Solutions for the problems of problem set 2
1. In general we have d(d − 1 − a) = (n − d − 1)b. Since in our case a = 0, b = 2
. Recall that
we have n = 1 + d(d+1)
2
(
)
1
2d + (n − 1)(a − b)
m± =
n−1∓ √
2
(a − b)2 + 4(d − b)
must be integers as they are the multiplicities of the non-trivial eigenvalues of
G. Note that 2d + (n − 1)(a − b) = 2(d + 1 − n) can only be 0 if G is the
complete graph Kn on n vertices, but then a = n√
− 2, so n = 3, d = 2 = 11 + 1.
If 2d + (n − 1)(a − b) = 2(d + 1 − n) ̸= 0, then (a − b)2 + 4(d − b) must be
rational, and if it’s rational then it is actually integer. Hence (a−b)2 +4(d−b) =
22 = 4 + 4(d − 2) = 4(d − 1) is a perfect square. Then d = r2 + 1 for some r.
2. (a) It is easy to see that this is a 2(n − 1)-regular graph, if two vertices are
adjacent then they have n−2 common neighbors (the vertices in their common
row or column), and if two vertices are not adjacent then they have 2 common
neighbors. So the parameters are (n2 , 2(n − 1), n − 2, 2).
By Theorem 3.3 of the lecture note, the eigenvalues of G are
2
2(n − 1), n − 2(2(n−1)) , −2((n−1) ) .
(b) Let Γ be an abelian group or order n, for instance the cyclic group Cn .
(Actually, it is not necessary that Γ should be abelian at all.) Let us consider
Γ × Γ and S = {(a, 0) | a ̸= 0} ∪ {(0, b) | b ̸= 0}. Then G = Cay(Γ × Γ, S).
(Actually, Γ1 × Γ2 would be just as good.) All characters of Γ × Γ can be
obtained as follows: let χi and χj be two character of Γ and let χij be a
function for which
χij (a, b) = χi (a)χj (b).
It’s easy to see that χij is a character of Γ × Γ, and since Γ × Γ has n2
characters and we just found n2 characters, it is enough to see that these
characters are pairwise different for different pairs of i and j which is indeed
true: it is easy ∑
to see that they are orthogonal with respect to the scalar
product ⟨f, g⟩ = x∈Γ×Γ f (x)g(x).
Now the eigenvalues of G is
∑
∑
∑
∑
∑
λij =
χij (s) =
χi (a)χj (0) +
χi (0)χj (b) =
χi (a) +
χj (b).
s∈S
a̸=0
b̸=0
a̸=0
Note that if χ is not the trivial character then
∑
∑
χ(a) =
χ(a) − χ(1) = 0 − 1 = −1.
a̸=0
a
1
b̸=0
2
while for the trivial character we have
∑
χ(a) = n − 1.
a̸=0
So if χi and χj are not the trivial characters then λij = −1 − 1 = −2, this
happens in (n − 1)2 cases. If exactly one of them is a trivial character then
λij = n − 1 − 1 = n − 2, this happens in 2(n − 1) cases. Finally, if both are
trivial characters then λij = 2(n − 1).
Remark 0.1. This result can be obtained from Practice Problem 2 of the first
problem set too as G = Kn Kn .
3. Let the Laplacian eigenvalues of G be λ1 ≥ · · · ≥ λn = 0. Similarly, let
the Laplacian eigenvalues of G be λ1 ≥ · · · ≥ λn = 0. We have seen that
λi = n − λn−i for i = 1, . . . , n − 1. Then
n−1
n−1
n
1∏
1∏
1 ∏
1
τ (G) =
λi =
(n − λn−i ) = 2
(n − λi ) = 2 L(G, n).
n i=1
n i=1
n i=1
n
4. The complement of G is Kk ∪ (n − k)K1 which has Laplacian-eigenvalues
k (k−1) , 0(n−k+1) . So G has Laplacian-eigenvalues n(n−k) , (n − k)(k−1) , 0. Hence
1∏
1
τ (G) =
λi = nn−k (n − k)k−1 = (n − k)k−1 nn−k−1 .
n i=1
n
n−1
–regular since if x2 ≡ y 2 ( mod p) then x ≡ ±y ( mod p).
5. Note that Pp is p−1
2
Now we need to determine the number of common neighbors of two vertices.
