Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. A simple combinatorics Le Monde mathematical puzzle:

N is a golden number if the sequence {1,2,…,N} can be reordered so that the sum of any consecutive pair is a perfect square. What are the golden numbers between 1 and 25?

Indeed, from an R programming point of view, all I have to do is to go over all possible permutations of {1,2,..,N} until one works or until I have exhausted all possible permutations for a given N. However, 25!=10²⁵ is a wee bit too large… Instead, I resorted once again to brute force simulation, by first introducing possible neighbours of the integers

```  perfect=(1:trunc(sqrt(2*N)))^2
friends=NULL
le=1:N
for (perm in 1:N){
amis=perfect[perfect>perm]-perm
amis=amis[amis<N]
le[perm]=length(amis)
friends=c(friends,list(amis)) #,recursive=TRUE)
}
```

and then proceeding to construct the permutation one integer at time by picking from its remaining potential neighbours until there is none left or the sequence is complete

```orderin=0*(1:N)
t=1
orderin=sample((1:N),1)
for (perm in 1:N){
friends[[perm]]=friends[[perm]]
[friends[[perm]]!=orderin]
le[perm]=length(friends[[perm]])
}
while (t<N){
if (length(friends[[orderin[t]]])==0)
break()
if (length(friends[[orderin[t]]])>1){
orderin[t+1]=sample(friends[[orderin[t]]],1)}else{
orderin[t+1]=friends[[orderin[t]]]
}
for (perm in 1:N){
friends[[perm]]=friends[[perm]]
[friends[[perm]]!=orderin[t+1]]
le[perm]=length(friends[[perm]])
}
t=t+1}
```

and then repeating this attempt until a full sequence is produced or a certain number of failed attempts has been reached. I gained in efficiency by proposing a second completion on the left of the first integer once a break occurs:

```while (t<N){
if (length(friends[[orderin]])==0)
break()
orderin[2:(t+1)]=orderin[1:t]
if (length(friends[[orderin]])>1){
orderin=sample(friends[[orderin]],1)}else{
orderin=friends[[orderin]]
}
for (perm in 1:N){
friends[[perm]]=friends[[perm]]
[friends[[perm]]!=orderin]
le[perm]=length(friends[[perm]])
}
t=t+1}
```

(An alternative would have been to complete left and right by squared taken at random…) The result of running this program showed there exist permutations with the above property for N=15,16,17,23,25,26,…,49.  Here is the solution for N=49:

25 39 10 26 38 43 21 4 32 49 15 34 30 6 3 22 42 7 9 27 37 12 13 23 41 40 24 1 8 28 36 45 19 17 47 2 14 11 5 44 20 29 35 46 18 31 33 16 48

As an aside, the authors of Le Monde puzzle pretended (in Tuesday, Nov. 12, edition) that there was no solution for N=23, while this sequence

22 3 1 8 17 19 6 10 15 21 4 12 13 23 2 14 11 5 20 16 9 7 18/p>

sounds fine enough to me… I more generally wonder at the general principle behind the existence of such sequences. It sounds quite probable that they exist for N>24. (The solution does not bring any light on this issue.)

Filed under: Books, Kids, R, Statistics Tagged: Le Monde, mathematical puzzle, perfect square, R  