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Oct 6 VH

 
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Clement



Joined: 24 Apr 2006
Posts: 1110
Location: Dar es Salaam Tanzania
PostPosted: Tue Oct 05, 2010 10:04 pm    Post subject: Oct 6 VH Reply with quote
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
PostPosted: Tue Oct 05, 2010 11:05 pm    Post subject: Reply with quote
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: 422
Location: Birmingham UK
PostPosted: Wed Oct 06, 2010 8:50 am    Post subject: Reply with quote
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
PostPosted: Thu Oct 07, 2010 2:16 am    Post subject: Reply with quote
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
PostPosted: Thu Oct 07, 2010 2:42 pm    Post subject: Reply with quote
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     |
 *-----------------------------------------------------------*

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
PostPosted: Thu Oct 07, 2010 3:46 pm    Post subject: Reply with quote
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     |
 *-----------------------------------------------------------*

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
PostPosted: Fri Oct 08, 2010 12:35 am    Post subject: Reply with quote
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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