The New Sudoku Players' Forum

[samer_sadek]

Here is the source
http://www.dailysudoku.com/sudoku/archive/2006/02/2006-02-12.shtml

This is how far I reached:

9-45-56-14-7-8-135-2-356
25-1-2578-3-6-29-4-578-589
3-247-2678-14-5-29-1789-678-689
245-8-257-6-3-1-2579-57-2459
24-2379-1-29-8-5-6-37-234
6-2359-235-29-4-7-2358-1-2358
1-235-235-8-9-46-235-46-7
8-235-4-7-1-36-235-9-2356
7-6-9-5-2-34-38-348-1

I'm not an expert, just started getting addicted, so this might be simple for all you pros...

[TKiel]

Samer_sadek,

You've got a naked pair in column 4 (two numbers confined to only two cells). Those two numbers can be excluded from all other cells in that same column.

As you are just getting started, I suggest you visit [url]angusj.com[/url] for an explanation of some techniques or the 'how to solve' link in this forum . I see some things in your puzzle that are pretty basic but I don't know if you don't know about them or you merely missed them in this puzzle, so I don't know if you just want hints or if you want explanations.

edit- My apologies for pointing out what you had already done. I didn't check the candidate list closely enough to realize that you had already made those exclusions. I believe my suggested reading assignment is pretty unnecessary also.

Tracy

[samer_sadek]

tracy, there are 2 naked apirs 1and4 and 2and9. I have already excluded all these from the columns and the square.
Am I missing something?

[vidarino]

Code: Select all

      9     45     56 |      14      7      8 |     135      2    356
     25      1   2578 |       3      6     29 |       4    578    589
      3    247   2678 |      14      5     29 |    1789    678    689
----------------------+-----------------------+----------------------
    245      8    257 |       6      3      1 |    2579     57   2459
     24   2379      1 |      29      8      5 |       6     37    234
      6   2359    235 |      29      4      7 |    2358      1   2358
----------------------+-----------------------+----------------------
      1    235    235 |       8      9     46 |     235     46      7
      8    235      4 |       7      1     36 |     235      9   2356
      7      6      9 |       5      2     34 |      38    348      1


There is a naked quad (four candidates spread over four cells) in column 7.

[TKiel]

Samer_sadek,

You could also look for an x-wing in 5.

Tracy

[samer_sadek]

I got the quad in 7, thanks vidarino, good eyes
Tracy, what's the x-wing in 5?

Thanks

[TKiel]

Samer_sadek,

An x-wing is two columns, each of which have only two cells that contain a certain candidate and those cells must align in the same two rows. Those four corners form a rectangle. Connect the opposite corners of the rectangle with a line, like an X. The cells on the end of each line will both either be true or false. The cells in each column are conjugate links, which means if one is true the other is false and, more importantly, if one is false the other is true. Those four corners are the only cells in the columns where the value can be placed. That means that all cells in either of the two columns that contain the value can have it excluded.

Code: Select all

 *--------------------------------------------------------------*
 | 9      45     56     | 14     7      8  | 135    2      356  |
 | 25X    1     -2578   | 3      6      29 | 4      578X  -589  |
 | 3      247    2678   | 14     5      29 | 1789   678    689  |
 |----------------------+----------------------+----------------|
 | 245X   8     -257    | 6      3      1  |-2579   57X   -2459 |
 | 24     2379   1      | 29     8      5  | 6      37     2349 |
 | 6      2359   235    | 29     4      7  | 23589  1      23589|
 |----------------------+--------------------+------------------|
 | 1      235    235    | 8      9      46 | 235    46     7    |
 | 8      235    4      | 7      1      36 | 235    9      2356 |
 | 7      6      9      | 5      2      34 | 38     348    1    |
 *--------------------------------------------------------------*


The four corners of the x-wing are marked with an X. The cells marked with - can have 5 excluded. If 5 is placed in any cell other than the four marked by the x-wing, one of the columns will end up with no 5. In other words, the 5 must go in one of those four cells and can go no where else in the row, therefore it can be excluded from all other cells in the row. If this isn't clear, visit the site mentioned in my post above for a much better explanation.

Tracy

[samer_sadek]

Got it, It is very clear and makes perfect sense.
thanks TKiel.

