随机推销员随机初始解决方案,优化算法返回意外结果

问题描述 投票:1回答:1

我知道旅行推销员是众所周知的,但我需要一些帮助来解释为什么我的优化算法会返回意外的结果。我通过随机顺序选择城市来创建初始解决方案。我还创建了一个带有构造函数的类,其中包含距离矩阵和初始解决方案作为参数。优化算法非常简单;它交换两个城市并检查路线距离是否有所改善,如果改进了,则应更新最佳解决方案。这持续了6次迭代。问题是,即使不满足覆盖它的条件,似乎最新的解决方案也会被更新和覆盖。我将添加一个显示测试运行结果的图像。

enter image description here

似乎变量bestSolution被覆盖但不是bestDistance。我必须拥有某种隧道视觉,因为即使代码非常简单,我也无法弄清楚这一点。有人可以请求为什么bestSolution被覆盖并返回意外结果?

代码示例如下:

package RandomMethod

import GreedyHeuristic
import java.util.*

fun main(args: Array<String>) {

                                           /*A  B  C*/
    val distances = arrayOf(/*A*/ intArrayOf(0, 2, 7),
                            /*B*/ intArrayOf(2, 0, 9),
                            /*C*/ intArrayOf(7, 9, 0))

    val initalSolution = findRandomRoute(distances)

    println("Initial solution: $initalSolution")
    println("Total distance: ${findTotalDistance(distances, initalSolution)}\n")

    val optimizedSolution = GreedyHeuristic(distances, initalSolution).optimize()

    println("\nOptimized solution with Greedy Heuristic: $optimizedSolution")
    println("Total distance: ${findTotalDistance(distances, optimizedSolution)}")

}

fun areAllCitiesVisited(isCityVisited: Array<Boolean>): Boolean {

    for (visited in isCityVisited) {
        if (!visited) return false
    }
    return true
}

fun findTotalDistance(distances: Array<IntArray>, orderToBeVisited: MutableList<Int>): Int {

    var totalDistance = 0

    for (i in 0..orderToBeVisited.size - 2) {
        val fromCityIndex = orderToBeVisited.get(i)
        val toCityIndex = orderToBeVisited.get(i + 1)
        totalDistance += distances[fromCityIndex].get(toCityIndex)
    }
    return totalDistance
}

fun findRandomRoute(distances: Array<IntArray>): MutableList<Int> {
    val visitedCities: Array<Boolean> = Array(distances.size, {i -> false})

    // Find starting city index. 0 = A, 1 = B, 2 = C .... N = X
    var currentCity = Random().nextInt(distances.size)
    val orderToBeVisited: MutableList<Int> = mutableListOf(currentCity)

    visitedCities[currentCity] = true

    while (!areAllCitiesVisited(visitedCities)) {

        currentCity = Random().nextInt(distances.size)

        if (!visitedCities[currentCity]) {
            orderToBeVisited.add(currentCity)
            visitedCities[currentCity] = true
        }
    }
    return orderToBeVisited
}

以及优化课程:

import java.util.*

class GreedyHeuristic(distances: Array<IntArray>, initialSoltion: MutableList<Int>) {

    val mInitialSolution: MutableList<Int> = initialSoltion
    val mDistances: Array<IntArray> = distances

    fun optimize(): MutableList<Int> {
        var bestSolution = mInitialSolution
        var newSolution = mInitialSolution
        var bestDistance = findTotalDistance(mDistances, bestSolution)
        var i = 0

        while (i <= 5) {
            println("best distance at start of loop: $bestDistance")

            var cityIndex1 = Integer.MAX_VALUE
            var cityIndex2 = Integer.MAX_VALUE

            while (cityIndex1 == cityIndex2) {
                cityIndex1 = Random().nextInt(mInitialSolution.size)
                cityIndex2 = Random().nextInt(mInitialSolution.size)
            }

            val temp = newSolution.get(cityIndex1)
            newSolution.set(cityIndex1, newSolution.get(cityIndex2))
            newSolution.set(cityIndex2, temp)

            val newDistance: Int = findTotalDistance(mDistances, newSolution)
            println("new distance: $newDistance\n")

            if (newDistance < bestDistance) {
                println("New values gived to solution and distance")
                bestSolution = newSolution
                bestDistance = newDistance
            }
            i++
        }
        println("The distance of the best solution ${findTotalDistance(mDistances, bestSolution)}")
        return bestSolution
    }

    fun findTotalDistance(distances: Array<IntArray>, orderToBeVisited: MutableList<Int>): Int {

        var totalDistance = 0

        for (i in 0..orderToBeVisited.size - 2) {
            val fromCityIndex = orderToBeVisited.get(i)
            val toCityIndex = orderToBeVisited.get(i + 1)
            totalDistance += distances[fromCityIndex].get(toCityIndex)
        }
        return totalDistance
    }

}
kotlin traveling-salesman
1个回答
1
投票

除非您特别要求,否则Kotlin(以及一般的JVM语言)不会复制值。这意味着,当你这样做时:

var bestSolution = mInitialSolution
var newSolution = mInitialSolution

你没有设置bestSolutionnewSolution来分离mInitialSolution的副本,而是让它们指向同一个MutableList,所以变异一个变异另一个。这就是说:你的问题不是bestSolution被覆盖,而是你每次修改newSolution时都会意外修改它。

然后,您可以在newSolution循环的每次迭代中重用while,而无需创建新列表。这导致我们做两件事:

  • 因为newSolution仍然使bestSolution别名,修改前者也会改变后者。
  • bestSolution = newSolution什么都不做。

正如评论中所提到的,解决这个问题的最简单方法是策略性地使用.toMutableList(),这将强制复制列表。您可以通过在顶部进行此更改来实现此目的:

var bestSolution = mInitialSolution.toMutableList()
var newSolution = mInitialSolution.toMutableList()

然后在循环内:

bestSolution = newSolution.toMutableList()

顺便说一下:作为一般规则,你应该返回并接受List而不是MutableList,除非你特别希望它成为你的函数合同的一部分,你将在原地改变它。在这个特殊的情况下,它会强迫你做一些不好的事情(比如不安全地投射mInitialSolutionMutableList,这应该会发出你头上的各种警告铃声),或者复制清单(这会让你朝着正确的答案)

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