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Menneskes

 
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Wed Feb 20, 2008 4:59 pm    Post subject: Menneskes Reply with quote

[Storm_norm asked of my last (v. difficult) Menneske what the original was. Following the conventions of this forum, I've prefaced the no. with an M, and added the (not very accurate) difficulty number after it. For the originals, just go into the Menneske site, click on Show puzzle number near top left, and type in the number without the M or the difficulty no.]

First, two VH standard puzzles for anyone fancying a relaxing puzzle, but (for me at anyrate) using different techniques - i.e. not X-, XY, XYZ - wings.

M6448613 (13). I needed 2 non-basic moves.

M3877484 (18.) Contains a type 3 triple UR (or UR loop?) - as basic as they get, I think. No doubt other ways to do it for UR-haters.

And a few notches above VH, but not needing serious AICs or Medusa. M4986282 (103) with the hidden trips etc. cleared up:-
Code:

+----------------+---------------+---------------+
| 9    8   15    | 145  3    7   | 2    56   146 |
| 237  257 12357 | 9    2458 6   | 3458 58   134 |
| 4    6   1235  | 1258 258  158 | 358  7    9   |
+----------------+---------------+---------------+
| 27   279 8     | 45   1    59  | 4567 3    246 |
| 5    3   4     | 6    7    2   | 9    1    8   |
| 1    279 6     | 3    4589 589 | 457  25   24  |
+----------------+---------------+---------------+
| 26   25  259   | 1258 2589 3   | 168  4    7   |
| 2367 1   23579 | 2578 2589 4   | 368  2689 236 |
| 8    4   2379  | 127  6    19  | 13   29   5   |
+----------------+---------------+---------------+

Play this puzzle online at the Daily Sudoku site

Quite long for me, but quite interesting, and rich enough for smart people to find smart moves no doubt.
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Wed Feb 20, 2008 6:58 pm    Post subject: Mennesse Reply with quote

A five-step xy-chain removes the 9 from R6C6 and solves the puzzle.
XY-chains may not be very sophisticated, but they work!

Earl


Last edited by Earl on Wed Feb 20, 2008 8:35 pm; edited 1 time in total
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Wed Feb 20, 2008 7:18 pm    Post subject: Reply with quote

Nice, suppose Johan will like it too.
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Thu Feb 21, 2008 4:47 pm    Post subject: Reply with quote

Quote:
Nice, suppose Johan will like it too.


Ah, those xy-chains Ravel, I do like them especially when the key is found to unlock them, and the chain-reaction they cause are sometimes very helpful for
analysing potential UR DP's,(in which we share a common interest I think, why not use them knowing that a valid sudoku has only one solution),when using bi-
value cells to find pincer cells for eliminating digits.
But in the puzzle posted by Victor I could only see an xy-chain elimination of digit <9> in R6C6, earlier meant by Earl, after the ER elimination of <4> in
R6C9, caused by the ER in Box 3, probably I missed something.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Thu Feb 21, 2008 5:47 pm    Post subject: Reply with quote

When i remember right, after the ER and an x-wing (xy-wing possible) you can use the UR 27 with a 4-cell-bivalue-chain (plus a 3-cell without the xy-wing) to get rid of two 3's.
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Thu Feb 21, 2008 7:51 pm    Post subject: Reply with quote

Quote:
When i remember right, after the ER and an x-wing (xy-wing possible) you can use the UR 27 with a 4-cell-bivalue-chain (plus a 3-cell without the xy-wing) to get rid of two 3's.


Ravel,

Now my penny dropped on the [27] UR, but when you transport digit <5> via R1C3 you don't need the pincer cells for eliminating both <3>'s. Those <5>'s in R2C23 are also part of an AIC which results in a contradiction in R2, but writing that down in eureka notation is like cracking the Navajo code for me.
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Thu Feb 21, 2008 11:57 pm    Post subject: Reply with quote

Johan wrote:
Those <5>'s in R2C23 are also part of an AIC which results in a contradiction in R2, but writing that down in eureka notation is like cracking the Navajo code for me.

