05/12/03 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : c9=1%col, g1=2%col, c1=9%cell, c5=8%col, g5=4%cell, g9=8%cell, c2=5%col, c3=2%col, d2=2%block lead to 28 filled cells.

2) Now :
Look at only possibles d3=9,f3=9 in their block. They forbid{i3=9, h3=9}.
Look at only possibles b2=3,b2=6 in their cell. Whether b2=6 (in which case a2=3%cell) or b2=3, in both cases, we have no more {e2=3, b1=3, h2=3, i2=3, a3=3}.
Look at only possibles b2=6,b2=3 in their cell. Whether b2=3 (in which case a2=6%cell) or b2=6, in both cases, we have no more {h2=6, e2=6, a3=6, i2=6, b1=6}.
Look at only possibles g7=1,g7=9 in their cell. Whether g7=9 (in which case g8=1%cell) or g7=1, in both cases, we have no more {i7=1, i8=1}.
Look at only possibles g7=9,g7=1 in their cell. Whether g7=1 (in which case g8=9%cell) or g7=9, in both cases, we have no more {h8=9, i8=9, i7=9}.

3) Now :
Look at only possibles i8=4,i8=6 in their cell. Whether i8=6 (in which case h8=4%cell) or i8=4, in both cases, we have no more {a8=4, h9=4, e8=4, i7=4, b8=4}.
Look at only possibles i8=6,i8=4 in their cell. Whether i8=4 (in which case h8=6%cell) or i8=6, in both cases, we have no more {h9=6, f8=6, e8=6}.
Look at only possibles e2=8,e2=1 in their cell. Whether e2=1 (in which case e8=8%cell) or e2=8, in both cases, we have no more {e4=8, e6=8, e7=8, e3=8}.
Look at only possibles e2=1,e2=8 in their cell. Whether e2=8 (in which case e8=1%cell) or e2=1, in both cases, we have no more {e6=1, e4=1, e7=1}.

4) Look at only possibles i5=7,f5=7 in their row. Whether i5=7 (in which case i7=3%cell) or f5=7 (in which case e6=3%cell), in both cases, we have no more {e7=3}.
Now easy fillings up to 41 filled cells. (If needed, e7=4%cell, d4=4%block, e4=6%cell, a5=6%block, a2=3%cell, b2=6%cell, b6=3%block, e6=7%cell, i5=7%block, h9=7%block, i7=3%block, e3=3%cell, h1=3%block)

5)Look at only possibles d7=8,d7=1 in their cell. Whether d7=1 (in which case e8=8%cell) or d7=8, in both cases, we have no more {f8=8, f7=8}.
Look at only possibles d7=1,d7=8 in their cell. Whether d7=8 (in which case e8=1%cell) or d7=1, in both cases, we have no more {f7=1, f8=1}.
Now easy fillings up to 47 filled cells. (If needed, f7=2%cell, f8=5%cell, d5=5%row, f5=3%block, d9=3%block, f9=6%block)

6)Look at only possibles e8=1,e2=1 in their col. Whether e8=1 (in which case d7=8%cell) or e2=1 (in which case f1=7%cell,b7=7%col), in both cases, we have no more {b7=8}.
Now easy fillings up to 81 filled cells. (If needed, b8=8%block, d7=8%block, e8=1%block, g7=1%block, g8=9%block, a7=9%block, b7=7%block, a3=7%block, f1=7%block, d1=1%block, i2=1%block, h2=9%block, a8=2%block, b4=2%block, h6=2%block, e2=8%cell, b1=4%block, i1=6%cell, h8=6%block, i8=4%block, h3=4%block, i3=8%block, h4=8%block, f6=8%block, f4=1%block, d6=9%block, f3=9%block, i4=9%block, a6=1%block, a9=4%block, b9=5%block, i6=5%block, a4=5%block, d3=6%block)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 9 in Be2, at 19 in g7g8, at 23 in a2b2, eliminate some possibles.
3) looking at 46 in h8i8, at 18 in e2e8, eliminate some possibles.
4) looking at 7 in R5, eliminate e7=3 (beware, heptagon). Now easy to 41 filled.
5)looking at 18 in d7e8, eliminate some possibles. Now easy to 47 filled.
6) looking at 1 in Ce, eliminate b7=8 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(d3=9)==(f3=9) forbids {h3=9, i3=9}
(b2=3)==(b2=6)--(a2=6)==(a2=3) forbids {i2=3, a3=3, b1=3, h2=3, e2=3}
(b2=6)==(b2=3)--(a2=3)==(a2=6) forbids {e2=6, b1=6, a3=6, i2=6, h2=6}
(g7=1)==(g7=9)--(g8=9)==(g8=1) forbids {i7=1, i8=1}
(g7=9)==(g7=1)--(g8=1)==(g8=9) forbids {i8=9, h8=9, i7=9}
(i8=4)==(i8=6)--(h8=6)==(h8=4) forbids {e8=4, i7=4, b8=4, h9=4, a8=4}
(i8=6)==(i8=4)--(h8=4)==(h8=6) forbids {e8=6, f8=6, h9=6}
(e2=8)==(e2=1)--(e8=1)==(e8=8) forbids {e4=8, e3=8, e7=8, e6=8}
(e2=1)==(e2=8)--(e8=8)==(e8=1) forbids {e6=1, e7=1, e4=1}
(i7=3)==(i7=7)--(i5=7)==(f5=7)--(e6=7)==(e6=3) forbids {e7=3}
Around 41 filled :
(d7=8)==(d7=1)--(e8=1)==(e8=8) forbids {f7=8, f8=8}
(d7=1)==(d7=8)--(e8=8)==(e8=1) forbids {f7=1, f8=1}
Around 47 filled :
(d7=8)==(d7=1)--(e8=1)==(e2=1)--(f1=1)==(f1=7)--(b1=7)==(b7=7) forbids {b7=8}

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 g3=4%col; in today's it becomes g1=2%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).