
SudoCue Users A forum for users of the SudoCue programs and the services of SudoCue.Net

View previous topic :: View next topic 
Author 
Message 
Sudtyro Hooked
Joined: 16 Jan 2007 Posts: 49

Posted: Thu Mar 15, 2007 5:00 pm Post subject: 12/16/06 Nightmare  Revisited 


Code:  1 2 567  49 3 49  56 67 8
8 C69 356  2 1 7  45 D36 469
C379 C79 4  6 5 8  1 237 A29
++
5 14 9  3 7 26  8 246 A126
26 3 68  5 48 1  9 246 7
267 14 678  48 9 26  3 5 A126
++
B69 5 1  7 2 49  B46 8 3
367 678 367  1 48 5  2 9 A46
4 89 2  89 6 3  7 1 5

Per previous posts on how to (manually) advance this puzzle:
1. Unkx80 pointed out the four cells (marked C & D above in boxes 1 & 3) that form an ALSXZ or the equivalent WXYZWing (C+D) that eliminates the 3 in r2c3.
Now, it also seems that one can apply Subset Counting to these same four cells in which candidates 6,7,9 each have a multiplicity of one, while candidate 3 has a multiplicity of two, giving a total subset multiplicity of 5. So, a 3 placed in r2c3 would eliminate both of the 3candidates in the subset and thereby reduce the total multiplicity to 3, which is one below the cell count. Hence, r2c3 <> 3, once again. These results clearly show that at least three different named solution techniques will eliminate the 3 in r2c3 using the same four cells. This is also in keeping with Ruud’s 4technique example grid used in the SudoCue Solving Guide’s sections on XYZWing, APE, ALSXZ, and Subset Counting.
2. Ruud pointed out a different elimination using another ALSXZ rule applied to the sets marked A and B above in c9 and r7, respectively. With X=4 as the restricted common, and Z=9, then r3c1 <> 9.
One would now hope that Subset Counting could also be applied to cells A+B, although it is not immediately obvious how to do this, since both the 9’s and 6’s have a multiplicity of two, and there are also six cells with five candidates.
So, the key question is...should Subset Counting and the ALSXZ rule both provide the same elimination when applied to the same subset of cells? And, if so, how would it work with the subset A+B above? 

Back to top 


Myth Jellies Hooked
Joined: 04 Apr 2006 Posts: 42

Posted: Sat Mar 17, 2007 7:25 am Post subject: Re: 12/16/06 Nightmare  Revisited 


Sudtyro wrote:  Code:  1 2 567  49 3 49  56 67 8
8 C69 356  2 1 7  45 D36 469
C379 C79 4  6 5 8  1 237 A29
++
5 14 9  3 7 26  8 246 A126
26 3 68  5 48 1  9 246 7
267 14 678  48 9 26  3 5 A126
++
B69 5 1  7 2 49  B46 8 3
367 678 367  1 48 5  2 9 A46
4 89 2  89 6 3  7 1 5

Per previous posts on how to (manually) advance this puzzle:
1. Unkx80 pointed out the four cells (marked C & D above in boxes 1 & 3) that form an ALSXZ or the equivalent WXYZWing (C+D) that eliminates the 3 in r2c3.
Now, it also seems that one can apply Subset Counting to these same four cells in which candidates 6,7,9 each have a multiplicity of one, while candidate 3 has a multiplicity of two, giving a total subset multiplicity of 5. So, a 3 placed in r2c3 would eliminate both of the 3candidates in the subset and thereby reduce the total multiplicity to 3, which is one below the cell count. Hence, r2c3 <> 3, once again. These results clearly show that at least three different named solution techniques will eliminate the 3 in r2c3 using the same four cells. This is also in keeping with Ruud’s 4technique example grid used in the SudoCue Solving Guide’s sections on XYZWing, APE, ALSXZ, and Subset Counting.
2. Ruud pointed out a different elimination using another ALSXZ rule applied to the sets marked A and B above in c9 and r7, respectively. With X=4 as the restricted common, and Z=9, then r3c1 <> 9.
One would now hope that Subset Counting could also be applied to cells A+B, although it is not immediately obvious how to do this, since both the 9’s and 6’s have a multiplicity of two, and there are also six cells with five candidates.
So, the key question is...should Subset Counting and the ALSXZ rule both provide the same elimination when applied to the same subset of cells? And, if so, how would it work with the subset A+B above? 
I find the easiest thing is to apply short AICs to these
(3&7&9=6)C  (6=3)D => any cell seeing all threes in sets C and D can be removed => r2c3 & r3c8 <> 3
(9&1&2&6=4)A  (4=6&9)B => r3c1 <> 9 in a similar fashion
Any ALS xz deductions, should have an equivalent subset counting deduction. The union of sets A & B has five digits (12469) filling six cells, with respective multiplicities of (11122) for a sum of seven. Any placement which prohibits all of the 9's or alternately all of the 6's from being placed in the union of these sets must be illegal as it would reduce the multiplicities to a value less than the number of cells. Thus r3c1 <> 9.
Note that the AIC implying one of the endpoints must be true says the same thing: that any 6 seeing all the 6's in A and B, or any 9's seeing all the 9's in A and B can be eliminated. 

Back to top 


Sudtyro Hooked
Joined: 16 Jan 2007 Posts: 49

Posted: Sat Mar 17, 2007 6:49 pm Post subject: Re: 12/16/06 Nightmare  Revisited 


Myth Jellies wrote: 
Any ALS xz deductions, should have an equivalent subset counting deduction. The union of sets A & B has five digits (12469) filling six cells, with respective multiplicities of (11122) for a sum of seven. Any placement which prohibits all of the 9's or alternately all of the 6's from being placed in the union of these sets must be illegal as it would reduce the multiplicities to a value less than the number of cells. Thus r3c1 <> 9.

MJ...yes, of course that's right! For some reason, I was thinking backward and gave up early...the extra cell makes it easier (not harder) to reduce the subset multiplicity below the cell count.
Also nice that you spotted the r3c8 <> 3 elimination...Unkx80 seems to have missed that one, too!
As always, your feedback is much appreciated! 

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
