QUALIFYING EXAM PROBLEMS
BENJAMIN IRIARTE
—————————————————————————————————————————————————
1. Let f (n) be the number we are studying for Sn . We have f (1) = 1 and f (2) = 2. Let n ≥ 3. Take w ∈ Sn
with w = a1 . . . an (written in word notation).
If an = n, then the standard representation of w will have the cycle (n) in last position, so ŵ = b1 . . . bn with
bn = n, so that w = ŵ iff the permutation u = a1 . . . an−1 in Sn−1 satisfies that u = û. Hence, the number of
w ∈ Sn with w(n) = n such that w = ŵ is f (n − 1).
Suppose now that w is such that w = ŵ but an 6= n. In the standard representation, w will have a last cycle
(nai . . . an ) with i ≤ n. But this means that w(n) = ai , so actually an = ai , so i = n, and the last cycle in the
standard form is precisely (nan ). Therefore, w(n − 1) = n and an = n − 1. Now, w = a1 . . . an−1 n(n − 1) satisfies
w = ŵ iff the permutation u = a1 . . . an−2 in Sn−1 satisfies u = û. Hence, the total number of w ∈ Sn with
w(n) 6= n such that w = ŵ is f (n − 2).
We obtain the recursion f (n) = f (n − 1) + f (n − 2) for all n ≥ 3, with f (1) = 1 = F2 and f (2) = 2 = F3 .
—————————————————————————————————————————————————
2. The problem is equivalent to proving the identity of formal power series
Y
Y 1
1
i
−
q
=
.
i
1−q
1 − qi
i≥1
i≥1
i6≡1,5 (mod 6)
Now,
1
1−q i
− qi =
1−q i +q 2i
1−q i
=
1+q 3i
(1−q i )(1+q i ) .
Multiplying both sides of the identity by
Y 1 + q 3i
=
1 + qi
i≥1
Y
Q
i≥1 (1
− q i ) we obtain
(1 − q i ).
i≥1
i≡1,5 (mod 6)
This is equivalent to having
Y
Y
(1 − q i )
i≥1
i≡1,5 (mod 6)
(1 + q k ) = 1.
k≥1
k≡1,2 (mod 3)
Every k ≡ r (mod 3) with r ∈ {1, −1} can be written uniquely as k = 2a b with b odd, a and b nonnegative integers.
We have b ≡ (−1)a r (mod 3) and hence, b ≡ (−1)a r (mod 6). We can then write
Y
Y
j
(1 + q k ) =
(1 + q 2 k ).
k≥1
k≡1,2 (mod 3)
k≥1
j≥1
k≡1,5 (mod 6)
However,
Y
i≥1
i≡1,5 (mod 6)
=
Y
i≥1
i≡1,5 (mod 6)
(1 − q i )
j
Y
(1 − q i )
(1 + q 2 k )
k≥1
j≥1
k≡1,5 (mod 6)
Y
j
(1 + q 2 i ) =
j≥1
Y
1 = 1.
i≥1
i≡1,5 (mod 6)
—————————————————————————————————————————————————
3. Suppose that we have a sequence w = a1 . . . ak such that inv(w) = n. Suppose that w has m 1’s and let
1 = i1 < · · · < im < k be the set of indices for which aij = 1. Let bj be the number of 0’s to the right of aij in
Pm
w. Then, b1 ≥ b2 ≥ · · · ≥ bm > 0 and i=1 bi = inv(w) = n, so b := (b1 , . . . , bm ) ` n. On the other hand, from
1
2
BENJAMIN IRIARTE
b ` n with bm > 0 but bm+1 = 0, let w = a1 . . . ak be the unique binary sequence with m 1’s, a1 = 1, ak = 0, and
such that there are bj 0’s in w to the right of the j-th 1 (from left to right) in w. This shows that f (n) = p(n),
the number of partitions of n. The number f (n, j) of such binary sequences w with inv(w) = n with exactly j 1’s
is just pj (n), the number of partitions of n with j nonzero parts. We have,
X
Y
1
.
1+
f (n)q n =
1 − qn
n≥1
n≥1
Also,
1+
X
n,j≥1
f (n, j)tj q n =
Y
n≥1
1
.
1 − tq n
—————————————————————————————————————————————————
¯ is just the number of maximal saturated chains of J(P ) which go through I. Every
4. For I ∈ J(P ), e(I)e(I)
such maximal chain has n + 1 elements and determines a unique linear extension of P . Furthermore, each such
maximal chain contains exactly one element I of rank i for 0 ≤ i ≤ n, and this rank is determined by ]I = i. These
observations suffice to write,
X
