Binary search: Difference between revisions
Line 23: | Line 23: | ||
'''Abstract view:''' | '''Abstract view:''' | ||
'''Correctnes:''' Obvious. | '''Correctnes:''' Obvious. |
Revision as of 06:04, 27 April 2016
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
The invariant is trivially fulfilled because [math]\displaystyle{ [\ell,\ldots,r] }[/math] is the entire index range of [math]\displaystyle{ A }[/math].
Induction step
Abstract view:
Correctnes: Obvious.
Complexity
Statement: Log(n)
Proff: Obvious.