05/11/27 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : b8=8%col, h2=3%col, b2=5%cell, b3=3%cell, b1=1%cell, b5=2%cell, h5=7%cell, h8=2%cell, d1=3%row lead to 28 filled cells.

2) Now :
Look at only possibles d3=5,f3=5 in their block. They forbid{g3=5, i3=5}.
Look at only possibles c1=6,c1=9 in their cell. Whether c1=9 (in which case a1=6%cell) or c1=6, in both cases, we have no more {a3=6, g1=6, e1=6, c2=6, i1=6}.
Look at only possibles c1=9,c1=6 in their cell. Whether c1=6 (in which case a1=9%cell) or c1=9, in both cases, we have no more {c2=9, i1=9, g1=9, e1=9, a3=9}.
Look at only possibles h7=8,h7=5 in their cell. Whether h7=5 (in which case h9=8%cell) or h7=8, in both cases, we have no more {i7=8, i9=8}.
Look at only possibles h7=5,h7=8 in their cell. Whether h7=8 (in which case h9=5%cell) or h7=5, in both cases, we have no more {i9=5, i7=5, g9=5}.

3) Now :
Look at only possibles i9=7,i9=9 in their cell. Whether i9=9 (in which case g9=7%cell) or i9=7, in both cases, we have no more {a9=7, c9=7, i7=7, g8=7, e9=7}.
Look at only possibles i9=9,i9=7 in their cell. Whether i9=7 (in which case g9=9%cell) or i9=9, in both cases, we have no more {f9=9, e9=9, g8=9}.
Look at only possibles e1=2,e1=8 in their cell. Whether e1=8 (in which case e9=2%cell) or e1=2, in both cases, we have no more {e3=2, e4=2, e6=2, e7=2}.
Look at only possibles e1=8,e1=2 in their cell. Whether e1=2 (in which case e9=8%cell) or e1=8, in both cases, we have no more {e4=8, e6=8, e7=8}.

4) Look at only possibles i5=4,f5=4 in their row. Whether i5=4 (in which case i7=6%cell) or f5=4 (in which case e4=6%cell), in both cases, we have no more {e7=6}.
Now easy fillings up to 41 filled cells. (If needed, e7=7%cell, e6=9%cell, d6=7%col, a5=9%row, a1=6%cell, c1=9%cell, c4=6%col, e4=4%cell, e3=6%cell, g8=4%col, g2=6%col, i7=6%col, i5=4%col)

5)Look at only possibles d7=2,d7=8 in their cell. Whether d7=8 (in which case e9=2%cell) or d7=2, in both cases, we have no more {f7=2, f9=2}.
Look at only possibles d7=8,d7=2 in their cell. Whether d7=2 (in which case e9=8%cell) or d7=8, in both cases, we have no more {f7=8, f9=8}.
Now easy fillings up to 47 filled cells. (If needed, f7=3%cell, f9=1%cell, f5=6%cell, d5=1%cell, d8=6%col, f8=9%cell)

6)Look at only possibles e9=8,e1=8 in their col. Whether e9=8 (in which case d7=2%cell) or e1=8 (in which case f2=4%cell,c7=4%col), in both cases, we have no more {c7=2}.
Now easy fillings up to 81 filled cells. (If needed, c9=2%col, c6=3%col, g4=3%col, a9=3%col, a7=5%col, h9=5%col, h7=8%col, a3=4%col, c2=7%cell, a8=7%col, f2=4%col, c7=4%col, e1=2%col, f3=5%cell, d3=9%cell, d2=8%cell, i2=9%cell, g1=5%cell, g6=2%cell, f6=8%cell, a6=1%cell, i6=5%cell, i4=1%cell, f4=2%cell, d4=5%cell, a4=8%cell, g3=7%cell, i3=2%cell, i1=8%cell, e9=8%col, i9=7%col, d7=2%col, c8=1%col, g9=9%cell)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 5 in Be2, at 69 in a1c1, at 58 in h7h9, eliminate some possibles.
3) looking at 79 in g9i9, at 28 in e1e9, eliminate some possibles.
4) looking at 4 in R5, eliminate e7=6 (beware, heptagon). Now easy to 41 filled.
5)looking at 28 in d7e9, eliminate some possibles. Now easy to 47 filled.
6) looking at 8 in Ce, eliminate c7=2 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(d3=5)==(f3=5) forbids {g3=5, i3=5}
(c1=6)==(c1=9)--(a1=9)==(a1=6) forbids {g1=6, i1=6, c2=6, a3=6, e1=6}
(c1=9)==(c1=6)--(a1=6)==(a1=9) forbids {i1=9, g1=9, c2=9, a3=9, e1=9}
(h7=8)==(h7=5)--(h9=5)==(h9=8) forbids {i9=8, i7=8}
(h7=5)==(h7=8)--(h9=8)==(h9=5) forbids {i7=5, g9=5, i9=5}
(i9=7)==(i9=9)--(g9=9)==(g9=7) forbids {g8=7, i7=7, e9=7, c9=7, a9=7}
(i9=9)==(i9=7)--(g9=7)==(g9=9) forbids {e9=9, g8=9, f9=9}
(e1=2)==(e1=8)--(e9=8)==(e9=2) forbids {e3=2, e4=2, e6=2, e7=2}
(e1=8)==(e1=2)--(e9=2)==(e9=8) forbids {e7=8, e6=8, e4=8}
(i7=6)==(i7=4)--(i5=4)==(f5=4)--(e4=4)==(e4=6) forbids {e7=6}
Around 41 filled :
(d7=2)==(d7=8)--(e9=8)==(e9=2) forbids {f9=2, f7=2}
(d7=8)==(d7=2)--(e9=2)==(e9=8) forbids {f7=8, f9=8}
Around 47 filled :
(d7=2)==(d7=8)--(e9=8)==(e1=8)--(f2=8)==(f2=4)--(c2=4)==(c7=4) forbids {c7=2}

Equivalence with 09/12

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/12 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/12 was c7=8%col; in today's it becomes b8=8%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 05/09/12's proof).