¯ = e(P )(n + 1).
e(I)e(I)
I∈J(P )
X n
¯ = 2n e(P ).
e(I)e(I)
]I
I∈J(P )
—————————————————————————————————————————————————
5. Call the chains considered in the sum good chains of P . We will show by induction on the size of posets that
f (P ) = ]P . For ]P = 1, the result f (P ) = 1 is immediate. Suppose it holds for all posets P with ]P < n for some
n ≥ 2, and let Q be a poset with ]Q = n. Let q be a minimal element of Q, and consider the induced subposets on
Q − {q} and on Uq := Vq − {q}. A good chain t1 < t2 < · · · < tn in Q can either contain q (in which case t1 = q)
or not (in which case t1 6= q). The former are in bijection with the good chains in Uq and the later are precisely
the good chains of Q − {q}. Furthermore,
X
1
1
= f (Uq )
.
(]V
−
1)
.
.
.
(]V
−
1)
(]V
− 1)
q
t
q
n−1
q<t2 <···<tn
tn maximal in Q
Note that ]Uq , ](Q − {q}) < n, so by induction and the above discussion we obtain,
1
f (Q) = f (Uq )
+ f (Q − {q}) = 1 + (n − 1) = n.
(]Vq − 1)
Therefore, we are done.
—————————————————————————————————————————————————
6. We use the Finite Field Method to compute the characteristic polynomials. Let fq (n) be the characteristic
polynomial for the n-dimensional case. We have fq (1) = q − 1, fq (2) = (q − 1)(q − 2) and fq (3) = (q − 1)(1 + (q −
3)(q − 2)). We first prove the recursion fq (n) = (q − 1)(q − 2)n−2 (q − 3) + fq (n − 2) for n ≥ 3. The case n = 3
holds by inspection.
Let n ≥ 4. Replacing the hyperplanes xn−1 + xn = 0 and xn + x1 = 0 in A with xn−1 − xn = 0 and xn − x1 = 0,
we obtain an hyperplane arrangement L1,n in Znq whose characteristic polynomial is exactly (by the Finite Field
Method) that of A, so χL1,n (q) = fq (n).
Let L2,n be the hyperplane arrangement in Znq given by xi = 0 for all i ∈ [n], xi + xi+1 = 0 for all i ∈ [n − 2],
and xn−1 − xn = 0. The number of points in Znq not in L2,n is clearly (q − 1)(q − 2)n−1 .
Now, the set of points (a1 , . . . , an ) ∈ (Z∗q )n not contained in L2,n but contained in L1,n is given by the conditions
ai + ai+1 6= 0 for i ∈ [n − 2], an−1 6= an , but a1 = an . Counting the number of such points is equivalent to counting
the number of points in Zn−1
not contained in the hyperplane arrangement L1,n−1 in Zn−1
given by equations
q
q
xi = 0 for all i ∈ [n − 1], xi + xi+1 = 0 for all i ∈ [n − 2], and xn−1 − x1 = 0.
Consider the hyperplane arrangement L2,n−1 in Zn−1
given by equations xi = 0 for all i ∈ [n−1], and xi +xi+1 = 0
q
n−1
for all i ∈ [n−2]. Clearly, the number of points in Zq not in L2,n−1 is (q −1)(q −2)n−2 . Now, the number of points
(a1 , . . . , an−1 ) ∈ (Z∗q )n−1 not contained L2,n−1 but contained in L1,n−1 is given by the conditions ai + ai+1 6= 0
for all i ∈ [n − 2], but an−1 = a1 . The number of such points is clearly equal to the number of points in Zn−2
not
q
QUALS PSET
3
contained in the hyperplane arrangement L1,n−2 in Zn−2
given by equations xi = 0 for all i ∈ [n − 2], xi + xi+1 = 0
q
for all i ∈ [n − 3], and xn−2 + x1 = 0, whose characteristic polynomial is fq (n − 2).
We obtain that
fq (n) = (q − 1)(q − 2)n−1 − ((q − 1)(q − 2)n−2 − fq (n − 2))
= (q − 1)(q − 2)n−2 (q − 3) + fq (n − 2)
for n ≥ 3, with fq (1) = q − 1 and fq (2) = (q − 1)(q − 2). Solving the recursion by working the geometric sums and
simplifying we obtain:
For n ≥ 1 odd,
fq (n) = (q − 2)n + 1.
For n ≥ 1 even,
fq (n) = (q − 2)n + q − 2.
—————————————————————————————————————————————————
