Prolog - 3 Prolog search trees + traces

Transcription

1 Prolog - 3 Prolog search trees + traces 1

2 Review member(a, [A B]). member(a, [B C]) :- member (A,C). goal-oriented semantics: can get value assignment for goal member(a,[b C]) by showing truth of subgoal member(a,c) and retaining value bindings of the variables procedural semantics: think of head of clause as procedure entry, terms as parameters. then body consists of calls within this procedure to do the calculation. variables bindings are like returned values. 2

3 Example?- member(a,[a,b]). yes?- member(a,[b,c]). no?- member(x,[a,b,c]). X = a ; X = b ; X = c ; no Invertability of Prolog arguments 1. member(a, [A B] ). 2. member(a, [B C]) :- member (A, C). 3

4 Example?- member(a,[b, c, X]). X= a ; no?- member(x,y). X = _123 Y = [ X _124] ) ; X = _123 Y = [_125, X _126] ; X = _123 Y = [ _127, _128, X _129] 1. member(a, [A B] ). 2. member(a, [B C]) :- member (A, C). Lazy evaluation of unbounded list structure. An unbound X as first element, second element, third element, etc. 4

5 Prolog Search Tree X=A=a,B=[b,c] member(x,[a,b,c]) X=a success X=A =b,b =[c] X=b success 1. member(a, [A B] ). 2. member(a, [B C]) :- member (A, C). X=A,B=a,C=[b,c] member(x,[b,c]) X=A =c,b =[ ] X=c success X=A,B =b,c =[c] member(x,[c]) X=A B =c, C =[ ] member(a,[ ]) fail fail 5

6 1. member(a, [A B] ). 2. member(a, [B C]) :- member (A, C).?- member(x, [a,b,c]). match rule 1. member(a, [A B] ) so X = A = a, B = [b,c] X = a ; match rule 2. member(a, [B C]) so X = A, B = a, C = [b,c] then evaluate subgoal member(x, [b,c]) match rule 1. member(a,[a B ]) so X = b, B = [c] X = b ; match rule 2. member(a,[b C ]) so X = A, B = b, C = [c] then evaluate subgoal member(x, [c]) match rule 1. member(a,[a B ]) so X=A = c, B =[ ] X = c ; match rule 2. member(a,[b C ]) so X=A, B =c, C =[ ], but member(x, [ ]) is unsatisfiable, no 6

7 Another Search Tree member(a, [b,c,x]) fail, a can t unify with b B = b, C = [c,x] member(a,[c, X]) B =c, C =[X] fail, a can t unify with c 1. member(a, [A B] ). 2. member(a, [B C]) :- member (A, C). X=a, B = [ ] success member(a,[ X]). fail, can t unify [ ] with a list X=B, C = [ ] member(a,[ ]) fail, ditto 7

8 Prolog Search Trees Really have built an evaluation tree for the query member(x,[a,b,c]). Search trees provide a formalism to consider all possible computation paths Leaves are success nodes or failures where computation can proceed no further By convention, to model Prolog, leftmost subgoal is tried first 8

9 Prolog Search Trees, cont. Label edges with variable bindings that occur by unification There can be more than one success node There can be infinite branches in the tree, representing non-terminating computations (performed lazily by Prolog); lazy evaluation implies only generate a node when needed. 9

10 Another Member_of Function Equivalent set of rules: don t care variables mem(a, [A _ ] ). mem(a, [ _ C]) :- mem(a,c). Can examine search tree and see the variables which have been excised were auxiliary variables in the clauses. 10

11 Calculus Example diff(x,1). diff(a,0):-const(a). diff(+(a,b), +(DA,DB)) :- diff(a,da),diff(b,db). diff(sin(a),*(cos(a),da)):- diff(a,da). const(a):-number(a). const(a):-atom(a),\+(a==x). 11

12 Backtracking eat(wolf, lamb). eat(lamb,grass). eat(lion,x):- eat(x,food),plant(food). 12

13 Backtracking eat(wolf, lamb). eat(lamb,grass). eat(lion,x):-eat(x,food),plant(food). plant(grass). Y = X X = wolf, Food = lamb eat(x, Food) eat(lion,y) plant(food) eat(wolf, lamb) fail lamb succeed 13

