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Sept. 28th Nightmare

 
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AZ Matt



Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA

PostPosted: Fri Sep 29, 2006 11:46 pm    Post subject: Sept. 28th Nightmare Reply with quote

This is yesterday's (Sept. 28th's) Nightmare:

482001000
063920000
000000000
730006900
001500000
000000305
000015000
390700001
000000083

I like this puzzle because it is so easy -- to a point.

It also has a solve that has helped me further develop a thought about strong links and what I will call anomalies, and it is in response to this quote from an earlier thread:

Myth Jellies, in a response to a defense of "double implication chains" by David Bryant, wrote:

Quote:
Since every basic method, as well as many coloring and chaining and ALS methods can be performed without making any assumption about any particular candidate being true or false; I continue to maintain that performing any step which requires such an assumption is performing brute force T&E.


So let's begin. It is fairly easy to get to here:
Code:
4 8  2 6  37 1   5  379 79
5 6  3 9  2  47  1  47  8
9 1  7 8  5  34  46 234 246

7 3  5 24 8  6   9  1   24
6 24 1 5  39 39  8  247 247
8 24 9 1  47 247 3  6   5

2 7  8 3  1  5   46 49  469
3 9  4 7  6  8   2  5   1
1 5  6 24 49 249 7  8   3


It easy to see the potential ripple effect in boxes 3, 4, 5, and 8, and without making any "assumptions" about what candidate goes in what cell.

There are many strong links on the {2}, {3}, {4}, {7}, and {9}, and the pattern of effect begins in the {24} in r5c2, goes through the center line of box 6, up to row 4 at r4c9, back to column 3 and box 5 at r4c4, down to box 8 and row 9 at r9c4. A second ripple from the {24} in r5c2 goes through the {24} in r6c2 and ends in box 5. And so (ah ha!), the ripples meet. This is a familiar spiral pattern that I see all the time. It is something like a remote pairs with weak links in this case, and it tells me this: It is quite likely that if you "assume," for example, that r5c2 is a 2, you will solve for all the cells in boxes 3, 4, 5, and 8 (though you might end up with a locked pair or two).

What I don't see is any "glitch" in the pattern that gives me a uniqueness problem. I haven't done it, but I am quite sure there are two ways to solve for just those those cells. And there is no reason to "assume" r5c2 is one or the other candidate. The puzzle's secret lies elswhere.

So I look at the remaining cells in boxes 2, 3, and 9. A lot of strong links (or soon to be revealed strong links) on the {3}, {4}, {6}, {7}, and {9}.

And here is the anomoly: There is an x-wing on the {6} in r37c79 and on the {9} in r17c89, so they are linked in r7c9, so however the numbers fall, if the {6} falls one way, the {9} must fall the other. Is there a number linking them? -- yes, the {4} directly in r7c7 and 8. And...??? Oh my gosh golly -- the {7} in r1c9 via the {47} in r2c8, via the {46} in r3c7.

Puzzle solved:
r3c7=6 ==> r7c8=9
r3c7=4 ==> r7c8=9
r7c8=9

You can call that brute force T&E all you want. And that's fine. But doesn't it "seem" like something more than that?
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sat Sep 30, 2006 1:46 pm    Post subject: 5-star constellation Reply with quote

That's a very nice analysis, Matt. I particularly like the way you noticed how the two "X-Wings" intersect each other ... I missed that entirely, although I did find the same "anomaly" when I worked this puzzle.

Oh -- I wouldn't call it "brute force", or "T&E". I'd call it a beautiful 5-star constellation in r3c6 (the "alpha star"), r7c6, r2c8, r5c8, and r7c8. It's very much like the familiar XY-Wing, except that it involves 5 cells instead of 4. dcb
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TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Sat Sep 30, 2006 3:11 pm    Post subject: Reply with quote

David Bryant wrote:
I'd call it a beautiful 5-star constellation in r3c6 (the "alpha star"), r7c6, r2c8, r5c8, and r7c8.


Are these cell locations correct? I've always thought of a constellation as an XY-chain (bivalue cells only) but this one passes through a cell that is solved (r7c6) and one that has 3 candidates (r5c8). Am also wondering if the 'alpha star' in a constellation would be similar to the 'stem' cell in an XY-wing or is it where the elimination takes place?


Last edited by TKiel on Sat Sep 30, 2006 8:46 pm; edited 1 time in total
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sat Sep 30, 2006 7:49 pm    Post subject: Terminology Reply with quote

TKiel wrote:
I've always thought of a constellation as an XY-chain (bivalue cells only) but this one passes through a cell that is solved (r7c6) and one that has 3 candidates (r5c8). Am also wondering if the 'alpha star' in a constellation would be similar to the 'stem' cell in an XY-wing or it it where the elimination takes place?

