fix R8a and add R8*.md and build them

2025-10-12 19:27:11 +09:00
parent 0a3ff3bbcb
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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 ")"
+```