Eulerian and Hamiltonian Graphs
There are many games and puzzles which can be analysed by graph theoretic concepts.
In fact, the two early discoveries which led to the existence of graphs arose from puz- zles, namely, the Konigsberg Bridge Problem and Hamiltonian Game, and these puzzles also resulted in the special types of graphs, now called Eulerian graphs and Hamiltonian graphs. Due to the rich structure of these graphs, they find wide use both in research and application.
3.1 Euler Graphs
A closed walk in a graphGcontaining all the edges ofGis called an Euler line inG. A graph containing an Euler line is called an Euler graph.
We know that a walk is always connected. Since the Euler line (which is a walk) contains all the edges of the graph, an Euler graph is connected except for any isolated vertices the graph may contain. As isolated vertices do not contribute anything to the understanding of an Euler graph, it is assumed now onwards that Euler graphs do not have any isolated vertices and are thus connected.
Example Consider the graph shown in Figure 3.1. Clearly,v1e1v2e2v3e3v4e4v5e5v3v6
e7v1in (a) is an Euler line, whereas the graph shown in (b) is non-Eulerian.
Fig. 3.1
The following theorem due to Euler [74] characterises Eulerian graphs. Euler proved the necessity part and the sufficiency part was proved by Hierholzer [115].
Theorem 3.1 (Euler) A connected graphGis an Euler graph if and only if all vertices ofGare of even degree.
Proof
Necessity LetG(V, E) be an Euler graph. ThusGcontains an Euler lineZ, which is a closed walk. Let this walk start and end at the vertexu∈ V. Since each visit of Z to an intermediate vertexvofZcontributes two to the degree ofvand sinceZtraverses each edge exactly once, d(v) is even for every such vertex. Each intermediate visit tou contributes two to the degree ofu, and also the initial and final edges ofZcontribute one each to the degree ofu. So the degreed(u)ofuis also even.
Sufficiency Let G be a connected graph and let degree of each vertex of G be even.
AssumeGis not Eulerian and letGcontain least number of edges. Sinceδ ≥ 2, Ghas a cycle. LetZbe a closed walk inGof maximum length. Clearly,G− E(Z)is an even degree graph. LetC1be one of the components ofG− E(Z). AsC1has less number of edges than G, it is Eulerian and has a vertexvin common withZ. LetZ0be an Euler line inC1. Then Z0∪ Zis closed inG, starting and ending atv. Since it is longer thanZ, the choice ofZ is contradicted. HenceGis Eulerian.
Second proof for sufficiency Assume that all vertices ofGare of even degree. We con- struct a walk starting at an arbitrary vertexvand going through the edges ofGsuch that no edge ofGis traced more than once. The tracing is continued as far as possible. Since every vertex is of even degree, we exit from the vertex we enter and the tracing clearly cannot stop at any vertex butv. Asvis also of even degree, we reach vwhen the tracing comes to an end. If this closed walkZwe just traced includes all the edges ofG, thenGis an Euler graph. If not, we remove fromGall the edges inZand obtain a subgraphZ0ofG formed by the remaining edges. Since bothGandZhave all their vertices of even degree, the degrees of the vertices ofZ0are also even. Also,Z0 touchesZat least at one vertex say u, becauseGis connected. Starting fromu, we again construct a new walk inZ0. As all the vertices ofZ0are of even degree, therefore this walk inZ0terminates at vertexu. This walk inZ0combined withZforms a new walk, which starts and ends at the vertexvand has more edges thanZ. This process is repeated till we obtain a closed walk that traces all the edges
ofG. HenceGis an Euler graph (Fig. 3.2) q
Fig. 3.2
3.2 Konigsberg Bridge Problem
Two islandsAandB formed by the Pregal river (now Pregolya) in Konigsberg (then the capital of east Prussia, but now renamed Kaliningrad and in west Soviet Russia) were connected to each other and to the banksCandDwith seven bridges. The problem is to start at any of the four land areas,A, B, C, orD, walk over each of the seven bridges exactly once and return to the starting point.
Euler modeled the problem representing the four land areas by four vertices, and the seven bridges by seven edges joining these vertices. This is illustrated in Figure 3.3.
Konigsberg Bridges C
A B
D
Underlying Graph C
A
D
B
Fig. 3.3
We see from the graphGof the Konigsberg bridges that not all its vertices are of even degree. ThusGis not an Euler graph, and implies that there is no closed walk inGcon- taining all the edges ofG. Hence it is not possible to walk over each of the seven bridges exactly once and return to the starting point.
Note Two additional bridges have been built since Euler’s day. The first has been built between land areasC andDand the second between the land areas AandB. Now in the graph of Konigsberg bridge problem with nine bridges, every vertex is of even degree and the graph is thus Eulerian. Hence it is now possible to walk over each of the nine bridges exactly once and return to the starting point (Fig 3.4).
Fig. 3.4
The following characterisation of Eulerian graphs is due to Veblen [254].
Theorem 3.2 A connected graphGis Eulerian if and only if its edge set can be decom- posed into cycles.
Proof LetG(V, E)be a connected graph and let Gbe decomposed into cycles. If kof these cycles are incident at a particular vertex v, thend(v) = 2k. Therefore the degree of every vertex ofGis even and henceGis Eulerian.
