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A griddy Le Monde mathematical puzzle:

1. On a 4×5 regular grid, find how many nodes need to be turned on to see all 3×4 squares to have at least one active corner in case of one arbitrary node failing.
2.  Repeat for a 7×9 grid.

The question is open to simulated annealing, as in the following R code:

n=3;m=4;np=n+1;mp=m+1

cvr=function(grue){
grud=grue
obj=(max(grue)==0)
for (i in (1:length(grue))[grue==1]){
grud[i]=0
obj=max(obj,max((1-grud[-1,-1])*(1-grud[-np,-mp])*
(1-grud[-np,-1])*(1-grud[-1,-mp])))
grud[i]=1}
obj=99*obj+sum(grue)
return(obj)}

dumban=function(grid,T=1e3,temp=1,beta=.99){
obj=bez=cvr(grid)
sprk=grid
for (t in 1:T){
grue=grid
if (max(grue)==1){ grue[sample(rep((1:length(grid))[grid==1],2),1)]=0
}else{ grue[sample(1:(np*mp),np+mp)]=1}
jbo=cvr(grue)
if (bez>jbo){ bez=jbo;sprk=grue}
if (log(runif(1))<(obj-jbo)/temp){
grid=grue;obj=cvr(grid)}
temp=temp*beta
}
return(list(top=bez,sol=sprk))}


>  dumban(grid,T=1e6,temp=100,beta=.9999)
$top [1] 8$sol
[,1] [,2] [,3] [,4] [,5]
[1,]    0    1    0    1    0
[2,]    0    1    0    1    0
[3,]    0    1    0    1    0
[4,]    0    1    0    1    0


which sounds like a potential winner.

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