05/12/02 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : b2=7%row, b8=8%cell, c8=7%cell, a6=7%block, a8=4%cell, e8=1%cell, e2=9%cell, h8=3%cell, h2=1%cell lead to 28 filled cells.

2) Now :
Look at only possibles c6=8,c4=8 in their block. They forbid{c3=8, c1=8}.
Look at only possibles a7=5,a7=6 in their cell. Whether a7=6 (in which case a9=5%cell) or a7=5, in both cases, we have no more {a1=5, b7=5, a5=5, a3=5, c9=5}.
Look at only possibles a7=6,a7=5 in their cell. Whether a7=5 (in which case a9=6%cell) or a7=6, in both cases, we have no more {a1=6, a5=6, a3=6, c9=6, b7=6}.
Look at only possibles g2=3,g2=8 in their cell. Whether g2=8 (in which case i2=3%cell) or g2=3, in both cases, we have no more {g1=3, i1=3}.
Look at only possibles g2=8,g2=3 in their cell. Whether g2=3 (in which case i2=8%cell) or g2=8, in both cases, we have no more {i3=8, i1=8, g1=8}.

3) Now :
Look at only possibles i1=9,i1=6 in their cell. Whether i1=6 (in which case i3=9%cell) or i1=9, in both cases, we have no more {h3=9, i9=9, i7=9, i5=9, g1=9}.
Look at only possibles i1=6,i1=9 in their cell. Whether i1=9 (in which case i3=6%cell) or i1=6, in both cases, we have no more {i4=6, i5=6, h3=6}.
Look at only possibles a5=1,a5=3 in their cell. Whether a5=3 (in which case i5=1%cell) or a5=1, in both cases, we have no more {g5=1, f5=1, d5=1, c5=1}.
Look at only possibles a5=3,a5=1 in their cell. Whether a5=1 (in which case i5=3%cell) or a5=3, in both cases, we have no more {f5=3, d5=3, g5=3}.

4) Look at only possibles e1=2,e4=2 in their col. Whether e1=2 (in which case g1=5%cell) or e4=2 (in which case f5=5%cell), in both cases, we have no more {g5=5}.
Now easy fillings up to 41 filled cells. (If needed, g5=9%cell, d6=9%block, d5=6%cell, e9=6%block, a7=6%block, a9=5%block, f7=5%block, f5=2%cell, e1=2%col, g1=5%cell, c5=5%cell, b3=5%block, h3=2%cell)

5)Look at only possibles g6=1,g6=3 in their cell. Whether g6=3 (in which case i5=1%cell) or g6=1, in both cases, we have no more {g4=1, i4=1}.
Look at only possibles g6=3,g6=1 in their cell. Whether g6=1 (in which case i5=3%cell) or g6=3, in both cases, we have no more {i4=3, g4=3}.
Now easy fillings up to 47 filled cells. (If needed, g4=7%cell, i4=4%cell, e6=4%col, e4=5%block, h6=5%block, h4=6%block)

6)Look at only possibles i5=3,a5=3 in their row. Whether i5=3 (in which case g6=1%cell) or a5=3 (in which case b4=2%cell,g7=2%row), in both cases, we have no more {g7=1}.
Now easy fillings up to 81 filled cells. (If needed, i7=1%row, i9=7%block, g9=8%block, i2=8%block, d7=7%block, f3=7%block, g2=3%block, i5=3%block, a5=1%row, a1=3%col, a3=8%block, g7=2%col, b4=2%col, b6=3%block, c6=6%block, c4=8%block, f6=8%block, d1=8%block, b1=6%block, i3=6%block, i1=9%block, c3=9%block, b7=9%block, h9=9%block, h7=4%block, f1=4%block, d9=4%block, f9=3%block, d4=3%block, c1=1%cell, g6=1%row, f4=1%row, d3=1%row, c9=2%col)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 8 in Bb5, at 83 in i2g2, at 65 in a7a9, eliminate some possibles.
3) looking at 69 in i1i3, at 13 in a5i5, eliminate some possibles.
4) looking at 2 in Ce, eliminate g5=5 (beware, heptagon). Now easy to 41 filled.
5)looking at 36 in g6i5, eliminate some possibles. Now easy to 47 filled.
6) looking at 3 in R5, eliminate g7=1 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(c6=8)==(c4=8) forbids {c3=8, c1=8}
(a7=5)==(a7=6)--(a9=6)==(a9=5) forbids {a1=5, b7=5, c9=5, a5=5, a3=5}
(a7=6)==(a7=5)--(a9=5)==(a9=6) forbids {a1=6, a3=6, a5=6, c9=6, b7=6}
(g2=3)==(g2=8)--(i2=8)==(i2=3) forbids {g1=3, i1=3}
(g2=8)==(g2=3)--(i2=3)==(i2=8) forbids {i1=8, g1=8, i3=8}
(i1=9)==(i1=6)--(i3=6)==(i3=9) forbids {h3=9, i7=9, g1=9, i5=9, i9=9}
(i1=6)==(i1=9)--(i3=9)==(i3=6) forbids {i5=6, h3=6, i4=6}
(a5=1)==(a5=3)--(i5=3)==(i5=1) forbids {f5=1, c5=1, g5=1, d5=1}
(a5=3)==(a5=1)--(i5=1)==(i5=3) forbids {d5=3, g5=3, f5=3}
(g1=5)==(g1=2)--(e1=2)==(e4=2)--(f5=2)==(f5=5) forbids {g5=5}
Around 41 filled :
(g6=1)==(g6=3)--(i5=3)==(i5=1) forbids {g4=1, i4=1}
(g6=3)==(g6=1)--(i5=1)==(i5=3) forbids {i4=3, g4=3}
Around 47 filled :
(g6=1)==(g6=3)--(i5=3)==(a5=3)--(b4=3)==(b4=2)--(b7=2)==(g7=2) forbids {g7=1}

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 h8=3%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).