## Fiendish #231

All about puzzles in newspapers, magazines, and books

### Fiendish #231

I'm at this stage:

Code: Select all
` 7 |4  | 3 3 8| 6 |  441 |   |   ---+---+---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.
p_court99

Posts: 9
Joined: 18 July 2005

Have a good look at row 2. You should find a candidate that can only go in one cell.
SteveF

Posts: 86
Joined: 26 March 2005

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. :|
p_court99

Posts: 9
Joined: 18 July 2005

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 | 3358| 6 |  441 |   |  5---+---+---132| 8 | 56765| 3 |1 984|516|372---+---+---5  |   | 66  | 9 |5 89 |6 5| 2 `

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

Posts: 9
Joined: 18 July 2005

You should see an a/b pair in box 3, then a bit of elimination will get you moving again.
Bernard Stay

Posts: 94
Joined: 22 March 2005

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

Posts: 9
Joined: 18 July 2005

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.
Bernard Stay

Posts: 94
Joined: 22 March 2005

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.
p_court99

Posts: 9
Joined: 18 July 2005

and a triple
gmc

Posts: 8
Joined: 03 July 2005

In that box? Can't see one.
Bernard Stay

Posts: 94
Joined: 22 March 2005

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).
george-no1

Posts: 150
Joined: 20 May 2005

1/8/9?
gmc

Posts: 8
Joined: 03 July 2005

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
george-no1

Posts: 150
Joined: 20 May 2005

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.
p_court99

Posts: 9
Joined: 18 July 2005