105 lines
2.7 KiB
Markdown
105 lines
2.7 KiB
Markdown
# 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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``` |