easy and uneasy riddles

On 15 January, The Riddler had both a straightforward and a challenging riddles. The first one was to optimise the choice of a real number d with the utility function U(d,θ)=d ℑ(θ>d), when θ is Uniform(0,100). Leading unsurprisingly to d=50…

The tough(er) one was to solve a form of sudoku where the 24 entries of a 8×3 table are integers in {1,…,9} and the information is provided by the row-wise and column-wise products of these integers. The vertical margins are

294, 216, 135, 98, 112, 84, 245, 40

and the horizontal margins are

8 890 560, 156 800, 55 566

After an unsuccessful brute-force (and pseudo-annealed) attempt achieving a minimum error of 127, although using the prime factor decompositions of these 11 margins, I realised that some entries were known: e.g., 7 at (1,2), 5 at (3,2), and 7 at (7,3), and (much later) that the (huge) product value for the first column implied that each term in that column had to be the maximal possible value for the corresponding rows, except for 5 on row 7. This leads to the starting grid

    [,1] [,2] [,3]
[1,]    7    7    6
[2,]    9    0    0
[3,]    9    5    3
[4,]    7    0    0
[5,]    8    0    0
[6,]    7    0    0
[7,]    5    7    7
[8,]    8    0    0

and an additional and obvious exclusion based on the absence of 3’s in the second column, of 5’s and 2’s in the third column shows there was a unique solution

    [,1] [,2] [,3]
[1,]    7    7    6
[2,]    9    8    3
[3,]    9    5    3
[4,]    7    2    7
[5,]    8    2    7
[6,]    7    4    3
[7,]    5    7    7
[8,]    8    5    1

as also demonstrated by a complete exploration with R:

Try it online!

Le Monde puzzle [#1092]

A Latin square Le Monde mathematical puzzle that I found rather dreary:

A hidden 3×3 board contains all numbers from 1 to 9. Anselm wants to guess the board and makes two proposals. Berenicke tells him how many entries are in the right rows and colums for each proposal, along with the information that no entry is at the right location. Anselm deduces the right board.

Which I solved by brute force and not even simulated annealing, first defining a target

ordoku1=ordoku2=matrix(1,9,2)
ordoku1[,1]=c(1,1,1,2,2,2,3,3,3)
ordoku1[,2]=rep(1:3,3)
ordoku2[,1]=c(3,2,3,1,2,3,2,1,1)
ordoku2[,2]=c(2,2,3,2,3,1,1,3,1)
fitz=function(ordo){
  (sum(ordo[c(1,4,7),2]==1)==1)+(sum(ordo[c(2,5,8),2]==2)==1)+
  (sum(ordo[c(3,6,9),2]==3)==0)+(sum(ordo[c(1,2,3),1]==1)==1)+
  (sum(ordo[c(4,5,6),1]==2)==1)+(sum(ordo[c(7,8,9),1]==3)==2)+
  (sum(ordo[c(6,7,9),2]==1)==2)+(sum(ordo[c(1,2,4),2]==2)==1)+
  (sum(ordo[c(3,5,8),2]==3)==2)+(sum(ordo[c(4,8,9),1]==1)==1)+
  (sum(ordo[c(7,2,5),1]==2)==1)+(sum(ordo[c(1,3,6),1]==3)==0)+
  (!(0%in%apply((ordo-ordoku1)^2,1,sum)))+(!(0%in%apply((ordo-ordoku2)^2,1,sum)))
}

on a 9×9 board entry reproducing all items of information given by Berenicke. If all constraints are met, the function returns 14. And then searched for a solution at random:

temp=1
randw=function(ordo){
  for (t in 1:1e6){
    chlg=sample(1:9,2)
    temp=ordo[chlg[1],]
    ordo[chlg[1],]=ordo[chlg[2],]
    ordo[chlg[2],]=temp
    if (fitz(ordo)==14){
      print(ordo);break()}}}

which produces the correct board

4 3 5
6 7 1
9 2 8

X entropy for optimisation

At Gregynog, with mounds of snow still visible in the surrounding hills, not to be confused with the many sheep dotting the fields(!), Owen Jones gave a three hour lecture on simulation for optimisation, which is a less travelled path when compared with simulation for integration. His second lecture covered cross entropy for optimisation purposes. (I had forgotten that Reuven Rubinstein and Dirk Kroese had put forward this aspect of their technique in the very title of their book. As “A Unified Approach to Combinatorial Optimization, Monte-Carlo Simulation and Machine Learning”.) The X entropy approaches pushes for simulations restricted to top values of the target function, iterating to find the best parameter in the parametric family used for the simulation. (Best to be understood in the Kullback sense.) Now, this is a wee bit like simulated annealing, where lots of artificial entities have to be calibrated in the algorithm, due to the original problem being unrelated to an specific stochastic framework. X entropy facilitates concentration on the highest values of the target, but requires a family of probability distributions that puts weight on the top region. This may be a damning issue in large dimensions. Owen illustrated the approach in the case of the travelling salesman problem, where the parameterised distribution is a Markov chain on the state space of city sequences. Further, if the optimal value of the target is unknown, avoiding getting stuck in a local optimum may be tricky. (Owen presented a proof of convergence for a temperature going to zero slowly enough that is equivalent to a sure exploration of the entire state space, in a discrete setting, which does not provide a reassurance in this respect, as the corresponding algorithm cannot be implemented.) This method falls into the range of methods that are doubly stochastic in that they rely on Monte Carlo approximations at each iteration of the exploration algorithm.

