## Le Monde puzzle [#13]

**T**his week, ** Le Monde** offers not one but three related puzzles:

- Is it possible to label the twelve edges of a cube by consecutive numbers such that the sum of the edge numbers at any of the eight nodes is constant?
- Is it possible to label the eight nodes of a cube by consecutive numbers such that the sums of the node numbers over the edges are all different?
- It possible to label the twelve edges of a cube by consecutive numbers except for two edges where the difference is two such that the sum of the edge numbers at any of the eight nodes is constant?

**A** first comment is that the lowest number amont the eight, twelve or fourteen can always be taken equal to zero. The problems are location invariant.

**F**or the first problem, since the edges are labeled from 0 to 11, the sum of the numbers is *1+…+11 = 11×12/2*. The sum of the values over all nodes is therefore *11×12*, which should also be equal to *8xs*, *s* being the constant sum at each node. Since 11×12 cannot be divided by 8, this is impossible.

**F**or the second problem, were there a solution, an R resolution would be

edges=rbind(c(1,1,1,2,2,3,3,4,5,5,6,7),c(2,4,5,3,6,4,7,8,6,8,7,8)) le=11 while (le<12){ node=sample(0:7) le=length(unique(node[edges[1,]]+node[edges[2,]]))}

However, the R code does not produce an answer after a long wait, implying this is not possible.

**F**or the third problem, there is a jump in the sequence of edge numbers at *a>-1*, then again at *10>b*>a. Therefore the sum of the edge numbers is *11×6 + (b-a)+2x(11-b)*, i.e *88-a-b,* which should be divisible by 4 when all the nodes have the same value. This occurs when *a+b* is divisible by 4, e.g. when *a=6, b=10*. This choice leads to the R code

b=sample(3:9,1) a=sample(c(4-b,8-b,12-b),1) while ((a<0)||(a>b)) a=sample(c(4-b,8-b,12-b),1) newedge=sample(c(0:a,(a+2):(b+1),(b+3):13)) le=NULL for (i in 1:8) le=c(le,sum(newedge[((edges[1,]!=i)*(edges[2,]!=i))==0])) while (length(unique(le))>1){ b=sample(3:9,1) a=sample(c(4-b,8-b,12-b),1) while ((a<0)||(a>b)) a=sample(c(4-b,8-b,12-b),1) newedge=sample(c(0:a,(a+2):(b+1),(b+3):13)) le=NULL for (i in 1:8) le=c(le,sum(newedge[((edges[1,]!=i)*(edges[2,]!=i))==0])) }

which produces a positive outcome

> le [1] 20 20 20 20 20 20 20 20 > a [1] 0 > b [1] 8 > newedge [1] 9 8 3 7 4 0 13 12 11 6 5 2

May 14, 2011 at 12:17 am

[…] week Le Monde puzzle (I have not received this week issue yet!) was about deriving an optimal strategy in less than 25 […]

April 14, 2011 at 5:39 am

Problem no.1. They don’t seem to indicate that the numbering has to start at 0 .

April 14, 2011 at 8:29 am

Yes, this is true, but the problem is invariant wrt the starting point.