The New Sudoku Players' Forum

[bennyd]

I solved this puzzle by guessing, but is there a way to solve it with logic? Note: my guess was that column 6 had to have a 3 and a 7; one way worked, the other didn't.

Also, is there a way to post puzzles here, and not have the columns get messed up in the posting?


9-- --- 82-
724 168 935
--8 9-2 74-

-46 8-9 -57
89- --- -62
-7- 6-1 498

587 396 214
632 714 589
419 285 673

Thanks for any and all hints.

[angusj]

is there a way to solve it with logic?

Naked Pair
Naked Triple
Multiple Colors
XY-Wing
then straightforward after that.

[bennyd]

NOTE (on edit):

Oops. I misread the definition of naked pair listed below, they were not saying that (6, 8) and (1,6,8) are naked pairs. Sorry 'bout that.
-----------------------------------------------------------
I don't think I understand naked pairs. I do understand that if 2 cells in a group contain only two numbers in each cell and the numbers are the same, then no other cells in that group can use those two numbers like (6, 8) and (6, 8). But, in an example I saw somewhere, the naked pair example is (6, 8) (1,6,8). Why doesn't that 1 mess up things? That is, a 6 could be in the first cell, a 1 in the 2nd cell and the 8 could be elsewhere. ????
-----------------------------------------------------------

[QBasicMac]

I don't think I understand naked pairs.

I hear you, man

The only naked pairs I can spot are
56 in box 1
16 in box 3
13 in box 6
45 in col 4
37 in col 6

But I can't see that any of those lead to the elimination of any candidate.

I found a locked candidate 5 in box 4 that eliminated 5 from r1c3
Then a hidden pair 47 in col 5 that eliminated 35 from those two cells.

But then you are stuck here

Code: Select all

+--------------+-------------+-----------+
| 9    56  13  | 45  47   37 | 8   2  16 |
| 7    2   4   | 1   6    8  | 9   3  5  |
| 13   56  8   | 9   35   2  | 7   4  16 |
+--------------+-------------+-----------+
| 123  4   6   | 8   23   9  | 13  5  7  |
| 8    9   135 | 45  47   37 | 13  6  2  |
| 23   7   35  | 6   235  1  | 4   9  8  |
+--------------+-------------+-----------+
| 5    8   7   | 3   9    6  | 2   1  4  |
| 6    3   2   | 7   1    4  | 5   8  9  |
| 4    1   9   | 2   8    5  | 6   7  3  |
+--------------+-------------+-----------+


So you must apply BAT (Boring Arcane Techniques) or T&E (Trial and Error) or get a book which has easier puzzles.

Regarding posting. Here is exactly what I posted above
{code}
+--------------+-------------+-----------+
| 9 56 13 | 45 47 37 | 8 2 16 |
| 7 2 4 | 1 6 8 | 9 3 5 |
| 13 56 8 | 9 35 2 | 7 4 16 |
+--------------+-------------+-----------+
| 123 4 6 | 8 23 9 | 13 5 7 |
| 8 9 135 | 45 47 37 | 13 6 2 |
| 23 7 35 | 6 235 1 | 4 9 8 |
+--------------+-------------+-----------+
| 5 8 7 | 3 9 6 | 2 1 4 |
| 6 3 2 | 7 1 4 | 5 8 9 |
| 4 1 9 | 2 8 5 | 6 7 3 |
+--------------+-------------+-----------+
{/code}

EXCEPT I used square brackets instead of curly brackes. Putting "code" and "/code" around a section of post will conserve spacing.

Mac

[bennyd]

QBasicMac:

I got exactly the same thing you did, except that in R1C5 I had 3457? How did you eliminate the 3 and the 5?

So does naked triples enter into this somehow?

[sweetbix]

So does naked triples enter into this somehow?

Yes in column 5 = 35, 23, 235

This is the same puzzle we just looked at another thread. Click here http://forum.enjoysudoku.com/viewtopic.php?p=16934#p16934 for hints.

