Question

我正在尝试将旅行商问题的2-opt实施转换为3-opt。据我所知，与2-opt相比，删除3个边缘并将其替换为希望获得更好的距离。我有问题找出改变/添加到我的2-opt交换的内容，使其成为三个选择。我遇到的主要问题是如何确保在单个交换函数中考虑所有8种交换。谢谢

2-opt代码：

private static int[] TwoOpt(int[] TSP) {
        double best = totalDistance;
        int numCities = TSP.length;
        int visited = 0;
        int current = 0;
        int[] newTour = TSP;
        while (visited < numCities) {
            for (int i = 0; i < numCities - 1; i++) {
                for (int j = i + 1; j < numCities; j++) {
                    int[] newerTour = Swap(i, j, newTour);
                    int newDistance = distance(newerTour);
                    if (newDistance < best) {
                        visited = 0;
                        best = newDistance;
                        newTour = newerTour;
                    }
                }
            }
            visited++;

        }
        return newTour;

    }

    private static int distance(int[] newTour) {
        int distance = 0;
        for (int i = 0; i < newTour.length - 1; i++) {
            distance += mstList.get(i).get(i + 1).p;
        }
        distance += mstList.get(newTour.length).get(0).p;
        return distance;
    }

    private static int[] Swap(int i, int j, int[] tour) {
        int size = tour.length;
        int[] newerTour = new int[tour.length];
        for (int c = 0; c <= i - 1; c++) {
            newerTour[c] = tour[c];
        }
        int change = 0;
        for (int d = i; d <= j; d++) {
            newerTour[d] = tour[d - change];
            change++;
        }
        for (int e = j + 1; e < size; e++) {
            newerTour[e] = tour[e];
        }
        return newerTour;

    }

这就是我对三选的实现，没有实现交换。

private static int[] ThreeOpt(int[] TSP) {
        double best = totalDistance;
        int numCities = TSP.length;
        int visited = 0;
        int current = 0;
        int[] newTour = TSP;
        while (visited < numCities) {
            for (int i = 0; i < numCities - 2; i++) {
                for (int j = i + 1; j < numCities - 1; j++) {
                    for (int k = j + 1; k < numCities; k++) {
                        int[] newerTour = Swap(i, j, k, newTour);
                        int newDistance = distance(newTour);
                        if (newDistance < best) {
                            visited = 0;
                            best = newDistance;
                            newTour = newerTour;
                        }
                    }
                }
            }
            visited++;

        }
        return newTour;
    }

Answer 1

实际上只有4种形式的3-opt。

将巡回演变为3部分A->，B-＆gt;，C-＆gt;

将B连接到A和C的两条边上的2-opt是：

A-＆GT;，＆LT; -B，C-＆GT;

即。只是扭转B的旅程。

现在为3-opt

A-＆GT;，＆LT; -B，＆LT; -C
A-＆GT;，C-＆GT;，B-＆GT;
A-＆GT;，C-＆GT;，＆LT; -B
A-＆GT;，＆LT; -C，B-＆GT;

注意A->，＆lt; -B，C-＆gt;和A->，B->，＆lt; -C和A->，＆lt; -C，＆lt; -B都是2-opt。

逆转很容易实现，你只需要考虑4个变量。

2 - 选择3-opt旅行推销员

1 个答案: