PLANAR GRAPHS. Plane graph (or embedded graph) A graph that is drawn on the plane without edge crossing, is called a Plane graph

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1 PLANAR GRAPHS Basc defntons Isomorphc graphs Two graphs G(V,E) and G2(V2,E2) are somorphc f there s a one-to-one correspondence F of ther vertces such that the followng holds: - u,v V, uv E, => F(u)F(v) E2 - x,y V, xy E => F(x)F(y) E2 Plane graph (or embedded graph) A graph that s drawn on the plane wthout edge crossng, s called a Plane graph Planar graph A graph s called Planar, f t s somorphc wth a Plane graph Phases A planar representaton of a graph dvdes the plane n to a number of connected regons, called faces, each bounded by edges of the graph. For every graph G, we denote n(g) the number of vertces, e(g) the number of edges, f(g) the number of faces. Degree We defne the degree of a face d(f), to be the number of edges boundng the face f. s The followng graphs are somorphc to 4 (the complete graph wth 4 vertces) f f2 G2 G f3 f4 f' f3' f2' f4' G3 Fgure : 3 somorphc graphs G2 and G3 are plane graphs, G s not. G s planar, as t s somorphc to G2 and G3. G2 and G3 are called planar embeddngs of K4

2 f, f2 and f3 (exteror phase) are phases of G2, wth degree d(f)=d(f2)=d(f3)=3. Theorem A graph s embeddable n the sphere f and only f t s embeddable n the plane. Proof. We show ths by usng a mappng known as stereographc projecton. Consder a sphercal surface S touchng a plane P at the pont SP (called south pole). The pont NP (called the pont of projecton or north pole) s on S and dametrcally opposte SP. Any pont z on P can be projected unquely onto S at z' by makng NP, z and z' collnear. In ths way any graph embedded n P can be projected onto S. Conversely, we can project any graph embedded n s onto P, choosng NP so as not to le an any vertex or edge of the graph. fgure 2: graph embeddable n sphere Theorem 2 A planar embeddng G' of a graph G can be transformed n to another embeddng such that any specfed face becomes the exteror face. Proof. Any face of G' s defned by the path whch forms ts boundary. Any such path, T, dentfed n a partcular planar representaton P of G, may be made to defne the exteror face of a dfferent planar representaton P' as follows. We form a sphercal embeddng P'' of P. P' s then formed by projectng P'' onto the plane n such a way that the pont of projecton les n the face defned by the mage of T on the sphere. a ) a b f c b c f f2 f3 f3 a f2 f3 f e d e d e d f2 b c fgure 3

3 Theorem 3 (Euler's formula) If G s a connected planar graph, for any embeddng G' the followng formula holds: n(g)+f(g) = e(g)+2 Proof. By nducton on f. For f(g) =, G s a tree. For every tree, e(g) = n(g)-, so n(g)+= e(g) + 2 n(g)+f(g) =e(g) +2 and the formula holds. Suppose t holds for all planar graphs wth less than f faces and suppose that G' has f 2 faces. Let (u,v) be an edge of G whch s not a cut-edge. Such an edge must exsts because G' has more than one face. The removal of (u.v) wll cause the two faces separated by (u,v) to combne, formng a sngle face. Hence (G-(u,v))' s a planar embeddng of a connected graph wth one less face than G', hence: f(g - (u,v)) = f(g) - n(g - (u,v) = n(g) e(g = (u,v)) = e(g) - But by the nducton hypothess: n(g- (u,v)) + f(g - (u,v)) = e(g - (u,v)) + 2 and so, by substtuton: n(g) + f(g) = e(g) + 2 Hence, by nducton, Euler's formula holds for all connected planar graphs. Lemma For any embeddng G' of any smple connected planar graph G, d( f ) 2e( G) = Proof. Each edge contrbutes to each face t s a bound, so t contrbutes 2 to the total sum. So the e(g) edges contrbutes 2e(G) to the total sum. Lemma 2 For any smple connected planar graph G, wth e(g) 3, the followng holds: e(g) 3n(G) - 6 Proof. Each face of any embeddng G' of G s bounded by at least three edges, hence: d( f ) 3 f ( G) Form the above lemma, d( f ) 2e( G), hence: 2e(G) 3f(G) => f(g) 3 2 e(g) = From Euler's formula, n(g)+f(g) = e(g)+2, so n(g) e(g) e(g) e(g) n(g) - 2 e(g) 3n(G) - 6 Defntons Bpartte graph A bpartte graph s a graph wth no cycles of odd number of edges. In a bpartte graph, the set of vertces can be parttoned to two dsjont not empty subsets V and V2, so that every edge of V connects a vertex of V wth a vertex of V2. Complete bpartte graph A complete bpartte graph, denoted as Km,n s a bpartte graph where V has m vertces, V2 has n vertces and every vertex of each subset s connected wth all other vertces of the other subset. Km,n haw m+n vertces and m*n edges.

