Consider a $A \in Mat_{n}(\{+1,-1\})$ (square matrix consisted of +1,-1).
Now we can make and operation majority , i.e. $a_{i,j} = $ median of his neighborhoods(closest elements around him, i.e. closest 3,5, or 8 elements).
If number of elements are equal , then majority return -1 , i.e.: $\begin{pmatrix} -1& -1 &1\\ -1& 1 &1\\ -1& 1& 1\\ \end{pmatrix} \to \begin{pmatrix} -1& 1 &1\\ -1& -1 &1\\ 1& 1& 1\\ \end{pmatrix}$ medium element go to $-1$, because of there are equal number of +1/-1.
After this we will have another matrix. Obviously there could be three situations (depends of initial positions of +1/-1): after some number of majority operations there will be 1 matrix,-1 matrix or matrix will cycled.
So my question: does there some estimates on number of $1$($-1$) to get 1 matrix (-1 matrix).
I've thought that it should be $O(n^{2})$ (actually I thought that $n(n-1)$ is enough for $n>3$).
For easier understanding I've left an example:
$\begin{pmatrix} -1& -1\\ 1& -1 \end{pmatrix} \to \begin{pmatrix} -1& -1\\ -1& -1 \end{pmatrix} $
Added example for $n=3$
$\begin{pmatrix} 1& 1& 1\\ -1& -1& 1\\ 1 & 1 &-1 \end{pmatrix} \to \begin{pmatrix} -1& 1& 1\\ 1& 1& 1\\ -1 & -1 &1 \end{pmatrix} \to \begin{pmatrix} 1& 1& 1\\ 1& 1& 1\\ 1 & 1 &1 \end{pmatrix}$
EDIT 1 :
I've tried to consider such sequence in OEIS. First of all I've found that for $n =2$ we have $f(2) = 3$, $n = 3$ we have $f(3) = $7 , if $n = 4$ we have $f(4) = 11$ and probably for $n = 5$ $f(5) = 17$.
I've considered some sequences but for some large $n$ this sequences became near the $n/2$. Which contradict with my logic.
With the clarified neighborhood definition:
Here is a probabilistic approach. Assume we start with a matrix that has each entry $1$ with probability $p$ and $-1$ with probability $1-p$. We assume the matrix is large so we don't care about the boundary. After one operation a cell becomes $1$ with probability $p^8+8p^7(1-p)+28p^6(1-p)^2+56p^5(1-p)^3$. If we want to go towards a $1$ matrix we need this to be greater than $p$. According to Alpha, this is true when $p \gt 0.604487$ so we need a bit more than $\frac 35$ of the cells to start out as $1$.
On the other hand, a matrix that has two neighboring rows and two neighboring columns filled with $-1$s at the start will go to $-1$, so you can have as many as $(n-2)^2\ 1$'s and still go to $-1$. I think a checkerboard matrix with a $3 \times 3$ block set to $1$ will eventually go to $1$, so you can have only half $1$s and get to a $1$ matrix.