Random graphs and positional games Tomasz uczak Adam Mickiewicz - - PowerPoint PPT Presentation
(Very) Nice workshop on Random Graphs Nice, 15th May 2014 Random graphs and positional games Tomasz uczak Adam Mickiewicz University, Pozna n, Poland Based on joint work with Magorzata Bednarska-Bzde ga, Danny Hefetz, and Michael
(Very) Nice workshop on Random Graphs Nice, 15th May 2014
Tomasz Łuczak Adam Mickiewicz University, Pozna´ n, Poland Based on joint work with Małgorzata Bednarska-Bzde ¸ga, Danny Hefetz, and Michael Krivelevich
◮ Maker-Breaker games ◮ Waiter-Client games
◮ Maker-Breaker games ◮ Waiter-Client games
Two players: Maker and Breaker
Two players: Maker and Breaker Board: the set of edges of Kn
Two players: Maker and Breaker Board: the set of edges of Kn In each round:
◮ Maker claims (color) 1 edge ◮ Breaker claims (color) q edges
Two players: Maker and Breaker Board: the set of edges of Kn In each round:
◮ Maker claims (color) 1 edge ◮ Breaker claims (color) q edges
Maker wins if his graph has property A,
Let us suppose that the property A is increasing .
Let us suppose that the property A is increasing (e.g. it is a property that a graph contains a triangle). .
Let us suppose that the property A is increasing (e.g. it is a property that a graph contains a triangle). The threshold bias ¯ q(n) = qA(n) is the maximum q so that Maker can win MB(n, q, A).
Let us suppose that the property A is increasing (e.g. it is a property that a graph contains a triangle). The threshold bias ¯ q(n) = qA(n) is the maximum q so that Maker can win MB(n, q, A). i.e. Maker has a winning strategy to build a graph with n
2
CLAIM FOLKLORE In MB(n, q, K3), when Maker tries to build a triangle, the threshold bias is Θ(
More specifically, Maker has a winning strategy if q < √n, and Breaker has a winning strategy if q > 2√n.
Two players: Waiter and Client
Two players: Waiter and Client Board: the set of edges of Kn
Two players: Waiter and Client Board: the set of edges of Kn In each round:
◮ Waiter offers Client q + 1 edges ◮ out of which Client selects 1 edge
Two players: Waiter and Client Board: the set of edges of Kn In each round:
◮ Waiter offers Client q + 1 edges ◮ out of which Client selects 1 edge
Waiter wins if Client’s graph has property A,
Let us suppose that the property A is increasing .
Let us suppose that the property A is increasing (e.g. it is a property that a graph contains a triangle). .
Let us suppose that the property A is increasing (e.g. it is a property that a graph contains a triangle). The threshold bias ¯ q(n) = qA(n) is the minimum q so that Waiter can win WC(n, q, A).
Let us suppose that the property A is increasing (e.g. it is a property that a graph contains a triangle). The threshold bias ¯ q(n) = qA(n) is the minimum q so that Waiter can win WC(n, q, A). i.e. Waiter has a winning strategy to force Client’s graph with n
2
CLAIM The threshold bias for WC(n, q, K3) is Θ(n). More specifically, Waiter can force a triangle if q < n/8, and Client can avoid it if q > n.
DEFINITION OF G(n, p)
G(n, p) is a random graph with vertex set {1, 2, . . . , n} in which each edge is generated with probability p, independently for each of n
2
DEFINITION OF G(n, M)
G(n, M) is a random graph with vertex set {1, 2, . . . , n} chosen uniformly at random from the family of all graphs with M edges.
DEFINITION A function M(n) = MA(n) is the threshold function for a increasing property A is the largest M such that Pr(G(n, p) has A) ≤ 1/2 .
OBSERVATION ERD ˝
OS, SELFRIDGE; BECK
If M(n) is the threshold function for A, then often the threshold bias qA(n) for MB(n, q, A) is roughly such that M(n) = n 2
i.e. quite often the result of MB(n, q, A) when both players play their optimal strategies is the same as the result of this game when both players play random strategies.
Connectivity: The threshold function: M(n) = n(log n + O(1))/2 = n
2
log n+O(1)
n
Erd˝
The threshold bias: q(n) = (1 + o(1))n/ log n Gebauer and Szab´
LEMMA ERD ˝
OS, SELFRIDGE; BECK
Let F be a family of graphs and EXF(n, q) be the expected number of graphs from F in G(n, n
2
Then there is a strategy for Breaker in the q-biased game which prevents Maker to create more then 2EXF copies of graphs from F.
