1.3. VERTEX DEGREES & COUNTING


 Michael Carr
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1 35 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 36 its eighbor o P. Note that P has at least three vertices. If G x v is coected, let y = v. Otherwise, a compoet cut off from P x v i G x v has at most oe vertex; call it w. The vertex w must be adjacet to v, sice otherwise we could build a loger path. I this case, let y = w A coected simple graph havig o 4vertex iduced subgraph that is a path or a cycle has a vertex adjacet to every other vertex. Cosider a vertex x of maximum degree. If x has a oeighbor y, let x, v, w be the begiig of a shortest path to y (w may equal y. Sice d(v d(x, some eighbor z of x is ot adjacet to v. If z w, the {z, x, v, w} iduce C 4 ; otherwise, {z, x, v, w} iduce P 4. Thus x must have o oeighbor The edges of a coected simple graph with 2k edges ca be partitioed ito paths of legth 2. The assumptio of coectedess is ecessary, sice the coclusio does ot hold for a graph havig compoets with a odd umber of edges. We use iductio o e(g; there is a sigle such path whe e(g = 2. For e(g > 2, let P = (x, y, z be a arbitrary path of legth two i G, ad let G = G {xy, yz}. If we ca partitio E(G ito smaller coected subgraphs of eve size, the we ca apply the iductio hypothesis to each piece ad combie the resultig decompositios. Oe way to do this is to partitio E(G ito coected subgraphs of eve size ad use P. Hece we are fiished uless G has two compoets of odd size (G caot have more tha three compoets, sice a edge deletio icreases the umber of compoets by at most oe. Each odd compoet cotais at least oe of {x, y, z}. Hece it is possible to add oe of xy to oe odd compoet ad yz to the other odd compoet to obtai a partitio of G ito smaller coected subgraphs VERTEX DEGREES & COUNTING A graph havig exactly two vertices of odd degree must cotai a path from oe to the other. The degree of a vertex i a compoet of G is the same as its degree i G. If the vertices of odd degree are i differet compoets, the those compoets are graphs with odd degree sum I a class with ie studets where each studet seds valetie cards to three others, it is ot possible that each studet seds to ad receives cards from the same people. The sedig of a valetie ca be represeted as a directed edge from the seder to the receiver. If each studet seds to ad receives cards from the same people, the the graph has x y if ad oly if y x. Modelig each opposed pair of edges by a sigle uorieted edge yields a 3regular graph with 9 vertices. This is impossible, sice every graph has a eve umber of vertices of odd degree If d(u + d(v = + k for a edge uv i a simple graph o vertices, the uv belogs to at least k triagles. This is the same as showig that u ad v have at least k commo eighbors. Let S be the eighbors of u ad T the eighbors of v, ad suppose S T = j. Every vertex of G appears i S or T or oe or both. Commo eighbors are couted twice, so S + T j = + k j. Hece j k. (Almost every proof of this usig iductio or cotradictio does ot eed it, ad is essetially just this coutig argumet The graph below is isomorphic to Q 4. It suffices to label the vertices with the ames of the vertices i Q 4 so that vertices are adjacet if ad oly if their labels differ i exactly oe place The kdimesioal cube Q k has ( k 2 k copies of P 3. Proof 1. To specify a particular subgraph isomorphic to P 3, the 3 vertex path, we ca specify the middle vertex ad its two eighbors. For each vertex of Q k, there are ( k ways to choose two distict eighbors, sice Q k is a simple kregular graph. Thus the total umber of P 3 s is ( k 2 k. Proof 2. We ca alteratively choose the startig vertex ad the ext two. There are 2 k ways to pick the first vertex. For each vertex, there are k ways to pick a eighbor. For each way to pick these vertices, there are k 1 ways to pick a third vertex completig P 3, sice Q k has o multiple edges. The product of these factors couts each P 3 twice, sice we build it from each ed. Thus the total umber of them is 2 k k(k 1/2. Q k has ( k 2 k 2 copies of C 4. Proof 1 (direct coutig. The vertices two apart o a 4cycle must differ i two coordiates. Their two commo eighbors each differ from each i exactly oe of these coordiates. Hece the vertices of a 4cycle
2 37 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 38 must use all 2tuples i two coordiates while keepig the remaiig coordiates fixed. All such choices yield 4cycles. There are ( k ways to choose the two coordiates that vary ad 2 k 2 ways to set a fixed value i the remaiig coordiates. Proof 2 (prior result. Every 4cycle cotais four copies of P 3, ad every P 3 cotais two vertices at distace 2 i the cube ad hece exteds to exactly oe 4cycle. Hece the umber of 4cycles is oefourth the umber of copies of P Coutig compoets. If G has k compoets ad H has l compoets, the G + H has k + l compoets. The maximum degree of G + H is max{ (G, (H} Largest bipartite subgraphs. P is already bipartite. C loses oe edge if is odd, oe if is eve. The largest bipartite subgraph of K is K /2, /2, which has 2 /4 edges The lists (5,5,4,3,2,2,2,1, (5,5,4,4,2,2,1,1, ad (5,5,5,3,2,2,1,1 are graphic, but (5,5,5,4,2,1,1,1 is ot. The aswers ca be obtaied from the HavelHakimi test; a list is graphic if ad oly if the list obtaied by deletig the largest elemet ad deletig that may extlargest elemets is graphic. Below are graphs realizig the first three lists, foud by the HavelHakimi algorithm. From the last list, we test (4, 4, 3, 1, 0, 1, 1, reordered to (4, 4, 3, 1, 1, 1, 0, the (3, 2, 0, 0, 1, 0. This is ot the degree list of a simple graph, sice a vertex of degree 3 requires three other vertices with ozero degree I a league with two divisios of 13 teams each, o schedule has each team playig exactly ie games agaist teams i its ow divisio ad four games agaist teams i the other divisio. If this were possible, the we could form a graph with the teams as vertices, makig two vertices adjacet if those teams play a game i the schedule. We are askig for the subgraph iduced by the 13 teams i a sigle divisio to be 9regular. However, there is o regular graph of odd degree with a odd umber of vertices, sice for every graph the sum of the degrees is eve If l, m, are oegative itegers with l + m = 1, the there exists a coected simple vertex graph with l vertices of eve degree ad m vertices of odd degree if ad oly if m is eve, except for (l, m, = (2, 0,. Sice every graph has a eve umber of vertices of odd degree, ad the oly simple coected graph with two vertices has both degrees odd, the coditio is ecessary. To prove sufficiecy, we costruct such a graph G. If m = 0, let G = C l (except G = K 1 if l = 1. For m > 0, we ca begi with K 1,m 1, which has m vertices of odd degree, ad the add a path of legth l beyod oe of the leaves. (Illustratio shows l = 3, m = 4. Alteratively, start with a cycle of legth l, ad add m vertices of degree oe with a commo eighbor o the cycle. That vertex of the cycle has eve degree because m is eve. May other costructios also work. It is also possible to prove sufficiecy by iductio o for 3, but this approach is loger ad harder to get right tha a explicit geeral costructio If C is a closed walk i a simple graph G, the the subgraph cosistig of the edges appearig a odd umber of times i C is a eve graph. Cosider a arbitrary vertex v V (G. Let S be the set of edges icidet to v, ad let f (e be the umber of times a edge e is traversed by C. Each time C passes through v it eters ad leaves. Therefore, e S f (e must be eve, sice it equals twice the umber of times that C visits v. Hece there must a eve umber of odd cotributios to the sum, which meas there are a eve umber of edges icidet to v that appear a odd umber of times i C. Sice we ca start a closed walk at ay of its vertices, this argumet holds for every v V (G If every vertex of G has eve degree, the G has o cutedge. Proof 1 (cotradictio. If G has a cutedge, deletig it leaves two iduced subgraphs whose degree sum is odd. This is impossible, sice the degree sum i every graph is eve. Proof 2 (costructio/extremality. For a edge uv, a maximal trail i G uv startig at u ca oly ed at v, sice wheever we reach a vertex we have use a odd umber of edges there. Hece a maximal such trail is a (u, vtrail. Every (u, vtrail is a (u, vwalk ad cotais a (u, vpath. Hece there is still a (u, vpath after deletio of uv, so uv is ot a cutedge. Proof 3 (prior results. Let G be a eve graph. By Propositio , G decomposes ito cycles. By the meaig of decompositio, every edge
