diff --git a/article.js b/article.js
index 9cf81087..99a74281 100644
--- a/article.js
+++ b/article.js
@@ -1,5 +1,5 @@
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e=this.getCurve(),t=this._segment;if(!t){var n=this.getParameter();0===n?t=e._segment1:1===n?t=e._segment2:null!=n&&(t=e.getPartLength(0,n)<e.getPartLength(n,1)?e._segment1:e._segment2),this._segment=t}return t},getCurve:function(){function e(e){var t=e&&e.getCurve();if(t&&null!=(r._parameter=t.getParameterOf(r._point)))return r._setCurve(t),r._segment=e,t}var t=this._curve,n=t&&t._path,r=this;return n&&n._version!==this._version&&(t=this._parameter=this._curve=this._offset=null),t||e(this._segment)||e(this._segment1)||e(this._segment2.getPrevious())},getPath:function(){var e=this.getCurve();return e&&e._path},getIndex:function(){var e=this.getCurve();return e&&e.getIndex()},getParameter:function(){var e=this.getCurve(),t=this._parameter;return e&&null==t?this._parameter=e.getParameterOf(this._point):t},getPoint:function(){return this._point},getOffset:function(){var e=this._offset;if(null==e){e=0;var t=this.getPath(),n=this.getIndex();if(t&&null!=n)for(var r=t.getCurves(),i=0;i<n;i++)e+=r[i].getLength();this._offset=e+=this.getCurveOffset()}return e},getCurveOffset:function(){var e=this.getCurve(),t=this.getParameter();return null!=t&&e&&e.getPartLength(0,t)},getIntersection:function(){return this._intersection},getDistance:function(){return this._distance},divide:function(){var e=this.getCurve(),t=null;return e&&(t=e.divide(this.getParameter(),!0),t&&this._setSegment(t._segment1)),t},split:function(){var e=this.getCurve();return e?e.split(this.getParameter(),!0):null},equals:function(e,t){var n=this===e,r=2e-7;if(!n&&e instanceof O&&this.getPath()===e.getPath()&&this.getPoint().isClose(e.getPoint(),r)){var i=this.getCurve(),a=e.getCurve(),o=Math.abs,s=o((i.isLast()&&a.isFirst()?-1:i.getIndex())+this.getParameter()-((a.isLast()&&i.isFirst()?-1:a.getIndex())+e.getParameter()));n=(s<4e-7||(s=o(this.getOffset()-e.getOffset()))<r||o(this.getPath().getLength()-s)<r)&&(t||!this._intersection&&!e._intersection||this._intersection&&this._intersection.equals(e._intersection,!0))}return n},toString:function(){var e=[],t=this.getPoint(),n=h.instance;t&&e.push("point: "+t);var r=this.getIndex();null!=r&&e.push("index: "+r);var i=this.getParameter();return null!=i&&e.push("parameter: "+n.number(i)),null!=this._distance&&e.push("distance: "+n.number(this._distance)),"{ "+e.join(", ")+" }"},isTouching:function(){var e=this._intersection;if(e&&this.getTangent().isCollinear(e.getTangent())){var t=this.getCurve(),n=e.getCurve();return!(t.isStraight()&&n.isStraight()&&t.getLine().intersect(n.getLine()))}return!1},isCrossing:function(){function e(e,t,n){return t<n?e>t&&e<n:e>t&&e<=c||e>=-c&&e<n}var t=this._intersection;if(!t)return!1;var n=this.getParameter(),r=t.getParameter(),i=4e-7,a=1-i;if(n>=i&&n<=a||r>=i&&r<=a)return!this.isTouching();var o=this.getCurve(),s=o.getPrevious(),l=t.getCurve(),u=l.getPrevious(),c=Math.PI;if(!s||!u)return!1;var h=s.getTangentAt(a,!0).negate().getAngleInRadians(),d=o.getTangentAt(i,!0).getAngleInRadians(),f=u.getTangentAt(a,!0).negate().getAngleInRadians(),p=l.getTangentAt(i,!0).getAngleInRadians();return e(f,h,d)^e(p,h,d)&&e(f,d,h)^e(p,d,h)},isOverlap:function(){return!!this._overlap}},s.each(A.evaluateMethods,function(e){var t=e+"At";this[e]=function(){var e=this.getParameter(),n=this.getCurve();return null!=e&&n&&n[t](e,!0)}},{preserve:!0}),new function(){function e(e,t,n){function r(n,r){for(var a=n+r;a>=-1&&a<=i;a+=r){var o=e[(a%i+i)%i];if(!t.getPoint().isClose(o.getPoint(),2e-7))break;if(t.equals(o))return o}return null}for(var i=e.length,a=0,o=i-1;a<=o;){var s,l=a+o>>>1,u=e[l];if(n&&(s=t.equals(u)?u:r(l,-1)||r(l,1)))return t._overlap&&(s._overlap=s._intersection._overlap=!0),s;var c=t.getPath(),h=u.getPath(),d=c===h?t.getIndex()+t.getParameter()-(u.getIndex()+u.getParameter()):c._id-h._id;d<0?o=l-1:a=l+1}return e.splice(a,0,t),t}return{statics:{insert:e,expand:function(t){for(var n=t.slice(),r=0,i=t.length;r<i;r++)e(n,t[r]._intersection,!1);return n}}}}),B=C.extend({_class:"PathItem",initialize:function(){},getIntersections:function(e,t,n,r){var i=this===e||!e,a=this._matrix.orNullIfIdentity(),o=i?a:(n||e._matrix).orNullIfIdentity();if(!i&&!this.getBounds(a).touches(e.getBounds(o)))return[];for(var s,e,l=this.getCurves(),u=i?l:e.getCurves(),c=l.length,h=i?c:u.length,d=[],f=[],p=0;p<h;p++)d[p]=u[p].getValues(o);for(var p=0;p<c;p++){var m=l[p],g=i?d[p]:m.getValues(a),v=m.getPath();v!==e&&(e=v,s=[],f.push(s)),i&&A._getSelfIntersection(g,m,s,{include:t,startConnected:1===c&&m.getPoint1().equals(m.getPoint2())});for(var w=i?p+1:0;w<h;w++){if(r&&s.length)return s;var y=u[w];A._getIntersections(g,d[w],m,y,s,{include:t,startConnected:i&&m.getPrevious()===y,endConnected:i&&m.getNext()===y})}}s=[];for(var p=0,b=f.length;p<b;p++)s.push.apply(s,f[p]);return s},getCrossings:function(e){return this.getIntersections(e,function(e){return e.isCrossing()})},_asPathItem:function(){return this},setPathData:function(e){function t(e,t){var n=+r[e];return s&&(n+=l[t]),n}function n(e){return new p(t(e,"x"),t(e+1,"y"))}var r,i,a,o=e.match(/[mlhvcsqtaz][^mlhvcsqtaz]*/gi),s=!1,l=new p,u=new p;this.clear();for(var c=0,h=o&&o.length;c<h;c++){var d=o[c],f=d[0],m=f.toLowerCase();r=d.match(/[+-]?(?:\d*\.\d+|\d+\.?)(?:[eE][+-]?\d+)?/g);var v=r&&r.length;switch(s=f===m,"z"!==i||/[mz]/.test(m)||this.moveTo(l=u),m){case"m":case"l":for(var w="m"===m,y=0;y<v;y+=2)this[0===y&&w?"moveTo":"lineTo"](l=n(y));a=l,w&&(u=l);break;case"h":case"v":for(var b="h"===m?"x":"y",y=0;y<v;y++)l[b]=t(y,b),this.lineTo(l);a=l;break;case"c":for(var y=0;y<v;y+=6)this.cubicCurveTo(n(y),a=n(y+2),l=n(y+4));break;case"s":for(var y=0;y<v;y+=4)this.cubicCurveTo(/[cs]/.test(i)?l.multiply(2).subtract(a):l,a=n(y),l=n(y+2)),i=m;break;case"q":for(var y=0;y<v;y+=4)this.quadraticCurveTo(a=n(y),l=n(y+2));break;case"t":for(var y=0;y<v;y+=2)this.quadraticCurveTo(a=/[qt]/.test(i)?l.multiply(2).subtract(a):l,l=n(y)),i=m;break;case"a":for(var y=0;y<v;y+=7)this.arcTo(l=n(y+5),new g(+r[y],+r[y+1]),+r[y+2],+r[y+4],+r[y+3]);break;case"z":this.closePath(!0)}i=m}},_canComposite:function(){return!(this.hasFill()&&this.hasStroke())},_contains:function(e){var t=this._getWinding(e,!1,!0);return!!("evenodd"===this.getWindingRule()?1&t:t)}}),z=B.extend({_class:"Path",_serializeFields:{segments:[],closed:!1},initialize:function(e){this._closed=!1,this._segments=[],this._version=0;var t=Array.isArray(e)?"object"==typeof e[0]?e:arguments:!e||e.size!==o||e.x===o&&e.point===o?null:arguments;t&&t.length>0?this.setSegments(t):(this._curves=o,this._selectedSegmentState=0,t||"string"!=typeof e||(this.setPathData(e),e=null)),this._initialize(!t&&e)},_equals:function(e){return this._closed===e._closed&&s.equals(this._segments,e._segments)},clone:function(e){var t=new z(C.NO_INSERT);return t.setSegments(this._segments),t._closed=this._closed,this._clockwise!==o&&(t._clockwise=this._clockwise),this._clone(t,e)},_changed:function e(t){if(e.base.call(this,t),8&t){var n=this._parent;if(n&&(n._currentPath=o),this._length=this._area=this._clockwise=this._monoCurves=o,16&t)this._version++;else if(this._curves)for(var r=0,i=this._curves.length;r<i;r++)this._curves[r]._changed()}else 32&t&&(this._bounds=o)},getStyle:function(){var e=this._parent;return(e instanceof D?e:this)._style},getSegments:function(){return this._segments},setSegments:function(e){var t=this.isFullySelected();this._segments.length=0,this._selectedSegmentState=0,this._curves=o,e&&e.length>0&&this._add(I.readAll(e)),t&&this.setFullySelected(!0)},getFirstSegment:function(){return this._segments[0]},getLastSegment:function(){
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Array(6),g=!0,v=[];if(0===f)return"";for(var w=0;w<f;w++)n(d[w]);return this._closed&&f>0&&(n(d[0],!0),v.push("z")),v.join("")}},{isEmpty:function(){return 0===this._segments.length},_transformContent:function(e){for(var t=new Array(6),n=0,r=this._segments.length;n<r;n++)this._segments[n]._transformCoordinates(e,t,!0);return!0},_add:function(e,t){for(var n=this._segments,r=this._curves,i=e.length,a=null==t,t=a?n.length:t,o=0;o<i;o++){var s=e[o];s._path&&(s=e[o]=s.clone()),s._path=this,s._index=t+o,s._selectionState&&this._updateSelection(s,0,s._selectionState)}if(a)n.push.apply(n,e);else{n.splice.apply(n,[t,0].concat(e));for(var o=t+i,l=n.length;o<l;o++)n[o]._index=o}if(r){var u=this._countCurves(),c=t+i-1===u?t-1:t,h=c,d=Math.min(c+i,u);e._curves&&(r.splice.apply(r,[c,0].concat(e._curves)),h+=e._curves.length);for(var o=h;o<d;o++)r.splice(o,0,new A(this,null,null));this._adjustCurves(c,d)}return this._changed(25),e},_adjustCurves:function(e,t){for(var n,r=this._segments,i=this._curves,a=e;a<t;a++)n=i[a],n._path=this,n._segment1=r[a],n._segment2=r[a+1]||r[0],n._changed();(n=i[this._closed&&0===e?r.length-1:e-1])&&(n._segment2=r[e]||r[0],n._changed()),(n=i[t])&&(n._segment1=r[t],n._changed())},_countCurves:function(){var e=this._segments.length;return!this._closed&&e>0?e-1:e},add:function(e){return arguments.length>1&&"number"!=typeof e?this._add(I.readAll(arguments)):this._add([I.read(arguments)])[0]},insert:function(e,t){return arguments.length>2&&"number"!=typeof t?this._add(I.readAll(arguments,1),e):this._add([I.read(arguments,1)],e)[0]},addSegment:function(){return this._add([I.read(arguments)])[0]},insertSegment:function(e){return this._add([I.read(arguments,1)],e)[0]},addSegments:function(e){return this._add(I.readAll(e))},insertSegments:function(e,t){return this._add(I.readAll(t),e)},removeSegment:function(e){return this.removeSegments(e,e+1)[0]||null},removeSegments:function(e,t,n){e=e||0,t=s.pick(t,this._segments.length);var 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t(t){if(0!==A.getLength(t)){var n=t[1],r=t[3],i=t[5],a=t[7];if(A.isStraight(t))e(t);else{var o=3*(r-i)-n+a,s=2*(n+i)-4*r,l=r-n,u=4e-7,c=1-u,h=[],f=d.solveQuadratic(o,s,l,h,u,c);if(0===f)e(t);else{h.sort();var p=h[0],m=A.subdivide(t,p);e(m[0]),f>1&&(p=(h[1]-p)/(1-p),m=A.subdivide(m[1],p),e(m[0])),e(m[1])}}}}var n,r=this._monoCurves;if(!r){r=this._monoCurves=[];for(var i=this.getCurves(),a=this._segments,o=0,s=i.length;o<s;o++)t(i[o].getValues());if(!this._closed&&a.length>1){var l=a[a.length-1]._point,u=a[0]._point,c=l._x,h=l._y,f=u._x,p=u._y;t([c,h,c,h,f,p,f,p])}if(r.length>0){var m=r[0],g=r[r.length-1];m.previous=g,g.next=m}}return r},getInteriorPoint:function(){var e=this.getBounds(),t=e.getCenter(!0);if(!this.contains(t)){for(var n=this._getMonoCurves(),r=[],i=t.y,a=[],o=0,s=n.length;o<s;o++){var l=n[o].values;if((1===n[o].winding&&i>=l[1]&&i<=l[7]||i>=l[7]&&i<=l[1])&&A.solveCubic(l,1,i,r,0,1)>0)for(var u=r.length-1;u>=0;u--)a.push(A.getPoint(l,r[u]).x);if(a.length>1)break}t.x=(a[0]+a[1])/2}return t},reorient:function(){return this.setClockwise(!0),this}}),D.inject({_getMonoCurves:function(){for(var e=this._children,t=[],n=0,r=e.length;n<r;n++)t.push.apply(t,e[n]._getMonoCurves());return t},reorient:function(){var e=this.removeChildren().sort(function(e,t){return t.getBounds().getArea()-e.getBounds().getArea()});if(e.length>0){this.addChildren(e);for(var t=e[0].isClockwise(),n=1,r=e.length;n<r;n++){for(var i=e[n].getInteriorPoint(),a=0,o=n-1;o>=0;o--)e[o].contains(i)&&a++;e[n].setClockwise(a%2===0&&t)}}return this}});var R=s.extend({
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r=t.index;r<=n.index;r++){var i=A.getPart(this.curves[r],r==t.index?t.value:0,r==n.index?n.value:1);r==t.index&&e.moveTo(i[0],i[1]),e.bezierCurveTo.apply(e,i.slice(2))}}},s.each(A.evaluateMethods,function(e){this[e+"At"]=function(t,n){var r=this.getParameterAt(t);return A[e](this.curves[r.index],r.value,n)}},{})),j=s.extend({initialize:function(e,t){for(var n,r=this.points=[],i=e._segments,a=0,o=i.length;a<o;a++){var s=i[a].point.clone();n&&n.equals(s)||(r.push(s),n=s)}e._closed&&(this.closed=!0,r.unshift(r[r.length-1]),r.push(r[1])),this.error=t},fit:function(){var e=this.points,t=e.length,n=this.segments=t>0?[new I(e[0])]:[];return t>1&&this.fitCubic(0,t-1,e[1].subtract(e[0]).normalize(),e[t-2].subtract(e[t-1]).normalize()),this.closed&&(n.shift(),n.pop()),n},fitCubic:function(e,t,n,r){if(t-e==1){var i=this.points[e],a=this.points[t],o=i.getDistance(a)/3;return void this.addCurve([i,i.add(n.normalize(o)),a.add(r.normalize(o)),a])}for(var s,l=this.chordLengthParameterize(e,t),u=Math.max(this.error,this.error*this.error),c=!0,h=0;h<=4;h++){var d=this.generateBezier(e,t,l,n,r),f=this.findMaxError(e,t,d,l);if(f.error<this.error&&c)return void this.addCurve(d);if(s=f.index,f.error>=u)break;c=this.reparameterize(e,t,l,d),u=f.error}var p=this.points[s-1].subtract(this.points[s]),m=this.points[s].subtract(this.points[s+1]),g=p.add(m).divide(2).normalize();this.fitCubic(e,s,n,g),this.fitCubic(s,t,g.negate(),r)},addCurve:function(e){var t=this.segments[this.segments.length-1];t.setHandleOut(e[1].subtract(e[0])),this.segments.push(new I(e[3],e[2].subtract(e[3])))},generateBezier:function(e,t,n,r,i){for(var a=1e-12,o=this.points[e],s=this.points[t],l=[[0,0],[0,0]],u=[0,0],c=0,h=t-e+1;c<h;c++){var d=n[c],f=1-d,p=3*d*f,m=f*f*f,g=p*f,v=p*d,w=d*d*d,y=r.normalize(g),b=i.normalize(v),_=this.points[e+c].subtract(o.multiply(m+g)).subtract(s.multiply(v+w));l[0][0]+=y.dot(y),l[0][1]+=y.dot(b),l[1][0]=l[0][1],l[1][1]+=b.dot(b),u[0]+=y.dot(_),u[1]+=b.dot(_)}var x,E,C=l[0][0]*l[1][1]-l[1][0]*l[0][1];if(Math.abs(C)>a){var k=l[0][0]*u[1]-l[1][0]*u[0],S=u[0]*l[1][1]-u[1]*l[0][1];x=S/C,E=k/C}else{var P=l[0][0]+l[0][1],T=l[1][0]+l[1][1];x=E=Math.abs(P)>a?u[0]/P:Math.abs(T)>a?u[1]/T:0}var N,M,I=s.getDistance(o),L=a*I;if(x<L||E<L)x=E=I/3;else{var A=s.subtract(o);N=r.normalize(x),M=i.normalize(E),N.dot(A)-M.dot(A)>I*I&&(x=E=I/3,N=M=null)}return[o,o.add(N||r.normalize(x)),s.add(M||i.normalize(E)),s]},reparameterize:function(e,t,n,r){for(var i=e;i<=t;i++)n[i-e]=this.findRoot(r,this.points[i],n[i-e]);for(var i=1,a=n.length;i<a;i++)if(n[i]<=n[i-1])return!1;return!0},findRoot:function(e,t,n){for(var r=[],i=[],a=0;a<=2;a++)r[a]=e[a+1].subtract(e[a]).multiply(3);for(var a=0;a<=1;a++)i[a]=r[a+1].subtract(r[a]).multiply(2);var o=this.evaluate(3,e,n),s=this.evaluate(2,r,n),l=this.evaluate(1,i,n),u=o.subtract(t),c=s.dot(s)+u.dot(l);return Math.abs(c)<1e-6?n:n-u.dot(s)/c},evaluate:function(e,t,n){for(var r=t.slice(),i=1;i<=e;i++)for(var a=0;a<=e-i;a++)r[a]=r[a].multiply(1-n).add(r[a+1].multiply(n));return r[0]},chordLengthParameterize:function(e,t){for(var n=[0],r=e+1;r<=t;r++)n[r-e]=n[r-e-1]+this.points[r].getDistance(this.points[r-1]);for(var r=1,i=t-e;r<=i;r++)n[r]/=n[i];return n},findMaxError:function(e,t,n,r){for(var i=Math.floor((t-e+1)/2),a=0,o=e+1;o<t;o++){var s=this.evaluate(3,n,r[o-e]),l=s.subtract(this.points[o]),u=l.x*l.x+l.y*l.y;u>=a&&(a=u,i=o)}return{error:a,index:i}}}),F=C.extend({_class:"TextItem",_boundsSelected:!0,_applyMatrix:!1,_canApplyMatrix:!1,_serializeFields:{content:null},_boundsGetter:"getBounds",initialize:function(e){this._content="",this._lines=[];var t=e&&s.isPlainObject(e)&&e.x===o&&e.y===o;this._initialize(t&&e,!t&&p.read(arguments))},_equals:function(e){return this._content===e._content},_clone:function e(t,n,r){return t.setContent(this._content),e.base.call(this,t,n,r)},getContent:function(){return this._content},setContent:function(e){this._content=""+e,this._lines=this._content.split(/\r\n|\n|\r/gm),this._changed(265)},isEmpty:function(){return!this._content},getCharacterStyle:"#getStyle",setCharacterStyle:"#setStyle",getParagraphStyle:"#getStyle",setParagraphStyle:"#setStyle"}),q=F.extend({_class:"PointText",initialize:function(){F.apply(this,arguments)},clone:function(e){return this._clone(new q(C.NO_INSERT),e)},getPoint:function(){var e=this._matrix.getTranslation();return new m(e.x,e.y,this,"setPoint")},setPoint:function(){var e=p.read(arguments);this.translate(e.subtract(this._matrix.getTranslation()))},_draw:function(e){if(this._content){this._setStyles(e);var t=this._style,n=this._lines,r=t.getLeading(),i=e.shadowColor;e.font=t.getFontStyle(),e.textAlign=t.getJustification();for(var a=0,o=n.length;a<o;a++){e.shadowColor=i;var s=n[a];t.hasFill()&&(e.fillText(s,0,0),e.shadowColor="rgba(0,0,0,0)"),t.hasStroke()&&e.strokeText(s,0,0),e.translate(0,r)}}},_getBounds:function(e,t){var n=this._style,r=this._lines,i=r.length,a=n.getJustification(),o=n.getLeading(),s=this.getView().getTextWidth(n.getFontStyle(),r),l=0;"left"!==a&&(l-=s/("center"===a?2:1));var u=new w(l,i?-.75*o:0,s,i*o);return t?t._transformBounds(u,u):u}}),V=s.extend(new function(){function e(e){var n,r=e.match(/^#(\w{1,2})(\w{1,2})(\w{1,2})$/);if(r){n=[0,0,0];for(var a=0;a<3;a++){var o=r[a+1];n[a]=parseInt(1==o.length?o+o:o,16)/255}}else if(r=e.match(/^rgba?\((.*)\)$/)){n=r[1].split(",");for(var a=0,s=n.length;a<s;a++){var o=+n[a];n[a]=a<3?o/255:o}}else{var l=i[e];if(!l){t||(t=ne.getContext(1,1),t.globalCompositeOperation="copy"),t.fillStyle="rgba(0,0,0,0)",t.fillStyle=e,t.fillRect(0,0,1,1);var u=t.getImageData(0,0,1,1).data;l=i[e]=[u[0]/255,u[1]/255,u[2]/255]}n=l.slice()}return n}var t,n={gray:["gray"],rgb:["red","green","blue"],hsb:["hue","saturation","brightness"],hsl:["hue","saturation","lightness"],gradient:["gradient","origin","destination","highlight"]},r={},i={},a=[[0,3,1],[2,0,1],[1,0,3],[1,2,0],[3,1,0],[0,1,2]],o={"rgb-hsb":function(e,t,n){var r=Math.max(e,t,n),i=Math.min(e,t,n),a=r-i,o=0===a?0:60*(r==e?(t-n)/a+(t<n?6:0):r==t?(n-e)/a+2:(e-t)/a+4);return[o,0===r?0:a/r,r]},"hsb-rgb":function(e,t,n){e=(e/60%6+6)%6;var r=Math.floor(e),i=e-r,r=a[r],o=[n,n*(1-t),n*(1-t*i),n*(1-t*(1-i))];return[o[r[0]],o[r[1]],o[r[2]]]},"rgb-hsl":function(e,t,n){var r=Math.max(e,t,n),i=Math.min(e,t,n),a=r-i,o=0===a,s=o?0:60*(r==e?(t-n)/a+(t<n?6:0):r==t?(n-e)/a+2:(e-t)/a+4),l=(r+i)/2,u=o?0:l<.5?a/(r+i):a/(2-r-i);return[s,u,l]},"hsl-rgb":function(e,t,n){if(e=(e/360%1+1)%1,0===t)return[n,n,n];for(var r=[e+1/3,e,e-1/3],i=n<.5?n*(1+t):n+t-n*t,a=2*n-i,o=[],s=0;s<3;s++){var l=r[s];l<0&&(l+=1),l>1&&(l-=1),o[s]=6*l<1?a+6*(i-a)*l:2*l<1?i:3*l<2?a+(i-a)*(2/3-l)*6:a}return o},"rgb-gray":function(e,t,n){return[.2989*e+.587*t+.114*n]},"gray-rgb":function(e){return[e,e,e]},"gray-hsb":function(e){return[0,0,e]},"gray-hsl":function(e){return[0,0,e]},"gradient-rgb":function(){return[]},"rgb-gradient":function(){return[]}};return s.each(n,function(e,t){r[t]=[],s.each(e,function(e,i){var a=s.capitalize(e),o=/^(hue|saturation)$/.test(e),l=r[t][i]="gradient"===e?function(e){var t=this._components[0];return e=U.read(Array.isArray(e)?e:arguments,0,{readNull:!0}),t!==e&&(t&&t._removeOwner(this),e&&e._addOwner(this)),e}:"gradient"===t?function(){return p.read(arguments,0,{readNull:"highlight"===e,clone:!0})}:function(e){return null==e||isNaN(e)?0:e};this["get"+a]=function(){return this._type===t||o&&/^hs[bl]$/.test(this._type)?this._components[i]:this._convert(t)[i]},this["set"+a]=function(e){this._type===t||o&&/^hs[bl]$/.test(this._type)||(this._components=this._convert(t),this._properties=n[t],this._type=t),this._components[i]=l.call(this,e),this._changed()}},this)},{_class:"Color",_readIndex:!0,initialize:function t(i){var a,o,s,l,u=Array.prototype.slice,c=arguments,h=0;Array.isArray(i)&&(c=i,i=c[0]);var d=null!=i&&typeof i;if("string"===d&&i in n&&(a=i,i=c[1],Array.isArray(i)?(o=i,s=c[2]):(this.__read&&(h=1),c=u.call(c,1),d=typeof i)),!o){if(l="number"===d?c:"object"===d&&null!=i.length?i:null){a||(a=l.length>=3?"rgb":"gray");var p=n[a].length;s=l[p],this.__read&&(h+=l===arguments?p+(null!=s?1:0):1),l.length>p&&(l=u.call(l,0,p))}else if("string"===d)a="rgb",o=e(i),4===o.length&&(s=o[3],o.length--);else if("object"===d)if(i.constructor===t){if(a=i._type,o=i._components.slice(),s=i._alpha,"gradient"===a)for(var m=1,g=o.length;m<g;m++){var v=o[m];v&&(o[m]=v.clone())}}else if(i.constructor===U)a="gradient",l=c;else{a="hue"in i?"lightness"in i?"hsl":"hsb":"gradient"in i||"stops"in i||"radial"in i?"gradient":"gray"in i?"gray":"rgb";var w=n[a],y=r[a];this._components=o=[];for(var m=0,g=w.length;m<g;m++){var b=i[w[m]];null==b&&0===m&&"gradient"===a&&"stops"in i&&(b={stops:i.stops,radial:i.radial}),b=y[m].call(this,b),null!=b&&(o[m]=b)}s=i.alpha}this.__read&&a&&(h=1)}if(this._type=a||"rgb",this._id=f.get(t),!o){this._components=o=[];for(var y=r[this._type],m=0,g=y.length;m<g;m++){var b=y[m].call(this,l&&l[m]);null!=b&&(o[m]=b)}}this._components=o,this._properties=n[this._type],this._alpha=s,this.__read&&(this.__read=h)},_serialize:function(e,t){var n=this.getComponents();return s.serialize(/^(gray|rgb)$/.test(this._type)?n:[this._type].concat(n),e,!0,t)},_changed:function(){this._canvasStyle=null,this._owner&&this._owner._changed(65)},_convert:function(e){var t;return this._type===e?this._components.slice():(t=o[this._type+"-"+e])?t.apply(this,this._components):o["rgb-"+e].apply(this,o[this._type+"-rgb"].apply(this,this._components))},convert:function(e){return new V(e,this._convert(e),this._alpha)},getType:function(){return this._type},setType:function(e){this._components=this._convert(e),this._properties=n[e],this._type=e},getComponents:function(){var e=this._components.slice();return null!=this._alpha&&e.push(this._alpha),e},getAlpha:function(){return null!=this._alpha?this._alpha:1},setAlpha:function(e){this._alpha=null==e?null:Math.min(Math.max(e,0),1),this._changed()},hasAlpha:function(){return null!=this._alpha},equals:function(e){var t=s.isPlainValue(e,!0)?V.read(arguments):e;return t===this||t&&this._class===t._class&&this._type===t._type&&this._alpha===t._alpha&&s.equals(this._components,t._components)||!1},toString:function(){for(var e=this._properties,t=[],n="gradient"===this._type,r=h.instance,i=0,a=e.length;i<a;i++){var o=this._components[i];null!=o&&t.push(e[i]+": "+(n?o:r.number(o)))}return null!=this._alpha&&t.push("alpha: "+r.number(this._alpha)),"{ "+t.join(", ")+" }"},toCSS:function(e){function t(e){return Math.round(255*(e<0?0:e>1?1:e))}var n=this._convert("rgb"),r=e||null==this._alpha?1:this._alpha;return n=[t(n[0]),t(n[1]),t(n[2])],r<1&&n.push(r<0?0:r),e?"