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Let's consider the following algorithm to multiply squares matrix: A is a matrix of NxN. r_i, r_j defines interval of rows. For example r_i = 2, r_j=3 means the second and the third rows. c_i, c_j means the same as r_i, r_j but for columns. We assume that N = 2^s for some s.

mul(A, B, r_i, r_j, c_i, c_j){
if(r_i != r_j){
   r_m = floor((r_i+r_j)/2)

   mul(A, B,  r_i, r_m, c_i, c_j) 
   mul(A, B, r_m+1, r_j, c_i, c_j)
} else if(c_i != c_j){   
   c_m = floor((c_i+c_j)/2)
   mul(A, B, r_i,r_j, c_i, c_m)
   mul(A, B, r_i, r_j, c_m+1, c_j)
}else{
 for i = 1 to N:
    C[r_i][c_i] += A[r_i][i] * B[i][c_i] 
 }

}

And the most important:

Complexity of that algorithm takes: T(n) = 4T(n/2) + n = O(n^2)

And it is not correct. The correct answer is O(n^3). Why my computation is incorrect.

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  • $\begingroup$ Duplicate? $\endgroup$ – Raphael Jul 25 '17 at 17:25
  • $\begingroup$ Note that O(n²) and O(n³) are not contradictory statements. Are you after $\Theta$? $\endgroup$ – Raphael Jul 25 '17 at 17:26
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    $\begingroup$ I'm voting to close this question as off-topic because "please check my work" questions are unlikely ever to be useful to anyone except the asker. $\endgroup$ – David Richerby Sep 23 '17 at 19:50
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Your calculation is not right... $$ U(R,C) = 2U(\frac{R}{2},C)\textrm{ or } 2U(R,\frac{C}{2})\\ U(1,1) = O(n) $$

Basically, you have $N^2$ unique positions (which you drive your way down to recursively, rather than just looping) - for each of those you do a loop of O(N) so you have $O(N^3)$

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    $\begingroup$ This is not the way the code works currently. It's only $2U(R/2,C)$ or only $2U(R,C/2)$. $\endgroup$ – Yuval Filmus Jun 25 '17 at 14:21
  • $\begingroup$ My algorithm is now correct? @Moti, can you edit your answer? $\endgroup$ – Logic Jun 25 '17 at 14:22
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Your algorithm can be unrolled to the usual $O(N^3)$ matrix multiplication algorithm. If you open up all the recursive steps you will get a recursion tree whose $N^2$ leaves are of the form mul(A, B, r, r, c, c), each of which has running time $\Theta(N)$, for a total of $\Theta(N^3)$.

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  • $\begingroup$ I know this. I would like to see how to to compute complexity for my algorithm. Because it is hard to me. $\endgroup$ – Logic Jun 25 '17 at 15:30
  • $\begingroup$ I have just shown you how to compute the complexity for your algorithm. $\endgroup$ – Yuval Filmus Sep 23 '17 at 20:22

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