3 39 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 40 of G is i a cycle. By Theorem , every edge i a cycle is ot a cutedge. Hece every edge of G is ot a cutedge. For k N, some (2k + 1regular simple graph has a cutedge. Costructio 1. Let H, H be copies of K 2k,2k with partite sets X, Y for H ad X, Y for H. Add a isolated edge vv disjoit from these sets. To H + H + vv, add edges from v to all of X ad from v to all of X, ad add k disjoit edges withi Y ad k disjoit edges withi Y. The resultig graph G k is (2k + 1regular with 8k + 2 vertices ad has vv as a cutedge. Below we sketch G 2 ; the graph G 1 is the graph i Example Y X v v X Y Costructio 2a (iductive. Let G 1 be the graph at the ed of Example (or i Costructio 1. This graph is 3regular with 10 vertices ad cutedge xy; ote that 10 = From a (2k 1regular graph G k 1 with 4k + 2 vertices such that G k 1 xy has two compoets of order 2k + 1, we form G k. Add two vertices for each compoet of G k 1 xy, adjacet to all the vertices of that compoet. This adds degree two to each old vertex, gives degree 2k + 1 to each ew vertex, ad leaves xy as a cutedge. The result is a (2k + 1regular graph G k of order 4k + 6 with cutedge xy. Costructio 2b (explicit. Form H k from K 2k+2 by removig k pairwise disjoit edges ad addig oe vertex that is adjacet to all vertices that lost a icidet edge. Now H k has 2k + 2 vertices of degree 2k + 1 ad oe of degree 2k. Form G k by takig two disjoit copies of H k ad addig a edge joiig the vertices of degree 2k. The graphs produced i Costructios 2a ad 2b are idetical Meetig o a moutai rage. A moutai rage is a polygoal curve from (a, 0 to (b, 0 i the upper halfplae; we start A ad B at opposite edpoits. Let P be a highest peak; A ad B will meet there. Let the segmets from P to (a, 0 be x 1,..., x r, ad let the segmets from P to (b, 0 be y 1,..., y s. We defie a graph to describe the positios; whe A is o x i ad B is o y j, the correspodig vertex is (i, j. We start at the vertex (r, s ad must reach (1, 1. We itroduce edges for the possible trasitios. We ca move from (i, j to (i, j + 1 if the commo edpoit of y j ad y j+1 has height betwee the heights of the edpoits of x i. Similarly, (i, j is adjacet to (i + 1, j if the commo edpoit of x i ad x i+1 has height betwee the heights of the edpoits of x j. To avoid triviality, we may assume that r + s > 2. We prove that (r, s ad (1, 1 are the oly vertices of odd degree i G. This suffices, because every graph has a eve umber of vertices of odd degree, which implies that (r, s ad (1, 1 are i the same compoet, coected by a path. The possible eighbors of (i, j are the pairs obtaied by chagig i or j by 1. Let X ad Y be the itervals of heights attaied by x i ad y j, ad let I = X Y. If the high ed of I is the high ed of exactly oe of X ad Y, the exactly oe eighborig vertex ca be reached by movig past the ed of the correspodig segmet. If it is the high ed of both, the usually oe or three eighborig vertices ca be reached, the latter whe both segmets reach peaks at their high eds. However, if (i, j = (1, 1, the the high ed of both segmets is P ad there is o eighbor of this type. Similarly, the low ed of I geerates oe or three eighbors, except that whe (i, j = (r, s there is o eighbor of this type. No eighbor of (i, j is geerated from both the low ed ad the high ed of I. Sice the cotributios from the high ad low ed of I to the degree of (i, j are both odd, each degree is eve, except for (r, s ad (1, 1, where exactly oe of the cotributios is odd Every simple graph with at least two vertices has two vertices of equal degree. The degree of a vertex i a vertex simple graph is i {0,..., 1}. These are distict values, so if o two are equal the all appear. However, a graph caot have both a isolated vertex ad a vertex adjacet to all others. This does ot hold for graphs allowig loops. I the 2vertex graph with oe loop edge ad oe oloop edge, the vertex degrees are 1 ad 3. This does ot hold for loopless graphs. I the 3vertex loopless graph with pairs havig multiplicity 0, 1, 2, the vertex degrees are 1, 3, Smallest kregular graphs. A simple kregular graph has at least k + 1 vertices, so K k+1 is the smallest. This is the oly isomorphism class of kregular graphs with k +1 vertices. With k +2 vertices, the complemet of a kregular graph must be 1regular. There is oe such class whe k is eve ((k + /2 isolated edges, oe whe k is odd. (Two graphs are isomorphic if ad oly if their complemets are isomorphic. With k + 3 vertices, the complemet is 2regular. For k 3, there are distict choices for such a graph: a (k + 3cycle or the disjoit uio of a 3cycle ad a kcycle. Sice these two 2regular graphs are oisomorphic, their complemets are oisomorphic kregular graphs with k + 3 vertices For k 2 ad g 2, there exists a kregular graph with girth g. We use strog iductio o g. For g = 2, take the graph cosistig of two vertices ad k edges joiig them. For the iductio step, cosider g > 2. Here we use iductio o k. For k = 2, a cycle of legth g suffices. For k > 2, the iductio hypothesis
4 41 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 42 provides a (k 1regular graph H with girth g. Sice g/2 < g, the global iductio hypothesis also provides a graph G with girth g/2 that is (H regular. Replace each vertex v i G with a copy of H; each vertex i the copy of H is made icidet to oe of the edges icidet to v i G. Each vertex i the resultig graph iherits k 1 icidet edges from H ad oe from G, so the graph is kregular. It has cycles of legth g i copies of H. A cycle C i G is cofied to a sigle copy of H or visits more tha oe such copy. I the first case, its legth is at least g, sice H has girth g. I the secod case, the copies of H that C visits correspod to a cycle i G, so C visits at least g/2 such copies. For each copy, C must eter o oe edge ad the move to aother vertex before leavig, sice the copy is etered by oly oe edge at each vertex. Hece the legth of such a cycle is at least 2 g/ Deletig a vertex of maximum degree caot icrease the average degree, but deletig a vertex of miimum degree ca reduce the average degree. Deletig ay vertex of a otrivial regular graph reduces the average degree, which proves the secod claim. For the first claim, suppose that G has vertices ad m edges, ad let a ad a be the average degrees of G ad G x, respectively. Sice G x has m d(x edges ad degree sum 2m 2d(x, we have a = a 2d(x ( a < a if d(x a > 0. Hece deletig a vertex of maximum degree i otrivial graph reduces the average 1 1 degree ad caot icrease it If k 2, the a kregular bipartite graph has o cutedge. Sice compoets of kregular graphs are kregular, it suffices to cosider a coected kregular X, Y bigraph. Let uv be a cutedge, ad let G ad H be the compoets formed by deletig uv. Let m = V (G X ad = V (G Y. By symmetry, we may assume that u V (G Y ad v V (H X. We cout the edges of G. The degree of each vertex of G i X is k, so G has mk edges. The degree of each vertex of G i Y is k except for d G (u = k 1, so G has k 1 edges. Hece m k = k 1, which is impossible because oe side is divisible by k ad the other is ot. The proof does t work if k = 1, ad the claim is false the. If vertex degrees k ad k + 1 are allowed, the a cutedge may exist. Cosider the example of 2K k,k plus oe edge joiig the two compoets A clawfree simple graph with maximum degree at least 5 has a 4 cycle. Cosider five edges icidet to a vertex v of maximum degree i such a graph G. Sice G has o iduced claw, the eighbors of v must iduce at least three edges. Sice these three edges have six edpoits amog the five eighbors of v, two of them must be icidet, say xy ad yz. Addig the edges xv ad zv to these two completes a 4cycle. There are arbitrarily large 4regular clawfree graphs with o 4cycles. Cosider a vertex v i such a graph G. Sice v has degree 4 ad is ot the ceter of a iduced claw ad does ot lie o a 4cycle, the subgraph iduced by v ad its eighbors cosists of two edgedisjoit triagles sharig v (a bowtie. Sice this happes at each vertex, G cosists of pairwise edgedisjoit triagles, with each vertex lyig i two of them. Hece each triagle has three eighborig triagles. Furthermore, two triagles that eighbor a give triagle i this way caot eighbor each other; that would create a 4cycle i the graph. Defie a graph H with oe vertex for each triagle i G; let vertices be adjacet i H if the correspodig triagles share a vertex i G. Now H is a 3regular graph with o 3cycles; a 3cycle i H would yield a 4cycle i G usig two edges from oe of the correspodig triagles. Also H must have o 4cycles, because a 4cycle i G could be built usig oe edge from each of the four triagles correspodig to the vertices of a 4cycle i H. Note that e(g = 2(G ad (H = e(g/3 = 2(G/3. O the other had, give ay 3regular graph H with girth at least 5, reversig the costructio yields G with the desired properties ad 3(H/2 vertices. Hece it suffices to show that there are arbitrarily large 3regular graphs with girth at least 5. Discoected such examples ca be formed by takig may copies of the Peterse graph as compoets. The graph G is coected if ad oly if H is coected. Coected istaces of H ca be obtaied from multiple copies of the Peterse graph by applyig 2switches (Defiitio Alteratively, arbitrarily large coected examples ca be costructed by takig two odd cycles (say legth 2m + 1 ad joiig the ith vertex o the first cycle to the 2ith vertex (modulo 2m + 1 o the secod cycle (this geeralizes the Peterse graph. We have costructed a coected 3regular graph. Sice we add disjoit edges betwee the cycles, there is o triagle. A 4cycle would have to alterate edges betwee the two odd cycles with oe edge of each, but the eighbors of adjacet vertices o the first cycle are two apart o the secod cycle K has ( 1!/2 cycles of legth, ad K, has!( 1!/2 cycles of legth 2. Each cycle i K is a listig of the vertices. These ca be listed i! orders, but we obtai the same subgraph o matter where we start the cycle ad o matter which directio we follow, so each cycle is listed 2 times. I K,, we ca list the vertices i order o a cycle (alteratig betwee the partite sets, i 2(! 2 ways, but by the same reasoig each cycle appears (2 2 times K m, has 6 ( m 3( 3 6cycles. To exted a edge i Km, to a 6cycle, we choose two more vertices from each side to be visited i order as we follow the cycle. Hece each edge i K, appears i (m 1( 1(m (
5 43 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 44 6cycles. Sice each 6cycle cotais 6 edges, we coclude that K, has m(m 1( 1(m ( /6 6cycles. Alteratively, each 6cycle uses three vertices from each partite set, which we ca choose i ( m 3( 3 ways. Each such choice of vertices iduces a copy of K 3,3 with 9 edges. There are 3! = 6 ways to pick three disjoit edges to be omitted by a 6cycle, so each K 3,3 cotais 6 6cycles Odd girth ad miimum degree i obipartite triaglefree  vertex graphs. Let k = δ(g, ad let l be the miimum legth of a odd cycle i G. Let C be a cycle of legth l i G. a Every vertex ot i V (C has at most two eighbors i V (C. It suffices to show that ay two eighbors of such a vertex v o C must have distace 2 o C, sice havig three eighbors would the require l = 6. Sice G is triaglefree, v does ot have cosecutive eighbors o C. If v has eighbors x ad y o C separated by distace more tha 2 o C, the the detour through v ca replace the x, ypath of eve legth o C to form a shorter odd cycle. b kl/2 (ad thus l 2/k. Sice C is a shortest odd cycle, it has o chords (it is a iduced cycle. Sice δ(g = k, each vertex of C thus has at least k 2 edges to vertices outside C. However, each vertex outside C has at most two eighbors o C. Lettig m be the umber of edges from V (C to V (G V (C, we thus have l(k m 2( l. Simplifyig the iequality yields kl/2. c The iequality of part (b is sharp whe k is eve. Form G from the cycle C l by replacig each vertex of C l with a idepedet set of size k/2 such that two vertices are adjacet if ad oly if the vertices they replaced were adjacet. Each vertex is ow adjacet to the vertices arisig from the two eighborig classes, so G is kregular ad has lk/2 vertices. Deletig the copies of ay oe vertex of C l leaves a bipartite graph, sice the partite sets ca be labeled alterately aroud the classes arisig from the rest of C l. Hece every odd cycle uses a copy of each vertex of C l ad has legth at least l, ad takig oe vertex from each class forms such a cycle Equivalet defiitios of the kdimesioal cube. I the direct defiitio of Q k, the vertices are the biary ktuples, with edges cosistig of pairs differig i oe place. The iductive defiitio gives the same graph. For k = 0 both defiitios specify K 1. For the iductio step, suppose k 1. The iductive defiitio uses two copies of Q k 1, which by the iductio hypothesis is the 1place differece graph of the biary (k 1tuples. If we apped 0 to the (k 1tuples i oe copy of Q k 1 ad 1 to the (k 1 tuples i the other copy, the withi each set we still have edges betwee the labels differig i exactly oe place. The iductive costructio ow adds edges cosistig of correspodig vertices i the two copies. This is also what the directio defiitio does, sice ktuples chose from the two copies differ i the last positio ad therefore differ i exactly oe positio if ad oly if they are the same i all other positios. e(q k = k2 k 1. By the iductive defiitio, e(q k = 2e(Q k 1 +2 k 1 for k 1, with e(q 0 = 0. Thus the iductive step for a proof of the formula is e(q k = 2(k 12 k k 1 = k k K 2,3 is the smallest simple bipartite graph that is ot a subgraph of the kdimesioal cube for ay k. Suppose the vectors x, y, a, b, c are the vertices of a copy of K 2,3 i Q k. Ay oe of a, b, c differs from x i exactly oe coordiate ad from y i aother (it ca t be the same coordiate, because the x = y. This implies that x ad y differ i two coordiate i, j. Paths from x to y i two steps ca be formed by chagig i ad the j or chagig j ad the i; these are the oly ways. I a cube two vertices have at most two commo eighbors. Hece K 2,3 is forbidde. Ay bipartite graph with fewer vertices or edges is cotaied i K 2,3 e or K 1,5, but K 2,3 e is a subgraph of Q 3, ad K 1,5 is a subgraph of Q 5, so K 2,3 is the smallest forbidde subgraph Every cycle of legth 2r i a hypercube belogs to a subcube of dimesio at most r, uiquely if r 3. Let C be a cycle of legth 2r i Q k ; V (C is a collectio of biary vectors of legth k. Let S be the set of coordiates that chage at some step while traversig the vectors i V (C. I order to retur to the first vector, each positio must flip betwee 0 ad 1 a eve umber of times. Thus traversig C chages each coordiate i S at least twice, but oly oe coordiate chages with each edge. Hece 2 S 2r, or S r. Outside the coordiates of S, the vectors of V (C all agree. Hece V (C is cotaied i a S dimesioal subcube. As argued above, at most two coordiates vary amog the vertices of a 4cycle; at least two coordiates vary, because otherwise there are ot eough vectors available to have four distict vertices. By the same reasoig, exactly three three coordiates vary amog the vertices of ay 6cycle; we caot fid six vertices i a 2dimesioal subcube. Thus the rdimesioal subcube cotaiig a particular cycle is uique whe r 3. Some 8cycles are cotaied i 3dimesioal subcubes, such as 000x, 001x, 011x, 010x, 110x, 111x, 101x, 100x, where x is a fixed vector of legth 3. Such a 8cycle is cotaied i 3 4dimesioal subcubes, obtaied by lettig some positio i x vary A 3dimesioal cube cotais 16 6cycles, ad the kdimesioal cube Q k cotais 16 ( k 3 2 k 3 6cycles. If we show that every 6cycle appears i exactly oe 3dimesioal subcube, the multiplyig the umber of 3 dimesioal subcubes by the umber of 6cycles i each subcube couts each 6cycle exactly oce.
6 45 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 46 For ay set S of vertices ot cotaied i a 3dimesioal subcube, there must be four coordiates i the correspodig ktuples that are ot costat withi S. A cycle through S makes chages i four coordiates. Completig the cycle requires returig to the origial vertex, so ay coordiate that chages must chage back. Hece at least eight chages are eeded, ad each edge chages exactly oe coordiate. The cycle has legth at least 8; hece 6cycles are cotaied i 3dimesioal subcubes. Furthermore, there are oly four vertices possible whe k 2 coordiates are fixed, so every 6cycle ivolves chages i three coordiates. Hece the oly 3dimesioal subcube cotaiig the 6cycle is the oe that varies i the same three coordiates as the 6cycle. By Example 1.3.8, there are ( k 3 2 k 3 3dimesioal subcubes, so it remais oly to show that Q 3 has 16 cycles of legth 6. We group them by the two omitted vertices. The two omitted vertices may differ i 1, 2, or 3 coordiates. If they differ i oe place (they are adjacet, the deletig them leaves a 6cycle plus oe edge joiig a pair of opposite vertices. Sice Q 3 has 12 edges, there are 12 6cycles of this type. Deletig two complemetary vertices (differig i every coordiate leaves oly a 6cycle. Sice Q 3 has four such pairs, there are four such 6cycles. The remaiig pairs differ i two positios. Deletig such a pair leaves a 4cycle plus two pedat edges, cotaiig o 6cycle. This cosiders all choices for the omitted vertices, so the umber of 6cycles i Q 3 is Properties of the middlelevels graph. Let G be the subgraph of Q 2k+1 iduced by vertices i which the umbers of 1s ad 0s differs by 1. These are the (2k + 1tuples of weight k ad weight k + 1, where weight deotes the umber of 1s. Each vertex of weight k has k + 1 eighbors of weight k + 1, ad each vertex of weight k + 1 has k + 1 eighbors of weight k. There are ( 2k+1 k vertices of each weight. Coutig edges by the DegreeSum Formula, e(g = (k + 1 (G = (k + 1 ( ( 2k+1 2 k+1 = (2k + 1 2k k. The graph is bipartite ad has o odd cycle. The 1s i two vertices of weight k must be covered by the 1s of ay commo eighbor of weight k +1. Sice the uio of distict ksets has size at least k + 1, there ca oly be oe commo eighbor, ad hece G has o 4cycle. O the other had, G does have a 6cycle. Give ay arbitary fixed vector of weight k 1 for the last 2k 2 positios, we ca form a cycle of legth six by usig 110, 100, 101, 001, 011, 010 successively i the first three positios Alterative descriptio of evedimesioal hypercubes. The simple graph Q k has vertex set {0, 1}k, with u v if ad oly if u ad v agree i exactly oe coordiate. Let the odd vertices be the vertices whose ame has a odd umber of 1s; the rest are eve vertices. Whe k is eve, Q k = Q k. To show this, reame all odd vertices by chagig 1s ito 0s ad 0s ito 1s. Sice k is eve, the resultig labels are still odd. Sice k is eve, every edge i Q k jois a eve vertex to a odd vertex. Uder the ew amig, it jois the eve vertex to a odd vertex that differs from it i oe coordiate. Hece the adjacecy relatio becomes precisely the adjacecy relatio of Q k. Whe k is odd, Q = k Q k, because Q k cotais a odd cycle ad hece is ot bipartite. Startig from oe vertex, form a closed walk by successively followig k edges where each coordiate is the coordiate of agreemet alog exactly oe of these edges. Hece each coordiate chages exactly k 1 times ad therefore eds with the value it had at the start. Thus this is a closed walk of odd legth ad cotais a odd cycle Automorphisms of Q k. a A subgraph H of Q k is isomorphic to Q l if ad oly if it is the subgraph iduced by a set of vertices agreeig i some set of k l coordiates. Let f be a isomorphism from H to Q l, ad let v be the vertex mapped to the vertex 0 of Q l whose coordiates are all 0. Let u 1,..., u l be the eighbors of v i H mapped to eighbors of 0 i Q l by f. Each u i differs from v i oe coordiate; let S be the set of l coordiates where these vertices differ from v. It suffices to show that vertices of H differ from v oly o the coordiates of S. This is immediate for l 1. For l 2, we prove that each vertex mapped by f to a vertex of Q l havig weight j differs from v i j positios of S, by iductio o j. Let x be a vertex mapped to a vertex of weight j i Q l. For j 1, we have already argued that x differs from v i j positios of S. For j 2, let y ad z be two eighbors of x whose images uder f have weight j 1 i Q l. By the iductio hypothesis, y ad z differ from v i j positios of S. Sice f (y ad f (z differ i two places, they have two commo eighbors i Q l, which are x ad aother vertex w. Sice w has weight j 2, the iductio hypothesis yields that w differs from v i j 1 positios of S. Sice the images of x, y, z, w iduce a 4cycle i Q l, also x, y, z, w iduce a 4cycle i H. The oly 4cycle i Q k that cotais all of y, z, w adds the vertex that differs from v i the j 2 positios of S where w differs, plus the two positios where y ad z differ from w. This completes the proof that x has the desired property. b The kdimesioal cube Q k has exactly 2 k k! automorphisms. (Part (a is uecessary. Form automorphisms of Q k by choosig a subset of the k coordiates i which to complemet 0 ad 1 ad, idepedetly, a permutatio of the k coordiates. There are 2 k k! such automorphisms.
