Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Lower bound for comparison based sorting algorithms

  • Difficulty Level : Easy
  • Last Updated : 06 Jul, 2021

The problem of sorting can be viewed as following. 

Input: A sequence of n numbers <a1, a2, . . . , an>. 
Output: A permutation (reordering) <a1, a2, . . . , an> of the input sequence such that a1 <= a2 ….. <= a’n

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

A sorting algorithm is comparison based if it uses comparison operators to find the order between two numbers.  Comparison sorts can be viewed abstractly in terms of decision trees. A decision tree is a full binary tree that represents the comparisons between elements that are performed by a particular sorting algorithm operating on an input of a given size. The execution of the sorting algorithm corresponds to tracing a path from the root of the decision tree to a leaf. At each internal node, a comparison ai <= aj is made. The left subtree then dictates subsequent comparisons for ai <= aj, and the right subtree dictates subsequent comparisons for ai > aj. When we come to a leaf, the sorting algorithm has established the ordering. So we can say following about the decision tree. 

1) Each of the n! permutations on n elements must appear as one of the leaves of the decision tree for the sorting algorithm to sort properly. 

2) Let x be the maximum number of comparisons in a sorting algorithm. The maximum height of the decision tree would be x. A tree with maximum height x has at most 2^x leaves. 

After combining the above two facts, we get following relation. 

      n!  <= 2^x

 Taking Log on both sides.
      log2(n!)  <= x

 Since log2(n!)  = Θ(nLogn),  we can say
      x = Ω(nLog2n)

Therefore, any comparison based sorting algorithm must make at least nLog2n comparisons to sort the input array, and Heapsort and merge sort are asymptotically optimal comparison sorts. 

Introduction to Algorithms, by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest and Clifford Stein

My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!