Logic Programming – Part 2• Lists• Backtracking Optimization (via the cut

operator)• Meta-Circular Interpreters

Lists – Basic Examples[] – The empty list[X,2,f(Y)] – A 3 element list[X|Xs] – A list starting with X. Xs

is a list as well.

Example - [3,5] may be written as [3|5|[]]

Lists – CFG with Prolog

Question: Which sentences can be constructed using this grammar?

s -> np vpnp -> det nvp -> v np | vdet -> a | then -> woman | manv -> shoots

Lists – CFG with PrologLets make relations out of it:

s(Z) :- np(X), vp(Y), append(X,Y,Z). np(Z) :- det(X), n(Y), append(X,Y,Z). vp(Z) :-  v(X), np(Y), append(X,Y,Z).vp(Z) :-  v(Z). det([the]).det([a]). n([woman]).n([man]). v([shoots]).

s -> np vpnp -> det nvp -> v np | vdet -> a | then -> woman | manv -> shoots

Lists – CFG with PrologWe can ask simple queries like:

Prolog generates entire sentences!

s([a,woman,shoots,a,man]).yes

?-s(X). X = [the,woman,shoots,the,woman] ;X = [the,woman,shoots,the,man] ;X = [the,woman,shoots,a,woman] ;X = [the,woman,shoots,a,man] ;X = [the,woman,shoots] …

?-s([the,man|X]). X = [the,man,shoots,the,woman] ;X = [the,man,shoots,the,man] ;X = [the,man,shoots,a,woman] …

Lists – CFG with PrologQuestion: Add a few rules to the grammar

What should we change in the code?Answer: we add the following code

s -> np vpnp -> det n | det adj nvp -> v np | vdet -> a | then -> woman | manv -> shootsadj -> vicious | marvelous

np(Z) :- det(X), adj(W), n(Y), append([X,W,Y],Z).adj([vicious]).adj([marvelous]).

Lists – The date RelationIn this example we’ll work with

datesWe assume, for simplicity that a

date comprises of a week day and an hour

We define the possible week days and hours with lists:

week_day(['Sun', 'Mon', 'Tue','Wed','Thu','Fri','Sat']).hour([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]).

Lists – The date Relation

Question: How can we tell if hour 2 is before hour 9?

Answer: 1. We can only do so by checking

precedence in the lists above2. A < relation isn’t really possible to

implement(There’s a more detailed answer in the PS document)

week_day(['Sun', 'Mon', 'Tue','Wed','Thu','Fri','Sat']).hour([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24]).

Lists – The date Relation

Some queries:

date([H,W]) :- hour(Hour_list), member(H, Hour_list), week_day(Weekday_list), member(W, Weekday_list). dateLT( date([_,W1]), date([_,W2]) ) :- week_day(Weekday_list),

precedes(W1,W2,Weekday_list).dateLT( date([H1,W]), date([H2,W]) ) :- hour(Hour_list),

precedes(H1,H2,Hour_list).date([1,'Sun']).true dateLT(date([5,'Mon']), date([1,'Tue'])).true

Lists – The date Relationprecedes is defined using append

/2

Notice that the first argument is a list of lists

This version of append is a strong pattern matcher

precedes(X,Y,Z) :- append( [_,[X],_,[Y],_] , Z).

Lists – Merging date ListsMerge 2 ordered date-lists

% Signature: merge(Xs, Ys, Zs)/3% purpose: Zs is an ordered list of dates obtained% by merging the ordered lists of dates Xs and Ys.merge([X|Xs] , [Y|Ys] , [X|Zs]) :- dateLT(X,Y), merge(Xs, [Y|Ys] ,Zs).merge([X|Xs] , [X|Ys] , [X,X|Zs]) :- merge(Xs, Ys, Zs).merge([X|Xs],[Y|Ys],[Y|Zs]) :- dateLT(Y,X), merge( [X|Xs] ,Ys, Zs). merge(Xs,[ ], Xs). merge([ ],Ys, Ys).?- merge( [date([5,'Sun']), date([5,'Mon'])], X,

[date([2, 'Sun']), date([5,'Sun']), date([5, 'Mon'])]).X = [date([2, 'Sun'])]

merge([d1,d3,d5],[d2,d3],Xs)

{X_1=d1,Xs_1=[d3,d5], Y_1=d2,Ys_1=[d3],Xs=[d1|Zs_1] } Rule 1 dateLT(d1,d2),

merge([d3,d5], [d2,d3] ,Zs_1)

merge([d3,d5], [d2,d3] ,Zs_1)

true

merge([d3,d5], [d3] ,Zs_2)

dateLT(d2,d3), merge([d3,d5], [d3] ,Zs_2)

Rule 2 – failure branch…Rule 1 – failure branch…

{ X_3=d3,Xs_3=[d5],Ys_3=[],Zs_2=[d3,d3|Zs_3] } Rule 2

Rule 1 – failure branch…

merge([d5], [] ,Zs_3)

{ Xs_4=[d5], Zs_3=[d5] } Fact 4

Rule 2 – failure branch…

Rule 3 – failure branch…

Rule 3 – failure branch…

{ X_2=d3,Xs_2=[d5],Y_2=d2,Ys_2=[d3],Zs_1=[d2|Zs_2]} Rule 3

Backtracking Optimization - CutThe cut operator (denoted ‘!’) allows to prune trees from unwanted branches.

