Let $A_1, A_2, A_3$, and $A_4$ be four matrices of dimensions $10 \times 5, 5 \times 20, 20 \times 10$, and $10 \times 5$, respectively. The minimum number of scalar multiplications required to find the product $A_1A_2A_3A_4$ using the basic matrix multiplication method is _______.

Let the sequence of dimensions be $p = \langle p_0, p_1, p_2, p_3, p_4 \rangle = \langle 10, 5, 20, 10, 5 \rangle$.

Using dynamic programming for Matrix Chain Multiplication, let $m[i, j]$ denote the minimum number of scalar multiplications needed to compute $A_i \dots A_j$:

Chains of length 1:

• $m[1, 1] = m[2, 2] = m[3, 3] = m[4, 4] = 0$

Chains of length 2:

• $m[1, 2] = p_0 p_1 p_2 = 10 \times 5 \times 20 = 1000$

• $m[2, 3] = p_1 p_2 p_3 = 5 \times 20 \times 10 = 1000$

• $m[3, 4] = p_2 p_3 p_4 = 20 \times 10 \times 5 = 1000$

Chains of length 3:

• For $m[1, 3]$:

  • Split at $k=1$: $m[1, 1] + m[2, 3] + p_0 p_1 p_3 = 0 + 1000 + 10 \times 5 \times 10 = 1500$

  • Split at $k=2$: $m[1, 2] + m[3, 3] + p_0 p_2 p_3 = 1000 + 0 + 10 \times 20 \times 10 = 3000$

  • $m[1, 3] = 1500$

• For $m[2, 4]$:

  • Split at $k=2$: $m[2, 2] + m[3, 4] + p_1 p_2 p_4 = 0 + 1000 + 5 \times 20 \times 5 = 1500$

  • Split at $k=3$: $m[2, 3] + m[4, 4] + p_1 p_3 p_4 = 1000 + 0 + 5 \times 10 \times 5 = 1250$

  • $m[2, 4] = 1250$

Chain of length 4 ($m[1, 4]$):

• Split at $k=1$: $m[1, 1] + m[2, 4] + p_0 p_1 p_4 = 0 + 1250 + 10 \times 5 \times 5 = 1250 + 250 = 1500$

• Split at $k=2$: $m[1, 2] + m[3, 4] + p_0 p_2 p_4 = 1000 + 1000 + 10 \times 20 \times 5 = 2000 + 1000 = 3000$

• Split at $k=3$: $m[1, 3] + m[4, 4] + p_0 p_3 p_4 = 1500 + 0 + 10 \times 10 \times 5 = 1500 + 500 = 2000$

The minimum cost is $\min(1500, 3000, 2000) = 1500$, corresponding to the parenthesization $A_1 ((A_2 A_3) A_4)$.