planar graphs graph colouringsisabela.dramnesc/gtc/lecture14... · 2018. 1. 9. · west university...

Planar graphsGraph colourings

Lecture 13Planar graphs.

Graph colourings. Chromatic polynomials

Isabela Dramnesc UVT

Computer Science Department,West University of Timisoara,

Romania

January 2018

Isabela Dramnesc UVT Graph Theory and Combinatorics – Lecture 13 1 / 31


Planar graphsDefinition and examples

A graph G is planar if it can be drawn in the plane such that pairsof edges intersect only at vertices, if at all. Such a drawing iscalled planar representation of G .

Example (Planar graphs)

. . . . ...

.. . .

..

. .

.

.

.

.

.



Auxiliary notions

Region of a planar representation of a graph G : maximal section of theplane in which any two points can be joined by a curve that does notintersect any part of G .

Example

. . .

.

.

.

. .

.

. .

determines 7 regions

R1R7

R2 R3R4

R5 R6

R7 is the outer region

Every region is delimited by edges.

An edge is in contact with one or two regions.

An edge borders a region R if it is in contact with R and withanother region.



Auxiliary notions

Region of a planar representation of a graph G : maximal section of theplane in which any two points can be joined by a curve that does notintersect any part of G .

Example

. . .

.

.

.

. .

.

. .

determines 7 regionsR1R7

R2 R3R4

R5 R6

R7 is the outer region

Every region is delimited by edges.

An edge is in contact with one or two regions.

An edge borders a region R if it is in contact with R and withanother region.



Regions and bound degrees

.

.

.

.

..

. .

e1

e2

e3

e4

e5

e 6e 7

e8

e9

S1

S2 S3

e1 is in contact only with S1

e2 and e3 are in contact only with S2

S1 is bordered by e7, e8, e9

S3 is bordered by e4, e5, e6

S2 is bordered by e4, e5, e6, e7, e8, e9

The bound degree b(S) of a region S is the number of edges thatborder S .

b(S1) = 3, b(S2) = 6, b(S3) = 3



Properties

Let G be a connected graph with n nodes, q edges, and a planarrepresentation of G with r regions.

n − q + r = 2 in all cases.



Properties of connected planar graphs

Theorem (Euler’s Formula)

If G is a connected planar graph with n nodes, q edges and rregions then n − q + r = 2.

Proof: Induction on q.Case 1: q = 0⇒ G = K1 and n = 1, q = 0, r = 1, thusn − q + r = 2.Case 2: G is a tree ⇒ q = n − 1 and r = 1, thusn − q + r = n − (n − 1) + 1 = 2.Case 3: G is a connected graph with a cycle. Let e be an edge ofthat cycle, and G ′ = G − e.G ′ is connected with n nodes, q − 1 edges, and r − 1 regions ⇒ byInduction Hypothesis: n − (q − 1) + (r − 1) = 2.Thus n − q + r = 2 holds in this case too.



Consequences of Euler’s Formula

Corollary 1

K3,3 is not planar.

Proof: K3,3 has n = 6 and q = 9 ⇒ if it were planar, it would

have r = q− n+ 2 = 5 regions Ri (1 ≤ i ≤ 5). Let C =5∑

i=1

b(Ri ).

Every edge is in contact with at most 2 regions⇒ C ≤ 2 q = 18.

K3,3 is bipartite ⇒ C3 is no subgraph of K3,3, thus b(Si ) ≥ 4for all i , therefore C ≥ 4 · 5 = 20

⇒ contradiction, thus K3,3 can not be planar.




Corollary 2

If G is a planar graph with n ≥ 3 nodes and q edges then q ≤ 3 n − 6.Moreover, if q = 3 n − 6 then b(S) = 3 for every region S of G .