7. WeP
use some standard facts from the theory of coxeter groups. We work with the left weak order. Let
Fq (n) = k≥0 f (n, k)q k for all n ≥ 0. We have Fq (0) = 1, Fq (1) = 1 and Fq (2) = 2 + q. We will prove the
recursion Fq (n) = nFq (n − 1) + n2 qFq (n − 2) for n ≥ 2. The case n = 2 holds by inspection.
Let n ≥ 3. Consider the symmetric group Sn . Let [w, z] ∼
= Bk be an interval of W (Sn ) with k ≥ 0. Define
si := (i, i + 1) for all i ∈ [n − 1]. To our generic choice of w and interval [w, z] above associate the sets I[w,z] :=
{i|si w ∈ [w, z]} and
Iw := {i|si occurs in a reduced expression for w}.
Then, either I[w,z] contains n − 1 or it does not. Let h(n, k) be the number of boolean intervals [w, z] ∼
= Bk in
W (Sn ) with n − 1 ∈ I[w,z] . Let g(n, k) be the number of boolean intervals [w, z] ∼
= Bk in W (Sn ) with n − 1 6∈ I[w,z] .
P
P
Define Hq (n) := k≥0 h(n, k)q k and Gq (n) := k≥0 g(n, k)q k . Clearly, Fq (n) = Gq (n) + Hq (n) for all n ≥ 0.
Case 1. Assume n − 1 6∈ I[w,z] . Then, either Iw contains n − 1 or it does not. If n − 1 6∈ Iw , then [w, z] can
be regarded as a uniquely determined interval of W (Sn−1 ). If n − 1 ∈ Iw , then among all reduced expressions of
w, consider one in which the first appearance from left to right of sn−1 is in its rightmost position. Let w = uv
be such an expression with sn−1 6∈ Iu but v = sn−1 v 0 . The simple reflection sn−1 cannot be pushed further to the
right in sn−1 v 0 . In particular, inv(si v) = inv(v) + 1 for all i ∈ [n − 2]. Hence, regarded as permutation in Sn , the
word expression for v satisfies that i appears to the left of i + 1 for all i ∈ [n − 2] but n appears to the left of
n − 1. In fact, any such v has all its reduced expressions of the form v = sn−1 v 0 . Now, the total number of such
v’s is n − 1, each uniquely determined by the position of n in its word expression: all positions are allowed except
the last one, and this determines the rest of the word expression. Thus, regarded as permutations in Sn , as u acts
only on the entries 1, 2, . . . , n − 1 of the word expression for v, then v is uniquely determined by w. More precisely,
given w with w(i) = n, v is the unique permutation as above with v(i) = n. On the other hand [u, zv −1 ] ∼
= Bk and,
consequently, it can be regarded as a uniquely determined boolean interval of W (Sn−1 ). Also, for every interval
[x, y] ∼
= Bk in W (Sn−1 ) and every element v ∈ Sn such that the word expression for v satisfies that i appears to the
left of i + 1 for all i ∈ [n − 2] but n appears to the left of n − 1, we have that (regarding now x, y as permutations
in Sn ) [xv, yv] ∼
= Bk in W (Sn ). For example, think of the word expression for all these permutations to see this.
This proves that Gq (n) = (1)Fq (n − 1) + (n − 1)Fq (n − 1) = nFq (n − 1).
Case 2. Assume n − 1 ∈ I[w,z] , so k ≥ 1. It is easy to see that for all i, j ∈ I[w,z] with i 6= j, si and sj commute.
This motivates the following argument. As n − 1 ∈ I[w,z] , me must have n − 2 6∈ I[w,z] . Now, either n − 2 is in Iw or
it is not. If n − 2 6∈ Iw , then sn−1 commutes with all simple reflections in a reduced expression for w, so actually
∼
n − 1 6∈ Iw . Hence, we regard [w, z] as being obtained from a uniquely determined interval [w, s−1
n−1 z] = Bk−1
in Sn−2 . Suppose now that n − 2 ∈ Iw . Among all reduced expressions for w, consider one in which the first
appearance of sn−2 is in its rightmost position, and say this expression is w = uv with n − 2 6∈ Iu and v = sn−2 v 0 .
Again, in this expression for v, sn−2 cannot be pushed further to the right, so as a permutation of Sn , v satisfies
that in its word expression, every i ∈ [n − 1]\{n − 2} is to left of i + 1, but n − 1 appears to the left of n − 2.
Notice that every such v is uniquely determined by the positions of n − 1 and n in its word expression: in this
expression, the numbers 1, 2, . . . , n − 2 must be ordered increasingly occupying some positions, and the remaining
two positions (which are then forced to be different from the last two positions) are filled with n − 1 and n, with
n − 1 to the left of n. This gives a total number n2 − 1 of these permutations. Now, as a permutation in Sn , v is
uniquely determined by w. We have that n − 2 6∈ Iu and hence, that n − 1 6∈ Iu (because n − 1 ∈ I[w,z] ), so the
word expression for u has the form a1 . . . an−2 (n − 1)n. Therefore, acting on v on the left by u to get w, i.e. w = uv
, we see that v is uniquely recovered from w by the positions of n − 1 and n in the word expression for w. More
precisely, given w with w(i) = n − 1 and w(j) = n, then v is the unique permutation as above for which v(i) = n − 1
4
BENJAMIN IRIARTE
and v(j) = n. Now, the interval [u, zv −1 ] ∼
= Bk can be regarded as being obtained from some uniquely determined
−1 ∼
[u, s−1
zv
]
B
in
W
(S
).
Moreover,
given an interval [x, y] ∼
= k−1
= Bk−1 in W (Sn−2 ) and a permutation v ∈ Sn
n−2
n−1
such that in its word expression every i ∈ [n − 1]\{n − 2} is to left of i + 1 but n − 1 appears to the left
of n − 2,
n
n
then [xv, sn−1 yv] ∼
B
in
W
(S
).
We
obtain
that
H
(n)
=
q(1)F
(n
−
2)
+
q
−
1
F
(n
−
2)
=
q
= k
n
q
q
2
2 Fq (n − 2).
q
n
We then obtain that Fq (n) = nFq (n − 1) + 2 qFq (n − 2) for all n ≥ 0, with initial values Fq (0) = 1 and
Fq (1) = 1. This recursion shows that
X
Fq (n)
n≥0
1
xn
=
.
n!
1 − x − 2q x2
—————————————————————————————————————————————————
8. Let u be a coatom of P . As rank(u) = d and P − {1} is simplicial, we know [0, u] ∼
= Bd , so [0, u] has d atoms.
As P has also d atoms, then all of them are in [0, u]. Therefore, for a fixed atom v of P , all coatoms of P are in
[v, 1].
We count the number N of edges between elements of rank d − 1 and elements of rank d in P which belong to
the interval [v, 1]. We do the counting in two different ways. For any coatom u of P , we know [v, u] ∼
= Bd−1 , so if
there are q coatoms of P , then N = (d − 1)q. On the other hand, for each w ∈ P of rank d − 1, the interval [w, 1]
is eulerian of rank 2, so [w, 1] ∼
= B2 . Let p be the number of elements of [v, 1] which are of rank d − 1 in P . Then,
N = 2p.
We obtain that (d − 1)q = 2p, which implies that q is even if d is even.
—————————————————————————————————————————————————
9. For part a), we can regard a tiling as an ordered concatenation of horizontal line segments of length 1 (whose
contribution to the generating function is ux) and of length 2 (whose contribution is x2 ). The tilings with n pieces
are then described by (ux + x2 )n , for n ≥ 1. Hence,
1+
X
n≥1
pn (u)xn =
X
(ux + x2 )n =
n≥0
1
.
1 − ux − x2
For part b), we want to understand the tilings of an m × m square, with m ∈ P, induced by picking a tiling of
