The New Sudoku Players' Forum

[sivodsi]

Hi there,
First post on the forum so hope I'm doing this right. The puzzle below is a 'fiendish' one scoped from the net (sorry don't have source, but I only choose them if they can be completed without guesswork).

I've gone as far as I can with it without guesswork and have exhausted the logical tricks at my disposal. To solve this kind of puzzle I need to get onto the next level with something new (unless I've overlooked something basic). Would really appreciate help with the logic necessary to solve this one.

8 7 4 | 2 _ 9 |5 _ _
_ _ _ | _ _ 7 | _ _ _
6 _ _ | _ 4 8 | _ _ 7

7 3 6 | 8 9 5 | _ _ 2
9 8 5 | 4 2 1 | 3 7 6
1 4 2 | 6 7 3 | 9 5 8

3 _ _ | _ 1 _ | _ _ 4
4 _ _ | _ _ _ | _ _ _
_ _ 9 | 7 _ 4 | 6 _ _


Let me know if there's any other information I can supply.
Cheers!

(edited for spelling)

[ravel]

Next step: box/box interaction (locked candidates 2): In row 1 the 1 must be in cols 8 or 9, so it cannot be in the rest of box 3. That gives you a single there.
It can be solved with pairs then.

Sites for basic techniques:
http://www.angusj.com/sudoku/hints.php
http://www.sadmansoftware.com/sudoku/techniques.htm
http://www.scanraid.com/BasicStrategies.htm

[QBasicMac]

Code: Select all

+---------------+---------------+---------------------+
| 8   7     4   | 2    36    9  | 5     136      13   |
| 25  1259  13  | 135  356   7  | 1248  1234689  139  |
| 6   1259  13  | 135  4     8  | 12    1239     7    |
+---------------+---------------+---------------------+
| 7   3     6   | 8    9     5  | 14    14       2    |
| 9   8     5   | 4    2     1  | 3     7        6    |
| 1   4     2   | 6    7     3  | 9     5        8    |
+---------------+---------------+---------------------+
| 3   256   78  | 59   1     26 | 278   289      4    |
| 4   1256  178 | 359  3568  26 | 1278  12389    1359 |
| 25  125   9   | 7    358   4  | 6     1238     135  |
+---------------+---------------+---------------------+


As ravel says, the only candidates 1 in row 1 are in columns 8 and 9 thus there can be no other 1's in box 3. That solves r3c7=2.

Now r2c9=r3c8=39, so all 3's and 9's can be erased from the rest of box 3. That solves r1c9=1 and thus also r1c8=6, r1c5=3

The only candidates 6 in column 6 are in rows 7 and 8. So 6 can be erased from r8c5 leaving another placement: r2c5=6, and r8c4=3 and finally, since r8c5=c9c5=58, r7c4 must be 9.

Actually, all singles from here on.

Mac

[sivodsi]

Thanks a lot for your help and explanations. Really appreciated.

What I dont get is that 3 and 9 are in r2c8 as well as r2c9 and r3c8, so how can 3 and 9 be only in those two cells but not r2c8? Why can you eliminate the '2' in r3c8? Surely it could still be there. What's the logical principle at work here?

Also, it seems that this level of Sudoku can only be worked out logically by marking up every vacant square; it gets really messy with paper and pen. I don't like doing Sudoku on a computer, so how can it be done neatly?

Cheers

[Crazy Girl]

Firstly all sudokus (with a unique solution) are solvable with logic, some basic logic some more advanced logic...

Lots of people solve sudoku's on paper without putting down the candidates for each cell, but when someone wants help on a puzzle in this forum if a candidate grid is shown, as below, then any helpful solver can see what one has done, and what one has not done, and help the person better.

Code: Select all

---+---------------+---------------+---------------------+
R1 | 8   7     4   | 2    36    9  | 5     136      13   |
R2 | 25  1259  13  | 135  356   7  | 1248  1234689  139  |
R3 | 6   1259  13  | 135  4     8  | 12    1239     7    |
---+---------------+---------------+---------------------+
R4 | 7   3     6   | 8    9     5  | 14    14       2    |
R5 | 9   8     5   | 4    2     1  | 3     7        6    |
R6 | 1   4     2   | 6    7     3  | 9     5        8    |
---+---------------+---------------+---------------------+
R7 | 3   256   78  | 59   1     26 | 278   289      4    |
R8 | 4   1256  178 | 359  3568  26 | 1278  12389    1359 |
R9 | 25  125   9   | 7    358   4  | 6     1238     135  |
---+---------------+---------------+---------------------+


Notice in Row 1 (R1 below) that the 1's are in the end box/block known as Block 3, so you can eliminate all 1's in Block 3 that are not in Row 1. so we get the situation below

