05/11/08 tough puzzle from sudoku.com.au

A complete proof

Filled cells are left blank on diagrams for easier reading.
First eliminations : [c5=5%block, b8=5%block, h2=5%block] lead to 23 filled cells.

Look at only possibles b3=1,b1=1 in their col. They forbid{a1=1, a2=1, c1=1, c2=1}.
Look at only possibles h7=9,h9=9 in their col. They forbid{i9=9, g8=9, g9=9, i8=9}.
Now easy fillings up to 46 filled cells. (If needed, d8=9%row, e1=9%block, g2=9%block, f2=1%row, e9=1%block, c8=1%block, e6=5%col, d6=8%row, c2=8%row, b4=8%block, a2=2%block, d2=7%row, a6=4%row, a5=1%block, a4=9%block, b7=9%block, h9=9%block, i5=9%block, c4=7%block, a1=7%block, c1=3%block, c7=2%row, c9=4%block)

Look at only possibles a7=3,a7=6 in their cell. Whether a7=6 (in which case h7=3%cell) or a7=3, in both cases, we have no more {e7=3, f7=3, d7=3}.
Look at only possibles a7=6,a7=3 in their cell. Whether a7=3 (in which case h7=6%cell) or a7=6, in both cases, we have no more {d7=6, e7=6, f7=6}.
Now easy fillings up to 51 filled cells. (If needed, d7=4%cell, f5=4%block, e5=7%block, f7=7%block, e7=8%row)

Look at only possibles d9=3,d3=3 in their col. Whether d9=3 (in which case d1=5%col) or d3=3 (in which case e3=6%cell), in both cases, we have no more {d1=6}.
Look at only possibles e4=3,f4=3 in their row. Whether e4=3 (in which case e3=6%cell) or f4=3 (in which case f8=6%cell), in both cases, we have no more {f1=6, f3=6}.
Now easy fillings up to 54 filled cells. (If needed, d1=5%cell, f9=5%block, f1=8%cell)

Now, Look at only possibles e3=6,d3=6 in their block. They forbid{h3=6, i3=6, g3=6}.
Now easy fillings up to 81 filled cells. (If needed, d1=5%cell, f9=5%block, f1=8%cell)

Short hints for a proof

One of the two heptagons is highlighted.

1) First eliminations to 23 filled cells.
2) Look at only possibles 9 in their column Ch. Look at only possibles 1 in their column Cb. Remove some possibles. Now easy eliminations take place again, leading to 46 filled cells.
3) Look at only possibles 36 in a7h7. Remove some possibles. Then simple eliminations take place again, up to 51 filled cells.
4) Looking at only possibles 3 in their column Cd, remove d1=6 ; looking at only possibles 3 in their row R4, remove f1=6,f3=6 (beware, heptagons).
5) Look then at only possibles 6 in their row R3. Remove some possibles. From here, simple eliminations lead to unique solution.

Forbidding-chain-like proof

One of the two heptagons is highlighted.

around 23 filled
(b3=1)==(b1=1) forbids {a2=1, c1=1, c2=1, a1=1}
(h7=9)==(h9=9) forbids {g9=9, i8=9, i9=9, g8=9}
around 46 filled
(a7=3)==(a7=6)--(h7=6)==(h7=3) forbids {d7=3, e7=3, f7=3}
(a7=6)==(a7=3)--(h7=3)==(h7=6) forbids {d7=6, e7=6, f7=6}
around 51 filled
(d1=5)==(d9=5)--(d9=3)==(d3=3)--(e3=3)==(e3=6) forbids {d1=6}
(e3=6)==(e3=3)--(e4=3)==(f4=3)--(f8=3)==(f8=6) forbids {f1=6, f3=6}
around 54 filled
(e3=6)==(d3=6) forbids {h3=6, g3=6, i3=6}

Equivalence with 09/24

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/24 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/24 was g5=1%block; in today's it becomes c5=5%block. 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 05/09/24's proof).