Prolog

173 SFS

?- sat([-20, 10, 0, -3, 13, 12],L). L = [-10,10,0,-3,10,10]

sat is short for saturate: these rules alter a list of positive integers so that any element whose abs. val. is > 10 keeps its sign but its abs. val is set to 10.

saturate([], []). % too big saturate([H|T], [X|Y]) :- H>10, X is 10, saturate(T,Y). % too small saturate([H|T], [X|Y]) :- H< -10, X is -10, saturate(T,Y). % everything else saturate([H|T], [X|Y]) :- X is H, saturate(T,Y). ...

Perhaps surprisingly, hitting ; here will find more copies of your solution, as prolog does the matches in a different order: the number of repetitions is is 2N, where N is the number of elements > 10 or < 10. This is a job for... a (green) CUT!!.

Yes, rule order matters. For instance, the base case has to be at the top, as usual.

More interestingly, if the last rule is moved up to be the second rule, the first answer generated is that the list is just copied with no changes since it's a default-copy rule and will be used first, before the non-default cases that need attention. Originally, it's the last resort after rules 1,2,3 fail. If it's moved to 2nd place, hitting ; generates 2N solutions, not all the same (!!) as backtracking selects different rules.

173 SFS + Answer

  1. Make a Prolog statement that asserts ``Namu is a mammal.''
  2. Make a Prolog rule that asserts ``If something is a mammal, it is an animal.''

mammal(namu). animal(Smthng) :- mammal(Smthng).

In Gprolog, Errors give line:column where compiler barfed. E.g.
Animal(X) :- ... gives
.../src/namu.pl:2:7: syntax error: . or operator expected after expression

Common Errors:

  1. Missing . at end of statement: Error found beginning of next line.
  2. Namu (Upper Case N) is a variable, not an individual: mistake.
  3. In MyPredicate(arg1, arg2), MyPredicate is a variable, error when see the (, as above.
  4. There is no ?- before facts or rules: that's the top-level, interactive prompt.
  5. For "if X then Y", or "X implies Y", the rule is Y :- X, not the other way around (think backwards arrow).

Prolog and Logic

Logic is one of the canonical formalisms for reasoning, thought, language, knowledge, mathematics, meta-mathematics etc. etc. Huge literature.

Prolog is for "programming in logic", and so inherits some of the paradigmicity of its formal inspiration. It also is a paradigmatical declarative language. That is, it has no control structure, at least ideally.

Clocksin and Mellish don't mention the logical basis of Prolog until the very last chapters of their text, so one doesn't need even to know formal logic exists to be a Prolog expert.

A prolog program states facts and rules, and asks a question. "It" (in the Zen sense) computes to produce an answer. Ideally the programmer doesn't care how. (Reality's another story, sadly). Under the hood there's an automatic theorem prover that does backtracking search to answer the question. That search involves another powerful search, a two-way pattern matcher that implements unification. Prolog statements are Horn Clauses, a slightly restricted but fast version of First Order Predicate Calculus.

SWI Prolog

As of 2011, the most recent SWI prolog is called swipl. An older version that may be around is pl. The former is recommended, so any stray pls in the overheads should be interpreted as swipl. /u/brown 560 swipl Welcome to SWI-Prolog (Multi-threaded, 32 bits, Version 5.7.11) . . . For help, use ?- help(Topic). or ?- apropos(Word). ?- help(nb_setval). [bounced to web page of help] true. halt. /u/brown 561

SWI Prolog Home

Reference Manual

Locally produced: Prologomena.

Another of many sets of Prolog lectures, tutorials, etc on the web: Prolog Introduction: Endriss, KCL.

Locally produced, highly telegraphic and probably no use Scattered basics.

GNU Prolog

Gnu Prolog

No Prolog lets you insert facts or rules at top level! Must create file, consult it. | ?- [tst]. compiling .... .. tst.pl compiled, 32 lines read .... (1 ms) yes | ?- dalter([i ,hate, you],X). X = [i,hate,i] yes | ?- halt.

No ?- help(foo).? Maybe not the most user-friendly documentation, either.

