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

Posted: Mon Nov 28, 2005 8:27 pm Post subject: A "5 Star Constellation" example 


Here's a puzzle I got from HollyDoll.
Code:  ... ... .73
4.8 ... ...
.31 45. ...
..7 ... 298
..5 2.7 3..
924 ... 6..
... .13 96.
... ... 7.5
84. ... ... 
After 32 moves you'll need to locate a "5 Star Constellation" to crack it. Enjoy! dcb
PS I looked it over again, and you can also solve it using "colors." But the "5 Star Constellation" is there, and fun to look for. 

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alanr555
Joined: 01 Aug 2005 Posts: 198 Location: Bideford Devon EX39

Posted: Tue Nov 29, 2005 3:06 pm Post subject: 


Code: 
This one seemed like a breeze. More and more Mandatory Pairs
came to light and cells resolved very easily  until I reached the
following position.
250 000 473
408 002 510
731 450 820
317 040 298
685 297 341
924 000 657
572 813 964
103 024 785
840 000 132
leaving 21 to go.
Rather than persevering with m/Pairs, I heeded the warning
that something interesting could be about to happen and so
moved to setting the candidate profiles. Column 6 and Row 2
needed normalising in terms of splitting the "Missing" profile.
Col 6 has (18) and (569) whilst row 2 has (37) and (69).
Overall inspection revealed that column 5 has a similar profile
to that identified in an earlier post on "Inferential Chains".
It has eight candidate occurrences in four cells  implying two
occurrences of each digit and two binary links from each cell.
This column looked good to contain the end point of the two
inferential chains that I was seeking  but where to start.
An inferential chain is one linking two cells where the end cell
must have the SAME value irrespective of the value of the
cell at the other end of the chain.
Experience suggests that this can be achieved only by making
use of the oneway link whereby a positive value X in one cell
implies 'notX' for everywhere else X is a candidate, whereas
the converse is not true (a cell being notX does NOT imply
value X in another cell  unless there are only TWO cells
containing X).
Thus row 1 seems to be interesting with 1,8,9 having binary links
but value 6 having the potential of a oneway link.
Thus, I guessed that r1c3 and r1c5 might be possible points for
the start and end of the inferential chain  and so it proved.
Starting from r1c5
r1c5=6
r1c3=9 (very short chain!)
r1c5=8 implies (not6) but this still leaves two possibilities
for the cell in row 1 where the 6 must be located and so is no use
to us in terms of forming an inferential chain. Thus: another route!
r1c5=8
r9c5=6
r9c4=7
r9c6=5
r4c6=6
r3c6=9
r3c9=6
r2c9=9
r2c2=6
r1c3=9
This (long!) chain was found visually by writing out the candidate
profiles and tracing the path using binary links. The latter part
is a trail of 69 alternations and so while seemingly complex is
really just the expansion of a simple logical thought. The real
work was making the connection from column 5 to the "right"
cell in the 69 chain so that one arrives at the end with value '9'
rather than with value '6'.
Now that the two chains have been identified, it is clear that r1c3
MUST take the value 9. This is a POSITIVE result and so does not
depend on the "allocate until contradicted" method.
Using r1c3=9 as a resolved cell, the remainder of the puzzle falls
into place to give the final solution.
Sadly it can take some time to find the two chains involved but
experience is bringing us some clues as to the more promising
places to look. Hopefully our prescience will improve over time
and these chains will start staring out at the less myopic among us.
Alan Rayner BS23 2QT



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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Tue Nov 29, 2005 3:30 pm Post subject: 


Hi,
For me a combination of the techniques:
 Hidden Pair (in Column)
 XWing (on Row) and
 XYWing
was enough to crack it.
I am almost sure that I got to the same position as David.
And from here, it can be done, like this:
Code: 
6 not in r1c6, Hidden Pair 1 8 in r1c6 and r6c6 (in Column)
9 not in r1c6, Hidden Pair 1 8 in r1c6 and r6c6 (in Column)
6 not in r1c4, it is in r1c3 r9c3 r1c5 r9c5 (XWing on Row)
6 not in r9c4, it is in r1c3 r9c3 r1c5 r9c5 (XWing on Row)
6 not in r9c6, it is in r1c3 r9c3 r1c5 r9c5 (XWing on Row)
5 not in r4c6, XYWing X=6 Y=9 in r8c4 X=6 Z=5 in r4c4 Y=9 Z=5 in r9c6
6 in r4c6  Sole Candidate ... 
Of course, nothing against a "5 stars constellation" ...
Or against finding the "chains" ...
There was a saying: "All roads are leading to Rome ...".
The empire falled apart, but the saying is still valid.
see u, 

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

Posted: Tue Nov 29, 2005 6:43 pm Post subject: I overlooked the "XWing" 


I could kick myself  I totally overlooked the "XWing" pattern in columns 3 & 5 ... with that, the "XYWing" pattern is relatively easy to spot. Good catch, someone!
Anyway, here's the "5 Star Constellation" I was talking about. Without noticing the XWing on "6" I had arrived at the following candidate table.
Code:  2 5 6/9 1/6/9 6/8 1/8 4 7 3
4 6/9 8 3/7 3/7 2 5 1 6/9
7 3 1 4 5 6/9 8 2 6/9
3 1 7 5/6 4 5/6 2 9 8
6 8 5 2 9 7 3 4 1
9 2 4 1/3 3/8 1/8 6 5 7
5 7 2 8 1 3 9 6 4
1 6/9 3 6/9 2 4 7 8 5
8 4 6/9 5/6/7/9 6/7 5/6/9 1 3 2 
The "Constellation" is in r8c4 (the "alpha star"), r4c4, r9c6, r4c6, and r3c6. One chain leads from r8c4 through r4c4 and r4c6 to r3c6  if r8c4 = 6 then r4c4 = 5, r4c6 = 6, and r3c6 = 9. The other chain leads from r8c4 through r9c6 & r4c6 to r3c6  if r8c4 = 9 then we have the {5, 6} pair in r4c6 & r9c6, so that r3c6 = 9 once again.
Interestingly, I found that I could also apply the "single chain" method starting from r8c4  assuming that r8c4 = 9 I deduce that r3c6 = 9 via r9c6 & r4c6, as before. But then there's a chain of {6, 9} pairs leading around the puzzle (r3c9, r2c9, r2c2, r8c2) that forces a contradiction, because it forces r8c2 = 9 (can't have two "9"s in the same row). dcb 

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someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Wed Nov 30, 2005 8:26 am Post subject: 


Hi,
I am starting for an alpha star that has 2 digits.
This idea is important, because it gives me a good probability that I will get to a contradiction. In this case I could eliminate (maybe) a digit of this star and mark as sure/fixed the other.
I could also get to a contradiction that would eliminate one digit from an other star (cell). And if I am lucky, this elimination would be an "important one" meaning that a digit can be found and set in the table.
The less lucky possibility is to find an contradiction and eliminate a digit which is not so relevant. Or not to find a contradiction at all.
If I would start from a cell with more than 2 digits, the above described probabilities would be smaller.
My problem now is what to do when there are no pairs or only a small numbers of pairs in the table. And I have a few such examples.
I am trying to think about what I call "entropic algorithm".
Any suggestions are welcomed.
P.S. I can supply some example that the techniques that starts from a pair and is looking for a "n" stars constellation is now working.
see u, 

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