Beck s Surplus Tic Tac Toe Game Exposition by William Gasarch

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1 1 Introduction Beck s Surplus Tic Tac Toe Game Exposition by William Gasarch (gasarch@cs.umd.edu) Consider the following game: Two players Mark (for Maker) and Betty (for Breaker) alternate (Mark going first) placing M s and B s on an n n checkerboard. Mark wins if he can get n M s in either the same row or the same column (getting n on the diagonal does not give him a win). Betty wins if she prevents him from doing this. The above game is stupid. Exercise 1 Show that for n 3 Betty wins the above game. Mark cannot win this game. But what if we lower our expectations? Consider the following game Definition 1.1 Let f : N N. The tic-tac-toe-f(n)-surplus game (henceforth ttt-f(n) game) is as follows. Two players Mark (for Maker) and Betty (for Breaker) alternate (Mark going first) placing M s and B s on an n n checkerboard. Mark wins if he can get n f(n) in either the same row or the same column (getting n f(n) on the diagonal does not give him a win). Betty wins if she prevents him from doing this. Question: For what value of f(n) does Mark have a winning strategy? For what value of f(n) does Betty have a winning strategy? We show that Ω( n) f(n) O( n log n). This is a result of Beck from [1]; however, we give a self contained proof. In addition, our exposition is online and hence available to anyone. Definition 1. RC is the set of rows and column. We assume that RC is ordered so that one can refer to the least element of RC such that.... Mark Can Achieve Surplus Ω( n) Theorem.1 There exists constants c R and n 0 N such that, for all n n 0, there is a strategy by which Mark can win the ttt-c n game. Proof: We define a potential function which will measure how well Mark is doing. Mark s strategy will be to (essentially)increase its value as much as possible. ( 1 Let 0 < ɛ < 1 be a parameter to be named later (it will be Θ n )). A turn is a pair of moves- one by Mark and the response by Betty. We will assume that n is even to avoid half-turns (Mark goes and there is no response from Betty since the game is over). Let t be how many turns have already been made. Let M t (A) be how many M s are in A after t turns. Let B t (A) be how many B s are in A after t turns. We define the potential function: 1

2 Φ t (1 ɛ) Mt(A) (1 ɛ) Bt(A). Here is the strategy for M. Assume that t turns have already occurred (t could be 0). Strategy for Mark: There are two possibilities. 1. There is some A RC such that M t (A) B t (A) c n. Let A be such that M t (A) B t (A) is maximized (if this A is non unique take the least such one). Place an M in A.. There is no such A. Play on an element of RC such that Φ t1 Φ t is maximized. (If there is a tie then use the least such element of RC.) Assume Mark has played this strategy. There are two cases; however, we will show that Case does not occur. Case 1: There is a stage t such that the first possibility of the strategy occurs. Let t 0 be the least such t. Let A be the element of RC that Mark places an M in during turn t. It is easy to see that Mark will play in A for the rest of the game. It is also easy to see that Since We have M n /(A) B n /(A) c n. M n /(A) B n /(A) n M n /(A) n c n. Case : There is no such stage t. Let be defined as Note that, for all t, for all A, max t,a M t (A) B t (A). M t (A) B t (A). We find a lower bound on Φ n / based on. We will then find an (easy) upper bound on Φ n /. We will use this upper and lower bound to get a lower bound on. The potential will decrease over time, but we need to show that it does not decrease too much. Let t 1 n. How big can Φ t1 Φ t be? We will need the following fact: Fact 1: 1. If Mark puts the M on the intersection of row A 1 and column A the potential function goes up by ɛ((1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) (1 ɛ) Mt(A ) (1 ɛ) Bt(A ) )

3 . If Mark puts the M on the intersection of row A 1 and column A, and then Betty puts the B on the intersection of row A 3 and column A 4, A 1 A 3, A A 4 then (1ɛ) Mt(A1) (1 ɛ) Bt(A1) (1ɛ) Mt(A) (1 ɛ) Bt(A) ) (1ɛ) Mt(A3) (1 ɛ) Bt(A3) (1ɛ) Mt(A4) (1 ɛ) Bt(A4) ) 3. If Mark puts the M on the intersection of row A 1 and column A, and then Betty puts the B on the intersection of row A 1 and column A 4, A A 4 then ( ) ɛ(1 ɛ) Mt(A) (1 ɛ) Bt(A) ɛ(1 ɛ) Mt(A4) (1 ɛ) Bt(A 4) Proof of Fact 1: 1) Marks move only affects the potential on row A 1 and column A. The potential goes up by (1ɛ) Mt(A1)1 (1 ɛ) Bt(A1) (1ɛ) Mt(A)1 (1 ɛ) Bt(A) (1ɛ) Mt(A1) (1 ɛ) Bt(A1) (1ɛ) Mt(A) (1 ɛ) Bt(A ) (1 ɛ) Mt(A1) (1 ɛ) Bt(A1) (1 ɛ 1) (1 ɛ) Mt(A) (1 ɛ) Bt(A) (1 ɛ 1) ɛ(1 ɛ) Mt(A1) (1 ɛ) Bt(A1) ) ɛ(1 ɛ) Mt(A) (1 ɛ) Bt(A) ) ɛ((1 ɛ) Mt(A1) (1 ɛ) Bt(A1) (1 ɛ) Mt(A) (1 ɛ) Bt(A) ) ) Since Mark s move maximizes potential it must create a bigger change of potential then the move that puts a marker at the intersection of row A 3 and column A 4. The inequality follows from this observation and Item 1. 3) This is a calculation similar to items 1 and above. End of Proof of Fact 1 Case 1: A 1 A 4 and A A 4. We look at Φ t Φ t1. We need only look at the parts of the sum that involve A 1, A, A 3, A 4 Φ t1 Φ t (1 ɛ) Mt(A 1)1 (1 ɛ) Bt(A 1) (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) (1 ɛ) Mt(A )1 (1 ɛ) Bt(A ) (1 ɛ) Mt(A ) (1 ɛ) Bt(A ) 3

