Graph Traversal Section 4.1–4.2 Dr. Mayfield and Dr. Lam Department of Computer Science James Madison University Oct 30, 2015
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Section 4.1–4.2 The following slides are by Steven Halim https://sites.google.com/site/stevenhalim/home/material
Graph Terms Graph Terms – Quick Review Quick Review – Vertices/Nodes / – (Strongly) Connected Component – Edges – Sub Graph S b G h – Un/Weighted – Complete Graph – Un/Directed – Directed Acyclic Graph Di d A li G h – In/Out Degree – Tree/Forest – Self ‐ Loop/Multiple Edges (Multigraph) vs Simple ( l h) l – Euler/Hamiltonian l / l Graph Path/Cycle – Sparse/Dense S /D – Bipartite Graph Bi tit G h – Path, Cycle – Isolated, Reachable I l t d R h bl CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
Depth ‐ First Search (DFS) Breadth ‐ First Search (BFS) Reachability Finding Connected Components Finding Connected Components Flood Fill Topological Sort Finding Cycles (Back Edges) Finding Articulation Points & Bridges Finding Strongly Connected Components Finding Strongly Connected Components GRAPH TRAVERSAL ALGORITHMS CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
Graph Traversal Algorithms Graph Traversal Algorithms • Given a graph, we want to traverse it! • There are 2 major ways: There are 2 major ways: – Depth First Search (DFS) • Usually implemented using recursion U ll i l t d i i • More natural • Most frequently used to traverse a graph M f l d h – Breadth First Search (BFS) • Usually implemented using queue (+ map), use STL • Can solve special case* of “shortest paths” problem! CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
Depth First Search Depth First Search – Template Template • O(V + E) if using Adjacency List • O(V 2 ) if using Adjacency Matrix typedef pair<int, int> ii; typedef vector<ii> vi; void dfs(int u) { // DFS for normal usage printf(" %d", u); // this vertex is visited dfs_num[u] = DFS_BLACK; // mark as visited for (int j = 0; j < (int)AdjList[u].size; j++) { ii v = AdjList[u][j]; // try all neighbors v of vertex u if (dfs_num[v.first] == DFS_WHITE) // avoid cycle dfs(v.first); // v is a (neighbor, weight) pair } } CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
Breadth First Search (using STL) Breadth First Search (using STL) • Complexity: also O(V + E) using Adjacency List map<int, int> dist; dist[source] = 0; <i t i t> di t di t[ ] 0 queue<int> q; q.push(source); // start from source while (!q.empty()) { hil (! t ()) { int u = q.front(); q.pop(); // queue: layer by layer! for (int j = 0; j < (int)AdjList[u].size(); j++) { ii v = AdjList[u][j]; // for each neighbours of u ii AdjLi t[ ][j] // f h i hb f if (!dist.count(v.first)) { dist[v.first] = dist[u] + 1; // unvisited + reachable q.push(v.first); // enqueue v.first for next steps h( fi t) // fi t f t t } } } CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
1 st Application: Connected Components 1 st Application: Connected Components • DFS (and BFS) can find connected components – A call of dfs( u ) visits only vertices connected to u ( ) y int numComp = 0; dfs_num.assign(V, DFS_WHITE); REP (i, 0, V - 1) // for each vertex i in [0..V-1] if (dfs num[i] == DFS WHITE) { // if not visited yet if (dfs_num[i] == DFS_WHITE) { // if not visited yet printf("Component %d, visit", ++numComp); dfs(i); // one component found printf("\n"); } printf("There are %d connected components\n", numComp); p ( p , p); CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
