Graph TraversalSection 4.1–4.2

Dr. Mayfield and Dr. Lam

Department of Computer ScienceJames Madison University

Oct 30, 2015

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Oct 30, 2015 Graph Traversal 3 of 18

Section 4.1–4.2

The following slides are by Steven Halim

https://sites.google.com/site/stevenhalim/home/material

Graph Terms Quick ReviewGraph Terms – Quick Review

/– Vertices/Nodes

– Edges

– (Strongly) Connected Component 

S b G h– Un/Weighted

– Un/Directed

– Sub Graph

– Complete Graph

Di d A li G h– In/Out Degree

– Self‐Loop/Multiple Edges ( l h) l

– Directed Acyclic Graph

– Tree/Forest

l / l(Multigraph) vs Simple Graph

S /D

– Euler/Hamiltonian Path/Cycle

Bi tit G h– Sparse/Dense

– Path, Cycle

I l t d R h bl

– Bipartite Graph

– Isolated, ReachableCS3233 ‐ Competitive Programming,

Steven Halim, SoC, NUS

Depth‐First Search (DFS)

Breadth‐First Search (BFS)Reachability

Finding Connected ComponentsFinding Connected Components 

Flood Fill

Topological Sort

Finding Cycles (Back Edges)

Finding Articulation Points & Bridges

Finding Strongly Connected Components

GRAPH TRAVERSAL ALGORITHMSFinding Strongly Connected Components

CS3233 ‐ Competitive Programming,Steven Halim, SoC, NUS

Graph Traversal AlgorithmsGraph Traversal Algorithms

• Given a graph, we want to traverse it!

• There are 2 major ways:There are 2 major ways:– Depth First Search (DFS)

U ll i l t d i i• Usually implemented using recursion

• More natural

M f l d h• Most frequently used to traverse a graph

– 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 TemplateDepth First Search – Template

• O(V + E) if using Adjacency List

• O(V2) if using Adjacency Matrix

typedef pair<int, int> ii; typedef vector<ii> vi;

void dfs(int u) { // DFS for normal usageprintf(" %d", u); // this vertex is visiteddfs_num[u] = DFS_BLACK; // mark as visitedfor (int j = 0; j < (int)AdjList[u].size; j++) {

ii v = AdjList[u][j]; // try all neighbors v of vertex uif (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

<i t i t> di t di t[ ] 0map<int, int> dist; dist[source] = 0;queue<int> q; q.push(source); // start from source

hil (! t ()) {while (!q.empty()) {int u = q.front(); q.pop(); // queue: layer by layer!for (int j = 0; j < (int)AdjList[u].size(); j++) {ii AdjLi t[ ][j] // f h i hb fii v = AdjList[u][j]; // for each neighbours of uif (!dist.count(v.first)) {dist[v.first] = dist[u] + 1; // unvisited + reachable

h( fi t) // fi t f t tq.push(v.first); // enqueue v.first for next steps}

}}}

CS3233 ‐ Competitive Programming,Steven Halim, SoC, NUS

1st Application: Connected Components1st 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 yetif (dfs_num[i] == DFS_WHITE) { // if not visited yetprintf("Component %d, visit", ++numComp);dfs(i); // one component foundprintf("\n");

}printf("There are %d connected components\n", numComp);p ( p , p);

CS3233 ‐ Competitive Programming,Steven Halim, SoC, NUS

Graph Traversal ComparisonGraph Traversal Comparison

• DFS

• Pros:

• BFS

• Pros:Pros:– Slightly easier to code

Pros:– Can solve SSSP on unweighted graphscode

– Use less memory

unweighted graphs (discussed later)

• Cons:– Cannot solve SSSP on

• Cons:– Slightly longer to Cannot solve SSSP on 

unweighted graphs

g y gcode

– Use more memoryUse more memoryCS3233 ‐ Competitive Programming,

Steven Halim, SoC, NUS

BFS (unweighted)

Dijk t ’ ( l )Dijkstra’s (non –ve cycle)

