Wednesday, January 13, 2016

Freivalds' algorithm


随机的力量(2) - 矩阵比较
最直接的想法,当然是先进行矩阵乘法得到矩阵D = A×B,然后再一一比较矩阵C和矩阵D是否相同?
比较两个n×n矩阵是否相同的时间复杂度为O(N^2),而矩阵乘法直接算法时间复杂度是O(N^3), 目前最快的矩阵乘法的时间复杂度是O(N^2.376)。综合来看,时间复杂度接近O(N^3)。

是否存在更快的方法呢?答案是肯定的,下面我们介绍一种技术 - Freivalds技术

先来看一个A*B=C的必要条件: 对于n×1向量r, A*B*r = C*r
注意,这并不是充分条件。如果A*B*r = C*r, 并不能推导出A*B一定等于C。但如果A*B*r = C*r不成立,那么A×B=C就一定不成立。

随机产生一个n×1向量r,r[i] = 0或1,A*B*r = C*r 但A*B不等于C的概率是多大呢?
可以证明,P{A*B*r = C*r, A*B != C} <= 1/2

Freivalds技术就是基于上述的必要条件和概率估计,采用Monto Carlo算法, 算法逻辑可以参考以下伪代码。

// 判断A*B=C是否成立
bool IsIdentical(A, B, C) {
for (int iter = 0; iter < K; iter++) {
r = RandomVector()
// 判断A*B*r = C*r是否成立
if(false == IsIdentical(A, B, C, r)) {
// 不满足必要条件,必定不成立
return false;
}
}
return true;
}

每次必要条件测试A*B*r = C*r的计算时间复杂度为O(N^2)。(计算A*B*r = A*(B*r)采取从右往左计算顺序)
每次测试的错误概率小于等于1/2, 那么采取Monto Carlo算法后,K次都错误的概率会下降到1/(2^K)。
综合来看,时间复杂度为O(K*N^2), 而准确概率提升到1 - 1/(2^K)

check whether the multiplication of matrix A and B equals the given matrix C. It does it by checking A*(B*r)-(C*r) where r is any random column vector consisting only 0/1 as its elements. If this value is zero algorithm prints Yes, No otherwise.
  1.         //random generation of the r vector containing only 0/1 as its elements
  2.         double [][]r = new double[n][1];
  3.         Random random = new Random();
  4.         for(int i=0; i<n; i++)
  5.         {
  6.             r[i][0] = random.nextInt(2);
  7.         }
  8.         //test A * (b*r) - (C*) = 0
  9.         double br[][] = new double[n][1];
  10.         double cr[][] = new double[n][1];
  11.         double abr[][] = new double[n][1];
  12.         br = multiplyVector(b, r, n);
  13.         cr = multiplyVector(c, r, n);
  14.         abr = multiplyVector(a, br, n);
  15.  
  16.         //check for all zeros in abr
  17.         boolean flag = true; 
  18.         for(int i=0; i<n; i++)
  19.         {
  20.             if(abr[i][0] == 0)
  21.                 continue;
  22.             else
  23.                 flag = false;
  24.         }
  25.         if(flag == true)
  26.             System.out.println("Yes");
  27.         else
  28.             System.out.println("No");
  29. 
    
  30.     public static double[][] multiplyVector(double[][] a, double[][] b, int n)
  31.     {
  32.         double result[][] = new double[n][1];
  33.         for (int i = 0; i < n; i++) 
  34.         {
  35.             for (int j = 0; j < 1; j++) 
  36.             {
  37.                 for (int k = 0; k < n; k++)
  38.                 {  
  39.                     result[i][j] = result[i][j] + a[i][k] * b[k][j];
  40.                 }
  41.             }
  42.         }
  43.         return result;
  44.     }


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