Can anyone help I am struggling to get head round naked doubles / triples in
the following puzzel
. . 2 | . . . | 1 . .
. . 1 | . . . | . 2 6
6 . 9 | 1 8 2 | . 7 .
-------+----------+------
. . 4 | . 7 . | 2 . 1
. . 6 | 2 . 1 | . 3 .
2 1 5 | . 4 . | . . .
-------+----------+------
1 . 7 | 3 . . | . . 5
. . 8 | . 1 . | . . .
. 6 3 | 7 . . | 9 1 .
Lost!
2 posts
• Page 1 of 1
by Karyobin » Mon Aug 08, 2005 11:29 pm
Right then, not sure about terminology but I'll give it a go...
Have a look at row 6 (2,1,5, etc...). The vacant cells in this row have the possible candidates {6,8,9} {3,6,8,9} {6,7,8} {6,8,9} {7,8,9}. There are two ways to approach this row from this point and they are:
1. Cells 4,7,8 and 9 can only contain the possible candidates {6,7,8,9}, which means they form a naked quad (I think!) So, discounting these four possibilities from the other cell means that it must be a 3.
2. Oh look! In row 6, only cell 6 can contain a 3, so cell 6 must be a 3!
Either way, you arrive at the same conclusion. Personally (and annoyingly) I find it much easier to look for pairs/triples/quads etc. than look for which cell in a row only contains one candidate. Perhaps it's just a matter of practice on my part.
Next bit:
If you look at box 9. you'll find that three cells contain the possible candidates {4,6 and 8 (not necessarily all three!)}, which again means that these candidates can be discounted from all the other cells in that box (three candidates in only three cells - that's the key - the same number of possibilities as cells). Which means it becomes pretty obvious where to place a '2' in box nine.
After this the whole puzzle pretty much 'opens up', as we clever chaps and chapesses like to say. Continue until complete. Or until tears prevent you seeing the screen anymore.
Sorry if any of this is unclear, the Stella-pixie is stamping on my bladder...
God, I love Tony Christie.
Have a look at row 6 (2,1,5, etc...). The vacant cells in this row have the possible candidates {6,8,9} {3,6,8,9} {6,7,8} {6,8,9} {7,8,9}. There are two ways to approach this row from this point and they are:
1. Cells 4,7,8 and 9 can only contain the possible candidates {6,7,8,9}, which means they form a naked quad (I think!) So, discounting these four possibilities from the other cell means that it must be a 3.
2. Oh look! In row 6, only cell 6 can contain a 3, so cell 6 must be a 3!
Either way, you arrive at the same conclusion. Personally (and annoyingly) I find it much easier to look for pairs/triples/quads etc. than look for which cell in a row only contains one candidate. Perhaps it's just a matter of practice on my part.
Next bit:
If you look at box 9. you'll find that three cells contain the possible candidates {4,6 and 8 (not necessarily all three!)}, which again means that these candidates can be discounted from all the other cells in that box (three candidates in only three cells - that's the key - the same number of possibilities as cells). Which means it becomes pretty obvious where to place a '2' in box nine.
After this the whole puzzle pretty much 'opens up', as we clever chaps and chapesses like to say. Continue until complete. Or until tears prevent you seeing the screen anymore.
Sorry if any of this is unclear, the Stella-pixie is stamping on my bladder...
God, I love Tony Christie.
- Karyobin
- Posts: 396
- Joined: 18 June 2005
2 posts
• Page 1 of 1