Bellman Ford’s (may have –ve cycle), not discussed

Floyd Warshall’s (all‐pairs

SHORTEST PATHSFloyd Warshall s (all pairs

CS3233 ‐ Competitive Programming,Steven Halim, SoC, NUS

BFS for Special Case SSSPBFS 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 vCheck how many nodes are un reachable from vor has distance > TTL from v

• i e length(shortest path(v node)) > TTLi.e. length(shortest_path(v, node)) > TTL

CS3233 ‐ Competitive Programming,Steven Halim, SoC, NUS

Example (1)0 1 2 3

4 765

8 Q = {5}D[5] = 0

9 10 11 12

5

0 1 2 3 Example (2)4 765

8 Q = {5}Q = {1, 6, 10} D[5] = 0

D[1] = D[5] + 1 = 19 10 11 12

D[1] = D[5] + 1 = 1D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1

1

5 6

10

0 1 2 3 Example (3)4 765

8 Q = {5}Q = {1, 6, 10}Q = {6 10 0 2}

D[5] = 0D[1] = D[5] + 1 = 1

9 10 11 12Q = {6, 10, 0, 2}Q = {10, 0, 2, 11}Q = {0, 2, 11, 9}

D[1] = D[5] + 1 = 1D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1D[0] = D[1] + 1 = 2

1 200 2D[0] = D[1] + 1 = 2D[2] = D[1] + 1 = 2D[11] = D[6] + 1 = 2D[9] = D[10] + 1 = 2

5 6D[9] D[10] + 1 2

10 119

0 1 2 3 Example (4)4 765

8 Q = {5}Q = {1, 6, 10}Q = {6 10 0 2}

D[5] = 0D[1] = D[5] + 1 = 1

9 10 11 12Q = {6, 10, 0, 2}Q = {10, 0, 2, 11}Q = {0, 2, 11, 9} Q = {2 11 9 4}

D[1] = D[5] + 1 = 1D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1D[0] = D[1] + 1 = 2

1 20Q = {2, 11, 9, 4}Q = {11, 9, 4, 3}Q = {9, 4, 3, 12} Q = {4, 3, 12, 8}

0 2 3D[0] = D[1] + 1 = 2D[2] = D[1] + 1 = 2D[11] = D[6] + 1 = 2D[9] = D[10] + 1 = 2

5 6Q {4, 3, 12, 8}

4D[9] D[10] + 1 2D[4] = D[0] + 1 = 3D[3] = D[2] + 1 = 3D[12] = D[11] + 1 = 3

8D[12] D[11] 1 3D[8] = D[9] + 1 = 3

10 119 12 77

0 1 2 3 Example (5)4 765

8 Q = {5}Q = {1, 6, 10}Q = {6 10 0 2}

D[5] = 0D[1] = D[5] + 1 = 1

9 10 11 12Q = {6, 10, 0, 2}Q = {10, 0, 2, 11}Q = {0, 2, 11, 9} Q = {2 11 9 4}

D[1] = D[5] + 1 = 1D[6] = D[5] + 1 = 1D[10] = D[5] + 1 = 1D[0] = D[1] + 1 = 2

1 20Q = {2, 11, 9, 4}Q = {11, 9, 4, 3}Q = {9, 4, 3, 12} Q = {4, 3, 12, 8}

0 2 3D[0] = D[1] + 1 = 2D[2] = D[1] + 1 = 2D[11] = D[6] + 1 = 2D[9] = D[10] + 1 = 2

5 6Q {4, 3, 12, 8}Q = {3, 12, 8}Q = {12, 8, 7}Q = {8, 7}

4 7D[9] D[10] + 1 2D[4] = D[0] + 1 = 3D[3] = D[2] + 1 = 3D[12] = D[11] + 1 = 3Q { , }

Q = {7}Q = {}8

D[12] D[11] 1 3D[8] = D[9] + 1 = 3D[7] = D[3] + 1 = 4

10 119 12 This is the BFS = SSSP spanning tree when BFS is started from vertex 5