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Jul 13 VH

 
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Clement



Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania

PostPosted: Mon Jul 13, 2015 12:40 am    Post subject: Jul 13 VH Reply with quote

Code:

+------------+------------+---------+
| 4   26  1  | 59  26 59  | 7 8  3  |
| 9   27  8  | 127 3  127 | 6 4  5  |
| 67  5   3  | 8   67 4   | 2 9  1  |
+------------+------------+---------+
| 1   8   6  | 4   5  27  | 3 27 9  |
| 5   9   27 | 3   8  6   | 4 1  27 |
| 27  3   4  | 279 1  279 | 5 6  8  |
+------------+------------+---------+
| 8   4   27 | 6   9  3   | 1 5  27 |
| 3   1   5  | 27  4  8   | 9 27 6  |
| 267 267 9  | 15  27 15  | 8 3  4  |
+------------+------------+---------+
Skyscraper on 7 along col 25; r3c46<>7; solves it.
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bat999



Joined: 09 Jul 2015
Posts: 55
Location: UK

PostPosted: Mon Jul 13, 2015 12:47 am    Post subject: Reply with quote

Code:
.--------------.---------------.-----------.
| 4    b26  1  | 59   c26  59  | 7  8   3  |
| 9    a27  8  | 127   3   127 | 6  4   5  |
| 67    5   3  | 8     67  4   | 2  9   1  |
:--------------+---------------+-----------:
| 1     8   6  | 4     5   27  | 3  27  9  |
| 5     9   27 | 3     8   6   | 4  1   27 |
| 27    3   4  | 279   1   279 | 5  6   8  |
:--------------+---------------+-----------:
| 8     4   27 | 6     9   3   | 1  5   27 |
| 3     1   5  | 27    4   8   | 9  27  6  |
| 267   6-7 9  | 15   d27  15  | 8  3   4  |
'--------------'---------------'-----------'
(7=2)r2c2 - (2=6)r1c2 - (6=2)r1c4 - (2=7)r9c5 => -7r9c2; stte
Cool


Last edited by bat999 on Thu Jul 23, 2015 5:48 pm; edited 1 time in total
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Clement



Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania

PostPosted: Mon Jul 13, 2015 6:44 am    Post subject: Jul 13 VH Reply with quote

bat999 completed the basics and an XY-Wing 27-6 pivoted in r9c5 comes out; r1c2<>6; stte
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bat999



Joined: 09 Jul 2015
Posts: 55
Location: UK

PostPosted: Mon Jul 13, 2015 9:53 am    Post subject: Re: Jul 13 VH Reply with quote

Clement wrote:
... an XY-Wing 27-6 pivoted in r9c5 comes out; r1c2<>6; stte

Yes, it makes a shorter chain. Very Happy
Code:
'.--------------.---------------.-----------.
| 4     2-6 1  | 59   a26  59  | 7  8   3  |
| 9     27  8  | 127   3   127 | 6  4   5  |
| 67    5   3  | 8     67  4   | 2  9   1  |
:--------------+---------------+-----------:
| 1     8   6  | 4     5   27  | 3  27  9  |
| 5     9   27 | 3     8   6   | 4  1   27 |
| 27    3   4  | 279   1   279 | 5  6   8  |
:--------------+---------------+-----------:
| 8     4   27 | 6     9   3   | 1  5   27 |
| 3     1   5  | 27    4   8   | 9  27  6  |
| 267  c67  9  | 15   b27  15  | 8  3   4  |
'--------------'---------------'-----------'
(6=2)r1c5 - (2=7)r9c5 - (7=6)r9c2 => -6r1c2; stte
Cool


Last edited by bat999 on Thu Jul 23, 2015 5:51 pm; edited 1 time in total
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jayagee



Joined: 26 Feb 2006
Posts: 13
Location: sunny north dakota

PostPosted: Mon Jul 13, 2015 7:12 pm    Post subject: Reply with quote

This was a funny situation. 27 in c3r5r7 and c9r5r7 means that there is NOT a unique solution to this puzzle. Simply put in a 2 or 7 in any of those 4 cells and it just falls apart.
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bat999



Joined: 09 Jul 2015
Posts: 55
Location: UK

PostPosted: Mon Jul 13, 2015 7:42 pm    Post subject: Reply with quote

jayagee wrote:
... there is NOT a unique solution to this puzzle...

