Matrix Multiplication: Difference between revisions

From NovaOrdis Knowledge Base
Jump to navigation Jump to search
Line 33: Line 33:


<font size='-1'>
<font size='-1'>
           │ AE+BG AF+BH │    │ P<sub>5</sub>+P<sub>4</sub>-P<sub>2</sub>+P<sub>6</sub>  P<sub>1</sub>+P<sub>2</sub>    
           │ AE+BG AF+BH │    │ P<sub>5</sub>+P<sub>4</sub>-P<sub>2</sub>+P<sub>6</sub>  P<sub>1</sub>+P<sub>2</sub>      
  M x N =  │            │  =  │                          │  
  M x N =  │            │  =  │                          │  
           │ CE+DG CF+DH │    │ P<sub>3</sub>+P<sub>4</sub>        P<sub>1</sub>+P<sub>5</sub>-P<sub>3</sub>-P<sub>7</sub> │       
           │ CE+DG CF+DH │    │ P<sub>3</sub>+P<sub>4</sub>        P<sub>1</sub>+P<sub>5</sub>-P<sub>3</sub>-P<sub>7</sub> │       

Revision as of 20:19, 20 September 2021

Internal

Overview

The straightforward iterative algorithm for matrix multiplication in case of two matrices M (mxn) and N (nxp) is m * n * p. For simplicity, we assume the matrices are square, with the number of rows and columns equal to n. Note that while usually we use "n" to denote the input size, here we use "n" to denote the matrix dimension, so the input size in this case is O(n2). In this case the obvious algorithm for multiplication is O(n3). That matches the intuition as the straightforward iterative algorithm is a triple nested loop over n.

One may think that a divide and conquer recursive algorithm could help, and the obvious divide and conquer algorithm would be to divide each matrix in equal blocks and perform recursive invocations multiplying the blocks:

     │ A  B │       │ E  F │
M =  │      │  N =  │      │   
     │ C  D │       │ G  H │

In this case final result is obtained by multiplying the blocks as follows:

         │ AE+BG AF+BH │
M x N =  │             │  
         │ CE+DG CF+DH │

However, applying Mater Method to infer the time complexity of the recursive multiplication algorithm for a = 8 (we invoke recursively eight times: AE, BG, AF, BH, CE, DG, CF and DH - none of the products re-occur, they are distinct), b = 2 (we multiply matrices twice as small, each matrix of size n/2 * n/2) and the combine step complexity is O(n2) so d = 2. a/bd is 8/22 = 2, so according to the Master Method, the complexity is bounded by O(nlogba) = O(n3), the same as the straightforward iterative method.

An improvement to this upper bound is provided by the Strassen method, which cleverly proposes only 7 recursive calls.

Strassen's Algorithm for Matrix Multiplication

The Strassen algorithm recursively computes the products of smaller (n/2) matrices, but it'll only be 7 of them:

P1=A(F-H), P2=(A+B)*H, P3=(C+D)E, P4=D(G-E), P5=(A+D)(E+H), P6=(B-D)(G+H), P7=(A-C)(E+F).

In the combine phase, it'll combine the results of the recursive calls, using only addition and subtraction (O(n2)):

         │ AE+BG AF+BH │     │ P5+P4-P2+P6   P1+P2       │
M x N =  │             │  =  │                          │ 
         │ CE+DG CF+DH │     │ P3+P4         P1+P5-P3-P7

The asymptotic complexity is Θ(nlog7).

Link to Gauss.

TODO CLRS page 75.