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Big-O Notation in Matrix Multiplication

Date: 12/10/2000 at 12:53:41
From: Peter D'Amore
Subject: Matrix multiplication done in big-O notation

Use the algorithm here to show that matrix multiplication can be done 
in O(n^k) where k < 3.

     [ Z11  Z12 ]   [ X11  X12 ]   [ Y11  Y12 ]
     [ Z21  Z22 ] = [ X21  X22 ] * [ Y21  Y22 ]

Simple algebraic manipulation shows that if we form the intermediate 

     W1 = (X11 + X22) * (Y11 + Y22)
     W2 = (X21 + X22) * (Y11)
     W3 = (X11) * (Y12 + Y22)
     W4 = (X22) * (Y11 + Y21)
     W5 = (X11 + X12) * (Y22)
     W6 = (X21 - X11) * (Y11 + Y12)
     W7 = (X12 - X22) * (Y21 + Y22)


     Z11 = W1+W4-W5+W7    
     Z12 = W3 + W5        
     Z21 = W2 + W4
     Z22 = W1 + W3 - W2 + W6

Please help me solve this question. I'm very familiar with matrix 
multiplication, but I've never worked with big-O notation, so any help 
would be greatly appreciated.


Date: 12/12/2000 at 11:26:13
From: Doctor Ian
Subject: Re: Matrix multiplication done in big-O notation

Hi Peter, 

Big-O notation is just a way of being able to compare expected run 
times without getting bogged down in detail.

Suppose you need to run an algorithm on a set of n inputs. Perhaps the 
actual number of 'operations' (however you choose to define that term) 
that would need to be carried out is something like:

     2n^2 + 3n - 5

This is really only interesting when n gets large, in which case the 
only term that really matters is the n^2 term, which dwarfs the 
others. And in fact, the constant coefficient isn't really that 
important, either. So we just say that the algorithm runs in 'order n 
squared' time, or O(n^2).

This notation, then, is used to compare algorithms in terms of speed: 
An O(n) algorithm is faster than an O(n^2) algorithm, but slower than 
an O(log n) algorithm.  

If you multiply two nxn matrices, A and B,

     [ a_1,1 a_1,2 ... a_1,n ]     [ b_1,1 b_1,2 ... b_1,n ]
     [ a_2,1 a_2,2 ... a_2,n ]  *  [ b_2,1 b_2,2 ... b_2,n ]  =   ?
         .     .         .            .     .         .
         .     .         .            .     .         .
     [ a_n,1 a_n,2 ... a_n,n ]     [ b_n,1 b_n,2 ... b_n,n ]

you end up with another nxn matrix, C, where:

     c_i,j = (row_i of A) . (col_j of B)

where '.' indicates a 'dot product', which is a sum of pairwise 
products. For example:

     (x, y, z) . (u, v, w) = x*u + y*v + z*w

So, to compute the first element of C,

     c_1,1 = (a_1,1 a_1,2 ... a_1,n) . (b_1,1 b_2,1 ... b_n,1)
           = a_1,1*b_1,1 + a_1,2*b_2,1 + ... + a_1,n*b_n,1

So, to construct each element of the output matrix in the most 
straightforward possible way requires n multiplications. (Well, more 
precisely, it would be 2n operations -- n multiplications, and n 
additions. But as I said, when constructing Big-O expressions, we 
generally ignore constant coefficients.)  

Since there are n^2 elements in the output matrix, that would require 
n*n^2 = n^3 operations to compute the entire matrix - that is, the 
speed of the most straightforward algorithm is O(n^k), where k = 3.

Your job is to show that it can be done in a more efficient way. You 
could do that by trying to invent a new algorithm by yourself, or you 
might just point to a paper like this one,

   Implementation of Strassen's Algorithm for Matrix Multiplication   

which discusses an algorithm that, at least for some matrices, reduces 
the exponent from 3 to 2.807, thus proving that it _can_ be done.

I hope this helps. Write back if you'd like to talk about this some 
more, or if you have any other questions. 

- Doctor Ian, The Math Forum   
Associated Topics:
College Algorithms
College Linear Algebra
High School Linear Algebra

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