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Consider the problem of sorting an n x n matrix (i.e. the rows and columns are in ascending order). I want to find the lower and upper bound of this problem.

I found that it is O(n^2 log n) by just sorting the elements and then outputting the first n elements as the first row, the next n elements as the second row, and so on. however i want to prove that it is also Omega(n^2 log n).

After trying smaller examples, I think I should prove that if I can solve this problem using less than n^2 log(n/e) comparisons, it would violates the log(m!) lower bound for comparisons needed to sort m elements.

Any ideas on how to prove that?

Jon 'links in bio' Ericson
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user2038833
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1 Answers1

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Have a look on http://en.wikipedia.org/wiki/Sorting_algorithm#Comparison_of_algorithms.

Your problem sound like you're just sorting n² elements instead of n, therefore 'O(n² log n²)' might be valid for mergesort for example.

If the first n elements in the first row don't have to be sorted themselves it might be faster, but not neccessarily, it depends on the algorithm.

Last but not least, trying some examples is no way prove something, especially small ones where statistics don't take effect(, they are not even indicating something)

Lord_Gestalter
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