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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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 Posted: Sun Nov 20, 2005 11:55 am Post subject: "5 stars constellation" - a generalization of Turb |
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Hi,
I would like to bring into discussion the "5 stars constellation" which a see as a generalization of the Turbot Fish (contradict me, if I am wrong).
I would in a very general mode describe it as:
1. it involves 5 stars (meaning 5 cells)
2. it has 2 starting digits, in the so called starting cell. We always find this alpha star, the bright one to start with.
3. we follow 2 chains, or pathes, to other stars/cells, forcing a digit TO BE or NOT TO BE in this cell.
4. the overall result is a chain of 5 stars, with an ending in some result, like for example:
- the 2 digits of the last star where both excluded, so there is no other digit to be set for this cell
- the same digit should be in to adiacent stars, which means same digit double in a row/column/3x3 block - and this is not acceptable.
- same digit to be in one cell, whatever digit we have choosen in the alpha star (this is the well know forcing chain - technique, involving 5 cells)
Sometimes you could use only one digit from the staring cell, sometimes, you could use one digit for one path and the other for the construction of the second path.
It does also not matter how much digit are in some cell, if we need only one digit to deal with for our cell.
The generalization, to the Turbot Fish is that it is not involving only one digit (if I got it right).
Any feedback, critics, dirty jokes, etc are wellcomed.
Examples can follow, on request.
P.S. it is "caming" from a more general technique of "double implication chains" that is starting from every digit of a pair from an "alpha" star. And this technique is cracking almost every nut that happens to have a sufficient number of pairs in the candidate table ...
see u, |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sun Nov 20, 2005 2:13 pm Post subject: Exmple of a "5 star constellation" |
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Hi,
Here a promissed example.
| Code: | 100000805
090300400
003005000
000609000
020080070
000507000
000900300
004001060
506000007 |
I worked myself up to:
| Code: | 3 in r1c8 5 in r2c3 5 in r5c7 5 in r8c5 - Unique Horizontal
9 in r1c5 5 in r4c2 5 in r7c8 - Unique Horizontal
7 in r3c7 - Unique Vertical
6 in r6c7 - Unique Vertical
6 in r5c1 9 in r8c1 - Unique in 3x3 block
2 in r8c7 - Sole Candidate
1 in r4c7 8 in r8c9 - Sole Candidate
7 in r8c4 9 in r9c7 - Sole Candidate
3 in r8c2 - Sole Candidate
7 in r2c5 - Unique Vertical
1 not in r3c8, it is in r2c8 or r2c9 (Row on 3x3 Block interaction)
1 not in r3c9, it is in r2c8 or r2c9 (Row on 3x3 Block interaction)
2 not in r7c5, it is in r9c4 or r9c5 or r9c6 (Row on 3x3 Block interaction)
2 not in r7c6, it is in r9c4 or r9c5 or r9c6 (Row on 3x3 Block interaction)
2 not in r3c5, it is in r4c5 or r6c5 of the 3x3 (3x3 Block on Row/Column interaction)
2 not in r9c5, it is in r4c5 or r6c5 of the 3x3 (3x3 Block on Row/Column interaction)
8 not in r4c1, XY-Wing X=7 Y=2 in r1c3 X=7 Z=8 in r4c3 Y=2 Z=8 in r2c1
8 not in r6c1, XY-Wing X=7 Y=2 in r1c3 X=7 Z=8 in r4c3 Y=2 Z=8 in r2c1 |
Here I started from pair: 2 & 7 in r1c3 for the:
| Code: | 1 467 27 24 9 246 8 3 5
. ... Aa .. . ... . . .
. ..a aA .. . ... . . .
28 9 5 3 7 268 4 12 126
.. . . . . ... . .. ...
.. . . . . ... . .. ...
248 468 3 1248 146 5 7 29 269
... ... . .... ... . . .. ...
... ... . .... ... . . .. ...
347 5 78 6 234 9 1 248 234
... . .. . ... . . ... ...
... . .. . ... . . ... ...
6 2 19 14 8 34 5 7 349
. . .. .b . .. . . ...
. . .. .. . .. . . ...
34 148 189 5 1234 7 6 2489 2349
.. ... ... . .... . . .... ....
.. ... ... . .... . . .... ....
