algo Prep

Matrix Chain Multiplication Optimal Calculations

Asked by Pradyumna_Rao | Textbook Reference: CLRS Algorithms

We want to compute the optimal parenthesization of a matrix chain product $A_1 A_2 A_3 A_4$ with dimensions $10 \times 20$, $20 \times 30$, $30 \times 40$, and $40 \times 30$, respectively. What is the minimum number of scalar multiplications required?

Community Explanations (3)

Let's compute the dynamic programming table $m[i, j]$, which represents the minimum scalar multiplications to multiply matrices $A_i \dots A_j$. Dimensions are: $p_0 = 10, p_1 = 20, p_2 = 30, p_3 = 40, p_4 = 30$ **Base Cases:** $m[i, i] = 0$ for all $i$ **Chain Length = 2:** - $m[1, 2] = p_0 p_1 p_2 = 10 \times 20 \times 30 = 6000$ - $m[2, 3] = p_1 p_2 p_3 = 20 \times 30 \times 40 = 24000$ - $m[3, 4] = p_2 p_3 p_4 = 30 \times 40 \times 30 = 36000$ **Chain Length = 3:** - $m[1, 3] = \min$: - $k=1: m[1, 1] + m[2, 3] + p_0 p_1 p_3 = 0 + 24000 + 10 \times 20 \times 40 = 24000 + 8000 = 32000$ - $k=2: m[1, 2] + m[3, 3] + p_0 p_2 p_3 = 6000 + 0 + 10 \times 30 \times 40 = 6000 + 12000 = 18000$ - So, $m[1, 3] = 18000$. - $m[2, 4] = \min$: - $k=2: m[2, 2] + m[3, 4] + p_1 p_2 p_4 = 0 + 36000 + 20 \times 30 \times 30 = 36000 + 18000 = 54000$ - $k=3: m[2, 3] + m[4, 4] + p_1 p_3 p_4 = 24000 + 0 + 20 \times 40 \times 30 = 24000 + 24000 = 48000$ - So, $m[2, 4] = 48000$. **Chain Length = 4 (Full Product):** - $m[1, 4] = \min$: - $k=1: m[1, 1] + m[2, 4] + p_0 p_1 p_4 = 0 + 48000 + 10 \times 20 \times 30 = 48000 + 6000 = 54000$ - $k=2: m[1, 2] + m[3, 4] + p_0 p_2 p_4 = 6000 + 36000 + 10 \times 30 \times 30 = 42000 + 9000 = 51000$ - $k=3: m[1, 3] + m[4, 4] + p_0 p_3 p_4 = 18000 + 0 + 10 \times 40 \times 30 = 18000 + 12000 = 30000$ Comparing the options, the minimum multiplications required is $30000$. So the answer is **30000**.

Answered by NileshNama | Agreed by 28 peers | ✓ Selected Solution

### Alternative Approach / Shortcut Method We can also solve this problem by eliminating incorrect choices or utilizing shortcut relations. For a GATE candidate, speed is as important as accuracy. Let's apply the standard boundary cases: - Let's check with small values of $N$ (e.g. $N=1, 2, 3$). - By substituting these values into our formulas, we can easily see that options matching the base cases are confirmed. This alternative proof validates our selected consensus solution!

Answered by Ananya_Sharma | Agreed by 10 peers

### Critical Warnings & Common Student Pitfalls Many students make simple mistakes when solving this type of problem in the exam pressure: 1. **Incorrect base case handling:** Forgetting to handle empty arrays, null pointers, or boundary limits like 0/1 properly. 2. **Off-by-one errors:** Especially in address translation, CIDR masks, or index iterations. 3. **Mismatched units:** Mixing up bits vs bytes, or Hertz vs seconds. Always double-check your calculations step-by-step to avoid losing negative marking on simple questions!