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swift - 如何修改递归分支定界树节点遍历以避免堆栈溢出?

转载 作者:行者123 更新时间:2023-11-28 15:55:59 24 4
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我创建了以下 Swift 3 程序来使用深度优先分支定界解决背包问题。不幸的是,它是递归的。它适用于多达 20 个项目。但是对于更多的项目(30个或更多),它会导致堆栈溢出。 (~5400 级深)。

如何改为非递归版本?

public enum BBState {
case created
case active
case death
}

public enum Direction {
case left
case right
}

public class BBTreeNode {
public var artifact: Artifact
public var nodeState: BBState
public var level: Int
public var path = [Direction]()
public var weightSoFar: Int = 0
public var optimumTotalValue: Int = 0
public var estimatedOptTotalValue: Double = 0
public weak var parentNode: BBTreeNode?
public var leftNode: BBTreeNode?
public var rightNode: BBTreeNode?
static var currentCompletedOptimumValues: Int = 0
static var currentCompletedEstimatedOptimumValues: Double = 0
static var currentOptimumPath = [Direction]()

// Initialization
public convenience init(artifact: Artifact) {
self.init(artifact: artifact, level: 0, path: [], parent: nil, left: nil, right: nil)
}

public init(artifact: Artifact, level: Int, path: [Direction], parent: BBTreeNode?, left: BBTreeNode?, right: BBTreeNode?) {
self.artifact = artifact
self.nodeState = .created
self.level = level
self.path = path
self.parentNode = parent
self.leftNode = left
self.rightNode = right
}


// BBTree
private func createRootAndInitiateBB(artifacts: [Artifact]) {
// return the optimum value from Depth-First branch and bound
let maxLevel = artifacts.count - 1
print("Max Level: \(maxLevel)")
// create dummy Root to start
let root = BBTreeNode(artifact: Artifact(value: 0, weight: 0))
root.estimatedOptTotalValue = Knapsack.calculateEstimatedOptimumValue(availableArtifacts: artifacts)
root.nodeState = .active
root.weightSoFar = Knapsack.capacity
// loop here for each artifact# - selected(left) / not(right) but going recursive to left until
// we have death node, then backtrack

func depthFirstTraversal(bbNode: BBTreeNode?, level: Int) {
guard let bbNode = bbNode // not nil
else { return }

guard level <= maxLevel
else { return }

print("Currently on path: \(bbNode.path)")

// switch to active is last state was created, else ignore
bbNode.nodeState = bbNode.nodeState == .created ? .active : bbNode.nodeState

// stop traverse down and traceback if calculated optimumValue < currentCompletedOptimumValue,
// or current weightSoFar is 0 or less than 0
// move up to parent
if bbNode.estimatedOptTotalValue < BBTreeNode.currentCompletedEstimatedOptimumValues ||
bbNode.weightSoFar < 0 {
print("Case for estimatedOptTotalValue: \(bbNode.estimatedOptTotalValue) < currentCompletedEstimatedOptimumValues: \(BBTreeNode.currentCompletedEstimatedOptimumValues)")
print("Or weight: \(bbNode.weightSoFar) < 0" )
bbNode.nodeState = .death
// remove references to children
bbNode.leftNode = nil
bbNode.rightNode = nil
depthFirstTraversal(bbNode: bbNode.parentNode, level: bbNode.level - 1)
} else if (bbNode.leftNode?.nodeState == .death &&
bbNode.rightNode?.nodeState == .death) || level == maxLevel {

