递归矩阵乘法 [英] Recursive matrix multiplication
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问题描述
我读算法导论由CLRS。本书展示了伪code简单的分而治之的矩阵乘法:
N = A.rows
令C是一个新的N×n矩阵
如果n == 1
C11 = A11 * B11
其他分区A,B和C
C11 = SquareMatrixMultiplyRecursive(A11,B11)
+ SquareMatrixMultiplyRecursive(A12,B21)
// ...
回复C
,其中,例如,A11是A尺寸N / 2 XN / 2的子矩阵。 作者还暗示我应该使用,而不是创造新的矩阵重新present子矩阵指数计算,所以我这样做:
的#include<的iostream>
#包括<载体>
模板<类T>
结构矩阵
{
矩阵(为size_t R,为size_t三)
{
Data.resize(C,性病::矢量< T>(R,0));
}
无效SetSubMatrix(const int的R,const int的C,const int的N,常量矩阵< T>&安培; A,常量矩阵< T>和B)
{
对于(INT _c = C; _c n种++ _三)
{
对于(INT _r = R; _r n种++ _ R)
{
数据[_c] [_ R] = A.Data [_c] [_ R] + B.Data [_c] [_ R];
}
}
}
静态矩阵< T> SquareMultiplyRecursive(矩阵< T>&安培; A,矩阵< T>和B,INT AR,诠释交流,诠释BR,公元前INT,INT N)
{
矩阵< T> C(N,N);
如果(正== 1)
{
C.Data [0] [0] = A.Data [AC] [AR] * B.Data [BC] [BR];
}
其他
{
C.SetSubMatrix(0,0,π/ 2,
SquareMultiplyRecursive(A,B,AR,交流,BR,公元前,N / 2),
SquareMultiplyRecursive(A,B,AR,AC +(N / 2),BR +(N / 2),BC,N / 2));
C.SetSubMatrix(0,N / 2,N / 2,
SquareMultiplyRecursive(A,B,AR,交流,BR,公元前+(N / 2),N / 2),
SquareMultiplyRecursive(A,B,AR,AC +(n / 2个峰),br +(N / 2),BC +(N / 2)中,n / 2));
C.SetSubMatrix(n / 2个,0,π/ 2,
SquareMultiplyRecursive(A,B,AR +(N / 2),交流,BR,公元前,N / 2),
SquareMultiplyRecursive(A,B,AR +(N / 2),交流+(N / 2),BR +(N / 2),BC,N / 2));
C.SetSubMatrix(N / 2,N / 2,N / 2,
SquareMultiplyRecursive(A,B,AR +(N / 2),交流,BR,公元前+(N / 2),N / 2),
SquareMultiplyRecursive(A,B,芳+(N / 2),AC +(n / 2个峰),br +(N / 2),BC +(N / 2)中,n / 2));
}
返回℃;
}
无效打印()
{
对于(INT C = 0;℃下Data.size(); ++ C)
{
对(INT R = 0; R<数据[0] .size(); ++ r)的
{
性病::法院<<数据[C] [R]<< ;
}
性病::法院<< \ N的;
}
性病::法院<< \ N的;
}
的std ::矢量<的std ::矢量< T> >数据;
};
诠释的main()
{
矩阵< INT> A(2,2);
矩阵< INT> B(2,2);
A.Data [0] [0] = 2;
A.Data [0] [1] = 1;
A.Data [1] [0] = 1;
A.Data [1] [1] = 2;
B.Data [0] [0] = 2;
B.Data [0] [1] = 1;
B.Data [1] [0] = 1;
B.Data [1] [1] = 2;
A.Print();
B.Print();
矩阵< INT> C(矩阵< INT> :: SquareMultiplyRecursive(A,B,0,0,0,0,2));
C.Print();
}
这给了我不正确的结果,寿我不知道我在做什么错了?
