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 Post subject: Twins NonConsecutive X
PostPosted: Tue Oct 23, 2012 5:06 pm 
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Twins NonConsecutive X


Twins: the cell is one of the two numbers.
Nonconsecutive: horizontally and vertically adjacent numbers cannot be consecutive.
X no repeats in the two main diagonals.

This is a hard one - assassin level.

The concept wellbeback pointed out in the previous puzzle will be a little help.



Image
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Candidates:
+----------------------------------+----------------------------------+----------------------------------+
| 12 123456789 123456789 | 123456789 123456789 123456789 | 123456789 123456789 34 |
| 123456789 23 123456789 | 123456789 123456789 123456789 | 123456789 123456789 123456789 |
| 123456789 123456789 34 | 123456789 123456789 123456789 | 78 123456789 123456789 |
+----------------------------------+----------------------------------+----------------------------------+
| 123456789 123456789 123456789 | 45 123456789 89 | 123456789 123456789 123456789 |
| 123456789 123456789 123456789 | 123456789 123456789 123456789 | 123456789 123456789 123456789 |
| 123456789 123456789 123456789 | 12 123456789 56 | 123456789 123456789 123456789 |
+----------------------------------+----------------------------------+----------------------------------+| 123456789 123456789 23 | 123456789 123456789 123456789 | 67 123456789 123456789 |
| 123456789 78 123456789 | 123456789 123456789 123456789 | 123456789 78 123456789 |
| 67 123456789 123456789 | 123456789 123456789 123456789 | 123456789 123456789 89 |
+----------------------------------+----------------------------------+----------------------------------+


Solution:
186375294
429618537
753942861
537429618
861753942
294186375
942861753
375294186
618537429


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PostPosted: Sat Dec 31, 2016 11:05 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
I first came across this puzzle about a year ago and did the easy early stages including the twin cells and routine NC steps. Then I forgot about this puzzle until recently.

The twins can easily be seen.:
R1C1 cannot be 2 because R2C28 = [32]would force 1 in R5C5 and R6C4 from the diagonals
and reached this position.

Image

After that things got a lot harder; it's definitely Assassin level.
Here are my remaining steps:
The obvious thought from the above position was T&E for the cells in N5 but that doesn't immediately lead anywhere, so I tried forcing chains with implied NC where appropriate.
R7C45 cannot be {89} which are consecutive -> R7C12 must contain one of 8,9 -> R8C1 = 3

2 in N3 only in R1C7 => 2 in N2 in R3C56, no 1 in R3C56 => 1 in R3 in R3C89 => R2C9 = 7
or 2 in R3C9, no 1 in R2C9
-> R2C9 = 7

7 in N1 only in R1C3 => 8 in R2C1 => 4 in R3C1
or 7 in R3C1 => R2C1 = 4
-> no 9 in R23C1 -> no 8 in R4C1

7 in N8 only in R9C4 => R5C4 = 8, R9C6 = 8, R1C6 = 7 => R2C5 = 8
or 7 in R9C6 => 8 in C6 in R12C6
-> no 8 in R1C45 + R2C4

R7C12 and R89C3 must each contain one of 8,9
R9C3 = 8 => R9C6 = 7
or R8C3 = 9 => R7C1 = 8, R23C1 = [47], R3C2 = 9, R3C4 = 6 => R2C4 = 9 => 8 in N2 in R12C6, locked for C6 => R9C6 = 7
-> R9C6 = 7, 8 in C6 only in R12C6, locked for N2

9 in N8 only in R78C45
R7C4 = 9 => R23C4 = [57] => R12C4 = [84] => R9C3 = 8 => no 9 in R8C3
or R7C5 = 9 => R5C4 = 8 => R9C3 = 8 => no 9 in R8C3
or 9 in R8C45
-> no 9 in R8C3, 9 in R7C12, locked for R7, R9C3 = 8, R9C2 = 1

R1C3 = {567} -> no 6 in R1C2, R8C3 = {45} -> no 5 in R8C4

R5C4 = 7
or R5C4 = 8 => R7C5 = 8, R8C4 = 9, R23C4 = [57]
-> no 7 in R1C4, 7 in R1 only in R1C35 -> no 6 in R1C4

R1C2 = 5 => R1C4 = {23}
or R1C2 = 8 => R1C3 = {56} => no 5 in R1C4
-> R1C4 = {23} -> no 2,3 in R1C5

R1C5 = {67} -> no 6 in R2C5

R1C3 = {56} => R2C3 = 9
or R1C3 = 7 => R1C2 = 5 => R2C3 = 9
-> R2C3 = 9, R2C4 = {56} -> no 6 in R3C4, R3C2 = {56} -> no 5,6 in R4C2

6 in N1 in R1C3+R3C2. R1C3 = 6
or R3C2 = 6 => R3C1 = 4 => R1C3 = 7
-> R1C3 = {67}, 5 in N1 only in R13C2, locked for C2

Naked pair {67} in R1C35, locked for R1 -> R1C8 = 9

3 in C2 only in R456C2 -> no 4 in R5C2, 4 in C2 only in R67C2 -> no 3 in R6C2, 7 in C8 only in R456C8 -> no 6 in R5C8

4 in N4 only in R5C3+R6C23 -> no 5 in R6C3

R8C3 = 5 -> R7C12 = [94], R6C2 = {89} -> no 8,9 in R5C2+R6C1 -> R6C2 = 9, R5C7 = 9

8 in N4 only in R4C2+R5C1 -> no 7 in R4C1

5 in C6 only in R123C6, locked for N2 -> R2C4 = 6, R1C35 = [67], R13C2 = [85]

R3C8 = 6, no 5,7 in R4C8

and the rest is naked singles without using NC


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PostPosted: Mon Jan 02, 2017 1:00 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
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Andrew et al

Happy New Year

Is this the longest first post solution on the site (at 5 years)?

Maurice


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