It is enough to determine the number of common neighbors of 0 and a, since
x and y has clearly the same number of vertices as 0 and x − y (we can rotate
everything with x − y). If c is a common neighbor of 0 and a, then c = u2 for
some u ̸= 0, and c − a = v 2 , so a = u2 − v 2 = (u + v)(u − v). Note that it
works backward too: if a = de then with u = (d + e)/2 and v = (d − e)/2 we
have a = u2 − v 2 , hence c = u2 is a common neighbor of 0 and a. Note that
(±u, ±v) will give the same solution. If d ̸= 0, then e = ad (here we use that Fp
is field). So there are p − 1 solutions for d and e, and this gives p−1
solution for
4
2 2
(u , v ). The only problem is that if u or v is 0. If u = 0 then −a = v 2 , it has
no solution if a is not a quadratic residue and has to solutions if a (implying
−a) is a quadratic residue. If v = 0 then a = u2 , it has no solution if a is not
a quadratic residue and has to solutions if a is a quadratic residue. Hence if
common neighbors. If a is
a is not a quadratic residue then 0 and a has p−1
4
3
not a quadratic residue then 0 and a has p−5
common neighbors. Hence Pp is
4
p−1 p−5 p−1
strongly regular with parameters (p, 2 , 4 , 4 ).
(b) The √eigenvalues of the adjacency matrix of Pp is p−1
with multiplicity 1
2
−1± p
p−1
and 2 with multiplicity 2 for both
signs. Since L(G) = p−1
I − A(G),
2
√
p± p
p−1
the Laplacian-eigenvalues of Pp are 2 with multiplicity 2 for both signs
and 0 with multiplicity 1. Then
(
(
)
√ )(p−1)/2 (
√ )
p − p (p−1)/2 1 p2 − p (p−1)/2
1 p+ p
τ (Pp ) =
=
.
p
2
2
p
4
Sketch of a second solution to part (a): First we show that G is strongly regular
without determining the parameters of G by considering its symmetries. Note
that for all a ∈ Fp , the map φa for which φa (x) = x + a is an automorphism
of the graph and since the orbit of Γ = {φa | a ∈ Fp } is the whole vertex set,
we get that Pp is regular. Note that if a = t2 for some t ̸= 0, then Ψa (x) = ax
is also an automorphism of Pp , since if u − v = x2 , then au − av = (tx)2 . Note
that any pair of adjacent (non-adjaccent) vertices can be mapped to any other
adjacent (non-adjaccent) pair of vertices by a composition of ϕa and Ψt2 , so
Pp is strongly regular.
Finally, if a is a quadratic non-residue then Ψa (x) = ax maps Pp to its
complement. So Pp and its complement are isomorphic. If G is a strongly
regular graph with parameters (n, d, a, b), then its complement is also strongly
regular with parameters (n, n − d − 1, n − 2d − 2 + b, n − 2d + a) (why?).
In our case n = p. This means that d = p−1
, and a = b − 1. Finally, from
2
d(d − 1 − a) = b(n − d − 1), one can deduce that b = p−1
. Hence the parameters
4
p−5 p−1
of Pp are (p, p−1
,
,
).
2
4
4
6. Note that by counting the number of√edges two different ways we get that
n1 d1 = n2 d2 . First we show that λ = d1 d2 is the largest eigenvalue of G.
√
√
Indeed, consider the vector v which takes the value n2 on A, and n1 on B,
then for a vertex a ∈ A we have
∑ √
√
√
√ ? ∑
λ · va = d1 d2 · n2 =
vb =
n1 = d1 n1 ,
b∈N (a)
b∈N (a)
which is indeed true, because n1 d1 = n2 d2 . Similarly, for b ∈ B we have
∑ √
√
√
√ ? ∑
va =
n2 = d2 n2
λ · vb = d1 d2 · n1 =
a∈N (b)
a∈N (b)
4
since n1 d1 = n2 d2 . Note that v is√a non-negative vector, so λ must be the
largest eigenvalue of G, i.e., µ1 = d1 d2 . Since G is bipartite, the vector v ′
√
√
which takes the value
n
on
A,
and
−
n1 on B is an eigenvector belonging
2
√
to µn = −λ = − d1 d2 .
From now on we repeat the argument which used to prove the expander
mixing lemma with slight modifications. Let χS and χT be the characteristic
vectors of the sets U and V : so χU (u) = 1 if u ∈ U and 0 otherwise. Observe
that
e(U, V ) = χTU AχV .
Let us write up χU and χV in the orthonormal basis u1 , . . . , un of eigenvectors,
where n = n1 + n2 . Note that we can choose u1 to be √2n11 n2 v, and un to be
√ 1
v ′ . Let
2n1 n2
n
∑
χU =
αi ui
i=1
and
χV =
n
∑
βi ui .
i=1
Then
χTU AχV =
n
∑
µi αi βi .
i=1
√
n2
|U |
=√
.
α1 = (χU , u1 ) = |U | √
2n1 n2
2n1
√
n2
|U |
=√
.
