Unsolvable Sudoku #16


The following illustrated proof for the almost diabolical & extreme Unsolvable #16 employs an interesting variety of techniques. If this is your first visit to this blog, WELCOME!!

I must apologize, as the terminology used here is not the standard terminolgy used in most other Sudoku sites. Therefor, previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries. Specifically, one may want to refer to the following pages:

The illustrations of forbidding chains, also called Alternating Inference Chains (AIC), shown in this proof will share this key:

  • black line = strong inference performed upon a set (strong link)
  • red line = weak inference performed upon a set (weak link)
  • black containers define a partioning of a strong set(s)
  • candidates crossed out in red = candidates proven false
Please be aware that, for me, strong and weak need not be mutually exclusive properties.

This puzzle is meant to be very difficult, but it is much easier than the previous unsolvable.


Unsolvable #16


Puzzle

Two Unique Possibilities are available here:

  • c3 = 9% column & box (hidden single)
  • i5 = 6% column & box
I like to look beyond the possibility matrix to find hidden sets at this time.


Hidden Pair 78, Hidden Triple 678, Locked candidate 4


3 steps

To save some space, I have chosen to illustrate a few steps simultaneously. The givens that are highlit above help one locate the hidden sets illustrated. Also, I have illustrated the Locked 4s eliminations. There are more locked sets eliminations available. This proof does not require them, so I have not bothered to detatil those.


Almost Hidden Pair 23


Almost Hidden Pair 23

The graph above attempts to illustrate the following chain:

  • b8=3 == a8=3 -- a3=3 == a3=2 -- g3=2 == g9=2 -- i8=2 == {Hidden Pair 23 at ab8}
    • => b8=23
Using the same strong inference sets, one can find a chain that also eliminates a15=3. Let me know if you find it! It is often instructive to see how the same strong inferences can yield two differing sets of eliminations. Usually some/most are common to both chains. In this case, the chain I choose to not illustrate is the stronger one. You may also find a Sue De Cox above. However, after solving c9=b1=4% column, a very short chain gets you to the same point.


My favorite Y wing style - also a Hub, Spoke Rim


Favorite Y Wing Style

Although b8=a3=23 is generally a marker for a Very Common Y Wing Style, in this case a different Y wing Style exists. Illustrated above is a continuous loop, written as:

  • a3=2 == a3=3 -- a8=3 == b8=3 -- b8=2 == a89=2 This proves:
    • a3=3 == a8=3 => a15≠3
    • a3=2 == a89=2 => a15≠2
    • b8=3 == b8=2 , but we already had that

Usually, the continuous loop Y Wing Style will use at least one grouped argument. This one uses a89=2. Also, this type of elimination and Sue De Cox tend to hang out together.

After making these eliminations, a naked pair 78 is revealed in row 1 => c1≠8


Hinged coloring with candidate 8


8s

It is a common trick to use a candidate that is limited to one each of a row and a column within a box. Above, this forbidding chain (AIC) on 8s is shown:

  • abc7 == a89 -- a1 == g1 => g7≠8
This same pattern could also be written:
  • bc7 == a789 -- a1 == g1
The key is the 8s must be in at least one of {row 7, column a}. Note here that it matters not even if 8 were in both row 7 and column a.


A continuous loop considering 6 strong inferences


deep continuous loop

Illustrated above is another continuous loop, or wrap around, chain. One way of many to write it is:

  • a1=7 == g1=7 -- g1=8 == g9=8 -- g9=2 == g3=2 -- a3=2 == c1=2 -- c56=2 == b5=2 -- b5=7 == a5=7 =>
    • g1=7 == g1=8, but we have that already
    • g9=8 == g9=2 => g9=28
    • g3=2 == a3=2 => fi3≠2
    • c1=2 == c56=2, but we also have that previously
    • b5=2 == b5=7 => b5=27
    • a1=7 == a5=7 => a2≠7
Continuous loops have been treated in some previous blog pages. If the reasons for all of the eliminations are not clear, perhaps review some of those pages.


A supefluous Y Wing Style


unneeded step

There are many Y wing styles available in this puzzle. I decided to illustrate this one primarily because it makes the following step easier. Just for fun, try to do the following step without executing this step. Hint: it is very similar, but is an Advanced Forbidding Chain.

The step above as a forbidding chain (AIC):

  • f9=7 == h9=7 -- g7=7 == g1=7 -- g1=8 == g9=8 => f9≠8


A standard loop considering 6 strong inferences


deep standard loop

Illustrated above is the following chain:

  • a3=2 == g3=2 -- g9=2 == g9=8 -- g1=8 == h2=8 -- h2=7 == h9=7 -- f9=7 == f9=1 -- f3=1 == f3=3
    • =>a1 ≠3
After making the indicated elimination, the puzzle cascades with Unique Possibilities (both naked and hidden singles) until 50 cells are solved. (UP 50).


