feat: Add Matrix Chain Multiplication in C++

Author: parrth20

CPP/dynamic_programming/Matrix_chain_mul.cpp

@@ -0,0 +1,86 @@
+/**
+ *  matrix_chain_multiplication.cpp
+ *  Implementation of the Matrix Chain Multiplication (MCM) problem.
+ *
+ * Description:
+ * This program finds the most efficient way to multiply a chain of matrices.
+ * The problem is not to perform the multiplications, but to decide the sequence
+ * of parenthesizations. We use a Dynamic Programming approach, specifically the
+ * Partition DP pattern.
+ *
+ * The state dp[i][j] represents the minimum number of scalar multiplications
+ * needed to compute the product of matrices from index i to j.
+ *
+ * The state transition is:
+ * dp[i][j] = min(dp[i][k] + dp[k+1][j] + dims[i-1]*dims[k]*dims[j])
+ * for all k from i to j-1.
+ *
+ * Complexity Analysis:
+ * - Time Complexity: O(n^3), where n is the number of matrices. Three nested loops are used.
+ * - Space Complexity: O(n^2) for the DP table.
+ */
+#include <iostream>
+#include <vector>
+#include <limits>
+
+using namespace std;
+
+/**
+ * @brief Solves the Matrix Chain Multiplication problem using tabulation.
+ * @param dims A vector where dims[i-1] and dims[i] are the dimensions of matrix i.
+ * @return The minimum number of scalar multiplications.
+ */
+int matrixChainOrder(const vector<int>& dims) {
+    // There are n-1 matrices.
+    int n = dims.size() - 1;
+    if (n < 2) {
+        return 0; // No multiplication needed if there's 0 or 1 matrix.
+    }
+
+    // dp[i][j] = Minimum number of multiplications to compute matrix M[i]...M[j]
+    // where M[i] has dimensions dims[i-1] x dims[i].
+    vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
+
+    // L is the chain length. It varies from 2 to n.
+    for (int L = 2; L <= n; ++L) {
+        // i is the starting matrix index.
+        for (int i = 1; i <= n - L + 1; ++i) {
+            // j is the ending matrix index.
+            int j = i + L - 1;
+            dp[i][j] = numeric_limits<int>::max();
+
+            // k is the partition point.
+            for (int k = i; k < j; ++k) {
+                // Cost to multiply M[i]..M[k] and M[k+1]..M[j]
+                // and then the final two resulting matrices.
+                int cost = dp[i][k] + dp[k + 1][j] + dims[i - 1] * dims[k] * dims[j];
+                if (cost < dp[i][j]) {
+                    dp[i][j] = cost;
+                }
+            }
+        }
+    }
+    return dp[1][n];
+}
+
+// Example usage and test cases
+int main() {
+    // Example: We have 4 matrices with dimensions:
+    // M1: 10x30, M2: 30x5, M3: 5x60, M4: 60x20
+    // The dimensions array will be {10, 30, 5, 60, 20}
+    vector<int> dims = {10, 30, 5, 60, 20};
+    cout << "The dimensions of the matrices are: ";
+    for(size_t i = 0; i < dims.size() -1; ++i) {
+        cout << dims[i] << "x" << dims[i+1] << (i == dims.size() - 2 ? "" : ", ");
+    }
+    cout << endl;
+
+    int min_multiplications = matrixChainOrder(dims);
+    cout << "Minimum number of scalar multiplications needed is: " << min_multiplications << endl; // Expected: 13500
+
+    vector<int> dims2 = {40, 20, 30, 10, 30};
+    cout << "\nFor dimensions {40, 20, 30, 10, 30}:" << endl;
+    cout << "Minimum number of scalar multiplications needed is: " << matrixChainOrder(dims2) << endl; // Expected: 26000
+
+    return 0;
+}