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feat: 2104 assignment

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Yadunand Prem
2023-11-13 18:55:36 +08:00
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38
labs/cs2102/Midterms/2110.sql
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-- Q1
SELECT c.name
FROM countries c
WHERE c.population > 100000000
AND c.continent = 'Africa';
SELECT c.continent, COUNT(*)
FROM countries c
WHERE NOT EXISTS (
SELECT 1
FROM airports a
WHERE c.iso2 = a.country_iso2
)
GROUP BY c.continent;
SELECT c.name, COUNT(*)
FROM borders b, countries c
WHERE c.iso2 = b.country1_iso2
GROUP BY c.iso2
ORDER BY COUNT(*) DESC
LIMIT 10;
SELECT b.country1_iso2, b.country2_iso2
FROM borders b, countries c1, countries c2
WHERE b.country1_iso2 = c1.iso2
AND b.country2_iso2 = c2.iso2
AND c1.continent = 'Asia' AND c2.continent = 'Europe';
SELECT c.name
FROM countries c
WHERE NOT EXISTS(
SELECT *
FROM routes r, airports a, countries c1
WHERE r.airline_code = 'SQ'
AND c1.continent = 'Asia'
AND r.to_code = a.code
AND c1.iso2 = a.country_iso2
) AND c.continent = 'Asia';

60
labs/cs2102/Midterms/2210.sql
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SELECT COUNT(*)
FROM subzones
WHERE population > 15000;
SELECT max_floor, COUNT(max_floor)
FROM hdb_blocks
GROUP BY max_floor
ORDER BY COUNT(max_floor) desc
LIMIT 1;
SELECT *
FROM areas a
WHERE region = 'central'
AND NOT EXISTS (
SELECT 1
FROM mrt_stations m,subzones s
WHERE m.subzone = s.name
AND s.area = a.name
);
SELECT sz.area, COUNT(stop.code) AS num_stops
FROM mrt_stops stop,
mrt_stations s,
subzones sz
WHERE stop.station = s.name
AND s.subzone = sz.name
GROUP BY sz.area
ORDER BY num_stops DESC
LIMIT 3;
SELECT DISTINCT m1.station
FROM mrt_connections c, mrt_stops m1, mrt_stops m2
WHERE c.from_code = m1.code
AND c.to_code = m2.code
AND m1.line = 'ew'
AND m2.line != 'ew';
SELECT sz.area, COUNT(DISTINCT so.line) AS num_lines
FROM mrt_stops so,
mrt_stations st,
subzones sz
WHERE so.station = st.name
AND st.subzone = sz.name
GROUP BY sz.area
HAVING COUNT(DISTINCT so.line) >= 3
ORDER BY num_lines DESC;
SELECT tab.area, COUNT(*) AS num_lines
FROM (
SELECT s.area, h.line
FROM mrt_stops h,
mrt_stations m,
subzones s
WHERE h.station = m.name
AND m.subzone = s.name
GROUP BY s.area, h.line
) tab
GROUP BY tab.area
HAVING COUNT (*) >= 3 ORDER BY num_lines DESC

65
labs/cs2102/Tutorials/tut04.sql
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-- How many loans involve an owner and a borrower from the same department?
SELECT COUNT(*)
FROM loan l,
student owner,
student borrower
WHERE l.owner = owner.email
AND l.borrower = borrower.email
AND owner.department = borrower.department;
-- For each faculty, print the number of loans that involve an owner and a borrower from this faculty?
SELECT owner.department, COUNT(*)
FROM loan l,
student owner,
student borrower
WHERE l.owner = owner.email
AND l.borrower = borrower.email
AND owner.department = borrower.department
GROUP BY owner.department;
-- What are the average and the standard deviation of the duration of a loan? Round the results to the nearest integer.
SELECT ROUND(AVG(l.returned - l.borrowed+1),0), ROUND(stddev_pop(l.returned - l.borrowed+1),0)
FROM loan l;
-- (a) Print the titles of the different books that have never been borrowed. Use a nested query.
SELECT b.title
FROM book b
WHERE b.isbn13 NOT IN (
SELECT l.book
FROM loan l
);
-- (b) Print the name of the different students who own a copy of a book that they have never lent to anybody.
SELECT s.name
FROM student s
WHERE s.email IN (
SELECT c.owner
FROM copy c
WHERE (c.owner, c.book,c.copy)NOT IN (
SELECT l.owner,l.book,l.copy
FROM loan l
) );
-- For each department, print the names of the students who lent the most.
SELECT s.name,s.department, COUNT(*)
FROM student s,loan l
WHERE l.owner = s.email
GROUP BY s.name, s.department
HAVING COUNT(*) >= ALL
(SELECT COUNT(*) FROM student s1, loan l1 WHERE l1.owner = s1.email AND s.department = s1.department GROUP BY s1.email);
-- Print email and name of different students who borrowed all books authored by adam smith
SELECT s.email, s.name
FROM student s
WHERE NOT EXISTS(
SELECT *
FROM book b
WHERE b.authors = 'Adam Smith'
AND NOT EXISTS(
SELECT *
FROM loan l
WHERE l.book = b.isbn13
AND l.borrower = s.email
)
)