Proof. Let R1, . . . ,Rr be the regions of G and C =r∑

i=1

b(Ri ). We

know that C ≤ 2 q and C ≥ 3r (because b(Ri ) ≥ 3 for all i). Therefore3 r ≤ 2 q ⇒ 3 (2 + q − n) ≤ 2 q ⇒ q ≤ 3n − 6.If the equality holds, then3 r = 2 q ⇒ C =

∑ri=1 b(Ri ) = 3 r ⇒ b(Ri ) = 3 for all regions Ri .

Corollary 3

K5 is not planar.

Proof: K5 has n = 5 nodes and q = 10 edges ⇒ 3 n − 6 = 9 < 10 = q

⇒ K5 can not be planar (Cf. Corollary 2).




Corollary 4

δ(G ) ≤ 5 for every planar graph G .

Proof: Suppose G has n nodes and q edges.Case 1: n ≤ 6⇒ every node has degree ≤ 5 ⇒ δ(G ) ≤ 5.

Case 2: n > 6. Let D =∑v∈V

deg(v). Then

D = 2 q (obvious)

≤ 2 (3 n − 6) (by Corollary 2)

= 6 n − 12.

If δ(G ) ≥ 6 then D =∑v∈V

deg(v) ≥∑v∈V

6 = 6 n, contradiction.

Thus δ(G ) ≤ 5 holds.



Subdivisions

Let G = (V ,E ) be an undirected graph, and (x , y) an edge.

A subdivision of (x , y) in G is a replacement of the edge(x , y) in G with a path from x to y through some newintermediate points.

A graph H is a subdivision of a graph G if H can be producedfrom G through a finite sequence of edge divisions.

Example

G :.

. .

.

. H :.

. .

.

...

.

.



Criteria to detect planar graphs

We say that a graph G contains a graph H if H can be producedby removing edges and nodes from G .

Remark

If H is a subgraph of G then G contains H. The converse is false:“G contains H” does not imply “H is a subgraph of G”.

H is a subgraph of G iff it can be produced from G by noderemovals.

Theorem (Kuratowski’s Theorem)

G is planar if and only if it contains no subdivisions of K3,3 and ofK5.



Kuratowski’s TheoremIllustrated examples

Apply Kuratowski’s Theorem to decide which of the followinggraphs are planar or not:

1. 1

7

23

4

5 6

No, because it contains a subdivision of K3,3:

1 7

3

5

2

4

6

2. 2

18

4

5

67

3

No, because it contains a subdivision of K3,3:

2

1

8

4

5

6

3

7





3. 1

23

4

5

67

8

No, because it contains a subdivision of K5:

2

35

6 4 87

1

4.

a1

a2

a3 a4

a5

1

2

3 4

5

No, because it contains a subdivision of K3,3:a4

1

a3

5

a1

2

4

3

a5 a2





3. 1

23

4

5

67

8

No, because it contains a subdivision of K5:

2

35

6 4 87

1

4.

a1

a2

a3 a4

a5

1

2

3 4

5No, because it contains a subdivision of K3,3:a4

1

a3

5

a1

2

4

3

a5 a2



Motivating problem

Alan, Bob, Carl, Dan, Elvis, Ford, Greg and John are senatorswhich comprise 7 committees:

C1 = {Alan, Bob, Carl}, C2 = {Carl, Dan, Elvis},C3 = {Dan,Ford}, C4 = {Adam, Greg}, C5 = {Elvis, John},C6 = {Elvis,Bob,Greg}, C7 = {John, Carl, Ford}.

Every committee must fix a meeting time. Since each senator mustbe present at each of his or her committee meetings, the meetingtimes need to be scheduled carefully.

Question: What is the minimum number of meeting times?



Answer to the motivating problem

Remarks:

Two committees Ci and Cj can not meet simultaneously ifand only if they have a common member (i.e., Ci ∩ Cj = ∅).

