Graph Theory in Networks

Lecture 4, 9/9/04EE 228A, Fall 2004

Rajarshi Gupta

University of California, Berkeley

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

2

Plan for Graph Segment

Lecture 3 – Tue (Sep 7, 2004) Paths and Routing Cycles and Protection Matching and Switching

Lecture 4 – Thu (Sep 9, 2004) Coloring and Capacity Trees and Broadcast, Multicast

Lecture 5 – Tue (Sep 14, 2004) Complete example: Capacity in Ad-Hoc

Networks Lectures 9 & 10

Student Presentations (have you signed up ?)

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

3

Coloring: History Map coloring: Color

all countries on a map using fewest colors

Model: Each country is a node, edge if share boundary

This forms a planar graph (edges don’t intersect)

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Graph Planarity Question in last class: How do I tell if graph is planar? Kuratowski’s Theorem: A graph is planar if and only if it has

no subgraph homeomorphic to K5 or K3,3 (See proof in Harary Thm 11.13)

Homeomorphic graphs: Can be obtained by sequence of subdivision of lines

K3,3: Complete bipartite graph with 3 nodes each Can be checked in O(n3)

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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4 & 5 Color Theorems Five color theorem:

Every planar graph is 5-colorable Heawood, 1890

Four Color Theorem: Every planar graph is 4-colorable Conjectured for many

years (since 1890) Proved by Appel and

Haken (1977)

Can do no better than 4 colors

K4 (complete graph in 4 vertices) is planar

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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5 Color Theorem (Lemma) Lemma: In a planar graph,

node w deg5 Recall Euler’s Formula

v-e+f=2 A planar graph with outside

face = polyhedron

Let d be avg node deg. So

dv=2e r be avg face deg. So rf=2e

Rewrite Euler But r3, So Replacing e=dv/2, Hence

For a finite graph, there must exist some node with degree < 6

f1 f2

f3

f4v=4

e=6

f=4

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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5 Color Theorem Proof Induction: If v<6, trivial Assume true for v-1

vertices In G, remove node v

with deg<6 Color rest of graph with

5 colors Can do this due to

induction hypothesis

v1

v5

v4v3

v2

v

Consider v Look at the 5 nodes around it If use < 4 colors, we are done

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

8 5 Color Theorem Proof (contd)

Else look at induced subgraph only with colors red and blue

If v1 and v3 not connected, interchange colors of v1-component s.t v1 is colored blue. And give red color to v.

Else: v1-v-v2-v1 topologically separates v2 from v4

Now repeat the procedure for v2 and v4

There cannot be path from v2 to v4 (with colors green and pink) since graph is planar

Hence can change color of v2

v1

v5

v4v3

v2

v

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Chromatic Number

Chromatic number Fewest number of colors required to color a

graph Few common graphs

Kn is n-colorable Bipartite graphs are 2-colorable Planar graphs are 4-colorable

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

10 (max degree)

Proof by algorithm Let the colors be 1,2,…k, where (k=1+) For node i = 1 to N

Pick the smallest color not used by any neighbor of i

Proof Suppose no such color exists for node i Neighbors of i have used up all k colors i has at least k neighbors deg(i) k > [Contradiction]

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

11 Theorem: Graph bi-colorable iff no odd cycles

‘’ Direction Proof by

contradiction Assume odd cycle

exists Color nodes on odd

cycle as R, B, R, B … Eventually last node

will have two neighbors with opposite colors

‘’ Direction Take any point v Make two sets V1 and V2

V1 = nodes odd dist from v V2 = nodes even dist frm v

Claim: Every edge joins V1 to V2

Suppose line xy joins two nodes in same set

{v…x} and {y…v} are both odd or both even

Then cycle {v…x}+{xy}+{y…v} is odd

By claim, color V1 red and V2 blue

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Independent Sets (IS) Set of nodes that do

not have an edge between them

Set of nodes that can have the same color

Maximal IS: not contained in any other IS

Result: Number of maximal IS in a graph is exponential

Graph coloring algorithms find IS

Finding the chromatic number of a graph is NP-hard

Many approximations exist

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Approximate Coloring Greedy Algorithm

Order nodes by degree Add first node to IS Discard all neighbors

into a Future bin Repeat till set empty Start with future bin

Correctness Are the sets generated

independent ? Are they maximal ?