43
labs/cs2102/Tutorials/tut05.sql
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-- How many loans involve
SELECT b.title
FROM book b
WHERE b.isbn13 != ALL (
SELECT l.book FROM loan l
);
-- 2b
SELECT s.name
FROM student s
WHERE s.email = ANY (
SELECT c.owner
FROM copy c
WHERE (c.owner, c.book, c.copy) NOT IN (
SELECT l.owner, l.book, l.copy
FROM loan l
)
);
-- 2c (When you see the most/the least, there's a standard way
-- of solving this
SELECT s.department, s.name, COUNT(*)
FROM student s, loan l
WHERE l.owner = s.email
GROUP BY s.department, s.name, s.email
HAVING COUNT(*) >= ALL (
SELECT COUNT(*)
FROM student s1, loan l1
WHERE l1.owner = s1.email
AND s.department = s1.department
GROUP BY s1.email
)
ORDER BY COUNT(*) DESC;
-- 2d
SELECT s.name, s.email
FROM student s
WHERE s.email = ALL (
SELECT c
FROM book b
WHERE b.authors LIKE '%Adam Smith%'
)

0
labs/cs2102/project/part3.sql Normal file
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311
labs/cs2104/prolog/assignment.pl Normal file
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%% In Prolog, atoms, numbers, and pairs (lists)
%% are data structures. Furthermore, We can construct a new
%% data structure using "function" symbols that are not
%% further interpreted. So
%% f(10, 20, 30)
%% is a data structure with the label 'f' and three components,
%% the numbers 10, 20, and 30.
%% A binary number tree is either null or
%% a node(T1, V, T2) where T1 is
%% a binary number tree, V is an number, and T2 is a
%% binary number tree.
%% Note that 'node' is the label, and T1, V, and T2 are
%% the components of a node data structure.
%% Examples:
tree1(
node(node(node(null,
1,
null),
2,
node(null,
3,
null)),
4,
node(node(null,
5,
null),
6,
node(null,
7,
null)))
).
tree2(null).
tree3(node(null,
4,
null)).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question 1, 5 points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Consider the following predicate leftmost_element:
leftmost_element(node(null, V, _), V).
leftmost_element(node(T1, _, _), W) :-
leftmost_element(T1, W).
test_leftmost_element :-
tree1(T1), leftmost_element(T1, 1),
tree2(T2), not(leftmost_element(T2,_)),
tree3(T3), leftmost_element(T3, 4).
%% Explain the behavior of the query
%% leftmost_element(X, 4) in three or four sentences:
%% T = node(null, 4, _) ;
%% T = node(node(null, 4, _), _, _) ;
%% T = node(node(node(null, 4, _), _, _), _, _) ;
%% T = node(node(node(node(null, 4, _), _, _), _, _), _, _)
%% Write your answer in comments here:
%%
%%
%%
%%
%%
%%
%%
%%
%%
%%
%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question 2, 10 points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% The depth of a tree is defined to be the longest path
%% from the root of the tree to any value, and 0 if there
%% are no values in the tree.
%% Write a predicate depth(T, X) that holds if X is the
%% depth of T.
%%%%%%%%%
%% your solution goes here
depth(null, 0).
depth(node(Left, _, Right), Res) :-
depth(Left, F1),
depth(Right, F2),
Res is max(F1, F2) + 1.
%%%%%%%%%
test_depth :-
tree1(T1), depth(T1, 3),
tree2(T2), depth(T2, 0),
tree3(T3), depth(T3, 1).
%% test your solution by
%% writing
%% ?- test_depth.
%% in the SWI-Prolog REPL.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question 3, 10 points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Write a predicate sum(T, S) that computes the sum S
%% of all elements of a tree T.
%%%%%%%%%%%%%%
%% your solution goes here
sum(null, 0).
sum(node(Left, X, Right), Res) :-
sum(Left, LeftRes),
sum(Right, RightRes),
Res is X + LeftRes + RightRes.
%%%%%%%%%%%%%%
test_sum :-
tree1(T1), sum(T1, 28),
tree2(T2), sum(T2, 0),
tree3(T3), sum(T3, 4).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question X, 0 points, just practice
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Write a predicate bt_member(T, V) that holds
%% if and only if V is a value in the tree T.
%%%%%%%%%%%%%%
%% solution at the end of this file
%%%%%%%%%%%%%%
test_bt_member :-
tree1(T1), bt_member(T1, 5),
tree2(T2), not(bt_member(T2, _)),
tree3(T3), bt_member(T3, 4).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question Y, 0 points, just practice
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Note that the given trees in tree1, tree2, and tree3
