06/01/07 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h8=5%col, i9=8%block, d9=5%row, f2=5%col, b3=5%col, h5=8%col lead to 27 filled cells.

2) Now :
Look at only possibles f6=1,f4=1 in their col. They forbid{e4=1, e6=1}.
Look at only possibles g9=4,g8=4 in their block. They forbid{g5=4, g2=4, g6=4, g1=4}.
Look at only possibles h2=1,h3=1 in their col. They forbid{g2=1}.
Look at only possibles g4=5,g6=5 in their col. Whether g6=5 (in which case g4=1%block) or g4=5, in both cases, we have no more {g4=3, g4=2}.
Look at only possibles g6=5,g4=5 in their col. Whether g4=5 (in which case g6=1%block) or g6=5, in both cases, we have no more {g6=4, g6=3}.
Now easy fillings up to 29 filled cells. (If needed, i6=4%block, d5=4%block)

3) Look at only possibles d6=7,e6=7 in their block. They forbid{c6=7, a6=7}.
Look at only possibles b2=3,b5=3 in their col. Whether b5=3 (in which case i4=3%block) or b2=3, in both cases, we have no more {i2=3}.

4) Here is the hardest step, so as suggested by _unknown, I highlighted the vertices of the 11-agon involved !
Look at only possibles e3=9,h3=9 in their row. Whether e3=9 (in which case d2=7%cell,i2=2%cell) or h3=9 (in which case h7=3%cell,i4=3%col), in both cases, we have no more {i4=2}.
Now easy fillings up to 50 filled cells. (If needed, i4=3%cell, g5=2%cell, f5=6%cell, d8=6%col, d6=3%col, e6=7%block, c4=6%block, f1=8%cell, d2=7%block, e4=8%col, f4=2%block, f8=9%cell, d4=9%block, f6=1%col, g4=1%row, g6=5%cell, a4=5%col, i2=2%block, e9=1%cell, b7=1%block, b9=6%cell)

5)Look at only possibles c6=9,c6=8 in their cell. Whether c6=9 (in which case c9=4%cell) or c6=8 (in which case a8=8%col), in both cases, we have no more {a8=4}.
Look at only possibles h7=9,h7=3 in their cell. Whether h7=9 (in which case e3=9%row) or h7=3 (in which case h1=4%cell,e1=6%cell), in both cases, we have no more {e3=6}.
Look at only possibles g9=9,c9=9 in their row. Whether g9=9 (in which case g2=3%cell) or c9=9 (in which case c6=8%cell,b2=8%row), in both cases, we have no more {b2=3}.
Now easy fillings up to 81 filled cells. (If needed, e1=6%col, i3=6%row, g1=7%block, i7=7%cell, b5=3%col, c5=7%block, a3=7%block, b8=7%block, b2=8%col, a1=4%col, c1=2%row, c2=3%col, c3=1%col, h1=3%block, h2=1%block, h3=4%block, e3=9%cell, g2=9%block, c9=9%row, a6=9%block, c8=4%block, c6=8%cell, e2=4%col, h7=9%block, g9=4%row, a8=8%row, a7=2%col, e8=2%row, e7=3%col, g8=3%col, g7=6%block)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 1 in column Cf, at 4 in block Bg8, at 1 in column Ch, at 15 in g4g6, eliminate some possibles. Then easy to 29 filled.
3) Looking at 7 in Be5, eliminate some possibles. Looking at 3 in Cb, eliminate i2=3 (beware, pentagon).
4) Looking at 9 in R3, eliminate i4=2 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at c6, eliminate a8=4 (heptagon). Looking at h7, eliminate e3=6 (nonagon). Looking at 9s in R9, eliminate b2=3(nonagon). Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(f6=1)==(f4=1) forbids {e6=1, e4=1}
(g9=4)==(g8=4) forbids {g5=4, g2=4, g1=4, g6=4}
(h2=1)==(h3=1) forbids {g2=1}
(g4=5)==(g6=5)--(g6=1)==(g4=1) forbids {g4=3, g4=2}
(g6=5)==(g4=5)--(g4=1)==(g6=1) forbids {g6=4, g6=3}
around 29 filled
(d6=7)==(e6=7) forbids {a6=7, c6=7}
(b2=3)==(b5=3)--(g5=3)==(i4=3) forbids {i2=3}
(i2=2)==(i2=7)--(d2=7)==(d2=9)--(e3=9)==(h3=9)--(h7=9)==(h7=3)--(i7=3)==(i4=3) forbids {i4=2}
around 50 filled
(c9=4)==(c9=9)--(c6=9)==(c6=8)--(a6=8)==(a8=8) forbids {a8=4}
(e3=9)==(h3=9)--(h7=9)==(h7=3)--(h1=3)==(h1=4)--(e1=4)==(e1=6) forbids {e3=6}
(g2=3)==(g2=9)--(g9=9)==(c9=9)--(c6=9)==(c6=8)--(c2=8)==(b2=8) forbids {b2=3}

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 h8=5%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 05/09/20's proof).