# Review 8-2

* Hajin Ju, 2024062806

## Problem 1

Fill in the blanks in the table below.

* $p[i]$: the price for a rod of length i
* $r[i]$: the maximum revenue for a rod of length i
* $s[i]$: the length of the leftmost piece when the revenue is maximum

### Solution 1

| i      | 0   | 1   | 2   | 3   | 4   | 5   | 6   | 7   | 8   | 9   | 10  |
| ------ | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| $p[i]$ | 0   | 1   | 5   | 8   | 9   | 10  | 17  | 17  | 20  | 24  | 30  |
| $r[i]$ | 0   | 1   | 5   | 8   | 10  | 13  | 17  | 18  | 22  | 25  | 30  |
| $s[i]$ | 0   | 1   | 2   | 3   | 2   | 2   | 6   | 1   | 2   | 3   | 10  |

## Problem 2

Fill in the blanks in the following pseudocode for `EXTENDED-BOTTOM-UP-CUT-ROD`.

### Solution 2

```text
EXTENDED-BOTTOM-UP-CUT-ROD (p, n)
    let r[0..n] and s[0..n] be new arrays
    r[0] = 0
    for j = 1 to n
        r[j] = -inf
        for i = 1 to j
            if r[j] < p[i] + r[j-i]
                r[j] = p[i] + r[j-i]
                s[j] = i
    return r, s
```

## Problem 3

Fill in the blanks in the following pseudocode for `PRINT-CUT-ROD-SOLUTION`.

### Solution 3

```text
PRINT-CUT-ROD-SOLUTION (p, n)
    (r, s) = EXTENDED-BOTTOM-UP-CUT-ROD(p, n)
    while n > 0
        print s[n]
        n = n - s[n]
```


## Problem 4

Fill in the blanks in the following pseudocode for `M-CUT-ROD`.

### Solution 4

```text
M-CUT-ROD (p, n)
    let r[0..n] be a new array
    for i = 0 to n
        r[i] = -inf
    return M-CUT-ROD-A(p, n, r)

M-CUT-ROD-A (p, n, r)
    if r[n] >= 0
        return r[n]
    if n == 0
        return 0
    else q = -inf
        for i = 1 to n
            q = max(q, p[i] + M-CUT-ROD-A(p, n-i, r))
    r[n] = q
    return q
```