05/11/01 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : b2=5%col, d1=5%block, b5=2%col, f8=5%col, h7=5%col, a1=2%block lead to 27 filled cells.

2) Now :
Look at only possibles f6=7,f4=7 in their col. They forbid{e6=7, e4=7}.
Look at only possibles c1=9,c2=9 in their block. They forbid{c5=9, c8=9, c6=9, c9=9}.
Look at only possibles b8=7,b7=7 in their col. They forbid{c8=7}.
Look at only possibles c4=5,c6=5 in their col. Whether c6=5 (in which case c4=7%block) or c4=5, in both cases, we have no more {c4=1, c4=6}.
Look at only possibles c6=5,c4=5 in their col. Whether c4=5 (in which case c6=7%block) or c6=5, in both cases, we have no more {c6=6, c6=9}.
Now easy fillings up to 29 filled cells. (If needed, d5=9%row, a6=9%row)

3) Look at only possibles d6=4,e6=4 in their block. They forbid{i6=4, g6=4}.
Look at only possibles h8=6,h5=6 in their col. Whether h5=6 (in which case a4=6%block) or h8=6, in both cases, we have no more {a8=6}.

4) Here is the hardest step, so as suggested by _unknown on comments of yesterday's puzzle, I highlighted the vertices of the 11-agon involved !
Look at only possibles e7=8,b7=8 in their row. Whether e7=8 (in which case d8=4%cell,a8=1%cell) or b7=8 (in which case b3=6%cell,a4=6%col), in both cases, we have no more {a4=1}. Now easy fillings up to 50 filled cells. (If needed, c5=1%block, a8=1%block, a4=6%cell, f5=3%cell, g4=3%block, d2=3%col, d6=6%col, e6=4%block, d8=4%block, d4=8%col, f2=8%col, e1=7%cell, h3=7%row, h1=3%block, f4=1%col, f6=7%col, f9=2%col, e4=2%col, c4=7%col, c6=5%block, i4=5%block)

5)Look at only possibles g6=8,g6=2 in their cell. Whether g6=8 (in which case g1=9%cell) or g6=2 (in which case i2=2%col), in both cases, we have no more {i2=9}.
Look at only possibles b3=8,b3=6 in their cell. Whether b3=8 (in which case e7=8%row) or b3=6 (in which case b9=9%cell,e9=3%cell), in both cases, we have no more {e7=3}.
Look at only possibles c1=8,g1=8 in their row. Whether c1=8 (in which case c8=6%cell) or g1=8 (in which case g6=2%cell,h8=2%row), in both cases, we have no more {h8=6}.
Now easy fillings up to 81 filled cells. (If needed, e9=3%block, a7=3%block, c9=4%block, a3=4%block, c3=3%block, h5=6%col, g5=4%block, i7=4%block, h2=4%block, h8=2%col, i9=9%col, g9=1%row, g8=6%col, g7=7%col, b8=7%col, c2=6%col, e3=6%col, i3=1%row, e2=1%row, b9=6%col, c1=9%col, g2=9%col, i2=2%row, g6=2%col, b7=9%col, e8=9%col, i6=8%col, g1=8%col, e7=8%col, c8=8%col, b3=8%col)

Short hints for a proof

The crux is at step 4 :

1) easy to 27 filled.
2) Looking at 7 in col Cf, at 9 in block Bb2, at 7 in col Cb, at 57 in c4c6, eliminate some possibles. Then easy to 29 filled.
3) Looking at 4 in Be5, eliminate some possibles. Looking at 6 in col Ch, eliminate a8=6 (beware, pentagon).
4) Looking at 8 in Row R7, eliminate a4=1 (beware, 11-agon ! but what a reward to get it by yourself...). Then easy to 50 filled.
5) Looking at g6, eliminate i2=9 (heptagon). Looking at b3, eliminate e7=3 (nonagon). Looking at 8s in row R1, eliminate h8=6(nonagon).Now easy to unique solution.

Forbidding-chain-like proof

A 11-agon is needed here :

Around 27 filled
(f6=7)==(f4=7) forbids {e6=7, e4=7}
(c1=9)==(c2=9) forbids {c5=9, c9=9, c6=9, c8=9}
(b8=7)==(b7=7) forbids {c8=7}
(c4=5)==(c6=5)--(c6=7)==(c4=7) forbids {c4=6, c4=1}
(c6=5)==(c4=5)--(c4=7)==(c6=7) forbids {c6=6, c6=9}
Around 29 filled
(d6=4)==(e6=4) forbids {i6=4, g6=4}
(h8=6)==(h5=6)--(c5=6)==(a4=6) forbids {a8=6}
(a8=1)==(a8=4)--(d8=4)==(d8=8)--(e7=8)==(b7=8)--(b3=8)==(b3=6)--(a3=6)==(a4=6) forbids {a4=1}
Around 50 filled
(g1=9)==(g1=8)--(g6=8)==(g6=2)--(i6=2)==(i2=2) forbids {i2=9}
(e7=8)==(b7=8)--(b3=8)==(b3=6)--(b9=6)==(b9=9)--(e9=9)==(e9=3) forbids {e7=3}
(c8=6)==(c8=8)--(c1=8)==(g1=8)--(g6=8)==(g6=2)--(g8=2)==(h8=2) forbids {h8=6}

Equivalence with 09/20

Now applying the same "dictionary" (renaming numbers, replacing cells) to proof of 05/09/20 you will mechanically get a proof of today's. E.g. first filled cell in 05/09/20 was a7=9%row; in today's it becomes b2=5%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 05/09/20's proof).

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