Number puzzle

[Guest]

Put the numbers from 1 to 14 into the square, such that no consecutive numbers (eg. 2 and 3) and consecutive parity numbers (e.g 2 and 4) touch (common edge or point).

[Guest]

nice

[Guest]

4 12 6 3

11 7 2 9 14 8

10 13 5 1

[Guest]

.. 4 12 6 3

11 7 2 9 14 8

.. 10 13 5 1

That is not correct. 4 and 2 are connected through a diagonal. The same reasoning with 10 and 11.

[bonanova]

If I understand the "parity" constraint, the idea is to keep neighbors from being nearest or next-nearest in value.

Neighboring numbers must differ by 3 or more in value.

I broke the problem down this way.

First note that 1 and 14 have 11 allowable neighboring numbers; 2 and 13 have 10; the others, 3 thru 12, have 9.

The 2 most central squares of the grid have 8 neighbors, and their outside neighbors have 6 neighbors.

The others have fewer: 3, 4 or 5.

So I start by placing 1-14 in the middle, and 13-2 as their outside neighbors.

The numbers with the most flexibility are placed in the most demanding squares.

So we start with this for the middle row:

x - 13 - 1 - 14 - 2 - x

Then I try to keep symmetry by placing the remaining numbers in pairs:

3-12, 4-11, 5-10, 6-9 and 7-8. Start with 3-12 in the outside squares:

3 - 13 - 1 - 14 - 2 - 12

and keep going, making sure the lines connecting the added pair pass thru the middle of the grid:

--------------11

3 - 13 - 1 - 14 - 2 - 12

---------4

----10-------11

3 - 13 - 1 - 14 - 2 - 12

---------4--------5

----10--6---11

3 - 13 - 1 - 14 - 2 - 12

---------4----9---5

----10---6--11---8

3 - 13 - 1 - 14 - 2 - 12

-----7---4----9---5

[Guest]

If I understand the "parity" constraint, the idea is to keep neighbors from being nearest or next-nearest in value.

Neighboring numbers must differ by 3 or more in value.

Exactly ...