Le Monde puzzle [#869]

A Le Monde mathematical puzzle once again in a Sudoku mode:

In an nxn table, all integers between 1 and n appear n times. If max denotes the maximum over the numbers of different integers on all rows and columns,  what is the minimum value of max when n=7? when n=11?

I tried to solve it by the following R code (in a pre-breakfast mode in my Reykjavik Airbnb flat!):

#pseudoku
n=7
T=10^4
vals=rep(1:n,n)
minmax=n
for (t in 1:T){
  psudo=matrix(sample(vals),ncol=n)
  maxc=maxr=max(sapply(apply(psudo,1,unique),length))
  if (maxc<minmax)
     maxr=max(sapply(apply(psudo,2,unique),length))
  minmax=min(minmax,max(maxc,maxr))
  }

but later realised that (a) the

sapply(apply(psudo,1,unique),length)

failed when all rows or all columns had the same number of unique terms and (b) I did not have to run the whole matrix:

vals=rep(1:n,n)
minmax=n
for (t in 1:T){
  psudo=matrix(sample(vals),ncol=n)
  maxc=max(length(unique(psudo[1,])),length(unique(psudo[,1])))
  i=1
  while((i<n)&(maxc<minmax)){
   i=i+1
   maxc=max(maxc,
        length(unique(psudo[i,])),
        length(unique(psudo[,i])))}
  minmax=min(minmax,maxc)
  }

gaining a factor of 3 in the R execution. With this random exploration, the minimum value returned was 2,2,2,3,4,5,5,6,7,8 for n=2,3,4,5,6,7,8,9,10,11. Half-hearted simulating annealing during one of the sessions of AISTATS 2014 did not show any difference…

3 Responses to “Le Monde puzzle [#869]”

  1. I think the answer may be ceiling(n/2) for n > 2. Consider n=6 and psudo=matrix(c(rep(rep(1:3, each=2), 3), rep(rep(4:6, each=2), 3)), nrow=6); For n=7, psudo=cbind(rbind(psudo, c(1,2,3,7,7,7)), c(4,7,5,7,6,7,7)).

  2. […] article was first published on Xi'an's Og » R, and kindly contributed to […]

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