a vertical side of the square, and then picking a tiling of an horizontal side of the square, where the nature of the
tilings on the sides is as described in part a). The u’s will describe the horizontal tilings and the v’s the vertical
tilings. Fix a tiling of an m × m square obtained in this way and draw the diagonal of the square. Consider the
points S in the diagonal where an horizontal and a vertical line in the tiling meet. There will be uniquely determined
right triangles to the right of the diagonal whose hypotenuses are given by the line segments between consecutive
points (along the diagonal) of S, and the ordered concatenation of these triangles determines completely the tiling
of the square. We call these good triangles. Notice that the catheti of a good triangle have equal length.
There are some possible choices of good triangles and we would now like to expose them. In order to do this,
we describe a good triangle by an ordered pair of ordered lists (A, B) := (a1 . . . ak , b1 . . . b` ) with k, ` ≥ 1, where
a1 , . . . , ak , b1 , . . . , b` ∈ {1, 2}. The list a1 . . . ak describes the tiling of the horizontal side of the triangle and the list
b1 . . . b` describes the vertical tiling of the triangle. In particular, we must have d := a1 + · · · + ak = b1 + · · · + b`
but a1 + · · · + ai 6= b1 + · · · + bj whenever i + j < k + `. An example is shown in the figure. Using this condition,
we see that the only possibilities are (1, 1), (2, 2), or one of the infinite families:
(11, 2), (121, 22), (1221, 222), (12221, 2222), . . .
(2, 11), (22, 121), (222, 1221), (2222, 12221), . . .
(12, 21), (122, 221), (1222, 2221), (12222, 22221), . . .
(21, 12), (221, 122), (2221, 1222), (22221, 12222), . . .
For the generic choice of pair above, let σ(A) be the number of 1’s in list A and let σ(B) be the number of 1’s
in B. Then, the contribution to the generating function of the pair (A, B) is uσ(A) v σ(B) xd .
QUALS PSET
(11, 2)
u2x2
(2, 11)
5
(1, 1)
2 2
v x
(2,2)
uvx
x2
The tilings under consideration of the square with n good triangles are then described by
n
uvx + x2 + (u2 + v 2 )(x2 + x4 + x6 + . . . ) + (2uv)(x3 + x5 + x7 + . . . )
n
x2
x3
2
2
2
.
= uvx + x + (u + v )
+ 2uv
1 − x2
1 − x2
Hence,
1+
X
pn (u)pn (v)xn =
n≥1
X
uvx + x2 + (u2 + v 2 )
n≥0
x2
x3
+ 2uv
2
1−x
1 − x2
n
1
.
x2
x3
1 − uvx − −
+ v 2 ) 1−x
2 − 2uv 1−x2
—————————————————————————————————————————————————
=
x2
(u2
10. We have
Fα (x) =
X
fα (n)xn
n≥0
n X
n−k
X n − k =
x
α =
(αx )
xn−2k
k
k
n≥0
k=0
k≥0
n≥2k
X
X n + k X (αx2 )k
=
(αx2 )k
xn =
(1 − x)k+1
k
n≥0
k≥0
k≥0
k
1
1
1 X αx2
=
·
=
αx2
1−x
1−x
1 − x 1 − 1−x
k≥0
X
n
k
X
2 k
1
.
1 − x − αx2
Hence, setting an := fα (n) for all n ≥ 0, we obtain the recursion
an = an−1
+
αan−2√for n ≥ 2, with a0 = a1 = 1.
√
1+ 1+4α
2
The characteristic polynomial of the recursion is t − t − α = t −
t − 1− 21+4α . If α 6= − 41 , then the
2
√
n √
n
solution to the recursion is an = 1+ 21+4α
+ 1+ 21+4α
for n ≥ 1. If α = − 14 , then the solution takes the
n
n
form an = 12 + n 12 for n ≥ 1.
—————————————————————————————————————————————————
=
P
11. For n ≥ 1, let h(n) = {B1 ,...,Bk }∈Πn f (]B1 ) . . . f (]Bk ). Define h(0) = 1. A rooted tree on vertices [n + 1] of
the kind described in the problem can be constructed by first picking a root i (for which there are n + 1 choices),
then picking a partition {B1 , . . . , Bk } in Π[n+1]\{i} ∼
