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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Fri Nov 19, 2010 5:42 pm Post subject: XY-Loop--yet again |
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Here I am after a UR and a couple of X-Wings:
| Code: |
+-------------+-----------+------------+
| 6 247 8 | 5 24 1 | 249 79 3 |
| 34 127 27 | 9 238 6 | 5 178 24 |
| 5 1234 9 | 23 7 48 | 248 18 6 |
+-------------+-----------+------------+
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
+-------------+-----------+------------+
| 7 2348 25 | 23 6 9 | 248 58 1 |
| 34 268 1 | 7 238 5 | 289 689 24 |
| 9 2468 256 | 1 248 48 | 3 568 7 |
+-------------+-----------+------------+
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Play this puzzle online at the Daily Sudoku site
I've done this a couple of times and ended with an invalidity and wonder if I'm messing up a loop. Consider this loop:
R7c3=5-->r7c8=8-->r3c8=1-->r2c8=7-->r2c3=2
First, is it still considered an XY-Loop if there's a non-bivalue cell, i.e., r2c8?
As to eliminations, I feel confident about eliminating the 2 from r9c3. What about the 8s from c8? Can I eliminate 8 from c8r289? I think I've got the eliminations down pretty well when all cells are bivalue, but I'm confused where there are polyvalue cells. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 19, 2010 10:09 pm Post subject: |
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Marty,
This is some kind of loop, but not an XY-loop. I don't know what to call it.
First, you are not done with basics. If you were, the puzzle reveals an XYZ-wing and then an XY-wing to solve it.
But, to return to your question. In the cells you quote, we have the following:
If R7C3 = 5: (Counterclockwise)
| Code: | +-------------+-----------+------------+
| 6 247 8 | 5 24 1 | 249 79 3 |
| 34 127 @2 | 9 238 6 | 5 @7 24 |
| 5 1234 9 | 23 7 48 | 248 @1 6 |
+-------------+-----------+------------+
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
+-------------+-----------+------------+
| 7 2348 *5 | 23 6 9 | 248 @8 1 |
| 34 268 1 | 7 238 5 | 289 689 24 |
| 9 2468 256 | 1 248 48 | 3 568 7 |
+-------------+-----------+------------+ |
If R7C3 = 2: (Clockwise)
| Code: | +-------------+-----------+------------+
| 6 247 8 | 5 24 1 | 249 79 3 |
| 34 127 @7 | 9 238 6 | 5 @18 24 |
| 5 1234 9 | 23 7 48 | 248 18 6 |
+-------------+-----------+------------+
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
+-------------+-----------+------------+
| 7 2348 *2 | 23 6 9 | 248 @5 1 |
| 34 268 1 | 7 238 5 | 289 689 24 |
| 9 2468 256 | 1 248 48 | 3 568 7 |
+-------------+-----------+------------+ |
Pincers on:
1: R23C8
2: R27C3
5: R7C38
7: R2C38
8: R237C8
and there are a number of eliminations. Unfortunately, they do not seem to progress the puzzle.
If you made an error, look closely at what you have for pincers on 8.
> Can I eliminate 8 from c8r289?
Not from c8r2.
I hope this helps.
Keith
PS: By the way, your grid after all basics:
| Code: |
+----------------+----------------+----------------+
| 6 247 8 | 5 24 1 | 249 79 3 |
| 34 127 27 | 9 238 6 | 5 178 24 |
| 5 1234 9 | 23 7 48 | 248 18 6 |
+----------------+----------------+----------------+
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
+----------------+----------------+----------------+
| 7 34 25 | 23 6 9 | 248 58 1 |
| 34 268 1 | 7 238 5 | 29 69 24 |
| 9 268 256 | 1 248 48 | 3 56 7 |
+----------------+----------------+----------------+ |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Nov 19, 2010 10:52 pm Post subject: |
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| Code: | Marty's Grid
+-------------+-----------+------------+
| 6 247 8 | 5 24 1 | 249 79 3 |
| 34 127 27 | 9 238 6 | 5 178 24 |
| 5 1234 9 | 23 7 48 | 248 18 6 |
+-------------+-----------+------------+
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
+-------------+-----------+------------+
| 7 2348 25 | 23 6 9 | 248 58 1 |
| 34 268 1 | 7 238 5 | 289 689 24 |
| 9 2468 256 | 1 248 48 | 3 568 7 |
+-------------+-----------+------------+
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This is not an XY-Loop. Depending on how you write the logic ...
| Code: | (2=5)r7c3 - (5=8)r7c8 - (18=7)r32c8 - (7=2)r2c3 - loop
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... you have a loop where the formation of the <18> pair in r23c8 must be taken into account in the clockwise direction.
