**Spoiler alert**: I will not hassle spoiler tagging the next wall of textual content: proceed at your individual threat.

The reply is that Alice wins for any finite $N$. In actual fact, the extra basic model holds:

Let $S$ be a finite set of factors on $mathbb R^2$. Alice and Bob take turns marking factors. Alice can solely mark a degree on her flip, whereas Bob can mark $N$ factors. They’re free to mark their factors wherever so long as they do not overwrite a earlier one. Then Alice can produce a congruent copy of $S$ inside her marked factors in finitely many strikes.

This can be a advanced theorem in combinatorial sport concept, and an entire proof will be present in József Beck’s *Combinatorial Video games: Tic-Tac-Toe Principle* (the place the above consequence happens as Theorem 2.3). The total argument is pretty concerned: I am going to briefly sketch the principle concepts within the proof.

Let’s work with the unit sq. case for simplicity. We’ll begin with the case $N=1$ after which point out how this generalizes to bigger values of $N$. The set of vertices of a unit sq. will be represented because the set of vectors $S={0,v_1,v_2,v_3=v_1+v_2}$ the place $v_1$ and $v_2$ are the unit vectors alongside the 2 coordinate axes. Step one is to select a lot of angles $theta_1,dots,theta_r$ in order that the vectors $v_{i,j}=v_i$ rotated by $theta_j$ are all *linearly unbiased over $mathbb Q$* (for these unfamiliar with linear algebra, this corresponds to taking plenty of rotated copies of the unit squares, and selecting the angles in a means in order that the one non-trivial “additive” relations between the vectors come from the apparent ones: $v_3=v_1+v_2$, and comparable relations maintain for every of the rotated copies, however no new relations pop up). Observe that it’s attainable to select such angles due to cardinality causes: linear dependence over rational “rule out” countably many selections of angles, however there are uncountably many choices.

The following step is to assemble a humongous “lattice” made out of those rotated copies that will likely be “wealthy” in copies of the unit sq.. Particularly, we select the set $$X(r,M)=left{left.sum_{i=0}^r (a_iv_{1,i}+b_iv_{2,i})proper| a_i,b_i textual content{ are integers within the vary }[-M,M]proper}.$$Right here $M$ is a big integer suitably chosen. Basically, this set incorporates plenty of rational linear combos of the rotated copies of $v_1$ and $v_2$, so the hope is that it’ll comprise plenty of copies of the unit sq..

Now one notes that $X$ has precisely $(2M+1)^{2(r+1)}$ factors (exactly the variety of methods to decide on the rational coordinates: linear independence ensures no two selections result in the identical level). Additional, one can rely that this has at the least $(2M+1-lambda)^{2(r+1)}(r+1)$ copies of the unit sq. ($lambda$ is a few fixed). Now let Alice all the time play on this set and attempt to kind certainly one of these copies of the unit sq.. If Bob performs exterior this set, that is to our benefit, so we are able to assume each gamers play on this set.

Thus now we have decreased our drawback to one thing about finite units: say now we have a big however finite set $X$, and sure subsets of $X$ of dimension $4$ are “successful”. If two gamers alternately occupy parts of $X$, then we wish to present that the primary participant can all the time occupy a successful set so long as there are “sufficiently many” successful units.

Fortuitously, there’s a helpful theorem that proves precisely this: see https://en.wikipedia.org/wiki/Maker-Breaker_game#A_winning_condition_for_Maker . It says that the primary participant wins so long as there are greater than $2d_2|X|$ successful units. (This $d_2$ is the so-called max-pair diploma of the corresponding “hypergraph”. I will not hassle defining any of these phrases right here, simply comment that it is easy to estimate $d_2$ in our case; the truth is, it is at most $6$, and, uh, issues work out.) An software of this theorem instantly solves our drawback; if we make $r$ and $M$ massive, our set does find yourself having sufficiently many successful units.

Now for the overall case the place $N>1$, one makes use of a biased model of the above consequence: we nonetheless use the identical set $X=X(r,M)$, however Bob is now allowed to occupy $N$ parts per transfer. Simplified for our particular case, this biased model tells us that Alice nonetheless wins so long as there are greater than $$N^2(1+N)d_2|X|$$ successful units (Theorem 2.2 in Beck). Thus the identical technique takes care of the overall case too.