Functional Programming in Java, Second Edition: p 148 "An optimization problem" & "Plain-Vanilla Recursion" problems

dtonhofer · 2023-04-26T12:58:59.270Z

On page 148, “An optimization problem” we read:

We’ll employ a solution for a company that buys rods at wholesale and sells them at retail. They figured that by cutting the rods into different sizes, they could maximize profits. The price that the company can charge for different lengths of rod changes often, so the company wants us to write a program to reveal what the maximum profit would be for a given size of rod.

The above is not computing the profit, but computing the revenue. The profit is revenue - expenses, but we don’t know the expenses, it might include manpower costs, machine costs etc.

The problem would also be more interesting if a rod of length 1 could only be sold at price 0 (i.e. it is wastage), at price 2 it’s too easy.

More seriously, on page 149, “Plain-Vanilla Recursion”, we read:

Continuing with this approach, we find that the maximum profit [revenue] for an arbitrary length n is the maximum of the profits [revenues] from each of the possible 2^(n-1) cuts length. That is, max(no cut, cut(1, n - 1), cut(2, n - 2), …), for a given length n.

It don’t understand the max() notation here, there should probably at least be revenue(.) of a cut schedule in there

In any case, the 2^(n-1) is imprecise Not considering symmetries, each cut point at marginal width 1 of which there are n-1, for example for width = 6:

≣|≣|≣|≣|≣|≣

can be switched on or off, giving us indeed 2^(n-1) “cut schedules.”

But considering all symmetries (to collapse similar “cut schedules”, consider only “cut schedules” where the width of a cut is monotonically (but not strictly) increasing from left to right), the number of possible “cut schedules” for width = n is then given by

[A000041 - OEIS] - the number of partitions of n (the partition numbers)

(I didn’t find this by myself, I first wrote the program to list the schedules, then duckduckgoed the sequence)

For example for width = 6, there are only 11 distinct ways to cut:

Number of ways of cutting for width = 6: 11
≣≣≣≣≣≣
≣|≣≣≣≣≣
≣≣|≣≣≣≣
≣≣≣|≣≣≣
≣|≣|≣≣≣≣
≣|≣≣|≣≣≣
≣≣|≣≣|≣≣
≣|≣|≣|≣≣≣
≣|≣|≣≣|≣≣
≣|≣|≣|≣|≣≣
≣|≣|≣|≣|≣|≣

Only increasing slowly:

|Width|Schedules|2^(n-1)|
|---|---|---|
|1|1|1|
|2|2|2|
|3|3|4|
|4|5|8|
|5|7|16|
|6|11|32|
|7|15|64|
|8|22|128|
|9|30|256|
|10|42|512|
|11|56|1024|
|12|77|2048|
|13|101|4096|
|14|135|8192|
|15|176|16384|
|16|231|32768|
|17|297|65536|
|18|385|131072|
|19|490|262144|
|20|627|524288|

Code to compute the above (unabashedly recursive, not memoizing/caching, slows down quickly with larger n. The SortedSet could be replaced by an array and “insertion sorting” if one wants “efficiency”)

import org.junit.jupiter.api.Test;

import java.util.*;
import java.util.stream.IntStream;

import static java.util.stream.Collectors.joining;

class CutSchedule implements Comparable<CutSchedule> {

    public List<Integer> increasingWidths = new ArrayList<>();

    public boolean verify() {
        if (increasingWidths.isEmpty()) {
            return false;
        }
        if (increasingWidths.get(0) <= 0) {
            return false;
        }
        for (int i = 1; i < increasingWidths.size(); i++) {
            if (increasingWidths.get(i - 1) > increasingWidths.get(i)) {
                return false;
            }
        }
        return true;
    }

    private static String toRodString(int width, char ch) {
        StringBuilder buf = new StringBuilder();
        IntStream.range(0, width).forEach(i -> buf.append(ch));
        return buf.toString();
    }

    public String toString(boolean numeric) {
        if (numeric) {
            return increasingWidths.stream().map(width -> Integer.toString(width)).collect(joining(","));
        } else {
            return increasingWidths.stream().map(width -> toRodString(width, '≣')).collect(joining("|"));
        }
    }

    public String toString() {
        return toString(false);
    }

    public int totalWidth() {
        return increasingWidths.stream().mapToInt(width -> width).sum();
    }

    public int cutCount() {
        return increasingWidths.size() - 1;
    }

    @Override
    public boolean equals(Object o) {
        if (o == null || !(o instanceof CutSchedule)) {
            return false;
        }
        return this.compareTo((CutSchedule) o) == 0;
    }