278 178 1278 9 46 468 3 5 14
Bbb ... .a.. . .. ... . . ..
.b. .B. .... . .. ... . . ..
9 3 4 7 5 1 2 6 8
. . . . . . . . .
. . . . . . . . .
5 18 6 248 34 2348 9 14 7
. .. . ... .. .... . .. .
. .. . ... .. .... . .. . |
Lets take a look at the "5 star constellation":
- the alpha star is in r1c3 and it is a double star with digits 2 & 7
- first path starting from number 2 is:
r7c3 <> 2 r7c1 = 2 r7c1 <> 7
- second path starting from number 7 is:
r1c2 <> 7 r7c2 = 7 r7c1 <> 7
Conclusion: number 7 can be excluded from r7c1
And this is cracking the nut!!!
| Code: | 7 in r4c1 - Unique Vertical
3 in r6c1 - Unique Vertical
4 in r3c1 4 in r6c2 - Unique Vertical
8 in r4c3 - Sole Candidate
8 in r6c8 - Unique Horizontal
9 in r3c8 - Unique Vertical
1 not in r7c3, it is in r7c2 or r9c2 (Column on 3x3 Block interaction)
2 not in r2c8, XY-Wing X=6 Y=8 in r3c2 X=6 Z=2 in r3c9 Y=8 Z=2 in r2c1
2 not in r2c9, XY-Wing X=6 Y=8 in r3c2 X=6 Z=2 in r3c9 Y=8 Z=2 in r2c1
2 not in r6c9, XY-Wing X=6 Y=1 in r3c5 X=6 Z=2 in r3c9 Y=1 Z=2 in r6c5
1 in r2c8 9 in r6c9 - Sole Candidate
6 in r2c9 1 in r6c3 4 in r9c8 - Sole Candidate
2 in r3c9 2 in r4c8 9 in r5c3 2 in r6c5 1 in r7c9 3 in r9c5 - Sole Candidate
4 in r4c5 - Sole Candidate
3 in r4c9 1 in r5c4 3 in r5c6 6 in r7c5 - Sole Candidate
8 in r3c4 1 in r3c5 4 in r5c9 - Sole Candidate
2 in r2c6 6 in r3c2 2 in r9c4 - Sole Candidate
7 in r1c2 4 in r1c4 8 in r2c1 8 in r9c6 - Sole Candidate
2 in r1c3 6 in r1c6 2 in r7c1 8 in r7c2 4 in r7c6 1 in r9c2 - Sole Candidate
7 in r7c3 - Sole Candidate |
and the final Sudoku table is:
| Code: | 1 7 2 4 9 6 8 3 5
8 9 5 3 7 2 4 1 6
4 6 3 8 1 5 7 9 2
7 5 8 6 4 9 1 2 3
6 2 9 1 8 3 5 7 4
3 4 1 5 2 7 6 8 9
2 8 7 9 6 4 3 5 1
9 3 4 7 5 1 2 6 8
5 1 6 2 3 8 9 4 7 |
P.S. the example is take from: KD-Super-Tough-Book-002-3
see u, |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sun Nov 20, 2005 11:41 pm Post subject: This is neat! |
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Thanks for the example, someone.
Oh -- should we call a 5-star constellation a Casseopeia?
I've taken the liberty of cleaning up the table you presented just a bit, and I've appended an explanation of the symbols you displayed to illustrate the "double-implication chain." If I made a mistake, please let me know.
| Code: | 1 467 27 24 9 246 8 3 5
. ..A Aa .. . ... . . .
. ..a aA .. . ... . . .
28 9 5 3 7 268 4 12 126
.. . . . . ... . .. ...
.. . . . . ... . .. ...
248 468 3 1248 146 5 7 29 269
... ... . .... ... . . .. ...
... ... . .... ... . . .. ...
347 5 78 6 234 9 1 248 234
... . .. . ... . . ... ...
... . .. . ... . . ... ...
6 2 19 14 8 34 5 7 349
. . .. .. . .. . . ...
. . .. .. . .. . . ...
34 148 189 5 1234 7 6 2489 2349
.. ... ... . .... . . .... ....
.. ... ... . .... . . .... ....
278 178 1278 9 46 468 3 5 14
Bbb .b. .aB. . .. ... . . ..
bb. .B. .Aa. . .. ... . . ..