解决方案
C语言//递归天真的矩阵乘法,而不是STRASSEN。
// 2013-FEB-15周五12点28分,在/ Gmail的moshahmed /
#包括< ASSERT.H>
#包括< stdio.h中>
#包括< stdlib.h中>
#包括< time.h中>
#定义M 2
#定义N(1<< M)
的typedef INT垫[N] [N]; //垫[2 **男,2 ** M]的分而治之MULT。
typedef结构{INT RA,RB,CA,CB; }角落; //用于跟踪行和列。
//设置A [A] = K
空集(垫A,角落,诠释K){
INT I,J;
对于(i = a.ra; I< a.rb;我++)
为(J = a.ca; J< a.cb; J ++)
A [1] [J] = K;
}
//设置A [A] = [随机(l..h)。
无效randk(垫A,角落,诠释L,INT高){
INT I,J;
对于(i = a.ra; I< a.rb;我++)
为(J = a.ca; J< a.cb; J ++)
A [1] [J] = L +兰特()%(H-1);
}
//打印A [一]
无效打印(垫A,角落,字符*名称){
INT I,J;
的printf(%S = {\ N,名);
对于(i = a.ra; I< a.rb;我++){
为(J = a.ca; J< a.cb; J ++)
的printf(%4d中,A [1] [J]);
的printf(\ N);
}
的printf(} \ N);
}
//返回矩阵的1/4:上/下,左/右。
无效find_corners(角落,诠释我,诠释J,角* B){
INT RM = a.ra +(a.rb - a.ra)/ 2;
INT厘米= a.ca +(a.cb - a.ca)/ 2;
* B = A;
如果(我== 0)B-> RB = RM; //前行
否则B-> RA = RM; // BOT行
如果(j == 0)B-> CB =厘米; //左COLS
否则B-> CA =厘米; //正确COLS
}
//天真乘法:A [A] * B [B] => C [C],递归。
无效MUL(垫A,板坯B,垫C,角落,角落B,角C){
角AII [2] [2],BII [2] [2],CII [2] [2];
INT I,J,M,N,P;
//检查:A [M N] * B [N P] = C [M P]
M = a.rb - a.ra;断言(M ==(c.rb-c.ra));
N = a.cb - a.ca;断言(N = =(b.rb-b.ra));
P = b.cb - b.ca;断言(P = =(c.cb-c.ca));
断言(米大于0);
如果(正== 1){
C [c.ra] [c.ca] + = A [a.ra] [a.ca] * B [b.ra] [b.ca]
返回;
}
//创建更小的矩阵:
对于(I = 0; I&2;我++){
为(J = 0; J&2; J ++){
find_corners(A,I,J,和放大器; AII [I] [J]);
find_corners(B,I,J,和放大器; BII [I] [J]);
find_corners(C,I,J,和放大器; CII [I] [J]);
}
}
//现在做8次矩阵乘法。
// C00 = A00 * B00 + A01 * B10
// C01 = A00 * B01 + A01 * B11
// C10 = A10 * B00 + A11 * B10
// C11 = A10 * B01 + A11 * B11
穆尔(甲,乙,丙,AII [0] [0],BII [0] [0],CII [0] [0]);
穆尔(甲,乙,丙,AII [0] [1],BII [1] [0],CII [0] [0]);
穆尔(甲,乙,丙,AII [0] [0],BII [0] [1],CII [0] [1]);
穆尔(甲,乙,丙,AII [0] [1],BII [1] [1],CII [0] [1]);
穆尔(甲,乙,丙,AII [1] [0],BII [0] [0],CII [1] [0]);
穆尔(甲,乙,丙,AII [1] [1],BII [1] [0],CII [1] [0]);
穆尔(甲,乙,丙,AII [1] [0],BII [0] [1],CII [1] [1]);
穆尔(甲,乙,丙,AII [1] [1],BII [1] [1],CII [1] [1]);
}
诠释的main(){
垫A,B,C;
角AI = {0,N,O,N};
角双向= {0,N,O,N};
角落CI = {0,N,O,N};
//设置(A,AI,2);
//设置(B,BI,2);
函数srand(时间(0));
randk(A,AI,0,2);
randk(乙,双,0,2);
组(C,CI,0); //设置为零多重峰之前。
打印(A,AI,A);
印刷(B,BI,B);
MUL(A,B,C,AI,BI,CI);
打印(C,CI,C);
返回0;
}
I am reading Introduction to Algorithms by CLRS. Book shows pseudocode for simple divide and conquer matrix multiplication:
n = A.rows
let c be a new n x n matrix
if n == 1
c11 = a11 * b11
else partition A, B, and C
C11 = SquareMatrixMultiplyRecursive(A11, B11)
+ SquareMatrixMultiplyRecursive(A12, B21)
//...
