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# Breadth First Search - PowerPoint PPT Presentation

Breadth First Search. Another graph search algorithm is Breadth First Search (bfs). In addition to finding connected components, the BFS approach enables us to calculate the distance from the start to every reachable vertex in its connected component.

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Presentation Transcript

Another graph search algorithm is Breadth First Search (bfs).

In addition to finding connected components, the BFS approach enables us to calculate the distance from the start to every reachable vertex in its connected component.

BFS uses a queue to hold vertices that have been visited but have not yet been used as a current vertex.

When searching from a vertex u, we look at every vertex adjacent to u,

one after another; for each of these vertices, if it is unvisited, we mark it

as visited and place it on the queue of pending vertices.

After inspecting all the vertices adjacent to u, we remove the next vertex

from the queue and explore from there.

We start by marking the start vertex as visited, in this algorithm by setting

its distance to 0; then we append it to the queue.

Then: while the queue is not empty:

Serve a vertex from the queue to be the current vertex;

for each vertex w adjacent from current:

if w has not been visited (distance is NIL):

mark w as visited;

make current the predecessor of w;

append w to the queue of pending vertices;

set the distance of w to be 1 + the distance of current;

We assume that we process the adjacent vertices of our current vertex in

alphabetical order. Our initial call will be bfs(a,…).

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

a

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

a

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

e

a

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

f e

a

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

f

e

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

d c b f

e

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

d c b

f

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

g d c b

f

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

g d c

b

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

g d

c

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

g

d

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

h g

d

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

h

g

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

h

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

j i

h

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

j

i

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

j

back

front

black graph vertices have distance -1

arrow points to the current vertex

teal vertices have been

completely explored and distance

is valid.

Queue of pending vertices

current

j

back

front

Code for BFS graph

• Input parameters: adj, start Output parameters: none

• bfs( adj, start ) { n = adj.last // number of vertices for i = 1 to n visit[i] = false visit[start] = true// q is an initially empty queue q.enqueue(start) while (!q.empty() ) { current = q.front( ) q.dequeue( ) trav = adj[current] // get pointer to first node of adjacency list of current while ( trav != null ) { v = trav.ver // get vertex from adjacency list of current if ( ! visit[v] ) { visit[v] = true q.enqueue(v) } trav = trav.next } }

• Assume the graph has n vertices and m edges

• The for-loop in bfs requires (n) steps

• Again, the worst-case for the nested while-loops is when each vertex in each adjacency list is visited.

• There are 2m such vertex occurrences

• Therefore BFS is (n+m)

• It is easy to use BFS, starting at vertex s, to determine if there is a path in graph G from s to v and, if so, to output a shortest path from s to v.

• The reason for the latter is that if vertex w is discovered from vertex v during BFS starting at s, the distance from s to w is 1 plus the distance from s to v.

• Thus we could have set dist[s] = 0 and after that, set dist[w] = 1+dist[v] whenever w is discovered from v.

• Theorem A graph G is bipartite if and only if G is 2-colorable

• Thus determining if G is bipartite is the same as determining if G is 2-colorable

• Theorem If G is a graph with a 2-coloring : V(G)  {R,B}, then the vertex colors on any path in G must alternate between the two colors R and B.

• Corollary If G is a 2-colorable graph, then G does not contain an odd-length cycle.

• Theorem If G does not contain an odd-length cycle, then G is 2-colorable

If the theorem is true for all connected graphs, then it is clearly true for arbitrary graphs, so we assume G = (V,E) is connected.

Choose a vertex u of G and define V0 = { v  V | dist(u,v) is even } and V1 = { v  V | dist(u,v) is odd}. Clearly V = V0  V1 and V0  V1 = .

Thus V0 ,V1 is a partition of V.

Claim: no edge of G connects vertices in the same partition set.

Proof Continued graph

• Claim: no edge of G connects vertices in the same partition set.

Suppose not and let v,w be adjacent vertices in G whose distances from u are either both odd or both even. Let Qv be a shortest path from u to v and Qw a shortest path to w from u.

Let x be the last vertex of Qw encountered when traversing Qv.

The length of the segments of the two paths from u to x must be the same as otherwise we could replace the longer by the shorter and get a shorter path to its target.

Pv

Pw

Let Pv and Pw be the segments of Qv and Qw from x to their targets.

Then the lengths of Pv and Pw are either both odd or both even, so the sum of their lengths is even. But this means that Pv Pw  {v,w} is an odd cycle.

• Determining if a connected graph is 2-colorable:

Run BFS with the following modifications:

Assign color WHITE to all vertices of G

Assign color RED to the start vertex s and put s in the queue

While the queue is not empty: u  dequeued element of the queue

for each vertex w adjacent to u: if color(w) == color(u) return FALSE

if color(w) == WHITE if color(u) = RED set color of w to BLUE else set color of w to RED

Enqueue w

return TRUE

Homework graph

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