使用算法查找组合位的奇怪行为

时间:2012-05-20 00:47:31

标签: java algorithm

我正在尝试获得所有独特的位组合(真正有趣......我知道......)。基本上它适用于小数字,如10位,3个唯一或100位和4个唯一。我正在大规模测试它,但是当我使用200位和30项时会发生奇怪的事情......我什么都没得到。我试图调试代码,看不出原因。

我正在使用所有多头所以我不确定为什么,比特有点限制或这是我的数学问题?另外,我正在为大量的比特做些什么,或者是否有一些限制(即在程序完成之前太阳烧掉了)?理想情况下,我希望从2000列表中获得30个唯一位。这种方法是唯一超过100但我不知道其限制的方法。

以下是代码:

import java.io.BufferedWriter;
import java.io.FileWriter;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Iterator;
import java.util.List;
import java.util.SortedMap;
import java.util.TreeMap;

public class Combinatorics {

    static final class CombinationsIterator implements Iterator<Long> {
        private int n;
        private int k;
        private long next;

        public CombinationsIterator(int n, int k) {
            this.n = n;
            this.k = k;
            next = (1L << k) - 1;
        }

        @Override
        public boolean hasNext() {
            return (next & (1L << n)) == 0;
        }

        @Override
        public Long next() {
            long result = next;
            long x = next;
            long u = x & -x;
            long v = u + x;
            x = v + (((v ^ x) / u) >> 2);
            next = x;
            return result;
        }

        @Override
        public void remove() { throw new UnsupportedOperationException(); }
    }

    static final class PermutationsIterator implements Iterator<List<Integer>> {
        int n;
        int k;
        int nPk;
        List<Integer> elements;
        int i;
        List<Integer> next;

        public PermutationsIterator(int n, int k) {
            this.n = n;
            this.k = k;

            nPk = permute(n, k);

            List<Integer> elements = new ArrayList<Integer>();
            for (int i = 0; i < n; i++)
                elements.add(i);
            this.elements = Collections.unmodifiableList(elements);
        }

        @Override
        public boolean hasNext() { return i < nPk; }

        @Override
        public List<Integer> next() {
            List<Integer> next = new ArrayList<Integer>();
            List<Integer> notNext = new ArrayList<Integer>(elements);
            int r = i;
            int np = nPk;
            for (int j = 0; j < k; j++) {
                np /= n - j;
                next.add(notNext.remove(r / np));
                r %= np;
            }
            i++;
            return next;
        }

        @Override
        public void remove() { throw new UnsupportedOperationException(); }
    }

    public static long toCombination(List<Integer> permutation) {
        long combination = 0;
        for (int i : permutation)
            combination |= (1L << i);
        return combination;
    }

    public static List<Integer> toPermutation(long combination) {
        List<Integer> permutation = new ArrayList<Integer>();
        long combinationRemaining = combination;
        int i = 0;
        while (combinationRemaining > 0) {
            if ((combinationRemaining & 1) > 0) {
                permutation.add(i);
            }
            combinationRemaining >>= 1;
            i++;
        }
        return permutation;
    }

    public static SortedMap<Integer, Integer> multiplicitiesOf(List<Integer> multiset) {
        SortedMap<Integer, Integer> multiplicities = new TreeMap<Integer, Integer>();
        for (Integer k : multiset) {
            Integer v = multiplicities.get(k);
            v = (v == null) ? 1 : (v + 1);
            multiplicities.put(k, v);
        }
        return multiplicities;
    }

    public static Iterator<Long> combinationsIterator(int n, int k) {
        return new CombinationsIterator(n, k);
    }

    public static Iterator<List<Integer>> permutationsIterator(int n, int k) {
        return new PermutationsIterator(n, k);
    }

    public static int factorial(int n) {
        int result = 1;
        for (int i = 1; i <= n; i++)
            result *= i;
        return result;
    }

    public static int permute(int n, int k) {
        int result = 1;
        for (int i = n - k + 1; i <= n; i++)
            result *= i;
        return result;
    }

    public static int choose(int n, int k) {
        return permute(n, k) / factorial(k);
    }

    public static void main(String[] args) throws IOException {
        System.out.println("Starting");
        for (Iterator<Long> it = combinationsIterator(200, 2); it.hasNext(); ) {
            long next = it.next();
            System.out.format("%d\t%10s\n", next, Long.toBinaryString(next));
        }

    }
}

main方法中,更改combinationsIterator(200, 2);会产生奇怪的行为(10,3会非常快速地给出正确的结果,甚至100,5 ..但随着数字越来越高,它就不会给出结果而不是花时间处理)

1 个答案:

答案 0 :(得分:4)

由于您使用的是long,因此限制为64位。但是,200选择30将会根据一些粗略的背后数学信息 - 在现代PC上迭代10亿亿美元。