动态规划之矩阵连乘

思考：

  三个矩阵A,B,C的阶分别是 $$a_0\ast a_1,a_1\ast a_2,a_2\ast a_3$$ ,从而(AxB)xC和Ax(BxC)的乘法次数是 $$a_0\ast a_1\ast a_2+a_0\ast a_2\ast a_3,a_1\ast a_2\ast a_3+a_0\ast a_1\ast a_3$$,二者一般是不相等的。则对于n个矩阵的连乘， $$A_1\ast A_2\cdots \ast A_n$$,如何改变计算次序，使得乘法次数计算量最小呢？

例如：

A1={30x35} ; A2={35x15} ;A3={15x5} ;A4={5x10} ;A5={10x20} ;A6={20x25} ;

结果为：((A1(A2A3))((A4A5)A6)) 最小的乘次为15125。

思路:

  设A[i:j]为矩阵$$A_i\ast A_{i+1}\cdots \ast A_j$$的连乘积，即从Ai到Aj的连乘积，其中，0 <= i <= j <= n-1

设m[i][j]为计算A[i:j]的最小乘次，所以原问题的最优值为m[0][n-1]。

• 当 i==j 时，单一矩阵，无需计算。$$m[i][i]=0，i=0,1\dots ,n-1$$
• 当 i < j 时，利用最优子结构，计算$$m[i][j]$$。即寻找断开位置k(i <= k < j)，使得$$m[i][k]+m[k+1][j]+P_i\ast P_{k+1}\ast P_{j+1}$$最小。

公式如下：

完整代码如下：

class Matrix:
  def __init__(self, row_num=0, col_num=0, matrix=None):
    if matrix != None:
      self.row_num = len(matrix)
      self.col_num = len(matrix[0])
    else:
      self.row_num = row_num
      self.col_num = col_num
    self.matrix = matrix
def matrix_chain(matrixs):
	matrix_num = len(matrixs)
	count = [[0 for j in range(matrix_num)] for i in range(matrix_num)]
	flag = [[0 for j in range(matrix_num)] for i in range(matrix_num)]
	for interval in range(1, matrix_num + 1):
		for i in range(matrix_num - interval):
			j = i + interval
			count[i][j] = count[i][i] + count[i + 1][j] + matrixs[i].row_num * matrixs[i + 1].row_num * matrixs[j].col_num
			flag[i][j] = i
			for k in range(i + 1, j):
				temp = count[i][k] + count[k + 1][j] + matrixs[i].row_num * matrixs[k + 1].row_num * matrixs[j].col_num
				if temp < count[i][j]:
					count[i][j] = temp	
					flag[i][j] = k
	traceback(0, matrix_num - 1, flag)
	return count[0][matrix_num - 1]

def traceback(i, j, flag):
	if i == j:
		return
	if j - i > 1:
		print(str(i + 1) + '~' + str(j + 1), end=': ')
		print(str(i + 1) + ":" + str(flag[i][j] + 1), end=',')
		print(str(flag[i][j] + 2) + ":" + str(j + 1))
	traceback(i, flag[i][j], flag)
	traceback(flag[i][j] + 1, j, flag)
	

def main():
	matrixs = [Matrix(30, 35), Matrix(35, 15), Matrix(15, 5), Matrix(5, 10), Matrix(10, 20), Matrix(20, 25)]
	result = matrix_chain(matrixs)
	print(result)

if __name__ == '__main__':
	main()

运行结果为：

1~6: 1:3,4:6
1~3: 1:1,2:3
4~6: 4:5,6:6
15125

表示计算顺序为 $$((A1(A2A3))((A4A5)A6))$$