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hughwill
Joined: 05 Apr 2010 Posts: 222 Location: Birmingham UK

Posted: Fri Jun 27, 2014 9:48 am Post subject: Jun 27 VH 


Could only do this in 3 steps:
Code: 
++++
 6 2 7  3 4 9  5 8 1 
 59 13 1359  16 156 8  2 4 7 
 8 4 15  157 2 157  9 6 3 
++++
 3 67 569  2 8 57  4 1 69 
 4 8 2  169 16 3  7 5 69 
 59 167 1569  5679 56 4  3 2 8 
++++
 2 36 36  15 9 15  8 7 4 
 1 9 4  8 7 2  6 3 5 
 7 5 8  4 3 6  1 9 2 
++++

Play this puzzle online at the Daily Sudoku site
(Select box to view)
Quote:  567 XYwing pivot r6c5 followed by Xwing on 5 in c15
leading to a 139 XYZwing pivot r2c3 setting r3c3 to 5 and finally
setting the puzzle 
If it needs all three different wings, I wonder how often this has happened
in VH Puzzles?
Hugh 

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arkietech
Joined: 31 Jul 2008 Posts: 1730 Location: Northwest Arkansas USA

Posted: Fri Jun 27, 2014 10:30 am Post subject: 


replaced
Last edited by arkietech on Fri Jun 27, 2014 10:39 am; edited 1 time in total 

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arkietech
Joined: 31 Jul 2008 Posts: 1730 Location: Northwest Arkansas USA

Posted: Fri Jun 27, 2014 10:38 am Post subject: 


arkietech wrote:  Nice puzzle! Here is a one step xywing solution using almost locked sets (ALS)
Code:  **
 6 2 7  3 4 9  5 8 1 
 59 b13 1359  16 156 8  2 4 7 
 8 4 b15  157 2 157  9 6 3 
++
 3 76 a569  2 8 57  4 1 a69 
 4 8 2  169 16 3  7 5 69 
 59 167 1569  5679 56 4  3 2 8 
++
 2 c36 36  15 9 15  8 7 4 
 1 9 4  8 7 2  6 3 5 
 7 5 8  4 3 6  1 9 2 
**
(6=5)r4c39(5=3)r3c3,r2c2(3=6)r7c2 => 6r4c2; ste 



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hughwill
Joined: 05 Apr 2010 Posts: 222 Location: Birmingham UK

Posted: Fri Jun 27, 2014 12:54 pm Post subject: 


Arkietech said:
Quote:  Nice puzzle! Here is a one step xywing solution using almost locked sets (ALS) 
Brilliant Dan, but how do you spot them?
Hugh 

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arkietech
Joined: 31 Jul 2008 Posts: 1730 Location: Northwest Arkansas USA

Posted: Fri Jun 27, 2014 5:55 pm Post subject: 


hughwill wrote: 
Brilliant Dan, but how do you spot them?
Hugh 
Software tools help a lot! Look first for bivalue cells then bivalue sets(pseudo cells).
xy wings are excellent puzzle breakers. 

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Marty R.
Joined: 12 Feb 2006 Posts: 5175 Location: Rochester, NY, USA

Posted: Fri Jun 27, 2014 11:40 pm Post subject: 


My solution was the same as Hugh's. When you stick to the standard VH steps, as I like to do, it's very, very seldom that a VH puzzle requires three steps. 

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rmireland
Joined: 21 Sep 2013 Posts: 15 Location: New Orleans

Posted: Sun Jun 29, 2014 5:24 pm Post subject: 


When Dan has a better solution, I think, "Why didn't I see that?" and then I focus on what Dan has done.
My question is, How is this solution different from a 569 XYZ in r4c3?
Is it an XY where the Y is the 69 locked set?
And if you do test it as an xyz, if you trace the 5r4c3 as Dan does, then indeed 6r4c2 but if you trace via r4c6 then 7r4c2 which contradiction means 5r4c3, leaving the 69 ALS to knock out the 6 from r4c2.
If someone has some spare time, could you please translate this into English?
(6=5)r4c39(5=3)r3c3,r2c2(3=6)r7c2 => 6r4c2
I really enjoy your site and Thanks,
Rick 

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arkietech
Joined: 31 Jul 2008 Posts: 1730 Location: Northwest Arkansas USA

Posted: Sun Jun 29, 2014 5:41 pm Post subject: 


rmireland wrote:  When Dan has a better solution, I think, "Why didn't I see that?" and then I focus on what Dan has done.
My question is, How is this solution different from a 569 XYZ in r4c3?
Is it an XY where the Y is the 69 locked set?
And if you do test it as an xyz, if you trace the 5r4c3 as Dan does, then indeed 6r4c2 but if you trace via r4c6 then 7r4c2 which contradiction means 5r4c3, leaving the 69 ALS to knock out the 6 from r4c2.
If someone has some spare time, could you please translate this into English?
(6=5)r4c39(5=3)r3c3,r2c2(3=6)r7c2 => 6r4c2
I really enjoy your site and Thanks,
Rick 
should be if r4c3 and r4c9 is not a 6 then it is 5 making r2c2 3 and r7c4 6 therefore r4c2 cannot be a 6
or
r4c39 is the same as a psuedo cell containing 56
r3c3,r2c2 is the same as a psuedo cell containing 35
r7c2 is a cell containing 36
These form an xywing removing 6 from r4c2
Last edited by arkietech on Mon Jun 30, 2014 2:41 am; edited 1 time in total 

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rmireland
Joined: 21 Sep 2013 Posts: 15 Location: New Orleans

Posted: Mon Jun 30, 2014 12:35 am Post subject: 


Hi Dan,
I thank you for your patience and willingness to help me understand.
I follow why you call it an XY with the psuedo cells, and I also follow why r4c2 cannot be a 6.
I also see why r2c2 with r3c3 is a psuedo cell with 35, because since they both can't be 1, one must be 3 or the other must be 5, or they can be 3 and 5 if r3c3 is a 1.
The part I don't presently see is the statement:
if r4c3 is not 6 then it is 5 ...
Is there some reason why not if r4c3 is not 6 then it could be 9? It seems that r4 can be resolved with a 9 there. I suspect it has to do with your saying
r4c39 is the same as a psuedo cell containing 56
so that if it's not a 6 then it must be a 5. And I see why it's a psuedo cell because 9 is forced to r4c3 or r4c9 and the only other outcome of the psuedo cell is a 5 or a 6. But unless I am stuck on some lower mental plane, I still don't see why real cell r4c3 can't ever be a 9?
Yours in confusion,
Rick 

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arkietech
Joined: 31 Jul 2008 Posts: 1730 Location: Northwest Arkansas USA

Posted: Mon Jun 30, 2014 2:23 am Post subject: 


rmireland wrote:  Hi Dan,
I thank you for your patience and willingness to help me understand.
I follow why you call it an XY with the psuedo cells, and I also follow why r4c2 cannot be a 6.
I also see why r2c2 with r3c3 is a psuedo cell with 35, because since they both can't be 1, one must be 3 or the other must be 5, or they can be 3 and 5 if r3c3 is a 1.
The part I don't presently see is the statement:
if r4c3 is not 6 then it is 5 ...
Is there some reason why not if r4c3 is not 6 then it could be 9? It seems that r4 can be resolved with a 9 there. I suspect it has to do with your saying
r4c39 is the same as a psuedo cell containing 56
so that if it's not a 6 then it must be a 5. And I see why it's a psuedo cell because 9 is forced to r4c3 or r4c9 and the only other outcome of the psuedo cell is a 5 or a 6. But unless I am stuck on some lower mental plane, I still don't see why real cell r4c3 can't ever be a 9?
Yours in confusion,
Rick 
if r4c3 is not 6 then it is 5 making r2c2 3 and r7c4 6 therefore r4c2 cannot be a 6
should be if r4c3 and r4c9 is not a 6
Sorry for the error and thanks for keeping me straight. 

