Undirected Graph G = (V,E)

CSE 421: Introduction to 1 2 Algorithms 10 3 11 Graph Traversal 12 4 8 Paul Beame 13 5 9 6

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Directed Graph G = (V,E) Graph Traversal

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Learn the basic structure of a graph
Walk from a fixed starting vertex s to 2 10 find all vertices reachable from s 3

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Three states of vertices 12 4 8
unvisited
visited/discovered 13
fully-explored 5 9 6

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Generic Graph Traversal Algorithm Generic Traversal Always Works

Claim: At termination R is the set of nodes Find: set R of vertices reachable from s∈∈∈V reachable from s
Proof ⊆ ∈∈∈ Reachable( s):
: For every node v R there is a path from s to v
⊇: Suppose there is a node w∉∉∉R reachable from s R←←← {s} via a path P ∉∉∉ While there is a ( u,v )∈∈∈E where u∈∈∈R and v∉∉∉R
Take first node v on P such that v R
Predecessor u of v in P satisfies Add v to R
u ∈∈∈ R

(u,v) ∈∈∈E

But this contradicts the fact that the algorithm exited the while loop.

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1 Breadth-First Search BFS(s)

Completely explore the vertices in order Global initialization: mark all vertices “unvisited” of their distance from s BFS( s) ← ← mark s “visited”; R {s}; layer L 0 {s}

while Li not empty ←←←∅∅∅ Li+1 ∈∈∈ For each u Li
Naturally implemented using a queue for each edge { u,v} if ( v is “unvisited”) mark v “visited”

Add v to set R and to layer Li+1 mark u “fully-explored” i ←←← i+1

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Properties of BFS (v) Properties of BFS

BFS( s) visits x if and only if there is a path in G from
On undirected graphs s to x.
All non-tree edges join vertices on the same or adjacent layers
Edges followed to undiscovered vertices define a “breadth first spanning tree" of G
Suppose not
Layer i in this tree, L
Then there would be vertices (x,y) such that i ∈∈∈ ∈∈∈ >>> x Li and y Lj and j i+1
those vertices u such that the shortest path in G from the root s is of length i.
Then, when vertices incident to x are considered in BFS y would be added to Li+1
On undirected graphs and not to Lj
All non-tree edges join vertices on the same or adjacent layers

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Graph Search Application: BFS Application: Shortest Paths Connected Components

Tree gives shortest 1 0
Want to answer questions of the paths from start vertex form: 2 3 1
Given : vertices u and v in G

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Is there a path from u to v? 6 Q: Why 7 not create 2
Idea: create array A such that 9 5 an array A[u] = smallest numbered vertex Path [u,v]? 11 3 that is connected to u 12 10
question reduces to whether A[u]= A[v]? 8 can label by distances from start 4 13 11 12

2 Graph Search Application: Connected Components DFS(u) – Recursive version

initial state: all v unvisited ← for s 1 to n do Global Initialization: mark all vertices "unvisited" if state( s) ≠ “fully-explored” then BFS( s): setting A[u] ←s for each u found DFS( u) (and marking u visited/fully-explored) mark u “visited” and add u to R endif for each edge { u,v} endfor if ( v is “unvisited”) DFS( v)
Total cost: O(n+m) end for
each vertex is touched once in this outer mark u “fully-explored” procedure and the edges examined in the different BFS runs are disjoint

works also with Depth First Search

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Properties of DFS(s) Non-tree edges

Like BFS( s):
All non-tree edges join a vertex and one
DFS( s) visits x if and only if there is a path in G from s to x of its descendents/ancestors in the DFS

Edges into undiscovered vertices define a "depth tree first spanning tree" of G
Unlike the BFS tree:

the DFS spanning tree isn't minimum depth
its levels don't reflect min distance from the root
No cross edges.
non-tree edges never join vertices on the same or adjacent levels
BUT…

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No cross edges in DFS on undirected Applications of Graph Traversal: graphs Bipartiteness Testing

Claim: During DFS(x) every vertex marked visited is
Easy: A graph G is not bipartite if it contains a descendant of x in the DFS tree T an odd length cycle
Claim: For every x,y in the DFS tree T, if (x,y) is an
WLOG: G is connected edge not in T then one of x or y is an ancestor of the
Otherwise run on each component other in T
Simple idea: start coloring nodes starting at a
Proof: given node s
One of x or y is visited first, suppose WLOG that x is visited first and therefore DFS(x) was called before DFS(y)
Color s red
Color all neighbors of s blue
During DFS(x), the edge (x,y) is examined

Since (x,y) is a not an edge of T, y was visited when the
Color all their neighbors red edge (x,y) was examined during DFS(x)
If you ever hit a node that was already colored

Therefore y was visited during the call to DFS(x) so y is a
the same color as you want to color it, ignore it descendant of x.
the opposite color, output error

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3 BFS gives Bipartiteness Why does it work?

s
Run BFS assigning all vertices from

layer Li the color i mod 2 u and v have a common ancestor
i.e. red if they are in an even layer, blue if L in an odd layer i

If there is an edge joining two vertices Cycle length 2(j-i)+1 from the same layer then output “Not Bipartite”

Lj Lj u v

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DFS(v) for a directed graph DFS(v)

1 1 tree edges 2 2 10 10 3 3 forward edges 11 11 12 back edges 12 4 8 4 8 13 13 5 9 5 9 ← cross edges 6 6 NO → cross edges 7 21 7 22

Properties of Directed DFS Directed Acyclic Graphs

Before DFS(s) returns, it visits all
A directed graph G=(V,E) is acyclic if it previously unvisited vertices reachable has no directed cycles via directed paths from s

Terminology: A directed acyclic graph is

Every cycle contains a back edge in the also called a DAG DFS tree

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4 Topological Sort Directed Acyclic Graph

Given: a directed acyclic graph (DAG) G=( V,E)
Output: numbering of the vertices of G with distinct numbers from 1 to n so edges only go from lower number to higher numbered vertices

Applications
nodes represent tasks
edges represent precedence between tasks
topological sort gives a sequential schedule for solving them

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In-degree 0 vertices Topological Sort

Every DAG has a vertex of in-degree 0
Can do using DFS
Proof: By contradiction

Suppose every vertex has some incoming edge

Consider following procedure:
Alternative simpler idea:

while (true) do
Any vertex of in-degree 0 can be given v←some predecessor of v number 1 to start

Remove it from the graph and then give a
After n+1 steps where n=| V| there will be a repeated vertex vertex of in-degree 0 number 2, etc.

This yields a cycle, contradicting that it is a DAG

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Topological Sort Topological Sort

1 1 2

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5 Topological Sort Topological Sort

1 2 1 2

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Topological Sort Topological Sort

1 2 1 2

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Topological Sort Topological Sort

1 2 1 2

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6 Topological Sort Topological Sort

1 2 1 2

4 4 3 3 8 8 5 5 6 6 9 9 10 7 7

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Topological Sort Topological Sort

1 2 1 2

4 4 3 3 8 8 5 5 6 6 9 10 9 10 7 7 11 11 12

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Topological Sort Topological Sort

1 2 1 2

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7 Implementing Topological Sort

Go through all edges, computing in-degree for each vertex O(m+n)

Maintain a queue (or stack) of vertices of in-degree 0

Remove any vertex in queue and number it

When a vertex is removed, decrease in- degree of each of its neighbors by 1 and add them to the queue if their degree drops to 0

Total cost O(m+n)

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