Some of my previous posts were made from my G1, so my options were practically somewhat limited. Now that I'm home again, it's easier to post, and to google as well as reference my math texts, so here goes...
If you look at the puzzle as having eight line segments, then the problem is solvable, and my solution is correct. My point is simple. Line segment intersections don't necessarily create new line segments, as a line segment is defined by endpoints, which can be any points on a line that you choose. I.e. the following picture has only two line segments, not four:
If you really want to get technical, since a point occupies no space, and there are an infinite number of points on a line between any two points, it then follows logically that if any two points on a line or line segment are chosen as endpoints, then there are an infinite number of line segments within any one line segment. For practical purposes I see two line segments in the picture above, and eight in John's diagram. If you do not accept that there are only eight line segments in the puzzle then the puzzle is not solvable:
http://en.wikipedia.org/wiki/Walls_&_Lines
Quote:
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The proof is as follows: "A continuous line that enters and leaves one of the rectangular rooms must of necessity cross two walls. Since the three bigger rooms have each an odd number of walls to be crossed, it follows that an end of a line must be inside each if all the 16 walls are crossed. But a continuous line has only two ends, so the puzzle is unsolvable."
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there are other "outside the box" methods of solving it, such as this (see #161):
http://www.puzzles.com/PuzzleHelp/Pu...ems157_168.htm
but again, they require some sort of outside the box thinking. Note also that the verbiage used in outlining the task the solver is to accomplish varies widely.