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**A** fairly simple puzzle in this weekend ** Le Monde** magazine:

*given five points on a line such that their pairwise distances are 1,2,4,…,14,18,20, find the respective positions of the five points over the line and deduce the missing distances*. Without loss of generality, we can set the first point at 0 and the fifth point at 20. The three remaining points are in between 1 and 19, but again without loss of generality we can choose the fourth point to be at 18. (Else the second point would be at 2.) Finding the three remaining points can be done by the R code

i=1 while (i==1){ prop=sort(c(0,sample(1:17,2),18,20)) a=sort(dist(prop)) if ((a[1]==1)&&(a[2]==2)&&(a[3]==4)&&(a[8]==14)) break() }

**R**emoving 18 from the above takes about the same time! A solution (modulo permutations) is *0, 13, 14, 18, 20*.

Filed under: R, Statistics Tagged: dist(), Le Monde, mathematical puzzle, R

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