# Must any "hourglass" touching the hexagon, in a Sudoku Hoshi, contain the same number twice?

I think I found a law for Sudoku Hoshi (also called Star Sudoku) puzzles but since I did the proof more or less in my head, please check.

In case you don't know the rules, I quote:

Fill the grid so that every framed triangle, every horizontal and every tilted row contains the digits 1 to 9, whereby the middle rows expand over the empty hexagon in the middle and the short rows on the edge are completed with the outer corners of the triangles.

Proclaimed Rule: See image - any "hourglass" touching the hexagon (red) must contain the same number twice. Please give a clean proof. (I also suspect that the other hourglasses across the border (yellow) will contain "a lot" of same numbers, but as you see at the top, the numbers may be different. Can you replace "a lot" by a number n<=12 such that any Hoshi will at least have n hourglasses with equal numbers?)

I'd like to add a simple, direct proof:

• Elegant but...too late. (Of course I at least upvoted!) Sep 7 at 7:27
• @HaukeReddmann loopy walt's argument is sufficiently better than my that I would suggest changing the accepted answer. Sep 7 at 15:29
• @JeremyDover: One can do that? (Done.) Sep 7 at 17:35
• That's a really useful thing to know in solving these... Sep 8 at 17:53
• This is a very satisfying answer. Great angle of attack, quick but understandable execution and appropriate visualization.
– mafu
Sep 19 at 18:10

The original claim that any "hourglass" contains the same pair of digits is true.

Proof:

Let's look at the hourglass consisting of the purple and red shaded cells in the image below:

First note that the purple cell must equal one of the other shaded cells, since the lower left triangle and the bottom "short row" overlap in four cells. Note that purple cannot be orange or yellow since they are in the same main diagonal row.

Suppose now that purple equals either green or blue; we'll use green to illustrate, but the argument is the same for both. Look in the upper left triangle. Since purple and green are equal, the upper right cell (1) in this triangle must also be green. Then the only possible place for green in the rightmost triangle is cell 2, which forces the green in the upper right triangle to be in cell 3. [Note: the placement for cell 2 works for both green and blue since the outer cell is added to this "short row"]. Cell 3 is in the same main diagonal row as the original green [blue] cell, which is our contradiction. Thus the purple cell must equal the red cell.

Image taken from https://nicksnels.gumroad.com/l/wrusE

• My original proof attempt was a bit different (I had to think about your firstmost statement a bit) but essentially amounts to the same contradiction. Sep 6 at 19:38