05/11/04 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : c3=6%row, a4=6%col, a1=7%block, e3=7%block, g6=6%block, h7=6%row lead to 27 filled cells.

2) Now :
Look at only possibles d6=4,f6=4 in their row. They forbid{f5=4, d5=4}.
Look at only possibles a2=5,c2=5 in their block. They forbid{i2=5, g2=5, d2=5, e2=5}.
Look at only possibles g3=4,h3=4 in their row. They forbid{g2=4}.
Look at only possibles f2=6,d2=6 in their row. Whether d2=6 (in which case f2=4%block) or f2=6, in both cases, we have no more {f2=8, f2=2}.
Look at only possibles d2=6,f2=6 in their row. Whether f2=6 (in which case d2=4%block) or d2=6, in both cases, we have no more {d2=5, d2=2}.
Now easy fillings up to 29 filled cells. (If needed, e4=5%col, d1=5%col)

3) Look at only possibles d4=3,d5=3 in their block. They forbid{d9=3, d8=3}.
Look at only possibles g7=2,e7=2 in their row. Whether e7=2 (in which case f1=2%block) or g7=2, in both cases, we have no more {g1=2}.

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 h5=1,h3=1 in their col. Whether h5=1 (in which case g4=3%cell,g1=8%cell) or h3=1 (in which case b3=2%cell,f1=2%row), in both cases, we have no more {f1=8}.
Now easy fillings up to 50 filled cells. (If needed, f1=2%cell, e2=8%cell, e6=9%cell, f8=9%col, i6=7%cell, f4=1%cell, c6=1%row, d6=4%cell, f2=4%col, f9=6%col, f5=7%col, f6=8%col, d2=6%col, g4=3%cell, d5=3%col, d4=2%cell, c4=9%cell, a5=4%cell, b7=4%col, a7=9%cell, g1=8%cell)

5)Look at only possibles d8=1,d8=7 in their cell. Whether d8=1 (in which case a8=5%cell) or d8=7 (in which case c9=7%row), in both cases, we have no more {c9=5}.
Look at only possibles b3=1,b3=2 in their cell. Whether b3=1 (in which case h5=1%col) or b3=2 (in which case i3=5%cell,i5=9%cell), in both cases, we have no more {h5=9}.
Look at only possibles a2=1,a8=1 in their col. Whether a2=1 (in which case g2=2%cell) or a8=1 (in which case d8=7%cell,g7=7%col), in both cases, we have no more {g7=2}.
Now easy fillings up to 81 filled cells. (If needed, h1=9%col, i5=9%col, b2=9%col, g7=7%cell, c7=3%cell, e8=3%col, h9=3%col, i2=3%col, b1=3%col, e7=2%col, i9=5%row, i8=8%col, i3=2%col, c2=2%row, c8=5%col, c9=7%col, c5=8%col, b9=8%col, d9=1%row, a8=1%row, g2=1%row, h5=1%row, b3=1%row, d8=7%col, h3=5%col, g5=5%col, a2=5%col, h8=4%col, g3=4%col, g8=2%col, b5=2%col)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 4 in row R6, at 5 in block Bb2, at 4 in row R3, at 46 in d2f2, eliminate some possibles. Then easy to 29 filled.
3) Looking at 3 in Be5, eliminate some possibles. Looking at 2 in R7, eliminate g1=2 (beware, pentagon).
4) Looking at 1 in Ch, eliminate h5=9 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at d8, eliminate c9=5 (heptagon). Looking at b3, eliminate h5=9 (nonagon). Looking at 1s in Ca, eliminate g7=2(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(d6=4)==(f6=4) forbids {d5=4, f5=4}
(a2=5)==(c2=5) forbids {i2=5, g2=5, e2=5, d2=5}
(g3=4)==(h3=4) forbids {g2=4}
(f2=6)==(d2=6)--(d2=4)==(f2=4) forbids {f2=2, f2=8}
(d2=6)==(f2=6)--(f2=4)==(d2=4) forbids {d2=2, d2=5}
around 29 filled
(d4=3)==(d5=3) forbids {d9=3, d8=3}
(g7=2)==(e7=2)--(e2=2)==(f1=2) forbids {g1=2}
(g1=8)==(g1=3)--(g4=3)==(g4=1)--(h5=1)==(h3=1)--(b3=1)==(b3=2)--(b1=2)==(f1=2) forbids {f1=8}
around 50 filled
(a8=5)==(a8=1)--(d8=1)==(d8=7)--(d9=7)==(c9=7) forbids {c9=5}
(h5=1)==(h3=1)--(b3=1)==(b3=2)--(i3=2)==(i3=5)--(i5=5)==(i5=9) forbids {h5=9}
(g2=2)==(g2=1)--(a2=1)==(a8=1)--(d8=1)==(d8=7)--(g8=7)==(g7=7) forbids {g7=2}

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 c3=6%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).