αn = (χU , un ) = |U | √
2n1 n2
2n1
√
n1
|V |
β1 = (χV , u1 ) = |T | √
=√
.
2n1 n2
2n2
√
n1
|V |
βn = (χV , un ) = −|T | √
= −√
.
2n1 n2
2n2
Here we used that U ⊆ A and V ⊆ B.
Hence
√
√
d1 |U ||V |
|U | |V |
d1 d2
√
= |U ||V | √
=
.
µ1 α1 β1 = d1 d2 √
2 n1 n2
2n2
2n1 2n2
Note that
and
√
√
d1 |U ||V |
|U | −|V |
d1 d2
√
µn αn βn = − d1 d2 √
= |U ||V | √
=
.
2 n1 n2
2n2
2n1 2n2
5
Hence
d1 |U ||V |
.
n2
µ1 α1 β1 + µn αn βn =
Hence
|S||T | ∑
e(S, T ) − d
=
µi αi βi .
n
i=2
n−1
n−1
n
∑
∑
e(S, T ) − d |S||T | = |αi ||βi |,
µi αi βi ≤ µ2
n Then
i=2
i=2
since µ2 ≥ µi ≥ µn−1 = −µ2 for 2 ≤ i ≤ n − 1. Now let us apply a CauchySchwartz inequality:
( n−1
)1/2 ( n−1
)1/2
n−1
∑
∑
∑
|αi ||βi | ≤
|αi |2
|βi |2
.
i=2
i=2
i=2
We will be a bit generous:
( n−1
)1/2 ( n−1
)1/2 ( n
)1/2 ( n
)1/2
∑
∑
∑
∑
|αi |2
|βi |2
≤
|αi |2
|βi |2
=
i=2
i=2
i=1
i=1
= ||χU || · ||χV || = |U |1/2 |V |1/2 .
Hence
√
d
|U
||V
|
1
e(U, V ) −
≤ µ2 |U ||V |.
n2
7. (a) We will use that A2 + (b − a)A − (d − b)I = bJ. On the other hand,
x2 + (b − a)x − (d − b) = (x − ϑ1 )(x − ϑ2 ), since ϑ1,2 were exactly the solutions
of this equation. Hence
(A − ϑ1 I)(A − ϑ2 I) = µJ.
Now let us multiply both sides with the vector 1. Now we get that
(A − ϑ1 I)(A − ϑ2 I)1 = (A − ϑ1 I)(d − ϑ2 )1 = (d − ϑ1 )(d − ϑ2 )1
and
bJ1 = bn1.
Hence
(d − ϑ1 )(d − ϑ2 ) = bn.
6
(b) First we show that if a strongly regular graph has a non-integer eigenvalue then its parameters are (n, n−1
, n−5
, n−1
); such a graph is called confer2
4
4
ence graph.
√
Let us first study the case when (a − b)2 + 4(d − b) is rational, so it is
actually integer. Then
)
√
1(
a − b ± (a − b)2 + 4(d − b)
ϑ1,2 =
2
rational and since these are the zeros of a monic polynomial, they must be
integers.
√
If (a − b)2 + 4(d − b) is not rational then
(
)
2d + (n − 1)(a − b)
1
n−1∓ √
m± =
2
(a − b)2 + 4(d − b)
can only be integers, in particular rationals if 2d + (n − 1)(a − b) = 0. Since
0 ≤ k ≤ v − 1, there are only three possibilities: d = 0, 21 (n − 1), n − 1. In the
first and last cases G is empty or the complete graphs and in these cases the
eigenvalues are integers. If d = 12 (n − 1) then b − a = 1. Now let us use that
(n − d − 1)b = d(d − a − 1). Now ew have n − d − 1 = d implying b = d − a − 1.
Hence 2b = (a + 1) + (d − b − 1) = d azaz b = 41 (n − 1) and a = 14 (n − 5).
Hence G is a conference graph.
Now we are ready to prove our claim. Let G be a strongly regular graph
with parameters (p, d, a, b) where p is a prime. If not all eigenvalues of G are
integers then G is a conference graph then we are done. We will show that if G
has only integer eigenvalues then G is the complete or empty graph. Since the
complement of a strongly regular graph is also strongly regular (why?) we only
need to prove that if d ≤ p−1
then G is the empty graph. (If a regular graph
2
has only integer eigenvalues then its complement has only integer eigenvalues
too. This follows from the solution of Problem 5 of the first problem set.) Now
let us use part (a): (d − ϑ1 )(d − ϑ2 ) = pb. Since everything is integer here, we
have:
p | (d − ϑ1 )(d − ϑ2 ).
On the other hand, |d − ϑ1 |, |d − ϑ2 | ≤ 2d ≤ (p − 1). So the only possibility
is that one of the terms (d − ϑ1 ), (d − ϑ2 ) is 0. Then (d − ϑ1 )(d − ϑ2 ) = pb
implies b = 0. But this can only happen if G is the union of some cliques of
the same size. Since p is a prime, this can occur only if G is Kp or pK1 . The
former case cannot occur since then the degree is p − 1 > (p − 1)/2. Hence G
must be the empty graph.
7
8. Recall Serre’s theorem: for every ε > 0, there exists a c = c(ε, d) such
√ that
for any d–regular graph G, the number of eigenvalues λ with λ ≥ (2−ε) d − 1
is at least c|V (G)|. In the proof of Serre’s theorem we have also seen that
√
√
1
2t0
−
(d
+
(2
−
ε)
(d
+
2
d
−
1)
d − 1)2t0
2
2(t +1)
√
c(ε, d) = 0
(2d)2t0 − (d + (2 − ε) d − 1)2t0
satisfies the conditions of the theorem if t0 satisfies
√
√
1
2t0
d
−
1)
−
(d
+
(2
−
ε)
d − 1)2t0 > 0.
(d
+
2
2
2(t0 + 1)
√
In our case √
d = 3 and clearly we would like to choose ε such that (2−ε) 2 >
2, say (2 − ε) 2 = 2.001. Then we need to choose t0 such that
√ 2t
1
(3
+
2
2) 0 − (3 + 2.001)2t0 > 0.
2(t0 + 1)2
Maple suggests that for instance t0 = 28 is good. Then c(ε, d) ≈ 0.00007989476237
If c|V (G)| > 1 then even the second largest eigenvalue is at least 2.001.
If |V (G)| > 12516.46504, then c|V (G)| > 1. Whence λ2 ≥ 2 implies that
|V (G)| ≤ 12516.
Remark 0.2. Once I saw a superoptimized computation which lead to the
bound 168. In this proof even the number of closed walks in the 3–regular
tree was not bounded, but the precise formula was used instead. On the other
hand, one can prove with other techniques that |V (G)| ≤ 30 and if |V (G)| = 30
then G is the so-called Tutte-Coxeter graph. See wikipedia or mathworld for
the definition of this graph. For this proof see the extra section.
9. Let us determine the Laplacian eigenvalues of the graph H. If λ is an
eigenvector of L(G) with eigenvector v then it is easy to see that λ is an
eigenvector of L(H) with eigenvector w, where w(u,i) = vu , where u ∈ V (G) and
i ∈ {0, 1}. This way we have already found n eigenvalues, where n = |V (G)|,
the eigenvalues λ1 ≥ . . . ≥ λn = 0 of L(G) are all eigenvalues of L(H).
Let’s try to find the remaining eigenvalues the following way: let w′ be a
′
′
vector such that w(u,i)
= −w(u,1−i)
for i ∈ {0, 1}. A little computation shows
′
that it is a Laplacian-eigenvector of H if and only if the vector w0′ = (w(u,0))
is an eigenvector of the following matrix L′ : L′vv = dv , the degree of vertex v,
and
{
1
if ((u, 0), (v, 0)) and ((u, 1), (v, 1)) ∈ E(H),
′
Luv =
−1 if ((u, 0), (v, 1)) and ((u, 1), (v, 0)) ∈ E(H).
8
Moreover, L(H)w′ = λ′ w′ if and only if L′ w0′ = λ′ w0′ . We can choose an
orthonormal eigenvectors of L = L(G), and also orthonormal eigenvectors of
L′ . Note that a vector w for which w(u,i) = w(u,1−i) is always orthogonal to a
′
′
vector w′ for which w(u,i)
= −w(u,1−i)
so we have found all eiganvalues of H.
Now we use the formula for the number spanning trees expressed by the
Laplacian-eigenvalues.
n−1
n
1
1 ∏ ∏ ′
λi
λi = τ (G) det(L′ ).
τ (H) =
2n i=1 i=1
2
The det(L′ ) is clearly an integer, we only need to prove that it is divisible by
2. This is true since over the field F2 we have L = L′ , whence
det(L′ ) ≡ det(L) = 0 (mod2).
We are done.
10. Let e1 , . . . , ek be the k disjoint edges. Let V be set of all spanning trees
of Kn . Let Ai be the set of spanning trees not containing the edge ei . This
means that the set of spanning trees containing e1 , . . . , ek is V \ ∪ki=1 Ai . By
the inculsion-exclusion formula we have
|V \
∪ki=1 Ai |
= |V | −
k
∑
i=1
|Ai | +
∑
|Ai ∩ Aj | − · · · + (−1)k |A1 ∩ · · · ∩ Ak |.