During a later talk, I tried to recycle one of my earlier R codes on simulated annealing for sudokus, but could not find a useful family of proposal distributions to reach the (unique) solution. Using a mere product of distributions on each of the free positions in the sudoku grid only led me to a penalty of 13 errors…

1    2    8    5    9    7    4    9    3
7    3    5    1    2    4    6    2    8
4    6    9    6    3    8    5    7    1
2    7    5    3    1    6    9    4    8
8    1    4    7    8    9    7    6    2
6    9    3    8    4    2    1    3    5
3    8    6    4    7    5    2    1    9
1    4    2    9    6    3    8    5    7
9    5    7    2    1    8    3    4    6

It is hard to consider a distribution on the space of permutations, 𝔖⁸¹.

Le Monde puzzle [#1001]

After a long lag (due to my missing the free copies distributed at Paris-Dauphine!), here is a Sudoku-like Le Monde mathematical puzzle:

A grid of size (n,n) holds integer values such that any entry larger than 1 is the sum of one term in the same column and one term in the same row. What is the maximal possible value observed in such a grid when n=3,4?

This can be solved in R by a random exploration of such possible grids in a simulated annealing spirit:

mat=matrix(1,N,N)
goal=1

targ=function(mat){ #check constraints
  d=0
  for (i in (1:(N*N))[mat>1]){
    r=(i-1)%%N+1;c=(i-1)%/%N+1
    d=d+(min(abs(mat[i]-outer(mat[-r,c],mat[r,-c],"+")))>0)} 
  return(d)}

cur=0
for (t in 1:1e6){
  i=sample(1:(N*N),1);prop=mat
  prop[i]=sample(1:(2*goal),1)
  d=targ(prop)
  if (10*log(runif(1))/t<cur-d){ 
      mat=prop;cur=d} 
  if ((d==0)&(max(prop)>goal)){
     goal=max(prop);maxx=prop}}

returning a value of 8 for n=3 and 37 for n=4. However, the method is quite myopic and I tried instead a random filling of the grid, using each time the maximum possible sum for empty cells:

goal=1
for (v in 1:1e6){
  mat=matrix(0,N,N)
  #one 1 per row/col
  for (i in 1:N) mat[i,sample(1:N,1)]=1
  for (i in 1:N) if (max(mat[,i])==0) mat[sample(1:N,1),i]=1
  while (min(mat)==0){
   parm=sample(1:(N*N)) #random order
   for (i in parm[mat[parm]==0]){
    r=(i-1)%%N+1;c=(i-1)%/%N+1
    if ((max(mat[-r,c])>0)&(max(mat[r,-c])>0)){
      mat[i]=max(mat[-r,c])+max(mat[r,-c])
      break()}}}
    if (goal<max(mat)){
    goal=max(mat);maxx=mat}}

which recovered a maximum of 8 for n=3, but reached 48 for n=4. And 211 for n=5, 647 for n=6… For instance, here is the solution for n=4:

[1,]    1    5   11   10
[2,]    2    4    1    5
[3,]   48    2   24    1
[4,]   24    1   22   11

Le Monde puzzle [#940]

A sudoku-like Le Monde mathematical puzzle:

On a 3×3 grid, all integers from 1 to 9 are present. Considering all differences between adjacent entries, the value of the grid is the minimum difference. What is the maximum possible value?

In a completely uninspired approach considering random permutations on {1,..,9}, the grid value can be computed as

neigh=c(1,2,4,5,7,8,1,4,2,5,3,6)
nigh=c(2,3,5,6,8,9,4,7,5,8,6,9)
perm=sample(9)
val<-function(perm){
min(abs(perm[neigh]-perm[nigh]))}

which produces a value of 3 for the maximal value. For a 4×4 grid

neigh=c(1:3,5:7,9:11,13:15,1+4*(0:2),2+4*(0:2),3+4*(0:2),4*(1:3))
nigh=c(2:4,6:8,10:12,14:16,1+4*(1:3),2+4*(1:3),3+4*(1:3),4*(2:4))
perm=sample(16)
val<-function(perm){
min(abs(perm[neigh]-perm[nigh]))}

the code returns 5. For the representation

[,1] [,2] [,3] [,4]
[1,] 8 13 3 11
[2,] 15 4 12 5
[3,] 9 14 6 16
[4,] 2 7 1 10