Post on 05-Jan-2016
The Ryjacek Closure
and
a Forbidden Subgraph
Akira Saito (Nihon University, Japan)
Liming Xiong (Beijing Institute of Technology, China)
Ryjacek Closure
(Preaching Buddhism to Buddha)
In 1997, Ryjacek discovered a new closure.
Ryjacek closure
• a locally-connected vertex• local completion
A vertex in a graph is said to be locally-connected if (the
neighborhood of ) induces a connected graph in .
Local completion at is the operation of joining every pair of non-
adjacent neighbors of by an edge.
A vertex in a graph is said to be eligible if is locally-connected
and does not induce a complete graph.
Ryjacek Closure
: a sequence of graphs
•
• is obtained from by local completion at an eligible vertex
in .
: a graph
If does not contain an eligible vertex, we call a (the) Ryjacek
closure and denote it by clR.
appears to depend on the order of eligible vertices chosen for
local completion at each step.
Ryjacek Closure
Theorem A ( Ryjacek 1997 )
1. The Ryjacek closure is uniquely determined, regardless of the order of
eligible vertices chosen for local completion.
2. If is a claw-free graph, then clR is a line graph.
3. If is a claw-free graph, then is hamiltonian if and only if clR is
hamiltonian.
(Note that the uniqueness of the Ryjacek closure is not limited to the
class of claw-free graphs.)
Ryjacek Closure
The Ryjacek closure is a powerful operation in the class of claw-free graphs.
Conjecture 1 (Matthews-Sumner Conjecture, 1984)
Every 4-connected claw-free graph is hamiltonian.
Conjecture 2 (Thomassen’s Conjecture, 1986)
Every 4-connected line graph is hamiltonian.
Corollary B (Ryjacek, 1997)
Conjecture 2 imlpies Conjecture 1.
Standard Approach
claw-free graph
line graph
pre-image
Ryjacek closure
Harary & Nash-Williams
super eulerian graphs and other techniques
Motivation
The Ryjacek closure is a powerful operation in the class of claw-free
graphs.
Is there any other class of graphs defined by a forbidden subgraph, in
which the Ryjacek closure works effectively?
: a connected graph of order at least three
𝑆 (𝐻 )=¿ “For every -free graph of sufficiently large order, is hamiltonian if
and only if clR is hamiltonian.”
Want : a graph which makes the statement true
Candidates
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
Proof (easy)
“For every -free graph of sufficiently large order, is hamiltonian if
and only if clR is hamiltonian.”
If holds, then a sufficiently large locally-connected -free graph is
hamiltonian.
Theorem 1
For a connected graph of order at least three, if the statment holds,
then H is , , , or .
(A graph is locally-connected if every vertex in is locally-connected.)
Proof (easy)
• non-hamiltonian
If holds, then a sufficiently large locally-connected -free graph is
hamiltonian.
• locally-connected
• can be arbitrarily large
This graph cannot be -free.
is an induced subgraph of this graph.
Proof (easy)
If holds, then a sufficiently large locally-connected -free graph is
hamiltonian.
𝐾 𝑠
Replace each edge in with the graph
on the right.
• non-hamiltonian
• locally-connected
• can be arbitrarily large
This graph cannot be -free.
is an induced subgraph of this graph.
Proof (easy)
is an induced subgraph of these graphs.
𝐾 𝑠
Replace each edge in with the graph
on the right.
H is , , , or .
Candidates
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
Check three graphs
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1Ryjacek’s result
Check three graphs
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
Check
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
Check
𝐾 1,2
“For every -free graph of sufficiently large order, is hamiltonian if
and only if clR is hamiltonian.”
What is a -free graph?
No pair of vertices of distance 2
Every component is a complete graph.
No eligible vertex
clR
is true, but trivial.
Check
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
Check
“For every -free graph of sufficiently large order, is hamiltonian if
and only if clR is hamiltonian.”
What is a -free graph?
The neighborhood of a vertex has no edge.
If is locally-connected, .
No eligible vertex
clR
is true, but trivial.
𝐾 3
Check
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
Check
“For every -free graph of sufficiently large order, is
hamiltonian if and only if clR is hamiltonian.”
What is a -free graph?
A graph is -free if and only if the neighborhood
of every vertex induces a -free graph
If is locally-connected, then induces a complete graph.
No eligible vertex clR
is true, but trivial.
𝐾 2+2𝐾 1
Check three graphs
Theorem 1
For a connected graph of order at least three, if the statement holds,
then H is , , , or .
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
These graphs make trivial.
Conclusion
𝐾 1,2 𝐾 3 𝐾 1 ,3 𝐾 2+2𝐾 1
𝑆 (𝐻 )=¿ “For every -free graph of sufficiently large order, is hamiltonian if
and only if clR is hamiltonian.”
Theorem 1
For a connected graph of order at least three, the statement holds if
and only if H is , , , or .
However, only gives a nontrivial statement.
Conclusion
𝑆 (𝐻 )=¿ “For every -free graph of sufficiently large order, is hamiltonian if
and only if clR is Hamiltonian.”
Theorem 1
For a connected graph of order at least three, the statement holds if
and only if H is , , , or .
The Ryjacek closure works in a non-trivial manner only in the class of
claw-free graphs.
Question
• The Ryjacek closure works in a non-trivial manner only in the class of
claw-free graphs.
But why is the Ryjacek closure so powerful in the class of claw-
free graphs, while it does now work effectively for other forbidden
subgraphs?
Is there any mechanism which makes the combination of the Ryjacek
closure and claw-free graphs so powerful?
• The Ryjacek closure works in a non-trivial manner only in the class of
claw-free graphs as long as we forbid one graph.
What happens if we forbid two or more graphs?
Děkuji mockrát !Thank you!