| View previous topic :: View next topic |
| Author |
Message |
Clement
Joined: 24 Apr 2006
Posts: 1110
Location: Dar es Salaam Tanzania
|
 Posted: Mon Jul 13, 2015 12:40 am Post subject: Jul 13 VH |
 |
|
| 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. |
|
| Back to top |
|
 |
bat999
Joined: 09 Jul 2015
Posts: 55
Location: UK
|
| Posted: Mon Jul 13, 2015 12:47 am Post subject: |
|
|
| 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
Last edited by bat999 on Thu Jul 23, 2015 5:48 pm; edited 1 time in total |
|
| Back to top |
|
|
Clement
Joined: 24 Apr 2006
Posts: 1110
Location: Dar es Salaam Tanzania
|
| Posted: Mon Jul 13, 2015 6:44 am Post subject: Jul 13 VH |
|
|
| bat999 completed the basics and an XY-Wing 27-6 pivoted in r9c5 comes out; r1c2<>6; stte |
|
| Back to top |
|
|
bat999
Joined: 09 Jul 2015
Posts: 55
Location: UK
|
| Posted: Mon Jul 13, 2015 9:53 am Post subject: Re: Jul 13 VH |
|
|
| Clement wrote: | | ... an XY-Wing 27-6 pivoted in r9c5 comes out; r1c2<>6; stte |
Yes, it makes a shorter chain. 
| 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
Last edited by bat999 on Thu Jul 23, 2015 5:51 pm; edited 1 time in total |
|
| Back to top |
|
|
jayagee
Joined: 26 Feb 2006
Posts: 13
Location: sunny north dakota
|
| Posted: Mon Jul 13, 2015 7:12 pm Post subject: |
|
|
| 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. |
|
| Back to top |
|
|
bat999
Joined: 09 Jul 2015
Posts: 55
Location: UK
|
| Posted: Mon Jul 13, 2015 7:42 pm Post subject: |
|
|
| jayagee wrote: | | ... there is NOT a unique solution to this puzzle... |
You have another solution? 
| 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 |
'---------'---------'---------' |
|
|
| Back to top |
|
|
dongrave
Joined: 06 Mar 2014
Posts: 564
|
| Posted: Mon Jul 13, 2015 10:49 pm Post subject: |
|
|
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. |
|
| Back to top |
|
|
jayagee
Joined: 26 Feb 2006
Posts: 13
Location: sunny north dakota
|
| Posted: Mon Jul 13, 2015 11:43 pm Post subject: |
|
|
| 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. |
|
| Back to top |
|
|
Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
|
| Posted: Mon Jul 13, 2015 11:50 pm Post subject: |
|
|
| 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). |
|
| Back to top |
|
|
jayagee
Joined: 26 Feb 2006
Posts: 13
Location: sunny north dakota
|
| Posted: Tue Jul 14, 2015 12:18 am Post subject: |
|
|
| 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 . |
|
| Back to top |
|
|
dongrave
Joined: 06 Mar 2014
Posts: 564
|
| Posted: Tue Jul 14, 2015 12:33 am Post subject: |
|
|
| 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! |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
