Fiendish #231

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Fiendish #231

Postby p_court99 » Mon Jul 18, 2005 1:33 pm

I'm at this stage:

Code: Select all
 7 |4  | 3
3 8| 6 |  4
41 |   |   
---+---+---
13 | 8 | 56
 65| 3 |1 
984|516|372
---+---+---
   |   | 6
6  | 9 |5 
 9 |6 5| 2


I've only ever done three other Su Doku's before - an easy, a mild and a difficult and they gave me no problems - do I need a new technique to solve this or am I missing something?

Thanks.:)
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Postby SteveF » Mon Jul 18, 2005 3:05 pm

Have a good look at row 2. You should find a candidate that can only go in one cell.
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Postby p_court99 » Mon Jul 18, 2005 3:46 pm

Haha, *oh yeah*. I don't know how I missed that, but thanks.:)

Edit: I must have been asleep - there were two other rows which had numbers that could be written straight in as well. :|
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Postby p_court99 » Mon Jul 18, 2005 4:05 pm

Hmm...I don't think I'm on much form today...can any point out what I'm stupidly missing now....you're allowed to laugh if you wish. :P

Code: Select all
27 |45 | 3
358| 6 |  4
41 |   |  5
---+---+---
132| 8 | 56
765| 3 |1
984|516|372
---+---+---
5  |   | 6
6  | 9 |5
89 |6 5| 2


I'm right in thinking that you don't need trial and error to solve these things, yeah?
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Postby Bernard Stay » Mon Jul 18, 2005 6:03 pm

You should see an a/b pair in box 3, then a bit of elimination will get you moving again.
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Postby p_court99 » Mon Jul 18, 2005 6:34 pm

Box three...I assume that's bottom left?

Okay, well in my bottom left I had this (the numbers in brackets are 'little' numbers, i.e. the possibles):

Code: Select all
5 (2/4) (1/2/3/7)
6 (2/4) (1/2/3/7)
8   9   (1/2/3/7)


I had already noticed the 2/4 pair (but thanks for the tip anyway), so I can reduce to this:

Code: Select all
5 (2/4) (1/3/7)
6 (2/4) (1/3/7)
8   9   (1/3/7)


Then, using the 4/7 pair in row nine, I can further reduce to this:

Code: Select all
5 (2/4) (1/3/7)
6 (2/4) (1/3/7)
8   9    (1/3)


So, that's where I was when I got stuck.

Just in case box three is top right, here's what I've got there:

Code: Select all
  (6/8/9)     3   (1/8/9)
  (2/7/9)   (1/9)    4
(2/6/7/8/9) (8/9)    5


If there's supposed to be an a/b pair there, then I've got something wrong.

Thanks.:)
p_court99
 
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Postby Bernard Stay » Mon Jul 18, 2005 6:39 pm

Yes, box 3 - top right. There's a 2/7 pair set, no?
Last edited by Bernard Stay on Mon Jul 18, 2005 2:40 pm, edited 1 time in total.
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Postby p_court99 » Mon Jul 18, 2005 6:40 pm

Ah yes, I didn't think to look for pair sets with other numbers...sorry, as I said, I'm pretty new to this. Thanks.:)
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Postby gmc » Mon Jul 18, 2005 6:41 pm

and a triple
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Postby Bernard Stay » Mon Jul 18, 2005 6:43 pm

In that box? Can't see one.
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Postby george-no1 » Mon Jul 18, 2005 6:45 pm

p_court99 wrote:I'm right in thinking that you don't need trial and error to solve these things, yeah?


Just to clarify, Pappocom puzzles never require trial and error (from the post topic I assume that this puzzle is from The Times, which uses Pappocom puzzles).
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Postby gmc » Mon Jul 18, 2005 6:46 pm

1/8/9?
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Postby george-no1 » Mon Jul 18, 2005 6:48 pm

The triple is (1,8,9) in that box but it is not really needed because spotting the (2,7) pair means that there is only one cell that can contain a 6.

George:)
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Postby p_court99 » Mon Jul 18, 2005 6:56 pm

Just to clarify, Pappocom puzzles never require trial and error (from the post topic I assume that this puzzle is from The Times, which uses Pappocom puzzles).

Thanks, that's good to know.:)

Thanks to everyone who helped me out, I've learnt something that I hadn't managed to figure out on my own and I've now solved this puzzle. I'll see if I get through a fiendish by myself next time.:)
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