Le Monde puzzle [#783]

In a political party, there are as many cells as there are members and each member belongs to at least one cell. Each cell has five members and an arbitrary pair of cells only shares one member. How many members are there in this political party?

Back to the mathematical puzzles of Le Monde (science leaflet of the weekend edition)! In addition to a tribune by Cédric Villani celebrating the 100th anniversary of the death of Henri Poincaré, this issue [now of last week] offers this interesting challenge. So much interesting that I could only solve it for three (instead of five) members and could not see the mathematical notion behind the puzzle…

Let us denote by n the number of both the cells and the number of members. Then, when picking an arbitrary order on the sets, if ij denotes the number of members in set j already seen in sets with lower indices, we have the following equality on the total number of members

n = 5n -i_2-\cdots-i_n

and the constraints are that i2<2, i3<3, i4<4, i5<5, and ij<6, for j>5. Hence, i2+i3+i4+i5+…+in≤5n-15, which implies n≥15.

Now, in terms of analytics, I could not go much further and thus turned to an R code to see if I could find a solution by brute force. Here is my code (where the argument a is the number of elements in each set):




 while (!stob){

  # ensuring intersections of one and one only
  for (i in 2:nrow(set)){

   if (chk>1){ #impossibile
   if (chk==0){
     #no common point yet but
     #can't increase the intersection size with previous sets
     if ((length(locprob)==0)){newset=1:(10*a)}else{newset=
   #else do nothing, restriction already satisfied

  if (length(newset)a)||(max(obj)>9*a)||(max(obj)==nrow(set)))

I build the sets and the collection of members by considering the constraints and stopping when (a) it is impossible to satisfy all constraints, (b) the current number of sets is the current number of members, or (c) there are too many members. (The R programming is very crude, witness the selection of possible values in the inside loop…)

Running the code for a=2 and a=3 led to the results

     [,1] [,2]
[1,]    1    2
[2,]    1    3
[3,]    2    3


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


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

which are indeed correct (note that the solution for a=3 is n=7 which allows for a symmetric allocation, rather than n=6 which is the simple upper bound). However, the code does not return a solution for a=5, which makes me wonder if the solution can be reached by brute force….

4 Responses to “Le Monde puzzle [#783]”

  1. Thanks for posting and translating this puzzle!

    N=3 SPOILER(?):

    For the n=3 case, the solution is beautifully represented by the Fano plane. Each sect is a line, each point a person (or visa-versa).

    I think the correct place to look for the n=5 solution will be the study of error-correcting codes, especially hamming codes. Eg. The Fano plane is hamming(7,4).

    Try representing each sect as a binary number, with a 1 in position n if it contains person n, and a 0 otherwise (or the other way round). This will coincide with the hamming code in the n=3 case.

    • Thanks, if you get a chance, I would be interested by the solution for n=5, obviously! As my computer code is still running…. Having used 40,000 CPU minutes so far!

      • You can stop!: The N=5 solution is (Check it!) 21.

        The k-th sect is defined as follows. Label the people i, i=1 to 20:

        Sect(k)={0, 2, 7, 8, 11}+k (mod 21).
        The set (with or without 21) is a golomb ruler:if s(i)-s(j)=n=s(l)-s(m), then i=l and j=m, that is, each difference can only be made one way.

        This means that if two sects have a person in common, then s(i)+n=s(l) for some n. Then there can be only one overlap: if s(j)+n=s(m), then rearranging, we get the Golomb condition applying. 

        There will always be an overlap because the Circle Golomb ruler ( or modular) is perfect: each number can be made in some way, as twice 5C2 is 20.

        You can also express the solution as a Finite Projective Plane, like the Fano plane. Here’s a visual solution:

        The finite projective plane has the same definition as the problem.

      • Thanks, terrific!!! (As it happens, my mainframe administrator killed my running program a few hours ago…)

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