Talk:Fujimura's problem
[Incomplete, just placing here what I have typed up so far.]
[math]\displaystyle{ \overline{c}^\mu_6 = 15 }[/math]
[math]\displaystyle{ \overline{c}^\mu_6 \geq 15 }[/math] from the bound for general n.
Note that there are ten extremal solutions to [math]\displaystyle{ \overline{c}^\mu_3 }[/math]:
Solution I: remove 300, 020, 111, 003
Solution II (and 2 rotations): remove 030, 111, 201, 102
Solution III (and 2 rotations): remove 030, 021, 210, 102
Solution III' (and 2 rotations): remove 030, 120, 012, 201
Also consider the same triangular lattice with the point 020 removed, making a trapezoid. Solutions based on I-III are:
Solution IV: remove 300, 111, 003
Solution V: remove 201, 111, 102
Solution VI: remove 210, 021, 102
Solution VI': remove 120, 012, 201
The on the 7x7x7 triangular lattice triangle 141-411-114 must have at least one point removed. Remove 141, noting by symmetry any logic that follows will also work for either of the other two points.
Suppose we can remove all equilateral triangles on our 7×7x7 triangular lattice with only 12 removals.
Here, "top triangle" means the top four rows of the lattice (with 060 at top) and "bottom trapezoid" means the bottom three rows.
At least 4 of those removals must come from the top triangle (the solutions of [math]\displaystyle{ \overline{c}^\mu_3 }[/math] mentioned above).
The bottom trapezoid includes the overlapping trapezoids 600-420-321-303 and 303-123-024-006. If the solutions of these trapezoids come from V, VI, or VI', then 6 points have been removed. Suppose the trapezoid 600-420-321-303 uses the solution IV (by symmetry the same logic will work with the other trapezoid). Then there are 3 disjoint triangles 402-222-204, 213-123-114, and 105-015-006. Then 6 points have been removed. Therefore at least six removals must come from the bottom trapezoid.
To make a total of 12 removals there must be either:
Case A: 4 removals from the top triangle and 8 from the bottom trapezoid.
Case B: 5 removals from the top triangle and 7 from the bottom trapezoid.
Case C: 6 removals from the top triangle and 6 from the bottom trapezoid.
- Suppose case C is true.
Suppose the trapezoid 600-420-321-303 used solution IV. There are three disjoint triangles 402-222-204, 213-123-114, and 105-015-006. The remainder of the points in the lower trapezoid (420, 321, 510, 501, 402, 312, 024) must be left open. 024 being open forces either 114 or 015 to be removed.
Suppose 114 is removed. Then 213 is open, and with 312 open that forces 222 to be removed. Then 204 is open, and with 024 that forces 006 to be removed. So the bottom trapezoid is a removal configuration of 600-411-303-222-114-006, and the rest of the points in the bottom trapezoid are open. All 10 points in the upper triangle form equilateral triangles with bottom trapezoid points, hence 10 removals in the upper triangle would be needed (more than the 6 allowed), so 114 being removed doesn't work.
Suppose 015 is removed. Then 006-024 forces 204 to be removed. Regardless of where the removal in 123-213-114, the points 420, 321, 222, 024, 510, 312, 501, 402, 105, and 006 must be open. This forces upper triangle removals at 330, 231, 042, 060, 051, 132, our remaining 6 removals on the top triangle. However, we have already removed 141, forcing one removal too many, so the trapezoid 600-420-321-303 doesn't use solution IV.
Suppose the trapezoid 600-420-321-303 uses solution VI. The trapezoid 303-123-024-006 can't be IV (already eliminated by symmetry) or VI' (leaves the triangle 402-222-204). Suppose the trapezoid 303-123-024-006 is solution VI. The removals from the lower trapezoid are then 420, 501, 312, 123, 204, and 015, leaving the remaining points in the lower trapezoid open. The remaining open points is forces 10 upper triangle removals, so the trapezoid 600-420-321-303 doesn't use solution VI. Therefore the trapezoid 303-123-024-006 is solution V. The removals from the lower trapezoid are then 420, 510, 312, 204, 114, and 105. The remaining points in the lower trapezoid are open, and force 9 upper triangle removals, hence the trapezoid 303-123-024-006 can't be V, and the solution for 600-420-321-303 can't be VI.
The solution VI' for the trapezoid 600-420-321-303 can be eliminated by the same logic by symmetry.
Therefore it is impossible for the bottom trapezoid to use only 6 removals.
- Suppose case A is true.
Because 141 is already removed, the solution to the upper triangle must remove either solution I (remove 060, 330, 033), solution II (remove 060, 231, 132), solution IIb (remove 033, 150, 240) or solution IIc (remove 330, 051, 042)
Suppose I is the solution for the upper triangle.
