I've found that luck is quite predictable. If you want more luck, take more chances. Be more active. Show up more often.

I'm taking Intro to Algorithms, which focuses mostly on graphs, and one of the first topics is about Eulerian paths and tours, you can easily find the theory of these topics in the course or on Wikipedia, let's write some code.

This is my brute force solution to the challenge in problem set 1: Find Eulerian tour:` ````
# Find Eulerian Tour
#
# Write a function that takes in a graph
# represented as a list of tuples
# and return a list of nodes that
# you would follow on an Eulerian Tour
#
# For example, if the input graph was
# [(1, 2), (2, 3), (3, 1)]
# A possible Eulerian tour would be [1, 2, 3, 1]
import copy
def contains_all_nodes(path, nodes):
unique = {}
for p in path:
unique[p] = True
for node in nodes:
if node not in unique:
return False
return True
def tour(u, graph, visited, path, nodes):
path.append(u)
for v in graph[u]:
e = (min(u, v), max(u, v))
if e not in visited:
visited_copy = copy.deepcopy(visited)
visited_copy[e] = True
path_copy = copy.deepcopy(path)
p = tour(v, graph, visited_copy, path_copy, nodes)
if contains_all_nodes(p, nodes):
return p
return path
def find_eulerian_tour(edges):
graph = {}
nodes = {}
for edge in edges:
u = edge[0]
v = edge[1]
nodes[u] = True
nodes[v] = True
if u not in graph:
graph[u] = []
graph[u].append(v)
if v not in graph:
graph[v] = []
graph[v].append(u)
for n in nodes:
path = tour(n, graph, {}, [], nodes)
if contains_all_nodes(path, nodes):
return path
return []
#edges = [(1, 2), (2, 3), (3, 1)]
edges = [(0, 1), (1, 5), (1, 7), (4, 5), (4, 8), (1, 6), (3, 7),
(5, 9), (2, 4), (0, 4), (2, 5), (3, 6), (8, 9)]
print find_eulerian_tour(edges)
```