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

Challenge Notebook

Problem: Find a build order given a list of projects and dependencies.


  • Is it possible to have a cyclic graph as the input?
    • Yes
  • Can we assume we already have Graph and Node classes?
    • Yes
  • Can we assume this is a connected graph?
    • Yes
  • Can we assume the inputs are valid?
    • Yes
  • Can we assume this fits memory?
    • Yes

Test Cases

  • projects: a, b, c, d, e, f, g
  • dependencies: (d, g), (f, c), (f, b), (f, a), (c, a), (b, a), (a, e), (b, e)
  • output: d, f, c, b, g, a, e

Note: Edge direction is down

    f     d
   /|\    |
  c | b   g
    a |

Test a graph with a cycle, output should be None


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 Dependency(object):

    def __init__(self, node_key_before, node_key_after):
        self.node_key_before = node_key_before
        self.node_key_after = node_key_after
In [ ]:
%run ../graph/
%load ../graph/
In [ ]:
class BuildOrder(object):

    def __init__(self, dependencies):
        # TODO: Implement me

    def find_build_order(self):
        # TODO: Implement me

Unit Test

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

In [ ]:
# %load
import unittest

class TestBuildOrder(unittest.TestCase):

    def __init__(self, *args, **kwargs):
        super(TestBuildOrder, self).__init__()
        self.dependencies = [
            Dependency('d', 'g'),
            Dependency('f', 'c'),
            Dependency('f', 'b'),
            Dependency('f', 'a'),
            Dependency('c', 'a'),
            Dependency('b', 'a'),
            Dependency('a', 'e'),
            Dependency('b', 'e'),

    def test_build_order(self):
        build_order = BuildOrder(self.dependencies)
        processed_nodes = build_order.find_build_order()

        expected_result0 = ('d', 'f')
        expected_result1 = ('c', 'b', 'g')
        self.assertTrue(processed_nodes[0].key in expected_result0)
        self.assertTrue(processed_nodes[1].key in expected_result0)
        self.assertTrue(processed_nodes[2].key in expected_result1)
        self.assertTrue(processed_nodes[3].key in expected_result1)
        self.assertTrue(processed_nodes[4].key in expected_result1)
        self.assertTrue(processed_nodes[5].key is 'a')
        self.assertTrue(processed_nodes[6].key is 'e')

        print('Success: test_build_order')

    def test_build_order_circular(self):
        self.dependencies.append(Dependency('e', 'f'))
        build_order = BuildOrder(self.dependencies)
        processed_nodes = build_order.find_build_order()
        self.assertTrue(processed_nodes is None)

        print('Success: test_build_order_circular')

def main():
    test = TestBuildOrder()

if __name__ == '__main__':

Solution Notebook

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