Logic Programming contd Lists in Prolog are represented

Logic Programming (cont’d): Lists in Prolog are represented by a functor (similar to cons in Scheme): 1. The empty list is represented by the constant - [ ]. 2. The term. (X , Xs) represents a list of which X is the 1 st element and Xs is the rest of its elements. 3. The functor ‘. ’ has an infix representation by ‘|’ (pipe symbol): [X | Xs] For example: [3 | [5 | [] ] ] 4. Known lists can be represented as a list of comma-separated expressions: For example: [ X, 2, f(Y) ] or [3 , 5]

Logic Programming (cont’d): Lists Example 1: Implementing a Context Free Grammar in PROLOG: S NP VP Det N V NP | V a | the woman | man saw Q: What language does this CFG represent? ? - s([a, woman, saw, a, man]). true ? -s(X). X = [the, woman, saw, the, woman]; X = [the, woman, saw, the, man]; X = [the, woman, saw, a, woman]; X = [the, woman, saw, a, man]; X = [the, woman, saw] … ? -s([the, man|X]). X = [saw, the, woman]; X = [saw, the, man]; X = [saw, a, woman] … Representation in PROLOG: 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([saw]).

Logic Programming (cont’d): Lists Example 1: Implementing a Context Free Grammar in PROLOG: S NP Adj VP Det N V NP VP Det N | Det Adj N vicious | marvelous V NP | V a | the woman | man shoots Q: How would we implement the modifications above? Representation in PROLOG: s(Z) : - np(X), vp(Y), append(X, Y, Z). np(Z): - det(X), n(Y), append(X, Y, Z). np(Z): - det(X), adj(W), n(Y), append([X, W, Y], Z]). vp(Z): - v(X), np(Y), append(X, Y, Z). vp(Z): - v(Z). adj([vicious]). adj([marvelous]). det([the]). det([a]). vp(Z): - v(X), np(Y), append(X, Y, Z). vp(Z): - v(Z). n([woman]). n([man]). det([the]). det([a]). v([shoots]). n([woman]). n([man]). v([shoots]).

Logic Programming (cont’d): Lists Example 2: Determining the order of specific times o In logic programming, symbols have no values: ‘ 2’ and ‘ 9’ are names and not numbers. o We cannot use the < relation to compare the order of numbers (it is infinite!). o A finite relation can be represented. We would like to implement an order relation between pairs of the form (hour, weekday). % Signature: weekday_list(List)/1 % Purpose: Holds the ordered list of week days. weekday_list([d('Sun'), d('Mon'), d('Tue'), d('Wed'), d('Thu'), d('Fri'), d('Sat')]). % Signature: hour_list(List)/1 % Purpose: Holds the ordered list of days hours. hour_list([h(0), h(1), h(2), h(3), h(4), h(5), h(6), h(7), h(8), h(9), h(10), h(11), h(12), h(13), h(14), h(15), h(16), h(17), h(18), h(19), h(20), h(21), h(22), h(23)]). % Signature: precedes(A, B, List)/1 % Purpose: The element A precedes the element B in the List. precedes(A, B, List) : - append([_, [A], _, [B], _], List).

Logic Programming (cont’d): Lists Example 2: Determining the order of specific times weekday_list([d('Sun'), d('Mon'), d('Tue'), d('Wed'), d('Thu'), d('Fri'), d('Sat')]). hour_list([h(0), h(1), h(2), h(3), h(4), h(5), h(6), h(7), h(8), h(9), h(10), h(11), h(12), h(13), h(14), h(15), h(16), h(17), h(18), h(19), h(20), h(21), h(22), h(23)]). A predicate to identify weekdays: % Signature: is_weekday(Day)/1 % Purpose: Day is a day of the week. is_weekday(d(D)) : - weekday_list(Weekday_list), member(d(D), Weekday_list). An order relation between weekdays: % Signature: weekday_order(Day 1, Day 2)/2 % Purpose: weekday Day 1 precedes the weekday Day 2 % in some weekday_order(d(D 1), d(D 2)) : - is_weekday(d(D 1)), is_weekday(d(D 2)), weekday_list(Weekday_list), precedes(d(D 1), d(D 2), Weekday_list).

