... 2006, 497–550.
the electronic journal of combinatorics 15 (2008), #N16 5
Necklace bisection with one cut less than needed
G´abor Simonyi
∗
Alfr´ed R´enyi Institute of Mathematics
Hungarian Academy ... specify
two disjoint sets D
1
, D
2
of the types of beads with D
1
∪ D
2
= ∅, it will always be
possible to cut the necklace (with k − 1 cuts) so that the first thief gets more...
... there
exists one- factorization of the complete graph K
2n
such that any two one- factors
do not induce a graph with a cycle of length k as a component. Moreover, some
infinite classes of one- factorizations, ... cycle of length k with all vertices in V
m
because one-
factorization induced by V
m
is the given one- factorization
˜
F of K
n+1
, which is k-cycle
free. Let C
l
be a cycle...
... oscillators gradually refining periodic movements executed on the
robot. Through this process, the user can grasp the rob otic arm and locally modify
the executed movement, which is learnt on-line, modulating ... current musical
parameters. After releasing the arm, the robot continues the execution of the
movement in consecutive loops. During the interaction, the impedance parameters
of our ro...
... to X
∗
, and K be
a nonempty closed convex subset of X. Suppose that T : Ω × K → K is
nonexpansive random mapping and f : Ω × K → K is a weakly contractive
random mapping with a function ϕ, then
(i) ... random fixed point.
5
for y ∈ X with y = x. It is well known that the above inequality is equivalent
to
lim sup
n→∞
x
n
− x < lim sup
n→∞
x
n
− y
for y ∈ X with y = x.
It is well...
... Definition of the algorithm
Assume that D is a skew shape with one box z removed (note that any skew
shape can be considered a skew shape with one box removed by simply attaching
a corner either on ... above
however, both terms have “degree” equal to three. Also, there is more than
one way to obtain n
i
for any i>1. This means that one generally cannot
reconstruct M by substit...
... of one of the
K+1
K
K-subsets of [K +1] with theset{K +2}.This
collection is itself followed by
K+2
K
sets, each of them being the union of one of the
K+2
K
K-subsets of [K +2]withtheset{K ... antichain with |I
K+1
A| ≥ K +1. Then V (A) <
U
n
.
Proof. If V (A) ≤ V
max2
(n)thenV (A) <U
n
and we are done. If V (A) >V
max2
(n)
then A must contain sets of size less...