14 Backtracking eat(wolf, lamb). eat(lamb,grass). eat(lion,x):-eat(x,food),plant(food). plant(grass). Y = X eat(x, Food) X = lamb, Food = grass eat(lion,y) eat(lamb, grass) grass plant(food) succeed succeed 14

15 Tracing in Sicstus Prolog?-[mem]. %loads file mem.pl?-trace. %turns on tracing facility [trace]?- memberof(x,[a,b,c]). (1) 1 Call: memberof(_6783,[a,b,c])? (1) 1 Exit: memberof(a,[a,b,c])? X = a ; (1) 1 Redo: memberof(a,[a,b,c])? (2) 2 Call: memberof(_6783,[b,c])? (2) 2 Exit: memberof(b,[b,c])? (1) 1 Exit: memberof(b,[a,b,c])? X = b ; 1. memberof(a, [A B] ). 2. memberof(a, [B C]) :- memberof (A, C). Matches memberof(a,[a b,c]) with rule #1. Asked for another answer, tries using rule #2. Tries memberof(x,[b,c]) and matches memberof(b,[b c]) with rule #1. 15

16 Tracing in Sicstus Prolog (1) 1 Redo: memberof(b,[a,b,c])? (2) 2 Redo: memberof(b,[b,c])? (3) 3 Call: memberof(_6783,[c])? (3) 3 Exit: memberof(c,[c])? (2) 2 Exit: memberof(c,[b,c])? (1) 1 Exit: memberof(c,[a,b,c])? X = c ; (1) 1 Redo: memberof(c,[a,b,c])? (2) 2 Redo: memberof(c,[b,c])? (3) 3 Redo: memberof(c,[c])? (4) 4 Call: memberof(_6783,[])? (4) 4 Fail: memberof(_6783,[])? (3) 3 Fail: memberof(_6783,[c])? (2) 2 Fail: memberof(_6783,[b,c])? (1) 1 Fail: memberof(_6783,[a,b,c])? no Asked for another answer, tries memberof(x,[b,c]) using rule #2. Tries memberof(x,[c]) and matches memberof(c,[c]) by rule #1. Finds no more choices to try 1. member(a, [A B] ). 2. member(a, [B C]) :- member (A, C). 16

17 Tracing in Prolog call 1.Memberof(A, [A B]). exit success fail 2.Memberof(A,[B C]):- Memberof(A,C) redo 17

18 Still More Append Generating an unbounded number of lists?- append(x,[b],y). X = [ ] Y = [b] ; X = [ _169] Y = [ _169, b] ; X = [ _169, _170 ] Y = [ _169, _170, b] ; etc. 18

19 Generate and Test Paradigm in which Prolog rules generate potential solutions and then test them for the desired properties Used often in simulations with lots of alternatives 19

20 n Queens Problem is given an n by n chessboard, place each of n queens on the board so that no queen can attack another in one move In chess, queens can move either vertically, horizontally or diagonally. This problem is a classic generate and test problem Code on lecture notes webpage 20

21 n Queens not(x):- X,!, fail. %same as saw in class not(_). in(h,[h _]). %same as our member_of in(h,[_ T]):- in(h,t). %%%nums generates a list of integers between two other numbers, L,H by putting the first number at the front of the list returned by a recursive call with a number 1 greater than the first. It only works when the first argument is bound to an integer. It stops when it gets to the higher number. nums(h,h,[h]). nums(l,h,[l R]):- L<H, N is L+1, nums(n,h,r). %%% The number of queens/size of board - use 4 queen_no(4). 21

22 n Queens %%% ranks and files generate the x and y axes of the chess board. Both are lists of numbers up to the number of queens; that is, ranks(l) binds L to the list [1,2,3,,#queens]. ranks(l):- queen_no(n), nums(1,n,l). files(l):- queen_no(n), nums(1,n,l). %%% R is a rank on the board; selects a particular rank R from the list of all ranks L. rank(r):- ranks(l), in(r,l). %%% F is a file on the board; selects a particular file F from the list of all files L. file(f):- files(l), in(f,l). 22

