05/11/14 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : i8=6%col, g7=8%block, i5=8%block, d7=6%block, f2=6%block, a1=6%col lead to 27 filled cells.

2) Now :
Look at only possibles f4=3,f6=3 in their col. They forbid{e4=3, e6=3}.
Look at only possibles h7=5,h8=5 in their block. They forbid{h4=5, h5=5, h2=5, h3=5}.
Look at only possibles i2=3,i1=3 in their col. They forbid{h2=3}.
Look at only possibles h6=6,h4=6 in their col. Whether h4=6 (in which case h6=3%block) or h6=6, in both cases, we have no more {h6=4, h6=2}.
Look at only possibles h4=6,h6=6 in their col. Whether h6=6 (in which case h4=3%block) or h4=6, in both cases, we have no more {h4=5, h4=4}.
Now easy fillings up to 29 filled cells. (If needed, g4=5%block, d5=5%block)

3) Look at only possibles d4=7,e4=7 in their block. They forbid{b4=7, c4=7}.
Look at only possibles a2=4,a5=4 in their col. Whether a5=4 (in which case g6=4%block) or a2=4, in both cases, we have no more {g2=4}.

4) Here is the hardest step, so as suggested by _unknown on comments of yesterday's puzzle, I highlighted the vertices of the 11-agon involved !
Look at only possibles e1=9,i1=9 in their row. Whether e1=9 (in which case d2=7%cell,g2=2%cell) or i1=9 (in which case i9=4%cell,g6=4%col), in both cases, we have no more {g6=2}.
Now easy fillings up to 50 filled cells. (If needed, h5=2%block, g6=4%block, g2=2%block, f5=1%cell, d8=1%col, d4=4%col, e4=7%block, d2=7%block, d6=9%cell, f3=8%cell, e6=8%block, f6=2%block, f4=3%block, h6=3%block, h4=6%block, c6=6%block, f8=9%cell, e7=3%cell, a9=3%block, a7=1%col, b6=1%col)

5)Look at only possibles b4=9,b4=8 in their cell. Whether b4=9 (in which case b7=5%cell) or b4=8 (in which case c8=8%col), in both cases, we have no more {c8=5}.
Look at only possibles i9=9,i9=4 in their cell. Whether i9=9 (in which case e1=9%row) or i9=4 (in which case i3=5%cell,e3=1%cell), in both cases, we have no more {e1=1}.
Look at only possibles h7=9,b7=9 in their row. Whether h7=9 (in which case h2=4%cell) or b7=9 (in which case b4=8%cell,a2=8%row), in both cases, we have no more {a2=4}.
Now easy fillings up to 81 filled cells. (If needed, a5=4%col, b5=7%block, a8=7%col, e3=1%col, h9=1%col, g9=7%block, h3=7%block, c1=7%block, c3=5%col, g1=1%col, i3=4%cell, h8=4%block, h7=5%block, i9=9%block, b7=9%block, c4=9%block, h2=9%block, e1=9%block, b4=8%block, c8=8%block, a2=8%block, b8=5%block, e2=5%block, i1=5%block, e9=4%block, b2=4%block, i2=3%block, b1=3%block, e8=2%block, c9=2%block, b3=2%col)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 3 in Cf, at 5 in block Bh8, at 3 in Ci, at 36 in h4h6, eliminate some possibles. Then easy to 29 filled.
3) Looking at 7 in Be5, eliminate some possibles. Looking at 4 in Ca, eliminate g2=4 (beware, pentagon).
4) Looking at 9 in R1, eliminate g6=2 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at b4, eliminate c8=5 (heptagon). Looking at i9, eliminate e1=1 (nonagon). Looking at 9s in R7, eliminate a2=4(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(f4=3)==(f6=3) forbids {e6=3, e4=3}
(h7=5)==(h8=5) forbids {h2=5, h5=5, h4=5, h3=5}
(i2=3)==(i1=3) forbids {h2=3}
(h6=6)==(h4=6)--(h4=3)==(h6=3) forbids {h6=4, h6=2}
(h4=6)==(h6=6)--(h6=3)==(h4=3) forbids {h4=5, h4=4}
around 29 filled
(d4=7)==(e4=7) forbids {c4=7, b4=7}
(a2=4)==(a5=4)--(h5=4)==(g6=4) forbids {g2=4}
(g2=2)==(g2=7)--(d2=7)==(d2=9)--(e1=9)==(i1=9)--(i9=9)==(i9=4)--(g9=4)==(g6=4) forbids {g6=2}
around 50 filled
(b7=5)==(b7=9)--(b4=9)==(b4=8)--(c4=8)==(c8=8) forbids {c8=5}
(e1=9)==(i1=9)--(i9=9)==(i9=4)--(i3=4)==(i3=5)--(e3=5)==(e3=1) forbids {e1=1}
(h2=4)==(h2=9)--(h7=9)==(b7=9)--(b4=9)==(b4=8)--(b2=8)==(a2=8) forbids {a2=4}

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 i8=6%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).