Conversely, letG be Eulerian. We showGcan be decomposed into cycles. To prove this, we use induction on the number of edges.
Sinced(v) ≥ 2for eachv∈ V, Ghas a cycleC. ThenG− E(C)is possibly a disconnected graph, each of whose componentsC1, C2, . . ., Ckis an even degree graph and hence Eule- rian. By the induction hypothesis, eachCiis a disjoint union of cycles. These together with
Cprovide a partition ofE(G)into cycles. q
The following result is due to Toida [244].
Theorem 3.3 IfW is a walk from vertex uto vertexv, thenW contains an odd number ofu− vpaths.
Proof LetW be a walk which we consider as a graph in itself, and not as a subgraph of some other graph. Letuandvbe initial and final vertices of the walkW. Clearly,d(u|W ) andd(v|W )are odd, andd(w|W )is even, for everyw∈ V (W ) − {u, v}. We count the number of distinctu− vwalks inW. These walks are the subgraphs ofW.
When we take a u− v walk by successively selecting the edges e1, e2, . . ., es, initial vertex ofe1beinguand terminal vertex ofesbeingv, for each edge there are an odd number of choices. The total number of such edges is the product of these odd numbers and is therefore odd. Now from these walks, we find theu− vpaths. If a u− v walkW1is not a path, then it contains one or more cycles. The traversal of these cycles in the two possible alternative directions (clockwise and anticlockwise) produces in all an even number of walks, all with the same edge set asW1. Omitting these even number of walks which are not paths from the total odd collection ofu− vwalks, gives an odd number ofu− vpaths.
q
Toida [244] proved the necessity part and McKee [157] the sufficiency part of the next characterisation. The second proof of this result can be found in Fleischner [79], [80].
Theorem 3.4 A connected graph is Eulerian if and only if each of its edges lies on an odd number of cycles.
Proof
Necessity LetGbe a connected Eulerian graph and let e= uvbe any edge ofG. Then G− eis au− vwalkW, and soG− e = W contains an odd number ofu− vpaths. Thus each of the odd number ofu− vpaths inW together withegives a cycle inGcontainingeand these are the only such cycles. Therefore there are an odd number of cycles inGcontaining e.
Sufficiency LetGbe a connected graph so that each of its edges lies on an odd number of cycles. Letvbe any vertex ofGandEv= {e1, . . ., ed}be the set of edges ofGincident on v, then|Ev| = d(v) = d. For eachi, 1 ≤ i ≤ d, letkibe the number of cycles ofGcontaining ei. By hypothesis, eachkiis odd. Letc(v)be the number of cycles ofGcontainingv. Then clearlyc(v) = 12 ∑d
i=1
kiimplying that2c(v) = ∑d
i=1
ki. Since2c(v)is even and eachki is odd,d is
even. HenceGis Eulerian. q
Corollary 3.1 The number of edge−disjoint paths between any two vertices of an Euler graph is even.
A consequence of Theorem 3.4 is the result of Bondy and Halberstam [37], which gives yet another characterisation of Eulerian graphs.
Corollary 3.2 A graph is Eulerian if and only if it has an odd number of cycle decom- positions.
Proof In one direction, the proof is trivial. IfGhas an odd number of cycle decomposi- tions, then it has at least one, and henceGis Eulerian.
Conversely, assume thatG is Eulerian. Lete∈ E(G) and let C1, . . ., Cr be the cycles containinge. By Theorem 3.4,ris odd. We proceed by induction on m= |E(G)|, withG being Eulerian.
IfGis just a cycle, then the result is true. Now assume thatGis not a cycle. This means that for eachi, 1 ≤ i ≤ r, by the induction assumption,Gi= G − E(Ci)has an odd number, saysi, of cycle decompositions. (IfGiis disconnected, apply the induction assumption to each of the nontrivial components ofGi). The union of each of these cycle decompositions ofGiandCiyields a cycle decomposition ofG. Hence the number of cycle decompositions ofGcontainingCiissi, 1 ≤ i ≤ r. Lets(G)denote the number of cycle decompositions of G. Then
s(G) ≡ ∑r
i=1
si≡ r(mod 2) (sincesi≡ 1(mod 2))
≡ 1(mod 2). q
Two examples of Euler graphs are shown in Figure 3.5.
Fig. 3.5
3.3 Unicursal Graphs
An open walk that includes (or traces) all edges of a graph without retracing any edge is called a unicursal line or open Euler line. A connected graph that has a unicursal line is called a unicursal graph. Figure 3.6 shows a unicursal graph.
Fig. 3.6
Clearly by adding an edge between the initial and final vertices of a unicursal line, we get an Euler line.
The following characterisation of unicursal graphs can be easily derived from Theorem 3.1.
Theorem 3.5 A connected graph is unicursal if and only if it has exactly two vertices of odd degree.
Proof LetGbe a connected graph and letGbe unicursal. ThenGhas a unicursal line, say fromutov, whereuandvare vertices ofG. Joinuandvto a new vertexwofGto get a graphH. ThenHhas an Euler line and therefore each vertex ofHis of even degree. Now, by deleting the vertexw, the degree of verticesuandveach get reduced by one, so thatu andvare of odd degree.
Conversely, letuandvbe the only vertices ofGwith odd degree. Joinuandvto a new vertexwto get the graphH. So every vertex ofH is of even degree and thusHis Eulerian.
