787d2daa8a
git-svn-id: https://yap.svn.sf.net/svnroot/yap/trunk@1388 b08c6af1-5177-4d33-ba66-4b1c6b8b522a
353 lines
8.9 KiB
HTML
353 lines
8.9 KiB
HTML
BEAM How-to-get-started
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This document gives you an introduction on how to use the EAM withing YAP.
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We start by explaining how to prepare YAP to use the EAM.
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Then we present some code examples that you can use to try
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out the EAM.
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WARNING:
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THE BEAM WITHING YAP IS STILL IN EARLY DEVELOPMENT,
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SO DON'T EXPECT IT TO RUN SMOOTHLY...
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WE ARE WORKING TO IMPROVE THE INTEGRATION OF BEAM WITHIN
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YAP, BUT THIS THINGS TAKE TIME. SO PLEASE BE PATIENT...
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---------------------------------------------------------------------------
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1. Compiling Yap to support the EAM.
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---------------------------------------------------------------------------
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If you want to use the BEAM prototype you must compile YAP using the
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flag --enable-eam
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tar -xvzf yapfile.tar.gz <- will extract yap to $YAPDIR
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mkdir tmp
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cd tmp
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../YAPDIR/configure --enable-eam <- prepare yap to compilation
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make <- compile yap
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su <- enter as root
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make install <- install yap
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You are now ready to try YAP-BEAM. Just run "yap" on the
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command line.
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---------------------------------------------------------------------------
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2. Quick start + Examples...
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---------------------------------------------------------------------------
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Quick example on how to run a Prolog program using the EAM.
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First you must enable the EAM using the comand
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?- eam.
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yes
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then you should load your program, and make your queries
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using ?- eam(query).
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You can write the query normally as in normal Prolog mode, but
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in this case you will only receive the first solution (or yes or no).
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Small example:
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Supose that you have the file Example.pl with the Prolog code:
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f(1).
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f(2).
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f(3).
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Here is an execution example:
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[user]$ ./yap
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% Restoring file /.../startup
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YAP version Yap-5.0.0
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?- eam.
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yes
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?- [t].
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% consulting /.../t.pl...
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% consulted /.../t.pl in module user, 1 msec 1328 bytes
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yes
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?- eam(f(X)).
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[ EAM execution started to solve f/1 ]
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X = 1 ? ;
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X = 2 ? ;
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X = 3 ? ;
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no
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?- f(X).
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[ EAM execution started to solve f/1 ]
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X = 1 ? ;
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no
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?-
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---------------------------------------------------------------------------
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A first example:
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You can try out the next example, the well-known benchmark scanner.pl
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that behaves badly in standard Prolog systems.
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Just run the demos... (demo1, demo2, demo3 or demo4).
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Remember to enable the EAM before loading the program.
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After running the examples, try restarting YAP and
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loading the program without the EAM enabled. Try to run
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the demo4... :)
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demo1:- demo(tiny).
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demo2:- demo(small).
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demo3:- demo(data).
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demo4:- demo(snake).
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demo(Data):-
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scannerdata(Data,R,C,D1,D2),
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write('Rows '), write(R),nl,
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write('Columns '), write(C),nl,
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write('Left diagonals '), write(D1),nl,
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write('Right diagonals '), write(D2),nl,nl,
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scanner(R,C,D1,D2,Image), !,
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displi(Image).
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sequence(Spec):-
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samples(Spec,Samples),
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cat(Samples,Images), !,
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displ(Images).
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samples([],Samples):- !, Samples=[].
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samples([S|Spec],Samples):- !,
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scannerdata(S,R,C,D1,D2),
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Samples=[sample(R,C,D1,D2)|Smpls],
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samples(Spec,Smpls).
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cat([],Images):- !, Images=[].
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cat([Sample|Samples], Images):- !,
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image(Sample,Image),
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Images=[Image|Imgs],
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cat(Samples,Imgs).
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image(sample(R,C,D1,D2), Image):-
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scanner(proc,R,C,D1,D2,I), !,
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Image=I.
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displ([]):- nl, nl.
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displ([I|Imgs]):- nl, displi(I), nl, displ(Imgs).
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displi([]):- nl .
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displi([R|Rws]):- write(' '),displr(R), displi(Rws).
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displr([]):- nl.
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displr([on|R]):- write('X '), displr(R).
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displr([off|R]):- write('_ '), displr(R).
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scannerdata(tiny, [1,1],[2,0],[1,1,0],[0,1,1]).
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scannerdata(small, [1,2,1],[2,1,1],[1,1,1,0,1],[0,0,3,1,0]).
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scannerdata(double,
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[2,2,3,2,2,1],
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[1,3,3,1,3,1],
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[0,1,1,2,1,2,3,1,1,0,0],
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[0,2,0,1,2,2,2,1,2,0,0]).
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scannerdata(snake,
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[4,2,6,2,4,4,3,2],
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[3,5,5,3,2,3,3,3],
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[1,2,2,1,1,2,5,2,2,3,3,2,1,0,0],
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[0,1,0,3,2,2,4,3,3,3,1,3,2,0,0]).
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scanner(RwData,ClData,D1Data, D2Data, Rws):-
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llength(RwData,R),
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llength(ClData,C),
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board(R,C,All,Rws,Cls,D1,D2),
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check(RwData,Rws),
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check(ClData,Cls),
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check(D1Data,D1),
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check(D2Data,D2).
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pixle(on).
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pixle(off).
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check([],[]).
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check([K|RwsD],[R|Rws]):-
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llength(R,L),
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line(K,L,R),
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check(RwsD,Rws).
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line(0,0,[]).
