06/02/07 tough puzzle from sudoku.com.au

A very hard puzzle. Beginners, it's at your own risks. Don't complain if you find it hot.

A complete proof

1) First eliminations : e6=8%block, i2=6%block, a7=5%block, d7=3%block, f7=6%block lead to 28 filled cells.

2)
Look at only possibles c1=1,a1=1 in their row. Whether a1=1 (in which case c1=6%row) or c1=1, in both cases, we have no more {c1=3, c1=9}.
Now easy fillings up to 29 filled cells. (If needed, c3=9%block)

3) First note that if d5 is not 6, then d6=6%block, f6=4%row, g6 or i6=1%row ; So one among d5=6,g6=1,i6=1 must be true.
Now look at these sets :
(1)a5=168 : possible values in a5
(2)c5=168 : possible values in a5
(3){d5=6,g6=1,i6=1} that we previously pointed
If g5 or i5=1, then d5=6 (set3),a5=8 (set1), c5=8, (set2) : too much 8s in row 5. So we eliminate gi5=1.

4)
Look at only possibles g6=1,i6=1 in their block. They forbid{f6=1}.
Look at only possibles b2=7,b3=7 in their block. They forbid{b9=7, b8=7}.
Look at only possibles f6=4,d6=4 in their row. They forbid{d4=4, e4=4, f4=4}.
Look at only possibles f5=5,i5=5 in their row. Whether i5=5 (in which case f3=5%row) or f5=5, in both cases, we have no more {f6=5, f4=5}.
Look at only possibles i3=5,f3=5 in their row. Whether f3=5 (in which case i5=5%row) or i3=5, in both cases, we have no more {i6=5}.
Look at only possibles a1=1,c1=1 in their row. Whether c1=1 (in which case a1=6%row) or a1=1, in both cases, we have no more {a1=2, a1=3}.
Now easy fillings up to 32 filled cells. (If needed, f6=4%cell, d3=4%col, g2=4%block)

5)
Look at only possibles d5=7,d4=7 in their col. They forbid{e4=7, f5=7, f4=7}.
Look at only possibles g1=3,h1=3 in their row. They forbid{g3=3, i3=3}.
Look at only possibles f5=5,f3=5 in their col. Whether f5=5 (in which case i5=2%cell) or f3=5 (in which case d1=2%block), in both cases, we have no more {d5=2}.

6)
Look at only possibles i3=2,i3=5 in their cell. Whether i3=5 (in which case i5=2%cell) or i3=2, in both cases, we have no more {i8=2, i9=2}.

7)
Look at only possibles g5=7,g5=2 in their cell. Whether g5=7 (in which case d4=7%col) or g5=2 (in which case i3=2%col,d1=2%block), in both cases, we have no more {d4=2}.
Look at only possibles b6=6,d6=6 in their row. Whether b6=6 (in which case h6=5%row,f5=5%row) or d6=6 (in which case d5=7%cell,g5=2%cell), in both cases, we have no more {f5=2}.
Now easy fillings up to 34 filled cells. (If needed, f4=2%block, d1=2%block)

8) Look at only possibles d6=9,d4=9 in their col. They forbid{e4=9}.

9)
Look at only possibles h4=9,h4=7 in their cell. Whether h4=9 (in which case i7=9%col) or h4=7 (in which case g7=7%block), in both cases, we have no more {g7=9}.
Look at only possibles e1=5,e1=9 in their cell. Whether e1=5 (in which case h6=5%block) or e1=9 (in which case h2=9%row), in both cases, we have no more {h6=9}.

10) Look at only possibles g5=7,g5=2 in their cell. Whether g5=7 (in which case g7=8%cell) or g5=2 (in which case g3=8%cell), in both cases, we have no more {g9=8}. Look at only possibles g7=8,g7=7 in their cell. Whether g7=8 (in which case g3=2%cell) or g7=7 (in which case g5=2%cell), in both cases, we have no more {g9=2}.
Now easy fillings up to 40 filled cells. (If needed, h8=2%block, b9=2%block, c9=8%row, a5=8%block, a2=2%row, a3=3%cell)

