Solution to "Set reduction"

Advanced methods and approaches for solving Sudoku puzzles

Solution to "Set reduction"

Postby clivem » Sat Jul 05, 2008 9:10 pm

I am new to Sudoku. I normally try to solve one of the Savant Registered daily puzzles> often come against what they term "Set reduction" which is defined or described as:"The total number candidates in a group of cells exactly matches the number of cells containing them."

What I cannot figure out is which of the involved candidates should be crossed off.

Can anyone help, please. If so, use "click by click" basic language.

clivem.
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Re: Solution to "Set reduction"

Postby ronk » Sat Jul 05, 2008 10:41 pm

clivem wrote:"The total number candidates in a group of cells exactly matches the number of cells containing them."

What I cannot figure out is which of the involved candidates should be crossed off.

In most places that's known as naked pairs, naked triples and naked quads ... and naked n-tuples or naked sets, in general. Angus Johnson's step-by-step guide should help.
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Postby Pat » Sun Jul 06, 2008 7:57 am

clivem wrote:---often come against what they term "Set reduction"
which is defined or described as
The total number candidates in a group of cells
exactly matches the number of cells containing them

What I cannot figure out is,
which of the involved candidates should be crossed off

when the "group of cells" is within one unit (a row, a column, or a box),
this "group of cells" will use up those candidates for that entire unit
-- so, remove those candidates elsewhere in that unit
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Postby wintder » Sun Jul 06, 2008 1:12 pm

In column 1 there is a naked pair on 67.
We have 2 cells and two candidates. "That uses up" all the 6s and 7s in that column.
Thus the 7s are not possible in r7c1 and r8c1.

Code: Select all
.---------------------.
| 2      7      68    |
| 9      4      68    |
| 5      3      1     |
:---------------------+
|#67     8      5     |
|#67     1      279   |
| 3      29     4     |
:---------------------+
| 148-7  6      279   |
| 48-7   29     3     |
| 18     5      29    |
'---------------------'


In the middle box there is a naked pair on 67.
There can be no other 6 or 7 in that box, thus r5c3 cannot be 7.

Code: Select all
.---------------------.
| 2      7      68    |
| 9      4      68    |
| 5      3      1     |
:---------------------+
|#67     8      5     |
|#67     1      29-7  |
| 3      29     4     |
:---------------------+
| 148    6      279   |
| 48     29     3     |
| 18     5      29    |
'---------------------'


It doesn't matter which order you do them in or whether you do them all "at once".

You could have looked at the last elimination differently, still valid, but it shows a slightly different type of elimination.

In the middle box only cells r5c3 and r6c2 have the digits 29.
That is a hidden pair and since we have 2 cells and 2 candidates there can be no other candidates in those 2 cells and we eliminate the 7 from r5c3

Code: Select all
.---------------------.
| 2      7      68    |
| 9      4      68    |
| 5      3      1     |
:---------------------+
| 67     8      5     |
| 67     1     #29-7  |
| 3     #29     4     |
:---------------------+
| 148    6      279   |
| 48     29     3     |
| 18     5      29    |
'---------------------'


Edited to correct typos, thanks clivem!
Last edited by wintder on Sat Jul 12, 2008 11:02 am, edited 2 times in total.
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Postby clivem » Sat Jul 12, 2008 1:35 pm

Thank you ronk, thank you Pat and thank you wintder. It is very helpful to know that one can get the solutions to my problem by looking at the involed numbers as naked pairs etc., something that I can understand.

wintder rote "In the middle box there is a naked pair on 67.
There can be no other 6 or 7 in that box, thus r3c5 cannot be 7." I think it should read "r5c3' but not "r3c5" Or is it a deliberate mistake for a novice like I?!

Pat's suggestion will probably to a quicker solution but I will have to think it out carefully before I can understand it fully and absorb it.

Angus Johnson's step-by-step guide is fine. It is now among my Favourites for future references.

Thanks again,

clivem.
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re(2): "naked" subset

Postby Pat » Sun Jul 13, 2008 6:11 am

clivem wrote:Pat's suggestion will probably to a quicker solution
but I will have to think it out carefully before I can understand it fully and absorb it.


hi clivem,

i wasn't making a separate suggestion,
merely explaining the logic of the "naked" subset
    and i thought it would be good to give the proper definition of the "naked" subset
    i.e. add that the "group of cells" must be within one unit
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Postby clivem » Tue Jul 15, 2008 7:20 pm

Hi Pat,

Thanks for your further clarification
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