7 47 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 48 We prove that every automorphism has this form. Let 0 be the all0 vertex. Let f be the iverse of a automorphism, ad let v be the vertex mapped to 0 by f. The eighbors of v must be mapped to the eighbors of 0. If these choices completely determie f, the f complemets the coordiates where v is ozero, ad the correspodece betwee the eighbors of 0 ad the eighbors of v determies the permutatio of the coordiates that expresses f as oe of the maps listed above. Suppose that x differs from v i coordiates r 1,..., r j. Let u 1,..., u j be the eighbors of v differig from v i these coordiates. We prove that f (x is the ktuple of weight j havig 1 i the coordiates where f (u 1,..., f (u j have 1. We use iductio o j. For j 1, the claim follows by the defiitio of u 1,..., u j. For j 2, let y ad z be two eighbors of x that differ from v i j 1 coordiates. Let w be the commo eighbor of y ad z that differs from v i j 2 coordiates. By the iductio hypothesis, f (y ad f (z have weight j 1 (i the appropriate positios, ad f (w has weight j 1. Sice f (x must be the other commo eighbor of f (y ad f (z, it has weight j, with 1s i the desired positios The Peterse graph has twelve 5cycles. Let G be the Peterse graph. We show first that each edge of G appears i exactly four 5cycles. For each edge e = xy i G, there are two other edges icidet to x ad two others icidet to y. Sice G has o 3cycles, we ca thus exted xy at both eds to form a 4vertex path i four ways. Sice G has o 4cycle, the edpoits of each such path are oadjacet. By Propositio , there is exactly oe vertex to add to such a path to complete a 5cycle. Thus e is i exactly four 5cycles. Whe we sum this cout over the 15 edges of G, we have couted 60 5cycles. However, each 5cycle has bee couted five times oce for each of its edges. Thus the total umber of 5cycles i G is 60/5 = 12. y x Combiatorial proofs with graphs. a For 0 k, ( ( = k ( 2 + k( k + k 2. Cosider the complete graph K, which has ( edges. If we partitio the vertices of K ito a k set ad a ( kset, the we ca cout the edges as those withi oe block of the partitio ad those choosig a vertex from each. Hece the total umber of edges is ( ( k 2 + k 2 + k( k. b If i =, the ( i (. Agai cosider the edges of K, ad partitio the vertices ito sets with i beig the size of the ith set. The left side of the iequality couts the edges i K havig both eds i the same S i, which is at most all of E(K For 1, there are 2 ( 1 2 simple eve graphs with a fixed vertex set of size. Let A be the set of simple eve graphs with vertex set v 1,..., v. Sice 2 ( 1 2 is the size of the set B of simple graphs with vertex set v 1,..., v 1, we establish a bijectio from A to B. Give a graph i A, we obtai a graph i B by deletig v. To show that each graph i B arises exactly oce, cosider a graph G B. We form a ew graph G by addig a vertex v ad makig it adjacet to each vertex with odd degree i G, as illustrated below. The vertices with odd degree i G have eve degree i G. Also, v itself has eve degree because the umber of vertices of odd degree i G is eve. Thus G A. Furthermore, G is the graph obtaied from G by deletig v, ad every simple eve graph i which deletig v yields G must have v adjacet to the same vertices as i G. Sice there is a bijectio from A to B, the two sets have the same size. G G v Triaglefree graphs i which every two oadjacet vertices have exactly two commo eighbors. (G = 1 + ( k+1 2, where k is the degree of a vertex x i G. For every pair of eighbors of x, there is exactly oe oeighbor of x that they have as a commo eighbor. Coversely, every oeighbor of x has exactly oe pair of eighbors of x i its eighborhood, because these are its commo eighbors with x. This establishes a bijective correspodece betwee the pairs i N(x ad the oeighbors of x. Coutig x, N(x, ad N(x, we have (G = 1 + k + ( ( k 2 = 1 + k+1 2. Sice this argumet holds for every x V (G, we coclude that G is kregular. Commet: Such graphs exist oly for isolated values of k. Uique graphs exist for k = 1, 2, 5. Viewig the vertices as x, N(x = [k], ad N(x = ( [k] 2, we have i adjacet to the pair { j, k} if ad oly if i { j, k}. The lack of triagles guaratees that oly disjoit pairs i ( [k] 2 ca be adjacet,
8 49 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 50 but each pair i ( ( [k] 2 must have exactly k 2 eighbors i [k] 2. For k = 5, this implies that N(x iduces the 3regular disjoitess graph of ( [5] 2, which is the Peterse graph. Sice the Peterse graph has girth 5 ad diameter 2, each itersectig pair has exactly oe commo eighbor i N(x i additio to its oe commo eighbor i N(x, so this graph has the desired properties. Numerical coditios elimiate k 3 (mod 4, because G would be regular of odd degree with a odd umber of vertices. There are stroger ecessary coditios. After k = 5, the ext possibility is k = 10, the 26, 37, 82, etc. A realizatio for k = 10 is kow to exist, but i geeral the set of realizable values is ot kow If G is a kitefree simple vertex graph such that every pair of oadjacet vertices has exactly two commo eighbors, the G is regular. Sice oadjacet vertices have commo eighbors, G is coected. Hece it suffices to prove that adjacet vertices x ad y have the same degree. To prove this, we establish a bijectio from A to B, where A = N(x N(y ad B = N(y N(x. Cosider u A. Sice u y, there exists v N(u N(y with v x. Sice G is kitefree, v x, so v B. Sice x ad v have commo eighbors y ad u, the vertex v caot be geerated i this way from aother vertex of A. Hece we have defied a ijectio from A to B. Iterchagig the roles of y ad x yields a ijectio from B to A. Sice these sets are fiite, the ijectios are bijectios, ad d(x = d(y If every iduced kvertex subgraph of a simple vertex graph G has the same umber of edges, where 1 < k < 1, the G is a complete graph or a empty graph. a If l k ad G is a graph o l vertices i which every iduced k vertex subgraph has m edges, the e(g = m ( ( l k / l 2 k. Coutig the edges i all the kvertex subgraphs of G yields m ( l k, but each edge appears i ( l 2 k of these subgraphs, oce for each kset of vertices cotaiig it. (Both sides of ( l 2 k 2 e(g = m ( l k cout the ways to pick a edge of G ad a kset of vertices i G cotaiig that edge. O the right, we pick the set first; o the left, we pick the edge first. b Uder the stated coditios, G = K or G = K. Give vertices u ad v, let A ad B be the sets of edges icidet to u ad v, respectively. The set of edges with edpoits u ad v is A B. We compute A B = e(g A B = e(g A B = e(g A B + A B. I this formula, A ad B are the edge sets of iduced subgraphs of order 1, ad A B is the edge set of a iduced subgraph of order 2. By part (a, the sizes of these sets do ot deped o the choice of u ad v The uique recostructio of the graph with vertexdeleted subgraphs below is the kite. Proof 1. A vertex added to the first triagle may be joied to 0,1,2, or 3 of its vertices. We elimiate 0 ad 1 because o vertexdeleted subgraph has a isolated vertex. We elimiate 3 because every vertexdeleted subgraph of K 4 is a triagle. Joiig it to 2 yields the kite. Proof 2. The graph G must have four vertices, ad by Propositio it has five edges. The oly such simple graph is the kite Retrievig a regular graph. Suppose that H is a graph formed by deletig a vertex from a regular graph G. We have H, so we kow (G = (H + 1, but we do t kow the vertex degrees i G. If G is d regular, the G has d(g/2 edges, ad H has d(g/2 d edges. Thus d = 2e(H/((G. Havig determied d, we add oe vertex w to H ad add d d H (v edges from w to v for each v V (H A graph with at least 3 vertices is coected if ad oly if at least two of the subgraphs obtaied by deletig oe vertex are coected. The edpoits of a maximal path are ot cutvertices. If G is coected, the the subgraphs obtaied by deleted such vertices are coected, ad there are at least of these. Coversely, suppose that at least two vertexdeleted subgraphs are coected. If G v is coected, the G is coected uless v is a isolated vertex. If v is a isolated vertex, the all the other subgraphs obtaied by deletig oe vertex are discoected. Hece v caot be isolated, ad G is coected Discoected graphs are recostructible. First we show that G is coected if ad oly if it has at least two coected vertexdeleted subgraphs. Necessity holds, because the edpoits of a maximal path caot be cutvertices. If G is discoected, the G v is discoected uless v is a isolated vertex (degree 0 i G ad G v is coected. This happes for at most oe vertex i G. After determiig that G is discoected, we obtai which discoected graph it is from its vertexdeleted subgraphs. We aim to idetify a coected graph M that is a compoet of G ad a vds i the deck that arises by deletig a specified vertex u of M. Replacig M u by M i that subgraph will recostruct G.
