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.ts
Replay: real traced execution (multi-file project)
const adj: Map<number, number[]> = new Map([
    [1, [2, 3]],
    [2, [1, 4]],
    [3, [1, 4]],
    [4, [2, 3, 5]],
    [5, [4, 6]],
    [6, [5]],
]);
const src = 1;
const dst = 6;
const dist: Map<number, number> = new Map([[src, 0]]);
const parent: Map<number, number> = new Map([[src, 0]]);
const queue: number[] = [src];
while (queue.length > 0) {
    const v = queue.shift() as number;
    for (const nb of adj.get(v)!) {
        if (!dist.has(nb)) {
            dist.set(nb, (dist.get(v) as number) + 1);
            parent.set(nb, v);
            queue.push(nb);
        }
    }
}
const path: number[] = [];
let node = dst;
while (node !== 0) {
    path.push(node);
    node = parent.get(node) as number;
}
path.reverse();
console.log(JSON.stringify(path));
console.log(dist.get(dst));

  1. dist ← {1: 0}

    10const dst = 6;11const dist: Map<number, number> = new Map([[src, 0]]);12const parent: Map<number, number> = new Map([[src, 0]]);
    values this step{1: 0}dist

  2. parent ← {1: null}

    11const dist: Map<number, number> = new Map([[src, 0]]);12const parent: Map<number, number> = new Map([[src, 0]]);13const queue: number[] = [src];
    values this step{1: null}parent

  3. dist ← {1: 0, 2: 1, 3: 1}, parent ← {1: null, 2: 1, 3: 1}, queue ← [2, 3]

    14while (queue.length > 0) {15    const v = queue.shift() as number;16    for (const nb of adj.get(v)!) {
    values this step{1: 0, 2: 1, 3: 1}dist{1: null, 2: 1, 3: 1}parent[2, 3]queue1dequeue

  4. dist ← {1: 0, 2: 1, 3: 1, 4: 2}, parent ← {1: null, 2: 1, 3: 1, 4: 2}

    14while (queue.length > 0) {15    const v = queue.shift() as number;16    for (const nb of adj.get(v)!) {
    values this step{1: 0, 2: 1, 3: 1, 4: 2}dist{1: null, 2: 1, 3: 1, 4: 2}parent[3, 4]queue2dequeue

  5. dist ← {1: 0, 2: 1, 3: 1, 4: 2}, parent ← {1: null, 2: 1, 3: 1, 4: 2}

    14while (queue.length > 0) {15    const v = queue.shift() as number;16    for (const nb of adj.get(v)!) {
    values this step{1: 0, 2: 1, 3: 1, 4: 2}dist{1: null, 2: 1, 3: 1, 4: 2}parent[4]queue3dequeue

  6. dist ← {1: 0, 2: 1, 3: 1, 4: 2, 5: 3}, parent ← {1: null, 2: 1, 3: 1, 4: 2, 5: 4}

    14while (queue.length > 0) {15    const v = queue.shift() as number;16    for (const nb of adj.get(v)!) {
    values this step{1: 0, 2: 1, 3: 1, 4: 2, 5: 3}dist{1: null, 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: null, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}

    14while (queue.length > 0) {15    const v = queue.shift() as number;16    for (const nb of adj.get(v)!) {
    values this step{1: 0, 2: 1, 3: 1, 4: 2, 5: 3, 6: 4}dist{1: null, 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: null, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}

    14while (queue.length > 0) {15    const v = queue.shift() as number;16    for (const nb of adj.get(v)!) {
    values this step{1: 0, 2: 1, 3: 1, 4: 2, 5: 3, 6: 4}dist{1: null, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}parent[]queue6dequeue

  9. path ← [1, 2, 4, 5, 6]

    24const path: number[] = [];25let node = dst;26while (node !== 0) {
    values this step[1, 2, 4, 5, 6]path{1: null, 2: 1, 3: 1, 4: 2, 5: 4, 6: 5}parent

  10. stdout ← [1, 2, 4, 5, 6]

    30path.reverse();31console.log(JSON.stringify(path));32console.log(dist.get(dst));
    values this step[1, 2, 4, 5, 6]stdout[1, 2, 4, 5, 6]path

  11. stdout ← 4

    31console.log(JSON.stringify(path));32console.log(dist.get(dst));
    values this step4stdout4dist[6]

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

    31console.log(JSON.stringify(path));32console.log(dist.get(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

  • TypeScript: a dist Map doubles as the visited check, parent records predecessors (0 marks the source), and queue.shift() dequeues.
  • 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.