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keith
Joined: 19 Sep 2005 Posts: 3182 Location: near Detroit, Michigan, USA

Posted: Mon Jun 30, 2014 4:09 pm Post subject: 


There is a solution using two wings: XY then XYZ: Code:  ++++
 6 2 7  3 4 9  5 8 1 
 59 13 1359  16 156 8  2 4 7 
 8 4 15  157 2 157  9 6 3 
++++
 3 #67 569  2 8 @57  4 1 69 
 4 8 2  169 16 3  7 5 69 
 59 167 1569  5679 #56 4  3 2 8 
++++
 2 36 36  15 9 15  8 7 4 
 1 9 4  8 7 2  6 3 5 
 7 5 8  4 3 6  1 9 2 
++++  Followed by Code:  ++++
 6 2 7  3 4 9  5 8 1 
 59 13 1359  16 56 8  2 4 7 
 8 4 #15  157 2 157  9 6 3 
++++
 3 67 56  2 8 57  4 1 9 
 4 8 2  9 1 3  7 5 6 
#59 17 @159  567 56 4  3 2 8 
++++
 2 36 36  15 9 15  8 7 4 
 1 9 4  8 7 2  6 3 5 
 7 5 8  4 3 6  1 9 2 
++++  Keith 

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rmireland
Joined: 21 Sep 2013 Posts: 15 Location: New Orleans

Posted: Sun Jul 06, 2014 1:36 am Post subject: 


Hi Dan,
I truly thank you for your help on this, but I think now I am even more confused.
When you say "if r4c3 and r4c9 is not a 6 ...", I don't know what you mean exactly.
Do you mean (1) r4c3 is not a 6 and at the same time r4c9 is not a 6, ie neither is a 6? or do you mean (2) r4c3 is possibly not a 6 and r4c9 is possibly not a 6, but exactly one of them must be a 6? or do you mean something I have not considered?
Please don't speak of links just yet. I am trying to understand at the logic level.
Thanks,
Rick
Three logicians walk into a bar. The bartender asks "Do all of you want a drink?"
The first logician says "I don't know."
The second logician says "I don't know."
The third logician says "Yes!" 

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arkietech
Joined: 31 Jul 2008 Posts: 1730 Location: Northwest Arkansas USA

Posted: Sun Jul 06, 2014 2:29 am Post subject: 


rmireland wrote:  Hi Dan,
I truly thank you for your help on this, but I think now I am even more confused.
When you say "if r4c3 and r4c9 is not a 6 ...", I don't know what you mean exactly.
Do you mean (1) r4c3 is not a 6 and at the same time r4c9 is not a 6, ie neither is a 6? or do you mean (2) r4c3 is possibly not a 6 and r4c9 is possibly not a 6, but exactly one of them must be a 6? or do you mean something I have not considered?
Please don't speak of links just yet. I am trying to understand at the logic level.
Thanks,
Rick
Three logicians walk into a bar. The bartender asks "Do all of you want a drink?"
The first logician says "I don't know."
The second logician says "I don't know."
The third logician says "Yes!" 
6=5)r4c39(5=3)r3c3,r2c2(3=6)r7c2 => 6r4c2
This is basically an xywing with two pinchers. 6 at the beginning and 6 at the end. If either one is a 6 then r4c2 cannot be a six.
This logic statement says that if r4c39 is not a 6 then r7c2 is a 6 therefore r4c2 can't be.
if r4c39 was a 6 we could quit there cause r4c2 could not be or we would have two 6's in r4
Hope this helps. I will have that drink. 

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Pat
Joined: 23 Feb 2010 Posts: 165

Posted: Thu Jul 10, 2014 3:02 pm Post subject: 


 "SIN":
r4c6 = 7
> (7) c2\r6
> (1) c2\b1
> r3c3 = 5
> (5) r4\c6 (curls back to bite itself)
 "forcing net" with 2 branches,
each short enough to follow easily:if r4c2 = 6,
this resolves both r4 and c2,
creating conflict at r3c3
Code: 
/> (7) r4\c6 > (5) r4\c3 \
r4c2 = 6 { }> r3c3 = none
\> (7) c2\r6 > (1) c2\b1 /

(similar to arkietech's path,
easier for me to follow) 

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