return C
Where for example, A11 is submatrix of A of size n/2 x n/2. Author also hints that I should use index calculations instead of creating new matrices to represent submatrices, so I did this:
#include <iostream>
#include <vector>
template<class T>
struct Matrix
{
Matrix(size_t r, size_t c)
{
Data.resize(c, std::vector<T>(r, 0));
}
void SetSubMatrix(const int r, const int c, const int n, const Matrix<T>& A, const Matrix<T>& B)
{
for(int _c=c; _c<n; ++_c)
{
for(int _r=r; _r<n; ++_r)
{
Data[_c][_r] = A.Data[_c][_r] + B.Data[_c][_r];
}
}
}
static Matrix<T> SquareMultiplyRecursive(Matrix<T>& A, Matrix<T>& B, int ar, int ac, int br, int bc, int n)
{
Matrix<T> C(n, n);
if(n == 1)
{
C.Data[0][0] = A.Data[ac][ar] * B.Data[bc][br];
}
else
{
C.SetSubMatrix(0, 0, n / 2,
SquareMultiplyRecursive(A, B, ar, ac, br, bc, n / 2),
SquareMultiplyRecursive(A, B, ar, ac + (n / 2), br + (n / 2), bc, n / 2));
C.SetSubMatrix(0, n / 2, n / 2,
SquareMultiplyRecursive(A, B, ar, ac, br, bc + (n / 2), n / 2),
SquareMultiplyRecursive(A, B, ar, ac + (n / 2), br + (n / 2), bc + (n / 2), n / 2));
C.SetSubMatrix(n / 2, 0, n / 2,
SquareMultiplyRecursive(A, B, ar + (n / 2), ac, br, bc, n / 2),
SquareMultiplyRecursive(A, B, ar + (n / 2), ac + (n / 2), br + (n / 2), bc, n / 2));
C.SetSubMatrix(n / 2, n / 2, n / 2,
SquareMultiplyRecursive(A, B, ar + (n / 2), ac, br, bc + (n / 2), n / 2),
SquareMultiplyRecursive(A, B, ar + (n / 2), ac + (n / 2), br + (n / 2), bc + (n / 2), n / 2));
}
return C;
}
void Print()
{
for(int c=0; c<Data.size(); ++c)
{
for(int r=0; r<Data[0].size(); ++r)
{
std::cout << Data[c][r] << " ";
}
std::cout << "\n";
}
std::cout << "\n";
}
std::vector<std::vector<T> > Data;
};
int main()
{
Matrix<int> A(2, 2);
Matrix<int> B(2, 2);
A.Data[0][0] = 2;
A.Data[0][1] = 1;
A.Data[1][0] = 1;
A.Data[1][1] = 2;
B.Data[0][0] = 2;
B.Data[0][1] = 1;
B.Data[1][0] = 1;
B.Data[1][1] = 2;
A.Print();
B.Print();
Matrix<int> C(Matrix<int>::SquareMultiplyRecursive(A, B, 0, 0, 0, 0, 2));
C.Print();
}
It gives me incorrect results, tho I am not sure what I'm doing wrong...
