Problem Statement
Explain the Matrix Chain Multiplication problem and how dynamic programming solves it optimally.
Explanation
Matrix Chain Multiplication is a classic optimization problem where you need to find the most efficient way to multiply a sequence of matrices. The order of multiplication matters because matrix multiplication is associative but the number of scalar multiplications depends on the parenthesization.
Given matrices A1, A2, dot dot dot, An with dimensions, we want to fully parenthesize the product to minimize the total number of scalar multiplications. For example, multiplying matrices A with dimensions 10 by 30, B with 30 by 5, and C with 5 by 60 can be done as either A times B times C which is 10 times 30 times 5 plus 10 times 5 times 60 equals 1500 plus 3000 equals 4500 operations, or A times B times C which is 30 times 5 times 60 plus 10 times 30 times 60 equals 9000 plus 18000 equals 27000 operations. The first order is much more efficient.
The problem exhibits optimal substructure. The optimal way to multiply matrices from i to j includes an optimal split at some k, where we optimally multiply matrices from i to k, matrices from k plus 1 to j, and then multiply the two results. We try all possible splits and choose the one with minimum cost.
The DP solution uses a 2D table where dp of i of j represents the minimum number of scalar multiplications needed to multiply matrices from i to j. The recurrence relation is: dp of i of j equals min over all k from i to j minus 1 of dp of i of k plus dp of k plus 1 of j plus cost of multiplying resulting matrices. The cost of multiplying two matrices with dimensions p times q and q times r is p times q times r.
Base cases are dp of i of i equals 0 for all i, as multiplying a single matrix costs nothing. The evaluation order is critical. We cannot fill the table row by row or column by column because dp of i of j depends on values in the same row to the left and same column below. Instead, we fill by chain length. First compute all chains of length 1, then length 2, and so on. This ensures dependencies are computed before needed.
The algorithm stores matrix dimensions in an array where matrix i has dimensions dimensions of i minus 1 by dimensions of i. For chain length len from 2 to n, for each starting position i from 1 to n minus len plus 1, compute ending position j equals i plus len minus 1. Try all split positions k from i to j minus 1. For each k, compute cost equals dp of i of k plus dp of k plus 1 of j plus dimensions of i minus 1 times dimensions of k times dimensions of j. Update dp of i of j to minimum cost found.
Time complexity is O of n cubed because we have O of n squared subproblems, each requiring O of n work to try all splits. Space complexity is O of n squared for the DP table.
To reconstruct the optimal parenthesization, we maintain an auxiliary table split of i of j that stores the optimal split point for matrices i to j. After filling the DP table, we can recursively build the parenthesization using these split points.
The algorithm can be extended to print the optimal solution as a string with parentheses. Starting from split of 1 of n, recursively construct left and right subproblems until reaching base cases.
Applications include optimizing database query execution where operations can be reordered, optimizing expression evaluation in compilers, and any scenario involving associative operations where order affects cost. The problem demonstrates the power of DP for problems with optimal substructure where trying all possibilities directly would be exponential.