LEMMA BECK
There is a strategy for Client in q-biased game so that, at the end of the game, his graph contains at most 2EXF copies of graphs from F.
LEMMA ERD ˝
OS, SELFRIDGE; BECK
Let F be a family of graphs and EXF(n, q) be the expected number of graphs from F in G(n, n
2
Then there is a strategy for Breaker in the q-biased game which prevents Maker to create more then 2EXF copies of graphs from F.
LEMMA BECK
There is a strategy for Client in q-biased game so that, at the end of the game, his graph contains at most 2EXF copies of graphs from F.
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2,
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2, so Client can avoid triangles in WC(n, n, K3) game,
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2, so Client can avoid triangles in WC(n, n, K3) game, so qK3(n) < n,
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2, so Client can avoid triangles in WC(n, n, K3) game, so qK3(n) < n, which, as we know, gives the right order of qK3(n) = Θ(n)!
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2,
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2, so Breaker can prohibit triangles in WC(n, n, K3) game,
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2, so Breaker can prohibit triangles in WC(n, n, K3) game, so qK3(n) < n,
The expected number of triangles in G(n, n
2
1/6 + o(1) < 1/2, so Breaker can prohibit triangles in WC(n, n, K3) game, so qK3(n) < n, but, as we know, the right order of qK3(n) is Θ(√n)!
If we apply Lemma for the family T of all spanning trees it gives qT (n) < n/2, (since nn−2(q + 1)−(n−1) < 1/2).
If we apply Lemma for the family T of all spanning trees it gives qT (n) < n/2, (since nn−2(q + 1)−(n−1) < 1/2). The real threshold is qT (n) = (1 + o(1))n/ log n and it agrees with the random graph heuristic!
If we apply Lemma for the family T of all spanning trees it gives qT (n) < n/2, (since nn−2(q + 1)−(n−1) < 1/2). The real threshold is qT (n) = (1 + o(1))n/ log n and it agrees with the random graph heuristic! Here the problem follows from the fact that the connectivity threshold for random graphs cannot be estimated by bounding the first moment of XT (n, q).
Go greedy!
Go greedy!
Idea: If only a very few graphs on n vertices and M = n
2
edges do not have property A, then Maker can win MB(n, q, A) playing greedily, i.e. he should always choose the edge which maximizes the chance that his final graph will have property A.
Go greedy!
Idea: If only a very few graphs on n vertices and M = n
2
edges do not have property A, then Maker can win MB(n, q, A) playing greedily, i.e. he should always choose the edge which maximizes the chance that his final graph will have property A. In a similar way, Waiter can win WC(n, q, A) proposing q + 1 edges which favor property A.
Go greedy!
Idea: If only a very few graphs on n vertices and M = n
2
edges do not have property A, then Maker can win MB(n, q, A) playing greedily, i.e. he should always choose the edge which maximizes the chance that his final graph will have property A. In a similar way, Waiter can win WC(n, q, A) proposing q + 1 edges which favor property A. The main problem here: in order this strategy to work the probability that G(n, M) has no property A must be very, very small, i.e. smaller than exp(−c0M) for some absolute constant c0.
LEMMA BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK For every c > 0 there exists (an explicit) 0 < b < 1 such that Maker [Waiter] has a strategy that in the first b n
2
has property A, provided only that Pr[G(n, M) has not A] ≤ exp(−cM), for M = b n
2
Let H be a given (small) subgraph and let m2(H) = max
F⊆H
e(F) − 1 v(F) − 2 .
Let H be a given (small) subgraph and let m2(H) = max
F⊆H
e(F) − 1 v(F) − 2 . If m2(H) = e(H)−1
v(H)−2 then H is 2-balanced.
Let H be a given (small) subgraph and let m2(H) = max
F⊆H
e(F) − 1 v(F) − 2 . If m2(H) = e(H)−1
v(H)−2 then H is 2-balanced.
Example: K3 is 2-balanced with m2(K3) = 2.
