Chapter 2. Algorithm Analysis

We can compare the efficiency of algorithms without implementing them. Our two most important tools are (1) the RAM model of computation and (2) the asymptotic analysis of worst-case complexity.

Machine-independent algorithm design depends upon a hypothetical computer called the Random Access Machine or RAM

Using the RAM model of computation, we can count how many steps our algorithm takes on any given input instance by executing it. However, to understand how good or bad an algorithm is in general, we must know how it works over all instances.

The formal definitions associated with the Big Oh notation are as follows:

Stop and Think: Back to the Definition

Complexity of algorithms

O(f (n)) + O(g(n)) → O(max(f (n), g(n)))
Ω(f (n)) + Ω(g(n)) → Ω(max(f (n), g(n)))
Θ(f (n)) + Θ(g(n)) → Θ(max(f (n), g(n)))

If f(n) = O(g(n)) and g(n) = O(h(n)), then f(n) = O(h(n)).

Problem: Substring Pattern Matching
Input: A text string t and a pattern string p.
Output: Does t contain the pattern p as a substring, and if so where?

Saying b^x = y is equivalent to saying that x = logby. Further, this definition is the same as saying b^(logby) = y.

A binary tree of height 1 can have up to 2 leaf nodes, while a tree of height two can have up to four leaves. What is the height h of a rooted binary tree with n leaf nodes? Note that the number of leaves doubles every time we increase the height by one. To account for n leaves, n = 2^h which implies that h = log2 n.

Logarithms arise whenever things are repeatedly halved or doubled.

As we have seen, stating b^x = y is equivalent to saying that x = logb y. The b term is known as the base of the logarithm. Three bases are of particular importance for mathematical and historical reasons:

Two implications of these properties of logarithms are important to appreciate from an algorithmic perspective:

The dominance relation between functions is a consequence of the theory of lim- its, which you may recall from Calculus. We say that f(n) dominates g(n) if limn→∞ g(n)/f(n) = 0. e.g. f(n) = n³ and g(n) = n².

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