A hexagonal sudoku variant

For fans of Killer Sudoku, Samurai Sudoku and other variants

A hexagonal sudoku variant

Postby jotti » Sun Oct 04, 2015 8:31 pm

I came up with this idea when I was creating a program which generated standard sudoku puzzles. Usually when I play games which contain a right angle grid, I start thinking of converting the game into a hexagonal grid. Later I found out that this same grid had already been used for one variant of sudoku. But my rules are a bit different.

Here's a rather easy puzzle for a starter:
hex25.png
hex25.png (35.5 KiB) Viewed 580 times


The rules are as follows. Use the numbers 1 to 9 to fill the whole grid. No number must occur twice or more in same round unit of 7 cells, same row of 7 cells or same row of 5 cells. Note that the rows run in three directions.

Things that differ from standard sudoku
The units consist of 7 or 5 cells, not 9. But this variation uses 9 numbers nevertheless. This is for a reason. With only 7 numbers, the only solution would always be that each round unit have an identical pattern of the 7 numbers. By adding numbers 8 and 9 this puzzle got very interesting!

The fact that not all numbers fit in one unit makes solving very different from standard sudoku. Some strategies you can use, some strategies are just not available here. I haven't even figured out very much special strategies.

Here are some more puzzles. Tell me what you think.
hex21.png
hex21.png (33.54 KiB) Viewed 580 times

hex17.png
hex17.png (33.17 KiB) Viewed 580 times
jotti
 
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Re: A hexagonal sudoku variant

Postby Smythe Dakota » Mon Oct 05, 2015 1:38 am

If you stare at one of your diagrams for a while, they will begin to look three-dimensional, with cubes arranged in various staircase patterns.

In fact, mathematically, this is essentially a three-dimensional puzzle. The horizontal rows constitute one dimension, the NE-SW diagonals another, and the NW-SE diagonals a third.

Bill Smythe
Smythe Dakota
 
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Re: A hexagonal sudoku variant

Postby jotti » Mon Oct 05, 2015 5:06 am

Haha, that's great! Maybe I should extend each cell into three significant squares of the cube. Each would have its own number.

But no! I stick with hexagons instead of 3D cubes. For time being.
jotti
 
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