9 3 4 7 5 1 2 6 8
. . . . . . . . .
. . . . . . . . .
5 18 6 248 34 2348 9 14 7
. .. . ... .. .... . .. .
. .. . ... .. .... . .. . |
We're starting the analysis with the {2, 7} pair in r1c3. The first assumption we make is that this cell contains a "2". So you have written the string "Aa" on the first line under the pair, to indicate (A) that 2 is true and also to indicate (a) that 7 is false.
Now we start to draw inferences. The string "A" appearing on the first line under r1c2 indicates that in this first case 7 must be true in this cell (because there only two ways to fit a "7" in the first row). You have also written the string "a" on the first line under r7c3 to indicate that 2 is false in that cell in this first case.
You move to the letter B for the second round of inferences. The string "Bbb" appearing on the first line under r7c1 is forced by the "a" under the 2 in r7c3 -- if r7c3 <> 2, then r7c1 =2, because there are only two ways the "2" can fit in row 7. And clearly, if r7c1 is a "2", then it cannot be a "7" or an "8" -- hence the "Bbb" string.
For completeness I have written in the "b" under the "7" appearing in r7c2 for this first case, and also the "B" under the "7" in r7c3 -- not because these values are important to the logical argument, but just to extend the illustration of the marking technique you are using.
Now we move to the second line underneath r1c3, where you have written "aA" to indicate that 2 is false and 7 is true this time through. This time we must write "a" under the "7" in r1c2, and "Aa" under r7c3.
The second round of inference is again marked by the letter "B" -- we get the "B" under the "7" in r7c2 because of the "a" in r1c2, and the "b" under the "7" in r7c1 because of the "B" in r7c2. And the fact that we now have two lower case "b"s appearing under the digit 7 informs us that we can safely eliminate that possibility in this cell.
This is pretty impressive, but I'm wondering if the proliferation of symbols A, a, B, b, ... might not be muddying the waters just a bit. Maybe the table would be improved by sticking to a simple "T" for true and "F" for false. Here's a version of the table incorporating that approach.
| Code: | 1 467 27 24 9 246 8 3 5
. ..T TF .. . ... . . .
. ..F FT .. . ... . . .
28 9 5 3 7 268 4 12 126
.. . . . . ... . .. ...
.. . . . . ... . .. ...
248 468 3 1248 146 5 7 29 269
... ... . .... ... . . .. ...
... ... . .... ... . . .. ...
347 5 78 6 234 9 1 248 234
... . .. . ... . . ... ...
... . .. . ... . . ... ...
6 2 19 14 8 34 5 7 349
. . .. .. . .. . . ...
. . .. .. . .. . . ...
34 148 189 5 1234 7 6 2489 2349
.. ... ... . .... . . .... ....
.. ... ... . .... . . .... ....
278 178 1278 9 46 468 3 5 14
TFF .F. .FT. . .. ... . . ..
.F. .T. .TF. . .. ... . . ..
9 3 4 7 5 1 2 6 8
. . . . . . . . .
. . . . . . . . .
5 18 6 248 34 2348 9 14 7
. .. . ... .. .... . .. .
. .. . ... .. .... . .. . |
What do you think -- is the table easier to interpret this way? dcb |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Mon Nov 21, 2005 2:38 am Post subject: Re: Exmple of a "5 star constellation" |
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| Code: |
This is a development of the general principle of finding two routes
through the grid from a start point to and end point using EITHER
a binary link OR a pseudo-binary link for each stage.
A binary link is as defined previously (ie where a digit appears as a
candidate exactly TWO times in a logical line). It works both ways
(ie If A then not B and also If not A then B)
A pseudo-binary link is a relaxation of this and works only one way.
It can have more than two members (here A,B,C,D ...) and states
if A then not B and not C and not D etc). There may be a better name
for this type of link!!!
If the two routes have differing parity (odd/even) in terms of the
number of links or steps in the routes then there is "new" information
and an elimination should be possible. Colouring is a special version
of this technique - using only true binary links.
Normally the parity of the routes between two cells will be the same
if composed of true binary links and using only rows/columns as the links.
In order to change the parity between routes it is necessary to bring in
a "distinguishing" feature. With the true binary paths used in the
colouring technique, this is use of a "diagonal" link (ie a link between
cells in a region that are NOT aligned by row or column) - or a number
of such links where the number of links has a different parity between
the two routes.
