![[Maple Plot]](images/10-141.gif)
05/10/14 tough puzzle from sudoku.com.au
This one is equivalent to 05/09/24. Proof :
1) Take 05/09/24 grid, exchange rows and columns : R1(bottom!)<-->Ca(left!), etc.
2) Now move rows 123456789 (1=bottom!) on new positions 213654798
3) Now move columns abcdefghi(a=left!) on new positions higdefcab. Do you begin to feel like déjà vu?
4) All that's left now is to rename last grid's values 123456789 on 623185479 and you get exactly 05/10/14's puzzle.
Here below is the "dictionary" that gets you from 05/09/24 to today's.
Now applying the same "dictionary" (changing rows, columns, values) to proof of
09/24 you will mechanically get a proof of today's. E.g. a3=1%col in 09/24 becomes g2=6%row in today's. Try it ! Or see the method explained step by step on 10/08 example.
See more illustrated examples of equivalence between puzzles : 09/26,09/29, 10/02, 10/03, 10/04.
In case that you be lazy enough not to find it by yourself, here is the proof automatically derived from 09/24's proof.
1) First eliminations : [e7=6%block, c8=6%block, g2=6%block] lead to 23 filled cells.
2) Look at only possibles i8=4,h8=4 in their row. They forbid {g7=4, i7=4, i9=4, g9=4}.
Look at only possibles a2=3,b2=3 in their row. They forbid {a1=3, a3=3, c3=3, c1=3}.
Now simple eliminations take place again, leading to 46 filled cells.
3) Look at only possibles b9=1,b9=8 in their cell. Whether b9=8 (in which case b2=1%cell) or b9=1, in both cases, we have no more {b5=1, b6=1, b4=1}.
Look at only possibles b9=8,b9=1 in their cell. Whether b9=1 (in which case b2=8%cell) or b9=8, in both cases, we have no more {b4=8, b6=8, b5=8, b7=8}.
Then simple eliminations take place again, up to 51 filled cells.
4) Look at only possibles h4=8,a4=8 in their row. Whether h4=8 (in which case h5=1%cell) or a4=8 (in which case i4=6%row), in both cases, we have no more {i4=1}.
Look at only possibles f5=8,f6=8 in their col. Whether f5=8 (in which case h5=1%row) or f6=8 (in which case c6=1%cell), in both cases, we have no more {h6=1, i6=1}.
Now, look at only possibles h5=1,h4=1 in their block. They forbid {h3=1, h2=1, h1=1}.
From here, simple eliminations lead to unique solution.