Suppose 222 is open. Then 420, 321, 123, and 024 must all be removed. This leaves five disjoint triangles which require removals in the bottom trapezoid (150-600-105, 051-501-006, 222-402-204, 231-411-213, 132-312-114); therefore the bottom trapezoid needs at least 9 removals, but we can only make 8, therefore 222 is closed.
Suppose 411 is open. Then 213 and 015 must be removed. This leaves five disjoint triangles such that each triangle must have exactly one removal (420-150-123, 321-051,024, 600-510-501, 402-312-303, 204-114-105), so the remaining point (006) must be open, forcing 501 to be removed. This makes 600 and 510 open, and based on the triangles 600-240-204 and 510-150-114 both 204 and 114 must both be removed, but 204 and 114 are on the same disjoint triangle, contradicting the statement that each triangle must have exactly one removal. So 411 is closed.
This leaves six disjoint triangles each which must have at least one removal (420-123-150, 321-024-051, 510-213-240, 312-015-042, 501-204-231, 402-105-132). This forces 600 and 006 to be open. Based on the triangles 006-501-051 and 600-204-240, this forces 501 and 204 to be open. But then there are no removals from the triangle 501-204-231, which is a contradiction. Therefore the solution of the upper triangle cannot be I.
Suppose II is the solution for the upper triangle.
There are seven disjoint triangles (150-600-105, 051-501-006, 222-402-204, 240-510-213, 042-312-015, 330-420-321, 033-123-024), therefore of the three points remaining in the bottom trapezoid (411, 303, 114) exactly one must be removed.
Suppose 411 is removed. Then 114 and 303 are open; 114 open forces 510 to be removed, forcing 213 to be open. 114 and 213 open force 123 to be closed, forcing 024 to be open. 024 open forces 222 to be closed, which forces 204 to be open, which leaves the triangle 213-303-204 open so we have a contradiction.
By symmetry 114 also can't be removed. Therefore we must remove 303. This leaves 411 and 114 open, forcing 510 and 015 closed. 510 and 015 closed forces 312 and 213 open, forcing 222 closed. 222 closed forces 402 and 204 open. 402 and 204 open force 501 and 105 closed. 510 and 105 closed force 600 and 006 open, leaving equilateral triangles 600-204-240 and 402-006-042 open. Therefore 303 can't be removed, and so the solution of the upper triangle can't be II.
Suppose IIb is the solution for the upper triangle.
Suppose 024 is open. This forces 420, 321, and 222 closed. Five disjoint triangles remain (510-600-501, 231-411-213, 132-312-114, 123-303-105, 024-204-006) so each must have exactly one point removed, and the remaining points in the lower trapezoid (402, 015) must be open. This forces 312, 411, 510, and 006 closed, which then force the the other points in their disjoint triangles open (600, 501, 231, 213, 132, 114, 204). The triangle 501-231-204 is therefore open, so we have a contradiction, therefore 024 is closed.
Suppose 006 is open. This forces 600, 501 and 402 to be closed, and leaves five disjoint triangles (510-420-411, 321-231-222, 312-132-114, 303-213-204, 105-015-006) and so 123 must be open. This forces 222 closed, which forces 321 open, which forces 420 closed, which forced 510 and 411 open, which forces 213 closed, which forces 303 and 204 open, which forces 105 closed, which forces 015 and 006 to be open, leaving an open triangle at 411-015-051. Therefore we have a contradiction, so 006 is closed.
Given 024 and 006 closed, now note there are six disjoint triangles (600-510-501, 402-312-303, 204-114-105, 420-330-321, 222-132-123, 411-231-213). Therefore the remaining point in the lower trapezoid 015 must be open, forcing 510, 411, and 312 to be closed. Using the disjoint triangles this forces 600, 501, 402, 303 and 213 to be open, which then forces 420 and 321 to be closed. Both 420 and 321 are on the same disjoint triangle, therefore we have a contradiction, so IIb can't be solution.
Note by symmetry, the same logic for IIb will apply for IIc. Therefore case A isn't true.
- Suppose case B is true.
The row 330-231-132-033 has ten possible solutions (excluding reflections, which by symmetry will be handled by the same logic):
Case Q: open-open-open-open Case R: closed-closed-closed-closed Case S: closed-open-open-open Case T: closed-open-closed-closed Case U: open-closed-closed-closed Case V: closed-open-closed-closed Case W: closed-closed-open-open Case X: closed-open-closed-open Case Y: open-closed-closed-open Case Z: closed-open-open-closed