#"+((1<<24)+(n[0]<<16)+(n[1]<<8)+n[2]).toString(16).slice(1):(4==n.length?"rgba(":"rgb(")+n.join(",")+")"},toCanvasStyle:function(e){if(this._canvasStyle)return this._canvasStyle;if("gradient"!==this._type)return this._canvasStyle=this.toCSS();var t,n=this._components,r=n[0],i=r._stops,a=n[1],o=n[2];if(r._radial){var s=o.getDistance(a),l=n[3];if(l){var u=l.subtract(a);u.getLength()>s&&(l=a.add(u.normalize(s-.1)))}var c=l||a;t=e.createRadialGradient(c.x,c.y,0,a.x,a.y,s)}else t=e.createLinearGradient(a.x,a.y,o.x,o.y);for(var h=0,d=i.length;h<d;h++){var f=i[h];t.addColorStop(f._rampPoint,f._color.toCanvasStyle())}return this._canvasStyle=t},transform:function(e){if("gradient"===this._type){for(var t=this._components,n=1,r=t.length;n<r;n++){var i=t[n];e._transformPoint(i,i,!0)}this._changed()}},statics:{_types:n,random:function(){var e=Math.random;return new V(e(),e(),e())}}})},new function(){var e={add:function(e,t){return e+t},subtract:function(e,t){return e-t},multiply:function(e,t){return e*t},divide:function(e,t){return e/t}};return s.each(e,function(e,t){this[t]=function(t){t=V.read(arguments);for(var n=this._type,r=this._components,i=t._convert(n),a=0,o=r.length;a<o;a++)i[a]=e(r[a],i[a]);return new V(n,i,null!=this._alpha?e(this._alpha,t.getAlpha()):null)}},{})}),U=s.extend({_class:"Gradient",initialize:function(e,t){this._id=f.get(),e&&this._set(e)&&(e=t=null),this._stops||this.setStops(e||["white","black"]),null==this._radial&&this.setRadial("string"==typeof t&&"radial"===t||t||!1)},_serialize:function(e,t){return t.add(this,function(){return s.serialize([this._stops,this._radial],e,!0,t)})},_changed:function(){for(var e=0,t=this._owners&&this._owners.length;e<t;e++)this._owners[e]._changed()},_addOwner:function(e){this._owners||(this._owners=[]),this._owners.push(e)},_removeOwner:function(e){var t=this._owners?this._owners.indexOf(e):-1;t!=-1&&(this._owners.splice(t,1),0===this._owners.length&&(this._owners=o))},clone:function(){for(var e=[],t=0,n=this._stops.length;t<n;t++)e[t]=this._stops[t].clone();return new U(e,this._radial)},getStops:function(){return this._stops},setStops:function(e){if(this.stops)for(var t=0,n=this._stops.length;t<n;t++)this._stops[t]._owner=o;if(e.length<2)throw new Error("Gradient stop list needs to contain at least two stops.");this._stops=W.readAll(e,0,{clone:!0});for(var t=0,n=this._stops.length;t<n;t++){var r=this._stops[t];r._owner=this,r._defaultRamp&&r.setRampPoint(t/(n-1))}this._changed()},getRadial:function(){return this._radial},setRadial:function(e){this._radial=e,this._changed()},equals:function(e){if(e===this)return!0;if(e&&this._class===e._class&&this._stops.length===e._stops.length){for(var t=0,n=this._stops.length;t<n;t++)if(!this._stops[t].equals(e._stops[t]))return!1;return!0}return!1}}),W=s.extend({_class:"GradientStop",initialize:function(e,t){if(e){var n,r;t===o&&Array.isArray(e)?(n=e[0],r=e[1]):e.color?(n=e.color,r=e.rampPoint):(n=e,r=t),this.setColor(n),this.setRampPoint(r)}},clone:function(){return new W(this._color.clone(),this._rampPoint)},_serialize:function(e,t){return s.serialize([this._color,this._rampPoint],e,!0,t)},_changed:function(){this._owner&&this._owner._changed(65)},getRampPoint:function(){return this._rampPoint},setRampPoint:function(e){this._defaultRamp=null==e,this._rampPoint=e||0,this._changed()},getColor:function(){return this._color},setColor:function(e){this._color=V.read(arguments),this._color===e&&(this._color=e.clone()),this._color._owner=this,this._changed()},equals:function(e){return e===this||e&&this._class===e._class&&this._color.equals(e._color)&&this._rampPoint==e._rampPoint||!1}}),G=s.extend(new function(){var e={fillColor:o,strokeColor:o,strokeWidth:1,strokeCap:"butt",strokeJoin:"miter",strokeScaling:!0,miterLimit:10,dashOffset:0,dashArray:[],windingRule:"nonzero",shadowColor:o,shadowBlur:0,shadowOffset:new p,selectedColor:o,fontFamily:"sans-serif",fontWeight:"normal",fontSize:12,font:"sans-serif",leading:null,justification:"left"},t={strokeWidth:97,strokeCap:97,strokeJoin:97,strokeScaling:105,miterLimit:97,fontFamily:9,fontWeight:9,fontSize:9,font:9,leading:9,justification:9},n={beans:!0},r={_defaults:e,_textDefaults:new s(e,{fillColor:new V}),beans:!0};return s.each(e,function(e,i){var a=/Color$/.test(i),l="shadowOffset"===i,u=s.capitalize(i),c=t[i],h="set"+u,d="get"+u;r[h]=function(e){var t=this._owner,n=t&&t._children;if(n&&n.length>0&&!(t instanceof D))for(var r=0,s=n.length;r<s;r++)n[r]._style[h](e);else{var l=this._values[i];l!==e&&(a&&(l&&(l._owner=o),e&&e.constructor===V&&(e._owner&&(e=e.clone()),e._owner=t)),this._values[i]=e,t&&t._changed(c||65))}},r[d]=function(e){var t,n=this._owner,r=n&&n._children;if(!r||0===r.length||e||n instanceof D){var t=this._values[i];if(t===o)t=this._defaults[i],t&&t.clone&&(t=t.clone());else{var u=a?V:l?p:null;!u||t&&t.constructor===u||(this._values[i]=t=u.read([t],0,{readNull:!0,clone:!0}),t&&a&&(t._owner=n))}return t}for(var c=0,h=r.length;c<h;c++){var f=r[c]._style[d]();if(0===c)t=f;else if(!s.equals(t,f))return o}return t},n[d]=function(e){return this._style[d](e)},n[h]=function(e){this._style[h](e)}}),C.inject(n),r},{_class:"Style",initialize:function(e,t,n){this._values={},this._owner=t,this._project=t&&t._project||n||a.project,t instanceof F&&(this._defaults=this._textDefaults),e&&this.set(e)},set:function(e){var t=e instanceof G,n=t?e._values:e;if(n)for(var r in n)if(r in this._defaults){var i=n[r];this[r]=i&&t&&i.clone?i.clone():i}},equals:function(e){return e===this||e&&this._class===e._class&&s.equals(this._values,e._values)||!1},hasFill:function(){return!!this.getFillColor()},hasStroke:function(){return!!this.getStrokeColor()&&this.getStrokeWidth()>0},hasShadow:function(){return!!this.getShadowColor()&&this.getShadowBlur()>0},getView:function(){return this._project.getView()},getFontStyle:function(){var e=this.getFontSize();return this.getFontWeight()+" "+e+(/[a-z]/i.test(e+"")?" ":"px ")+this.getFontFamily()},getFont:"#getFontFamily",setFont:"#setFontFamily",getLeading:function e(){var t=e.base.call(this),n=this.getFontSize();return/pt|em|%|px/.test(n)&&(n=this.getView().getPixelSize(n)),null!=t?t:1.2*n}}),H=new function(){function e(e,t,n,r){for(var i=["","webkit","moz","Moz","ms","o"],a=t[0].toUpperCase()+t.substring(1),o=0;o<6;o++){var s=i[o],l=s?s+a:t;if(l in e){if(!n)return e[l];e[l]=r;break}}}return{getStyles:function(e){var t=e&&9!==e.nodeType?e.ownerDocument:e,n=t&&t.defaultView;return n&&n.getComputedStyle(e,"")},getBounds:function(e,t){var n,r=e.ownerDocument,i=r.body,a=r.documentElement;try{n=e.getBoundingClientRect()}catch(e){n={left:0,top:0,width:0,height:0}}var o=n.left-(a.clientLeft||i.clientLeft||0),s=n.top-(a.clientTop||i.clientTop||0);if(!t){var l=r.defaultView;o+=l.pageXOffset||a.scrollLeft||i.scrollLeft,s+=l.pageYOffset||a.scrollTop||i.scrollTop}return new w(o,s,n.width,n.height)},getViewportBounds:function(e){var t=e.ownerDocument,n=t.defaultView,r=t.documentElement;return new w(0,0,n.innerWidth||r.clientWidth,n.innerHeight||r.clientHeight)},getOffset:function(e,t){return H.getBounds(e,t).getPoint()},getSize:function(e){return H.getBounds(e,!0).getSize()},isInvisible:function(e){return H.getSize(e).equals(new g(0,0))},isInView:function(e){return!H.isInvisible(e)&&H.getViewportBounds(e).intersects(H.getBounds(e,!0))},getPrefixed:function(t,n){return e(t,n)},setPrefixed:function(t,n,r){if("object"==typeof n)for(var i in n)e(t,i,!0,n[i]);else e(t,n,!0,r)}}},X={add:function(e,t){for(var n in t)for(var r=t[n],i=n.split(/[\s,]+/g),a=0,o=i.length;a<o;a++)e.addEventListener(i[a],r,!1)},remove:function(e,t){for(var n in t)for(var r=t[n],i=n.split(/[\s,]+/g),a=0,o=i.length;a<o;a++)e.removeEventListener(i[a],r,!1)},getPoint:function(e){var t=e.targetTouches?e.targetTouches.length?e.targetTouches[0]:e.changedTouches[0]:e;return new p(t.pageX||t.clientX+document.documentElement.scrollLeft,t.pageY||t.clientY+document.documentElement.scrollTop)},getTarget:function(e){return e.target||e.srcElement},getRelatedTarget:function(e){return e.relatedTarget||e.toElement},getOffset:function(e,t){return X.getPoint(e).subtract(H.getOffset(t||X.getTarget(e)))},stop:function(e){e.stopPropagation(),e.preventDefault()}};X.requestAnimationFrame=new function(){function e(){for(var t=i.length-1;t>=0;t--){var o=i[t],s=o[0],l=o[1];(!l||("true"==u.getAttribute(l,"keepalive")||a)&&H.isInView(l))&&(i.splice(t,1),s())}n&&(i.length?n(e):r=!1)}var t,n=H.getPrefixed(window,"requestAnimationFrame"),r=!1,i=[],a=!0;return X.add(window,{focus:function(){a=!0},blur:function(){a=!1}}),function(a,o){i.push([a,o]),n?r||(n(e),r=!0):t||(t=setInterval(e,1e3/60))}};var K=s.extend(l,{_class:"View",initialize:function e(t,n){function r(e){return n[e]||parseInt(n.getAttribute(e),10)}function i(){var e=H.getSize(n);return e.isNaN()||e.isZero()?new g(r("width"),r("height")):e}this._project=t,this._scope=t._scope,this._element=n;var a;this._pixelRatio||(this._pixelRatio=window.devicePixelRatio||1),this._id=n.getAttribute("id"),null==this._id&&n.setAttribute("id",this._id="view-"+e._id++),X.add(n,this._viewEvents);var o="none";if(H.setPrefixed(n.style,{userSelect:o,touchAction:o,touchCallout:o,contentZooming:o,userDrag:o,tapHighlightColor:"rgba(0,0,0,0)"}),u.hasAttribute(n,"resize")){var s=this;X.add(window,this._windowEvents={resize:function(){s.setViewSize(i())}})}if(this._setViewSize(a=i()),u.hasAttribute(n,"stats")&&"undefined"!=typeof Stats){this._stats=new Stats;var l=this._stats.domElement,c=l.style,h=H.getOffset(n);c.position="absolute",c.left=h.x+"px",c.top=h.y+"px",document.body.appendChild(l)}e._views.push(this),e._viewsById[this._id]=this,this._viewSize=a,(this._matrix=new b)._owner=this,this._zoom=1,e._focused||(e._focused=this),this._frameItems={},this._frameItemCount=0},remove:function(){return!!this._project&&(K._focused===this&&(K._focused=null),K._views.splice(K._views.indexOf(this),1),delete K._viewsById[this._id],this._project._view===this&&(this._project._view=null),X.remove(this._element,this._viewEvents),X.remove(window,this._windowEvents),this._element=this._project=null,this.off("frame"),this._animate=!1,this._frameItems={},!0)},_events:s.each(["onResize","onMouseDown","onMouseUp","onMouseMove"],function(e){this[e]={install:function(e){this._installEvent(e)},uninstall:function(e){this._uninstallEvent(e)}}},{onFrame:{install:function(){this.play()},uninstall:function(){this.pause()}}}),_animate:!1,_time:0,_count:0,_requestFrame:function(){var e=this;X.requestAnimationFrame(function(){e._requested=!1,e._animate&&(e._requestFrame(),e._handleFrame())},this._element),this._requested=!0},_handleFrame:function(){a=this._scope;var e=Date.now()/1e3,t=this._before?e-this._before:0;this._before=e,this._handlingFrame=!0,this.emit("frame",new s({delta:t,time:this._time+=t,count:this._count++})),this._stats&&this._stats.update(),this._handlingFrame=!1,this.update()},_animateItem:function(e,t){var n=this._frameItems;t?(n[e._id]={item:e,time:0,count:0},1===++this._frameItemCount&&this.on("frame",this._handleFrameItems)):(delete n[e._id],0===--this._frameItemCount&&this.off("frame",this._handleFrameItems))},_handleFrameItems:function(e){for(var t in this._frameItems){var n=this._frameItems[t];n.item.emit("frame",new s(e,{time:n.time+=e.delta,count:n.count++}))}},_update:function(){this._project._needsUpdate=!0,this._handlingFrame||(this._animate?this._handleFrame():this.update())},_changed:function(e){1&e&&(this._project._needsUpdate=!0)},_transform:function(e){this._matrix.concatenate(e),this._bounds=null,this._update()},getElement:function(){return this._element},getPixelRatio:function(){return this._pixelRatio},getResolution:function(){return 72*this._pixelRatio},getViewSize:function(){var e=this._viewSize;return new v(e.width,e.height,this,"setViewSize")},setViewSize:function(){var e=g.read(arguments),t=e.subtract(this._viewSize);t.isZero()||(this._viewSize.set(e.width,e.height),this._setViewSize(e),this._bounds=null,this.emit("resize",{size:e,delta:t}),this._update())},_setViewSize:function(e){var t=this._element;t.width=e.width,t.height=e.height},getBounds:function(){return this._bounds||(this._bounds=this._matrix.inverted()._transformBounds(new w(new p,this._viewSize))),this._bounds},getSize:function(){return this.getBounds().getSize()},getCenter:function(){return this.getBounds().getCenter()},setCenter:function(){var e=p.read(arguments);this.scrollBy(e.subtract(this.getCenter()))},getZoom:function(){return this._zoom},setZoom:function(e){this._transform((new b).scale(e/this._zoom,this.getCenter())),this._zoom=e},isVisible:function(){return H.isInView(this._element)},scrollBy:function(){this._transform((new b).translate(p.read(arguments).negate()))},play:function(){this._animate=!0,this._requested||this._requestFrame()},pause:function(){this._animate=!1},draw:function(){this.update()},projectToView:function(){return this._matrix._transformPoint(p.read(arguments))},viewToProject:function(){return this._matrix._inverseTransform(p.read(arguments))}},{statics:{_views:[],_viewsById:{},_id:0,create:function(e,t){return"string"==typeof t&&(t=document.getElementById(t)),new Q(e,t)}}},new function(){function e(e){var t=X.getTarget(e);return t.getAttribute&&K._viewsById[t.getAttribute("id")]}function t(e,t){return e.viewToProject(X.getOffset(t,e._element))}function n(){if(!K._focused||!K._focused.isVisible())for(var e=0,t=K._views.length;e<t;e++){var n=K._views[e];if(n&&n.isVisible()){K._focused=o=n;break}}}function r(e,t,n){e._handleEvent("mousemove",t,n);var r=e._scope.tool;return r&&r._handleEvent(c&&r.responds("mousedrag")?"mousedrag":"mousemove",t,n),e.update(),r}var i,a,o,s,l,u,c=!1,h=window.navigator;h.pointerEnabled||h.msPointerEnabled?(s="pointerdown MSPointerDown",l="pointermove MSPointerMove",u="pointerup pointercancel MSPointerUp MSPointerCancel"):(s="touchstart",l="touchmove",u="touchend touchcancel","ontouchstart"in window&&h.userAgent.match(/mobile|tablet|ip(ad|hone|od)|android|silk/i)||(s+=" mousedown",l+=" mousemove",u+=" mouseup"));var d={"selectstart dragstart":function(e){c&&e.preventDefault()}},f={mouseout:function(e){var n=K._focused,i=X.getRelatedTarget(e);!n||i&&"HTML"!==i.nodeName||r(n,t(n,e),e)},scroll:n};d[s]=function(n){var r=K._focused=e(n),a=t(r,n);c=!0,r._handleEvent("mousedown",a,n),(i=r._scope.tool)&&i._handleEvent("mousedown",a,n),r.update()},f[l]=function(s){var l=K._focused;if(!c){var u=e(s);u?(l!==u&&r(l,t(l,s),s),a=l,l=K._focused=o=u):o&&o===l&&(l=K._focused=a,n())}if(l){var h=t(l,s);(c||l.getBounds().contains(h))&&(i=r(l,h,s))}},f[u]=function(e){var n=K._focused;if(n&&c){var r=t(n,e);c=!1,n._handleEvent("mouseup",r,e),i&&i._handleEvent("mouseup",r,e),n.update()}},X.add(document,f),X.add(window,{load:n});var p={mousedown:{mousedown:1,mousedrag:1,click:1,doubleclick:1},mouseup:{mouseup:1,mousedrag:1,click:1,doubleclick:1},mousemove:{mousedrag:1,mousemove:1,mouseenter:1,mouseleave:1}};return{_viewEvents:d,_handleEvent:function(){},_installEvent:function(e){var t=this._eventCounters;if(t)for(var n in p)t[n]=(t[n]||0)+(p[n][e]||0)},_uninstallEvent:function(e){var t=this._eventCounters;if(t)for(var n in p)t[n]-=p[n][e]||0},statics:{updateFocus:n}}}),Q=K.extend({_class:"CanvasView",initialize:function(e,t){if(!(t instanceof HTMLCanvasElement)){var n=g.read(arguments,1);if(n.isZero())throw new Error("Cannot create CanvasView with the provided argument: "+[].slice.call(arguments,1));t=ne.getCanvas(n)}if(this._context=t.getContext("2d"),this._eventCounters={},this._pixelRatio=1,!/^off|false$/.test(u.getAttribute(t,"hidpi"))){var r=window.devicePixelRatio||1,i=H.getPrefixed(this._context,"backingStorePixelRatio")||1;this._pixelRatio=r/i}K.call(this,e,t)},_setViewSize:function(e){var t=this._element,n=this._pixelRatio,r=e.width,i=e.height;if(t.width=r*n,t.height=i*n,1!==n){if(!u.hasAttribute(t,"resize")){var a=t.style;a.width=r+"px",a.height=i+"px"}this._context.scale(n,n)}},getPixelSize:function(e){var t,n=a.browser;if(n&&n.firefox){var r=this._element.parentNode,i=document.createElement("div");i.style.fontSize=e,r.appendChild(i),t=parseFloat(H.getStyles(i).fontSize),r.removeChild(i)}else{var o=this._context,s=o.font;o.font=e+" serif",t=parseFloat(o.font),o.font=s}return t},getTextWidth:function(e,t){var n=this._context,r=n.font,i=0;n.font=e;for(var a=0,o=t.length;a<o;a++)i=Math.max(i,n.measureText(t[a]).width);return n.font=r,i},update:function(e){var t=this._project;if(!t||!e&&!t._needsUpdate)return!1;var n=this._context,r=this._viewSize;return n.clearRect(0,0,r.width+1,r.height+1),t.draw(n,this._matrix,this._pixelRatio),t._needsUpdate=!1,!0}},new function(){function e(e,t,n,r,i,a){function o(e){if(e.responds(t)&&(s||(s=new J(t,n,r,i,a?r.subtract(a):null)),e.emit(t,s)&&s.isStopped))return n.preventDefault(),!0}for(var s,l=i;l;){if(o(l))return!0;l=l.getParent()}return!!o(e)}var t,n,r,i,a,o,s,l,u;return{_handleEvent:function(c,h,d){if(this._eventCounters[c]){var f=this._project,p=f.hitTest(h,{tolerance:0,fill:!0,stroke:!0}),m=p&&p.item,g=!1;switch(c){case"mousedown":for(g=e(this,c,d,h,m),l=a==m&&Date.now()-u<300,i=a=m,t=n=r=h,s=!g&&m;s&&!s.responds("mousedrag");)s=s._parent;break;case"mouseup":g=e(this,c,d,h,m,t),s&&(n&&!n.equals(h)&&e(this,"mousedrag",d,h,s,n),m!==s&&(r=h,e(this,"mousemove",d,h,m,r))),!g&&m&&m===i&&(u=Date.now(),e(this,l&&i.responds("doubleclick")?"doubleclick":"click",d,t,m),l=!1),i=s=null;break;case"mousemove":s&&(g=e(this,"mousedrag",d,h,s,n)),g||(m!==o&&(r=h),g=e(this,c,d,h,m,r)),n=r=h,m!==o&&(e(this,"mouseleave",d,h,o),o=m,e(this,"mouseenter",d,h,m))}return g}}}}),Y=s.extend({_class:"Event",initialize:function(e){this.event=e},isPrevented:!1,isStopped:!1,preventDefault:function(){this.isPrevented=!0,this.event.preventDefault()},stopPropagation:function(){this.isStopped=!0,this.event.stopPropagation()},stop:function(){this.stopPropagation(),this.preventDefault()},getModifiers:function(){return $.modifiers}}),Z=Y.extend({_class:"KeyEvent",initialize:function(e,t,n,r){Y.call(this,r),this.type=e?"keydown":"keyup",this.key=t,this.character=n},toString:function(){return"{ type: '"+this.type+"', key: '"+this.key+"', character: '"+this.character+"', modifiers: "+this.getModifiers()+" }"}}),$=new function(){function e(n,i,c,h){var d,f=c?String.fromCharCode(c):"",p=r[i],m=p||f.toLowerCase(),g=n?"keydown":"keyup",v=K._focused,w=v&&v.isVisible()&&v._scope,y=w&&w.tool;if(u[m]=n,n?l[i]=c:delete l[i],p&&(d=s.camelize(p))in o){o[d]=n;var b=a.browser;if("command"===d&&b&&b.mac)if(n)t={};else{for(var _ in t)_ in l&&e(!1,_,t[_],h);t=null}}else n&&t&&(t[i]=c);y&&y.responds(g)&&(a=w,y.emit(g,new Z(n,m,f,h)),v&&v.update())}var t,n,r={8:"backspace",9:"tab",13:"enter",16:"shift",17:"control",18:"option",19:"pause",20:"caps-lock",27:"escape",32:"space",35:"end",36:"home",37:"left",38:"up",39:"right",40:"down",46:"delete",91:"command",93:"command",224:"command"},i={9:!0,13:!0,32:!0},o=new s({shift:!1,control:!1,option:!1,command:!1,capsLock:!1,space:!1}),l={},u={};return X.add(document,{keydown:function(t){var a=t.which||t.keyCode;a in r||o.command?e(!0,a,a in i||o.command?a:0,t):n=a},keypress:function(t){null!=n&&(e(!0,n,t.which||t.keyCode,t),n=null)},keyup:function(t){var n=t.which||t.keyCode;n in l&&e(!1,n,l[n],t)}}),X.add(window,{blur:function(t){for(var n in l)e(!1,n,l[n],t)}}),{modifiers:o,isDown:function(e){return!!u[e]}}},J=Y.extend({_class:"MouseEvent",initialize:function(e,t,n,r,i){Y.call(this,t),this.type=e,this.point=n,this.target=r,this.delta=i},toString:function(){return"{ type: '"+this.type+"', point: "+this.point+", target: "+this.target+(this.delta?", delta: "+this.delta:"")+", modifiers: "+this.getModifiers()+" }"}}),ee=Y.extend({_class:"ToolEvent",_item:null,initialize:function(e,t,n){this.tool=e,this.type=t,this.event=n},_choosePoint:function(e,t){return e?e:t?t.clone():null},getPoint:function(){return this._choosePoint(this._point,this.tool._point)},setPoint:function(e){this._point=e},getLastPoint:function(){return this._choosePoint(this._lastPoint,this.tool._lastPoint)},setLastPoint:function(e){this._lastPoint=e},getDownPoint:function(){return this._choosePoint(this._downPoint,this.tool._downPoint)},setDownPoint:function(e){this._downPoint=e},getMiddlePoint:function(){return!this._middlePoint&&this.tool._lastPoint?this.tool._point.add(this.tool._lastPoint).divide(2):this._middlePoint},setMiddlePoint:function(e){this._middlePoint=e},getDelta:function(){return!this._delta&&this.tool._lastPoint?this.tool._point.subtract(this.tool._lastPoint):this._delta;
