Background

DPV Ch 6.5
If $A$ is a matrix of dimension $p\times q$ and $B$ is a matrix of dimension $q\times r$, then the matrix product $A\cdot B$ is well-defined. It is the matrix $C$ of dimension $p\times r$ such that entry $c_{ij} = \sum_{k=1}^q a_{ik}b_{kj}$ for all $1\le i\le p$ and $1\le j\le r$.
The naive algorithm for multiplying two matrices using the above definition requires $pqr$ scalar multiplications.
Matrix multiplication is associative. Given a sequence (chain) of matrices any two consecutive ones of which are compatible for multiplication, we may compute the product of the whole sequence of matrices by repeatedly replacing any two consecutive matrices by their product, until only one matrix remains.

Parenthesization

We can specify this sequence of choices of multiplication by adding to the original sequence of matrices a pair of parentheses, for each choice of multiplication.
For example, one possible way to compute the product $A_1A_2A_3A_4A_5$ is as follows.
- [Step 1] Multiply $A_1$ by $A_2$.
- [Step 2] Multiply $A_3$ by $A_4$.
- [Step 3] Multiply the result matrix from Step 2 by $A_5$.
- [Step 4] Multiply the result matrix from Step 1 by the result matrix from Step 3.
This corresponds to the fully parenthesized expression $((A_1A_2)((A_3A_4)A_5))$.
Note that one full parenthesization may correspond to more than one sequence of choices.
It’s clear that computing the product $A_1A_2\cdots A_n$ according to different full parenthesizations may require different numbers of scalar multiplications.

Given a sequence of $n$ matrices $A_1, A_2, \ldots, A_n$, fully parenthesize it to give the fewest number of scalar multiplications used in computing the product $A_1A_2\cdots A_n$.
Note that we are not trying to compute the matrix product $A_1A_2\cdots A_n$. We are just figuring out how to parenthesize the matrix chain so that if we were to compute the matrix product using the naive method, the number of scalar multiplications required would be fewest.
Therefore, the entries of each matrix $A_i$ are irrelevant. Only their dimensions are relevant.
Assume the dimensions of the $A$ matrices are input as an array $p$ so that, for all $1\le i\le n$, the dimension of $A_i$ is $p_{i-1}\times p_i$.

For $1\le i\le j\le n$, define $m(i,j)$ to be the fewest number of scalar multiplications required to compute the product $A_iA_{i+1}\cdots A_j$.
We seek $m(1,n)$.

If $i=j$, the product $A_iA_{i+1}\cdots A_j$ is simply $A_i$ itself and it requires no scalar multiplication to compute it. So, $m(i,j)=0$ when $i=j$.
Suppose $i<j$. Let the very last matrix multiplication to compute the product $A_iA_{i+1}\cdots A_j$ optimally (i.e., using the fewest number of scalar multiplications) be the multiplication of the product $A_iA_{i+1}\cdots A_k$ by the product $A_{k+1}A_{k+2}\cdots A_j$. Then the product $A_iA_{i+1}\cdots A_k$ and the product $A_{k+1}A_{k+2}\cdots A_j$ must have been computed optimally too. Thus, $m(i,j) = m(i, k) + m(k+1, j) + p_{i-1}p_kp_j$.
We don't know what the value of $k$ is, but we know that $i\le k< j$.
The above reasoning gives the recurrence. \[ m(i, j) = \left\{ \begin{array}{ll} 0, & \mbox{if $i=j$} \\ \min \{\,m(i,k)+m(k+1,j)+p_{i-1}p_kp_j : i\le k< j\,\}, & \mbox{if $i<j$.} \end{array} \right. \]

MISSING PICTURE OF m(i,j) TABLE

[Step 1] Fill in the table $m(\cdot,\cdot)$, plus a companion table of minimizers.
[Step 2] Find an optimal full parenthesization by using the minimizer as a pointer to two subproblems. Since each problem has 2 subproblems, we program this recursively.

Suppose we fill in the $m(\cdot,\cdot)$ table row-by-row or column-by-column. Explain why the row index must be decreasing and the column index increasing.

We find an optimal parenthesization in time $O(n^3)$ because
- step 1 fills in each table entry in time $O(n)$, in time $O(n^3)$ total, and
- step 2 finds each pair of matrices to multiply in time $O(1)$, $O(n)$ time total.
This problem illustrates 2D-dynamic programming where $m(i,j)$ depends on $O(n)$ smaller values and the time is $O(n^3)$.