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TKiel
Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI
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 Posted: Mon Oct 08, 2007 10:57 am Post subject: Oct 8 VH |
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| The only thing I remember that seemed like an advanced step was simple coloring. Definitely not one of the harder Very Hards. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Mon Oct 08, 2007 12:03 pm Post subject: |
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For me an almost identical Empty Rectangle solution to the last VH. Note the two 3s in R8 and the L hinge in Box 3 which take out the 3 in R1C2.
| Code: |
+-------+-------+-------+
| 3 3 . | . . . | . 3 . |
| 3 . . | . . . | . . 3 |
| . . . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | . . . |
| 3 . 3 | . . . | . . . |
| . . . | . . . | . . . |
+-------+-------+-------+
| 3 3 3 | . . . | . 3 . |
| . 3 . | . . . | . . 3 |
| . . . | . . . | . . . |
+-------+-------+-------+
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sdq_pete
Joined: 30 Apr 2007
Posts: 119
Location: Rotterdam, NL
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| Posted: Mon Oct 08, 2007 12:30 pm Post subject: |
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I shall have to look at this ER business when I get around to it. For me it was one simple XY wing 139 R7C7.
Peter |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Mon Oct 08, 2007 1:43 pm Post subject: Oct 8 VH |
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An x-wing and an xy-wing did it for me.
Earl |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Oct 08, 2007 3:39 pm Post subject: |
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| Couple of rectangles did it for me. I noticed them before I started looking for Wings. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Mon Oct 08, 2007 4:37 pm Post subject: |
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| An xy-wing with pivot in R7C3 was the final step for me. |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Mon Oct 08, 2007 6:55 pm Post subject: Daily Sudoku Oct 8-2007 |
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After normal scanning and reducing by pairs I have the following:-
49 349 8 5 1 6 2 39 7
13 2 5 8 7 9 4 6 13
6 7 19 4 3 2 159 1589 89
7 6 2 1 5 4 3 89 89
13 5 13 6 9 8 7 2 4
89 89 4 7 2 3 15 15 6
489 13489 39 2 6 7 19 1349 5
2 13 6 9 4 5 8 7 13
5 49 7 3 8 1 6 49 2
An XY Wing with {1,9} as pivot in r7c7 and {1,3}r8c9: {3,9}r7c3 eliminating the 3 in r8c2 solves the puzzle. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Oct 08, 2007 7:09 pm Post subject: |
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There is also a UR on {15} in R36C78 (Type 6 I believe it is: the <5>s form an X-Wing). The <1>s in R3C78 are eliminated and the puzzle solved.
PS:
Craig's ER coincides with a Skyscraper in R38. Also, it recently occurred to me that a Skyscraper is a form of Sashimi X-Wing (a "double" one, in fact) ... just to keep that terminology pot stirred up! |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Mon Oct 08, 2007 9:35 pm Post subject: |
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| Quote: | | There is also a UR on {15} in R36C78 (Type 6 I believe it is: the <5>s form an X-Wing). |
Asellus: I couldn't find ref. to Type 6 URs in my Sudoku guide at www.sudocue.net/guide.php. However, I thought at least two corners of a UR had to be pairs (this was pointed out to me in a post some weeks ago). Here the rectangle you mention has three 159's and one 1598 . I'm a bit lost mate -but am very interested cos this pattern pops up a lot. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Oct 09, 2007 12:07 am Post subject: |
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| I don't have any intermediate positions, but I don't recall the 15 being a Type 6, which I thought had the bivalue cells on the diagonals. With a Type 6 you actually solve two cells to preclude the deadly pattern. Eliminating a candidate from each of the two roof cells is typical of a Type 4, which involves one of the deadly candidates being a strong link, so the other one gets removed. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Oct 09, 2007 4:03 am Post subject: |
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Okay...
First, let's make sure we are referring to the same grid. I agree with Clement's grid after basic methods. I'll repost it here so that it is easier to scan:
| Code: | +--------------+-------+-------------+
| 49 349 8 | 5 1 6 | 2 39 7 |
| 13 2 5 | 8 7 9 | 4 6 13 |
| 6 7 19 | 4 3 2 | 159 1589 89 |
+--------------+-------+-------------+
| 7 6 2 | 1 5 4 | 3 89 89 |
| 13 5 13 | 6 9 8 | 7 2 4 |
| 89 89 4 | 7 2 3 | 15 15 6 |
+--------------+-------+-------------+
| 489 13489 39 | 2 6 7 | 19 1349 5 |
| 2 13 6 | 9 4 5 | 8 7 13 |
| 5 49 7 | 3 8 1 | 6 49 2 |
+--------------+-------+-------------+ |
Craig,
As you can see, there are two {15} cells sharing a side (in R6). Because the <5>s form an X-Wing, neither of the cells R3C78 can contain <1>. To see this: If R6C7 is <5>, then R6C8 is <1> and R3C8 is <5> due to the X-Wing (which means it isn't <1>) and R3C7 cannot be <1> (to avoid the DP). If R6C8 is <5>, the same sort of reasoning says that R3C7 must be <5> (i.e. not <1>) and R3C8 cannot be <1>. Either way, R3C78 aren't <1>. [As I note below, this elimination would still be valid even if there were another <5> candidate in R6. The essential requirement is that the <5>s in R3 are strongly linked.]
