TITLE: Degree of vertex in left part of a bipartite graph with distance less than 3 in right part.
QUESTION [5 upvotes]: A friend and I have worked on the following problems and would appreciate your help generalizing the answer for $n\ge5$:
Suppose participants $P$ at a conference speak one of more of the langages in the set $L$ and that each pair can communicate in at least one language. We already proved that if $|L|=3$ and $|P| \ge10$ then one language has to be spoken by at least $2/3$ of participants. That ratio is $3/5$ if $|L|=4$ and $|P|$ is large enough to dodge low participants irregularity.
We generalize easily to the following problem and would appreciate your take :
Let $P$ and $L$ be the two sides of a bipartite graph such that for any pair $(x,y)$ of distinct vertices in $P$ then $dist(x,y)=2$. What is the lowest value for
$$p(n)=\max_{\ell\in L} \frac{\deg(\ell)}{n}$$
where $|L|=n$.

REPLY [3 votes]: An other incomplete answer. I assume that the number of participants is large, so we can continuously divide them into groups. With this method I was able to calculate $p(n)$ for $n$ up to 6.
If we have $n$ languages, then we can divide the people into $2^n$ groups. Each group does or doesn't speak certain languages.
We can name the languages $0$ to $n-1$ and name the groups $0$ to $2^n-1$. Group $i$ speaks language $j$ if the bit $2^j$ is set in the binary notation of $i$. For example, if $n=4$, then group $3 = 0011_2$ speaks languages 0 and 1, but not 2 and 3.
Let the variables $a_0$ up to $a_{2^n-1}$ indicate the size of the groups. The following restrictions apply:

$a_0 = 0$
$\displaystyle \forall_i \ \ a_i \ge 0$
$\displaystyle \sum_{i=0}^{2^n-1} a_i = 1$
For groups $i$ and $j$: $\quad\displaystyle i\ \&\ j = 0\ \ \implies\ \ a_i = 0\ \vee\ a_j = 0$
Where $\&$ is the bitwise and.

We can simplify the problem a bit. For $n>2$ we can assume that:

No one speaks only one language, so $a_i=0$ when $i$ is a power of $2$.
If anyone speaks only two languages, then without loss of generality we can assume that this will be languages $0$ and $1$. So for all groups $i$ that can speak only two languages, and neither of them is $0$ or $1$, we have $a_i = 0$.

We have to calculate:
$$
\min_{\displaystyle a_i}\ \max\ \left\{\sum \{a_i\ |\ i\ \&\ 2^j \neq 0\} \ \ \middle|\ \ j=0,1,\dots n-1\right\}
$$
We can enter the formulas in Mathematica and use the Minimize function to find the minimum for each $n$ and get the corresponding values $a_i$. This can be used to look for patterns that may also be applicable to higher $n$. The following code works for $n \ge 2$:
solve[n_] := Module[{vari, or, max},
    vari = Select[Range[0, 2^n - 1], Total[IntegerDigits[#, 2]] > 1 && (Total[IntegerDigits[#, 2]] > 2 || BitAnd[#, 3] != 0)&];
    or = Or @@ (a[#] == 0 & /@ #) & /@ Select[Tuples[vari, 2], Less @@ # && BitAnd @@ # == 0 &];
    max = Table[(Select[vari, BitGet[#, i] == 1 &] // a // Thread // Total) <= x, {i, 0, n - 1}];
    Minimize[{x, a[#] >= 0 & /@ vari, Total[a /@ vari] == 1, or, max}, Join[a /@ vari, {x}]]
];

Below the results. Note that these solutions are probably not unique.
$$
\begin{array}{|c|c|c|}
\hline
 n & p(n) & a_i \\
\hline
 1 & 1 & 
\begin{array}{c|c}
 0 & a_i \\
\hline
 1 & 1 \\
\end{array}
 \\
\hline
 2 & 1 & 
\begin{array}{cc|c}
 0 & 1 & a_i \\
\hline
 1 & 1 & 1 \\
\end{array}
 \\
\hline
 3 & 2/3 & 
\begin{array}{ccc|c}
 0 & 1 & 2 & a_i \\
\hline
 0 & 1 & 1 & 1/3 \\
 1 & 0 & 1 & 1/3 \\
 1 & 1 & 0 & 1/3 \\
\end{array}
 \\
\hline
 4 & 3/5 & 
\begin{array}{cccc|c}
 0 & 1 & 2 & 3 & a_i \\
\hline
 0 & 0 & 1 & 1 & 1/5 \\
 0 & 1 & 1 & 0 & 1/5 \\
 1 & 0 & 1 & 0 & 1/5 \\
 1 & 1 & 0 & 1 & 2/5 \\
\end{array}
 \\
\hline
 5 & 5/9 & 
\begin{array}{ccccc|c}
 0 & 1 & 2 & 3 & 4 & a_i \\
\hline
 0 & 0 & 1 & 1 & 1 & 2/9 \\
 0 & 1 & 0 & 1 & 1 & 2/9 \\
 1 & 0 & 0 & 0 & 1 & 1/9 \\
 1 & 0 & 0 & 1 & 0 & 1/9 \\
 1 & 1 & 1 & 0 & 0 & 1/3 \\
\end{array}
 \\
\hline
 6 & 1/2 & 
\begin{array}{cccccc|c}
 0 & 1 & 2 & 3 & 4 & 5 & a_i \\
\hline
 0 & 1 & 0 & 1 & 1 & 0 & 1/4 \\
 0 & 1 & 1 & 0 & 0 & 1 & 1/4 \\
 1 & 0 & 0 & 0 & 1 & 1 & 1/4 \\
 1 & 0 & 1 & 1 & 0 & 0 & 1/4 \\
\end{array}
 \\
\hline
\end{array}
$$
Unfortunately the time to calculate increases very fast, so it is not feasible to calculate this for $n \ge 7$. Unless we can add more restrictions about some $a_i$ being $0$.
Because of Misha's answer, the output for $n = 7$ would look something like this:
$$
\begin{array}{|c|c|c|}
\hline
n & p(n) & a_i \\
\hline
 7 & 3/7 & 
\begin{array}{ccccccc|c}
 0 & 1 & 2 & 3 & 4 & 5 & 6 & a_i \\
\hline
 0 & 0 & 1 & 0 & 1 & 1 & 0 & 1/7 \\
 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1/7 \\
 0 & 1 & 0 & 0 & 0 & 1 & 1 & 1/7 \\
 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1/7 \\
 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1/7 \\
 1 & 0 & 0 & 1 & 0 & 1 & 0 & 1/7 \\
 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1/7 \\
\end{array}
 \\
\hline
\end{array}
$$