05/12/10 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : c8=8%col, a7=9%block, c5=9%block, f7=8%block, d2=8%block, g1=8%col lead to 27 filled cells.

2) Now :
Look at only possibles d6=1,d4=1 in their col. They forbid{e4=1, e6=1}.
Look at only possibles b7=2,b8=2 in their block. They forbid{b6=2, b5=2, b3=2, b2=2}.
Look at only possibles c2=1,c1=1 in their col. They forbid{b2=1}.
Look at only possibles b4=8,b6=8 in their col. Whether b6=8 (in which case b4=1%block) or b4=8, in both cases, we have no more {b4=4, b4=5}.
Look at only possibles b6=8,b4=8 in their col. Whether b4=8 (in which case b6=1%block) or b6=8, in both cases, we have no more {b6=2, b6=4}.
Now easy fillings up to 29 filled cells. (If needed, f5=2%row, a6=2%row)

3) Look at only possibles f6=6,e6=6 in their block. They forbid{i6=6, h6=6}.
Look at only possibles g2=4,g5=4 in their col. Whether g5=4 (in which case a4=4%block) or g2=4, in both cases, we have no more {a2=4}.

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 e1=7,c1=7 in their row. Whether e1=7 (in which case f2=6%cell,a2=5%cell) or c1=7 (in which case c9=4%cell,a4=4%col), in both cases, we have no more {a4=5}.
Now easy fillings up to 50 filled cells. (If needed, b5=5%block, a4=4%block, d5=3%cell, h4=3%block, d3=9%cell, a2=5%block, f4=7%cell, d6=1%cell, b6=8%cell, i4=8%block, d4=5%cell, e4=9%cell, b4=1%cell, f2=6%cell, e6=6%block, f6=4%block, f8=3%col, d8=7%col, e7=1%cell, g9=1%row, g7=3%block)

5)Look at only possibles h6=7,h6=9 in their cell. Whether h6=7 (in which case h7=2%cell) or h6=9 (in which case i8=9%col), in both cases, we have no more {i8=2}.
Look at only possibles c9=7,c9=4 in their cell. Whether c9=7 (in which case e1=7%row) or c9=4 (in which case c3=2%cell,e3=3%cell), in both cases, we have no more {e1=3}.
Look at only possibles b7=7,h7=7 in their row. Whether b7=7 (in which case b2=4%cell) or h7=7 (in which case h6=9%cell,g2=9%row), in both cases, we have no more {g2=4}.
Now easy fillings up to 81 filled cells. (If needed, e3=3%block, a1=3%block, b9=3%block, b3=6%block, a9=6%block, g2=9%cell, g5=4%col, h5=6%block, i1=6%block, g8=6%block, i3=2%col, h1=1%cell, c3=4%cell, h2=4%block, b2=7%cell, c1=2%cell, e1=7%cell, c9=7%block, h7=7%block, h8=2%block, b7=2%block, i8=9%block, h6=9%block, i6=7%block, i9=5%block, e8=5%block, h3=5%block, e9=4%block, b8=4%block, e2=2%cell, c2=1%cell)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 1 in column Cd, at 2 in block Bb8, at 1 in column Cc, at 18 in b4b6, eliminate some possibles. Then easy to 29 filled.
3) Looking at 6 in Be5, eliminate some possibles. Looking at 4 in Cg, eliminate a2=4 (beware, pentagon).
4) Looking at 7 in R1, eliminate a4=5 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at h6, eliminate i8=2 (heptagon). Looking at c9, eliminate e1=3 (nonagon). Looking at 7s in R7, eliminate g2=4(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

The 11-agon is needed here :

around 27 filled
(d6=1)==(d4=1) forbids {e4=1, e6=1}
(b7=2)==(b8=2) forbids {b5=2, b6=2, b3=2, b2=2}
(c2=1)==(c1=1) forbids {b2=1}
(b4=8)==(b6=8)--(b6=1)==(b4=1) forbids {b4=5, b4=4}
(b6=8)==(b4=8)--(b4=1)==(b6=1) forbids {b6=4, b6=2}
around 29 filled
(f6=6)==(e6=6) forbids {i6=6, h6=6}
(g2=4)==(g5=4)--(b5=4)==(a4=4) forbids {a2=4}
(a2=5)==(a2=6)--(f2=6)==(f2=7)--(e1=7)==(c1=7)--(c9=7)==(c9=4)--(a9=4)==(a4=4) forbids {a4=5}
around 50 filled
(h7=2)==(h7=7)--(h6=7)==(h6=9)--(i6=9)==(i8=9) forbids {i8=2}
(e1=7)==(c1=7)--(c9=7)==(c9=4)--(c3=4)==(c3=2)--(e3=2)==(e3=3) forbids {e1=3}
(b2=4)==(b2=7)--(b7=7)==(h7=7)--(h6=7)==(h6=9)--(h2=9)==(g2=9) forbids {g2=4}

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 c8=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 05/09/20's proof).