Therefore,G= H − whas au− vunicursal line so thatGis unicursal. q
The following result is the generalisation of Theorem 3.5.
Theorem 3.6 In a connected graphGwith exactly2kodd vertices, there exists kedge disjoint subgraphs such that they together contain all edges ofGand that each is a unicursal graph.
Proof LetGbe a connected graph with exactly2k odd vertices. Let these odd vertices be namedv1, v2, . . ., vk; w1, w2, . . ., wkin any arbitrary order. Addkedges toGbetween the vertex pairs(v1, w1), (v2, w2), . . ., (vk, wk)to form a new graphH, so that every vertex ofHis of even degree. ThereforeH contains an Euler lineZ.
Now, if we remove fromZthekedges we just added (no two of these edges are incident on the same vertex), then Z is divided intok walks, each of which is a unicursal line.
The first removal gives a single unicursal line, the second removal divides that into two unicursal lines, and each successive removal divides a unicursal line into two unicursal
lines, until there arekof them. Hence the result. q
3.4 Arbitrarily Traceable Graphs
An Eulerian graphGis said to be arbitrarily traceable (or randomly Eulerian) from a vertex vif every walk with initial vertexvcan be extended to an Euler line ofG. A graph is said to be arbitrarily traceable if it is arbitrarily traceable from every vertex (Fig. 3.7).
Fig. 3.7
The following characterisation of arbitrarily traceable graphs is due to Ore [174]. Such graphs were also characterised by Chartrand and White [56] .
Theorem 3.7 An Eulerian graphGis arbitrarily traceable from a vertexvif and only if every cycle ofGpasses throughv.
Proof
Necessity Let the Eulerian graphGbe arbitrarily traceable from a vertexv. Assume there is a cycleCnot passing throughv. LetH= G − E(C). Then every vertex ofHhas an even degree and the component ofH containingv is Eulerian. This component of H can be traversed as an Euler lineZ, starting and ending withvand contains all those edges ofG which are incident atv. Clearly, thisv− vwalk cannot be extended to contain the edges of Calso, contradicting thatGcontainsv. Thus every cycle inGcontainsv.
Sufficiency Let every cycle of the Eulerian graphGpass through the vertexvofG. We show thatGis arbitrarily traceable fromv. Assume, on the contrary, thatGis not arbitrarily traceable from v. Then there is av− vclosed walkW ofGcontaining all the edges ofG incident withvand yet not containing all the edges ofG. Let one such edge be incident at a vertex u onW. So every vertex of H= G − E(W )is of even degree andvis an isolated vertex ofHanduis not. The component ofHcontaininguis therefore Eulerian subgraph ofGnot passing throughv, contradicting the assumption. Hence the result follows. q
Corollary 3.3 Cycles are the only arbitrarily traceable graphs.
3.5 Sub-Eulerian Graphs
A graphGis said to be sub-Eulerian if it is a spanning subgraph of some Eulerian graph.
The following characterisation of sub-Eulerian graphs is due to Boesch, Suffel and Tin- dell [28].
Theorem 3.8 A connected graphGis sub-Eulerian if and only ifGis not spanned by a complete bipartite graph.
Proof
Necessity We prove that no spanning supergraphHof an odd complete bipartite graphG is Eulerian. LetV1∪V2be the bipartition of the vertex set ofG. Since degree of each vertex ofGis odd, andGis complete bipartite, therefore|V1|and|V2|are odd. IfH1is the induced subgraph ofH onV1, then at least one vertex, sayv, ofV1has even degree inH1, since|V1| is odd. But thend(v|H) = d(v|H) + |V2|, which is odd. ThereforeHis not Eulerian.
Sufficiency Refer Boesch et. al., [28]. q
Super-Eulerian graphs
A non-Eulerian graphGis said to be super-Eulerian if it has a spanning Eulerian subgraph.
The following sufficient conditions for super-Eulerian graphs are due to Lesniak-Foster and Williams [148].
Theorem 3.9 If a graphGis such thatn≥ 6,δ ≥ 2andd(u) + d(v) ≥ n − 1, for every pair of non-adjacent verticesuandv, thenGis super-Eulerian.
The following result is due to Balakrishnan and Paulraja [12].
Theorem 3.10 IfGis any connected graph and if each edge ofGbelongs to a triangle inG, thenGhas a spanning Eulerian subgraph.
Proof SinceGhas a triangle,Ghas a closed walk. LetW be the longest closed walk in G. ThenW must be a spanning Eulerian subgraph ofG. If not, there exists a vertexv∈ W/ andvis adjacent to a vertex uofW. By hypothesis,uv belongs to a triangle, sayuvw. If none of the edges of this triangle is inW, thenW∪ {uv, vw, wu}yields a closed walk longer thanW (Fig. 3.8). Ifuw∈ W, then(W − uw) ∪ {uv, vw}would be a closed walk longer than W. This contradiction proves thatW is a spanning closed walk inG. q
Fig. 3.8
3.6 Hamiltonian Graphs
A cycle passing through all the vertices of a graph is called a Hamiltonian cycle. A graph containing a Hamiltonian cycle is called a Hamiltonian graph. A path passing through all the vertices of a graph is called a Hamiltonian path and a graph containing a Hamiltonian path is said to be traceable. Examples of Hamiltonian graphs are given in Figure 3.9.
