5. planar graphs

8/4/2019 5. Planar Graphs

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Dr.Gangaboraiah, PhD

Department of Community MedicineKempegowda Institute of Medical Sciences

Banashankari 2nd Stage, Bangalore-70

Mobile: 98451 28875

E-mail: [email protected]

Planar graphs


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IntroductionEarlier it was indicated that a graphcan be represented by more than one

figure (geometrical drawing). In somefigures which are representing graphsthe edges intersect (cross over) at

points, which are not vertices of thegraph and in some others the edgesmeet only at the vertices.


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A graph which can be represented byat least one plane drawing, in whichthe edges meet only at the vertices is

called a planar graph. A graph whichcannot be represented by a planedrawing, in which edges meet only at

the vertices is called a nonplanargraph.


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Thus, in every plane drawing of anonplanar graph at least two edges ofthe graph intersect at a point which is

not a vertex of the graph.


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Planar graphA graph G is said to be a planar graphif it can be represented by a figuredrawn on a plane (of the paper, blackboard etc.,) in such a way that no twoof its edges intersect (cross overexcept at a vertex on which both areincident).


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Nonplanar graphA graph G is said to be a nonplanargraph that cannot be represented by a

figure drawn on a plane without across over between its edges.


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EmbeddingA figure representing a graph drawnon any surface in such a way that no

two edges intersect is calledembedding. An embedding of a planargraph is called plane representation.


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For example, consider the followingthree drawings representing thecomplete graph of four vertices, viz.,

K4.


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Observe that in the first of thedrawings, two edges intersect and inthe second and third no two edges

intersect.


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Therefore, the first drawing is not anembedding of K4 while the other twoare embeddings.

Further, the second and third drawingsillustrate that K4 can be drawn on a

plane in such a way that no two of itsedges intersect. Therefore, K4 is aplanar graph.


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The second and third drawings are itsplane representations.

Two standard examples of nonplanargraphs are the complete graph K5 andthe complete bipartite graph K3, 3.


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Theorem:The complete graph of five vertices K5(Kuratowskis first graph) is nonplanar.

Proof:Recall that in the graph K5 there are fivevertices and every vertex is joined to allother four vertices by an edge so thatthe graph contains ten edges.


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Let us name the vertices as v1, v2, v3,v4, and v5 and the edges as e1, e2, ,e10 as specified below.


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Consider the edge e7=(v2,v5). Thisedge can bedrawn either inside oroutside the pentagon shown in figure

2 (slide no. 8). Consider the edgee2=(v1, v3) and e3=(v1, v4). If we drawthese edges inside the pentagon, they

will intersect e7; therefore, let us drawthem outside. Therefore, we have todraw inside the pentagon.


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If we draw the edge e9 outside thepentagon, it intersects the edge e3.Therefore, we have to draw it inside

the pentagon.Thus, both of the edges e6 and e9 areto be drawn inside the pentagon, but

then they themselves intersect.


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This demonstrates that in everypossible plane drawing of K5 at leasttwo edges of K5 intersect at a point

which is not a vertex of K5. Hence, K5is nonplanar.


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Theorem 2Kuratowskis second graph isnonplanar.

Proof:Be definition, K3, 3 is the graph in

which the vertex set is partitioned intotwo sets V1 and V2 each containingthree vertices such that every vertex in


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V1 is joined to every vertex in V2 by anedge and vice-versa.

Let us name the vertices in V1 as v1, v2,v3 and the vertices in V2 as v4, v5, v6.


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e1

=(v1

, v4

) e2

=(v1

, v5

) e3

=(v1

, v6

)e4=(v2, v4) e5=(v2, v5) e6=(v2, v6)e7=(v3, v4) e8=(v3, v5) e9=(v3, v6)


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Observe that the edgese1=(v1, v4), e4=(v4, v2), e5=(v2, v5),

e8=(v5, v3), e9=(v3, v6), e3=(v6, v1)

form a hexagonal circuit.


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Consider the edge e6=(v2, v6). Thisedge can be drawn either wholly insidethe hexagon or wholly outside it. Let

us draw it inside- the other case issimilar.

Consider the edge e2=(v

1, v

5). If we

draw this edge inside the hexagon, itintersects the edge e6.


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Therefore, let us draw it outside thehexagon.

Now, consider the edge e7=(v3, v4). Ifthis edge is drawn inside the hexagon,it crosses the edge e6 and if it is drawn

outside the hexagon, it crosses theedge e2.


