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Topic: "Logic puzzle" logic
Replies: 15   Last Post: Dec 11, 1999 6:57 PM

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Quentin Grady

Posts: 291
Registered: 12/6/04
Re: "Logic puzzle" logic
Posted: Dec 1, 1999 2:18 AM
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On Wed, 01 Dec 1999 00:46:47 GMT, DaleRush@planetnet.com (Dale Rush)
wrote:

>On Wed, 01 Dec 1999 Quentin Grady <quentin@inhb.co.nz> wrote:
>>1. Patricia lives in Brussels. She's one year older than John's
>> mother who lives in Eittligen.
>>2. Stasik's mother is Nona. She's one year older than Bina whose son
>> isn't Alex.
>>3. The woman in Kirov (who isn't Maria) is younger than the worman in
>> Odessa.
>>4. Nicholas's mother (no Nina) is 43.
>>5. The woman living in Chittagong is 45.
>>The women's ages are 41, 42, 43, 45, 46.
>>Their names are Bina, Maria, Nina, Nona, Patricia.
>>They live in Brussels, Chittagong,Ettlingen, Kirov and Odessa.
>>How old is each woman, who is her son and where do they live.

>
>Does each woman have a son? I see only four sons specified (John,
>Stasik, Alex, Nicholas). Anyway, by inspection, I'd say
>
> Brussels Chittagong Ettlingen Kirov Odessa
> Particia Bina Maria Nina Nona
> Nicholas ???? John Alex Stasik
> 43 45 42 41 46
>
>I suppose what you're asking about is some sort of formal technique
>for solving problems like this. Maybe if we thought about the thought
>process we use to solve them "by hand", and then think about how we
>might program a computer to do the same thing, the process could be
>formalized.
>
>The most useful information seems to be the ages and their relations
>to each other. There are two distinct pairs of women whose ages
>differ by only one, and the list of ages contains only three such
>possibilities. Furthermore, we know John's mother isn't 45 (from her
>location), so we must have either [Pat=42,John's=41] or else [Pat=43,
>John's=42]. Both of these involve the age 42, so we have Nona=46
>and Bina=45, hence Bina is in Chittagong. Also, we must put Nona
>in Odessa to make its resident older than the woman in Kirov, which
>must be Nina. Then we see Nicholas must be Pat's son, so she is 43,
>and the rest falls into place.


I admire the facility with which you do this. I had been endeavouring
to find a way of meaningfully representing the "Either {Pat = 42,
John's = 41} or else {Pat = 43, John's=42) situation on the 5 x 5
grids. It seems to me that at this critical decision stage the grids
aren't helpful. At least I haven't discovered how to make them work.

>On Wed, 01 Dec 1999 Quentin Grady <quentin@inhb.co.nz> wrote:
>>There might be something else. A sort of Booleanoid algebra that
>>allows information found in one 5 x 5 grid to be used in another 5 x 5
>>grid. Perhaps there isn't. Is there such a Booleanoid algebra?
>>If there is, what are its rules?

>
>Well, there are many kinds of algebras, and there are multi-valued
>logics (i.e., instead of just True and False they have three or more
>truth values), but I'm not sure how any of those would apply here.
>This kind of problem seems like ordinary Boolean logic. I suppose you
>could formalize it with some kind of Boolean vectors or matrices,
>essentially like the 5x5 arrays in your puzzle book, and then define
>rules for "intersecting" multiple arrays of information, allowing the
>individual arrays to be incomplete, i.e., entries may contain True,
>False, or Undetermined.


Yes. This is what I am looking for. A grid algebra to allow
information in one 5x5 grid to be used in another.

Thanks,







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