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fix R8a and add R8*.md and build them

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# Review 7
# Review 8-1
* Hajin Ju, 2024062806

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# 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
```

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# 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)
```

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# Review 8-4
* Hajin Ju, 2024062806
## Problem 1
What is the dimension of the matrix product $AB$ if $A$ is a $p\times q$ matrix and $B$ is a $q\times r$ matrix?
### Solution 1
$p\times r$
## Problem 2
Count the number of scalar multiplications to multiply $A$ and $B$ where $A$ is a $p\times q$ matrix and $B$ is a $q\times r$ matrix.
### Solution 2
$pqr$
## Problem 3
Count the number of scalar multiplications where the dimensions of $A_1$, $A_2$, and $A_3$$ are $10\times 100$, $100\times 5$ and $5\times 50$, respectively.
1. $(A_1A_2)A_3$
2. $A_1(A_2A_3)$
### Solution 3
1. $5000 + 2500 = 7500$
2. $25000 + 50000 = 75000$
## Problem 4
Fully parenthesize the product $A_1A_2A_3A_4$. (There are five distinct ways.)
### Solution 4
1. $(A_1A_2)(A_3A_4)$
2. $(A_1(A_2A_3))A_4$
3. $A_1(A_2(A_3A_4))$
4. $A_1((A_2A_3)A_4)$
5. $((A_1A_2)A_3)A_4$

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# 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 ")"
```