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朴素矩阵相乘算法，思想明了，编程实现简单。时间复杂度是Θ(n^3)。伪码如下
1 for i ← 1 to n 2 do for j ← 1 to n 3 do c[i][j] ← 0 4 for k ← 1 to n 5 do c[i][j] ← c[i][j] + a[i][k]⋅ b[k][j]
矩阵乘法中采用分治法，第一感觉上应该能够有效的提高算法的效率。如下图所示分治法方案，以及对该算法的效率分析。有图可知，算法效率是Θ(n^3)。算法效率并没有提高。下面介绍下矩阵分治法思想：
鉴于上面的分治法方案无法有效提高算法的效率，要想提高算法效率，由主定理方法可知必须想办法将2中递归式中的系数8减少。Strassen提出了一种将系数减少到7的分治法方案，如下图所示。
效率分析如下：
伪码如下：
1 Strassen (N,MatrixA,MatrixB,MatrixResult) 2 3 //splitting input Matrixes, into 4 submatrices each. 4 for i <- 0 to N/2 5 for j <- 0 to N/2 6 A11[i][j] <- MatrixA[i][j]; //a矩阵块 7 A12[i][j] <- MatrixA[i][j + N / 2]; //b矩阵块 8 A21[i][j] <- MatrixA[i + N / 2][j]; //c矩阵块 9 A22[i][j] <- MatrixA[i + N / 2][j + N / 2];//d矩阵块 10 11 B11[i][j] <- MatrixB[i][j]; //e 矩阵块 12 B12[i][j] <- MatrixB[i][j + N / 2]; //f 矩阵块 13 B21[i][j] <- MatrixB[i + N / 2][j]; //g 矩阵块 14 B22[i][j] <- MatrixB[i + N / 2][j + N / 2]; //h矩阵块 15 //here we calculate M1..M7 matrices . 17 //递归求M1 18 HalfSize <- N/2 19 AResult <- A11+A22 20 BResult <- B11+B22 21 Strassen( HalfSize, AResult, BResult, M1 ); //M1=(A11+A22)*(B11+B22) p5=(a+d)*(e+h) 22 //递归求M2 23 AResult <- A21+A22 24 Strassen(HalfSize, AResult, B11, M2); //M2=(A21+A22)B11 p3=(c+d)*e 25 //递归求M3 26 BResult <- B12 - B22 27 Strassen(HalfSize, A11, BResult, M3); //M3=A11(B12-B22) p1=a*(f-h) 28 //递归求M4 29 BResult <- B21 - B11 30 Strassen(HalfSize, A22, BResult, M4); //M4=A22(B21-B11) p4=d*(g-e) 31 //递归求M5 32 AResult <- A11+A12 33 Strassen(HalfSize, AResult, B22, M5); //M5=(A11+A12)B22 p2=(a+b)*h 34 //递归求M6 35 AResult <- A21-A11 36 BResult <- B11+B12 37 Strassen( HalfSize, AResult, BResult, M6); //M6=(A21-A11)(B11+B12) p7=(c-a)(e+f) 38 //递归求M7 39 AResult <- A12-A22 40 BResult <- B21+B22 41 Strassen(HalfSize, AResult, BResult, M7); //M7=(A12-A22)(B21+B22) p6=(b-d)*(g+h) 42 43 //计算结果子矩阵 44 C11 <- M1 + M4 - M5 + M7; 45 46 C12 <- M3 + M5; 47 48 C21 <- M2 + M4; 49 50 C22 <- M1 + M3 - M2 + M6; 51 //at this point , we have calculated the c11..c22 matrices, and now we are going to 52 //put them together and make a unit matrix which would describe our resulting Matrix. 53 for i <- 0 to N/2 54 for j <- 0 to N/2 55 MatrixResult[i][j] <- C11[i][j]; 56 MatrixResult[i][j + N / 2] <- C12[i][j]; 57 MatrixResult[i + N / 2][j] <- C21[i][j]; 58 MatrixResult[i + N / 2][j + N / 2] <- C22[i][j];
Strassen.h
1 #ifndef STRASSEN_HH 2 #define STRASSEN_HH 3 template<typename T> 4 class Strassen_class{ 5 public: 6 void ADD(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize ); 7 void SUB(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize ); 8 void MUL( T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize );//朴素算法实现 9 void FillMatrix( T** MatrixA, T** MatrixB, int length);//A,B矩阵赋值 10 void PrintMatrix(T **MatrixA,int MatrixSize);//打印矩阵 11 void Strassen(int N, T **MatrixA, T **MatrixB, T **MatrixC);//Strassen算法实现 12 }; 13 template<typename T> 14 void Strassen_class<T>::ADD(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize ) 15 { 16 for ( int i = 0; i < MatrixSize; i++) 17 { 18 for ( int j = 0; j < MatrixSize; j++) 19 { 20 MatrixResult[i][j] = MatrixA[i][j] + MatrixB[i][j]; 21 } 22 } 23 } 24 template<typename T> 25 void Strassen_class<T>::SUB(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize ) 26 { 27 for ( int i = 0; i < MatrixSize; i++) 28 { 29 for ( int j = 0; j < MatrixSize; j++) 30 { 31 MatrixResult[i][j] = MatrixA[i][j] - MatrixB[i][j]; 32 } 33 } 34 } 35 template<typename T> 36 void Strassen_class<T>::MUL( T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize ) 37 { 38 for (int i=0;i<MatrixSize ;i++) 39 { 40 for (int j=0;j<MatrixSize ;j++) 41 { 42 MatrixResult[i][j]=0; 43 for (int k=0;k<MatrixSize ;k++) 44 { 45 MatrixResult[i][j]=MatrixResult[i][j]+MatrixA[i][k]*MatrixB[k][j]; 46 } 47 } 48 } 49 } 50 51 /* 52 c++使用二维数组，申请动态内存方法 53 申请 54 int **A; 55 A = new int *[desired_array_row]; 56 for ( int i = 0; i < desired_array_row; i++) 57 A[i] = new int [desired_column_size]; 58 59 释放 60 for ( int i = 0; i < your_array_row; i++) 61 delete [] A[i]; 62 delete[] A; 63 64 */ 65 template<typename T> 66 void Strassen_class<T>::Strassen(int N, T **MatrixA, T **MatrixB, T **MatrixC) 67 { 68 69 int HalfSize = N/2; 70 int newSize = N/2; 71 72 if ( N <= 64 ) //分治门槛，小于这个值时不再进行递归计算，而是采用常规矩阵计算方法 73 { 74 MUL(MatrixA,MatrixB,MatrixC,N); 75 } 76 else 77 { 78 T** A11; 79 T** A12; 80 T** A21; 81 T** A22; 82 83 T** B11; 84 T** B12; 85 T** B21; 86 T** B22; 87 88 T** C11; 89 T** C12; 90 T** C21; 91 T** C22; 92 93 T** M1; 94 T** M2; 95 T** M3; 96 T** M4; 97 T** M5; 98 T** M6; 99 T** M7; 100 T** AResult; 101 T** BResult; 102 103 //making a 1 diminsional pointer based array. 104 A11 = new T *[newSize]; 105 A12 = new T *[newSize]; 106 A21 = new T *[newSize]; 107 A22 = new T *[newSize]; 108 109 B11 = new T *[newSize]; 110 B12 = new T *[newSize]; 111 B21 = new T *[newSize]; 112 B22 = new T *[newSize]; 113 114 C11 = new T *[newSize]; 115 C12 = new T *[newSize]; 116 C21 = new T *[newSize]; 117 C22 = new T *[newSize]; 118 119 M1 = new T *[newSize]; 120 M2 = new T *[newSize]; 121 M3 = new T *[newSize]; 122 M4 = new T *[newSize]; 123 M5 = new T *[newSize]; 124 M6 = new T *[newSize]; 125 M7 = new T *[newSize]; 126 127 AResult = new T *[newSize]; 128 BResult = new T *[newSize]; 129 130 int newLength = newSize; 131 132 //making that 1 diminsional pointer based array , a 2D pointer based array 133 for ( int i = 0; i < newSize; i++) 134 { 135 A11[i] = new T[newLength]; 136 A12[i] = new T[newLength]; 137 A21[i] = new T[newLength]; 138 A22[i] = new T[newLength]; 139 140 B11[i] = new