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This demonstrates that in everypossible plane drawing of K3, 3 at leasttwo edges of K

3, 3

intersect at a pointwhich is not a vertex of K3, 3.

Hence K3, 3

is a nonplanar graph.


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Common properties of K5 and K3, 3By the virtue of the definitions andTheorems 1 and 2 the Kuratowskis

graphsK5 and K3, 3 share some commonproperties.

1. Both are regular graphs, where asK5 is 5-regular, K3, 3 is 3-regular.


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2. Both are nonplanar.3. Removal of one edge makes each aplanar graph .

(In theorem 1, the drawing would havecontained only non intersecting edgesif e9 was not there. In theorem 2, the

drawing would have contained onlynonintersecting edges if e7 was notthere.


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4. Removal of one vertex makes each aplanar graph (Note that K4 and K2, 3are planar.

5. K5 is the nonplanar graph with thesmallest number of vertices, and

K3,3 is the nonplanar graph withthe smallest number of edges.


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Homeomorphic graphsTwo graphs G1 and G2 are said to behomeomorphic if one of these can be obtainedfrom the other by insertion of new vertices of

degree two into its edges or by the merger ofedges in series.


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Observe that the second graph isobtained from the first graph byinserting the vertex v which of degree

2 into an edge. Hence, the two graphsare homeomorphic.

The following theorem known as theKuratowskis theorem is fundamental

in the study of planar graphs.


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31
http://en.wikipedia.org/wiki/File:Graph_homeomorphism_example_3.svghttp://en.wikipedia.org/wiki/File:Graph_homeomorphism_example_2.svghttp://en.wikipedia.org/wiki/File:Graph_homeomorphism_example_1.svghttp://en.wikipedia.org/wiki/File:Graph_subdivision_step1.svghttp://en.wikipedia.org/wiki/File:Graph_subdivision_step2.svghttp://en.wikipedia.org/wiki/File:Graph_subdivision_step2.svghttp://en.wikipedia.org/wiki/File:Graph_subdivision_step1.svg


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Theorem 3A necessary and sufficient conditionfor a graph G to be a planar is that G

does not contain either K5 or K3, 3 assubgraphs or any subgraphhomeomorphic to either of these.

Note: If a graph G contains K5 or K3, 3 as a subgraph, thenin view of Kuratowskis theorem we may infer that G is

nonplanar.


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Example 1Show that(i) The graph of order 5 and size 8

(ii) The graph of order 6 and size 12 areplanar

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SolutionTo show that a graph is planar, it isenough if we draw one plane diagram

representing the graph in which notwo edges cross each other. Thefigures in slide 34 shows that they are

planar.

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Example 2:If n4, show that thecomplete graph Kn is planar.

Solution: The diagrams represent

graphs K1, K2, K3, and K4.

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Solution: The diagrams representgraphs K1, K2, K3, and K4. In none ofthese diagrams, no two edges cross

each other. Therefore, K1, K2, K3, andK4 are all planar graphs. Since K1, K2,K3, andK4are planar graphs and K5 isa nonplanar graph, it may be infer thatK5 is the complete nonplanar graph ofthe smallest order.

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Example 3Show that the graphs K2, 2 and K2, 3 areplanar graphs.

SolutionIn K2, 2 the vertices set is partitioned

into two sets of vertices say {v1, v2}and {v3, v4}.

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SolutionIn K2, 2 the vertices set is partitionedinto two sets of vertices say {v1, v2}

and {v3, v4}.

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respectively and there is an edgejoining every vertex in V1 with vertex V2

and vice-versa. The first figure in slide

no. 36 represents this graph. Evidently,

in this graph, no two edges cross each

other. Therefore, K2, 2 is planar.

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In K2,3, the vertex set is partitioned intotwo sets V1, andV2, withV2, containingtwo vertices v1, v2, and V2 containingthree vertices, say, v3, v4, v5 and there

is an edge joining every vertex in Vwith every vertex in V2 and vice-versa.The figure 2 in slide no. 39 represents

this graph. Observe that in this figure,no two edges cross each other.Therefore, K2,3 is planar.

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Note: The graph K2, 3 can be obtained by removingone vertex from K3, 3. Thus, the nonplanar graphK3, 3 becomes planar if one vertex is removed fromit.

