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24 Dec, 2005 "Nightmare"

 
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David Bryant
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Joined: 20 Jan 2006
Posts: 86
Location: Denver, Colorado

PostPosted: Thu Feb 09, 2006 4:01 pm    Post subject: 24 Dec, 2005 "Nightmare" Reply with quote

I've been going back through the archives to see what I can learn from Ruud's excellent puzzles. I
had to assume the solution for this puzzle is unique in order to solve it -- I'm wondering if someone else
can find the solution without making that assumption.

Here's how I attacked it. After completing 15 cells and making a few eliminations by means of simple
coloring I arrived at this position.

Code:
  4    268   26    68     1     7     3     5     9
 356    7    356   346    9     2     1     8    46
 368    9     1   3468   48     5     2    47    467
3569    4   3569   59    27    36     8     1    27
  1    268  2569   589  2478   46    579   479    3
23589  238    7   1459   248   134   59     6    24
  7     1     8     2     6     9     4     3     5
 269    5    269   14     3    14    679   279    8
2369   236    4     7     5     8    69    29     1


Not seeing any simple patterns I reasoned as follows.

-- Assume that the solution is unique.

-- If the "8" in column 4 lies in the top center 3x3 box there will be a "non-unique rectangle" fully
formed in r6c4, r6c6, r8c4, & r8c6, with the crucial digit "3" lying in r6c6.

-- In particular, r1c4 <> 8. Assume the converse. Then we have

r1c4 = 8 ==> r3c5 = 4 ==> r3c8 = 7 ==> r5c8 = 4 ==> r5c6 = 6 ==> r4c6 = 3

and this contradicts the uniquity assumption.

Therefore r1c4 = 6 and the rest is just an exercise. dcb
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keith
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Joined: 07 Feb 2006
Posts: 35
Location: near Detroit, Michigan, USA

PostPosted: Fri Feb 10, 2006 3:10 am    Post subject: Unique puzzles? Reply with quote

David,

I too have been rather quiet in the forums, for I have been learning by solving Ruud's excellent puzzles with pencil and paper.

I have noticed a curious side effect. I don't think I am much better at Ruud's puzzles, but I can solve my daily newspaper puzzles in a few minutes.

Ruud has stated that all his puzzles have a unique solution. Your refusal to use techniques that utilize or exploit this fact is interesting.

Keith
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David Bryant
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Posts: 86
Location: Denver, Colorado

PostPosted: Fri Feb 10, 2006 11:51 am    Post subject: I'm not refusing ... Reply with quote

Keith wrote:
Ruud has stated that all his puzzles have a unique solution. Your refusal to use techniques that utilize or exploit this fact is interesting.

Hi, Keith!

I'm not refusing to use such techniques -- I just used a non-unique rectangle to solve this puzzle. I
have an idea, though, that there will always be a logical way to find the solution without using
the "uniquity" assumption, so when I hit an apparent counterexample my curiousity is piqued. dcb
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lac
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Joined: 02 Jan 2006
Posts: 43
Location: Göteborg, Sweden

PostPosted: Sat Feb 11, 2006 11:46 pm    Post subject: Dec 24 nightmare Reply with quote

Here is another way to solve this. Looking for placing of the 4s, you
can prove that r5c6 can not be 4. If r2c4 is 4, then r8c4 must be 1,
thus r8c6 is 4, and r5c6 is 4. But if r2c4 is not 4, then r2c9 must
be 4, r3c8 must be 7 and r5c8 must be 4. Again, r5c4 cannot be 4, so it
is 6.

Code:
 
 
  4    268   26    68     1     7     3     5     9
  5     7     3    46     9     2     1     8    46
  68    9     1   3468    48    5     2    47    467
  69    4    569   59     27    3     8     1    27
  1    28    259   589   2478   6    579   479    3
2389   238    7   14589   248    14   59     6    24
  7     1     8     2     6     9     4     3     5
 269    5    269   14     3    14    679   279    8
2369   236    4     7     5     8    69    29     1



(How did you know there was no 8 in r5c4? I am assuming this is a typo
and putting it back in.)

At this point I used the the unique rectangles, and eliminated the 1 4
from r6c4. At this point you can prove that r3c5 must be 8. If r5c8
== 4, r6c9 == 2, r6c4 must be 4, so r3c5 must not be 4. On the other
hand, if r5c8 != 4, the r3c8 == 4, so again r3c5 must not be 4.

So the question is, can you prove that r6c4 must not be 4 some other
way? If you just put it in and see what happens, you get 2 6s row 9,
but this is hardly a satisfying way to solve things. Does this help
you find a different way?

Laura
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David Bryant
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Joined: 20 Jan 2006
Posts: 86
Location: Denver, Colorado

PostPosted: Sun Feb 12, 2006 1:08 am    Post subject: Thanks for the ideas, Laura Reply with quote

Laura wrote:
(How did you know there was no 8 in r5c4? I am assuming this is a typo and putting it back in.)

I think you meant to say r6c4. Laura, because I had an "8" (589} at r5c4 in my original post, but no "8" at r6c4.

Anyway, there are two "8"s in column 1, and 2 "8"s in row 1, and the way they're lined up there's either an "8"
at r6c1 or else there's an "8" at r1c4. So r6c4 can't be "8".

Thanks for the suggestions about another way to look at it. That will give me something else to work on
on a rainy day! dcb
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lac
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Joined: 02 Jan 2006
Posts: 43
Location: Göteborg, Sweden

PostPosted: Sun Feb 12, 2006 9:53 am    Post subject: typo, meant r6c4 Reply with quote

Yes. meant r6c4. Sorry about that.

Laura
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