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.scala
Replay: real traced execution (multi-file project)
import scala.collection.mutable.{HashMap, Queue, ArrayBuffer}
object Main {
	def main(args: Array[String]): Unit = {
		val adj = HashMap.empty[Int, List[Int]]
		adj(1) = List(2, 3)
		adj(2) = List(1, 4)
		adj(3) = List(1, 4)
		adj(4) = List(2, 3, 5)
		adj(5) = List(4, 6)
		adj(6) = List(5)
		val src = 1
		val dst = 6
		val dist = HashMap.empty[Int, Int]
		val parent = HashMap.empty[Int, Int]
		dist(src) = 0
		parent(src) = 0
		val queue = Queue.empty[Int]
		queue.enqueue(src)
		while (queue.nonEmpty) {
			val v = queue.dequeue()
			for (nb <- adj(v)) {
				if (!dist.contains(nb)) {
					dist(nb) = dist(v) + 1
					parent(nb) = v
					queue.enqueue(nb)
				}
			}
		}
		val path = ArrayBuffer.empty[Int]
		var node = dst
		while (node != 0) {
			path += node
			node = parent(node)
		}
		println(path.reverse.mkString("[", ", ", "]"))
		println(dist(dst))
	}
}

  1. dist ← {1: 0}

    14val parent = HashMap.empty[Int, Int]15dist(src) = 016parent(src) = 0
    values this step{1: 0}dist

  2. parent ← {1: null}

    15dist(src) = 016parent(src) = 017val queue = Queue.empty[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]

    19while (queue.nonEmpty) {20	val v = queue.dequeue()21	for (nb <- 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}

    19while (queue.nonEmpty) {20	val v = queue.dequeue()21	for (nb <- 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}

    19while (queue.nonEmpty) {20	val v = queue.dequeue()21	for (nb <- 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}

    19while (queue.nonEmpty) {20	val v = queue.dequeue()21	for (nb <- 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}

    19while (queue.nonEmpty) {20	val v = queue.dequeue()21	for (nb <- 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}

    19while (queue.nonEmpty) {20	val v = queue.dequeue()21	for (nb <- 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]

    29val path = ArrayBuffer.empty[Int]30var node = dst31while (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]

    34}35println(path.reverse.mkString("[", ", ", "]"))36println(dist(dst))
    values this step[1, 2, 4, 5, 6]stdout[1, 2, 4, 5, 6]path

  11. stdout ← 4

    35	println(path.reverse.mkString("[", ", ", "]"))36	println(dist(dst))37}
    values this step4stdout4dist[6]

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

    35	println(path.reverse.mkString("[", ", ", "]"))36	println(dist(dst))37}
    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

  • Scala: a dist map doubles as the visited check, parent records predecessors (0 marks the source), and a Queue 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.