The New Sudoku Players' Forum

[Jasper32]

A couple of months ago, I wrote about help with APE’s. Your replies has helped me find (2) APE’s since then. Now I came up with what I think is an APE but it is certainly complex in comparison to any APE I have ever seen before. I solved the puzzle below using this APE BUT I am not certain the APE I created is a valid one. That is why I would appreciate your help in either telling me I have indeed a valid APE or showing me the error of my ways. It is hard for me to believe that this is a valid APE but I checked it (3) times and it seems to meet the criteria.

The APE is in box (3) and the cells are shown by the letter (a) and (b). There is a reason for my showing them as (a) and (b), which will become apparent as we go along. All the other cells, (6) of them, are identified with the letter (c). If you look at the puzzle, you will see what I mean.

After going through the elimination’s, I came up with the fact that (1) could be eliminated from (a) and both (3) and (4) could be eliminated from (b). Now my actual question is whether my logic is correct or did I make lucky guesses.

Your help, will all be graciously received. Thank you.

Code: Select all

 
 
 
 *--------------------------------------------------------------------*
 | 14     6      8      | 249    3      29     | 5      7      124c    |
 | 1349c   159    1345   | 78     24c     78     | 1369a   169c    12349b  |
 | 2      479    347    | 6      1      5      | 39c     8      349c    |
 |----------------------+----------------------+----------------------|
 | 5      2      6      | 39     8      139    | 4      19     7      |
 | 8      3      14     | 5      47     79     | 19     2      6      |
 | 7      14     9      | 24     26     126    | 8      3      5      |
 |----------------------+----------------------+----------------------|
 | 36     57     2      | 1      9      4      | 367    56     8      |
 | 13469  14579  13457  | 28     267    268    | 13679  1569   139    |
 | 169    8      17     | 37     5      36     | 2      4      19     |
 *--------------------------------------------------------------------*



[udosuk]

Unfortunately, I don't think your deduction is correct APE, and it's not even logically sound (i.e. probably just lucky guesses).

Have a read at these links:

http://forum.enjoysudoku.com/viewtopic.php?t=3882
http://forum.enjoysudoku.com/viewtopic.php?t=4755
http://www.sudopedia.org/wiki/Aligned_Pair_Exclusion

To make the elimination r2c7<>1, you have to prove that all of the cases r2c79=[12|13|14|19] leads to easy-to-see contradictions, which unfortunately isn't true. Ditto for r2c9<>{34}.

Here I will show you several moves similar to APE (e.g UPE), which help to crack this puzzle:

Code: Select all

+----------------------+----------------------+----------------------+
| 14     6      8      | 249    3      29     | 5      7      124    |
| 1349   159    1345   | 78     24     78     | 1369   169    12349  |
| 2      479    347    | 6      1      5      | 39     8      349    |
+----------------------+----------------------+----------------------+
| 5      2      6      | 39     8      139    | 4      19     7      |
| 8      3      14     | 5      47     79     | 19     2      6      |
| 7      14     9      | 24     26     126    | 8      3      5      |
+----------------------+----------------------+----------------------+
|*36    #57     2      | 1      9      4      | 367   #56     8      |
| 13469  14579  13457  | 28     267    268    | 13679  1569   139    |
|#169    8     *17     | 37     5      36     | 2      4     #19     |
+----------------------+----------------------+----------------------+

Possible pairings of r7c1+r9c3: [31|37|61|67]
r7c28 can't be [55] => r7c1+r9c3 can't be [67]
r9c19 can't be [99] => r7c1+r9c3 can't be [61]
=> r7c1+r9c3 must be [31|37]
=> r7c1=3
=> 6 @ r7,n9 locked @ r7c78

Possible pairings of r1c6+r6c4: [22|24|92|94]
r5c56 can't be [77] => r1c6+r6c4 can't be [94]
=> r1c6+r6c4 must be [22|24|92], must have 2
=> r1c4+r6c6, seeing r1c6+r6c4, can't have 2

Possible pairings of r1c1+r9c3: [11|17|41|47]
r4c4 from {39} => r19c4 can't be [93]
=> r1c1+r9c3 can't be [47]
=> r1c1+r9c3 must be [11|17|41], must have 1
=> r2c3+r89c1, seeing r1c1+r9c3, can't have 1

Turbot fish (skyscraper):
1 @ r5 locked @ r5c37
1 @ r9 locked @ r9c39
r59c3 can't both be 1
=> at least one of r5c7+r9c9 must be 1
=> r8c7, seeing r5c7+r9c9, can't be 1

Possible pairings of r5c5+r8c7: [43|47|49|73|77|79]
r5c37 can't be [11] => r5c5+r8c7 can't be [49]
r3c7+r5c37 can't be [911] => r5c5+r8c7 can't be [43]
=> r5c5+r8c7 must be [47|73|77|79], must have 7
=> r8c5, seeing r5c5+r8c7, can't be 7

The rest is easy stuff, including a simple xy-wing @ r28c37.

[Luke]

Sorry, Jasper, no cigar.

