Biased MakerBreaker Hamilton cycle game

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Biased Maker-Breaker Hamilton cycle game

• Jozsef Beck

Slides by Pavel Anissimov and Sela Ferdman
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Game description

• Game is on complete graph Kn
• Maker takes one edge, Breaker b edges
• Maker wins if succeeds to build Hamiltonian
  cycle.

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Games breaking point
Maker wins Breaker wins
b

• We will prove lower bound for breaking point

by showing winning strategy for Maker.
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Makers Strategy

• Build graph with certain properties, using
  potential blocking strategy (from
  Erdos-Selftridge theorem)
• Apply several times Longer-Path Argument

After performing both steps, Makers graph will
contain Hamiltonian cycle.
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Longer-Path Argument

• We will need several definitions

- set of vertices adjacent to at least one vertex
of S
Pósa-deformation Let P (v0,v1, ,vm) path in
graph G. If vi, vm?G then we say that the path
(v0, , vi, vm, vm-1, , vi1) arises by
Pósa-deformation from P. End(G, P, v0) set of
all endpoints of paths arising by repeated
Posa-deformations from P, keeping the starting
point v0 fixed.
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Pósa-deformations example
Endpoints
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Pósa Lemma

• Lemma Given graph G. Assume for every
• vertex-set ,G satisfies .
• Let P (v0,v1, ,vm) maximum length path in G
  which starts from v0.
• Then .

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Pósa Lemma example
k 1 - each vertex has at least 2 neighbors
Maximal length path
Path after Pósa-deformation
2 endpoints k1
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Proof Idea

• Well show that satisfy
• But assumption is that every vertex-set
  satisfies
• It implies

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Contd

• Why satisfy ?

• Neighbor of endpoint must be vertex on P
  (otherwise well have path longer than P)
• Moreover it can be shown that this neighbor has
  to be close to some another endpoint.

Neighbors Endpoints
v0
vm
Formally Its clear that
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Longer-Path Argument

• Suppose were in the middle of the play and
  Makers graph G has two properties
• a) For every
• b) G is connected on n vertices
• Well show how maker can enlarge maximum path in
  G.
• Repeating this several times well have path of
  length V - Hamiltonian Cycle.

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Longer-Path Argument

• Assume Makers graph G satisfies properties (a)
  and (b), and P(v0,v1, ,vm) maximum length path
  in G.
• We define P(xi) - path arising from P by
  Posa-deformations such that xi is its endpoint.
• Let
• Graph Close(G, P) consists of closing edges
  edges which turn some maximum path aroused from
  P(x) into a cycle.
• Notice that

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Example of Close(G,P)
G
P(x3)
P(x2)
x0
x0
x3
x2
x0
x0
x0
x3
x2
x0
x1
x0
x3
x2
x0
x1
x0
x1
x3
x3
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Makers Move Case 1

• Some edges of Close(G, P) arent occupied yet,
  and Maker doesnt own them.

Makers move is to take edge e from Close(G,
P). After the move, Makers graph will contain
cycle C of length m1 (m is length of maximum
path P). If m1 n then C is Hamiltonian
Cycle. Otherwise there is a vertex w outside of
C, which is connected to C via some path (because
G is connected). Combining this path with C we
obtain new path longer than P.
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Example
P
C
C
w
Egde from Close(G,P)
P
w
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Makers Move Case 2

• Maker already owns an edge from Close(G,P).

• This case is possible only when Makers graph
  already contains Hamiltonian cycle.
• Indeed, Makers graph already contains cycle C of
  length m1. If m1 lt n it follows from previous
  case that there is a path P longer than P.
  Contradiction to the fact that P is maximum path.

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Makers Move Case 3

• All edges in Close(G,P) are occupied by Breaker.

In this case Breaker already owns at least
edges. This require at least moves.
Thus If Maker can guarantee that, after
moves,
his graph satisfies properties (a) and (b), then
in at most n extra moves he will always own
Hamiltonian cycle.
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Transversal Hypergraph Method

• Now, in order to force properties (a) and (ß) we
  will use the Transversal Hypergraph Method.
• Let be the set of all complete bipartite
  graphs of type
• If Maker can occupy at least one edge from each
  during the first t moves,
  then property (a) is satisfied.

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Why property (a) is satisfied?

• Assume property (a) is NOT satisfied. Meaning,
• Let ,which leaves
• vertices not connected with S.
• But, we can build a complete bipartite graph
• with these vertices and we assumed that Maker
  has occupied at least one edge in every such
  graph.
• Contradiction.

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Why property (ß) is satisfied?

• Notice that because of the way we satisfied
  property (a), it implies property (ß).
• Property (ß) says that G is connected on n
  vertices.
• Assume (ß) is NOT satisfied. Then, there exists a
  complete bipartite graph
  without Maker edge in it.
• But, (with an
  appropriate choice
• of ) and we assumed that Maker
  has occupied an edge at each .
  
• Contradiction.

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Constructing Hypergraph

• We want to translate our set of all complete
  bipartite graphs to hypergraph .
• Vertices of are the edges of .
• Hyperedge is edges of .
• So, when Maker occupies an edge at ,
  he actually chooses a vertex at the
  Hypergraph, which blocks hyperedge.

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Maker builds via blocking

• As we have seen, Makers goal is to occupy at
  least one edge at each graph. It is
  equal to blocking each hyperedge at .

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The Trick of Fake Moves

• Makers goal is, therefore, to block the
  hypergraph during first t moves.
• Maker can use the Potential technique
  (Erdös-Selfridge), but this technique can
  guarantee blocking only at the end.
• To overcome this difficulty we employ a large
  number of fake moves.

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Fake Moves Play Definition

• Consider the play.
• Breaker starts, he has 1 move. Maker has b moves.
• Breaker defines a fake play and uses the
  Potential technique for this play.
• Maker plays the original game, so with a correct
  choice of moves by Breaker the game will end when
  the Maker owns points. When is
  the size of the board.

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Correct choice of moves.

• Breaker virtually adds some points to each
  Makers move, so that at every turn Maker will
  virtually take points.
• Maker is not aware of the added points by
  Breaker, so he can choose his new points from the
  points Breaker consider to be taken already. Then
  Breaker will add more points to create Maker
  move.
• Notice that applying this trick weakens the
  number of moves allowed to Maker.

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When can Maker achieve this early stage blocking?

• Applying the criterion
  for our play we obtain that if
• then the first player can block every
  when the second player owns points.
  

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Specifying k and b

• Let
• Define integer
  .

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Proof of

• It suffices to check
• where

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Contd

• By using the fact we can
  estimate

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Contd

• There exists such that

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Summary

• We proved that Maker can construct graph with
  properties (a) and (b) in the early enough stage
  of the game.

• Then it applies Longer-Path Argument at most n
  times, and obtains Hamiltonian cycle.