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Example 4 (Three utility problem)Suppose there are three houses andthree utility points (electricity, water,

sewage, say) which are such that eachutility joint is joined to each house.Can the lines of joining be such that

no two lines cross each other?

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Since each house is joined to each utility

point, the graph has to be K3, 3. This graphis nonplanar and therefore at least two ofits edges cross each other. As such, it is

not possible to have the lines joining thehouses and the utility points such that notwo line cross each other.

Example 5Verify that the following graphs arehomeomorphic but not isomorphic.

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SolutionEach graph can be obtained from the otherby adding or removing appropriate vertices.Therefore, they are homeomorphic. They

are not isomorphic is evident if we observethe incident relationship which is notidentical.

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Eulers formulaIf G is a planar graph, then G can berepresented by a diagram in a plane inwhich no two edges intersect (cross over).

Such a diagram divides the plane into anumber of parts called regions (or meshes,or windows or faces) of which exactly one

(part) is unbounded. The number of edgeson the boundary of a region (face) is called

the degree of that region*.

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*If a pendant edge is a part of theboundary of a region, by conventionthis edge is counted twice while

determining the degree of the region.

For example, in the diagram of a

planar graph shown in the abovefigure, the diagram divides the planeinto 6 regions R1, R2, R3, R4, R5, and R6.

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Observe that whereas each of theregions R1 to R5 is bounded, the region

R6 is unbounded. That is, R1 to R5 arein the interior of the graph while R6 isin the exterior.

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Further observe that, in the abovefigure, the boundary of the region R1 ismade up of two edges. Therefore, the

degree of R1 is 2 and we write d(R1)=2.

The boundary of each of the regions R2

and R4 is made up of 3 edges,therefore, d(R2)=d(R4)=3.

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The boundary of the regions R3 consistsof 4 edges of which one is a pendantedge. Therefore, d(R3)=5.

The region R5 is bounded by a singleedge (self-loop), therefore, d(R5)=1.

The boundary of the exterior region R6

consists of six edges; therefore, d(R6)=6.

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Note that,d(R1)+d(R2)+d(R3)+d(R4)+d(R5)+d(R6)=20

which is twice the number of edges inthe graph. This property is analogousto the handshaking property and is

true for all planar graphs.

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In the figure, two trivial interestingcases are depicted. The first of thesefigures corresponds to the null graph,

N1, with one vertex and no edges.

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Evidently, the diagram determines onlyone region R-the unbounded region.The second figure corresponds to atree. In this case also, the diagramdetermines only one region R

-the

unbounded region.

It should be pointed out that the regionsare determined by a diagram of a planargraph and not by the graph itself.

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This means that if we change the diagramof the graph, the regions determined by theold one in the sense that the unboundedregion in the old. However, the interesting

fact is that the total number of regions inthe two diagrams remains the same.

Theorem 1: A connected planar graph Gwith n vertices and m edges has exactlym-n+2 regions in every one of its diagrams.

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ProofLet r denote the number of regions in adiagram of G. The theorem states thatr = m n + 2 or nm + r = 2 (1)

The proof is by induction.If m = 0, then n must be equal to 1.Because, if n>1, then G will have at least

two vertices and there must be an edgeconnecting them (because G is connected)so that m 0.

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If n = 1, a diagram of G will have only one

region in it- the entire plane region (asshown in figure one of slide no. 39). Thus,if m = 0, we have n = 1 and r = 1 so that

n m + r = 2. This verifies the theorem form = 0.

Now, assume that the theorem holds for

all graphs with m = k number of edges,where k is a specified non-negativeinteger.

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Consider a graph Gk+1

with k+1 edges andn vertices. If Gk+1 is a tree, then Gk+1 willhave n-1 edges and a diagram of Gk+1 willdetermine only one region-the entire plane

region (as shown in figure 2 in slide no.39). Thus, for Gk+1, we have k + 1 = n 1and r = 1, so that

n (k+1) + r = 2.This means that the result (1) is true whenm = k + 1 as well if the graph is a tree.

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Suppose Gk+1

is not tree and let r be thenumber of regions which a diagram ofGk+1 has. Since Gk+1 is not a tree, itcontains at least one circuit. Consider an

edge e in a circuit and remove it from Gk+1.The resulting graph, Gk+1- e, will have nvertices and k edges and a diagram of the

graph will have r-1 regions. Since Gk+1- ehas k edges, the theorem holds for thisgraph (by induction assumption made).

