Avoider

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Avoider Enforcer games
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Same Boards, Different goal.
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Hyper Graph G (V, E)
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Simple illustration
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Biased Game (p, q, H)

• Avoider on each move, selects p vertices
• Enforcer on each move, selects q vertices

• P
• q

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Are Avoider Enforcer andMaker Breaker
games equivalent ?

• Makers goal is to fully occupy a hyperedge
• Avoiders goal is NOT fully occupying an hyperedge

• Breakers goal is failing Maker
• Enforcers goal is failing Avoider

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A remainder a game of (p,q,H)

• Maker Breaker a sufficient condition for
  Breaker to win
• Was derived (by Beck) by constructing a
  potential function, taking after ErdosSelfridge.

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More about Maker Breaker

• Tight bound
• Monotone
• if Maker wins a game of (p, q, H), then he surely
  wins a game of (p1, q, H) and of (p, q-1, H).
• If Breaker wins a game of (p, q, H), then he
  surely wins a game of (p, q1, H) and of (p-1, q,
  H).

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Avoider Enforcer - There are some similarities

• A sufficient condition for Avoider to win a game
• of (1, 1, H) is
• Which is the same as the last condition for
  Breaker to win for all
• (p, q, H).
• However, in general there is much difference.

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Avoider Enforcer A sufficient condition for
Avoider

• In a game of (p, q, H) -
• if
• then Avoider has a winning strategy.
• If Enforcer has the last move, this condition can
  be relaxed to

Based upon Beck ErdosSelfridge
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Avoider Enforcer A sufficient condition for
Avoider

• Notice the condition is independent of q
  (Enforcers bias) not a tight bound.
• Although, for any constant q, it is not far.
• (One of the important open questions)
• Not monotone in general !
• as demonstrated next in two games victory is
  accomplished through parity!

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1. A game of (1, q, H)
EVEN

• H
• For t large enough
• Enforcer wins
• q is even

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1. A game of (1, q, H)
ODD

• H
• For t large enough
• Enforcer wins
• q is even

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2. A game of (p, 1, H)

• Vertices of H vertices of
• Hyperedge of H set of a vertex of each level
• For t large enough
• Avoider wins
• p is even
• Exactly the same method as the last game

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Avoider Enforcer games

• Are considerably different from Maker Breaker
• Not monotone
• Different sufficient condition (for Avoider)
• The sufficient condition is not tight
• Next, we study more closely 2 quite natural
  Avoider Enforcer (1, q, H) games
• Perfect Matching
• Connectivity
• Searching for a sufficient condition for Enforcer
  to win (1,q,H)
• Maker Breaker threshold of order n/log(n).

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Perfect Matching over the edges of (1, q, K2n
)
U V

• H is a complete graph on 2n vertices.
• Let be a copy of
  in .
• Enforcer enforces Avoider to select edges so that
  at the end of the game Avoider has a complete
  Matching.
• Take t edges of E and let E1E \ those t edges
• t is the minimal integer such that (q1)
  n(n-1)t
• Let E2 denote all the remaining edges except E1

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Perfect Matching over the edges of (1, q, H)

• Each time Avoider picks an edge from E2, Enforcer
  picks q edges from E2. this is always possible as
  E2 is divisible by q1 (choosing t
  appropriately)
• We will focus on the edges of E1
• Each time Avoider picks an edge from E1, Enforcer
  picks q edges from E1 (this is always possible,
  except maybe once).

U V
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Perfect Matching over the edges of (1, q, H)
U V

• In order to show a complete Matching of (U, V,
  E1), we will see that at the end of this game,
  the set of edges picked by Avoider satisfies
  Halls condition
• Let (X,Y,E) be a bipartite graph
• for every X U,
• N(X) X
• X has a perfect Matching.

X
N(X)
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The trick A transposition
U V

• Define a hypergraph
• as following
• 1. the hyper vertices of
• are the edges of E1.
• 2. The hyperedges of
• are all the sets of edges between vertices of X
  U and Y V so that
• XY n1.

X
Y
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The trick A transposition
U V

• In the new graph, Enforcer
• plays as Halls-Avoider,
• and Avoider plays as Halls-Enforcer.
• The new game (q, 1, )
• Now - Halls-Avoider
• applies the sufficient condition for winning the
  new game thus, not fully occupying any
  hyperedge of .
• Later, we examine the size of the appropriate q

X
Y
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The trick A transposition
U V

• Now well see that if Halls-Avoider
  doesnt fully occupy any hyperedge, Halls
  condition is satisfied by Halls-Enforcer.
• At the end of the new game, take any subset X of
  U.
• Look at N(X) which were picked by
  Halls-Enforcer.
• This implies all the other edges (the new hyper
  vertices) were picked by Halls-Avoider.

N(X)
X
V-N(X)
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The trick A transposition
U V

• As Halls-Avoider wins, it is impossible
  that
• Xn-N(X) n1
• (otherwise Halls-Avoider would have lost), and
  so
• Xn-N(X) n
• N(X) X
• and so Hall condition is satisfied by
  Halls-Enforcer.
• Now, retranspose
• Halls-Enforcer is our Avoider

N(X)
X
V-N(X)
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About that q

• Showing that Halls-Avoiders bias in the (q,1,
  ) game satisfies Avoiders sufficient condition
  at the start of the new game

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About that q
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About that q
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Remember The condition
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Perfect Matching over the edges of (1, q, H)
U V

• We saw for
• Enforcer has a winning strategy!
• We dont know whether this is a tight bound!
• We dont know whether it is monotone!
• Next, we play connectivity

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Connectivity on (1, q, H)

• Where H contains q1 pairwise edge disjoint
  spanning trees(or more).
• Avoider and Enforcer pick edges.
• We will see that Enforcer can enforce Avoider to
  pick a spanning tree.

