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 05/09/12 was c7=8%col; in today's it becomes g9=4%row. 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 are hints for today's proof (automatically derived from 05/09/12's proof).

1) First simple eliminations : [c1=8%row, g9=4%row, e9=1%row, a9=5%row, b9=8%row, a4=8%block, c9=9%row, e1=6%cell, g1=1%cell] lead to 28 filled cells.
2) Look at only possibles b4=9,b6=9 in their block. They forbid {b2=9, b3=9}. Look at only possibles a8=3,a8=2 in their cell. Whether a8=2 (in which case a7=3%cell) or a8=3, in both cases, we have no more {c8=3, a5=3, b7=3, a2=3, a3=3}. Look at only possibles a8=2,a8=3 in their cell. Whether a8=3 (in which case a7=2%cell) or a8=2, in both cases, we have no more {b7=2, c8=2, a2=2, a3=2, a5=2}. Look at only possibles h1=4,h1=9 in their cell. Whether h1=9 (in which case i1=4%cell) or h1=4, in both cases, we have no more {i3=4, h3=4}. Look at only possibles h1=9,h1=4 in their cell. Whether h1=4 (in which case i1=9%cell) or h1=9, in both cases, we have no more {i3=9, i2=9, h3=9}. Look at only possibles i3=6,i3=2 in their cell. Whether i3=2 (in which case i2=6%cell) or i3=6, in both cases, we have no more {h3=6, i5=6, i8=6, i7=6, g2=6}. Look at only possibles i3=2,i3=6 in their cell. Whether i3=6 (in which case i2=2%cell) or i3=2, in both cases, we have no more {i5=2, i6=2, g2=2}. Look at only possibles a5=1,a5=4 in their cell. Whether a5=4 (in which case i5=1%cell) or a5=1, in both cases, we have no more {h5=1, f5=1, d5=1, b5=1}. Look at only possibles a5=4,a5=1 in their cell. Whether a5=1 (in which case i5=4%cell) or a5=4, in both cases, we have no more {f5=4, h5=4, d5=4}.
3) Look at only possibles e3=7,e6=7 in their col. Whether e3=7 (in which case h3=3%cell) or e6=7 (in which case f5=3%cell), in both cases, we have no more {h5=3}. This drives us by simple eliminations to 41 filled cells.
4) Look at only possibles h4=1,h4=4 in their cell. Whether h4=4 (in which case i5=1%cell) or h4=1, in both cases, we have no more {i6=1, h6=1}. Look at only possibles h4=4,h4=1 in their cell. Whether h4=1 (in which case i5=4%cell) or h4=4, in both cases, we have no more {i6=4, h6=4}. Now simple eliminations to 47 filled cells.
5) Look at only possibles i5=4,a5=4 in their row. Whether i5=4 (in which case h4=1%cell) or a5=4 (in which case c6=7%cell,h8=7%row), in both cases, we have no more {h8=1}. Now easy eliminations lead to unique solution.

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