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Added to 'why use worst case'. Formatting changes.

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Jake Prather
2015-02-01 11:27:32 -06:00
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asymptotic-notation.html.markdown
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@@ -38,8 +38,9 @@ You can label a function, or algorithm, with an Asymptotic Notation in many diff
Some examples are, you can describe an algorithm by it's best case, worse case, or equivalent case. Some examples are, you can describe an algorithm by it's best case, worse case, or equivalent case.
The most common is to analyze an algorithm by it's worst case. This is because if you determine an The most common is to analyze an algorithm by it's worst case. This is because if you determine an
algorithm's run time or time complexity, by it's best case, what if it's best case is only obtained algorithm's run time or time complexity, by it's best case, what if it's best case is only obtained
given at a low, unrealistic, input size? It is equivalent to having a 5 meter sprinting race. given at a low, unrealistic, input size? It is equivalent to having a 5 meter sprinting race to find
That isn't the best measurement. the fastest sprinter on earth. Or testing the 40 to 50MPH time of a car to determine the fastest car
in the world. The measurement loses meaning because it doesn't represent the problem well.
### Types of functions, limits, and simplification ### Types of functions, limits, and simplification
``` ```
@@ -85,9 +86,13 @@ g(n) = log n
is f(n) O(g(n))? is f(n) O(g(n))?
is 3 log n + 100 O(log n)? is 3 log n + 100 O(log n)?
Let's look to the definition of Big-Oh. Let's look to the definition of Big-Oh.
```
3log n + 100 <= c * log n 3log n + 100 <= c * log n
```
Is there some constant c that satisfies this for all n? Is there some constant c that satisfies this for all n?
```
3log n + 100 <= 150 * log n, n > 2 (undefined at n = 1) 3log n + 100 <= 150 * log n, n > 2 (undefined at n = 1)
```
Yes! The definition of Big-Oh has been met therefore f(n) is O(g(n)). Yes! The definition of Big-Oh has been met therefore f(n) is O(g(n)).
*Example 2* *Example 2*
@@ -99,7 +104,9 @@ g(n) = n
is f(n) O(g(n))? is f(n) O(g(n))?
is 3 * n^2 O(n)? is 3 * n^2 O(n)?
Let's look at the definition of Big-Oh. Let's look at the definition of Big-Oh.
```
3 * n^2 <= c * n 3 * n^2 <= c * n
```
Is there some constant c that satisfies this for all n? Is there some constant c that satisfies this for all n?
No there isn't, f(n) is NOT O(g(n)). No there isn't, f(n) is NOT O(g(n)).