We give a sufficient condition for a distance-regular graph to be Hamiltonian. In particular, the Petersen graph is the only connected non-Hamiltonian strongly regular graph on fewer than 99 vertices.Distance-regular graphs;Hamilton cycles JEL-code

Brouwer, A.E.

Haemers, W.H.

English

Research Papers in Economics

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
No. 2008–28 
 
 
HAMILTONIAN STRONGLY REGULAR GRAPHS 
 
 
By A.E. Brouwer, W.H. Haemers  
 
 
 
March 2008 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
ISSN 0924-7815 Hamiltonian strongly regular graphs
A. E. Brouwer & W. H. Haemers
Abstract
We give a suﬃcient condition for a distance-regular graph to be Hamil-
tonian. In particular, the Petersen graph is the only connected non-
Hamiltonian strongly regular graph on fewer than 99 vertices.
Keywords: Distance-regular graphs, Hamilton cycles. JEL-code C0.
1 Introduction
For a graph Γ, let α(Γ) denote the independence number (the maximum
size of an independent set) and let κ(Γ) denote the vertex connectivity
(the minimum number of vertices one has to remove from Γ in order to
disconnect it; note that the graph without vertices is not connected).
Chv´ atal & Erd˝ os [5] proved that every graph Γ on at least three ver-
tices, with α(Γ) ≤ κ(Γ) is Hamiltonian. Hoﬀman showed (cf. [1], 1.3.2)
that every regular graph Γ on n vertices with valency k (k > 0) and
smallest eigenvalue s satisﬁes α(Γ) ≤ −ns/(k − s). It follows that every
k-regular k-connected graph on n > 2 vertices with −ns/(k − s) < k + 1
is Hamiltonian.
However, for many interesting k-regular k-connected graphs we have
−ns/(k − s) = k + 1. This includes the Petersen graph (which has
(n,k,s) = (10,3,−2)). Here we show:
Theorem 1 Let Γ be a k-regular k-connected graph with n vertices and
smallest eigenvalue s which is not the Petersen graph. If k > 1 and
−ns ≤ (k + 1)(k − s) then Γ is Hamiltonian.
Brouwer & Koolen [3] proved that every non-complete distance-regular
graph of degree k is k-connected. Hence
Corollary 2 Let Γ be a k-regular distance-regular graph with n vertices
and smallest eigenvalue s which is not the Petersen graph. If k > 1 and
−ns ≤ (k + 1)(k − s) then Γ is Hamiltonian.
Several distance-regular graphs satisfy this inequality (and some with
equality). For connected strongly regular graphs (i.e., distance regular
graphs of diameter 2), this becomes
Corollary 3 Let Γ be a connected strongly regular graph with parameters
(n,k,λ, ) and smallest eigenvalue s. If s is not integral, or if −s ≤  +1,
then Γ is Hamiltonian.
(Here   is the number of common neighbors of two nonadjacent vertices.)
This corollary covers most admissible parameter sets for strongly regular
graphs. We conjecture that the Petersen graph is the only connected
strongly regular graph that is not Hamiltonian.
An important ingredient of the proof of the theorem is the following
characterization of the case of equality in Hoﬀman’s bound (cf. [1], 1.3.2).
1Lemma 4 Let Γ be a k-regular graph (k > 0) with n vertices and smallest
eigenvalue s. Then α(Γ) ≤ −ns/(k − s). A coclique S satisﬁes the bound
with equality, if and only if every vertex not in S is adjacent to exactly
−s vertices of S.
2 Rearranging cycles
Suppose the graph Γ is connected and non-Hamiltonian, with at least one
cycle. Let C be a longest cycle. Provide C with a cyclic order, and write
x
+ (resp. x
−) for the successor (predecessor) of x on C. Write D
+ (resp.
D
−) for {u
+ | u ∈ D} (resp. {u
