05/11/02 tough puzzle from sudoku.com.au

A complete proof

1) First eliminations : h8=4%col, b2=5%col, b5=7%col, b9=8%col, b7=4%col, d9=4%row, b8=2%col, h5=9%cell, h2=7%cell lead to 28 filled cells.

2) Now :
Look at only possibles d7=2,f7=2 in their block. They forbid{i7=2, g7=2}.
Look at only possibles c9=6,c9=1 in their cell. Whether c9=1 (in which case a9=6%cell) or c9=6, in both cases, we have no more {i9=6, e9=6, g9=6, a7=6, c8=6}.
Look at only possibles c9=1,c9=6 in their cell. Whether c9=6 (in which case a9=1%cell) or c9=1, in both cases, we have no more {c8=1, i9=1, e9=1, a7=1, g9=1}.
Look at only possibles h3=5,h3=2 in their cell. Whether h3=2 (in which case h1=5%cell) or h3=5, in both cases, we have no more {i1=5, i3=5}.
Look at only possibles h3=2,h3=5 in their cell. Whether h3=5 (in which case h1=2%cell) or h3=2, in both cases, we have no more {i1=2, i3=2, g1=2}.

3) Now :
Look at only possibles i1=9,i1=1 in their cell. Whether i1=1 (in which case g1=9%cell) or i1=9, in both cases, we have no more {a1=9, i3=9, e1=9, g2=9, c1=9}.
Look at only possibles i1=1,i1=9 in their cell. Whether i1=9 (in which case g1=1%cell) or i1=1, in both cases, we have no more {g2=1, f1=1, e1=1}.
Look at only possibles e9=7,e9=5 in their cell. Whether e9=5 (in which case e1=7%cell) or e9=7, in both cases, we have no more {e6=7, e3=7, e7=7, e4=7}.
Look at only possibles e9=5,e9=7 in their cell. Whether e9=7 (in which case e1=5%cell) or e9=5, in both cases, we have no more {e4=5, e3=5, e6=5}.

4) Look at only possibles i5=3,f5=3 in their row. Whether i5=3 (in which case i3=6%cell) or f5=3 (in which case e4=6%cell), in both cases, we have no more {e3=6}.
Now easy fillings up to 41 filled cells. (If needed, e3=9%cell, d6=9%row, e6=1%cell, a5=1%block, c9=1%row, a9=6%row, c4=6%col, e7=6%col, e4=3%col, i5=3%row, g2=3%col, g8=6%col, i3=6%col)

5)Look at only possibles d3=7,d3=5 in their cell. Whether d3=5 (in which case e1=7%cell) or d3=7, in both cases, we have no more {f1=7, f3=7}.
Look at only possibles d3=5,d3=7 in their cell. Whether d3=7 (in which case e1=5%cell) or d3=5, in both cases, we have no more {f3=5, f1=5}.
Now easy fillings up to 47 filled cells. (If needed, f3=4%cell, f1=8%cell, d5=8%row, f5=6%row, d2=6%row, f2=1%row)

6)Look at only possibles e1=5,e9=5 in their col. Whether e1=5 (in which case d3=7%cell) or e9=5 (in which case f8=3%cell,c3=3%col), in both cases, we have no more {c3=7}.
Now easy fillings up to 81 filled cells. (If needed, d3=7%row, h3=5%row, a3=2%row, h1=2%row, c3=3%row, f8=3%row, a7=3%row, e1=5%row, i9=5%row, g9=2%row, d8=5%row, i8=1%row, d7=1%row, f7=2%row, i6=2%row, d4=2%row, c1=7%row, a1=4%row, g4=4%row, c6=4%row, e9=7%row, a6=8%row, f6=5%row, a4=5%row, g6=7%row, f4=7%row, c2=8%row, i7=7%row, i4=8%row, a2=9%row, c8=9%row, g7=9%row, i1=9%row, g1=1%row)

Short hints for a proof

Beware, 9-agon needed here:

1) easy to 28 filled.
2) looking at 2 in Be8, at 16 in a9c9, at 25 in h1h3, eliminate some possibles.
3) looking at 19 in g1i1, at 57 in e1e9, eliminate some possibles.
4) looking at 3 in R5, eliminate e3=6 (beware, heptagon). Now easy to 41 filled.
5)looking at 57 in e1d3, eliminate some possibles. Now easy to 47 filled.
6) 5 in Ce, eliminate c3=7 (beware, nonagon). then easy to unique solution.

Forbidding-chain-like proof

The 9-agon is needed here :

Around 28 filled :
(d7=2)==(f7=2) forbids {i7=2, g7=2}
(c9=6)==(c9=1)--(a9=1)==(a9=6) forbids {c8=6, i9=6, a7=6, e9=6, g9=6}
(c9=1)==(c9=6)--(a9=6)==(a9=1) forbids {a7=1, e9=1, c8=1, g9=1, i9=1}
(h3=5)==(h3=2)--(h1=2)==(h1=5) forbids {i3=5, i1=5}
(h3=2)==(h3=5)--(h1=5)==(h1=2) forbids {g1=2, i1=2, i3=2}
(i1=9)==(i1=1)--(g1=1)==(g1=9) forbids {e1=9, i3=9, a1=9, g2=9, c1=9}
(i1=1)==(i1=9)--(g1=9)==(g1=1) forbids {e1=1, g2=1, f1=1}
(e9=7)==(e9=5)--(e1=5)==(e1=7) forbids {e6=7, e7=7, e4=7, e3=7}
(e9=5)==(e9=7)--(e1=7)==(e1=5) forbids {e6=5, e4=5, e3=5}
(i3=6)==(i3=3)--(i5=3)==(f5=3)--(e4=3)==(e4=6) forbids {e3=6}
Around 41 filled :
(d3=7)==(d3=5)--(e1=5)==(e1=7) forbids {f3=7, f1=7}
(d3=5)==(d3=7)--(e1=7)==(e1=5) forbids {f3=5, f1=5}
Around 47 filled :
(d3=7)==(d3=5)--(e1=5)==(e9=5)--(f8=5)==(f8=3)--(c8=3)==(c3=3) forbids {c3=7}

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 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 05/09/12's proof).

Go : back to sandbox