# Final Page of Diabolic Proof for 03/11/07

The first page of this proof is here.

### Puzzle at UP 36

At this point, assuming one does not use the AUR 69 at ab38, the puzzle is resistant to simple forbidding chains. There are some possible, but I could not advance the puzzle very much with any of those that I found. For this reason, I have pulled out the elephant gun on the next step.

The Alternating Inference Chain (AIC), or forbidding chain illustrated above is:

• Complex, as it uses a chain as a Boolean variable
• A partial wrap around or partial continuous nice loop, as f3=3 -- f7=3
The chain can be written many ways. Here is my favorite way to write this one:

f2=3 == e3=3 -- g3=3 ==1 {h7=6 == h7=9 -- h1=9 == h1=5 -- g3=5 ==1 g3=6 -- g79=6 == h7=6} -- f7=6 == f7=3

Although the macro chain is a continuous nice loop since f2=3 and f7=3 are in conflict with each other, the bracketed chain used as a Boolean is not, as h7=6 is not in conflict with h7=6. Therefor, the deductions one can make from the chain are best found by making the following substitutions in the chain and considering only the remaining weak links to be strong:

• f2=3 == e3=3 -- g3=3 ==non-native link h7=6 -- f7=6 == f7=3
Therefor, we have the following proven non-native strong sets:
1. f2=3 == f7=3 => f8≠3
2. e3=3 == g3=3 => no possible eliminations here
3. h7=6 == f7=6 => eg7≠6

The puzzle is advanced a bit further:

• f8 = 2% cell
• c8 = 3% cell
• c9 = 2% row, column & box
• c3 = 4% column
• Thus UP 40

How to find this one? Much experience! This one is not easy to find, but the column f, with the 3s, told me that there existed a wrap around chain somewhere on the puzzle grid. Much experience in finding Y wing styles led me find a conditional Y wing style that led me to find this chain. It would not be a truly diabolical puzzle if all the chains were easy to find!

### A much simpler Advanced Forbidding Chain

Illustrated above the Almost Locked Set (ALS) 589 at ef5 is used in a chain as follows:

• i6=9 == i6=6 -- f6=6 == f6=5 -- f5=5 == {pair 89 at ef5} => i5≠9
• => i5 = 3% cell, i2 = 6% cell, i6 = 9% cell, a4 = 3% row, column & box
• Thus we arive at UP 44

### One more depth 4 chain

Illustrated above is one more forbidding chain using 4 native strong sets:

• a5=5 == a3=5 -- g3=5 == g3=3 -- e3=3 == e3=9 -- e5=9 == f5=9 => f5≠5
• => f6 = 5% box & column
• thus UP 45
• => Pair 89 at ef5 => h5 ≠8
• => h4 = 8% box & column
• thus UP 46

### Final step: Simple coloring

Above, as a forbidding chain on 5s:

• a3 == g3 -- g4 == b4 => a5≠5
• => a3 = 5% column
• => %cells, or naked singles, to the end

### Proof

1. Start at 22 filled - the given puzzle. Unique Possibilities to 27 filled. (UP 27).
1. Locked 2's at b46 forbids b789=2
2. Locked 3's at a45 forbids a8=3
3. Pair 69 at ab8 forbids ab9b7fg8=6, b7g9=9 UP 28
1. Locked 8's at d13 forbids e13=8
2. Locked 2's at d23 forbids d3f2=2
3. Hidden pair 28 at dh3 forbids d3=6, h3=4569
4. Locked 1's at c12 forbids c79=1
5. Locked 2's at hi7 forbids cef7=2
6. Locked 7's at g79 forbids h7=7
7. Locked 6's at d12 forbids e13f2=6
8. Hidden pair 69 at d38 forbids d3=45
9. Y style: i2=3 == i5=3 -- i5=8 == i1=8 -- h3=8 == h3=2 forbids i2=2 UP 29
2. e3=3 == e7=3 -- e7=1 == b7=1 -- b7=5 == c7=5 -- c3=5 == c3=4 forbids e3=4 UP 35
1. Locked 9's at ghi1 forbids g23i2=9
2. i5=3 == i2=3 -- i2=6 == i6=6 -- gh4=6 == e4=6 -- e4=8 == h4=8 UP 36
3. f2=3 == e3=3 -- g3=3 ==1{h1=9 == h1=5 -- g3=5 ==1 g3=6 -- g79=6 == h7=6} -- h7=9 == h7=6 -- f7=6 == f7=3 forbids f8=3, eg7=6 UP 40
4. i6=9 == i6=6 -- f6=6 == f6=5 -- f5=5 == {f45=pair 89} forbids i5=9 UP 44
5. a5=5 == a3=5 -- g3=5 == g3=3 -- e3=3 == e3=9 -- e5=9 == f5=9 forbids f5=5 UP 45
6. pair 89 at ef5 forbids h5=8 UP 46
7. fc on 5's: a3 == g3 -- g4 == b4 forbids a5=5 UP 81
• Sets: 9(1) + 5(2) + 3 + 4(4) + 6 = 44
• Max depth 6 at step 6
• Rating: 9(.01) + 5(.03) + .07 + 4(.15) + .63 = 1.54