05/11/15 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : i1=1%row, g6=1%col, a4=1%block, e1=7%row, g3=7%row, b9=1%row lead to 27 filled cells.

2) Now :
Look at only possibles f4=8,d4=8 in their row. They forbid{f5=8, d5=8}.
Look at only possibles g2=3,i2=3 in their block. They forbid{e2=3, c2=3, f2=3, a2=3}.
Look at only possibles a1=8,b1=8 in their row. They forbid{a2=8}.
Look at only possibles d2=1,f2=1 in their row. Whether f2=1 (in which case d2=8%block) or d2=1, in both cases, we have no more {d2=4, d2=5}.
Look at only possibles f2=1,d2=1 in their row. Whether d2=1 (in which case f2=8%block) or f2=1, in both cases, we have no more {f2=3, f2=5}.
Now easy fillings up to 29 filled cells. (If needed, e6=3%col, f3=3%col)

3) Look at only possibles f6=6,f5=6 in their block. They forbid{f8=6, f7=6}.
Look at only possibles a9=5,e9=5 in their row. Whether e9=5 (in which case d3=5%block) or a9=5, in both cases, we have no more {a3=5}.

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 !. Thanks to Peter who let us know that a 11-agon is called a hendekagon.
Look at only possibles b5=2,b1=2 in their col. Whether b5=2 (in which case a6=6%cell,a3=4%cell) or b1=2 (in which case h1=5%cell,d3=5%row), in both cases, we have no more {d3=4}.
Now easy fillings up to 50 filled cells. (If needed, a3=4%row, e2=4%row, d3=5%block, e4=9%cell, i6=9%row, f6=5%row, f5=6%col, a6=6%row, d6=2%row, i4=2%row, d4=4%row, f4=8%row, d2=8%row, d7=1%col, f2=1%col, c4=7%row, d5=7%row, d8=9%col, g5=8%cell, g9=9%cell, h9=8%col)

5)Look at only possibles f8=2,f8=7 in their cell. Whether f8=2 (in which case g8=3%cell) or f8=7 (in which case i7=7%row), in both cases, we have no more {i7=3}.
Look at only possibles h1=2,h1=5 in their cell. Whether h1=2 (in which case b5=2%col) or h1=5 (in which case c1=3%cell,c5=9%cell), in both cases, we have no more {b5=9}.
Look at only possibles g2=2,g8=2 in their col. Whether g2=2 (in which case a2=5%cell) or g8=2 (in which case f8=7%cell,a9=7%col), in both cases, we have no more {a9=5}.
Now easy fillings up to 81 filled cells. (If needed, c5=9%row, h2=9%row, b3=9%row, h3=6%row, c2=6%row, e9=5%row, e8=6%col, b7=6%col, i9=6%col, a9=7%row, c7=3%row, c8=4%col, c1=5%col, h5=5%col, i2=5%col, a8=5%col, h7=4%col, i5=4%col, i8=3%col, i7=7%col, f8=7%col, g2=3%col, h1=2%col, g8=2%col, f7=2%col, b5=2%col, b1=3%col, a5=3%col, a2=2%col, b8=8%row, a1=8%row)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 8 in row R4, at 3 in block Bg2, at 8 in row R1, at 18 in d2f2, eliminate some possibles. Then easy to 29 filled.
3) Looking at 6 in Be5, eliminate some possibles. Looking at 5 in R9, eliminate a3=5 (beware, pentagon).
4) Looking at 2 in Cb, eliminate d3=4 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at f8, eliminate i7=3 (heptagon). Looking at h1, eliminate b5=9 (nonagon). Looking at 2s in Cg, eliminate a9=5(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(f4=8)==(d4=8) forbids {d5=8, f5=8}
(g2=3)==(i2=3) forbids {c2=3, e2=3, a2=3, f2=3}
(a1=8)==(b1=8) forbids {a2=8}
(d2=1)==(f2=1)--(f2=8)==(d2=8) forbids {d2=4, d2=5}
(f2=1)==(d2=1)--(d2=8)==(f2=8) forbids {f2=5, f2=3}
around 29 filled
(f6=6)==(f5=6) forbids {f8=6, f7=6}
(a9=5)==(e9=5)--(e2=5)==(d3=5) forbids {a3=5}
(a3=4)==(a3=6)--(a6=6)==(a6=2)--(b5=2)==(b1=2)--(h1=2)==(h1=5)--(h3=5)==(d3=5) forbids {d3=4}
around 50 filled
(g8=3)==(g8=2)--(f8=2)==(f8=7)--(f7=7)==(i7=7) forbids {i7=3}
(b5=2)==(b1=2)--(h1=2)==(h1=5)--(c1=5)==(c1=3)--(c5=3)==(c5=9) forbids {b5=9}
(a2=5)==(a2=2)--(g2=2)==(g8=2)--(f8=2)==(f8=7)--(a8=7)==(a9=7) forbids {a9=5}

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 i1=1%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).