Very Hard puzzle - no solution by logic?

Post the puzzle or solving technique that's causing you trouble and someone will help


Postby Guest » Sat Jun 04, 2005 9:02 pm

i got a solution in about 45 minutes (had to start over once)
Posts: 312
Joined: 25 November 2005

The {1 or 6} elimination: does is stand up to rigorous exami

Postby laidback_tyke » Sat Jun 04, 2005 10:50 pm

Anonymous wrote:
whohe wrote:
laidback_tyke wrote:
Animator wrote:Look at row 2. What cells can have the number 1 or the number 6? there are only two such cells...

OK, I see that principle (thanks for that - I wasn't familiar with it before).
However, R2C2 can also be 6 (or 2,5,9), which would leave R2C1 & R2C6 to have a 1 in them.
We could have 2 or 5 in R2C1, R2C2 and R2C8 ... and so on.

Is this a case of having to chase it down, repeatedly applying the same principle to eliminate possibilities?

Hi there.

R2C2 can't be a 6 because Column 1 must have a 6 in it, and the 6 can only go in R1C1 or R2C1, therefore no other cell in the top left block of 9 could be a 6.

Ahh, yes, now that's coming in from a different direction - thanks for the reminder not to rely on only one principle.
I'd still like to pin down the "1-6 pair" thing. Is this a hidden thing? I'll have to do some more thinking about this before I make any more input:)

... and I thought I was good at logic! I need to get some sleep (not staying up until 0200 hrs working on "fiendish" sudoku!)

Thanks folks, for the stimulation:)

Hi again - I've had a further look at this 1-6 pair assertion and I think I've found a hole in it: see what you think ...
It's really focussing on the logic relating to the line AS GIVEN, i.e. not influenced by what else has been found (e.g. what "whohe" noted above).

With what was originally given, the possiblities for row2 are
1256 / 2569 / 8 / 579 / 59 / 167 / 3 / 25 / 4
To see if I could disprove the previous "1-6" pair assertions, I put 1 in C1, then of the remaining possibilites (6 or 7) in C6 I put 7.
Then 6 can go in only C2.
This forces C8 to be 2, and C4 can be 5 and C5 can be 9.
Hence an arrangement is possible that doesn't have a 6 in C6.

I know this arrangement probably doesn't fit with the rest of the matrix, but that's not the point. This all about the tools that we use to eliminate certain possibilities.

Looking forward to you comments,
Posts: 3
Joined: 29 May 2005

Re: The {1 or 6} elimination: does is stand up to rigorous e

Postby scrose » Sat Jun 04, 2005 11:03 pm

laidback_tyke wrote:With what was originally given, the possiblities for row2 are
1256 / 2569 / 8 / 579 / 59 / 167 / 3 / 25 / 4

You should eliminate your candidate 6 from r2c2.

There are no candidate 6's in column 1 of box[2,1] or box[3,1]. Thus the 6 in box[1,1] must be in column 1. Therefore you can eliminate 6 as a candidate for r2c2. Hence you are left with only two cells in row 2 that contain {1,6}.
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Joined: 31 May 2005


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