Very Common Y Wing Style


Very Common Y Wing Style

There are many easy ways to finish this puzzle from this point. The most efficient way that I found to complete the puzzle with just one easy elimination is illustrated above:

  • c4=8 == c4=6 -- d4=6 == d7=6 -- b7=6 == b7=8 => c7≠8
Again, it may bear repeating that this type of Y Wing Style should be considered standard issue. The puzzle is reduced to naked singles to the end.


Solution


Solution


PROOF

  1. Start with the given puzzle. Unique Possibilities to 25 filled. (UP 25)
    1. Hp 78 at g1h2 => g1h2=78
    2. Ht 678 at abh2 => ab2=678
    3. Locked 4s at gi3 =>i12≠4
    4. alternate b8=2 == {fc on 2s:a8 == i8 -- g9 == g3} -- a3=2 == a3=3-- a8=3 == b8=3
      • => b8≠48, a15≠3
    UP 27
    1. Y style: a3=2 == a3=3 -- a8=3==b8=3 -- b8=2 ==a89 =2 => a15≠2
    2. pair 78 at ag1 => c1≠8/li>
    3. fc on 8's: bc7 == a789 -- a1 == g1 => g7≠8/li>
    4. g3=2 == g9=2 -- g9=8 == g1=8 -- g1=7 == a1=7 -- a5=7 == b5=7 -- b5=2 == b8=2 == a89=2 == a3=2
      • => if3≠2, g9≠17, a2≠7, b5≠38
    5. a3=2 == a3=3 -- f3=3 == f3=1 -- f9=1 == {VCY style f9=8 == f9=7 -- h9=7 == h2=7 -- g1=7 == g1=8} -- g9=8 == g9=2
      • => g3≠2
    UP 50
  2. vcystyle: c4=8 == c4=6 -- d4=6 == d7=6 -- b7=6 == b7=8 => c7≠8 UP 81
  • Sets: 2 + 3 + 1 + 4 + 3 + 2 + 2 + 6 + 6 + 3 = 32
  • Rating: .01 + 3(.03) + 3(.07) + .15 + 2(.63) = 1.72
  • Max depth 6 at steps 3d,3e.


NOTES

This puzzle does not quite deserve the label, unsolvable. I am fairly certain that the path that I illustrated is not the most efficient path. Unlike unsolvable 13, there was never a point in this puzzle that I had to look for chains. Therefor, I did not take much time to optimize my proof. However, this is an excellent puzzle for study, as it has some nice and heavily indicated chains to find. Hopefully, many of you are able to chain straight through!




7 Comments
Indicate which comments you would like to be able to see

It looks like there were some good puzzles this week on the tough page. I am sorry that I missed them. I have been trying to educate myself further in sudoku. Thus, I have been spending much time reading sudoku forums. I have even posted a bit there.

Since the puzzle grid labels that we use More...
20/Apr/07 6:04 AM
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Thanks Steve,
I enjoyed reviewing this proof. I have noted that most Sudoku puzzles start out very symmetrical in the provided numbers. Both #13 and #16 that you presented don't start symmetrical. Is this a Sudoku requirement for a starting point or is this up to the generator to make their More...
22/Apr/07 5:19 AM
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Hi jm!
There seems to be a prejudice towards puzzles that have the symmetrical givens. I believe this to be entirely an aesthetics thing.

Generally, I only post the eliminations required for my proof. Many other possible eliminations are left untaken in the proof. Generally, my first More...
22/Apr/07 5:04 PM
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change e5 to 3 and then take e8 and swap it with c8 and then put a 6 in b3 and then the puzzle hould start making sence
try it sometime
29/Apr/07 11:02 PM
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Possible elimination of a15=3 using hidden pair 23: ?
a3=3 == a3=2 -- g3=2 == g9=2 -- i8=2 == {Hidden Pair 23 at ab8} -- a8=158 == {Pair 23 at a38}, forbids a15=3.
30/Apr/07 9:54 PM
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Very nice Dave.

If anyone understands the wizard's comment, please help me out!
01/May/07 8:45 AM
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I think what the wizard was trying to propose is modifying the original puzzle to make one that was easier to solve. Trouble is (I just tried) is that the changes are impossible. Since h3 is 6, b3 can't be 6, as he proposes. Even if the changes did make sense, that wasn't the point anyway.
21/May/07 8:10 AM
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