-},setDelta:function(e){this._delta=e},getCount:function(){return/^mouse(down|up)$/.test(this.type)?this.tool._downCount:this.tool._count},setCount:function(e){this.tool[/^mouse(down|up)$/.test(this.type)?"downCount":"count"]=e},getItem:function(){if(!this._item){var e=this.tool._scope.project.hitTest(this.getPoint());if(e){for(var t=e.item,n=t._parent;/^(Group|CompoundPath)$/.test(n._class);)t=n,n=n._parent;this._item=t}}return this._item},setItem:function(e){this._item=e},toString:function(){return"{ type: "+this.type+", point: "+this.getPoint()+", count: "+this.getCount()+", modifiers: "+this.getModifiers()+" }"}}),te=(c.extend({_class:"Tool",_list:"tools",_reference:"tool",_events:["onActivate","onDeactivate","onEditOptions","onMouseDown","onMouseUp","onMouseDrag","onMouseMove","onKeyDown","onKeyUp"],initialize:function(e){c.call(this),this._firstMove=!0,this._count=0,this._downCount=0,this._set(e)},getMinDistance:function(){return 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strict";var r=n(0),i=n(2),a=new i,o="aligning",s=r.createClass({displayName:"Aligning",getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t)},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t)},align:function(e,t){var n=t.p1.x,r=t.p1.y,i=-Math.atan2(t.p2.y-r,t.p2.x-n),a=Math.cos,o=Math.sin,s=function(e){return{x:(e.x-n)*a(i)-(e.y-r)*o(i),y:(e.x-n)*o(i)+(e.y-r)*a(i)}};return e.map(s)},draw:function(e,t){e.setPanelCount(2),e.reset(),e.drawSkeleton(t),e.drawCurve(t);var n=t.points,r={p1:n[0],p2:n[n.length-1]},i=this.align(n,r),a=new e.Bezier(i),o=e.getPanelWidth(),s=e.getPanelHeight(),l={x:o,y:0};e.setColor("black"),e.drawLine({x:0,y:0},{x:0,y:s},l),l.x+=o/4,l.y+=s/2,e.setColor("grey"),e.drawLine({x:0,y:-s/2},{x:0,y:s/2},l),e.drawLine({x:-o/4,y:0},{x:o,y:0},l),e.setFill("grey"),e.setColor("black"),e.drawSkeleton(a,l),e.drawCurve(a,l)},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(14),a=n(9),o=n(15),s=Math.atan2,l=Math.PI,u=2*l,c=Math.cos,h=Math.sin,d=r.createClass({displayName:"Introduction",statics:{keyHandlingOptions:{propName:"error",values:{38:.1,40:-.1},controller:function(e){e.error<.1&&(e.error=.1)}}},getDefaultProps:function(){return{title:"Approximating Bézier curves with circular arcs"}},setupCircle:function(e){var t=new e.Bezier(70,70,140,40,240,130);e.setCurve(t)},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t)},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t),e.error=.5},getCCenter:function(e,t,n,r){var i,a=n.x-t.x,o=n.y-t.y,d=r.x-n.x,f=r.y-n.y,p=a*c(l/2)-o*h(l/2),m=a*h(l/2)+o*c(l/2),g=d*c(l/2)-f*h(l/2),v=d*h(l/2)+f*c(l/2),w=(t.x+n.x)/2,y=(t.y+n.y)/2,b=(n.x+r.x)/2,_=(n.y+r.y)/2,x=w+p,E=y+m,C=b+g,k=_+v,S=e.utils.lli8(w,y,x,E,b,_,C,k),P=e.utils.dist(S,t),T=s(t.y-S.y,t.x-S.x),N=s(n.y-S.y,n.x-S.x),M=s(r.y-S.y,r.x-S.x);return T<M?((T>N||N>M)&&(T+=u),T>M&&(i=M,M=T,T=i)):M<N&&N<T?(i=M,M=T,T=i):M+=u,S.s=T,S.e=M,S.r=P,S},drawCircle:function(e,t){e.reset();var n=t.points,r=this.getCCenter(e,n[0],n[1],n[2]);e.setColor("grey"),e.drawCircle(r,e.utils.dist(r,n[0])),e.setColor("black"),n.forEach(function(t){return e.drawCircle(t,3)});var i;e.setColor("blue"),e.drawLine(n[0],n[1]),i={x:(n[0].x+n[1].x)/2,y:(n[0].y+n[1].y)/2},e.drawLine(i,{x:r.x+(r.x-i.x),y:r.y+(r.y-i.y)}),e.setColor("red"),e.drawLine(n[1],n[2]),i={x:(n[1].x+n[2].x)/2,y:(n[1].y+n[2].y)/2},e.drawLine(i,{x:r.x+(r.x-i.x),y:r.y+(r.y-i.y)}),e.setColor("green"),e.drawLine(n[2],n[0]),i={x:(n[2].x+n[0].x)/2,y:(n[2].y+n[0].y)/2},e.drawLine(i,{x:r.x+(r.x-i.x),y:r.y+(r.y-i.y)}),e.setColor("black"),e.drawPoint(r),e.setFill("black"),e.text("Intersection point",r,{x:-25,y:10})},drawSingleArc:function(e,t){e.reset();var n=t.arcs(e.error);e.drawSkeleton(t),e.drawCurve(t);var r=n[0];e.setColor("red"),e.setFill("rgba(255,0,0,0.2)"),e.debug=!0,e.drawArc(r),e.setFill("black"),e.text("Arc approximation with total error "+e.utils.round(e.error,1),{x:10,y:15})},drawArcs:function(e,t){e.reset();var n=t.arcs(e.error);e.drawSkeleton(t),e.drawCurve(t),n.forEach(function(t){e.setRandomColor(.3),e.setFill(e.getColor()),e.drawArc(t)}),e.setFill("black"),e.text("Arc approximation with total error "+e.utils.round(e.error,1)+" per arc segment",{x:10,y:15})},render:function(){return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"Let's look at doing the exact opposite of the previous section: rather than approximating circular arc using Bézier curves, let's approximate Bézier curves using circular arcs."),r.createElement("p",null,"We already saw in the section on circle approximation that this will never yield a perfect equivalent, but sometimes you need circular arcs, such as when you're working with fabrication machinery, or simple vector languages that understand lines and circles, but not much else."),r.createElement("p",null,'The approach is fairly simple: pick a starting point on the curve, and pick two points that are further along the curve. Determine the circle that goes through those three points, and see if it fits the part of the curve we\'re trying to approximate. Decent fit? Try spacing the points further apart. Bad fit? Try spacing the points closer together. Keep doing this until you\'ve found the "good approximation/bad approximation" boundary, record the "good" arc, and then move the starting point up to overlap the end point we previously found. Rinse and repeat until we\'ve covered the entire curve.'),r.createElement("p",null,"So: step 1, how do we find a circle through three points? That part is actually really simple. You may remember (if you ever learned it!) that a line between two points on a circle is called a ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Chord_%28geometry%29"},"chord"),", and one property of chords is that the line from the center of any chord, perpendicular to that chord, passes through the center of the circle."),r.createElement("p",null,"So: if we have have three points, we have three (different) chords, and consequently, three (different) lines that go from those chords through the center of the circle. So we find the centers of the chords, find the perpendicular lines, find the intersection of those lines, and thus find the center of the circle."),r.createElement("p",null,"The following graphic shows this procedure with a different colour for each chord and its associated perpendicular through the center. You can move the points around as much as you like, those lines will always meet!"),r.createElement(i,{preset:"simple",title:"Finding a circle through three points",setup:this.setupCircle,draw:this.drawCircle}),r.createElement("p",null,"So, with the procedure on how to find a circle through three points, finding the arc through those points is straight-forward: pick one of the three points as start point, pick another as an end point, and the arc has to necessarily go from the start point, over the remaining point, to the end point."),r.createElement("p",null,"So how can we convert a Bezier curve into a (sequence of) circular arc(s)?"),r.createElement("ul",null,r.createElement("li",null,"Start at ",r.createElement("em",null,"t=0")),r.createElement("li",null,"Pick two points further down the curve at some value ",r.createElement("em",null,"m = t + n")," and ",r.createElement("em",null,"e = t + 2n")),r.createElement("li",null,"Find the arc that these points define"),r.createElement("li",null,"Determine how close the found arc is to the curve:",r.createElement("ul",null,r.createElement("li",null,"Pick two additional points ",r.createElement("em",null,"e1 = t + n/2")," and ",r.createElement("em",null,"e2 = t + n + n/2"),"."),r.createElement("li",null,"These points, if the arc is a good approximation of the curve interval chosen, should lie ",r.createElement("em",null,"on")," the circle, so their distance to the center of the circle should be the same as the distance from any of the three other points to the center."),r.createElement("li",null,"For point points, determine the (absolute) error between the radius of the circle, and the",r.createElement("em",null,"actual")," distance from the center of the circle to the point on the curve."),r.createElement("li",null,"If this error is too high, we consider the arc bad, and try a smaller interval.")))),r.createElement("p",null,"The result of this is shown in the next graphic: we start at a guaranteed failure: s=0, e=1. That's the entire curve. The midpoint is simply at ",r.createElement("em",null,"t=0.5"),", and then we start performing a ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Binary_search_algorithm"},"Binary Search"),"."),r.createElement("ol",null,r.createElement("li",null,"We start with ",1),r.createElement("li",null,"That'll fail, so we retry with the interval halved: ",.5),r.createElement("ul",null,r.createElement("li",null,"If that arc's good, we move back up by half distance: ",.75,"."),r.createElement("li",null,"However, if the arc was still bad, we move ",r.createElement("em",null,"down")," by half the distance: ",.25,".")),r.createElement("li",null,"We keep doing this over and over until we have two arcs found in sequence of which the first arc is good, and the second arc is bad. When we find that pair, we've found the boundary between a good approximation and a bad approximation, and we pick the former")),r.createElement("p",null,"The following graphic shows the result of this approach, with a default error threshold of 0.5, meaning that if an arc is off by a ",r.createElement("em",null,"combined")," half pixel over both verification points, then we treat the arc as bad. This is an extremely simple error policy, but already works really well. Note that the graphic is still interactive, and you can use your up and down arrow keys keys to increase or decrease the error threshold, to see what the effect of a smaller or larger error threshold is."),r.createElement(i,{preset:"simple",title:"Arc approximation of a Bézier curve",setup:this.setupCubic,draw:this.drawSingleArc,onKeyDown:this.props.onKeyDown}),r.createElement("p",null,"With that in place, all that's left now is to \"restart\" the procedure by treating the found arc's end point as the new to-be-determined arc's starting point, and using points further down the curve. We keep trying this until the found end point is for ",r.createElement("em",null,"t=1"),", at which point we are done. Again, the following graphic allows for up and down arrow key input to increase or decrease the error threshold, so you can see how picking a different threshold changes the number of arcs that are necessary to reasonably approximate a curve:"),r.createElement(i,{preset:"simple",title:"Arc approximation of a Bézier curve",setup:this.setupCubic,draw:this.drawArcs,onKeyDown:this.props.onKeyDown}),r.createElement("p",null,'So... what is this good for? Obviously, If you\'re working with technologies that can\'t do curves, but can do lines and circles, then the answer is pretty straight-forward, but what else? There are some reasons why you might need this technique: using circular arcs means you can determine whether a coordinate lies "on" your curve really easily: simply compute the distance to each circular arc center, and if any of those are close to the arc radii, at an angle betwee the arc start and end: bingo, this point can be treated as lying "on the curve". Another benefit is that this approximation is "linear": you can almost trivially travel along the arcs at fixed speed. You can also trivially compute the arc length of the approximated curve (it\'s a bit like curve flattening). The only thing to bear in mind is that this is a lossy equivalence: things that you compute based on the approximation are guaranteed "off" by some small value, and depending on how much precision you need, arc approximation is either going to be super useful, or completely useless. It\'s up to you to decide which, based on your application!'))}});e.exports=o(d)},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="arclength",s=Math.sin,l=2*Math.PI,u=r.createClass({displayName:"Arclength",getDefaultProps:function(){return{title:a.getTitle(o)}},setup:function(e){var t,n=e.getPanelWidth(),r=e.getPanelHeight();this.generator||(t=function(e,t){return t=t||1,{x:e*n/l,y:t*s(e)}},t.start=0,t.end=l,t.step=.1,t.scale=r/3,this.generator=t)},drawSine:function(e,t){var n=e.getPanelWidth(),r=e.getPanelHeight(),i=this.generator;i.dheight=t,e.setColor("black"),e.drawLine({x:0,y:r/2},{x:n,y:r/2}),e.drawFunction(i,{x:0,y:r/2})},drawSlices:function(e,t){var n=e.getPanelWidth(),r=e.getPanelHeight(),i=n/l,a=0,o=t<=25?1:0;e.reset(),e.setColor("transparent"),e.setFill("rgba(150,150,255, 0.4)");for(var 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i,o="arclengthapprox",s=n(15),l=r.createClass({displayName:"ArclengthApprox",statics:{keyHandlingOptions:{propName:"steps",values:{38:1,40:-1},controller:function(e){e.steps<1&&(e.steps=1)}}},getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t),e.steps=10},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t),e.steps=16},draw:function(e,t){e.reset(),e.drawSkeleton(t);for(var n,r=t.getLUT(e.steps),i=1/e.steps,a=t.points[0],o=i;o<1+i;o+=i)n=t.get(Math.min(o,1)),e.setColor("red"),e.drawLine(a,n),a=n;for(var s,l,u,c=t.length(),h=0,d=0;d<r.length-1;d++)a=r[d],s=r[d+1],l=s.x-a.x,u=s.y-a.y,h+=Math.sqrt(l*l+u*u);h=(100*h|0)/100,c=(100*c|0)/100,e.text("Approximate length, "+e.steps+" steps: "+h+" (true: "+c+")",{x:10,y:15})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s(l)},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="boundingbox",s=r.createClass({displayName:"BoundingBox",getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t)},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t)},draw:function(e,t){e.reset(),e.setColor("#00FF00"),e.drawbbox(t.bbox()),e.setColor("black"),e.drawSkeleton(t),e.drawCurve(t),e.setColor("red"),t.extrema().values.forEach(function(n){e.drawCircle(t.get(n),3)})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,setup:function(){this.size(600,300),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,setup:function(){this.size(400,400);for(var e=2*Math.PI,t=0;t<e;t+=e/9)this.points.push({x:this.width/2+100*Math.cos(t),y:this.height/2+100*Math.sin(t)});this.knots=this.formKnots(this.points);var n=0|Math.round(this.points.length/2);this.knots[n+0]=this.knots[n],this.knots[n+1]=this.knots[n],this.knots[n+2]=this.knots[n];for(var r=n+3;r<this.knots.length;r++)this.knots[r]=this.knots[r-1]+1;this.props.controller&&this.props.controller(this,this.knots),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";var r=n(0),i=n(107),a=n(9),o=n(111),s=n(116),l=r.createClass({displayName:"BoundingBox",getDefaultProps:function(){return{title:"B-Splines"}},bindKnots:function(e,t,n){this.refs[n].bindKnots(e,t)},bindWeights:function(e,t,n,r){this.refs[r].bindWeights(e,t,n)},render:function(){var e=this;return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"No discussion on Bézier curves is complete without also giving mention of that other beast in the curve design space: B-Splines. Easily confused to mean Bézier splines, that's not actually what they are; they are \"basis function\" splines, which makes a lot of difference, which we'll be looking at in this section. We're not going to dive as deep into B-Splines as we have for Bézier curves (that would be an entire primer on its own) but we'll be looking at how B-Splines work, what kind of maths is involved in computing them, and how to draw them based on a number of parameters that you can pick for individual B-Splines."),r.createElement("p",null,"First off: B-Splines are ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Piecewise"},'"piecewise polynomial interpolation curves"'),', where the "single curve" is built by performing polynomial interpolation over a set of points, using a sliding window of a fixed number of points. For instance, a "cubic" B-Spline defined by twelve points will have its curve built by evaluating the polynomial interpolation of four points, and the curve can be treated as a lot of different sections, each controlled by four points at a time, such that the full curve consists of smoothly connected sections defined by points ',"{","1,2,3,4","}",", ","{","2,3,4,5","}",", ..., ","{","8,9,10,11","}",", and finally ","{","9,10,11,12","}",", for eight sections."),r.createElement("p",null,"What do they look like? They look like this! .. okay that's an empty graph, but simply click to place some point, with the stipulation that you need at least four point to see any curve. More than four points simply draws a longer B-Spline curve:"),r.createElement(i,{sketch:n(125)}),r.createElement("p",null,"The important part to notice here is that we are ",r.createElement("strong",null,"not"),' doing the same thing with B-Splines that we do for poly-Béziers or Catmull-Rom curves: both of the latter simply define new sections as literally "new sections based on new points", so a 12 point cubic poly-Bézier curve is actually impossible, because we start with a four point curve, and then add three more points for each section that follows, so we can only have 4, 7, 10, 13, 16, etc point Poly-Béziers. Similarly, while Catmull-Rom curves can grow by adding single points, this addition of a single point introduces three implicit Bézier points. Cubic B-Splines, on the other hand, are smooth interpolations of ',r.createElement("em",null,"each possible curve involving four consecutive points"),", such that at any point along the curve except for our start and end points, our on-curve coordinate is defined by four control points."),r.createElement("p",null,"Consider the difference to be this:"),r.createElement("ul",null,r.createElement("li",null,"for Bézier curves, the curve is defined as an interpolation of points, but:"),r.createElement("li",null,"for B-Splines, the curve is defined as an interpolation of ",r.createElement("em",null,"curves"),".")),r.createElement("p",null,"In order to make this interpolation of curves work, the maths is necessarily more complex than the maths for Bézier curves, so let's have a look at how things work."),r.createElement("h2",null,"How to compute a B-Spline curve: some maths"),r.createElement("p",null,"Given a B-Spline of degree ",r.createElement("code",null,"d")," and thus order ",r.createElement("code",null,"k=d+1")," (so a quadratic B-Spline is degree 2 and order 3, a cubic B-Spline is degree 3 and order 4, etc) and ",r.createElement("code",null,"n")," control points ",r.createElement("code",null,"P",r.createElement("sub",null,"0"))," through ",r.createElement("code",null,"P",r.createElement("sub",null,"n-1")),", we can compute a point on the curve for some value ",r.createElement("code",null,"t")," in the interval [0,1] (where 0 is the start of the curve, and 1 the end, just like for Bézier curves), by evaluting the following function:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/329da80e737b0005f4dbe4c84ff868bde5dfaee0.svg",width:"177.79999999999998rem",height:"43.4rem"})),r.createElement("p",null,'Which, honestly, doesn\'t tell us all that much. All we can see is that a point on a B-Spline curve is defined as "a mix of all the control points, weighted somehow", where the weighting is achieved through the ',r.createElement("em",null,"N(...)")," function, subscipted with an obvious parameter ",r.createElement("code",null,"i"),", which comes from our summation, and some magical parameter ",r.createElement("code",null,"k"),". So we need to know two things: 1. what does N(t) do, and 2. what is that ",r.createElement("code",null,"k"),"? Let's cover both, in reverse order."),r.createElement("p",null,"The parameter ",r.createElement("code",null,"k"),' represents the "knot interval" over which a section of curve is defined. As we learned earlier, a B-Spline curve is itself an interpoliation of curves, and we can treat each transition where a control point starts or tops influencing the total curvature as a "knot on the curve". Doing so for a degree ',r.createElement("code",null,"d")," B-Spline with ",r.createElement("code",null,"n")," control point gives us ",r.createElement("code",null,"d + n + 1")," knots, defining ",r.createElement("code",null,"d + n")," intervals along the curve, and it is these intervals that the above ",r.createElement("code",null,"k")," subscript to the N() function applies to."),r.createElement("p",null,"Then the N() function itself. What does it look like?"