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JV
Joined: 09 Jan 2011 Posts: 24 Location: Devon, England
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Posted: Sat Feb 26, 2011 3:08 pm Post subject: Magic bullet |
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Menneske 2170801
Code: |
+------------------+----------------------+----------------+
| 8 345 1 | 235 2345 9 | 6 235 7 |
| 9 345 2 | 3567 34567 356 | 1 35 8 |
| 356 7 36 | 8 1235 1235 | 234 9 245 |
+------------------+----------------------+----------------+
| 2367 1 5 | 236 236 4 | 23789 78 269 |
| 2346 239 3469 | 12356 8 7 | 23 235 1256 |
| 2367 8 367 | 9 12356 12356 | 237 4 1256 |
+------------------+----------------------+----------------+
| 12457 259 8 | 1257 12579 125 | 2479 6 3 |
| 123457 2359 3479 | 123567 1235679 12356 | 24789 78 249 |
| 237 6 379 | 4 2379 8 | 5 1 29 |
+------------------+----------------------+----------------+
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Play this puzzle online at the Daily Sudoku site
This looks very difficult (as it should - it's a 'Super hard +', rating 906). But there is a magic bullet - easy to follow, hard to spot (for me anyway).
(I used two moves in the bottom right, and eventually found a third that starts there.) |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sun Feb 27, 2011 5:56 pm Post subject: |
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Every puzzle should have a solution. I don't know about JV's "magic bullet", but ...
I don't recall my solver's network logic previously leading me to a 5-cell ALS.
Code: | (4)r3c9 = r3c7|r8c9 - r7c7 = r7c1 - (4=5)r54693c1 => r3c9<>5
+-----------------------------------------------------------------------------------------+
| 8 345 1 | 235 2345 9 | 6 235 7 |
| 9 345 2 | 3567 34567 356 | 1 35 8 |
| e356 7 36 | 8 1235 1235 | b234 9 a24-5 |
|-----------------------------+-----------------------------+-----------------------------|
| e2367 1 5 | 236 236 4 | 23789 78 269 |
| e2346 239 3469 | 12356 8 7 | 23 235 1256 |
| e2367 8 367 | 9 12356 12356 | 237 4 1256 |
|-----------------------------+-----------------------------+-----------------------------|
| d12457 259 8 | 1257 12579 125 | c2479 6 3 |
| 123457 2359 3479 | 123567 1235679 12356 | 24789 78 b249 |
| e237 6 379 | 4 2379 8 | 5 1 29 |
+-----------------------------------------------------------------------------------------+
# 143 eliminations remain
Basics complete the puzzle.
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Sun Feb 27, 2011 6:12 pm Post subject: |
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What a move
I really doubt that a person would ever search for such an als but life is always got a surprise for us.
Ted |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sun Feb 27, 2011 6:29 pm Post subject: |
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Danny, nice!
I don't understand why you need to worry about r8c9?
Some might see it easier from the other angle...
Code: | (5)r3c1=hp(15)r78c1 - (4)r7c1=r7c7 - r3c7=r3c9 |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Sun Feb 27, 2011 8:17 pm Post subject: |
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peterj wrote: | Danny, nice!
I don't understand why you need to worry about r8c9?
Some might see it easier from the other angle...
Code: | (5)r3c1=hp(15)r78c1 - (4)r7c1=r7c7 - r3c7=r3c9 |
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I just missed the HP() when trying to unravel my solver's network.
As for r8c9, I just threw that in to show that an alternate path was available from r3c9 to r7c7.
Thanks for making the obvious obvious.
Regards, Danny |
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JV
Joined: 09 Jan 2011 Posts: 24 Location: Devon, England
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Posted: Sun Feb 27, 2011 8:22 pm Post subject: |
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Thanks Danny & Peter. Amazing Eurekas! By 'magic bullet' I did indeed mean deleting the 5 in r3c9.
I don't know how other folk see things: I mean do you think in Eureka? (I do for an ordinary chain, but something like this is too complex for me.) I simply saw, as you no doubt did, that r8c9 = 4 => r7c1 = 4 => r8c1 = 1 => r3c1 = 5 => r3c9 <> 5.
Another test if you'd like one, how one expresses an APE in EUreka: there's one in r78c7. (I'd already removed the 7s in r47c7 because of the UR 78, but I don't think that makes any difference.) |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sun Feb 27, 2011 9:43 pm Post subject: |
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JV, I have never spent the time to get to grips with APE. Partly because I never see anyone use them (other than some computer solvers) and my understanding is that they can usually be represented by xyz+-wings and als-xz techniques...
There is an wxyz-wing (I think)... which is hard to represent in Eureka other than a kraken presentation...
Code: | APE/wxyz-wing(2479) r7c7 ; r8c7<>2
(2)r7c7
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(9)r7c7 - (9=2)r9c9
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(4)r7c7 - als(4=23)r35c7
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(7)r7c7 - als(7=23)r56c7 |
In this case you can see it as an "almost naked quad" move also...
Code: | anq(2347=9)r3567 - (9=2)r9c9 ; r8c7<>2 |
I think if you consider all the possible combinations of values in the "aligned pairs" r7c7 and r8c7 you will find that all the ones with (2)r8c7 are not valid - I haven't tried this! Life is too short... |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sun Feb 27, 2011 10:20 pm Post subject: |
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Peter, three years ago we started discussing APE here and this post by Keith probably put an end to it:
"All,
I did a search on APE, and was surprised to find my footsteps all over the discussion.
http://www.sudoku.com/boards/viewtopic.php?t=3882
As I now recall, Ruud proposed that APE was a missing method from the usual arsenal.
I did an analysis that showed APE is another (less general) form of XY, XYZ, and WXYZ-wings. In other words, APE makes one exclusion at a time, XY.. wings make multiple exclusions.
