The New Sudoku Players' Forum

by **Guest** » Wed Mar 30, 2005 2:14 pm

I must admit that I solved this yesterday lunch/afternoon in between phone calls.

The key is column 1 and the top middle cube and then column 4.

Column four solves your 1/7 dilema.

It did take me two goes though.

by **Guest** » Wed Mar 30, 2005 2:16 pm

Hmm. Maybe cryptic isn't what I need after all! I think janders solution counts as T&E, and I don't understand Bernard at all! I don't think box 2 can be resolved without T&E can it? I have possibilities of (67) (36) (37), which can be either 6 3 7 or 7 6 3. Likewise, I don't thing row 3 can be reduced from (578) (578) (37) (36) (68). My head hurts :o(

by **shakers** » Wed Mar 30, 2005 2:19 pm

Nigel's suggestion will work, but it smacks of T&E to me!

I'd love Wayne to comfirm whether this would count as a logical deduction, or T&E...

by **Guest** » Wed Mar 30, 2005 2:21 pm

Nigel - our posts crossed over...

How does this work without Trial and error? I can't see anyway to resolve col 1, box 2 or col 4, each of which have two valid resolutions, nor are there any interesting intersections.

Like I said - my head hurts!

by **Guest** » Wed Mar 30, 2005 2:24 pm

shakers wrote:Nigel's suggestion will work, but it smacks of T&E to me!

I'd love Wayne to comfirm whether this would count as a logical deduction, or T&E...

If using pencil marks to eliminate possibilities counts as T&E then so be it.

I did establish what was impossible and therefore reached a logical conclusion.

Looking at a completed one and trying to remember where you started makes your head hurt.

I will have to print it again if you are still stuck.

by **Bernard Stay** » Wed Mar 30, 2005 2:25 pm

Shakers: the version you showed had a 1 in box 2 at r3/c6. My options are 3/7, 6/3/7, 1. I contended that the 6 doesn't actually affect the 3/7 pairs and thus has to go in r1,c6. May be quite wrong, but looks logical to me.

by **shakers** » Wed Mar 30, 2005 2:30 pm

The way the puzzle stands is that
r1c6 could be (6,7)
r3c4 could be (3,7)
r3c5 could be (3,6)

r1c6 does have to be 6 (given by Nigel's T&E), but I still can't see how you made the leap...

by **Guest** » Wed Mar 30, 2005 2:32 pm

Bernard - I think you ment IJ, not Shakers, but anyway. You can eliminate 7 from r3c5 because all the 7s in box 5 are in col 5, leaving (37) and (36) - so these are not pairs.

by **Bernard Stay** » Wed Mar 30, 2005 2:47 pm

Yes, I meant you, sorry. But that's all right then, as 6 has gone to r1c6 we now have 3/7, 3, 1 along the bottom - thus 7,3,1 and all's well!

by **shakers** » Wed Mar 30, 2005 2:56 pm

All's well... apart from the fact there was no pair to allow you to put 6 in r1c6 which makes it a lucky placing!

My, and I don't think I'm alone, contention is that to put the 6 in r1c6 you have to use T&E...

by **Guest** » Wed Mar 30, 2005 3:05 pm

sorry to go back loads.... but how can you not have a 3 in r8c4?

by **shakers** » Wed Mar 30, 2005 3:10 pm

Because the other possibilities for r8c4 included the 1/7 pair, so anything but the 1/7 can be eliminated.

by **Guest** » Wed Mar 30, 2005 3:11 pm

Indeed, you are not alone Shakers!

Bernard - How did you get that 6?

Lil miss - 1 & 7 only appear on row 8 in cols 1 and 4, so 1 & 7 can be eliminated from the rest of the row, and those two cells (r8c1 & r8c4) can be reduced to just (17) - this is how you get the 4 in r8c5 (see the response to Tim's question at the start of this thread)

by **Guest** » Wed Mar 30, 2005 3:12 pm

oooooooooooo ok makes sense...... thank you!

by **Guest** » Wed Mar 30, 2005 3:16 pm

Well I have started again, to try and help you and now I can't do it.

Must have hit brain fade time.

Is it normal to blank like this

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