Lecture 1

Teleport to Lecture 1

BACKTRACKING AND INSTANTIATION

?- append([1,2], [3,4,5], Z). Z = [1, 2, 3, 4, 5] ; No ?- append([1,2], Y, [1,2,3,4,5]). Y = [3, 4, 5] ; No ?- append(X, Y, [1,2,3,4,5]). X = [] Y = [1, 2, 3, 4, 5] ; X = [1] Y = [2, 3, 4, 5] ; X = [1, 2] Y = [3, 4, 5] ; X = [1, 2, 3] Y = [4, 5] ; X = [1, 2, 3, 4] Y = [5] ; X = [1, 2, 3, 4, 5] Y = [] ; No

MORE BACKTRACKING AND INSTANTIATION

?- append([1], Y, Z). Z = [1|Y] ; No ?- append(X, [1], Z). X = [] Z = [1] ; X = [_G271] Z = [_G271, 1] ; X = [_G271, _G277] Z = [_G271, _G277, 1] ; ?- append(X, Y, Z). X = [] Y = Z ; X = [_G277] Z = [_G277|Y] ; X = [_G277, _G283] Z = [_G277, _G283|Y] ; ...

Recursion Example: Children, Descendents

ch(b,a). desc(X,Y) :- ch(X,Y). ch(c,b). desc(X,Y) :- ch(X,Mid), desc(Mid,Y). ch(d,c). /* not a legal file!*/ consult('rec.pl'). ?- desc(d,a). Call: (7) desc(d, a) ? creep Call: (8) ch(d, a) ? creep Fail: (8) ch(d, a) ? creep Redo: (7) desc(d, a) ? creep Call: (8) ch(d, _L192) ? creep Exit: (8) ch(d, c) ? creep Call: (8) desc(c, a) ? creep Call: (9) ch(c, a) ? creep Fail: (9) ch(c, a) ? creep Redo: (8) desc(c, a) ? creep Call: (9) ch(c, _L203) ? creep Exit: (9) ch(c, b) ? creep Call: (9) desc(b, a) ? creep Call: (10) ch(b, a) ? creep Exit: (10) ch(b, a) ? creep Exit: (9) desc(b, a) ? creep Exit: (8) desc(c, a) ? creep Exit: (7) desc(d, a) ? creep Yes

Lecture 2a

Teleport to Lecture

The Cut Lecture!

Pretty Darn Good Alternative Lecture

THE CUT !

Not a logic concept: it controls backtracking search. So you need to have a good picture of how Prolog control is operating behind the scenes. Sad! The whole ideal of insulating us from "It" crumbles into dust....

Cut is a goal whose effects go back to the parent goal, the rule with the cut on its RHS. Its common uses are:

Example: Negation as Failure

The cut-fail trick shows the principle of "Negation as Failure:" If P can be shown, first rule is applicable, second otherwise. If P can't be proved, backtracking gets to the 2nd rule. not(P) :- P, !, fail. not(P).

Without the cut in the rule, get: | ?- not(true). yes | ?- not(fail). yes Without the cut, if P is true, the first rule applies, and fails (as it should). But instead of returning fail and giving up (due to the cut), Prolog looks for another rule or fact to apply and finds the 2nd line. Oops.

Generally: ``to prove ans if you can prove condgoal then prove thengoal else prove elsegoal.'' is
ans:- condgoal, !, thengoal. ans:- elsegoal.

A Previous Example

We want h(A,B): "if A<3 set B=0 else set B=2". Two candidates, f/2 and g/2. 


f(X,0) :- X<3, !. % if f's first arg <3, 
                  % 2nd arg is 0. 
                  % NO MORE SEARCH
f(X,2).   % f's 2nd arg is 2
          %  for any 1st arg.

g(X,0) :- X<3.    % as above 
                  % but will look for 
                  % another rule
                  % if fail (X>=3).

g(X,2).           % like f(X,2).
-------------
*unix*:swipl
?- [cut].
Warning:  Singleton variables: [X]
% cut compiled 0.00 sec, 6 clauses
true.

?- f(1,X).
X = 0.

?- f(2,X).
X = 0.

?- f(3,X).
X = 2.

