% Exercitiul 1 [rezolvat] % Fie un predicat employee_salary/2 care configureaza un nume de angajat si salariul acestuia. % De exemplu, employee_salary(peter, 1000) % Primind o lista de elemente employee_salary, determinati numele tuturor angajatilor care castiga o suma minima configurata. % veti scrie un predicat employee_with_good_salary(InputList, MinimumSalary, ResultList). % de exemplu % ?- employee_with_good_salary([employee_salary(peter, 1000), employee_salary(oliver, 1200), employee_salary(sam, 700), employee_salary(ben, 15000), employee_salary(sandra, 4500)], 1200, Res). % Res = [oliver, ben, sandra] employee_with_good_salary(InputList, MinSalary, Result) :- findall(Name, (member(employee_salary(Name, Salary), InputList), Salary >= MinSalary), Result). % Exercitiul 2 [rezolvat] % Primind o lista identica celei de la exercitiul 1, dar cu aparitii multiple ale unui angajat, % (considerand ca fiecare intrare reprezinta salariul intr-o anumita luna) % determinati salariul mediu al fiecarui angajat. % de exemplu, daca peter apare employee_salary(peter, 1000), respectiv employee_salary(peter, 1200), in lista de rezultat va fi doar employee_salary(peter, 1100) % ?- employee_average_salary([employee_salary(peter, 1000), employee_salary(oliver, 1200), employee_salary(sam, 700), employee_salary(oliver, 800), employee_salary(sam, 900), employee_salary(sandra, 4500), employee_salary(peter, 1200), employee_salary(oliver, 400)], Result). % Result = [(peter, 1100), (oliver, 800), (sam, 800), (sandra, 4500)] sum_and_count_by_name(_, [], 0, 0) :- !. sum_and_count_by_name(Name, [employee_salary(Name, CurrentSalary) | Tail], Sum, Count) :- sum_and_count_by_name(Name, Tail, SumTail, CountTail), Sum is SumTail + CurrentSalary, Count is CountTail + 1, !. sum_and_count_by_name(Name, [_ | Tail], Sum, Count) :- sum_and_count_by_name(Name, Tail, Sum, Count). remove_by_name(_, [], []). remove_by_name(Name, [employee_salary(Name, _) | Tail], TR) :- remove_by_name(Name, Tail, TR), !. remove_by_name(Name, [H | T], [H | TR]) :- remove_by_name(Name, T, TR). employee_average_salary([], []) :- !. employee_average_salary([employee_salary(Name, Salary) | Tail], [(Name, Average) | TR]) :- sum_and_count_by_name(Name, [employee_salary(Name, Salary) | Tail], Sum, Count), Average is Sum / Count, remove_by_name(Name, [employee_salary(Name, Salary) | Tail], NewList), employee_average_salary(NewList, TR). % Exercitiul 3 [rezolvat] % Primind o lista identica celei de la exercitiul 2, in lista de rezultat ordonati angajatii dupa salariul mediu castigat, in ordine descrescatoare. % ?- sort_employee_average_salary([employee_salary(peter, 1000), employee_salary(oliver, 1200), employee_salary(sam, 700), employee_salary(oliver, 800), employee_salary(sam, 900), employee_salary(sandra, 4500), employee_salary(peter, 1200), employee_salary(oliver, 400)], Result). % Result = [(sandra, 4500), (peter, 1100), (oliver, 800), (sam, 800)] % vreau sa fie descrescator my_cmp(X, (_, Salary1), (_, Salary2)) :- compare(X, Salary2, Salary1). sort_employee_average_salary(InputList, SortedList) :- employee_average_salary(InputList, TempList), predsort(my_cmp, TempList, SortedList). % Exercitiul 4 - Dutch Royal Family [rezolvat] mother(wilhelmina,juliana). mother(juliana,beatrix). mother(juliana,christina). mother(juliana,irene). mother(juliana,margriet). mother(beatrix,friso). mother(beatrix,alexander). mother(beatrix,constantijn). mother(emma,wilhelmina). father(hendrik,juliana). father(bernard,beatrix). father(bernard,christina). father(bernard,irene). father(bernard,margriet). father(claus,friso). father(claus,alexander). father(claus,constantijn). father(willem,wilhelmina). % citim mother(beatrix, alexander) ca "beatrix este mama lui alexander". aceeasi interpretare la father queen(beatrix). queen(juliana). queen(wilhelmina). queen(emma). king(willem). % 4a) definiti urmatoarele predicate % parent/2 - parent(X, Y). sa fie true cand X este parinte al lui Y % ruler(X) - sa fie true cand X este conducator parent(X, Y) :- mother(X, Y). parent(X, Y) :- father(X, Y). ruler(X) :- king(X). ruler(X) :- queen(X). % 