Oops! I described it incorrectly ... the legs of the chain run r3c7 -> r7c7 -> r7c8, and r3c7 -> r2c8 -> r1c9 -> r7c8.

A. r3c7 = 6 ==> r7c7 = 4 ==> r7c8 = 9
B. r3c7 = 4 ==> r2c8 = 7 ==> r1c9 = 9 ==> r7c8 = 9

Sorry about that ... it was too early for me to think straight. Rolling Eyes

Oh -- when "someone_somewhere" introduced the "constellation" idea last fall, he also named the "alpha star" -- it would correspond to the "stem cell" in an XY-Wing. And the "double-implication chains" in the "constellations" don't have to pass through bi-valued cells only. Sometimes we can find the implication because there are only two positions in a row/column/3x3 box in which a particular value can fall. For instance, in this case, the last step

r1c9 = 9 ==> r7c8 = 9

works because there are only two spots in box 9 where a "9" can possibly fit. dcb
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TKiel



Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI

PostPosted: Sat Sep 30, 2006 8:51 pm    Post subject: Reply with quote

Thanks dcb.

The thing about only using bivalue cells in a constellation must have been an assumption on my part, maybe because most of the other ones I've seen described did that.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Sat Sep 30, 2006 9:07 pm    Post subject: Re: Sept. 28th Nightmare Reply with quote

Code:
4 8  2 6  37 1   5  379 79
5 6  3 9  2  47  1  47  8
9 1  7 8  5  34  46 234 246

7 3  5 24 8  6   9  1   24
6 24 1 5  39 39  8  247 247
8 24 9 1  47 247 3  6   5

2 7  8 3  1  5   46 49  469
3 9  4 7  6  8   2  5   1
1 5  6 24 49 249 7  8   3
I think the "alpha star" is only known in this forum. Is it the same as a double implication chain?

In the other forum it is more common to write it as a contradiction loop like this (which i know is displeasing for some people):
r7c9=9 => r1c9=>7 => r2c8 =4 => r3c7=6 => r7c7=4 => r7c8=9
=> r7c8=9
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Sat Sep 30, 2006 10:02 pm    Post subject: Reply with quote

Code:
 *-----------*
 |482|..1|...|
 |.63|92.|...|
 |...|...|...|
 |---+---+---|
 |73.|..6|9..|
 |..1|5..|...|
 |...|...|3.5|
 |---+---+---|
 |...|.15|...|
 |39.|7..|..1|
 |...|...|.83|
 *-----------*
 *--------------------------------------------------*
 | 4    8    2    | 6    37   1    | 5    379  79   |
 | 5    6    3    | 9    2    47   | 1    47   8    |
 | 9    1    7    | 8    5    34   | 46   234  246  |
 |----------------+----------------+----------------|
 | 7    3    5    | 24   8    6    | 9    1    24   |
 | 6    24   1    | 5    39   39   | 8    247  247  |
 | 8    24   9    | 1    47   247  | 3    6    5    |
 |----------------+----------------+----------------|
 | 2    7    8    | 3    1    5    | 46   49   469  |
 | 3    9    4    | 7    6    8    | 2    5    1    |
 | 1    5    6    | 24   49   249  | 7    8    3    |
 *--------------------------------------------------*

AZ Matt, there are quite a few ways to solve this from here without assuming any cell is any particular value.

If you think uniqueness arguments are okay, the easiest thing might be to note that r89c35 is in danger of being filled in with all 4's and 6's; therefore r9c5 = 9

There is also the xy-wing
(9=7)r1c9 - (7=4)r2c8 - (4=9)r7c8
one or both of the endpoint 9's must be true, therefore any cell which can see both of them cannot be a nine, so r1c8 and r7c9 <> 9

And there is the chain that uses the same candidates that you used in your deduction:
(4=6)r7c7 - (6=4)r3c7 - (4=7)r2c8 - (7=9)r1c9 - (9)r7c9 = (9)r7c8

This alternating inference chain tells you that either r7c7 = 4 or r7c8 = 9 or both. These endpoints see each other, therefore one endpoint cannot assume the other endpoints value. Thus r7c8 <> 4.
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AZ Matt



Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA

PostPosted: Mon Oct 02, 2006 6:19 pm    Post subject: Thank you Reply with quote

Myth Jellies

I had not spotted a UR like that before. Thanks.

And I see I should have spotted the xy-wing. It was right inside of the numbers I was looking at for the solve (in fact, it is probably the more logical way of decribing my solve). I am not much of a technical solver. I get distracted by the patterns of effects in the strong links. I don't like looking for xy and xyz-wings either. Sometimes they are fairly easy to spot, but, in general, it feels like random searching.
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