nxnxn Dots Puzzle in 3D and its generalization to k-dimensions!

This post it to announce that I have just solved the nxnxnx...xn dots puzzle, in any dimensions amount you like (n>=2)... the problem is that, for 3 dimensions and more, I can provide only a lower plus an upper bound (considering the straight lines you need to connect nxnx...xn dots).

In 3D, the upper bound of the number of consecutive straight lines you need is given by the formula SL:=2(n^2)-n-1, for n>2 (if n=2, SL=7... it is trivial).
The solving pattern is the same one I have just shown in 2D, connecting every plane using one more line (so, there are n-1 lines more).

2D-->2(n-1) SL.
We have to reproduce the “square spiral” n-times, connecting every plane we have considered... so there are “n-1” lines more.
Thus, 2(n-1)*n+(n-1)=2n*n-2n+n-1 (Q.E.D.).

Square spiral method in k-dimensions (general solving method):

The total straight lines amount, in k-dimensions (nxnx...xn where “n” appears k-times), would be:

This is a general result, it is an upper bound for any nxnx...xn problem!


Now, let’s focus ourselves on the 3D generalization.
My result/proof is as follows:
Lower Bound (Proof - by definition): k(l)>=[(n^3-n)/(n-1)]+1=n(n+1)+1.
Upper Bound (Square Spiral Proof): k(m)=2(n^2)-n-1.
Corrected Upper Bound: for any n>3, k(m)=2(n^2)-n-2.

Thus, the GAP (by proofs) is given by 2(n^2)-n-2-(n^2+n+1)=n^2-2n-3.

Considering n=4, I can solve the puzzle using only 26 straight lines, instead of 2(4^2)-4-1=27.



The table below refers to the Upper Bound (by the Square Spiral) above, while the Lower Bound is based on the consideration that you cannot connect more than "n" dots using the first line and then the maximum number is "n-1" for any additional line:



The 3D approximated upper-bound pattern, implies that, for n>2, SL(n+1)=SL(n)+5+4(n-1).


Here is the Corrected Upper Bound solution for n=5 (43 straight lines only):



P.S.
Corrected Upper Bound in k dimensions. Let "h" denote the straight lines number you need to fit every dot:
h=(2*n-1)*n^(k-2)-1.

Lower Bound in k dimensions. Let "h" denote the straight lines number you need to fit every dot:
n^k:≤n+(h-1)*(n-1)-->(n^k-n)/(n-1)=h-1-->h=(n^k-1)/(n-1).