05/11/13 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h8=9%col, h4=6%col, g6=9%col, h5=8%col, c8=2%block lead to 26 filled cells.

2) Now :
Look at only possibles b8=4,b7=4 in their block. They forbid{b5=4, b2=4, b3=4}.
Look at only possibles a9=3,c9=3 in their block. They forbid{g9=3, e9=3, d9=3, i9=3}.
Look at only possibles e4=8,e6=8 in their block. They forbid{e9=8, e8=8, e1=8, e2=8}.
Look at only possibles i4=4,i4=5 in their cell. Whether i4=5 (in which case g4=4%cell) or i4=4, in both cases, we have no more {c4=4, a4=4, g5=4}.
Look at only possibles i4=5,i4=4 in their cell. Whether i4=4 (in which case g4=5%cell) or i4=5, in both cases, we have no more {i6=5, g5=5, a4=5, e4=5, c4=5}.
Look at only possibles c1=8,f1=8 in their row. Whether f1=8 (in which case c1=9%row) or c1=8, in both cases, we have no more {c1=7, c1=1}.
Look at only possibles f1=8,c1=8 in their row. Whether c1=8 (in which case f1=9%row) or f1=8, in both cases, we have no more {f1=6, f1=4, f1=1}.
Now easy fillings up to 39 filled cells. (If needed, a1=6%row, b6=6%col, b2=8%col, b9=9%col, c1=9%cell, f2=9%col, f1=8%cell, c2=7%block)

3)
Look at only possibles c3=5,c3=1 in their cell. Whether c3=1 (in which case b3=5%cell) or c3=5, in both cases, we have no more {h3=5, a2=5, a3=5}.
Look at only possibles c3=1,c3=5 in their cell. Whether c3=5 (in which case b3=1%cell) or c3=1, in both cases, we have no more {f3=1, d3=1}.
Look at only possibles a9=5,a9=3 in their cell. Whether a9=3 (in which case c9=5%cell) or a9=5, in both cases, we have no more {b7=5, b8=5, g9=5, i9=5, e9=5}.
Now easy fillings up to 54 filled cells. (If needed, e9=1%cell, f5=1%col, d2=1%col, b3=1%col, c4=1%col, b5=5%col, a9=5%col, c3=5%col, a6=3%col, c9=3%col, c6=8%col, e4=8%col, i9=8%cell, d8=8%col, e6=5%cell)

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=4 in their cell. Whether e8=3 (in which case d7=7%cell) or e8=4 (in which case b8=7%cell), in both cases, we have no more {b7=7}.
Now easy fillings up to 81 filled cells. (If needed, b8=7%col, b7=4%col, f7=5%cell, h2=5%col, e2=3%row, d7=3%col, i8=3%col, i4=5%col, g8=5%col, g9=6%col, d3=6%col, f8=6%col, i1=4%col, i7=1%col, g1=1%col, g4=4%col, f3=4%col, e8=4%col, a2=4%col, a3=2%col, h3=3%col, e1=2%col, h7=2%col, g2=2%col, h1=7%col, g7=7%col, d9=7%col)

Short hints for a proof

Beware, 6-FC needed here:

1) easy to 26 filled.
2) Looking at 4 in Bb8, at 3 in Bb8, at 8 in Be5, at 45 in g4i4, at 89 in c1f1, eliminate some possibles, then easy to 39 filled.
3) Looking at 15 in b3c3, at 35 in a9c9, eliminate some possibles, then easy to 54 filled.
4) Looking at e8, eliminate b7=7 (beware, heptagon). Then easy to unique solution.

Forbidding-chain-like proof

The 6-FC is needed here :

around 26 filled
(b8=4)==(b7=4) forbids {b5=4, b2=4, b3=4}
(a9=3)==(c9=3) forbids {i9=3, d9=3, e9=3, g9=3}
(e4=8)==(e6=8) forbids {e9=8, e1=8, e2=8, e8=8}
(i4=4)==(i4=5)--(g4=5)==(g4=4) forbids {c4=4, g5=4, a4=4}
(i4=5)==(i4=4)--(g4=4)==(g4=5) forbids {c4=5, i6=5, a4=5, e4=5, g5=5}
(c1=8)==(f1=8)--(f1=9)==(c1=9) forbids {c1=7, c1=1, c1=4}
(f1=8)==(c1=8)--(c1=9)==(f1=9) forbids {f1=4, f1=6, f1=1}
around 39 filled
(c3=5)==(c3=1)--(b3=1)==(b3=5) forbids {a3=5, h3=5, a2=5}
(c3=1)==(c3=5)--(b3=5)==(b3=1) forbids {f3=1, d3=1}
(a9=5)==(a9=3)--(c9=3)==(c9=5) forbids {b7=5, b8=5, i9=5, e9=5, g9=5}
around 54 filled
(d7=7)==(d7=3)--(e8=3)==(e8=4)--(b8=4)==(b8=7) forbids {b7=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 f8=6%col; in today's it becomes g6=9%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).