解决方案
// Recursive naive matrix multiplication in C, not strassen.
// 2013-Feb-15 Fri 12:28 moshahmed/at/gmail
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define M 2
#define N (1<<M)
typedef int mat[N][N]; // mat[2**M,2**M] for divide and conquer mult.
typedef struct { int ra, rb, ca, cb; } corners; // for tracking rows and columns.
// set A[a] = k
void set(mat A, corners a, int k){
int i,j;
for(i=a.ra;i<a.rb;i++)
for(j=a.ca;j<a.cb;j++)
A[i][j] = k;
}
// set A[a] = [random(l..h)].
void randk(mat A, corners a, int l, int h){
int i,j;
for(i=a.ra;i<a.rb;i++)
for(j=a.ca;j<a.cb;j++)
A[i][j] = l + rand()% (h-l);
}
// Print A[a]
void print(mat A, corners a, char *name) {
int i,j;
printf("%s = {\n",name);
for(i=a.ra;i<a.rb;i++){
for(j=a.ca;j<a.cb;j++)
printf("%4d, ", A[i][j]);
printf("\n");
}
printf("}\n");
}
// Return 1/4 of the matrix: top/bottom , left/right.
void find_corners(corners a, int i, int j, corners *b) {
int rm = a.ra + (a.rb - a.ra)/2 ;
int cm = a.ca + (a.cb - a.ca)/2 ;
*b = a;
if (i==0) b->rb = rm; // top rows
else b->ra = rm; // bot rows
if (j==0) b->cb = cm; // left cols
else b->ca = cm; // right cols
}
// Naive Multiply: A[a] * B[b] => C[c], recursively.
void mul(mat A, mat B, mat C, corners a, corners b, corners c) {
corners aii[2][2], bii[2][2], cii[2][2];
int i, j, m, n, p;
// Check: A[m n] * B[n p] = C[m p]
m = a.rb - a.ra; assert(m==(c.rb-c.ra));
n = a.cb - a.ca; assert(n==(b.rb-b.ra));
p = b.cb - b.ca; assert(p==(c.cb-c.ca));
assert(m>0);
if (n==1) {
C[c.ra][c.ca] += A[a.ra][a.ca] * B[b.ra][b.ca];
return;
}
// Create the smaller matrices:
for(i=0;i<2;i++) {
for(j=0;j<2;j++) {
find_corners(a, i, j, &aii[i][j]);
find_corners(b, i, j, &bii[i][j]);
find_corners(c, i, j, &cii[i][j]);
}
}
// Now do the 8 sub matrix multiplications.
// C00 = A00*B00 + A01*B10
// C01 = A00*B01 + A01*B11
// C10 = A10*B00 + A11*B10
// C11 = A10*B01 + A11*B11
mul( A, B, C, aii[0][0], bii[0][0], cii[0][0] );
mul( A, B, C, aii[0][1], bii[1][0], cii[0][0] );
mul( A, B, C, aii[0][0], bii[0][1], cii[0][1] );
mul( A, B, C, aii[0][1], bii[1][1], cii[0][1] );
mul( A, B, C, aii[1][0], bii[0][0], cii[1][0] );
mul( A, B, C, aii[1][1], bii[1][0], cii[1][0] );
mul( A, B, C, aii[1][0], bii[0][1], cii[1][1] );
mul( A, B, C, aii[1][1], bii[1][1], cii[1][1] );
}
int main() {
mat A, B, C;
corners ai = {0,N,0,N};
corners bi = {0,N,0,N};
corners ci = {0,N,0,N};
//set(A,ai,2);
//set(B,bi,2);
srand(time(0));
randk(A,ai, 0, 2);
randk(B,bi, 0, 2);
set(C,ci,0); // set to zero before mult.
print(A, ai, "A");
print(B, bi, "B");
mul(A,B,C, ai, bi, ci);
print(C, ci, "C");
return 0;
}
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