15.1 Rod cutting

15.1-1

Show that equation \(\text{(15.4)}\) follows from equation \(\text{(15.3)}\) and the initial condition \(T(0) = 1\).

• For \(n = 0\), this holds since \(2^0 = 1\).

• For \(n > 0\), substituting into the recurrence, we have

  \[ \begin{aligned} T(n) & = 1 + \sum_{j = 0}^{n - 1} 2^j \\\\ & = 1 + (2^n - 1) \\\\ & = 2^n. \end{aligned} \]

15.1-2

Show, by means of a counterexample, that the following "greedy" strategy does not always determine an optimal way to cut rods. Define the density of a rod of length \(i\) to be \(p_i / i\), that is, its value per inch. The greedy strategy for a rod of length \(n\) cuts off a first piece of length \(i\), where \(1 \le i \le n\), having maximum density. It then continues by applying the greedy strategy to the remaining piece of length \(n - i\).

The counterexample:

15.1-3

Consider a modification of the rod-cutting problem in which, in addition to a price \(p_i\) for each rod, each cut incurs a fixed cost of \(c\). The revenue associated with a solution is now the sum of the prices of the pieces minus the costs of making the cuts. Give a dynamic-programming algorithm to solve this modified problem.

We can modify \(\text{BOTTOM-UP-CUT-ROD}\) algorithm from section 15.1 as follows:

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MODIFIED-CUT-ROD(p, n, c)
    let r[0..n] be a new array
    r[0] = 0
    for j = 1 to n
        q = p[j]
        for i = 1 to j - 1
            q = max(q, p[i] + r[j - i] - c)
        r[j] = q
    return r[n]

We need to account for cost \(c\) on every iteration of the loop in lines 5-6 but the last one, when \(i = j\) (no cuts).

We make the loop run to \(j - 1\) instead of \(j\), make sure \(c\) is subtracted from the candidate revenue in line 6, then pick the greater of current best revenue \(q\) and \(p[j]\) (no cuts) in line 7.

15.1-4

Modify \(\text{MEMOIZED-CUT-ROD}\) to return not only the value but the actual solution, too.

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MEMOIZED-CUT-ROD(p, n)
    let r[0..n] and s[0..n] be new arrays
    for i = 0 to n
        r[i] = -∞
    (val, s) = MEMOIZED-CUT-ROD-AUX(p, n, r, s)
    print "The optimal value is" val "and the cuts are at" s
    j = n
    while j > 0
        print s[j]
        j = j - s[j]

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MEMOIZED-CUT-ROD-AUX(p, n, r, s)
    if r[n] ≥ 0
        return r[n]
    if n == 0
        q = 0
    else q = -∞
        for i = 1 to n
            (val, s) = MEMOIZED-CUT-ROD-AUX(p, n - i, r, s)
            if q < p[i] + val
                q = p[i] + val
                s[n] = i
    r[n] = q
    return (q, s)

15.1-5

The Fibonacci numbers are defined by recurrence \(\text{(3.22)}\). Give an \(O(n)\)-time dynamic-programming algorithm to compute the nth Fibonacci number. Draw the subproblem graph. How many vertices and edges are in the graph?

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FIBONACCI(n)
    let fib[0..n] be a new array
    fib[0] = 1
    fib[1] = 1
    for i = 2 to n
        fib[i] = fib[i - 1] + fib[i - 2]
    return fib[n]

There are \(n + 1\) vertices in the subproblem graph, i.e., \(v_0, v_1, \dots, v_n\).

• For \(v_0, v_1\), each has \(0\) leaving edge.
• For \(v_2, v_3, \dots, v_n\), each has \(2\) leaving edges.

Thus, there are \(2n - 2\) edges in the subproblem graph.

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