Logic Programming (cont’d): Lists Example 2: Determining the order of specific times weekday_list([d('Sun'), d('Mon'), d('Tue'), d('Wed'), d('Thu'), d('Fri'), d('Sat')]). hour_list([h(0), h(1), h(2), h(3), h(4), h(5), h(6), h(7), h(8), h(9), h(10), h(11), h(12), h(13), h(14), h(15), h(16), h(17), h(18), h(19), h(20), h(21), h(22), h(23)]). A predicate to identify hours: % Signature: is_hour(Hour)/1 % Purpose: Hour is an hour of the day. is_hour(h(H)) : - hour_list(Hour_list), member(h(H), Hour_list). An order relation between hours: % Signature: hour_order(H 1, H 2)/2 % Purpose: hour H 1 precedes the hour H 2 in some day. hour_order(h(H 1), h(H 2)) : - is_hour(h(H 1)), is_hour(h(H 2)), hour_list(Hour_list), precedes(h(H 1), h(H 2), Hour_list).

Logic Programming (cont’d): Lists Example 2: Determining the order of specific times A predicate to identify times: % Signature: is_time(T)/1 % Purpose: T is an hour of the week. is_time(h(H), d(D))) : - is_hour(h(H)), is_weekday(d(D)). An order relation between times: % Signature: time_order(T 1, T 2)/1 % Purpose: The time T 1 precedes the time T 2 in the week. time_order(time(h(H 1), d(D 1)), time(h(H 2), d(D 2))) : - is_time(h(H 1), d(D 1))), is_time(h(H 2), d(D 2))), weekday_order(d(D 1), d(D 2)). time_order(time(h(H 1), d(D)), time(h(H 2), d(D))) : - is_time(h(H 1), d(D))), is_time(h(H 2), d(D))), hour_order(h(H 1), h(H 2)). ? - is_time(h(1), d('Sun'))). true ? - time_order(time(h(5), d('Mon')), time(h(1), d('Tue'))). true

Logic Programming (cont’d): Lists Example 3: Deleting all occurrences of an element from a list. % Signature: delete(List, X, Has. No. Xs)/3 % Purpose: The list Has. No. Xs is the result of removing % all occurrences of X from the list List. % Example: ? - delete([2, 3, 2, 4, 5, 2, 4], 2, X). % X = [3, 4, 5, 4] delete([], _, []). delete([X|Xs], Z, [X|Ys]) : - X = Z, delete(Xs, Z, Ys). delete([X|Xs], X, Ys) : - delete(Xs, X, Ys).

Logic Programming (cont’d): Lists Example 5: Deleting one occurrence of an element from a list. % Signature: select(X, Has. Xs, One. Less. Xs)/3 % Purpose: The list One. Less. Xs is the list X without % one occurrence of X. select(Element, Has. Xs, One. Less. Xs) : - append([Prefix, [Element], Suffix], Has. Xs), append(Prefix, Suffix, One. Less. Xs). ? - select(4, [2, 3, 2, 4, 5, 2, 4], X). X = [2, 3, 2, 5, 2, 4]; X = [2, 3, 2, 4, 5, 2]; false

Logic Programming (cont’d): Lists Example 4: Merging sorted lists % Signature: lt(Obj 1, Obj 2)/2 % Purpose: Obj 1 precedes Obj 2 by some comparison criteria. lt(Time 1, Time 2) : - time_order(Time 1, Time 2). % Signature: merge(Xs, Ys, Zs)/3 % Purpose: Zs is the sorted merge of the sorted lists Xs and Ys. We assume that % there is a predicate "lt" of order between the elements of Xs and Ys. merge([X|Xs], [Y|Ys], [X|Zs]) : - lt(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]) : - lt(Y, X), merge([X|Xs], Ys, Zs). merge(Xs, [], Xs). merge([], Ys). %1 %2 %3 %4 %5 ? - merge([time(h(5), d('Sun')), time(h(5), d('Mon'))], X, [time(h(2), d('Sun')), time(h(5), d('Mon'))]). X = [time(h(2), d( 'Sun'))] ? - merge([time(h(1), d('Sun')), time(h(3), d('Wed')), time(h(5), d('Sat'))], [time(h(2), d('Sun')), time(h(3), d('Wed'))], Xs). Xs = [time(h(1), d('Sun')), time(h(2), d('Sun')), time(h(3), d('Wed')), time(h(5), d('Sat'))]; false