Fig. 3.9
If the last edge of a Hamiltonian cycle is dropped, we get a Hamiltonian path. However, a non-Hamiltonian graph can have a Hamiltonian path, that is, Hamiltonian paths cannot always be used to form Hamiltonian cycles. For example, in Figure 3.10,G1has no Hamil- tonian path, and so no Hamiltonian cycle;G2has the Hamiltonian pathv1v2v3v4, but has no Hamiltonian cycle, whileG3has the Hamiltonian cyclev1v2v3v4v1.
Fig. 3.10
Hamiltonian graphs are named after Sir William Hamilton, an Irish Mathematician (1805−1865), who invented a puzzle, called the Icosian game, which he sold for25guineas to a game manufacturer in Dublin. The puzzle involved a dodecahedron on which each of the20vertices was labelled by the name of some capital city in the world. The aim of the game was to construct, using the edges of the dodecahedron a closed walk of all the cities which traversed each city exactly once, beginning and ending at the same city. In other words, one had essentially to form a Hamiltonian cycle in the graph corresponding to the dodecahedron. Figure 3.11 shows such a cycle.
Fig. 3.11
Clearly, then-cycleCnwithndistinct vertices (andnedges) is Hamiltonian. Now, given any Hamiltonian graphG, the supergraph G0 (obtained by adding in new edges between non-adjacent vertices ofG) is also Hamiltonian. This is because any Hamiltonian cycle in Gis also a Hamiltonian cycle ofG0. For instance,Knis a supergraph of ann-cycle and so Knis Hamiltonian.
A multigraph or general graph is Hamiltonian if and only if its underlying graph is Hamiltonian, because if G is Hamiltonian, then any Hamiltonian cycle in G remains a Hamiltonian cycle in the underlying graph ofG. Conversely, if the underlying graph of a graphGis Hamiltonian, thenGis also Hamiltonian.
LetG be a graph withn vertices. Clearly, Gis a subgraph of the complete graph Kn. FromG, we construct step by step supergraphs ofGto getKn, by adding an edge at each step between two vertices that are not already adjacent (Fig. 3.12).
Fig. 3.12
Now, let us start with a graphGwhich is not Hamiltonian. Since the final outcome of the procedure is the Hamiltonian graphKn, we change from a non-Hamiltonian graph to a Hamiltonian graph at some stage of the procedure. For example, the non-Hamiltonian
graphG1above is followed by the Hamiltonian graphG2. Since supergraphs of Hamilto- nian graphs are Hamiltonian, once a Hamiltonian graph is reached in the procedure, all the subsequent supergraphs are Hamiltonian.
Definition: A simple graphGis called maximal non-Hamiltonian if it is not Hamiltonian and the addition of an edge between any two non-adjacent vertices of it forms a Hamilto- nian graph. For example, G1 above is maximal non-Hamiltonian. Figure 3.13 shows a maximal non-Hamiltonian graph.
Fig. 3.13
It follows from the above procedure that any non-Hamiltonian graph withn-vertices is a subgraph of a maximal non-Hamiltonian graph withnvertices.
The above procedure is used to prove the following sufficient conditions due to Dirac [68].
Theorem 3.11 (Dirac) IfGis a graph withnvertices, wheren≥ 3andd(v) ≥ n/2, for every vertexvofG, thenGis Hamiltonian.
Proof Assume that the result is not true. Then for some value n≥ 3, there is a non- Hamiltonian graphH in which d(v) ≥ n/2, for every vertex of H. In any spanning super graphK (i.e., with the same vertex set) of H, d(v) ≥ n/2for every vertex ofK, since any proper supergraph of this form is obtained by adding more edges. Thus there is a maximal non-Hamiltonian graphGwithnvertices andd(v) ≥ n/2for everyvinG. Using thisG, we obtain a contradiction.
Clearly,G6= Kn, asKnis Hamiltonian. Therefore there are non-adjacent verticesu and vinG. LetG+ uvbe the supergraph of Gby adding an edge betweenu andv. SinceG is maximal non-Hamiltonian,G+ uvis Hamiltonian. Also, ifCis a Hamiltonian cycle of G+ uv, thenCcontains the edgeuv, since otherwiseCis a Hamiltonian cycle ofG, which is not possible. Let this Hamiltonian cycleCbeu= v1, v2, . . ., vn= v, u.
Now, letS= {vi∈ C :there is an edge fromutovi+1inG}andT = {vj∈ C :there is an edge fromvtovjinG}.
Thenvn∈ T/ , since otherwise there is an edge fromvtovn= v, that is a loop, which is impossible.
Alsovn∈ S/ , (takingvn+1 asv1), since otherwise we again get a loop from u tov1= u. Therefore,vn∈ S ∪ T(Fig. 3.14).
Fig. 3.14
Let |S|, |T | and |S ∪ T | be the number of elements inS, T and S∪ T respectively. So
|S ∪ T | < n. Also, for every edge incident with u, there corresponds one vertex vi in S. Therefore,|S| = d(u). Similarly,|T | = d(v).
