The New Sudoku Players' Forum

[GregM]

This puzzle appeared Saturday April 1 2006 in Knight/Ridder papers.

Code: Select all

2   6  135|9   57  157|37     4     8
4   59  15|3    8  157|2     79     6
7   39   8|6    4    2|1      5    39
-------------------------------------
8  257  25|4  2357 579|6     2379   1
3   27   9|1    6    8|5       27   4
6    1   4|25 2357 579|3789 23789 379
-------------------------------------
5    8   6|7    9    3|4       1    2
9   23   7|258  1    4|38      6   35
1    4  23|258  25   6|3789 3789 3579


This is as far as I can get it. When I run the pointing pair algorithm, I have a pair of 5's in [R4C2] and [R4C3] that eliminates the other 5's in [R4C5] and [R4C6]. X-wing, Y-wing don't reveal anything new.
Is there a guru who can offer me a suggestion for a possible next move?
Are there puzzles where a "guess" is required, or should all puzzles be solvable with logic?
Thanks,

[GregM]

Looking over the rules in "How to ask for help" I see that I have posted this in the wrong forum. I will format it per the instructions and repost in the correct forum.
GregM

[Ruud]

You may want to look here again to read this reply

A finned X-Wing is found for digit 3.

Code: Select all

.------------------.------------------.------------------.
| 2     6    *135  | 9     57    157  |*37    4     8    |
| 4     59    15   | 3     8     157  | 2     79    6    |
| 7     39    8    | 6     4     2    | 1     5     39   |
:------------------+------------------+------------------:
| 8     257   25   | 4     237   79   | 6     2379  1    |
| 3     27    9    | 1     6     8    | 5     27    4    |
| 6     1     4    | 25    2357  579  | 3789  23789 379  |
:------------------+------------------+------------------:
| 5     8     6    | 7     9     3    | 4     1     2    |
| 9     23    7    | 258   1     4    |-38    6     35   |
| 1     4    *23   | 258   25    6    |*3789 #3789 #3579 |
'------------------'------------------'------------------'

This eliminates 3 from R8C7, leaving a single 8.

A regular X-Wing for digit 3 eliminates 2 more candidates.

Code: Select all

.---------------.---------------.---------------.
| 2    6    135 | 9    57   157 | 37   4    8   |
| 4    59   15  | 3    8    157 | 2    79   6   |
| 7   *39   8   | 6    4    2   | 1    5   *39  |
:---------------+---------------+---------------:
| 8    257  25  | 4    37   79  | 6    2379 1   |
| 3    27   9   | 1    6    8   | 5    27   4   |
| 6    1    4   | 25   2357 579 | 379  8   -379 |
:---------------+---------------+---------------:
| 5    8    6   | 7    9    3   | 4    1    2   |
| 9   *23   7   | 25   1    4   | 8    6   *35  |
| 1    4    23  | 8    25   6   | 379  379 -3579|
'---------------'---------------'---------------'

There are 2 XY-Wings in a row, rooted in R3C9 and the next in R1C7.

The final blow is a sashimi swordfish for digit 7.

Code: Select all

.---------------.---------------.---------------.
| 2    6    135 | 9    57   157 | 37   4    8   |
| 4   *59   15  | 3    8   *157 | 2   *79   6   |
| 7    39   8   | 6    4    2   | 1    5    39  |
:---------------+---------------+---------------:
| 8   *257  25  | 4   #37  *9   | 6   *237  1   |
| 3   *27   9   | 1    6   *8   | 5   *27   4   |
| 6    1    4   | 25   2357-57  | 39   8    79  |
:---------------+---------------+---------------:
| 5    8    6   | 7    9    3   | 4    1    2   |
| 9    23   7   | 25   1    4   | 8    6    35  |
| 1    4    23  | 8    25   6   | 379  379  579 |
'---------------'---------------'---------------'

From this point on, singles is all you need.

Ruud.

