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java - Java 中 TSP 的动态编程方法

转载 作者:行者123 更新时间:2023-12-02 09:49:03 26 4
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我是一名初学者,我正在尝试使用动态规划方法编写一个旅行推销员问题。

这是我的计算函数的代码:

public static int compute(int[] unvisitedSet, int dest) {
if (unvisitedSet.length == 1)
return distMtx[dest][unvisitedSet[0]];

int[] newSet = new int[unvisitedSet.length-1];
int distMin = Integer.MAX_VALUE;

for (int i = 0; i < unvisitedSet.length; i++) {

for (int j = 0; j < newSet.length; j++) {
if (j < i) newSet[j] = unvisitedSet[j];
else newSet[j] = unvisitedSet[j+1];
}

int distCur;

if (distMtx[dest][unvisitedSet[i]] != -1) {
distCur = compute(newSet, unvisitedSet[i]) + distMtx[unvisitedSet[i]][dest];

if (distMin > distCur)
distMin = distCur;
}
else {
System.out.println("No path between " + dest + " and " + unvisitedSet[i]);
}
}
return distMin;
}

代码没有给我正确的答案,我正在尝试找出错误发生的位置。我认为当我添加以下内容时会发生错误:distCur = 计算(newSet, unvisitedSet[i]) + distMtx[unvisitedSet[i]][dest]; 所以我一直在搞乱那部分,在返回 distMin 之前将添加内容移到最后,依此类推......但我无法弄清楚。

虽然我确信可以从代码中推断出来,但我将陈述以下事实来澄清。

distMtx 存储所有城际距离,距离是对称的,也就是说如果从城市 A 到城市 B 的距离是 3,那么从城市 B 到城市 A 的距离也是 3。另外,如果两个城市没有直接路径,距离值为-1。

任何帮助将不胜感激!谢谢!

编辑:

主函数从文本文件中读取城际距离。因为我假设城市数量始终小于 100,所以全局 int 变量 distMtx 为 [100][100]。

一旦矩阵填充了必要的信息,就会创建一个包含所有城市的数组。城市的名称基本上都是数字。因此,如果我有 4 个城市,set[4] = {0, 1, 2, 3}

在主函数中,创建distMtxset后,首先调用compute():

int optRoute = compute(set, 0);
System.out.println(optRoute);

示例输入:

-1 3 2 7
3 -1 10 1
2 10 -1 4
7 1 4 -1

预期输出:

10

最佳答案

这是一个使用动态规划的 TSP 迭代解决方案。让您的生活更轻松的是将当前状态存储为位掩码而不是数组。这样做的优点是状态表示形式紧凑并且可以轻松缓存。

我做了一个video Youtube 上详细介绍了该问题的解决方案,敬请欣赏!代码取 self 的 github repo

/**
* An implementation of the traveling salesman problem in Java using dynamic
* programming to improve the time complexity from O(n!) to O(n^2 * 2^n).
*
* Time Complexity: O(n^2 * 2^n)
* Space Complexity: O(n * 2^n)
*
**/

import java.util.List;
import java.util.ArrayList;
import java.util.Collections;

public class TspDynamicProgrammingIterative {

private final int N, start;
private final double[][] distance;
private List<Integer> tour = new ArrayList<>();
private double minTourCost = Double.POSITIVE_INFINITY;
private boolean ranSolver = false;

public TspDynamicProgrammingIterative(double[][] distance) {
this(0, distance);
}

public TspDynamicProgrammingIterative(int start, double[][] distance) {
N = distance.length;

if (N <= 2) throw new IllegalStateException("N <= 2 not yet supported.");
if (N != distance[0].length) throw new IllegalStateException("Matrix must be square (n x n)");
if (start < 0 || start >= N) throw new IllegalArgumentException("Invalid start node.");

this.start = start;
this.distance = distance;
}

// Returns the optimal tour for the traveling salesman problem.
public List<Integer> getTour() {
if (!ranSolver) solve();
return tour;
}