Now, ifvk is a vertex belonging to bothSandT, there is an edgeejoiningutovk+1and an edge f joiningvtovk. This implies thatC0= v1, vk+1, vk+2, . . ., vn, vk, vk−1, . . ., v2, v1is a Hamiltonian cycle inG, which is a contradiction as Gis non-Hamiltonian. This shows that there is no vertexvkinS∩ T, so thatS∩ T = Φ.
Thus|S ∪ T | = |S| + |T| − |S ∩ T |gives|S| + |T| = |S ∪ T|, so thatd(u) + d(v) < n. This is a contradiction, becaused(u) ≥ n/2 for all u inG, and so d(u) + d(v) ≥ n/2 + n/2 giving
d(u) + d(v) ≥ n. Hence the theorem follows. q
The following result is due to Ore [176].
Theorem 3.12 (Ore) LetGbe a graph withn vertices and letu andvbe non-adjacent vertices inGsuch thatd(u) + d(v) ≥ n. LetG+ uvdenote the super graph ofGobtained by joininguandvby an edge. ThenGis Hamiltonian if and only ifG+ uvis Hamiltonian.
Proof LetG be a graph withn vertices and supposeuand vare non-adjacent vertices inGsuch thatd(u) + d(v) ≥ n. LetG+ uvbe the super graph ofGobtained by adding the edgeuv. Let Gbe Hamiltonian. Then obviouslyG+ uv is Hamiltonian. Conversely, let G+ uvbe Hamiltonian. We have to show thatGis Hamiltonian. Then, as in Theorem 3.11, we getd(u) + d(v) < n, which contradicts the hypothesis that d(u) + d(v) ≥ n. Hence Gis
Hamiltonian. q
The following is the proof of Bondy [35] of Theorem 3.12, and this proof bears a close resemblance to the proof of Dirac’s theorem given by Newman [170], but is more direct.
Proof (Bondy [35]) Consider the complete graphK on the vertex set ofGin which the edges ofGare coloured blue and the remaining edges ofK are coloured red. LetC be a
Hamiltonian cycle ofKwith as many blue edges as possible. We show that every edge of Cis blue, in other words, thatCis Hamiltonian cycle ofG.
Suppose to the contrary,Chas a red edgeuu−(whereu−is the successor ofuonC). Con- sider the setSof vertices joined touby blue edges (that is, the set of neighbours ofuinG).
The successoru−ofuonCmust be joined by a blue edge to some vertexv−ofS−, because ifu−is adjacent inConly to verticesV− (S−U{u−}),dG(u) + dG(u−) = |NG(u)| + |NG(u−)| ≤
|S| + (|V | − |S−| − 1) = |V (G)| − 1, contradicting the hypothesis thatdG(u) + dG(u−) ≥ |V (G)|, uandu−being non-adjacent inG. But now the cycleCobtained fromCby exchanging the edgesuu−andvv−has more blue edges thanC, which is a contradiction. q
Definition: Let Gbe a graph withn vertices. If there are two non-adjacent verticesu1
andv1inGsuch thatd(u1) + d(v1) ≥ n, joinu1andv1by an edge to form the super graph G1. Now, if there are two non-adjacent verticesu2andv2inG1such thatd(u2) + d(v2) ≥ n, joinu2andv2by an edge to form supergraphG2. Continue in this way, recursively joining pairs of non-adjacent vertices whose degree sum is at least nuntil no such pair remains.
The final supergraph thus obtained is called the closure ofGand is denoted byc(G). The example in Figure 3.15 illustrates the closure operation.
Fig. 3.15
We observe in this example that there are different choices of pairs of non-adjacent verticesuandvwithd(u) + d(v) ≥ n. Therefore the closure procedure can be carried out in several different ways and each different way gives the same result.
In the graph shown in Figure 3.16,n= 7andd(u) + d(v) < 7, for any pairu, vof adjacent vertices. Therefore,c(G) = G.
G
Fig. 3.16
The importance ofc(G)is given in the following result due to Bondy and Chvatal [36].
Theorem 3.13 A graphGis Hamiltonian if and only if its closurec(G)is Hamiltonian.
Proof Letc(G)be the closure of the graphG. Sincec(G)is a supergraph ofG, therefore, ifGis Hamiltonian, thenc(G)is also Hamiltonian.
Conversely, letc(G)be Hamiltonian. LetG, G1, G2, . . ., Gk−1, Gk= c(G)be the sequence of graphs obtained by performing the closure procedure onG. Sincec(G) = Gkis obtained fromGk−1by settingGk= Gk−1+ uv, whereu, vis a pair of non adjacent vertices inGk−1 withd(u) + d(v) ≥ n, therefore it follows thatGk−1is Hamiltonian. SimilarlyGk−2, soGk−3,
. . .,G1and thusGis Hamiltonian. q
Corollary 3.4 LetGbe a graph withnvertices withn≥ 3. Ifc(G)is complete, thenGis Hamiltonian.
There can be more than one Hamiltonian cycle in a given graph, but the interest lies in the edge-disjoint Hamiltonian cycles. The following result gives the number of edge-disjoint Hamiltonian cycles in a complete graph with odd number of vertices.
Theorem 3.14 In a complete graph with n vertices there are (n − 1)/2 edge-disjoint Hamiltonian cycles, ifnis an odd number,n≥ 3.
