05/11/26 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h8=1%row, i6=1%col, b4=1%row, e8=8%row, i7=8%row, c2=1%row lead to 27 filled cells.

2) Now :
Look at only possibles f4=6,d4=6 in their row. They forbid{d5=6, f5=6}.
Look at only possibles i9=4,h9=4 in their block. They forbid{e9=4, f9=4, a9=4, b9=4}.
Look at only possibles b8=6,c8=6 in their row. They forbid{b9=6}.
Look at only possibles d9=1,f9=1 in their row. Whether f9=1 (in which case d9=6%block) or d9=1, in both cases, we have no more {d9=9, d9=3}.
Look at only possibles f9=1,d9=1 in their row. Whether d9=1 (in which case f9=6%block) or f9=1, in both cases, we have no more {f9=4, f9=3}.
Now easy fillings up to 29 filled cells. (If needed, e6=4%col, f7=4%col)

3) Look at only possibles f6=7,f5=7 in their block. They forbid{f3=7, f1=7}.
Look at only possibles b2=3,e2=3 in their row. Whether e2=3 (in which case d7=3%block) or b2=3, in both cases, we have no more {b7=3}.

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 c5=2,c8=2 in their col. Whether c5=2 (in which case b6=7%cell,b7=9%cell) or c8=2 (in which case g8=3%cell,d7=3%row), in both cases, we have no more {d7=9}.
Now easy fillings up to 50 filled cells. (If needed, d7=3%cell, e9=9%block, e4=5%cell, a4=8%cell, d6=2%cell, f4=6%cell, d4=9%cell, d5=8%cell, d1=5%cell, d3=1%cell, h4=2%cell, b6=7%cell, b7=9%cell, f6=3%cell, f5=7%cell, h6=5%cell, i5=6%cell, i2=5%cell, f9=1%col, d9=6%col, g2=6%col)

5)Look at only possibles f1=2,f1=8 in their cell. Whether f1=2 (in which case i1=4%cell) or f1=8 (in which case h3=8%row), in both cases, we have no more {h3=4}.
Look at only possibles g8=2,g8=3 in their cell. Whether g8=2 (in which case c5=2%col) or g8=3 (in which case a8=4%cell,a5=5%cell), in both cases, we have no more {c5=5}.
Look at only possibles i9=2,i1=2 in their col. Whether i9=2 (in which case b9=3%cell) or i1=2 (in which case f1=8%cell,b2=8%col), in both cases, we have no more {b2=3}.
Now easy fillings up to 81 filled cells. (If needed, b2=8%cell, h2=7%cell, e2=3%cell, e1=7%cell, c7=5%col, g7=7%cell, a5=5%col, g9=5%col, a9=7%row, c3=7%row, a3=4%row, c1=6%cell, b1=3%cell, a1=9%cell, b8=6%col, b5=4%col, c5=2%cell, b9=2%col, c8=4%col, g8=2%row, g3=9%cell, h3=8%cell, h1=4%cell, i1=2%cell, f1=8%cell, f3=2%cell, g5=3%cell, h5=9%cell, h9=3%col, i9=4%col, a8=3%cell)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 6 in row R4, at 4 in block Bh8, at 6 in row R8, at 16 in d9f9, eliminate some possibles. Then easy to 29 filled.
3) Looking at 7 in Be5, eliminate some possibles. Looking at 3 in R2, eliminate b7=3 (beware, pentagon).
4) Looking at 2 in Cc, eliminate d7=9 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at f1, eliminate h3=4 (heptagon). Looking at g8, eliminate c5=5 (nonagon). Looking at 2s in Ci, eliminate b2=3(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(f4=6)==(d4=6) forbids {d5=6, f5=6}
(i9=4)==(h9=4) forbids {f9=4, e9=4, b9=4, a9=4}
(b8=6)==(c8=6) forbids {b9=6}
(d9=1)==(f9=1)--(f9=6)==(d9=6) forbids {d9=9, d9=3}
(f9=1)==(d9=1)--(d9=6)==(f9=6) forbids {f9=3, f9=4}
around 29 filled
(f6=7)==(f5=7) forbids {f1=7, f3=7}
(b2=3)==(e2=3)--(e9=3)==(d7=3) forbids {b7=3}
(b7=9)==(b7=7)--(b6=7)==(b6=2)--(c5=2)==(c8=2)--(g8=2)==(g8=3)--(g7=3)==(d7=3) forbids {d7=9}
around 50 filled
(i1=4)==(i1=2)--(f1=2)==(f1=8)--(f3=8)==(h3=8) forbids {h3=4}
(c5=2)==(c8=2)--(g8=2)==(g8=3)--(a8=3)==(a8=4)--(a5=4)==(a5=5) forbids {c5=5}
(b9=3)==(b9=2)--(i9=2)==(i1=2)--(f1=2)==(f1=8)--(b1=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=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).