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

- 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

- 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

Refer to the Solution Notebook. If you are stuck and need a hint, the solution notebook's algorithm discussion might be a good place to start.

In [ ]:

```
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)
```

In [ ]:

```
class Knapsack(object):
def fill_knapsack(self, input_items, total_weight):
# TODO: Implement me
pass
```

**The following unit test is expected to fail until you solve the challenge.**

In [ ]:

```
# %load 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()
```

Review the Solution Notebook for a discussion on algorithms and code solutions.