Le Monde puzzle [1]

Following the presentation of the first Le Monde puzzle of the year, I tried a simulated annealing solution on an early morning in my hotel room. Here is the R code, which is unfortunately too rudimentary and too slow to be able to tackle n=1000.

#minimise \sum_{i=1}^I x_i
#for 1\le x_i\le 2n+1, 1\e i\le I
#    I\ge n, x_i \ne x_j
#    a=x_i,b=x_j,i,j\in I implies a+b=x_k for a k\in I
n=6
m=2*n+1

complete=function(inde){

 len=length(inde)
 comp=outer(inde,inde,"+")
 diag(comp)=inde
 comp=sort(unique(comp[comp<m+1]))

while ((length(comp)>len)&&(length(comp)<m)){
 inde=comp
 len=length(inde)
 comp=outer(inde,inde,"+")
 diag(comp)=inde
 comp=sort(unique(comp[comp<m+1]))
 }

 comp
 }

move=function(inde,tempe){

 ind=inde

 # movin

 if (length(ind)<m)

 off=sample(ind,1,prob=ind)
 inn=sample((1:m)[-ind],1)
 newinde=sort(c(ind[ind!=off],inn))
 newinde=complete(newinde)

 if (tempe*log(runif(1))<(sum(ind)-sum(newinde)))
 ind=newinde
 }

# prunnin

 if (length(ind)>n){

 off=sample(ind,1,prob=ind)
 newinde=sort(ind[ind!=off])
 newinde=complete(newinde)

 if (tempe*log(runif(1))<sum(ind)-sum(newinde))
 ind=newinde
 }

 ind
 }

T=10^4
fact=0.1
tpt=fact*seq(1,log(1+T),le=T)

inde=complete(sample(1:m,n))

recor=list(ind=inde,val=sum(inde))
for (t in 1:T){

 inde=move(inde,tpt[t])
 if (sum(inde)<recor$val){

 recor=list(ind=inde,val=sum(inde))
 }
 }

print(recor)

The solution to the puzzle (given in the next Le Monde issue) is to take only the even digits, resulting in a minimum sum equal to n(n+1).