?- g(1,X).
X = 0 ;          %aha!
X = 2.

?- g(2,X).
X = 0 ;
X = 2.

?- g(3,X).
X = 2.
?- halt.

BUILT-IN PREDICATES

Check SWI documentation first!

new clauses: consult(S), reconsult(X), also
?- [f1, -f2, 'fred'].
true, fail
classifying terms: var(X), novar(X), atom(X), integer(X), atomic(X)
clauses as terms: listing(A), clause(X,Y), asserta(X), assertz(X), retract(X)
complex structures: functor(T,F,N), arg(N,T,A), name(A,L)
backtracking: repeat.
complex goals: X,Y [and], X;Y [or], call(X), not(X) equality: X=Y, X \= Y, X == Y, X \== Y, X =:= Y, X =\= Y
I/O: get0(X), get(X), skip(X), read(X), put(X), nl, tab(X), write(X), display(X), op(X,Y,Z)
files: see(X), seeing(X), seen, tell(X), telling(X), told
arithmetic: ``X is Y'', +,-,*,/,mod, =, \=, <, >, >=, =<
spying: trace, notrace, spy P, debugging, nodebug, nospy

PROLOG SAMPLES: VIET-NAM

hanoi

hanoi(N) :- move(N, left, center, right). move(0,_,_,_) := !. % nothing to do, quit. move(N,A,B,C) :- M is N-1, move(M,A,C,B), inform(A,B), move(M,C,B,A). inform(X,Y) :- write([move, disc, from, X, peg, to, Y, peg]),nl.

PROLOG SAMPLES: LISTS

Note _ the anonymous ("don't care") variable.

last(X,[X]). last(X,[ _ | Y] :- last(X,Y). nextto(X,Y, [X,Y | _ ]). nextto(X,Y, [ _ | Z ]) :- nextto(X,Y,Z). append([],L,L). append([X | L1], L2, [X | L3]) :- append (L1, L2, L3). Fun with instantiation: last(E, List) :- append(_,[E], List). nextto(E1, E2, List) :- append( _, [E1, E2 | _ ], List). member(E, List) := append(_, [E1 | _ ], List).

PROLOG SAMPLES: MORE LISTS

Recursive: rev([],[]). rev([H | T] , L) :- rev(T,Z), append(Z, [H],L). Accumulator: 2nd arg is ans so far rev2(L1, L2) := revzap(L1, [], L2). revzap([X|L],L2,L3) :- revzap(L,[X|L2],L3). revzap([],L,L). at end, copy 2nd to 3rd arg.

Delete one matching elt: efface(_, [],[]). efface(A, [A|L], L) :- !. efface(A, [B|L], [B|M]) :- efface(A,L,M). Delete all matching elts: new base case (keep on deleting!). mydelete(_,[],[]). mydelete(X, [X|L], M) :- ! , mydelete(X,L,M). mydelete(X, [Y|L], [Y|M]) :- mydelete(X,L,M).

PROLOG SAMPLES: Executions

efface(A, [A|L], L) :- ! . /*base case*/ efface(A, [B|L], [B|M]) :- efface(A,L,M). efface(_,[],[]). ?- efface(a, [b,a,c,a,d],X). Call: (8) efface(a, [b, a, c, a, d], _G332) ? creep Call: (9) efface(a, [a, c, a, d], _G416) ? creep Exit: (9) efface(a, [a, c, a, d], [c, a, d]) ? creep Exit: (8) efface(a, [b, a, c, a, d], [b, c, a, d]) ? creep X = [b, c, a, d] ; Redo: (8) efface(a, [b, a, c, a, d], _G332) ? creep No

BUG WITH NO CUT !

