在背包 0/1 代码中实现回溯 table

Question

我正在尝试在我的背包 0/1 问题的动态规划算法中实施回溯 table。在我的代码中，我设置了一个回溯 table 矩阵，但我不确定如何根据我的程序填写这些值。

def optimizeInvestments(invstmt, money):
""" knapsack problem """
n = len(invstmt)
val = []
name = []
roi = []
traceback = [[0 for i in range(n)] for i in range(n)]

for i in invstmt:
    name.append(i[0])
    val.append(i[-1])
    roi.append(i[1])

K = [[0 for x in range(money + 1)] for x in range(n + 1)]
I = [[0 for x in range(money + 1)] for x in range(n + 1)]

for i in range(n + 1):
    for w in range(money + 1):
        if i == 0 or w == 0:
            K[i][w] = 0
            I[i][w] = ""

        elif roi[i - 1] <= w:

            if (val[i - 1] + K[i - 1][w - roi[i - 1]] > K[i - 1][w]):
                K[i][w] = val[i - 1] + K[i - 1][w - roi[i - 1]]
                if len(I[i - 1][w - roi[i - 1]]) > 0:
                    I[i][w] = name[i - 1] + " & " + I[i - 1][w - roi[i - 1]]
                else:
                    I[i][w] = name[i - 1]

            else:
                K[i][w] = K[i - 1][w]
                I[i][w] = I[i - 1][w]

        else:
            K[i][w] = K[i - 1][w]
            I[i][w] = I[i - 1][w]

portfolio = 'With $'+ str(money) + ", invest in " + str(I[n][money]) + " for a ROI of $" + str(K[n][money])
return portfolio

Answer 1

想出如何添加回溯 table。这是我的代码：

def optimizeInvestments(invstmt, money):
""" knapsack problem """
n = len(invstmt)
val = []
name = []
roi = []

for i in invstmt:
    name.append(i[0])
    val.append(i[-1])
    roi.append(i[1])

K = [[0 for x in range(money + 1)] for x in range(n + 1)]
I = [[0 for x in range(money + 1)] for x in range(n + 1)]
optimal = [[None for x in range(money + 1)] for x in range(n + 1)]
traceback = [[False for x in range(money + 1)] for x in range(n + 1)]


for i in range(n + 1):
    for w in range(money + 1):
        optimal[i][w] = 0
        if i == 0 or w == 0:
            K[i][w] = float(0)
            I[i][w] = ""
            optimal[i][w] = 0
            traceback[i][w] = False

        elif roi[i - 1] <= w:

            if (val[i - 1] + K[i - 1][w - roi[i - 1]] > K[i - 1][w]):
                K[i][w] = val[i - 1] + K[i - 1][w - roi[i - 1]]
                optimal[i][w] = val[i - 1] + optimal[i - 1][w - roi[i - 1]]
                traceback[i][w] = True

                if len(I[i - 1][w - roi[i - 1]]) > 0:
                    I[i][w] = name[i - 1] + " & " + I[i - 1][w - roi[i - 1]]
                else:
                    I[i][w] = name[i - 1]

            else:
                K[i][w] = K[i - 1][w]
                I[i][w] = I[i - 1][w]
                optimal[i][w] = optimal[i-1][w]
                traceback[i][w] = False




        else:
            K[i][w] = K[i - 1][w]
            I[i][w] = I[i - 1][w]
            optimal[i][w] = optimal[i-1][w]
            traceback[i][w] = False

    print('optimal: ')
    for row in optimal:
        print(row)
    print("traceback: ")
    for row in traceback:
        print(row)

在背包 0/1 代码中实现回溯 table

Implementing traceback table in knapsack 0/1 code

python

algorithm

dynamic-programming

knapsack-problem

traceback