1
$\begingroup$

So each matrix has $N^{2}$ elements, and so just by comparing each element we would be doing $O(N^{2})$ operations. Is there any other way to compare these two matrices such that the number of operations is less than $O(N^{2})$ or is the matrix comparison lower bound also $\Omega(N^2)$?

$\endgroup$
1
  • 2
    $\begingroup$ What kind of comparison are you looking at? $\endgroup$
    – Raphael
    Feb 22, 2016 at 13:43

1 Answer 1

4
$\begingroup$

A very simple adversary argument shows that when comparing two vectors of length $M$ (in your case, $M = N^2$), you must query (in the worst case) all positions of both vectors to know whether they are equal. That takes time $\Omega(M)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.