Proof A complete graphGofnvertices hasn(n − 1)/2edges and a Hamiltonian cycle in Gcontainsnedges. Therefore the number of edge-disjoint Hamiltonian cycles inGcannot exceed(n − 1)/2. Whenn is odd, we show there are(n − 1)/2edge-disjoint Hamiltonian cycles.
The subgraph of a complete graph withn vertices shown in Figure 3.17 is a Hamiltonian cycle. Keeping the vertices fixed on a circle, rotate the polygonal pattern clockwise by
360
n−1, 2.n−1360, . . ., n−32 .n−1360 degrees. We see that each rotation produces a Hamiltonian cycle that has no edge in common with any of the previous ones. Therefore, there are(n − 3)/2 new Hamiltonian cycles, all disjoint from the one in Figure 3.17, and also edge-disjoint among themselves. Thus there are(n − 1)/2edge disjoint Hamiltonian cycles. q
Fig. 3.17
The next result involving degrees give the sufficient conditions for a graph to be Hamil- tonian.
Theorem 3.15 LetD= [di]n1be a degree sequence of a graphG= (V, E), d1≤ d2≤ . . . ≤ dn. Each of the following gives the sufficient conditions forGto be Hamiltonian.
A. 1≤ k ≤ n ⇒ dk≥n
2 (Dirac [68]) B. uv∈ E ⇒ d(u) + d(v) ≥ n/ (Ore [176]) C. 1≤ k ≤n2⇒ dk> k(Posa [210]).
D. j< k, dj≤ janddk≤ k − 1 ⇒ dj+ dk≥ n(Bondy [33]) E. dk≤ k < n2⇒ dn−k≥ n − k(Chvatal [59])
F. For every iand j with 1≤ i ≤ n, 1 ≤ j ≤ n, i + j ≥ n, vivj∈ E/ , d(vi) ≤ i andd(vj) ≤ j− 1 ⇒ d(vi) + d(vj) ≥ n(Las Vergnas [256].
G. c(G)is complete (Bondy and Chvatal [36]).
Proof We first prove that A⇒ B(i) (ii)⇒C(iii)⇒ D(iv)⇒ E⇒ F(v) (vi)⇒ G. i. This can be easily established.
ii. Assume that (C) is not true, so that there exists akwith1≤ k <n2 anddk≤ k. Then the induced subgraph on the verticesv1, v2, . . ., vkis a complete graph. For, if there are verticesiand jwith1≤ i < j ≤ kandvivj∈ E/ , thendi+ dj≤ 2dk< n, contradicting (B). Sincedk≤ k, eachvi, 1 ≤ i ≤ k, is adjacent to at most onevj, k + 1 ≤ j ≤ n. Also, n− k > k, becausek<n2. Therefore there is a vertexvj, k + 1 ≤ j ≤ nnot adjacent to any of the verticesv1, v2, . . ., vk. For thisvj, we havedj≤ n − k − 1. But thendj+ dk≤ (n − k − 1) + k = n − 1. Thus there is avjvk∈ E/ withdj+ dk≤ n − 1, contradicting (B).
Hence proving (ii).
iii. Assume that (D) is not true, so that there exist jandkwith j< k, dj≤ j, dk≤ k −1and dj+ dk< n. This givesi= dj<n2. But thendj≤ j givesddj ≤ dj, since the sequence is non-decreasing. Therefore, di≤ dj= i. Thus there is an i, 1 ≤ i ≤ n2 withdi≤ i, contradicting (C). This proves (iii).
iv. If(E)is not true, there is akwithdk≤ k <n2anddn−k≤ n −k −1. Thendk+dn−k≤ n −1. Settingn− k = j, we havek< j, dk≤ k, dj≤ j − 1anddj+ dk≤ n − 1. This contradicts (D) and so (iv) is proved.
v. Assume that (F) is not true, so that there is a pair of vertices vi and vj, i < j with vivj∈ E/ and violating (F). Choosei to be the least such possible integer. Then by minimality ofi, di−1> i − 1. Thusdi≥ di−1≥ iand sincedi≤ i, we obtaindi= i. If
i≥ n2, we get di+ dj≥ 2di≥ n, contradicting the violation of (F). Therefore, i<n2. Thus there is ani, 1 ≤ i <n2 withdi= i. Now, if (E) is satisfied, we havedn−i≥ n − i and since j≥ n − i, we obtaindj≥ dn−i≥ n − i. By minimality ofi, di= iand we have dj+ di≥ (n − i) + i = n, again contradicting the violation of (F). Thus negation of (F) implies negation of (E) and (V) is established.
vi. Assume thatc(G) = His not complete. Let vi andvj be non-adjacent vertices inH such that (a) j is as large as possible and (b) iis as large as possible subject to (a).
Theni< j, and sinceHis the closure ofG, therefore
d(vi|H) + d(vj|H) ≤ n − 1, (3.15.1)
d(vi|G) + d(vj|G) ≤ n − 1.
By the choice of j, vi is adjacent inHto allvkwithk> j, so that
d(vi|H) ≥ n − j. (3.15.2)
Again, by the choice ofi, vj is adjacent inHto allvkwithk> i, k 6= j, so that
d(vj|H) ≥ n − i − 1. (3.15.3)
From (3.15.1) and (3.15.2), we have
d(vj|G) ≤ d(vj|H) ≤ (n − 1) − (n − j) = j − 1. From (3.15.1) and (3.15.3), we have
d(vj|G) ≤ d(vi|H) ≤ (n − 1) − (n − i − 1) = i. From (3.15.2) and (3.15.3), we have
i+ j ≥ (2n − 1) − d(vi|H) − d(vj|H) ≥ n. (using (1)) Therefore i and j contradict the given conditions. Thus H= c(G) is complete. This proves (vi).