Keep in mind the basic concept, obvious as it seems:
There cannot be two candidates for three cells.
There cannot be three candiates for four cells.
When you have identified such a situation, then you have an APE.

Look at one of your eliminations, r2c7 <> 1. What you're saying here is that you have identified an aligned pair for every possible combination of [ab] where a=1. Is that the case?

Code: Select all

 *--------------------------------------------------------------------*
 | 14     6      8      | 249    3      29     | 5      7      124c   |
 | 1349c  159    1345   | 78     24c    78     | 1369a  169c   2349b  |
 | 2      479    347    | 6      1      5      | 39c    8      349c   |
 |----------------------+----------------------+----------------------|
 | 5      2      6      | 39     8      139    | 4      19     7      |
 | 8      3      14     | 5      47     79     | 19     2      6      |
 | 7      14     9      | 24     26     126    | 8      3      5      |
 |----------------------+----------------------+----------------------|
 | 36     57     2      | 1      9      4      | 367    56     8      |
 | 13469  14579  13457  | 28     267    268    | 13679  1569   139    |
 | 169    8      17     | 37     5      36     | 2      4      19     |
 *--------------------------------------------------------------------*

If a=1, then the possibilities for [ab] are [12], [13], [14], and [19]. If there's an APE, all four of these pairs must exist outside [ab] in cells that can see [ab]. That's a pretty tall order. The "aligned pair" can either be a bivalue cell, or part of a triple (three candidates in two cells.)

In this case, not one of the four possibilities exists as an aligned pair. I suspect you may have looked at [24] and [124] as a triple, in which case I could see you taking out [12] and [14]. Unfortunately, [24] cannot "see" [124] and thus does not qualify as a triple. Even if it did there's still the issue of what to do with [13] and [19].

You did identify a legitimate triple in box 3, [39] and [349]. This triple can be used to eliminate [34],[39],[93] and [94] as possible configurations for [ab], but this alone is not enough to allow an exclusion.

In general, the greater the number of candidates in [ab], the more difficult it will be to find an APE. Also, when you're looking for [c] cells don't try to include any four-digit cells like r2c1. Look for just bivalue cells and triples.

udosuk's "UPE" (Unaligned Pair Exclusion?) twist is new to me. A very interesting and creative approach. The first two examples make sense to me as something I might be able to use right away. Examples 3 and 5 also make sense but definitely don't leap off the grid at me. I think spotting those might require some practice (or genius) I haven't cultivated.

[ronk]

Code: Select all

 *--------------------------------------------------------------------*
 | 14     6      8      | 249    3      29     | 5      7      124c   |
 | 1349c  159    1345   | 78     24c    78     | 1369a  169c   12349b |
 | 2      479    347    | 6      1      5      | 39c    8      349c   |
 |----------------------+----------------------+----------------------|
 | 5      2      6      | 39     8      139    | 4      19     7      |
 | 8      3      14     | 5      47     79     | 19     2      6      |
 | 7      14     9      | 24     26     126    | 8      3      5      |
 |----------------------+----------------------+----------------------|
 | 36     57     2      | 1      9      4      | 367    56     8      |
 | 13469  14579  13457  | 28     267    268    | 13679  1569   139    |
 | 169    8      17     | 37     5      36     | 2      4      19     |
 *--------------------------------------------------------------------*

Starting with the fills (clues and placements) you provided:

Code: Select all

.68.3.57..........2..615.8.526.8.4.783.5...267.9...835..2194..8..........8..5.24.

SSTS gets one to:

Code: Select all

 149   6     8     | 249   3     29    | 5     7     1249
 1349  1459  1345  | 24789 247   2789  | 1369  169   12349
 2     479   347   | 6     1     5     | 39    8     349
-------------------+-------------------+------------------
 5     2     6     | 39    8     139   | 4     19    7
 8     3     14    | 5     47    179   | 19    2     6
 7     14    9     | 24    246   126   | 8     3     5
-------------------+-------------------+------------------
 36    57    2     | 1     9     4     | 367   56    8
 13469 14579 13457 | 2378  267   23678 | 13679 1569  139
 1369  8     137   | 37    5     367   | 2     4     139
 

Do you recall how you made the 9s eliminations in the top band (band 1), the 3s elims in band 3, and r5c6<>1?

[Jasper32]

Many thanks to "udosuk" for your reply and telling me where I went wrong in my thinking. I have read the references you listed and little by little, I am beginning to grasp what should be a simple concept but has caused me problems. Luke451, thanks for the advice and your clarity made it clear what you were writing about. Ronk, I am sorry I can't answer your questions about how I made the eliminations you asked about. I "played" with this puzzle for a loooooong time before submitting my problem to the forum. I just can't remember.

To all of you many thanks for your help. I learn a lot from this forum and appreciate your unselfishness in helping me.

[udosuk]

Jasper32, I also appreciate your kind words, wish you a happy new year!

Luke451, all the APE/UPE moves I applied are in fact just different presentations of ALS-xz or ALS-xy-wing moves. I think experts such as ronk should have no trouble seeing how they work. In practice I think most chain-using players should be able to spot these eliminations quite easily but probably would not present their moves that way.