With the current technique, the distinguishing feature can involve
using a 'pseudo-binary' link. However using such a link means that
the chain is "ONE WAY ONLY" and the inference applies only to the
END-cell - whereas using only true binary links gives an inference
to the cells at EACH end - and possibly also to some in the middle.
+++
1 467 27; - - - ; - - -
28 9 5 ; - - - ; - - -
248 468 3 ; - - - ; - - -
347 5 78; - - - ; - - -
6 2 19; - - - ; - - -
34 148 189; - - - ; - - -
278 178 1278; - - - ; - - -
9 3 4 ; - - - ; - - - ;
5 18 6 ; - - - ; - - -
- the alpha star is in r1c3 and it is a double star with digits 2 & 7
- first path starting from number 2 is:
r1c3 = 2
r7c3 <> 2
r7c1 = 2 [r7c1 <> 7]
This route has TWO steps.
- second path starting from number 7 is:
r1c3 = 7
r1c2 <> 7
r7c2 = 7
r7c1 <> 7
This route has THREE steps
Conclusion:
There are two routes from r1c3 to r7c1 with differing parity.
The parity difference arises because the longer route has a 'dog-leg'
from the start cell to row 7 by going via r1c2 instead of directly to
row 7 as is the case with the first route.
Thus new information is gleaned.
We need a general rule.
"If a cell with exactly two candidates can be linked to another cell
in the grid by two forcing chains (which may contain non-binary one-way
links) which have a different number of steps in them AND the odd-even
parity of the numbers of such steps differs AND each chain uses the SAME
digit for its final step then that digit is either eliminated or confirmed in
the end-cell according to the nature of that final link."
Using our example:
The two cells are r1c3 and r7c1.
The forcing chains are given above.
They EACH using a non-binary link
(and so the inference is ONE-WAY only)
They EACH use digit 7 as the final link.
The end link is "Not 7" in each case and so digit 7 can be excluded
from r7c1.
The definition is extended to include the "nature of the last link" as
we could have plotted two routes from r1c3 to r4c1. These would be
as above to r7c1 and then by binary link to set r4c1 actually to be '7'.
The last link would be "must be 7" in each case. In practice, one would
probably not do that (as r7c1 would be the target) but the definition
must include the possibility of a positive setting rather than just an
elimination.
+++
The rule above is then totally independent of the specific values in
the start or end-cells. It is constrained by being one-way but allows
two-way inferences in special cases (ie where only true binary links
are used). It is NOT 'trial and error' as it uses only information already
on the grid, does not allocate values until completion of the computation
and has no possibility of 'back-tracking'.
It seems to be a satisfactory rule. All that is then needed is a method
of finding the things!! The good news is that searching for these would
also find possible usages of the colouring techniques.
I trust that this assists. There is excellent work here - but it should
not be called the five-star constellation. It can work for multi-starred
constellations (not perhaps for 81-stars but maybe the 48 that appear
on a national flag?? Is there any real limit?).
+++
I like the T/F ascription as it clarifies the demonstration of the logic -
but, unfortunately, it loses the picture of the chain. In my method
above the sequence of the chain is important - but only to count the
number of steps and to check the nature of the last link. The values
held in the 'en route' cells are irrelevant - once one has confidence
in having the method as a solution tool. That said, one does still have
to find and to plot the routes. If they do happen to work on purely
binary links and using the same digit for each step then one has a
forcing chain of the 'colouring' type and then the intermediate
values may be of relevance. Otherwise, sadly not - apart from any
common path to the end-cell where the cells on it could be treated
as the end-point for deriving an inference.
Alan Rayner BS23 2QT
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Mon Nov 21, 2005 10:11 am Post subject: |
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Hi,
Thank you, David & Alan.
You have refined the "4 star constellation" and the "5 star constellation".
I will supply soome "6 star constellation" examples. Of course that this ones should not contain any "4" or "5" (otherwise, why should we look for a "6" ?).
If I find other "Start Constellation" examples, I will post them.
(The nice part is when 2 pathes are needed !!!)
