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andras
Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales
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 Posted: Tue Mar 18, 2008 9:22 am Post subject: March 18 VH |
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An interesting one - I found an X-wing on 2, an x-y on 2,6,8 and finally an x-y-z also on 2,6,8 were needed to solve it, though I suspect there are easier ways. 
John |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Tue Mar 18, 2008 11:52 am Post subject: March 18 VH |
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For openers
Earl
| Code: |
+------------+------------+---------------+
| 5 48 469 | 1 7 246 | 269 268 3 |
| 3 1 69 | 26 89 5 | 4 7 268 |
| 2 478 4679 | 3 89 46 | 569 1 568 |
+------------+------------+---------------+
| 9 3 1 | 4 26 8 | 57 26 57 |
| 7 6 2 | 5 1 3 | 8 9 4 |
| 8 45 45 | 9 26 7 | 1 3 26 |
+------------+------------+---------------+
| 1 2 37 | 268 5 9 | 2367 4 2678 |
| 4 9 8 | 7 3 26 | 26 5 1 |
| 6 57 357 | 28 4 1 | 37 28 9 |
+------------+------------+---------------+
|
Play this puzzle online at the Daily Sudoku site
For closers: an x-wing (2), an xy-wing (268), then an xyz-wing (268).
Last edited by Earl on Tue Mar 18, 2008 4:57 pm; edited 3 times in total |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Tue Mar 18, 2008 12:01 pm Post subject: |
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| Code: |
+------------+-----------+-------------+
| 5 48 469 | 1 7 246 | 269 268 3 |
| 3 1 69 | 26 89 5 | 4 7 268 |
| 2 478 4679 | 3 89 46 | 569 1 568 |
+------------+-----------+-------------+
| 9 3 1 | 4 26 8 | 57 26 57 |
| 7 6 2 | 5 1 3 | 8 9 4 |
| 8 45 45 | 9 26 7 | 1 3 26* |
+------------+-----------+-------------+
| 1 2 37 | 68 5 9 | 367 4 678 |
| 4 9 8 | 7 3 26 | 26* 5 1 |
| 6 57 357 | 28 4 1 | 37 28 9 |
+------------+-----------+-------------+
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Well, the *d 2s are conjugate, and therefore the 6s, which eliminates the 6 in r7c9. The 26 in r2 is also conjugate to these, but I can't see any use for that. Maybe somebody else can make something of all those pairs.
Otherwise same as Andras: X-wing removes 2 from r1c8, which exposes XYZ-wing pivoted on r2c9. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Tue Mar 18, 2008 1:09 pm Post subject: |
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I used the two 2s in R9 and the ER in Box to remove the 2 in R1C8. This left a 268 xyz wing that removed the 6 from R3C9. Another 268 xy wing removed the 6’s from Box3C7. This left a triple in R3 which produced another triple in C2 which removed the 4 in R1C2.
Sure - someone will probably come up with a one step xy wing – but where’s the poetry – where’s the elegance. It’s like comparing a dirty limerick with a Shakesperian sonnet. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Mar 18, 2008 3:56 pm Post subject: |
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| I had to work for this one, presumably because there was something I missed. First was a Type 6 UR on 37, then a skyscraper on 2 and an XY-Wing on 28-26-68. Stuck at this point until I eliminated a 6 from r3c9 as a result of an APE. Knowing that something else is usually lurking when there's an APE, I found the same exclusion could be made by the XYZ-Wing which I hadn't noticed. That exposed a triple in row 3 which solved the puzzle. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Tue Mar 18, 2008 7:57 pm Post subject: March 18 VH |
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Help !
The HINT on Draw/Play says that the next move in this grid is 8 in R2C5.
Why?
Earl
| Code: |
+------------+-----------+------------+
| 5 48 469 | 1 7 246 | 269 68 3 |
| 3 1 69 | 26 89 5 | 4 7 268 |
| 2 478 4679 | 3 89 46 | 569 1 568 |
+------------+-----------+------------+
| 9 3 1 | 4 26 8 | 57 26 57 |
| 7 6 2 | 5 1 3 | 8 9 4 |
| 8 45 45 | 9 26 7 | 1 3 26 |
+------------+-----------+------------+
| 1 2 37 | 68 5 9 | 367 4 678 |
| 4 9 8 | 7 3 26 | 26 5 1 |
| 6 57 357 | 28 4 1 | 37 28 9 |
+------------+-----------+------------+
|
Play this puzzle online at the Daily Sudoku site |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Tue Mar 18, 2008 8:51 pm Post subject: |
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| Quote: | | First was a Type 6 UR on 37, |
I'm not sure what a Type 6 UR is - my reference clippings only go up to a Type 5. I do see the 37's in 4 corners of R79 - but how do you treat the extra 5 and 6.
