Welcome back!

This proof is multi-page. Some interesting stuff was analyzed on the
First Page, and on the
Second Page.

This page is considerably less complex than the preceding pages. Nevertheless, this
puzzle is still fertile ground for studying Sudoku. Again, this page begins with some easy
eliminations.

### Naked Pair 68

The Pair 68 at gh1 eliminates the indicated 6s, allowing two more cell solutions: e1=7, f3=6.

### Some Locked Candidates

Several Locked Candidate eliminations are available. Illustrated above are Locked candidates
involving 7s, 4s, 6s and 9s. The proven eliminations help to reveal the next step.

### Y Wing Style on 45

Above, the newly strong 4s in row 3 uncover:

- h3=4 == d3=4 -- d6=4 == f6=4 -- f6=5 == g5=5 => g3≠5

This elimination can be found easily. The typical markers for a very common Y wing style, the
cells g3h6 = 58, tell one to focus here. Often, when the common Y Wing Style elimination is
not available, a coloring link will eliminate a candidate in the marker cells. Clearly, one more
cell is solved.

### A Continuous Loop

Continuous Loops are always my favorite, as they illustrate the power of AIC. Most proofs
by contradiction will only allow one elimination at a time. Continuous loops often provide
seemingly unrelated eliminations. Illustrated above is one such loop:

- d2=4 == d6=4 -- g6=4 == g6=5 -- g3=5 == g3=3 -- d3=3 == d2=3 => d2≠1, g2≠5

### A Standard Forbidding Chain using only bivalues and bilocations

Above, we finally have a rather non-descript standard chain, with no almosting, except perhaps
almost boring....

- g2=9 == g2=3 -- d2=3 == d2=4 -- f2=4 == f2=2 -- f5=2 == f5=9 -- e6=9 == e9=9 => g9≠9

There are many alternative ways to get this same elimination.

### A Standard Forbidding Chain using only bivalues and bilocations - Again

Another elimination in the same cell with many possible alternatives. One of them is:

- e9=6 == d8=6 -- d8=7 == d5=7 -- f4=7 == f4=4 -- g4=4 == g4=8 -- g1=8 == g1=6 => g9≠6

One more cell is solved, but to me it is a key cell!

### Almost Hidden Pair 36 used in a chain

This elimination has been available for quite some time. There are alternative ways to prove
this one also, some of them using an almost Y wing. However, this is what I saw first:

- {Hidden Pair 36 at ci4} == e4=6 -- d5=6 == d5=7 -- f4=7 == f4=4 -- g4=4 == g4=8 => c4≠8

Although this 8 seems innocuous enough, it does create a bilocation situation with the 8s both in
column c and in row 4.

### Almost Hidden Pair 36 used in a chain - again

Once again, this one has been lurking about for a while. It did not seem like it was worth
executing until that 7 solved at g9. The chain could be written as:

- b6=6 == {Hidden Pair 36 at ci6} -- i6=9 == e6=9 -- e9=9 == e9=6 => b9≠6

Ok, I know one cannot see both of these Hidden Pair 36 arguments without also knowing that a
Almost Unique Rectangle argument must exist. I see no real value, however, in using it to
eliminate the 6 from c4.

After b9≠6, one clearly has a naked pair 58 at ab9 => c9≠58. Then, c2=8% column.

### X wing on 5s

The X wing on 5s illustrated above was revealed by the last step, plus the cell solution.
The X wing can be written as a continuous loop forbidding chain on 5s:

- c3 == c6 -- g6 == g3 => ab6,b3≠5

### Very Common Y Wing Style

Perhaps it is a fitting end to this puzzle that the final step be one of my pet projects, a
**Very Common Y Wing Style**. The simplicity both in logic and in finding this
elimination technique should make it standard issue in everyone's bag of Sudoku solving tricks!
BTW, there is at least one other Very Common Y Wing Styles available here, but this one serves
to unlock the puzzle:

- a6=2 == a6=1 -- d6=1 == e4=1 -- e2=1 == e2=2 => e6≠2

Now the puzzle is reduced to a cascade of naked singles to the end!

### Unsolvable 13 - Solved

### Unsolvable 13 - An AIC based Proof

- Given 23 cells at Puzzle start. UP 25 (d7=8% box and f7=3% box)
- Hidden pair 12 at i13 => i1≠46, i3≠346, h123≠1, g123≠2
- Hidden pair 29 at e6f5 => e6≠16, f5≠67
- Naked pair 67 at d58 => d36≠6, d3≠7
- Hidden pair 67 at e1f3 => e1≠12, f3≠24
- Locked 2s at ef2 => bc2≠2
- Locked 3s at g23 => g46≠3
- Let A = {e9=6 == e9=9 -- f8=9 == f5=9 -- i5=9 == i5=6 -- d5=6 == d8=6 -- e9=6 == e9=9}
- thus, A is remote pairs 69 at both e9,i5

- e9=7 == A -- i9=69 == i9=4 -- a9=4 == a1=4 -- gh1=4 == pair 68 at gh1 -- e1=6 == e1=7
- => f8≠6, e4≠7, bg9≠4, ac1≠8, gh2≠8

- fc on 9s: e6 == e9 -- i9 == i56 => g6≠9
- Let A = {a56=5 and a45=8}, Let B = {Hidden Pair 58 at ah5}
- i9=4 == a9=4 --a9=58 == A -- b5=58 == B -- h5=9 == i56=9 => i9≠9

- Use the same A and B the previous step
- i456=6 == i9=6 -- i9=4 == a9=4 -- a9=58 == A -- b5=58 == B => h5≠6