T[newLength]; 141 B12[i] = new T[newLength]; 142 B21[i] = new T[newLength]; 143 B22[i] = new T[newLength]; 144 145 C11[i] = new T[newLength]; 146 C12[i] = new T[newLength]; 147 C21[i] = new T[newLength]; 148 C22[i] = new T[newLength]; 149 150 M1[i] = new T[newLength]; 151 M2[i] = new T[newLength]; 152 M3[i] = new T[newLength]; 153 M4[i] = new T[newLength]; 154 M5[i] = new T[newLength]; 155 M6[i] = new T[newLength]; 156 M7[i] = new T[newLength]; 157 158 AResult[i] = new T[newLength]; 159 BResult[i] = new T[newLength]; 160 161 162 } 163 //splitting input Matrixes, into 4 submatrices each. 164 for (int i = 0; i < N / 2; i++) 165 { 166 for (int j = 0; j < N / 2; j++) 167 { 168 A11[i][j] = MatrixA[i][j]; 169 A12[i][j] = MatrixA[i][j + N / 2]; 170 A21[i][j] = MatrixA[i + N / 2][j]; 171 A22[i][j] = MatrixA[i + N / 2][j + N / 2]; 172 173 B11[i][j] = MatrixB[i][j]; 174 B12[i][j] = MatrixB[i][j + N / 2]; 175 B21[i][j] = MatrixB[i + N / 2][j]; 176 B22[i][j] = MatrixB[i + N / 2][j + N / 2]; 177 178 } 179 } 180 181 //here we calculate M1..M7 matrices . 182 //M1[][] 183 ADD( A11,A22,AResult, HalfSize); 184 ADD( B11,B22,BResult, HalfSize); //p5=(a+d)*(e+h) 185 Strassen( HalfSize, AResult, BResult, M1 ); //now that we need to multiply this , we use the strassen itself . 186 187 188 //M2[][] 189 ADD( A21,A22,AResult, HalfSize); //M2=(A21+A22)B11 p3=(c+d)*e 190 Strassen(HalfSize, AResult, B11, M2); //Mul(AResult,B11,M2); 191 192 //M3[][] 193 SUB( B12,B22,BResult, HalfSize); //M3=A11(B12-B22) p1=a*(f-h) 194 Strassen(HalfSize, A11, BResult, M3); //Mul(A11,BResult,M3); 195 196 //M4[][] 197 SUB( B21, B11, BResult, HalfSize); //M4=A22(B21-B11) p4=d*(g-e) 198 Strassen(HalfSize, A22, BResult, M4); //Mul(A22,BResult,M4); 199 200 //M5[][] 201 ADD( A11, A12, AResult, HalfSize); //M5=(A11+A12)B22 p2=(a+b)*h 202 Strassen(HalfSize, AResult, B22, M5); //Mul(AResult,B22,M5); 203 204 205 //M6[][] 206 SUB( A21, A11, AResult, HalfSize); 207 ADD( B11, B12, BResult, HalfSize); //M6=(A21-A11)(B11+B12) p7=(c-a)(e+f) 208 Strassen( HalfSize, AResult, BResult, M6); //Mul(AResult,BResult,M6); 209 210 //M7[][] 211 SUB(A12, A22, AResult, HalfSize); 212 ADD(B21, B22, BResult, HalfSize); //M7=(A12-A22)(B21+B22) p6=(b-d)*(g+h) 213 Strassen(HalfSize, AResult, BResult, M7); //Mul(AResult,BResult,M7); 214 215 //C11 = M1 + M4 - M5 + M7; 216 ADD( M1, M4, AResult, HalfSize); 217 SUB( M7, M5, BResult, HalfSize); 218 ADD( AResult, BResult, C11, HalfSize); 219 220 //C12 = M3 + M5; 221 ADD( M3, M5, C12, HalfSize); 222 223 //C21 = M2 + M4; 224 ADD( M2, M4, C21, HalfSize); 225 226 //C22 = M1 + M3 - M2 + M6; 227 ADD( M1, M3, AResult, HalfSize); 228 SUB( M6, M2, BResult, HalfSize); 229 ADD( AResult, BResult, C22, HalfSize); 230 231 //at this point , we have calculated the c11..c22 matrices, and now we are going to 232 //put them together and make a unit matrix which would describe our resulting Matrix. 