05/11/25 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : i3=5%col, a3=8%cell, a1=5%cell, a2=1%cell, a5=4%cell, i5=2%cell, a7=9%cell, i7=4%cell, f2=5%row lead to 28 filled cells.

2) Now :
Look at only possibles f1=8,d1=8 in their block. They forbid{h1=8, g1=8}.
Look at only possibles b2=6,b2=7 in their cell. Whether b2=7 (in which case c2=6%cell) or b2=6, in both cases, we have no more {g2=6, e2=6, h2=6, c1=6, b3=6}.
Look at only possibles b2=7,b2=6 in their cell. Whether b2=6 (in which case c2=7%cell) or b2=7, in both cases, we have no more {e2=7, c1=7, h2=7, g2=7, b3=7}.
Look at only possibles i9=9,i9=8 in their cell. Whether i9=8 (in which case i8=9%cell) or i9=9, in both cases, we have no more {g9=9, g8=9}.
Look at only possibles i9=8,i9=9 in their cell. Whether i9=9 (in which case i8=8%cell) or i9=8, in both cases, we have no more {g9=8, g8=8, h8=8}.

3) Now :
Look at only possibles g8=2,g8=7 in their cell. Whether g8=7 (in which case h8=2%cell) or g8=2, in both cases, we have no more {c8=2, b8=2, e8=2, g9=2, h7=2}.
Look at only possibles g8=7,g8=2 in their cell. Whether g8=2 (in which case h8=7%cell) or g8=7, in both cases, we have no more {d8=7, h7=7, e8=7}.
Look at only possibles e2=4,e2=9 in their cell. Whether e2=9 (in which case e8=4%cell) or e2=4, in both cases, we have no more {e1=4, e4=4, e6=4, e9=4}.
Look at only possibles e2=9,e2=4 in their cell. Whether e2=4 (in which case e8=9%cell) or e2=9, in both cases, we have no more {e9=9, e6=9, e4=9}.

4) Look at only possibles g5=3,d5=3 in their row. Whether g5=3 (in which case g9=6%cell) or d5=3 (in which case e4=6%cell), in both cases, we have no more {e9=6}.
Now easy fillings up to 41 filled cells. (If needed, e9=2%cell, e6=7%cell, b2=7%col, c2=6%cell, c5=7%col, b4=6%col, e4=3%cell, e1=6%cell, h7=3%col, h3=6%col, g9=6%col, g5=3%col, f6=2%col)

5)Look at only possibles f9=4,f9=9 in their cell. Whether f9=9 (in which case e8=4%cell) or f9=4, in both cases, we have no more {d8=4, d9=4}.
Look at only possibles f9=9,f9=4 in their cell. Whether f9=4 (in which case e8=9%cell) or f9=9, in both cases, we have no more {d9=9, d8=9}.
Now easy fillings up to 47 filled cells. (If needed, d9=5%cell, d8=1%cell, d5=6%cell, f5=1%cell, f7=6%col, d7=7%cell)

6)Look at only possibles e8=9,e2=9 in their col. Whether e8=9 (in which case f9=4%cell) or e2=9 (in which case d3=3%cell,b9=3%col), in both cases, we have no more {b9=4}.
Now easy fillings up to 81 filled cells. (If needed, b8=4%col, b6=5%col, h4=5%col, c8=5%col, e2=4%col, h2=8%cell, g2=9%cell, b7=1%col, b3=2%col, g3=7%cell, f3=9%cell, d3=3%cell, d1=8%cell, f1=7%cell, c1=3%cell, h8=7%col, b9=3%col, h1=2%col, g1=4%cell, h6=4%col, d6=9%cell, c6=1%cell, g6=8%cell, g4=1%cell, d4=4%cell, f4=8%cell, c4=9%cell, f9=4%col, g8=2%col, c7=2%col, c9=8%cell, i9=9%cell, i8=8%cell, e8=9%cell)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 8 in Be2, at 67 in b2c2, at 89 in i8i9, eliminate some possibles.
3) looking at 27 in g8h8, at 49 in e2e8, eliminate some possibles.
4) looking at 3 in R5, eliminate e9=6 (beware, heptagon). Now easy to 41 filled.
5)looking at 49 in e8f9, eliminate some possibles. Now easy to 47 filled.
6) looking at 9 in Ce, eliminate b9=4 (beware, nonagon). Then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(f1=8)==(d1=8) forbids {g1=8, h1=8}
(b2=6)==(b2=7)--(c2=7)==(c2=6) forbids {e2=6, g2=6, h2=6, b3=6, c1=6}
(b2=7)==(b2=6)--(c2=6)==(c2=7) forbids {e2=7, g2=7, h2=7, b3=7, c1=7}
(i9=9)==(i9=8)--(i8=8)==(i8=9) forbids {g9=9, g8=9}
(i9=8)==(i9=9)--(i8=9)==(i8=8) forbids {h8=8, g8=8, g9=8}
(g8=2)==(g8=7)--(h8=7)==(h8=2) forbids {c8=2, h7=2, e8=2, g9=2, b8=2}
(g8=7)==(g8=2)--(h8=2)==(h8=7) forbids {h7=7, e8=7, d8=7}
(e2=4)==(e2=9)--(e8=9)==(e8=4) forbids {e9=4, e6=4, e1=4, e4=4}
(e2=9)==(e2=4)--(e8=4)==(e8=9) forbids {e6=9, e4=9, e9=9}
(g9=6)==(g9=3)--(g5=3)==(d5=3)--(e4=3)==(e4=6) forbids {e9=6}
Around 41 filled :
(f9=4)==(f9=9)--(e8=9)==(e8=4) forbids {d9=4, d8=4}
(f9=9)==(f9=4)--(e8=4)==(e8=9) forbids {d9=9, d8=9}
Around 47 filled :
(f9=4)==(f9=9)--(e8=9)==(e2=9)--(d3=9)==(d3=3)--(b3=3)==(b9=3) forbids {b9=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 g3=4%col; in today's it becomes i3=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 05/09/12's proof).