05/11/09 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : i1=9%row, e1=4%row, c1=1%row, a9=8%row, b1=8%row, a1=3%row, c6=8%col, e9=6%cell, i9=4%cell lead to 28 filled cells.

2) Now :
Look at only possibles b6=3,b4=3 in their block. They forbid{b7=3, b8=3}.
Look at only possibles c2=5,c2=2 in their cell. Whether c2=2 (in which case c3=5%cell) or c2=5, in both cases, we have no more {b3=5, a2=5, c8=5, c5=5, c7=5}.
Look at only possibles c2=2,c2=5 in their cell. Whether c2=5 (in which case c3=2%cell) or c2=2, in both cases, we have no more {c7=2, b3=2, a2=2, c8=2, c5=2}.
Look at only possibles h9=9,h9=3 in their cell. Whether h9=3 (in which case g9=9%cell) or h9=9, in both cases, we have no more {g7=9, h7=9}.
Look at only possibles h9=3,h9=9 in their cell. Whether h9=9 (in which case g9=3%cell) or h9=3, in both cases, we have no more {g8=3, h7=3, g7=3}.

3) Now :
Look at only possibles g7=6,g7=2 in their cell. Whether g7=2 (in which case g8=6%cell) or g7=6, in both cases, we have no more {i8=6, g5=6, h7=6, g2=6, g3=6}.
Look at only possibles g7=2,g7=6 in their cell. Whether g7=6 (in which case g8=2%cell) or g7=2, in both cases, we have no more {g4=2, g5=2, i8=2}.
Look at only possibles c5=4,c5=9 in their cell. Whether c5=9 (in which case g5=4%cell) or c5=4, in both cases, we have no more {d5=4, b5=4, h5=4, f5=4}.
Look at only possibles c5=9,c5=4 in their cell. Whether c5=4 (in which case g5=9%cell) or c5=9, in both cases, we have no more {h5=9, f5=9, d5=9}.

4) Look at only possibles e7=7,e4=7 in their col. Whether e7=7 (in which case h7=5%cell) or e4=7 (in which case d5=5%cell), in both cases, we have no more {h5=5}.
Now easy fillings up to 41 filled cells. (If needed, h5=6%cell, f6=6%col, f5=2%cell, c2=2%row, d2=5%row, b5=5%row, c3=5%row, e3=2%row, d5=7%row, e7=7%col, i8=7%row, a8=5%row, h7=5%row)

5)Look at only possibles h6=4,h6=9 in their cell. Whether h6=9 (in which case g5=4%cell) or h6=4, in both cases, we have no more {g4=4, h4=4}.
Look at only possibles h6=9,h6=4 in their cell. Whether h6=4 (in which case g5=9%cell) or h6=9, in both cases, we have no more {h4=9, g4=9}.
Now easy fillings up to 47 filled cells. (If needed, h4=8%cell, g4=1%cell, e6=1%col, i6=5%row, e4=5%row, i4=2%block)

6)Look at only possibles g5=9,c5=9 in their row. Whether g5=9 (in which case h6=4%cell) or c5=9 (in which case a4=7%cell,h2=7%row), in both cases, we have no more {h2=4}.
Now easy fillings up to 81 filled cells. (If needed, g2=4%row, c5=4%row, c7=9%col, g3=8%col, h3=3%row, g9=3%row, g5=9%col, h9=9%col, d8=8%col, f2=8%col, i2=1%row, h2=7%col, a4=7%col, a6=9%col, b6=2%row, d6=3%row, h6=4%row, f7=3%row, c8=3%row, b4=3%row, g8=2%row, a7=2%row, d7=1%row, b7=4%row, f8=4%row, d4=4%row, f3=1%row, f4=9%col, d3=9%col, b3=7%col, i3=6%col, g7=6%col, b8=6%row, a2=6%row)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 3 in Bb5, at 52 in c2c3, at 39 in g9h9, eliminate some possibles.
3) looking at 26 in g7g8, at 49 in c5g5, eliminate some possibles.
4) looking at 7 in Ce, eliminate h5=5 (beware, heptagon). Now easy to 41 filled.
5)looking at 49 in g5h6, eliminate some possibles. Now easy to 47 filled.
6) looking at 9 in R5, eliminate c3=7 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(b6=3)==(b4=3) forbids {b7=3, b8=3}
(c2=5)==(c2=2)--(c3=2)==(c3=5) forbids {c5=5, c7=5, c8=5, b3=5, a2=5}
(c2=2)==(c2=5)--(c3=5)==(c3=2) forbids {c8=2, a2=2, c5=2, c7=2, b3=2}
(h9=9)==(h9=3)--(g9=3)==(g9=9) forbids {g7=9, h7=9}
(h9=3)==(h9=9)--(g9=9)==(g9=3) forbids {h7=3, g7=3, g8=3}
(g7=6)==(g7=2)--(g8=2)==(g8=6) forbids {g2=6, g3=6, g5=6, h7=6, i8=6}
(g7=2)==(g7=6)--(g8=6)==(g8=2) forbids {g5=2, g4=2, i8=2}
(c5=4)==(c5=9)--(g5=9)==(g5=4) forbids {b5=4, h5=4, f5=4, d5=4}
(c5=9)==(c5=4)--(g5=4)==(g5=9) forbids {f5=9, h5=9, d5=9}
(h7=5)==(h7=7)--(e7=7)==(e4=7)--(d5=7)==(d5=5) forbids {h5=5}
Around 41 filled :
(h6=4)==(h6=9)--(g5=9)==(g5=4) forbids {h4=4, g4=4}
(h6=9)==(h6=4)--(g5=4)==(g5=9) forbids {h4=9, g4=9}
Around 47 filled :
(h6=4)==(h6=9)--(g5=9)==(c5=9)--(a4=9)==(a4=7)--(a2=7)==(h2=7) forbids {h2=4}

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 i1=9%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).