This notebook was prepared by Donne Martin. Source and license info is on GitHub.

# Solution Notebook¶

## Constraints¶

• Can we replace the items once they are placed in the knapsack?
• Yes, this is the unbounded knapsack problem
• Can we split an item?
• No
• Can we get an input item with weight of 0 or value of 0?
• No
• Do we need to return the items that make up the max total value?
• No, just the total value
• Can we assume the inputs are valid?
• No
• Are the inputs in sorted order by val/weight?
• Yes
• Can we assume this fits memory?
• Yes

## Test Cases¶

• items or total weight is None -> Exception
• items or total weight is 0 -> 0
• General case
total_weight = 8
items
v | w
0 | 0
a 1 | 1
b 3 | 2
c 7 | 4

max value = 14


## Algorithm¶

We'll use bottom up dynamic programming to build a table.

Taking what we learned with the 0/1 knapsack problem, we could solve the problem like the following:

v = value
w = weight

j
-------------------------------------------------
| v | w || 0 | 1 | 2 | 3 | 4 | 5 |  6 |  7 |  8  |
-------------------------------------------------
| 0 | 0 || 0 | 0 | 0 | 0 | 0 | 0 |  0 |  0 |  0  |
a | 1 | 1 || 0 | 1 | 2 | 3 | 4 | 5 |  6 |  7 |  8  |
i b | 3 | 2 || 0 | 1 | 3 | 4 | 6 | 7 |  9 | 10 | 12  |
c | 7 | 4 || 0 | 1 | 3 | 4 | 7 | 8 | 10 | 11 | 14  |
-------------------------------------------------

i = row
j = col



However, unlike the 0/1 knapsack variant, we don't actually need to keep space of O(n * w), where n is the number of items and w is the total weight. We just need a single array that we update after we process each item:

    -------------------------------------------------
| v | w || 0 | 1 | 2 | 3 | 4 | 5 |  6 |  7 |  8  |
-------------------------------------------------

-------------------------------------------------
a | 1 | 1 || 0 | 1 | 2 | 3 | 4 | 5 |  6 |  7 |  8  |
-------------------------------------------------

-------------------------------------------------
b | 3 | 2 || 0 | 1 | 3 | 4 | 6 | 7 |  9 | 10 | 12  |
-------------------------------------------------

-------------------------------------------------
c | 7 | 4 || 0 | 1 | 3 | 4 | 7 | 8 | 10 | 11 | 14  |
-------------------------------------------------

if j >= items[i].weight:
T[j] = max(items[i].value + T[j - items[i].weight],
T[j])



Complexity:

• Time: O(n * w), where n is the number of items and w is the total weight
• Space: O(w), where w is the total weight

## Code¶

### Item Class¶

In [1]:
class Item(object):

def __init__(self, label, value, weight):
self.label = label
self.value = value
self.weight = weight

def __repr__(self):
return self.label + ' v:' + str(self.value) + ' w:' + str(self.weight)


### Knapsack Bottom Up¶

In [2]:
class Knapsack(object):

def fill_knapsack(self, items, total_weight):
if items is None or total_weight is None:
raise TypeError('items or total_weight cannot be None')
if not items or total_weight == 0:
return 0
num_rows = len(items)
num_cols = total_weight + 1
T = [0] * (num_cols)
for i in range(num_rows):
for j in range(num_cols):
if j >= items[i].weight:
T[j] = max(items[i].value + T[j - items[i].weight],
T[j])
return T[-1]


## Unit Test¶

In [3]:
%%writefile test_knapsack_unbounded.py
import unittest

class TestKnapsack(unittest.TestCase):

def test_knapsack(self):
knapsack = Knapsack()
self.assertRaises(TypeError, knapsack.fill_knapsack, None, None)
self.assertEqual(knapsack.fill_knapsack(0, 0), 0)
items = []
items.append(Item(label='a', value=1, weight=1))
items.append(Item(label='b', value=3, weight=2))
items.append(Item(label='c', value=7, weight=4))
total_weight = 8
expected_value = 14
results = knapsack.fill_knapsack(items, total_weight)
total_weight = 7
expected_value = 11
results = knapsack.fill_knapsack(items, total_weight)
self.assertEqual(results, expected_value)
print('Success: test_knapsack')

def main():
test = TestKnapsack()
test.test_knapsack()

if __name__ == '__main__':
main()

Overwriting test_knapsack_unbounded.py

In [4]:
%run -i test_knapsack_unbounded.py

Success: test_knapsack