06/02/04 tough puzzle from sudoku.com.au

Today's puzzle is considerably tougher than the last ones.

Want to see the whole thing? A complete proof
Just stuck somewhere and willing to have still work to do ? Short hints for a proof
Studied enough forbidding chains to appreciate this Forbidding-chain-like proof ?

A complete proof

1) First eliminations : i7=2%row, a3=5%block, b9=9%block lead to 25 filled cells.

2) Now some basic eliminations :
Look at only possibles d7=1,f7=1 in their row. They forbid{e8=1, d8=1, f8=1}.
Look at only possibles d3=8,f3=8 in their row. They forbid{d2=8, e2=8, f2=8}.
Look at only possibles c8=3,c9=3 in their block. They forbid{c1=3, c4=3, c6=3}.
Look at only possibles i4=9,i6=9 in their col. They forbid{g4=9, g6=9}.
Look at only possibles d4=9,f6=9 in their block. Whether f6=9 (in which case d4=4%block) or d4=9, in both cases, we have no more {d4=3, d4=8, d4=1}.
Look at only possibles f6=4,d4=4 in their block. Whether d4=4 (in which case f6=9%block) or f6=4, in both cases, we have no more {f6=1, f6=7, f6=3, f6=2}.
Look at only possibles g2=7,g1=7 in their block. Whether g1=7 (in which case g2=9%block) or g2=7, in both cases, we have no more {g2=6, g2=1, g2=2}.
Look at only possibles g1=9,g2=9 in their block. Whether g2=9 (in which case g1=7%block) or g1=9, in both cases, we have no more {g1=2, g1=1}.
Now easy fillings up to 26 filled cells. (If needed, h1=1%block)

3) another pair leads to 29 filled cells :
Look at only possibles i8=5,g8=5 in their block. Whether g8=5 (in which case i8=1%row) or i8=5, in both cases, we have no more {i8=6, i8=7}.
Now easy fillings up to 29 filled cells. (If needed, i9=7%col, h9=4%block, i3=4%col)

4) basic eliminations + one pentagon (color on 8's) :
Look at only possibles i5=6,i6=6 in their col. They forbid{g6=6, h6=6, h5=6}.
Look at only possibles e4=8,g4=8 in their row. Whether g4=8 (in which case h8=8%col) or e4=8, in both cases, we have no more {e8=8}.
Look at only possibles f9=8,f9=3 in their cell. Whether f9=3 (in which case f3=8%cell) or f9=8, in both cases, we have no more {f5=8, f8=8, f7=8}.
Look at only possibles f3=3,f3=8 in their cell. Whether f3=8 (in which case f9=3%cell) or f3=3, in both cases, we have no more {f8=3, f5=3}.

5) Here is the hardest step, we'll manage it using intermediate sets of possibles, as Steve introduced some days ago.
Build three "Steve sets" in which at least one element must be true:
Steve0 : {e8=7,c9=3}. Because if e8 is not 7, e8=3%cell, c9=3%row.
Steve1 : {a6=7,c9=3}. Because if a6 is not 7, e6=7%row, c9=3%Steve0.
Steve2 : {i6=9,c9=3}. Because if i6 is not 9, f6=9%row, f8=7%cell, c9=3%Steve0.
Then consider possible 6s in row 6 : aci6=6. Now you need no more looking at the diagram :
Whether a6=6 or i6=6(giving c9=3 by Steve1 or Steve2), or c6=6 : in any case, c9 is not 6.

Short hints for a proof

Tough elimination of c9=6 here


1) Easy fillings to 25 filled cells.
2) Looking at 1s in row7, at 8s in row3, at 3s in Bb8, at 9s in Col i, at 49 in d4f6, at 79 in g1g2, eliminate some possibles. Then easy to 26 filled cells.
3) Looking at 15 in g8i8, eliminate some possibles. Then easy to 29 filled cells.
4) Looking at 6s in col i, at 38 in f3f9, eliminate some possibles. Looking at 8s in row4, eliminate e8=8 (pentagon).
5) Very hard ! looking at 3s in col c, at e8, at 7 in row 6, at f8, at 9s in row6, at 6s in row6, eliminate c9=6. Then easy to the end.

Forbidding-chain-like proof

Tough elimination of c9=6 here


around 25 filled
(d7=1)==(f7=1) forbids {d8=1, e8=1, f8=1}
(d3=8)==(f3=8) forbids {f2=8, e2=8, d2=8}
(c8=3)==(c9=3) forbids {c6=3, c4=3, c1=3}
(i4=9)==(i6=9) forbids {g6=9, g4=9}
(d4=9)==(f6=9)--(f6=4)==(d4=4) forbids {d4=8, d4=1, d4=3}
(f6=4)==(d4=4)--(d4=9)==(f6=9) forbids {f6=3, f6=2, f6=7, f6=1}
(g2=7)==(g1=7)--(g1=9)==(g2=9) forbids {g2=6, g2=1, g2=2}
(g1=9)==(g2=9)--(g2=7)==(g1=7) forbids {g1=1, g1=2}
around 26 filled
(i8=5)==(g8=5)--(g8=1)==(i8=1) forbids {i8=6, i8=7}
around 29 filled
(i5=6)==(i6=6) forbids {h6=6, h5=6, g6=6}
(e4=8)==(g4=8)--(h5=8)==(h8=8) forbids {e8=8}
(f9=8)==(f9=3)--(f3=3)==(f3=8) forbids {f5=8, f8=8, f7=8}
(f3=3)==(f3=8)--(f9=8)==(f9=3) forbids {f5=3, f8=3}
Then non-fc step (see above) forbids c9=6.

That's all for today, folks...