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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)
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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