矩阵链乘法问题描述(Matrix-chain multiplication)

矩阵链乘法问题描述(Matrix-chain multiplication)

flyfish 2015-9-13

矩阵链乘法问题：给定n个要相乘的矩阵构成序列<A1,A2,...,An>,计算乘积A1,A2...AnA_1,A_2...A_n。

Matrix-chain multiplication：We are given a sequence(chain)<A1,A2,...,An> or n matices to be multiplied,and we wish to compute the product A1,A2...AnA_1,A_2...A_n。

product

n乘积

multiply

英 [‘mʌltɪplaɪ] 美 [‘mʌltɪplaɪ]

vt. 乘；使增加；使繁殖；使相乘

vi. 乘；繁殖；增加

过去式 multiplied

过去分词 multiplied

现在分词 multiplying

为了计算，可以将两个矩阵相乘的算法作为一个子程序，根据括号的次序做全部的矩阵乘法。

We can evaluate the express using the stand algorithm for multiplying paires of matrices as a subroutine once we have parenthesized it to resolve all ambiguities in how the matrices are multiplied together.

parenthesize

英 [pə’renθɪsaɪz] 美 [pə’rɛnθə,saɪz]

vt. 加插入语于…；将…加上括弧

过去式 parenthesized

过去分词 parenthesized

现在分词 parenthesizing

ambiguity

英 [æmbɪ’gjuːɪtɪ] 美 [,æmbɪ’ɡjuəti]

n. 含糊；不明确；暧昧；模棱两可的话

复数 ambiguities

如果矩阵链为<A1,A2,A3,A4>乘积A1A2A3A4A_1A_2A_3A_4可用5种不同方式加全部括号:

if the chain of matrices is <A1,A2,A3,A4>,then we can fully parenthesized the product A1A2A3A4A_1A_2A_3A_4 in five distinct ways:

(A1(A2(A3A4)))(A_1 (A_2 (A_3 A_4 )))

(A1((A2A3)A4))(A_1 ((A_2 A_3 )A_4 ))

((A1A2)(A3A4))((A_1 A_2 )(A_3 A_4 ))

((A1(A2A3))A4)((A_1 (A_2 A_3 ))A_4 )

(((A1A2)A3)A4)(((A_1 A_2 )A_3 )A_4 )

只有当矩阵A的列数与矩阵B的行数相等时A×B才有意义

如果A是p*q矩阵，B是q*r矩阵，那么结果的矩阵C就是p*r矩阵

We can multiple tow metrices A and B only if they are compatible:the number of colums of A must equal the number of rows of B.

if A is a p*q matrix and

B is a q*r matrix，

the resulting matrix C is a p*r matrix

考虑3个矩阵的链<A1，A2，A3>，假设维数分别为10X100，100X5，5X50。

如果按照(A1A2)A3次序计算，那么需要10x100x5 + 10x5x50 = 7500次乘法运算；

如果按照A1(A2A3)次序计算，那么需要100x5x50 + 10x100x50 = 75000次乘法运算；因此按照第一种次序计算要快10倍。

consider the problem of chain<A1，A2，A3> of three matrices,Suppose that the dimensions of the matrices are 10x100,100x5,and 5x50,

respectively,If we multiply according to the parenthesization((A1A2)A3)((A_1 A_2) A_3),we perform 10x100x5=5000 scalar multiplications to compute the 10x5 matrix product A1A2A_1 A_2 , plus another 10 x 5 x50 = 2500 scalar multiplications to multiply this matrix by A3A_3 , for a total of 7500 scalar multiplications.

respective

英 [rɪ’spektɪv] 美 [rɪ’spɛktɪv]

adj. 分别的，各自的

respectively

英 [rɪ’spektɪvlɪ] 美 [rɪ’spɛktɪvli]

adv. 分别地；各自地，独自地

If instead we multiply according to the parenthesization(A1(A2A3))(A_1 (A_2 A_3)) we perform 100 x 5 x 50 = 25,000 scalar multiplications to compute the 100 x 50 matrix product A2A3A_2 A_3 , plus another 10 x 100 x 50 = 50,000 scalar multiplications to multiply A1A_1 by this matrix, for a total of 75,000 scalar multiplications. Thus, computing the product according tothe first parenthesization is 10 times faster.

假设矩阵 A B

A 2行3列

B 3行2列

A={142536},B=⎧⎩⎨⎪⎪123456⎫⎭⎬⎪⎪\begin{equation*}A=\begin{Bmatrix}
1 & 2 & 3\\
4 & 5& 6
\end{Bmatrix},
B=\begin{Bmatrix}
1 & 4\\
2 & 5\\
3 & 6
\end{Bmatrix}
\end{equation*}

乘积AB是

AB={1×1+2×2+3×34×1+5×2+6×31×4+2×5+3×64×4+5×5+6×6}\begin{equation*}AB=\begin{Bmatrix}
1\times 1+2\times 2+3\times 3 & 1\times 4+2\times 5+3\times 6 \\
4\times 1+5\times 2+6\times 3 & 4\times 4+5\times 5+6\times 6
\end{Bmatrix}
\end{equation*}

2*3*2 理解为2行 每行中一个元素需要3个乘，一共2列

A1A2A_1 A_2 结果是10行5列 每个元素100个乘.

A1A2A_1 A_2 的结果 乘以 A3A_3 结果是10行50列 每个元素5个乘.

所以是10x100x5 + 10x5x50 = 7500

矩阵链乘法问题可表述为：给定n个矩阵构成的一个链<A1，A2，A3>，其中i=1,2,3,4…..,n，矩阵AiA_i的维数为Pi1×PiPi_1 ×Pi，对乘积A1A2A3.....AnA_1A_2A_3.....A_n，以一种最小标量乘法次数的方式进行加全部括号。

We state the matrix-chain multiplication problem as follows: given a chain<A1，A2，A3> of n matrices, where for i=1,2,3,4…..,n matrix AiA_i has dimension Pi1×PiPi_1 ×Pi , fully parenthesize the product A1A2A3.....AnA_1A_2A_3.....A_n in a way that minimizes the number of scalar multiplications.

注意在矩阵链相乘题目中，实际上并没有把矩阵相乘，目的仅是确定一个具有最小代价的矩阵相乘次序

Note that in the matrix-chain multiplication problem, we are not actually multiplying matrices. Our goal is only to determine an order for multiplying matrices that has the lowest cost。