1≤i<j≤k
Next we determine |Ai1 ∩ · · · ∩ Aij |. This is the number of spanning trees
not containing the edges ei1 , . . . , eij , in other words the number of spanning
trees of G = Kn − {ei1 , . . . , eij }. First we determine the Laplacian eigenvalues
of G. Note that the complement of G consists of j edges and n − 2j isolated
vertices, hence its Laplacian eigenvalues are 2(j) , 0(n−j ). Hence G has Laplacian
eigenvalues (n − 2)(j) , n(n−j−1) , 0. Hence
τ (G) =
Hence
|V
\∪ki=1 Ai |
1
(n − 2)j nn−j−1 = (n − 2)j nn−j−2 .
n
( )
k
=
(−1)
(n−2)j nn−j−2 = (n−(n−2))k nn−k−2 = 2k nn−k−2 .
j
j=0
k
∑
j
11. Let s(G) be the smallest eigenvalue of G. First we show that if H is an
induced subgraph of G then s(H) ≥ s(G). Indeed, if x is a vector in R|V (H)|
9
then we can consider x′ = (x, 0) ∈ R|V (G)| , hence
s(H) = min xT A(H)x = min
x′T A(G)x′ ≥ min xT A(G)x = s(G).
′
||x||=1
||x ||=1
||x||=1
Our strategy will be to find small graphs H with smallest eigenvalue at most
−2, then we know that G cannot contain H. For instance, the spectrum of the
4-cycle C4 is {2, 0(2) , −2}, so s(C4 ) = −2. Hence G cannot contain an induced
4-cycle.
We distinguish two cases according to G contain a triangle or not.
Case 1: G does not contain triangle. Let 1 and 2 be two adjacent vertices. Let 3, 4 be the other two neighbors of 1, and let 5, 6 be the other two
neighbors of 2. Note that 3, 4, 5, 6 are distinct as G does not contain triangle. G not containing a triangle also implies that G does not contain the edges
(1, 5), (1, 6), (2, 3), (2, 4), (3, 4) and (5, 6). But then G cannot contain the edges
(3, 5), (3, 6), (4, 5), (4, 6) since then G would contain an induced 4-cycle. Hence
the induced subgraph of G on the vertices 1, 2, 3, 4, 5, 6 is exactly the graph T
appearing in Problem 1 in the first problem set where we saw that its spectrum is {2, 1, 0(2) , −1, −2}, so s(T ) = −2. But then s(G) ≤ s(T ) = −2, a
contradiction. So G must contain a triangle.
Case 2: G contains a triangle. First we show that if G contains two triangles
which have a common edge then G must be K4 . First of all, if G contains a
K4 then in fact G = K4 , because G is connected and K4 is 3–regular. So we
can assume that G does not contain a K4 , but it contains two triangles which
have a common edge. It means that it contains 4 vertices, 1, 2, 3, 4 such that it
contains every edges except one, say (2, 4). Now let 5 be the last neighbor of 2.
Note that the vertex 1 is incident to two triangles, so by the vertex-transitivty
this must be true for the vertex 2 too. It means that 5 have to be adjacent to
1 or 3, but this is impossible since the vertices 1 and 3 already have the three
neighbors. Hence if G is not K4 then it cannot contain two triangles which
have a common edge. Note that G cannot contain two triangles having exactly
one common vertex since this vertex would have degree at least 4.
This means that if G contains a triangle, then by the vertex-transitivity one
can partition the vertex set of G into disjoint triangles. Between two triangles
there cannot be at most one edge, since otherwise G would either contain a
vertex of degree at least 4 or an induced 4 cycle. Now contract all triangles
in G, the obtained graph H is also 3–regular. H must contain a cycle (why?),
let us consider the smallest one: Ck . Since this is the smallest cycle, Ck must
be an induced subgraph of H. This means that G contains an induced cycle
C2k . But C2k is a 2-regular bipartite graph hence s(C2k ) = −2. This means
that s(G) ≥ −2, a contradiction.
10
Remark 0.3. Surprisingly, it turns out that in the second case s(G) = −2
exactly for every graph G. It is because G is a so-called line graph. Indeed, let
us consider the graph H again, which was obtained by contracting all triangles
in G. Next, let us subdivide all edges of H: this means that we put a vertex
to every edge of H, let S(H) denote the obtained graph. If we have a graph
K, we can always condider its line graph L(K) which is obtained as follows:
the vertices of L(K) are the edges of K, and two vertices of L(K) (i.e. two
edges of K) are adjacent if the original edges had a common end vertex. Now
observe that in our case, G is nothing else than the line graph of S(H), hence
G = L(S(H)).
