Feb 05 Nightmare

[lac]

Nice to see one that doesn't depend on uniqueness. I found 2
forcing chains. Anybody solve it differently? I suspect there is more
than one way to solve this.

Laura

[Steve]

Your suspicions are well founded.
This is just about as tough as I like them: elementary logic plus x- and xy-wing will resolve it.

Steve

[David Bryant]

This puzzle is interesting because of multiple paths to the one solution.
After completing just ten squares I arrived at this position.

Code: Select all

 569    3   4569  1456   25     8    259    7    19
 567    8     1     9    257   567   245   245    3
  2    47    459  1345   357  1357    6   1459    8
  8     5    349   37     1     2   3479   349    6
 79    47     2     8     6    35     1    35    49
 167   16    36    357    4     9    357    8     2
 146   126    7    136    8    136  2349  12349   5
  3    126   56   156     9     4     8    12     7
 145    9     8     2    357  1357   34     6    14

The first thing we have to notice here is an X-Wing on the "9"s in columns
1 and 9, allowing us to eliminate "9" at r1c3 and r1c7. Now if we look
closely at the top right 3x3 box we can spot a hidden pair {1, 9} in r1c9
& r3c8. And there's an X-Wing on "7", in columns 4 & 7, that lets us
eliminate the "7" at r6c1. Now the grid looks like this.

Code: Select all

 69     3    46    146   25     8    25     7    19
 567    8     1     9    257   567   245   245    3
  2    47    459  1345   357  1357    6    19     8
  8     5    49    37     1     2   3479   349    6
 79    47     2     8     6    35     1    35    49
 16    16     3    57     4     9    57     8     2
 146   126    7    136    8    136  2349  12349   5
  3    126   56    156    9     4     8    12     7
 145    9     8     2    357  1357   34     6    14

At this point I actually found two different ways to proceed. The first method
is based on a "double-implication chain" rooted in r1c1:

A. r1c1 = 6 ==> r5c1 = 9 ==> r4c3 = 4 ==> r1c3 = 6

So we can set r1c1 = 9, and make quite a bit of progress, as follows:

Code: Select all

  9     3    46    46    25     8    25     7     1
 56     8     1     9    257   567   245   45     3
  2     7    45   1345   35    135    6     9     8
  8     5     9    37     1     2    47    34     6
  7     4     2     8     6    35     1    35     9
 16    16     3    57     4     9    57     8     2
  4    126    7    136    8    136    9    12     5
  3    126   56    156    9     4     8    12     7
 15     9     8     2    57    157    3     6     4

Now, if we're willing to assume the solution to this puzzle is unique, we can
obtain the solution immediately by observing that "6" must lie in r7c2 or r8c2,
forcing a "1" at r6c2. Or, if we prefer not to make that assumption, we can
trace a double-implication chain rooted in r9c1:

A. r9c1 = 5 ==> r8c3 = 6 ==> {1, 2} in r7c2, r8c2 ==> {3, 6} in r7c4, r7c6
B. r9c1 = 5 ==> r9c5 = 7 ==> r9c6 = 1 ==> {3, 5} in r3c6, r5c6 ==> r7c6 = 6

But now it's impossible to complete column 4, as the sole candidate at r8c4 is
a "5", and the sole candidate at r7c4 is a "3". Therefore r9c1 = 1, and the
rest of the solution is the same as if we had set r6c2 = 1 based on uniquity.

* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

The second method for solving this puzzle relies on two binary chains of "5"s
that can be linked together.

Code: Select all

 69     3    46    146   25     8    25     7    19
5+67    8     1     9    257   567   245  245=    3
  2    47    459  1345   357  1357    6    19     8
  8     5    49    37     1     2   3479   349    6
 79    47     2     8     6    35=    1    35~   49
 16    16     3    5~7     4    9    5=7    8     2
 146   126    7    136    8    136  2349  12349   5
  3    126   5+6  15-6    9     4     8    12     7
145-    9     8     2    357  1357   34     6    14

The two chains appear to be disjoint. But we have an "=" across from a "+" in
row 1, and a "~" opposite a "-" in column 4. So we can equate the "=" with the
"-", and we can also equate the "+" with the "~" -- this allows us to eliminate
"5" at r2c5, c6, & c7, and at r3c4.

We also have an XY-Wing in r1c3 which allows us to set r5c1 = 7. After following
up on the results of this move we reach the following position. (Note that the chain
of "5"s has to be extended a little to get {1, 7} in r9c6.)

