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The card game Set lead to the following question. Lets call a subset $A$ of $(\mathbb{Z}/3)^n$ dependent, if there is $\{x,y,z\}\subset A$ with $x+y+z=0$. (So unlike the case of linear dependence we are not allowing any coefficients here).

Let $f(n)$ denote the maximal size of a independent subset of $(\mathbb{Z}/3)^n$. Is there an explicit expression / a recursion for $f(n)$? Can anything be said about its asymptotic behaviour (like $f\in O(c^n)$ for some minimal $c$)?

As $\{0;1\}^n$ is independent, we know that $f(n)\ge 2^n$. And so $c\ge 2$. The card game set deals with the case $n=4$ and $f(4)=20$ if i remember correctly.

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You may be interested in some of the discussion around Polymath1, which I believe tackles similar questions. – Qiaochu Yuan Oct 6 '10 at 12:01
$f(4)=20$ is correct--that is, 20 is the minimum number of SET cards needed to guarantee a "set" in the context of the game. – Isaac Oct 6 '10 at 16:23
I am a bit confused. Both answers I got are contradicting each other, as the given recursion does not satisfy the claimed lower bound c>2. Furthermore In this presentation is sounded like its unknown,whether this recursion holds (it is in the chapter further work and it is followed by a ?-sign). Anyhow I am a bit confused right now. – HenrikRueping Oct 7 '10 at 0:16
You claimed c≥2, which the formulas I gave do satisfy (as it is O(2^n), so c=2). On a second look, you're right that the presentation to which I linked does not quite make it clear whether or not these formulas (which are not recursive) are correct, but that was how I read it and the formulas do seem to work for n=2, 3, 4, 5, and 6. – Isaac Oct 7 '10 at 0:40
"We have a lower bound of order c^n for some c > 2" I think this statement in harrisons answer contradicts the suggested formula. – HenrikRueping Oct 7 '10 at 9:37
up vote 3 down vote accepted

Mathematicians call these "independent sets" capsets, and they're fairly widely studied. But surprisingly, we don't know much about their asymptotic behavior! We have a lower bound of order c^n for some c > 2; and we have an upper bound of order about 3^n/n (although this is not in the least trivial to prove!) Most mathematicians seem to suspect that the upper bound can't be improved to c^n for c strictly less than 3, but no one's sure how to prove this.

For more information (perhaps a little out-of-date, but I don't think much has changed) see Terry Tao's discussion.

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Based on this presentation (PPT; Google-based HTML conversion), I think a formula for your $f(n)$ is: $$f(n)=\begin{cases} \frac{2^{n+2}-4}{3} & \text{ if }n\text{ is even} \\ \frac{2^{n+2}-5}{3} & \text{ if }n>1\text{ is odd} \end{cases}$$

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