06/02/11 tough puzzle from sudoku.com.au

A medium puzzle. The hardest step is 3sets-deep and unicolored.

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 : a7=3%row, b8=9%block, i1=5%row lead to 26 filled cells.

2)
Look at only possibles g8=8,g8=2 in their cell. Whether g8=2 (in which case g9=8%cell) or g8=8, in both cases, we have no more {g2=8, h7=8, g1=8, i7=8, h8=8, i9=8}.
Look at only possibles g9=2,g9=8 in their cell. Whether g9=8 (in which case g8=2%cell) or g9=2, in both cases, we have no more {g2=2, i9=2}.

3)
Look at only possibles g1=3,g1=1 in their cell. Whether g1=1 (in which case e1=3%cell) or g1=3, in both cases, we have no more {c1=3}.
Look at only possibles i9=4,i9=7 in their cell. Whether i9=7 (in which case c9=4%cell) or i9=4, in both cases, we have no more {d9=4, a9=4}.
Look at only possibles c9=7,c9=4 in their cell. Whether c9=4 (in which case i9=7%cell) or c9=7, in both cases, we have no more {d9=7, a9=7}.
Look at only possibles e1=1,e1=3 in their cell. Whether e1=3 (in which case g1=1%cell) or e1=1, in both cases, we have no more {c1=1, a1=1, f1=1}.
Now easy fillings up to 30 filled cells. (If needed, c2=3%block, a3=1%block, c4=1%block, c8=6%col)

4)
Look at only possibles a6=2,a4=2 in their col. They forbid{b4=2, b6=2}.
Look at only possibles e2=1,e2=6 in their cell. Whether e2=6 (in which case g2=1%cell) or e2=1, in both cases, we have no more {h2=1, i2=1}.
Now easy fillings up to 41 filled cells. (If needed, b4=5%cell, d6=5%block, d4=3%block, h4=8%cell, h2=7%cell, h3=3%cell, g6=3%col, g2=6%col, e2=1%row, e1=3%block, g1=1%cell)

5)
Look at only possibles f8=7,a8=7 in their row. Whether a8=7 (in which case c1=7%col) or f8=7, in both cases, we have no more {f1=7}.

6)
Look at only possibles d2=8,d2=4 in their cell. Whether d2=4 (in which case f1=8%cell) or d2=8, in both cases, we have no more {d3=8, f3=8}.

7) Look at 8s in col a. Whether a1=8 (giving f7 or f8=8%col), a8=8 (giving g9=8%col) or a9=8, d9 is never 8.

Now easy fillings up to 81 filled cells. (If needed, d9=9%cell, d5=6%cell, a5=4%cell, d3=7%cell, e3=9%col, f3=6%block, e7=6%row, h7=5%row, h8=4%cell, i9=7%cell, i6=4%row, i4=6%block, a4=2%cell, i5=9%col, h6=1%block, f4=9%block, c1=7%col, b2=4%block, f1=4%block, d2=8%block, f5=1%block, b3=2%block, i2=2%block, i3=8%row, a1=8%block, b7=8%block, f7=7%cell, b6=7%col, f8=8%col, g9=8%block, g8=2%cell, a6=6%col, a8=7%block, a9=5%col, e8=5%col, c9=4%row, f6=2%cell, e9=2%cell, d7=4%col, i7=1%cell)

Short hints for a proof

3sets color move eliminates d9=8.


1) easy fillings to 26 filled.
2) basic eliminations (locked candidates, pairs... total 9 sets) to 41 filled.
3) eliminate f1=7 (causes trouble in Bb8, 2sets, color)
4) eliminate df3=8 (2sets, pairs)
5) looking at 8s in col a, eliminate d9=8 (3 sets) then easy fillings to the end.

Total sets to examine : 16, max depth : 3 (at step 5)

Forbidding-chain-like proof

3sets color move eliminates d9=8.


Here are, ordered, the eliminations needed. The rest is only easy fillings.
(g8=8)==(g8=2)--(g9=2)==(g9=8) forbids {g2=8, h7=8, i7=8, g1=8, i9=8, h8=8}
(g9=2)==(g9=8)--(g8=8)==(g8=2) forbids {g2=2, i9=2}
(g1=3)==(g1=1)--(e1=1)==(e1=3) forbids {c1=3}
(i9=4)==(i9=7)--(c9=7)==(c9=4) forbids {a9=4, d9=4}
(c9=7)==(c9=4)--(i9=4)==(i9=7) forbids {a9=7, d9=7}
(e1=1)==(e1=3)--(g1=3)==(g1=1) forbids {a1=1, f1=1, c1=1}
(a6=2)==(a4=2) forbids {b6=2, b4=2}
(e2=1)==(e2=6)--(g2=6)==(g2=1) forbids {i2=1, h2=1}
(f8=7)==(a8=7)--(c9=7)==(c1=7) forbids {f1=7}
(d2=8)==(d2=4)--(f1=4)==(f1=8) forbids {f3=8, d3=8}
Sets a189=8, g89=8, f178=8 forbid d9=8.

That's all for today, folks...