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That is, we haver-1 = k n + 2, or n- (k + 1 ) + r = 2.

This means that the result (1) is true when

m = k + 1 as well it is true for m = k and thegraph is not a tree. Thus, the result (1) if itis true for m = k + 1, it is true for m = k 0,

whether the graph is a tree or not. Hence,by induction, it follows that the result (1) istrue for all non-negative integers m.

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RemarksThe formula (1) which gives the number ofregions (which a diagram of a connectedplanar graph has) in terms of the number

of vertices and the number of edges isknown as the Eulers formula.

Corollary 1If G is a connected simple planar graphwith n ( 3) vertices, m (>2) edges

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and r regions, then(i) m (3/2) r and

(ii) m 3n-6

Further, if G is a triangle-free, then(iii) m 2n-4.

Proof

Since the graph G is simple, it has noparallel edges and no self-loops. As such,every region must be bounded

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by three or more edges. Therefore, thetotal number of edges that bound allregions is greater than or equal to 3r. Onthe other hand, an edge is in theboundary of at most two regions.Therefore, the total number of edges thatbound all regions is less than or equal to

2m. Thus, 3r 2m or m (3/2) r. This isrequired result (i).

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Now, substituting for r from the Eulers

formula in the result just proved, we getm (3/2) (m n + 2)

Which simplifies to m 3n 6. This is the

required result (ii).

If G is triangle-free, every region must bebounded by four or more edges, so thatthe total number of edges that bound all isgreater than or equal to 4r. Consequently,we have 4r 2m.

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Substituting for r from the Eulers formulain this, we get m 2 (m n + 2) orm 2n-4. This is the required result (iii).

Corollary 2Kuratowskis first graph, K5, is nonplanar.Proof

The graph K5 is simple, connected and hasn=5 vertices and m=10 edges.

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If this graph is planar, then from the result(ii) of Corollary 1, we should have m 3n-6,i.e., 10 15-6, which is not true. Therefore,K5 is nonplanar.

Corollary 3Kuratowskis second graph K3,3 is nonplanarProof:Note that K3,3 is simple, connected and hasn=6 vertices and m=9 edges.

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Suppose K3, 3 is planar. By examiningthe K3, 3 figure we note that K3, 3 has notriangles. (In fact no bipartite graph has

a circuit of odd length). Therefore, byresult (iii) of Corollary 1 we shouldhave m 2n- 4, i.e., 9 12- 4, which is

not true. Hence K3, 3 is nonplanar.

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Corollary 4Every connected simple planar graph Gcontains a vertex of degree at most 5.

ProofSuppose each vertex of G is of degreegreater than or equal to 6. Then, if d1, d2,

, dn are the degrees of the n vertices ofG, we have d1 6, d2 6, d3 6,, dn 6.

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Adding these, we get

d1+d2++dn 6n.

By handshaking property, the lefthand side of this inequality is equal

to 2m where m is the number of

edges in G. Thus, 2m 6n or 3n m.

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By result (ii) of Corollary 1, we shouldhave m 3n 6. Thus, 3n 3n 6.

This cannot be true. Therefore, G

must have a vertex of degree at most

5.

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ExampleVerify Eulers formula for the graph

shown in slide no. 45

Solution:Here n=6, m=10 and r=6Therefore,

n m + r = 6 10 + 6 = 2Here the Eulers formula is verified.

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ExampleWhat is the maximum number ofedges possible in a simple connected

planar graph with eight vertices?

Solution:

m 3n-6 = 18 ( Here n=8).Thus the maximum number of edgespossible is 18.

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ExampleWhat is the minimum number of verticesnecessary for a simple connected graphwith 11 edges to be a planar?

SolutionFor a simple connected planar (n, m)graph, we should have by Corollary 1

m 3n 6 or n (m +6)/3When m = 11, we get n 17/3. Thus, the

required minimum number of vertices is 6.

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ExampleShow that the condition m 3n 6 is not asufficient condition for a connected simplegraph with n vertices and m edges is not aplanar.

Solution:Consider the graph K3,3 which is simple andconnected and which has n=6 vertices andm=9 edges. We check that for this graph,m 3n-6. But the graph is nonplanar. Thiscounter example proves the required result.