233 //组合小矩阵到一个大矩阵 234 for (int i = 0; i < N/2 ; i++) 235 { 236 for (int j = 0 ; j < N/2 ; j++) 237 { 238 MatrixC[i][j] = C11[i][j]; 239 MatrixC[i][j + N / 2] = C12[i][j]; 240 MatrixC[i + N / 2][j] = C21[i][j]; 241 MatrixC[i + N / 2][j + N / 2] = C22[i][j]; 242 } 243 } 244 245 // 释放矩阵内存空间 246 for (int i = 0; i < newLength; i++) 247 { 248 delete[] A11[i];delete[] A12[i];delete[] A21[i]; 249 delete[] A22[i]; 250 251 delete[] B11[i];delete[] B12[i];delete[] B21[i]; 252 delete[] B22[i]; 253 delete[] C11[i];delete[] C12[i];delete[] C21[i]; 254 delete[] C22[i]; 255 delete[] M1[i];delete[] M2[i];delete[] M3[i];delete[] M4[i]; 256 delete[] M5[i];delete[] M6[i];delete[] M7[i]; 257 delete[] AResult[i];delete[] BResult[i] ; 258 } 259 delete[] A11;delete[] A12;delete[] A21;delete[] A22; 260 delete[] B11;delete[] B12;delete[] B21;delete[] B22; 261 delete[] C11;delete[] C12;delete[] C21;delete[] C22; 262 delete[] M1;delete[] M2;delete[] M3;delete[] M4;delete[] M5; 263 delete[] M6;delete[] M7; 264 delete[] AResult; 265 delete[] BResult ; 266 267 }//end of else 268 269 } 270 271 template<typename T> 272 void Strassen_class<T>::FillMatrix( T** MatrixA, T** MatrixB, int length) 273 { 274 for(int row = 0; row<length; row++) 275 { 276 for(int column = 0; column<length; column++) 277 { 278 279 MatrixB[row][column] = (MatrixA[row][column] = rand() %5); 280 //matrix2[row][column] = rand() % 2;//ba hazfe in khat 50% afzayeshe soorat khahim dasht 281 } 282 283 } 284 } 285 template<typename T> 286 void Strassen_class<T>::PrintMatrix(T **MatrixA,int MatrixSize) 287 { 288 cout<<endl; 289 for(int row = 0; row<MatrixSize; row++) 290 { 291 for(int column = 0; column<MatrixSize; column++) 292 { 293 294 295 cout<<MatrixA[row][column]<<"\t"; 296 if ((column+1)%((MatrixSize)) == 0) 297 cout<<endl; 298 } 299 300 } 301 cout<<endl; 302 } 303 #endif
Strassen.cpp
1 #include <iostream> 2 #include <ctime> 3 #include <Windows.h> 4 using namespace std; 5 #include "Strassen.h" 6 7 int main() 8 { 9 Strassen_class<int> stra;//定义Strassen_class类对象 10 int MatrixSize = 0; 11 12 int** MatrixA; //存放矩阵A 13 int** MatrixB; //存放矩阵B 14 int** MatrixC; //存放结果矩阵 15 16 clock_t startTime_For_Normal_Multipilication ; 17 clock_t endTime_For_Normal_Multipilication ; 18 19 clock_t startTime_For_Strassen ; 20 clock_t endTime_For_Strassen ; 21 srand(time(0)); 22 23 cout<<"\n请输入矩阵大小(必须是2的幂指数值(例如:32,64,512,..): "; 24 cin>>MatrixSize; 25 26 int N = MatrixSize;//for readiblity. 