9 51 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 52 Amog all compoets of all graphs i the deck, let M be oe with maximum order. Sice every compoet H of a potetial recostructio G appears as a compoet of some G v, M caot belog to ay larger compoet of G. Hece M is a compoet of G. Let L be a fixed coected subgraph of M obtaied by deletig a leaf u of some spaig tree of M. The L is a compoet of G u. We wat to recostruct G by substitutig M for L i G u; we must idetify G u. There may be several isomorphic copies of G u. As i the discoected graph G show above, M may appear as a compoet of every vds G v. However, sice M caot be created by a vertex deletio, a vds with the fewest copies of M must arise by deletig a vertex of M. Amog these, we seek a subgraph with the most copies of L as compoets, because i additio to occurreces of L as a compoet of G, we obtai a additioal copy if ad oly if the deleted vertex of M ca play the role of u. This idetifies G u, ad we obtai G by replacig oe of its compoets isomorphic to L with a compoet isomorphic to M Largest graphs of specified types. a Largest vertex simple graph with a idepedet set of size a. Proof 1. Sice there are o edges withi the idepedet set, such a graph has at most ( ( a ( 2 edges, which equals a 2 + ( aa. This boud is achieved by the graph cosistig of a copy H of K a, a idepedet set S of size a, ad edges joiig each vertex of H to each vertex of S. Proof 2. Each vertex of a idepedet set of size a has degree at most a. Each other vertex has degree at most 1. Thus d(v a( a+ ( a( 1. By the DegreeSum Formula, e(g ( a( 1 + a/2. This formula equals those above ad is achieved by the same graph, sice this graph achieves the boud for each vertex degree. ( b The maximum size of a vertex simple graph with k compoets is k+1 2. The graph cosistig of K k+1 plus k 1 isolated vertices has k compoets ad ( k+1 2 edges. We prove that other vertex graphs with k compoets do t have maximum size. Let G be such a graph. If G has a compoet that is ot complete, the addig edges to make it complete does ot chage the umber of compoets. Hece we may assume that every compoet is complete. If G has compoets with r ad s vertices, where r s > 1, the we move oe vertex from the sclique to the rclique. This deletes s 1 edges ad creates r edges, all icidet to the moved vertex. The other edges remai the same, so we gai r s + 1 edges, which is positive. Thus the umber of edges is maximized oly whe every compoet is a complete graph ad oly oe compoet has more tha oe vertex. c The maximum umber of edges i a discoected simple vertex graph is ( 1 2, with equality oly for K1 + K 1. Proof 1 (usig part (b. The maximum over graphs with k compoets is ( k+1 2, which decreases as k icreases. For discoected graphs, k 2. We maximize the umber of edges whe k = 2, obtaiig ( 1 2. Proof 2 (direct argumet. Give a discoected simple graph G, let S be the vertex set of oe compoet of G, ad let t = S. Sice o edges joi S ad S, e(g ( t( t. This boud is weakest whe t( t is smallest, which for 1 t 1 happes whe t {1, 1}. Thus always e(g ( ( 2 1( 1 = 1 2, ad equality holds whe G = K1 + K 1. Proof 3 (iductio o. Whe = 2, the oly simple graph with e(g > ( 1 2 = 1 is K2, which is coected. For > 2, suppose e(g > ( 1 2. If (G = 1, the G is coected. Otherwise, we may select v with d(v 2. The e(g v > ( ( = 2 2. By the iductio hypothesis, G v is coected. Sice e(g > ( 1 2 ad G is simple, we have d(v > 0, so there is a edge from v to G v, ad G is also coected. Proof 4 (complemetatio. If G is discoected, the G is coected, so e(g 1 ad e(g ( ( 2 ( 1 = 1 2. I fact, G must cotai a spaig complete bipartite subgraph, which is as small as 1 edges oly whe G = K 1, 1 ad G = K 1 + K Every vertex simple graph with maximum degree /2 ad miimum degree /2 1 is coected. Let x be a vertex of maximum degree. It suffices to show that every vertex ot adjacet to x has a commo eighbor with x. Choose y / N(x. We have N(x = /2 ad N(y /2 1. Sice y x, we have N(x, N(y V (G {x, y}. Thus N(x N(y = N(x + N(y N(x N(y /2 + /2 1 ( = Strogly idepedet sets. If S is a idepedet set with o commo eighbors i a graph G, the the vertices of S have pairwisedisjoit closed eighborhoods of size at least δ(g + 1. Thus there are at most (G/(δ(G + 1 of them. Equality is achievable for the 3dimesioal cube usig S = {000, 111}. Equality is ot achievable whe G = Q 4, sice with 16 vertices ad miimum degree 4 it requires three parwisedisjoit closed eighborhoods of size 5. If v S, the o vertex differig from v i at most two places is i S. Also, at most oe vertex differig from v i at least three places is i
10 53 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 54 S, sice such vertices differ from each other i at most two places. Thus oly two disjoit closed eighborhoods ca be foud i Q Every simple graph has a vertex whose eighbors have average degree as large as the overall average degree. Let t(w be the average degree of the eighbors of w. I the sum w V (G t(w = w V (G y N(w d(y/d(w, we have the terms d(u/d(v ad d(v/d(u for each edge uv. Sice x/y + y/x 2 wheever x, y are positive real umbers (this is equivalet to (x y 2 0, each such cotributio is at least 2. Hece t(w d(u uv E(G d(v + d(v 2e(G. Hece the average of the eighborhood average degrees is at least the average degree, ad the pigeohole priciple d(u yields the desired vertex. It is possible that every average eighborhood degree exceeds the average degree. Let G be the graph with 2 vertices formed by addig a ( matchig betwee a complete graph ad a idepedet set. Sice G has 2 + edges ad 2 vertices, G has average degree ( + 1/2. For each vertex of the clique, the eighborhood average degree is 1 + 1/. For each leaf, the eighborhood average degree is Subgraphs with large miimum degree. Let G be a loopless graph with average degree a. a If x V (G, the G = G x has average degree at least a if ad oly if d(x a/2. Let a be the average degree of G, ad let be the order of G. Deletig x reduces the degree sum by 2d(x, so ( 1a = a 2d(x. Hece ( 1(a a = a 2d(x. For > 1, this implies that a a if ad oly if d(x a/2. Alterative presetatio. The average degree of G is 2e(G/(G. Sice G has e(g d(x edges, the average degree is at least a if ad oly if 2[e(G d(x] (G 1 a. Sice e(g = (Ga/2, we ca rewrite this as (Ga 2d(x = 2e(G 2d(x a(g a. By cacelig (Ga, we fid that the origial iequality is equivalet to d(x a/2. b If a > 0, the G has a subgraph with miimum degree greater tha a/2. Iteratively delete vertices with degree at most half the curret average degree, util o such vertex exists. By part (a, the average degree ever decreases. Sice G is fiite, the procedure must termiate. It eds oly by fidig a subgraph where every vertex has degree greater tha a/2. c The result of part (b is best possible. To prove that o fractio of a larger tha 1 2 a ca be guarateed, let G be a vertex tree. We have a(g = 2( 1/ = 2 2/, but subgraphs of G have miimum degree at most 1. Give β > 1 2, we ca choose large eough so that 1 βa(g Bipartite subgraphs of the Peterse graph. a Every edge of the Peterse graph is i four 5cycles. I every 5cycle through a edge e, the edge e is the middle edge of a 4vertex path. Such a path ca be obtaied i four ways, sice each edge exteds two ways at each edpoit. The eighbors at each edpoit of e are distict ad oadjacet, sice the girth is 5. Sice the edpoits of each such P 4 are oadjacet, they have exactly oe commo eighbor. Thus each P 4 yields oe 5cycle, ad each 5cycle through e arises from such a P 4, so there are exactly four 5cycles cotaiig each edge. b The Peterse graph has twelve 5cycles. Sice there are 15 edges, summig the umber of 5cycles through each edge yields 60. Sice each 5cycle is couted five times i this total, the umber of 5cycles is 12. c The largest bipartite subgraph has twelve edges. Proof 1 (breakig odd cycles. Each edge is i four 5cycles, so we must delete at least 12/4 edges to break all 5cycles. Hece we must delete at least three edges to have a bipartite subgraph. The illustratio shows that deletig three is eough; the Peterse graph has a bipartite subgraph with 12 edges (see also the cover of the text. A B A B A A B B A A Proof 2 (study of bipartite subgraphs. The Peterse graph G has a idepedet set of size 4, cosistig of the vertices {ab, ac, ad, ae} i the structural descriptio. The 12 edges from these four vertices go to the other six vertices, so this is a bipartite subgraph with 12 edges. Let X ad Y be the partite sets of a bipartite subgraph H. If X 4, the e(h 12, with equality oly whe X is a idepedet 4set i G. Hece we eed oly cosider the case X = Y = 5. To obtai e(g > 10, some vertex x X must have three eighbors i Y. The two oeighbors of x i Y have commo eighbors with x, ad these must lie i N(x, which is cotaied i Y. Hece e(g[y ] 2. Iterchagig X ad Y i the argumet shows that also e(g[x] 2. Hece e(h Whe the algorithm of Theorem is applied to a bipartite graph, it eed ot fid the bipartite subgraph with the most edges. For the bipartite graph below, the algorithm may reach the partitio betwee the upper vertices ad lower vertices.
