# Review 8-3

* Hajin Ju, 2024062806

## Problem 1

Fill in the blanks in the following LCS computation

### Solution 1

$$c[i][j] = \text{The length of an LCS of the subsequences} \,X_i\, \text{and}\, Y_j.$$

$$c[i][j] = \begin{cases}
0 & \text{if}\, i = 0 \,\text{or}\, j = 0\\
c[i-1][j-1] + 1 &\text{if}\, i,j > 0 \,\text{and}\, x_i = y_j\\
max(c[i][j-1], c[i-1][j]) &\text{if}\, i, j > 0 \,\text{and}\, x_i \neq y_j
\end{cases}$$

| *     | $y_j$ | $B$          | $D$           | $C$           | $A$            | $B$            | $A$            |
| ----- | ----- | ------------ | ------------- | ------------- | -------------- | -------------- | -------------- |
| $x_i$ | 0     | 0            | 0             | 0             | 0              | 0              | 0              |
| $A$   | 0     | $\uparrow 0$ | $\uparrow 0$  | $\uparrow 0$  | $\nwarrow 1$   | $\leftarrow 1$ | $\nwarrow 1$   |
| $B$   | 0     | $\nwarrow1$  | $\leftarrow1$ | $\leftarrow1$ | $\uparrow 1$   | $\nwarrow 2$   | $\leftarrow 2$ |
| $C$   | 0     | $\uparrow 1$ | $\uparrow 1$  | $\nwarrow 2$  | $\leftarrow 2$ | $\uparrow 2$   | $\uparrow 2$   |
| $B$   | 0     | $\nwarrow 1$ | $\uparrow 1$  | $\uparrow 2$  | $\uparrow2$    | $\nwarrow3$    | $\leftarrow3$  |
| $D$   | 0     | $\uparrow1$  | $\nwarrow2$   | $\uparrow2$   | $\uparrow2$    | $\uparrow3$    | $\uparrow3$    |
| $A$   | 0     | $\uparrow1$  | $\uparrow2$   | $\uparrow2$   | $\nwarrow 3$   | $\uparrow 3$   | $\nwarrow4$    |
| $B$   | 0     | $\nwarrow1$  | $\uparrow2$   | $\uparrow2$   | $\uparrow3$    | $\nwarrow4$    | $\uparrow4$    |

## Problem 2

Fill in the blanks in the following multiple LCS computation.

### Solution 2

| *     | $y_j$ | $B$                    | $D$                    | $C$                    | $A$                    | $B$                    | $A$                    |
| ----- | ----- | ---------------------- | ---------------------- | ---------------------- | ---------------------- | ---------------------- | ---------------------- |
| $x_i$ | 0     | 0                      | 0                      | 0                      | 0                      | 0                      | 0                      |
| $A$   | 0     | $\leftarrow\uparrow 0$ | $\leftarrow\uparrow 0$ | $\leftarrow\uparrow 0$ | $\nwarrow 1$           | $\leftarrow 1$         | $\nwarrow 1$           |
| $B$   | 0     | $\nwarrow1$            | $\leftarrow1$          | $\leftarrow1$          | $\leftarrow\uparrow 1$ | $\nwarrow 2$           | $\leftarrow 2$         |
| $C$   | 0     | $\uparrow 1$           | $\leftarrow\uparrow 1$ | $\nwarrow 2$           | $\leftarrow 2$         | $\leftarrow\uparrow 2$ | $\leftarrow\uparrow 2$ |
| $B$   | 0     | $\nwarrow 1$           | $\leftarrow\uparrow 1$ | $\uparrow 2$           | $\uparrow2$            | $\nwarrow3$            | $\leftarrow3$          |
| $D$   | 0     | $\uparrow1$            | $\nwarrow2$            | $\leftarrow\uparrow2$  | $\uparrow2$            | $\uparrow3$            | $\uparrow3$            |
| $A$   | 0     | $\uparrow1$            | $\uparrow2$            | $\uparrow2$            | $\nwarrow 3$           | $\leftarrow\uparrow 3$ | $\nwarrow4$            |
| $B$   | 0     | $\nwarrow1$            | $\uparrow2$            | $\leftarrow\uparrow2$  | $\uparrow3$            | $\nwarrow4$            | $\leftarrow\uparrow4$  |


## Problem 3

Fill in the blanks in the following pseudocode for `LCS-LENGTH`.

### Solution 3

```text
LCS-LENGTH(X, Y)
    m = X.length
    n = Y.length
    let b[1..m, 1..n] and c[1..m, 1..n] be new tables
    for i = 1 to m
        c[i][0] = 0
    for j = 1 to n
        c[0][j] = 0
    for i = 1 to m
        for j = 1 to n
            if X[i] = Y[j]
                c[i][j] = c[i - 1][j - 1] + 1
                b[i][j] = \nwarrow
            else if c[i-1][j] >= c[i][j-1]
                c[i][j] = c[i-1][j]
                b[i][j] = \uparrow
            else
                c[i][j] = c[i][j-1]
                b[i][j] = \leftarrow
    return c, b
```

## Problem 4

Fill in the blanks in the following pseudocode for `PRINT-LCS`.

### Solution 4

```text
PRINT-LCS(b, X, i, j)
    if i = 0 or j = 0
        return
    if b[i][j] == \nwarrow
        PRINT-LCS(b, X, i-1, j)
        print X[i]
    else if b[i][j] == \uparrow
        PRINT-LCS(b, X, i-1, j)
    else
        PRINT-LCS(b, X, i, j-1)
```