| Quote: | | There's an X-Wing in c37. |
Don't see that either |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Mar 18, 2008 9:17 pm Post subject: |
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| Quote: | | I'm not sure what a Type 6 UR is - my reference clippings only go up to a Type 5. I do see the 37's in 4 corners of R79 - but how do you treat the extra 5 and 6. |
A Type 6 is when the bivalue cells are diagonal and one of the digits (in this case, 3) form an X-Wing. Both the bivalue cells are solved with that digit and the move normally doesn't get you too far.
There are also some digits that can be eliminated when it's only a potential Type 6, based on the presence of strong links, but I don't know where the explanation is. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Tue Mar 18, 2008 9:38 pm Post subject: |
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| Marty: Sorry but looking at Earl's grid I still can't see how and why anything can be eliminated from the 4 cells with 37, 357, 367, 37 (assuming that is the UR you are refering to) |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Wed Mar 19, 2008 6:18 am Post subject: |
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x-wing on 2 removes a 2 in box 3
xy-wing 2,6,8 removes two 6's in box 3
xyz-wing removes another 6 in box 3
moral of the story... use box 3. |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Wed Mar 19, 2008 8:09 am Post subject: Re: March 18 VH |
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| Earl wrote: | Help !
The HINT on Draw/Play says that the next move in this grid is 8 in R2C5.
Why?
Earl
| Code: |
+------------+-----------+------------+
| 5 48 469 | 1 7 246 | 269 68 3 |
| 3 1 69 | 26 89 5 | 4 7 268 |
| 2 478 4679 | 3 89 46 | 569 1 568 |
+------------+-----------+------------+
| 9 3 1 | 4 26 8 | 57 26 57 |
| 7 6 2 | 5 1 3 | 8 9 4 |
| 8 45 45 | 9 26 7 | 1 3 26 |
+------------+-----------+------------+
| 1 2 37 | 68 5 9 | 367 4 678 |
| 4 9 8 | 7 3 26 | 26 5 1 |
| 6 57 357 | 28 4 1 | 37 28 9 |
+------------+-----------+------------+
|
Play this puzzle online at the Daily Sudoku site |
One way of getting the 8- although not elegant is to postulate that an 8 goes into r2c9 - analysis of line 2 and the 8s in boxes 9 and 8 quickly gives 2 in two places in col 4 therefore the original postulate is false so r2c5 must be 8 |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Wed Mar 19, 2008 11:59 am Post subject: |
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| cgordon wrote: | | Marty: Sorry but looking at Earl's grid I still can't see how and why anything can be eliminated from the 4 cells with 37, 357, 367, 37 (assuming that is the UR you are refering to) |
I do not remember the typecasting but my explanation goes like this:
a) in row 7, if r7c3=7 (the "other" possibility, since the x-wing is in "3") r7c7 must be 3 (strong link)
b) then r9c8 must be 7 (bi-value), then r9c3 must be 3 (strong link)
c) we have now a situation where 3 and 7 form a deadly pattern, thus
our assumption (a) must be false and r7c3=3. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Mar 19, 2008 12:19 pm Post subject: |
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| Sorry Nataraj but if this is a Type 6 UR the creators should have stopped at Type 5. To me it looks like guessing to artificially reduce 4 cells to a deadly pattern. But this has been argued before - so I'll forever hold my piece. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Mar 19, 2008 4:31 pm Post subject: |
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| cgordon wrote: | | Sorry Nataraj but if this is a Type 6 UR the creators should have stopped at Type 5. To me it looks like guessing to artificially reduce 4 cells to a deadly pattern. But this has been argued before - so I'll forever hold my piece. |
Craig, I don't understand what you don't like about the Type 6. As far as I know, and I'm no theoretician, but all the logic of URs is based on eliminating deadly patterns, and I don't know why Type 6 is any different. Stated simply, the two bivalue cells are on the diagonals and one of the digits (3, in this case) forms an X-Wing.