11) Look at only possibles h7=7,g7=7 in their block. They forbid{g9=7, c7=7}.
Now easy fillings up to 81 filled cells. (If needed, c8=7%block, b8=3%block, a8=6%block, c1=6%block, a1=1%block, b6=6%block, d5=6%block, d4=7%block, d6=9%block, h4=9%block, i7=9%block, g1=9%block, e2=9%block, g5=7%block, h7=7%block, g7=8%block, h2=8%block, b3=8%block, f3=7%block, e9=7%block, b2=7%block, b4=5%block, f5=5%block, e4=1%block, f9=1%block, i8=1%block, g6=1%block, c5=1%block, h6=5%block, i3=5%block, e1=5%block, i9=4%block, e8=4%block, c7=4%block, a4=4%block, g9=3%block, i6=3%block, c4=3%block, h1=3%block, i5=2%block, g3=2%block)

Short hints for a proof

Max depth 5 needed here, to eliminate f5=2

Well, there's work to be done...
1) easy to 28.
2) eliminate c1=9 (pair,2sets), then easy to 29.
3) eliminate gi5=1 (hard : 5sets! first thicklink d5=6 to gi6=1, then look at a5&c5)
4) basic eliminations (max depth 2, pairs, xwing... total 7 sets) then easy to 32.
5) looking at 5s in col f, eliminate d5=2 (3 sets)+basic eliminations depth 1 or 2, total 4 sets.
6) eliminate d4=2 (gives pb for 2s in row 5, 4 sets), f5=2 (look at 6s in row 6, 5 sets), then easy to 34.
7) eliminate b2=2 (color, 2sets), h4=35 (pair, 2 sets), g9=7 (gives pb in i3, 4 sets)
8) eliminate g7=9 (look at h4,3sets), h6=9 (look at e1, 3 sets)
9) eliminate g9=28, (look at g7, 3 sets), then easy to 40 filled.
10) 1set elimination then easy to the end.
Total sets used :50, max depth :5.

Forbidding-chain-like proof

Max depth 5 needed here, to eliminate f5=2.

Here are, ordered, the eliminations needed. The rest is only easy fillings.
(c1=1)==(a1=1)--(a1=6)==(c1=6) forbids {c1=3, c1=9}
Sets a5=168, c5=168, d56=6, df6=4, fgi6=1 forbid gi5=1
(g6=1)==(i6=1) forbids {f6=1}
(b2=7)==(b3=7) forbids {b8=7, b9=7}
(f6=4)==(d6=4) forbids {d4=4, f4=4, e4=4}
(f5=5)==(i5=5)--(i3=5)==(f3=5) forbids {f6=5, f4=5}
(i3=5)==(f3=5)--(f5=5)==(i5=5) forbids {i6=5}
(a1=1)==(c1=1)--(c1=6)==(a1=6) forbids {a1=2, a1=3}
(d5=7)==(d4=7) forbids {e4=7, f4=7, f5=7}
(g1=3)==(h1=3) forbids {g3=3, i3=3}
(i5=2)==(i5=5)--(f5=5)==(f3=5)--(f3=2)==(d1=2) forbids {d5=2}
(i3=2)==(i3=5)--(i5=5)==(i5=2) forbids {i9=2, i8=2}
(d4=7)==(d5=7)--(g5=7)==(g5=2)--(i5=2)==(i3=2)--(f3=2)==(d1=2) forbids {d4=2}
(f5=5)==(i5=5)--(h6=5)==(b6=5)--(b6=6)==(d6=6)--(d5=6)==(d5=7)--(g5=7)==(g5=2) forbids {f5=2}
(d6=9)==(d4=9) forbids {e4=9}
(i7=9)==(i6=9)--(h4=9)==(h4=7)--(g5=7)==(g7=7) forbids {g7=9}
(h6=5)==(h1=5)--(e1=5)==(e1=9)--(e2=9)==(h2=9) forbids {h6=9}
(g7=8)==(g7=7)--(g5=7)==(g5=2)--(g3=2)==(g3=8) forbids {g9=8}
(g3=2)==(g3=8)--(g7=8)==(g7=7)--(g5=7)==(g5=2) forbids {g9=2}
(h7=7)==(g7=7) forbids {g9=7, c7=7}

That's all for today, folks...