## APE Problem

Post the puzzle or solving technique that's causing you trouble and someone will help

### APE Problem

I am having problems with the puzzle shown below. I got to this point in the puzzle and the Scanraid Solver showed the next step to be an APE. The Solver’s description of how it arrived at this conclusion leaves me completely baffled.

Scanraid write:
APE: Row pair D8 / D9 reduced from 2/3/9-> 2/3/9 and
2/3/4/5/9->3/4/5/9
--Pair combination 2/9 found in F8
--Triple Combinations 3/8 D2 + 2/8 D5
My comment: Why isn’t the 4/5 in D6 mentioned?

Rod Hagglund wrote:
“Any two cells aligned on a row or column within the same box cannot duplicate the contents of any two-candidate cell they both see.”

In the puzzle below, the two aligned cells are marked (a) and (b). The two-candidate cells are marked (c).

I show the following possibilities as possible for the two aligned cells after reduction.

2-3, 2-4, 2-5, 2-9, 3-4, 3-5, 3-9, 4-9.\

Now with that being written, I hope someone can give me a simple explanation as to what qualifies the (2) in (b) to be eliminated. What is the reconciliation between what Scanraid says and what Hagglund says? I can't make any sense out of Scanraid or Hagglund on this particular puzzle.

Code: Select all
` *--------------------------------------------------------------------* | 3      9      57     | 4      28     258    | 6      178    18     | | 1      4      57     | 358    36     568    | 2      789    89     | | 8      26     26     | 9      7      1      | 34     5      34     | |----------------------+----------------------+----------------------| | 2479   38c    1      | 6      28c     45     | 34579  239a    23459b  | | 247    5      38     | 278    9      248    | 1347   6      1234   | | 24679  267    246    | 57     1      3      | 8      29c     2459   | |----------------------+----------------------+----------------------| | 24     1      2348   | 238    5      7      | 39     2389   6      | | 257    2378   9      | 1238   36     268    | 135    4      12358  | | 256    2368   2368   | 1238   4      9      | 135    1238   7      | *--------------------------------------------------------------------*`
Jasper32

Posts: 60
Joined: 04 January 2008

I don't know APE, so I went to Sudopedia. Consider these cell "b+a" combinations:

Code: Select all
`2+2                  => invalid!2+3 => D2=8 and D5=8 => invalid!2+9 => F8=empty      => invalid!`

All combinations containing a 2 in cell b are invalid. The candidate 2 can be removed from cell b. D6 does not figure into the solution!
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

Jasper32 Another way of looking at it based on Scanraid's solving notes is:

D8 and D9 can together contain {23},{24},{25},{29},{34},{35},{39},{49},{59}
{23} is blocked by D2 and D5 combined.
{29} is blocked by F8.
Leaves {24},{25},{34},{35},{39},{49},{59}.
The only remaining options for digit 2 require 4 or 5 which can only be found in D9. Therefore D9 cannot be the 2.
Glyn

Posts: 357
Joined: 26 April 2007

### Re: APE Problem

Jasper32 wrote:I hope someone can give me a simple explanation as to what qualifies the (2) in (b) to be eliminated.
Code: Select all
` *--------------------------------------------------------------------* | 3      9      57     | 4      28     258    | 6      178    18     | | 1      4      57     | 358    36     568    | 2      789    89     | | 8      26     26     | 9      7      1      | 34     5      34     | |----------------------+----------------------+----------------------| | 2479   38c    1      | 6      28c     45    | 34579  239a   23459b | | 247    5      38     | 278    9      248    | 1347   6      1234   | | 24679  267    246    | 57     1      3      | 8      29c    2459   | |----------------------+----------------------+----------------------| | 24     1      2348   | 238    5      7      | 39     2389   6      | | 257    2378   9      | 1238   36     268    | 135    4      12358  | | 256    2368   2368   | 1238   4      9      | 135    1238   7      | *--------------------------------------------------------------------*`

Here's a graphic way to reveal any aligned pair exclusion, but esp ones with triples. It's easier than it reads!

The candidates in the cells under consideration (a and b) contain 239 and 23459 respectively. If you write down the possibilities for the two cells you'd end up with this:
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`a = 239b = 2345923  32  9224  34  9325  35  9429  39  95`

One of the cells you've designated "c" contains <29> and it can see both of the cells under consideration (<29> becomes an "aligned pair.") This means you can eliminate <29> (and <92>) because you can't have 3 cells with only 2 candidates. Your possibilities are now these:
Code: Select all
`a = 239b = 2345923  32  X24  34  9325  35  94X   39  95`

Now, examine any triple (2 cells w/a total of 3 candidates between them) that can see both cells under consideration. The cells in row 4 designated "c" form a triple with <38> and <28>. The only possible combinations for these cells are <32>, <38> and <82>. This means you can eliminate <32> (and of course <23>) because you can't have 4 cells with only 3 candidates. Your possibilities are now these:
Code: Select all
`a = 239b = 23459X   X   X24  34  9325  35  94X   39  95`

Now there remains no combination with <2> as the second candidate, so it can be excluded from r4c9. The "aligned pairs" have forced an "exclusion."

I like this process for APE's because it's immediately obvious when an elimination becomes available. Go APE!

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

### APE

Thanks to all of you who replied to my post. It was sure helpful and appreciated.
Jasper32

Posts: 60
Joined: 04 January 2008