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line(K,L,[on|R]):-
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K > 0,
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K1 is K - 1,
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L1 is L - 1,
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line(K1,L1,R).
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line(K,L,[off|R]):-
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L > K,
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L1 is L - 1,
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line(K,L1,R).
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board(0,C,[],[],Cls,D,D):-
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C > 0,
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C1 is C - 1,
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seed(C,Cls),
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seed(C1,D).
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board(R,C,All,[Row|Rws],Cols,Rdiag,Ldiag):-
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R > 0,
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R1 is R - 1,
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row(C, Row),
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all2(Row,Al,All),
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column(Row,Cls,Cols),
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diagonal(Row,Rdg,Rdiag),
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rev(Row,[],Rev),
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diagonal(Rev,Ldg,Ldiag),
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board(R1,C,Al,Rws,Cls,Rdg,Ldg).
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seed(0,[]).
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seed(C,[[]|S1]):- C > 0, C1 is C -1 , seed(C1,S1).
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all2([],Al,Al).
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all2([H|Row],Al,[H|All]):- all2(Row,Al,All).
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row(0,[]).
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row(C,[_|R]):- C > 0, C1 is C -1, row(C1,R).
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column([],X,X).
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column([H|R],[Cl|Cls],[[H|Cl]|Columns]):- column(R,Cls,Columns).
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diagonal([H|Row],Dg, [[H]|Diag]):- column(Row,Dg,Diag).
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rev([],Y,Y).
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rev([H|T],Y,Z):- rev(T,[H|Y],Z).
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llength([],0).
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llength([A|R],N):- llength(R,M), N is M+1.
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---------------------------------------------------------------------------
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Another example:
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The famours queens...
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Try to evaluate in normal Prolog que query:
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?- queens([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],X).
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Now quit YAP and enable eam before loading the queens program
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Here is the code for you to try it:
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demo1:- queens([1,2,3,4,5,6,7,8,9],L), write(L), nl, fail.
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demo2:- queens([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15],L), write(L), nl.
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queens(L,C):-
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perm(L,P),
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pair(L,P,C),
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safe([],C).
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perm([],[]).
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perm(Xs,[Z|Zs]):-
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select(Z,Xs,Ys),
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perm(Ys,Zs).
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select(X,[X|Xs],Xs).
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select(X,[Y|Ys],[Y|Zs]):-
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select(X,Ys,Zs).
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pair([],[],[]).
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pair([X|Y],[U|V],[p(X,U)|W]):-
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pair(Y,V,W).
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safe(X,[]).
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safe(X,[Q|R]):-
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test(X,Q),
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safe([Q|X],R).
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test([],X).
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test([p(C1,R1)|S],p(C2,R2)):-
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test(S,p(C2,R2)),
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nd(p(C1,R1),p(C2,R2)).
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nd(p(C1,R1),p(C2,R2)):-
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wait_while_var([C1,C2,R1,R2]),
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C is C1-C2,
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R is R1-R2,
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C=\=R,
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NR is R2-R1,
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C=\=NR.
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Note that on the nd predicate, we have used wait_while_var
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to force the EAM to wait while C1, C2, R1, R2 are not bound,
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because the operations in this predicate can't be done with
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those variables unbound.
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---------------------------------------------------------------------------
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3. Some notes...
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---------------------------------------------------------------------------
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- BEAM create the indexing code (try, retry and trust) only
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considering the first argument and only when the predicates are first called.
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So the first time you run a query there can be a slowdown, because the
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EAM is indexing the code.
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---------------------------------------------------------------------------
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- A lot of builtins/code are not yet supported...
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For example var(X), not , ; (or), ...
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You will have a internal compiler error for these cases.
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and the clause that uses the builtin/code not supported will always fail.
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Example: consider the code:
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tst(X):- var(X), X=1.
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tst(2).
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You will receive a warning that there is unsupported code.
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Although you can still use the tst predicate, the first alternative
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will always fail...
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?- tst(X).
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[ EAM execution started to solve tst/1 ]
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X = 2 ? ;
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no
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---------------------------------------------------------------------------
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- The EAM prefers deterministic instead of non-deterministic.
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and thus can change the order of goals of your code to delay
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non-deterministic bindings.
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So, you must be careful when using builtins that have side-effects,
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or that may expect variables to be bound.
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For example, supose that you have the following Prolog code:
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f(1).
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f(2).
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tst(Y):- f(X), Y is X+1.
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Normal prolog would be fine, but the EAM can not execute this
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code correctly because since f(X) is non-deterministic,
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Y is X+1 is executed before X being bound. The result would be:
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?- tst(X).
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[ EAM execution started to solve tst/1 ]
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% INSTANTIATION ERROR- in arithmetic at user:tst/1 (clause 1): expected bound value
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The solution for this case is to force the EAM to wait for X to be bound.
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So the code correct code would be:
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tst(Y):- f(X), skip_while_var(X), Y is X+1.
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That means that the code Y is X+1 should be skipped
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while X is var. Note that in this case there is no more
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after Y is X+1. If there were, execution would continue
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on that code.
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Another alternative is to use:
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tst(Y):- f(X), wait_while_var(X), Y is X+1.
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that means what execution can not proceed while X is not bound.
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---------------------------------------------------------------------------
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- To support the EAM within the YAP the WAM compilation was
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specially the classification of permanent variables.
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This code was adapted from the initial BEAM implementation and is not
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yet completed.
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I've have already discover some code examples where the variables
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should are not being classified as permanent, and as result the BEAM
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returns false solutions. We are also working on this problem...
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---------------------------------------------------------------------------
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Expect more info soon...
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...
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