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tlanglet · Posted: Sun Oct 25, 2009 7:20 pm Post subject: Eureka Notation Question

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:

ttt · Joined: 06 Dec 2008 Posts: 42 Location: vietnam

tlanglet · Posted: Mon Oct 26, 2009 1:39 pm Post subject:

Thanks for the feedback ttt.

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

Ted

daj95376 · Joined: 23 Aug 2008 Posts: 3854

tlanglet · Posted: Mon Oct 26, 2009 6:17 pm Post subject:

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

Anyway, I learned more about using the Eureka notation.

Thanks Danny........

Ted

daj95376 · Joined: 23 Aug 2008 Posts: 3854

Asellus · Posted: Mon Nov 02, 2009 8:12 am Post subject:

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.

tlanglet · Posted: Tue Nov 03, 2009 3:01 pm Post subject:

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