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/c10575fb591062784484357356796a4c0be4f83e.svg",width:"588rem",height:"44.8rem"})),r.createElement("p",null,"So this is where we see the interpolation: N(t) for an (i,k) pair (that is, for a step in the above summation, on a specific knot interval) is a mix between N(t) for (i,k-1) and N(t) for (i+1,k-1), so we see that this is a recursive iteration where ",r.createElement("code",null,"i")," goes up, and ",r.createElement("code",null,"k")," goes down, so it seem reasonable to expect that this recursion has to stop at some point; obviously, it does, and specifically it does so for the following ",r.createElement("code",null,"i"),"/",r.createElement("code",null,"k")," values:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/6664a4fc5832059bbc68eaa8068a4b2577e1d96a.svg",width:"251.99999999999997rem",height:"42rem"})),r.createElement("p",null,"And this function finally has a straight up evaluation: if a ",r.createElement("code",null,"t")," value lies within a knot-specific interval once we reach a ",r.createElement("code",null,"k=1"),' value, it "counts", otherwise it doesn\'t. We did cheat a little, though, because for all these values we need to scale our ',r.createElement("code",null,"t")," value first, so that it lies in the interval bounded by ",r.createElement("code",null,"knots[d]")," and ",r.createElement("code",null,"knots[n]"),", which are the start point and end point where curvature is controlled by exactly ",r.createElement("code",null,"order")," control points. For instance, for degree 3 (=order 4) and 7 control points, with knot vector [1,2,3,4,5,6,7,8,9,10,11], we map ",r.createElement("code",null,"t")," from [the interval 0,1] to the interval [4,8], and then use that value in the functions above, instead."),r.createElement("h2",null,"Can we simplify that?"),r.createElement("p",null,"We can, yes."),r.createElement("p",null,"People far smarter than us have looked at this work, and two in particular —",r.createElement("a",{href:"http://www.npl.co.uk/people/maurice-cox"},"Maurice Cox")," and ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Carl_R._de_Boor"},"Carl de Boor"),"— came to a mathematically pleasing solution: to compute a point P(t), we can compute this point by evaluating ",r.createElement("em",null,"d(t)")," on a curve section between knots ",r.createElement("em",null,"i")," and ",r.createElement("em",null,"i+1"),":"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/3780a420cd9b1bc59bec2c49bbd29f5e58497a3c.svg",width:"295.4rem",height:"22.4rem"})),"This is another recursive function, with ",r.createElement("em",null,"k")," values decreasing from the curve order to 1, and the value ",r.createElement("em",null,"α")," (alpha) defined by:",r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/ba3b25cd54993b4601d8f415bc4cde73af4fc460.svg",width:"267.4rem",height:"40.599999999999994rem"})),r.createElement("p",null,'That looks complicated, but it\'s not. Computing alpha is just a fraction involving known, plain numbers and once we have our alpha value, computing (1-alpha) is literally just "computing one minus alpha". Computing this d() function is thus simply a matter of "computing simple arithmetics but with recursion", which might be computationally expensive because we\'re doing "a lot of" steps, but is also computationally cheap because each step only involves very simple maths. Of course as before the recursion has to stop:'),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/1405067abebab73574934e3e69d7a7158106c744.svg",width:"387.79999999999995rem",height:"42rem"})),r.createElement("p",null,"So, we see two stopping conditions: either ",r.createElement("code",null,"i")," becomes 0, in which case d() is zero, or ",r.createElement("code",null,"k"),' becomes zero, in which case we get the same "either 1 or 0" that we saw in the N() function above.'),r.createElement("p",null,"Thanks to Cox and de Boor, we can compute points on a B-Spline pretty easily: we just need to compute a triangle of interconnected values. For instance, d() for i=3, k=3 yields the following triangle:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/a0a1069b001c75a1fab7f40ffa8bc403e1408f0d.svg",width:"438.2rem",height:"242.2rem"})),r.createElement("p",null,"That is, we compute d(3,3) as a mixture of d(2,3) and d(2,2): d(3,3) = a(3,3) x d(2,3) + (1-a(3,3)) x d(2,2)... and we simply keep expanding our triangle until we reach the terminating function parameters. Done deal!"),r.createElement("p",null,"One thing we need to keep in mind is that we're working with a spline that is contrained by its control points, so even though the ",r.createElement("code",null,"d(..., k)")," values are zero or one at the lowest level, they are really \"zero or one, times their respective control point\", so in the next section you'll see the algorithm for running through the computation in a way that starts with a copy of the control point vector and then works its way up to that single point: that's pretty essential!"),r.createElement("p",null,'If we run this computation "down", starting at d(3,3), then without special code in place we would be computing quite a few terms multiple times at each step. On the other hand, we can also start with that last "column", we can generate the terminating d() values first, then compute the a() constants, perform our multiplcations, generate the previous step\'s d() values, compute their a() constants, do the multiplications, etc. until we end up all the way back at the top. If we run our computation this way, we don\'t need any explicit caching, we can just "recycle" the list of numbers we start with and simply update them as we move up the triangle. So, let\'s implement that!'),r.createElement("h2",null,"Cool, cool... but I don't know what to do with that information"),r.createElement("p",null,"I know, this is pretty mathy, so let's have a look at what happens when we change parameters here. We can't change the maths for the interpolation functions, so that gives us only one way to control what happens here: the knot vector itself. As such, let's look at the graph that shows the interpolation functions for a cubic B-Spline with seven points with a uniform knot vector (so we see seven identical functions), representing how much each point (represented by one function each) influences the total curvature, given our knot values. And, because exploration is the key to discovery, let's make the knot vector a thing we can actually manipulate. Normally a proper knot vector has a constraint that any value is strictly equal to, or larger than the previous ones, but screw it this is programming, let's ignore that hard restriction and just mess with the knots however we like."),r.createElement("div",{className:"two-column"},r.createElement(o,{ref:"interpolation-graph"}),r.createElement(i,{sketch:n(128),controller:function(t,n){return e.bindKnots(t,n,"interpolation-graph")}})),r.createElement("p",null,"Changing the values in the knot vector changes how much each point influences the total curvature (with some clever knot value manipulation, we can even make the influence of certain points disappear entirely!), so we can see that while the control points define the hull inside of which we're going to be drawing a curve, it is actually the knot vector that determines the actual ",r.createElement("em",null,"shape")," of the curve inside that hull."),r.createElement("p",null,"After reading the rest of this section you may want to come back here to try some specific knot vectors, and see if the resulting interpolation landscape makes sense given what you will now think should happen!"),r.createElement("h2",null,"Running the computation"),r.createElement("p",null,"Unlike the de Casteljau algorithm, where the ",r.createElement("code",null,"t")," value stays the same at every iteration, for B-Splines that is not the case, and so we end having to (for each point we evaluate) run a fairly involving bit of recursive computation. The algorithm is discussed on ",r.createElement("a",{href:"http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/de-Boor.html"
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Determine the circle that goes through those three points, and see if it fits the part of the curve we\'re trying to approximate. Decent fit? Try spacing the points further apart. Bad fit? Try spacing the points closer together. Keep doing this until you\'ve found the "good approximation/bad approximation" boundary, record the "good" arc, and then move the starting point up to overlap the end point we previously found. Rinse and repeat until we\'ve covered the entire curve.'),r.createElement("p",null,"So: step 1, how do we find a circle through three points? That part is actually really simple. You may remember (if you ever learned it!) that a line between two points on a circle is called a ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Chord_%28geometry%29"},"chord"),", and one property of chords is that the line from the center of any chord, perpendicular to that chord, passes through the center of the circle."),r.createElement("p",null,"So: if we have have three points, we have three (different) chords, and consequently, three (different) lines that go from those chords through the center of the circle. So we find the centers of the chords, find the perpendicular lines, find the intersection of those lines, and thus find the center of the circle."),r.createElement("p",null,"The following graphic shows this procedure with a different colour for each chord and its associated perpendicular through the center. You can move the points around as much as you like, those lines will always meet!"),r.createElement(i,{preset:"simple",title:"Finding a circle through three points",setup:this.setupCircle,draw:this.drawCircle}),r.createElement("p",null,"So, with the procedure on how to find a circle through three points, finding the arc through those points is straight-forward: pick one of the three points as start point, pick another as an end point, and the arc has to necessarily go from the start point, over the remaining point, to the end point."),r.createElement("p",null,"So how can we convert a Bezier curve into a (sequence of) circular arc(s)?"),r.createElement("ul",null,r.createElement("li",null,"Start at ",r.createElement("em",null,"t=0")),r.createElement("li",null,"Pick two points further down the curve at some value ",r.createElement("em",null,"m = t + n")," and ",r.createElement("em",null,"e = t + 2n")),r.createElement("li",null,"Find the arc that these points define"),r.createElement("li",null,"Determine how close the found arc is to the curve:",r.createElement("ul",null,r.createElement("li",null,"Pick two additional points ",r.createElement("em",null,"e1 = t + n/2")," and ",r.createElement("em",null,"e2 = t + n + n/2"),"."),r.createElement("li",null,"These points, if the arc is a good approximation of the curve interval chosen, should lie ",r.createElement("em",null,"on")," the circle, so their distance to the center of the circle should be the same as the distance from any of the three other points to the center."),r.createElement("li",null,"For point points, determine the (absolute) error between the radius of the circle, and the",r.createElement("em",null,"actual")," distance from the center of the circle to the point on the curve."),r.createElement("li",null,"If this error is too high, we consider the arc bad, and try a smaller interval.")))),r.createElement("p",null,"The result of this is shown in the next graphic: we start at a guaranteed failure: s=0, e=1. That's the entire curve. The midpoint is simply at ",r.createElement("em",null,"t=0.5"),", and then we start performing a ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Binary_search_algorithm"},"Binary Search"),"."),r.createElement("ol",null,r.createElement("li",null,"We start with ",1),r.createElement("li",null,"That'll fail, so we retry with the interval halved: ",.5),r.createElement("ul",null,r.createElement("li",null,"If that arc's good, we move back up by half distance: ",.75,"."),r.createElement("li",null,"However, if the arc was still bad, we move ",r.createElement("em",null,"down")," by half the distance: ",.25,".")),r.createElement("li",null,"We keep doing this over and over until we have two arcs found in sequence of which the first arc is good, and the second arc is bad. When we find that pair, we've found the boundary between a good approximation and a bad approximation, and we pick the former")),r.createElement("p",null,"The following graphic shows the result of this approach, with a default error threshold of 0.5, meaning that if an arc is off by a ",r.createElement("em",null,"combined")," half pixel over both verification points, then we treat the arc as bad. This is an extremely simple error policy, but already works really well. Note that the graphic is still interactive, and you can use your up and down arrow keys keys to increase or decrease the error threshold, to see what the effect of a smaller or larger error threshold is."),r.createElement(i,{preset:"simple",title:"Arc approximation of a Bézier curve",setup:this.setupCubic,draw:this.drawSingleArc,onKeyDown:this.props.onKeyDown}),r.createElement("p",null,"With that in place, all that's left now is to \"restart\" the procedure by treating the found arc's end point as the new to-be-determined arc's starting point, and using points further down the curve. We keep trying this until the found end point is for ",r.createElement("em",null,"t=1"),", at which point we are done. Again, the following graphic allows for up and down arrow key input to increase or decrease the error threshold, so you can see how picking a different threshold changes the number of arcs that are necessary to reasonably approximate a curve:"),r.createElement(i,{preset:"simple",title:"Arc approximation of a Bézier curve",setup:this.setupCubic,draw:this.drawArcs,onKeyDown:this.props.onKeyDown}),r.createElement("p",null,'So... what is this good for? Obviously, If you\'re working with technologies that can\'t do curves, but can do lines and circles, then the answer is pretty straight-forward, but what else? There are some reasons why you might need this technique: using circular arcs means you can determine whether a coordinate lies "on" your curve really easily: simply compute the distance to each circular arc center, and if any of those are close to the arc radii, at an angle betwee the arc start and end: bingo, this point can be treated as lying "on the curve". Another benefit is that this approximation is "linear": you can almost trivially travel along the arcs at fixed speed. You can also trivially compute the arc length of the approximated curve (it\'s a bit like curve flattening). The only thing to bear in mind is that this is a lossy equivalence: things that you compute based on the approximation are guaranteed "off" by some small value, and depending on how much precision you need, arc approximation is either going to be super useful, or completely useless. It\'s up to you to decide which, based on your application!'))}});e.exports=o(d)},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="arclength",s=Math.sin,l=2*Math.PI,u=r.createClass({displayName:"Arclength",getDefaultProps:function(){return{title:a.getTitle(o)}},setup:function(e){var t,n=e.getPanelWidth(),r=e.getPanelHeight();this.generator||(t=function(e,t){return t=t||1,{x:e*n/l,y:t*s(e)}},t.start=0,t.end=l,t.step=.1,t.scale=r/3,this.generator=t)},drawSine:function(e,t){var n=e.getPanelWidth(),r=e.getPanelHeight(),i=this.generator;i.dheight=t,e.setColor("black"),e.drawLine({x:0,y:r/2},{x:n,y:r/2}),e.drawFunction(i,{x:0,y:r/2})},drawSlices:function(e,t){var n=e.getPanelWidth(),r=e.getPanelHeight(),i=n/l,a=0,o=t<=25?1:0;e.reset(),e.setColor("transparent"),e.setFill("rgba(150,150,255, 0.4)");for(var s,u,c,h=l/t,d=h/2;d<l+h/2;d+=h)s=this.generator(d),u={x:s.x-i*h/2+o,y:0},c={x:s.x+i*h/2-o,y:s.y*this.generator.scale},o||e.setFill("rgba(150,150,255,"+(.4+.3*Math.random())+")"),e.drawRect(u,c,{x:0,y:r/2}),a+=h*Math.abs(s.y*this.generator.scale);e.setFill("black");var f=(400*r/3|0)/100,p=(100*a|0)/100;e.text("Approximating with "+t+" strips (true area: "+f+"): "+p,{x:10,y:r-15})},drawCoarseIntegral:function(e){e.reset(),this.drawSlices(e,10),this.drawSine(e)},drawFineIntegral:function(e){e.reset(),this.drawSlices(e,24),this.drawSine(e)},drawSuperFineIntegral:function(e){e.reset(),this.drawSlices(e,99),this.drawSine(e)},setupCurve:function(e){var t=e.getDefaultCubic();e.setCurve(t)},drawCurve:function(e,t){e.reset(),e.drawSkeleton(t),e.drawCurve(t);var n=t.length();e.setFill("black"),e.text("Curve length: "+n+" pixels",{x:10,y:15})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=u},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="arclengthapprox",s=n(15),l=r.createClass({displayName:"ArclengthApprox",statics:{keyHandlingOptions:{propName:"steps",values:{38:1,40:-1},controller:function(e){e.steps<1&&(e.steps=1)}}},getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t),e.steps=10},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t),e.steps=16},draw:function(e,t){e.reset(),e.drawSkeleton(t);for(var n,r=t.getLUT(e.steps),i=1/e.steps,a=t.points[0],o=i;o<1+i;o+=i)n=t.get(Math.min(o,1)),e.setColor("red"),e.drawLine(a,n),a=n;for(var s,l,u,c=t.length(),h=0,d=0;d<r.length-1;d++)a=r[d],s=r[d+1],l=s.x-a.x,u=s.y-a.y,h+=Math.sqrt(l*l+u*u);h=(100*h|0)/100,c=(100*c|0)/100,e.text("Approximate length, "+e.steps+" steps: "+h+" (true: "+c+")",{x:10,y:15})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s(l)},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="boundingbox",s=r.createClass({displayName:"BoundingBox",getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t)},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t)},draw:function(e,t){e.reset(),e.setColor("#00FF00"),e.drawbbox(t.bbox()),e.setColor("black"),e.drawSkeleton(t),e.drawCurve(t),e.setColor("red"),t.extrema().values.forEach(function(n){e.drawCircle(t.get(n),3)})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,setup:function(){this.size(600,300),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,setup:function(){this.size(400,400);for(var e=2*Math.PI,t=0;t<e;t+=e/9)this.points.push({x:this.width/2+100*Math.cos(t),y:this.height/2+100*Math.sin(t)});this.knots=this.formKnots(this.points);var n=0|Math.round(this.points.length/2);this.knots[n+0]=this.knots[n],this.knots[n+1]=this.knots[n],this.knots[n+2]=this.knots[n];for(var r=n+3;r<this.knots.length;r++)this.knots[r]=this.knots[r-1]+1;this.props.controller&&this.props.controller(this,this.knots),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";var r=n(0),i=n(107),a=n(9),o=n(111),s=n(116),l=r.createClass({displayName:"BoundingBox",getDefaultProps:function(){return{title:"B-Splines"}},bindKnots:function(e,t,n){this.refs[n].bindKnots(e,t)},bindWeights:function(e,t,n,r){this.refs[r].bindWeights(e,t,n)},render:function(){var e=this;return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"No discussion on Bézier curves is complete without also giving mention of that other beast in the curve design space: B-Splines. Easily confused to mean Bézier splines, that's not actually what they are; they are \"basis function\" splines, which makes a lot of difference, which we'll be looking at in this section. We're not going to dive as deep into B-Splines as we have for Bézier curves (that would be an entire primer on its own) but we'll be looking at how B-Splines work, what kind of maths is involved in computing them, and how to draw them based on a number of parameters that you can pick for individual B-Splines."),r.createElement("p",null,"First off: B-Splines are ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Piecewise"},'"piecewise polynomial interpolation curves"'),', where the "single curve" is built by performing polynomial interpolation over a set of points, using a sliding window of a fixed number of points. For instance, a "cubic" B-Spline defined by twelve points will have its curve built by evaluating the polynomial interpolation of four points, and the curve can be treated as a lot of different sections, each controlled by four points at a time, such that the full curve consists of smoothly connected sections defined by points ',"{","1,2,3,4","}",", ","{","2,3,4,5","}",", ..., ","{","8,9,10,11","}",", and finally ","{","9,10,11,12","}",", for eight sections."),r.createElement("p",null,"What do they look like? They look like this! .. okay that's an empty graph, but simply click to place some point, with the stipulation that you need at least four point to see any curve. More than four points simply draws a longer B-Spline curve:"),r.createElement(i,{sketch:n(125)}),r.createElement("p",null,"The important part to notice here is that we are ",r.createElement("strong",null,"not"),' doing the same thing with B-Splines that we do for poly-Béziers or Catmull-Rom curves: both of the latter simply define new sections as literally "new sections based on new points", so a 12 point cubic poly-Bézier curve is actually impossible, because we start with a four point curve, and then add three more points for each section that follows, so we can only have 4, 7, 10, 13, 16, etc point Poly-Béziers. Similarly, while Catmull-Rom curves can grow by adding single points, this addition of a single point introduces three implicit Bézier points. Cubic B-Splines, on the other hand, are smooth interpolations of ',r.createElement("em",null,"each possible curve involving four consecutive points"),", such that at any point along the curve except for our start and end points, our on-curve coordinate is defined by four control points."),r.createElement("p",null,"Consider the difference to be this:"),r.createElement("ul",null,r.createElement("li",null,"for Bézier curves, the curve is defined as an interpolation of points, but:"),r.createElement("li",null,"for B-Splines, the curve is defined as an interpolation of ",r.createElement("em",null,"curves"),".")),r.createElement("p",null,"In order to make this interpolation of curves work, the maths is necessarily more complex than the maths for Bézier curves, so let's have a look at how things work."),r.createElement("h2",null,"How to compute a B-Spline curve: some maths"),r.createElement("p",null,"Given a B-Spline of degree ",r.createElement("code",null,"d")," and thus order ",r.createElement("code",null,"k=d+1")," (so a quadratic B-Spline is degree 2 and order 3, a cubic B-Spline is degree 3 and order 4, etc) and ",r.createElement("code",null,"n")," control points ",r.createElement("code",null,"P",r.createElement("sub",null,"0"))," through ",r.createElement("code",null,"P",r.createElement("sub",null,"n-1")),", we can compute a point on the curve for some value ",r.createElement("code",null,"t")," in the interval [0,1] (where 0 is the start of the curve, and 1 the end, just like for Bézier curves), by evaluting the following function:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/329da80e737b0005f4dbe4c84ff868bde5dfaee0.svg",width:"177.79999999999998rem",height:"43.4rem"})),r.createElement("p",null,'Which, honestly, doesn\'t tell us all that much. All we can see is that a point on a B-Spline curve is defined as "a mix of all the control points, weighted somehow", where the weighting is achieved through the ',r.createElement("em",null,"N(...)")," function, subscipted with an obvious parameter ",r.createElement("code",null,"i"),", which comes from our summation, and some magical parameter ",r.createElement("code",null,"k"),". So we need to know two things: 1. what does N(t) do, and 2. what is that ",r.createElement("code",null,"k"),"? Let's cover both, in reverse order."),r.createElement("p",null,"The parameter ",r.createElement("code",null,"k"),' represents the "knot interval" over which a section of curve is defined. As we learned earlier, a B-Spline curve is itself an interpoliation of curves, and we can treat each transition where a control point starts or tops influencing the total curvature as a "knot on the curve". Doing so for a degree ',r.createElement("code",null,"d")," B-Spline with ",r.createElement("code",null,"n")," control point gives us ",r.createElement("code",null,"d + n + 1")," knots, defining ",r.createElement("code",null,"d + n")," intervals along the curve, and it is these intervals that the above ",r.createElement("code",null,"k")," subscript to the N() function applies to."),r.createElement("p",null,"Then the N() function itself. What does it look like?"