## Le Monde puzzle [1]

**F**ollowing 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)

**T**he 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)*.

February 4, 2011 at 9:10 am

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