# Unsolvable Sudoku #13 Page Last

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

1. Given 23 cells at Puzzle start. UP 25 (d7=8% box and f7=3% box)
1. Hidden pair 12 at i13 => i1≠46, i3≠346, h123≠1, g123≠2
2. Hidden pair 29 at e6f5 => e6≠16, f5≠67
3. Naked pair 67 at d58 => d36≠6, d3≠7
4. Hidden pair 67 at e1f3 => e1≠12, f3≠24
5. Locked 2s at ef2 => bc2≠2
6. 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
7. fc on 9s: e6 == e9 -- i9 == i56 => g6≠9
8. 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
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
10. 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
2. Pair 68 at gh1 => e1≠6, e1=7, f3=6 UP 30
1. Locked 4s at h23 => g23≠4
2. Locked 6s at i456 => g46≠6
3. Locked 7s at df8 => cg8≠7
4. Locked 9s at i56 => h5≠9
5. Y wing style: h3=4 == d3=4 -- d6=4 == g6=4 -- g6=5 == h5=5 => h3≠5 h3=4 (UP 31)
1. d2=4 == d6=4 -- g6=4 == g6=5 -- g3=5 == g3=3 -- d3=3 == d2=3 => d2≠1, g2≠5
2. e9=9 == f8=9 -- f8=7 == f4=7 -- f4=4 == f2=4 -- d2=4 == d2=3 -- g2=3 == g2=9 => g9≠9
3. 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
1. {Hidden pair 36 at ci4} == e4=6 -- e4=1 == d6=1 -- d6=4 == f4=4 -- g4=4 == g4=8 => c4≠8
2. b6=6 == {Hidden pair 36 at ci6} -- i6=9 == e6=9 -- e9=9 == e9=6 => b9≠6
3. pair 58 at ab9 => c9≠58 c2=8 UP 33
1. X wing on 5s at cg36 => b3, ab6≠5
2. 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:

• Puzzle forbids something
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! 