4 (1 ɛ) Mt(A 3) (1 ɛ) Bt(A 3)1 (1 ɛ) Mt(A 3) (1 ɛ) Bt(A 3) (1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4)1 (1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4) (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) (1 ɛ 1) (1 ɛ) Mt(A ) (1 ɛ) Bt(A ) (1 ɛ 1) (1 ɛ) Mt(A 3) (1 ɛ) Bt(A 3) (1 ɛ 1) (1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4) (1 ɛ 1) ɛ(1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) ɛ(1 ɛ) Mt(A ) (1 ɛ) Bt(A ) ɛ(1 ɛ) Mt(A 3) (1 ɛ) Bt(A 3) ɛ(1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4) ( ɛ (1 ɛ) Mt(A1) (1 ɛ) Bt(A1) ɛ(1 ɛ) Mt(A) (1 ɛ) Bt(A ) ) (1 ɛ) Mt(A3) (1 ɛ) Bt(A3) (1 ɛ) Mt(A4) (1 ɛ) Bt(A4) ) This quantity is 0 by Fact 1. Case : A 1 A 3. Only rows A 1, A 3 and column A 4 are affected by the turn. Φ t1 Φ t (1 ɛ) Mt(A 1)1 (1 ɛ) Bt(A 1)1 (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) 4

5 (1 ɛ) Mt(A )1 (1 ɛ) Bt(A ) (1 ɛ) Mt(A ) (1 ɛ) Bt(A ) (1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4)1 (1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4) (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) ((1 ɛ)(1 ɛ) 1) (1 ɛ) Mt(A ) (1 ɛ) Bt(A ) (1 ɛ 1) (1 ɛ) Mt(A 4) (1 ɛ) Bt(A 4) (1 ɛ 1) ɛ (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) ( ) ɛ(1 ɛ) Mt(A) (1 ɛ) Bt(A) ɛ(1 ɛ) Mt(A4) (1 ɛ) Bt(A 4) The expression in big parenthesis must be 0 by Fact 1.. So we have Φ t1 Φ t ɛ (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) Φ t1 Φ t ɛ (1 ɛ) Mt(A 1) (1 ɛ) Bt(A 1) We want to get this in terms of. By the definition of Let Z t Mt(A 1)B t(a 1 ). Then Hence M t (A 1 ) B t (A 1 ) M(A 1 ) Z t B(A 1 ) Z t ( ) ( Φ t1 Φ t ɛ (1 ɛ) Zt (1 ɛ) Zt Φ t ɛ 1ɛ 1 ɛ (1 ɛ ) Zt Φ t ɛ We use an approximation to simplify this expression. Note that 1ɛ 1 ɛ ) 5

6 Hence we have 1 ɛ 1 ɛ 1 ɛ 1 ɛ eɛ. Φ t1 Φ t ɛ e ɛ Hence Φ 0 (1 ɛ) M0(A) (1 ɛ) B0(A) (1 ɛ) 0 (1 ɛ) 0 n. Φ n / n ɛ ɛ n e n eɛ ɛ n. We will now pick ɛ though it will be in terms of another constant. Let ɛ β/n where β will be chosen later. Hence we have We also have an upper bound on Φ n /. By the definition of Φ n / Φ n / n e 8β/n βn. (1 ɛ) M n / (A) (1 ɛ) B n / (A) Hence M n /(A) B n /(A) n n Φ n / (1 ɛ)m n / (A) (1 ɛ) B n / (A) (1 ɛ)n/ (1 ɛ) n/ n(1 ɛ) n/ (1 ɛ) ( ) n/ n (1 ɛ ) n/ 1ɛ 1 ɛ We use two approximations to simplify this expression. From above we have We also use 1 ɛ 1 ɛ 1 ɛ 1 ɛ eɛ. 1 ɛ e ɛ Hence we have 6

