# A game on a graph

Alice and Bob play a game on a complete graph ${G}$ with $2014$ vertices. They take moves in turn with Alice beginning. At each move Alice directs one undirected edge of $G$. At each move Bob chooses a positive integer number $m,\: 1\leq m \leq 1000$ and after that directs $m$ undirected edges of $G$. The game ends when all edges are directed. If there is some directed cycle in $G$ Alice wins. Determine whether Alice has a winning strategy.

This problem is from the Turkey JBMO TST 2014. Could someone help? I have not gone anywhere with it. Thanks a lot.

• What the... this is a funny problem! Jun 21, 2014 at 16:33
• I do hope by funny you mean something that interests you. Thanks anyways. :) Jun 21, 2014 at 16:34
• If Alice wins then she has a triangle. Jun 23, 2014 at 1:17
• @Bananarama: No, she could have a directed hexagon (for example) with no other edges. Jun 24, 2014 at 13:20
• The game ends when all edges are directed. At the end there are two cases: the tournament is transitive (contains no cycle). The tournament is not transitive(contains at least one triangle). Jun 24, 2014 at 13:24

This is not a full solution, but this may help:

The name of the structure we get at the end is tournament.

A transitive tournament is one where $a\to b$ and $b\to c$ implies $a\to c$, thus they give us a total ordering of the vertices.

Tournaments are only acyclic if they are transitive, in particular a tournament $T$ has no $3$-cycle if and only if it is transitive. Notice a transitive graph has no cycles. Now suppose a graph is not transitive, then there exists $a,b,c$ such that $a\to b, b\to c$ and $c\to a$.

So If Bob wants to win he has to build a transitive graph. Notice any tournament with outdegree sequence $(0,1,2\dots ,n-1)$ is transitive. Can Bob build this graph?

• Your solution seems clear, but I can't seem to find a way to draw the graph with outdegree sequence $(0,1,2,\cdots,n-1)$. Could you shed some light on it? I apologise for my ignorance. Thanks a lot. Jun 23, 2014 at 3:44
• number the vertices $0,1,2,3\dots 2014$ and make all arrows point towards the smaller number, then the vertex with number $0$ will have $0$ edges coming out of it, the vertex with number $1$ shall have $1$ vertex on it and so on. I repeat this isn't a full solution, but at least we know if Bob wins all the resulting graphs are isomorphic. Jun 23, 2014 at 3:59

A sketch showing Alice can win. Alice makes a directed chain, extending it by one step (or two if there is a stray edge she can use) at each turn. When the chain has $n$ steps, Bob needs to have directed a total of $\frac 12(n)(n-1)$ edges or Alice can win the next time. We therefore need $\frac 12(n)(n-1) \gt 1000n$, which happens at $n=2001$, so Alice can win by the $2004$th turn

Added: When Alice's chain is 2 edges long, there is only one edge that can complete a cycle. Bob has to orient that one the wrong way to prevent Alice winning on the next move. When her chain is $n=4$ edges long, there are $\frac 12(4-1)(4-2)=6$ edges that could win- three from the head end of the chain, to from the next point down, and one from the next to bottom, so Bob has to orient all $6$ of these by his third turn. As Alice's chain gets longer, the number of edges Bob has to orient grows quadratically. The number of edges Bob can orient only grows linearly, so eventually (unless she runs out of points) she will win.

• could you complete your sketch? Thanks in advance. I don't really get your solution. My apologies for not understanding it, forgive my ignorance. I also don't understand how the inequality comes in play. Thanks for all the help. Jun 24, 2014 at 13:30
• Is it obvious that Alice can always extend the chain? What if all 14 vertices Alice has yet to go to when $n=2000$ each already have 1014+ directed edges? Then Bob can prevent Alice from extending the chain. Also I think you should have $\binom{n+1}{2} - n = \frac{1}{2}n(n-1)$. Jun 24, 2014 at 14:28
• @PerryIverson: Alice doesn't have to go to a point that doesn't have edges directed. The idea is that Bob can't account for enough of them. If those $14$ already have $1014$ directed edges, Bob has wasted a lot of moves. Now he has at most $2,000,000-14*1014$ to block the $\frac 12(2000)(1999)$ threats and he has already lost. I agree with your off by one and have updated. Thanks Jun 24, 2014 at 14:40
• @RossMillikan Right, I just wanted to point out that for a full solution, you would need to prove that Bob cannot possibly block Alice from extending the chain. If at turn $n$, Bob blocks all of Alice's moves, and directs at least 1014 edges out of all $(2014 - (n+1))$ vertices Alice has yet to visit, he will be able to stop Alice from extending the chain. This would require $\frac{1}{2}n(n-1) + 1014(2013-n) \leq 1000n$, which has no solutions. It's actually a narrow margin - if Bob had been allowed 1007 edges each turn, it's possible that he could block Alice on turn 1952. Jun 24, 2014 at 15:01
• @PerryIverson: this solution requires twice as many vertices as edges Bob can play in one turn. I believe that it hurts Bob to play edges, and that with a tighter analysis we can show Alice wins on 1003. Bob can clearly win on 1002-take the outgoing vertex of Alice's first play and mark all the other edges from there outgoing. Then keep up, needing to do one less edge each turn. I'm thinking on it. For a start, I am seeing if Alice can win on 8 vertices with Bob limited to 5 plays. I believe she does. Jun 24, 2014 at 16:09

Alice has the winning strategy. We first prove that on each of her turns, if the length of the longest directed chain isn't already $$2014$$, Alice can always extend it by one. We then use this to give an explicit winning strategy.

To prove our lemma, suppose otherwise. Let $$(v_1, v_2, \ldots, v_n)$$ be the longest directed chain as of yet, and let $$S\neq\varnothing$$ be the set of vertices not on the chain. Every single point of $$S$$ must point to $$v_n$$ – otherwise, the chain could be trivially extended by Alice. Let $$i be the least integer such that not every single point of $$S$$ points to $$v_i$$. Let $$a\in S$$ such that $$a$$ does not point to $$v_i$$. Alice can then connect $$v_i$$ to $$a$$, thus creating the new longest chain $$(v_1, \ldots, v_i, a, v_{i+1}, \ldots, v_n)$$, as we wanted to prove.

Alice thus has the following winning strategy. She builds an ever-increasing chain by the method above. By turn $$n=2002$$, Alice will have created a chain of length at least $$2002$$. Since $$\binom{2002}{2}>1000\cdot2002$$, after Bob's turn, two vertices of the chain will remain unconnected. Alice can then form a cycle with them. $$\blacksquare$$