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Avoider Enforcer games

Same Boards, Different goal.

Hyper Graph G (V, E)

Simple illustration

Biased Game (p, q, H)

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

- P
- q

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

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.

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).

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.

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

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!

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

EVEN

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

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

ODD

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

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

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).

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

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

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)

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

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

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)

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)

About that q

- Showing that Halls-Avoiders bias in the (q,1,

) game satisfies Avoiders sufficient condition

at the start of the new game

About that q

About that q

Remember The condition

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

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.

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.

Enforcers strategy

- Maintains in memory q1 acyclic graphs G1, G2,

, Gq1. - In the beginning, Gi Ti (i1.q1) which are

the original spanning trees.

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.

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.

An example

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).

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.

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!

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.

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.

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

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

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

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

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

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

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.

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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