# Knapsack Problem (0-1 solution) - Dynamic Algorithm

Posted on Sep 27, 2016

I’ve recently dug up old code from my University days, which I thought I’d share for the benefit/misfortune of others.

There’s a common problem in programming called the knapsack problem. Here was my solution based on the dynamic algorithm.

# Math is used to floor/round the floats to an interger and back! import math

# Total allowed weight

totalWeight = 2392;

# Define the items name, Weight, Profit

items = ((“Weapon and Ammunition”, 4.13, 1.4), (“Water”, 2.13, 2.74), (“Pith Helmet”, 3.03, 1.55), (“Sun Cream”, 2.26, 0.82), (“Tent”, 3.69, 2.38), (“Flare Gun”, 3.45, 2.93), (“Olive Oil”, 1.09, 1.77), (“Firewood”, 2.89, 0.53), (“Kendal Mint Cake”, 1.08, 2.77), (“Snake Repellant Spray”, 3.23, 4.29), (“Bread”, 2.29, 2.85), (“Pot Noodles”, 0.55, 0.34), (“Software Engineering Textbook”, 2.82, -0.45), (“Tinned Food”, 2.31, 2.17), (“Pork Pie”, 1.63, 1.62))

# Get the total of bagged items.

def totalvalue(comb): totwt = totval = 0 for item, wt, val in comb: totwt += wt totval += val return (totval, -totwt) if totwt <= totalWeight else (0, 0)

# The knapsack 0-1 Dynamic Programming Algorithm.

def knapsack(items, limit): table = [[0 for w in range(limit + 1)] for j in xrange(len(items) + 1)]

for j in xrange(1, len(items) + 1):
item, wt, val = items$j-1$
wt = int(wt \* 100)
val = int(val \* 100)
for w in xrange(1, limit + 1):
if wt > w:
table$j$$w$ = table$j-1$$w$
else:
table$j$$w$ = max(table$j-1$$w$,
table$j-1$$w-wt$ + val)

result = 
w = limit
for j in range(len(items), 0, -1):
was\_added = table$j$$w$ != table$j-1$$w$

item, wt, val = items$j-1$
wt = int(wt \* 100)
val = int(val \* 100)
result.append(items$j-1$)
w -= wt

return result


# Bag the items

bagged = knapsack(items, totalWeight)

# Print the results

print(“Bagged the following items\n " + ‘\n ‘.join(sorted(item for item,_,_ in bagged))) val, wt = totalvalue(bagged)

print(“Total Value: %f - Total Weight: %f” % (float(val), float(-wt)))