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Connectivity on (1, q, H)

• Let T1, T2, , Tq1 be pairwise disjoint spanning
  trees of G(V,E).
• Let W be the union (of the edges) of the q1
  spanning trees.
• Let L E\W.

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Enforcers strategy

• Maintains in memory q1 acyclic graphs G1, G2,
  , Gq1.
• In the beginning, Gi Ti (i1.q1) which are
  the original spanning trees.

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Enforcers strategy

• Whenever Avoider picks an edge e of some Gj,
  Enforcer picks one edge fi of each Gi (i J),
  hence a total of q edges.
• For all i j If Gi U e contains a (unique)
  cycle then fi is chosen as some unclaimed edge on
  this cycle. If Gi U e doesnt contains a cycle
  then fi is chosen arbitrarily.
• In any case, Enforcer replaces Gi with Gi U e \
  fi.

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Enforcers strategy

• If Avoider picks an edge of L then Enforcer picks
  q edges of L. if there are not enough edges in L,
  Enforcer picks the rest of his acyclic graphs
  one edge of a different Gi.
• If Enforcer starts the game, he would also start
  on the edges of L.
• In any case, Enforcer removes his picks from the
  Gi.

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An example
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A winning strategy

• Every unclaimed edge of W is in exactly one Gi,
  and every edge of W claimed by Avoider is in
  every Gi.
• (an edge is added to Gi it was chosen
  by Avoider).
• Every edge claimed by Enforcer is in no Gi.
• (an edge is removed from Gi it was
  chosen by Enforcer).

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A winning strategy

• After each round, each Gi is either a spanning
  tree, or a panning tree minus one edge.
• It is obviously true at the beginning.
• Suppose it is true after the Kth round.
• If Avoider, on his K1 move, picks an edge e of
  some Gj, then Enforcer, on his K1 move, picks an
  edge fi of the other Gi.
• If Gi U e is acyclic, then Gi U e \ fi is a
  spanning tree minus one edge.
• If Gi U e contains a cycle then removing an
  edge fi from this cycle makes Gi U e \ fi a
  spanning tree over the same vertices as it was on
  the Kth round, thus it is a spanning tree or a
  spanning tree minus one edge.

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A winning strategy

• In the end, each Gi has no unclaimed edges.
• thus, all the edges of Gi are those which were
  picked by Avoider (as Enforcers picks are
  removed of each Gi and Avoiders picks are in
  each Gi).
• So every Gi where is the set of
  edges picked by Avoider.
• Gi in the end is a spanning tree or a spanning
  tree minus one edge.
• We know that V - 1
  because there are
• (q1)(n-1) edges in W, so each player plays at
  least (n-1) rounds.
• Thats why Gi must be a spanning tree, and
  Enforcer has won!

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Connectivity on (1, q, H)

• We saw that if H contains q1 edge disjoint
  spreading trees, Enforcer has a winning strategy.
• If H does not contain q1 edge disjoint spanning
  trees, and there is some limit on the number of
  total edges and the identity of the first player
  then, Avoider has a winning strategy.

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Connectivity on (1, q, H)

• For maker Breaker generally, the existence of
  q1 edge disjoint spanning trees does not ensures
  Makers win.
• There exists a polynomial time algorithm for
  finding q1 edge disjoint spanning trees, so
  Enforcers strategy is realistic.
• Next we apply this outcome on special cases.

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Connectivity on complete graphs

• On complete graphs, the game of (1, q, Kn ) has a
  threshold of
• Kn contains edge disjoint spanning trees.
• By what we have seen, Enforcer would win.
• If q almost always Enforcer wins

Kn
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Connectivity on complete graphs

• If q then each round total of
  are colored.
• For Avoider to lose, the game must last at least
  n-1 rounds.
• So total of edges.
• If n is even then this is more than we have.
• So Avoider wins.

Kn
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Connectivity on complete graphs

• If n is odd and the graph has
  total edges.
• In case Enforcer is the first player, we need
• total edges.
• which is still too much, so Avoider wins.

Kn
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Connectivity on complete graphs

• If Avoider is the first player, we need
• Which is exactly what we have.
• Avoider wins only if he completes some cycle.
• In an analogues game of Maker Breaker, it could
  be shown that Breaker wins.
• So Avoider loses in this case.

Kn
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Connectivity on complete graphs

• We saw the threshold is
• except for special cases where n is odd and
  Avoider starts, and then the threshold is
• Another way of putting it is that this game is
  monotone.

Kn
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Connectivity on complete graphs

• Another special game is the game of (1,1 Kn).
• Enforcer can enforce Avoider to build
  spanning trees.
• The strategy playing
• separate games, each on a graph with 2 edge
  disjoint spanning trees.

Kn
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Summery

• Avoider Enforcer games are not generally
  monotone.
• A sufficient condition for Avoider to win in
  (p,q,H), which was not tight.
• We studied 2 special games
• Perfect Matching in (1,q,Kn,n), where Enforcer
  has a winning strategy for q of order n/log2n. We
  dont know whether there exists a threshold, and
  if so, of what ordeer.

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Summery

• Connectivity we saw a general sufficient
  condition for Enforcer to enforce a spanning tree
  in (1,q,H), where H contains q1 edge disjoint
  spanning trees.
• Connectivity in (1,q,Kn), which is a special case
  of a monotone game with a threshold of
• Connectivity in (1,1,Kn), in which Enforcer can
  enforce Avoider to build edge disjoint
  spanning trees.