[mercury-users] Constraint puzzles - understanding how to go from Prolog to Mercury

[Vladimir Gubarkov]

Hi,

You can also check this out
http://www.mercury.csse.unimelb.edu.au/mailing-lists/mercury-users/mercury-users.201106/0029.html


Sincerely yours,
Vladimir

On Tue, Sep 6, 2011 at 4:34 AM, Daniel Patterson <lists.mercury at dbp.mm.st>wrote:

> Hi again,
>
> An example used in The Art of Prolog (1986, page 216) is a puzzle that
> follows the similar pattern of "There are N people, each who have a
> different X, Y, and Z." followed by a series of clues that should completely
> describe a unique solution (because of the constraints about uniqueness).
> The prolog code they showed consisted of a series of logical statements that
> amounted to that clue, and then further logical statements to extract the
> desired information (the solution to the puzzle).
>
> In trying to translate this to Mercury, I ran into problems due to the
> instantiatedness of some of the parts. I looked through the section of the
> reference manual on Solvers and the corresponding paper by Becket et al from
> 2006, but I couldn't tell if that was even talking about the type of problem
> I was trying to solve. I also looked through the Prolog -> Mercury guide,
> but didn't notice anything that pertained to these kind of problems.
>
> Either way, the code I wrote (which was a simplified translation of what
> was written in The Art of Prolog), was the following. This problem involves
> the ranking of a programming competition between three individuals from
> different countries who prefer different sports. I've left out my attempts
> at writing the modes because none of them worked!
>
> :- type nationality ---> american; australian; israeli.
> :- type sport ---> tennis; basketball; cricket.
> :- type name ---> michael; richard; simon.
> :- type friend ---> friend(name, nationality, sport).
>
> :- pred did_better(friend,friend,list(friend)).
> did_better(A,B,[A,B,_]).
> did_better(A,C,[A,_,C]).
> did_better(B,C,[_,B,C]).
>
> :- pred name(friend,name).
> name(friend(A,_,_),A).
> :- pred nationality(friend,nationality).
> nationality(friend(_,B,_),B).
> :- pred sport(friend,sport).
> sport(friend(_,_,C),C).
>
> main(!IO) :-
>  (
>    if
>    (
>      Friends = [friend(_,_,_),friend(_,_,_),friend(_,_,_)],
>      % clue 1
>      did_better(Man1Clue1,Man2Clue1,Friends),
>      name(Man1Clue1,michael), sport(Man1Clue1,basketball),
>      nationality(Man2Clue1,american),
>      % clue 2
>      did_better(Man1Clue2,Man2Clue2,Friends),
>      name(Man1Clue2,simon),nationality(Man1Clue2,israeli),
>      sport(Man2Clue2,tennis),
>      % clue 3
>      first(Friends,ManClue3),sport(ManClue3,cricket),
>      % queries
>      list.member(Q1,Friends),name(Q1,Name),nationality(Q1,australian),
>      member(Q2,Friends),name(Q2,richard),sport(Q2,Sport)
>    )
>    then
>      % results
>      print(Name,!IO),
>      print(Sport,!IO)
>    else
>      print("Could not solve\n", !IO)
>  ).
>
> The main problems seemed to arise from the fact that all of the predicates
> were partially filling in the friends, so, for example, it would say that
> the final instantiatedness was friend(ground,free,free), for example.
>
> Are these types of problems solvable, and if so, how? Or does someone have
> an example of the way that they have done it (it could be any similar
> problem, not just this one)?
>
> Thanks in advance,
> Daniel
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