LOGO-L> Logical Puzzle

["Olga Tuzova" <olgatu@hotmail.com>]

Hello everybody,

I'm intruding on your recursion discussion with very different problem. 
I haven't seen this kind of problems there, but may be somebody is 
interested, or has had an experience in solving them.
The problem is from a "logical field".

For example let's formulate it as follows.

Three experts are having discussion about an ancient cup.

Expert A: This is a China cup from the 5th century,
Expert B: This is a Japan cup from the third century,
Expert C: This is not a China cup, it was made in the 4th century.

It turned out, only one of two statements of each expert was right, the 
other was wrong.

What country and what century does the cup belong?
----------------

I think, the main idea of the solution is standart.

Le's denote with C the statement "the cup is from China",
            J - "from Japan",
            T - "it was made in the third century",
            F - " in the 4th century",
            V - " in the 5th century".

We should find such boolean values of C J T F V, which make the
following statements  "true":
    or (and C not V)(and not C V)
    or (and J not T)(and not J T)
    or (and not not C F)(and not C not F)
Also:
    not and T F
    not and T V
    not and F V
    not and C J.

The solutions that I've seen, all use looking over(?) all the possible 
combinations of values of C, J, T, F, V. It doesn't look great. I tried 
to present these combination as a binary code (binary representation of 
the number from 0 through 31), but not sure it improves the solution 
much.
Would be greatful for any ideas and suggestions. Also, does anybody work 
on such problems this way or other with the school children?

The solution presented below, says, that the cup was from Japan and was 
made in the 5th century.

Thanks,
Olga.
---------------------------------
to logic.main
for [i 0 31][if (and bool1 c :i v :i ~
                     bool1 j :i t :i ~
                     bool2 c :i f :i ~
                     bool3 c :i j :i ~
                     bool3 t :i f :i ~
                     bool3 t :i v :i ~
                     bool3 f :i v :i)~
            [(pr c :i j :i t :i  f :i v :i)]]
end

to c :i
op int :i/16
end

to j :i
op remainder int :i/8 2
end

to t :i
op remainder int :i/4 2
end

to f :i
op remainder int :i/2 2
end

to v :i
op remainder :i 2
end

to bool1 :c :v
output or (and l :c not l :v) (and l :v not l :c)
end

to bool2 :c :f
output or (and not not l :c l :f)(and not l :c not l :f)
end

to bool3 :c :j
output not and l :c l :j
end

to l :a
ifelse :a=1 [op "true][op "false]
end
-------------------------------

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