dg-publish: true

Q1: Head and Tail Extraction

اكتب ناتج (Output) كل استعلام من الاستعلامات دي عشان تفصل الـ Head عن الـ Tail:

?- [H|T] = [a, b, c, d].
?- [H|T] = [a].
?- [H1, H2|T] = [a, b, c, d].
?- [H|T] = [].
?- [A, B, C|T] = [a, b, c, d, e].
?- [A|T] = [a].
?- [H, _|T] = [1, 2, 3].
?- [_, _|T] = [x, y, z].

Solution (Head and Tail Extraction)

1) H = a, T = [b, c, d].
2) H = a, T = [].
3) H1 = a, H2 = b, T = [c, d].
4) false. % (Empty list has no Head or Tail)
5) A = a, B = b, C = c, T = [d, e].
6) A = a, T = [].
7) H = 1, T = [3].
8) T = [z].

Q2: List Unification

اكتب الناتج للاستعلامات دي (unification):

?- [X, 2, Y] = [1, 2, 3].
?- [X, 2|Y] = [1, 2, 3, 4].
?- [X, Y|Z] = [a].
?- [X|Y] = [1, 2, 3].
?- [X, Y, Z] = [a, b, c|T].
?- [a,B,c,D]=[A,b,C,d].
?- [[X,a]]=[b,Y].
?- [(a+X),(Y+b)]=[(W+c),(d+b)].
?- [[a],[B,c],[]]=[X,[b,c],Y].

Solution (List Unification)

1) X = 1, Y = 3.
2) X = 1, Y = [3, 4].
3) false. % (Left side expects at least 2 elements, right side has 1)
4) X = 1, Y = [2, 3].
5) X = a, Y = b, Z = c, T = [].
6) A = a, B = b, C = c, D = d.
7) false. % (Length mismatch inside the lists)
8) W = a, X = c, Y = d.
9) B = b, X = [a], Y = [].

Q3: Translate a List

استخدم الـ Facts دي اللي بترجم الأرقام لكلمات:

tran(1,one).
tran(2,two).
tran(3,three).
...

اكتب دالة listtran تاخد List فيها أرقام وترجع List فيها الكلمات المقابلة ليها.
وجربها على الاستعلام ده: ?- listtran([1, 2, 3], L).

Solution (Translate a List)

listtran([], []).
listtran([Hn|Tn], [He|Te]) :- 
    tran(Hn, He), 
    listtran(Tn, Te).

Query Output:

?- listtran([1, 2, 3], L).
L = [one, two, three].

Q4: List Membership

اكتب الـ Rules بتاعة الـ member وجرب الاستعلامات دي وطلع ناتجها:

?- member(c, [a, b, c, d]).
?- member(x, [a, b, c]).
?- member(X, [1, 2, 3]).

Solution (List Membership)

member(X, [X|_]).
member(X, [_|T]) :- member(X, T).

Queries:

1) true.
2) false.
3) X = 1 ; X = 2 ; X = 3.

Q5: List Concatenation (Append)

اكتب ناتج الاستعلامات دي باستخدام دالة append:

?- append([a, b, c], [1, 2, 3], L).
?- append([a, [b, c], d], [a, [], b], L).
?- append(L1, L2, [a, b, c]).
?- append(L, _, [1, 2, 3, 4, 5, 6, 7, 8]), length(L, 5).
?- append([A], L1, [1, 2, 3, 4]).

Solution (Append)

1) L = [a, b, c, 1, 2, 3].
2) L = [a, [b, c], d, a, [], b].
3) L1 = [], L2 = [a, b, c] ;
   L1 = [a], L2 = [b, c] ;
   L1 = [a, b], L2 = [c] ;
   L1 = [a, b, c], L2 = [] ;
   false.
4) L = [1, 2, 3, 4, 5].
5) A = 1, L1 = [2, 3, 4].

Q6: Adding and Deleting Items

اكتب الـ Rule الخاصة بإضافة عنصر (Add) والـ Rule الخاصة بمسح عنصر (Delete).
وهات ناتج الـ Queries دي:

?- add(5, [1, 2], L).
?- del(c, [a, b, c, d], L).
?- del(a, [a, b, a, c, a], L).

Solution (Adding and Deleting)

% Add relation
add(X, L, [X|L]).

% Delete relation
del(X, [X|Tail], Tail).
del(X, [Y|Tail], [Y|Tail1]) :- del(X, Tail, Tail1).

Queries:

1) L = [5, 1, 2].
2) L = [a, b, d].
3) L = [b, a, c, a] ;
   L = [a, b, c, a] ;
   L = [a, b, a, c]. % (Notice how it deletes one occurrence at a time!)

Q7: Insert an item

استخدم الـ del بالعكس عشان تعمل Predicate اسمها insert(X, L, BiggerList) وظيفتها إنها تضيف العنصر X في أي مكان في الـ List.
وجرب الاستعلام: ?- insert(a, [c, d], B).

Solution (Insert)

insert(X, L, BiggerList) :- del(X, BiggerList, L).

Query Output:

?- insert(a, [c, d], B).
B = [a, c, d] ;
B = [c, a, d] ;
B = [c, d, a].

Q8: Sublist and Permutations

هات النواتج:

?- sublist([c, e], [a, b, c, d, e, f]).
?- sublist([a, b], [c, a, b, d]).
?- sublist(S, [a, b, d]).
?- perm([a, b, c], P).

Solution (Sublist and Permutations)

1) false. % (They are not consecutive)
2) true.
3) S = [] ; S = [a] ; S = [a, b] ; S = [a, b, d] ; S = [] ; S = [b] ; S = [b, d] ; S = [] ; S = [d] ; S = [].
4) P = [a, b, c] ;
   P = [a, c, b] ;
   P = [b, a, c] ;
   P = [b, c, a] ;
   P = [c, a, b] ;
   P = [c, b, a].

