View previous topic :: View next topic 
Author 
Message 
AZ Matt
Joined: 03 Nov 2005 Posts: 63 Location: Hiding under my desk in Phoenix AZ USA

Posted: Fri Sep 29, 2006 11:46 pm Post subject: Sept. 28th Nightmare 


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 xwing 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? 

Back to top 


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

Posted: Sat Sep 30, 2006 1:46 pm Post subject: 5star constellation 


That's a very nice analysis, Matt. I particularly like the way you noticed how the two "XWings" 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 5star constellation in r3c6 (the "alpha star"), r7c6, r2c8, r5c8, and r7c8. It's very much like the familiar XYWing, except that it involves 5 cells instead of 4. dcb 

Back to top 


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

Posted: Sat Sep 30, 2006 3:11 pm Post subject: 


David Bryant wrote:  I'd call it a beautiful 5star 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 XYchain (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 XYwing 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 

Back to top 


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

Posted: Sat Sep 30, 2006 7:49 pm Post subject: Terminology 


TKiel wrote:  I've always thought of a constellation as an XYchain (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 XYwing 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.
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 XYWing. And the "doubleimplication chains" in the "constellations" don't have to pass through bivalued 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 

Back to top 


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

Posted: Sat Sep 30, 2006 8:51 pm Post subject: 


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. 

Back to top 


ravel
Joined: 21 Apr 2006 Posts: 536

Posted: Sat Sep 30, 2006 9:07 pm Post subject: Re: Sept. 28th Nightmare 


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 

Back to top 


Myth Jellies
Joined: 27 Jun 2006 Posts: 64

Posted: Sat Sep 30, 2006 10:02 pm Post subject: 


Code:  **
482..1...
.6392....
.........
++
73...69..
..15.....
......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 xywing
(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. 

Back to top 


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

Posted: Mon Oct 02, 2006 6:19 pm Post subject: Thank you 


Myth Jellies
I had not spotted a UR like that before. Thanks.
And I see I should have spotted the xywing. 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 xyzwings either. Sometimes they are fairly easy to spot, but, in general, it feels like random searching. 

Back to top 




You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum

Powered by phpBB © 2001, 2005 phpBB Group
