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

A complete proof

1) First eliminations : a4=2%block, i3=6%block, c3=2%block, c6=9%col, c5=8%col lead to 26 filled cells.

2) Now :
Look at only possibles g3=7,g1=7 in their block. They forbid{g7=7, g9=7, g5=7}.
Look at only possibles h2=5,i2=5 in their block. They forbid{b2=5, f2=5, e2=5, a2=5}.
Look at only possibles e6=8,e4=8 in their block. They forbid{e2=8, e3=8, e7=8, e8=8}.
Look at only possibles b6=7,b6=1 in their cell. Whether b6=1 (in which case a6=7%cell) or b6=7, in both cases, we have no more {h6=7, i6=7, a5=7}.
Look at only possibles b6=1,b6=7 in their cell. Whether b6=7 (in which case a6=1%cell) or b6=1, in both cases, we have no more {a5=1, h6=1, i6=1, e6=1, b4=1}.
Look at only possibles i8=8,d8=8 in their row. Whether d8=8 (in which case i8=2%row) or i8=8, in both cases, we have no more {i8=3, i8=4}.
Look at only possibles d8=8,i8=8 in their row. Whether i8=8 (in which case d8=2%row) or d8=8, in both cases, we have no more {d8=7, d8=4, d8=9}.
Now easy fillings up to 39 filled cells. (If needed, h8=9%row, g4=9%block, g7=8%col, d8=8%block, d7=2%col, i8=2%block, g2=2%block, i7=3%block)

3)
Look at only possibles i9=1,i9=4 in their cell. Whether i9=4 (in which case g9=1%cell) or i9=1, in both cases, we have no more {h9=1, c9=1, h7=1}.
Look at only possibles i9=4,i9=1 in their cell. Whether i9=1 (in which case g9=4%cell) or i9=4, in both cases, we have no more {d9=4, f9=4}.
Look at only possibles h2=1,h2=5 in their cell. Whether h2=5 (in which case i2=1%cell) or h2=1, in both cases, we have no more {b2=1, g1=1, e2=1, g3=1, a2=1}.
Now easy fillings up to 54 filled cells. (If needed, e2=4%cell, d5=4%col, g9=4%col, i9=1%row, h2=1%block, e4=1%row, e6=8%block, i4=8%col, g5=1%cell, b2=8%cell, f3=8%block, i6=4%block, h4=5%block, i2=5%block, f7=4%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 e3=5,e3=7 in their cell. Whether e3=5 (in which case f1=3%cell) or e3=7 (in which case g3=3%cell), in both cases, we have no more {g1=3}.
Now easy fillings up to 81 filled cells. (If needed, g3=3%col, g1=7%block, d1=1%cell, c7=1%col, e7=5%row, f1=5%col, e8=6%col, c9=5%col, h9=6%row, a7=6%row, c1=6%col, c8=3%col, h7=7%block, f9=9%cell, d9=7%cell, d3=9%block, a2=9%block, f2=3%block, a1=3%block, b1=4%block, b3=5%block, a3=1%block, e3=7%block, a8=4%block, a6=7%col, b8=7%block, b6=1%block)

Short hints for a proof

Beware, 6-FC needed here:

1) easy to 26 filled.
2) Looking at 7 in Bg2, at 5 in Bg2, at 8 in Be5, at 17 in a6b6, at 28 in i8d8, eliminate some possibles, then easy to 39 filled.
3) Looking at 15 in h2i2, at 14 in g9i9, eliminate some possibles, then easy to 54 filled.
4) Looking at e3, eliminate g1=3 (beware, heptagon). Then easy to unique solution.

Forbidding-chain-like proof

The 6-FC is needed here :

around 26 filled
(g3=7)==(g1=7) forbids {g9=7, g5=7, g7=7}
(h2=5)==(i2=5) forbids {e2=5, b2=5, a2=5, f2=5}
(e6=8)==(e4=8) forbids {e3=8, e7=8, e2=8, e8=8}
(b6=7)==(b6=1)--(a6=1)==(a6=7) forbids {a5=7, i6=7, h6=7}
(b6=1)==(b6=7)--(a6=7)==(a6=1) forbids {b4=1, e6=1, i6=1, a5=1, h6=1}
(i8=8)==(d8=8)--(d8=2)==(i8=2) forbids {i8=4, i8=7, i8=3}
(d8=8)==(i8=8)--(i8=2)==(d8=2) forbids {d8=4, d8=7, d8=9}
around 39 filled
(i9=1)==(i9=4)--(g9=4)==(g9=1) forbids {h9=1, h7=1, c9=1}
(i9=4)==(i9=1)--(g9=1)==(g9=4) forbids {d9=4, f9=4}
(h2=1)==(h2=5)--(i2=5)==(i2=1) forbids {g3=1, g1=1, e2=1, b2=1, a2=1}
around 54 filled
(f1=3)==(f1=5)--(e3=5)==(e3=7)--(g3=7)==(g3=3) forbids {g1=3}

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 c3=2%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 05/09/28's proof).