- Let
- A = {i5=6 == i5=9 -- f5=9 == f8=9 -- c8=9 == c9=9 -- eg9=9 == {{eg9=pair67}}
- B = fc on 6's: i5 == d5 -- d8 == e9
- C = fc on 6s: B == b5 -- b7 == gh7

A == c7=9 -- c7=6 == C => i9≠6 UP 28 i9, a1, b8=4

- Pair 68 at gh1 => e1≠6, e1=7, f3=6 UP 30
- Locked 4s at h23 => g23≠4
- Locked 6s at i456 => g46≠6
- Locked 7s at df8 => cg8≠7
- Locked 9s at i56 => h5≠9
- Y wing style: h3=4 == d3=4 -- d6=4 == g6=4 -- g6=5 == h5=5 => h3≠5 h3=4 (UP 31)

- d2=4 == d6=4 -- g6=4 == g6=5 -- g3=5 == g3=3 -- d3=3 == d2=3 => d2≠1, g2≠5
- e9=9 == f8=9 -- f8=7 == f4=7 -- f4=4 == f2=4 -- d2=4 == d2=3 -- g2=3 == g2=9 => g9≠9
- e9=6 == d8=6 -- d5=6 == d5=7 -- f4=7 == f4=4 -- g4=4 == g4=8 -- g1=8 == g1=6 => g9≠6

g9=7 UP 32
- {Hidden pair 36 at ci4} == e4=6 -- e4=1 == d6=1 -- d6=4 == f4=4 -- g4=4 == g4=8 => c4≠8
- b6=6 == {Hidden pair 36 at ci6} -- i6=9 == e6=9 -- e9=9 == e9=6 => b9≠6
- pair 58 at ab9 => c9≠58 c2=8 UP 33

- X wing on 5s at cg36 => b3, ab6≠5
- Very common Y Wing style: a6=2 == a6=1 -- d6=1 == e4=1 -- e2=1 == e2=2 => e6≠2

Naked singles to end UP 81

Proof statistics:

- Sets
- 4(2) + 2(1) + 9 + 2+ 6 + 6 + 8 + 2 + 4(1) + 3 + 4 + 3(5) + 4 + 2(2) + 3
- 6(1) + 8(2) + 2(3) + 2(4) + 3(5) + 2(6) + 8 + 9 = 80

- Maximum depth: 9 at step 2.7
- Rating: .06 + .24 + .14 + .3 + .93 + 1.26 + 2.55 + 5.11 = 10.59

### Notes

This puzzle is certainly difficult. A rating of about 10, using my rating scale, is roughly
equivalent to one step requiring 10 simulataneous native strong inferences. As such, it borders
upon, in my opinion, being opaque to the human mind. Of course, that is a completely arbitrary
opinion.

Unsolvable 13 is fertile with possible examples of grouping native strong inference constraints
within **forbidding chains, or alternating inference chains (AIC)**.
This is equivalent to what I call **Advanced Forbidding Chains**.
Applying a tiny fraction of creativity, and some elementary logic, across such chains can
be a simple, but powerful, tool to prove the occasional vexing sudoku puzzle.

This brings me to my personal point of view involving sudoku solving. Technique based solving
is the general goal. To achieve that goal, I believe the sharpest instrument is: identifying
the shortest possible path to achieve a particular elimination. By analyzing such a path, one
can derive patterns, or techniques, which may or not repeat themselves in other puzzles.
Eventually, combining disparite ideas together becomes a technique that is useful.
The analysis needs some sort of foundation. I use forbidding chains as the foundation.
The analysis needs some parameters for determining *shortest possible path*. I use
native strong sets as that parameter. I use native strong sets as that parameter because puzzles
are defined by setting some cells to given values. This means, puzzles are defined by limiting
some of the native strong sets to singular values. Native weak sets are, in my opinion, an
irrevelent parameter. Why? A solved sudoku puzzle is a puzzle such that all 81(4) of the native
strong sets have been reduced to singular values. The native weak sets are actually not reduced
at all! Forbidding chains and native strong inference sets work together adequately, as forbidding
chains are based upon the interaction between strong inferences, which are puzzle specific
in sudoku, and weak inferences, which are puzzle neutral in sudoku.

Finally, when a potentially useful idea is found, I am in favor of viewing the idea broadly.
For example, the concept of a hinge in sudoku is more broadly recognized as the potential for
the interaction between a row in a box with a column in a box. There is no need to limit that
concept to singular candidates, nor to certain specific small groups of candidates. Instead,
the base inspiration, in my opinion, is merely: a specific type of interaction by location
(hinge) is useful. A further generalization is still possible: The interaction between
any subset of locations in a box with any other subset of locations within a box is useful. This
begs a further generalization. The interaction between any subset of locations within any large
house with any other subset of locations within any large house is useful. Probably, I
generalize to the point of uselessness too often. Nevertheless, recognition of the base concept
is important, as failure to recognize it will blind one to potential patterns. Thus, there is
much to be gained by recognizing small specific patterns, but it is important, in my opinion, to
keep in mind the larger generalities lurking behind every small specific pattern.

There is an important underlying prejudice to my point of view. Clearly, if one wishes to
solve just one puzzle, the quickest route, after perhaps some easy techniques are exhausted,
is to invoke *Trial and Error*. My prejudice is: I am not particularly interested
in solving just one specific sudoku puzzle! I am much more interested in discovering the
underlying patterns that allow deductive reasoning to solve puzzles. Thus, my prejudice
gravitates towards the elimination of candidates from cells based upon reasoned deductions.

Having said all that, sudoku analysis of candidate eliminations (for me) boils down to:

Now, find the smallest part of what one knows about Puzzle that still forbids that something.
Express those smallest parts as a logical construct. Repeat as necessary. Catalog discoveries.
Have fun!

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