How Does Matrix Chain Multiplication Work?
Learn the O(n cubed) interval DP solution for optimal matrix chain multiplication parenthesization.
Expected Interview Answer
Matrix chain multiplication finds the cheapest order to parenthesize a chain of matrix multiplications — not the product itself — using dynamic programming in O(n³) time and O(n²) space, where dp[i][j] stores the minimum scalar multiplications needed to compute the product of matrices i through j by trying every possible split point k between them.
Multiplying matrices is associative, so (A×B)×C and A×(B×C) give the same result matrix, but the number of scalar multiplications performed can differ enormously depending on the matrix dimensions involved, making the parenthesization order a genuine optimization problem. The recurrence fills dp[i][j] as the minimum over every split point k (i <= k < j) of dp[i][k] + dp[k+1][j] + cost of multiplying the resulting two matrices, where that multiplication cost is p[i-1] * p[k] * p[j] given a dimensions array p. The table is filled by increasing chain length, since dp[i][j] depends on shorter subchains, giving the classic O(n³) interval DP pattern: O(n²) subproblems, each trying up to O(n) split points. A companion table tracks which split k achieved the minimum at each [i][j], enabling reconstruction of the actual optimal parenthesization after the DP table is filled.
- O(n³) time, O(n²) space via interval DP
- Dramatically reduces scalar multiplication count versus naive left-to-right order
- Split-point table reconstructs the actual optimal parenthesization
- Classic template for interval/range DP problems generally
AI Mentor Explanation
Matrix chain multiplication is like deciding the order to combine a series of partnership stats across an innings — combining stats pairwise costs different amounts of "work" depending on which pairs you merge first, even though the final combined total is the same regardless of order. The scorer tries every possible split point in the sequence of partnerships, computing the cheapest way to merge everything to the left of the split and everything to the right, then adding the cost of merging those two merged results together. Building this up from the shortest sub-sequences to the full innings, always reusing already-solved shorter merges, is what keeps the search from re-deriving the same merge costs repeatedly. The final answer is the cheapest total merge cost, plus a record of exactly which split points achieved it.
Step-by-Step Explanation
Step 1
Define the dimensions array
p[0..n] where matrix i has dimensions p[i-1] x p[i], for a chain of n matrices.
Step 2
Fill dp by increasing chain length
dp[i][j] = min over k of dp[i][k] + dp[k+1][j] + p[i-1]*p[k]*p[j], starting from length-2 chains.
Step 3
Track the optimal split point
Store the k that achieved the minimum at each [i][j] in a separate split table.
Step 4
Reconstruct the parenthesization
Recursively read the split table from dp[1][n] to print the actual optimal grouping.
What Interviewer Expects
- Clarify this optimizes parenthesization order, not the multiplication itself
- State O(n³) time, O(n²) space and explain the O(n²) subproblems × O(n) split points
- Give the correct cost formula: dp[i][k] + dp[k+1][j] + p[i-1]*p[k]*p[j]
- Mention the split-point table for reconstructing the actual optimal order
Common Mistakes
- Confusing this with actually multiplying the matrices (it only counts scalar operations)
- Forgetting matrix multiplication is associative but not commutative — order changes cost, not the result
- Getting the dimensions array indexing wrong (off-by-one on p[i-1] vs p[i])
- Omitting the split-point table, leaving no way to reconstruct the actual grouping
Best Answer (HR Friendly)
“Multiplying a chain of matrices always gives the same final result no matter how you group the multiplications, but grouping them differently can take wildly different amounts of computation. I would use dynamic programming to try every possible way of splitting the chain in two, reusing the best answer for each smaller piece, to find the grouping that costs the least — and I would keep track of exactly which split points won so I can reconstruct the actual best order.”
Code Example
def matrix_chain_order(p):
n = len(p) - 1 # number of matrices
dp = [[0] * (n + 1) for _ in range(n + 1)]
split = [[0] * (n + 1) for _ in range(n + 1)]
for length in range(2, n + 1):
for i in range(1, n - length + 2):
j = i + length - 1
dp[i][j] = float("inf")
for k in range(i, j):
cost = dp[i][k] + dp[k + 1][j] + p[i - 1] * p[k] * p[j]
if cost < dp[i][j]:
dp[i][j] = cost
split[i][j] = k
return dp[1][n], split
def print_parens(split, i, j):
if i == j:
return f"M{i}"
k = split[i][j]
return f"({print_parens(split, i, k)}{print_parens(split, k + 1, j)})"Follow-up Questions
- How would you reconstruct the actual optimal parenthesization from the split table?
- Why is matrix multiplication associative but not commutative, and how does that matter here?
- How does this problem generalize to other interval DP problems like optimal BST?
- What is the space complexity and how would you reduce it if only the minimum cost is needed?
MCQ Practice
1. What does matrix chain multiplication actually optimize?
The product is identical regardless of grouping; only the number of scalar multiplications performed changes.
2. What is the time complexity of the standard DP solution?
There are O(n²) subproblems (i, j pairs), each trying up to O(n) split points, giving O(n³).
3. What does the split table store?
The split table records the optimal split point at each subproblem so the actual parenthesization can be rebuilt recursively.
Flash Cards
What does matrix chain multiplication find? — The cheapest parenthesization order for multiplying a chain of matrices, not the product itself.
What is the time and space complexity of the DP solution? — O(n³) time, O(n²) space.
What is the cost recurrence for dp[i][j]? — min over k of dp[i][k] + dp[k+1][j] + p[i-1]*p[k]*p[j].
Why is matrix multiplication order-sensitive in cost but not in result? — It is associative (grouping does not change the result) but the scalar multiplication count varies hugely with grouping.