7 Φ n / ne ɛ e ɛ n/ We now use the definition of ɛ to obtain Φ n / ne 8β/n e β Combining the upper and lower bounds on Φ n / we obtain the following. n e 8β/n βn ne 8β/n e β e 8β/n β e 8β/n e β e 8β/n β e 8β/n e β e 8β/n (β e β ) q 8β e n β e β ( ) 8β ln β e β n ( ) n 8β ln β e β 1 8β ln ( β e β ) n Pick β so that the constant in front of the n is positive. The reader may want to pick β so as to maximize the constant. Let d be such that d n. By the definition of there exists a t and an A such that Let c d. Hence M t (A) B t (A) d n. M t (A) B t (A) c n. At this stage t the first part of the strategy will happen. Hence this case, case, cannot occur. 7

8 3 Betty Can Make Sure Surplus O( n ln n) (This section will only sketch the proof.) Theorem 3.1 There exist constants c R and n N such that, for all n n 0, there is a strategy by which Betty can win the ttt-c n ln n game. Proof: We let Betty go first in this game. This will only affect the constants in the asymptotics. We define a potential function which will measure how well Mark is doing. Betty s strategy will be to decrease its value as much as possible. ( ) ln n Let 0 < ɛ < 1 be a parameter to be named later (it will be Θ n ). A turn is a pair of moves- one by Betty and the response by Mark. We will assume that n is even to avoid half-turns (Betty goes and there is no response from Mark since the game is over). Let t be how many turns have already been made. Let M t (A) be how many M s are in A after t turns. Let B t (A) be how many B s are in A after t turns. We define the potential function: Φ t (1 ɛ) Mt(A) ((1ɛ)n/) (1 ɛ) Bt(A) ((1 ɛ)n/). Strategy for Betty: Assume that t turns have already occurred (t could be 0). Play on an element of RC such that Φ t Φ t1 is maximized. (If there is a tie then use the least such element of RC.) Claim 1: If Betty plays the strategy above then the potential always either stays the same or decreases. Proof of Claim 1: This is a calculation that we will omit. (This is what I meant when I said we would sketch the proof) End of Proof of Claim 1 Using (1 ɛ) e ɛ and (1 ɛ) e ɛ we obtain the following. Φ 0 (1 ɛ) ((1ɛ)n/) (1 ɛ) ((1 ɛ)n/) n(1 ɛ) ((1ɛ)n/) (1 ɛ) ((1 ɛ)n/) ne ɛ n β ln n We will now set ɛ though it will depend on a constant. Let ɛ n. Note that ne ɛn ne β ln n n n β n 1 β. For the next few equations let M(A) M n /(A) and B(A) B n /(A). Recall that Φ n / Also recall that Φ n / Φ 0. (1 ɛ) M(A) ((1ɛ)n/) (1 ɛ) B(A) ((1 ɛ)n/). 8

9 If there is an A such that M(A) n 1 ɛ n ( ) log n Θ n that summand will be greater than 1. Hence if Φ n / < 1 then there can be no such A. Since Φ n / Φ 0 all we need is Φ 0 < 1. We can accomplish that by taking β < 1. 4 This section has a proof of something I know is false- Help me if you can When I was trying to derive what ɛ should be I got results that did not make sense. It may be an arithmetic mistake or I may have a fundamental misunderstanding of something. If you can tell me whats wrong I will be enlightened. Assume Betty has played this strategy. Let be defined as max M n /(A) B n /(A). We use the approximation (1 x) e x on Φ n /. Φ n / (1 ɛ)m(a) ((1ɛ)n/) (1 ɛ) B(A) ((1 ɛ)n/) eɛ(m(a) ((1ɛ)n/)) e ɛ(b(a) ((1 ɛ)n/)) eɛ(m(a) B(A)) ɛ n Let A be the row where M(A) B(A). summands. Hence Since Φ n / Φ 0 we have Φ n / e ɛ ɛ n e ɛ ɛn Φ n / Φ 0 ne ɛ n The sum above is greater than one of its KEY: the e ɛn cancel out. This leads to results that do not make sense. e ɛ n Gee, I could just take ɛ to be (say) 1 ln(n) This contradicts Theorem.1. ɛ ln(n) ln(n) ɛ and get that O((log n) ). 9

10 One possible Fix: I cheated a bit by letting Betty go first. What if Mark goes first? The proof would start the potential after Mark s first move. Φ 0 does start out a little bigger, but this did not help since I still got all of the e ɛ to cancel out. Also, for large n, the player who goes first shouldn t matter. Another possible fix: The proof that the potential never increases did that impose bounds on ɛ. The proof by Beck didn t seem to. Another possible fix: Do the approximation more carefully. This wouldn t help anything since the approximation I am using is correct for large n. References [1] J. Beck. Surplus of graphs and the local Lovasz lemma. In Building bridges between math and computer science, pages , New York, 008. Springer. Bolyai society mathematical studies number

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