On Tuesday, Eric Chi and Kenneth Lange arXived a paper on a comparison of numerical techniques for solving sudokus. (The very Kenneth Lange who wrote this fantastic book on numerical analysis.) One of these techniques is the simulated annealing approach I had played with a long while ago.  They seem to use the same penalisation function as mine, i.e., the number of constraint violations, but the moves are different in that they pick pairs of cells without clues (i.e., not constrained) and swap their contents. The pairs are not picked at random but with probability proportional to exp(k), if k is the number of constraint violations. The temperature decreases geometrically and the simulated annealing program stops when the zero cost is achieved or when a maximum 10⁵ iterations are reached. The R program I wrote while visiting SAMSI had more options, but it was also horrendously slow! The CPU time reported by the authors is far far lower, almost in the range of the backtracking solution that serves as their reference. (Of course, it is written in Fortran 95, not in R…) As in my case, the authors mentioned they sometimes get stuck in a local minimum with only 2 cells with constraint violations.

So I reprogrammed an R code following (as much as possible) their scheme. However, I do not get a better behaviour than with my earlier code, and certainly no solution within seconds, if any. For instance, the temperature decrease in 100(.99)^t seems too steep to manage 10⁵ steps. So, either I am missing a crucial element in the code, or my R code is very poor and clever Fortran programming does the trick! Here is my code

target=function(s){
tar=sum(apply(s,1,duplicated)+apply(s,2,duplicated))
for (r in 1:9){
bloa=(1:3)+3*(r-1)%%3
blob=(1:3)+3*trunc((r-1)/3)
tar=tar+sum(duplicated(as.vector(s[bloa,blob])))
}
return(tar)
}

cost=function(i,j,s){
#constraint violations in cell (i,j)
  cos=sum(s[,j]==s[i,j])+sum(s[i,]==s[i,j])
  boxa=3*trunc((i-1)/3)+1;
  boxb=3*trunc((j-1)/3)+1;
  cos+sum(s[boxa:(boxa+2),boxb:(boxb+2)]==s[i,j])
  }

entry=function(){
  s=con
  pop=NULL
  for (i in 1:9)
    pop=c(pop,rep(i,9-sum(con==i)))
  s[s==0]=sample(pop)
  return(s)
  }

move=function(tau,s,con){
  pen=(1:81)
  for (i in pen[con==0]){
    a=((i-1)%%9)+1
    b=trunc((i-1)/9)
    pen[i]=cost(((i-1)%%9)+1,trunc((i-1)/9),s)
    }
  wi=sample((1:81)[con==0],2,prob=exp(pen[(1:81)[con==0]]))
  prop=s
  prop[wi[1]]=s[wi[2]]
  prop[wi[2]]=s[wi[1]]

  if (runif(1)<exp((target(s)-target(prop)))/tau)
    s=prop
  return(s)
  }

#Example:
s=matrix(0,ncol=9,nrow=9)
s[1,c(1,6,7)]=c(8,1,2)
s[2,c(2:3)]=c(7,5)
s[3,c(5,8,9)]=c(5,6,4)
s[4,c(3,9)]=c(7,6)
s[5,c(1,4)]=c(9,7)
s[6,c(1,2,6,8,9)]=c(5,2,9,4,7)
s[7,c(1:3)]=c(2,3,1)
s[8,c(3,5,7,9)]=c(6,2,1,9)

con=s
tau=100
s=entry()
for (t in 1:10^4){
  for (v in 1:100) s=move(tau,s,con)
  tau=tau*.99
  if (target(s)==0) break()
  }