Complexity Each iteration adds one

node, so O(m) Discarding neighbors O(n) Total: O(mn)

Followup: How far is this from optimal coloring Arbitrary or Bounded

Many approx algos give coloring within 4X, 6X of optimal

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Vertex/Node Cover (NC) Recall Lecture Graph1: Nodes NC of G such that

every edge of G has at least one node in NC Theorem: S is NC (V-S) is IS Proof ‘=>’

Suppose not Then edge with both end-points in (V-S) Then S does not cover this edge (Contradiction)

Proof ‘<=’ Suppose not. Let S’ be an IS but V-S’ is not a NC Then we have an edge with both end-points not in V-

S’ i.e. both end-points are in S’ S’ is not independent (Contradiction)

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Cliques

Observe Each node in a clique

needs a different color So Clique is complement of

IS

A

B C

E F

D

Maximal Cliques:

ABC, BCEF, CDF

Definitions Clique = Complete

Subgraph Maximal Clique = Clique

not a subset of any other Clique Number = size

of largest clique

= 4

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Perfect Graphs A graph is said to be perfect when its

Chromatic Number Clique Number Are equal For all induced subgraphs

Strong Perfect Graph Theorem Chudnovsky, Robertson, Seymour, Thomas (2002) A graph is perfect if and only if it has no odd holes or

odd anti-holes Odd-hole is a odd cycle with no chord Odd anti-hole is complement of odd-hole

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Duplication Lemma Lovasz (1972)

Take a vertex v in G Add v’ s.t. v’ is neighbor

to every neighbor of v If G is perfect, then G’ is

perfect Proof

Consider induced subgraph H’ of G’

If v’H’, then (H’)=(H’) If v’H’ but vH’, then H’

isomorphic to corres. H

If v,v’H’ Consider H=H’\{v’} Since v and v’ are not

connected, can give color of v to v’ (same neighbors)

So (H’)= (H) Also (H’)=(H)

Clique w/o v is not enlarged Clique with v is not enlarged

as v and v’ not connected But (H)=(H) as G perfect Hence (H’)=(H’) and G’ is

also perfect

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

18

Trees

Define a tree G=(V,E) |V| = |E| + 1 Connected

Terminology Height of a tree Depth of a node In-degree Out-degree Leaves

r

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Useful Trees

Types of Trees Binary Tree

Each non-leaf node has exactly 2 children

Ternary Tree Height Balanced

Tree Root chosen s.t.

height is minimized All trees are

Bipartite Graphs

Induced Trees Formed by choosing a

subset of V,E from G Rooted Tree

Rooted at particular node

Spanning Tree Covers every node

‘Subset Spanning Tree’ Covers every node in a

chosen subset

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Trees in Networks

Broadcast Tree Search Trees

BFS DFS

Spanning Tree Protocol (STP) in Bridged Networks (e.g. Ethernet)

Multicast Tree And many, many other places …

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Trees and Ethernet

Tree Feature Connected No cycles

Removing any edge makes it disconnected

Adding any edge forms a cycle

Every pair of nodes has unique path

Ethernet Behavior Need all LANs to talk Loops cause

broadcast storms Any bridge/port

failure needs recomputation

Need to disable all redundant ports

Guarantees path with STP

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Concrete Example

Weight balanced Spanning Tree (WBST) Each edge has cost

e.g. 10/100/1000 Mbps Choose spanning tree that has min difference

between min and max edge cost Want to minimize buffering requirements as

packets move through the network. Also control jitter.

Complexity

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

23 Recall Spanning Tree Algos

Kruskal’s Greedy Algorithm to compute Spanning Tree 1. Order the edges in term of weight 2. Add lowest cost edge (provided no loops) 3. Check if all vertices is connected 4. If not, return to step 2

Any search (BFS/DFS) can check for connectedness

BFS/DFS may be performed in O(m)

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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WBST : Algorithm Sort edges in increasing order of weight c1<c2<…<cm

Initialize: diff = cm-c1, low = 1, high = 1 Take Glow,high = G (V, {ci: lowihigh}) if Glow,high is connected

if diff > (chigh – clow) diff := chigh – clow Remember low, high

low := low + 1 repeat

else high := high + 1 repeat

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

25

WBST : Correctness Spanning Tree exists

By checking Glow,high is connected => spanning tree must exist

Do this using DFS. Also finds the WBST Why balanced

Keep adding low cost edges till graph connected

Then try to discard as many low cost edges Move both ‘low’ and ‘high’ pointers to

consider other ranges Remember best diff

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

26

WBST : Complexity

In each iteration, we either increment low, or increment high

So max number of iterations = 2m Each iteration needs to check

connectedness: DFS O(m) So entire algo is O(m2)

Think: How would we make this algorithm distributed

Lecture 4, 9/9/04

EE 228A, Fall 2004

Rajarshi Gupta

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Lessons from this lecture

Coloring Cliques Independent Sets Useful in capacity, scheduling

Trees Bridged Networks Used often when want to ensure unique

path etc