Le Monde puzzle [#1087]

A board-like Le Monde mathematical puzzle in the digit category:

Given a (k,m) binary matrix, what is the maximum number S of entries with only one neighbour equal to one? Solve for k=m=2,…,13, and k=6,m=8.

For instance, for k=m=2, the matrix

\begin{matrix} 0 &0\\ 1 &1\\ \end{matrix}

is producing the maximal number 4. I first attempted a brute force random filling of these matrices with only a few steps of explorations and got the numbers 4,8,16,34,44,57,… for the first cases. Since I was convinced that the square k² of a number k previously exhibited to reach its maximum as S=k² was again perfect in this regard, I then tried another approach based on Gibbs sampling and annealing (what else?):

gibzit=function(k,m,A=1e2,N=1e2){
  temp=1 #temperature
  board=sole=matrix(sample(c(0,1),(k+2)*(m+2),rep=TRUE),k+2,m+2)
  board[1,]=board[k+2,]=board[,1]=board[,m+2]=0 #boundaries
  maxol=counter(board,k,m) #how many one-neighbours?
  for (t in 1:A){#annealing
    for (r in 1:N){#basic gibbs steps
      for (i in 2:(k+1))
        for (j in 2:(m+1)){
          prop=board
          prop[i,j]=1-board[i,j]
          u=runif(1)
          if (log(u/(1-u))<temp*(counter(prop,k,m)-val)){ 
             board=prop;val=counter(prop,k,m) 
             if (val>maxol){
               maxol=val;sole=board}}
      }}
    temp=temp*2}
  return(sole[-c(1,k+2),-c(1,m+2)])}

which leads systematically to the optimal solution, namely a perfect square k² when k is even and a perfect but one k²-1 when k is odd. When k=6, m=8, all entries can afford one neighbour exactly,

> gibzbbgiz(6,8)
[1] 48
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]    1    0    0    1    1    0    0    1
[2,]    1    0    0    0    0    0    0    1
[3,]    0    0    1    0    0    1    0    0
[4,]    0    0    1    0    0    1    0    0
[5,]    1    0    0    0    0    0    0    1
[6,]    1    0    0    1    1    0    0    1

but this does not seem feasible when k=6, m=7, which only achieves 40 entries with one single neighbour.

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