# Two [n] to [n] function families

$\bf Note.$ This question had a bounty, so at the end I accepted the best (and only) answer but in fact it is still open. It is (hopefully) equivalent to this question, if you have any ideas, please post them there.

$\bf Question.$ Fix n. We are interested in the biggest t for which there exist two families of functions, $P_i,Q_i$, of size t from [n] to [n] such that for any $i,j$ whenever we consider the infinite sequence $P_i(Q_j(P_i(Q_j\ldots P_i(3))\ldots)$ (where the number of iterations tends to infinity), it contains no 2's and infinitely many 1's if $i=j$ and it contains no 1's and infinitely many 2's if $i\ne j$.

$\bf A lower bound.$ I know a construction that shows that $t\ge 2^{\frac n2-O(1)}.$ For every subset $S$ of [n] that contains exactly one of $2k$ and $2k+1$ for $2\le k\le \frac n2-2$ we construct a pair of functions, $P_S,Q_S$ as follows. For any number m denote by $m^+$ the smallest element of $S$ that is bigger than m or if all elements of $S$ are at most m then define it to be 1. $P_S(1)=1, P_S(2)=2$ and for bigger $m$'s $P_S(m)=m^+$, while $Q_S(1)=1, Q_S(2)=2$ and for bigger $m$'s $Q_S(m)=m$ if $m\in S$ and $Q_S(m)=2$ if $m\notin S$. This way we go through all the elements of S and end in 1 if the functions have the same index, but we are pushed to 2 if they differ.

$\bf Upper bound.$ It is of course true that $t\le n^n$. So can you do better than $2^n$?

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To get bold, you can wrap the text with pairs of asterisks; see daringfireball.net/projects/markdown/syntax#em for a general reference of the notation available. – Mariano Suárez-Alvarez Feb 15 '10 at 4:09
I did not manage but I guess it's fine like this... – domotorp Feb 21 '10 at 7:14
Surely, you don't mean "$Q_S(m)=2$ if $m\in S$" - that does not define a permutation. – Igor Pak Feb 22 '10 at 4:14
You are completely right, I am talking about functions and not permutations everywhere, so I changed permutations to functions, I donno how I could be so stupid. Btw, now I wonder if the original question makes any sense... – domotorp Feb 22 '10 at 8:25
Please forgive me, but I am only a hobbyist, s I have to ask: what is [n] meaning. Probably it is obvious but I do not know and it is difficult to goggle such symbol... There is at least several meaning which may fit,see: en.wikipedia.org/wiki/Table_of_mathematical_symbols – kakaz Feb 22 '10 at 10:30

Okay, so I tried to see how this could possibly work. After some thinking I decided that one may as well take $P_i=Q_i$, so that the orbit of 3 (under the action of $P_i$) is a cycle containing 1. If you take the length of this cycle to be roughly $n/2$, send $2\to 3$ and everything else to 2, that's not a bad idea except that it doesn't work for all possible $n/2$-subsets; otherwise we would have roughly $\binom{n}{n/2}$ possible $i$, as you wanted to begin with. If you now look at the orbit of 3 under $(P_iP_j)$ in this setting you pretty quickly conclude that there is an inherent "even-town theorem" (see Babai-Frankl's book) and thus $2^{n/2}$ is really the best possible. Of course, in the full generality weird things might be possible - I have no intuition for this, but this doesn't look good and unless the difference is really really important for some applications I wouldn't recommend working on this problem.
As I said, I considered possible pairs of subsets A,B for which $P_A$ and $P_B$ defined as above satisfy $P_AP_B$ has a 2 orbit cycle. When you do a calculation, this shows $|A\cap B|$ must be even (among other things). All of this is under certain "natural" assumptions, so I don't have any kind of theorem. That would take quite a bit of work, I imagine. Sorry I can't be more clearer and resolve the whole problem... – Igor Pak Feb 25 '10 at 18:01
Umm, kind of. First, you actually have a freedom of choice which cycle $3\to A\to 1\to 3$ to take - there are many ordering of A. Say you fix one $P_A$. Now you need to take a another, say $P_B$, and try what conditions does it have to satisfy so that $[P_AP_B]^k(3)$ has no 1. I remember concluding that this must satisfy some kind of parity conditions and really similar in form to what you have in the question. But as I said, I didn't formally prove anything - I simply convinced myself that there is no better construction, but you might try going along this route and see where it takes you. – Igor Pak Feb 26 '10 at 23:47