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’90
For every H and a constant a > 0 there exists a constant c > 0 such that Pr(G(n, M) ⊇ H) ≤ exp(−cM), provided M ≥ an2−m2(H).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’90
For every 2-balanced H and a constant a > 0 there exist constants b > 0 and c > 0 such that the probability that G(n, M) contains fewer than bM edge disjoint copies of H is less than exp(−cM), provided M ≥ an2−m2(H).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’90
For every H and a constant a > 0 there exists a constant c > 0 such that Pr(G(n, M) ⊇ H) ≤ exp(−cM), provided M ≥ an2−m2(H).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’90
For every 2-balanced H and a constant a > 0 there exist constants b > 0 and c > 0 such that the probability that G(n, M) contains fewer than bM edge disjoint copies of H is less than exp(−cM), provided M ≥ an2−m2(H).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI
For every 2-balanced H and a constant a > 0 there exist constants b > 0 and c > 0 such that the probability that G(n, M) contains fewer than bM edge disjoint copies of H is less than exp(−cM), provided M ≥ an2−m2(H).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI
For every 2-balanced H and a constant a > 0 there exist constants b > 0 and c > 0 such that the probability that G(n, M) contains fewer than bM edge disjoint copies of H is less than exp(−cM), provided M ≥ an2−m2(H). Remark 1: If we want the expected number of H to be of the
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI
For every 2-balanced H and a constant a > 0 there exist constants b > 0 and c > 0 such that the probability that G(n, M) contains fewer than bM edge disjoint copies of H is less than exp(−cM), provided M ≥ an2−m2(H). Remark 1: If we want the expected number of H to be of the
Remark 2: Typically, c cannot be made arbitrarily large, even for large a. For instance, the probability that G(n, M) ⊇ K3 is larger than (1/2 + o(1))M, whenever M = o(n2).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’89
If M ≥ 10n3/2 then with probability at least exp(−M/5) G(n, M) contains at least M/100 edge disjoint triangles.
LEMMA BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK’14+
Maker has a strategy that in the first 0.01 n
2
graph has property A, provided Pr[G(n, M) has not A] ≤ exp(−M/5), for M = 0.01 n
2
COROLLARY
If M = n
2
Maker can win MB(n, q, K3).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’89
If M ≥ 10n3/2 then with probability at least exp(−M/5) G(n, M) contains at least M/100 edge disjoint triangles.
LEMMA BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK’14+
Maker has a strategy that in the first 0.01 n
2
graph has property A, provided Pr[G(n, M) has not A] ≤ exp(−M/5), for M = 0.01 n
2
COROLLARY
If M = n
2
Maker can win MB(n, q, K3).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’89
If M ≥ 10n3/2 then with probability at least exp(−M/5) G(n, M) contains at least M/100 edge disjoint triangles.
LEMMA BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK’14+
Maker has a strategy that in the first 0.01 n
2
graph has property A, provided Pr[G(n, M) has not A] ≤ exp(−M/5), for M = 0.01 n
2
COROLLARY
If M = n
2
Maker can win MB(n, q, K3).
THEOREM JANSON, ŁUCZAK, RUCI ´
NSKI’89
If M ≥ 10n3/2 then with probability at least exp(−M/5) G(n, M) contains at least M/100 edge disjoint triangles.
LEMMA BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK’14+
Maker has a strategy that in the first 0.01 n
2
graph has property A, provided Pr[G(n, M) has not A] ≤ exp(−M/5), for M = 0.01 n
2
In fact, Maker can win MB(n, q, K3) quite early, when fewer than 1% of all pairs have been used, and it can create at least 0.1n2/3 triangles.
For every Maker-Breaker game MB(n, q, H), when Maker tries to build a copy of graph H, this method gives the right (lower) bound for the threshold bias! THEOREM BEDNARSKA, ŁUCZAK’00 For each graph H there are constants C > c > 0 such that:
◮ if q < cnm2(H), then Maker wins MB(n, q, H); ◮ if q > Cnm2(H), then Breaker wins MB(n, q, H).
For every Maker-Breaker game MB(n, q, H), when Maker tries to build a copy of graph H, this method gives the right (lower) bound for the threshold bias! THEOREM BEDNARSKA, ŁUCZAK’00 For each graph H there are constants C > c > 0 such that:
◮ if q < cnm2(H), then Maker wins MB(n, q, H); ◮ if q > Cnm2(H), then Breaker wins MB(n, q, H).
Remark: Note that for q < cnm2(H) Maker can quickly (!) create Ω n
2
For Waiter-Client game MB(n, q, H), this method gives the same lower bound for the threshold bias, but typically (in fact almost always) it is far off the right value. Example: In the case of H = K3, we have n ≫ √n.
