fix R8a and add R8*.md and build them
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# Review 7
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# Review 8-1
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* Hajin Ju, 2024062806
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# Review 8-2
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* Hajin Ju, 2024062806
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## Problem 1
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Fill in the blanks in the table below.
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* $p[i]$: the price for a rod of length i
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* $r[i]$: the maximum revenue for a rod of length i
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* $s[i]$: the length of the leftmost piece when the revenue is maximum
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### Solution 1
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| i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
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| ------ | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
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| $p[i]$ | 0 | 1 | 5 | 8 | 9 | 10 | 17 | 17 | 20 | 24 | 30 |
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| $r[i]$ | 0 | 1 | 5 | 8 | 10 | 13 | 17 | 18 | 22 | 25 | 30 |
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| $s[i]$ | 0 | 1 | 2 | 3 | 2 | 2 | 6 | 1 | 2 | 3 | 10 |
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## Problem 2
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Fill in the blanks in the following pseudocode for `EXTENDED-BOTTOM-UP-CUT-ROD`.
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### Solution 2
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```text
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EXTENDED-BOTTOM-UP-CUT-ROD (p, n)
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let r[0..n] and s[0..n] be new arrays
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r[0] = 0
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for j = 1 to n
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r[j] = -inf
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for i = 1 to j
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if r[j] < p[i] + r[j-i]
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r[j] = p[i] + r[j-i]
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s[j] = i
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return r, s
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```
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## Problem 3
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Fill in the blanks in the following pseudocode for `PRINT-CUT-ROD-SOLUTION`.
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### Solution 3
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```text
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PRINT-CUT-ROD-SOLUTION (p, n)
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(r, s) = EXTENDED-BOTTOM-UP-CUT-ROD(p, n)
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while n > 0
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print s[n]
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n = n - s[n]
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```
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## Problem 4
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Fill in the blanks in the following pseudocode for `M-CUT-ROD`.
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### Solution 4
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```text
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M-CUT-ROD (p, n)
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let r[0..n] be a new array
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for i = 0 to n
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r[i] = -inf
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return M-CUT-ROD-A(p, n, r)
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M-CUT-ROD-A (p, n, r)
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if r[n] >= 0
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return r[n]
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if n == 0
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return 0
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else q = -inf
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for i = 1 to n
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q = max(q, p[i] + M-CUT-ROD-A(p, n-i, r))
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r[n] = q
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return q
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```
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# Review 8-3
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* Hajin Ju, 2024062806
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## Problem 1
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Fill in the blanks in the following LCS computation
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### Solution 1
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$$c[i][j] = \text{The length of an LCS of the subsequences} \,X_i\, \text{and}\, Y_j.$$
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$$c[i][j] = \begin{cases}
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0 & \text{if}\, i = 0 \,\text{or}\, j = 0\\
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c[i-1][j-1] + 1 &\text{if}\, i,j > 0 \,\text{and}\, x_i = y_j\\
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max(c[i][j-1], c[i-1][j]) &\text{if}\, i, j > 0 \,\text{and}\, x_i \neq y_j
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\end{cases}$$
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| * | $y_j$ | $B$ | $D$ | $C$ | $A$ | $B$ | $A$ |
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| ----- | ----- | ------------ | ------------- | ------------- | -------------- | -------------- | -------------- |
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| $x_i$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
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| $A$ | 0 | $\uparrow 0$ | $\uparrow 0$ | $\uparrow 0$ | $\nwarrow 1$ | $\leftarrow 1$ | $\nwarrow 1$ |
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| $B$ | 0 | $\nwarrow1$ | $\leftarrow1$ | $\leftarrow1$ | $\uparrow 1$ | $\nwarrow 2$ | $\leftarrow 2$ |
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| $C$ | 0 | $\uparrow 1$ | $\uparrow 1$ | $\nwarrow 2$ | $\leftarrow 2$ | $\uparrow 2$ | $\uparrow 2$ |
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| $B$ | 0 | $\nwarrow 1$ | $\uparrow 1$ | $\uparrow 2$ | $\uparrow2$ | $\nwarrow3$ | $\leftarrow3$ |
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| $D$ | 0 | $\uparrow1$ | $\nwarrow2$ | $\uparrow2$ | $\uparrow2$ | $\uparrow3$ | $\uparrow3$ |
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| $A$ | 0 | $\uparrow1$ | $\uparrow2$ | $\uparrow2$ | $\nwarrow 3$ | $\uparrow 3$ | $\nwarrow4$ |
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| $B$ | 0 | $\nwarrow1$ | $\uparrow2$ | $\uparrow2$ | $\uparrow3$ | $\nwarrow4$ | $\uparrow4$ |
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## Problem 2
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Fill in the blanks in the following multiple LCS computation.