27 28 //申请内存 29 MatrixA = new int *[MatrixSize]; 30 MatrixB = new int *[MatrixSize]; 31 MatrixC = new int *[MatrixSize]; 32 33 for (int i = 0; i < MatrixSize; i++) 34 { 35 MatrixA[i] = new int [MatrixSize]; 36 MatrixB[i] = new int [MatrixSize]; 37 MatrixC[i] = new int [MatrixSize]; 38 } 39 40 stra.FillMatrix(MatrixA,MatrixB,MatrixSize); //矩阵赋值 41 42 //*******************conventional multiplication test 43 cout<<"朴素矩阵算法开始时钟: "<< (startTime_For_Normal_Multipilication = clock()); 44 45 stra.MUL(MatrixA,MatrixB,MatrixC,MatrixSize);//朴素矩阵相乘算法 T(n) = O(n^3) 46 47 cout<<"\n朴素矩阵算法结束时钟: "<< (endTime_For_Normal_Multipilication = clock()); 48 49 cout<<"\n矩阵运算结果... \n"; 50 stra.PrintMatrix(MatrixC,MatrixSize); 51 52 //*******************Strassen multiplication test 53 cout<<"\nStrassen算法开始时钟: "<< (startTime_For_Strassen = clock()); 54 55 stra.Strassen( N, MatrixA, MatrixB, MatrixC ); //strassen矩阵相乘算法 56 57 cout<<"\nStrassen算法结束时钟: "<<(endTime_For_Strassen = clock()); 58 59 60 cout<<"\n矩阵运算结果... \n"; 61 stra.PrintMatrix(MatrixC,MatrixSize); 62 63 cout<<"矩阵大小 "<<MatrixSize; 64 cout<<"\n朴素矩阵算法: "<<(endTime_For_Normal_Multipilication - startTime_For_Normal_Multipilication)<<" Clocks.."<<(endTime_For_Normal_Multipilication - startTime_For_Normal_Multipilication)/CLOCKS_PER_SEC<<" Sec"; 65 cout<<"\nStrassen算法:"<<(endTime_For_Strassen - startTime_For_Strassen)<<" Clocks.."<<(endTime_For_Strassen - startTime_For_Strassen)/CLOCKS_PER_SEC<<" Sec\n"; 66 system("Pause"); 67 return 0; 68 69 }
输出：
| 矩阵大小 | 朴素矩阵算法（秒） | Strassen算法（秒） |
| 32 | 0.003 | 0.003 |
| 64 | 0.004 | 0.004 |
| 128 | 0.021 | 0.071 |
| 256 | 0.09 | 0.854 |
| 512 | 0.782 | 6.408 |
| 1024 | 8.908 | 52.391 |
1）采用Strassen算法作递归运算，需要创建大量的动态二维数组，其中分配堆内存空间将占用大量计算时间，从而掩盖了Strassen算法的优势
2）于是对Strassen算法做出改进，设定一个界限。当n<界限时，使用普通法计算矩阵，而不继续分治递归。需要合理设置界限，不同环境（硬件配置）下界限不同
3）矩阵乘法一般意义上还是选择的是朴素的方法，只有当矩阵变稠密，而且矩阵的阶数很大时，才会考虑使用Strassen算法。
分析原因：（网上总结的说法）
http://blog.csdn.net/handawnc/article/details/7987107
仔细研究后发现，采用Strassen算法作递归运算，需要创建大量的动态二维数组，其中分配堆内存空间将占用大量计算时间，从而掩盖了Strassen算法的优势。于是对Strassen算法做出改进，设定一个界限。当n<界限时，使用普通法计算矩阵，而不继续分治递归。
改进后算法优势明显，就算时间大幅下降。之后，针对不同大小的界限进行试验。在初步试验中发现，当数据规模小于1000时，下界S法的差别不大，规模大于1000以后，n取值越大，消耗时间下降。最优的界限值在32～128之间。
因为计算机每次运算时的系统环境不同（CPU占用、内存占用等），所以计算出的时间会有一定浮动。虽然这样，试验结果已经能得出结论Strassen算法比常规法优势明显。使用下界法改进后，在分治效率和动态分配内存间取舍，针对不同的数据规模稍加试验可以得到一个最优的界限。
http://www.cppblog.com/sosi/archive/2010/08/30/125259.html
时间复杂度就马上降下来了。。但是不要过于乐观。
从实用的观点看，Strassen算法通常不是矩阵乘法所选择的方法：
1 在Strassen算法的运行时间中，隐含的常数因子比简单的O(n^3)方法常数因子大
2 当矩阵是稀疏的时候，为稀疏矩阵设计的算法更快
3 Strassen算法不像简单方法那样子具有数值稳定性
4 在递归层次中生成的子矩阵要消耗空间。
所以矩阵乘法一般意义上还是选择的是朴素的方法，只有当矩阵变稠密，而且矩阵的阶数>20左右，才会考虑使用Strassen算法。
【1】http://blog.csdn.net/xyd0512/article/details/8220506
【2】http://blog.csdn.net/zhuangxiaobin/article/details/36476769
【3】http://blog.csdn.net/handawnc/article/details/7987107
【4】http://www.xuebuyuan.com/552410.html
【5】http://blog.csdn.net/chenhq1991/article/details/7599824
标签：des style blog http color os io 使用 ar
原文地址：http://www.cnblogs.com/zhoutaotao/p/3963048.html