bugefface(A, [A|L], L). % no cut bugefface(A, [B|L], [B|M]) :- bugefface(A,L,M). bugefface(_,[],[]). ?- bugefface(a, [b, a,c, a, d],X). Call: (8) bugefface(a, [b, a, c, a, d], _G332) ? creep Call: (9) bugefface(a, [a, c, a, d], _G413) ? creep Exit: (9) bugefface(a, [a, c, a, d], [c, a, d]) ? creep Exit: (8) bugefface(a, [b, a, c, a, d], [b, c, a, d]) ? creep X = [b, c, a, d] ; Redo: (9) bugefface(a, [a, c, a, d], _G413) ? creep Call: (10) bugefface(a, [c, a, d], _G416) ? creep Call: (11) bugefface(a, [a, d], _G419) ? creep Exit: (11) bugefface(a, [a, d], [d]) ? creep Exit: (10) bugefface(a, [c, a, d], [c, d]) ? creep Exit: (9) bugefface(a, [a, c, a, d], [a, c, d]) ? creep Exit: (8) bugefface(a, [b, a, c, a, d], [b, a, c, d]) ? creep X = [b, a, c, d] ; Redo: (11) bugefface(a, [a, d], _G419) ? creep Call: (12) bugefface(a, [d], _G422) ? creep Call: (13) bugefface(a, [], _G425) ? creep Exit: (13) bugefface(a, [], []) ? creep Exit: (12) bugefface(a, [d], [d]) ? creep Exit: (11) bugefface(a, [a, d], [a, d]) ? creep Exit: (10) bugefface(a, [c, a, d], [c, a, d]) ? creep Exit: (9) bugefface(a, [a, c, a, d], [a, c, a, d]) ? creep Exit: (8) bugefface(a, [b, a, c, a, d], [b, a, c, a, d]) ? creep X = [b, a, c, a, d] ; Redo: (12) bugefface(a, [d], _G422) ? creep Redo: (11) bugefface(a, [a, d], _G419) ? creep Redo: (10) bugefface(a, [c, a, d], _G416) ? creep Redo: (9) bugefface(a, [a, c, a, d], _G413) ? creep Redo: (8) bugefface(a, [b, a, c, a, d], _G332) ? creep No

PROLOG SAMPLES: Mapping

Given a structure (like a list or tree), traverse it and produce a new, similar structure with elements that are transformations of the original elements.

Example with lists: alter a list by changing certain items in it by a set of rules.

  1. Make the head of the output list be the changed version of the head of the input list.
  2. Make the tail of the output list be the altered version of the tail of the input list.
  3. If at end of input list, terminate output list with empty list [].

Or... Altering a list with head H and tail T gives a list with head X and tail Y if:
changing item H gives item X, and
altering the list T gives the list Y.

If we want to substitute terms (which happen to be spelled like English words) in a list, we could see: %change rules: change(you,i). change(are, [am, not]). change(french, dutch). change(do, no). change(X,X). /* the catchall rule...no change */ % the list-altering program: alter([],[]). alter([H|T], [X|Y]) :- change(H,X), alter(T,Y).

So get -? alter([pretend you are french],Z). Z = [pretend i [am not] dutch]

PROLOG SAMPLES: SETS

member(X, [X | _ ]). member(X, [ _ | Y]) :- member(X,Y). subset([],Y). subset([A|X], Y) :- member(A,Y), subset(X,Y). disjoint(X,Y) :- not (( member(Z,X), member(Z,Y))). No duplicated elts in set. intersection([],X,[]). intersection ([X|R], Y, [X|Z]) :- member(X,Y), !, intersection(R,Y,Z). intersection([X|R],Y,Z):- intersection(R,Y,Z).

PERMUTATION

perm1([],[]). perm1(List, [Head | Tail]) :- append(V, [Head|U], List), % find me a (new) V, U such that...(!!) append(V,U,W), perm1(W,Tail).

LHS: To generate a permutation, successively (recursively) generate the next element to be head of the "output" list at this level. RHS: (1) To do that, pick it from somewhere in the list -- there is a leading sublist V in front of it and a trailing sublist U behind it. (2) Compose V and U into the new list from which to pick the next "head" element (in the next recursive level). (3) Permute that list, and put it at the tail of the "head" element at this level of recursion.

If you don't like this permutation, just hit ";" and Prolog will go find a different V (thus) U at all levels... N! times. "Nondeterministic programming".