23 n Queens %%% Squares on the board are (rank,file) coordinates. attacks decides if a queen on the square at rank R1, file F1 attacks the square at rank R2, file F2 or vice versa. A queen attacks every square on the same rank, the same file, or the same diagonal. attacks((r,_),(r,_)). attacks((_,f),(_,f)). %a Prolog tuple attacks((r1,f1),(r2,f2)):- diagonal((r1,f1),(r2,f2)). %%%can decompose a Prolog tuple by unification (X,Y)=(1,2) results in X=1,Y=2; tuples have fixed size and there is not head-tail type construct for tuples x same diagonal, diagonal same rank safe placement same file 23

24 n Queens %%% Two squares are on the same diagonal if the slope of the line between them is 1 or -1. Since / is used, real number values for 1 and -1 are needed. diagonal((x,y),(x,y)). %degenerate case, 0 length diag diagonal((x1,y1),(x2,y2)):- N is Y2-Y1,D is X2-X1, Q is N/D, Q is 1.0E+00. %diagonal needs bound args diagonal((x1,y1),(x2,y2)):- N is Y2-Y1, D is X2-X1, Q is N/D, Q is -1.0E+00. %%%because of use of is, diagonal is NOT invertible. 24

25 n Queens %%%placement can be used as a generator. If placement is called with a free variable, it will construct every possible list of squares on the chess board. The first predicate will allow it to establish the empty list as a list of squares on the board. The second predicate will allow it to add any (R,F) pair onto the front of a list of squares if R is a rank of the board and F is a file of the board. placement first generates all 1 element lists, then all 2 element lists, etc. Switching the order of predicates in the second clause will cause it to try varying the length of the list before it varies the squares added to the list placement([]). placement([(r,f) P]):- placement(p), rank(r), file(f). 25

26 n Queens %%%these two routines check the placement of the next queen %%%Checks a list of squares to see that no queen on any of them would attack any other. does by checking that position j doesn t conflict with positions (j+1),(j+2) etc. ok_place([]). ok_place([(r,f) P]):- no_attacks((r,f),p),ok_place(p). %%% Checks that a queen at square (R,F) doesn't attack any square (rank,file pair) in list L; uses attacks predicate defined previously no_attacks(_,[]). no_attacks((r,f),[(r2,f2) P]):- not(attacks((r,f),(r2,f2))), no_attacks((r,f),p). 26

27 n Queens %%% This solution works by generating every list of squares, such that the length of the list is the same as the number of queens, and then checks every list generated to see if it represents a valid placement of queens to solve the N queens problem; assume list length function queens(p):- queen_no(n), length(p,n), placement(p), ok_place(p). generate code given first test code follows 27

28 Unification, Informally Intuitively, unification between 2 Prolog terms tries to assign values to the variables so that the resulting trees, representing the terms, are isomorphic (including matching labels) To use a Prolog rule, we must unify the head of the rule with the subgoal to be proved, matching term by term 28

29 Unification, Informally Given a subgoal <functor>(<term>{, <term>}) how to unify it with a clause head? Rule and subgoal have same name Any uninstantiated variable matches any term If term is also an uninstantiated variable, this means if either takes on a value, they both do Integer and symbolic constants match themselves, only A structured term matches another term iff Has same relation name Has same number of components and corresponding components match 29

30 Unification Unification looks for the most general (or least restrictive) value to assign A substitution (s ) is a finite map from variables to terms in the language append([a B],Y,[A Z]):-...?- append([a,b],[c],w) s: A=a, B=[b], Y=[c], W=[a Z] A term U is an instance of another term T, if there is a substitution s such that U = T s. 30

31 Unification Two terms S,T unify if they have a common instance U (that is, S s 1 = T s 2 = U) Note: if variable X is contained in both S and T, then s 1 and s 2 both must have the same substitution for X. If two terms unify, they can be made identical under some substitution 31

32 Unification There may be more than one substitution to unify two terms times(z,times(y,7)) and times(4,w) s 1 : Z=4, Y=plus(3,5), W=times(plus(3,5),7) s 2 : Z=4, W=times(Y,7) Which substitution is simpler? less restrictive on the values of the variables? s 2 32

33 Most General Unifier We say g is the most general unifier (mgu) of two terms T, W iff for all other unifiers s of T,W, T s is an instance of T g. therefore, s can be obtained by a substitution d applied to g, s = g d?- member(a,b) returns A=_123, B=[A _ ] when it could return A= _123, B=[ A,b] or A=_123, B= [A, c, d] etc. Note, the 2nd and 3rd B values are obtainable from the mgu by additional substitutions 33