05/11/12 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : a3=7%col, i7=6%col, i5=4%col, i2=1%col, i1=7%col, d2=7%block, i3=2%col, a5=3%cell, a7=4%cell lead to 28 filled cells.

2) Now :
Look at only possibles d1=2,f1=2 in their block. They forbid{c1=2, b1=2}.
Look at only possibles g2=5,g2=9 in their cell. Whether g2=9 (in which case h2=5%cell) or g2=5, in both cases, we have no more {h1=5, b2=5, c2=5, e2=5, g3=5}.
Look at only possibles g2=9,g2=5 in their cell. Whether g2=5 (in which case h2=9%cell) or g2=9, in both cases, we have no more {g3=9, h1=9, c2=9, e2=9, b2=9}.
Look at only possibles a9=6,a9=2 in their cell. Whether a9=2 (in which case a8=6%cell) or a9=6, in both cases, we have no more {b9=6, b8=6}.
Look at only possibles a9=2,a9=6 in their cell. Whether a9=6 (in which case a8=2%cell) or a9=2, in both cases, we have no more {b9=2, b8=2, c8=2}.

3) Now :
Look at only possibles b8=3,b8=9 in their cell. Whether b8=9 (in which case c8=3%cell) or b8=3, in both cases, we have no more {b9=3, h8=3, g8=3, c7=3, e8=3}.
Look at only possibles b8=9,b8=3 in their cell. Whether b8=3 (in which case c8=9%cell) or b8=9, in both cases, we have no more {c7=9, e8=9, f8=9}.
Look at only possibles e2=4,e2=6 in their cell. Whether e2=6 (in which case e8=4%cell) or e2=4, in both cases, we have no more {e6=4, e1=4, e4=4, e9=4}.
Look at only possibles e2=6,e2=4 in their cell. Whether e2=4 (in which case e8=6%cell) or e2=6, in both cases, we have no more {e4=6, e9=6, e6=6}.

4) Look at only possibles b5=8,f5=8 in their row. Whether b5=8 (in which case b9=5%cell) or f5=8 (in which case e6=5%cell), in both cases, we have no more {e9=5}.
Now easy fillings up to 41 filled cells. (If needed, e9=3%cell, d4=3%block, e4=9%cell, h5=9%block, g2=9%block, h2=5%block, g6=5%block, e1=5%col, e6=8%col, b5=8%block, c7=8%block, b9=5%block, c3=5%block)

5)Look at only possibles d9=4,d9=6 in their cell. Whether d9=6 (in which case e8=4%cell) or d9=4, in both cases, we have no more {f8=4, f9=4}.
Look at only possibles d9=6,d9=4 in their cell. Whether d9=4 (in which case e8=6%cell) or d9=6, in both cases, we have no more {f9=6, f8=6}.
Now easy fillings up to 47 filled cells. (If needed, f9=7%cell, f8=1%cell, d5=1%row, f5=5%block, d7=5%block, f7=9%block)

6)Look at only possibles e8=6,e2=6 in their col. Whether e8=6 (in which case d9=4%cell) or e2=6 (in which case f3=8%cell,g9=8%col), in both cases, we have no more {g9=4}.
Now easy fillings up to 81 filled cells. (If needed, g8=4%block, h8=7%block, g4=7%block, c6=7%block, d9=4%block, e8=6%block, a9=6%block, h9=2%block, g9=8%block, h1=8%block, f3=8%block, d3=6%block, d1=9%block, b3=9%block, c8=9%block, b2=6%block, b8=3%block, g3=3%block, h7=3%block, c1=3%block, a8=2%block, f1=2%block, e2=4%block, b1=4%block, c4=4%block, f6=4%block, f4=6%block, h6=6%block, d6=2%block, b4=2%block, c2=2%block, g7=1%block, h4=1%block, b6=1%block)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 2 in Be2, at 59 in g2h2, at 26 in a8a9, eliminate some possibles.
3) looking at 39 in b8c8, at 46 in e2e8, eliminate some possibles.
4) looking at 8 in R5, eliminate e9=5 (beware, heptagon). Now easy to 41 filled.
5)looking at 46 in d9e8, eliminate some possibles. Now easy to 47 filled.
6) looking at 6 in Ce, eliminate g9=4 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(d1=2)==(f1=2) forbids {c1=2, b1=2}
(g2=5)==(g2=9)--(h2=9)==(h2=5) forbids {h1=5, g3=5, c2=5, e2=5, b2=5}
(g2=9)==(g2=5)--(h2=5)==(h2=9) forbids {h1=9, e2=9, b2=9, c2=9, g3=9}
(a9=6)==(a9=2)--(a8=2)==(a8=6) forbids {b8=6, b9=6}
(a9=2)==(a9=6)--(a8=6)==(a8=2) forbids {c8=2, b9=2, b8=2}
(b8=3)==(b8=9)--(c8=9)==(c8=3) forbids {h8=3, e8=3, g8=3, b9=3, c7=3}
(b8=9)==(b8=3)--(c8=3)==(c8=9) forbids {f8=9, c7=9, e8=9}
(e2=4)==(e2=6)--(e8=6)==(e8=4) forbids {e1=4, e4=4, e6=4, e9=4}
(e2=6)==(e2=4)--(e8=4)==(e8=6) forbids {e9=6, e4=6, e6=6}
(b9=5)==(b9=8)--(b5=8)==(f5=8)--(e6=8)==(e6=5) forbids {e9=5}
Around 41 filled :
(d9=4)==(d9=6)--(e8=6)==(e8=4) forbids {f8=4, f9=4}
(d9=6)==(d9=4)--(e8=4)==(e8=6) forbids {f8=6, f9=6}
Around 47 filled :
(d9=4)==(d9=6)--(e8=6)==(e2=6)--(f3=6)==(f3=8)--(g3=8)==(g9=8) forbids {g9=4}

Equivalence with 09/12

Now applying the same "dictionary" (renaming numbers, replacing cells) 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 i7=6%col. Try it ! Or see the method explained step by step on 10/08 example.
In case that you be lazy enough not to find it by yourself, see hints for today's proof (automatically derived from 05/09/12's proof).