− | u ∈ D}) for any subset D of C. When
describing subpaths of C, use u...v for the path from u to v in the cyclic
order of C, and u,,,v for the path from u to v in the opposite direction.
For a set of vertices X, let N(X) be the set of vertices not in X but with
a neighbor in X. Let H be a component of Γ \ C. Let K = N(H). Then
K ⊆ C.
(i) If u,v ∈ K and v = u
+, then uHv...u is a longer cycle, if uHv
denotes a path joining u and v via vertices (at least one) in H. Contra-
diction. So, K ∩ K
− = K ∩ K
+ = ∅ and K is a vertex cut of Γ and
|K| = |K
+| = |K
−| ≥ κ(Γ).
(ii) If u,v ∈ K and u
+ is adjacent to v
+, then uHv,,,u
+v
+...u is a
longer cycle, contradiction. So K
+ (and K
−) are cocliques nonadjacent
to any point of H, and α(Γ) ≥ |K| + α(H). This argument proves the
Chv´ atal-Erd˝ os theorem.
Now assume that α(Γ) = κ(Γ)+1. Then κ(Γ) = |K| and α(H) = 1, so H
is a clique. Assume moreover that whenever S is a coclique of size α(Γ),
each vertex outside S has at least two neighbors in S.
Claim: |C| = n − 1.
Proof. Suppose h,h
′ / ∈ C, with h ∈ H, and consider the coclique
S = K
+∪{h} of size α(Γ). Since h
′ / ∈ S the vertex h
′ has at least
two neighbours in S. Since K ∩ K
+ = ∅, it follows that h
′ belongs to a
component H
′  = H of Γ \ C and has two neighbors u
+ and v
+ in K
+.
But now uhv,,,u
+h
′v
+...u is a longer cycle. Contradiction. 2
Assume that Γ is not a complete bipartite graph Kt,t+1. Put H = {h}.
Claim: K
+ ∩ K
− = ∅.
Proof. Suppose a = u
+ = v
− with u,v ∈ K. Then C
′ = hv...uh is
another maximum cycle, missing a. Let A be the set of neighbors of a.
Assume there exists a w ∈ K \ A. The set S = A
− ∪ {h} is a coclique of
size α(Γ), and w
− / ∈ S, and w
− / ∈ K, so w
− has at least one neighbor b in
A
− \{a}. If b lies between a and w, then hw...ab
+...w
−b,,,vh is a longer
cycle. If b lies between w and a, then hw...bw
−,,,ab
+...uh is a longer
cycle. So there is no such vertex w, and hence K = A.
If K
+ = K
− then κ(Γ) = (n − 1)/2. Since each vertex has at least
κ(Γ) neighbours, Γ is Kt,t+1, which was excluded.
Hence we may assume that v was chosen such that v
+ ∈ K
+ \ K
−.
Let w be the ﬁrst vertex in C after v such that w
− ∈ K
− \ K
+. Now
w
− has at least two neighbors in K
+∪{h}, so certainly one neighbor
x
+ ∈ K
+ is diﬀerent from v
+ and hence lies between w and a. But now
hw...xa,,,x
+w
−,,,vh is a longer cycle, contradiction. 2
Now assume that whenever S is a coclique of size α(Γ), each vertex outside
S has at least three neighbors in S.
2Call an edge {a,b} ∈ E(Γ) a special cord of C if a,b ∈ C, {a,b} / ∈ E(C),
and a or b lies in K
+ ∪ K
−. The length of a special cord {a,b} is the
length of the shortest (undirected) path between a and b via C. Let {a,b}
be a special cord of minimum length for a cycle C, where the minimum is
taken over all possible special cords and maximum cycles. Assume that
C is oriented from a to b.
If a
+ / ∈ K
+ then there is a vertex u
+ ∈ K
+ \ {a,a
++}, adjacent to
a
+. Since {a,b} is a minimal special cord, u
+ is not in a...b. Similarly,
if b
− / ∈ K
+ there is a vertex v
+ ∈ K
+ \ {b
−−,b} adjacent to b
−, but not
in a...b. Now hu,,,v
+b
−,,,a
+u
+...ab...vh or hv,,,u
+a
+...b
−v
+...ab...uh
is a longer cycle.
In a similar way we ﬁnd a longer cycle if a
+ / ∈ K
− and b
− / ∈ K
−. It
is impossible that a
+ ∈ K
+ and b
− ∈ K
− because K ∩ (K
+ ∪ K
−) = ∅.
Therefore a
+ ∈ K
−, b
− ∈ K
+ and a ∈ K
+ or b ∈ K
−. By symmetry, we
may assume a ∈ K
+.
Now C
