Python 背包问题分支限界算法

Question

我已经花了一周的时间编写这个背包问题的分支限界代码，并查阅了许多相关文章和书籍。然而，当我运行我的代码时，得到的结果并不是我期望的那样。输入来自于文本文件，像这样：

12
4 1
1 1
3 2
2 3

第一行表示背包容量，随后每一行都是价值/重量对。使用这个文件得到的结果是“8”，而不是“10”（除非我错了，所有物品都无法装入背包）。以下是我的代码：

#!/usr/bin/python
# -*- coding: utf-8 -*-

import Queue
from collections import namedtuple
Item = namedtuple("Item", ['index', 'value', 'weight', 'level', 'bound', 'contains'])

class Node:
    def __init__(self, level, value, weight, bound, contains):
         self.level = level
         self.value = value
         self.weight = weight
         self.bound = bound
         self.contains = contains

def upper_bound(u, k, n, v, w):
    if u.weight > k:
        return 0

    else:
        bound = u.value
        wt = u.weight
        j = u.level + 1

        while j < n and wt + w[j] <= k:
            bound += v[j]
            wt += w[j]
            j += 1

    # fill knapsack with fraction of a remaining item
            if j < n:
                bound += (k - wt) * (v[j] / w[j])

            return bound

def knapsack(items, capacity):
    item_count = len(items)
    v = [0]*item_count
    w = [0]*item_count

# sort items by value to weight ratio
    items = sorted(items, key=lambda k: k.value/k.weight, reverse = True)

    for i,item in enumerate(items, 0):
        v[i] = int(item.value)
        w[i] = int(item.weight)

    q = Queue.Queue()

    root = Node(0, 0, 0, 0.0, [])
    root.bound = upper_bound(root, capacity, item_count, v, w)
    q.put(root)

    value = 0
    taken = [0]*item_count
    best = set()

    while not q.empty():
        c = q.get()

        if c.bound > value:
            level = c.level+1

    # check 'left' node (if item is added to knapsack)
        left = Node(c.value + v[level], c.weight + w[level], level, 0.0, c.contains[:])
        left.contains.append(level)

        if left.weight <= capacity and left.value > value:
            value = left.value
            best |= set(left.contains)

        left.bound = upper_bound(left, capacity, item_count, v, w)

        if left.bound > value:
            q.put(left)

        # check 'right' node (if items is not added to knapsack)
        right = Node(c.value, c.weight, level, 0.0, c.contains[:])
        right.contains.append(level)
        right.bound = upper_bound(right, capacity, item_count, v, w)

        if right.bound > value:
            q.put(right)

    for b in best:
        taken[b] = 1

    value = sum([i*j for (i,j) in zip(v,taken)])

    return str(value)

我的索引有误吗？我没有正确遍历树或计算边界吗？

- wikenator

---

你可能想看一下http://rosettacode.org/wiki/Knapsack_Problem/Python - Dietrich

我正在努力将背包算法应用于包含10,000个以上项目的数据集。我已经成功地在较小的数据集上实现了DP背包，但是在某一点上，内存成为了一个问题，这就是为什么我转而使用分支定界法的原因。 - wikenator

所以我用 bound += v[j] 替换了 bound += (k - wt) * (v[j] / w[j])，显然前者会导致算法过早停止。当我初始化左右节点时，也有一些反向变量（现在已经修复）。现在我卡在了标记哪些物品被取走的问题上。在更大的数据集上，我要么得到 [0,1,3,...] 的模式，要么得到 [0,2,4,...] 的模式。'contains' 数组更新/修改的位置是否正确？ - wikenator