11 55 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 56 This bipartite subgraph with eight edges has more tha half of the edges at each vertex, ad o further chages are made. However, the bipartite subgraph with the most edges is the full graph Every otrivial loopless graph G has a bipartite subgraph cotaiig more tha half its edges. We use iductio o (G. If (G = 2, the G cosists of copies of a sigle edge ad is bipartite. For (G > 2, choose v V (G that is ot icidet to all of E(G (at most two vertices ca be icidet to all of E(G. Thus e(g v > 0. By the iductio hypothesis, G v has a bipartite subgraph H cotaiig more tha e(g/2 edges. Let X, Y be a bipartitio of H. If X cotais at least half of N G (v, the add v to Y ; otherwise add v to X. The augmeted partitio captures a bipartite subgraph of G havig more tha half of E(G v ad at least half of the remaiig edges, so it has more tha half of E(G. Commet. The statemet ca also be proved without iductio. By Theorem , G has a bipartite subgraph H with at least e(g/2 edges. By the proof of Theorem , equality holds oly if d H (v = d G (v/2 for every v V (G. Give a edge uv, each of u ad v has exactly half its eighbors i its ow partite set. Switchig both to the opposite set will capture those edges while retaiig the edge uv, so the ew bipartite subgraph has more edges No fractio of the edges larger tha 1/2 ca be guarateed for the largest bipartite subgraph. If G is the complete graph K 2, the e(g = = (2 1, ad the largest bipartite subgraph is K,, which has 2 ( 2 2 edges. Hece lim f (G /e(g = lim = 1. For large eough 2, the fractio of the edges i the largest bipartite subgraph is arbitrarily close to 1/2. (I fact, i every graph the largest bipartite subgraph has more tha half the edges Every loopless graph G has a spaig kpartite subgraph H such that e(h (1 1/ke(G. Proof 1 (local chage. Begi with a arbitrary partitio of V (G ito k parts V 1,..., V k, ad cosider the kpartite subgraph H cotaiig all edges of G cosistig of two vertices from distict parts. Give a partitio of V (G, let V (x deote the part cotaiig x. If i G some vertex x has more eighbors i V j tha i some other part, the shiftig x to the other part icreases the umber of edges captured by the kpartite subgraph. Sice G has fiitely may edges, this shiftig process must termiate. It termiates whe for each x V (G the umber N(x V i is miimized by V i = V (x. The d G (x = i N G(x V i k N G (x V (x. We coclude that N G (x V (x (1/kd G (x, ad hece d H (x (1 1/kd G (x for all x V (G. By the degreesum formula, e(h (1 1/ke(G. Proof 2 (iductio o. We prove that whe G is otrivial, some such H has more tha (1 1/ke(G edges. This is true whe = 2. We procede by iductio for > 2. Choose a vertex v V (G. By the iductio hypothesis, G v has a spaig kpartite subgraph with more tha (1 1/ke(G v edges. This subgraph partitios V (G v ito k partite sets. Oe of these sets cotais at most 1/k eighbors of v. Add v to that set to obtai the desired kpartite subgraph H. Now e(h > (1 1/ke(G v + (1 1/kd G (v = (1 1/ke(G For 3, the miimum umber of edges i a coected vertex graph i which every edge belogs to a triagle is 3( 1/2. To achieve the miimum, we eed oly cosider simple graphs. Say that coected graphs with each edge i a triagle are good graphs. For = 3, the oly such graph is K 3, with three edges. Whe is odd, a costructio with the claimed size cosists of ( 1/2 triagles sharig a commo vertex. Whe is eve, add oe vertex to the costructio for 1 ad make it adjacet to both edpoits of oe edge. For the lower boud, let G be a smallest vertex good graph. Sice G has fewer tha 3/2 edges (by the costructio, G has a vertex v of degree 2. Let x ad y be its eighbors. Sice each edge belogs to a triagle, x y. If > 3, the we form G by deletig v ad, if xy have o other eighbor, cotractig xy. Every edge of G belogs to a triagle that cotaied it i G. The chage reduces the umber of vertices by 1 or 2 ad reduces the umber of edges by at least 3/2 times the reductio i the umber of vertices. By the iductio hypothesis, e(g 3((G 1/2, ad hece the desired boud holds for G Let G be a simple vertex graph. v V (G a e(g = e(g v. If we cout up all the edges i all the subgraphs 2 obtaied by deletig oe vertex, the each edge of G is couted exactly 2 times, because it shows up i the 2 subgraphs obtaied by deletig a vertex other tha its edpoits. b If 4 ad G has more tha 2 /4 edges, the G has a vertex whose deletio leaves a graph with more tha ( 1 2 /4 edges. Sice G has more tha 2 /4 edges ad e(g is a iteger, we have e(g ( 2 + 4/4 whe is eve ad e(g ( 2 + 3/4 whe is odd (sice (2k = 4k 2 + 4k + 1, every square of a odd umber is oe more tha a multiple of 4. Thus always we have e(g ( 2 + 3/4. By part (a, we have e(g v v V (G ( 2 + 3/4. I the sum we have 2 terms. Sice the largest umber i a set is at least the average, there is a vertex v such that e(g v We rewrite this as 2 4 e(g v (2 + 3( = =
12 57 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 58 Whe 4, the last term is positive, ad we obtai the strict iequality e(g v > ( 1 2 /4. c Iductive proof that G cotais a triagle if e(g > 2 /4. We use iductio o. Whe 3, they oly simple graph with more tha 2 /4 edges is whe = 3 ad G = K 3, which ideed cotais a triagle. For the iductio step, cosider 4, ad let G be a vertex simple graph with more tha 2 /4 vertices. By part (b, G has a subgraph G v with 1 vertices ad more tha ( 1 2 /4 edges. By the iductio hypothesis, G v therefore cotais a triagle. This triagle appears also i G K /2, /2 is the oly vertex triaglefree graph of maximum size. As i the proof of Matel s result, let x be a vertex of maximum degree. Sice N(x is a idepedet set, x ad its oeighbors capture all the edges, ad we have e(g ( (G (G. If equality holds, the summig the degrees i V (G N(x couts each edge exactly oce. This requires that V (G N(x also is a idepedet set, ad hece G is bipartite. If G is bipartite ad has ( (G (G edges, the G = K ( (G, (G. Hece e(g is maximized oly by K /2, / The bridge club with 14 members (o game ca be played if two of the four people table have previously bee parters: If each member has played with four others ad the six additioal games have bee played, the the arrival of a ew member allows a game to be played. We show that the ew player yields a set of four people amog which o two have bee parters. This is true if ad oly if the previous games must leave three people (i the origial 14 amog which o two have bee parters. The graph of pairs who have NOT bee parters iitially is K 14. For each game played, two edges are lost from this graph. At the breakpoit i the sessio, each vertex has lost four icidet edges, so 28 edges have bee deleted. I the remaiig six games, 12 more edges are deleted. Hece 40 edges have bee deleted. Sice e(k 14 = 91, there remai 51 edges for pairs that have ot yet bee parters. By Matel s Theorem (Theorem , the maximum umber of edges i a simple 14vertex graph with o triagle is 14 2 /4. Sice 51 > 49, the graph of remaiig edges has a triagle. Thus, whe the 15th perso arrives, there will be four people of whom oe have partered each other The miimum umber of triagles t(g i a vertex graph G ad its complemet. a t(g = ( ( 3 ( e + d(v v V (G 2. Let d1,..., d deote the vertex degrees. We prove that the right side of the formula assigs weight 1 to the vertex triples that iduce a triagle i G or G ad weight 0 to all other triples. Amog these terms, ( 3 couts all triples, ( e couts those determied by a edge of G ad a vertex off that edge, ad ( d i 2 couts 1 for each pair of icidet edges. I the table below, we group these cotributios by how may edges the correspodig triple iduces i G. ( t(g i G ( 3 ( e di edges edges edge edges b t(g ( 1( 5/24. Begi with the formula for k 3 (G + k 3 (G from part (a. Usig the covexity of quadratic fuctios, we get a lower boud for the sum o the right by replacig the vertex degrees by the average degree 2e/. The boud is ( ( 3 ( e + 2e/ 2, which reduces to ( ( 3 2e( e/. As a fuctio of e, this is miimized whe e = 1 2(. This substitutio ad algebraic simplificatio produce t(g ( 1( 5/24. Commet. The proof of part (b uses two miimizatios. These imply that equality ca hold oly for a regular graph (d i = 2e/ for all i with e = 2( 1. There is such a regular graph if ad oly if is odd ad ( 1/2 is eve. Thus we eed = 4k + 1 ad G is 2kregular Maximum size with o iduced P 4. a If G is a simple coected graph ad G is discoected, the e(g (G 2, with equality oly for K (G, (G. Sice G is discoected, (G (G/2, with equality oly if G = K (G, (G. Thus e(g = d i /2 (G (G/2 (G 2. As observed, equality whe G is discoected requires G = K (G, (G. b If G is a simple coected graph with maximum degree D ad o iduced subgraph isomorphic to P 4, the e(g D 2. It suffices by part (a to prove that G is discoected whe G is coected ad P 4 free. We use iductio o (G for (G 2; it is immediate whe (G = 2. For the iductio step, let v be a ocutvertex of G. The graph G = G v is also P 4 free, so its complemet is discoected, by the iductio hypothesis. Thus V (G v has a vertex partitio X, Y such that all of X is adjacet to all of Y i G. Sice G is coected, v has a eighbor z X Y ; we may assume be symmetry that z Y. If G is coected, the G has a v, zpath. Let y be the vertex before z o this path; ote that y Y. Also G coected requires x X such that vx E(G. Now {v, z, x, y} iduces P 4 i G Iductive proof that for d i eve there is a multigraph with vertex degrees d 1,..., d. Proof 1 (iductio o d i. If d i = 0, the all d i are 0, ad the vertex graph with o edges has degree list d. For the iductio step, suppose d i > 0. If oly oe d i is ozero, the it must be eve, ad the
13 59 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 60 graph cosistig of 1 isolated vertices plus d i /2 loops at oe vertex has degree list d (multigraphs allow loops. Otherwise, d has at least two ozero etries, d i ad d j. Replacig these with d i 1 ad d j 1 yeilds a list d with smaller eve sum. By the iductio hypothesis, some graph G with degree list d. Form G by addig a edge with edpoits u ad v to G, where d G (u = d i 1 ad d G (v = d j 1. Although u ad v may already be adjacet i G, the resultig multigraph G has degree list d. Proof 2 (iductio o. For = 1, put d 1 /2 loops at v 1. If d is eve, put d /2 loops at v ad apply the iductio hypothesis. Otherwise, put a edge from v to some other vertex correspodig to positive d i (which exists sice d i is eve ad proceed as before A tuple of oegative itegers with largest etry k is graphic if the sum is eve, k <, ad every etry is k or k 1. Let A( be the set of  tuples satisfyig these coditios. Let B( be the set of graphic tuples. We prove by iductio o that tuples i A( are also i B(. Whe = 1, the oly list i A( is (0, ad it is graphic. For the iductio step, let d be a tuple i A(, ad let k be its largest elemet. Form d from d by deletig a copy of k ad subtractig 1 from k largest remaiig elemets. The operatio is doable because k <. To apply the iductio hypothesis, we eed to prove that d A( 1. Sice we delete a istace of k ad subtract oe from k other values, we reduce the sum by 2k to obtai d from d, so d does have eve sum. Let q be the umber of copies of k i d. If q > k + 1, the d has ks ad (k 1s. If q = k + 1, the d has oly (k 1s. If q < k + 1, the d has (k 1s ad (k s. Also, if k = 1, the the first possibility caot occur. Thus d has legth 1, its largest value is less tha 1, ad its largest ad smallest values differ by at most 1. Thus d A( 1, ad we ca apply the iductio hypothesis to d. The iductio hypothesis (d A( 1 (d B( 1 tells us that d is graphic. Now the HavelHakimi Theorem implies that d is graphic. (Actually, we use oly the easy part of the HH Theorem, addig a vertex joied to vertices with desired degrees If d is a oicreasig list of oegative itegers, ad d is obtaied by deletig d k ad subtractig 1 from the k largest other elemets, the d is graphic if ad oly if d is graphic. The proof is like that of the Havel Hakimi Theorem. Sufficiecy is immediate. For ecessity, let w be a vertex of degree d k i a simple graph with degree sequece d. Alter G by 2switches to obtai a graph i which w has the d k highestdegree other vertices as eighbors. The argumet to fid a 2switch icreasig the umber of desired eighbors of w is as i the proof of the Havel Hakimi Theorem The list d = (d 1,..., d 2k with d 2i = d 2i 1 = i for 1 i k is graphic. This is the degree list for the bipartite graph with vertices x 1,..., x k ad y 1,..., y k defied by x r y s if ad oly if r +s > k. Sice the eighborhood of x r is {y k, y k 1,..., y k r+1 }, the degree of x r is r. Thus the graph has two vertices of each degree from 1 to k Necessary ad sufficiet coditios for a list d to be graphic whe d cosists of k copies of a ad k copies of b, with a b 0. Sice the degree sum must be eve, the quatity ka + ( kb must be eve. I additio, the iequality ka k(k 1 + ( k mi{k, b} must hold, sice each vertex with degree b has at most mi{k, b} icidet edges whose other edpoit has degree a. We costruct graphs with the desired degree sequece whe these coditios hold. Note that the iequality implies a 1. Case 1: b k ad a k. Begi with K k, k, havig partite sets X of size k ad Y of size k. If k(a + k ad ( k(b k are eve, the add a (a + kregular graph o X ad a (b kregular graph o Y. To show that this is possible, ote first that 0 a + k k 1 ad 0 b k a k k 1. Also, whe pq is eve, a qregular graph o p vertices i a circle ca be costructed by makig each vertex adjacet to the q/2 earest vertices i each directio ad also to the opposite vertex if q is odd (sice the p is eve. Note that k(a +k ad ( k(b k have the same parity, sice their differece ak ( kb differs from the give eve umber ka + ( kb by a eve amout. If they are both odd, the we delete oe edge from K k, k, ad ow oe vertex i the subgraph o X should have degree a + k + 1 ad oe i the subgraph o Y should have degree b k +1. Whe pq is odd, such a graph o vertices v 0,..., v p 1 i a circle (qregular except for oe vertex of degree q + 1 ca be costructed by makig each vertex adjacet to the (q 1/2 earest vertices i each directio ad the addig the edges {v i v i+(p 1/2 : 0 i (p 1/2. Note that all vertices are icidet to oe of the added edges, except that v (p 1/2 is icidet to two of them. Case 2: k 1 a < k. Begi by placig a complete graph o a set S of k vertices. These vertices ow have degree k 1 ad will become the vertices of degree a, which is okay sice a b. Put a set T of k additioal vertices i a circle. For each vertex i S, add a k +1 cosecutive eighbors i T, startig the ext set immediately after the previous set eds. Sice a 1, each vertex i S is assiged a k + 1 distict eighbors i T. Sice k(a k + 1 ( kb ad the edges are distributed early equally to vertices of T, there is room to add these edges. For the subgraph iduced by T, we eed a graph with k vertices ad [( kb k(a k + 1]/2 edges ad degrees differig by at most 1. The desired umber of edges is itegral, sice ka + ( kb is eve, ad it
14 61 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 62 is oegative, sice k(a k + 1 ( kb. The largest degree eeded is ( kb k(a k + 1 k. This is at most b, which is less tha k sice b a < k. The desired graph ow exists by Exercise Case 3: b < k ad a k. Put the set S of size k i a circle. For each vertex i the set T of size k, assig b cosecutive eighbors i S; these are distict sice b < k. Sice a k, o vertex of S receives too may edges. O S we put a almostregular graph with k vertices ad [ak b( k]/2 edges. Agai, this umber of edges is itegral, ad i the case specified it is oegative. Existece of such a graph requires a b( k/k k 1, which is equivalet to the give iequality k(a k + 1 ( kb. Now agai Exercise provides the eeded graph. Case 4: b < k ad a < mi{k 1, k}. Sice a < k, also b < k. Therefore, we ca use the idea of Case 1 without the complete bipartite graph. Agai take disjoit vertex sets X of size k ad Y of size k. If ka ad ( kb are eve, the we use a aregular graph o X ad a bregular graph o Y. As observed before, these exist. Note that ka ad ( kb have the same parity, sice their sum is give to be eve. If they are both odd, the we put mi{k, k} disjoit edges with edpoits i both X ad Y. We ow complete the graph with a regular graph of eve degree o oe of these sets ad a almostregular graph guarateed by Exercise o the other If G is a selfcomplemetary vertex graph ad is odd, the G has a vertex of degree ( 1/2. Let d 