Because of the X-Wing, the 3s have to be in the upper left and lower right corners of the rectangle or upper right and lower left. If the 3s go in the polyvalue cells, you have the DP, so the 3s must go in the bivalue cells and thus eliminate the 3s on the other two corners. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Mar 19, 2008 5:21 pm Post subject: |
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Marty: I dunno. I had an image of deadly patterns as 4 equal pairs - not 4 bivalue singles. But in any event, if you put the 3's in the polyvalue cells you get this:
| Code: |
+-------+-------+------+
| . . 7 | . . . | 3 . .|
| . . . | . . . | . . .|
| . . 3 | . . . | 7 . .|
+-------+-------+------+
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What's Deadly about that?? There aren't two options (I don't think)
. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Mar 19, 2008 5:50 pm Post subject: |
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| cgordon wrote: | Marty: I dunno. I had an image of deadly patterns as 4 equal pairs - not 4 bivalue singles. But in any event, if you put the 3's in the polyvalue cells you get this:
| Code: |
+-------+-------+------+
| . . 7 | . . . | 3 . .|
| . . . | . . . | . . .|
| . . 3 | . . . | 7 . .|
+-------+-------+------+
|
What's Deadly about that?? There aren't two options (I don't think)
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Craig, consider this:
| Code: | +------------+-----------+------------+
| 1 2 37 | 68 5 9 | 367 4 678 |
| 4 9 8 | 7 3 26 | 26 5 1 |
| 6 57 357 | 28 4 1 | 37 28 9 |
+------------+-----------+------------+ |
Either r9c3 must = 5 or r7c7 must = 6 or else there will be four 37s. That's standard UR logic. But you can take it further in this type because of the X-Wing. Once you use either of those values, you have removed a 3 from both a row and a column, leaving no choice but to have the 3s occupy the two corners with the bivalues, because they are now the only remaining 3s in the column and row. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Mar 19, 2008 7:06 pm Post subject: |
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| Quote: | | Either r9c3 must = 5 or r7c7 must = 6 or else there will be four 37s. That's standard UR logic. |
I'm probably gonna get excommunicated from this Forum - BUT - if I make r9c3 = 5, I don't automatically end up with a three 37's. I get a 37,37,367 (similarly if r7c7 = 6, I get 37,37, 357). Seems I MUST use the X-wing to solve it. Is that what a Type 6 is - a URXwing? I'll keep an eye out for the next one - but I wonder how often they show up.
[/i] |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Wed Mar 19, 2008 7:41 pm Post subject: |
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| cgordon wrote: | Marty: I dunno. I had an image of deadly patterns as 4 equal pairs - not 4 bivalue singles. But in any event, if you put the 3's in the polyvalue cells you get this:
| Code: |
+-------+-------+------+
| . . 7 | . . . | 3 . .|
| . . . | . . . | . . .|
| . . 3 | . . . | 7 . .|
+-------+-------+------+
|
What's Deadly about that?? There aren't two options (I don't think)
. |
Craig, I think I can shed some light here.
Once we arrive at your position, it is clear, that in rows 7 and 9 the "7" and "3" are confined to cols 3 and 7.
Same goes for columns 3 and 7: the "3" and "7" are confined to rows 7 and 9.
What's to prevent us from xx the two - thus creating an alternate solution?
| Code: |
+-------+-------+------+
| . . 3 | . . . | 7 . .|
| . . . | . . . | . . .|
| . . 7 | . . . | 3 . .|
+-------+-------+------+
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What I am trying to say here, is that a position like this
_+_
a|b
b|a
-+-
is a deadly pattern already.
But if we set r7c3=3 then the deadly pattern can be avoided. r7c3 and r9c7 are still "3", but 7 does not have to go in the other corners of the rectangle.
I hope this helps. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Wed Mar 19, 2008 10:55 pm Post subject: |
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tsk tsk, I am away for a day, and look what happens! 
Yes, a "Type 6" is a Unique X-wing. It is really quite simple.
| Code: | +------------+-----------+------------+
| 1 2 37@ | 68 5 9 |367# 4 678 |
| 4 9 8 | 7 3 26 | 26 5 1 |
| 6 57 357# | 28 4 1 | 37@ 28 9 |
+------------+-----------+------------+ |
There is an X-wing on <3>. Either the cells marked @ are <3>, or the cells marked # are <3>.
But, <3> in # forces <7> in @, which is a non-unique (deadly) solution. So, the cells @ must be <3>.
Keith |
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