4 Corollary A smple connected planar bpartte graph, has each face wth even degree. Proof. Each face s a cycle and the graph s bpartte, so each face must has even number of vertces. Lemma 3 For any smple connected bpartte graph G, wth e(g) 3. the followng holds: e(g) 2n(G) - 4 Proof. G s bpartte, so each face of every embeddng G' has at least 4 edges, hence d( f ) 4 f ( G) For every smple connected planar graph, d( f ) 2e( G), hence: 2e(G) 4f(G) => f(g) 2 e(g) = From Euler's formula, n(g)+f(g) = e(g)+2, hence n(g) + 2 e(g) e(g) e(g) n(g) - 2 e(g) 2n(G) - 4 s a) K5 s not planar. K5 n(k5) = 5 5 e(k5) = = 0 2 3n(K5) - 6 = 5-6 = 9. fgure 4 e(k5) > 3n(K5) - 6, hence K5 s not planar. b) K3,3 s not planar. K3,3 n(k3,3) = 6 e(k3,3) = 3*3=9 2n(K3,3) - 4 = 2-4 = 8 e(k3,3) > 2n(K3,3) - 4, hence K3,3 s not planar. c) fgure 5 The formula e(g) 3n(G) - 6 does not holds for e(g) < 3. K has n(k)=, e(k)=0 and e(k) >3n(K) - 6 K2 has n(k2)=2, e(k2)= and e(k2) > 3n(K2) - 6 Defntons Genus For any non-negatve nteger g, we can construct a surface n whch t s possble to embed g non-ntersectng closed curves wthout separatng the surface nto two regons. If for the same surface (g+) closed curves always cause a separaton, then the surface s sad to have a genus equal to g. A graph that can be embedded n a surface of genus g, but not on a surface of genus (g-) s called a graph of genus g. Crossng number The Crossng number of a graph s the mnmum number of crossngs of edges for the graph drawn n the plane.

5 Thckness The thckness T (G) of a graph G s the mnmum number of planar sub-graphs of G whose unon s G. Corollary 2 The thckness T of a smple graph G satsfes e( G) T(G) 3n( G) 6 Proof. Each planar sub-graph wll contan at most 3n(G)- edges. s fgure 6: a sphere of genus 0 fgure 7: a sphere wth a handle fgure 8: a torus, are both genus. fgure 9: a sphere wth two handles fgure 0: a double torus, are both genus 2

6 fgure : a graph of genus. Fgure 2: K5 has crossng number. fgure 3: K9 s the unon of three planar graphs, hence T(K9) 3 Theorem 4 If G s a connected graph wth genus g, then: f(g) = e(g) n(g) + 2 2g Proof. By nducton on g. For g=0 the graph s planar and we have f(g) = e(g) n(g) + 2 2g from Euler s formula. We assume that the theorem s true for all graphs wth genus (g-) These graphs may be drawn on a sphercal surface wth (g-) handles and nclude all those graphs obtaned by deletng those edges passng over a sngle handle n any graph of genus g. We construct G wth genus g on a surface o genus g by addng a sngle edge to G, requrng an addtonal handle. Usng prme letters for G, we have by the nducton hypothess: F(G ) = e(g ) n(g ) + 2 2g But e(g)=e(g )+, g = g +, n-n(g ) + Also f(g) = f(g ) because the handle connects two dstnct faces n G makng A sngle face n G. Hence by substtuton: F(G) = e(g) n(g) g. And so by nducton the theorem s proved. Corollary 3 If G s a connected graph wth genus n(g) 4 then g 6 (e(g) 3n(G)) + Proof. From the above theorem we get: g= 2 (e(g) n(g) - f(g)) +

7 Every face of an embeddng of the graph s bound by at least three edges of whch separates two faces, therefore 3f(G) 2e(G), and so the result follows by substtuton. Defnton Homeomorphc graphs Two graphs are sad to be homeomorphc f one can be made somorphc to the other by the addton or the deleton of vertces of degree two, n the followng manner: By dvdng an edge nto two edges n seres by the nserton of a vertex of degree 2, or by reverse of ths process;. Fgure 4: homeomorphc graphs Theorem 5 (Kuratowsk) A graph s planar f and only f t has no sub-graph homeomorphc to K5 or to K3,3. Proof. The proof s out of the subject of ths course. For a proof you can look at Alan Gbbons book, Algorthmc graph theory, page fgure 5: graphs G, G2 and G3 Graph G s not planar, snce t has a sub-graph (G2) homeomorphc to G3, Whch s somorphc to K3,3 (The partton of G3 vertces s{,8,9} and {2,5,6}) Defntons Colorng A colorng of the vertces of a graph s a mappng of any vertex of the graph to a color such that any vertces connected wth an edge have dfferent colors. The mnmum number of colors requred for a graph colorng s called colorng number of the graph. Theorem 6 (Four Color theorem) A planar graph has colorng number 4 The proof was gven n 976 by Appel and Haken, usng computers and analyzng the problem to 936 dfferent cases.