[GregM]

Ruud,
Thanks for the rapid reply
It appears I need to take my game to the next level.
GregM

[ravel]

Just to give you alternatives for Ruud's eliminations:
Instead of the finned x-wing you can use the 2 strong links r8c2-r9c3 and r1c3-r1c7 to eliminate 3 from r8c7.
Instead of the sashimi swordfish you can use the following chain
r1c5=7, r12c6<>7, r6c6=7, r6c9=9, r3c9=3, r1c7=7 (contradiction) => r1c5=5
(r6c6<>7 follows)

[CathyW]

Instead of the finned x-wing you can use the 2 strong links r8c2-r9c3 and r1c3-r1c7 to eliminate 3 from r8c7.

Presume you actually mean r9c7! I made the same elimination using colouring (which amounts to the same thing).
I didn't keep track of all the steps but this puzzle is solvable without very advanced techniques - simple and multiple colouring and an xy-wing did the job for me.

Edit: Ravel was right! The elimination is at r8c7. See below.

[ravel]

Sorry, dont know what you mean, Cathy.
(In Ruuds first candidate list above)
I see 2 possibilities for the strong links to eliminate 3 in r8c7:
r8c2-r9c3 and r1c3-r1c7
r8c2-r3c2 and r1c3-r1c7
Also 3 in r9c9 could be eliminated by one of those 2 strong links (but the x-wing does it anyway):
r9c3-r1c3 and r1c7-r3c9
r9c3-r8c2 and r3c2-r3c9
All of them are simple coloring too (the 2 strong links are linked strongly also).

But i cannot see any coloring in Ruuds last diagram.
[Edit: ah, i mixed advanced and multiple coloring, the latter should do it and of course grouped coloring also]
A possible xy-chain is (using Speds notation):
3-r1c7-7-r1c5-5-r9c5-2-r8c4-5-r8c9-3, which eliminates 3 from r3c9 and r9c7.
(i.e. either r1c7=3 or, if r1c7=7, through the chain follows r8c9=3)

[GregM]

Referring to Ruud's first reply, where can I find information on the "finned X-wing"? What I see violates the X-wing rule, but the additional 3s are outside the rectangle formed by the X. Is that all there is to it?

[Ruud]

Finned fish are introduced here:

http://forum.enjoysudoku.com/viewtopic.php?t=2793

There is also a sticky topic with an index in the advanced solving techniques forum.

Ruud.

[CathyW]

Sorry, dont know what you mean, Cathy.

Sorry!! Getting my wires crossed and getting muddled up with my colours. I still didn't need a finned x-wing or any variety of fish to solve the puzzle.

Basic steps (naked & hidden singles & locked candidates) from the original puzzle take you to this point:

Code: Select all

 *-----------*
 |...|9..|..8|
 |...|3..|2.6|
 |7.8|6..|.5.|
 |---+---+---|
 |...|...|6.1|
 |3.9|1.8|5.4|
 |6.4|...|...|
 |---+---+---|
 |.8.|..3|4.2|
 |9.7|..4|...|
 |1..|..6|...|
 *-----------*
 *--------------------------------------------------------------------*
 | 2      6      135B   | 9      57     157    | 37A    4      8      |
 | 4      59     15     | 3      8      157    | 2      79     6      |
 | 7      39A    8      | 6      4      2      | 1      5      39     |
 |----------------------+----------------------+----------------------|
 | 8      257    25     | 4      237    79     | 6      2379   1      |
 | 3      27     9      | 1      6      8      | 5      27     4      |
 | 6      1      4      | 25     2357   579    | 3789   23789  379    |
 |----------------------+----------------------+----------------------|
 | 5      8      6      | 7      9      3      | 4      1      2      |
 | 9      23B    7      | 258    125    4      | 38-    6      35     |
 | 1      4      23A    | 258    25     6      | 3789   3789   3579   |
 *--------------------------------------------------------------------*


Next step is simple colouring on 3s to eliminate at r8c7 as you correctly stated Ravel, which gives you the 8s in box 6 and 8 as well. Having made that elimination you can also eliminate 3 at r6c9 and r9c9.