Figure 1. Here H = K4 and the red graph is the original graph
G. The curvy edges must be red, but they don’t seem to be for
some reason.
The importance of this observation lies in the fact that the smallest eigenvalue of a line graph L(K) is always at least −2, and if the original graph
K had more edges than vertices then it is exactly −2. This is because if X
denotes the edge-vertex incidence matrix of K then the adjacency matrix of
the line graph A(L(K)) = XX T − 2I, and XX T is positive semidefinite and
it is singular if v(K) < e(K).
∑
12. Let L(G, x) = det(xI − L(G)) = nk=1 (−1)n−k ak xk . We will prove that
∑
ak =
b(S1 )b(S2 ) . . . b(Sk ),
S1 ∪···∪Sk =V (G)
Si ∩Sj =∅
where b(Si ) = |Si |τ (G[Si ]). We will prove it in three steps.
(a) Let us consider the coefficients of xk in det(xI − L(G)). We get this
coefficient by first choosing k vertices and choose x from the corresponding
diagonal elements and multiply it by the determinant of the remaining (n −
k) × (n − k) submatrix of L(G).
∑
(−1)n−k ak =
(−1)n−k det(L(G)I ),
|I|=k
11
where L(G)I is the matrix obtained from L(G) be deleting the rows and
columns corresponding to G. Now observe that L(G)I can be obtained as
follows: we conract all vertices of I into one vertex thus obtaining a graph
G/I, and then we delete the row and column corresponding to this new vertex. But then by Kichhoff’s theorem we get that det(L(G)I ) = τ (G/I).
(b) Now let us count the following set in two different ways
{(F, I) | F ∈ Fk , F = T1 ∪ · · · ∪ Tk , |I ∩ V (Ti )| = 1 (i = 1, . . . , k)},
where Fk is the set of spanning forests of the graph G with exactly k components. For fixed I we can choose F in τ (G/I) ways since by contracting I
we get a spanning tree of G/I and if we have a spanning tree of G/I then by
decomposing the vertex corresponding to I we ∏
get an element of F ∈ Fk . If
we fix F then then we get I in exactly γ(F ) = k=1 |V (Ti )| ways: we need to
choose one vertex from each components of F . Hence
∑
∑
ak =
τ (G/I) =
γ(F ).
F ∈Fk
I⊂V (G)
|I|=k
(c) Let us group together the elements of Fk according that the components
of F induces which sets S1 , . . . , Sk . For given a partition S1 ∪ · · · ∪ Sk we
can choose such a spanning forest in τ (G[S1 ]) . . . τ (G[Sk ]) ways. On the other
∏
∏
hand, for every such F we have γ(F ) = ki=1 |V (Ti )| = ki=1 |Si |. Hence
∑
∑
ak =
γ(F ) =
b(S1 )b(S2 ) . . . b(Sk ).
F ∈Fk
S1 ∪···∪Sk =V (G)
Si ∩Sj =∅
Now we are ready to prove the claim of the problem. Let Pk (S) denote the
set of partitions of S into exactly k non-empty sets. Then
∑
L(S1 , x)L(S2 , y) =
S1 ⊎S2 =V (G)
S1 ⊎S2 =V (G)
=
( n
∑
∑


n
∑

(−1)|S1 |−k
∑
S1 ⊎S2 =V (G)
k=1


n
∑
·  (−1)|S2 |−ℓ 
ℓ=1
)(
(−1)|S1 |−k ak (S1 )xk
k=1
∑
{T1 ,T2 ,...,Tℓ }∈Pℓ (S2 )
ℓ=1
)
(−1)|S2 |−ℓ aℓ (S2 )y ℓ


b(R1 )b(R2 ) . . . b(Rk ) xk  ·
{R1 ,R2 ,...,Rk }∈Pk (S1 )
∑
n
∑


b(T1 )b(T2 ) . . . b(Tℓ ) y ℓ  =
=
12
=
n
∑

∑
(−1)n−r 
{Q1 ,Q2 ,...,Qr }∈Pr (G)
r=1
=
n
∑
(
)
r ( )
∑
r
b(Q1 ) . . . b(Qr )
xi y r−i =
k
k=0
(−1)n−r ar (G)(x + y)r = L(G, x + y).
r=1
Remark 0.4. This is the rare case when I can explain how anyone can discover
such an identity. First of all, I conjectured the statement because for the
complete graph this identity reduces to an identity of Abel:
n ( )
∑
n
x(x − k)k−1 y(y − (n − k))n−k−1 = (x + y)(x + y − n)n−1 .
k
k=0
(Actually, Abel’s identity is generally with + signs, but if you apply it to −x
and −y and you multiply it with (−1)n then you can get the above identity
and vice versa.)