A cut prunes all the goals below itA cut prunes all alternative solutions of

goals to the left of itA cut does not affect the goals to it’s

right

The cut operator is a goal that always succeeds

Example - Merge with Cut In the merge example, only 1 of the 3 first

rules can be true. There is no reason to try to others.

Modify rule 1:merge([X|Xs] ,[Y|Ys], [X|Zs]) :- dateLT(X,Y), !, merge (Xs, [Y |Ys],Zs).

merge([d1,d3,d5],[d2,d3],Xs)

dateLT(d1,d2),!, merge([d3,d5], [d2,d3] ,Zs_1)

merge([d3,d5], [d2,d3] ,Zs_1)

!, merge([d3,d5], [d2,d3] ,Zs_1)

Rule 2 – failure branch…

Rule 3 – failure branch…

Another ExampleHow many results does this query

return?

Why does this happen?The query fits both rules 4 and 5How can we avoid this?Add cut to rule 4

?- merge([],[],X) .

merge(Xs, [ ],Xs) :- !.

X = [];X = [];No

Meta-Circular InterpretersWe have seen 3 different

interpreters in class

Version 1 is trivial

We can’t control the computation this way

solve( A ) :- A.

Interpreter Version 2

clause finds the first rule unifying with A with body B

% Signature: solve(Goal)/1% Purpose: Goal is true if it is true when posed to the original program P.

solve(true).solve( (A, B) ) :- solve(A), solve(B).solve(A) :- A\=true, clause(A, B), solve(B).

?- clause( parent(X,isaac),Body).X = abraham Body = true

?- clause(ancestor(abraham, P),Body).P = Y, Body = parent(abraham, Y) ;P = Z, Body = parent(abraham, Y), ancestor(Y, Z)

{<A_1 = ancestor(abraham, P)>}Rule 3 solve

solve(ancestor(abraham, P))

clause(ancestor(abraham, P), B_1), solve(B_1) { <B_1 = parent(abraham, P)>,

<X_2 = abraham>, <Y_2 = P> }Rule 1 ancestor

solve(parent(abraham, P))

{<A_3 = parent(abraham, P)>}Rule 3 solve  

{<P = issac>, <B_3 =true>}Fact 1 parent  

solve(parent(abraham,Y_2), ancestor(Y_2, P))

{ <B_1 = parent(abraham,Y_2), ancestor(Y_2, P)> }Rule 2 ancestor 

{<A_3 = parent(abraham,Y_2)><B_3 = ancestor(Y_2, P>}Rule 2 solve 

clause(parent(abraham, P), B_3), solve(B_3).

solve(true)

true

solve( parent(abraham,Y_2)), solve(ancestor(Y_2, P))

clause(parent(abraham, Y_2), B_4), solve(B_4)

solve(ancestor(Y_2, P))Fact 1 solve  

{<P = issac>}

 

{<A_4 = parent(abraham,Y_2)>}Rule 3 solve 

{<Y_2 = issac>, <B_4 =true>}Fact 1 parent  solve(true) ,

solve(ancestor(issac, P))

Interpreter Version 3In this version we control the

goal selection order by using a stack of goals

Preprocessing – The given program is converted into a program with a single predicate rule

Queries are checked against the new program

Interpreter Version 3

Sample converted program:

% Signature: solve(Goal)/1% Purpose: Goal is true if it is true when posed to the original program P.1. solve(Goal) :- solve(Goal, []). % Signature: solve(Goal, Rest_of_goals)/21. solve([],[]).2. solve([],[G | Goals]):- solve(G, Goals).3. solve([A|B],Goals):- append(B, Goals, Goals1), solve(A, Goals1).4. solve( A, Goals) :- rule(A, B), solve(B, Goals).

%rule (Head, BodyList)/21. rule( member(X, [X|Xs] ), [] ).2. rule( member(X, [Y|Ys] ), [member(X, Ys)] ).

{ <X_3=X>,<Y_3= a>,<Ys_3=[b, c]>,<B_2 = [member(X, [b,c])] > } Rule 2 rule

{ <A_4= member(X, [b,c])>, <B_4=[]>, <Goals_4=[]> } Rule 3 rule

{ <Goals1_4=[]>}Rule of append

{ <A_5=member(X,[b,c])>, <Goals_5=[]>}Rule 4 solve

{ <X=b>,<X_6 = b>, <Xs_6=[c]>,<B_5 = []> }Rule 1 rule

{<Goal_1 = member(X, [a, b, c])>}Rule 1 solve

solve(member(X, [a, b, c]))

solve(member(X, [a, b, c]), []){ <A_2 = member(X, [a, b, c]>,<Goals_1 = []> }Rule 4 solve rule(member(X, [a, b, c], B_2),

solve(B_2, [])

solve([],[])

{ <X=a>,<X_3 = a>,<Xs_3=[b, c]>,<B_2 = []> }Rule 1 rule

true

{<X=a>}

Rule 1 solvesolve([member(X, [b,c])], [])

append([], [], Goals1_4), solve(member(X, [b,c]), Goals1_4).

solve(member(X, [b,c]), []).

rule(member(X,[b,c]), B_5), solve(B_5, [])

solve([],[])

true

{<X=b>}

Rule 1 solve