⇒ we can consider the undirected graph G with

nodes = the committees C1,C2,C3,C4,C5,C6,C7

edges (Ci ,Cj) if Ci and Cj share a member (i.e., Ci ∩ Cj = ∅)We color every node Ci with a colour representing its meetingtime Ci

⇒ the problem is reduced to: what is the minimum number ofcolours that can be assigned to the nodes of G , such that noedge has endpoints with the same colours?



Answer to the motivating problem

G : C1

C2C3

C4

C5

C6C7

C1

C2C3

C4

C5

C6C7

K(C1) = K(C3) = K(C5) = 1

K(C2) = K(C4) = 2

K(C6) = K(C7) = 3

⇒ the minimum number is 3.

(we need 3 colours)

Definition (node colouring, chromatic number)

A k-colouring of the nodes of a graph G = (V ,E ) is a mapK : V → {1, . . . , k} such that K (u) 6= K (v) if (u, v) ∈ E .The chromatic number χ(G ) of a graph G is the minimum valueof k ∈ N for which there exists a k-colouring of G .



Chromatic polynomials

The computation of χ(G ) is a hard problem (NP-complete).

Birkhoff (≈ 1900) found a method to compute a polynomialcG (z) for any graph G , called the chromatic polynomial of G ,such that

cG (k) = the number of k-colourings of the nodes of G

⇒ χ(G ) = niminum value of k for which cG (k) > 0.

We will present

1 simple formulas of cG (z) for some special graphs G .

2 two recursive algorithms for the computation of cG (z) for anygraph G .



Chromatic polynomials for special graphs

1 The empty graph En: v1 v2 . . . vn

every node can be coloured with any of the z available colours:⇒ cEn(z) = zn and χ(En) = 1

2 Tree Tn with n nodes:z alternatives to colour the root nodeany other node can be coloured with any colour different fromthe colour of the parent node ⇒ z − 1 alternatives to colour it

⇒ cTn(z) = z · (z − 1)n−1 and χ(Tn) =

{1 if n = 1,2 if n > 1.

3 Special case: the graph Pn (path with n nodes) is a special

tree with n nodes: v1 v2 . . . vn

⇒ cPn(z) = z · (z − 1)n−1 and χ(Pn) =

{1 if n = 1,2 if n > 1.

4 Complete graph Kn:cKn(z) = z · (z − 1) · . . . · (z − n + 1) and χ(Kn) = n.



The computation of chromatic polynomialsSpecial operations on graphs

LEt G = (V ,E ) be an undirected graph, and e = (x , y) an edgefrom E

I G − e is the graph produced from G by removing e

I G/e is the graph produced G as follows:Collapse x and y into one node, whose neighbours are theprevious neighbours of x and y .

Example

G :

• •

•••• •

w za

b c

x y

G − (b, c):

G/(b, c):

• •

•••• •w za

b c

x y

• •

••• •w za

b&c

x y



The computation of chromatic polynomialsRecursive formulas

Note, that, for every e ∈ E : cG (z) = cG−e(z)− cG/e(z)⇒ two algorithms for the recursive computation of the chromaticpolynomial:

1 Reduce G by eliminating edges e ∈ E one by one:

cG (z) = cG−e(z)− cG/e(z)

until we reach special polynomials En or Tn:

Base cases: cEn(z) = zn and cTn(z) = z · (z − 1)n−1

2 Extend G by adding edges that are missing from G :

cG (z) = cG (z) + cG/e(z)

where e is an edge missing from G , and G = G + e

Base case: cKn(z) = z · (z − 1) · . . . · (z − n + 1)



The computation of the chromatic polynomial by reductionIllustrated example

G : cG (z) = cG1(z)− cG2

(z), where

a b

cd

e

G1 = G − (a, b):

a b

cd

eG2 = G/(a, b):

a&b

cd

e

cG1(z) = cG11

(z)− cG12(z) and cG2

(z) = cG21(z)− cG22

(z), where

G11:

a b

cd

eG12:

a b&c

de

G21:

a&b

cd

eG22:

a&b&c

de

The following graphs are isomorphic: G12 ≡ G21 and G22 = K3,thus:

cG (z) = cG11(z)− 2 · cG12(z) + z(z − 1)(z − 2)︸︷︷︸cK3

(z)



