Algorithm Runtime Analysis

• Correctness (proof)
• Running time, depends on:
  • input size
  • computer
  • implementation

1. Unit-time random access machine model
2. Asymptotic notation

RAM model

Unit-cost model.excalidraw

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fährt elementare Operationen aus

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NOTE

Elementary operations:

• read/write one number
• arithmetic: +, -, *, /
• comparisons: $<$, $=$, $\geq$

Running time in the RAM model = number of elementary operations. This corresponds to practical running time up to a constant factor.

Asymptotic notation

Definition

NOTE

$N=\{n_0,n_0+1,\dots \}$ is the set of possible input sizes. For $f:N\to {\mathbb{R}}^{+}$, $O(f)=\{g:N\to {\mathbb{R}}^{+}\mid \text{there exists }C>0\text{ such that }g(n)\le C\cdot f(n)\text{ for all }n\in N\}$.

$O(f)$: order of $f$. $g\in O(f)\leftrightarrow \frac{g(n)}{f(n)}\le C$ (equivalently: the ratio is bounded).

Example 1

$f(n)=n^2$, $g(n)=4n^2$

$f\in O(g)$ because $n^2\le 4n^2$ with $C=1$.

$g\in O(f)$ because $4n^2\le 4n^2$ with $C=4$.

In general, for constant $a\in {\mathbb{R}}^{+}$:

$$O(af(n))=O(f(n))$$

Example 2

$f(n)=100n$, $g(n)=n^2$

$f\in O(g)$ because $100n\le 100n^2$ with $C=100$, $n\ge 1$.

NOTE

$O(n^2)$ is the set of all functions that grow at most quadratically.

Useful tools

• ${\lim }_{n\to \infty }\frac{g(n)}{f(n)}=\infty ⟹g\notin O(f)$
• ${\lim }_{n\to \infty }\frac{g(n)}{f(n)}\in {\mathbb{R}}^{+}⟹g\in O(f)$

Example 1

$f(n)=n^{1.6}+3n+4\in O(n^{1.6})$

$\frac{f(n)}{n^{1.6}}=1+3n^{-0.6}+4n^{-1.6}<1+3+4=8$.

Example 2

Let $a>0$.

$f(n)=\ln n\in O(n^a)$, because

$$\underset{n\to \infty }{\lim }\frac{\ln n}{n^a}=\underset{n\to \infty }{\lim }\frac{\frac{1}{n}}{a{n}^{a-1}}=\underset{n\to \infty }{\lim }\frac{1}{a{n}^a}=0.$$