QUICKSORT REVIEW

Quicksort is O(n log n)

(9 1 6 2 5 8 3 7 4 0)

pivot V (9 8 7) 6 (1 2 5 3 4 0) V V qsort qsort || \/ (9 8 7 6 5 4 3 2 1 0)

PERMUTATION SORTS

Bubble Sort: bubble (L,S) :- append(U,[A,B|V],L), B < A, !, append(U, [B,A|V],M), % !! find me some U, V such that... bubble(M,S). % above succeeds only if need swap bubble(L,L). % so need this(!)

Quicksort: qsort1 ([H|T],S) :- split(H,T,U1, U2), qsort1(U1, V1), qsort1(U2,V2), append (V1, [H | V2], S). qsort1([],[]). Split maps L = [H|T] to U1, U2. U1's elts ≤ H, U2's > H, order in same as in L. split(H,[H1|T], [H1|U1],U2) :- H1 < H, split(H,T, U1, U2). split(H,[H1|T], U1, [H1|U2]) :- H1 > H, split(H,T, U1, U2). split(_,[],[],[]).

WARREN'S APPEND-FREE QSORTS

Beware, from Prolog by Example, VERY typo-ridden book. Very! qsort2(L,S) :- sort2(L,S,[]). sort2([H|T],S,X) :- split(H,T,U1,U2), sort2(U1,S, [H|Y]), sort2(U2, Y, X). sort2([],X,X). Use list difference: qsort3 ([H|T],S-X) :- split(H,T,U1,U2), qsort3(U1, S-[H|Y]), qsort3(U2, Y-X). qsort3([],X-X). split(H,[H1|T], [H1|U1],U2) :- H1 < H, split(H,T, U1, U2). split(H,[H1|T], U1, [H1|U2]) :- H1 > H, split(H,T, U1, U2). split(_,[],[],[]). }

Using Accumulators

Accumulators can reduce copying and structure-building in recursive routines (see the assignment). They're common. Straightforward recursive List length: listlen([],0). listlen([H|T],N) :- listlen(T,Nb4), N is Nb4 +1. To use accumulator, always introduce second functor with extra argument that accumulates a partial answer: when done its other args match args of the original functor you wanted. Here: listlen(L,N) :- lenacc(L,0,N). /* new 3-arg func*/ lenacc([],A,A). /*copy accum. to 'output' (2nd arg to 3rd)*/ lenacc([H|T],A,N) :- A1 is A + 1, lenacc(T, A1,N).

N just gets carried along and is instantiated w/ special 'copy 2nd arg to 3rd' rule, then 2nd arg tossed out by first rule. Subgoals look like lenacc([a,b,c],0,N) /* all N's co-refer*/ lenacc([a,b],1,N) lenacc([a],2,N) lenacc([],3,N)

More!

Some More Advanced Features... structures, globals, cut, etc...

STRUCTURES 1

STRUCTURES 2

THE S AND P PUZZLE

There are two numbers M and N, 1 < M, N, < 100. I tell Mr. S their sum, and I tell Mr. P their product. S knows P knows the product, P knows S knows the sum.

Mr. P: I don't know the numbers.

Mr. S: Yeah, I knew you didn't. Neither do I.

Mr. P: Oh, now I know them!

Mr. S: Oh yeah? Well then so do I.

. . .
So, what are M and N?

Also very much to the point... How translate this English into Prolog?

THE S AND P PUZZLE

Home-Grown Operators

Declaring Operators:

Operators have:

Operator has descriptive atom: for infix: xfx, xfy, yfx, yfy, for prefix: fx, fy, for postfix: xf, yf. f is the operator, and x,y are the arguments. In the absence of brackets, the y argument can have operators of the same or lower precedence class than operator f; x means any operators in the argument must have strictly lower precedence than f. So with + declared as yfx, you can't have a + b + c interpreted as {a + (b + c) since the arg after + constains arg of same precedence: thus yfx means left associativity, and xfy means right associativity. E.g.---. :- op(650,yfx,'::'). :- op(675,xfy,'::='). Z ::= X :: Y :- append(X,Y,Z). ?- MyList ::= [a,b] :: [x,y,z]. MyList = [a, b, x, y, z] Yes

A HAT PUZZLE

Hat Puzzles

Prisoner and Hat Puzzles