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Dec 16 VH

 
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kragzy



Joined: 01 May 2007
Posts: 112
Location: Australia

PostPosted: Mon Dec 15, 2008 10:40 pm    Post subject: Dec 16 VH Reply with quote

A very obvious UR (79) or, if you're like me and prefer to avoid uniqueness solutions, an xyz wing (678). No XY wings that I could see - nice for a change!
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Tue Dec 16, 2008 3:28 am    Post subject: Dec 16 VH Reply with quote

I needed two steps: an x-wing (5) which made possible an xyz-wing (678). The UR did not help me in the solution. Any one-steppers out there?

Earl
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Tue Dec 16, 2008 4:39 am    Post subject: Reply with quote

I also needed to remove the digit <5> from r4c3 to open the xyz <678> and complete the puzzle. Either a x-wing <5> or a Type 4 UR <58> deletes the <5>.

The only one step solution I found was a chain.

Ted
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Dec 16, 2008 4:49 am    Post subject: Reply with quote

Quote:
Any one-steppers out there?

I think the first poster is reporting a one-stepper. I also needed the X-Wing to expose the XYZ.

However, I generally look for X-Wings last and the first time I did the puzzle, it took me 5 or 6 steps, ending with a BUG+2 followed by a BUG+1. Sequence of steps makes a huge difference.
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Tue Dec 16, 2008 9:03 am    Post subject: Reply with quote

Marty R. wrote:
Quote:
Any one-steppers out there?

I think the first poster is reporting a one-stepper. I also needed the X-Wing to expose the XYZ.

However, I generally look for X-Wings last and the first time I did the puzzle, it took me 5 or 6 steps, ending with a BUG+2 followed by a BUG+1. Sequence of steps makes a huge difference.


umm, don't think that UR79 does it in one step, but since someone asked.

this xy-chain does...

Code:
.------------------.------------------.------------------.
| 6     49    45   | 79    3     1    | 2     789   589  |
| 39    7     2    | 4     8     5    | 6     19    139  |
| 35    8     1    | 79    2     6    | 35    79    4    |
:------------------+------------------+------------------:
|*59   *69   56-78 | 3     1     4    |*57    2     568  |
| 1     2    *67   | 5     9     8    | 37    4     36   |
| 4     3     58   | 6     7     2    | 19    189   1589 |
:------------------+------------------+------------------:
| 8     146   46   | 2     5     7    | 149   3     19   |
| 7     14    3    | 8     6     9    | 14    5     2    |
| 2     5     9    | 1     4     3    | 8     6     7    |
'------------------'------------------'------------------'

(7=6)r5c3 - (6=9)r4c2 - (9=5)r4c1 - (5=7)r4c7; r4c3 <> 7
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George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Tue Dec 16, 2008 11:48 am    Post subject: A kind of 1 stepper Reply with quote

If you consider the 3s in col 9, one of then demands a 9 at r2c1, and the other a 9 at r4c2. But the 9s in boxes 1 and 4 have two possible states, and only one of them satisfies the above condition so r4c2 is 9!

any idea what this tecnique is formally called?
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eddieg



Joined: 12 Jan 2006
Posts: 47
Location: San Diego, CA USA

PostPosted: Tue Dec 16, 2008 3:53 pm    Post subject: Reply with quote

Great puzzle today. I use only basic moves (stay away from skycrappers and coloring, etc) and almost always am able to solve every puzzle.

You guys are way better than I will ever be.
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Tue Dec 16, 2008 3:56 pm    Post subject: Re: A kind of 1 stepper Reply with quote

George Woods wrote:
If you consider the 3s in col 9, one of then demands a 9 at r2c1, and the other a 9 at r4c2. But the 9s in boxes 1 and 4 have two possible states, and only one of them satisfies the above condition so r4c2 is 9!

any idea what this tecnique is formally called?


George, I am still trying to make sense of these types of conditions so I am definitely not the person to answer your question. But I do have a question and an observation which I hope is valid.

Question: What exactly was the "technique" you used to determine that only one of the <3s> in col 9 satisfies the conditions?

Observation: Starting with the <3> in r5c9, TWO CONFLICTING chains/paths exist to determine the value <9> in box 4.
(a) r5c9=3 => r5c3=6 => r4c2=9 (this is as you noted)
(b) r5c9=3 => r5c7=7 => r4c7=5 => r4c1=9 which is a conflict.

Maybe the best way to identify this is as a conflicting chain that deletes <3> in r5c9.

Ted
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Dec 16, 2008 4:46 pm    Post subject: Reply with quote

Quote:
umm, don't think that UR79 does it in one step, but since someone asked.

