I don't understand this.

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I don't understand this.

Postby Gee » Mon Nov 10, 2008 3:07 am

I have solved the puzzle to this point but I could go no further. Even after 90 minutes, there was no place for me to turn. Finally, I ran the puzzle through the Scanraid Solver and next step shown was :
APE: Row Pair B4 / B6 reduced from 3/4/6/8->3/6/8
and 4/5/8->4/5/8
- PAIR combination 4/5 found in B1
- TRIPLE combinations 4/7/8 A5 + 7/8 C5

APE: Row Pair B5 / B6 reduced
from 4/6/7/8->6/7/8 and 4/5/8->4/5/8
- PAIR combination 4/5 found in B1
- PAIR combination 7/8 found in B2
- TRIPLE combinations 4/7/8 A5 + 7/8 C5


I don't have the slighest idea how to interpet what is shown above nor how the eliminations were derived and made. I know it is an APE, which I don't fully understand, but try as I might, I can not logically figure it out even after researching APE's on several web sites including Scanraid's explanation. This seems, to me, like a very difficult puzzle to say the least. I hope I can get some help from somebody to explain what I need to understand this. Simply, I hope.:) ANY help will be appreciated. DEFINITELY!!!

Code: Select all
230109060001000920600002041000730000700020008000096000857003204062000500040257080


Code: Select all
 
 *-----------*
 |23.|1.9|.6.|
 |..1|...|92.|
 |6..|..2|.41|
 |---+---+---|
 |...|73.|...|
 |7..|.2.|..8|
 |...|.96|...|
 |---+---+---|
 |857|..3|2.4|
 |.62|...|5..|
 |.4.|257|.8.|
 *-----------*

 
 *--------------------------------------------------------------------*
 | 2      3      458    | 1      478    9      | 78     6      57     |
 | 45     78     1      | 3468   4678   458    | 9      2      357    |
 | 6      789    589    | 358    78     2      | 38     4      1      |
 |----------------------+----------------------+----------------------|
 | 459    1289   45689  | 7      3      1458   | 146    159    269    |
 | 7      19     34569  | 45     2      145    | 1346   1359   8      |
 | 345    128    3458   | 458    9      6      | 1347   1357   237    |
 |----------------------+----------------------+----------------------|
 | 8      5      7      | 69     16     3      | 2      19     4      |
 | 139    6      2      | 489    148    48     | 5      379    379    |
 | 139    4      39     | 2      5      7      | 136    8      369    |
 *--------------------------------------------------------------------*
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Postby Glyn » Mon Nov 10, 2008 3:41 am

Gee Hope this helps I'll use row and column numbers rather than Andrew's chess notation.
If you look at r2c4 and r2c6 these are 8 possible combinations for the pair.
[34],[35],[38],[45],{48},[64],[65],[68] (NOTE {} is a reversible pair, [] is an ordered pair)
Now if we had [45] we would eliminate all the candidates for r2c1.
If we had {48} we get left with two 7's in Box 2. (In Andrews description forms the triple (478)r2c46+r3c5 killing all candidates at r1c5.)
So we are now left with [34][35][38][64][65][68].
Notice how there is only one combo with a 4 and that it must have a 6 with it which can only come from r2c4. => r2c4<>4
I can do the other one as well but perhaps this is enough to put you on track.

Well I'm on a roll so here's the other one in r2c56 possible pairs [45],{48},[64],[65],[68],[74],[75],[78],[85].
[45],{48} blocked as for the other APE also [78] blocked by r3c5 leaving
[64],[65],[68],[74],[75],[85]. Only combo with 4 needs a 6 or 7 from r2c5. => r2c5<>4.

Now the bad news: it still leaves loads to do.:(
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Postby Gee » Mon Nov 10, 2008 5:18 am

Wow, Thanks for the prompt reply. I will be printing it out and going over it as you suggested in your email. I will let you know later how I made out. Just want to let you know that I appreciate your reply.
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Postby Luke » Mon Nov 10, 2008 5:44 am

Click here to see last month's discussion on APEs with triples.
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Postby Glyn » Mon Nov 10, 2008 5:49 am

Thanks Luke451 I completely forgot we'd answered one of these recently. Looks like I'm down to 80 brain cells already.:(
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Re: I don't understand this.