%% are binary search trees: For every node, the values in
%% the left subtree of the node are all smaller than the
%% value, and the values in the right subtree of the tree
%% are all larger than the value.
%% Write a predicate bst_member(T, V) that exploits this fact
%% and that holds if and only if V is a value in the
%% binary search tree T. The number of recursive calls
%% should be limited by the depth of the binary search tree.
%%%%%%%%%%%%%%
%% solution at the end of this file
%%%%%%%%%%%%%%
test_bst_member :-
tree1(T1), bst_member(T1, 5),
tree2(T2), not(bst_member(T2, _)),
tree3(T3), bst_member(T3, 4).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question Z, 0 points, just practice
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Write a predicate make_bst(N, T) that makes a binary search
%% tree that contains the integers 1,..,N.
%% Hint: Remember Question 1
%%%%%%%%%%%%%%
%% solution at the end of this file
%%%%%%%%%%%%%%
test_make_bst :-
make_bst(10, T1), bst_member(T1, 10),
not(bst_member(T1, 11)),
make_bst(100, T2), bst_member(T2, 100),
not(bst_member(T2, 101)).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question 4, 10 points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Here is a length predicate for lists.
len([], 0).
len([_|T], N) :-
len(T, N1), N is N1 + 1.
test_length :-
len([1,2,3,4], 4),
len([], 0),
len([1,2,3,4,5,6,7,8,9,10], 10).
%% Write a predicate nth(I, L, V) that holds
%% if and only if V is the nth element of list L,
%% assuming we start counting at 0.
%%
%%%%%%%%%%%%%%
%% your solution goes here
%% The 0th element of any list is V
nth(0, [V|_], V).
nth(Idx, [_|Tail], Elem) :-
Idx > 0, %% Catch for negative indices
Idx1 is Idx - 1,
nth(Idx1, Tail, Elem).
%%%%%%%%%%%%%%
test_nth :-
nth(0, [1,2,3,4], 1),
not(nth(0, [], 0)),
nth(9, [1,2,3,4,5,6,7,8,9,10], 10).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question 5, 10 points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Most likely, your nth predicate is not going
%% to work backwards. It doesn't allow you to
%% identify the position of a given value in
%% a list: nth(I, [10,20,30,40], 30) will not bind
%pos(
%% I to the value 2. Do you see why? (no submission)
%%
%% Write a predicate pos(X, L, I) that holds when
%% I is a position at which L has the value X.
%% Thus pos(30, [10,20,30,40], I) should bind
%% I to the value 2.
%% Make sure that your predicate will hold for
%% all positions of the given value X, not just
%% the first one. Thus pos(30, [10, 30, 30, 40], I)
%% should bind I to 1, and upon backtracking, it
%% should bind I to the value 2.
%%%%%%%%%%%%%%
%% your solution goes here
pos(Target, List, Index) :- pos_helper(Target, List, 0, Index).
pos_helper(Target, [Target|_], Acc, Acc).
pos_helper(Target, [_|Tail], Acc, Index) :-
Acc1 is Acc + 1,
pos_helper(Target, Tail, Acc1, Index).
%%%%%%%%%%%%%%
test_pos :-
pos(10, [10,20,30,40], 0),
not(pos(0, [], _)),
not(pos(0, [1,2,3,4], _)),
pos(20, [10,20,20,40], 2),
pos(10, [1,2,3,4,5,6,7,8,9,10], 9).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Question 6, 10 points
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Write a predicate all_pos(X, L, I) that
%% computes the list of all positions at which
%% L has the value X. Thus
%% all_pos(30, [10, 30, 30, 40], X) should bind
%% X to the list [1, 2].
%%%%%%%%%%%%%%
%% your solution goes here
all_pos(Target, List, Res) :- all_pos_helper(Target, List,0, Res).
all_pos_helper(_, [],_, []).
all_pos_helper(Target, [Target|Tail], Acc, [Acc|Rest]) :-
Acc1 is Acc + 1,
all_pos_helper(Target, Tail, Acc1, Rest).
all_pos_helper(Target, [_|Tail], Acc, Rest) :-
Acc1 is Acc + 1,
all_pos_helper(Target, Tail, Acc1, Rest).
%%%%%%%%%%%%%%
test_all_pos :-
all_pos(0, [1,2,3,4], []),
all_pos(0, [], []),
all_pos(10, [1,2,3,4,5,6,7,8,9,10], [9]),
all_pos(4, [1,2,3,4,4,4,4,4,9,10], [3,4,5,6,7]).
%% Solution Question X
bt_member(node(_, V,_), V).
bt_member(node(T1,_,_), V) :- bt_member(T1, V).
bt_member(node(_,_,T2), V) :- bt_member(T2, V).
%% Solution Question Y
bst_member(node(_, V, _), V).
bst_member(node(T1,V,_), W) :- W < V, bst_member(T1, W).
bst_member(node(_,V,T2), W) :- W > V, bst_member(T2, W).
%% Solution Question Z
make_bst(0, null).
make_bst(N, node(T1, N, null)) :-
N > 0, N1 is N - 1, make_bst(N1, T1).