Answer 1

def upper_bound(u, k, n, v, w):
        if u.weight > k:
            return 0
        else:
            bound = u.value
            wt = u.weight
            j = u.level 
            while j < n and wt + w[j] <= k:
                bound += v[j]
                wt += w[j]
                j += 1
            # fill knapsack with fraction of a remaining item
            if j < n:
                bound += (k - wt) * float(v[j])/ w[j]
            return bound


def knapsack(items, capacity):
        item_count = len(items)
        v = [0]*item_count
        w = [0]*item_count
        # sort items by value to weight ratio
        items = sorted(items, key=lambda k: float(k.value)/k.weight, reverse = True)
        for i,item in enumerate(items, 0):
            v[i] = int(item.value)
            w[i] = int(item.weight)
        q = Queue.Queue()
        root = Node(0, 0, 0, 0.0,[])
        root.bound = upper_bound(root, capacity, item_count, v, w)
        q.put(root)
        value = 0
        taken = [0]*item_count
        best = set()
        while not q.empty():
            c = q.get()
            if c.bound > value:
                level = c.level+1
            # check 'left' node (if item is added to knapsack)
            left = Node(level,c.value + v[level-1], c.weight + w[level-1], 0.0, c.contains[:])
            left.bound = upper_bound(left, capacity, item_count, v, w)
            left.contains.append(level)
            if left.weight <= capacity:
                if left.value > value:
                    value = left.value
                    best = set(left.contains)
                if left.bound > value:
                    q.put(left)
                # check 'right' node (if items is not added to knapsack)   
            right = Node(level,c.value, c.weight, 0.0, c.contains[:])
            right.bound = upper_bound(right, capacity, item_count, v, w)
            if right.weight <= capacity:
                if right.value > value:
                    value = right.value
                    best = set(right.contains)
                if right.bound > value:
                    q.put(right)
        for b in best:
            taken[b-1] = 1
        value = sum([i*j for (i,j) in zip(v,taken)])
        return str(value)

- pbellot

Answer 2

import functools
class solver():
    def __init__(self, Items, capacity):
        self.sortedItems = list(filter(lambda x: x.value > 0, Items))
        self.sortedItems = sorted(self.sortedItems, key=lambda    x:float(x.weight)/float(x.value))
    self.numItems = len(Items)
    self.capacity = capacity
    self.bestSolution = solution(0, self.capacity)

def isOptimisitcBetter(self, sol, newItemIdx):
    newItem = self.sortedItems[newItemIdx]
    rhs = (sol.value + (sol.capacity/newItem.weight)*newItem.value)
    return rhs > self.bestSolution.value

def explore(self, sol, itemIndex):
    if itemIndex < self.numItems:
        if self.isOptimisitcBetter(sol, itemIndex):
            self.exploreLeft(sol, itemIndex)
            self.exploreRight(sol, itemIndex)

def exploreLeft(self, sol, itemIndex):
    newItem = self.sortedItems[itemIndex]
    thisSol = sol.copy()
    if thisSol.addItem(newItem):
        if thisSol.value > self.bestSolution.value:
            self.bestSolution = thisSol
        self.explore(thisSol, itemIndex+1)

def exploreRight(self, sol, itemIndex):
    self.explore(sol, itemIndex+1)

def solveWrapper(self):
    self.explore(solution(0, self.capacity), 0)


class solution():
    def __init__(self, value, capacity, items=set()):
    self.value, self.capacity = value, capacity
    self.items = items.copy()

def copy(self):
    return solution(self.value,  self.capacity, self.items)

def addItem(self, newItem):
    remainingCap = self.capacity-newItem.weight
    if remainingCap < 0:
        return False
    self.items.add(newItem)
    self.capacity = remainingCap
    self.value+=newItem.value
    return True


solver = solver(items, capacity)
solver.solveWrapper()
bestSol = solver.bestSolution

- Adel

Answer 3

我认为你只需要在不拿走物品时计算边界。如果你拿走了物品，那意味着你的边界仍然可以达到。如果你不拿走，你就必须重新调整你的期望值。

- user3088454

---

P.S. 很高兴你正在参加Coursera的离散优化课程 :) - user3088454