Postby Luke » Mon Nov 10, 2008 10:18 pm

At the risk of being redundant..........
Code: Select all
 *--------------------------------------------------------------------*
 | 2      3      458    | 1      478    9      | 78     6      57     |
 | 45     78     1      |a3468   4678  b458    | 9      2      357    |
 | 6      789    589    | 358    78     2      | 38     4      1      |
 |----------------------+----------------------+----------------------|
 | 459    1289   45689  | 7      3      1458   | 146    159    269    |
 | 7      19     34569  | 45     2      145    | 1346   1359   8      |
 | 345    128    3458   | 458    9      6      | 1347   1357   237    |
 |----------------------+----------------------+----------------------|
 | 8      5      7      | 69     16     3      | 2      19     4      |
 | 139    6      2      | 489    148    48     | 5      379    379    |
 | 139    4      39     | 2      5      7      | 136    8      369    |
 *--------------------------------------------------------------------*
The candidates in the cells under consideration (a and b) contain 3468 and 458 respectively. If you write down the possibilities for the two cells you'd end up with this:
Code: Select all
a = 3468
b = 458

34  35  38
    45  48
64  65  68
84  85

"PAIR combination 4/5 found in B1" refers to [45] in r2c1. It can see both of the cells under consideration so [45] becomes an "aligned pair." This means you can eliminate [45] because you can't have 3 cells with only 2 candidates. Your possibilities are now these:
Code: Select all
a = 3468
b = 458

34  35  38
    X   48
64  65  68
84  85

Now, examine any "TRIPLE combinations" (2 cells w/a total of 3 candidates between them) that can see both cells under consideration. The cells in row 5 form a triple with [78] and [478]. The only possible combinations for these cells are [47],[48], and [78]. This means you can eliminate [48] (and of course [84]) because you can't have 4 cells with only 3 candidates. Your possibilities are now these:
Code: Select all
a = 3468
b = 458

34  35  38
    X   X
64  65  68
X   85

Now there remains no combination with [4] as the first candidate, so it can be excluded from r2c4. The "aligned pairs" have forced an "exclusion."
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Postby eleven » Mon Nov 10, 2008 11:20 pm

Interesting method, but a bit cumbersome.

In this case the w-wing 45 (r2c1, r5c4) with strong link for 5 in row 3 (or column 6) makes the same elimination.
Code: Select all
 *--------------------------------------------------------------------*
 | 2      3      458    | 1     *478    9      |*78     6      57     |
 |#45     78     1      | 3-468  4678   458    | 9      2      357    |
 | 6      789   @589    |@358   *78     2      |*38     4      1      |
 |----------------------+----------------------+----------------------|
 | 459    1289   45689  | 7      3      1458   | 146    159    269    |
 | 7      19     34569  |#45     2      145    | 1346   1359   8      |
 | 345    128    3458   | 458    9      6      | 1347   1357   237    |
 |----------------------+----------------------+----------------------|
 | 8      5      7      | 69     16     3      | 2      19     4      |
 | 139    6      2      | 489    148    48     | 5      379    379    |
 | 139    4      39     | 2      5      7      | 136    8      369    |
 *--------------------------------------------------------------------*

Also from the (diagonal) UR 78 in r13c57 (the 7 in r3c7 already missing) you can eliminate 8 from r1c5.

But then i need chains.
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Postby Gee » Tue Nov 11, 2008 10:41 am

In looking at Luke471 reply I have a question:

luke471 wrote:
Now, examine any "TRIPLE combinations" (2 cells w/a total of 3 candidates between them) that can see both cells under consideration. The cells in row 5 form a triple with [78] and [478]. The only possible combinations for these cells are [47],[48], and [78]. This means you can eliminate [48] (and of course [84]) because you can't have 4 cells with only 3 candidates. Your possibilities are now these:Code:
a = 3468
b = 458

34 35 38
X X
64 65 68
X 85

Now there remains no combination with [4] as the first candidate, so it can be excluded from r2c4. The "aligned pairs" have forced an "exclusion."