′ = ha
−,,,ba...b
−−h is another maximum cycle, missing b
−. If
a
++  = b
−−, then C
′ has a special cord {a
++,h}, which is shorter than
the special cord {a,b} of C, contradiction. So a
++ = b
−−. (Note that the
Petersen graph has this structure.) Let E be the set of neighbors of b
−
distinct from b,b
−−. Then S = E
− ∪ {b
−,a,h} is a coclique of size α(Γ).
We have a
+ ∈ K
−, so a
+ has a neighbor u
+ in K
+ \ {a}. Now
C
′′ = ha
−,,,u
+a
+ab...uh is a circuit of length n−2 missing b
−− and b
−.
Since b
−− / ∈ S, it follows that b
−− has a neighbor c
− ∈ E
−. Now
P = cb
−b
−−c
− is a path of length 3. Since b
− ∈ K
+ we have c  = u
+ so
{c
−,c} is an edge of C
′′. Replacing this edge by P gives a Hamiltonian
circuit. This proves:
Proposition 5 Suppose that α(Γ) ≤ κ(Γ) + 1, and that each vertex out-
side any coclique S of size α(Γ) has at least 3 neighbours in S. Then Γ is
Hamiltonian, or Γ is Kt,t+1 (with t ≥ 2). 2
Corollary 6 Let Γ be a k-regular and k-connected graph with smallest
eigenvalue s < −2. If −ns/(k − s) ≤ k + 1, then Γ is Hamiltonian. 2
3 Smallest eigenvalue −2
Suppose Γ is connected and k-regular with smallest eigenvalue −2. Then Γ
is a line graph or belongs to a ﬁnite list of 187 so called regular exceptional
graphs (see [4] or [6]). These graphs and their spectra are known ([6],
Table A3). For these graphs Hoﬀman’s coclique bound equals 2n/(k +2),
and for only ﬁve cases this number is equal to k +1. All ﬁve have n = 10
and k = 3 (see [6], p.218 for the drawings of these ﬁve graphs). One is
the Petersen graph, but the other four are easily seen to be Hamiltonian.
So we can conclude:
Proposition 7 The Petersen graph is the only non-Hamiltonian k-regular
exceptional graph with a coclique of size k + 1.
Next we deal with the case that Γ is a k-regular line graph.
Proposition 8 A k-connected k-regular line graph Γ on n (n ≥ 3) ver-
tices satisfying α(Γ) = 2n/(k + 2) ≤ k + 1, is Hamiltonian.
Proof. Let Γ be the line graph of Γ
′ with n
′ vertices. Note that a coclique
in Γ corresponds to a matching in Γ
′. It is well-known that Γ
′ is either
regular or bipartite biregular. Assume Γ
′ is bipartite biregular. Let n
′
1
3and n
′
2 (n
′
1 ≤ n
′
2) be the sizes of the two parts with degrees k
′
1 and k
′
2,
respectively. Then k = k
′
1+k
′
2−2, n = n
′
1k
′
1 = n
′
2k
′
2, and k
′
1 ≥ k
′
2. Clearly,
a matching in Γ
′ has at most n
′
1 edges, so α(Γ) ≤ n
′
1. On the other hand,
2n/(k + 2) = 2n
′
1k
′
1/(k
′
1 + k
′
2) ≥ n
′
1 with equality only if k
′
1 = k
′
2. This
shows that Γ
′ is regular (of degree k
′ say).
A sequence (e1,...,eℓ) of ℓ distinct edges of a graph is called sequential
if ei and ei+1 intersect for i = 1,...,ℓ−1. A graph is called sequential if it
admits a sequential sequence (e1,...,em) consisting of all edges, such that
e1 intersects em. It is clear that a graph is sequential if and only if its line
graph is Hamiltonian. We will show that Γ
′ is sequential. If k
′ is even,
then Γ
′ is Eulerian and therefore sequential. So assume that k
′ is odd.
Note that n = n
′k
′/2 and k = 2(k
′ −1). Since α(Γ) = 2n/(k +2) = n
′/2,
Γ
′ has a perfect matching M (say). Consider the graph Γ
′′, obtained from
Γ
′ by deleting M. If Γ
′′ is connected, Γ
′′ is Eulerian, and Γ
′ is easily seen
to be sequential. If Γ
′′ is not connected, it consists of two disjoint (k
′−1)-
regular components, both with n
′/2 vertices, and all edges of M meet both
components. (Indeed, since Γ is k-connected, and n
′/2 = α(Γ) ≤ k + 1,