print("Case for no more path available. Need to backtrack. ")
print("Current level: \(level)")
// stop, record and traceback if at maxLevel or when both children are death
if level == maxLevel && bbNode.estimatedOptTotalValue > BBTreeNode.currentCompletedEstimatedOptimumValues {
BBTreeNode.currentCompletedEstimatedOptimumValues = bbNode.estimatedOptTotalValue
BBTreeNode.currentCompletedOptimumValues = bbNode.optimumTotalValue
BBTreeNode.currentOptimumPath = bbNode.path
print("blah...")
print("Candidate for optimum: \(bbNode.path)")
print("Completed optimum path: \(BBTreeNode.currentCompletedOptimumValues)")
print("Estimated optimum value: \(BBTreeNode.currentCompletedEstimatedOptimumValues)")
}
bbNode.nodeState = .death
// remove references to children
bbNode.leftNode = nil
bbNode.rightNode = nil
let _ = path.popLast()
depthFirstTraversal(bbNode: bbNode.parentNode, level: bbNode.level - 1)
} else if bbNode.leftNode == nil {
// create left child
print("create left child node")
let childLeftNode = createBBNode(leftChild: true, parent: bbNode, path: bbNode.path + [.left])
bbNode.leftNode = childLeftNode
depthFirstTraversal(bbNode: childLeftNode, level: childLeftNode.level)
} else if bbNode.rightNode == nil {
// create right child
print("create right child node")
let childRightNode = createBBNode(leftChild: false, parent: bbNode, path: bbNode.path + [.right])
bbNode.rightNode = childRightNode
depthFirstTraversal(bbNode: childRightNode, level: childRightNode.level)
}

}

func createBBNode(leftChild: Bool, parent: BBTreeNode, path: [Direction]) -> BBTreeNode {
let level = parent.level + 1
let artifact = artifacts[level]
let newBBNode = BBTreeNode(artifact: artifact, level: level, path: path, parent: parent, left: nil, right: nil )
if leftChild {
newBBNode.optimumTotalValue = parent.optimumTotalValue + artifact.value
newBBNode.estimatedOptTotalValue = parent.estimatedOptTotalValue
newBBNode.weightSoFar = parent.weightSoFar - artifact.weight
} else {
// right direction, we don't pick this item
// Artifact is a struct, artifacts is array of Artifact, so we don't need to write deep copy
var artifactsWithItemsRemoved = artifacts

print("Current artifacts before any removal: \(artifactsWithItemsRemoved)")
print("for path \(newBBNode.path) we are going to remove artifacts...")
// first remove the dummy artifact
artifactsWithItemsRemoved.remove(at: 0)
// now the real artifacts
var indexOfItemForRemoval = [Int]()
for (index,direction) in path.enumerated() {
if direction == .right {
indexOfItemForRemoval.append(index)
}
}
// actual removal, need to reverse the array index to avoid out of range index
for idx in indexOfItemForRemoval.reversed() {
artifactsWithItemsRemoved.remove(at: idx)
}

print("Artifacts with items removed: \(artifactsWithItemsRemoved)")

newBBNode.optimumTotalValue = parent.optimumTotalValue
newBBNode.estimatedOptTotalValue = Knapsack.calculateEstimatedOptimumValue(availableArtifacts: artifactsWithItemsRemoved)
print("New estimatedOptValue for \(newBBNode.path) is \(newBBNode.estimatedOptTotalValue)")
// no weight deduction if we don't pick
newBBNode.weightSoFar = parent.weightSoFar
}
return newBBNode
}

depthFirstTraversal(bbNode: root, level: 0)
}

public func findOptimumValueAndPath(artifacts: [Artifact]) -> (Int,[Direction], Double) {
createRootAndInitiateBB(artifacts: artifacts)
return (BBTreeNode.currentCompletedOptimumValues, BBTreeNode.currentOptimumPath, BBTreeNode.currentCompletedEstimatedOptimumValues)
}

}

====更新===

我设法将递归深度限制在某个限制,例如 4000,一旦计数器达到限制,我将 currentNode 信息返回给调用者。然后使用 currentNode 再次调用“depthFirstTraversal”,但使用全新的堆栈。

这是我的做法:

  1. 将 createRootAndInitiateBB 和 findOptimumValueAndPath 移至调用方。
  2. 更新 depthFirstTraversal 以具有如下函数签名:

    static func depthFirstTraversal(bbNode: BBTreeNode, level: Int, recursiveCount: Int, complete completeFlag: inout Bool, currentProcessedNode: inout BBTreeNode)

  3. 其他一些重构工作可保持计数器计数并修复一些错误。 (例如 parentNode 不应该是弱的,否则在返回到调用者之后,currentNode 的父节点变为 nil,我们失去回溯到更高节点的能力)。

最佳答案

关于swift - 如何修改递归分支定界树节点遍历以避免堆栈溢出?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41736058/

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