CLAIM
For q < O(nm2(H)) Waiter can quickly create Ω n
2
For Waiter-Client game MB(n, q, H), this method gives the same lower bound for the threshold bias, but typically (in fact almost always) it is far off the right value. Example: In the case of H = K3, we have n ≫ √n. However this method gives the following statement.
CLAIM
For q < O(nm2(H)) Waiter can quickly create Ω n
2
We know that greedy method gives lower bounds for the threshold bias for both MB(n, q, H) and WC(n, q, H) which is of the correct order for the former one but does poorly for the latter one.
We know that greedy method gives lower bounds for the threshold bias for both MB(n, q, H) and WC(n, q, H) which is of the correct order for the former one but does poorly for the latter one. The derandomization method provides upper bounds for the threshold bias for both MB(n, q, H) and WC(n, q, H) which typically, is very much off the former one but does quite well for the latter one (at least for H = K3).
CONJECTURE BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
For every q and H, in Waiter-Client q-biased game Waiter can always force at least Ω
F⊆H EXH
n 2
THEOREM BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
The above conjecture is true for many ‘nice’ graphs H (such as, for instance, complete graphs Kk).
CONJECTURE BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
For every q and H, in Waiter-Client q-biased game Waiter can always force at least Ω
F⊆H EXH
n 2
THEOREM BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
The above conjecture is true for many ‘nice’ graphs H (such as, for instance, complete graphs Kk).
CONJECTURE BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
For every q and H, in Waiter-Client q-biased game Waiter can always force at least Ω(f(n)) copies of H, where f(n) = min
F⊆H EXH
n 2
CLAIM BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
By the greedy method, the above conjecture is true, when q is small enough. In fact, when the expected number f(n) is of the
n
2
least Ω(n) copies of H in Client’s graph.
Clearly, EXH(n, n 2
where v(H) and e(H) denote the number of vertices and edges in H respectively, while ‘≍’ means up to constant factor which may depend on H.
EXH ≍ nv(H)(q + 1)−e(H).
EXH ≍ nv(H)(q + 1)−e(H). Let H(e1, . . . , ek) denote the graph obtained from H by removing k of its edges: e1, . . . , ek. Then, EXH(e1,...,ek) ≍ EXH(q + 1)k.
THEOREM BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
If H is ‘nice’ Waiter can force in Client’s graph roughly as many copies of H as EXH(n, n
2
THEOREM BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
If H is ‘nice’ Waiter can force in Client’s graph roughly as many copies of H as EXH(n, n
2
Proof We know (by the greedy method) that the above is true for 2-balanced H such that EXH is at least as large as n
2
THEOREM BEDNARSKA-BZDE ¸ GA, HEFETZ, ŁUCZAK
If H is ‘nice’ Waiter can force in Client’s graph roughly as many copies of H as EXH(n, n
2
Proof We know (by the greedy method) that the above is true for 2-balanced H such that EXH is at least as large as n
2
Delete from H some of its edges such that H(e1, . . . , ek) is 2-balanced and EXH(e1,...,ek) is at least as large as n
2
(one can do if H is ‘nice’ enough).
CLAIM If H is ‘nice’ enough then, for some α > 0, Waiter can quickly force in Client’s graph a family H(e1, . . . , ek) of α′EXH(e1,...,ek) ≥ αEXH(q + 1)k copies of H(e1, . . . , ek) such that: (A) Each of the missing pairs e1, . . . , ek belongs to just
(B) None of the missing pairs e1, . . . , ek has been offered by Waiter yet.
|H(e1, . . . , ek−1, ek)| ≥ αEXH(q + 1)k
|H(e1, . . . , ek−1, ek)| ≥ αEXH(q + 1)k |H(e1, . . . , ek−1)| ≥ αEXH(q + 1)k−1
|H(e1, . . . , ek−1, ek)| ≥ αEXH(q + 1)k |H(e1, . . . , ek−1)| ≥ αEXH(q + 1)k−1 . . . |H| ≥ αEXH
Even ‘good-looking’ graphs may not be ‘nice’.
Even ‘good-looking’ graphs may not be ‘nice’. Example: H = K5 − e.
Even ‘good-looking’ graphs may not be ‘nice’. Example: H = K5 − e. EXK5−e = n5(q + 1)−9 ≍ 1. Thus, q ≍ n5/9 and M = n
2
Consequently, we have to remove from K5 − e three edges so that the expectation of the resulting graph jumps to n15/9 ≫ n13/9.
