05/10/15 tough puzzle from sudoku.com.au

This one is equivalent to 05/09/12. Proof :

1) Take 05/09/12 grid.

2) Now move rows 123456789 (1=bottom!) on new positions 798456213

3) Now move columns abcdefghi(a=left!) on new positions cbadefihg. Do you begin to feel like déjà vu?

4) All that's left now is to rename last grid's values 123456789 on 692814357 and you get exactly 05/10/15's puzzle.

Here is the dictionary which takes you from former puzzle to today's.

Now applying the same "dictionary" (changing rows, columns, values) to proof of 05/09/12 you will mechanically get a proof of today's. E.g. first filled cell in 09/12 was "c7=8%col". It becomes "a2=5%col". 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 09/12's proof.

1) First simple eliminations : [i8=8%col, a2=5%col, a5=9%col, a7=3%col, a9=8%col, a8=7%col, i5=2%cell, d7=8%block, i2=9%cell] lead to 28 filled cells.

2) Look at only possibles d9=7,f9=7 in their block. They forbid {h9=7, g9=7}.

Look at only possibles c7=6,c7=4 in their cell. Whether c7=4 (in which case b7=6%cell) or c7=6, in both cases, we have no more {c8=6, h7=6, g7=6, e7=6, b9=6}.

Look at only possibles c7=4,c7=6 in their cell. Whether c7=6 (in which case b7=4%cell) or c7=4, in both cases, we have no more {g7=4, e7=4, h7=4, b9=4, c8=4}.

Look at only possibles i1=5,i1=7 in their cell. Whether i1=7 (in which case i3=5%cell) or i1=5, in both cases, we have no more {h1=5, h3=5}.

Look at only possibles i1=7,i1=5 in their cell. Whether i1=5 (in which case i3=7%cell) or i1=7, in both cases, we have no more {h3=7, g3=7, h1=7}.

Look at only possibles h3=2,h3=4 in their cell. Whether h3=4 (in which case g3=2%cell) or h3=2, in both cases, we have no more {g2=2, b3=2, e3=2, c3=2, h1=2}.

Look at only possibles h3=4,h3=2 in their cell. Whether h3=2 (in which case g3=4%cell) or h3=4, in both cases, we have no more {e3=4, f3=4, g2=4}.

Look at only possibles e7=9,e7=5 in their cell. Whether e7=5 (in which case e3=9%cell) or e7=9, in both cases, we have no more {e6=9, e9=9, e1=9, e4=9}.

Look at only possibles e7=5,e7=9 in their cell. Whether e7=9 (in which case e3=5%cell) or e7=5, in both cases, we have no more {e1=5, e4=5, e6=5}.

3) Look at only possibles h5=1,f5=1 in their row. Whether h5=1 (in which case h1=6%cell) or f5=1 (in which case e4=6%cell), in both cases, we have no more {e1=6}.

This drives us by simple eliminations to 41 filled cells.

4) Look at only possibles d1=9,d1=5 in their cell. Whether d1=5 (in which case e3=9%cell) or d1=9, in both cases, we have no more {f1=9, f3=9}.

Look at only possibles d1=5,d1=9 in their cell. Whether d1=9 (in which case e3=5%cell) or d1=5, in both cases, we have no more {f1=5, f3=5}.

Now simple eliminations to 47 filled cells.

5) Look at only possibles e3=5,e7=5 in their col. Whether e3=5 (in which case d1=9%cell) or e7=5 (in which case f8=1%cell,c1=1%col), in both cases, we have no more {c1=9}.

Now easy eliminations lead to unique solution.