Code: Select all

  9     3    46     46   25     8    25     7     1
 56     8     1     9    27    67    24    45     3
  2     7    45    14     3    15     6     9     8
  8     5     9    37     1     2    47    34     6
  7     4     2     8     6    35     1    35     9
 16    16     3    57     4     9    57     8     2
  4    126    7    136    8    136    9    12     5
  3    126   56    156    9     4     8    12     7
 15     9     8     2    57    17     3     6     4

And now another "XY-Wing", this one rooted in r9c6, cracks the puzzle wide open
by allowing us to set r1c5 = 2. dcb

[lac]

I got to the same place where David started his double implication chain.
And I too started with r1c1. But what I showed was that no matter
which value you set r1c1 to, r4c3 must be 9. If 6, then r1c3 is 4,
thus r4c3 is 9. If 9, then r5c1 is 7, r5c2 is 4¸ and again r4c3 is 9.

This however, immediately gets you the r1c1 to be 9, so we ought to
be solving the same way.... and indeed we get to the same place.
But there is a third path to take from there.

No matter what value you place in r9c1, r3c3 must be 4.
If set to 1, then r2c1 must be 5, and thus r3c3 must be 4. If set to 5,
r8c6 must be 6, then r1c3 must be 1, and r3c6 must be 5, and once
again, r3c3 must be 4.

And this cracks it wide open.

I find it fascinating that each time we found the same weak cell, but
then did different things with it. I wonder if this is a coincidence, or
if there is some proof that we could construct that double implication
chains implies a cell whose value is forced? or more likely could be
forced if there are only 2 candidates in the cell ...

pondering, pondering ....

Laura

[Steve]

My way of getting past the hurdle in David's second matrix was to put r5c1 = 7 using the XY-Wing for (46) pivoted on r1c3 with pincers r1c1 and r4c3.

The second hurdle (= David's third matrix) took me a long time to resolve. I eventually noticed that there were only two places for 5 in row 3 and in column 1. This fork configuration excludes 5 from r9c6 and produces the XY-Wing for (17) pivoted on that cell which David also used.

Finally, I have just checked the puzzle on a solver after prohibiting the use of XY-Wings and chains. It was forced to a guess at the first hurdle. There is at least one further way round the second but it is too tedious to record.

From now on I have promised to stick to puzzles that I can solve within 30 minutes.

Steve

[David Bryant]

From now on I have promised to stick to puzzles that I can solve within 30 minutes.

That's nice. But isn't it hard to know which puzzles those are?

It reminds me of what Will Rogers said about the stock market: "I only buy stocks that go up. If they're going to go down, I don't buy them!" dcb

[David Bryant]

I find it fascinating that each time we found the same weak cell, but
then did different things with it. I wonder if this is a coincidence, or
if there is some proof ...

One thing I've noticed in a lot of puzzles with the "non-unique rectangles" is that there is often a close
connection between the value we can infer by assuming that the solution is unique and a fruitful
starting point for the double-implication chain technique.

Here's an example from the Nightmare for 4 Feb, 2006. I made the obvious moves to get to this point.

Code: Select all

  4     1     8    23    23     5     9     7     6
  3     6     7    14     9    148   18     5     2
  5     2     9     6     7    18     4    18     3
 26     9    46    13    13     7     5    24     8
 28     5    13    89     4     6    137   123   79
  7    48    13     5    28    29    13     6    49
 89     7     2    48     5     3     6    489    1
  1    48     5    29     6    29    378   38    47
 689    3    46     7    18    14     2    489    5

Now there's a non-unique rectangle starting up in rows 5 & 6, columns 3 & 7.
If we're willing to assume the solution is unique we can set r5c7 = 7 and
go merrily on our way.

What if we don't want to make that assumption? Well, there are several forcing
chain arguments one can make to demonstrate that r5c7 = 7. All of them seem
to involve the cells r9c5 & r9c6 in some way or another.

Simplest and most direct is the assumption that r9c3 = 4:

A. r9c3 = 4 ==> r9c6 = 1
B. r9c3 = 4 ==> r8c2 = 8 ==> r6c5 = 8 ==> r9c5 = 1

Notice that this leads directly to the proper placement of the "7" in row 5:

r9c3 <> 4 ==> r8c2 = 4 ==> r8c9 = 7 ==> r5c7 = 7

We can also start by putting the "7" in the wrong place.

C. r5c9 = 7 ==> r7c9 = 4 ==> r9c3 = 4
D. r8c7 = 7 ==> r7c9 = 4 ==> r9c3 = 4

So the contradiction we reached seems to be pretty directly related to the
non-unique rectangle. I wonder -- is this always the case? dcb

PS I found a lot of additional examples among the Nightmares for December, and will post a few of those another day.