The first step is to directly verify that the following strategies must be feasible.
1 1 1, 2 12, 3 13, 4 14
122, 223, 324, 42 1
133, 234, 33 1, 432
144, 24 1, 342, 443
555, 656, 757, 858
566, 667, 768, 865
577, 678, 775, 876
588, 685, 786, 887
In the second step, the geometric model can be used to describe it more concisely, and the lattice points {1, 2, 3, 4, 5, 6, 7, 8} 3 in R 3 can be investigated. Each attempt of a point is equivalent to 22 points on three straight lines parallel to the coordinate axis that have been verified.
If only 3 1 inspection points are selected and eight planes perpendicular to the Z axis are investigated, it is suggested to set z= 1 as the least inspection points.
If there are no more than two detection points on z= 1, then at least 64 points on the plane should be covered by the points on the upper 7 layers, which is contradictory. So there are three inspection points on this plane.
At least 25 points on z= 1 are not on the straight line generated by these three points (marked as Class A points) and need to be covered by points in the above seven-layer plane (25 points that meet the conditions are marked as Class B points), and the remaining three free points are marked as Class C points.
The straight line generated by Class A points covers 64-25+3*7=60 points at most;
The straight line generated by Class B points covers at most 25*(8+6)=350 points;
The straight line generated by class C points covers at most 3*2 1=63 points out of z=0 plane (all points on z=0 have been counted).
These points are not enough to cover all 5 12 points.