Logic Programming (cont’d): Lists Example 5: Merging sorted lists merge([X|Xs], [Y|Ys], [X|Zs]) : - lt(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]) : - lt(Y, X), merge([X|Xs], Ys, Zs). merge(Xs, [], Xs). merge([], Ys). merge( [ time(h(1), d('Sun')), time(h(3), d('Wed')), time(h(5), d('Sat'))], [ time(h(2), d('Sun')), time(h(3), d('Wed'))], Xs) merge([t 1, t 3, t 5], [t 2, t 3], Xs) {X_1=t 1, Xs_1=[t 3, t 5], Y_1=t 2, Ys_1=[t 3], Xs=[t 1|Zs_1]} 1 lt(t 1, t 2), merge([t 3, t 5], [t 2, t 3], Zs_1) 2 – failure branch… 3 – failure branch… * merge([t 3, t 5], [t 2, t 3], Zs_1) 1 – failure branch… 2 – failure branch… {X_2=t 3, Xs_2=[t 5], Y_2=t 2, Ys_2=[t 3], Zs_1=[t 2|Zs_2]} 3 lt(t 2, t 3), merge([t 3, t 5], [t 3], Zs_2) * merge([t 3, t 5], [t 3], Zs_2) 1 – failure branch… {X_3=t 3, Xs_3=[t 5], Y_3=t 3, Ys_3=[], Zs_2=[t 3, t 3|Zs_3]} 2 merge([t 5], [], Zs_3) {Xs_4=[t 5], Ys_4=[], Zs_3=[t 5]} 4 true 1 – failure branch…

Logic Programming (cont’d): Backtracking optimization The cut operator (!): Used to avoid redundant computations. Rule k: A : - B 1, …Bi, !, Bi+1, …, Bn. Rule k+1: A : - …. … Rule k Following selections of A rules are cut. ? - B 1’, …, Bi, !, Bi+1, …Bn’, Q 1’, …Qn’ * Further substitutions for goals before ‘!’ are cut. ? - !, Bi+1, …Bn’, Q 1’, …Qn’ ? - Bi+1, …Bn’, Q 1’, …Qn’ …

Logic Programming (cont’d): Backtracking optimization Example 7: Optimize the procedure merge. o When none of the first two params is [], only one of the rules can be valid, o Since either lt(Y, X), or lt(X, Y) or X==Y holds, exclusively. o Once one of the above holds, we can skip further rule selections. merge([X|Xs], [Y|Ys], [X|Zs]) : - lt(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]) : - lt(Y, X), merge([X|Xs], Ys, Zs). merge([t 1, t 3, t 5], [t 2, t 3], Xs) lt(t 1, t 2), !, merge([t 3, t 5], [t 2, t 3], Zs_1) 3 – failure branch… 2 – failure branch… !, merge([t 3, t 5], [t 2, t 3], Zs_1)

Logic Programming (cont’d): Backtracking optimization Example 8: red VS green cuts. o Green cut: Helps optimizing the program by avoiding redundant computations. o Red cut: Omits possible solutions. o Red cuts are undesired, unless specifically required (E. g. , to provide only the first solution). Q: How many possible solutions does the following query have? ? - merge([], X). X = []; X = [] The query matches both facts 4 and 5: merge(Xs, [], Xs). merge([], Ys). To avoid this, we add ‘!’ to fact 4, rewriting it as a rule: merge(Xs, [], Xs): - !. merge([], Ys). A red cut! The number of solutions has changed!