By Theorem 3.13 it follows that if (G) holds, thenGis Hamiltonian. q
The next result is due to Nash-Williams [168].
Theorem 3.16 (Nash-Williams) Everyk-regular graph on2k+ 1vertices is Hamilto- nian.
Proof LetGbe a k-regular graph on2k+ 1vertices. Add a new vertex wand join it by an edge to each vertex ofG. The resulting graphH on2k+ 2vertices hasδ= k + 1. Thus by Theorem 3.15 (A),His Hamiltonian. RemovingwfromH, we get a Hamiltonian path, sayv0v1. . .v2k.
Assume thatGis not Hamiltonian, so that (a) ifv0vi∈ E, thenvi−1v2k∈ E/ , (b) ifv0vi∈ E/ , thenvi−1v2k∈ E, sinced(v0) = d(v2k) = k.
The following cases arise.
Case (i) v0 is adjacent tov1, v2, . . ., vk, andv2k is adjacent tovk, vk+1, . . ., v2k−1. Then there is aniwith1≤ i ≤ ksuch thatvi is not adjacent to somevj for0≤ j ≤ k( j 6= i). But d(vi) = k. Soviis adjacent tovj for some jwithk+ 1 ≤ j ≤ 2k − 1. Then the cycleCgiven byvivi−1. . .v0vi+1. . .vj−1v2kv2k+1. . .vjis a Hamiltonian cycle ofG(Fig 3.18).
Fig. 3.18
Case (ii) There is an i with 1≤ i ≤ 2k − 1 such thatvi+1v0∈ E, but viv0∈ E/ . Then by (b), vi−1v2k ∈ E. Thus G contains the2k-cycle vi−1vi−2. . .v0vi+1. Renaming the 2k-cycle C as u1u2. . .u2k and letu0 be the vertex of G not on C. Then u0 cannot be adjacent to two consecutive vertices onCand hence u0is adjacent to every second vertex onC, say u1, u3, . . ., u2k−1. Replacingu2ibyu0, we obtain another maximum cycleC0ofGand hence u2imust be adjacent tou1, u3, . . ., u2k−1. But thenu1is adjacent tou0, u2, . . ., u2k, implying d(u1) ≥ k + 1. This is a contradiction and henceGis Hamiltonian. q
3.7 Pancyclic Graphs
Definition: A graphGof ordern(≥ 3)is pancyclic ifGcontains all cycles of lengths from 3 ton. Gis called vertex-pancyclic if each vertexvofGbelongs to a cycle of every length`, 3 ≤ ` ≤ n.
Example Clearly, a vertex-pancyclic graph is pancyclic. However, the converse is not true. Figure 3.19 displays a pancyclic graph that is not vertex-pancyclic.
Fig. 3.19
The result of pancyclic graphs was initiated by Bondy [34], who showed that Ore’s sufficient condition for a graphGto be Hamiltonian (Theorem 6.2.5) actually implies much more. Note that ifδ ≥n2, thenm≥n22. The proof of the following result due to Thomassen can be found in Bollobas [29].
Theorem 3.17 Let G be a simple Hamiltonian graph on n vertices with at least hn22
i
edges. ThenGis either pancyclic or else is the complete bipartite graphKn
2,n2. In particular, ifGis Hamiltonian andm>n42, thenGis pancyclic.
Proof The result can easily be verified forn= 3. We may therefore assume thatn≥ 4. We apply induction onn. Suppose the result is true for all graphs of order at mostn− 1(n ≥ 4), and letGbe a graph of ordern.
First, assume thatGhas a cycleC= v0v1. . .vn−2v0of lengthn− 1. Letvbe the (unique) vertex of Gnot belonging toC. Ifd(v) ≥n2, vis adjacent to two consecutive vertices on Cand henceGhas a cycle of length 3. Suppose for somer, 2 ≤ r ≤n−12 , Chas no pair of verticesu andwonCadjacent tovinGwithdC(u, w) = r. Then ifvi1, vi2, . . .vid(v) are the vertices ofCthat are adjacent tovinG(recall thatCcontains all the vertices ofGexceptv), thenvi1+r, vi2+r, . . ., vid(v)+r are nonadjacent tovinG, where the suffixes are taken modulo (n− 1). Thus,2d(v) ≤ n − 1, a contradiction. Hence, for eachr, 2 ≤ r ≤n−12 , Chas a pair of verticesuandwonCadjacent tovinGwithdC(u, w) = r. Thus for eachr, 2 ≤ r ≤n−12 , G has a cycle of lengthr+ 2as well as a cycle of lengthn− 1 − r + 2 = n − r + 1(Fig. 3.20).
ThusGis pancyclic.
Fig. 3.20
Ifd(v) ≤n−12 , thenG[V (C)], the subgraph ofGinduced byV(C)has at least n42−n−12 >
(n−1)2
4 edges. So, by the induction assumption,G[V (C)]is pancyclic and henceGis pan- cyclic. (By hypothesis,Gis Hamiltonian).
