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

Challenge Notebook

Problem: Implement the Towers of Hanoi with 3 towers and N disks.

Constraints

  • Can we assume we already have a stack class that can be used for this problem?
    • Yes
  • Can we assume the inputs are valid?
    • No
  • Can we assume this fits memory?
    • Yes

Test Cases

  • None tower(s) -> Exception
  • 0 disks -> None
  • 1 disk
  • 2 or more disks

Algorithm

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.

Code

In [ ]:
%run ../../stacks_queues/stack/stack.py
%load ../../stacks_queues/stack/stack.py
In [ ]:
class Hanoi(object):

    def move_disks(self, num_disks, src, dest, buff):
        # TODO: Implement me
        pass

Unit Test

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

In [ ]:
# %load test_hanoi.py
import unittest


class TestHanoi(unittest.TestCase):

    def test_hanoi(self):
        hanoi = Hanoi()
        num_disks = 3
        src = Stack()
        buff = Stack()
        dest = Stack()

        print('Test: None towers')
        self.assertRaises(TypeError, hanoi.move_disks, num_disks, None, None, None)

        print('Test: 0 disks')
        hanoi.move_disks(num_disks, src, dest, buff)
        self.assertEqual(dest.pop(), None)

        print('Test: 1 disk')
        src.push(5)
        hanoi.move_disks(num_disks, src, dest, buff)
        self.assertEqual(dest.pop(), 5)

        print('Test: 2 or more disks')
        for disk_index in range(num_disks, -1, -1):
            src.push(disk_index)
        hanoi.move_disks(num_disks, src, dest, buff)
        for disk_index in range(0, num_disks):
            self.assertEqual(dest.pop(), disk_index)

        print('Success: test_hanoi')


def main():
    test = TestHanoi()
    test.test_hanoi()


if __name__ == '__main__':
    main()

Solution Notebook

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