Logic Programming (cont’d): Meta circular interpreter Recall the second version of the meta-circular interpreter (seen in class): o Operates in two phases: 1. Preprocessing: The given program, P, is translated into a new program P’, which has a single predicate, rule, and consists of facts only. 2. Queries are applied to the new program using the procedure solve. % Signature: solve(Exp, Proof)/2 % Purpose: Interpret and associate proofs with success answers solve(true, true). %1 solve([], []). %2 solve([A|B], [Proof. A|Proof. B]) : - solve(A, Proof. A), %3 solve(B, Proof. B). solve(A, node(A, Proof)) : - rule(A, B), %4 solve(B, Proof). % Signature: parent(Father, Son)/2 % Purpose: Father is parent of Son parent(abraham, isaac). parent(isaac, jacob). % Signature: ancestor(Ancestor, Descendant)/2 % Purpose: Ancestor is ancestor of % Descendant ancestor(X, Y) : - parent(X, Y). ancestor(X, Z) : - parent(X, Y), ancestor(Y, Z). rule(parent(abraham, isaac), true). rule(parent(isaac, jacob), true). rule(ancestor(X, Y), [parent(X, Y)]). rule(ancestor(X, Z), [parent(X, Y), ancestor(Y, Z)]).

Logic Programming (cont’d): Meta circular interpreter % Signature: solve(Exp, Proof)/2 % Purpose: Interpret and associate proofs with success answers solve(true, true). %1 solve([], []). %2 solve([A|B], [Proof. A|Proof. B]) : - solve(A, Proof. A), %3 solve(B, Proof. B). solve(A, node(A, Proof)) : - rule(A, B), %4 solve(B, Proof). % Signature: parent(Father, Son)/2 % Purpose: Father is parent of Son parent(abraham, isaac). parent(isaac, jacob). % Signature: ancestor(Ancestor, Descendant)/2 % Purpose: Ancestor is ancestor of % Descendant ancestor(X, Y) : - parent(X, Y). ancestor(X, Z) : - parent(X, Y), ancestor(Y, Z). rule(parent(abraham, isaac), true). rule(parent(isaac, jacob), true). rule(ancestor(X, Y), [parent(X, Y)]). rule(ancestor(X, Z), [parent(X, Y), ancestor(Y, Z)]).

Logic Programming (cont’d): Meta circular interpreter Example 10: Applying the meta circular interpreter. member(X, [X|Xs]). member(X, [Y|Ys]) : - member(X, Ys ). % Signature: rule(Head, Body. List)/2 rule(member(X, [X|Xs]), []). rule(member(X, [Y|Ys]), [member(X, Ys)]). Preprocessing % Signature: solve(Exp)/1 % Purpose: Interpret with explicit management % of a waiting list of goals. solve(Goal) : - solve(Goal, []). %1 solve(true, Goals) : - solve([], Goals). %1 solve([], []). %2 solve([], [G|Goals]) : - solve(G, Goals). %3 solve([A|B], Goals) : - %4 extend_waiting_list([A|B], Goals, New. Goals), solve([], New. Goals). solve(A, Goals) : - rule(A, B), %5 solve(B, Goals).

solve(member(X, [a, b, c])) {Goal_1 = member(X, [a, b, c])} 1 solve(member(X, [a, b, c]), []) {A_2 = member(X, [a, b, c], Goals_1 = []} 4 rule(member(X, [a, b, c], B_2), solve(B_2, []) {X_3 = X, Y_3 = a, Ys_3 = [b, c], B_2 = [member(X, [b, c])]} {X = a, X_3 = a, Xs_3 = [b, c], B_2 = []} 1 solve([], []) 1 true {X = a} solve([member(X, [b, c])] , []) Rule 2 rule append([], Goals 1_4), solve(member(X, [b, c]), Goals 1_4) {A_4 = member(X, [b, c]), B_4 = [], Goals_4 = []} 3 {Goals 1_4 = []} Rule of append solve(member(X, [b, c]), []) {A_5 = member(X, [b, c]), Goals_5 = []} 4 rule(member(X, [b, c]), B_5), solve(B_5, []) {X = b, X_6 = b, Xs_6 = [c], B_5 = []} 1 solve([], []) 1 {X = b} true …