Now, back to USSA, to work!!!
see u, |
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alanr555
Joined: 01 Aug 2005
Posts: 198
Location: Bideford Devon EX39
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| Posted: Mon Nov 21, 2005 10:49 am Post subject: Re: Exmple of a "5 star constellation" |
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| Code: |
Sorry!
My previous definition was incorrect. The error arose because I was
attempting to conflate the solutions to two different situations.
The revised definition is
"If ANY cell can be linked to another cell in the grid by as many forcing
chains (which may contain non-binary one-way links) as the original cell
has candidates AND each chain starts with a different digit from that
candidate set AND the final link in each chain EITHER uses the SAME digit
(X) as every other such chain OR can be construed to have the same
"NOT X" implication as the final link in every other chain
THEN
that digit (X) is either eliminated or confirmed in the end-cell according
to the nature of the common element of that final link."
Using our example:
The two cells are r1c3 and r7c1.
Cell r1c3 is the start cell. It has TWO candidates
(We could choose a cell with three candidates but the likelihood of
finding three chains obeying the rules would be considerably less)
Thus, we need to find TWO chains.
The candidates for r1c3 are 2 and 7.
Thus one chain must start with 2 and the other with 7.
The end cell is r7c1
One of the "final links" is "Not-7"
(If ANY of the final links is a "Not" then ALL of them need to be converted
to a "not" format. It is less likely - but possible - that the same positive
value will result from the chains)
Thus the other chain setting a specific value implies "not-7"
With all (two) chains resulting in 'not-7' that MUST apply and the
'7' digit may be excluded from the profile for r7c1.
There are non-binary one-way links and so NO inference can be made
about the start cell or any intermediate cells.
The forcing chains were given previously.
They EACH using a non-binary link
(and so the inference is ONE-WAY only)
They EACH use digit 7 as the final link (or inference).
The end link infers "Not 7" in each case and so
digit 7 can be excluded from r7c1.
The definition has been amended but the consideration of what is
needed to form a chain (binary links and one-way links) still applies.
However, consideration of the parity is irrelevant here.
The definition is extended to include the "nature of the last link" as
we could have plotted two routes from r1c3 to r4c1. These would be
as above to r7c1 and then by binary link to set r4c1 actually to be '7'.
The last link would be "must be 7" in each case. In practice, one would
probably not do that (as r7c1 would be the target) but the definition
must include the possibility of a positive setting rather than just an
elimination.
+++
The rule above is then totally independent of the specific values in
the start or end-cells. It is constrained by being one-way but allows
two-way inferences in special cases (ie where only true binary links
are used). It is NOT 'trial and error' as it uses only information already
on the grid, does not allocate values until completion of the computation
and has no possibility of 'back-tracking'.
It seems to be a satisfactory rule. All that is then needed is a method
of finding the things!!
+++
I do apologise for the misleading earlier post.
In view of the generality, I would suggest that the rule be called
"Inferential Chaining"
Its use is likely to limited to situations where other techniques fail to
unlock the puzzle and there might be a temptation to apply some form
of trial and error. By being a rule that does not depend upon allocation
and back-tracking, Inferential Chaining would seem to fit the bill of
avoiding the use of 'trial and error'.
Alan Rayner BS23 2QT
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sat Nov 26, 2005 8:36 am Post subject: Example of a "5 star constellation" |
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Hi,
I would like to supply an other example of a "5 star constellation".
You can check that this is not a "Open Chain" and not an "Forcing chain" and also does not correspond to the definition of "Turbot fish".
It starts 2 chais from an alpha (double) star. But one of the stars of the double star is the main one. Getting to a contradiction, we will eliminate (explode) this main star.
But now to business: I took the USATODAY from 30 sept 2005 puzzle.
| Code: | Initial SuDoku Table
1 2 3 - 4 5 - - -
- - - - - - - - -
6 - - 7 - - 2 - -
- 7 - - 8 - - 3 -
- 4 - - 2 - - 6 -
- 5 - - 1 - - 9 -
- - 9 - - 3 - - 8
- - - - - - - - -
- - - 2 9 - 7 1 4 |
And worked it out ...