= Πn , then constructing a rooted tree of the same kind on each
of the blocks B1 , . . . , Bk , and finally appending the roots of these k trees to i. All the resulting trees are pairwise
distinct, so for n ≥ 1, we have f (n + 1) = (n + 1)h(n). Consider
X
X
xn+1
xn+1
xey =
h(n)
=
(n + 1)h(n)
n!
(n + 1)!
n≥0
=x+
n≥0
X
n≥1
As f (1) = 2, we have that x + xey = x(1 + ey ) = y.
f (n + 1)
xn+1
.
(n + 1)!
6
BENJAMIN IRIARTE
From the recursion we obtain x =
y
1+ey ,
so y =
x
1+ex
h−1i
. From the Lagrange Inversion Formula, we know
Pn
)
in−1
= n1 [xn−1 ](1 + ex )n . But (1 + ex )n = i=0 ni eix and [xn−1 ]eix = (n−1)!
, so [xn ]y =
that [xn ]y = n1 [x−1 ] (1+e
xn
P
P
n
n
n
n
1
1 n−1
n−1
](1 + ex )n = n!
. As y = Ef (x), this proves the formula f (n) = i=0 i in−1 .
i=0 i i
n [x
—————————————————————————————————————————————————
x n
12. Suppose that D := D(V, E) with |V | = n ≥ 2 is a digraph with exactly one Eulerian tour. We know that D
is balanced and connected, that outdeg(v) is either 1 or 2 and that τ (D, v) = 1 for all v ∈ V .
Let u, v ∈ V with u 6= v. We prove that there is exactly one diwalk (directed simple walk) from u to v. Clearly
there is at least one such diwalk. Suppose there are two distinct diwalks from u to v, say (v0 , v1 , . . . , vi ) and
(u0 , u1 , . . . , uj ) with u = v0 = u0 and v = vi = uj . WLOG we can assume that these walks meet only at u and
v. Consider the unique spanning tree T of D with root v, and assume (v0 , v1 , . . . , vi ) is the diwalk given by T . In
T , delete the edge (v0 , v1 ) and then add the edge (u0 , u1 ). This gives a different spanning tree of D with root v,
contradicting the uniqueness of such a tree.
Consider the dicycles (directed simple cycles) of D, say they are C1 , . . . , Cm . Now, consider any v ∈ V and
directed edge e with init(e) = v. Let (v0 , v1 , . . . , vk ) be the Eulerian tour of D, with v = v0 = vk . Consider the
segment of the tour between two consecutive occurrences of v whose first edge is e. In this segment, there are either
repeated vertices or none. If there are no repeated vertices, we obtain a dicycle of D containing v and e. If there
are repeated vertices, let u be the first repeated vertex. Pasting the subsegment from v to the first occurrence of
u to the subsegment from the last occurrence of u to v, we obtain again a dicycle containing v and e. Hence, the
dicycles C1 , . . . , Cm contain all vertices V and directed edges E. In particular, m ≥ 1.
Suppose that Ci and Cj meet, for some i 6= j. If they meet at two distinct vertices u and v, then either we
obtain two distinct diwalks from u to v or two distinct walks from v to u, neither of which can be true. Therefore,
Ci and Cj meet only in one vertex. Hence, every vertex in V belongs to at most two dicycles.
Consider the graph G(D) with vertex set {c1 , . . . , cm } for which vertices ci and cj with i 6= j are connected by
an edge iff the dicycles Ci and Cj meet. Suppose G(D) has a simple cycle of length ≥ 3, say (c1 , c2 , . . . , ck , c1 ) for
ease of notation. Pick two different vertices u and v in C1 , say u = V (C1 ) ∩ V (C2 ) and v = V (C1 ) ∩ V (Ck ). There
is a diwalk from u to v given by directed edges in C1 , but also one coming from directed edges in C2 ∪ · · · ∪ Ck ,
and these diwalks are distinct, which is not possible. Hence, G(D) is a tree.