[TKiel]

Samer_sadek,

The concept also entends to three rows, three cells, three columns (called a swordfish), four of each (called a jellyfish) and 5 of each (called a starfish or a squirmbag).

Tracy

[samer_sadek]

I know, or at least so I heard.
I read about the swordFish pattern, and I understand it when I see an example in the explanation, but I was never able to see it while solving a puzzle.

BTW I found this particular puzzle being discussed in another forum with another way of solving it.
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=452

[TKiel]

Samer_sadek,

Not sure to which of the techniques mentioned in the other forum you are referring. The thing about the 9's (by geoff h) is called either locked candidates or row-box interaction. Basically the only place for a 9 in row 4 is in box 6. If any cell in box other than ones in that row are assigned 9, then row 4 will have no 9. Therefore no cells in the box except those in row 4 can have 9 as a candidate. The same eliminations can be made with an x-wing in column 2 & 4, row 5 & 6. If you're referring to the post by David Bryant, the technique is a variety of colouring, in which conjugate links are labeled to indicate opposite states of being. A conjugate link is one in which if one is true the other is false and if one is false the other is true. This one involves two seperate conjugate chains linked on candidate 7.

Code: Select all

 
 *-------------------------------------------------------*
 | 9     45    56    | 14    7     8  | 135   2     356 |
 | 25    1     278b  | 3     6     29 | 4     578B  89  |
 | 3     247   2678  | 14    5     29 | 1789  678   689 |
 |-------------------+-------------------+--------------|
 | 245   8     27A   | 6     3     1  | 279   57    249 |
 | 24    2379a 1     | 29    8     5  | 6     37A   234 |
 | 6     2359  235   | 29    4     7  | 2358  1     2358|
 |-------------------+-------------------+--------------|
 | 1     235   235   | 8     9     346| 235   346   7   |
 | 8     235   4     | 7     1     36 | 235   9     2356|
 | 7     6     9     | 5     2     34 | 38    348   1   |
 *------------------------------------------------------*
 

I've labeled one chain with A-a and the other with B-b. In column 8, B & A share a group. In column 3, b & A share a group. If A is true, then neither B nor b can be true, which means row 2 will not have a 7. Therefore A is false and a must be true.

If you are referring to the 'Glassmans pan' referred to by dotdot, I have no idea what that means.

Tracy

[tarek]

Hi there samer,

there is also a hidden double in column 7:

Code: Select all

*--------------------------------------------------------*
| 9     45    56   | 14    7     8    | 135   2     356  |
| 25    1     2578 | 3     6     29   | 4     578   589  |
| 3     247   2678 | 14    5     29   |*1789  678   689  |
|------------------+------------------+------------------|
| 245   8     257  | 6     3     1    |*2579  57    2459 |
| 24    2379  1    | 29    8     5    | 6     37    234  |
| 6     2359  235  | 29    4     7    | 2358  1     2358 |
|------------------+------------------+------------------|
| 1     235   235  | 8     9     46   | 235   46    7    |
| 8     235   4    | 7     1     36   | 235   9     2356 |
| 7     6     9    | 5     2     34   | 38    348   1    |
*--------------------------------------------------------*
r3c7 Must only have 79 as valid Candidates (79 is a Hidden Double in Column 7)
r4c7 Must only have 79 as valid Candidates (79 is a Hidden Double in Column 7)

which solves the puzzle as it allows for the 1 in row 7 to be a hidden single.

Tarek

[samer_sadek]

tarek, again, good eyes, thanks. I do not know how you guys see these things.
Tracy, I'm very interested in what you said, but I did not understand it well. I understand the concept of conjugate links, but you said
'If A is true, then neither B nor b can be true'. I do not get that part. Why can't A be true and b be true? The way I understand it, A and b can be true, or a and B can be true
A second question: I see now the 9 x-wing in column 2 & 4, row 5 & 6, but I do not know how it will help me. there are already no 9s in these 2 rows or columns. In other words, now that I see it, what can I eleminate?

Thanks

[Animator]

'If A is true, then neither B nor b can be true'.

If A is true then r4c3 and r5c8 are the number 7.

Now, in row 2 there are only two cells that can contain the number 7: r2c3 (b) and r2c8 (B).

If A is true then there is already a 7 in column 3 and column 8. In that case there is no way that you can place a 7 on row 2.

[samer_sadek]

got it
when you explain it it seems so obvious