From charlesreid1

(Created page with "=Notes= ==Goodrich chapter 4== Recursion: binary search analysis. '''Proposition:''' The binary search algorithm runs in O(log n) time for a sorted sequence with n elements...")
 
Line 14: Line 14:


<math>
<math>
(mid - 1) - low + 1 = \floor \dfrac{low+high}{2} - low \leq \dfrac{high - low + 1}{2}
(mid - 1) - low + 1 = floor( \dfrac{low+high}{2} ) - low \leq \dfrac{high - low + 1}{2}
</math>
</math>

Revision as of 00:37, 12 July 2017

Notes

Goodrich chapter 4

Recursion: binary search analysis.

Proposition: The binary search algorithm runs in O(log n) time for a sorted sequence with n elements.

Justification: To prove this claim, a crucial fact is that within each recursive call the number of candidate entries still to be searched is given by the value

high - low + 1

Moreover, the number of remaining candidates is reduced by at least one half with each recursive call. Specifically, from the definition of mid, the number of remaining candidates is either

$ (mid - 1) - low + 1 = floor( \dfrac{low+high}{2} ) - low \leq \dfrac{high - low + 1}{2} $