As a next step, one characterize those families of polynomials which satisfies such an identity. Note that the actual form of the function b(.) was not
important at all, and one can prove that a family of polynomials satisfies such
an identity if and only if such a function b(.) exists. It’s quite easy to prove.
Last step is to prove that the only candidate for the function b(.), namely
b(S) = |S|τ (G[S]) is good. (It is the only candidate as b(G) must be a1 (G).)
Here I was in a good position: I knew part (b), it appears in the literature, so
I only needed to realize that it is equivalent with part (c).
So the way I wrote down this solution is a bit counter intuitive, in reality
one works backward.
1. Problem 8 revisited
In this part we prove that a 3–regular graph with second largest eigenvalue
at most 2 has size at most 30, and the only 3-regular graph with second largest
eigenvalue 2 on 30 vertices is the so-called Tutte–Coxeter graph. In fact, we
will only prove that the extremal graph has at most 32 vertices and its shortest
cycle is of size at least 7, then a computer can check the remaining cases.
The Tutte–Coxeter graph is a bipartite graph on 30 vertices which can be
constructed
as follows: let A be the 2-element subsets of {1, 2, 3, 4, 5, 6}. Then
()
A = 62 = 15. Let B be the partitions of {1, 2, 3, 4, 5, 6} into three sets of
size 2, for instance {1, 4}{2, 6}{3, 5} is such a partition. Then |B| = 15. Now
connect an element of A to an element of B if the set of size 2 appears in the
partition. This is a 3–regular bipartite graph on 30 vertices. Its shortest cycle
13
has length 8, this is the largest possible size. In fact, it is easy to prove that
if a 3–regular graph has girth 8 (girth is the length of the shortest cycle) then
the graph has at least 30 vertices. The spectrum of the Tutte–Coxeter graph
is {3, 2(9) , 0(10) , −2(9) , −2}. Let me mention that there is only one 3–regular
non-bipartite graph on 28 vertices with second largest eigenvalue at most 2
and it has girth 7, it is called Coxeter-graph. It is again a very symmetric
graph. There is a general
√ observation that regular graphs with small second
eigenvalue, say λ2 ≤ 2 d − 1 has only a few short cycles. This claim has some
quantitative forms, most of them not very satisfactory.
You can read more about the Coxeter and Coxeter-Tutte graphs at http://
en.wikipedia.org/wiki/Coxeter_graph and http://en.wikipedia.org/wiki/
Tutte-Coxeter_graph.
For sake of simplicity, we will say that the graph G has property R if G
is 3-regular graph with second largest eigenvalue at most 2. In what follows,
let H[S] be the induced subgraph of H induced by the vertex set S ⊆ V (H).
In all our proofs, the strategy will be the following: assume that the graph
H has many vertices, then we can decompose the vertex set V (H) as follows:
V (H) = A∪B∪C such that H[A] induces a small graph with largest eigenvalue
bigger than 2, and the neighbors N (A) = B, and finally the number of edges
of H[C] satisfies e(C) > |C| (or some other property) which ensures that
the largest eigenvalue of H[C] is also bigger than 2. Note that the largest
eigenvalue of a graph is at least the average degree, so e(C) > |C| indeed
implies that the largest eigenvalue of H[C] is bigger than 2. Then H[A ∪ C] =
H[A] ∪ H[C] since there is no edg between A and C since B = N (A). Note
that H[A ∪ C] = H[A] ∪ H[C] has two eigenvalues bigger than 2, and so H
has a second largest eigenvalue bigger than 2 by interlacing: from Practice
Problem 4 it follows that λk (H) ≥ λk (H ′ ) for any induced subgraph H ′ of H.
In particular, if k = 2, then we get thet λ2 (H) ≥ λ2 (H[A ∪ C]) > 2.
Lemma 1.1. Assume that the graph G has property R and its shortest cycle
has length g. Then G has at most 4(g + 1) vertices.
Proof. We prove the claim by contradcition. Assume that the graph G has at
least 4g + 6 vertices (bigger than 4(g + 1) and it should be even). Let D be
the vertex set of a shortest cycle. Then G[D] cannot contain a chord. Let u
be an arbitrary vertex adjacent to one of the vertices of D. Let A = D ∪ {u}.
Then the largest eigenvalue of G[A] is strictly bigger than 2, since the largest
eigenvalue of a cycle is 2, and G[A] is a connected graph containing a cycle. Let
B the set of neighbors of A. Then |B| ≤ |A| = g + 1. Let C = V (G) \ (A ∪ B).
14
Let us estimate the number of edges of G[C]:
1
1
e(C) = (3|C| − e(B, C)) ≥ (3|C| − 2|B|).