| Code: | 3 in r3c5 - Sole Candidate
6 in r2c5 - Sole Candidate
2 in r2c6 - Unique Horizontal
7 not in r2c8, it is in r1c8 or r1c9 (Row on 3x3 Block interaction)
7 not in r2c9, it is in r1c8 or r1c9 (Row on 3x3 Block interaction)
3 not in r8c1, it is in r9c1 or r9c2 (Row on 3x3 Block interaction)
3 not in r8c2, it is in r9c1 or r9c2 (Row on 3x3 Block interaction)
5 not in r7c1, it is in r9c1 or r9c3 (Row on 3x3 Block interaction)
5 not in r8c1, it is in r9c1 or r9c3 (Row on 3x3 Block interaction)
5 not in r8c3, it is in r9c1 or r9c3 (Row on 3x3 Block interaction)
3 in r9c2 7 in r1c8 - Unique Vertical
1 not in r8c3, it is in r7c2 or r8c2 (Column on 3x3 Block interaction)
6 not in r8c3, it is in r7c2 or r8c2 (Column on 3x3 Block interaction)
6 not in r9c3, it is in r7c2 or r8c2 (Column on 3x3 Block interaction)
9 not in r2c1, it is in r2c2 or r3c2 (Column on 3x3 Block interaction)
5 not in r7c4, it is in r7c5 or r8c5 (Column on 3x3 Block interaction)
5 not in r8c4, it is in r7c5 or r8c5 (Column on 3x3 Block interaction)
7 not in r8c6, it is in r7c5 or r8c5 (Column on 3x3 Block interaction)
2 not in r8c9, it is in r7c8 or r8c8 (Column on 3x3 Block interaction)
4 not in r2c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
8 not in r1c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
8 not in r2c7, it is in r2c8 or r3c8 (Column on 3x3 Block interaction)
8 in r1c4 6 in r9c6 - Unique Horizontal
8 in r8c6 - Unique Vertical
1 in r3c6 - Unique Vertical
9 in r2c4 - Unique in 3x3 block
9 in r3c2 - Unique in 3x3 block
8 in r2c2 5 in r3c9 - Sole Candidate
4 in r2c8 4 in r3c3 - Sole Candidate
8 in r3c8 - Sole Candidate
4 not in r4c4, it is in r4c6 or r6c6 (Column on 3x3 Block interaction)
4 not in r6c4, it is in r4c6 or r6c6 (Column on 3x3 Block interaction)
5 not in r7c7, it is in r7c8 or r8c8 (Column on 3x3 Block interaction)
5 not in r8c7, it is in r7c8 or r8c8 (Column on 3x3 Block interaction)
6 in r7c7 - Sole Candidate
9 in r1c7 1 in r7c2 - Sole Candidate
6 in r1c9 4 in r7c4 6 in r8c2 3 in r8c7 - Sole Candidate
1 in r2c7 1 in r8c4 9 in r8c9 - Sole Candidate
3 in r2c9 - Sole Candidate
4 in r8c1 - Unique Horizontal |
So, what's left is:
| Code: | 1 2 3 8 4 5 9 7 6
57 8 57 9 6 2 1 4 3
6 9 4 7 3 1 2 8 5
29 7 126 56 8 49 45 3 12
389 4 18 35 2 79 58 6 17
238 5 268 36 1 47 48 9 27
27 1 9 4 57 3 6 25 8
4 6 27 1 57 8 3 25 9
58 3 58 2 9 6 7 1 4 |
5 star constellation that I found was:
Alpha star c4r1. The other: r4c6, c4c9, r5c6, r5c9
| Code: | c1 c6 c9
r4 *2/9 4/9* *1/2
r5 *7/9 1/7* |
Does not look like a "open chain", "forcing chain" or "Turbot Fish".
After excluding 2 from r4c1, the rest is not a problem:
| Code: | 9 in r4c1 - Sole Candidate
4 in r4c6 - Sole Candidate
5 in r4c7 7 in r6c6 - Sole Candidate
6 in r4c4 9 in r5c6 8 in r5c7 2 in r6c9 - Sole Candidate
1 in r4c9 3 in r5c1 1 in r5c3 3 in r6c4 4 in r6c7 - Sole Candidate
2 in r4c3 5 in r5c4 7 in r5c9 8 in r6c1 - Sole Candidate
6 in r6c3 7 in r8c3 5 in r9c1 - Sole Candidate
7 in r2c1 5 in r2c3 2 in r7c1 5 in r8c5 8 in r9c3 - Sole Candidate
7 in r7c5 5 in r7c8 2 in r8c8 - Sole Candidate |
and
| Code: | Final SuDoku Table
1 2 3 8 4 5 9 7 6
7 8 5 9 6 2 1 4 3
6 9 4 7 3 1 2 8 5
9 7 2 6 8 4 5 3 1
3 4 1 5 2 9 8 6 7
8 5 6 3 1 7 4 9 2
2 1 9 4 7 3 6 5 8
4 6 7 1 5 8 3 2 9
5 3 8 2 9 6 7 1 4 |
see u, |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sat Nov 26, 2005 9:37 am Post subject: |
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Hi,
5 stars constellation can be so nice.
Here an other example:
| Code: | - alpha star in r1c7
- rest of stars: r1c8, r1c9, r2c8, r5c8
c7 c8 c9
r1 2/7* 3/7/8*/9 *2/3/7/8#/9
r2 *6/7
r5 6/8*
7 can be eliminated from r1c7 because of
contracdiction in column 8 for number 8 |
Now, the Sudoku where I found it:
| Code: | 5 - - 4 - 6 - - -
- - - 3 9 - 5 - 4
- - 2 - - - - - -
9 6 - - 3 5 - - -
3 - 5 - - 1 - - -
- - - 8 - - - - -
7 - - - 4 - 6 2 -
- - - 5 - - 3 - -
- - - - - - 8 - - |
A couple of moves to get to the difficult point:
| Code: | 2 in r2c6 5 in r3c5 - Unique Horizontal
4 in r6c6 - Unique Vertical
9 in r5c4 - Unique in 3x3 block
1 in r7c4 - Sole Candidate
7 in r3c4 - Sole Candidate
8 in r3c6 2 in r4c4 - Sole Candidate
1 in r1c5 6 in r9c4 - Sole Candidate
8 in r8c5 - Unique Vertical
8 in r2c1 2 in r9c5 - Unique Vertical
4 not in r4c8, it is in r4c7 or r5c7 (Column on 3x3 Block interaction)
4 not in r5c8, it is in r4c7 or r5c7 (Column on 3x3 Block interaction)
1 not in r6c8, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
1 not in r6c9, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
2 not in r6c9, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
6 not in r6c8, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
6 not in r6c9, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
7 not in r6c8, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
7 not in r6c9, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
9 not in r6c8, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
9 not in r6c9, Hidden Pair 3 5 in r6c8 and r6c9 (in Row)
6 in r6c5 9 in r6c7 - Unique Horizontal
7 in r5c5 - Unique Vertical
1 in r3c7 - Sole Candidate
1 not in r4c3, it is in r6c1 or r6c2 or r6c3 (Row on 3x3 Block interaction)
2 not in r5c2, it is in r6c1 or r6c2 (Row on 3x3 Block interaction)
7 not in r4c3, it is in r6c2 or r6c3 (Row on 3x3 Block interaction)
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And here the candidate table where we have to look for the "5 stars
constellation":
| Code: | 5 379 379 4 1 6 27 3789 23789
8 17 167 3 9 2 5 67 4
46 349 2 7 5 8 1 369 369
9 6 48 2 3 5 47 178 178
3 48 5 9 7 1 24 68 268
12 127 17 8 6 4 9 35 35
7 3589 389 1 4 39 6 2 59
1246 1249 1469 5 8 79 3 1479 179
14 13459 1349 6 2 379 8 14579 1579 |
see u, |
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someone_somewhere
Joined: 07 Aug 2005
Posts: 275
Location: Munich
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| Posted: Sat Nov 26, 2005 10:19 am Post subject: |
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Hi,
Now a "Schmankerl", how the austrian would say.
