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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Sun May 30, 2010 4:27 pm    Post subject: Puzzle 10/05/30: (C) Advanced |   |  
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				| Probably a BBDB as well. 
 
  	  | Code: |  	  | +-----------------------+ | 6 . . | . 3 . | . . . |
 | . 5 . | 8 . . | 6 . . |
 | . . 3 | . . 6 | 7 . . |
 |-------+-------+-------|
 | . 4 . | . 5 . | 8 . . |
 | 7 . . | 4 2 . | . . 5 |
 | . . 5 | . . 7 | . . . |
 |-------+-------+-------|
 | . 9 4 | 3 . . | . 7 . |
 | . . . | . . . | 2 . 4 |
 | . . . | . 7 . | . 6 1 |
 +-----------------------+
 
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 Play this puzzle online at the Daily Sudoku site
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		| peterj 
 
 
 Joined: 26 Mar 2010
 Posts: 974
 Location: London, UK
 
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				|  Posted: Sun May 30, 2010 7:12 pm    Post subject: |   |  
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				| First time through as played - four steps with a nice UR 
  	  | Quote: |  	  | x-wing(2); r1c6<>2, r369c1<>2 type 6 UR(49); r2c1<>4, r3c5<>4
 kite(9); r4c6<>9
 xy-chain (9=3)r4c1-(3=1)r4c6-(1=2)r7c6-(2=9)r2c6; r2c1<>9
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 Went back to see if my last step could be used earlier - two steps..
 
  	  | Quote: |  	  | x-wing(2); r1c6<>2, r369c1<>2 AIC (9=3)r4c1-r6c1=r6c7-(3=1)r5c7-(1=8)r5c3-(8=2)r9c3-(2=9)r1c3; r23c1<>9
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Mon May 31, 2010 7:18 am    Post subject: |   |  
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				|  	  | Quote: |  	  | A different 2 step: 
 1. Saw the swordfish which eliminates six 2's
 2. (8)r8c6=(8-3)r5c6=r5c7-r6c7=(3-8)r6c1=r6c5-r8c5=(8)r8c6; r8c6 is 8
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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Mon May 31, 2010 8:30 am    Post subject: |   |  
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				| Sometimes, I can't get to sleep and come up with crazy thoughts that probably should have been forgotten. An example ... 
 Consider the following four rows and a way to use ANP() to get a strong link on two different values in two different cells.
 
 
  	  | Code: |  	  | +-----------------------------------------------+ |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   23  .   |   .   23  .   |   .   23  .   |   3 cells and 2 values
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |---------------+---------------+---------------|
 |   .  123  .   |   .   23  .   |   .   23  .   |   3 cells and 3 values
 |   .   .   .   |   .   .   .   |   .   .   .   |
 |   .   23  .   |   .   23  .   |   .   234 .   |   3 cells and 3 values
 |---------------+---------------+---------------|
 |   .  123  .   |   .   23  .   |   .   234 .   |   3 cells and 4 values where
 |   .  123  .   |   .   23  .   |   .   24  .   |   two values appear only once
 |   .  123  .   |   .   23  .   |   .   34  .   |   and in different cells
 +-----------------------------------------------+
 
 row 2: invalid
 
 row 4: (  23)r4c58 - (23=1)r4c2  =>  Naked Pair forces a Naked Single
 
 row 6: (  23)r6c25 - (23=4)r6c8  =>  Naked Pair forces a Naked Single
 
 row 7: (1=23)r7c25 - (23=4)r7c8  =>  SL equivalent: (1)r7c2 = (4)r7c8
 row 8: (1=23)r8c25 - (2 =4)r8c8  =>  SL equivalent: (1)r8c2 = (4)r8c8
 row 9: (1=23)r9c25 - ( 3=4)r9c8  =>  SL equivalent: (1)r9c2 = (4)r9c8
 
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 Now, why would anyone want the SL equivalent when the original chain segment will always work? Someone who has a chain() routine in their solver that only allows for two single-digit values to be stored for a strong link.
  _ 
 BTW: I used this logic to crack Puzzle 10/05/30: (C) Advanced in one step.
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		| peterj 
 
 
 Joined: 26 Mar 2010
 Posts: 974
 Location: London, UK
 
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				|  Posted: Mon May 31, 2010 9:40 am    Post subject: |   |  
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				| Danny, I tried hard to find a one-step here so I had to go back to look after reading your post... Is this it?! Very nice!
 
 
  	  | Quote: |  	  | (8=19)r45c3-(9=3)r4c1-(3=19)r4c6,r6c4-(19=8)r6c5; r5c6<>8, r6c12<>8 
 or in Danny chain() solver terms...
 
 (8)r5c3=(3)r4c1-(3)r4c6=(8)r6c5
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 [Edit(daj): added missing cells to first chain.]
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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Mon May 31, 2010 3:42 pm    Post subject: |   |  
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				| Peter: Congratulations! You almost matched my results exactly. I think yours is more interesting because you used the approach twice. (note: I added 2x missing cells to your first chain.) Nice going on the second version of the chain. 
 
  	  | Code: |  	  | (8=19)r45c3 - (9=3)r4c1 - r4c6 = (3)r5c6  =>  r5c6<>8 (8   )r 5c3 = (  3)r4c1 - r4c6 = (3)r5c6  =>  r5c6<>8   SL equivalent version
 +--------------------------------------------------------------+
 |  6     128   1289  |  7     3     1259  |  4     58    29    |
 |  249   5     7     |  8     49    29    |  6     1     3     |
 |  2489  128   3     |  1259  149   6     |  7     58    29    |
 |--------------------+--------------------+--------------------|
 |  39    4     19    |  6     5     139   |  8     2     7     |
 |  7     6     189   |  4     2     1389  |  139   39    5     |
 |  2389  1238  5     |  19    189   7     |  139   4     6     |
 |--------------------+--------------------+--------------------|
 |  12    9     4     |  3     6     12    |  5     7     8     |
 |  135   7     6     |  159   189   1589  |  2     39    4     |
 |  2358  238   28    |  259   7     4     |  39    6     1     |
 +--------------------------------------------------------------+
 # 70 eliminations remain
 
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Mon May 31, 2010 5:57 pm    Post subject: ALMOST ALMOST ALS |   |  
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				| I had that <189> ANP but didn't follow that particular path round! Nice one chaps. That ANP proves that r5c3 is 8, however, I was looking at it from a different structure - the almost ALS XZ. (see diagram with apologies for the double "almost" but that's not new!  ) 
 Everything hinges around that same 8 in r5c3
 
 
   
 X ={1,2,3,8,9} shown in green and in block 4.
 Z ={1,3,8,9} shown in orange and in block 5.
 
 a) IF that 8 at r5c3 is false then the ALS XZ is TRUE and restricted common is 8 and the common candidate is the 3 in r4c1 which is eliminated (in blue) and puzzle solved
 
 b) IF the 8 at r5c3 is true puzzle is solved
 
 
 PS - Yes I know that the contradiction in scenario (a) is that a locked pair <19> is formed if  r5c3 <> 8 which in turn should mean that r4c1 <> 9.  Either way, there is an inherent contradiction between the ALS hypothesis and the locked pair which indicates that the hypothesis r5c3 <> 8 is also  false!!
  
 Last edited by Mogulmeister on Tue Jun 01, 2010 5:39 am; edited 2 times in total
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Tue Jun 01, 2010 4:51 am    Post subject: |   |  
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				| This is another example of the AALS that Ronk has referred to previously? |  | 
	
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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Tue Jun 01, 2010 6:15 am    Post subject: |   |  
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				| Forcing Network: r5c6=8 r4c6=3 ... leaving contradiction in cells r4c1, r4c3, and r5c3 
 
  	  | Code: |  	  | +--------------------------------------------------------------+ |  6     128   1289  |  7     3     1259  |  4     58    29    |
 |  249   5     7     |  8     49    29    |  6     1     3     |
 |  2489  128   3     |  1259  149   6     |  7     58    29    |
 |--------------------+--------------------+--------------------|
 |  9     4     19    |  6     5     3     |  8     2     7     |
 |  7     6     19    |  4     2     8     |  139   39    5     |
 |  2389  1238  5     |  19    19    7     |  139   4     6     |
 |--------------------+--------------------+--------------------|
 |  12    9     4     |  3     6     12    |  5     7     8     |
 |  135   7     6     |  159   189   159   |  2     39    4     |
 |  2358  238   28    |  259   7     4     |  39    6     1     |
 +--------------------------------------------------------------+
 
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 Too bad it qualifies as T&E. However, it does show the critical ALS cells that I used.
 
 Last edited by daj95376 on Tue Jun 01, 2010 6:29 am; edited 1 time in total
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Tue Jun 01, 2010 6:26 am    Post subject: |   |  
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				| Danny, even when the contradiction is expressed as: 
 (8-3)r5c6=r4c6-(3=19)r4c13-(19=8)r5c3-(8)r5c6; contradiction so r5c6 <>8 ?
 
 Excuse notation but It looks like another discontinuous loop....
 
 [edit to include missing cell ref]
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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Tue Jun 01, 2010 6:50 am    Post subject: |   |  
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				|  	  | Mogulmeister wrote: |  	  | Danny, even when the contradiction is expressed as: 
 (8-3)r5c6=r4c6-(3=19)r4c13-(19=8 )r5c3-(8)r5c6; contradiction so r5c6 <>8 ?
 
 It looks like another discontinuous loop....
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 Yes, and that's the hook that draws everyone into using ALS and ANP().
 
 However, if you examine the (3=19)r4c13-(19=8 )r5c3 logic closely, you'll see that the subset operation (=19)r4c13-(19)r5c3 is really a forcing network in disguise. All subset operations are really forcing network logic. So, what you have is an embedded forcing network in the middle of what looks like a discontinuous chain.
 
 I've learned to accept the use of subsets in chains because they are a natural extension to chain logic. I also accept ALS and UR segments, even though I consider them less natural.
 
 Last edited by daj95376 on Tue Jun 01, 2010 7:17 am; edited 1 time in total
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Tue Jun 01, 2010 7:13 am    Post subject: |   |  
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				| Understood - I watched many debates unfold on T&E (due to my background I wondered what time and expenses had to do with anything) and stayed out of them as some of the raciocination in the debates descended to the level of "angels on a pinhead". 
 Like you, I am happy to use the interesting ANPs, ALSs but I won't be worrying if I am not restoring the table back to the natural wood.
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		| daj95376 
 
 
 Joined: 23 Aug 2008
 Posts: 3854
 
 
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				|  Posted: Tue Jun 01, 2010 7:34 am    Post subject: |   |  
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				| Everyone: Please forgive my soap-box rantings on chains vs. forcing networks. 
 When I wrote my first Sudoku solver, I missed an important constraint in the definition of a chain. What resulted, after much effort, was a useless chain() routine that was no better than forcing network logic. Since then, I've been very sensitive to what's being called a chain.
 
 Regards, Danny
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		| peterj 
 
 
 Joined: 26 Mar 2010
 Posts: 974
 Location: London, UK
 
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				|  Posted: Tue Jun 01, 2010 7:55 am    Post subject: |   |  
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				| For me it all boils down to the mental attitude I had when I went looking for a 'chain'. If I spot a pattern that looks like a ANP/ANT/ALS etc or a useful crossover between strong links on single digits and then see that I can use it in a chain to disprove something useful... I'm happy. If however I have descended into mentally plugging numbers into cells in promising parts of the grid and following the consequences ... it's really T&E.
 I'm sure I am not alone in having sufficient self-awareness to know when I am really just guessing!
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Tue Jun 01, 2010 8:40 am    Post subject: |   |  
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				| Speaking purely personally, it's always good to hear your POV Danny and those of the others here.  From what I've seen thus far I'd say this is a pretty broad church which is not always true at some other places I have visited. |  | 
	
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		| ronk 
 
 
 Joined: 07 May 2006
 Posts: 398
 
 
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				|  Posted: Tue Jun 01, 2010 11:28 am    Post subject: |   |  
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				|  	  | Mogulmeister wrote: |  	  | This is another example of the AALS that Ronk has referred to previously? | 
 Not in this case. Even when candidate <8> exists at r5c3, there are only five candidate values in your four cells.
 
 I see your deduction as an ALS-xy-wing with endpoint overlap, where the overlap is illustrated below with bicolor cells. Using some of this forum's AN[PTQ] lingo
  ... 
 ANP(19=3)r4c13 - ANT(3=189)[r4c6,r6c45] - ANQ(8=1239)[r4c13,r6c12] --> r5c3<>19
 
 
   
 It avoids such awkwardness as ...
 a) IF that 8 at r5c3 is false then the .... puzzle solved
 b) IF the 8 at r5c3 is true puzzle is solved
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		| tlanglet 
 
 
 Joined: 17 Oct 2007
 Posts: 2468
 Location: Northern California Foothills
 
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				|  Posted: Tue Jun 01, 2010 1:23 pm    Post subject: |   |  
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				| A two step solution without chains or forcing networks........ 
 w-wing 39 r4c1 & r5c8 SL3 r45c6; r5c3<>9
 Flightless xy-wing -128 with vertex 28 in r9c3 plus transport: (1)r5c3 - r4c3 = (1)r4c6; r7c6<>1
 
 Ted
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		| peterj 
 
 
 Joined: 26 Mar 2010
 Posts: 974
 Location: London, UK
 
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				|  Posted: Tue Jun 01, 2010 1:52 pm    Post subject: |   |  
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				| Ted, lovely xy-wing - a satisfying find! |  | 
	
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		| Mogulmeister 
 
 
 Joined: 03 May 2007
 Posts: 1151
 
 
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				|  Posted: Tue Jun 01, 2010 2:46 pm    Post subject: |   |  
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				|  	  | Mogulmeister wrote: |  	  | This is another example of the AALS that Ronk has referred to previously? | 
 
 
  	  | Ronk wrote: |  	  | Not in this case. Even when candidate <8> exists at r5c3, there are only five candidate values in your four cells. | 
 
 Thanks - it remains an ALS and the extra "almost" not required. It is the presence or otherwise of the 8 that determines if the ALS XZ works.
 
 
 V Interesting - see what you mean. 	  | Ronk wrote: |  	  | I see your deduction as an ALS-xy-wing with endpoint overlap, where the overlap is illustrated below with bicolor cells. | 
 
 ANP(19=3)r4c13 - ANT(3=189)[r4c6,r6c45] - ANQ(8=1239)[r4c13,r6c12] --> r5c3<>19
 
 I was also interested in the inherent logical contradiction that says:
 
 The ALS XZ when true (ie without an 8 in r5c3) implies that r4c1 <>3 AND that r4c1 <> 9 due to the locked pair. This contradiction implies that the putative ALS XZ is False and so 8 must be in r5c3 solving the puzzle.
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		| Marty R. 
 
 
 Joined: 12 Feb 2006
 Posts: 5770
 Location: Rochester, NY, USA
 
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				|  Posted: Tue Jun 01, 2010 10:54 pm    Post subject: |   |  
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				| I finished up with Ted's flightless XY-Wing, but before that I had difficulties. X-Wing (2), ERs (1,1), Multi-coloring twice (2,2) and Type 6 UR (49) before the ending XY-Wing.
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