add review 14 15

notes/midterm.md

@@ -0,0 +1,44 @@
+# Alogorithm Midterm Access
+
+## Sorting Problem
+
+Comparison Sorts(Lower bound is $\Omega(n \lg n)$ because decision-tree model)
+
+* Insertion Sort
+* Selection Sort
+* Merge Sort
+* Quick Sort
+* Heap Sort
+
+Sorting In Linear Time
+
+* Counting Sort
+* Radix Sort
+
+## Asymtotic Notation
+
+* $Theta$-notation
+* $O$-notation
+* $\Omega$-notation
+
+Transitivity, Reflexivity, Symmetry, Transpose Symmetry
+
+Three methods for solving recurrences:
+1. Substitution method
+2. Recursion-tree method
+3. Master method
+
+**Getting asymtotic notation by recursion tree is important**
+
+## Dynamic Programming
+
+* Assembly-line scheduling
+* Rod cutting
+* Longest-Common Subsequence
+* Matrix-chain Multiplication
+
+## Greedy Algorithm
+
+* An Activity Selection Problem
+* Elements of the greedy strategy
+* Huffman codes

reviews/R7.md

@@ -194,8 +194,5 @@ col4.append(create_radix_table(soup,
         "TAR",], [0, 1, 2]
 ))
 
-
-
-
 print(soup)
 ```

reviews/R8a.md

@@ -1,4 +1,4 @@
-# Review 7
+# Review 8-1
 
 * Hajin Ju, 2024062806
 

reviews/R8b.md

@@ -0,0 +1,78 @@
+# 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
+```

reviews/R8c.md

@@ -0,0 +1,94 @@
+# 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)
+```

reviews/R8d.md

@@ -0,0 +1,43 @@
+# 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. $((A_1A_2)A_3)A_4$

reviews/R8e.md

@@ -0,0 +1,105 @@
+# 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 ")"
+```  

reviews/__pyfunc__/r7.py

@@ -17,6 +17,9 @@ TH_STYLE_TEMPLATE = "font-weight: normal; border: none; width: 1.2em; height: 0.
 def td_style_with_gray(style):
     return style + "background-color: lightgray;"
 
+def td_style_with_no_left_border(style):
+    return style + "border-left: none;"
+
 def td_style_with_no_right_border(style):
     return style + "border-right: none;"
 
@@ -68,6 +71,8 @@ def create_radix_table(soup: BeautifulSoup,
         tr = soup.new_tag("tr")
         for i, char in enumerate(word):
             style = TD_STYLE_TEMPLATE
+            if i != 0:
+                style = td_style_with_no_left_border(style)
             if i != len(word) - 1:
                 style = td_style_with_no_right_border(style)
             if highlighted and i in highlighted: