06/01/10 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : c8=3%row, g4=3%row, h2=3%row, e8=9%row, a7=9%row, a6=3%row lead to 27 filled cells.

2) Now :
Look at only possibles d4=5,f4=5 in their row. They forbid{d5=5, f5=5}.
Look at only possibles a9=6,c9=6 in their block. They forbid{e9=6, i9=6, d9=6, g9=6}.
Look at only possibles g8=5,h8=5 in their row. They forbid{g9=5}.
Look at only possibles f9=3,d9=3 in their row. Whether d9=3 (in which case f9=5%block) or f9=3, in both cases, we have no more {f9=2, f9=1}.
Look at only possibles d9=3,f9=3 in their row. Whether f9=3 (in which case d9=5%block) or d9=3, in both cases, we have no more {d9=6, d9=1}.
Now easy fillings up to 29 filled cells. (If needed, d7=6%block, e6=6%block)

3) Look at only possibles d6=7,d5=7 in their block. They forbid{d1=7, d3=7}.
Look at only possibles g2=1,e2=1 in their row. Whether e2=1 (in which case f7=1%block) or g2=1, in both cases, we have no more {g7=1}.

4) Here is the hardest step, so as suggested by _unknown, I highlighted the vertices of the 11-agon involved !
Look at only possibles h5=8,h8=8 in their col. Whether h5=8 (in which case g6=7%cell,g7=2%cell) or h8=8 (in which case b8=1%cell,f7=1%row), in both cases, we have no more {f7=2}.
Now easy fillings up to 50 filled cells. (If needed, f7=1%cell, e9=2%block, g7=2%block, e4=4%cell, i4=9%cell, f1=4%col, c6=4%row, d6=1%row, d5=7%col, f5=9%row, f4=2%block, d4=5%row, d9=3%cell, f3=3%col, f6=8%block, f9=5%col, c4=8%row, a5=5%cell, b2=5%col, a2=4%block, g6=7%block)

5) Here is a possible non-fc move, involving 3 triples, that leads to end. You'll find a pure fc solution in the fc proof section below.
One of c1=9, d1=9, g1=9 is true.
If c1=9 then i1=2 so g1=1 or g2=1.
if g1=9 then g2=1.
if d1=9 then a1=8.
So one of g1=1, g2=1, a1=8 is true.
But if g1=1 or g2=1 then g9=8 so one of g9=8 ,a1=8 is true, eliminating a9=8.
Now easy fillings up to 81 filled cells. (If needed, h7=4%col, b9=4%col, i5=4%block, g2=9%cell, c2=7%cell, b7=7%block, e1=7%col, h3=7%col, i3=6%row, h1=5%cell, g1=1%cell, g9=8%cell, h5=8%row, h8=6%col, g5=6%col, g8=5%cell, i1=2%cell, a1=8%col, b3=2%cell, b5=1%cell, i8=1%row, c9=1%row, a9=6%row, c3=9%cell, d3=8%cell, d1=9%cell, b8=8%col, i9=7%col, c1=6%col, e2=1%block, c5=2%block)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 5 in row R4, at 6 in block Bb8, at 5 in row R8, at 35 in d9f9, eliminate some possibles. Then easy to 29 filled.
3) Looking at 7 in Be5, eliminate some possibles. Looking at 1 in R2, eliminate g1=2 (beware, pentagon).
4) Looking at 8 in Ch, eliminate f7=2 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at 9 in R1, eliminate a9=8 (beware, nonfc move, involving 3 triples). Then easy to end.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(d4=5)==(f4=5) forbids {f5=5, d5=5}
(a9=6)==(c9=6) forbids {e9=6, i9=6, d9=6, g9=6}
(g8=5)==(h8=5) forbids {g9=5}
(f9=3)==(d9=3)--(d9=5)==(f9=5) forbids {f9=1, f9=2}
(d9=3)==(f9=3)--(f9=5)==(d9=5) forbids {d9=1, d9=6}
around 29 filled
(d6=7)==(d5=7) forbids {d1=7, d3=7}
(g2=1)==(e2=1)--(e9=1)==(f7=1) forbids {g7=1}
(g7=2)==(g7=7)--(g6=7)==(g6=8)--(h5=8)==(h8=8)--(b8=8)==(b8=1)--(b7=1)==(f7=1) forbids {f7=2}
around 50 filled
(a1=6)==(a1=8)--(d1=8)==(d1=9)--(d3=9)==(c3=9) forbids {c3=6}
(h5=8)==(h8=8)--(b8=8)==(b8=1)--(i8=1)==(i8=6)--(i5=6)==(i5=4) forbids {h5=4}
(g9=1)==(g9=8)--(a9=8)==(a1=8)--(d1=8)==(d1=9)--(g1=9)==(g2=9) forbids {g2=1}

Equivalence with 09/20

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/20 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/20 was a7=9%row; in today's it becomes c8=3%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 05/09/20's proof).