05/11/30 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : b1=2%row, h9=6%row, h1=9%cell, g1=6%cell, i1=1%cell, e1=5%cell, i4=6%col, e9=3%cell, b9=5%cell lead to 28 filled cells.

2) Now :
Look at only possibles g4=9,g6=9 in their block. They forbid{g7=9, g8=9}.
Look at only possibles i3=4,i3=7 in their cell. Whether i3=7 (in which case i2=4%cell) or i3=4, in both cases, we have no more {i8=4, i7=4, g2=4, i5=4, h3=4}.
Look at only possibles i3=7,i3=4 in their cell. Whether i3=4 (in which case i2=7%cell) or i3=7, in both cases, we have no more {g2=7, i5=7, i7=7, i8=7, h3=7}.
Look at only possibles c9=2,c9=9 in their cell. Whether c9=9 (in which case a9=2%cell) or c9=2, in both cases, we have no more {c8=2, a8=2}.
Look at only possibles c9=9,c9=2 in their cell. Whether c9=2 (in which case a9=9%cell) or c9=9, in both cases, we have no more {a8=9, a7=9, c8=9}.

3) Now :
Look at only possibles a8=3,a8=7 in their cell. Whether a8=7 (in which case a7=3%cell) or a8=3, in both cases, we have no more {a3=3, a5=3, a2=3, c8=3, b7=3}.
Look at only possibles a8=7,a8=3 in their cell. Whether a8=3 (in which case a7=7%cell) or a8=7, in both cases, we have no more {a6=7, b7=7, a5=7}.
Look at only possibles i5=5,i5=2 in their cell. Whether i5=2 (in which case a5=5%cell) or i5=5, in both cases, we have no more {g5=5, f5=5, c5=5, d5=5}.
Look at only possibles i5=2,i5=5 in their cell. Whether i5=5 (in which case a5=2%cell) or i5=2, in both cases, we have no more {d5=2, c5=2, f5=2}.

4) Look at only possibles e8=8,e6=8 in their col. Whether e8=8 (in which case c8=4%cell) or e6=8 (in which case f5=4%cell), in both cases, we have no more {c5=4}.
Now easy fillings up to 41 filled cells. (If needed, c5=3%cell, d5=7%cell, e2=7%block, i3=7%block, i2=4%cell, d4=3%row, f3=4%block, f5=8%cell, e8=8%block, b7=8%block, c8=4%cell, g5=4%cell, h7=4%col)

5)Look at only possibles c4=5,c4=2 in their cell. Whether c4=2 (in which case a5=5%cell) or c4=5, in both cases, we have no more {a6=5, c6=5}.
Look at only possibles c4=2,c4=5 in their cell. Whether c4=5 (in which case a5=2%cell) or c4=2, in both cases, we have no more {c6=2, a6=2}.
Now easy fillings up to 47 filled cells. (If needed, c6=6%cell, a6=1%cell, e6=4%cell, b6=7%cell, e4=1%cell, b4=4%cell)

6)Look at only possibles a5=2,i5=2 in their row. Whether a5=2 (in which case c4=5%cell) or i5=2 (in which case h6=8%cell,c3=8%row), in both cases, we have no more {c3=5}.
Now easy fillings up to 81 filled cells. (If needed, c3=8%cell, g2=8%block, h6=8%block, h3=3%cell, b3=1%cell, d3=6%cell, f7=6%block, a3=5%cell, a5=2%cell, i5=5%cell, i7=9%cell, d7=5%cell, g6=9%cell, f4=9%block, d8=9%block, f8=1%cell, d6=2%cell, f6=5%cell, g4=7%cell, h8=7%block, a7=7%block, a8=3%cell, g7=3%cell, h4=2%cell, c4=5%cell, f2=2%cell, d2=1%cell, c2=9%cell, a9=9%block, b2=3%cell, a2=6%cell, c9=2%row, i8=2%row, g8=5%col)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 9 in Bg8, at 74 in i2i3, at 29 in a9c9, eliminate some possibles.
3) looking at 37 in a7a8, at 25 in a5i5, eliminate some possibles.
4) looking at 8 in col Ce, eliminate c5=4 (beware, heptagon). Now easy to 41 filled.
5)looking at 25 in c4c5, eliminate some possibles. Now easy to 47 filled.
6) looking at 2 in Row R5, eliminate c3=5 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(g4=9)==(g6=9) forbids {g7=9, g8=9}
(i3=4)==(i3=7)--(i2=7)==(i2=4) forbids {i8=4, i7=4, i5=4, g2=4, h3=4}
(i3=7)==(i3=4)--(i2=4)==(i2=7) forbids {i7=7, h3=7, g2=7, i8=7, i5=7}
(c9=2)==(c9=9)--(a9=9)==(a9=2) forbids {a8=2, c8=2}
(c9=9)==(c9=2)--(a9=2)==(a9=9) forbids {a7=9, a8=9, c8=9}
(a8=3)==(a8=7)--(a7=7)==(a7=3) forbids {a5=3, a2=3, a3=3, c8=3, b7=3}
(a8=7)==(a8=3)--(a7=3)==(a7=7) forbids {b7=7, a5=7, a6=7}
(i5=5)==(i5=2)--(a5=2)==(a5=5) forbids {c5=5, f5=5, d5=5, g5=5}
(i5=2)==(i5=5)--(a5=5)==(a5=2) forbids {f5=2, d5=2, c5=2}
(c8=4)==(c8=8)--(e8=8)==(e6=8)--(f5=8)==(f5=4) forbids {c5=4}
Around 41 filled :
(c4=5)==(c4=2)--(a5=2)==(a5=5) forbids {c6=5, a6=5}
(c4=2)==(c4=5)--(a5=5)==(a5=2) forbids {c6=2, a6=2}
Around 47 filled :
(c4=5)==(c4=2)--(a5=2)==(i5=2)--(h6=2)==(h6=8)--(h3=8)==(c3=8) forbids {c3=5}

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 b1=2%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 05/09/12's proof).