Notes

Sometimes it's a pain to figure something out. When documentation is lacking or I think it can be articulated more clearly, it gets a note here. I've also made a series of simple notes (largely hello world samples) for a collection of programming languages to help programmers remember syntax. These notes are not for learning the language, they are intended to help polyglot programmers 'get in the mode'.

Prolog

A few quick and dirty Prolog notes for getting up and running.

Data types

1, -1, 1.11 % constants
a 'a b c' % atoms
X The_end % variables start with capitals
a(1,2) functionname(x,y) % compound terms

Interacting with prolog

  • Start prolog: pl
  • Quit prolog: halt.
  • Start entering facts: [user].
  • Stop entering facts: Control+D
  • Find another solution: ; (or enter to just accept current answer)
  • See full result: w

To create your own predicates: save them in a file as
whatever.pl. To load them in Prolog, type [whatever].

Solving problems

  • See as Prolog sees: write_canonical([1,2,3]).
  • Print: print('hello world').
  • Debugging: trace,program(A,B,C). or guitracer,program(A,B,C).
  • Help files: help.

Example code

% Write a predicate replace(OldCharacter, NewCharacter, InputList, OutputList)
% that replaces all the occurrences of OldCharacter in InputList with
% NewCharacter.

replace(_,_,[],[]).
replace(Old,New,[Old|TOld],[New|TNew]) :-
        replace(Old,New,TOld,TNew).
replace(Old,New,[Different|TOld],[Different|TNew]) :-
        not(Old = Different),
        replace(Old,New,TOld,TNew).

% Write a predicate textify(ListToReplace, NewCharacter, InputList, OutputList)
% that replaces all the occurrences of ListToReplace in InputList with the
% character NewCharacter. Assume that ListToReplace always has exactly three
% characters.
% textify([a,t,e],8,[s, e, e,' ',y,o,u,' ',l,a,t,e,r,' ',k,a,t,e],Answer).

textify(_,_,[],[]).
textify(_,_,[_],[]).
textify(_,_,[_,_],[]).
textify([Old1,Old2,Old3],New,[Old1,Old2,Old3|TOld],[New|TNew]) :-
    textify([Old1,Old2,Old3],New,TOld,TNew).
textify([Old1,Old2,Old3],New,[Different|TOld],[Different|TNew]) :-
    not(Old1=Different),
    textify([Old1,Old2,Old3],New,TOld,TNew).

% Describe how you would modify textify to deal with lists of any length.

% replace first rule with:
textify(ListToReplace, _, InputList, InputList) :-
        length(ListToReplace,L), length(InputList,I), I < L.
% replace second rule with:
textify([Old|OldT],New,[Old|TOld],[New|TNew]) :-
    TOld = [OldT|TOld2],
    textify([Old|OldT],New,TOld2,TNew).

% Given an input tree T, write a Prolog program that constructs a tree of the
% same shape as T, but in which the value of each node has been set to the value
% of the maximum value node in T.

% readable solution:
max(X,X,X) :- !.
max(X,Y,X) :- X > Y.
max(X,Y,Y) :- X < Y.

setMaxTree(n(V,[],[]), V, Max, n(Max,[],[])).
setMaxTree(n(V,L,[]), CurrMax, Max, n(Max,LNew,[])) :-
    setMaxTree(L, V2, Max, LNew),
    max(V, V2, CurrMax).
setMaxTree(n(V,[],R), CurrMax, Max, n(Max,[],RNew)) :-
    setMaxTree(R, V2, Max, RNew),
    max(V, V2, CurrMax).
setMaxTree(n(V,L,R), CurrMax, Max, n(Max,LNew,RNew)) :-
    setMaxTree(L,V2,Max,LNew),
    setMaxTree(R,V3,Max,RNew),
    max(V,V2,VMax1),
    max(VMax1,V3,CurrMax).

applyMax(T,TNew) :- setMaxTree(T,Max,Max,TNew).

% condensed to four clauses:
setMaxTree(n(V,[],[]), V, Max, n(Max,[],[])).
setMaxTree(n(V,L,[]), CurrMax, Max, n(Max,LNew,[])) :-
    setMaxTree(L, V2, Max, LNew),
    (V>=V2 -> CurrMax=V; CurrMax=V2).
setMaxTree(n(V,[],R), CurrMax, Max, n(Max,[],RNew)) :-
    setMaxTree(R, V2, Max, RNew),
    (V>=V2 -> CurrMax=V; CurrMax=V2).
setMaxTree(n(V,L,R), CurrMax, Max, n(Max,LNew,RNew)) :-
    setMaxTree(L,V2,Max,LNew),
    setMaxTree(R,V3,Max,RNew),
    (V>=V2 -> VMax1=V; VMax1=V2),
    (VMax1>=V3 -> CurrMax=VMax1; CurrMax=V3).
% 'simply' call with setMaxTree(Tree,M,M,X).
% to produce X as the new tree with all the nodes set to max

% Write a Prolog program to return a list of all the 0’s and a list of all the
% 1’s in a given tree. For example, the goal enum(n(n(0, 1, 0), 1, 0), X, Y)
% should instantiate X to [0, 0, 0] and Y to [1, 1]. The program should use
% difference lists.

% without difference lists:
flatten(0,[0]).
flatten(1,[1]).
flatten(n(X,Y,Z),List) :-
    flatten(X,XL),flatten(Y,YL),flatten(Z,ZL),
    append(XL,YL,List1),
    append(List1,ZL,List).

enum(T,X,Y) :- flatten(T,FlatT),getN(0,FlatT,X),getN(1,FlatT,Y).

getN(X,[X|TIn],[X|TOut]) :- getN(X,TIn,TOut).
getN(X,[Y|TIn],TOut) :- not(X=Y),getN(X,TIn,TOut).
getN(_,[],[]).

% with difference lists:
flatten(0,[0|T]-T).
flatten(1,[1|T]-T).
flatten(n(X,Y,Z),List) :-
    flatten(X,XL),flatten(Y,YL),flatten(Z,ZL),
    dapp(XL,YL,List1),
    dapp(List1,ZL,List).

dapp(A-B,B-C,A-C).

getN(_,A-B,A-B) :- unify_with_occurs_check(A,B), !.
getN(X,[X|TIn]-T1,[X|TOut]-T2) :- getN(X,TIn-T1,TOut-T2).
getN(X,[Y|TIn]-T1,TOut-T2) :- not(X=Y),getN(X,TIn-T1,TOut-T2).

enum(T,X,Y) :- flatten(T,FlatT),getN(0,FlatT,X),getN(1,FlatT,Y).

% A terminal node of the trinary tree is said to be of odd parity if the number
% of its 1 components is an odd number. For example, n(1, 1, 1) is of odd
% parity, and n(1, 0, 1) is not of odd parity. Write a Prolog program to count
% the number of terminal nodes in a tree that have odd parity. For example,
% the goal odd(n(n(0, 1, 0), 1, 0), X) should instantiate X to 1.

isodd(n(0,0,1)).
isodd(n(0,1,0)).
isodd(n(1,0,0)).
isodd(n(1,1,1)).

odd(1,0).
odd(0,0).
odd(n(X,Y,Z),1) :- isodd(n(X,Y,Z)),!.
odd(n(X,Y,Z),Count) :- odd(X,C1),odd(Y,C2),odd(Z,C3), Count is C1+C2+C3.

% note to self: would be better to use the 99 prolog challenges rather than old supervision work, so have removed questions that do not show anything new

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