Graphs

Shortest Path (Unweighted, via BFS)

BFS explores a graph layer by layer, so the first time it reaches a vertex is along a shortest path. Track dist[v] and parent[v] while exploring, then walk parents back from the target to reconstruct the route.

Algorithm

On the canonical graph from graph-adjacency-list, the shortest path from 1 to 6 is [1, 2, 4, 5, 6] with distance 4. The path is rebuilt from parent: 6 -> 5 -> 4 -> 2 -> 1, reversed.

layers equal distance BFS order equals distance in an unweighted graph.

Basic Implementation

basic.py
Replay: real traced execution (multi-file project)
from collections import deque

adj = {
    1: [2, 3],
    2: [1, 4],
    3: [1, 4],
    4: [2, 3, 5],
    5: [4, 6],
    6: [5],
}

src, dst = 1, 6
dist = {src: 0}
parent = {src: None}
queue = deque([src])
while queue:
    v = queue.popleft()
    for nb in adj[v]:
        if nb not in dist:
            dist[nb] = dist[v] + 1
            parent[nb] = v
            queue.append(nb)

path = []
node = dst
while node is not None:
    path.append(node)
    node = parent[node]
path.reverse()
print(path)
print(dist[dst])

  1. dist ← {1: 0}

    12src, dst = 1, 613dist = {src: 0}14parent = {src: None}
    values this step{1: 0}dist

  2. parent ← {1: None}

    13dist = {src: 0}14parent = {src: None}15queue = deque([src])
    values this step{1: None}parent

  3. dist ← {1: 0, 2: 1, 3: 1}, parent ← {1: None, 2: 1, 3: 1}, queue ← [2, 3]

    16while queue:17    v = queue.popleft()18    for nb in adj[v]:
    values this step{1: 0, 2: 1, 3: 1}dist{1: None, 2: 1, 3: 1}parent[2, 3]queue1dequeue

  4. dist ← {1: 0, 2: 1, 3: 1, 4: 2}, parent ← {1: None, 2: 1, 3: 1, 4: 2}

    16while queue:17    v = queue.popleft()18    for nb in adj[v]:
    values this step{1: 0, 2: 1, 3: 1, 4: 2}dist{1: None, 2: 1, 3: 1, 4: 2}parent[3, 4]queue2dequeue

  5. dist ← {1: 0, 2: 1, 3: 1, 4: 2}, parent ← {1: None, 2: 1, 3: 1, 4: 2}

    16while queue:17    v = queue.popleft()18    for nb in adj[v]:
    values this step{1: 0, 2: 1, 3: 1, 4: 2}dist{1: None, 2: 1, 3: 1, 4: 2}parent[4]queue3dequeue

  6. dist ← {1: 0, 2: 1, 3: 1, 4: 2, 5: 3}, parent ← {1: None, 2: 1, 3: 1, 4: 2, 5: 4}

    16while queue:17    v = queue.popleft()18    for nb in adj[v]:
    values this step{1: 0, 2: 1, 3: 1, 4: 2, 5: 3}dist{1: None, 2: 1, 3: 1, 4: 2, 5: 4}parent[5]queue4dequeue

  7. dist ← {1: 0, 2: 1, 3: 1, 4: 2, 5: 3, 6: 4}, parent ← {1: None, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}

    16while queue:17    v = queue.popleft()18    for nb in adj[v]:
    values this step{1: 0, 2: 1, 3: 1, 4: 2, 5: 3, 6: 4}dist{1: None, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}parent[6]queue5dequeue

  8. dist ← {1: 0, 2: 1, 3: 1, 4: 2, 5: 3, 6: 4}, parent ← {1: None, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}

    16while queue:17    v = queue.popleft()18    for nb in adj[v]:
    values this step{1: 0, 2: 1, 3: 1, 4: 2, 5: 3, 6: 4}dist{1: None, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}parent[]queue6dequeue

  9. path ← [1, 2, 4, 5, 6]

    29path.reverse()30print(path)31print(dist[dst])
    values this step[1, 2, 4, 5, 6]path{1: None, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}parent

  10. stdout ← [1, 2, 4, 5, 6]

    30print(path)31print(dist[dst])
    values this step[1, 2, 4, 5, 6]stdout[1, 2, 4, 5, 6]path

  11. stdout ← 4

    30print(path)31print(dist[dst])
    values this step4stdout4dist[6]

  12. BFS path ← 1 -> 2 (1 edge, cost 10), cheaper weighted path ← 1 -> 3 -> 2 (2 edges, cost 2)

    30print(path)31print(dist[dst])
    values this step1 -> 2 (1 edge, cost 10)BFS path1 -> 3 -> 2 (2 edges, cost 2)cheaper weighted pathuse Dijkstra with a priority queueweighted algorithm1->2 weight 10, 1->3 weight 1, 3->2 weight 1edge weights

Complexity

  • Time: O(V + E)
  • Space: O(V)

Implementation notes

  • Python: a dist dict doubles as the visited check (a vertex is discovered once dist has it), and parent records the predecessor.
  • The replay shows dist, parent, and the queue filling in, then the reconstructed path. It also contrasts that unweighted result with a weighted graph where Dijkstra with a priority queue is required.