Transcript Proof of the middle levels conjecture

Proof of the middle
levels conjecture
Torsten Mütze
Hamilton cycles
• Hamilton cycle = cycle that visits every vertex exactly once
Hamilton cycles
• Problem: Given a graph, does it have a Hamilton cycle?
• fundamental problem with many applications
(special case of travelling salesman problem)
• computational point of view:
• no efficient algorithm known (NP-complete [Karp 72]),
i.e. brute-force approach essentially best possible
• what about particular families of graphs?
The middle layer graph
Consider the cube
11...1
level 3
111
level 2 110
101
011
level 1 100
010
001
level 0
000
00...0
Middle layer graph
The middle layer graph
Consider the cube
11...1
level 3
111
level 2 110
101
011
level 1 100
010
001
level 0
000
00...0
Middle layer graph
The middle layer graph
Middle layer of
11100 11010
11001
10110
10101
10011
01110
01101
01011
00111
11000 10100
10010
10001
01100
01010
01001
00110
00101
00011
Question:
Does the middle layer of
for every
?
have a Hamilton cycle
The middle layer graph
Middle layer of
11100 11010
11001
10110
10101
10011
01110
01101
01011
00111
11000 10100
10010
10001
01100
01010
01001
00110
00101
00011
• bipartite, connected
• number of vertices:
• degree:
The middle layer graph
Middle layer of
•
•
•
•
11100 11010
11001
10110
10101
10011
01110
01101
01011
00111
11000 10100
10010
10001
01100
01010
01001
00110
00101
00011
bipartite, connected
number of vertices:
degree:
automorphisms:
bit permutation + inversion,
The middle layer graph
Middle layer of
•
•
•
•
11100 11010
11001
10110
10101
10011
01110
01101
01011
00111
11000 10100
10010
10001
01100
01010
01001
00110
00101
00011
bipartite, connected
number of vertices:
degree:
automorphisms:
bit permutation + inversion,
The middle layer graph
Middle layer of
•
•
•
•
11100 11010
11001
10110
10101
10011
01110
01101
01011
00111
11000 10100
10010
10001
01100
01010
01001
00110
00101
00011
bipartite, connected
number of vertices:
degree:
automorphisms:
bit permutation + inversion,
• vertex-transitive
The middle levels conjecture
Conjecture:
The middle layer graph has a Hamilton cycle for every
• also known as „revolving door conjecture“
.
The middle levels conjecture
Conjecture:
The middle layer graph has a Hamilton cycle for every
.
• probably first mentioned in [Havel 83], [Buck, Wiedemann 84]
• also (mis)attributed to Dejter, Erdős, Trotter [Kierstead, Trotter 88]
and various others
• exercise (!!!) in [Knuth 05]
The middle levels conjecture
Conjecture:
The middle layer graph has a Hamilton cycle for every
.
Motivation:
• Gray codes (applications e.g. in digital communication)
111
111
101
011
110
001
100
010
000
000
010
110
111
011
001
101
100
101
011
110
001
100
010
000
001
011
010
110
100
101
The middle levels conjecture
Conjecture:
The middle layer graph has a Hamilton cycle for every
.
Motivation:
• Conjecture [Lovász 70]: Every connected vertex-transitive graph
has a Hamilton cycle (apart from five exceptions).
History of the conjecture
Numerical evidence:
The conjecture holds for all
[Moews, Reid 99], [Shields, Savage 99],
[Shields, Shields, Savage 09], [Shimada, Amano 11]
History of the conjecture
Asymptotic results:
The middle layer graph has a cycle of length
•
[Savage 93]
•
[Felsner, Trotter 95]
•
[Savage, Winkler 95]
•
[Johnson 04]
History of the conjecture
Other relaxations and partial results:
[Kierstead, Trotter 88]
[Duffus, Sands, Woodrow 88]
[Dejter, Cordova, Quintana 88]
[Duffus, Kierstead, Snevily 94]
[Hurlbert 94]
[Horák, Kaiser, Rosenfeld, Ryjácek 05]
[Gregor, Škrekovski 10]
…
Our results
Theorem 1:
The middle layer graph has a Hamilton cycle for every
Theorem 2:
The middle layer graph has
.
different Hamilton cycles.
Remarks:
number of automorphisms is only
,
so Theorem 2 is not an immediate consequence of Theorem 1
Our results
Theorem 1:
The middle layer graph has a Hamilton cycle for every
Theorem 2:
The middle layer graph has
.
different Hamilton cycles.
Remarks:
number of Hamilton cycles is at most
so Theorem 2 is best possible
,
Our results
Theorem 1:
The middle layer graph has a Hamilton cycle for every
Theorem 2:
The middle layer graph has
.
different Hamilton cycles.
Remarks:
proof is constructive  algorithm to compute the cycle (inefficient)
[M., Nummenpalo 15+]:
algorithm to compute next cycle vertex in time
on average
Proof ideas
Step 1: Build a 2-factor in the middle layer graph
Step 2: Connect the cycles in the 2-factor to a single cycle
Structure of the middle layer graph
Structure of the middle layer graph
Structure of the middle layer graph
A Hamilton cycle
Catalan numbers
Structure of the middle layer graph
A Hamilton cycle
Structure of the middle layer graph
A Hamilton cycle
Structure of the middle layer graph
A Hamilton cycle
Structure of the middle layer graph
A Hamilton cycle
Step 1: Build a 2-factor
Construction from [M., Weber 12]
???
isomorphism
(bit permutation + inversion)
Step 1: Build a 2-factor
Construction from [M., Weber 12]
2-factor
isomorphism
(bit permutation + inversion)
Step 1: Build a 2-factor
Construction from [M., Weber 12]
• parametrizing yields
different 2-factors
• essentially only one can be analyzed:
= plane trees with edges
2-factor
Fundamental problem:
varying changes
globally
Step 2: Connect the cycles
New ingredient: Flippable pairs
2-factor
is a flippable pair,
if there is a flipped pair
such that
Step 2: Connect the cycles
New ingredient: Flippable pairs
2-factor
is a flippable pair,
if there is a flipped pair
such that
Step 2: Connect the cycles
New ingredient: Flippable pairs
2-factor
flippable pairs
yield
different 2-factors
+ very precise local control
…we can construct many
flippable pairs
Step 2: Connect the cycles
New ingredient: Flippable pairs
2-factor
Auxiliary graph
Step 2: Connect the cycles
Lemma 1: If
is connected, then the middle layer
graph contains a Hamilton cycle.
Lemma 2: If
contains
the middle layer graph contains
2-factor
different spanning trees, then
different Hamilton cycles.
Auxiliary graph
The crucial reduction
Prove that
middle layer graph
has a Hamilton cycle
(many Hamilton cycles)
Prove that
is connected
(has many spanning trees)
Analysis of
2 leaves
6 leaves
5 leaves
4 leaves
= plane trees with
3 leaves
edges
Analysis of
2 leaves
6 leaves
5 leaves
4 leaves
= plane trees with
3 leaves
edges
Thank you!