Another mathematical puzzle from Le Monde that relates to a broken calculator (skipping the useless tale):

Given a pair of arbitrary positive integers (x,y) a calculator can either substract the same integer [lesser than min(x,y)] from both x and y or multiply either x or y by 2. Is it always possible to obtain equal entries by iterating calls to this calculator?

While the solution provided in this weekend edition of Le Monde is to keep multiplying x=min(x,y) by 2 until it is larger than or equal to y=max(x,y)/2,  at which stage subtracting 2x-y leads to (y-x,2y-2x) which is one multiplication away from equality, I wrote a simple R code that blindly searches for a path to equality, using as a target function exp{x²+y²+(x-y)²}. I did not even include a simulated annealing schedule as the optimal solution is known. Here is the R code:

#algorithm that brings two numbers (x,y) to be equal by
#operations x=2*x and (x,y)=(x,y)-(c,c)
emptied=function(a,b){

 mab=min(a,b)-1
 a=a-mab
 b=b-mab
 prop=matrix(0,3,2)
 targ=rep(0,3)
 targ0=a^2+b^2+(a-b)^2

 stop=(a==b)
 while (!stop){

   prop[1,]=c(a,b)-sample(0:(min(a,b)-1),1)
   targ[1]=sum(prop[1,]^2)+diff(prop[1,])^2
   prop[2,]=c(2*a,b)
   targ[2]=sum(prop[2,]^2)+diff(prop[2,])^2
   prop[3,]=c(a,2*b)
   targ[3]=sum(prop[3,]^2)+diff(prop[3,])^2

   i=sample(1:3,1,prob=exp(targ0-targ))
   a=prop[i,1];b=prop[i,2];targ0=targ[i]
   stop=(a==b)

   print(c(a,b))
   }
}

For instance,

> emptied(39,31)
[1] 9 2
[1] 8 1
[1] 8 2
[1] 7 1
[1] 7 2
[1] 7 4
[1] 6 3
[1] 5 2
[1] 4 1
[1] 4 2
[1] 3 1
[1] 3 1
[1] 3 1
[1] 3 1
[1] 3 1
[1] 3 2
[1] 2 1
[1] 2 1
[1] 2 1
[1] 2 1
[1] 2 1
[1] 2 2