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/c10575fb591062784484357356796a4c0be4f83e.svg",width:"588rem",height:"44.8rem"})),r.createElement("p",null,"So this is where we see the interpolation: N(t) for an (i,k) pair (that is, for a step in the above summation, on a specific knot interval) is a mix between N(t) for (i,k-1) and N(t) for (i+1,k-1), so we see that this is a recursive iteration where ",r.createElement("code",null,"i")," goes up, and ",r.createElement("code",null,"k")," goes down, so it seem reasonable to expect that this recursion has to stop at some point; obviously, it does, and specifically it does so for the following ",r.createElement("code",null,"i"),"/",r.createElement("code",null,"k")," values:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/6664a4fc5832059bbc68eaa8068a4b2577e1d96a.svg",width:"251.99999999999997rem",height:"42rem"})),r.createElement("p",null,"And this function finally has a straight up evaluation: if a ",r.createElement("code",null,"t")," value lies within a knot-specific interval once we reach a ",r.createElement("code",null,"k=1"),' value, it "counts", otherwise it doesn\'t. We did cheat a little, though, because for all these values we need to scale our ',r.createElement("code",null,"t")," value first, so that it lies in the interval bounded by ",r.createElement("code",null,"knots[d]")," and ",r.createElement("code",null,"knots[n]"),", which are the start point and end point where curvature is controlled by exactly ",r.createElement("code",null,"order")," control points. For instance, for degree 3 (=order 4) and 7 control points, with knot vector [1,2,3,4,5,6,7,8,9,10,11], we map ",r.createElement("code",null,"t")," from [the interval 0,1] to the interval [4,8], and then use that value in the functions above, instead."),r.createElement("h2",null,"Can we simplify that?"),r.createElement("p",null,"We can, yes."),r.createElement("p",null,"People far smarter than us have looked at this work, and two in particular —",r.createElement("a",{href:"http://www.npl.co.uk/people/maurice-cox"},"Maurice Cox")," and ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Carl_R._de_Boor"},"Carl de Boor"),"— came to a mathematically pleasing solution: to compute a point P(t), we can compute this point by evaluating ",r.createElement("em",null,"d(t)")," on a curve section between knots ",r.createElement("em",null,"i")," and ",r.createElement("em",null,"i+1"),":"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/3780a420cd9b1bc59bec2c49bbd29f5e58497a3c.svg",width:"295.4rem",height:"22.4rem"})),"This is another recursive function, with ",r.createElement("em",null,"k")," values decreasing from the curve order to 1, and the value ",r.createElement("em",null,"α")," (alpha) defined by:",r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/ba3b25cd54993b4601d8f415bc4cde73af4fc460.svg",width:"267.4rem",height:"40.599999999999994rem"})),r.createElement("p",null,'That looks complicated, but it\'s not. Computing alpha is just a fraction involving known, plain numbers and once we have our alpha value, computing (1-alpha) is literally just "computing one minus alpha". Computing this d() function is thus simply a matter of "computing simple arithmetics but with recursion", which might be computationally expensive because we\'re doing "a lot of" steps, but is also computationally cheap because each step only involves very simple maths. Of course as before the recursion has to stop:'),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/1405067abebab73574934e3e69d7a7158106c744.svg",width:"387.79999999999995rem",height:"42rem"})),r.createElement("p",null,"So, we see two stopping conditions: either ",r.createElement("code",null,"i")," becomes 0, in which case d() is zero, or ",r.createElement("code",null,"k"),' becomes zero, in which case we get the same "either 1 or 0" that we saw in the N() function above.'),r.createElement("p",null,"Thanks to Cox and de Boor, we can compute points on a B-Spline pretty easily: we just need to compute a triangle of interconnected values. For instance, d() for i=3, k=3 yields the following triangle:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/a0a1069b001c75a1fab7f40ffa8bc403e1408f0d.svg",width:"438.2rem",height:"242.2rem"})),r.createElement("p",null,"That is, we compute d(3,3) as a mixture of d(2,3) and d(2,2): d(3,3) = a(3,3) x d(2,3) + (1-a(3,3)) x d(2,2)... and we simply keep expanding our triangle until we reach the terminating function parameters. Done deal!"),r.createElement("p",null,"One thing we need to keep in mind is that we're working with a spline that is contrained by its control points, so even though the ",r.createElement("code",null,"d(..., k)")," values are zero or one at the lowest level, they are really \"zero or one, times their respective control point\", so in the next section you'll see the algorithm for running through the computation in a way that starts with a copy of the control point vector and then works its way up to that single point: that's pretty essential!"),r.createElement("p",null,'If we run this computation "down", starting at d(3,3), then without special code in place we would be computing quite a few terms multiple times at each step. On the other hand, we can also start with that last "column", we can generate the terminating d() values first, then compute the a() constants, perform our multiplcations, generate the previous step\'s d() values, compute their a() constants, do the multiplications, etc. until we end up all the way back at the top. If we run our computation this way, we don\'t need any explicit caching, we can just "recycle" the list of numbers we start with and simply update them as we move up the triangle. So, let\'s implement that!'),r.createElement("h2",null,"Cool, cool... but I don't know what to do with that information"),r.createElement("p",null,"I know, this is pretty mathy, so let's have a look at what happens when we change parameters here. We can't change the maths for the interpolation functions, so that gives us only one way to control what happens here: the knot vector itself. As such, let's look at the graph that shows the interpolation functions for a cubic B-Spline with seven points with a uniform knot vector (so we see seven identical functions), representing how much each point (represented by one function each) influences the total curvature, given our knot values. And, because exploration is the key to discovery, let's make the knot vector a thing we can actually manipulate. Normally a proper knot vector has a constraint that any value is strictly equal to, or larger than the previous ones, but screw it this is programming, let's ignore that hard restriction and just mess with the knots however we like."),r.createElement("div",{className:"two-column"},r.createElement(o,{ref:"interpolation-graph"}),r.createElement(i,{sketch:n(128),controller:function(t,n){return e.bindKnots(t,n,"interpolation-graph")}})),r.createElement("p",null,"Changing the values in the knot vector changes how much each point influences the total curvature (with some clever knot value manipulation, we can even make the influence of certain points disappear entirely!), so we can see that while the control points define the hull inside of which we're going to be drawing a curve, it is actually the knot vector that determines the actual ",r.createElement("em",null,"shape")," of the curve inside that hull."),r.createElement("p",null,"After reading the rest of this section you may want to come back here to try some specific knot vectors, and see if the resulting interpolation landscape makes sense given what you will now think should happen!"),r.createElement("h2",null,"Running the computation"),r.createElement("p",null,"Unlike the de Casteljau algorithm, where the ",r.createElement("code",null,"t")," value stays the same at every iteration, for B-Splines that is not the case, and so we end having to (for each point we evaluate) run a fairly involving bit of recursive computation. The algorithm is discussed on ",r.createElement("a",{href:"http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/spline/de-Boor.html"
 },"this Michigan Tech")," page, but an easier to read version is implemented by ",r.createElement("a",{href:"https://github.com/thibauts/b-spline/blob/master/index.js#L59-L71"},"b-spline.js"),", so we'll look at its code."),r.createElement("p",null,"Given an input value ",r.createElement("code",null,"t"),", we first map the input to a value from the domain [0,1] to the domain [knots[degree], knots[knots.length - 1 - degree]. Then, we find the section number ",r.createElement("code",null,"s")," that this mapped ",r.createElement("code",null,"t")," value lies on:"),r.createElement("pre",null,"\n        for(s=domain[0]; s < domain[1]; s++) {\n          if(knots[s] <= t && t <= knots[s+1]) break;\n        }"),r.createElement("p",null,"after running this code, ",r.createElement("code",null,"s")," is the index for the section the point will lie on. We then run the algorithm mentioned on the MU page (updated to use this description's variable names):"),r.createElement("pre",null,"\n        let v = copy of control points\n\n        for(let L = 1; L <= order; L++) {\n          for(let i=s; i > s + L - order; i--) {\n            let numerator = t - knots[i]\n            let denominator = knots[i - L + order] - knots[i]\n            let alpha = numerator / denominator\n            let v[i] = alpha * v[i] + (1-alpha) * v[i-1]\n          }\n        }"),r.createElement("p",null,'(A nice bit of behaviour in this code is that we work the interpolation "backwards", starting at ',r.createElement("code",null,"i=s")," at each level of the interpolation, and we stop when ",r.createElement("code",null,"i = s - order + level"),", so we always end up with a value for ",r.createElement("code",null,"i")," such that those ",r.createElement("code",null,"v[i-1]")," don't try to use an array index that doesn't exist)"),r.createElement("h2",null,"Open vs. closed paths"),r.createElement("p",null,"Much like poly-Béziers, B-Splines can be either open, running from the first point to the last point, or closed, where the first and last point are ",r.createElement("em",null,"the same point"),". However, because B-Splines are an interpolation of curves, not just point, we can't simply make the first and last point the same, we need to link a few point point: for an order ",r.createElement("code",null,"d")," B-Spline, we need to make the last ",r.createElement("code",null,"d")," point the same as the first ",r.createElement("code",null,"d")," points. And the easiest way to do this is to simply append ",r.createElement("code",null,"points.splice(0,d)")," to ",r.createElement("code",null,"points"),". Done!"),r.createElement("p",null,'Of course if we want to manipulate these kind of curves we need to make sure to mark them as "closed" so that we know the coordinate for ',r.createElement("code",null,"points[0]")," and ",r.createElement("code",null,"points[n-k]")," etc. are the same coordinate, and manipulating one will equally manipulate the other, but programming generally makes this really easy by storing references to coordinates (or other linked values such as coordinate weights, discussed in the NURBS section) rather than separate coordinate objects."),r.createElement("h2",null,"Manipulating the curve through the knot vector"),r.createElement("p",null,"The most important thing to understand when it comes to B-Splines is that they work ",r.createElement("em",null,"because"),' of the concept of a knot vector. As mentioned above, knots represent "where individual control points start/stop influencing the curve", but we never looked at the ',r.createElement("em",null,"values")," that go in the knot vector. If you look back at the N() and a() functions, you see that interpolations are based on intervals in the knot vector, rather than the actual values in the knot vector, and we can exploit this to do some pretty interesting things with clever manipulation of the knot vector. Specifically there are four things we can do that are worth looking at:"),r.createElement("ol",null,r.createElement("li",null,"we can use a uniform knot vector, with equally spaced intervals,"),r.createElement("li",null,"we can use a non-uniform knot vector, without enforcing equally spaced internvals,"),r.createElement("li",null,'we can collapse sequential knots to the same value, locally lowering curve complexity using "null" intervals, and'),r.createElement("li",null,"we can form a special case non-uniform vector, by combining (1) and (3) to for a vector with collapsed start and end knots, with a uniform vector in between.")),r.createElement("h3",null,"Uniform B-Splines"),r.createElement("p",null,"The most straightforward type of B-Spline is the uniform spline. In a uniform spline, the knots are distributed uniformly over the entire curve interval. For instance, if we have a knot vector of length twelve, then a uniform knot vector would be [0,1,2,3,...,9,10,11]. Or [4,5,6,...,13,14,15], which defines ",r.createElement("em",null,"the same intervals"),", or even [0,2,3,...,18,20,22], which also defines ",r.createElement("em",null,"the same intervals"),", just scaled by a constant factor, which becomes normalised during interpolation and so does not contribute to the curvature."),r.createElement("div",{className:"two-column"},r.createElement(o,{ref:"uniform-spline"}),r.createElement(i,{sketch:n(131),controller:function(t,n){return e.bindKnots(t,n,"uniform-spline")}})),r.createElement("p",null,"This is an important point: the intervals that the knot vector defines are ",r.createElement("em",null,"relative")," intervals, so it doesn't matter if every interval is size 1, or size 100 - the relative differences between the intervals is what shapes any particular curve."),r.createElement("p",null,"The problem with uniform knot vectors is that, as we need ",r.createElement("code",null,"order"),' control points before we have any curve with which we can perform interpolation, the curve does not "start" at the first point, nor "ends" at the last point. Instead there are "gaps". We can get rid of these, by being clever about how we apply the following uniformity-breaking approach instead...'),r.createElement("h3",null,"Reducing local curve complexity by collapsing intervals"),r.createElement("p",null,"By collapsing knot intervals by making two or more consecutive knots have the same value, we can reduce the curve complexity in the sections that are affected by the knots involved. This can have drastic effects: for ever interval collapse, the curve order goes down, and curve continuity goes down, to the point where collapsing ",r.createElement("code",null,"order"),' knots creates a situation where all continuity is lost and the curve "kinks".'),r.createElement("div",{className:"two-column"},r.createElement(o,{ref:"center-cut-bspline"}),r.createElement(i,{sketch:n(126),controller:function(t,n){return e.bindKnots(t,n,"center-cut-bspline")}})),r.createElement("h3",null,"Open-Uniform B-Splines"),r.createElement("p",null,"By combining knot interval collapsing at the start and end of the curve, with uniform knots in between, we can overcome the problem of the curve not starting and ending where we'd kind of like it to:"),r.createElement("p",null,"For any curve of degree ",r.createElement("code",null,"D")," with control points ",r.createElement("code",null,"N"),", we can define a knot vector of length ",r.createElement("code",null,"N+D+1")," in which the values ",r.createElement("code",null,"0 ... D+1")," are the same, the values ",r.createElement("code",null,"D+1 ... N+1"),' follow the "uniform" pattern, and the values ',r.createElement("code",null,"N+1 ... N+D+1"),' are the same again. For example, a cubic B-Spline with 7 control points can have a knot vector [0,0,0,0,1,2,3,4,4,4,4], or it might have the "identical" knot vector [0,0,0,0,2,4,6,8,8,8,8], etc. Again, it is the relative differences that determine the curve shape.'),r.createElement("div",{className:"two-column"},r.createElement(o,{ref:"open-uniform-bspline"}),r.createElement(i,{sketch:n(129),controller:function(t,n){return e.bindKnots(t,n,"open-uniform-bspline")}})),r.createElement("h3",null,"Non-uniform B-Splines"),r.createElement("p",null,'This is essentialy the "free form" version of a B-Spline, and also the least interesting to look at, as without any specific reason to pick specific knot intervals, there is nothing particularly interesting going on. There is one constraint to the knot vector, and that is that any value ',r.createElement("code",null,"knots[k+1]"),"should be equal to, or greater than ",r.createElement("code",null,"knots[k]"),"."),r.createElement("h2",null,"One last thing: Rational B-Splines"),r.createElement("p",null,'While it is true that this section on B-Splines is running quite long already, there is one more thing we need to talk about, and that\'s "Rational" splines, where the rationality applies to the "ratio", or relative weights, of the control points themselves. By introducing a ratio vector with weights to apply to each control point, we greatly increase our influence over the final curve shape: the more weight a control point carries, the close to that point the spline curve will lie, a bit like turning up the gravity of a control point.'),r.createElement("div",{className:"two-column"},r.createElement(s,{ref:"rational-uniform-bspline-weights"}),r.createElement(i,{scrolling:!0,sketch:n(130),controller:function(t,n,r,i){e.bindWeights(t,r,i,"rational-uniform-bspline-weights")}})),r.createElement("p",null,'Of course this brings us to the final topic that any text on B-Splines must touch on before calling it a day: the NURBS, or Non-Uniform Rational B-Spline (NURBS is not a plural, the capital S actually just stands for "spline", but a lot of people mistakenly treat it as if it is, so now you know better). NURBS are an important type of curve in computer-facilitated design, used a lot in 3D modelling (as NURBS surfaces) as well as in arbitrary-precision 2D design due to the level of control a NURBS curve offers designers.'),r.createElement("p",null,"While a true non-uniform rational B-Spline would be hard to work with, when we talk about NURBS we typically mean the Open-Uniform Rational B-Spline, or OURBS, but that doesn't roll off the tongue nearly as nicely, and so remember that when people talk about NURBS, they typically mean open-uniform, which has the useful property of starting the curve at the first control point, and ending it at the last."),r.createElement("h2",null,"Extending our implementation to cover rational splines"),r.createElement("p",null,"The algorithm for working with Rational B-Splines is virtually identical to the regular algorithm, and the extension to work in the control point weights is fairly simple: we extend each control point from a point in its original number of dimensions (2D, 3D, etc) to one dimension higher, scaling the original dimensions by the control point's weight, and then assigning that weight as its value for the extended dimension."),r.createElement("p",null,"For example, a 2D point ",r.createElement("code",null,"(x,y)")," with weight ",r.createElement("code",null,"w")," becomes a 3D point ",r.createElement("code",null,"(w * x, w * y, w)"),"."),r.createElement("p",null,"We then run the same algorithm as before, which will automaticall perform weight interpolation in addition to regular coordinate interpolation, because all we've done is pretended we have coordinates in a higher dimension. The algorithm doesn't really care about how many dimensions it needs to interpolate."),r.createElement("p",null,'In order to recover our "real" curve point, we take the final result of the point generation algorithm, and "unweigh" it: we take the final point\'s derived weight ',r.createElement("code",null,"w'")," and divide all the regular coordinate dimensions by it, then throw away the weight information."),r.createElement("p",null,"Based on our previous example, we take the final 3D point ",r.createElement("code",null,"(x', y', w')"),", which we then turn back into a 2D point by computing ",r.createElement("code",null,"(x'/w', y'/w')"),". And that's it, we're done!"))}});e.exports=l},function(e,t,n){"use strict";var r=["#C00","#CC0","#0C0","#0CC","#00C","#C0C","#600","#660","#060","#066","#006","#606"];e.exports={degree:3,activeDistance:9,cache:{N:[]},setup:function(){this.size(600,300),this.points=[{x:0,y:0},{x:100,y:-100},{x:200,y:100},{x:300,y:-100},{x:400,y:100},{x:500,y:0}],this.knots=this.formKnots(this.points),this.props.controller&&this.props.controller(this,this.knots),this.draw()},draw:function(){this.clear();var e=25;this.grid(e),this.stroke(0),this.line(e,0,e,this.height);var t=this.height-e;this.line(0,t,this.width,t);for(var n=this.degree,r=this.points.length||4,i=0;i<r+1+n;i++)this.drawN(i,n,e,(this.width-e)/(2*(r+2)),this.height-2*e)},drawN:function(e,t,n,i,a){this.stroke(r[e]);this.knots;this.beginPath();for(var o=e-1,s=o,l=.1,u=e+t+1;s<u;s+=l){var c=n+e*i+s*i,h=this.height-n-this.N(e,t,s)*a;this.vertex(c,h)}this.endPath()},N:function(e,t,n){var r=this.knots[e],i=this.knots[e+1],a=this.knots[e+t-1],o=this.knots[e+t];if(1===t)return r<=n&&n<=i?1:0;var s=n-r,l=a-r,u=0===l?0:s/l,c=o-n,h=o-i,d=0===h?0:c/h,f=0;if(0!==u){var p=this.ensureN(e,t-1,n);f=void 0===p?this.N(e,t-1,n):p}var m=0;if(0!==d){var g=this.ensureN(e+1,t-1,n);m=void 0===g?this.N(e+1,t-1,n):g}return this.cacheN(e,t,n,u*f+d*m),this.cache.N[e][t][n]},ensureN:function(e,t,n){this.cache.N||(this.cache.N=[]);var r=this.cache.N;return r[e]||(r[e]=[]),r[e][t]||(r[e][t]=[]),r[e][t][n]},cacheN:function(e,t,n,r){this.ensureN(e,t,n),this.cache.N[e][t][n]=r}}},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,setup:function(){this.size(400,400);for(var e=2*Math.PI,t=0;t<e;t+=e/10)this.points.push({x:this.width/2+100*Math.cos(t),y:this.height/2+100*Math.sin(t)});this.knots=this.formKnots(this.points,!0),this.props.controller&&this.props.controller(this,this.knots),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,weights:[],setup:function(){this.size(400,400);for(var e=2*Math.PI,t=this.width/3,n=0;n<6;n++)this.points.push({x:this.width/2+t*Math.cos(n/6*e),y:this.height/2+t*Math.sin(n/6*e)});this.points=this.points.concat(this.points.slice(0,3)),this.closed=this.degree,this.knots=this.formKnots(this.points),this.weights=this.formWeights(this.points),this.props.controller&&this.props.controller(this,this.knots,this.weights,this.closed),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";e.exports={degree:3,activeDistance:9,setup:function(){this.size(400,400);for(var e=2*Math.PI,t=0;t<e;t+=e/10)this.points.push({x:this.width/2+100*Math.cos(t),y:this.height/2+100*Math.sin(t)});this.knots=this.formKnots(this.points),this.props.controller&&this.props.controller(this,this.knots),this.draw()},draw:function(){var e=this;this.clear(),this.grid(25);var t=this.points[0];this.points.forEach(function(n){e.stroke(200),e.line(n.x,n.y,t.x,t.y),t=n,e.stroke(0),e.circle(t.x,t.y,4)}),this.drawSplineData()},drawSplineData:function(){if(!