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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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 Posted: Thu Aug 26, 2010 10:10 pm Post subject: Aug 27 VH |
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| XY-Wing 34 47 37 with Pivot in r5c5 removing 3 in r9c6 solves it. |
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kuskey
Joined: 10 Dec 2008
Posts: 141
Location: Pembroke, NH
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| Posted: Fri Aug 27, 2010 4:43 am Post subject: 27 Aug VH |
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| Barring a mistake, basics did it for me - very easy basics at that. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Aug 27, 2010 4:58 am Post subject: Re: 27 Aug VH |
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| kuskey wrote: | | Barring a mistake, basics did it for me - very easy basics at that. |
Based on experience, both personal and otherwise, it's a good idea to try it again under these circumstances.
I also needed the XY-Wing or, alternatively, looking at the implications of the potential Type 6 UR on 79. |
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kuskey
Joined: 10 Dec 2008
Posts: 141
Location: Pembroke, NH
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| Posted: Fri Aug 27, 2010 5:29 am Post subject: 27 Aug VH |
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| That "phantom" mistake must have hiding behind the 347 xy-wing I came across. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Aug 27, 2010 3:17 pm Post subject: |
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| Quote: | | I also needed the XY-Wing or, alternatively, looking at the implications of the potential Type 6 UR on 79. |
Marty:
I tried applying Type 6 UR rules and removed 7's and 9's from the QUAD in R1C6 based on there being only two 7's in C9 and two 9's in R3. But it didn't work. So I guess Type 6 UR's only work when there are triples and not QUADs. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Aug 27, 2010 3:31 pm Post subject: |
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| cgordon wrote: | | Quote: | | I also needed the XY-Wing or, alternatively, looking at the implications of the potential Type 6 UR on 79. |
Marty:
I tried applying Type 6 UR rules and removed 7's and 9's from the QUAD in R1C6 based on there being only two 7's in C9 and two 9's in R3. But it didn't work. So I guess Type 6 UR's only work when there are triples and not QUADs. |
Craig,
A true, full Type 6 has an X-Wing on one of the candidates and that number can be placed in the two bivalue cells. When you have a strong link on a number you can remove that number from the polyvalue cell in the other line. Because of the 9 strong link in r1, you remove the 9 from r3c9. Because of the strong link on 7 in r3 you can remove the 7 from r1c6.
What I did was test the possibilities of the DP killers and found that 28 in r1c6 led to an invalidity. With the removal of those it then became a Type 1 UR, with a 4 in r3c9. Hope this helps. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Fri Aug 27, 2010 3:57 pm Post subject: |
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Marty:
I can only do things by rote and my understanding of Type 6 UR’s is that the pairs are diagonally located as in the unrelated example below - and that in one of the rows or columns there are only TWO of the numbers.
Thus if there are only TWO 3’s in R1 – remove the 3 from the opposite triple in R3C9. Or, if say there are only two 9’s in C9 – remove the 9 from the triple in R1C1.
| Code: |
+---------+-------+--------+
| 369 . . | . . . | . . 39 |
| . . . | . . . | . . . |
| 39 . . | . . . | . . 379|
+---------+-------+--------+
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This has always worked before, but I have no idea if there are wings involved or whether this is really a Type 6.
If it isn’t – I demand it be given a name !!! |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Aug 27, 2010 4:43 pm Post subject: |
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| cgordon wrote: | Marty:
I can only do things by rote and my understanding of Type 6 UR’s is that the pairs are diagonally located as in the unrelated example below - and that in one of the rows or columns there are only TWO of the numbers.
Thus if there are only TWO 3’s in R1 – remove the 3 from the opposite triple in R3C9. Or, if say there are only two 9’s in C9 – remove the 9 from the triple in R1C1.
| Code: |
+---------+-------+--------+
| 369 . . | . . . | . . 39 |
| . . . | . . . | . . . |
| 39 . . | . . . | . . 379|
+---------+-------+--------+
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This has always worked before, but I have no idea if there are wings involved or whether this is really a Type 6.
If it isn’t – I demand it be given a name !!! |
Not enough here to tell if it's a true Type 6. But if either 3 or 9 is an X-Wing, then both bivalue cells can be solved with that number. The eliminations described in your last two sentences are correct. In your first post there were two strong links but you eliminated both from the same cell rather than eliminating each from the polyvalue cell in the other line. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Fri Aug 27, 2010 10:12 pm Post subject: |
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The last time I tried to help someone my phraseology seem to just make things worse but hopeful not this time.
Craig, my interpretation of your rules for Type 6 UR is that you are correct, but I am not sure you applied then correctly for this puzzle. Here is my code after basics with the UR marked #:
| Code: |
*-----------------------------------------------------------*
| 37 4 137 | 28 6 #2789 | 5 18 #79 |
| 9 17 5 | 38 37 4 | 2 18 6 |
| 6 2 8 | 5 1 #79 | 49 3 #479 |
|-------------------+-------------------+-------------------|
| 28 3 4 | 126 5 12 | 69 7 89 |
| 1 9 267 | 2468 47 278 | 46 5 3 |
| 78 5 67 | 346 9 37 | 1 2 48 |
|-------------------+-------------------+-------------------|
| 5 17 17 | 9 8 6 | 3 4 2 |
| 234 6 23 | 7 34 5 | 8 9 1 |
| 34 8 9 | 134 2 13 | 7 6 5 |
*-----------------------------------------------------------* |
Note the strong links (only two instances ) on 7 in row3 & col9
Using your rule and considering row3, we can remove 7 in r1c6.
And why is that valid? If r1c6=7, then r1c9=9, then r3c9 must be 7 due to the strong link on 7 in col9, and then r3c6=9 thereby creating the deadly pattern; thus r1c6 can not be 7.
If we repeat the process using col9 instead of row3 we get the same result; r1c6<>7
Now lets look at the 9. The strong links are in row1 & col6. Again using your rule for either the row or column, we get r3c9<>9. And again, why is that deletion valid? If r3c9=9, then r1c9=7, then r1c6 must be 9 due to the strong link on 9 in row1, and then r3c6=7 which gives us the deadly pattern; thus r3c9<>9
So, using your rule, I get r1c6<>7 and r3c9<>9 whereas I understood you post to indicate that you deleted both 7 & 9 from r1c6.
Also note that your rule works independently of the number of extra digits in the polyvalue cells.
Hope this is helpful......
Ted |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sat Aug 28, 2010 4:11 pm Post subject: |
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| Quote: | So, using your rule, I get r1c6<>7 and r3c9<>9 whereas I understood you post to indicate that you deleted both 7 & 9 from r1c6.
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Ted: I see where I made a basic error. There are three 9's in R3 (not two) - so I erred by removing 9 from R1C6. However you answered my question that there can be more than three nos. in the 'elimination' cell.
As I mentioned I don't get into the reasons for things (eg DP's). Being naturally lazy, I just have simple and basic rules and do things by rote.
What a Philistine eh!
Cheers |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Aug 29, 2010 1:00 am Post subject: |
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| cgordon wrote: |
As I mentioned I don't get into the reasons for things (eg DP's). Being naturally lazy, I just have simple and basic rules and do things by rote.
What a Philistine eh!
Cheers |
And I, being a mathematician, like to know why 2+2=4......
Ted |
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