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### Solution 2
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| * | $y_j$ | $B$ | $D$ | $C$ | $A$ | $B$ | $A$ |
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| ----- | ----- | ---------------------- | ---------------------- | ---------------------- | ---------------------- | ---------------------- | ---------------------- |
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| $x_i$ | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
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| $A$ | 0 | $\leftarrow\uparrow 0$ | $\leftarrow\uparrow 0$ | $\leftarrow\uparrow 0$ | $\nwarrow 1$ | $\leftarrow 1$ | $\nwarrow 1$ |
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| $B$ | 0 | $\nwarrow1$ | $\leftarrow1$ | $\leftarrow1$ | $\leftarrow\uparrow 1$ | $\nwarrow 2$ | $\leftarrow 2$ |
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| $C$ | 0 | $\uparrow 1$ | $\leftarrow\uparrow 1$ | $\nwarrow 2$ | $\leftarrow 2$ | $\leftarrow\uparrow 2$ | $\leftarrow\uparrow 2$ |
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| $B$ | 0 | $\nwarrow 1$ | $\leftarrow\uparrow 1$ | $\uparrow 2$ | $\uparrow2$ | $\nwarrow3$ | $\leftarrow3$ |
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| $D$ | 0 | $\uparrow1$ | $\nwarrow2$ | $\leftarrow\uparrow2$ | $\uparrow2$ | $\uparrow3$ | $\uparrow3$ |
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| $A$ | 0 | $\uparrow1$ | $\uparrow2$ | $\uparrow2$ | $\nwarrow 3$ | $\leftarrow\uparrow 3$ | $\nwarrow4$ |
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| $B$ | 0 | $\nwarrow1$ | $\uparrow2$ | $\leftarrow\uparrow2$ | $\uparrow3$ | $\nwarrow4$ | $\leftarrow\uparrow4$ |
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## Problem 3
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Fill in the blanks in the following pseudocode for `LCS-LENGTH`.
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### Solution 3
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```text
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LCS-LENGTH(X, Y)
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m = X.length
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n = Y.length
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let b[1..m, 1..n] and c[1..m, 1..n] be new tables
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for i = 1 to m
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c[i][0] = 0
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for j = 1 to n
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c[0][j] = 0
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for i = 1 to m
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for j = 1 to n
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if X[i] = Y[j]
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c[i][j] = c[i - 1][j - 1] + 1
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b[i][j] = \nwarrow
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else if c[i-1][j] >= c[i][j-1]
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c[i][j] = c[i-1][j]
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b[i][j] = \uparrow
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else
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c[i][j] = c[i][j-1]
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b[i][j] = \leftarrow
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return c, b
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```
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## Problem 4
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Fill in the blanks in the following pseudocode for `PRINT-LCS`.
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### Solution 4
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```text
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PRINT-LCS(b, X, i, j)
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if i = 0 or j = 0
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return
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if b[i][j] == \nwarrow
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PRINT-LCS(b, X, i-1, j)
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print X[i]
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else if b[i][j] == \uparrow
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PRINT-LCS(b, X, i-1, j)
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else
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PRINT-LCS(b, X, i, j-1)
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```
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# Review 8-4
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* Hajin Ju, 2024062806
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## Problem 1
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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?
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### Solution 1
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$p\times r$
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## Problem 2
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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.
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### Solution 2
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$pqr$
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## Problem 3
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Count the number of scalar multiplications where the dimensions of $A_1$, $A_2$, and $A_3$$ are $10\times 100$, $100\times 5$ and $5\times 50$, respectively.
