python 3.x - Why does my binary search use so many comparisons? -


here code:

def binary_search(sortedlist,target,comparisons): start = 0 end = len(sortedlist) - 1   while(end - start >= 0):     mid = (start + end)//2     comparisons += 2     if(target < sortedlist[mid]):         comparisons -= 1         end = mid - 1     elif(target > sortedlist[mid]):         start = mid + 1     else:         return target, comparisons return false, comparisons

it same every other post on here binary searching reason uses way many comparisons.

here code after fixed myself

from classes import genelist ## uncomment following line able make own testing genes # classes import gene, genome  def binary_search(sortedlist, target, comparisons):     reducedlist = sortedlist      while len(reducedlist) > 1:         mid = len(reducedlist) // 2         comparisons += 1         if target < reducedlist[mid]:             reducedlist = reducedlist[:mid]         else:             reducedlist = reducedlist[mid:]      comparisons += 1     if reducedlist[0] == target:         return reducedlist[0], comparisons     else:         return false, comparisons   def genetic_similarity_binary(first_genome, second_genome):     """ function takes 2 genome objects, , returns genelist     , integer. genelist of genes common     between first_genome , second_genome, while integer     how many comparisons took find similar genes.     hint: might pay define helper function.     """     comparisons = 0     similar_list = genelist()     target in first_genome:         result, comparisons = binary_search(second_genome, target, comparisons)         if result:             similar_list.append(result)     return similar_list, comparisons

you dont have middle check

for optimisation purposes, should check termination condition first. try way,

def binary_search(sortedlist,target,comparisons): start = 0 end = len(sortedlist) - 1   while(end - start >= 0):     mid = (start + end)//2     comparisons += 2     if(target == sortedlist[mid]):         comparisons -= 1         return target, comparisons     elif(target > sortedlist[mid]):         start = mid + 1     else:         end = mid - 1 return false, comparisons

as upon execution of binary_search([1,2,3,4,5,6,7,8,9,10],9,0)

for code gives outcome (9, 6) , above mentioned changes give (9, 5)


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