We stop briefly the argument to introduce a definition. Let T := T (V, F, v) be a labeled tree on V with
distinguished vertex v and with set of directed edges F . Let T 0 := T 0 (V, F 0 , v) be the rooted tree (with root v)
obtained from T by orienting all its edges out of v. Suppose that T is binary. Furthermore, suppose that for every
(x, y), (x, z) ∈ F 0 , either (x, y), (z, x) ∈ F or (y, x), (x, z) ∈ F . Then, we call T a di-binary tree with root v.
Continuing with the problem, pick a vertex v ∈ V . Assume C1 contains v. We construct a di-binary tree Tv
on V with root v. We construct Tv from D by deleting exactly one edge from every dicycle Ci to obtain a diwalk
Wi . Consider G(D) to be rooted with root c1 , all edges pointing out of c1 . Color the vertices of G(D) red or blue.
Let c1 be red, and complete the coloring of G(D) by requiring that no two adjacent vertices have the same color.
For (ci , cj ) a directed edge in G(D), let vj := V (Ci ) ∩ V (Cj ), and set v1 := v. Now, for every 1 ≤ i ≤ m and
Ci = (vi , u1 , . . . , uk , vi ), let Wi = (vi , u1 , . . . , uk ) if ci is red and let Wi = (u1 , . . . , uk , vi ) if ci is blue. This yields
Tv after all cycles have been considered.
There are two key things to notice now. First, D can be uniquely recovered from Tv . In Tv , consider all maximal
diwalks W := (v0 , v1 , . . . , vk ) for which v 6= v1 , . . . , vk−1 (so when k = 1 the condition is satisfied trivially), and call
them good diwalks. For every such good diwalk W , add the directed edge (vk , v0 ) to W to complete a dicycle. The
set of good diwalks of Tv gives a partition of the set of directed edges of Tv . Completing the dicycle as above for
all good diwalks in Tv , we obtain back D. Now, if we start from a coloring of G(D) for which c1 is blue instead of
red, we obtain a di-binary tree Tv0 with root v from which, analogously, D can be uniquely recovered by completing
the good diwalks. But Tv is distinct from Tv0 as di-binary trees. Hence, there are exactly two di-binary trees on
V with root v to which we associate D, and both of them are uniquely associated to D. Furthermore, from every
di-binary tree T on V with root v, the above process produces a digraph on V with exactly one Eulerian tour. For
example, this can be checked by induction on the number of good diwalks in T .
Second, note that di-binary trees on [n] with root n are equivalent to labeled binary (undirected) trees on [n] with
root n (which explains the somehow artificial definition of di-binary tree). The number of binary trees on n vertices
is Cn , so the number of labeled binary trees on [n] with root n is then (n − 1)!Cn . Hence, the number of di-binary
trees on V with root v is also (n − 1)!Cn . As discussed above, to every D we associate two distinct di-binary trees
with root v, so the total number of digraphs on n vertices with exactly one Eulerian tour is 21 (n − 1)!Cn .
QUALS PSET
7
—————————————————————————————————————————————————
P −1
P
P
P
−1
13. We have
λ zλ pλ (x)pλ (y) =
λ sλ (x)sλ (y). In particular,
λ`n zλ pλ (x)pλ (y) =
λ`n sλ (x)sλ (y),
P
−1 2
m1 m2
mn
2
2m1 m2
so yn = λ`n zλ pλ . For λ = (1 2 . . . n ) ` n, let λ ` 2n be the partition (1
2 . . . n2mn ). Hence,
P
−1
yn = λ`n zλ pλ2 and
X zλ2
hyn , yn i =
z2
λ`n λ
X 2m1 (λ)2m2 (λ) 2mn (λ)
=
...
.
m1 (λ)
m2 (λ)
mn (λ)
λ`n
n
P
√ 1
Now, n≥0 2n
. Clearly, defining hy0 , y0 i = 1, then
n q =
1−4q