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Dual graph:A dual graph of a given planar graphGis agraph which has a vertex for each planeregion of G, and an edge for each edge

in Gjoining two neighboring regions, for acertain embedding of G. The term "dual" isused because this property is symmetric,

meaning that if His a dual of G, then Gis adual of H(if Gis connected).

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http://en.wikipedia.org/wiki/Planar_graphhttp://en.wikipedia.org/wiki/Graph_embeddinghttp://en.wikipedia.org/wiki/Duality_(mathematics)http://en.wikipedia.org/wiki/Symmetric_functionhttp://en.wikipedia.org/wiki/Symmetric_functionhttp://en.wikipedia.org/wiki/Duality_(mathematics)http://en.wikipedia.org/wiki/Graph_embeddinghttp://en.wikipedia.org/wiki/Planar_graph


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http://en.wikipedia.org/wiki/File:Dualgraphs.png


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PropertiesThe dual of a planar graph is a

planar multigraph - a graph that may

have loops and multiple edges. If G is a connected graph and if G is

a dual of Gthen Gis a dual of G.

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http://en.wikipedia.org/wiki/Planar_graphhttp://en.wikipedia.org/wiki/Multigraphhttp://en.wikipedia.org/wiki/Multigraphhttp://en.wikipedia.org/wiki/Planar_graph


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Example

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Example: Prove that the Petersen


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Example: Prove that the Petersengraph is a non-planar graph.

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Solution: Recall that the Petersengraph is a 3-regular graph of order 10and size 15. The graph is repeated in

figure 2 with vertices as A, B, C etc.Now consider the graph shown infigure 2. Verify that this graph is

another representation of the Petersengraph.

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Suppose we delete the edges {C, D} &{Q, T} along with their end verticesfrom this graph. The resulting graph

is K3, 3. Thus the Petersen graphcontains K3, 3 as a subgraph.Therefore, the Petersen graph is non-

planar (by Kurotowskis theorem).

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Detection of planarity:Given a graph G, the determination ofits planarity or otherwise is an

important problem. This problem canbe tackled by employing what isknown as elementary reduction. The

steps involved in this reduction are asfollows:

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Step 1:Given a graph G, determine the setA = {G1, G2, G3, , Gk}

Where G1, G2, G3, , Gk are subgraphsof G, every pair of which has exactlyone vertex in common (such graphs

are called blocks)

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Step 2:Remove all loops from all of Gis.Step 3:Remove all but one edge between every

pair of vertices joined by multiple edges(if any).Step 4:

Eliminate all vertices of degree 2 bymerging the edges incident on thesevertices.

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Step 5:Repeat step 3 and 4 repeatedly untileach block Gi is reduced to a new

graph Hi, which will be one of thefollowing:(1) A graph with a single edge.

(2) A complete graph of order four.

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If Hi is the first or the second of theabove possible forms, we concludethat Hi is a planar graph.

Consequently, each Gi with which westarted is planar and therefore G is

planar.

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Example:Carry out the elementary reductionprocess for the following graph

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Step 1:The given graph is a single block,hence, contains only G

Step 2:Remove the loops e9

Step 3:

Remove one of the two parallel edgese1,e8, say e8.

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The resulting graph left out after thefirst three steps is

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Step 4:Eliminate the vertices of degree 2 bymerging the edges incident on these

vertices. Thus, we merge(i) The edges e1 and e2 into an edge,say e10

(ii) The edges e6 and e7 into an edge,say e11

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The resulting graph is

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As per step 3, let us remove one of theparallel edges e5 and e10 and one ofthe parallel edges e3 and e11. Theresulting graph is

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As per step 4, let us merge the edgese3 and e4 into an edge e12 to get thefollowing graph

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As per step 3, let us remove one oftwo parallel edges e5 and e12, say theedge e12 and thus, we get the graph

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This graph is the final graph obtainedby the process of elementaryreduction applied to the graph given in

slide no. 86.

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Example: Check the planarity (or


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otherwise) of the following graph

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Step 1: Splitting G into blocks


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p p g

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Step 2: Removing loops and


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p g peliminating multiple edges

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Step 3: Merging the edges incident


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on vertices of degree 2

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Step 4: Eliminating the parallel


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edges

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The reduction is now complete. Thefinal reduced graph in slide no.88 hasthree block of which the first and thethird (which are single edges) areobviously planar. The second one isevidently the complete graph K5,

which is non-planar. Thus, the givengraph contains K5 as a subgraph andis therefore non-planar.

5. planar graphs

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