Introduction to Graphs And Breadth First Search

Graphs: what are they? Representations of pairwise relationships Collections of objects under some specified relationship Representations of pairwise relationships Collections of objects under some specified relationship

Graphs: what are they mathematically? A graph G is a pair (V,E) V is a set of vertices (nodes) E is a set of pairs (a,b), a,b  V V is the set of relatable objects E is the set of relationships A graph G is a pair (V,E) V is a set of vertices (nodes) E is a set of pairs (a,b), a,b  V V is the set of relatable objects E is the set of relationships

A Visual Example G = ( {1,2,3,4,5}, {(1,2), (1,4), (2,3), (2,4), (1,5)} )

Directed Graphs In a directed graph (a,b)  E does not imply (b,a)  E Undirected graphs are a subset (a,b)  E if and only if (b,a)  E Visually, directed graphs are drawn with arrows In a directed graph (a,b)  E does not imply (b,a)  E Undirected graphs are a subset (a,b)  E if and only if (b,a)  E Visually, directed graphs are drawn with arrows

Directed Graph Example G = ( {1,2,3,4,5}, { (1,5), (2,1), (2,3), (2,4), (3,2), (4,1) } )

Weighted Graphs Have weights associated with edges Can be directed or undirected Can have pairs, in a directed graph, where the weights from (a,b) have no relationship on the weights from (b,a) Have weights associated with edges Can be directed or undirected Can have pairs, in a directed graph, where the weights from (a,b) have no relationship on the weights from (b,a)

Weighted Graph Example G = ( {1,2,3,4,5}, { (1,5,-5), (2,1,3.2), (2,3,42), (3,2, π ), (2,4,777), (4,1,666) } ) 3.2 π

Graph Representation How to represent in memory? Two common ways: Adjacency Lists Adjacency Matrix How to represent in memory? Two common ways: Adjacency Lists Adjacency Matrix

Adjacency Lists Compact usage in sparse graphs where |E| << |V| 2 Stores graph as array of |V| lists Each v has a list of adjacent v in G Compact usage in sparse graphs where |E| << |V| 2 Stores graph as array of |V| lists Each v has a list of adjacent v in G

Undirected Graph Example G = ( {1,2,3,4,5}, {(1,2), (1,4), (2,3), (2,4), (1,5)} )

Directed Graph Example G = ( {1,2,3,4,5}, { (1,5), (2,1), (2,3), (2,4), (3,2), (4,1) } )

Adjacency Lists Wrap Up Sum of list lengths for undirected 2|E| For some apps, could optimize to |E| Sum of list lengths for directed |E| Weighted graphs: left as exercise Sum of list lengths for undirected 2|E| For some apps, could optimize to |E| Sum of list lengths for directed |E| Weighted graphs: left as exercise

Adjacency Matrix Often less memory for dense graphs Faster check for edge existence Mathematically: M is a |V|*|V| matrix Dimensions represent vertices M(i,j)=1 if (i,j)  E, 0 otherwise Often less memory for dense graphs Faster check for edge existence Mathematically: M is a |V|*|V| matrix Dimensions represent vertices M(i,j)=1 if (i,j)  E, 0 otherwise

Undirected Graph Example G = ( {1,2,3,4,5}, {(1,2), (1,4), (2,3), (2,4), (1,5)} )

Directed Graph Example G = ( {1,2,3,4,5}, { (1,5), (2,1), (2,3), (2,4), (3,2), (4,1) } )

Adjacency Matrix Wrap Up Size is always |V| 2 If |E| close to |V|, can be more efficient because edge is 1 bit instead of a 4 bytes for a pointer Weighted graphs: use weight instead of 0’s and 1’s Size is always |V| 2 If |E| close to |V|, can be more efficient because edge is 1 bit instead of a 4 bytes for a pointer Weighted graphs: use weight instead of 0’s and 1’s

Breadth-first Search Problem: For a given graph G, and a specified s in the graph, find all vertices v that are reachable from s and determine the shortest path in G from s to v.

How BFS works Constructs a breadth first tree Root is s Path from s to v is shortest path from s to v in G Constructs a breadth first tree Root is s Path from s to v is shortest path from s to v in G

The BFS algorithm Assigns a color to each node white = vertex has not been reached gray = vertex is in the BFS frontier) black = vertex and ALL of its neighbors have been processed. Assigns a color to each node white = vertex has not been reached gray = vertex is in the BFS frontier) black = vertex and ALL of its neighbors have been processed.

BFS algorithm (cont.) Computes d[v] for each v Shortest distance from s to v in G Computes p[v] for each v Predecessor of v in the breadth-first tree Computes d[v] for each v Shortest distance from s to v in G Computes p[v] for each v Predecessor of v in the breadth-first tree

BFS Pseudo Code (initialization) 1.for each vertex v in V 2. color[v] = white 3. d[v] = INFINITY 4. p[v] = NULL 5.color[s] = gray 6.d[s] = 0 7.Queue.clear() 8.Queue.put(s) 1.for each vertex v in V 2. color[v] = white 3. d[v] = INFINITY 4. p[v] = NULL 5.color[s] = gray 6.d[s] = 0 7.Queue.clear() 8.Queue.put(s)

BFS Pseudo Code (tree construction) 9.while (!Queue.empty()) 10. u = Queue.get() 11. for each v adjacent to u 12. if (color[v] == white) 13. color[v] = gray 14. d[v] = d[u] p[v] = u 16. Queue.put(v) 17. color[u] = black 9.while (!Queue.empty()) 10. u = Queue.get() 11. for each v adjacent to u 12. if (color[v] == white) 13. color[v] = gray 14. d[v] = d[u] p[v] = u 16. Queue.put(v) 17. color[u] = black

