Question

是否可以将函数go转换为非递归函数？一些提示或启动草图将非常有用

public static TSPSolution solve(CostMatrix _cm, TSPPoint start, TSPPoint[] points, long seed) {
    TSPSolution sol = TSPSolution.randomSolution(start, points, seed, _cm);
    double t = initialTemperature(sol, 1000);
    int frozen = 0;
    System.out.println("-- Simulated annealing started with initial temperature " + t + " --");
    return go(_cm, sol, t, frozen);
}

private static TSPSolution go(CostMatrix _cm, TSPSolution solution, double t, int frozen) {
    if (frozen >= 3) {
        return solution;
    }
    i++;

    TSPSolution bestSol = solution;
    System.out.println(i + ": " + solution.fitness() + " " + solution.time() + " "
            + solution.penalty() + " " + t);
    ArrayList<TSPSolution> nHood = solution.nHood();

    int attempts = 0;
    int accepted = 0;

    while (!(attempts == 2 * nHood.size() || accepted == nHood.size()) && attempts < 500) {
        TSPSolution sol = nHood.get(rand.nextInt(nHood.size()));
        attempts++;

        double deltaF = sol.fitness() - bestSol.fitness();
        if (deltaF < 0 || Math.exp(-deltaF / t) > Math.random()) {
            accepted++;
            bestSol = sol;
            nHood = sol.nHood();
        }
    }

    frozen = accepted == 0 ? frozen + 1 : 0;

    double newT = coolingSchedule(t);

    return go(_cm, bestSol, newT, frozen);

}

Answer 1

这很简单，因为它是尾递归的：递归调用和放大之间没有代码。函数返回什么。因此，一旦循环结束，您可以将go的正文包裹在循环while (frozen<3)和return solution中。并使用参数的赋值替换递归调用：solution=bestSol; t=newT;。

Answer 2

你需要考虑两件事：

每一步都有哪些变化？
算法什么时候结束？

答案答案应该是

bestSol（solution），newT（t），frozen（frozen）
当frozen >= 3为真时

因此，最简单的方法就是将整个函数包含在

while (frozen < 3) {
    ...
    ...
    ...

    frozen = accepted == 0 ? frozen + 1 : 0;
    //double newT = coolingSchedule(t);
    t = coolingSchedule(t);
    solution = bestSol;
}

Answer 3

根据经验，迭代递归函数的最简单方法是将第一个元素加载到Stack上，而不是调用递归，而是将结果添加到Stack中。

例如：

public Item recursive(Item myItem)
{
    if(myItem.GetExitCondition().IsMet()
    {
        return myItem;
    }
    ... do stuff ...
    return recursive(myItem);
}

会变成：

public Item iterative(Item myItem)
{
    Stack<Item> workStack = new Stack<>();
    while (!workStack.isEmpty())
    {
        Item workItem = workStack.pop()
        if(myItem.GetExitCondition().IsMet()
        {
            return workItem;
        }
        ... do stuff ...
        workStack.put(workItem)
    }
    // No solution was found (!).
    return myItem;
}

此代码未经测试，可能（读取：确实）包含错误。它可能甚至不能编译，但应该给你一个大致的想法。

将递归函数转换为非递归函数

3 个答案: