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hughwill
Joined: 05 Apr 2010
Posts: 424
Location: Birmingham UK
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 Posted: Fri Jun 27, 2014 9:48 am Post subject: Jun 27 VH |
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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 |
+-------------+--------------+--------+
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Play this puzzle online at the Daily Sudoku site
(Select box to view)
| Quote: | 567 XY-wing pivot r6c5 followed by X-wing on 5 in c15
leading to a 139 XYZ-wing 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: 1834
Location: Northwest Arkansas USA
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| Posted: Fri Jun 27, 2014 10:30 am Post subject: |
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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: 1834
Location: Northwest Arkansas USA
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| Posted: Fri Jun 27, 2014 10:38 am Post subject: |
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| arkietech wrote: | Nice puzzle!  Here is a one step xy-wing 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 7-6 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: 424
Location: Birmingham UK
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| Posted: Fri Jun 27, 2014 12:54 pm Post subject: |
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Arkietech said:
| Quote: | | Nice puzzle! Here is a one step xy-wing 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: 1834
Location: Northwest Arkansas USA
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| Posted: Fri Jun 27, 2014 5:55 pm Post subject: |
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| 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: 5770
Location: Rochester, NY, USA
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| Posted: Fri Jun 27, 2014 11:40 pm Post subject: |
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| 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: 33
Location: New Orleans
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| Posted: Sun Jun 29, 2014 5:24 pm Post subject: |
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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: 1834
Location: Northwest Arkansas USA
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| Posted: Sun Jun 29, 2014 5:41 pm Post subject: |
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| 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 xy-wing 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: 33
Location: New Orleans
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| Posted: Mon Jun 30, 2014 12:35 am Post subject: |
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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: 1834
Location: Northwest Arkansas USA
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| Posted: Mon Jun 30, 2014 2:23 am Post subject: |
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| 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: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Jun 30, 2014 4:09 pm Post subject: |
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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 1-67 15-69 | 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: 33
Location: New Orleans
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| Posted: Sun Jul 06, 2014 1:36 am Post subject: |
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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: 1834
Location: Northwest Arkansas USA
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| Posted: Sun Jul 06, 2014 2:29 am Post subject: |
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| 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 xy-wing 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: 207
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| Posted: Thu Jul 10, 2014 3:02 pm Post subject: |
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- "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 -/
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(similar to arkietech's path,
easier for me to follow) |
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