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@ -76,27 +76,31 @@ for a given function. Say f(n) is your algorithm runtime, and g(n) is an arbitra
you are trying to relate to your algorithm. f(n) is O(g(n)), if for any real constant c (c>0), you are trying to relate to your algorithm. f(n) is O(g(n)), if for any real constant c (c>0),
f(n) <= c g(n) for every input size n (n>0). f(n) <= c g(n) for every input size n (n>0).
Example 1 *Example 1*
f(n) = 3log n + 100 ```
f(n) = 3log n + 100
g(n) = log n 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*
f(n) = 3*n^2 ```
f(n) = 3*n^2
g(n) = n 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)).
### Big-Omega ### Big-Omega