I believe you are just looking at the XY Wing from the point of view of the "victims."
Code:

+-----------------+-----------------+--------------+
| 9     8    15   | 145   3     7   | 2    6   14  |
| 237   257  1257 | 9     245   6   | 45   8   134 |
| 4     6    1235 | 1258  258   158 | 35   7   9   |
+-----------------+-----------------+--------------+
| 27    279  8    | 45    1     59  | 467  3   46  |
| 5     3    4    | 6     7     2   | 9    1   8   |
| 1     79   6    | 3     489   89  | 47   5   2   |
+-----------------+-----------------+--------------+
| 26    25   259  | 1258  2589  3   | 168  4   7   |
| 2367  1    2579 | 2578  2589  4   | 68   29  36  |
| 8     4    2379 | 127   6     19  | 13   29  5   |
+-----------------+-----------------+--------------+

The <5>s in r2c23 are the victims of the {145} XY Wing that pivots in r1c9.

Meanwhile, those <3> eliminations using the {27} UR provide a nice example for Eureka notation... which is actually not at all difficult to learn. First, let's consider those r2c23 <5>s to be gone. The {27} UR creates a strong link between the <3> in r2c1 and the <9> in r4c2: one or both of them must be true (they cannot both be false or we have the DP). This is an inferential strong link (not a conjugate link). Using the Eureka "=" symbol for a strong link, it is:
(3)r2c1=(9)r4c2

We then have the chain of bivalues {79}, {47}, {45} and {35}. The digits within each bivalue cell have a strong (conjugate) link. For instance, the first bivalue cell can be written in Eureka as:
(7=9)r6c2

The bivalue cells are joined together by weak links. So, the first two bivalues are joined using the Eureka "-" symbol for a weak link:
(9=7)r6c2-(7=4)r6c7

I reversed the order of the digits in the two bivalue notations so that it is clear that it is the weak link on <7> that joins the two bivalues. I.e., the <7>s are "next to each other" on either side of the "-" weak link.

Now, we can write out the entire chain:
(3)r2c1=(9)r4c2-(9=7)r6c2-(7=4)r6c7-(4=5)r2c7-(5=3)r3c7

All of the links alternate strong-weak-strong-weak-etc., as required. You might object that since there are only two <9>s in c2 (or only two <7>s in r6, etc.) that these are strong links, not weak links. If so, the answer is that they are conjugate (one must be true and the other false) links and are thus both strong and weak for inference purposes. We can use them either way. However, since there are three <4>s in c7, then (4)r6c7-(4)r2c7 can only be seen as a weak link.

The <3>s on the ends of the chain are both attached by strong links. Thus, they have a strong inference between them: one or both must be true. This means that they can serve as pincers and eliminate any <3>s that are weakly linked to both of them ("common peers").

All that remains for the proper complete Eureka notation is to add those weakly linked victims to the beginning and end of the chain and then write the conclusion that they cannot be <3>:

(3)r3c3|r2c9-(3)r2c1=(9)r4c2-(9=7)r6c2-(7=4)r6c7-(4=5)r2c7-(5=3)r3c7-(3)r3c3|r2c9; r3c3|r2c9<>3

The "|" character means "and" and is used to group things together. Here, "(3)r3c3|r2c9" means "the <3>s in r3c3 and r2c9." And "<>" means "are not equal to" or "do not contain."

Now, just for the heck of it, let's revisit that XY Wing in Eureka notation. It is:

(5)r2c23-(5=1)r1c3-(1=4)r1c9-(4=5)r2c7-(5)r2c32; r2c23<>5

So, maybe Eureka isn't quite so difficult as it may appear?

[Edit to correct a typo.]


Last edited by Asellus on Fri Feb 22, 2008 10:25 am; edited 1 time in total
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storm_norm



Joined: 18 Oct 2007
Posts: 1741

PostPosted: Fri Feb 22, 2008 1:02 am    Post subject: Reply with quote

Quote:
So, maybe Eureka isn't quite so difficult as it may appear?


for me, its always hard to remember that the equal sign means strong inference and the hyphen is weak inference.

I am always trying to read it correctly in my mind and therefore not really paying attention to all that meaning packed into the notation itself.
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Johan



Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium

PostPosted: Fri Feb 22, 2008 9:43 am    Post subject: Reply with quote

Asellus,

Many thanks for taking your time explaining this matter on Eureka notation.
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Fri Feb 22, 2008 1:37 pm    Post subject: Reply with quote

Asellus says:
Quote:
maybe Eureka isn't quite so difficult as it may appear?

That's true I suppose, but it's not that easy either! (It's like my schoolboy French for me: I can make sense of written French but prefer English of course, and I find it difficult to write grammatically in either French or Eureka.) In short: it gains in conciseness, buit perhaps loses in clarity.

Earl, guess it's ironic or something like that: you solved this puzzle easily with XY-chains whereas I took a convoluted road, but I solved one of yours fairly easily with XY-chains. Evens!
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