// Returns the minimal tour cost.
public double getTourCost() {
if (!ranSolver) solve();
return minTourCost;
}

// Solves the traveling salesman problem and caches solution.
public void solve() {

if (ranSolver) return;

final int END_STATE = (1 << N) - 1;
Double[][] memo = new Double[N][1 << N];

// Add all outgoing edges from the starting node to memo table.
for (int end = 0; end < N; end++) {
if (end == start) continue;
memo[end][(1 << start) | (1 << end)] = distance[start][end];
}

for (int r = 3; r <= N; r++) {
for (int subset : combinations(r, N)) {
if (notIn(start, subset)) continue;
for (int next = 0; next < N; next++) {
if (next == start || notIn(next, subset)) continue;
int subsetWithoutNext = subset ^ (1 << next);
double minDist = Double.POSITIVE_INFINITY;
for (int end = 0; end < N; end++) {
if (end == start || end == next || notIn(end, subset)) continue;
double newDistance = memo[end][subsetWithoutNext] + distance[end][next];
if (newDistance < minDist) {
minDist = newDistance;
}
}
memo[next][subset] = minDist;
}
}
}

// Connect tour back to starting node and minimize cost.
for (int i = 0; i < N; i++) {
if (i == start) continue;
double tourCost = memo[i][END_STATE] + distance[i][start];
if (tourCost < minTourCost) {
minTourCost = tourCost;
}
}

int lastIndex = start;
int state = END_STATE;
tour.add(start);

// Reconstruct TSP path from memo table.
for (int i = 1; i < N; i++) {

int index = -1;
for (int j = 0; j < N; j++) {
if (j == start || notIn(j, state)) continue;
if (index == -1) index = j;
double prevDist = memo[index][state] + distance[index][lastIndex];
double newDist = memo[j][state] + distance[j][lastIndex];
if (newDist < prevDist) {
index = j;
}
}

tour.add(index);
state = state ^ (1 << index);
lastIndex = index;
}

tour.add(start);
Collections.reverse(tour);

ranSolver = true;
}

private static boolean notIn(int elem, int subset) {
return ((1 << elem) & subset) == 0;
}

// This method generates all bit sets of size n where r bits
// are set to one. The result is returned as a list of integer masks.
public static List<Integer> combinations(int r, int n) {
List<Integer> subsets = new ArrayList<>();
combinations(0, 0, r, n, subsets);
return subsets;
}

// To find all the combinations of size r we need to recurse until we have
// selected r elements (aka r = 0), otherwise if r != 0 then we still need to select
// an element which is found after the position of our last selected element
private static void combinations(int set, int at, int r, int n, List<Integer> subsets) {

// Return early if there are more elements left to select than what is available.
int elementsLeftToPick = n - at;
if (elementsLeftToPick < r) return;

// We selected 'r' elements so we found a valid subset!
if (r == 0) {
subsets.add(set);
} else {
for (int i = at; i < n; i++) {
// Try including this element
set |= 1 << i;

combinations(set, i + 1, r - 1, n, subsets);

// Backtrack and try the instance where we did not include this element
set &= ~(1 << i);
}
}
}

public static void main(String[] args) {
// Create adjacency matrix
int n = 6;
double[][] distanceMatrix = new double[n][n];
for (double[] row : distanceMatrix) java.util.Arrays.fill(row, 10000);
distanceMatrix[5][0] = 10;
distanceMatrix[1][5] = 12;
distanceMatrix[4][1] = 2;
distanceMatrix[2][4] = 4;
distanceMatrix[3][2] = 6;
distanceMatrix[0][3] = 8;

int startNode = 0;
TspDynamicProgrammingIterative solver = new TspDynamicProgrammingIterative(startNode, distanceMatrix);

// Prints: [0, 3, 2, 4, 1, 5, 0]
System.out.println("Tour: " + solver.getTour());

// Print: 42.0
System.out.println("Tour cost: " + solver.getTourCost());
}
}

关于java - Java 中 TSP 的动态编程方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33527127/

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