Matrix Chain Multiplication
Find optimal way to parenthesize chain of matrix multiplications to minimize total scalar multiplications. Given matrices A1, A2, ..., An with dimensions, there are exponentially many ways to parenthesize, but DP finds optimal in O(n³) time. Classic example of optimal substructure in DP.
Visualization
Interactive visualization for Matrix Chain Multiplication
Matrix Chain Multiplication
Matrices: A1(30×35), A2(35×15), A3(15×5), A4(5×10), A5(10×20), A6(20×25)
• Time: O(n³)
• Space: O(n²)
• Finds optimal parenthesization
Interactive visualization with step-by-step execution
Implementation
1function matrixChainOrder(dims: number[]): number {
2 const n = dims.length - 1;
3 const dp: number[][] = Array(n).fill(0).map(() => Array(n).fill(0));
4
5 for (let len = 2; len <= n; len++) {
6 for (let i = 0; i < n - len + 1; i++) {
7 const j = i + len - 1;
8 dp[i][j] = Infinity;
9
10 for (let k = i; k < j; k++) {
11 const cost = dp[i][k] + dp[k + 1][j] + dims[i] * dims[k + 1] * dims[j + 1];
12 dp[i][j] = Math.min(dp[i][j], cost);
13 }
14 }
15 }
16
17 return dp[0][n - 1];
18}Deep Dive
Theoretical Foundation
For matrices A[i..j], optimal cost = min over all split points k of: cost(A[i..k]) + cost(A[k+1..j]) + cost of multiplying resulting matrices. Use 2D DP table where dp[i][j] = minimum multiplications for A[i..j]. Fill diagonally by chain length.
Complexity
Time
O(n³)
O(n³)
O(n³)
Space
O(n²)
Applications
Use Cases
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