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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Tue Oct 05, 2010 10:04 pm Post subject: Oct 6 VH |
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XY-Wing 18 16 68 with Pivot 16 in r3c4 removing 8 in r1c2 solves it. |
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kragzy
Joined: 01 May 2007 Posts: 112 Location: Australia
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Posted: Tue Oct 05, 2010 11:05 pm Post subject: |
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Also a 468 XY wing centred on R1C4 provides a solution. Since after basics this puzzle comprises only 1s, 4s, 6s, and 8s, I suspect that there are several solutions. I'll leave it to others to find them and report.
Cheers |
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hughwill
Joined: 05 Apr 2010 Posts: 424 Location: Birmingham UK
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Posted: Wed Oct 06, 2010 8:50 am Post subject: |
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UR 68 leaves R3C8 a 1- all else follows. Very hard?? |
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prakash
Joined: 02 Jan 2008 Posts: 67 Location: New Jersey, USA
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Posted: Thu Oct 07, 2010 2:16 am Post subject: |
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I took the UR route since it seemed so obvious. Didn't care to look for anything else. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Thu Oct 07, 2010 2:42 pm Post subject: |
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hughwill & prakash,
Here is my code after basics:
Code: | *-----------------------------------------------------------*
| 2 68 5 | 146 18 7 | 148 3 9 |
| 9 3 1 | 2 5 48 | 6 7 48 |
| 68 4 7 | 16 3 9 | 2 18 5 |
|-------------------+-------------------+-------------------|
| 5 68 3 | 9 7 2 | 148 1468 148 |
| 4 1 2 | 5 68 68 | 3 9 7 |
| 68 7 9 | 3 4 1 | 5 68 2 |
|-------------------+-------------------+-------------------|
| 1 2 46 | 8 9 46 | 7 5 3 |
| 3 9 468 | 7 16 5 | 148 2 148 |
| 7 5 48 | 14 2 3 | 9 148 6 |
*-----------------------------------------------------------*
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I do not believe that this code contains a valid UR. A valid UR must contain the same two digits in a square pattern of four cells that occupy two rows, two columns and two boxes. Here is a link to an excellent post detailing URs.
Ted |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Thu Oct 07, 2010 3:46 pm Post subject: |
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tlanglet wrote: | hughwill & prakash,
Here is my code after basics:
Code: | *-----------------------------------------------------------*
| 2 68 5 | 146 18 7 | 148 3 9 |
| 9 3 1 | 2 5 48 | 6 7 48 |
| 68 4 7 | 16 3 9 | 2 18 5 |
|-------------------+-------------------+-------------------|
| 5 68 3 | 9 7 2 | 148 1468 148 |
| 4 1 2 | 5 68 68 | 3 9 7 |
| 68 7 9 | 3 4 1 | 5 68 2 |
|-------------------+-------------------+-------------------|
| 1 2 46 | 8 9 46 | 7 5 3 |
| 3 9 468 | 7 16 5 | 148 2 148 |
| 7 5 48 | 14 2 3 | 9 148 6 |
*-----------------------------------------------------------*
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I do not believe that this code contains a valid UR. A valid UR must contain the same two digits in a square pattern of four cells that occupy two rows, two columns and two boxes. Here is a link to an excellent post detailing URs.
Ted |
Certainly not one of the defined URs. If one wants to monkey around in boxes 69 with the potential 18 DP, it means one of the four cells must be=4. The solution to box 2 is then forced, or so it looked to me. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Fri Oct 08, 2010 12:35 am Post subject: |
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Marty,
Using internal analysis for AUR(18)r48c79, one of the four cells, r48c79, must equal 4.
If r48c7=4 then r1c7<>4, then r1c4=4
If r48c9=4 then r2c9<>4, then r2c6=4
This results in two pincers equal 4 in box 2, but I not see a forced result. What path did you utilize to force a solution in box2?
I did locate a deletion for AUR(18)r48c79 using an external analysis as follows
External SIS:
Digit 1: col7 => r1c7=1; col9 is empty
Digit 8: box6 => r46c8=8; box9 => r9c8=8
Thus, we have four conditions to satisfy: r1c7=1,r4c8=8,r6c8=8,r9c8=8
If r469c8=8 then r3c8=1, then r9c8<>1, then r9c4=1
We now have two pincers for digit 1 that deletes 1 in r1c4
Another possiblity is AUR(48)r47c79
External SIS:
Digit 4: box6 => r4c8=4; box9 => r9c8=4
Digit 8: col7 => r1c7=8; col9 => r2c9=8
If r4c8=4, then r4c8<>6, then r4c2=6, then r1c2=8, then r1c5=1
If r9c8=4, then r9c4=1, then r8c5<>1, then r1c5=1
If r1c7=8, then r1c5=1
If r2c9=8, then r3c8=1, then r9c8<>1, then r9c4=1, then r8c5<>1, then r1c5=1
Thus all four inferences result in r1c5=1 to complete the puzzle in one step.
So, my initial statement that this puzzle does not contain a valid UR is incorrect; I should have stated that this puzzle does not contain a predefined "Type n UR".
For those that are interested, a post on using external inferences for AURs is available here.
Ted |
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