Correctness of BFS Definition 1. b(s,v) is the min number of edges in any path from s to v. If there is no path from s to v then b(s,v) = INFINITY. b(s,v) is the shortest-path distance. Lemma 1. Let G=(V,E), v in V. For any edge (u,v) in E b(s,v) <= b(s,u) + 1 Definition 1. b(s,v) is the min number of edges in any path from s to v. If there is no path from s to v then b(s,v) = INFINITY. b(s,v) is the shortest-path distance. Lemma 1. Let G=(V,E), v in V. For any edge (u,v) in E b(s,v) <= b(s,u) + 1

Proof of Lemma 1 If u is reachable from s, so is v. The shortest path from s to v cannot be more than the shortest path from s to u plus the edge (u,v), thus the inequality holds. If u is not reachable then b(s,u) = INFINITY so the inequality holds

Lemma 2 Upon termination, the BFS algorithm computes d[v] for every vertex and d[v] >= b(s,v)

Proof of Lemma 2 By induction on the number i of enqueue operations. For i = 1 (s is enqueued), d[s]=[0]=b(s,s) d[v]=INFINITY>=b(s,v) for all v != s For i = n, consider white v discovered from u. By induction, d[u]>=b(s,u). Since d[v]=d[u]+1 >= b(s,u)+1 >= b(s,v) By induction on the number i of enqueue operations. For i = 1 (s is enqueued), d[s]=[0]=b(s,s) d[v]=INFINITY>=b(s,v) for all v != s For i = n, consider white v discovered from u. By induction, d[u]>=b(s,u). Since d[v]=d[u]+1 >= b(s,u)+1 >= b(s,v)

Lemma 3 At all times during execution of BFS the queue contains vertices (v1, v2, … vr) such that d[v1] <= d[v2]…<=d[vr] d[vr] <= d[v1] + 1 At all times during execution of BFS the queue contains vertices (v1, v2, … vr) such that d[v1] <= d[v2]…<=d[vr] d[vr] <= d[v1] + 1

Proof of Lemma 3 By induction on number i of queue op’s. For i=1, queue only has s, hypothesis holds For i=n After dequeueing v1: d[vr]<=d[v1]+1 and d[v1]<=d[v2], then d[vr]<=d[v2]+1, so hypothesis holds After enqueueing v r+1 : D[v r+1 ] = d[v1]+1 >= d[vr] D[v r+1 ] = d[v1]+1 <= d[v2]+1, since d[v1]<=d[v2] Since v2 is the new head of queue, hypothesis holds By induction on number i of queue op’s. For i=1, queue only has s, hypothesis holds For i=n After dequeueing v1: d[vr]<=d[v1]+1 and d[v1]<=d[v2], then d[vr]<=d[v2]+1, so hypothesis holds After enqueueing v r+1 : D[v r+1 ] = d[v1]+1 >= d[vr] D[v r+1 ] = d[v1]+1 <= d[v2]+1, since d[v1]<=d[v2] Since v2 is the new head of queue, hypothesis holds

Corollary (4) to Lemma 3 If vertices u and v are enqueued during execution of BFS and u is enqueued before v, then d[u] <= d[v]

Theorem 5 Given G=(V,E) and s BFS discovers every v reachable from s Upon termination, d[v]=b(s,v) Moreover, for v reachable from s One of the shortests paths from s to v is a path followed from s to p[v], followed by edge (p[v],v). Given G=(V,E) and s BFS discovers every v reachable from s Upon termination, d[v]=b(s,v) Moreover, for v reachable from s One of the shortests paths from s to v is a path followed from s to p[v], followed by edge (p[v],v).

Proof of Theorem 5 By contradiction. Assume v assigned d[v] != b(s,v). By lemma 2, d[v]>=b(s,v), so d[v] > b(s,v). v must be reachable, else b(s,v)>=d[v] Let u be predecessor on path to v b(s,v) = b(s,u)+1 = d[u]+1 This would mean d[v] > d[u]+1 By contradiction. Assume v assigned d[v] != b(s,v). By lemma 2, d[v]>=b(s,v), so d[v] > b(s,v). v must be reachable, else b(s,v)>=d[v] Let u be predecessor on path to v b(s,v) = b(s,u)+1 = d[u]+1 This would mean d[v] > d[u]+1

Proof completion d[v] > d[u]+1 cannot happen! Look at when BFS dequeues u v is either white, black, or gray If v is black, already removed from queue, and by corollary 4, d[v]<=d[u] If v is gray, it was made gray when other vertex w was dequeued, so d[v]=d[w]+1 <= d[u]+1 (by corollary 4) If v is white, then the code sets d[v] d[v] = d[u] + 1 d[v] > d[u]+1 cannot happen! Look at when BFS dequeues u v is either white, black, or gray If v is black, already removed from queue, and by corollary 4, d[v]<=d[u] If v is gray, it was made gray when other vertex w was dequeued, so d[v]=d[w]+1 <= d[u]+1 (by corollary 4) If v is white, then the code sets d[v] d[v] = d[u] + 1

BFS Wrap Up So, d[v]=b(s,v) for all v in V All reachable vertices discovered, else d = INFINITY If p[v]=u, then d[v]=d[u]+1, so one of the shortest paths from s to v takes path from s to u then (u,v) So, d[v]=b(s,v) for all v in V All reachable vertices discovered, else d = INFINITY If p[v]=u, then d[v]=d[u]+1, so one of the shortest paths from s to v takes path from s to u then (u,v)

Applications for Graphs Link structure of a website Problems in travel, biology, etc. Network representation Solution space: EXAMPLE: Sudoku Link structure of a website Problems in travel, biology, etc. Network representation Solution space: EXAMPLE: Sudoku