r
Y
X
Y X 2n
1
in 
n

q
=
hyn , yn iq =
n
1 − 4q i
n≥0
n≥0
i≥1
s
=
i≥1
p
1
= P (q, 4).
i
1 − 4q
Y
i≥1
—————————————————————————————————————————————————
14. We have,
a(m, n) =
X X X
f µ f ν f λ cλµν
µ`m ν`n−m λ`n
*

X
=

!
X
µ
f sµ 
µ`m
ν
f sν
+
,
ν`n−m
X
λ
f sλ
λ`n
n−m n
= hpm
, p1 i = hp1m p1n−m , p1n i = hp1n , p1n i
1 p1
= z1n = n!.
And,
X X X
b(m, n) =
f µ f ν cλµν
µ`m ν`n−m λ`n
*
=

X

!
X
µ
f sµ 
µ`m
=
+
p1n ,
=
sλ
=
pω
ω`n
p1n ,
zω−1 χλ (ω)pω
λ`n ω`n
zω−1 χλ (ω)
*
!
X
p1n , z1−1
n
=
λ`n
=
sλ
λ`n
XX
!+
p1n ,
,
+
λ`n
X
+
X
*
*
X
f sν
ν`n−m
*
X
ν
λ
+
n
χ (1 ) p1n
λ`n
X
χλ (1n ) =
λ`n
X
f λ = t(n).
λ`n
And,
X X X
c(m, n) =
f ν f λ cλµν
µ`m ν`n−m λ`n
*
=

X

sµ 
µ`m
*
=

X

sµ  p1n−m , p1n
*
=
ν
f sν
=
X


z1−1
m
µ`m
λ
f sλ
λ`n
pω 

X
,
X

*
ω`m

+
ν`n−m
+
µ`m
!
X
χµ (1m ) p1m p1n−m , p1n

X
+
zω−1 χµ (ω) p1n−m , p1n
µ`m
+


= z1−1
m z1n

X
µ`m
f µ =
n!
t(m)
m!
8
BENJAMIN IRIARTE
Also,
X X X
d(m, n) =
f λ cλµν
µ`m ν`n−m λ`n
*

X
=

sµ 
µ`m
*
=

pa 
a`m
*
=
λ
f sλ
λ`n
!!
X
za−1 χµ (a)

X
pb
b`n−m
+
, p1n
!
X
µ m 
z1−1
) p1n−m
m χ (1
+
ν n−m
z1−1
) , p1n
n−m χ (1
ν`n−m

−1
z1n z1−1
m z n−m
1
zb−1 χν (b)
ν`n−m

X
µ`m
=
,
X

X
µ`m
p1m 
sν
+
ν`n−m

X
!
X

X

µ`m
f
µ
!
X
ν`n−m
f
ν
=
n
t(m)t(n − m).
m
—————————————————————————————————————————————————