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.
An Advanced Forbidding Chain
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:
- f2=3 == f7=3 => f8≠3
- e3=3 == g3=3 => no possible eliminations here
- 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
Proved solution
Proof
- Start at 22 filled - the given puzzle. Unique Possibilities to 27 filled. (UP 27).
- Locked 2's at b46 forbids b789=2
- Locked 3's at a45 forbids a8=3
- Pair 69 at ab8 forbids ab9b7fg8=6, b7g9=9 UP 28
- Locked 8's at d13 forbids e13=8
- Locked 2's at d23 forbids d3f2=2
- Hidden pair 28 at dh3 forbids d3=6, h3=4569
- Locked 1's at c12 forbids c79=1
- Locked 2's at hi7 forbids cef7=2
- Locked 7's at g79 forbids h7=7
- Locked 6's at d12 forbids e13f2=6
- Hidden pair 69 at d38 forbids d3=45
- Y style: i2=3 == i5=3 -- i5=8 == i1=8 -- h3=8 == h3=2 forbids i2=2 UP 29
- e3=3 == e7=3 -- e7=1 == b7=1 -- b7=5 == c7=5 -- c3=5 == c3=4 forbids e3=4 UP 35
- Locked 9's at ghi1 forbids g23i2=9
- i5=3 == i2=3 -- i2=6 == i6=6 -- gh4=6 == e4=6 -- e4=8 == h4=8 UP 36
- 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
- i6=9 == i6=6 -- f6=6 == f6=5 -- f5=5 == {f45=pair 89} forbids i5=9 UP 44
- a5=5 == a3=5 -- g3=5 == g3=3 -- e3=3 == e3=9 -- e5=9 == f5=9 forbids f5=5 UP 45
- pair 89 at ef5 forbids h5=8 UP 46
- 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