Hidden pairs?

Advanced methods and approaches for solving Sudoku puzzles

Hidden pairs?

Postby markb » Thu Jan 19, 2006 7:28 pm

Hello again

I've had some advice on doing 'fiendish' (in The Times) level puzzles. Crazy Girl etc said that the most difficult aspect of these would be a hidden pair.

Why can't I see one in this puzzle??

* 7 9 | * * * | 2 * *
* * * | 2 * 1 | * * *
6 8 4 | * 2 7 | * 3 1
7 2 1 | * * 3 |6 4 *
3 9 5 | 4 1 6 | 7 8 2
9 5 * | * 4 * | * * *
* * * | 3 * 9 | * **
* * 6 | * * * | 4 9 *

Thanks for any help

Mark
markb
 
Posts: 12
Joined: 27 December 2005

Postby ronk » Thu Jan 19, 2006 7:34 pm

Hi markb, your puzzle is missing one row, Ron
ronk
2012 Supporter
 
Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Postby markb » Thu Jan 19, 2006 7:45 pm

Maybe that's why it's so difficult!

Sorry, should have been -

* 7 9 | * * * | 2 * *
* * * | 2 * 1 | * * *
* * * | * 3 * | * 5 *
6 8 4 | * 2 7 | * 3 1
7 2 1 | * * 3 |6 4 *
3 9 5 | 4 1 6 | 7 8 2
9 5 * | * 4 * | * * *
* * * | 3 * 9 | * **
* * 6 | * * * | 4 9 *

Cheers
Mark
markb
 
Posts: 12
Joined: 27 December 2005

Postby Carcul » Thu Jan 19, 2006 8:00 pm

Hi Markb.

You have an hidden single in r1c9 (which allows the determination of the remaining positions of the candidate in the grid) and then a naked quad in box 2 that solve the puzzle.

Regards, Carcul
Carcul
 
Posts: 724
Joined: 04 November 2005

Postby markb » Thu Jan 19, 2006 8:26 pm

Ok I can now see the single 3 but don''t get the hidden quad

??

Can you explain please?

Thanks
Mark
markb
 
Posts: 12
Joined: 27 December 2005

Postby Carcul » Thu Jan 19, 2006 8:34 pm

Hi Markb.

It's a naked quad: four cells populated only with the same four candidates (but not necessarily with all those four candidates in each of the four cells). In box 2 you have cells r1c4,r1c5,r1c6,r3c6 populated just with the numbers 4,5,6,8: therefore, these candidates can be eliminated from r2c5 and r3c4.

Hope this help.

Regards, Carcul
Last edited by Carcul on Thu Jan 19, 2006 4:35 pm, edited 1 time in total.
Carcul
 
Posts: 724
Joined: 04 November 2005

Postby Animator » Thu Jan 19, 2006 8:35 pm

It's either a naked quad or a hidden pair. (It is not a hidden quad.)

Hidden pair: 7 and 9 in r2c5 and r3c4.

Naked quad: 4, 5, 6, 8 in r1c4, r1c5, r1c6, r3c6.
Animator
 
Posts: 469
Joined: 08 April 2005

Postby markb » Thu Jan 19, 2006 8:43 pm

Sorry I meant naked quad.


It's a naked quad: four cells populated only with the same four candidates (but not necessarily with all those four candidates in each of the four cells). In box 2 you have cells r1c4,r1c5,r1c6,r3c6 populated just with the numbers 4,5,6,8: therefore, these candidates can be eliminated from r2c5 and r3c4


But 5, 6 & 8 are in r2 c5 AND 4 & 8 are in r3 c6

???

Mark
markb
 
Posts: 12
Joined: 27 December 2005

Postby Animator » Thu Jan 19, 2006 9:52 pm

That's why it's a naked quad and not a hiddden quad.

Naked quad = there are 4 cells that have only those 4 candidates. (read: there are 4 cells in which no other number then those 4 are possible)

If you fill in 5, 6, 8 in r2c5 then you end up with an empty cell. (Feel free to try it).
The same is true for r3c6.
Animator
 
Posts: 469
Joined: 08 April 2005

Re: Hidden pairs?

Postby QBasicMac » Thu Jan 19, 2006 11:12 pm

markb wrote:Crazy Girl etc said that the most difficult aspect of these would be a hidden pair.


Hunhh? What's all the talk about quads?

First place the hidden singles in row 1, col 7, row 9 and row 2.
Then see the hidden pair 79 in box 2, right?

Then after placing the 5 in row 2, there are locked candidates: 6 in box 2 and 8 in row 2.

After that, singles to the end.

As Crazy Girl, etc. said: nothing more difficult than one hidden pair.

Mac
QBasicMac
 
Posts: 441
Joined: 13 July 2005

Postby TKiel » Thu Jan 19, 2006 11:43 pm

All the candidate 3's can be placed as singles.

X-wing in 1's.

Naked triple in row 3. (Which for my eyes is usually easier to spot that a hidden anything.)

Singles after that.

Tracy
TKiel
 
Posts: 209
Joined: 05 January 2006

OK thanks everyone!

Postby markb » Fri Jan 20, 2006 11:57 am

OK thanks everyone!

Mark
markb
 
Posts: 12
Joined: 27 December 2005


Return to Advanced solving techniques