Determining a "spaghetti boundary" 0f formal $n$-squares Consider a formal $n$-square, i.e., the set $\{ 0, 1, .., n-1 \}^2$ of positions (column number, row number) of a square $n$-matrix. A cyclic rotation of the $j$-th row $\{ (i,j) : i = 0 .. n-1 \}$ by an integer amount $r$ maps $(i,j)$ to $(i + r \text{ mod } n, j)$ for each $i$. A horizontal move consists of a cyclic rotation of some rows by some amount (which may be different for each involved row).
Let $S$ be a set of $n$ positions, distributed over a number of rows. The more rows containing positions of $S$, the less elements on individual rows. The "spaghetti effect" occurs when it is possible to apply a horizontal move shifting the positions of $S$ into all different columns. Considering a horizontal move as "exercising pressure" on $S$ and moving positions row by row into distinct columns as "crumbling", some intuition gets explained: a set of $n$ positions involving many rows is like a string of spaghetti, having small (rows) sections. With the appropriate pressure, it should crumble completely.
The spaghetti boundary of a formal $n$-square is the least number $s(n)$ such that all sets of $n$ positions involving more than $s(n)$ rows can have their row sections moved apart. In my paper "shuffled equi-$n$-squares", available at
http://arxiv.org/abs/1701.02325
a method was developed to estimate spaghetti boundaries for various sizes $n$. The method relies on results about rotating sets apart in a regular $n$-gon. The results for $8 \leq n \leq 50$ can be found in table 5 of the cited paper.
Only a few values are proven sharp. They are listed here in the format $(n, s(n))$ for $n \geq 4$:
$$(4,2), (5,2), (6,3), (7,3), (8,4), (9,5), (12,7).$$
For $n = 2, 3$ the spaghetti effect occurs all the time whence $s(2) = s(3) = 0$. For a few other $n$ we have a value that is proven sharp up to one unit.
Question. Find sharp spaghetti boundaries for other $n$.
Added 04/29/2017: s(10)=5 is sharp too.
Added 19/05/2017: I finally managed to write a (rather lengthy) inductive proof showing that Hagen von Eitzen's proposal for $s(n)$ is correct for all $n$. I don't see a way to attach a pdf-file here.

Theorem. Let $n\ge 4$. Then
$$ s(n)=n+2-\min_{ab\ge n}(a+b)=n+2-\lceil\sqrt n\,\rceil-\left\lceil\frac{n}{\lceil\sqrt n\,\rceil}\right\rceil.$$
Proof.
First we show the second equality: If we want to minimize $a+b$ under the condition $ab\ge n$ (and wlog. $a\ge b$), we certainly need $a\ge\sqrt n$. And if $a=\lceil\sqrt n\,\rceil+r$ with $r\ge 0$, then we need $b\ge\frac na$, i.e., we need to choose $r$ to minimize $r+\left\lceil\frac{n}{\lceil\sqrt n\,\rceil+r}\right\rceil$, but increasing $r$ by one decreases the second summand by at most one, so that the minimum is attained already for $r=0.
Next we show $s(n)\ge m$ if $m=n-a-b+2$ with $ab\ge n\ge a+b$ (and so $m\ge2$). We may assume wlog. that $a(b-1)<n$.
Let $$\begin{align}A_1&=\{0,1,\ldots,a-1\},\\A_2&=\{0,a,2a,\ldots,(b-1)a\},\\ A_k&=\{0\}\qquad\text{for }3\le k\le m.\end{align}$$ (All viewed as subsets of $\Bbb Z/n\Bbb Z$). Then $\sum_{k=1}^m|A_k|=a+b+(m-2)=n$. Assume there are integers $c_1,\ldots,c_m$ such that $\bigcup(A_i+c_i)=\Bbb Z/n\Bbb Z$, or equivalently $(A_i+c_i)\cap (A_j+c_j)=\emptyset$ for $i\ne j$. We may assume wlog $c_1=a(b-1)$. Then $0\le c_2<a$ leads to a conflict because $(b-1)a\in A_2$, $a\le c_2<2a$ does the same because $(b-2)a\in A_2$, and so on, so that $0\le c_2<ab$ leads to a conflict. By assumption, this covers all residue classes $\bmod n$. Hence $(c_1,\ldots,c_m)$ as desired cannot exist, thus showing $s(n)\ge m$ and ultimately
$$s(n)\ge  n+2-\lceil\sqrt n\,\rceil-\left\lceil\frac{n}{\lceil\sqrt n\,\rceil}\right\rceil.$$
The other direction follows from the claim below, where we work with $d=n-m$ and thus without all one-element sets $A_i$, which never pose a problem. $\square$
Claim. 
Let $n\ge 4$ and let $d$ be a non-negative integer $<\lceil\sqrt n\,\rceil-\left\lceil\frac{n}{\lceil\sqrt n\,\rceil}\right\rceil$.
Let $k\in\Bbb N$ and $a_1\ge a_2\ge \ldots\ge a_k\ge 2$ and $a_1+\ldots+a_k\le k+d$. Then for any choice of sets $A_1,\ldots,A_k\subseteq\Bbb Z/n\Bbb Z$ with $|A_i|=a_i$, there exist integers $c_1,\ldots,c_k$ with $A_i+c_i\cap A_j+c_j=\emptyset$.
Proof [by induction on $k$].
The case $k=1$ is trivial.
Assume $k\ge2$ and the claim is already correct for smaller $k$.
Let $A_1,\ldots,A_k$ as in the claim be given.
From $a_1+\ldots+a_{k-1}\le(k-1)+d$ and the induction hypothesis, we find suitable $c_1,\ldots,c_{k-1}$. Of the $n$ possible choices for $c_k\bmod n$, this prohibits at most
$$\tag1 a_k\cdot(a_1+\ldots+a_{k-1})\le a_k(k+d-a_k)\le\frac1k\cdot\frac{k-1}{k}\cdot (k+d)^2$$
(using $ka_k\le a_1+\ldots+a_k\le k+d$ and that$x\mapsto x(a-x)$ is increasing on $[0,\tfrac a2]$). 
As $k\le (a_1-1)+\ldots+(a_k-1)\le d<2\sqrt n$, this is $<\frac{4n}k$, which already $<n$ and hence sufficient if $k\ge4$.
If $k=3$, our estimate $(1)$ is $< \frac29\cdot 4n<n$ and we are done again. 
Remains the case $k=2$.
But then the conditions of the claim give us
$$a_1+a_2<\lceil\sqrt n\,\rceil-\left\lceil\frac{n}{\lceil\sqrt n\,\rceil}\right\rceil=\min_{ab\ge n}(a+b)$$
so that we conclude $a_1a_2<n$, i.e., again there are less than $n$ choices prohibited. $\square$