The computation of the chromatic polynomial by reductionIllustrated example (continued)

cG (z) = cG11(z)− 2 · cG12(z) + z(z − 1)(z − 2)

G11:

a b

cd

eG12:

a b&c

de

Note that

cG11(z) = cT5(z)− cT4(z) = z(z − 1)4 − z(z − 1)3

cG12(z) = cT4(z)− cT3(z) = z(z − 1)3 − z(z − 12)

⇒ cG (z) =z(z − 1)4 − z(z − 1)3 − 2(z(z − 1)3 − z(z − 1)2)

+ z(z − 1)(z − 2)

= z5 − 7 z4 + 18 z3 − 20 z2 + 8 z



The computation of the chromatic polynomial by extensionIllustrated example

G : cG (z) = cG1(z) + cG2

(z), where

a b

cd

e

G1 :

a b

cd

eG2 :

a b

c&ed

cG2(z) = z(z − 1)(z − 2)(z − 3) because G2 ≡ K4, and

cG1(z) = cG11

(z) + cG12(z) where G11 :

a b

cd

eG12 :

a b&e

cd

cG11(z) = cG111

(z) + cG112(z)

= cK5(z) + cK4

(z) whereG111 :

a b

cd

e

G111 ≡ K5

G112 :

b

a&cd

e

G112 ≡ K4



The computation of the chromatic polynomial by extensionIllustrated example (continued)

cG (z) = cG1 (z) + cG2 (z) = (cG11 (z) + cG12 (z)) + cK4 (z)

= cK5 (z) + cK4 (z) + cG12 (z) + cK4 (z)

where G12 :

a b&e

cd

cG12(z) = cG121

(z) + cG122(z) = cK4

(z) + cK3(z) where

where G121 :

a b&e

cd

G121 ≡ K4

where G122 :

a&c b&e

d

G122 ≡ K3

⇒ cG (z) = cK5(z) + 3cK4

(z) + cK3(z) = z5 − 7 z4 + 18 z3 − 20 z2 + 8 z



Properties of the chromatic polynomial

If G = (V ,E ) is an undeirected graph with n nodes and q edgesthen the chromatic polynomial cG (z) satisfies the followingconditions:

I It has degree n.

I The coefficient of zn is 1.

I Its coefficients have alternating signs.

I The constant term is 0.

I The coefficient of zn−1 is −q.

Example

G :n = 5, q = 7

cG (z) = z5 − 7 z4 + 18 z3 − 20 z2 + 8 z

a b

cd

e



Remarkable resultsMaps and planar graphs

Every country from a planar map is represented by a node (apoint inside it)Two nodes get connected if and only if their respectivecountries share a nontrivial border (mode than just a dot).

⇒ undirected graph GH corresponding to a map H. For example:

Remark: H is a planar map if and only if GH is a planargraph.



Remarkable results4-colourings of a map

The countries of a planar map H can be coloured with 4colours, such that no two neighbouring countries have the

same colour.

Remarks

1 This is one of the most famous problems from Graph Theory

Extremely long, tedious, and complex proofThe problem was proposed in 1858; first proof was given in1976 (Appel & Haken)The problem is equivalent with the statement that the planargraph GH is 4-colorable.

2 This theorem is equivalent with the statement:

χ(G ) ≤ 4 for every planar graph G .



Remarkable results4-colourings of a map

The countries of a planar map H can be coloured with 4colours, such that no two neighbouring countries have the

same colour.

Remarks1 This is one of the most famous problems from Graph Theory

Extremely long, tedious, and complex proofThe problem was proposed in 1858; first proof was given in1976 (Appel & Haken)The problem is equivalent with the statement that the planargraph GH is 4-colorable.