A "5 stars constellation" that needs 3 pathes to explode the alpha star from the starting double star.
| Code: | c1 c2 c3 c7
r1 1/2 1/2/5/8 1/5 1/2/8
T F T T F F
r6 2/5 2/8
T F F T |
5 stars constellation with 3 pathes:
- alpha star r6c2 with 2 the starting digit
- the other stars: r6c7, r1c7; r1c1, r1c2, r1c3
the pathes to follow:
Path-1: r6c2 = 2, r6c2 <> 5, r1c2 = 5, r1c3 <> 5, r1c3 = 1
Path-2: r6c2 = 2, r1c2 <> 2 & r2c2 <> 2, r1c1 = 2
(only place left for in the 3x3 block), r1c7 <> 2
Path-3: r6c2 = 2, r6c7 <> 2, r6c7 = 8, r1c7 <> 8
from results of path-2 and path-3 we conclude r1c7 = 1 (the only one left)
and this contradicts with result of path 1 (digit 1 should be unnique in row 1)
This means that we can expolde (eliminate) digit 2 from the r6c2 (alpha star)
And now the entire puzzle, for all who would like to try to solve it in one way or the other:
| Code: | Initial SuDoku Table
- - - 9 7 - - - 4
6 3 - 4 2 - - - -
- - - - - - - - -
9 - 8 - - 2 7 3 -
- - 6 - 3 5 4 - -
- - - - - - - 1 -
- 7 2 - 4 - 3 - -
5 - - - - - 6 - 1
- - - - - - - - - |
The warm-up moves:
| Code: | 3 in r1c6 3 in r3c9 4 in r4c2 4 in r6c6 - Unique Horizontal
6 in r1c8 5 in r4c9 - Unique Horizontal
6 in r9c2 6 in r6c9 - Unique Vertical
9 in r6c5 - Unique in 3x3 block
8 in r8c5 - Sole Candidate
9 in r8c2 - Sole Candidate
7 in r8c6 - Sole Candidate
1 not in r5c4, it is in r4c4 or r4c5 (Row on 3x3 Block interaction)
8 not in r1c1, it is in r1c2 or r3c2 (Column on 3x3 Block interaction)
8 not in r3c1, it is in r1c2 or r3c2 (Column on 3x3 Block interaction)
8 not in r3c4, it is in r2c6 or r3c6 (Column on 3x3 Block interaction)
5 not in r3c2, it is in r3c4 or r3c5 of the 3x3 (3x3 Block on Row/Column interaction)
5 not in r3c3, it is in r3c4 or r3c5 of the 3x3 (3x3 Block on Row/Column interaction)
5 not in r3c7, it is in r3c4 or r3c5 of the 3x3 (3x3 Block on Row/Column interaction)
5 not in r3c8, it is in r3c4 or r3c5 of the 3x3 (3x3 Block on Row/Column interaction)
1 not in r9c4, Hidden Pair 2 3 in r8c4 and r9c4 (in Column)
5 not in r9c4, Hidden Pair 2 3 in r8c4 and r9c4 (in Column)
1 not in r2c3, Hidden Triple 7 4 9 in r2c3 r3c1 r3c3
1 not in r3c1, Hidden Triple 7 4 9 in r2c3 r3c1 r3c3
1 not in r3c3, Hidden Triple 7 4 9 in r2c3 r3c1 r3c3
2 not in r3c1, Hidden Triple 7 4 9 in r2c3 r3c1 r3c3
5 not in r2c3, Hidden Triple 7 4 9 in r2c3 r3c1 r3c3
5 not in r1c7, it is in r2c7 or r2c8 (Row on 3x3 Block interaction) |
And the obtained candidate table to work on:
| Code: | 12 1258 15 9 7 3 128 6 4
6 3 79 4 2 18 1589 5789 789
47 128 479 156 156 168 1289 2789 3
9 4 8 16 16 2 7 3 5
127 12 6 78 3 5 4 289 289
237 25 357 78 9 4 28 1 6
18 7 2 156 4 169 3 589 89
5 9 34 23 8 7 6 24 1
1348 6 134 23 15 19 2589 245789 2789 |
see u, |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sun Nov 27, 2005 12:10 am Post subject: Another way to crack this nut |
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| Someone_Somewhere wrote: | 5 stars constellation with 3 pathes:
- alpha star r6c2 with 2 the starting digit
- the other stars: r6c7, r1c7; r1c1, r1c2, r1c3 |
This is certainly an interesting position. Here's another way to analyze it.
-- The possible "5"s at r1c2, r1c3, r6c2, & r6c3 form an "X-Wing" formation.
-- Assuming r6c2 = 2 forces r6c7 = 8, which creates the {1, 2} pair in r1c1 & r1c7.
-- This in turn forces r1c3 = 5, leaving no way to complete the X-Wing. dcb |
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