05/12/11 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h1=9%col, h5=7%col, h8=8%col, b5=4%cell, b1=7%cell, b9=3%col, h9=5%cell, h7=3%cell, d8=3%block lead to 28 filled cells.

2) Now :
Look at only possibles d7=5,f7=5 in their block. They forbid{c7=5, a7=5}.
Look at only possibles i8=6,i8=2 in their cell. Whether i8=2 (in which case g8=6%cell) or i8=6, in both cases, we have no more {i9=6, g7=6, e8=6, a8=6, c8=6}.
Look at only possibles i8=2,i8=6 in their cell. Whether i8=6 (in which case g8=2%cell) or i8=2, in both cases, we have no more {i9=2, e8=2, c8=2, a8=2, g7=2}.
Look at only possibles b3=9,b3=5 in their cell. Whether b3=5 (in which case b2=9%cell) or b3=9, in both cases, we have no more {c3=9, c2=9}.
Look at only possibles b3=5,b3=9 in their cell. Whether b3=9 (in which case b2=5%cell) or b3=5, in both cases, we have no more {c2=5, c3=5, a2=5}.

3) Now :
Look at only possibles c2=4,c2=2 in their cell. Whether c2=2 (in which case a2=4%cell) or c2=4, in both cases, we have no more {i2=4, c3=4, a1=4, e2=4, g2=4}.
Look at only possibles c2=2,c2=4 in their cell. Whether c2=4 (in which case a2=2%cell) or c2=2, in both cases, we have no more {f2=2, e2=2, a1=2}.
Look at only possibles e8=7,e8=9 in their cell. Whether e8=9 (in which case e2=7%cell) or e8=7, in both cases, we have no more {e6=7, e7=7, e3=7, e4=7}.
Look at only possibles e8=9,e8=7 in their cell. Whether e8=7 (in which case e2=9%cell) or e8=9, in both cases, we have no more {e3=9, e4=9, e6=9}.

4) Look at only possibles c5=1,f5=1 in their row. Whether c5=1 (in which case c3=6%cell) or f5=1 (in which case e6=6%cell), in both cases, we have no more {e3=6}.
Now easy fillings up to 41 filled cells. (If needed, e3=4%cell, d4=4%block, e4=2%cell, g5=2%block, i8=2%block, g8=6%block, i6=6%block, e6=1%cell, c5=1%block, a1=1%block, c3=6%block, e7=6%cell, a9=6%block)

5)Look at only possibles d3=7,d3=9 in their cell. Whether d3=9 (in which case e2=7%cell) or d3=7, in both cases, we have no more {f2=7, f3=7}.
Look at only possibles d3=9,d3=7 in their cell. Whether d3=7 (in which case e2=9%cell) or d3=9, in both cases, we have no more {f2=9, f3=9}.
Now easy fillings up to 47 filled cells. (If needed, f3=3%cell, f2=8%cell, d5=8%row, f5=6%block, d1=6%block, f1=2%block)

6)Look at only possibles e2=9,e8=9 in their col. Whether e2=9 (in which case d3=7%cell) or e8=9 (in which case f9=1%cell,i3=1%col), in both cases, we have no more {i3=7}.
Now easy fillings up to 81 filled cells. (If needed, i2=7%block, d3=7%block, e2=9%block, b3=9%block, b2=5%block, g3=5%block, g2=3%block, i4=3%block, a6=3%block, i3=1%block, g7=1%block, i9=4%block, g1=4%block, i1=8%block, f9=1%block, d9=9%block, c8=9%block, a8=5%block, d7=2%block, f7=5%block, e8=7%block, d6=5%block, c4=5%block, c6=8%block, g4=8%block, g6=9%block, f4=9%block, f6=7%block, a4=7%block, c7=7%block, a7=4%block, c2=4%block, c9=2%block, a2=2%block)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 5 in Be8, at 26 in i8g8, at 59 in b2b3, eliminate some possibles.
3) looking at 24 in a2c2, at 79 in e2e9, eliminate some possibles.
4) looking at 1 in R5, eliminate e3=6 (beware, heptagon). Now easy to 41 filled.
5)looking at 79 in d3e2, eliminate some possibles. Now easy to 47 filled.
6) looking at 9 in Ce, eliminate i3=7 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(d7=5)==(f7=5) forbids {a7=5, c7=5}
(i8=6)==(i8=2)--(g8=2)==(g8=6) forbids {e8=6, g7=6, i9=6, a8=6, c8=6}
(i8=2)==(i8=6)--(g8=6)==(g8=2) forbids {e8=2, g7=2, i9=2, a8=2, c8=2}
(b3=9)==(b3=5)--(b2=5)==(b2=9) forbids {c3=9, c2=9}
(b3=5)==(b3=9)--(b2=9)==(b2=5) forbids {c2=5, c3=5, a2=5}
(c2=4)==(c2=2)--(a2=2)==(a2=4) forbids {e2=4, c3=4, i2=4, g2=4, a1=4}
(c2=2)==(c2=4)--(a2=4)==(a2=2) forbids {f2=2, e2=2, a1=2}
(e8=7)==(e8=9)--(e2=9)==(e2=7) forbids {e7=7, e6=7, e3=7, e4=7}
(e8=9)==(e8=7)--(e2=7)==(e2=9) forbids {e6=9, e3=9, e4=9}
(c3=6)==(c3=1)--(c5=1)==(f5=1)--(e6=1)==(e6=6) forbids {e3=6}
Around 41 filled :
(d3=7)==(d3=9)--(e2=9)==(e2=7) forbids {f2=7, f3=7}
(d3=9)==(d3=7)--(e2=7)==(e2=9) forbids {f3=9, f2=9}
Around 47 filled :
(d3=7)==(d3=9)--(e2=9)==(e8=9)--(f9=9)==(f9=1)--(i9=1)==(i3=1) forbids {i3=7}

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 h1=9%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).