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 More Than One Solution? (Posted on 2017-10-26)
Every week the Parade magazine supplement to many American newspapers publishes a puzzle called Numbrix. Below is a sample with the answer worked out.

The bold-face numbers are the ones given; the non-bolded numbers are the solution. The puzzle is always 9x9 and the idea is to produce a path of numbers from 1 through 81 so that the sixteen given numbers fit into the sequence without change. Each week the sixteen numbers have been in the same positions as the bolded numbers here. Each step from one number to the next must be either vertical or horizontal--never diagonal.

The question for you is: If the author of one of these puzzles first produces an arbitrary path, starting anywhere and ending anywhere (not necessarily on the edge of the tableau), and the numbers in the bolded positions revealed as the puzzle, will the puzzle always be solvable.

Another way of phrasing the question is: We know that some sets of numbers in the bolded positions have no solutions and would never be presented. Obviously also, some sets of numbers can be solved as they have only one solution; samples of this are published every week. The question becomes, Can a set of numbers be placed in the bolded positions so that there'd be two or more possible paths that would fit, and therefore be two or more solutions?

 67 66 63 62 59 58 1 2 3 68 65 64 61 60 57 6 5 4 69 70 53 54 55 56 7 8 9 72 71 52 37 36 13 12 11 10 73 74 51 38 35 14 15 16 17 76 75 50 39 34 33 32 19 18 77 48 49 40 41 30 31 20 21 78 47 46 45 42 29 26 25 22 79 80 81 44 43 28 27 24 23
Sample puzzles can be found at

Note that some have more than the specified positions marked. Are the extra positions needed?

 No Solution Yet Submitted by Charlie Rating: 3.0000 (1 votes)

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 Solution (spoiler) | Comment 1 of 2
Yes, a set of numbers be can placed in the bolded positions so that there'd be two or more possible paths that would fit, and therefore be two or more solutions.

For example, consider the following path:  Place the 1 in the center and start spiraling clockwise.  The center becomes:

9  2  3
8  1  4
7  6  5  etc.

until the entire perfectly spiral path is complete.

Form a different path, a nearly perfect spiral, by reversing the 1 and the 3.  Now the center is

9  2  1
8  3  4
7  6  5  etc.

The two paths are identical except for the location of the 1 and 3.  So, at a minimum, a problem with a unique solution would require either the 1 or the 3 to be bolded.

This also applies to spirals that start on the outside.

 Posted by Steve Herman on 2017-10-26 15:33:36

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