The New Sudoku Players' Forum

[Noakesy97]

stuck on this puzzle and can't work out a way to solve it without guessing a number can anyone show me what to do next

Black numbers are the starting numbers
Blue are the ones I have filled in (I'm pretty certain they are right)

[JasonLion]

There is a naked pair in row 4 and a naked triple in block 4.

After that things get a little more complex. I would use a Finned X-Wing and a Sashimi X-Wing, though there are several other approaches.

Code: Select all

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

[daj95376]

JasonLion's X-Wings can be replaced by a finned Swordfish. (probably one of the alternatives he was referencing)

_

[pjb]

Code: Select all

 3456789 2       345678 | 1      59     589    | 35789  b5789   36789 
 35789   59      3578   | 4      6      589    | 135789 b125789 123789
*5689    1      *568    | 3      2      7      |a589     4      689   
------------------------+----------------------+----------------------
 28      59      1      | 27     3      59     | 6      78     4     
 268     4       268    | 27     59     1      | 5789    3      789   
 39-5    7       3-5    | 8      4      6      | 2      c159    19     
------------------------+----------------------+----------------------
 124     8       9      | 5      7      34     | 134     6      123   
*2457    6      *2457   | 9      1      34     | 3478    278    2378   
 147     3       47     | 6      8      2      | 1479    179    5     


This is getting more complicated, but I couldn't resist posting this cute one-stepper:
There is a finned (Kraken) X-wing of 5s at r3c137, r8c13 (fin r3c7)
If 5 at r3c7 is false, then X-wing of r38c13 => -5 r6c13
if 5 at r3c7 is true, then 5s at r12c8 are false and therefore 5 at r6c8 is true => -5 r6c13
Either way, r6c13 <> 5 which solves puzzle with singles.
Phil

PS One doesn't have to use fish, the following grouped chain of 5s does the same: (5)r4c2 = r2c2 - r3c13 = r3c7 - r12c8 = r6c8 => -5 r6c13; stte

[Leren]

Code: Select all

*--------------------------------------------------------------------------------*
| 3456789 2       345678   | 1       59      589      | 35789   5789    36789    |
| 35789  d59      3578     | 4       6       589      | 135789  125789  123789   |
|e5689    1      e568      | 3       2       7        |f589     4       689      |
|--------------------------+--------------------------+--------------------------|
| 28     c59      1        | 27      3      b59       | 6       78      4        |
| 268     4       268      | 27     a59      1        | 789-5   3       789      |
| 359     7       35       | 8       4       6        | 2       159     19       |
|--------------------------+--------------------------+--------------------------|
| 124     8       9        | 5       7       34       | 134     6       123      |
| 2457    6       2457     | 9       1       34       | 3478    278     2378     |
| 147     3       47       | 6       8       2        | 1479    179     5        |
*--------------------------------------------------------------------------------*

After the naked pair and triple mentioned by Jason you should be at this pencilmark situation.

Now follow the cells I've marked abcdef. First the standard notation (gobldegook) for this move:

(5=9) r5c5 - r4c6 = r4c2 - (9=5) r2c2 - r3c13 = (5) r3c7 => - 5 r5c7

Since you don't say what your level of experience is I'll explain this in words.

Suppose r5c5 (cell a) was not 5, then it would be 9. So r4c6 (b) would not be 9 and r4c2 (c) would be 9 (since there are only 2 9's in Row 4).

So r2c2 (d) would not be 9, so it would be 5. So r3c1 and r3c3 (cells marked e) would both not be 5 and r3c7 (f) would be 5 (since there are only 3 5's in Row 3).

You can reverse this process, start at cell f, assume it is not 5 and follow an analagous chain of reasoning using cells fedcba to deduce that r5c5 would be 5.

What this all means is that at least one of r5c5 and r3c7 must be 5. They might both be 5 but they can't both be not 5.

Since r5c7 can see both of these cells it can't be 5. Removing the 5 from r5c7 you'll find that the puzzle solves easily from there, since this shows that both r5c5 and r6c8 must both be 5, and you're away.

Leren