... interpretation.
This interpretation is in line with the rest of this art icle, counting permutations having an
initial descending segment and λ-coloured fixed points to the right of the initial segment.
In addition, ... n,
generalising the results with λ = 0 from [9]. In our proofs, we will use the following
convent io ns. If λ is a positive integer, and (π,C) is a permutation w...
... lozenge in exactly one-third of the tilings.
Proof. We compare the result of setting m = n in Corollary 12 with the result of
setting m = n in 15 and evaluating the sum by Lemma 16.
8.
Computing ... Aztec diamond of
order 3.
We are interested in tilings of hexagons with lozenges, which are rhombi with unit
sides and angles of 120 and 60 degrees, and tilings of Azte...
... monotonic uniformly continuous function of x and y
on the unit square. In fact, it is increasing in x if and only if α begins with d and
is increasing in y if and only if α ends with u.
(b) f(x, y, ... in (8.1), but it is needed now because we normalize
φ to have
1
0
φ = 1 and we incorporate a factor of λ in C.)
One can interpret finding the eigenfunction φ in terms of...
... A perfect match-
ing of a graph G is a set o f independent edges of G covering all vertices of G. Problems
involving enumeration of perfect mat chings of a graph were first examined by chemists
and ... Enumeration of perfect matchings of Q
m,n,r
3.1 The sign of pfB(x
1
, x
2
)
In order to decide the sign of pfaffians of B(x
1
, x
2
)(x
1
, x
2
= ±1), we distinguish the
perf...
... q-enumeration of ASM’s, which consists in
counting ASM’s (or classes of symmetry of ASM’s) with respect to their number of −1
entries (to the number of orbits of −1 entries in the case of symmetric ... symmetric
in all its variables.
Stroganov proved this surprising symmetry property by a study of Izergin-Korepin
determinant. A proof only involving Yang-Baxter equation...
... theorem.
Furthermore, letting x = 0 in (10), we deduce that the number of permutation triples
of S
n
with “RRR” forbidden satisfies the following binomial relation
n
k=0
(−1)
k
n
k
3
C
k
= 0 with n > ... 6
Enumeration of Restricted
Permutation Triples
Xiaojing Chen
School of Mathematics and Computational Science
China University of Petroleum
Dongying 257061, P. R. Chin...
... choice of the origin p
0
, a pin representation is not associated
with a unique pin word, but with at most 8 pin words (see Figure 4). The set of pin words
is the set of all encodings of pin-permutations. ... by keeping in the same order points p
i
that form an
occurrence of π in σ.
Instead of random patterns of a pin-permutat io n σ, we will often be interested in
pat...
... defect. Interest in such graphs is part of the general study of the
∗
mirka.miller @newcastle.edu.au
†
This research was supported by a Marie Curie International Incoming Fellowship within the ... graphs of defect 2. In such a graph G any vertex v can reach within D steps either two
vertices (called repeats of v) in two different ways each, or one vertex (called double repeat
of...
... element, as in a
classical structure, but a subset of H. One of the interests of the researchers in the field
of hyperstructur es is to construct new hyperoperations using g raphs [18], binary relations
[2, ... examples of join spaces have
been constructed in connection with binary relations, graphs and lattices. In order to de-
fine a join space, we need the following notat...