You have another solution? Confused
Code:
.---------.---------.---------.
| 4  2  1 | 5  6  9 | 7  8  3 |
| 9  7  8 | 2  3  1 | 6  4  5 |
| 6  5  3 | 8  7  4 | 2  9  1 |
:---------+---------+---------:
| 1  8  6 | 4  5  2 | 3  7  9 |
| 5  9  7 | 3  8  6 | 4  1  2 |
| 2  3  4 | 9  1  7 | 5  6  8 |
:---------+---------+---------:
| 8  4  2 | 6  9  3 | 1  5  7 |
| 3  1  5 | 7  4  8 | 9  2  6 |
| 7  6  9 | 1  2  5 | 8  3  4 |
'---------'---------'---------'
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Mon Jul 13, 2015 10:49 pm    Post subject: Reply with quote

I just had to chime in here being a traditionalist that knows that these VHs can be solved using just basics and wings!

Clement's original grid missed the fact that the 2 can be eliminated from r9c2 because of the locked 2's in box 1. Then the 'standard' solution (for us traditionalists) is to simply use one of the two XY wings that are available now that the 2 has been removed from r9c2. Clement's 2nd post mentioned the 27-6 XY-wing with pivot at r9c5 which forces r1c2<>6 and there's also a 67-2 XY-wing with pivot at r9c2 which forces r1c5<>2.

I don't have any idea what jayagee is talking about! I'm pretty sure he's mistaking the 27's that he's referring to as a non-UR (which they obviously aren't). It's an easy mistake to make. They're in the same row and column but they're not in the same box.
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jayagee



Joined: 26 Feb 2006
Posts: 13
Location: sunny north dakota

PostPosted: Mon Jul 13, 2015 11:43 pm    Post subject: Reply with quote

It's my understanding that each Sudoku MUST have one unique solution. This puzzle; with 2&7 in four boxes forming a rectangle indicates that there is NOT a unique solution. You can plug in either a 2 or a 7 in any one of the four cells (c3r5,r7 or c9r5,r7) and solve the puzzle.
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Mon Jul 13, 2015 11:50 pm    Post subject: Reply with quote

jayagee wrote:
This was a funny situation. 27 in c3r5r7 and c9r5r7 means that there is NOT a unique solution to this puzzle. Simply put in a 2 or 7 in any of those 4 cells and it just falls apart.


If Keith were reading this, He could explain it much better than I could. I think there is a unique solution because there's no deadly pattern visible. With a DP, there are a certain number of identical bivalue cells. You can plug numbers into the cells and complete the puzzle and arrive at a valid solution. Then you could reverse those numbers and arrive at another valid solution that is not identical. In this case, after entering the numbers in the four 27 cells make it look like a deadly pattern because the rows, columns and boxes each have 27, making for a valid solution. But after solving the four 27 cells, start making eliminations in other cells and the grids will be different, only one of which will be valid. It looks like a UR, but it isn't. The principles of URs apply only when the four cells are in two boxes which are in the same chute (band or stack).
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jayagee



Joined: 26 Feb 2006
Posts: 13
Location: sunny north dakota

PostPosted: Tue Jul 14, 2015 12:18 am    Post subject: Reply with quote

OOPS!! I stand corrected. I happened to plug a 2 in c9r5 and the thing just fell apart, but when I went back and tried a 7 at that location it wouldn't solve so I guess I have misunderstood the rectangular pattern of two numbers alike. Guess i'll have to do more study on that .
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Tue Jul 14, 2015 12:33 am    Post subject: Reply with quote

jayagee wrote:
It's my understanding that each Sudoku MUST have one unique solution. This puzzle; with 2&7 in four boxes forming a rectangle indicates that there is NOT a unique solution. You can plug in either a 2 or a 7 in any one of the four cells (c3r5,r7 or c9r5,r7) and solve the puzzle.


Holy crow! I don't know how I could make it any clearer than that Marty. There IS only one solution jayagee! That's what we're trying to tell you. If you'd take the time to plug a 2 into r5c3, you'll see that it doesn't work! In order for there be what Marty called a deadly pattern (and what I referred to as a 'non-UR'), the numbers have to be in the same row, same column, AND same box!

Go ahead, plug a 2 in r5c3 and try to fill out the entire grid and you'll see. And if you think you do have a solution with a 2 plugged in r5c3, then post it for us (I won't stay up late though because I know it doesn't exist.) As I said, it's an easy mistake to make! It looks logical at first - but it's not right. It's not just the same row and column! They also have to be in the same box.

BUT - don't confuse that rule with the X-wings! They only have to be in the same row and column!
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