# Review 8-5

* Hajin Ju, 2024062806

## Problem 1

The minimum number of scalar multiplications for computing $A_iA_{i+1}\dots A_j$, denoted by $m[i, j]$, is as follows. Fill in the blanks.

### Solution 1

$$m[i,j] = \begin{cases}
0 & \text{if}\; i = j\\
\min_{i\leq k < j}{(m[i][k] + m[k+1][j] + p_{i-1}p_{k}p_j)} & \text{if}\;{i < j}
\end{cases}$$

## Problem 2

Compute (a)$m [2, 5]$ and (b)$s[2, 5]$ in the following example and parenthesize (c) the prodct $A_1A_2A_3A_4A_5A_6$ fully to minimize the number of scalar multiplications.

|  $m$ |    1 |     2 |    3 |    4 |     5 |     6 |
| :---: | ---: | ----: | ---: | ---: | ----: | ----: |
|    1 |    0 | 15750 | 7875 | 9375 | 11875 | 15125 |
|    2 |      |     0 | 2625 | 4375 |   (a) | 10500 |
|    3 |      |       |    0 |  750 |  2500 |  5375 |
|    4 |      |       |      |    0 |  1000 |  3500 |
|    5 |      |       |      |      |     0 |  5000 |
|    6 |      |       |      |      |       |     0 |

| $s$ | 2   | 3   | 4   | 5   | 6   |
| :---: | --- | --- | --- | --- | --- |
| 1   | 1   | 1   | 3   | 3   | 3   |
| 2   |     | 2   | 3   | (b) | 3   |
| 3   |     |     | 3   | 3   | 3   |
| 4   |     |     |     | 4   | 5   |
| 5   |     |     |     |     | 5   |
| 6   |     |     |     |     |     |

| name  | matrix dimension |
| :-----: | ---------------- |
| $A_1$ | $30\times 35$    |
| $A_2$ | $35\times 15$    |
| $A_3$ | $15\times 5$     |
| $A_4$ | $5\times 10$     |
| $A_5$ | $10\times 20$    |
| $A_6$ | $20\times 25$    |

### Solution 2

| index | p   |
| ----- | --- |
| 0     | 30  |
| 1     | 35  |
| 2     | 15  |
| 3     | 5   |
| 4     | 10  |
| 5     | 20  |
| 6     | 25  |

(a). $$m[2, 5] = \left(\min\begin{cases}m[2][2] + m[3][5] + p[1][2][3] &= 13000\\
m[2][3] + m[4][5] + p[1][3][5] &= 7125\\
m[2][4] + m[5][5] + p[1][4][5] &= 11375\\
\end{cases}\right)=7125$$

(b). therefore $s[2, 5] = 3$

(c). $$(A_1 (A_2A_3))((A_4A_5)A_6)$$


## Problem 3

Fill in the blanks in the following pseudocode for `MATRIX-CHAIN-ORDER`.

### Solution 3

```text
MATRIX-CHAIN-ORDER (p)
    let m[1..n, 1..n] and s[1..(n-1), 2..n] be new tables
    for i = 1 to n
        m[i, i] = 0
    for l = 2 to n
        for i = 1 to n - l + 1
            j = i + l - 1
            m[i, j] = inf
            for k = i to j - 1
                q = m[i][k] + m[k + 1][j] + p[i-1] * p[k] * p[j]
                if q < m[i, j]
                    m[i, j] = q
                    s[i, j] = k
    return m and s
```

## Problem 4
Fill in the blanks in the following pseudocode for `PRINT-OPTIMAL-PARENS`.

### Solution 4

```text
PRINT-OPTIMAL-PARENS (s, i, j)
    if i == j
        print "A_i"
    else print "("
        PRINT-OPTIMAL-PARENS(s, i, s[i, j])
        PRINT-OPTIMAL-PARENS(s, s[i, j] + 1, j)
        print ")"
```