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A generalization of the well-known "Nine Dots Problem", where the regular axis-aligned bounding box (RAABB:=[0, n] X [0, n] X [0, n]) has been declared.

From Marco Ripà, Aug 10 2020: (Start)

In particular, if we loosen the constraint on the allowed AABB, covering paths characterized by a shorter link-length can be found, such as 5 <= a(2) <= 6, where the aforementioned upper bound has been discovered by Koki Goma in August 2021, providing the self-crossing covering path (0,0,0)-(2,2,0)-(1/2,1/2,3/2)-(2,-1,0)-(0,1,0)-(0,1,1)-(0,0,1).

Moreover, the above pattern suggests different uncrossing covering paths of the same link-length, such as (1,0,0)-(0,0,0)-(2,2,2)-(1/2,-1,1/2)-(-1/2,1,3/2)-(1,1,0)-(1,1,0) and also the (self-crossing) covering path (1,0,0)-(0,0,0)-(0,1,0)-(3/2,1,3/2)-(1/2,-1,1/2)-(-1/2,1,3/2)-(1,1,0) which is entirely contained inside a box of 1.5 X 2 X 2 units^3 but which does not match the RAABB. (End)

Except for the first term a duplicate of A003462.

This is an n-dimensional generalization of the well-known "Nine Dots Problem".

Except for n < 2, the a(n) represent "outside the box" solutions, but (for any n) the minimal covering trail C(n) is still inside a box of (hyper)volume 3^n units^n. - Marco Ripà, Jul 19 2020

The most natural mathematical generalization of the well-known "Nine Dots Problem" by Sam Loyd (published in Cyclopedia of puzzles, p. 301, in 1914) is an NP-hard challenge with no AABB, no constraint about visiting any vertex more than once or self-crossing lines, no restriction on the turning angles available.

This problem has been solved for n < 4 (see links), while it has been proved that a(4) is 21, 22, or 23, since a covering trail (for the n = 4 case) having 23 links is given by (3,3,1)-(1,3,1)-(-2,0,1)-(4,0,1)-(3,0,3)-(3,3,3)-(0,0,0)-(3,0,0)-(0,3,3)-(0,0,3)-(3,3,0)-(-1,3,0)-(2,3,3)-(2,-1,3)-(-1,2,0)-(4,2,0)-(1,-1,3)-(1,4,3)-(4,1,0)-(-1,1,0)-(3,3,2)-(3,-2,2)-(0,7,2)-(0,0,2).

A covering trail for the n = 5 case with a link-length of 36 is (2,3,3)-(-1,0,3)-(4,0,3)-(0,4,3)-(5,4,3)- (3,2,1)-(-1,0,1)-(4,5,1)-(4,0,1)-(0,0,1)-(0,4,1)-(5,-1,1)-(3,3,3)-(0,-3,0)-(0,5,0)-(4,1,4)- (-1,1,4)-(3,5,0)-(3,0,0)-(-1,4,4)-(4,4,4)-(4,0,0)-(4,4,0)-(0,0,4)-(5,0,4)-(1,4,0)-(1,-1,0)- (5,3,4)-(0,3,4)-(2,-1,0)-(2,4,0)-(4,2,4)-(0,2,4)-(4,0,2)-(0,0,2)-(0,4,2)-(4,4,2).

A generalization of the well-known "Nine Dots Problem".

For any n > 3, an upper bound for this problem is given by U(n) := (t + 1)*n^(n - 3) - 1, where "t" is the best known solution for the corresponding n X n X n case, and (for any n > 5) t = floor((3/2)*n^2) - floor((n - 1)/4) + floor((n + 1)/4) - floor((n - 2)/4) + floor(n/2) + n - 2 (e.g., U(4) = 95, since 23 is the current upper bound for the 4 X 4 X 4 problem). In particular, it is easily possible to prove the existence of an Hamiltonian path without self crossing such that U(4) = 95 (in fact, an Hamiltonian path with link-length 23 for the 4 X 4 X 4 problem was explicitly shown in June 2020).

A (trivial) lower bound is given by B(n):= (n^n - 1)/(n - 1). - Marco Ripà, Aug 25 2020

For any n > 2, is it possible to construct an "inside the box" covering tree for any n X n X n set of points consisting of n^2 + n lines if n is even and n^2 + n + 1 lines if n is odd.

In the special case of any 2 X 2 X ... X 2 points problem (k-times n), every optimal covering tree has (exactly) 2^(k-1) rectilinear segments, thus 2^(3-1) = 4 lines for the 2 x 2 x 2 case.

If n = 3, then the (outside the box) solution is a(3) = 12, since such a connected arrangement of line segments has already been shown to exist, and this explicit upper bound matches 3^3 + 3. - Marco Ripà, Aug 25 2020

Let n >= 5, we know that it is impossible to cover more than n^3 points with n^2 segments (trivial), and we need at least n segments to obtain a connected graph (otherwise, we cannot join more than n + h*(n - 1) points with h + 1 connected lines). Thus, assuming n > 2, the general lower bound is confirmed to be n^2 + n, therefore a(n even) = 4, 20, 42, 72, ...

This sequence describes the optimal solution of the 2D generalization of the well-known nine dots problem, published in Loyd’s Cyclopedia of Puzzles (1914), p. 301.

Since Solomon Golomb constructively proved that, for any n >= 3, the minimum-link polygonal chain covering a given {0,1,...,n-1} X {0,1,...,n-1} grid consists of (exactly) 2*(n - 1) line segments, we only need to find the shortest trail satisfying the constraint above.

In detail, if n = 2, the trivial spanning path (0,1)-(0,0)-(1,0)-(1,1) is optimal. If n = 3, we have the classic solution of the nine dots problem (0,1)-(0,3)-(3,0)-(0,0)-(2,2). Now, if n > 3, a valid upper bound is given by n^2 + 5*sqrt(2) - 3, but it is possible to improve this solution for the n = 5 case by providing the trail.

(2,3)-(4,3)-(1,0)-(1,3)-(4,0)-(0,0)-(0,4)-(4,4)-(4,1), whose total Euclidean length is 20 + 6*sqrt(2). In the end, assuming n > 5, we can recycle the mentioned solution, then extend the last line segment to reach (4,-1), and finally apply the square spiral pattern to the (extended) ending segment of the {0,1,...,n-1} X {0,1,...,n-1} grid solution in order to get the solution for the {0,1,...,n} X {0,1,...,n} case, joining 2*n + 1 more points by spending two additional line segments having a combined length of 2*n (and this is an iterative strategy which is optimal for any n > 5).