(this.points.length<=this.degree)){var e=this.points.map(function(e){return[e.x,e.y]});this.drawCurve(e),this.drawKnots(e)}}}},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="canonical",s=r.createClass({displayName:"Canonical",getDefaultProps:function(){return{title:a.getTitle(o)}},setup:function(e){var t=e.getDefaultCubic();e.setCurve(t),e.reset(),e._map_loaded=!1},draw:function(e,t){var n=400,r=n,i=this.unit,a={x:n/2,y:r/2};e.setSize(n,r),e.setPanelCount(2),e.reset(),e.drawSkeleton(t),e.drawCurve(t),e.offset.x+=400,e._map_loaded?e.image(e._map_image):setTimeout(function(){this.drawBase(e,t),this.draw(e,t)}.bind(this),100),e.drawLine({x:0,y:0},{x:0,y:r});var o=[{x:0,y:0},{x:0,y:i},{x:i,y:i},this.forwardTransform(t.points,i)],s=new e.Bezier(o);e.setColor("blue"),e.drawCurve(s,a),e.drawCircle(o[3],3,a)},forwardTransform:function(e,t){t=t||1;var n=e[0],r=e[1],i=e[2],a=e[3],o=-n.x+a.x-(-n.x+r.x)*(-n.y+a.y)/(-n.y+r.y),s=-n.x+i.x-(-n.x+r.x)*(-n.y+i.y)/(-n.y+r.y),l=t*o/s,u=t*(-n.y+a.y)/(-n.y+r.y),c=t-t*(-n.y+i.y)/(-n.y+r.y),h=c*o/s,d=u+h;return{x:l,y:d}},drawBase:function(e,t){e.reset();var n=400,r=n,i=this.unit=n/5,a={x:n/2,y:r/2};e.setSize(n,r),e.setColor("lightgrey");for(var o=0;o<n;o+=i/2)e.drawLine({x:o,y:0},{x:o,y:r});for(var s=0;s<r;s+=i/2)e.drawLine({x:0,y:s},{x:n,y:s});e.setColor("black"),e.drawLine({x:n/2,y:0},{x:n/2,y:r}),e.drawLine({x:0,y:r/2},{x:n,y:r/2}),e.setColor("green"),e.drawLine({x:-n/2,y:i},{x:n/2,y:i},a),e.setColor("black"),e.setFill("black"),e.drawCircle({x:0,y:0},4,a),e.text("(0,0)",{x:5+a.x,y:15+a.y}),e.drawCircle({x:0,y:i},4,a),e.text("(0,1)",{x:5+a.x,y:i+15+a.y}),e.drawCircle({x:i,y:i},4,a),e.text("(1,1)",{x:i+5+a.x,y:i+15+a.y}),e.setWeight(1.5),e.setColor("#FF0000"),e.setFill(e.getColor());var l=[],u=1,c=1;for(o=-10;o<=1;o+=.01)s=(-o*o+2*o+3)/4,o>-10&&(l.push({x:i*u,y:i*c}),e.drawLine({x:i*u,y:i*c},{x:i*o,y:i*s},a)),u=o,c=s;l.push({x:i*u,y:i*c}),e.text("Curve form has cusp →",{x:n/2-2*i,y:r/2+i/2.5}),e.setColor("#FF00FF"),e.setFill(e.getColor());var h=Math.sqrt;for(o=1;o>=0;o-=.005)l.push({x:i*u,y:i*c}),s=.5*(h(3)*h(4*o-o*o)-o),e.drawLine({x:i*u,y:i*c},{x:i*o,y:i*s},a),u=o,c=s;for(l.push({x:i*u,y:i*c}),e.text("← Curve forms a loop at t = 1",{x:n/2+i/4,y:r/2+i/1.5}),e.setColor("#3300FF"),e.setFill(e.getColor()),o=0;o>-n;o-=.01)l.push({x:i*u,y:i*c}),s=(-o*o+3*o)/3,e.drawLine({x:i*u,y:i*c},{x:i*o,y:i*s},a),u=o,c=s;l.push({x:i*u,y:i*c}),e.text("← Curve forms a loop at t = 0",{x:n/2-i+10,y:r/2-1.25*i}),e.setColor("transparent"),e.setFill("rgba(255,120,100,0.2)"),e.drawPath(l,a),l=[{x:-n/2,y:i},{x:n/2,y:i},{x:n/2,y:r},{x:-n/2,y:r}],e.setFill("rgba(0,200,0,0.2)"),e.drawPath(l,a),e.setColor("black"),e.setFill(e.getColor()),e.text("← Curve form has one inflection →",{x:n/2-i,y:r/2+1.75*i}),e.text("← Plain curve ↕",{x:n/2+i/2,y:r/6}),e.text("↕ Double inflection",{x:10,y:r/2-10}),e._map_image=e.toImage(),e._map_loaded=!0},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="catmullconv",s=r.createClass({displayName:"CatmullRomConversion",getDefaultProps:function(){return{title:a.getTitle(o)}},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="catmullmoulding",s=n(15),l=r.createClass({displayName:"CatmullRomMoulding",statics:{keyHandlingOptions:{propName:"distance",values:{38:1,40:-1}}},getDefaultProps:function(){return{title:a.getTitle(o)}},setup:function(e){e.setPanelCount(3),e.lpts=[{x:56,y:153},{x:144,y:83},{x:188,y:185}],e.distance=0},convert:function(e,t,n,r){var i=.5;return[t,{x:t.x+(n.x-e.x)/(6*i),y:t.y+(n.y-e.y)/(6*i)},{x:n.x-(r.x-t.x)/(6*i),y:n.y-(r.y-t.y)/(6*i)},n]},draw:function(e){e.reset(),e.setColor("lightblue"),e.drawGrid(10,10);var t=e.lpts;e.setColor("black"),e.setFill("black"),t.forEach(function(t,n){e.drawCircle(t,3),e.text("point "+(n+1),t,{x:10,y:7})});var n=e.getPanelWidth(),r=e.getPanelHeight(),i={x:n,y:0};e.setColor("lightblue"),e.drawGrid(10,10,i),e.setColor("black"),e.drawLine({x:0,y:0},{x:0,y:r},i),t.forEach(function(t,n){e.drawCircle(t,3,i)});var a=t[0],o=t[1],s=t[2],l=s.x-a.x,u=s.y-a.y,c=Math.sqrt(l*l+u*u);l/=c,u/=c,e.drawLine(a,s,i);var h={x:a.x+(s.x-o.x)-e.distance*l,y:a.y+(s.y-o.y)-e.distance*u},d={x:a.x+(s.x-o.x)+e.distance*l,y:a.y+(s.y-o.y)+e.distance*u},f=e.utils.lli4(a,s,o,{x:(h.x+d.x)/2,y:(h.y+d.y)/2});e.setColor("blue"),e.drawCircle(f,3,i),e.drawLine(t[1],f,i),e.setColor("#666"),e.drawLine(f,h,i),e.drawLine(f,d,i),e.setFill("blue"),e.text("p0",h,{x:-20+i.x,y:i.y+2}),e.text("p4",d,{x:10+i.x,y:i.y+2}),e.setColor("red"),e.drawCircle(h,3,i),e.drawLine(o,h,i),e.drawLine(a,{x:a.x+(o.x-h.x)/5,y:a.y+(o.y-h.y)/5},i),e.setColor("#00FF00"),e.drawCircle(d,3,i),e.drawLine(o,d,i),e.drawLine(s,{x:s.x+(d.x-o.x)/5,y:s.y+(d.y-o.y)/5},i);var p=new e.Bezier(this.convert(h,a,o,s)),m=new e.Bezier(this.convert(a,o,s,d));e.setColor("lightgrey"),e.drawCurve(p,i),e.drawCurve(m,i),i.x+=n,e.setColor("lightblue"),e.drawGrid(10,10,i),e.setColor("black"),e.drawLine({x:0,y:0},{x:0,y:r},i),e.drawCurve(p,i),e.drawCurve(m,i),e.drawPoints(p.points,i),e.drawPoints(m.points,i),e.setColor("lightgrey"),e.drawLine(p.points[0],p.points[1],i),e.drawLine(p.points[2],m.points[1],i),e.drawLine(m.points[2],m.points[3],i)},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s(l)},function(e,t,n){"use strict";var r=n(0),i=n(14),a=n(9),o=Math.sin,s=Math.cos,l=r.createClass({displayName:"Circles",getDefaultProps:function(){return{title:"Circles and quadratic Bézier curves"}},setup:function(e){e.w=e.getPanelWidth(),e.h=e.getPanelHeight(),e.pad=20,e.r=e.w/2-e.pad,e.mousePt=!1,e.angle=0;var t={x:e.w-e.pad,y:e.h/2};e.setCurve(new e.Bezier(t,t,t))},draw:function(e,t){e.reset(),e.setColor("lightgrey"),e.drawGrid(1,1),e.setColor("red"),e.drawCircle({x:e.w/2,y:e.h/2},e.r),e.setColor("transparent"),e.setFill("rgba(100,255,100,0.4)");var n={x:e.w/2,y:e.h/2,r:e.r,s:e.angle<0?e.angle:0,e:e.angle<0?0:e.angle};e.drawArc(n),e.setColor("black"),e.drawSkeleton(t),e.drawCurve(t)},onMouseMove:function(e,t){var n=e.offsetX-t.w/2,r=e.offsetY-t.h/2,i=Math.atan2(r,n),a=t.curve.points,l=t.r,u=(s(i)-1)/o(i);a[1]={x:t.w/2+l*(s(i)-u*o(i)),y:t.w/2+l*(o(i)+u*s(i))},a[2]={x:t.w/2+t.r*s(i),y:t.w/2+t.r*o(i)},t.setCurve(new t.Bezier(a)),t.angle=i},render:function(){return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"Circles and Bézier curves are very different beasts, and circles are infinitely easier to work with than Bézier curves. Their formula is much simpler, and they can be drawn more efficiently. But, sometimes you don't have the luxury of using circles, or ellipses, or arcs. Sometimes, all you have are Bézier curves. For instance, if you're doing font design, fonts have no concept of geometric shapes, they only know straight lines, and Bézier curves. OpenType fonts with TrueType outlines only know quadratic Bézier curves, and OpenType fonts with Type 2 outlines only know cubic Bézier curves. So how do you draw a circle, or an ellipse, or an arc?"),r.createElement("p",null,"You approximate."),r.createElement("p",null,"We already know that Bézier curves cannot model all curves that we can think of, and this includes perfect circles, as well as ellipses, and their arc counterparts. However, we can certainly approximate them to a degree that is visually acceptable. Quadratic and cubic curves offer us different curvature control, so in order to approximate a circle we will first need to figure out what the error is if we try to approximate arcs of increasing degree with quadratic and cubic curves, and where the coordinates even lie."),r.createElement("p",null,"Since arcs are mid-point-symmetrical, we need the control points to set up a symmetrical curve. For quadratic curves this means that the control point will be somewhere on a line that intersects the baseline at a right angle. And we don't get any choice on where that will be, since the derivatives at the start and end point have to line up, so our control point will lie at the intersection of the tangents at the start and end point."),r.createElement("p",null,"First, let's try to fit the quadratic curve onto a circular arc. In the following sketch you can move the mouse around over a unit circle, to see how well, or poorly, a quadratic curve can approximate the arc from (1,0) to where your mouse cursor is:"),r.createElement(i,{preset:"arcfitting",title:"Quadratic Bézier arc approximation",setup:this.setup,draw:this.draw,onMouseMove:this.onMouseMove}),r.createElement("p",null,"As you can see, things go horribly wrong quite quickly; even trying to approximate a quarter circle using a quadratic curve is a bad idea. An eighth of a turns might look okay, but how okay is okay? Let's apply some maths and find out. What we're interested in is how far off our on-curve coordinates are with respect to a circular arc, given a specific start and end angle. We'll be looking at how much space there is between the circular arc, and the quadratic curve's midpoint."),r.createElement("p",null,"We start out with our start and end point, and for convenience we will place them on a unit circle (a circle around 0,0 with radius 1), at some angle ",r.createElement("i",null,"φ"),":"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/ef34ab8f466ed3294895135a346b55ada05d779d.svg",width:"183.39999999999998rem",height:"42rem"})),r.createElement("p",null,"What we want to find is the intersection of the tangents, so we want a point C such that:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/5660e8512b07dbac7fcf04633de8002fa25aa962.svg",width:"298.2rem",height:"42rem"})),r.createElement("p",null,"i.e. we want a point that lies on the vertical line through S (at some distance ",r.createElement("i",null,"a")," from S) and also lies on the tangent line through E (at some distance ",r.createElement("i",null,"b")," from E). Solving this gives us:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/d16e7a1c1e9686e1afb82f4ffcec07078d264565.svg",width:"229.6rem",height:"42rem"})),r.createElement("p",null,"First we solve for ",r.createElement("i",null,"b"),":"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/3128b31a874166ebe4479d3002d70f280de375a1.svg",width:"588rem",height:"18.2rem"})),r.createElement("p",null,"which yields:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/02b158f9ef2191b970dc2fe69c0903eba2b1f8b5.svg",width:"106.39999999999999rem",height:"40.599999999999994rem"})),r.createElement("p",null,"which we can then substitute in the expression for ",r.createElement("i",null,"a"),":"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/3bd9c2d6740ff530aabcbe60252742032af816e9.svg",width:"242.2rem",height:"204.39999999999998rem"})),r.createElement("p",null,"A quick check shows that plugging these values for ",r.createElement("i",null,"a")," and ",r.createElement("i",null,"b")," into the expressions for C",r.createElement("sub",null,"x")," and C",r.createElement("sub",null,"y"),' give the same x/y coordinates for both "',r.createElement("i",null,"a"),' away from A" and "',r.createElement("i",null,"b")," away from B\", so let's continue: now that we know the coordinate values for C, we know where our on-curve point T for ",r.createElement("i",null,"t=0.5")," (or angle φ/2) is, because we can just evaluate the Bézier polynomial, and we know where the circle arc's actual point P is for angle φ/2:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/0b80423188012451e0400f473c19729eb2bad654.svg",width:"197.39999999999998rem",height:"33.599999999999994rem"})),r.createElement("p",null,"We compute T, observing that if ",r.createElement("i",null,"t=0.5"),", the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/bc50559ff8bd9062694a449aae5f6f85f91de909.svg",width:"264.59999999999997rem",height:"36.4rem"})),r.createElement("p",null,"Which, worked out for the x and y components, gives:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/c7fca7664a3acb855eeaaf412aa2331202f41097.svg",width:"428.4rem",height:"81.19999999999999rem"})),r.createElement("p",null,"And the distance between these two is the standard Euclidean distance:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/3251cd91a1cffc27a1695ece4c13cc651d7007fb.svg",width:"418.59999999999997rem",height:"161rem"})),r.createElement("p",null,"So, what does this distance function look like when we plot it for a number of ranges for the angle φ, such as a half circle, quarter circle and eighth circle?"),r.createElement("table",null,r.createElement("tbody",null,r.createElement("tr",null,r.createElement("td",null,r.createElement("p",null,r.createElement("img",{src:"images/arc-q-pi.gif",height:"190px"})),r.createElement("p",null,"plotted for 0 ≤ φ ≤ π:")),r.createElement("td",null,r.createElement("p",null,r.createElement("img",{src:"images/arc-q-pi2.gif",height:"187px"})),r.createElement("p",null,"plotted for 0 ≤ φ ≤ ½π:")),r.createElement("td",null,this.props.showhref?"http://www.wolframalpha.com/input/?i=plot+sqrt%28%281%2F4+*+%28sin%28x%29+%2B+2tan%28x%2F2%29%29+-+sin%28x%2F2%29%29%5E2+%2B+%282sin%5E4%28x%2F4%29%29%5E2%29+for+0+%3C%3D+x+%3C%3D+pi%2F4":null,r.createElement("p",null,r.createElement("img",{src:"images/arc-q-pi4.gif",height:"174px"})),r.createElement("p",null,"plotted for 0 ≤ φ ≤ ¼π:"))))),r.createElement("p",null,"We now see why the eighth circle arc looks decent, but the quarter circle arc doesn't: an error of roughly 0.06 at ",r.createElement("i",null,"t=0.5")," means we're 6% off the mark... we will already be off by one pixel on a circle with pixel radius 17. Any decent sized quarter circle arc, say with radius 100px, will be way off if approximated by a quadratic curve! For the eighth circle arc, however, the error is only roughly 0.003, or 0.3%, which explains why it looks so close to the actual eighth circle arc. In fact, if we want a truly tiny error, like 0.001, we'll have to contend with an angle of (rounded) 0.593667, which equates to roughly 34 degrees. We'd need 11 quadratic curves to form a full circle with that precision! (technically, 10 and ten seventeenth, but we can't do partial curves, so we have to round up). That's a whole lot of curves just to get a shape that can be drawn using a simple function!"),r.createElement("p",null,"In fact, let's flip the function around, so that if we plug in the precision error, labelled ε, we get back the maximum angle for that precision:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/61a938fa10b77e8c41c3c064ed39bd1145d6bbcc.svg",width:"259rem",height:"56rem"})),r.createElement("p",null,"And frankly, things are starting to look a bit ridiculous at this point, we're doing way more maths than we've ever done, but thankfully this is as far as we need the maths to take us: If we plug in the precisions 0.1, 0.01, 0.001 and 0.0001 we get the radians values 1.748, 1.038, 0.594 and 0.3356; in degrees, that means we can cover roughly 100 degrees (requiring four curves), 59.5 degrees (requiring six curves), 34 degrees (requiring 11 curves), and 19.2 degrees (requiring a whopping nineteen curves). "),r.createElement("p",null,"The bottom line? ",r.createElement("strong",null,"Quadratic curves are kind of lousy")," if you want circular (or elliptical, which are circles that have been squashed in one dimension) curves. We can do better, even if it's just by raising the order of our curve once. So let's try the same thing for cubic curves."))}});e.exports=l},function(e,t,n){"use strict";var r=n(0),i=n(14),a=n(9),o=Math.sin,s=Math.cos,l=Math.tan,u=r.createClass({displayName:"CirclesCubic",getDefaultProps:function(){return{title:"Circles and cubic Bézier curves"
 }},setup:function(e){e.setSize(400,400),e.w=e.getPanelWidth(),e.h=e.getPanelHeight(),e.pad=80,e.r=e.w/2-e.pad,e.mousePt=!1,e.angle=0;var t={x:e.w-e.pad,y:e.h/2};e.setCurve(new e.Bezier(t,t,t,t))},guessCurve:function(e,t,n){var r={x:(e.x+n.x)/2,y:(e.y+n.y)/2},i={x:t.x+(t.x-r.x)/3,y:t.y+(t.y-r.y)/3},a=(n.x-e.x)/4,o=(n.y-e.y)/4,s={x:t.x-a,y:t.y-o},l={x:t.x+a,y:t.y+o},u={x:i.x+2*(s.x-i.x),y:i.y+2*(s.y-i.y)},c={x:i.x+2*(l.x-i.x),y:i.y+2*(l.y-i.y)},h={x:e.x+2*(u.x-e.x),y:e.y+2*(u.y-e.y)},d={x:n.x+2*(c.x-n.x),y:n.y+2*(c.y-n.y)};return[h,d]},draw:function(e,t){e.reset(),e.setColor("lightgrey"),e.drawGrid(1,1),e.setColor("rgba(255,0,0,0.4)"),e.drawCircle({x:e.w/2,y:e.h/2},e.r),e.setColor("transparent"),e.setFill("rgba(100,255,100,0.4)");var n={x:e.w/2,y:e.h/2,r:e.r,s:e.angle<0?e.angle:0,e:e.angle<0?0:e.angle};e.drawArc(n);var r={x:e.w/2+e.r*s(e.angle/2),y:e.w/2+e.r*o(e.angle/2)},i=t.points[0],a=t.points[3],l=this.guessCurve(i,r,a),u=new e.Bezier([i,l[0],l[1],a]);e.setColor("rgb(140,140,255)"),e.drawLine(u.points[0],u.points[1]),e.drawLine(u.points[1],u.points[2]),e.drawLine(u.points[2],u.points[3]),e.setColor("blue"),e.drawCurve(u),e.drawCircle(u.points[1],3),e.drawCircle(u.points[2],3),e.drawSkeleton(t),e.setColor("black"),e.drawLine(t.points[1],t.points[2]),e.drawCurve(t)},onMouseMove:function(e,t){var n=e.offsetX-t.w/2,r=e.offsetY-t.h/2;if(!(n>t.w/2)){var i=Math.atan2(r,n);i<0&&(i=2*Math.PI+i);var a=t.curve.points,u=t.r,c=4*l(i/4)/3;a[1]={x:t.w/2+u,y:t.w/2+u*c},a[2]={x:t.w/2+t.r*(s(i)+c*o(i)),y:t.w/2+t.r*(o(i)-c*s(i))},a[3]={x:t.w/2+t.r*s(i),y:t.w/2+t.r*o(i)},t.setCurve(new t.Bezier(a)),t.angle=i}},drawCircle:function(e){e.setSize(325,325),e.reset();var t=e.getPanelWidth(),n=e.getPanelHeight(),r=60,i=t/2-r,a=.55228,o={x:-r/2,y:-r/4},s=new e.Bezier([{x:t/2+i,y:n/2},{x:t/2+i,y:n/2+a*i},{x:t/2+a*i,y:n/2+i},{x:t/2,y:n/2+i}]);e.setColor("lightgrey"),e.drawLine({x:0,y:n/2},{x:t+r,y:n/2},o),e.drawLine({x:t/2,y:0},{x:t/2,y:n+r},o);var l=s.points;e.setColor("red"),e.drawPoint(l[0],o),e.drawPoint(l[1],o),e.drawPoint(l[2],o),e.drawPoint(l[3],o),e.drawCurve(s,o),e.setColor("rgb(255,160,160)"),e.drawLine(l[0],l[1],o),e.drawLine(l[1],l[2],o),e.drawLine(l[2],l[3],o),e.setFill("red"),e.text(l[0].x-t/2+","+(l[0].y-n/2),{x:l[0].x+7,y:l[0].y+3},o),e.text(l[1].x-t/2+","+(l[1].y-n/2),{x:l[1].x+7,y:l[1].y+3},o),e.text(l[2].x-t/2+","+(l[2].y-n/2),{x:l[2].x+7,y:l[2].y+7},o),e.text(l[3].x-t/2+","+(l[3].y-n/2),{x:l[3].x,y:l[3].y+13},o),l.forEach(function(e){e.x=-(e.x-t)}),e.setColor("blue"),e.drawCurve(s,o),e.drawLine(l[2],l[3],o),e.drawPoint(l[2],o),e.setFill("blue"),e.text("reflected",{x:l[2].x-r/2,y:l[2].y+13},o),e.setColor("rgb(200,200,255)"),e.drawLine(l[1],l[0],o),e.drawPoint(l[1],o),l.forEach(function(e){e.y=-(e.y-n)}),e.setColor("green"),e.drawCurve(s,o),l.forEach(function(e){e.x=-(e.x-t)}),e.setColor("purple"),e.drawCurve(s,o),e.drawLine(l[1],l[0],o),e.drawPoint(l[1],o),e.setFill("purple"),e.text("reflected",{x:l[1].x+10,y:l[1].y+3},o),e.setColor("rgb(200,200,255)"),e.drawLine(l[2],l[3],o),e.drawPoint(l[2],o),e.setColor("black"),e.setFill("black"),e.drawLine({x:t/2,y:n/2},{x:t/2+i-2,y:n/2},o),e.drawLine({x:t/2,y:n/2},{x:t/2,y:n/2+i-2},o),e.text("r = "+i,{x:t/2+i/3,y:n/2+10},o)},render:function(){return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"In the previous section we tried to approximate a circular arc with a quadratic curve, and it mostly made us unhappy. Cubic curves are much better suited to this task, so what do we need to do?"),r.createElement("p",null,'For cubic curves, we basically want the curve to pass through three points on the circle: the start point, the mid point at "angle/2", and the end point at "angle". We then also need to make sure the control points are such that the start and end tangent lines line up with the circle\'s tangent lines at the start and end point.'),r.createElement("p",null,'The first thing we can do is "guess" what the curve should look like, based on the previously outlined curve-through-three-points procedure. This will give use a curve with correct start, mid and end points, but possibly incorrect derivatives at the start and end, because the control points might not be in the right spot. We can then slide the control points along the lines that connect them to their respective end point, until they effect the corrected derivative at the start and end points.  