blue skies over Batignoles [jatp]

a bar for MCMC algorithms [jatp]

Le Monde puzzle (rainy Sunday!)

On October 14, the weekend edition of Le Monde had the following puzzle: consider four boxes that contain all integers between 1 and 9999, in such a way that for any N, N, 2N, 3N, and 4N are in four different boxes. If 1,2,3 and 4 are in boxes labelled 1,2,3 and 4, respectively, in which box is 972 located? The direct resolution of the puzzle is that, since N and 6N are always in the same box (proof below), 972=2²x3⁵ is the same box as 162 and 27. Filling the boxes by considering the multiples of 2,…,9 leads to 27 being in the second box. Here is however a brute force resolution in R, filling the boxes by looking at the three multiples of each value met so far.

boxin=function(N){ #gives the box N is in

 sum(N==box1)+2*sum(N==box2)+3*sum(N==box3)+4*sum(N==box4)
 }

nomul=function(N,j){ #check for no multiple in the same box

 if (j==1) nopb=((sum(N==2*box1)+sum(N==3*box1)+sum(N==4*box1))==0)&&
                ((sum(2*N==box1)+sum(3*N==box1)+sum(4*N==box1))==0)
 if (j==2) nopb=((sum(N==2*box2)+sum(N==3*box2)+sum(N==4*box2))==0)&&
                ((sum(2*N==box2)+sum(3*N==box2)+sum(4*N==box2))==0)
 if (j==3) nopb=((sum(N==2*box3)+sum(N==3*box3)+sum(N==4*box3))==0)&&
                ((sum(2*N==box3)+sum(3*N==box3)+sum(4*N==box3))==0)
 if (j==4) nopb=((sum(N==2*box4)+sum(N==3*box4)+sum(N==4*box4))==0)&&
                ((sum(2*N==box4)+sum(3*N==box4)+sum(4*N==box4))==0)
 nopb
 }

box1=c(1)
box2=c(2)
box3=c(3)
box4=c(4)
N=1

while (N<972){

  allbox=c(box1,box2,box3,box4)
  N=min(allbox[allbox>N])
  ndx=rep(0,4)
  ndx[1]=boxin(N)
  for (t in 2:4){

    if (sum(t*N==allbox)>0){
      ndx[t]=boxin(t*N)}
    }
  if (sum(ndx==0)==1){ #no choice
    ndx[ndx==0]=(1:4)[-ndx[ndx>0]]
  }else{
    while (min(ndx)==0){

     pro=ndx
     pro[ndx==0]=sample((1:4)[-ndx[ndx>0]])
     okk=nomul(N,pro[1])&&nomul(2*N,pro[2])&&nnomul(3*N,pro[3])&&nomul(4*N,pro[4])
     if (okk) ndx=pro
     }
  }
  for (t in 2:4){

   if (ndx[t]==1) box1=unique(c(box1,t*N))
   if (ndx[t]==2) box2=unique(c(box2,t*N))
   if (ndx[t]==3) box3=unique(c(box3,t*N))
   if (ndx[t]==4) box4=unique(c(box4,t*N))
   }
  }
boxin(972)

To prove that N and 6N are in the same box, consider that the numbers whose box is set are of the form 2ⁱ3^j. Considering 2ⁱ3^j.in box 1, say, and 2ⁱ⁺¹3^j, 2ⁱ3^j+1, and 2ⁱ⁺²3^j.in boxes 2, 3, and 4 respectively, 6×2ⁱ3^j=2ⁱ⁺¹3^j+1 cannot be in boxes 2 and 3. Furthermore, since 2ⁱ⁺²3^j=2×2ⁱ⁺¹3^j, is in box 4, 2ⁱ⁺¹3^j+1=3×2ⁱ⁺¹3^j cannot be in box 4 either. Ergo, it is in box 1.