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Eureka Notation Question

 
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Sun Oct 25, 2009 7:20 pm    Post subject: Eureka Notation Question Reply with quote

I started working on the puzzle "MM 1568, 10-25-09" posted in the "Other Forum" and my first step raised the issue of how to describe it properly.

After basics:
Code:

*-----------------------------------------------------------*
 | 79    8     1     | 3     4     5     | 679   2     679   |
 | 5     4     6     | 7     289   29    | 389   39    1     |
 | 2     3     79    | 1     89    6     | 789   5     4     |
 |-------------------+-------------------+-------------------|
 | 789   6     5     | 2     3     1     | 79    4     789   |
 | 4     1     3789  | 58    59    79    | 3579  6     2     |
 | 3789  2     3789  | 458   6     479   | 1     39    5789  |
 |-------------------+-------------------+-------------------|
 | 1     7     4     | 56*   25    3     | 2569  8     569*  |
 | 6     5     2     | 9     7     8     | 4     1     3     |
 | 38    9     38    | 456*  1     24    | 256   7     56*   |
 *-----------------------------------------------------------*

A type 6 UR56 (marked with *) is in r79c49; either r7c9=9 or r9c4=4 to prevent the deadly pattern. So,

UR56[(9)r7c9 = (4)r9c4] - (4=2)r9c6 - (2=5)r7c5 - (5=6)r7c4.

At this stage, r7c4=6 and r7c5=5 forcing r7c9=9 which satisfies the constrain of the original UR. Thus we can set r7c9=9.

How can this condition be notated in Eureka? How about something like:

UR56[(9)r7c9 = (4)r9c4] - (4=2)r9c6 -(2=56)r7c45 - (56=9)r7c9

Ted
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ttt



Joined: 06 Dec 2008
Posts: 42
Location: vietnam

PostPosted: Mon Oct 26, 2009 4:18 am    Post subject: Re: Eureka Notation Question Reply with quote

tlanglet wrote:
Code:

 *-----------------------------------------------------------*
 | 79    8     1     | 3     4     5     | 679   2     679   |
 | 5     4     6     | 7     289   29    | 389   39    1     |
 | 2     3     79    | 1     89    6     | 789   5     4     |
 |-------------------+-------------------+-------------------|
 | 789   6     5     | 2     3     1     | 79    4     789   |
 | 4     1     3789  | 58    59    79    | 3579  6     2     |
 | 3789  2     3789  | 458   6     479   | 1     39    5789  |
 |-------------------+-------------------+-------------------|
 | 1     7     4     | 56*   25    3     | 2569  8     569*  |
 | 6     5     2     | 9     7     8     | 4     1     3     |
 | 38    9     38    | 456*  1     24    | 256   7     56*   |
 *-----------------------------------------------------------*

A type 6 UR56 (marked with *) is in r79c49; either r7c9=9 or r9c4=4 to prevent the deadly pattern. So,
.......
...we can set r7c9=9.
How can this condition be notated in Eureka? How about something like:
UR56[(9)r7c9 = (4)r9c4] - (4=2)r9c6 -(2=56)r7c45 - (56=9)r7c9

Nice find!
IMO, your presenting is correct or you can present as: UR56[(9)r7c9 = (4)r9c4] - (4=2)r9c6 -(2=56)r9c79 => r7c9<>56

ttt
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Mon Oct 26, 2009 1:39 pm    Post subject: Reply with quote

Thanks for the feedback ttt.

The sad part of the story is that the step did nothing obvious towards solving the puzzle.

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Oct 26, 2009 5:44 pm    Post subject: Reply with quote

tlanglet wrote:
Thanks for the feedback ttt.

The sad part of the story is that the step did nothing obvious towards solving the puzzle.

Cells {42} {25} {56} form an ALS, so I'm sure someone would write your Eureka sequence with an embedded ALS entry. Personally, I use forcing chain notation for situations like yours because many ALS sequences simply disguise network sections of a solution anyway.

My original 2-step solution to this puzzle.

Code:
 <56> UR r79c49
                                 r7c9=9
 r9c4=4 r9c6=2 ( r7c5=5 r7c4=6 ) r7c9=9   (same as Ted)   -or-
 r9c4=4 r9c6=2      r9c79=56     r7c9=9   (same as ttt)
 +--------------------------------------------------------------+
 |  79    8     1     |  3     4     5     |  679   2     679   |
 |  5     4     6     |  7     289   29    |  389   39    1     |
 |  2     3     79    |  1     89    6     |  789   5     4     |
 |--------------------+--------------------+--------------------|
 |  789   6     5     |  2     3     1     |  79    4     789   |
 |  4     1     3789  |  58    59    79    |  3579  6     2     |
 |  3789  2     3789  |  458   6     479   |  1     39    5789  |
 |--------------------+--------------------+--------------------|
 |  1     7     4     | *56    25    3     |  2569  8    *56+9  |
 |  6     5     2     |  9     7     8     |  4     1     3     |
 |  38    9     38    | *56+4  1     24    |  256   7    *56    |
 +--------------------------------------------------------------+
 # 58 eliminations remain

Incorrect !!!