| Code: | Counter-clockwise: r7c3=5
*-----------------------------------------------------------*
| 6 47 8 | 5 24 1 | 249 9 3 |
| 34 1 d2 | 9 38 6 | 5 c7 4 |
| 5 34 9 | 23 7 48 | 248 c1 6 |
|-------------------+-------------------+-------------------|
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
|-------------------+-------------------+-------------------|
| 7 234 a5 | 23 6 9 | 24 b8 1 |
| 34 268 1 | 7 238 5 | 29 69 24 |
| 9 2468 6 | 1 248 48 | 3 56 7 |
*-----------------------------------------------------------*
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| Code: | Clockwise: r7c3=2
*-----------------------------------------------------------*
| 6 24 8 | 5 24 1 | 249 79 3 |
| 34 12 b7 | 9 238 6 | 5 c18 24 |
| 5 1234 9 | 23 7 48 | 24 c18 6 |
|-------------------+-------------------+-------------------|
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 6 | 4 9 3 | 1 2 5 |
|-------------------+-------------------+-------------------|
| 7 348 a2 | 3 6 9 | 48 d5 1 |
| 34 68 1 | 7 238 5 | 289 69 24 |
| 9 468 56 | 1 248 48 | 3 6 7 |
*-----------------------------------------------------------*
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| Code: | Eliminations: r9c3<>2, r89c8<>8 & r2c2<>7 thanks to Keith
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Last edited by daj95376 on Fri Nov 19, 2010 11:10 pm; edited 2 times in total |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Nov 19, 2010 11:06 pm Post subject: |
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| daj95376 wrote: |
| Code: | Eliminations: r9c3<>2, r89c8<>8
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How about r2c2<>7?
Keith |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Nov 19, 2010 11:11 pm Post subject: |
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| keith wrote: | | daj95376 wrote: |
| Code: | Eliminations: r9c3<>2, r89c8<>8
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How about r2c2<>7?
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You caught me sleeping. _  _
I updated my post. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Nov 19, 2010 11:57 pm Post subject: |
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Thanks Keith and Danny. My invalid elimination of 8 from r2c8 was based on what Danny told me in another thread:
"Instead of getting technical, I'm going to provide a "do this" answer.
* Look at each assignment and its corresponding elimination(s) in the loop.
* Delete that value from any cells that see the assignment and elimination cells."
I considered the 8 to be an assignment and elimination. Danny did go on to say additional instructions would be needed in loops more complicated than XY-Loops, which this obviously is.
| Quote: | | First, you are not done with basics. |
I'm sure I'll kick myself, but even knowing I'm not done, I'm still not seeing it. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Nov 20, 2010 12:25 am Post subject: |
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| Marty R. wrote: | My invalid elimination of 8 from r2c8 was based on what Danny told me in another thread:
"Instead of getting technical, I'm going to provide a "do this" answer.
* Look at each assignment and its corresponding elimination(s) in the loop.
* Delete that value from any cells that see the assignment and elimination cells."
I considered the 8 to be an assignment and elimination. |
In defense of my instructions in the other thread, it was based on the assumption of a chain. You stepped outside that assumption when you used r7c8=8 and r3c8=1 to perform r2c8<>18=7. You were then incorporating network logic because r2c8=7 does not follow from r3c8=1 alone. The easiest way for me to get around your network was to alter the representation. I did this through an ALS in the chain that I wrote above.
Regards, Danny
FWIW: When I wrote my first Sudoku solver, I spent what seemed like countless hours on the chain() routine... only to discover that it was useless because I had implemented network logic by mistake.
Last edited by daj95376 on Sat Nov 20, 2010 12:47 am; edited 1 time in total |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Nov 20, 2010 12:47 am Post subject: |
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| Marty R. wrote: |
| Quote: | | First, you are not done with basics. |
I'm sure I'll kick myself, but even knowing I'm not done, I'm still not seeing it. |
Marty, your grid:
| Code: | +----------------+----------------+----------------+
| 6 247 8 | 5 24 1 | 249 79 3 |
| 34 127 27 | 9 238 6 | 5 178 24 |
| 5 1234 9 | 23 7 48 | 248 18 6 |
+----------------+----------------+----------------+
| 2 5 4 | 8 1 7 | 6 3 9 |
| 1 9 3 | 6 5 2 | 7 4 8 |
| 8 67 67 | 4 9 3 | 1 2 5 |
+----------------+----------------+----------------+
| 7 2348 25 | 23 6 9 | 248 58 1 |
| 34 268 1 | 7 238 5 | 289 689 24 |
| 9 2468 256 | 1 248 48 | 3 568 7 |
+----------------+----------------+----------------+ |
R9C2 is not 4.
R7C2 r8C1 are a hidden pair 34.
R89B8 are not 8.
That completes the basics. There is an XYZ-wing -249 in R3C7, leading to an XY-wing -234 in R2C3.
None of this changes the discussion above.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Nov 20, 2010 1:56 am Post subject: |
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| Quote: | R9C2 is not 4.
R7C2 r8C1 are a hidden pair 34.
R89B8 are not 8.
That completes the basics. |
I hate to be a pest, but I'm not seeing it. Of course, if r9c2<>4, then I see the hidden pair or, more appropriate for me, a 2568 quad. But I don't know why r9c2<>4 nor why r89b8<>8. I've looked and looked and see nothing in subsets or locked candidates.  |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Nov 20, 2010 2:07 am Post subject: |
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Marty,
The 4 in R9 is in B8 ...
Then, after the 34 pair, the 8 in R7 is in B9.
(Or, the 4 in B8 is in R9, and the 8 in B9 is in R7.)
I am going from the grid you posted, quoted above.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Nov 20, 2010 2:21 am Post subject: |
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| How could I miss those 4s in box 8? Thanks again. |
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