Graph Traversal Comparison Graph Traversal Comparison • DFS • BFS • Pros: Pros: • Pros: Pros: – Slightly easier to – Can solve SSSP on code code unweighted graphs unweighted graphs (discussed later) – Use less memory • Cons: • Cons: – Slightly longer to g y g – Cannot solve SSSP on Cannot solve SSSP on code unweighted graphs – Use more memory Use more memory CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
BFS (unweighted) Dijk t Dijkstra’s (non –ve cycle) ’ ( l ) Bellman Ford’s (may have –ve cycle), not discussed Floyd Warshall’s (all ‐ pairs Floyd Warshall s (all pairs SHORTEST PATHS CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
BFS for Special Case SSSP BFS for Special Case SSSP • SSSP is a classical problem in Graph theory: – Find shortest paths from one source to the rest^ p • Special case: UVa 336 (A Node Too Far) • Problem Description: – Given an un ‐ weighted & un ‐ directed Graph, g p , a starting vertex v , and an integer TTL – Check how many nodes are un ‐ reachable from v Check how many nodes are un reachable from v or has distance > TTL from v • i e length( shortest path (v node)) > TTL i.e. length( shortest_path (v, node)) > TTL CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS
Example (1) D[5] = 0 Q = {5} 12 3 7 11 2 6 10 1 5 5 0 4 8 9
0 1 2 3 Example (2) 4 5 6 7 8 Q = {5} Q = {1, 6, 10} D[5] = 0 D[1] = D[5] + 1 = 1 D[1] = D[5] + 1 = 1 9 10 11 12 D[6] = D[5] + 1 = 1 D[10] = D[5] + 1 = 1 1 5 6 10
0 1 2 3 Example (3) 4 5 6 7 8 Q = {5} Q = {1, 6, 10} D[5] = 0 Q = {6 10 0 2 } Q = {6, 10, 0, 2 } D[1] = D[5] + 1 = 1 D[1] = D[5] + 1 = 1 9 10 11 12 Q = {10, 0, 2, 11 } D[6] = D[5] + 1 = 1 Q = {0, 2, 11, 9 } D[10] = D[5] + 1 = 1 D[0] = D[1] + 1 = 2 D[0] = D[1] + 1 = 2 D[2] = D[1] + 1 = 2 0 1 20 2 D[11] = D[6] + 1 = 2 D[9] = D[10] + 1 = 2 D[9] D[10] + 1 2 5 6 11 9 10
0 1 2 3 Example (4) 4 5 6 7 8 Q = {5} Q = {1, 6, 10} D[5] = 0 Q = {6 10 0 2 } Q = {6, 10, 0, 2 } D[1] = D[5] + 1 = 1 D[1] = D[5] + 1 = 1 9 10 11 12 Q = {10, 0, 2, 11 } D[6] = D[5] + 1 = 1 Q = {0, 2, 11, 9 } D[10] = D[5] + 1 = 1 Q = {2 11 9 4 } Q = {2, 11, 9, 4 } D[0] = D[1] + 1 = 2 D[0] = D[1] + 1 = 2 Q = {11, 9, 4, 3 } D[2] = D[1] + 1 = 2 0 1 20 2 3 Q = {9, 4, 3, 12 } D[11] = D[6] + 1 = 2 Q {4, 3, 12, 8 } Q = {4, 3, 12, 8 } D[9] = D[10] + 1 = 2 D[9] D[10] + 1 2 D[4] = D[0] + 1 = 3 4 5 6 D[3] = D[2] + 1 = 3 D[12] = D[11] + 1 = 3 D[12] D[11] 1 3 D[8] = D[9] + 1 = 3 8 11 12 77 9 10
0 1 2 3 Example (5) 4 5 6 7 8 Q = {5} Q = {1, 6, 10} D[5] = 0 Q = {6, 10, 0, 2 } Q = {6 10 0 2 } D[1] = D[5] + 1 = 1 D[1] = D[5] + 1 = 1 9 10 11 12 Q = {10, 0, 2, 11 } D[6] = D[5] + 1 = 1 Q = {0, 2, 11, 9 } D[10] = D[5] + 1 = 1 Q = {2 11 9 4 } Q = {2, 11, 9, 4 } D[0] = D[1] + 1 = 2 D[0] = D[1] + 1 = 2 Q = {11, 9, 4, 3 } D[2] = D[1] + 1 = 2 0 1 20 2 3 Q = {9, 4, 3, 12 } D[11] = D[6] + 1 = 2 Q = {4, 3, 12, 8 } Q {4, 3, 12, 8 } D[9] = D[10] + 1 = 2 D[9] D[10] + 1 2 Q = {3, 12, 8} D[4] = D[0] + 1 = 3 4 5 6 7 Q = {12, 8, 7 } D[3] = D[2] + 1 = 3 Q Q = {8, 7} { , } D[12] D[11] 1 3 D[12] = D[11] + 1 = 3 Q = {7} D[8] = D[9] + 1 = 3 8 Q = {} D[7] = D[3] + 1 = 4 This is the BFS = SSSP spanning 11 12 9 10 tree when BFS is started from vertex 5
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