Code:
 XYZ-Wing r1c7/r1c9+r4c7 => r3c7<>7
 +--------------------------------------------------------------+
 |  79    8     1     |  3     4     5     | *679   2    *67    |
 |  5     4     6     |  7     289   29    |  389   39    1     |
 |  2     3     79    |  1     89    6     |  89-7  5     4     |
 |--------------------+--------------------+--------------------|
 |  789   6     5     |  2     3     1     | *79    4     78    |
 |  4     1     3789  |  58    59    79    |  3579  6     2     |
 |  3789  2     3789  |  458   6     479   |  1     39    578   |
 |--------------------+--------------------+--------------------|
 |  1     7     4     |  56    25    3     |  256   8     9     |
 |  6     5     2     |  9     7     8     |  4     1     3     |
 |  38    9     38    |  456   1     24    |  256   7     56    |
 +--------------------------------------------------------------+
 # 52 eliminations remain
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Mon Oct 26, 2009 6:17 pm    Post subject: Reply with quote

What can I say!!! I unknowingly came close but no brass ring.

Anyway, I learned more about using the Eureka notation.

Thanks Danny........

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3855

PostPosted: Mon Oct 26, 2009 6:34 pm    Post subject: Reply with quote

tlanglet wrote:
What can I say!!! I unknowingly came close but no brass ring.

Anyway, I learned more about using the Eureka notation.

Thanks Danny........

Ted,

Anyone who manually plows through these puzzles like you and others do, automatically earns an honorary brass ring in my book!!!

The best I can do is let my solver point out bottlenecks and see if I get lucky enough to find an alternate path that it doesn't know about. That was the case in this puzzle. First, I found the 2-step solution; and then I concentrated on folding the <9> elimination in [r1c9] into the subsequent XYZ-Wing. That's how I came up with my 1-step solution that I posted with the original puzzle.

Regards, Danny
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Asellus



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

PostPosted: Mon Nov 02, 2009 8:12 am    Post subject: Reply with quote

Ted,

Just now perusing this...

I would use the strong inference within one of the ALS, as Danny mentions. There is no need to think in terms of forcing (as your description of how you found the relationship suggests, and as some of the discussion above does explicitly). Rather, the key is to "see" ALS in a "new" way that reveals the inherent strong inferences they contain. The essential principle is this: All instances of any two digits within an ALS have a strong inference between them. By "all instances" I mean grouped digits.

This principle should be obvious since it is not possible for all instances of any two digits within an ALS both to be false.

So, there are a couple of options for using such an ALS inference in your AIC. We can proceed to the 25 bivalue and then consider the 569 ALS. (This is most likely how one would see it.) The inference within this ALS that we are interested in is:
ALS[(5)r7c49=(9)r7c9]

So, the AIC is:
UR56[(9)r7c9 = (4)r9c4] - (4=2)r9c6 - (2=5)r7c5 - ALS[(5)r7c49=(9)r7c9]; r7c9=9

Another way is to consider the 3-cell 2569 ALS. In this case, we are interested in this inference:
ALS[(2)r7c5=(9)r7c9]r7c459
I place the complete ALS cell reference outside the brackets since the references inside do not fully describe it.

So now, the AIC is:
UR56[(9)r7c9 = (4)r9c4] - (4=2)r9c6 - ALS[(2)r7c5=(9)r7c9]r7c459; r7c9=9

Note 1: In either case, I have used only the inference that is useful to your AIC. But these are not the only inherent strong inferences in these ALS. For instance, in the 569 ALS we might under different circumstances want to use (6)r7c49=(9)r7c9 instead. (The inference (5)r7c49=(6)r7c49 also exists though is not, I suspect, likely to be useful.)

Note 2: I do not believe that these ALS strong inferences are "macros" or "shorthands" for longer inference sequences. True, it is often possible to construct equivalent longer sequences (or always possible with enough creative notations). But these inferences are immediately evident once the ALS is recognized without thinking of any such longer sequences. They are an immediate consequence of the logic of ALS, as explained above.

This is true in much the same way as the UR inference used above is an immediate consequence of the logic of the UR and not a shorthand for some equivalent longer sequence of inferences.
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tlanglet



Joined: 17 Oct 2007
Posts: 2461
Location: Northern California Foothills

PostPosted: Tue Nov 03, 2009 3:01 pm    Post subject: Reply with quote

Asellus,

As always, thanks for your comments. I have tried understanding/using ALS a couple of times in the past without any success. However I now see a flicker of light after this post by you, both as inferences for AICs and also as a basis for directly making deletions. My challenge now is in finding and using ALS given my better understanding.

I greatly appreciate you taking the time and energy to provide this type of info to those of us still on the learning curve.

Ted
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