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15 Nov 2014 12:14:48

shiro
Joined: 8 Nov 2014
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What is next logical move?
This message was edited 1 time. Last update was at 15 Nov 2014 12:15:26



16 Nov 2014 05:37:25

Stickywulf
Joined: 22 Nov 2013
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shiro wrote:What is next logical move?
Look at Row 5 Column 9 (R5C9). It must be either 1 or 4. If we pretend that it does equal 4 then we find out quickly that this can't be true. Follow the logical steps below:
Let R5C9 = 4
Then R4C8=1
Then R7C8=9
Then R7C2=1
Then R5C2=4 .... but R5C2 can't equal 4 since we let R5C9 equal 4.
Therefore R5C9 can't be 4, and so R5C9=1
I think this might be known as a "jellyfish" if you want to Google more information.
This message was edited 1 time. Last update was at 16 Nov 2014 05:38:26



24 Nov 2014 02:17:24

rdwells
Joined: 29 Jul 2013
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shiro wrote:What is next logical move?
Check out R1C3. If that's a 6, then there are 4 squares in a rectangle (R2C3, R2C5, R3C3, and R3C5) that are all either 1 or 4; this would lead to multiple solutions, which would mean the puzzle is invalid. So R1C3 must be 4.
Question for others: Is the assumption that there are no multiple solutions valid? It should be true of any wellcreated puzzle, and I use often it in situations like the one here, but it just kinda feels wrong.



28 Nov 2014 05:42:05

GreenLantern
Joined: 28 Nov 2014
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A Unique Rectangle at r23c35 allows us to conclude that either r2c3=6 or r3c3=6 and therefore, r1c3<>6, and the puzzle is solved.
Some solvers like to avoid using uniqueness assumptions in their solution, but I don't see the point in that since by definition, a valid sudoku must have a single unique solution. Why limit the number of solving techniques in your toolbox?



28 Nov 2014 15:15:58

jeka
(Forum Admin)
Joined: 12 Jul 2010
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rdwells wrote:
Check out R1C3. If that's a 6, then there are 4 squares in a rectangle (R2C3, R2C5, R3C3, and R3C5) that are all either 1 or 4; this would lead to multiple solutions, which would mean the puzzle is invalid. So R1C3 must be 4.
If you keep solving the puzzle with this assumption, i.e. that R1C3 is 6, you will come to a point where you'll realize that, in fact, no solution exists, i.e. putting a 6 in R1C3 is wrong. So, no, this will not lead to multiple solutions, but rather to no solution at all.
rdwells wrote:
Question for others: Is the assumption that there are no multiple solutions valid? It should be true of any wellcreated puzzle, and I use often it in situations like the one here, but it just kinda feels wrong.
This assumption is valid, however, this approach is still wrong.
Now about solving this particular puzzle, notice that R7C2 and R8C9 will always have the same value (either a 1 or a 9) that is enforced by R7C8, however, that value cannot be 1 because that would prevent both R5C2 and R5C9 from having a 1. Therefore, the value in both R7C2 and R8C9 must be a 9.
This message was edited 1 time. Last update was at 1 Dec 2014 22:53:59



28 Nov 2014 15:52:30

GreenLantern
Joined: 28 Nov 2014
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Actually, I was too hasty. r1c3<>6 only advances the puzzle to this point:
And here, the puzzle is finally solved by making another uniqueness deduction. This is a BUG+1 grid which allows us to immediately conclude r8c2=1, and the puzzle is reduced to singles.