4b) definiti un predicat predecessor/2, astfel incat predecessor(X, Y) sa ne spuna ierarhia de predecesori pentru un conducator X % (vrem sa raspundem, de fapt, la cine precede pe cine) predecessor(X, Y) :- ruler(X), ruler(Y), parent(Y, X). predecessor(X, Y) :- ruler(X), parent(Z, X), ruler(Z), predecessor(Z, Y). % Exercitiul 5 [rezolvat] % Fie L = (R, F, C, ari) o structura de ordinul I, unde F = {f, g, h}, C = {u, v, w} iar ari(f) = 2, ari(g) = 1, ari(h) = 3 % Considerand a, b, c, d variabile, care dintre urmatorii termeni se pot unifica? % a) f(g(a), h(b, c, v)) cu f(g(u), h(g(w), w, g(a))) % b) f(g(a), h(b, c, d)) cu f(g(u), h(g(w), w, g(a))) % se va scrie in terminal % ?- unify_with_occurs_check(f(g(A), h(B, C, v)), f(g(u), h(g(w), w, g(A)))). % raspunsul este false, pentru ca termenii nu unifica. v/0 iar g/1, se va incerca unificarea v = g(A) => esec % ?- unify_with_occurs_check(f(g(A), h(B, C, D)), f(g(u), h(g(w), w, g(A)))). % raspunde A = u, B = g(w), C = w, D = g(u) % Exercitiul 6 [rezolvat] % Fie predicatele succ/2 si obj/1 % Adaptati algoritmul BFS astfel incat sa pastram drumurile cele mai scurte care ajung de la un nod de start la un nod obiectiv. % Vom pastra doar cele mai scurte drumuri din lista rezultat, plecand de la un nod de start catre obiectivul deja setat. % Daca rulati solve(1, R), veti gasi R = [[5, 3, 4, 2, 1], [5, 2, 1], [5, 3, 1], [5, 3, 4, 1]] % vrem sa pastram doar drumurile de lungime minima, adica de lungime 3 % scriem o functie de minim min([], 0) :- !. min([X], X) :- !. min([H | T], MinTail) :- min(T, MinTail), MinTail =< H, !. min([H | T], H) :- min(T, MinTail), H =< MinTail. minimiumPaths(Start, Result) :- solve(Start, AllPaths), % gasim toate drumurile posibile % vrem acum sa mapam fiecare lista din lista de drumuri la lungimea ei findall(Length, (member(Path, AllPaths), length(Path, Length)), Lengths), % acum, Lengths este o lista de lungimi de forma [5, 3, 3, 4] % adica primul drum are lungime 5, al doilea are lungime 3, al treilea lungime 3, ultimul lungime 4 min(Lengths, MinLength), % am aflat ca drumul de lungime minima are lungime 3 (minim 3) findall(Path, (member(Path, AllPaths), length(Path, Length), Length =:= MinLength), Result). % am pus in Result toate Path care au lungimea minima calculata % Exercitiul 7 [rezolvat] % verificati daca un sir de caractere primit ca intrare, si reprezentat ca un atom, este palindrom sa nu % palindrome(prolog). atunci false % palindrome(ele). atunci true palindrome(Str) :- string_chars(Str, ListChars), reverse(ListChars, ReversedList), ListChars == ReversedList. % Exercitiul 8 [rezolvat] % Fie urmatoarea baza de cunostinte, definita de predicatele % employee_info(name, department_number, scale) % department(department_number, department_name) % salary(scale, amount) employee_info(mcardon,1,5). employee_info(treeman,2,3). employee_info(chapman,1,2). employee_info(claessen,4,1). employee_info(petersen,5,8). employee_info(cohn,1,7). employee_info(duffy,1,9). department(1,board). department(2,human_resources). department(3,production). department(4,technical_services). department(5,administration). salary(1,1000). salary(2,1500). salary(3,2000). salary(4,2500). salary(5,3000). salary(6,3500). salary(7,4000). salary(8,4500). salary(9,5000). % a) determinati toti angajatii din departamentul 1 si care au scale > 2 % Exemplu pentru rezolvare % ?- employee_info(Name, Department_Number, Scale), Department_Number = 1, Scale > 2 % acum, puneti toate aceste rezultate intr-o singura lista employees_from_department_eq_1_with_scale_gt_2(List) :- findall(Name, (employee_info(Name, Department_Number, Scale), Department_Number = 1, Scale > 2), List). % b) determinati toti angajatii dintr-un anume departament employees_from_department(DepNumber, List) :- findall(Name, (employee_info(Name, Department_Number, _), Department_Number = DepNumber), List). % c) selectati name si scale al angajatilor din departamentul 1, si scale > 3 employees_from_department_eq_1_with_scale_gt_3(List) :- findall((Name, Scale), (employee_info(Name, Department_Number, Scale), Department_Number = 1, Scale > 3), List).