1,..., d be the degree list of G i oicreasig order. The degree list of G i oicreasig order is 1 d,..., 1 d 1. Sice G = G, the lists are the same. Sice is odd, the cetral elemets i the list yield d (+1/2 = 1 d (+1/2, so d (+1/2 = ( 1/ Whe is cogruet to 0 or 1 modulo 4, there is a vertex simple graph G with 1 2( edges such that (G δ(g 1. This is satisfied by the costructio give i the aswer to Exercise More geerally, let G be ay 2kregular simple graph with 4k + 1 vertices, where = 4k + 1. Such a graph ca be costructed by placig 4k + 1 vertices aroud a circle ad joiig each vertex to the k closest vertices i each directio. By the DegreeSum Formula, e(g = (4k + 1(2k/2 = 1 2(. For the case where = 4k, delete oe vertex from the graph costructed above to form G. Now e(g = e(g 2k = (4k 1(2k/2 = 1 2( The oegative itegers d 1 d are the vertex degrees of a loopless graph if ad oly if d i is eve ad d 1 d d. Necessity. If such a graph exists, the d i couts two edpoits of each edge ad must be eve. Also, every edge icidet to the vertex of largest degree has its other ed couted amog the degrees of the other vertices, so the iequality holds. Sufficiecy. Specify vertices v 1,..., v ad costruct a graph so that d(v i = d i. Iductio o has problems: It is ot eough to make d edges joi v 1 ad v degrees ad apply the iductio hypothesis to (d 1 d, d 2,..., d 1. Although d 1 d d d 1 holds, d 1 d may ot be the largest of these umbers. Proof 1 (iductio o d i. The basis step is d i = 0, realized by a idepedet set. Suppose that d i > 0; we cosider two cases. If d 1 = i=2 d i, the the desired graph cosists of d 1 edges from v 1 to v 2,..., v. If d 1 < i=2 d i, the the differece is at least 2, because the total degree sum is eve. Also, at least two of the values after d 1 are ozero, sice d 1 is the largest. Thus we ca subtract oe from each of the last two ozero values to obtai a list d to which we ca apply the iductio hypothesis (it has eve sum, ad the largest value is at most the sum of the others. To the resultig G, we add oe edge joiig the two vertices whose degrees are the reduced values. (This ca also be viewed as iductio o ( i=2 d i d 1. Proof 2 (iductio o d i. Basis as above. Cosider d i > 0. If d 1 > d 2, the we ca subtract 1 from d 1 ad from d 2 to obtai d with smaller sum. Still d 1 1 is a largest value i d ad is bouded by the sum of the other values. If d 1 = d 2, the we subtract 1 from each of the two smallest values to form d. If these are d 1 ad d 2, the d has the desired properties, ad otherwise i=2 d i exceeds d 1 by at least 2, ad agai d has the desired properties. I each case, we ca apply the iductio hypothesis to d ad complete the proof as i Proof 1. Proof 3 (local chage. Every oegative iteger sequece with eve sum is realizable whe loops ad multiple edges are allowed. Give such a realizatio with a loop, we chage it to reduce the umber of loops without chagig vertex degrees. Elimiatig them all produces the desired realizatio. If we have loops at distict vertices u ad v, the we replace two loops with two copies of the edge uv. If we have loops oly at v ad have a edge xy betwee two vertices other tha v, the we replace oe loop ad oe copy of xy by edges vx ad vy. Such a edge xy must exist because the sum of the degrees of the other vertices is as large as the degree of v A simple graph with degree sequece d 1 d 2 d is coected if d j j for all j such that j 1 d. Let V (G = {v 1,..., v }, with d(v i = d i, ad let H be the compoet of G cotaiig v ; ote that H has at least 1 + d vertices. If G is ot coected, the G has aother compoet H. Let j be the umber of vertices i H. Sice H has at least 1 + d vertices, we have j 1 d. By the hypothesis, d j j. Sice H has j vertices, its maximum degree is at least d j. Sice d j j, there are at
15 63 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 64 least j + 1 vertices i H, which cotradicts the defiitio of j. Hece G is i fact coected If D = {a i } is a set of distict positive itegers, with 0 < a 1 < < a k, the there is a simple graph o a k + 1 vertices whose set of vertex degrees (repetitio allowed is D. Proof 1 (iductive costructio. We use iductio o k. For k = 1, use K a1 +1. For k = 2, use the joi K a1 K a2 a That is, G cosists of a clique Q with a 1 vertices, a idepedet set S with a 2 a vertices, ad all edges from Q to S. The vertices of S have degree a 1, ad those of Q have degree a 2. For k 2, take a clique Q with a 1 vertices ad a idepedet set S with a k a k 1 vertices. Each vertex of S has eighborhood Q, ad each vertex of Q is adjacet to all other vertices. Other vertices have a 1 eighbors i Q ad oe i S, so the degree set of G Q S should be {a 2 a 1,..., a k 1 a 1 }. By the iductio hypthesis, there is a simple graph H with a k 1 a vertices havig this degree set (the degree set is smaller by. Usig H for G Q S completes G as desired. Proof 2 (iductio ad complemetatio. Agai use iductio o k, usig K a1 +1 whe k = 1. For k > 1 ad 0 < a 1 < a k, the complemet of the desired graph with a vertices has degree set {a k a 1,..., a k a k 1, 0}. By the iductio hypothesis, there is a graph of order a k a with degree set {a k a 1,, a k a k 1 }. Add a 1 isolated vertices ad take the complemet to obtai the desired graph G Costructio of cubic graphs ot obtaiable by expasio aloe. A simple cubic graph G that caot be obtaied from a smaller cubic graph by the expasio operatio is the same as a cubic graph o which o erasure ca be performed, sice ay erasure yieldig a smaller H from G could be iverted by a expasio to obtai G from H. A edge caot be erased by this operatio if ad oly if oe of the subsequet cotractios produces a multiple edge. This happes if the other edges icidet to the edge beig erased belog to a triagle, or i oe other case, as idicated below. ad Fially, we eed oly provide a simple cubic graph with 4k vertices where every edge is oerasable i oe of these two ways. To do this place copies of G 1,..., G k of K 4 e (the uique 4vertex graph with 5 edges aroud i a rig, ad for each cosecutive pair G i, G i+1 add a edge joiig a pair of vertices with degree two i the subgraphs, as idicated below, where the wraparoud edge has bee cut Costructio of 3regular simple graphs a A 2switch ca be performed by performig a sequece of expasios ad erasures. We achieve a 2switch usig two expasios ad the two erasures as show below. If the 2switch deletes xy ad zw ad itroduces xw ad yz, the the first expasio places ew vertices u ad v o xy ad zw, the secod itroduces s ad t o the resultig edges ux ad vz, the first erasure deletes su ad its vertices, ad the secod erasure deletes tv ad its vertices. The resultig vertices are the same as i the origial graph, the erasures were legal because they created oly edges that were ot preset origially, ad we have deleted xy ad zw ad itroduced xw ad yz. y z x w y z y z s t u v u v x w x w y z v t x w y z x w b Every 3regular simple graph ca be obtaied from K 4 by a sequece of expasios ad erasures. Erasure is allowed oly if o multiple edges result. Suppose H is the desired 3regular graph. Every 3regular graph has a eve umber of vertices, at least four. Ay expasio of a 3regular graph is a 3regular graph with two more vertices. Hece successive expasios from K 4 produce a 3regular graph G with (H vertices. Sice G ad H have the same vertex degrees, there is a sequece of 2switches from G to H. Sice every 2switch ca be produced by a sequece of expasios ad erasures, we ca costruct a sequece of expasios ad erasures from K 4 to H by goig through G If G ad H are X, Y bigraphs, the d G (v = d H (v for all v X Y if ad oly if there is a sequece of 2switches that trasforms G ito H without ever chagig the bipartitio. The coditio is sufficiet, sice 2 switches do ot chage vertex degrees. For ecessity, assume that d G (v = d H (v for all v. We build a sequece of 2switches trasformig G to H. Proof 1 (iductio o X. If X = 1, the already G = H, so we may assume that X > 1. Choose x X ad let k = d(x. Let S be a selectio of k vertices of highest degree i Y. If N(x S, choose y S ad y Y S so that x y ad x y. Sice d(y d(y, there exists x X so that y x ad y x. Switchig xy, x y for xy, x y icreases
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