**Xi'an's Og » R**, and kindly contributed to R-bloggers)

**T**he puzzle in the weekend edition of Le Monde this week can be expressed as follows:

Consider four integer sequences

(x, and_{n}), (y_{n}), (z_{n})(wsuch that_{n}),and, if

u=(x_{n},y_{n},z_{n},wfor_{n}),i=1,…,4,

if

uis not the maximum of_{i}uandotherwise. Find the first return time

n(if any) such thatx. Find the value of_{n}=0(ythat minimises this return time._{0},z_{0},w_{0})

The difficulty stands with the constraint that the sequences only take integer values, which eliminates a lot of starting values. I wrote an R code that corresponds to this puzzle:

library(schoolmath) nodd=TRUE while (nodd){ suite=start=c(0,sort(2*sample(1:23,3))) clock=seq(0,48,le=49)*2*pi/48 clock=rbind(sin(clock),cos(clock)) plot(clock[1,],clock[2,],type="l", axes=F,xlab="",ylab="") radii=c("gold","sienna","steelblue","tomato") for (t in 1:10^5){ for (j in 1:4){ if (suite[j]<max(suite)){ suite[j]=((suite[j]+min(suite[suite>suite[j]]))/2)%%48 }else{ suite[j]=((suite[j]+48+min(suite[-j]))/2)%%48} } plot(clock[1,],clock[2,],type="l", axes=F,xlab="",ylab="") for (j in 1:4) lines(c(0,sin(2*pi*suite[j]/48)),c(0,cos(2*pi*suite[j]/48)),col=radii[j]) if ((suite[1]==0)||(max(is.decimal(suite))==1)){ nodd=(max(is.decimal(suite))==1) print(t) break()} }}

but it fails to produce a result, always bumping into unacceptable starting values after one or two (rarely three) iterations! So either the starting conditions are very special out of the 23*22*21/6=1771 possible values of sort(2*sample(1:23,3)) or…I missed a point in the original puzzle.

Filed under: R Tagged: Le Monde, mathematical puzzle, R

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