2
2
In the final step we used the fact that all vertices of B has 1 neighbor in A,
so there are at most 2 edges going to C from every vertex of B. Note that
|B| ≤ g + 1, and |C| ≥ 2g + 4. Hence
e(C) > |C|.
This shows that G[C] has largest eigenvalue bigger than 2. But then G[A ∪ C]
has at least two eigenvalues bigger than 2. By interlacing we get that G has
second largest eigenvalue bigger than 2, a contradiction.
Lemma 1.2. Assume that the graph G has property R. Then G has at most
34 vertices.
Proof. We prove the claim by contradcition. Assume that the graph G has at
least 36 vertices. The girth of G must be at least 8 by the previous lemma.
We will only use the fact that it is at least 5, so it must contain the following
graph T as an induced subgraph.
u
1
1
2
2
1
1
The tree T has largest eigenvalue bigger than 2. Indeed, if we erase the
vertex u, then its largest eigenvalue is exactly 2, and the elements of the
corresponding eigenvector are given in the figure. Since the tree is connected,
its largest eigenvalue is strictly greater than 2. Let A ⊂ V (G) be the vertex
set of this tree.
Let B the set of neighbors of A. Then |B| ≤ 9. Let C = V (G) \ (A ∪ B).
Let us estimate the number of edges of C:
1
1
1
e(C) = (3|C| − e(B, C)) ≥ (3|C| − 2|B|) = |C| + (|C| − 2|B|) > |C|,
2
2
2
since |C| ≥ 36 − (7 + 9) = 20, while |B| ≤ 9. This shows that G[C] has largest
eigenvalue bigger than 2. But then G[A∪C] has at least two eigenvalues bigger
than 2. By interlacing we get that G has second largest eigenvalue bigger than
2, a contradiction.
15
Now we improve on Lemma 1.2 a bit.
Lemma 1.3. Assume that the graph G has property R. Then G has at most
32 vertices.
Proof. We prove the claim by contradiction. Assume that the graph G has
at least 34 vertices. The girth of G must be at least 7 or 8 by the previous
two lemmas. (If the girth is at least 9, then the graph has at least 46 vertices
by the Moore-bound which we already excluded.) We distinguish two cases,
according to the girth is 7 or 8.
Case 1: girth is 7. Let D be a 7-cycle, and N (D) be the set of neighbors of D.
Then D∪N (D) contain the same tree T which we used in Lemma 1.2 such that
H[D] and the tree contain 4 common edges, and consequently |V (T )∩N (D)| =
2. We make the same partition as before with A being this tree, B = N (A)
and C = V (G) \ (A ∪ B). The only new ingredient we use that B contains an
edge (of the 7-cycle), so e(B, C) ≤ 2|B| − 2. But then
1
1
1
e(C) = (3|C|−e(B, C)) ≥ (3|C|−(2|B|−2)) = |C|+ (|C|−2|B|+2) > |C|,
2
2
2
since |C| ≥ 34 − (7 + 9) = 18, while |B| ≤ 9. We are done as before.
Case 2: girth is 8. Let D be a 8-cycle, and N (D) be the set of neighbors of D.
Then D ∪ N (D) contain the same tree which we used in Lemma 1.2 such that
H[D] and the tree contain 4 common edges, and consequently |V (T )∩N (D)| =
2. We make the same partition as before with A being this tree, B = N (A)
and C = V (G) \ (A ∪ B). As before e(B, C) ≤ 2|B|, and
1
1
1
e(C) = (3|C| − e(B, C)) ≥ (3|C| − 2|B|) = |C| + (|C| − 2|B|) ≥ |C|,
2
2
2
If there is no inequality, then we are done as before. Equality can hold if
|C| = 18, |B| = 9, and e(B, C) = 18, e(C) = 18. If the largest eigenvalue of
G[C] is bigger than 2, then we are done as before. If it is at most 2, then it
must be disjoint union of cycles. So every vertex of G[C] has degree 2. But
the vertex of D ∩ C has already 2 neighbors in B, so it would have degree at
least 4, contradiction.
We got contradiction in both cases, so we are done.
Remark 1.4. By combining the ideas of Case 1 and Case 2, one can also
exclude the case of g = 7 and v = 32. This would save a little work for the
computer. The case of g = 8, v = 32 is surprisingly non-existent, there is no
such 3-regular graph. Here we give the number of 3-regular graphs of g = 7, 8
and v = 30, 32:
16
g=7 g=8
v = 30 546
1
v = 32 30368
0
By checking all these graphs, we get that v(3, 2) = 30 and the only graph
on 30 vertices with property R is the Tutte–Coxeter graph.