Archive for linear programming

Le Monde puzzle [#1021]

Posted in Books, Kids, R with tags , , , , , on September 17, 2017 by xi'an

A puzzling Le Monde mathematical puzzle for which I could find no answer in the allotted time!:

A most democratic electoral system allows every voter to have at least one representative by having each of the N voters picking exactly m candidates among the M running candidates and setting the size n of the representative council towards this goal, prior to the votes. If there are M=25 candidates, m=10 choices made by the voters, and n=10 representatives, what is the maximal possible value of N? And if N=55,555 and M=33, what is the minimum value of n for which m=n is always possible?

I tried a brute force approach by simulating votes from N voters at random and attempting to find the minimal number of councillors for this vote, which only provides an upper bound of the minimum [for one vote], and a lower bound in the end [over all votes]. Something like

for (i in 1:N) votz[i,]=sample(1:M,n)
#exploration by majority
  remz=1:N;conz=NULL
  while (length(remz)>0){
    seatz=order(-hist(votz[remz,],
    breaks=(0:M)+0.5,plot=FALSE)$density)[1]
    conz=c(conz,seatz);nuremz=NULL
    for (v in remz)
      if (!(seatz%in%votz[v,])) nuremz=c(nuremz,v)
    remz=nuremz}
  solz=length(conz)
#exploration at random
   kandz=matrix(0,N,M)
   for (i in 1:N) kandz[i,votz[i,]]=1
   for (t in 1:1e3){
#random choice of councillors
    zz=sample(c(0,1),M,rep=TRUE)
    while (min(kandz%*%zz)!=1)
      zz=sample(c(0,1),M,rep=TRUE)
    solz=min(solz,sum(zz))
#random choice of remaining councillor per voter
    remz=1:N;conz=NULL
    while (length(remz)>0){
      seatz=sample(votz[remz[1],],1)
      conz=c(conz,seatz);nuremz=NULL
      for (i in remz)
        if (!(seatz%in%votz[i,])) nuremz=c(nuremz,i)
      remz=nuremz}
    solz=min(solz,length(conz))}
maxz=max(solz,maxz)}

which leads to a value near N=4050 for the first question, with 0% confidence… Obviously, the problem can be rephrased as a binary integer linear programming problem of the form

n= \max_A \min_{c;\,\min Ac=1}\mathbf{1}^\text{T}c

where A is the NxM matrix of votes and c is the vector of selected councillors. But I do not see a quick way to fix it!

Le Monde puzzle [#1006]

Posted in Kids, R with tags , , , , , , , on May 3, 2017 by xi'an

Once the pseudo-story [noise] removed, a linear programming Le Monde mathematical puzzle:

For the integer linear programming problem

max 2x¹+2x²+x³+…+x¹⁰

under the constraints

x¹>x²+x³, x²>x³+x⁴, …, x⁹>x¹⁰+x¹, x¹⁰>x¹+x²

find a solution with the maximal number of positive entries.

Expressed this way, it becomes quite straightforward to solve with the help of a linear programming R code like lpSolve. Indeed, a simple iteration of the constraints shows that positive entries are necessarily bracketed by negative entries, since, e.g.,

x²<-88x¹/55, x¹⁰<-33x¹/55

(which applies to all consecutive triplets since the constraints are invariant by transposition). Hence there are at most five positive entries but calling lpSolve with this option

> lp (direction="max", 
objective.in=c(2,2,rep(1,8)),
const.mat=A, 
const.dir=rep(">=",10), 
const.rhs=rep(1,10)+A%*%c(rep(c(20,-1),5)), 
all.int=TRUE) 
Error: no feasible solution found

shows this is not possible. (The added vector is my way of getting around the constraint that lpSolve only considers positive entries. I therefore moved the negative entries by 20, meaning they are assumed to be larger than -20. Using the larger bound 50 does not change the outcome.) From there, there are 10 possible versions of vectors with four positive entries and a simple loop returns

> masume
[1] -90
> topast
 [1] -11 1 -13 1 -15 1 -17 1 -19 -9

as the maximal criterion and argument of this maximum with four positive entries.

As an aside, the previous Le Monde puzzle [#1005] was also provided by a linear system: given 64 cubes, each of the 384 faces being painted in one of four colours, with exactly 40 of these cubes exhibiting the four colours,  the question was about the number of cubes that could be bicolour so that a mono-colour super-cube could be reconstituted for all colours.  Which amounted to solve the four equations

4a+2b=24,4c+2d=40, b+c=8,d+3a=24,

leading to 40 quadri-colour, 16 tri-colour, and 8 bi-colour cubes.

%d bloggers like this: