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Added BFS and DFS (graph algorithms) #408

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Feb 11, 2019
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39 changes: 39 additions & 0 deletions Graphs/BFS.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,39 @@
"""pseudo-code"""

"""
BFS(graph G, start vertex s):
// all nodes initially unexplored
mark s as explored
let Q = queue data structure, initialized with s
while Q is non-empty:
remove the first node of Q, call it v
for each edge(v, w): // for w in graph[v]
if w unexplored:
mark w as explored
add w to Q (at the end)

"""

import collections


def bfs(graph, start):
explored, queue = set(), [start] # collections.deque([start])
explored.add(start)
while queue:
v = queue.pop(0) # queue.popleft()
for w in graph[v]:
if w not in explored:
explored.add(w)
queue.append(w)
return explored


G = {'A': ['B', 'C'],
'B': ['A', 'D', 'E'],
'C': ['A', 'F'],
'D': ['B'],
'E': ['B', 'F'],
'F': ['C', 'E']}

print(bfs(G, 'A'))
36 changes: 36 additions & 0 deletions Graphs/DFS.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,36 @@
"""pseudo-code"""

"""
DFS(graph G, start vertex s):
// all nodes initially unexplored
mark s as explored
for every edge (s, v):
if v unexplored:
DFS(G, v)
"""


def dfs(graph, start):
"""The DFS function simply calls itself recursively for every unvisited child of its argument. We can emulate that
behaviour precisely using a stack of iterators. Instead of recursively calling with a node, we'll push an iterator
to the node's children onto the iterator stack. When the iterator at the top of the stack terminates, we'll pop
it off the stack."""
explored, stack = set(), [start]
explored.add(start)
while stack:
v = stack.pop() # the only difference from BFS is to pop last element here instead of first one
for w in graph[v]:
if w not in explored:
explored.add(w)
stack.append(w)
return explored


G = {'A': ['B', 'C'],
'B': ['A', 'D', 'E'],
'C': ['A', 'F'],
'D': ['B'],
'E': ['B', 'F'],
'F': ['C', 'E']}

print(dfs(G, 'A'))