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Jun 27 VH

 
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hughwill



Joined: 05 Apr 2010
Posts: 214
Location: Birmingham UK

PostPosted: Fri Jun 27, 2014 9:48 am    Post subject: Jun 27 VH Reply with quote

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 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: 1727
Location: Northwest Arkansas USA

PostPosted: Fri Jun 27, 2014 10:30 am    Post subject: Reply with quote

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: 1727
Location: Northwest Arkansas USA

PostPosted: Fri Jun 27, 2014 10:38 am    Post subject: Reply with quote

arkietech wrote:
Nice puzzle! Very Happy 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: 214
Location: Birmingham UK

PostPosted: Fri Jun 27, 2014 12:54 pm    Post subject: Reply with quote

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: 1727
Location: Northwest Arkansas USA

PostPosted: Fri Jun 27, 2014 5:55 pm    Post subject: Reply with quote

hughwill wrote:


Brilliant Dan, but how do you spot them?

Hugh


Software tools help a lot! Very Happy 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: 5164
Location: Rochester, NY, USA

PostPosted: Fri Jun 27, 2014 11:40 pm    Post subject: Reply with quote

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

PostPosted: Sun Jun 29, 2014 5:24 pm    Post subject: Reply with quote

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: 1727
Location: Northwest Arkansas USA

PostPosted: Sun Jun 29, 2014 5:41 pm    Post subject: Reply with quote

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: 15
Location: New Orleans

PostPosted: Mon Jun 30, 2014 12:35 am    Post subject: Reply with quote

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: 1727
Location: Northwest Arkansas USA

PostPosted: Mon Jun 30, 2014 2:23 am    Post subject: Reply with quote

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: 3178
Location: near Detroit, Michigan, USA

PostPosted: Mon Jun 30, 2014 4:09 pm    Post subject: Reply with quote

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: 15
Location: New Orleans

PostPosted: Sun Jul 06, 2014 1:36 am    Post subject: Reply with quote

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: 1727
Location: Northwest Arkansas USA

PostPosted: Sun Jul 06, 2014 2:29 am    Post subject: Reply with quote

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: 165

PostPosted: Thu Jul 10, 2014 3:02 pm    Post subject: Reply with quote


  • "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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