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.kt
Replay: real traced execution (multi-file project)
fun main() {
	val adj = HashMap<Int, List<Int>>()
	adj[1] = listOf(2, 3)
	adj[2] = listOf(1, 4)
	adj[3] = listOf(1, 4)
	adj[4] = listOf(2, 3, 5)
	adj[5] = listOf(4, 6)
	adj[6] = listOf(5)
	val src = 1
	val dst = 6
	val dist = HashMap<Int, Int>()
	val parent = HashMap<Int, Int>()
	dist[src] = 0
	parent[src] = 0
	val queue = ArrayDeque<Int>()
	queue.addLast(src)
	while (queue.isNotEmpty()) {
		val v = queue.removeFirst()
		for (nb in adj[v]!!) {
			if (!dist.containsKey(nb)) {
				dist[nb] = dist[v]!! + 1
				parent[nb] = v
				queue.addLast(nb)
			}
		}
	}
	val path = mutableListOf<Int>()
	var node = dst
	while (node != 0) {
		path.add(node)
		node = parent[node]!!
	}
	path.reverse()
	println(path.joinToString(prefix = "[", postfix = "]"))
	println(dist[dst])
}

  1. dist ← {1: 0}

    12val parent = HashMap<Int, Int>()13dist[src] = 014parent[src] = 0
    values this step{1: 0}dist

  2. parent ← {1: null}

    13dist[src] = 014parent[src] = 015val queue = ArrayDeque<Int>()
    values this step{1: null}parent

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

    17while (queue.isNotEmpty()) {18	val v = queue.removeFirst()19	for (nb in adj[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}

    17while (queue.isNotEmpty()) {18	val v = queue.removeFirst()19	for (nb in adj[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}

    17while (queue.isNotEmpty()) {18	val v = queue.removeFirst()19	for (nb in adj[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}

    17while (queue.isNotEmpty()) {18	val v = queue.removeFirst()19	for (nb in adj[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}

    17while (queue.isNotEmpty()) {18	val v = queue.removeFirst()19	for (nb in adj[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}

    17while (queue.isNotEmpty()) {18	val v = queue.removeFirst()19	for (nb in adj[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]

    27val path = mutableListOf<Int>()28var node = dst29while (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]

    33path.reverse()34println(path.joinToString(prefix = "[", postfix = "]"))35println(dist[dst])
    values this step[1, 2, 4, 5, 6]stdout[1, 2, 4, 5, 6]path

  11. stdout ← 4

    34	println(path.joinToString(prefix = "[", postfix = "]"))35	println(dist[dst])36}
    values this step4stdout4dist[6]

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

    34	println(path.joinToString(prefix = "[", postfix = "]"))35	println(dist[dst])36}
    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

  • Kotlin: a dist map doubles as the visited check, parent records predecessors (0 marks the source), and an ArrayDeque gives FIFO order.
  • 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.