APE is a monkey that you should not worry about.
I welcome any example of an APE exclusion that is not an XY... wing.
Keith" |
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JV
Joined: 09 Jan 2011 Posts: 24 Location: Devon, England
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Posted: Sun Feb 27, 2011 10:38 pm Post subject: |
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Thanks Peter, but I think I'm sorry I asked!
Right-hand boxes with 7s gone (because of the UR78):
Code: |
+---------------+
| 6 235 7 |
| 1 35 8 |
| 234 9 245 |
+---------------+
| 2389 78 269 |
| 23 235 1256 |
| 237 4 1256 |
+---------------+
| 2479 6 3 |
| 2489 78 249 |
| 5 1 29 |
+---------------+
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Clearly, (7)r7c7 = (8)r8c7 . (It doesn't matter but they're in fact strictly conjugate , since their only other occurence in the box is as a naked pair.
If you have 7 in r7c7 then there's a locked set 234,23, 237, 7 in the column: no 2 or 4 in r8c7.
Or, r8c7= 8; no 2 or 4.
I know it's not very common, but it's also very easy. (The combination of 7 & 8 in box 9 hint's at the existence of an APE, since that's at least one pair disallowed.) |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sun Feb 27, 2011 10:50 pm Post subject: |
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Here is the corrected link:
http://forum.enjoysudoku.com/aligned-pair-exclusion-ape-no-dj-t3882.html
Keith
Marty R. wrote: | Peter, three years ago we started discussing APE here and this post by Keith probably put an end to it:
"All,
I did a search on APE, and was surprised to find my footsteps all over the discussion.
http://www.sudoku.com/boards/viewtopic.php?t=3882
As I now recall, Ruud proposed that APE was a missing method from the usual arsenal.
I did an analysis that showed APE is another (less general) form of XY, XYZ, and WXYZ-wings. In other words, APE makes one exclusion at a time, XY.. wings make multiple exclusions.
APE is a monkey that you should not worry about.
I welcome any example of an APE exclusion that is not an XY... wing.
Keith" |
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JV
Joined: 09 Jan 2011 Posts: 24 Location: Devon, England
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Posted: Sun Feb 27, 2011 10:54 pm Post subject: |
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Marty, I've just read your post. I'm sure you're right, that APEs are a subset of more general methods. (And as I've said they're probably not common.) But . . . they're easy for some of us to spot where more complex methods might not be.
Isn't it the case with many common methods that they're special cases, used because they're easy to find? We like skyscrapers, but they're just one special type of single digit chain. Or W-wings, or whatever.
In this puzzle I was scratching round for something to do. I saw the APE simply because I find them fairly easy to see. I don't have the ability to see in any other way the two deletions I made. |
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JV
Joined: 09 Jan 2011 Posts: 24 Location: Devon, England
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Posted: Sun Feb 27, 2011 11:02 pm Post subject: |
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Now I've read Keith's post, including the link to the APE discussion, but that hasn't left me much wiser. I can see vaguely that the deletions I made are essentially because of an intersection of ALS in c7 and box 9. And if I were pressed I just might be able to formulate that. But so what? - I'd never have been able to see this in any other way. |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sun Feb 27, 2011 11:17 pm Post subject: |
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JV wrote: | Now I've read Keith's post, including the link to the APE discussion, but that hasn't left me much wiser. I can see vaguely that the deletions I made are essentially because of an intersection of ALS in c7 and box 9. And if I were pressed I just might be able to formulate that. But so what? - I'd never have been able to see this in any other way. |
JV, can you be more explicit about what you are calling an APE? Which cells, what eliminations?
Thank you,
Keith |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Mon Feb 28, 2011 2:42 am Post subject: |
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I have started an APE thread in the "Solving Techniques" forum. Please post further discussion and examples of Aligned Pair Exclusions there.
Keith |
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Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
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Posted: Mon Feb 28, 2011 3:20 am Post subject: |
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tlanglet wrote: | I really doubt that a person would ever search for such an als but life is always got a surprise for us. |
I probably wouldn't have noticed its usefulness in this case, but it's not so hard to spot such an ALS. Whenever you have a conjugate pair within a house, the other unsolved cells of the house form an ALS. This should be evident. You have n candidates distributed in n cells. There exists one candidate that is limited to two of those cells. The other n-2 cells thus contain n-1 candidates and form an ALS. (I'm assuming that the grid has been properly reduced and that there are no hidden sets.)
In this case, c1 had 7 unsolved cells with conjugate <1>s in r78c1. The remaining 5 cells, r34569c1, form the ALS with the 6 remaining candidates, 234567.
peterj's wing is another example. It uses the 5-digit ALS (almost naked quad, which is just an ALS) formed by excluding the conjugate <8>s in c7. His notation is fine, though I would notate it as:
ALS[(2)r3567c7=(9)r7c7] - (9=2)r9c9 ; r8c7<>2
It's just another way to write the same thing.
ALS are definitely a powerful solving technique and very useful in AICs. Spotting them is the trick. |
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JV
Joined: 09 Jan 2011 Posts: 24 Location: Devon, England
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Posted: Mon Feb 28, 2011 10:00 am Post subject: |
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Thanks Asellus.
Keith, I do see that it's not necessary to use an APE, but that's how I saw it. If you write down the possible pairs for r78c7 you'll find that none of them includes a 2 or a 4 in r78c7, thus deleting them.
And I have to admit that Asellus is right, that it's not too diffcult to spot once you know to look. I suppose the question is how to know where to look for something. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Mon Feb 28, 2011 2:34 pm Post subject: |
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Asellus,
Once again you have pointed out another insight into the sudoku world. I, for one, have not considered the simple, obvious perspective for spotting an ALS that you presented.
Thanks...........
Ted |
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