Binary search

Binary search

Algorithmic problem: Finding an element in a sorted array

Type of algorithm: recursion

Abstract view

The subroutine does nothing but calling another, recursive subroutine with the same output and the following input: array [math]\displaystyle{ A }[/math], comparison [math]\displaystyle{ cmp }[/math], element [math]\displaystyle{ s }[/math] and, in addition, two indices of [math]\displaystyle{ A }[/math], [math]\displaystyle{ \ell }[/math] and [math]\displaystyle{ r }[/math], such that [math]\displaystyle{ \ell\lt r }[/math]. In the original call to this recursive subroutine, [math]\displaystyle{ \ell }[/math] is the first and [math]\displaystyle{ r }[/math] the last index of [math]\displaystyle{ A }[/math].

Invariant: For every recursive call: If [math]\displaystyle{ s }[/math] is present at least once in [math]\displaystyle{ A }[/math], then all indices of [math]\displaystyle{ A }[/math] where [math]\displaystyle{ s }[/math] is present are in the interval [math]\displaystyle{ [\ell,\ldots,r] }[/math].

Variant: The value [math]\displaystyle{ r-\ell }[/math] is roughly halved in every descent in the recursion tree.

Break condition: Either an occurrence of [math]\displaystyle{ s }[/math] is found or it is [math]\displaystyle{ \ell\gt r }[/math]

Induction basis

Abstract view: Set [math]\displaystyle{ p:= }[/math] n/2.

Implementation: Obvious.

Proof: Nothing to show.

Induction step

Abstract view:

If p points to a node but not with key K, p descends in the appropriate direction, left or right.

Implementation:

1. If [math]\displaystyle{ p.key = K }[/math], terminate the algorithm and return true.
2. Otherwise:
   1. If [math]\displaystyle{ K \lt p.key }[/math], set [math]\displaystyle{ p }[/math] to the sequence element [math]\displaystyle{ k_{i+1}=k_i-n/2^i }[/math].
   2. If [math]\displaystyle{ K \gt p.key }[/math], set [math]\displaystyle{ p }[/math]to the sequence element [math]\displaystyle{ k_{i+1}=k_i-n/2^i }[/math].

Correctnes: Obvious.

Complexity

Statement: Log(n)

Proff: Obvious.