Down Under Upside Down - a Sudoku puzzle

Post puzzles for others to solve here.

Postby champagne » Wed Dec 24, 2008 2:38 am

ronk wrote:Huh? Unicity, a term no one else on this forum uses, means unique, does it not?

Thank to udosuk and you for improvement of my poor English:D


ronk wrote:
udosuk wrote:If r2c37 were to have 3, then r37c8 would have 2 by 90 symmetry, causing an empty cell @ r2c8.
Therefore r2c37 can't have 3 (and you can copy this elimination 3 times using 90 symmetry).

Thanks udosuk, I obvously missed the chapter explaining that 90-degree symmetry uses two cyclic permutations of four values each ... (1234)(6789) for this example.



You got it and in that situation, your discontinued nice loop does the same.

You can see the effect of that 90° cyclic permutation looking at the layer B,b.

At the end, you have for example 4r2c2.b;4r8c2.B the typical status for a strong link in column 2.

champagne
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Postby udosuk » Tue Dec 30, 2008 2:32 am

I've just developed a new simple technique for Block Symmetry.:) I call this one Bludo:D :

Suppose a grid has Block Symmetry e.g. b159,b267,b348 with digit cycles (123|456|789).

I'll call 3 cells "along the flow" if these 3 cells occupy the exact same spot of one of the above 3 groups of 3 blocks.
For example, r1c1+r4c4+r7c7, r2c9+r5c3+r8c6 are "along the flow" cells.
By definition if a grid has Block Symmetry then all sets of 3 "along the flow" cells must contain a digit cycle.
E.g. r1c1+r4c4+r7c7=[123], r2c9+r5c3+r8c6=[897].

Now, I'll call 3 cells "against the flow" if these 3 cells occupy the exact same spot of 3 blocks "anti-diagonally aligned".
For example, with the 3 groups of blocks above, the anti-diagonally aligned groups of blocks would be b168,b249,b357.
So examples of 3 "against the flow" cells would be r1c1+r4c7+r7c4 or r3c8+r6c5+r9c2.

Here is the definition of the Bludo technique:

Between every pair of cells among 3 "against the flow" cells, they can't hold two different members of a digit cycle.
E.g. r1c1+r4c7 can't be {12|13|23|45|46|56|78|79|89}, but they can have 2 equal digits or one each from 2 digit cycles.

Proof

Just look at the following grid:

Code: Select all
A..|...|...
...|...|...
...|...|...
---+---+---
...|B..|*..
...|...|...
...|...|...
---+---+---
...|*..|C..
...|...|...
...|...|...

Here, suppose (ABC) is a digit cycle, with r1c1=A.
Then r4c4+r7c7=[BC], so r4c7+r7c4 can't have B or C.
Therefore the "against the flow" cells r1c1+r4c7+r7c4 can't contain {AB} or {AC} or {BC} together.

And then you can generalise this case to all "against the flow" cells. (Q.E.D.)



Corollary

3 "against the flow" cells must contain either:
- 3 identical digits, or
- 2 identical digits plus another from a different digit cycle, or
- 1 digit each from all 3 digit cycles.



I admit, that this technique is not as powerful as Gludo, but it does help quite a bit when solving Block Symmetry puzzles, especially when most cells in the grid are left with few candidates.

I'll demonstrate its application in solving one of the puzzles from eleven in this thread.

:idea:
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