At this point you can use xy-wing to eliminate 7 at r6c7 (though I didn't spot it at this point previously!):

Code: Select all

 
 *-----------------------------------------------------------*
 | 2     6     135   | 9     57    157   | *37   4     8     |
 | 4     59    15    | 3     8     157   | 2     79    6     |
 | 7     39    8     | 6     4     2     | 1     5     *39   |
 |-------------------+-------------------+-------------------|
 | 8     257   25    | 4     37    79    | 6     2379  1     |
 | 3     27    9     | 1     6     8     | 5     27    4     |
 | 6     1     4     | 25    2357  579   | -379  8     *79   |
 |-------------------+-------------------+-------------------|
 | 5     8     6     | 7     9     3     | 4     1     2     |
 | 9     23    7     | 25    1     4     | 8     6     35    |
 | 1     4     23    | 8     25    6     | 379   379   579   |
 *-----------------------------------------------------------*


Then multiple colouring on 7s allows elimination of 7 at r6c6.
Another xy-wing - this time r1c7, r2c8 and r6c7 to eliminate 9 at r4c8 and it falls apart after that.

[Myth Jellies]

Here is a slightly more complex, but quicker path that cracks the puzzle...

Code: Select all

2   6  135|9  *57  157|*37     4     8
4   59  15|3    8  157| 2     79     6
7   39   8|6    4    2| 1      5    39
-------------------------------------
8  257  25|4  2357 579| 6     2379   1
3   27   9|1    6    8| 5       27   4
6    1   4|25 2357 579| 3789 23789 379
-------------------------------------
5    8   6|7    9    3| 4       1    2
9   23   7|258  1    4| 38      6   35
1    4 *23|258 *25   6| 3789 3789 3579


The starred cells form a [3=7]-[7=5]-[5=2]-[2=3] candidate chain. This means that the cells, r1c3 & r9c7, which see both ends of the chain cannot be a three. That leaves us with...

Code: Select all

2   6  *15|9   57 *157| 37     4     8
4   59 *15|3    8 *157| 2     79     6
7   39   8|6    4    2| 1      5    39
-------------------------------------
8  257  25|4  2357 579| 6     2379   1
3   27   9|1    6    8| 5       27   4
6    1   4|25 2357 579| 3789 23789 379
-------------------------------------
5    8   6|7    9    3| 4       1    2
9   23   7|258  1    4| 38      6   35
1    4  23|258  25   6| 789  3789 3579


The 15's in the starred cells are a deadly pattern which must be avoided to preserve a unique solution. Since the ones are locked in those four squares, the sevens must push out the fives (Uniqueness type 3). This leaves...

Code: Select all

2   6   15|9   57   17| 37     4     8
4   59  15|3    8   17| 2     79     6
7   39   8|6    4    2| 1      5    39
-------------------------------------
8  257  25|4  2357 579| 6     2379   1
3   27   9|1    6    8| 5       27   4
6    1   4|25 2357 579| 3789 23789 379
-------------------------------------
5    8   6|7    9    3| 4       1    2
9   23   7|258  1    4| 38      6   35
1    4  23|258  25   6| 789  3789 3579


...and the resulting naked pairs and locked candidates result in a fairly trivial end game.

[re'born]

Myth Jellies,

Your solution is quite elegant and elementary. Kudos! At the end, however, I'm not sure why you need to employ the unique rectangle. In fact, after removing the 3 from r1c3, it appears to me that the puzzle solves with singles.

[ravel]

The starred cells form a [3=7]-[7=5]-[5=2]-[2=3] candidate chain. This means that the cells, r1c3 & r9c7, which see both ends of the chain cannot be a three.

Just to say it, this is a classic xy-chain, in Speds notation:
3-r1c7-7-r1c5-5-r9c5-2-r9c3-3

[Myth Jellies]

...I'm not sure why you need to employ the unique rectangle. In fact, after removing the 3 from r1c3, it appears to me that the puzzle solves with singles.

Heh, removing the 3 in row one unlocked an obvious (to me) uniqueness rectangle, so I didn't even bother to check if the puzzle was already cracked . UR's, because they are often so easy to spot in candidate lists, tend to be part of my basic techniques arsenal.

[re'born]

Heh, removing the 3 in row one unlocked an obvious (to me) uniqueness rectangle, so I didn't even bother to check if the puzzle was already cracked . UR's, because they are often so easy to spot in candidate lists, tend to be part of my basic techniques arsenal.

Generally speaking, I do the same thing. Once we agree to overlook that the puzzle is already solved by your first move, then another nice point about your UR can be made. You mention that the 7's push out the 5's, but even more is true, the 7's push out the other 7's in box 2 and column 6 which, for instance, implies r1c5 = 5.