Please help MODIFY this (0-1) knapsack code to non (0-1)? I have this bounded knapsack code that given capacity, weight, value and max number of items u can take it generates a table where the last value is the greatest value possible that can be generated.BUT it only allows one item to be taken once?Please modify it so that one item can be taken multiple times.I think it has something to do with the max?

def solve3d (capacity, value, weight, maxitems):

"""

solve the 3d-knapsack problem specified in its parameters: capacity is the

overall capacity of the knapsack and the ith position of the arrays value

and weight specify the value and weight of the ith item. This is called the

3d-knapsack not because it refers to a cuboid but because it also considers

a maximum number of items to insert which is given in the last parameter

"""

# initialization - the number of items is taken from the length of any array

# which shall have the same length

nbitems = len (value)

# we use dynamic programming to solve this problem. Thus, we'll need a table

# that contains (N, M) cells where 0<=N<=capacity and 0<=M<=nbitems

table=dict ()

# initialize the contents of the dictionary for all capacities and number of

# items to another dictionary which returns 0 by default

for icapacity in range (0, 1+capacity):

table [icapacity]=dict ()

for items in range (0, 1+nbitems):

table [icapacity][items] = defaultdict (int)

# now we are ready to start, ... for the first j items

for j in range (0, nbitems):

# for all capacities ranging from 1 to the maximum overall capacity

for i in range (1, 1+capacity):

# and for all cardinalities of items from 1 until the maximum

# allowed

for k in range (1, 1+maxitems):

# if this item can not be inserted

if (weight [j] > i):

table [i][1+j][k] = table [i][j][k] # just do not add it

# otherwise, consider inserting it

else:

# if this is item is known to fit the knapsack (because its

# weight does not exceed the current capacity and adding it

# creates a set with a cardinality less or equal than the

# cardinality currently considered), then compute the

# optimal value as usual but considering sets of the same

# cardinality (k)

if (j

table [i][1+j][k] = max (table[i][j][k],

table[i-weight[j]][j][k]+value[j])

else:

prev = []

# retrieve the optimal solution for all values of

# (i-weight[j], kappa [0 .. j-1], k-1)

for kappa in range (0, 1+j):

prev.append (table[i-weight[j]][kappa][k-1])

# and consider inserting this item taking into account

# the optimal solution from the preceding cardinality

# considering all sets of items

table [i][1+j][k] = max (table[i][j][k],

max (prev) + value[j])

# return the table computed so far

return table