05/12/08 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : a4=9%block, b8=9%block, i8=5%block, b6=4%col, b5=2%col lead to 26 filled cells.

2) Now :
Look at only possibles h8=1,h9=1 in their block. They forbid{h2=1, h1=1, h5=1}.
Look at only possibles g7=3,i7=3 in their block. They forbid{f7=3, c7=3, a7=3, e7=3}.
Look at only possibles e6=2,e4=2 in their block. They forbid{e2=2, e3=2, e8=2, e7=2}.
Look at only possibles c6=1,c6=8 in their cell. Whether c6=8 (in which case a6=1%cell) or c6=1, in both cases, we have no more {i6=1, g6=1, a5=1}.
Look at only possibles c6=8,c6=1 in their cell. Whether c6=1 (in which case a6=8%cell) or c6=8, in both cases, we have no more {e6=8, i6=8, c4=8, a5=8, g6=8}.
Look at only possibles i3=2,d3=2 in their row. Whether d3=2 (in which case i3=9%row) or i3=2, in both cases, we have no more {i3=6, i3=7}.
Look at only possibles d3=2,i3=2 in their row. Whether i3=2 (in which case d3=9%row) or d3=2, in both cases, we have no more {d3=7, d3=4, d3=1}.
Now easy fillings up to 39 filled cells. (If needed, g3=4%row, h4=4%block, h2=2%col, d3=2%block, i2=6%block, i3=9%block, d2=9%block, h7=9%block)

3)
Look at only possibles i1=8,i1=7 in their cell. Whether i1=7 (in which case h1=8%cell) or i1=8, in both cases, we have no more {g1=8, g2=8, b1=8}.
Look at only possibles i1=7,i1=8 in their cell. Whether i1=8 (in which case h1=7%cell) or i1=7, in both cases, we have no more {f1=7, d1=7}.
Look at only possibles g7=8,g7=3 in their cell. Whether g7=3 (in which case i7=8%cell) or g7=8, in both cases, we have no more {a7=8, c7=8, h9=8, e7=8, h8=8}.
Now easy fillings up to 54 filled cells. (If needed, e7=7%cell, f2=7%block, d5=7%block, i6=7%block, h1=7%block, i1=8%block, e4=8%block, e6=2%block, h5=8%block, g7=8%block, i7=3%block, i4=2%block, g4=3%block, c7=2%cell, f8=2%block)

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 e8=3,e8=1 in their cell. Whether e8=3 (in which case f9=6%cell) or e8=1 (in which case h8=6%cell), in both cases, we have no more {h9=6}.
Now easy fillings up to 81 filled cells. (If needed, h8=6%block, h9=1%block, d9=8%cell, b2=8%col, e2=3%row, f9=3%block, c8=3%block, b1=3%row, a8=8%block, c6=8%block, f1=4%cell, d8=4%block, a7=4%block, e3=5%block, a2=5%block, g1=5%block, g2=1%cell, b3=6%cell, b9=5%block, f7=6%block, a9=6%block, c9=7%block, a3=7%block, c3=1%cell, d1=1%cell, a6=1%block, e8=1%block)

Short hints for a proof

Beware, 6-FC needed here:

1) easy to 26 filled.
2) Looking at 1 in Bh8, at 3 in Bh8, at 2 in Be5, at 18 in a6c6, at 29 in i3d3, eliminate some possibles, then easy to 39 filled.
3) Looking at 78 in h1i1, at 38 in g7i7, eliminate some possibles, then easy to 54 filled.
4) Looking at e8, eliminate h9=6 (beware, heptagon). Then easy to unique solution.

Forbidding-chain-like proof

The 6-FC is needed here :

around 26 filled
(h8=1)==(h9=1) forbids {h1=1, h2=1, h5=1}
(g7=3)==(i7=3) forbids {f7=3, c7=3, a7=3, e7=3}
(e6=2)==(e4=2) forbids {e3=2, e2=2, e7=2, e8=2}
(c6=1)==(c6=8)--(a6=8)==(a6=1) forbids {a5=1, g6=1, i6=1}
(c6=8)==(c6=1)--(a6=1)==(a6=8) forbids {c4=8, i6=8, e6=8, a5=8, g6=8}
(i3=2)==(d3=2)--(d3=9)==(i3=9) forbids {i3=1, i3=7, i3=6}
(d3=2)==(i3=2)--(i3=9)==(d3=9) forbids {d3=7, d3=1, d3=4}
around 39 filled
(i1=8)==(i1=7)--(h1=7)==(h1=8) forbids {g1=8, g2=8, b1=8}
(i1=7)==(i1=8)--(h1=8)==(h1=7) forbids {d1=7, f1=7}
(g7=8)==(g7=3)--(i7=3)==(i7=8) forbids {e7=8, h8=8, h9=8, a7=8, c7=8}
around 54 filled
(f9=6)==(f9=3)--(e8=3)==(e8=1)--(h8=1)==(h8=6) forbids {h9=6}

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 h2=8%block; in today's it becomes i8=5%block. 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).