8 Lemma 4 The four color theorem s equvalent to the followng lemma: The vertces of a planar graph can be parttoned n two sets, V,V2 such that G/V, G/V2 are both bpartte graphs. Proof. a) 4 Color Theorem => Lemma I can get mmedately the two bpartte graphs by choosng 2 of the four colors for one of the bpartte graphs. b) Lemma => 4 Color theorem I can get the two bpartte graphs and put 2 dfferent colors on each graph. Defnton Dual graph. Let G be a planar graph and a planar embeddng G A dual graph G* has a vertex for each face of G and an edge between two faces f and f2 f and only f f and f2 share an edge. Dual graph s not always a smple graph. fgure 6 We can see that the dual of the frst planar representaton of the graph, s not somorphc to the dual of the second representaton. The dual graph refers to of a planar representaton and not to the graph. Lemma 5 A graph s planar f and only f the dual graphs of t s planar embeddngs are planar For a proof you can look at Alan Gbbons book, Algorthmc graph theory, page 83. Corollary 4 Usng dual graphs, w can see that the four color theorem s equvalent to: We can color the faces of a graph such that two faces who share an edge have dfferent color, usng at most four colors. Planarty Testng A sophstcated algorthm gven by Hopcroft and Tarjan, tests planarty n O(n) tme. 3 Here we wll see a smple algorthm that test planarty O( n ) tme.

9 As a frst applcaton of the dvde and conquer prncple, we observe that: A graph s planar f and only f all ts connected components are planar A connected graph s planar f and only f all ts bconnected components are planar Thus, va prelmnary decomposton nto connected and bconnected components, we can restrct our attenton to the problem of testng the planarty of a bconnected graph. Defntons Pece Let G be a bconnected graph. Let C be a cycle n G. A pece s: a) an edge wth vertces on C b) a connected component after removng the edges of C Separatng cycle A cycle C of G s sad to be separatng f t has at least two peaces, and s called nonseparatng f t has one peace. Of course, f G = C, then C has no peces. In any bconnected graph there s a separatng cycle and from any non-separatng cycle, I can always get a separatng cycle. fgure 7: a bconnected graph G and a cycle C. fgure 8: the peces of G wth respect to C

10 Lemma 5 Let G be a bconnected graph and let C be a nonseparatng cycle of G wth pece P.\If P s not a path, then G has a separatng cycle C consstng of a subpath of C plus a path of P between two attachments. Proof. Let u and v be two attachments of P that are consecutve n the crcular orderng, and let g be a subpath of C between u and v that does not contan any attachment of C. Snce P s connected, there s a path p, between u and v. Let C be the cycle obtaned from C by replacng g wth p. We have that g s a pece of G wth respect to C. If P s not a path, let e be an edge of P not n p. There s a peces of C dstnct from g contanng e. Thus, f P s not a path, then C has at least two peces and s thus a separatng cycle of G. fgure 9: a nonseparatng cycle C and a separatng cycle C obtaned from C as shown n the proof of the Lemma. Defnton The nterlacement graph If the graph G s planar, then n any planar drawng of G each peace s drawn ether nsde C or entrely outsde c. We say that two peces of G, wth respect to C, nterlace f they cannot be drawn on the same sde of C wthout volatng planarty. The nterlacement graph of the peces of G, wth respect to C, s the graph whose vertces are the peces of G and whose edges are the pars of peces that nterlace. fgure 20 A planar drawng of the above graph, where peces P, P3 and P6 are drawn nsde cycle C and the other peces are drawn outsde and the nterlacement graph I of the peces of F wth respect to cycle C

11 Theorem 7 A bconnected graph G wth a cycle C s planar f and only f the followng condtons hold: For each pece P of G wth respect to C, the graph obtaned by addng P to C s planar The nterlacement graph of the peces of G, wth respect to C, s bpartte. Algorthm The correctness of the algorthm s based on the above lemma and theorem and on the fact that the graph P obtaned by addng a pece P to C s bconnected. Input: a bconnected graph, G, wth n vertces and at most 3n-6 edges, and a separatng cycle C of G. Output: and ndcaton of whether G s planar.. Compute the peces of G wth respect to C. 2. for each pece P of G that s not a path (of one or more edges): (a) let P be the graph obtaned by addng P to C (b) let C be the cycle of P obtaned from C by replacng the porton of C between two consecutve attachments wth a path of P between them (c) apply the algorthm recursvely to graph P and cycle C. If P s non planar, return nonplanar 3. Compute the nterlacement graph I of the peces. 4. Test whether I s bpartte. If s not bpartte, return nonplanar 5. Return planar Tme analyss Step : O(n) Step 2: O(n)) 2 Step 3: O( n ) 2 Step 4: O( n ) 2 So a recursve nvocaton of algorthm of planarty testng takes O( n ) tme. We observe that number of recursve nvocatons as O(n) by assocatng wth each nvocaton a dstnct edge of G.. We conclude that the runnng tme of algorthm Planarty Testng s 3 O( n )??µ?t???????d?? 945 G?????? G?a????d?ßa?d?? 960

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