## What can I work out from here... ?

Everything about Sudoku that doesn't fit in one of the other sections

### What can I work out from here... ?

Hi all,

I've got a puzzle here where one of the blocks looks like this:

Code: Select all
`  5            4    {7, 8}  {3, 6, 8}    2    {6, 8}  {3, 7, 8}    1    9`

Where {x, y .. } indicates that the cell must contain one of x, y, ...

(I didn't see a standard notation, so I've come up with this in the hope that it's not too confusing).

Let us refer to:
the {7, 8} choice as A
the {3, 6, 8} choice as B
the {6, 8} choice as C
the {3, 7, 8} choice as D

Now, given an analysis of (what I understand of) naked pairs, we get:
- A and D are 7 and 8
- B and C are 6 and 8
- one of B and D MUST be 3.

If the first statement holds true, then the 3 choice can be removed from B. This means that B must hold 3, which means that C holds 6, which leaves A and D. However, the same logic holds for the second statement, with A swapped for C and B swapped for D.

The question I have is - have I missed anything about this situation? Is there anything else I can work out with the information given? It feels like I should...
rik

Posts: 3
Joined: 14 June 2006

### while we're at it...

I have a boxrow that reads:

Code: Select all
`{1, 2}  {1, 3}  {1, 2, 3}`

Where {1, 2} indicates that only 1 and 2 are the valid choices for this cell.

Can I use the hidden/naked pairs rules on this to say that the boxrow should be 2, 1 3, in this case?
rik

Posts: 3
Joined: 14 June 2006

### Re: while we're at it...

rik wrote:I have a boxrow that reads:

Code: Select all
`{1, 2}  {1, 3}  {1, 2, 3}`

(...)
Can I use the hidden/naked pairs rules on this to say that the boxrow should be 2, 1 3, in this case?

As it stands, that is a naked triple, i.e, 3 candidates for 3 cells. If it were reducible, there would be either a naked single (with a complementary hidden pair) ... or a hidden single (with a complementary naked pair). If either a naked or hidden single existed ...

digit 1 would exist in only one cell, or
digit 2 would exist in only one cell, or
digit 3 would exist in only one cell.

Since neither is true, your naked triple is irreducible. By a similar though more complex argument, it can be shown that your first example is not reducible either.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ahh. i understand.

thank you.
rik

Posts: 3
Joined: 14 June 2006

rik,

Code: Select all
`5            4    {7, 8} {3, 6, 8}    2    {7, 8} {3, 7, 8}    1    9`

Then, {7, 8} are indeed a naked pair. You can reduce this to
Code: Select all
`5         4    {7, 8} {3, 6}    2    {7, 8} {3}       1    9`

etc.

Keith
keith
2017 Supporter

Posts: 216
Joined: 03 April 2006