It's quite a hard puzzle and my solution needed chains and colours but don't be put off by QBasic Mac because I didn't know what these methods were a few weeks ago and I don't think they're that hard to understand.

[bennyd]

Thanks, I'll look at that other thread. I'm quite new to this too, but quite stubborn. So there is hope yet!

[sweetbix]

Stubborn is good!

angusj, your reply included multiple colours for this puzzle. I understand simple colours but could you explain how you used multiples. Thanks.

[angusj]

angusj, your reply included multiple colours for this puzzle. I understand simple colours but could you explain how you used multiples. Thanks.

http://www.setbb.com/phpbb/viewtopic.php?p=2575&mforum=sudoku#2575
http://www.setbb.com/phpbb/viewtopic.php?p=1412&mforum=sudoku#1412

[Jeff]

This grid can be easily solved using BUG from here.

Code: Select all

 *--------------------------------------------------*
 | 9    56   13   | 45   47   37   | 8    2    16   |
 | 7    2    4    | 1    6    8    | 9    3    5    |
 | 13   56   8    | 9    35   2    | 7    4    16   |
 |----------------+----------------+----------------|
 | 12+3 4    6    | 8    23   9    | 13   5    7    |
 | 8    9    15+3 | 45   47   37   | 13   6    2    |
 | 23   7    35   | 6    25+3 1    | 4    9    8    |
 |----------------+----------------+----------------|
 | 5    8    7    | 3    9    6    | 2    1    4    |
 | 6    3    2    | 7    1    4    | 5    8    9    |
 | 4    1    9    | 2    8    5    | 6    7    3    |
 *--------------------------------------------------*

The BUG principle suggests that at least one of the non-BUG candidates is true.

Since r6c1=3 will eliminate all non-BUG candidates, therefore r6c1<>3 => r6c1=2.
Since r6c3=3 will eliminate all non-BUG candidates, therefore r6c3<>3 => r6c3=5.

[QBasicMac]

QBasicMac: I got exactly the same thing you did, except that in R1C5 I had 3457? How did you eliminate the 3 and the 5?

Then a hidden pair 47 in col 5 that eliminated 35 from those two cells.

To explain further:
In column 5, there are only two occurrences of 4.
In addition, there are only two occurrences of 7.
These occur in the same two cells: r1c5 and r5c5.
Therefore those two cells are restricted to containing 4 and 7.
That is called a "hidden pair".

So I erased 35 from r1c5, since it can contain only 47.

Now I'm curious how you eliminated 35 from r5c5 without simultaneously clearing r1c5.

So does naked triples enter into this somehow?

Sorry, I got lazy and didn't look for triples. Other replies answer your question.

Good luck

Mac

[ihope127]

Why doesn't that 1 mess up things? That is, a 6 could be in the first cell, a 1 in the 2nd cell and the 8 could be elsewhere. ????

If you know that the 8 can't be elsewhere, and the 6 can't either, then they both have to be in the pair, meaning the 1 can't be.

(By the way, sticking an empty pair of bold tags inside a tag will prevent that tag from parsing: [code] can be written as [co[b][/b]de].)

[QBasicMac]

(By the way, sticking an empty pair of bold tags inside a tag will prevent that tag from parsing: [code] can be written as [co[b][/b]de].)

[code]
This is a test
[/code]

Good trick for answering the reappearing question of how to post code.

Thanks!

Mac

[sweetbix]

Thanks for the colouring links angusj. I understand the theory but I guess the hard part is putting it into practice.

[9X9]

Jeff

Although I'm on a sabbatical, I risk losing the BUG plot if I don't pitch in now. Although I can grasp part of it, the long thread developing and elucidating BUG, contributed to by genuine experts like yourself, is mainly just so much "number soup" for me.

Any chance of a succint and crystal clear "dummies" version?