Any comparison based algorithm in which at every step the output is decided by comparison result of 2 keys at a time and 2 outcome is there based on which key is minimum and which one in maximum. So in decision tree where every vertex represent comparison between 2 keys and there are 2 branches corresponding to 2 output and finally the result is obtained when we are at the leaf node.
Since n elements can be arranged in n! ways, so the decision tree will have n! number of leafs corresponding to different arrangement. Since decision tree is a binary tree with every node represent comparison between 2 keys and every edge represent the comparison result. So a binary tree with n! leaf node will have minimum height
Since the number of comparisons is equal to number of vertices in path from root node to leaf of decision tree. Therefore the height of decision tree is equal to number of comparisons needed to get the output. Since minimum height of decision tree is , therefore the lower bound of comparison based sorting is .
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3. (20 points) Prove the best case performance of comparison-based sortings is S2(nlgn) by using the...
not A Question 5 In the worst case, the very best that a comparison based sorting algorithm can do when sorting n records is Q (n^2) Q(log (n!)) Q (logn) (n)
In the worst case, the very best that a comparison based sorting algorithm can do when sorting n records is 2 (n^2) (log (n!)) (logn) (n)
(30 points) Prove or disprove the following statement: There exists a comparison-based sorting algorithm whose running time is linear for at least a fraction of 1/2" of the n! possible input instances of length n.
In the worst case, the very best that a comparison based sorting algorithm can do when sorting n records is Q (n^2) Q(log (n!)) (log n) O Q (n)
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