If I'm thinking of the thread you alluded to, the point wasn't that two cells of a UR had to be bivalues in order to perform any elimination. The point was that the four cells had to be contained within two Boxes.
If one of the digits involved in a UR also forms an X-Wing, then some sort of elimination is always possible provided that at least one of the UR cells is a bivalue. Here are the possiblities:
(1) One corner is a bivalue: The non X-Wing digit can be removed from the corner diagonally opposite.
(2) Two cells on a side are bivalues: The non X-Wing digit can be removed from the other two cells. (This is the case above.)
(3) Two diagonally opposite cells are bivalues: They can be fixed with the X-Wing digit. (This, it seems, is the real Type 6 McCoy! But, don't quote me!) Note that this is true even if 3 of the cells are bivalues: the two that are diagonally opposite are still set to the X-Wing digit. However, in this last case, the non X-Wing digit is removed from the non bivalue corner (this is more easily seen as a simple Type 1 elimination that triggers the X-Wing).
Got it?
PS: It might be possible, under the right circumstances, to perform some sort of UR elimination even if none of the four cells is a bivalue. Probably, this would depend upon an overlapping ALS or something. But, that is just a hunch; I don't have any examples.
Marty,
I knew I shouldn't have mentioned a Type number! That's why I qualified it with "I believe." I'm no good with remembering "Type" numbers. I just remember the reasonings. To me, Type 6 is associated with the overlapping X-Wing, not the diagonal bivalue configuration. My mistake. The Type 4 only requires the strong link on the non bivalue "roof" side, if I'm figuring it correctly. Clearly, the overlapping X-Wing provides this (and is thus a Type 4) but isn't required for the Type 4. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Tue Oct 09, 2007 7:38 pm Post subject: |
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| Quote: | | To see this: If R6C7 is <5>, then R6C8 is <1> and R3C8 is <5> due to the X-Wing |
Hey: I'm just back from the 19th hole and am having tremendous difficulty relating this UR relationship to an X wing. But if it works, a Type whatever UR should become a definitive tecnhique. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Oct 10, 2007 6:40 pm Post subject: |
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| Quote: | | Because the <5>s form an X-Wing, neither of the cells R3C78 can contain <1>. |
Asellus: I must be missing something here. Why can't R3C78 contain an 8 or 9 - or even a 1.
Craig |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Oct 10, 2007 9:34 pm Post subject: |
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| cgordon wrote: | | Quote: | | Because the <5>s form an X-Wing, neither of the cells R3C78 can contain <1>. |
Asellus: I must be missing something here. Why can't R3C78 contain an 8 or 9 - or even a 1.
Craig |
Craig,
I ain't Asellus, that's for sure. R3c8 can't be 8 or 9 because it would cause the "deadly pattern" with r3c9, r4c8 and r4c9. This is a "Type 1" rectangle, i.e., three corners have the same bivalue.
R3c7 and r3c8 can't be a 1 because one of the cells has to be a 5 and if the other contained a 1 there would be a deadly pattern with r6c7 and r6c8. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed Oct 10, 2007 10:00 pm Post subject: |
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That's funny! I didn't even see the {89} UR. 
Once I saw the {15} UR, I became blind to the {89} thing starting me in the face.
Craig,
As Marty said, it is the strong link on <5> in R3C78 together with the {15} pair in R6C78 that remove the <1>s. I believe we established that this is Type 4. (The X-Wing isn't necessary, but does provide the required strong link.) My main point was that I routinely check for an X-Wing in one of the potential UR digits since that results in some sort of elimination. |
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George Woods
Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
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| Posted: Sat Oct 13, 2007 11:18 am Post subject: Many many ways to finish |
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Was away from home for this one - so first chance to comment!
But look at the number of different ways of solving from the situation shown above
1 XY wing 39-91-13 Making r8c2 1
2 Colour the 3s starting at r8c2 showing it can't be 3
3 Anti DR in 89 in boxes 3 and 6 make r3 c8 15 ; followed by DR logic in 15 in same two boxes to make r3c7 9
4 DR properties described by Asellus (2 of the 4 cells must be 5(X Wing) and one of them 1 with Anti DR denying 1 to fourth cell hence r2c9 is 1
5 W wing in 13s in boxes 3 and 7 operating on box1 means that r8c9 is 3
6 This W wing logic is identical to what I think of as "gift wrapping" (The three 13s define a rectangle with the fourth corner in Box1 so if 3 is not to be denied in box1 then r2c8 and r8c2 cannot both be 3 and must be 1
As it happens I refused the DR logic and used method 6 (gift wrapping!) |
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