    @Override
    public int compareTo(CutSchedule o) {
        assert o != null;
        int widthDelta = this.totalWidth() - o.totalWidth();
        if (widthDelta != 0) {
            // if total width is smaller, the CutSchedule is "smaller"
            return widthDelta;
        }
        int cutCountDelta = this.cutCount() - o.cutCount();
        if (cutCountDelta != 0) {
            // if cut count is smaller, the CutSchedule is "smaller"
            return cutCountDelta;
        }
        for (int i = 0; i < cutCount(); i++) {
            int deltaCutWidth = this.increasingWidths.get(i) - o.increasingWidths.get(i);
            if (deltaCutWidth != 0) {
                // the first having a smaller cut at position i is "smaller"
                return deltaCutWidth;
            }
        }
        return 0;
    }
}

public class RodCuttingOptimization {

    private static void extendToFullWidthAndCollect(final SortedSet<CutSchedule> csSetForSmallerWidth, final int width, final int firstCutWidth, final Set<CutSchedule> res) {
        for (CutSchedule subCs : csSetForSmallerWidth) {
            assert subCs.verify();
            assert subCs.totalWidth() == width - firstCutWidth;
            CutSchedule cs = new CutSchedule();
            cs.increasingWidths.add(firstCutWidth);
            cs.increasingWidths.addAll(subCs.increasingWidths);
            res.add(cs);
        }
    }

    private static SortedSet<CutSchedule> generateAllCutsSchedulesForGivenNumCutsAndWidth(final int numCuts, final int width, final int minCutWidth) {
        assert numCuts >= 0;
        assert width > 0;
        assert minCutWidth > 0;
        SortedSet<CutSchedule> res = new TreeSet<>();
        if (numCuts == 0) {
            CutSchedule cs = new CutSchedule();
            cs.increasingWidths.add(width);
            res.add(cs);
        } else {
            // Make the first cut at increasingly larger positions. It must be the smallest cut made!
            IntStream.rangeClosed(minCutWidth, width / 2).forEach(firstCutWidth -> {
                SortedSet<CutSchedule> csSetForSmallerWidth =
                        Collections.unmodifiableSortedSet(
                                generateAllCutsSchedulesForGivenNumCutsAndWidth(
                                        numCuts - 1,
                                        width - firstCutWidth,
                                        firstCutWidth
                                ));
                extendToFullWidthAndCollect(csSetForSmallerWidth, width, firstCutWidth, res);
            });
        }
        return res;
    }

    private static void verifyAll(final Set<CutSchedule> csSet, int width, final Set<CutSchedule> mustNotContain) {
        csSet.stream().forEach(cs -> {
            assert cs.verify();
            assert cs.totalWidth() == width;
            assert !mustNotContain.contains(cs);
        });
    }

    private static SortedSet<CutSchedule> tryingAllCutsForWidth(final int width) {
        final int minNumCuts = 0;
        final int maxNumCuts = width - 1;
        SortedSet<CutSchedule> res = new TreeSet<>();
        IntStream.rangeClosed(minNumCuts, maxNumCuts).forEach(numCuts -> {
            Set<CutSchedule> csSetForWidth = generateAllCutsSchedulesForGivenNumCutsAndWidth(numCuts, width, 1);
            verifyAll(csSetForWidth, width, res);
            res.addAll(csSetForWidth);
        });
        return res;
    }

    private final static boolean withPrintout = false;

    @Test
    public void loopOverWidths() {
        final int minWidth = 1;
        final int maxWidth = 100;
        IntStream.rangeClosed(minWidth, maxWidth).forEach(width -> {
            SortedSet<CutSchedule> all = tryingAllCutsForWidth(width);
            System.out.println("Number of ways of cutting for width = " + width + ": " + all.size());
            if (withPrintout) {
                all.stream().forEach(cs -> System.out.println(cs.toString(false)));
            }
        });
    }

}

venkats · 2023-05-31T20:03:25.291Z

We can assume the given values are profit instead of revenue. The exponential time complexity also comes from the worst case scenario.