However, if you look back at the section on fitting curves through three points, the rules used were such that they optimized for a near perfect hemisphere, so using the same guess won\'t be all that useful: guessing the solution based on knowing the solution is not really guessing.'),r.createElement("p",null,'So have a graphical look at a "bad" guess versus the true fit, where we\'ll be using the bad guess and the description in the second paragraph to derive the maths for the true fit:'),r.createElement(i,{preset:"arcfitting",title:"Cubic Bézier arc approximation",setup:this.setup,draw:this.draw,onMouseMove:this.onMouseMove}),r.createElement("p",null,'We see two curves here; in blue, our "guessed" curve and its control points, and in grey/black, the true curve fit, with proper control points that were shifted in, along line between our guessed control points, such that the derivatives at the start and end points are correct.'),r.createElement("p",null,'We can already seethat cubic curves are a lot better than quadratic curves, and don\'t look all that wrong until we go well past a quarter circle; ⅜th starts to hint at problems, and half a circle has an obvious "gap" between the real circle and the cubic approximation. Anything past that just looks plain ridiculous... but quarter curves actually look pretty okay!'),r.createElement("p",null,'So, maths time again: how okay is "okay"? Let\'s apply some more maths to find out.'),r.createElement("p",null,"Unlike for the quadratic curve, we can't use ",r.createElement("i",null,"t=0.5")," as our reference point because by its very nature it's one of the three points that are actually guaranteed to lie on the circular curve. Instead, we need a different ",r.createElement("i",null,"t")," value. If we run some analysis on the curve we find that the actual ",r.createElement("i",null,"t"),' value at which the curve is furthest from what it should be is 0.211325 (rounded), but we don\'t know "why", since finding this value involves root-finding, and is nearly impossible to do symbolically without pages and pages of math just to express one of the possible solutions.'),r.createElement("p",null,"So instead of walking you through the derivation for that value, let's simply take that ",r.createElement("i",null,"t")," value and see what the error is for circular arcs with an angle ranging from 0 to 2π:"),r.createElement("table",null,r.createElement("tbody",null,r.createElement("tr",null,r.createElement("td",null,r.createElement("p",null,r.createElement("img",{src:"images/arc-c-2pi.gif",height:"187px"})),r.createElement("p",null,"plotted for 0 ≤ φ ≤ 2π:")),r.createElement("td",null,r.createElement("p",null,r.createElement("img",{src:"images/arc-c-pi.gif",height:"187px"})),r.createElement("p",null,"plotted for 0 ≤ φ ≤ π:")),r.createElement("td",null,r.createElement("p",null,r.createElement("img",{src:"images/arc-c-pi2.gif",height:"187px"})),r.createElement("p",null,"plotted for 0 ≤ φ ≤ ½π:"))))),r.createElement("p",null,"We see that cubic Bézier curves are much better when it comes to approximating circular arcs, with an error of less than 0.027 at the two \"bulge\" points for a quarter circle (which had an error of 0.06 for quadratic curves at the mid point), and an error near 0.001 for an eighth of a circle, so we're getting less than half the error for a quarter circle, or: at a slightly lower error, we're getting twice the arc. This makes cubic curves quite useful!"),r.createElement("p",null,'In fact, the precision of a cubic curve at a quarter circle is considered "good enough" by so many people that it\'s generally considered "just fine" to use four cubic Bézier curves to fake a full circle when no circle primitives are available; generally, people won\'t notice that it\'s not a real circle unless you also happen to overlay an actual circle, so that the difference becomes obvious.'),r.createElement("p",null,'So with the error analysis out of the way, how do we actually compute the coordinates needed to get that "true fit" cubic curve? The first observation is that we already know the start and end points, because they\'re the same as for the quadratic attempt:'),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/ef34ab8f466ed3294895135a346b55ada05d779d.svg",width:"183.39999999999998rem",height:"42rem"})),r.createElement("p",null,"But we now need to find two control points, rather than one. If we want the derivatives at the start and end point to match the circle, then the first control point can only lie somewhere on the vertical line through S, and the second control point can only lie somewhere on the line tangent to point E, which means:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/4df65dae78bc5a0e6c5f23a2faae9a9d7a8b39b3.svg",width:"118.99999999999999rem",height:"42rem"})),r.createElement("p",null,'where "a" is some scaling factor, and:'),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/cb32f8f9c3ae2b264a48003c237a798d02dc8935.svg",width:"170.79999999999998rem",height:"42rem"})),r.createElement("p",null,'where "b" is also some scaling factor.'),r.createElement("p",null,"Starting with this information, we slowly maths our way to success, but I won't lie: the maths for this is pretty trig-heavy, and it's easy to get lost if you remember (or know!) some of the core trigonoetric identities, so if you just want to see the final result just skip past the next section!"),r.createElement("div",{className:"note"},r.createElement("h2",null,"Let's do this thing."),r.createElement("p",null,"Unlike for the quadratic case, we need some more information in order to compute ",r.createElement("i",null,"a")," and ",r.createElement("i",null,"b"),", since they're no longer dependent variables. First, we observe that the curve is symmetrical, so whatever values we end up finding for C",r.createElement("sub",null,"1")," will apply to C",r.createElement("sub",null,"2")," as well (rotated along its tangent), so we'll focus on finding the location of C",r.createElement("sub",null,"1")," only. So here's where we do something that you might not expect: we're going to ignore for a moment, because we're going to have a much easier time if we just solve this problem with geometry first, then move to calculus to solve a much simpler problem."),r.createElement("p",null,"If we look at the triangle that is formed between our starting point, or initial guess C",r.createElement("sub",null,"1"),"and our real C",r.createElement("sub",null,"1"),", there's something funny going on: if we treat the line ","{","start,guess","}"," as our opposite side, the line ","{","guess,real","}"," as our adjacent side, with ","{","start,real","}"," our hypothenuse, then the angle for the corner hypothenuse/adjacent is half that of the arc we're covering. Try it: if you place the end point at a quarter circle (pi/2, or 90 degrees), the angle in our triangle is half a quarter (pi/4, or 45 degrees). With that knowledge, and a knowledge of what the length of any of our lines segments are (as a function), we can determine where our control points are, and thus have everything we need to find the error distance function. Of the three lines, the one we can easiest determine is ","{","start,guess","}",", so let's find out what the guessed control point is. Again geometrically, because we have the benefit of an on-curve ",r.createElement("i",null,"t=0.5")," value."),r.createElement("p",null,"The distance from our guessed point to the start point is exactly the same as the projection distance we looked at earlier. Using ",r.createElement("i",null,"t=0.5"),' as our point "B" in the "A,B,C" projection, then we know the length of the line segment ',"{","C,A","}",", since it's d",r.createElement("sub",null,"1")," = ","{","A,B","}"," + d",r.createElement("sub",null,"2")," = ","{","B,C","}",":"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/b15a274c1e0a6aeeaf517b5d2c8ee0a7997dd617.svg",width:"417.2rem",height:"42rem"})),r.createElement("p",null,"So that just leaves us to find the distance from ",r.createElement("i",null,"t=0.5")," to the baseline for an arbitrary angle φ, which is the distance from the centre of the circle to our ",r.createElement("i",null,"t=0.5")," point, minus the distance from the centre to the line that runs from start point to end point. The first is the same as the point P we found for the quadratic curve:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/0b80423188012451e0400f473c19729eb2bad654.svg",width:"197.39999999999998rem",height:"33.599999999999994rem"})),r.createElement("p",null,"And the distance from the origin to the line start/end is another application of angles, since the triangle ","{","origin,start,C","}"," has known angles, and two known sides. 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 draw:this.draw,onKeyDown:this.props.onKeyDown}))}});e.exports=o(s)},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="inflections",s=r.createClass({displayName:"ABC",getDefaultProps:function(){return{title:a.getTitle(o)}},setupCubic:function(e){var t=new e.Bezier(135,25,25,135,215,75,215,240);e.setCurve(t)},draw:function(e,t){e.reset(),e.drawSkeleton(t),e.drawCurve(t),e.setColor("red"),t.inflections().forEach(function(n){e.drawCircle(t.get(n),5)})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="intersections",s=Math.min,l=Math.max,u=r.createClass({displayName:"Intersections",getDefaultProps:function(){return{title:"Intersections"}},setupLines:function(e){var t=new e.Bezier([50,50,150,110]),n=new e.Bezier([50,250,170,170]);e.setCurve(t,n)},drawLineIntersection:function(e,t){e.reset();var n=e.utils.lli4,r=n(t[0].points[0],t[0].points[1],t[1].points[0],t[1].points[1]),i=0;t.forEach(function(t){if(e.drawSkeleton(t),e.setColor("black"),r){var n=t.points,a=s(n[0].x,n[1].x),o=s(n[0].y,n[1].y),u=l(n[0].x,n[1].x),c=l(n[0].y,n[1].y);a<=r.x&&o<=r.y&&u>=r.x&&c>=r.y&&(e.setColor("#00FF00"),i++)}e.drawCurve(t)}),r&&(e.setColor(i<2?"red":"#00FF00"),e.drawCircle(r,3))},setupQuadratic:function(e){var t=e.getDefaultQuadratic(),n=new e.Bezier([15,250,220,20]);e.setCurve(t,n)},setupCubic:function(e){var t=new e.Bezier([100,240,30,60,210,230,160,30]),n=new e.Bezier([25,260,230,20]);e.setCurve(t,n)},draw:function(e,t){e.reset(),t.forEach(function(t){e.drawSkeleton(t),e.drawCurve(t)});var n=e.utils,r={p1:t[1].points[0],p2:t[1].points[1]},i=n.align(t[0].points,r),a=new e.Bezier(i),o=n.roots(a.points);o.forEach(function(n){var r=t[0].get(n);e.drawCircle(r,3),e.text("t = "+n,{x:r.x+5,y:r.y+10})})},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=u},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="introduction",s=r.createClass({displayName:"Introduction",getDefaultProps:function(){return{title:a.getTitle(o)}},drawQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t)},drawCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t)},drawCurve:function(e,t){e.reset(),e.drawSkeleton(t),e.drawCurve(t)},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="matrix",s=r.createClass({displayName:"Matrix",getDefaultProps:function(){return{title:a.getTitle(o)}},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="matrixsplit",s=r.createClass({displayName:"MatrixSplit",getDefaultProps:function(){return{title:a.getTitle(o)}},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="moulding",s=Math.abs,l=r.createClass({displayName:"Moulding",getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){e.setPanelCount(3);var t=e.getDefaultQuadratic();t.points[2].x-=30,e.setCurve(t)},setupCubic:function(e){e.setPanelCount(3);var t=new e.Bezier([100,230,30,160,200,50,210,160]);t.points[2].y-=20,e.setCurve(t),e.lut=t.getLUT(100)},saveCurve:function(e,t){t.t&&(t.setCurve(t.newcurve),t.t=!1,t.redraw())},findTValue:function(e,t){var n=t.curve.on({x:e.offsetX,y:e.offsetY},7);return!(n<.05||n>.95)&&n},markQB:function(e,t){if(t.t=this.findTValue(e,t),t.t){var n=t.t,r=2*n,i=r*n-r,a=i+1,o=s(i/a),l=t.curve,u=t.A=l.points[1],c=t.B=l.get(n);t.C=t.utils.lli4(u,c,l.points[0],l.points[2]),t.ratio=o}},markCB:function(e,t){if(t.t=this.findTValue(e,t),t.t){var n=t.t,r=1-n,i=n*n*n,a=r*r*r,o=i+a,l=o-1,u=s(l/o),c=t.curve,h=c.hull(n),d=t.A=h[5],f=t.B=c.get(n);t.db=c.derivative(n),t.C=t.utils.lli4(d,f,c.points[0],c.points[3]),t.ratio=u}},drag:function(e,t){if(t.t){var n=t.newB={x:e.offsetX,y:e.offsetY};t.newA={x:n.x-(t.C.x-n.x)/t.ratio,y:n.y-(t.C.y-n.y)/t.ratio}}},dragQB:function(e,t){t.t&&(this.drag(e,t),t.update=[t.newA])},dragCB:function(e,t){if(t.t){this.drag(e,t);var n=t.curve,r=n.hull(t.t),i=t.B,a=r[7],o=r[8],s={x:a.x-i.x,y:a.y-i.y},l={x:o.x-i.x,y:o.y-i.y},u=n.points,c={x:t.newB.x+s.x,y:t.newB.y+s.y},h={x:t.newA.x-(t.newA.x-c.x)/(1-t.t),y:t.newA.y-(t.newA.y-c.y)/(1-t.t)},d={x:t.newB.x+l.x,y:t.newB.y+l.y},f={x:t.newA.x+(d.x-t.newA.x)/t.t,y:t.newA.y+(d.y-t.newA.y)/t.t},p={x:u[0].x+(h.x-u[0].x)/t.t,y:u[0].y+(h.y-u[0].y)/t.t},m={x:u[3].x-(u[3].x-f.x)/(1-t.t),y:u[3].y-(u[3].y-f.y)/(1-t.t)};t.p1=c,t.p2=d,t.sc1=h,t.sc2=f,t.nc1=p,t.nc2=m,t.update=[p,m]}},drawMould:function(e,t){e.reset(),e.drawSkeleton(t),e.drawCurve(t);var n=e.getPanelWidth(),r=e.getPanelHeight(),i={x:n,y:0},a=e.utils.round;if(e.setColor("black"),e.drawLine({x:0,y:0},{x:0,y:r},i),e.drawLine({x:n,y:0},{x:n,y:r},i),e.t){e.drawCircle(t.get(e.t),3),e.npts=[t.points[0]].concat(e.update).concat([t.points.slice(-1)[0]]),e.newcurve=new e.Bezier(e.npts),e.setColor("lightgrey"),e.drawCurve(e.newcurve);var o=e.drawHull(e.newcurve,e.t,i);if(e.drawLine(e.npts[0],e.npts.slice(-1)[0],i),e.drawLine(e.newA,e.newB,i),e.setColor("grey"),e.drawCircle(e.newA,3,i),e.setColor("blue"),e.drawCircle(e.B,3,i),e.drawCircle(e.C,3,i),e.drawCircle(e.newB,3,i),e.drawLine(e.B,e.C,i),e.drawLine(e.newB,e.C,i),e.setFill("black"),e.text("A'",e.newA,{x:i.x+7,y:i.y+1}),e.text("start",t.get(0),{x:i.x+7,y:i.y+1}),e.text("end",t.get(1),{x:i.x+7,y:i.y+1}),e.setFill("blue"),e.text("B'",e.newB,{x:i.x+7,y:i.y+1}),e.text("B, at t = "+a(e.t,2),e.B,{x:i.x+7,y:i.y+1}),e.text("C",e.C,{x:i.x+7,y:i.y+1}),3===t.order){var s=t.hull(e.t);e.drawLine(s[7],s[8],i),e.drawLine(o[7],o[8],i),e.drawCircle(o[7],3,i),e.drawCircle(o[8],3,i),e.text("e1",o[7],{x:i.x+7,y:i.y+1}),e.text("e2",o[8],{x:i.x+7,y:i.y+1})}i.x+=n,e.setColor("lightgrey"),e.drawSkeleton(e.newcurve,i),e.setColor("black"),e.drawCurve(e.newcurve,i)}else i.x+=n,e.drawCurve(t,i)},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=l},function(e,t,n){"use strict";var r=n(0),i=n(14),a=n(9),o=n(15),s=r.createClass({displayName:"Offsetting",statics:{keyHandlingOptions:{propName:"distance",values:{38:1,40:-1}}},getDefaultProps:function(){return{title:"Curve offsetting"}},setup:function(e,t){e.setCurve(t),e.distance=20},setupQuadratic:function(e){var t=e.getDefaultQuadratic();this.setup(e,t)},setupCubic:function(e){var t=e.getDefaultCubic();this.setup(e,t)},draw:function(e,t){e.reset(),e.drawSkeleton(t);var n=t.reduce();n.forEach(function(t){e.setRandomColor(),e.drawCurve(t),e.drawCircle(t.points[0],1)});var r=n.slice(-1)[0];e.drawPoint(r.points[3]||r.points[2]),e.setColor("red");var i=t.offset(e.distance);i.forEach(function(t){e.drawPoint(t.points[0]),e.drawCurve(t)}),r=i.slice(-1)[0],e.drawPoint(r.points[3]||r.points[2]),e.setColor("blue"),i=t.offset(-e.distance),i.forEach(function(t){e.drawPoint(t.points[0]),e.drawCurve(t)}),r=i.slice(-1)[0],e.drawPoint(r.points[3]||r.points[2])},render:function(){return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"Perhaps you are like me, and you've been writing various small programs that use Bézier curves in some way or another, and at some point you make the step to implementing path extrusion. But you don't want to do it pixel based, you want to stay in the vector world. You find that extruding lines is relatively easy, and tracing outlines is coming along nicely (although junction caps and fillets are a bit of a hassle), and then decide to do things properly and add Bézier curves to the mix. Now you have a problem."),r.createElement("p",null,"Unlike lines, you can't simply extrude a Bézier curve by taking a copy and moving it around, because of the curvatures; rather than a uniform thickness you get an extrusion that looks too thin in places, if you're lucky, but more likely will self-intersect. The trick, then, is to scale the curve, rather than simply copying it. But how do you scale a Bézier curve?"),r.createElement("p",null,"Bottom line: ",r.createElement("strong",null,"you can't"),". So you cheat. We're not going to do true curve scaling, or rather curve offsetting, because that's impossible. Instead we're going to try to generate 'looks good enough' offset curves."),r.createElement("div",{className:"note"},r.createElement("h2",null,'"What do you mean, you can\'t. Prove it."'),r.createElement("p",null,'First off, when I say "you can\'t" what I really mean is "you can\'t offset a Bézier curve with another Bézier curve". not even by using a really high order curve. You can find the function that describes the offset curve, but it won\'t be a polynomial, and as such it cannot be represented as a Bézier curve, which',r.createElement("strong",null,"has")," to be a polynomial. Let's look at why this is:"),r.createElement("p",null,"From a mathematical point of view, an offset curve ",r.createElement("i",null,"O(t)")," is a curve such that, given our original curve",r.createElement("i",null,"B(t)"),", any point on ",r.createElement("i",null,"O(t)")," is a fixed distance ",r.createElement("i",null,"d")," away from coordinate ",r.createElement("i",null,"B(t)"),". So let's math that:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/3aff5cef0028337bbb48ae64ad30000c4d5e238f.svg",width:"113.39999999999999rem",height:"16.799999999999997rem"})),r.createElement("p",null,"However, we're working in 2D, and ",r.createElement("i",null,"d")," is a single value, so we want to turn it into a vector. If we want a point distance ",r.createElement("i",null,"d"),' "away" from the curve ',r.createElement("i",null,"B(t)")," then what we really mean is that we want a point at ",r.createElement("i",null,"d"),' times the "normal vector" from point ',r.createElement("i",null,"B(t)"),', where the "normal" is a vector that runs perpendicular ("at a right angle") to the tangent at ',r.createElement("i",null,"B(t)"),". Easy enough:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/2cf48e2f8525258a3fa0fe4f10ec2acef67104b3.svg",width:"158.2rem",height:"16.799999999999997rem"})),r.createElement("p",null,"Now this still isn't very useful unless we know what the formula for ",r.createElement("i",null,"N(t)")," is, so let's find out.",r.createElement("i",null,"N(t)")," runs perpendicular to the original curve tangent, and we know that the tangent is simply",r.createElement("i",null,"B'(t)"),", so we could just rotate that 90 degrees and be done with it. However, we need to ensure that ",r.createElement("i",null,"N(t)")," has the same magnitude for every ",r.createElement("i",null,"t"),", or the offset curve won't be at a uniform distance, thus not being an offset curve at all. The easiest way to guarantee this is to make sure",r.createElement("i",null,"N(t)")," always has length 1, which we can achieve by dividing ",r.createElement("i",null,"B'(t)")," by its magnitude:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/4941ecbff4c50732ba66fec53307456fc605f032.svg",width:"125.99999999999999rem",height:"42rem"})),r.createElement("p",null,"Determining the length requires computing an arc length, and this is where things get Tricky with a capital T. First off, to compute arc length from some start ",r.createElement("i",null,"a")," to end ",r.createElement("i",null,"b"),', we must use the formula we saw earlier. Noting that "length" is usually denoted with double vertical bars:'),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/f6d8c2965b02363e092acb00bbc1398cfbb170a4.svg",width:"177.79999999999998rem",height:"37.8rem"})),r.createElement("p",null,"So if we want the length of the tangent, we plug in ",r.createElement("i",null,"B'(t)"),", with ",r.createElement("i",null,"t = 0")," as start and",r.createElement("i",null,"t = 1")," as end:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/1f024282044316a9e4b3de2c855d2ceb96aff056.svg",width:"219.79999999999998rem",height:"37.8rem"})),r.createElement("p",null,"And that's where things go wrong. It doesn't even really matter what the second derivative for ",r.createElement("i",null,"B(t)"),"is, that square root is screwing everything up, because it turns our nice polynomials into things that are no longer polynomials."),r.createElement("p",null,"There is a small class of polynomials where the square root is also a polynomial, but they're utterly useless to us: any polynomial with unweighted binomial coefficients has a square root that is also a polynomial. Now, you might think that Bézier curves are just fine because they do, but they don't; remember that only the ",r.createElement("strong",null,"base")," function has binomial coefficients. That's before we factor in our coordinates, which turn it into a non-binomial polygon. The only way to make sure the functions stay binomial is to make all our coordinates have the same value. And that's not a curve, that's a point. We can already create offset curves for points, we call them circles, and they have much simpler functions than Bézier curves."),r.createElement("p",null,"So, since the tangent length isn't a polynomial, the normalised tangent won't be a polynomial either, which means ",r.createElement("i",null,"N(t)")," won't be a polynomial, which means that ",r.createElement("i",null,"d")," times ",r.createElement("i",null,"N(t)")," won't be a polynomial, which means that, ultimately, ",r.createElement("i",null,"O(t)")," won't be a polynomial, which means that even if we can determine the function for ",r.createElement("i",null,"O(t)")," just fine (and that's far from trivial!), it simply cannot be represented as a Bézier curve."),r.createElement("p",null,"And that's one reason why Bézier curves are tricky: there are actually a ",r.createElement("i",null,"lot")," of curves that cannot be represent as a Bézier curve at all. They can't even model their own offset curves. They're weird that way. So how do all those other programs do it? Well, much like we're about to do, they cheat. We're going to approximate an offset curve in a way that will look relatively close to what the real offset curve would look like, if we could compute it.")),r.createElement("p",null,'So, you cannot offset a Bézier curve perfectly with another Bézier curve, no matter how high-order you make that other Bézier curve. However, we can chop up a curve into "safe" sub-curves (where safe means that all the control points are always on a single side of the baseline, and the midpoint of the curve at ',r.createElement("i",null,"t=0.5")," is roughly in the centre of the polygon defined by the curve coordinates) and then point-scale those sub-curves with respect to the curve's scaling origin (which is the intersection of the point normals at the start and end points)."),r.createElement("p",null,"A good way to do this reduction is to first find the curve's extreme points, as explained in the earlier section on curve extremities, and use these as initial splitting points. After this initial split, we can check each individual segment to see if it's \"safe enough\" based on where the center of the curve is. If the on-curve point for ",r.createElement("i",null,"t=0.5")," is too far off from the center, we simply split the segment down the middle. Generally this is more than enough to end up with safe segments."),r.createElement("p",null,"The following graphics show off curve offsetting, and you can use your up and down arrow keys to control the distance at which the curve gets offset. The curve first gets reduced to safe segments, each of which is then offset at the desired distance. Especially for simple curves, particularly easily set up for quadratic curves, no reduction is necessary, but the more twisty the curve gets, the more the curve needs to be reduced in order to get segments that can safely be scaled."),r.createElement(i,{preset:"simple",title:"Offsetting a quadratic Bézier curve",setup:this.setupQuadratic,draw:this.draw,onKeyDown:this.props.onKeyDown}),r.createElement(i,{preset:"simple",title:"Offsetting a cubic Bézier curve",setup:this.setupCubic,draw:this.draw,onKeyDown:this.props.onKeyDown}),r.createElement("p",null,"You may notice that this may still lead to small 'jumps' in the sub-curves when moving the curve around. This is caused by the fact that we're still performing a naive form of offsetting, moving the control points the same distance as the start and end points. If the curve is large enough, this may still lead to incorrect offsets."))