3,170 research outputs found
The Complexity of Change
Many combinatorial problems can be formulated as "Can I transform
configuration 1 into configuration 2, if certain transformations only are
allowed?". An example of such a question is: given two k-colourings of a graph,
can I transform the first k-colouring into the second one, by recolouring one
vertex at a time, and always maintaining a proper k-colouring? Another example
is: given two solutions of a SAT-instance, can I transform the first solution
into the second one, by changing the truth value one variable at a time, and
always maintaining a solution of the SAT-instance? Other examples can be found
in many classical puzzles, such as the 15-Puzzle and Rubik's Cube.
In this survey we shall give an overview of some older and more recent work
on this type of problem. The emphasis will be on the computational complexity
of the problems: how hard is it to decide if a certain transformation is
possible or not?Comment: 28 pages, 6 figure
The complexity of Free-Flood-It on 2xn boards
We consider the complexity of problems related to the combinatorial game
Free-Flood-It, in which players aim to make a coloured graph monochromatic with
the minimum possible number of flooding operations. Our main result is that
computing the length of an optimal sequence is fixed parameter tractable (with
the number of colours present as a parameter) when restricted to rectangular
2xn boards. We also show that, when the number of colours is unbounded, the
problem remains NP-hard on such boards. This resolves a question of Clifford,
Jalsenius, Montanaro and Sach (2010)
Guessing Numbers of Odd Cycles
For a given number of colours, , the guessing number of a graph is the
base logarithm of the size of the largest family of colourings of the
vertex set of the graph such that the colour of each vertex can be determined
from the colours of the vertices in its neighbourhood. An upper bound for the
guessing number of the -vertex cycle graph is . It is known that
the guessing number equals whenever is even or is a perfect
square \cite{Christofides2011guessing}. We show that, for any given integer
, if is the largest factor of less than or equal to
, for sufficiently large odd , the guessing number of with
colours is . This answers a question posed by
Christofides and Markstr\"{o}m in 2011 \cite{Christofides2011guessing}. We also
present an explicit protocol which achieves this bound for every . Linking
this to index coding with side information, we deduce that the information
defect of with colours is for sufficiently
large odd . Our results are a generalisation of the case which was
proven in \cite{bar2011index}.Comment: 16 page
Towards on-line Ohba's conjecture
The on-line choice number of a graph is a variation of the choice number
defined through a two person game. It is at least as large as the choice number
for all graphs and is strictly larger for some graphs. In particular, there are
graphs with whose on-line choice numbers are larger
than their chromatic numbers, in contrast to a recently confirmed conjecture of
Ohba that every graph with has its choice number
equal its chromatic number. Nevertheless, an on-line version of Ohba conjecture
was proposed in [P. Huang, T. Wong and X. Zhu, Application of polynomial method
to on-line colouring of graphs, European J. Combin., 2011]: Every graph
with has its on-line choice number equal its chromatic
number. This paper confirms the on-line version of Ohba conjecture for graphs
with independence number at most 3. We also study list colouring of
complete multipartite graphs with all parts of size 3. We prove
that the on-line choice number of is at most , and
present an alternate proof of Kierstead's result that its choice number is
. For general graphs , we prove that if then its on-line choice number equals chromatic number.Comment: new abstract and introductio
An analogue of Ryser's Theorem for partial Sudoku squares
In 1956 Ryser gave a necessary and sufficient condition for a partial latin
rectangle to be completable to a latin square. In 1990 Hilton and Johnson
showed that Ryser's condition could be reformulated in terms of Hall's
Condition for partial latin squares. Thus Ryser's Theorem can be interpreted as
saying that any partial latin rectangle can be completed if and only if
satisfies Hall's Condition for partial latin squares.
We define Hall's Condition for partial Sudoku squares and show that Hall's
Condition for partial Sudoku squares gives a criterion for the completion of
partial Sudoku rectangles that is both necessary and sufficient. In the
particular case where , , , the result is especially simple, as
we show that any partial -Sudoku rectangle can be completed
(no further condition being necessary).Comment: 19 pages, 10 figure
Strongly representable atom structures of relation algebras
Accepted versio
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