Q9: Delete the last 3 elements

اكتب استعلام (Query) باستخدام الـ append بيمسح آخر 3 عناصر من أي List ويسيب الباقي في متغير جديد, وجربها على [a, b, c, d, e, f].

Solution (Delete last 3 elements)

الفكرة إن الـ List الأصلية هي عبارة عن دمج (Append) للباقي اللي عايزينه، مع 3 عناصر في الآخر ملهمش لازمة [_, _, _].

?- append(NL, [_, _, _], [a, b, c, d, e, f]).
NL = [a, b, c].

Q10: Mathematical Functions

اكتب Prolog Predicates تنفذ المعادلات دي:

y = (x + 3)^2 + 3x
الدالة المنطقية F = AB' + C

Solution (Mathematical Functions)

1. Mathematical Equation:

calculate_y(X, Y) :-
    Y is (X + 3) * (X + 3) + 3 * X.

2. Boolean Function:

% Define boolean operations first
not(1, 0).
not(0, 1).

and(0, 0, 0).
and(0, 1, 0).
and(1, 0, 0).
and(1, 1, 1).

or(0, 0, 0).
or(0, 1, 1).
or(1, 0, 1).
or(1, 1, 1).

% The function
bool_f(A, B, C, F) :-
    not(B, B_Not),
    and(A, B_Not, Term1),
    or(Term1, C, F).

Q11: One_more Predicate

اكتب Predicate اسمها one_more(L1, L2) بتاخد List من الأرقام، وتطلع List تانية كل عنصر فيها بيزيد بواحد عن العنصر المقابل ليه في الـ List الأولى.
one_more([1, 2, 3], L). يخلي L = [2, 3, 4].

Solution (One_more Predicate)

one_more([], []).
one_more([H1|T1], [H2|T2]) :-
    H2 is H1 + 1,
    one_more(T1, T2).

Q12: True/False (From Lecture)

حدّد إذا كانت الجمل دي صح ولا غلط مع تصحيح الغلط:

?-X is 9//2. Prolog will respond to this query as: X=4.5
?-X = 9/3. Prolog will respond to this query as: X=3
?-X is 9/2. Prolog will respond to this query as: X=4.5
member( b, [a,b,c]).
?- append(L, [C], [1,3,4]). Prolog will respond to this query as: L=[1,3], C=4
?- L=[1,2], L1=[5|L]. Prolog will respond to this query as: L1 = [5, 1, 2].
?- L=[1,2], L1=[5|L]. Prolog will respond to this query as: L1 = [1, 2, 5].

Solution (True/False)

Q13: MCQs (From Lecture)

1. ?- L=[a, b, c, d, e], append([ _ ], [A1,A2| _ ], L).

a) A1 = c, A2 = d.
b) A1 = b, A2 = c.
c) A1 = d, A2 = e.
d) A1 = a, A2 = b.

2. Write a goal to delete the first two elements from a list L producing list L2.

a) ?-append( _ , _ , L2, L)
b) ?-append([ _ , _ ], L2, L)
c) ?-append(L, [ _ , _ ], L2)
d) ?-append([ _ , _ ], L, L2)

3. To find the output of the function X = A^2 + AB. In prolog, we can write:

a) f(A,B,X):- X = A^2+A*B.
b) f(A,B,X):- X = A*A+A*B.
c) f(A,B,X):- X is A^2+AB.
d) f(A,B,X):- X is A^2+A*B.

4. Write a rule to find the last 2 elements of the list.

a) last2(L,A,B):- append(_,A,B,L).
b) last2(L,A,B):- append(_,(A1,A2),L).
c) last2(L,A,B):- append(L,_,[A,B]).
d) last2(L,A,B):- append(_,[A,B],L).

5. Write a goal to find the last 2 elements of the list.

a) ?-append( L, A, B)
b) ?-append(_,[A, B],L)
c) ?-append(_, (A,B], L)
d) ?-append([A,B], _, L)

6. To find the output of the Boolean function F = B' + A. In prolog, we can write:

a) fun(A, B, F):- not(B, X ), and(X, A, F).
b) fun(A, B, F):- or(B', A ).
c) fun(A, B, F):- not(B, X ), or(X, A, F).
d) fun(A, B, F):- not(B ), or(A, B, F).

Solution (MCQs)

Q14: List Built-in / Arithmetic Operations

استنتج نواتج الدوال الجاهزة والعمليات الحسابية دي

?- length([a,b,c,d,e,[a,x],t], X).
?- sumlist([1,2,3,4], N).
?- even([a,b,c,d]).
?- reverse([1,2,3], L).
?- msort([a,c,d,b], A).

Solution (Operations)

1) X = 7. % (الـ list اللي جوه `[a,x]` بتتحسب كعنصر واحد).
2) N = 10. % (مجموع 1+2+3+4).
3) true. % (عدد العناصر 4، وده عدد زوجي).
4) L = [3, 2, 1]. % (عكس الترتيب).
5) A = [a, b, c, d]. % (ترتيب أبجدي عادي).

Chapter 5 (Part 2) - Practice

Q1: Head and Tail Extraction

Q2: List Unification

Q3: Translate a List

Q4: List Membership

Q5: List Concatenation (Append)

Q6: Adding and Deleting Items

Q7: Insert an item

Q8: Sublist and Permutations

Q9: Delete the last 3 elements

Q10: Mathematical Functions

Q11: One_more Predicate

Q12: True/False (From Lecture)

Q13: MCQs (From Lecture)

Q14: List Built-in / Arithmetic Operations

Nour Eldeen Mahmoud