}});e.exports=o(s)},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="pointcurves",s=Math.abs,l=r.createClass({displayName:"PointCurves",getDefaultProps:function(){return{title:a.getTitle(o)}},setup:function(e){e.lpts=[{x:56,y:153},{x:144,y:83},{x:188,y:185}]},onClick:function(e,t){3==t.lpts.length&&(t.lpts=[]),t.lpts.push({x:e.offsetX,y:e.offsetY}),t.redraw()},getQRatio:function(e){var t=2*e,n=t*e-t,r=n+1;return s(n/r)},getCRatio:function(e){var t=1-e,n=e*e*e,r=t*t*t,i=n+r,a=i-1;return s(a/i)},drawQuadratic:function(e,t){var n=["start","t=0.5","end"];if(e.reset(),e.setColor("lightblue"),e.drawGrid(10,10),e.setFill("black"),e.setColor("black"),e.lpts.forEach(function(t,r){e.drawCircle(t,3),e.text(n[r],t,{x:5,y:2})}),3===e.lpts.length){var r=e.lpts[0],i=e.lpts[2],a=e.lpts[1],o={x:(r.x+i.x)/2,y:(r.y+i.y)/2};e.setColor("blue"),e.drawLine(r,i),e.drawLine(a,o),e.drawCircle(o,3);var s=this.getQRatio(.5),l={x:a.x+(a.x-o.x)/s,y:a.y+(a.y-o.y)/s};t=new e.Bezier([r,l,i]),e.setColor("lightgrey"),e.drawLine(l,a),e.drawLine(l,r),e.drawLine(l,i),e.setColor("black"),e.drawCircle(l,1),e.drawCurve(t)}},drawCubic:function(e,t){var n=["start","t=0.5","end"];if(e.reset(),e.setFill("black"),e.setColor("black"),e.lpts.forEach(function(t,r){e.drawCircle(t,3),e.text(n[r],t,{x:5,y:2})}),e.setColor("lightblue"),e.drawGrid(10,10),3===e.lpts.length){var r=e.lpts[0],i=e.lpts[2],a=e.lpts[1],o={x:(r.x+i.x)/2,y:(r.y+i.y)/2};e.setColor("blue"),e.drawLine(r,i),e.drawLine(a,o),e.drawCircle(o,1);var s=this.getCRatio(.5),l={x:a.x+(a.x-o.x)/s,y:a.y+(a.y-o.y)/s},u=e.utils.dist(r,i),c=u/8,h=e.utils.dist(a,o),d=4,f=c+h/d,p=f*(i.x-r.x)/u,m=f*(i.y-r.y)/u,g={x:a.x-p,y:a.y-m},v={x:a.x+p,y:a.y+m},w={x:l.x+2*(g.x-l.x),y:l.y+2*(g.y-l.y)},y={x:l.x+2*(v.x-l.x),y:l.y+2*(v.y-l.y)},b={x:r.x+2*(w.x-r.x),y:r.y+2*(w.y-r.y)},_={x:i.x+2*(y.x-i.x),y:i.y+2*(y.y-i.y)};t=new e.Bezier([r,b,_,i]),e.drawLine(g,v),e.setColor("lightgrey"),e.drawLine(l,o),e.drawLine(l,w),e.drawLine(l,y),e.drawLine(r,b),e.drawLine(i,_),e.drawLine(b,_),e.setColor("black"),e.drawCircle(l,1),e.drawCircle(b,1),e.drawCircle(_,1),e.drawCurve(t)}},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=l},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="pointvectors",s=r.createClass({displayName:"PointVectors",getDefaultProps:function(){return{title:a.getTitle(o)}},setupQuadratic:function(e){var t=e.getDefaultQuadratic();e.setCurve(t)},setupCubic:function(e){var t=e.getDefaultCubic();e.setCurve(t)},draw:function(e,t){e.reset(),e.drawSkeleton(t);var n,r,i,a,o,s,l=20;for(n=0;n<=10;n++)r=n/10,i=t.get(r),a=t.derivative(r),s=Math.sqrt(a.x*a.x+a.y*a.y),a={x:a.x/s,y:a.y/s},o=t.normal(r),e.setColor("blue"),e.drawLine(i,{x:i.x+a.x*l,y:i.y+a.y*l}),e.setColor("red"),e.drawLine(i,{x:i.x+o.x*l,y:i.y+o.y*l}),e.setColor("black"),e.drawCircle(i,3)},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(14),a=n(9),o=Math.atan2,s=Math.sqrt,l=Math.sin,u=Math.cos,c=r.createClass({displayName:"PolyBezier",getDefaultProps:function(){return{title:"Forming poly-Bézier curves"}},setupQuadratic:function(e){var t=e.getPanelWidth(),n=e.getPanelHeight(),r=t/2,i=n/2,a=40,o=[{x:r,y:a},{x:t-a,y:a},{x:t-a,y:i},{x:t-a,y:n-a},{x:r,y:n-a},{x:a,y:n-a},{x:a,y:i},{x:a,y:a}];e.lpts=o},setupCubic:function(e){var t=e.getPanelWidth(),n=e.getPanelHeight(),r=t/2,i=n/2,a=40,o=(t-2*a)/2,s=.55228,l=s*o,u=[{x:r,y:a},{x:r+l,y:a},{x:t-a,y:i-l},{x:t-a,y:i},{x:t-a,y:i+l},{x:r+l,y:n-a},{x:r,y:n-a},{x:r-l,y:n-a},{x:a,y:i+l},{x:a,y:i},{x:a,y:i-l},{x:r-l,y:a}];e.lpts=u},movePointsQuadraticLD:function(e,t){for(var n,r,i,a=1;a<4;a++)n=t+(2*a-2)+e.lpts.length,n=e.lpts[n%e.lpts.length],r=t+(2*a-1),r=e.lpts[r%e.lpts.length],i=t+2*a,i=e.lpts[i%e.lpts.length],i.x=r.x+(r.x-n.x),i.y=r.y+(r.y-n.y);i=t+6,i=e.lpts[i%e.lpts.length],e.problem=i},movePointsCubicLD:function(e,t){var n,r;t%3===1?(r=t-1,r+=r<0?e.lpts.length:0,n=t-2,n+=n<0?e.lpts.length:0):(r=(t+1)%e.lpts.length,n=(t+2)%e.lpts.length),r=e.lpts[r],n=e.lpts[n],n.x=r.x+(r.x-e.mp.x),n.y=r.y+(r.y-e.mp.y)},linkDerivatives:function(e,t){if(t.mp){var n=8===t.lpts.length,r=t.mp_idx;n?r%2!==0&&this.movePointsQuadraticLD(t,r):r%3!==0&&this.movePointsCubicLD(t,r)}},movePointsQuadraticDirOnly:function(e,t){var n,r,i;[-1,1].forEach(function(a){n=e.mp,r=t+a+e.lpts.length,r=e.lpts[r%e.lpts.length],i=t+2*a+e.lpts.length,i=e.lpts[i%e.lpts.length];var c=o(r.y-n.y,r.x-n.x),h=i.x-r.x,d=i.y-r.y,f=s(h*h+d*d);i.x=r.x+f*u(c),i.y=r.y+f*l(c)}),i=t+4,i=e.lpts[i%e.lpts.length],e.problem=i},movePointsCubicDirOnly:function(e,t){var n,r;t%3===1?(r=t-1,r+=r<0?e.lpts.length:0,n=t-2,n+=n<0?e.lpts.length:0):(r=(t+1)%e.lpts.length,n=(t+2)%e.lpts.length),r=e.lpts[r],n=e.lpts[n];var i=o(r.y-e.mp.y,r.x-e.mp.x),a=n.x-r.x,c=n.y-r.y,h=s(a*a+c*c);n.x=r.x+h*u(i),n.y=r.y+h*l(i)},linkDirection:function(e,t){if(t.mp){var n=8===t.lpts.length,r=t.mp_idx;n?r%2!==0&&this.movePointsQuadraticDirOnly(t,r):r%3!==0&&this.movePointsCubicDirOnly(t,r)}},bufferPoints:function(e,t){t.bpts=JSON.parse(JSON.stringify(t.lpts))},moveQuadraticPoint:function(e,t){this.moveCubicPoint(e,t),[-1,1].forEach(function(n){var r=t-n+e.lpts.length;r=e.lpts[r%e.lpts.length];var i=t-2*n+e.lpts.length;i=e.lpts[i%e.lpts.length];var a=t-3*n+e.lpts.length;a=e.lpts[a%e.lpts.length];var c=o(i.y-r.y,i.x-r.x),h=a.x-i.x,d=a.y-i.y,f=s(h*h+d*d);a.x=i.x+f*u(c),a.y=i.y+f*l(c)});var n=t+4;n=e.lpts[n%e.lpts.length],e.problem=n},moveCubicPoint:function(e,t){var n=e.bpts[t],r=e.lpts[t],i=r.x-n.x,a=r.y-n.y,o=e.lpts.length,s=t-1+o,l=t+1,u=e.bpts[s%o],c=e.bpts[l%o],h=e.lpts[s%o],d=e.lpts[l%o];return h.x=u.x+i,h.y=u.y+a,d.x=c.x+i,d.y=c.y+a,{x:i,y:a}},modelCurve:function(e,t){if(t.mp){var n=8===t.lpts.length,r=t.mp_idx;n?r%2!==0?this.movePointsQuadraticDirOnly(t,r):this.moveQuadraticPoint(t,r):r%3!==0?this.movePointsCubicDirOnly(t,r):this.moveCubicPoint(t,r)}},draw:function(e,t){e.reset();var n=e.lpts,r=8===n.length,i=r?new e.Bezier(n[0],n[1],n[2]):new e.Bezier(n[0],n[1],n[2],n[3]);e.drawSkeleton(i,!1,!0),e.drawCurve(i);var a=r?new e.Bezier(n[2],n[3],n[4]):new e.Bezier(n[3],n[4],n[5],n[6]);e.drawSkeleton(a,!1,!0),e.drawCurve(a);var o=r?new e.Bezier(n[4],n[5],n[6]):new e.Bezier(n[6],n[7],n[8],n[9]);e.drawSkeleton(o,!1,!0),e.drawCurve(o);var s=r?new e.Bezier(n[6],n[7],n[0]):new e.Bezier(n[9],n[10],n[11],n[0]);e.drawSkeleton(s,!1,!0),e.drawCurve(s),e.problem&&(e.setColor("red"),e.drawCircle(e.problem,5))},render:function(){return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"Much like lines can be chained together to form polygons, Bézier curves can be chained together to form poly-Béziers, and the only trick required is to make sure that:"),r.createElement("ol",null,r.createElement("li",null,"the end point of each section is the starting point of the following section, and"),r.createElement("li",null,"the derivatives across that dual point line up.")),r.createElement("p",null,"Unless, of course, you want discontinuities; then you don't even need 2."),r.createElement("p",null,"We'll cover three forms of poly-Bézier curves in this section. First, we'll look at the kind that just follows point 1. where the end point of a segment is the same point as the start point of the next segment. This leads to poly-Béziers that are pretty hard to work with, but they're the easiest to implement:"),r.createElement(i,{preset:"poly",title:"Unlinked quadratic poly-Bézier",setup:this.setupQuadratic,draw:this.draw}),r.createElement(i,{preset:"poly",title:"Unlinked cubic poly-Bézier",setup:this.setupCubic,draw:this.draw}),r.createElement("p",null,'Dragging the control points around only affects the curve segments that the control point belongs to, and moving an on-curve point leaves the control points where they are, which is not the most useful for practical modelling purposes. So, let\'s add in the logic we need to make things a little better. We\'ll start by linking up control points by ensuring that the "incoming" derivative at an on-curve point is the same as it\'s "outgoing" derivative:'),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/37740bb1a0b7b1ff48bf3454e52295fc717cacbb.svg",width:"130.2rem",height:"18.2rem"})),r.createElement("p",null,"We can effect this quite easily, because we know that the vector from a curve's last control point to its last on-curve point is equal to the derivative vector. If we want to ensure that the first control point of the next curve matches that, all we have to do is mirror that last control point through the last on-curve point. And mirroring any point A through any point B is really simple:"),r.createElement("p",null,r.createElement("img",{className:"LaTeX SVG",src:"images/latex/ce6e3939608c4ed0598107b06543c2301b91bb7f.svg",width:"319.2rem",height:"42rem"})),r.createElement("p",null,"So let's implement that and see what it gets us. The following two graphics show a quadratic and a cubic poly-Bézier curve again, but this time moving the control points around moves others, too. However, you might see something unexpected going on for quadratic curves..."),r.createElement(i,{preset:"poly",title:"Loosely connected quadratic poly-Bézier",setup:this.setupQuadratic,draw:this.draw,onMouseMove:this.linkDerivatives}),r.createElement(i,{preset:"poly",title:"Loosely connected cubic poly-Bézier",setup:this.setupCubic,draw:this.draw,onMouseMove:this.linkDerivatives}),r.createElement("p",null,"As you can see, quadratic curves are particularly ill-suited for poly-Bézier curves, as all the control points are effectively linked. Move one of them, and you move all of them. Not only that, but if we move the on-curve points, it's possible to get a situation where a control point's positions is different depending on whether it's the reflection of its left or right neighbouring control point: we can't even form a proper rule-conforming curve! This means that we cannot use quadratic poly-Béziers for anything other than really, really simple shapes. And even then, they're probably the wrong choice. Cubic curves are pretty decent, but the fact that the derivatives are linked means we can't manipulate curves as well as we might if we relaxed the constraints a little."),r.createElement("p",null,"So: let's relax the requirement a little."),r.createElement("p",null,"We can change the constraint so that we still preserve the ",r.createElement("em",null,"angle")," of the derivatives across sections (so transitions from one section to the next will still look natural), but give up the requirement that they should also have the same ",r.createElement("em",null,"vector length"),". Doing so will give us a much more useful kind of poly-Bézier curve:"),r.createElement(i,{preset:"poly",title:"Loosely connected quadratic poly-Bézier",setup:this.setupQuadratic,draw:this.draw,onMouseMove:this.linkDirection}),r.createElement(i,{preset:"poly",title:"Loosely connected cubic poly-Bézier",setup:this.setupCubic,draw:this.draw,onMouseMove:this.linkDirection}),r.createElement("p",null,"Cubic curves are now better behaved when it comes to dragging control points around, but the quadratic poly-Bézier still has the problem that moving one control points will move the control points and may ending up defining \"the next\" control point in a way that doesn't work. Quadratic curves really aren't very useful to work with..."),r.createElement("p",null,"Finally, we also want to make sure that moving the on-curve coordinates preserves the relative positions of the associated control points. With that, we get to the kind of curve control that you might be familiar with from applications like Photoshop, Inkscape, Blender, etc."),r.createElement(i,{preset:"poly",title:"Loosely connected quadratic poly-Bézier",setup:this.setupQuadratic,draw:this.draw,onMouseDown:this.bufferPoints,onMouseMove:this.modelCurve}),r.createElement(i,{preset:"poly",title:"Loosely connected cubic poly-Bézier",setup:this.setupCubic,draw:this.draw,onMouseDown:this.bufferPoints,onMouseMove:this.modelCurve}),r.createElement("p",null,'Again, we see that cubic curves are now rather nice to work with, but quadratic curves have a new, very serious problem: we can move an on-curve point in such a way that we can\'t compute what needs to "happen next". Move the top point down, below the left and right points, for instance. There is no way to preserve correct control points without a kink at the bottom point. Quadratic curves: just not that good...'),r.createElement("p",null,'A final improvement is to offer fine-level control over which points behave which, so that you can have "kinks" or individually controlled segments when you need them, with nicely well-behaved curves for the rest of the path. Implementing that, is left as an excercise for the reader.'))}});e.exports=c},function(e,t,n){"use strict";var r=n(0),i=n(2),a=new i,o="preface",s=r.createClass({displayName:"Preface",getDefaultProps:function(){return{title:a.getTitle(o)}},render:function(){return r.createElement("section",null,a.getContent(o,this))}});e.exports=s},function(e,t,n){"use strict";var r=n(0),i=n(14),a=n(9),o=r.createClass({displayName:"Projections",getDefaultProps:function(){return{title:"Projecting a point onto a Bézier curve"}},setup:function(e){e.setSize(320,320);var t=new e.Bezier([{x:248,y:188},{x:218,y:294},{x:45,y:290},{x:12,y:236},{x:14,y:82},{x:186,y:177},{x:221,y:90},{x:18,y:156},{x:34,y:57},{x:198,y:18}]);e.setCurve(t),e._lut=t.getLUT()},findClosest:function(e,t,n){var r,i,a=e.length,o=n(e[0],t),s=0;for(r=1;r<a;r++)i=n(e[r],t),i<o&&(s=r,o=i);return s/(a-1)},draw:function(e,t){if(e.reset(),e.drawSkeleton(t),e.drawCurve(t),e.mousePt){e.setColor("red"),e.setFill("red"),e.drawCircle(e.mousePt,3);var n=this.findClosest(e._lut,e.mousePt,e.utils.dist),r=t.get(n);e.drawLine(r,e.mousePt),e.drawCircle(r,3),e.text("t = "+e.utils.round(n,2),r,{x:10,y:3})}},onMouseMove:function(e,t){t.mousePt={x:e.offsetX,y:e.offsetY},t._lut=t.curve.getLUT()},render:function(){return r.createElement("section",null,r.createElement(a,this.props),r.createElement("p",null,"Say we have a Bézier curve and some point, not on the curve, of which we want to know which ",r.createElement("i",null,"t")," value on the curve gives us an on-curve point closest to our off-curve point. Or: say we want to find the projection of a random point onto a curve. How do we do that?"),r.createElement("p",null,"If the Bézier curve is of low enough order, we might be able to ",r.createElement("a",{href:"http://jazzros.blogspot.ca/2011/03/projecting-point-on-bezier-curve.html"},"work out the maths for how to do this"),", and get a perfect ",r.createElement("i",null,"t")," value back, but in general this is an incredibly hard problem and the easiest solution is, really, a numerical approach again. We'll be finding our ideal ",r.createElement("i",null,"t")," value using a ",r.createElement("a",{href:"https://en.wikipedia.org/wiki/Binary_search_algorithm"},"binary search"),". First, we do a coarse distance-check based on ",r.createElement("i",null,"t"),' values associated with the curve\'s "to draw" coordinates (using a lookup table, or LUT). This is pretty fast. Then we run this algorithm:'),r.createElement("ol",null,r.createElement("li",null,"with the ",r.createElement("i",null,"t")," value we found, start with some small interval around ",r.createElement("i",null,"t")," (1/length_of_LUT on either side is a reasonable start),"),r.createElement("li",null,"if the distance to ",r.createElement("i",null,"t ± interval/2")," is larger than the distance to ",r.createElement("i",null,"t"),", try again with the interval reduced to half its original length."),r.createElement("li",null,"if the distance to ",r.createElement("i",null,"t ± interval/2")," is smaller than the distance to ",r.createElement("i",null,"t"),", replace ",r.createElement("i",null,"t")," with the smaller-distance value."),r.createElement("li",null,"after reducing the interval, or changing ",r.createElement("i",null,"t"),", go back to step 1.")),r.createElement("p",null,"We keep repeating this process until the interval is small enough to claim the difference in precision found is irrelevant for the purpose we're trying to find ",r.createElement("i",null,"t")," for. In this case, I'm arbitrarily fixing it at 0.0001."),r.createElement("p",null,'The following graphic demonstrates the result of this procedure.Simply move the cursor around, and if it does not lie on top of the curve, you will see a line that projects the cursor onto the curve based on an iteratively found "ideal" ',r.createElement("i",null,"t")," value."),r.createElement(i,{
diff --git a/en-GB/index.html b/en-GB/index.html
index 12d23c27..0eb99107 100644
--- a/en-GB/index.html
+++ b/en-GB/index.html
@@ -4,6 +4,23 @@
     <meta charset="utf-8">
     <title>A Primer on Bézier Curves</title>
 
+    <script>
+      (function() {
+        // force https except during local testing (e.g. not on port 80)
+        let loc = window.location.toString();
+        if (loc.indexOf(":8080") === -1 && loc.indexOf("http:") === 0) {
+          window.location = loc.replace("http:","https:");
+        }
+        // if we're still here, load google analytics
+        (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
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diff --git a/index.html b/index.html
index 866b8c6d..c7495025 100644
--- a/index.html
+++ b/index.html
@@ -5,6 +5,23 @@
     <meta charset="utf-8">
     <title>A Primer on Bézier Curves</title>
 
+    <script>
+      (function() {
+        // force https except during local testing (e.g. not on port 80)
+        let loc = window.location.toString();
+        if (loc.indexOf(":8080") === -1 && loc.indexOf("http:") === 0) {
+          window.location = loc.replace("http:","https:");
+        }
+        // if we're still here, load google analytics
+        (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
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diff --git a/index.template.html b/index.template.html
index 53db7784..f977694f 100644
--- a/index.template.html
+++ b/index.template.html
@@ -4,6 +4,23 @@
     <meta charset="utf-8">
     <title>A Primer on Bézier Curves</title>
 
+    <script>
+      (function() {
+        // force https except during local testing (e.g. not on port 80)
+        let loc = window.location.toString();
+        if (loc.indexOf(":8080") === -1 && loc.indexOf("http:") === 0) {
+          window.location = loc.replace("http:","https:");
+        }
+        // if we're still here, load google analytics
+        (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
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+        })(window,document,'script','https://www.google-analytics.com/analytics.js','ga');
+        ga('create', 'UA-69475841-3', 'auto');
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diff --git a/ja-JP/index.html b/ja-JP/index.html
index 7593e859..efc03e34 100644
--- a/ja-JP/index.html
+++ b/ja-JP/index.html
@@ -4,6 +4,23 @@
     <meta charset="utf-8">
     <title>A Primer on Bézier Curves</title>
 
+    <script>
+      (function() {
+        // force https except during local testing (e.g. not on port 80)
+        let loc = window.location.toString();
+        if (loc.indexOf(":8080") === -1 && loc.indexOf("http:") === 0) {
+          window.location = loc.replace("http:","https:");
+        }
+        // if we're still here, load google analytics
+        (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
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+        })(window,document,'script','https://www.google-analytics.com/analytics.js','ga');
+        ga('create', 'UA-69475841-3', 'auto');
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diff --git a/zh-CN/index.html b/zh-CN/index.html
index 20e1d2ac..d60c5e26 100644
--- a/zh-CN/index.html
+++ b/zh-CN/index.html
@@ -4,6 +4,23 @@
     <meta charset="utf-8">
     <title>A Primer on Bézier Curves</title>
 
+    <script>
+      (function() {
+        // force https except during local testing (e.g. not on port 80)
+        let loc = window.location.toString();
+        if (loc.indexOf(":8080") === -1 && loc.indexOf("http:") === 0) {
+          window.location = loc.replace("http:","https:");
+        }
+        // if we're still here, load google analytics
+        (function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){
+        (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o),
+        m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m)
+        })(window,document,'script','https://www.google-analytics.com/analytics.js','ga');
+        ga('create', 'UA-69475841-3', 'auto');
+        ga('send', 'pageview');
+      }());
+    </script>
+
     <!-- optional <base> element to make sure localised content still uses the base location -->
     <base href="..">