The New Sudoku Players' Forum

[koushanejad74]

"if two crossing 4 cell compartments have exactly the same range, the puzzle is unsolvable.", much better

another corollary=> crossing 4 cell compartments:

if the common cell is not 5, then either both compartments are LO or both compartments are HI

[Leren]

I think my last corollary is correct and easy to show, and in line with what you just said. You could also have said that if the common cell is 5, one must be LO and one must be HI, otherwise they would have exactly the same range.

As you said, if the common cell is not 5, then both compartments are LO or both compartments are HI. But they can't have exactly the same range, so one must hold a 5 and one must not.

Considering the LO common cell case, one is 1234 / 5678 and the grey cell is 9, and the other is 2345 / 6789 and the grey cell is 1.
Considering the HI common cell case, one is 5678 / 1234 and the grey cell is 9, and the other is 6789 / 2345 and the grey cell is 1.

Considering the 5 crossing case, one is 5678 / 1234 and the grey cell is 9, and the other is 2345 / 6789 and the grey cell is 1.

I've Indicated the possible common cell digits in red to make this short proof easier to read. The net result is that you can eliminate 19 from the common cell and r5c5, right at the start of the puzzle.

Leren

[Leren]

I can make a few more individual eliminations using the ammunition we now have.

Suppose r123c1 was HI and r3c1 was 8 or 9. Then r3c123456 would be HI, r3c78 LO, r1234c8 = 2345 r6789c8 HI and does not contain a 5, so r9c6789 is HI, r9c1234 is LO, r89c3 LO, r7c12 HI, r1c56789HI, so r3c1 <> 89.

I was trying to find a solution to the puzzle that does not rely on a contradiction. No success yet, but good practice in handling HI/LO chains to prove eliminations.

Leren