I don't think so either, but I interpreted the first poster as saying the XYZ-Wing was one step. I, and others, needed an X-Wing to open up the XYZ.
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Tue Dec 16, 2008 6:00 pm    Post subject: Re: A kind of 1 stepper Reply with quote

George Woods wrote:
If you consider the 3s in col 9, one of then demands a 9 at r2c1, and the other a 9 at r4c2. But the 9s in boxes 1 and 4 have two possible states, and only one of them satisfies the above condition so r4c2 is 9!

any idea what this tecnique is formally called?

Your information about the 9s in [b1] and [b4] is extraneous.

For me, you are describing a Double Implication Chain, but it goes back to a very old definition that isn't used anymore. I think it would now be called a Forcing Net ... that can be reorganized as an AIC. Note: your initial strong link on (3) is still present between the 2nd and 3rd cell in the AIC.

Code:
(9=3)r2c1 - (3)r2c9 = (3-6)r5c9 = (6)r4c9 - (6=9)r4c2 => [r1c2],[r4c1]<>9

This is a perfect example of the claim that one or both ends of an AIC must be true. After the eliminations above are performed, we are left with Hidden Singles: [r2c1]=9 and [r4c2]=9.
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Clement



Joined: 24 Apr 2006
Posts: 1110
Location: Dar es Salaam Tanzania

PostPosted: Tue Dec 16, 2008 8:40 pm    Post subject: Daily Sudoku: Tue 16-Dec-2008 VH Reply with quote

I used 3 UR's 1) {7,9}r13c48
2) {1,4}r78c27 and choosing 4 in r1c3 leads to another UR{5,8} in r46c39. Therefore, r1c3 must be 5, which solves the puzzle.
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George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK

PostPosted: Tue Dec 16, 2008 11:58 pm    Post subject: Re: A kind of 1 stepper Reply with quote

Another description of what I did shows this solution to fairly barbaric.

It can be described as an XY chain of the form 96 67 73 3X X3 39, the two ends being at r4c2 and r2c1 with the strongly linked 3s in col 9 shown as 3X this knocks out both the 9s in r1c2 and r4c1.

I guess I ought to learn the algebraic notaion used by you experts!
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Wendy W



Joined: 04 Feb 2008
Posts: 144

PostPosted: Wed Dec 17, 2008 1:29 am    Post subject: Reply with quote

Red-letter day! My first UR spotted!
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Wed Dec 17, 2008 2:22 am    Post subject: Reply with quote

Quote:
the two ends being at r4c2 and r2c1 with the strongly linked 3s in col 9 shown as 3X

George Woods,

think about that for a minute. that sounds very familiar because this is the same structure as a W-wing. you are using the strongly linked 3's in column 9 to connect two pair of candidates, just like in a W-wing...
Code:
normal w-wing...(x=y) - (y) = (y) - (y=x)

 this w-wing... (9=3) - (3) = (3) - (3=7) - plus the two cell xy-chain extension


the pair (3,9) in r2c1 and the pair (3,7) in r5c7 are connected like a W-wing via those strong links in column 9... makes this chain
(9=3)r2c1 - (3)r2c9 = (3)r5c9 - (3=7)r5c7

then it is extended from the (3,7) cell through the 6's in box 5

hope this image helps...


what is important to notice is the black lines that is essentially a w-wing structure... except for one little difference. the ending candidates aren't the same, we have on one side 9 is true, or on the other side 7 is true Exclamation Idea

now, the extensions, like you pointed out, are just simple to see xy-chains...

(7=6)r5c3 - (6=9)r4c2
put it all together...
(9=3)r2c1 - (3)r2c9 = (3)r5c9 - (3=7)r5c7 - (7=6)r5c3 - (6=9)r4c2;
means that the 9 in r4c1 is toast, burnt and crispy.
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Wed Dec 17, 2008 5:00 am    Post subject: Reply with quote

I'm coming in a bit late here - but being xy or xyz wing challenged, I used all manner of 'pattern' solutions here - an x-wing, two UR's and an ER. Then I saw the xyz. Oh well.
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kragzy



Joined: 01 May 2007
Posts: 112
Location: Australia

PostPosted: Thu Dec 18, 2008 12:08 am    Post subject: Reply with quote

Quote:
I don't think so either, but I interpreted the first poster as saying the XYZ-Wing was one step. I, and others, needed an X-Wing to open up the XYZ.


Marty, I did need an X wing. It's just that I often find X-wings as part of the box-line checking process by following the pattern of the numbers through the puzzle. As a result, I am tending to regard X wings as a cleaning up step rather than an advanced technique. Hope that doesn't sound arrogant - I'm no expert. Colouring, skyscapers, W wings, strong links ???? ... all foreign language to me.

Cheers
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