algorithm - Recurrence relation of number of comparisons in binary search -


i have doubt recurrence relation number of comparisons in binary search.

i read recurrence can written t(n) = t(n/2) + 1 in website http://faculty.simpson.edu/lydia.sinapova/www/cmsc250/ln250_weiss/l14-recrel.htm

according me should t(n) = t(n/2) + 2, in worst case element might not present in array , end doing 2 comparisons in each pass.

please tell me whether right or wrong.

i think right.

imho, compare b means know a=b, a>b or a<b @ same time. say, 1 comparison may have 3 different results.

but programming languages. have use 2 comparisons.

if mid == x:      found it!          # 1st else if mid < x:  search right       # 2nd else:             search left

you mean == , < 2 comparisons.

it not affect result though. because use big o notation represent complexity. matter of constant, o don't care it.

according master theorem. either +1 or +2 result same complexity o(log n).

what want limit (big-o), not mathematical equation precise result.

i think matters here 1 , 2 both constant time. can split ==, > machine instructions, , may greater 2. or maybe programming languages or cpu utilize comparison, cost 1 comparison. not matter here when doing asymptotic analysis.

  1. https://en.wikipedia.org/wiki/master_theorem
  2. https://en.wikipedia.org/wiki/asymptotic_analysis

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