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

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

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 ")"

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