05/11/10 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : c2=6%block, c7=2%row, d8=2%row, f7=8%row, e7=1%row lead to 26 filled cells.

2) Now :
Look at only possibles c3=4,b3=4 in their block. They forbid{g3=4, e3=4, h3=4}.
Look at only possibles a1=9,a2=9 in their block. They forbid{a4=9, a5=9, a8=9, a9=9}.
Look at only possibles f5=1,d5=1 in their block. They forbid{c5=1, i5=1, g5=1, a5=1}.
Look at only possibles f9=4,f9=5 in their cell. Whether f9=5 (in which case f8=4%cell) or f9=4, in both cases, we have no more {e8=4, f1=4, f2=4}.
Look at only possibles f9=5,f9=4 in their cell. Whether f9=4 (in which case f8=5%cell) or f9=5, in both cases, we have no more {f5=5, d9=5, f2=5, f1=5, e8=5}.
Look at only possibles i2=1,i6=1 in their col. Whether i6=1 (in which case i2=2%col) or i2=1, in both cases, we have no more {i2=3, i2=7}.
Look at only possibles i6=1,i2=1 in their col. Whether i2=1 (in which case i6=2%col) or i6=1, in both cases, we have no more {i6=4, i6=3, i6=8}.
Now easy fillings up to 39 filled cells. (If needed, i1=8%col, d3=8%block, g3=1%row, g2=7%block, a3=2%row, i2=2%row, i6=1%cell, g6=2%row)

3)
Look at only possibles h2=5,h2=3 in their cell. Whether h2=3 (in which case h3=5%cell) or h2=5, in both cases, we have no more {g1=5, h7=5, h1=5}.
Look at only possibles h2=3,h2=5 in their cell. Whether h2=5 (in which case h3=3%cell) or h2=3, in both cases, we have no more {h6=3, h4=3}.
Look at only possibles a1=5,a1=9 in their cell. Whether a1=9 (in which case a2=5%cell) or a1=5, in both cases, we have no more {a9=5, b3=5, c3=5, a8=5, a5=5}.
Now easy fillings up to 54 filled cells. (If needed, a5=3%cell, e6=3%row, g4=3%row, h3=3%row, f2=3%row, d2=1%row, d1=9%block, a2=9%block, f5=1%row, e3=5%row, h2=5%row, a1=5%row, d5=5%block, a9=1%cell, c4=1%row)

Next chain was found by mark from ny.
4) Here is the hardest step, so as suggested by _unknown, I highlighted the vertices of the heptagon involved !
Look at only possibles c5=9,c5=4 in their cell. Whether c5=9 (in which case b4=7%cell) or c5=4 (in which case c3=7%cell), in both cases, we have no more {b3=7}.
Now easy fillings up to 81 filled cells. (If needed, c3=7%block, b3=4%row, b6=5%cell, g7=5%row, h7=9%block, g5=9%block, i5=6%block, g8=6%block, b7=6%block, h1=6%block, b4=9%block, a4=7%block, c6=8%block, h6=4%cell, h4=8%block, b8=7%block, i7=7%block, c9=9%block, a8=8%block, c5=4%row, g1=4%row, i8=3%row, f8=4%row, f9=5%block, b9=3%row, c8=5%row, i9=4%row)

Short hints for a proof

Beware, 6-FC needed here:

1) easy to 27 filled.
2) Looking at 4 in Bb2, at 9 in Bb2, at 1 in Be5, at 45 in f8f9, at 12 in i2i6, eliminate some possibles, then easy to 39 filled.
3) Looking at 23 in h2h3, at 59 in a1a2, eliminate some possibles, then easy to 54 filled.
4) Looking at c5, eliminate b3=7 (beware, heptagon). Then easy to unique solution.

Forbidding-chain-like proof

The 6-FC is needed here :

around 27 filled
(c3=4)==(b3=4) forbids {h3=4, g3=4, e3=4}
(a1=9)==(a2=9) forbids {a9=9, a4=9, a5=9, a8=9}
(f5=1)==(d5=1) forbids {c5=1, g5=1, i5=1, a5=1}
(f9=4)==(f9=5)--(f8=5)==(f8=4) forbids {f1=4, f2=4, e8=4}
(f9=5)==(f9=4)--(f8=4)==(f8=5) forbids {f2=5, e8=5, f1=5, d9=5, f5=5}
(i2=1)==(i6=1)--(i6=2)==(i2=2) forbids {i2=3, i2=4, i2=7}
(i6=1)==(i2=1)--(i2=2)==(i6=2) forbids {i6=4, i6=8, i6=3}
around 39 filled
(h2=5)==(h2=3)--(h3=3)==(h3=5) forbids {h1=5, h7=5, g1=5}
(h2=3)==(h2=5)--(h3=5)==(h3=3) forbids {h6=3, h4=3}
(a1=5)==(a1=9)--(a2=9)==(a2=5) forbids {c3=5, b3=5, a8=5, a5=5, a9=5}
around 54 filled
(b4=7)==(b4=9)--(c5=9)==(c5=4)--(c3=4)==(c3=7) forbids {b3=7}

Equivalence with 2005/09/28

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/28 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/28 was h9=6%row; in today's it becomes c7=2%row. 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 05/09/28's proof).