Re: Re: Re: Re: LOGO-L> Logical Puzzle

["Dreyfuss Pierre-Andre" <p.a.dreyfuss@mailexcite.com>]

Solving the puzzle using sets.



Building a microworld for the set  theory.

Then the puzzle is just a math problem.

I don't know the exact english names so corrct the names if necessary.

Union : the set of elements  member of :a or member of :b Each element only
once.


to union  :A :B
op remdup se :A :B
end

Intersection: The set of elements member of :A and member of :B

to inter :A :B
op filter [memberp ? :A] :B
end

Complement: set of elements of :R not in :A

to compl :A :R
op filter [not memberp ?  :A] :R
end

Product:  All couples made of an element of :a and an element: of :B

to setproduct :A :B
op (crossmap [list ?1 ?2]  :A :B)
end

And now let play. You can try writing commands in the commander.


The world of the puzzle.

make "cup [China Japan]
make "century [3 4 5]

What the experts are saying.

make "cupa [China]
make "cupb [Japan]
make "cupc compl [China] :cup

show :cupc
[Japan]

make "centurya[5]
make "centuryb[3]
make "centuryc[4]

If The cup is true The Century is false .

make "expa1 setproduct :cupa compl :centurya :century
show :expa1
[[China 3] [China 4]]
make "expb1 setproduct :cupb compl :centuryb :century
show :expb1
[[Japan 4] [Japan 5]]
make "expc1 setproduct :cupc compl :centuryc :century
show :expc1
[[Japan 3] [Japan 5]]

If the cup is false the century is true.

make "expa2 setproduct compl :cupa :cup :centurya
show :expa2
[[Japan 5]]
make "expb2 setproduct compl :cupb :cup :centuryb
show :expb2
[[China 3]]
make "expc2 setproduct compl :cupc :cup :centuryc
show :expc2
[[China 4]]

All each expert can say.
make "expa union :expa1 :expa2
show :expa
[[China 3] [China 4] [Japan 5]]
make "expb union :expb1 :expb2
show :expb
[[Japan 4] [Japan 5] [China 3]]
make "expc union :expc1 :expc2
show :expc
[[Japan 3] [Japan 5] [China 4]]


The truth should be sayed by all experts.
show inter :expa inter :expb :expc
[[Japan 5]]

Copy all that in the editor , keep what is necessary to build the program.


Sorry that set theory is no more a part of the curriculum of our pupils (12
- 15)
in Geneva ...

I don't know about yours.

Original :

>> To:            logo-l@gsn.org

>> Date:          Sun, 18 Jan 1998 23:38:34 -0700
>> From:          "Dreyfuss Pierre-Andre" <p.a.dreyfuss@mailexcite.com>
>> Cc:            olgatu@hotmail.com

>> Subject:       : Re: LOGO-L> Logical Puzzle
>> Organization:  MailExcite  (http://www.mailexcite.com)
>> Reply-to:      "Dreyfuss Pierre-Andre" <p.a.dreyfuss@mailexcite.com>
>
>Thank  Erich Neuwirth and Pierre-Andre  for introducing me to such an
>interesting approach to logic puzzle solving. Especially I like the
>one-line solution:
> print filter[memberp ? :experta]filter[memberp ? :expertb] :expertc
>
>(BTW, Pierre-Andre, there was a minor typo in this line. It was
>obvious, I just want to show, I investigated the message carefully.
>:-))
>
>And, I find it suitable for educational aims, that for this solution
>half of the work should be done in the student's head. But, for this
>approach I should introduce students to maps and filters. I'm not
>sure, it's trivial. At least, I have had no experience of that kind
>by far. I think, the most difficult part of it might be templates. By
>now I have no ideas how it could be done on the "school level". Any
>suggestions?
>
>Regards,
>Olga.
>
>> Hello Olga,
>>
>> Here is a solution that is not too difficult for your pupils.
>>
>> The basic I Idea is that we can build a set of affirmations that could
be true
for
>> each expert.
>>
>> Each affirmation is a list of country century
>> Expert A :[ China 4] [China 3] [Japan 5]
>>
>> Expert B :  [ Japan 4] [Japan 5] [China 3]
>>
>> Expert C :(changing not China into  Japan)
>>
>>             [Japan 3] [Japan 5] [China 4]
>>
>> make "experta [[China 4] [China 3] [Japan 5]]
>> make "expertb [[Japan 4] [Japan 5] [China 3]]
>> make "expertc [[Japan 3] [Japan 5] [China 4]]
>>
>> Now the problem is to find which affirmation is the same for all experts.

>>
>> It is the intersection of the three lists.
>>
>> The library primitive FILTER does the job.
>>
>> print filter[memberp ? :experta]filter[memberp ? :expertb] :experta Will
give
the
>> solution. [Japan 5]
>>
>> If you want the computer to build the affirmations then use the procedure
assert

>> below.
>>
>> to  assert :name :cupexp :centuryexp
>>  make :name  ~
>>  se map  [list :cupexp  ? ] ~
>>  filter [not ? = :centuryexp] :century ~
>>  map  [list ? :centuryexp] ~
>>  filter [not ? = :cupexp] :cup
>> end
>>
>> to solution
>>   local "expa
>>   local "expb
>>   local "expc
>>
>>   assert "expa "china 5
>>   assert "expb "japan 3
>>   assert "expc "japan 4
>>   print  filter [memberp ? :expa] ~
>>   filter [memberp ? :expb] :expc
>> end
>>
>> You can disgard the local declarations.
>>
>> Sincerly,
>> Pierre-AndrE Dreyfuss
>> Olga wrote .
>> >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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