So we know we will need at least five segments. Can it be done in four lines? Any four-line solution would of course also cover one of the 3-by-3 sub-squares of points in the 3-by-4 grid, and we’ve already discarded the only 3-by-3 solution. Hence, we need a completely new solution for the 3-by-4 grid. The problem is that solution, which is unique, also inherently involves one of the grid points lying at the junction of two of the line segments, and hence not satisfying the conditions of this problem (which require that each grid point lie on exactly one line segment). The obvious first thing to try is to build on the standard 3-by-3 solution after all, it does “go outside the box” and hit other grid points. So the puzzle comes down to finding five segments that go through the remaining 3-by-4 grid. The extra condition actually provides a bit of a hint - the first line segment must just go through four of the dots along one edge of the square.
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