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Avoider

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Tight bound. Monotone: ... Notice the condition is independent of q (Enforcer's bias) not a tight bound. ... Avoider to win in (p,q,H), which was not tight. ... – PowerPoint PPT presentation

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Title: Avoider


1
Avoider Enforcer games
2
Same Boards, Different goal.
3
Hyper Graph G (V, E)
4
Simple illustration
5
Biased Game (p, q, H)
  • Avoider on each move, selects p vertices
  • Enforcer on each move, selects q vertices
  • P
  • q

6
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

7
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.

8
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).

9
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.

10
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
11
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!

12
1. A game of (1, q, H)
EVEN
  • H
  • For t large enough
  • Enforcer wins
  • q is even

13
1. A game of (1, q, H)
ODD
  • H
  • For t large enough
  • Enforcer wins
  • q is even

14
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

15
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).

16
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

17
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
18
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)
19
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
20
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
21
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)
22
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)
23
About that q
  • Showing that Halls-Avoiders bias in the (q,1,
    ) game satisfies Avoiders sufficient condition
    at the start of the new game

24
About that q
25
About that q
26
Remember The condition
27
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

28
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.

29
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.

30
Enforcers strategy
  • Maintains in memory q1 acyclic graphs G1, G2,
    , Gq1.
  • In the beginning, Gi Ti (i1.q1) which are
    the original spanning trees.

31
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.

32
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.

33
An example
34
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).

35
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.

36
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!

37
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.

38
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.

39
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
40
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
41
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
42
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
43
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
44
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
45
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.

46
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.
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