Next, assume thatGhas no cycle of lengthn− 1. ThenGis not pancyclic. In this case, we show thatGisKn
2,n2.
LetC= v0v1v2. . .vn−1v0be a Hamilton cycle ofG. We claim that of the two pairsvivkand vi+1vk+2 (where suffixes are taken modulon), at most only one of them can be an edge of G. Otherwise,vkvk−1vk−2. . .vi+1vk+2vk+3vk+4. . .vivk is an(n − 1)-cycle inG, a contradiction.
Hence, ifd(vi) = r, then there arervertices adjacent tovi inGand hence at leastrvertices (includingvi+1 sincevivi−1∈ E(G)) that are nonadjacent tovi+1. Thus,d(vi+1) ≤ n − r and d(vi) + d(vi+1) ≤ n.
Summing the last inequality overifrom 0 ton− 1, we get4m≤ n2. But by hypothesis, 4m≥ n2. Hence,m=n42 and sonmust be even.
This givesd(vi) + d(vi+1) = nfor eachi, and thus for eachiandk, exactly one ofvivkand
vi+1vk+2is an edge ofG. (3.17.1)
Thus, ifG6= Kn
2 ,n
2, then certainly there existiandjsuch thatvivj∈ E andi≡ j(mod 2).
Hence for somej, there exists an even positive integerssuch thatvj+1vj+1+s∈ E. Chooses to be the least even positive integer with the above property. Thenvjvj+1+s∈ E/ . Hence,s≥ 4 (ass= 2would mean thatvjvj+1∈ E/ ). Again, by (3.17.1),vj−1vj+s−3= vj−1vj−1+(s−2)∈ E(G) contradicting the choice of s. Thus,G= Kn
2 ,n2. The last part follows from the fact that
|E(Kn
2 ,n2, )| =n42. q
Theorem 3.18 LetG6= Kn
2 ,n2, be a simple graph withn≥ 3vertices and letd(u)+d(v) ≥ n for every pair of non-adjacent vertices ofG. ThenGis pancyclic.
Proof By Ore’s Theorem (Theorem 3.12), Gis Hamiltonian. We show that Gis pan- cyclic by first proving thatm≥n42 and then invoking Theorem 3.17. This is true ifδ ≥n2 (as2m= ∑n
i=1
di≥δn≥ n2/2). So assume thatδ <n2.
LetS be the set of vertices of degreeδ inG. For every pair(u, v)of vertices of degree δ, d(u) + d(v) < n2+n2= n. Hence by hypothesis,Sinduces a clique ofGand|S| ≤δ+ 1. If
|S| =δ+ 1, thenGis disconnected withG[S]as a component, which is impossible (asGis Hamiltonian). Thus,|S| ≤δ. Further, ifv∈ S, vis nonadjacent ton− 1 −δ vertices ofG. If uis such a vertex,d(v) + d(u) ≥ nimplies thatd(u) ≥ n −δ. Further,vis adjacent to at least one vertexw∈ S/ andd(w) ≥δ+ 1, by the choice ofS. These facts give that
2m=
n
∑
i=1
di≥ (n −δ− 1)(n −δ) +δ2+ (δ+ 1),
where the last (δ+ 1) comes out of the degree ofw. Thus, 2m≥ n2− n(2δ+ 1) + 2δ2+ 2δ+ 1,
which implies that
4m≥ 2n2− 2n(2δ+ 1) + 4δ2+ 4δ+ 2
= (n − (2δ+ 1)2+ n2+ 1
≥ n2+ 1, sincen> 2δ. Consequently,m>n2
4 , and by Theorem 3.17,Gis pancyclic. q
3.8 Exercises
1. Prove that the wheelWnis Hamiltonian for everyn≥ 2, andn-cubeQnis Hamiltonian for eachn≥ 2.
2. IfGis ak-regular graph with2k− 1vertices, then prove thatGis Hamiltonian.
3. Show that if a cubic graphGhas a spanning closed walk, thenGis Hamiltonian.
4. IfG= G(X, Y )is a bipartite Hamiltonian graph, then show that|X|=|Y |.
5. Prove that for each n≥ 1, the complete tripartite graph Kn, 2n, 3n is Hamiltonian, but Kn, 2n, 3n+1is not Hamiltonian.
6. How many spanning cycles are there in the complete bipartite graphsK3, 3andK4, 3? 7. Prove that a graphGwithn≥ 3vertices is arbitrarily traceable if and only if it is one
of the graphs Cn, Knor Kn, nwithn= 2p.
8. Prove that a graph G with n≥ 3 vertices is randomly traceable if and only if it is randomly Hamiltonian.
9. Find the closure of the graph given in Figure 3.2. Is it Hamiltonian?
10. Does there exist an Eulerian graph with
i. an even number of vertices and an odd number of edges, ii. and odd number of vertices and an even number of edges.
Draw such a graph if it exists.
11. Characterise graphs which are both Eulerian and Hamiltonian.
12. Characterise graphs which possess Hamiltonian paths but not Hamiltonian cycles.
13. Characterise graphs which are unicursal but not Eulerian.
14. Give an example of a graph which is neither pancyclic nor bipartite, but whose n- closure is complete.