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1. $(A_1A_2)A_3$
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2. $A_1(A_2A_3)$
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### Solution 3
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1. $5000 + 2500 = 7500$
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2. $25000 + 50000 = 75000$
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## Problem 4
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Fully parenthesize the product $A_1A_2A_3A_4$. (There are five distinct ways.)
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### Solution 4
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1. $(A_1A_2)(A_3A_4)$
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2. $(A_1(A_2A_3))A_4$
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3. $A_1(A_2(A_3A_4))$
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4. $A_1((A_2A_3)A_4)$
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5. $((A_1A_2)A_3)A_4$
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# Review 8-5
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* Hajin Ju, 2024062806
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## Problem 1
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The minimum number of scalar multiplications for computing $A_iA_{i+1}\dots A_j$, denoted by $m[i, j]$, is as follows. Fill in the blanks.
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### Solution 1
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$$m[i,j] = \begin{cases}
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0 & \text{if}\; i = j\\
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\min_{i\leq k < j}{(m[i][k] + m[k+1][j] + p_{i-1}p_{k}p_j)} & \text{if}\;{i < j}
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\end{cases}$$
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## Problem 2
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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.
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| $m$ | 1 | 2 | 3 | 4 | 5 | 6 |
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| :---: | ---: | ----: | ---: | ---: | ----: | ----: |
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| 1 | 0 | 15750 | 7875 | 9375 | 11875 | 15125 |
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| 2 | | 0 | 2625 | 4375 | (a) | 10500 |
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| 3 | | | 0 | 750 | 2500 | 5375 |
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| 4 | | | | 0 | 1000 | 3500 |
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| 5 | | | | | 0 | 5000 |
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| 6 | | | | | | 0 |
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| $s$ | 2 | 3 | 4 | 5 | 6 |
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| :---: | --- | --- | --- | --- | --- |
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| 1 | 1 | 1 | 3 | 3 | 3 |
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| 2 | | 2 | 3 | (b) | 3 |
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| 3 | | | 3 | 3 | 3 |
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| 4 | | | | 4 | 5 |
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| 5 | | | | | 5 |
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| 6 | | | | | |
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| name | matrix dimension |
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| :-----: | ---------------- |
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| $A_1$ | $30\times 35$ |
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| $A_2$ | $35\times 15$ |
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| $A_3$ | $15\times 5$ |
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| $A_4$ | $5\times 10$ |
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| $A_5$ | $10\times 20$ |
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| $A_6$ | $20\times 25$ |
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### Solution 2
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| index | p |
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| ----- | --- |
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| 0 | 30 |
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| 1 | 35 |
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| 2 | 15 |
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| 3 | 5 |
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| 4 | 10 |
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| 5 | 20 |
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| 6 | 25 |
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(a). $$m[2, 5] = \left(\min\begin{cases}m[2][2] + m[3][5] + p[1][2][3] &= 13000\\
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m[2][3] + m[4][5] + p[1][3][5] &= 7125\\
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m[2][4] + m[5][5] + p[1][4][5] &= 11375\\
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\end{cases}\right)=7125$$
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(b). therefore $s[2, 5] = 3$
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(c). $$(A_1 (A_2A_3))((A_4A_5)A_6)$$
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## Problem 3
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Fill in the blanks in the following pseudocode for `MATRIX-CHAIN-ORDER`.
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### Solution 3
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```text
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MATRIX-CHAIN-ORDER (p)
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let m[1..n, 1..n] and s[1..(n-1), 2..n] be new tables
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for i = 1 to n
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m[i, i] = 0
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for l = 2 to n
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for i = 1 to n - l + 1
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j = i + l - 1
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m[i, j] = inf
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for k = i to j - 1
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q = m[i][k] + m[k + 1][j] + p[i-1] * p[k] * p[j]
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if q < m[i, j]
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m[i, j] = q
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s[i, j] = k
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return m and s
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```
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## Problem 4
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Fill in the blanks in the following pseudocode for `PRINT-OPTIMAL-PARENS`.
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### Solution 4
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```text
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PRINT-OPTIMAL-PARENS (s, i, j)
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if i == j
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print "A_i"
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else print "("
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PRINT-OPTIMAL-PARENS(s, i, s[i, j])
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PRINT-OPTIMAL-PARENS(s, s[i, j] + 1, j)
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print ")"
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```
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