05/11/03 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : f7=9%col, g9=9%block, g3=4%block, d9=7%row, e9=5%row lead to 26 filled cells.

2) Now :
Look at only possibles g1=1,i1=1 in their block. They forbid{c1=1, e1=1, a1=1}.
Look at only possibles h2=8,h3=8 in their block. They forbid{h4=8, h8=8, h5=8, h7=8}.
Look at only possibles d5=5,f5=5 in their block. They forbid{b5=5, g5=5, c5=5, h5=5}.
Look at only possibles d8=1,d8=6 in their cell. Whether d8=6 (in which case d7=1%cell) or d8=1, in both cases, we have no more {e7=1, d2=1, d3=1}.
Look at only possibles d8=6,d8=1 in their cell. Whether d8=1 (in which case d7=6%cell) or d8=6, in both cases, we have no more {d2=6, d3=6, f8=6, e7=6, d5=6}.
Look at only possibles b3=5,b6=5 in their col. Whether b6=5 (in which case b3=9%col) or b3=5, in both cases, we have no more {b3=2, b3=3}.
Look at only possibles b6=5,b3=5 in their col. Whether b3=5 (in which case b6=9%col) or b6=5, in both cases, we have no more {b6=7, b6=1, b6=2}.
Now easy fillings up to 39 filled cells. (If needed, b2=7%col, f1=7%col, c1=5%row, b6=5%col, b3=9%col, h1=9%col, c6=9%col, c3=3%block)

3)
Look at only possibles a3=6,a3=2 in their cell. Whether a3=2 (in which case a1=6%cell) or a3=6, in both cases, we have no more {a2=6, c2=6, a9=6}.
Look at only possibles a3=2,a3=6 in their cell. Whether a3=6 (in which case a1=2%cell) or a3=2, in both cases, we have no more {a6=2, a4=2}.
Look at only possibles h2=6,h2=8 in their cell. Whether h2=8 (in which case h3=6%cell) or h2=6, in both cases, we have no more {h8=6, i1=6, h5=6, h7=6, g1=6}.
Now easy fillings up to 54 filled cells. (If needed, h5=2%cell, e6=2%row, c4=2%row, a1=2%row, d3=2%row, f3=5%row, f2=8%col, h3=8%col, h2=6%col, f5=6%col, e1=6%col, a3=6%col, d5=5%col, h8=5%cell, g4=5%col)

4) Here is the hardest step, so as suggested by _unknown, I highlighted the vertices of the heptagon involved !
Look at only possibles c5=8,c9=8 in their col. Whether c5=8 (in which case g8=8%col) or c9=8 (in which case i9=6%row), in both cases, we have no more {g8=6, i9=8}.
Now easy fillings up to 81 filled cells. (If needed, g8=8%cell, i4=8%col, h4=3%row, a9=8%col, c5=8%col, b5=4%row, g5=1%row, a6=1%row, c2=1%row, a2=4%row, i1=1%row, g1=3%row, h7=7%col, g6=7%col, a4=7%col, g7=6%col, i6=6%col, d8=6%col, b8=1%row, i8=2%row, b7=2%row, i7=3%row, c7=4%row, i9=4%row, b9=3%row, d7=1%row, c9=6%col)

Short hints for a proof

Beware, 6-FC needed here:

1) easy to 26 filled.
2) Looking at 1 in Bg2, at 8 in Bh2, at 5 in Be5, at 16 in d7d8, at 59 in b3b6, eliminate some possibles, then easy to 39 filled.
3) Looking at 26 in a1a3, at 68 in h2h3, eliminate some possibles, then easy to 54 filled.
4) Looking at 8 in Cc, eliminate {g8=6, i9=8} (beware, heptagons). Then easy to unique solution.

Forbidding-chain-like proof

The 6-FC is needed here :

around 26 filled
(g1=1)==(i1=1) forbids {a1=1, e1=1, c1=1}
(h2=8)==(h3=8) forbids {h5=8, h7=8, h4=8, h8=8}
(d5=5)==(f5=5) forbids {h5=5, b5=5, g5=5, c5=5}
(d8=1)==(d8=6)--(d7=6)==(d7=1) forbids {e7=1, d2=1, d3=1}
(d8=6)==(d8=1)--(d7=1)==(d7=6) forbids {d2=6, d3=6, f8=6, e7=6, d5=6}
(b3=5)==(b6=5)--(b6=9)==(b3=9) forbids {b3=2, b3=3, b3=1}
(b6=5)==(b3=5)--(b3=9)==(b6=9) forbids {b6=1, b6=2, b6=7}
around 39 filled
(a3=6)==(a3=2)--(a1=2)==(a1=6) forbids {a2=6, c2=6, a9=6}
(a3=2)==(a3=6)--(a1=6)==(a1=2) forbids {a6=2, a4=2}
(h2=6)==(h2=8)--(h3=8)==(h3=6) forbids {i1=6, h5=6, h7=6, h8=6, g1=6}
around 50 filled
(g8=8)==(g5=8)--(c5=8)==(c9=8)--(c9=6)==(i9=6) forbids {g8=6, i9=8}

Equivalence with 2005/09/28

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/28 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/28 was f8=6%col; in today's it becomes f7=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 05/09/28's proof).

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