Code: Select all

---+---------------+---------------+---------------------+
R1 | 8   7     4   | 2    36    9  | 5     136      13   |
R2 | 25  1259  13  | 135  356   7  | 248   234689   39   |
R3 | 6   1259  13  | 135  4     8  | *2    239      7    |
---+---------------+---------------+---------------------+
R4 | 7   3     6   | 8    9     5  | 14    14       2    |
R5 | 9   8     5   | 4    2     1  | 3     7        6    |
R6 | 1   4     2   | 6    7     3  | 9     5        8    |
---+---------------+---------------+---------------------+
R7 | 3   256   78  | 59   1     26 | 278   289      4    |
R8 | 4   1256  178 | 359  3568  26 | 1278  12389    1359 |
R9 | 25  125   9   | 7    358   4  | 6     1238     135  |
---+---------------+---------------+---------------------+


Now in Row 3 Column 7 there is a single value, known as a Naked Single, so one can eliminate all other 2's from Block 3, Row 3 and Column 7. After you do this there is two cells in Block 3 that only have candidates 3 and 9, so either R2C9=3 and R3C8=9 or vice versa, but no other candidate can occupy these two cells, so we can eliminate all other 3's and 9's in Block 3, This is known as a Naked Pair.

From this you should be able to solve the rest of the puzzle, also i recommend taking a look at the links in Ravel's post to explain some more basic techniques.

If you are still stuck, do not hesitate to post again

[ravel]

.. it gets really messy with paper and pen. I don't like doing Sudoku on a computer, so how can it be done neatly?

This is, how you could do it without filling out all candidates:
Lets start from your situation above:

Code: Select all

 8 7 4 | 2 . 9 | 5 . .
 . . . | . . 7 | . . .
 6 . . | . 4 8 | . . 7

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

 3 . . | . 1 . | . . 4
 4 . . | . . . | . . .
 . . 9 | 7 . 4 | 6 . .


Lets look for pairs:
Obviously are those in column 1 and 6.
Now hidden pairs are easier to spot.
Since 78 are in box 1, they have to be in rows 78 of column 3. Then 13 are left for r23c3.
48 must be in r2c78, because rows 1,3 and column 9 both have 48.

Then look, where only 3 cells are remaining:
In the first row only 136 are left, 36 for c5, 13 for c9.

Now lets count the remaining candidates, but stop, where more than 2 are left.

We get here (on paper i give small numbers to the candidates):

Code: Select all

 8  7  4  |  2  36 9  |  5  .  13
 25 .  13 |  .  .  7  |  48 48 .
 6  .  13 |  .  4  8  |  12 .  7
----------------------------------
 7  3  6  |  8  9  5  |  14 14 2
 9  8  5  |  4  2  1  |  3  7  6
 1  4  2  |  6  7  3  |  9  5  8
---------------------------------
 3  .  78 |  59 1  26 |  .  .  4
 4  .  78 |  .  .  26 |  .  .  .
 25 .  9  |  7  .  4  |  6  .  .


Now we can notice, what i said above. Since the 1 has to be either in column 8 or 9 of row 1, it cannot be in r3c7 (of the same box). So r3c7=2. Lets look, what it helps in the same row/col/box:
Having in mind, that the 1 in box 3 is reserved to row 1 (and the 48-pair), we can see, that only 39 remain for both r2c9 and r3c8 (and 78 remains in r7c6). This is the hardest step (but filling in all remaining candidates of the box would show it anyway).
The 39-pair gives us a 1 in r1c9, 6 in r1c8 (then r2c5) and 3 in r1c5 (then r8c4).
[Edit: An advanced solver would look for unique rectangles here: From the 13-pair in box 1/column 3 we know that not both 1 and 3 can be in r23 of column 4. Since the only free cell in the column is r8c4 and we already have a 1 in this box, r8c4 must be 3 - and r7c4=9]

Code: Select all

 8  7  4  |  2  3  9  |  5  6  1
 25 .  13 |  .  6  7  |  48 48 39
 6  .  13 |  .  4  8  |  2  39 7
----------------------------------
 7  3  6  |  8  9  5  |  14 14 2
 9  8  5  |  4  2  1  |  3  7  6
 1  4  2  |  6  7  3  |  9  5  8
---------------------------------
 3  .  78 |  59 1  26 |  78 .  4
 4  .  78 |  3  .  26 |  .  .  .
 25 .  9  |  7  .  4  |  6  .  .


With the 3 in r8c4 we get 9 in r7c4, 5 in r7c2, 2 in r7c8 and r9c1 and so on ...

[QBasicMac]

This is, how you could do it without filling out all candidates:

Nice!

Mac

[tarek]

after exhausting box-line interactions, the naked triple (139) in box 3 seems to be the only non single technique needed [although several lesser techniques are present]

tarek