The New Sudoku Players' Forum

[Bernard Stay]

The Times today:

*7*|864|*3*
*2*|***|*9*
**1|***|5**
–––––––––––
***|7*3|***
***|*4*|***
***|619|***
________
*14|*3*|72*
2**|*8*|**6
8*3|2*5|9*1

Easy to:

975|864|132
32*|157|*94
**1|392|5*7
__________
**9|723|**5
*3*|548|**9
*5*|619|**3
_________
514|936|728
297|481|356
863|275|941

From there there is one X-wing (6s) but what then?

Thanks.

[aran]

975|864|132
32*|157|*94
**1|392|5*7
__________
**9|723|**5
*3*|548|**9
*5*|619|**3
_________
514|936|728
297|481|356
863|275|941

From there there is one X-wing (6s) but what then?

Thanks.

Bernard
Simplest from there if you know and approve of uniqueness techniques :
after X-wing elims, only cell not a bivalue is r6c7 (248).
Had that cell only two candidates there would be at least two solutions to the puzzle (see later)
So the answer lies in that cell : and the quetion to ask is :
which candidate if false would reduce the puzzle to bi-values only ? ie a puzzle with two solutions. That candidate must therefore be true.
So which one is it ?
Rule of thumb : that candidate which occurs three times in the box : 8
ie 8 can be placed in r6c7.
Logic for that : a mental exercice : if 8 were false in r6c7, then each remaining candidate on the grid (1 2 4 6 7 8) would occur an even number of times. Assigning any candidate anywhere will...necessarily (well after thinking about it)...give rise to a solution=>impossible.

[Bernard Stay]

Thanks aran. I sort of follow that, though it's beyond the techniques normally required of us. For the more simple minded of us, now you draw attention to the tri-value cell is trial & error. 2 & 4 lead to contradictions, so it must be 8. Not really satisfying though.

[udosuk]

So much fuss over such an easy puzzle...

Method 1:

Code: Select all

975|864|132
32#|157|@94
..1|392|5@7
---+---+---
..9|723|..5
.3.|548|..9
.5#|619|.%3
---+---+---
514|936|728
297|481|356
863|275|941

8 of column 3 is locked in r2c3 & r6c3 (marked #)
8 of box 3 is locked in r2c7 & r3c8 (marked @)
Now r2c3 & r2c7 can't be both 8 (same row)
=> one or both of r6c3 & r3c8 must be 8
=> r6c8 (marked %), seeing r6c3 & r3c8, can't be 8
(This is called a turbot fish/skyscraper.)

Method 2:

Code: Select all

975|864|132
32.|157|.94
..1|392|5.7
---+---+---
.#9|723|%%5
.3.|548|..9
*5%|619|.@3
---+---+---
514|936|728
297|481|356
863|275|941

r4c2 (marked #) must be 4 or 8
r6c8 (marked @) must be 7 or 8
Now r6c1 (marked *) must be 4 or 7
=> one or both of r4c2 & r6c8 must be 8
=> r4c7+r4c8+r6c3 (marked %), seeing r4c2 & r6c8, can't be 8
(This is called an xy-wing.)

[aran]

Thanks aran. I sort of follow that, though it's beyond the techniques normally required of us. For the more simple minded of us, now you draw attention to the tri-value cell is trial & error. 2 & 4 lead to contradictions, so it must be 8. Not really satisfying though.

Or a nice bit of logic, depending on your point of view.
An alternative that you might find "legitimate" (after the Xwing elims) :
1. either r6c27 is a pair 28 : in which case : r6c8=7
or
2. r6c7=4 : in which case r4c7=8 and : r6c8=7.
But then you might regard that as testing both sides of a True/False and consider as "doubtful".

In that case just write it as a chain :
28r6c27=4r6c7-(4=8)r4c7 : <8>r6c7.

[Luke]

Bernard, click here for more info on BUG+1, the technique aran demonstrated. While there are many ways to go, the BUG+1 is instantly recognizable, and it is not trial and error. While you're figuring out the logic behind it, you can use the perfunctory approach (as aran mentioned):

Code: Select all

 *--------------------------------------------------*
 | 9    7    5    | 8    6    4    | 1    3    2    |
 | 3    2    68   | 1    5    7    | 68   9    4    |
 | 46   48   1    | 3    9    2    | 5    68   7    |
 |----------------+----------------+----------------|
 | 16   48   9    | 7    2    3    | 48   16   5    |
 | 17   3    26   | 5    4    8    | 26   17   9    |
 | 47   5    28   | 6    1    9    | 248  78   3    |
 |----------------+----------------+----------------|
 | 5    1    4    | 9    3    6    | 7    2    8    |
 | 2    9    7    | 4    8    1    | 3    5    6    |
 | 8    6    3    | 2    7    5    | 9    4    1    |
 *--------------------------------------------------*

Only one cell has three candidates, (248) r6c7.
Of the three candidates, find which one appears three times in the same row (or col or box) as the triple.
Place that candidate in r6c7, and the puzzle solves.

Here's the one from Sudopedia:

Code: Select all

 *--------------------------------------------------*
 | 3    2    19   | 59   579  6    | 4    8    17   |
 | 79   6    8    | 4    79   1    | 2    3    5    |
 | 17   4    5    | 8    2    3    | 6    9    17   |
 |----------------+----------------+----------------|
 | 8    5    4    | 2    6    7    | 9    1    3    |
 | 69   3    69   | 1    8    5    | 7    4    2    |
 | 2    1    7    | 39   39   4    | 5    6    8    |
 |----------------+----------------+----------------|
 | 4    7    16   | 36   13   2    | 8    5    9    |
 | 16   8    2    | 56   15   9    | 3    7    4    |
 | 5    9    3    | 7    4    8    | 1    2    6    |
 *--------------------------------------------------*

Same drill:
Only one cell has three candidates.
Of the three candidates, find which one appears three times in the same row (or col or box) as the triple.
Place that candidate in the cell that holds the triple.

Simple, and "satisfying," if immediately solving the puzzle brings any satisfaction. Also worth mentioning is that this is the first stepping stone to many other useful BUGgy techniques.

[Pat]

The Times today:

Code: Select all

 . 7 . | 8 6 4 | . 3 .
 . 2 . | . . . | . 9 .
 . . 1 | . . . | 5 . .
-------+-------+------
 . . . | 7 . 3 | . . .
 . . . | . 4 . | . . .
 . . . | 6 1 9 | . . .
-------+-------+------
 . 1 4 | . 3 . | 7 2 .
 2 . . | . 8 . | . . 6
 8 . 3 | 2 . 5 | 9 . 1



hi Bernard Stay,

thanks for posting the puzzle!

The Times has posted the article on www,
but without the puzzles

Su Doku battle goes a little off the wall in search for world champ