2 This theorem is equivalent with the statement:

χ(G ) ≤ 4 for every planar graph G .



5-colourings of planar maps

The countries of a planar map H can be coloured with 5colours, such that no two neighbouring countries have thesame colour. or, equivqlently: χ(G ) ≤ 5 for every planar graph G .

Proof: Induction on n = the number of nodes of G .The statement is obvious for n ≥ 5, thus we assume n ≥ 6.δ(G ) ≤ 5 by Corollary 4, thus G has a node v with deg(v) ≤ 5.Let G ′ be the graph produced by removing v from G ⇒ G ′ hasn − 1 nodes, thus χ(G ′) ≤ 5 by Inductive Hypothesis. Therefore,we can assume G ′ has a 5-colouring with colours 1,2,3,4,5.Case 1: deg(G ) = d ≤ 4. Let v1, . . . , vd be the neighbours of v ,with colours c1, . . . , cd .

v

v1

v2

vdc1

c2

cdfor v we can choose any colourc ∈ {1, 2, 3, 4, 5} − {c1, . . . , cd}⇒ G is 5-colourable.



5-colourings of planar mapsProof (continued)

Case 2: deg(v) = 5, thus v has 5 neighbours v1, v2, v3, v4, v5,which we assume to be coloured with c1, c2, c3, c4, c5, respectively.

1 If {c1, c2, c3, c4, c5} 6= {1, 2, 3, 4, 5}, we can colour v with anycolour c ∈ {1, 2, 3, 4, 5} − {c1, c2, c3, c4, c5} ⇒ G is5-colourable.

2 If {c1, c2, c3, c4, c5} = {1, 2, 3, 4, 5}, we can assumec1 = 1, c2 = 2, c3, c4 = 4, c5 = 5.

v

v1

v2v3

v4 v5

1

2

3

4 5

Main idea: We will rearrange the colours of G ′ in order to makepossible a colouring of v .




v

v1

v2v3

v4 v5

1

2

3

4 5

We consider all nodes of G ′ which are coloured with 1 (red) and 3 (green).Case 2.1. G ′ has no path from v1 to v3 coloured only with 1 and 3.Let H be the subgraph of G ′ made of all paths starting from v1 which are colouredonly with 1 (red) and 3 (green).

v

..

. .

. ..

. ..

.

colour inter-

change in H

v

..

. .

. ..

. ..

.

V [v3] ∩ V (H) = ∅, that is, neither v3 nor any of its neighbours is a node of H.We can interchange colours 1 and 3 in H, and afterwards assign colour 1 (red)to v ⇒ G is 5-colourable.




v

v1

v2v3

v4 v5

1

2

3

4 5


v

..

. .

. ..

. ..

.

colour inter-

change in H

v

..

. .

. ..

. ..

.

V [v3] ∩ V (H) = ∅, that is, neither v3 nor any of its neighbours is a node of H.

We can interchange colours 1 and 3 in H, and afterwards assign colour 1 (red)to v ⇒ G is 5-colourable.




v

v1

v2v3

v4 v5

1

2

3

4 5


v

..

. .

. ..

. ..

.

colour inter-

change in H

v

..

. .

. ..

. ..

.





v

v1

v2v3

v4 v5

1

2

3

4 5


v

..

. .

. ..

. ..

.colour inter-

change in H

v

..

. .

. ..

. ..

.





v

v1

v2v3

v4 v5

12

3

4 5

Case 2.2. G ′ has a path from v1 to v3 coloured only with colours 1 and 3 ⇒ we arein one of the following two situations:

v

v2v3

v4 v5

v1

orv

v2v3

v4 v5

v1

In both cases, there can be no path from v2 to v4 coloured only with 2 and 4 ⇒ case

2.1 is applicable to nodes v2 and v4 ⇒ G is 5-colourable in this case too.


planar graphs graph colouringsisabela.dramnesc/gtc/lecture14... · 2018. 1. 9. · west university...

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