Thus, H(e1, e2, e3) is C5 ∪ e. But C5 ∪ e is not 2-balanced!
Thus, H(e1, e2, e3) is C5 ∪ e. But C5 ∪ e is not 2-balanced! Indeed, m2(C5 ∪ e) = e(C5 ∪ e) − 1 v(C5 ∪ e) − 2 = 4 3 but m2(C3) = 2 > 4 3 .
Thus, H(e1, e2, e3) is C5 ∪ e. But C5 ∪ e is not 2-balanced! Indeed, m2(C5 ∪ e) = e(C5 ∪ e) − 1 v(C5 ∪ e) − 2 = 4 3 but m2(C3) = 2 > 4 3 . Note that EXC5∪e ≍ n5q−6 ≍ n15/9 but EXC3 ≍ n3q−3 = n12/9 .
Let C(n, q) be the longest cycle Waiter can force in Client’s graph. THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (A) If q > 1.3n then C(n, q) = 0. (B) If q > (1 + ǫ)n, then C(n, q) = Oǫ(1). (C) If q < (1 − ǫ)n, then C(n, q) > ǫn/6. (D) If q < 10−9n, then Waiter can make Client’s graph to be pancyclic.
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
(A) If q > 1.3n then C(n, q) = 0. (B) If q > (1 + ǫ)n for some ǫ > 0, then C(n, q) = Oǫ(1).
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
(A) If q > 1.3n then C(n, q) = 0. (B) If q > (1 + ǫ)n for some ǫ > 0, then C(n, q) = Oǫ(1). Proof The expected number of cycles in G(n, n
2
bounded from above by
n
n k (k − 1)! 2 1 q + 1 ≤
∞
1 2k n q k .
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
(A) If q > 1.3n then C(n, q) = 0. (B) If q > (1 + ǫ)n for some ǫ > 0, then C(n, q) = Oǫ(1). Proof The expected number of cycles in G(n, n
2
bounded from above by
n
n k (k − 1)! 2 1 q + 1 ≤
∞
1 2k n q k . For q > 1.3n it is smaller than 1, for each q > (1 + ǫ)n, the series is convergent, so the assertion follows from the derandomization result.
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (A) If q > 1.3n then C(n, q) = 0. (B) If q > (1 + ǫ)n, then C(n, q) = Oǫ(1). Possibly, the following holds.
CONJECTURE BB, HEFETZ, Ł, KRIVELEVICH
If q > (1 + ǫ)n for some constant ǫ > 0, then C(n, q) = 0 for n large enough.
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (A) If q > 1.3n then C(n, q) = 0. (B) If q > (1 + ǫ)n, then C(n, q) = Oǫ(1). Possibly, the following holds.
CONJECTURE BB, HEFETZ, Ł, KRIVELEVICH
If q > (1 + ǫ)n for some constant ǫ > 0, then C(n, q) = 0 for n large enough.
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (D) If q < 10−9n, then Waiter can make Client’s graph to be pancyclic.
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (D) If q < 10−9n, then Waiter can make Client’s graph to be pancyclic. The crucial step in the proof is to show that for q = Ω(n) Waiter can quickly force Client to construct a good expander.
Let L(n, q) be the size of the largest component Waiter can force in Client’s graph. THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (A) If q > en then L(n, q) = O(log n). (B) If en ≥ q > (1 + ǫ)n, then L(n, q) = Oǫ(√n). (C) If q < (1 − ǫ)n, then L(n, q) > (2ǫ + o(1))n. (D) If q < n/2 − 1, then L(n, q) = n.
THEOREM BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (A) If q > en then L(n, q) = O(log n). (B) If en ≥ q > (1 + ǫ)n, then L(n, q) = Oǫ(√n). (C) If q < (1 − ǫ)n, then L(n, q) > (2ǫ + o(1))n. (D) If q < n/2 − 1, then L(n, q) = n.
CONJECTURE BB, HEFETZ, Ł, KRIVELEVICH
Let ǫ > 0. (AB) If q > (1 + ǫ)n then L(n, q) = O(log n). (C) If q < (1 − ǫ)n, then L(n, q) = (2ǫ + o(1))n.
QUESTION
QUESTION
QUESTION
It is certainly smaller than n/k (since Client’s graphs must have at least nk/2 edges) and larger than n/(3k) (then it is easy to construct tripartite graph where each vertex from Vi has at least k neighbours in Vi+1).