Le Monde puzzle [#817]

April 18, 2013
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(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)

The weekly Le Monde puzzle is (again) a permutation problem that can be rephrased as follows:

Find

\min_{\sigma\in\mathfrak{S}_{10}} \max_{0\le i\le 10}\ \sigma_i+\sigma_{i+1}

where \mathfrak{S}_{10} denotes the set of permutations on {0,…,10} and \sigma_i is defined modulo 11 [to turn {0,...,10} into a torus]. Same question for

\min_{\sigma\in\mathfrak{S}_{10}} \max_{0\le i\le 10}\ \sigma_i+\sigma_{i+1}+\sigma_{i+2}

and for

\min_{\sigma\in\mathfrak{S}_{10}} \max_{0\le i\le 10}\ \sigma_i+\cdots+\sigma_{i+5}

This is rather straightforward to code if one adopts a brute-force approach::

perminmax=function(T=10^3){
  mins=sums=rep(500,3)
  permin=matrix(0:10,ncol=11,nrow=3,byrow=TRUE)

  for (t in 1:T){
    perms=sample(0:10)
    adde=perms+perms[c language="(11,1:10)"][/c]
    sums[1]=max(adde)
    adde=adde+perms[c language="(10,11,1:9)"][/c]
    sums[2]=max(adde)
    adde=adde+perms[c language="(9:11,1:8)"][/c]+perms[c language="(8:11,1:7)"][/c]
    sums[3]=max(adde)
    for (j in 1:3)
     if (sums[j]<mins[j]){
       mins[j]=sums[j];permin[j,]=perms}
    }

  print(mins)
  print(permin)
  }

(where I imposed the first term to be zero because of the invariance by permutation), getting the solutions

> perminmax(10^5)
[1] 11 17 28
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,]    0   10    1    6    5    4    7    3    8     2     9
[2,]    0   10    4    3    5    1    9    6    2     8     7
[3,]    0    2    9    6    7    3    1    4   10     8     5

for 2, 3, and 5 terms.  (Since 10! is a mere 3.6 million, I checked with an exhaustive search, using the function permutation from the gtools package.)


Filed under: Books, Kids, R Tagged: gtools, Le Monde, mathematical puzzle, permutations, R

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