tvl-depot/scratch/facebook/graph-coloring.py
William Carroll 9549dbb266 Update BFS impls
I've subtly been implementing breadth-first traversals in graphs
incorrectly. The change is subtle, but updating `seen` needs to happen
immediately after queuing an item.

The results will remain the same, but the runtimes will differ dramatically. I
didn't notice this until I attempted to complete LeetCode's "count islands"
challenge, and LeetCode rejected my solution because it could not finish before
timing out. After looking at other candidates' solutions and comparing them to
mine, I couldn't see any difference... except for this subtle difference.

This SO answer provides a helpful explanation:
https://stackoverflow.com/questions/45623722/marking-node-as-visited-on-bfs-when-dequeuing

The take-away lesson here is to always call `seen.add(..)` immediately after
enqueuing.
2020-11-23 23:21:20 +00:00

60 lines
1.4 KiB
Python

from collections import deque
class Palette(object):
def __init__(self, n):
self.i = 0
self.colors = list(range(n))
def get(self):
return self.colors[self.i]
def advance(self):
self.i += 1 % len(self.colors)
class GraphNode(object):
def __init__(self, label):
self.label = label
self.neighbors = set()
self.color = None
def __repr__(self):
result = []
xs = deque()
xs.append(self)
seen = set()
while xs:
node = xs.popleft()
result.append('{} ({})'.format(node.label, str(node.color)))
for c in node.neighbors:
if c.label not in seen:
xs.append(c)
seen.add(node.label)
return ', '.join(result)
def color_graph(graph, d):
seen = set()
start = graph
xs = deque()
palette = Palette(d + 1)
xs.append((start, palette.get()))
while xs:
x, color = xs.popleft()
x.color = color
for c in x.neighbors:
if c.label not in seen:
palette.advance()
xs.append((c, palette.get()))
seen.add(x.label)
a = GraphNode('a')
b = GraphNode('b')
c = GraphNode('c')
a.neighbors.add(b)
b.neighbors.add(a)
b.neighbors.add(c)
c.neighbors.add(b)
print(a)
color_graph(a, 3)
print(a)