Γ
′ cannot be separated by removing fewer than n
′/2 edges.) The two
components Γ1 and Γ2 of Γ
′′ are Eulerian. Consider two edges {w,x}
and {y,z} from M such that {x,y} is an edge in Γ2. From an Eulerian
cycle of Γ2 one easily constructs a sequential sequence S of Γ
′, such that
S contains all edges of Γ2 and M, starts with {w,x} and ends with {y,z}.
If {w,z} is an edge of Γ1 we can insert S in an Eulerian tour in Γ1 right
before or after {w,z} and ﬁnd that Γ
′ is sequential. If w and z are not
adjacent, they have a common neighbor v in Γ1 (since 2k
′ > n
′/2). Now
consider the graph obtained from Γ2 by deleting the edges {v,w} and
{v,z} and adding the edges {w,x} and {x,y} and {y,z}. Now we can
insert S in an Eulerian tour of this latter graph, minus the three added
edges, and also insert the two removed edges again, and ﬁnd that Γ
′ is
sequential. 2
The two propositions above together with the theorem of Chv´ atal and
Erd˝ os prove the main theorem if s = −2. If s = −1, then Γ = Kn, and
the result is trivial. With Corollary 6, this completes the proof of the
main theorem.
4 Examples
Several distance-regular graphs of small diameter satisfy the inequality
of Corollary 2. An interesting inﬁnite family consists of the collinearity
graphs of generalized quadrangles GQ(r,t) from which a spread has been
deleted (see [1], p.385). These graphs have diameter 3 and (n,k,s) =
((r + 1)(rt + 1),rt,−t). They satisfy the condition (with equality), and
hence are Hamiltonian. The smallest of these has r = t = 2, and is
isomorphic to the line graph of the Petersen graph.
The distance-regular graphs with diameter 2 are precisely the con-
nected strongly regular graphs. For these graphs the connectivity was
determined already in [2]. Only few parameter sets for connected strongly
regular graphs have −s >   + 1. For n ≤ 100 there are only ﬁve cases:
4# (v,k,λ, ) s name Hamiltonian
1 (50,7,0,1) −3 Hoﬀman-Singleton graph yes
2 (56,10,0,2) −4 Gewirtz graph yes
3 (77,16,0,4) −6 M22 graph yes
4 (99,14,1,2) −4 ?
5 (100,22,0,6) −8 Higman-Sims graph yes
In cases 1, 2, 3 and 5 there is a unique strongly regular graph with the
given parameters, and the graph is Hamiltonian by direct veriﬁcation. (In
case 5 Hamiltonicity also follows from the fact that the smallest eigenvalue
has multiplicity 22, so that α ≤ 22.)
In case 4, existence of a strongly regular graph is open. Thus, all con-
nected strongly regular graphs on fewer than 99 vertices are Hamiltonian.
We conjecture that the Petersen graph is the only connected non-
Hamiltonian strongly regular graph.
References
[1] A. E. Brouwer, A. M. Cohen & A. Neumaier, Distance-Regular Graphs,
Springer-Verlag, 1989.
[2] A. E. Brouwer & D. M. Mesner, The connectivity of strongly regular
graphs, Europ. J. Combin. 6 (1985) 215–216.
[3] A. E. Brouwer & J. H. Koolen, The vertex-connectivity of a distance-
regular graph, Europ. J. Combin. (to appear).
[4] F. C. Bussemaker, D. Cvetkovi´ c & J. J. Seidel, Graphs related to ex-
ceptional root systems, T. H. Report 76-WSK-05, Technical University
Eindhoven, 1976.
[5] V. Chv´ atal & P. Erd˝ os, A note on Hamiltonian circuits, Discr. Math.
2 (1972) 111–113.
[6] D. Cvetkovi´ c, P. Rowlinson & S. Simi´ c, Spectral Generalizations of
Line Graphs: On graphs with least eigenvalue −2, Cambridge Univ.
Press, 2004.
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Hamiltonian Strongly Regular Graphs

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