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labs/cs2104/prolog/family.pl Normal file
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parent(a, b).
parent(a, c).
parent(b, d).
parent(b, e).
parent(c, f).
parent(c, g).
parent(f, h).
parent(f, i).
ancestor(X, Y) :- parent(X, Y).
ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y).
factorial(1, 1).
factorial(N, F) :- N > 1, N1 is N - 1, factorial(N1, F1), F is F1 * N.
faciter(1, Res, Res).
faciter(N, Acc, Res) :- N > 1, N1 is N -1, Acc1 is N * Acc, faciter(N1, Acc1, Res).
fib(0, 0).
fib(1, 1).
fib(N, F) :- N > 1, N1 is N-1, N2 is N-2, fib(N1, F1), fib(N2, F2), F is F1 + F2.
fibiter(X, X, F1, F2, F1).
fibiter(X, N, F1, F2, Result) :- X < N, X1 is X + 1, Aux is F1 + F2, fibiter(X1, N, F2, Aux, Result).
head([H|_], H).
tail([_|T], T).
member(X, [X|_]).
member(X, [_|T]) :- member(X, T).
append([], L, L).
append([H|T], L, [H|R]) :- append(T, L, R).
reverse([], []).
reverse([H|T], R) :-
reverse(T, R1), append(R1, [H], R).

44
labs/cs2104/prolog/likes.pl Normal file
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%% Demo coming from http://clwww.essex.ac.uk/course/LG519/2-facts/index_18.html
%%
%% Please load this file into SWI-Prolog
%%
%% Sam's likes and dislikes in food
%%
%% Considering the following will give some practice
%% in thinking about backtracking.
%%
%% You can also run this demo online at
%% http://swish.swi-prolog.org/?code=https://github.com/SWI-Prolog/swipl-devel/raw/master/demo/likes.pl&q=likes(sam,Food).
/** <examples>
?- likes(sam,dahl).
?- likes(sam,chop_suey).
?- likes(sam,pizza).
?- likes(sam,chips).
?- likes(sam,curry).
*/
likes(sam,Food) :-
indian(Food),
mild(Food).
likes(sam,Food) :-
chinese(Food).
likes(sam,Food) :-
italian(Food).
likes(sam,chips).
indian(curry).
indian(dahl).
indian(tandoori).
indian(kurma).
mild(dahl).
mild(tandoori).
mild(kurma).
chinese(chow_mein).
chinese(chop_suey).
chinese(sweet_and_sour).
italian(pizza).
italian(spaghetti).