I follow exactly what Luke471 wrote but my question is this.
I can see there are no combinations beginning with a {4} but what about {34} or {64} ? Can't they be reversed to {43} and {46}? I thought so. Maybe this is the cause of my confusion. When working with "Simple Solver" and I have an aligned pair, it always show the pair as {37} and {37} ...never as {37} and {73}. Evidentlly the order of the pair makes a difference as Luke471 mentions. Also,I have never seen the combinations displayed in a hortizontal order as opposed to a vertical display. I like that. Seems easier.

Thanks to Glynn and Luke471 for your insights and help. I think I am on the right now and will grasp the concept with just a little more help.
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Postby DonM » Tue Nov 11, 2008 12:31 pm

Gee, interesting puzzle for more than one reason. Would be interested in knowing where it's from. Thanks.
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Postby Glyn » Tue Nov 11, 2008 2:28 pm

Gee You have no 3 or 6 in r2c6 so you can not get 43 or 46. Have a look at my post again where I use {} for a reversible pair [] for an ordered pair.
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Postby Gee » Tue Nov 11, 2008 6:03 pm

Don M
The puzzle is from last Sunday's Daily Telegraph weekly "Extreme" puzzle.

Glyn
I guess I didn't understand what reversible pair meant. I assume it is, as in this case, 48 could also mean 84 and an ordered pair is as 34 which has to be 34 and cannot also be interpreted as 43. In all my reading, I have never heard of a “reversible pair” or an “ordered pair”. Now, the question is what makes (48) reversible and (34) ordered? I assume this is of paramount importance in working with APE’s.

If I saw something before as 39 and then later 93, I assumed it was the same thing so I would eliminate one or the other from my list. Evidently I was wrong...very wrong! How do I know when I have an "ordered pair" and when do I know it is a "reversible pair"?????
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Postby DonM » Tue Nov 11, 2008 7:54 pm

Gee wrote:Don M
The puzzle is from last Sunday's Daily Telegraph weekly "Extreme" puzzle.


I thought it looked rather familiar. It's #112 which is this week's puzzle which I've already solved. Can't say much about the solution because the solution or much about it isn't supposed to be discussed until the deadline of Saturday midnight eastern U.S. time. However, fwiw: Although I know it's not really the point of the discussion, as it turns out, I had a short chain eliminating 4 from r2c4 and beside it had written 'not too useful' & as it turned out, it wasn't.:)
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Postby Glyn » Wed Nov 12, 2008 3:07 am

Gee By reversible {34} I mean that you could have r2c4=3 and r2c6=4 OR r2c4=4 and r2c6=3.
By ordered [34] I mean that you can only have r2c4=3 and r2c6=4.
Now is it possible to have r2c6=3? Look at the grid and decide whether {} or [] applies here and what that implies for r2c4.

As it's this weeks Extreme you'll have to wait for DonM or ttt to post their walkthroughs over in the Eureka Forum. Watch that space!
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Postby Gee » Wed Nov 12, 2008 6:25 am

If I start in (b4) with a (3) then go to (b6) I will get a (4) . If I start at (b6) with a (4) and go to (b4) I get a (3) ….so it appears that [34] is structured as ordered. Since (48) appears in both (b4) and (b6) it is {4/8} and reversible. Correct???

If we look at (b4) and (b5) (48) is also {48} reversible since those numbers are in both b4) and (b6) Correct?

If we looked at E7 and E8, we could say {13} is reversible but [35] would be ordered. Correct?

I hope I got that right. If so no wonder I was having problems.

I do sincerely apologize to the group and especially donM and ttt for my posting an “open” puzzle. I knew the deadline for posting was midnight the following Saturday. It never occurred to me that I should wait before posting a problem to an “open’ puzzle. It will never happen again and thank you for letting me know. I will try to locate the Eureka Forum and use it. These “extreme” puzzles do cause me problems. Some I solve….some I don’t. In this case I ran it through Scanraid and didn’t understand their explanation for the elimination. Again, I am sorry.
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Postby Glyn » Wed Nov 12, 2008 6:40 am

Gee The point I was trying to get home was that r2c6 cannot be 3 so there is only one way to have the pair 34 in r2c46. The notation I used is often applied to permutation analysis for Killer Sudoku so I borrowed it for the discussion of APE.

For Extreme help (after the cutoff date) it is probably still best to ask here unless it specifically relates to a walkthrough posted on Eureka.
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