
Re: "Logic puzzle" logic
Posted:
Dec 1, 1999 2:18 AM


This post not CC'd by email 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 multivalued >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,

