My first crack at a logical paradigm programming language. If you already know java or C++, I’d highly recommend learning prolog because it will make you a better problem solver in imperative languages.

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/* Homework Assignment 2
Programming Languages
Spring 2023
*/
/* Instructions */
/*
This section deals with general submission instructions. Use this
source file. You will be
able to code in and run the file in the prolog interpreter directly.
We will be using swipl for our prolog environment:
To load/reload this file, cd to its directory and run swipl. Then, in the prompt, type [hw2].
On the lab computers, you can run swipl by typing `prolog`.
cd PATH_TO_FILE
prolog
[hw2].
From then on you may execute queries (goals) in the prompt.
You should provide your answers in the designated spot. Once you have
added some code to the file, rerun [hw2]. in the swipl prompt to
reload.
In addition, there are unit tests for each problem. These are there to
help you better understand what the question asks for, as well as
check your code. They are included in our knowledge base as queries
and are initially commented out -- % is a Prolog line comment.
Lines marked as SUCCEED should be satisfied by your defined relation(s).
Lines marked as FAIL should not be satisfied.
%:- member_times(4,[3,3,2,3],0). % SUCCEED
%:- member_times(4,[1,2,3],3)-> fail ; true. % FAIL
After you have finished a problem and are ready to test, remove the
initial % for each test for the associated problem and reload the
assignment file ([hw2].). Each line should silently load and
succeed. If any line throws a WARNING, then you solution is not
correct. If you pass the tests there is a good chance that your code
is correct, but not guaranteed; the tests are meant as guided feedback
and are not a check for 100% correctness.
This file is named hw2.txt because BrightSpace blocks uploading of .pl files.
You will likely need to rename it to hw2.pl in order to load it in a Prolog
interpreter, and then rename it back to hw2.txt in order to submit.
*/
/* Submission */
/*
For this assignment -- and the remaing Prolog assignments -- you must
submit only the source file. Non-code answers may be written in comments.
Coding should be done directly in hw2.pl.
*/
/* Homework 2 */
/* Due: Next Wednesday, 11:59 PM */
/*
Purpose: To get comfortable with Logic programming, and get a good
grasp on list manipulation in Prolog.
*/
/* Problem 1a:
Programming with matching. A line can be defined by 2 points. A point has an
x and y coordinate. A line is vertical if both points have the same x value.
A line is horizontal if both points have the same y values. The following
is a knowledge base which specify what is meant for a line to be vertical
or horizontal respectively. This example is due to Ivan Bratko.
*/
vertical(line(point(X,_),point(X,_))).
horizontal(line(point(_,Y),point(_,Y))).
/* 1. Name the clauses, predicates, rules, and facts.
2. Name the atoms, variables, and data structure constructors. */
% 1.) There are only 2 clauses, which are both facts, and they are the given 2 lines above.
% There are 2 predicates: veritcal and horizontal. There are no rules.
%
% 2.) The atoms are the underscores. The variables are X and Y.
% The data structure constructors are point, which takes two numbers.
% There's also a line constructor, which is built with two points.
/* Problem 1b:
A way of writing numerals, which is sometimes used in mathematical logic, makes
use of just four symbols: 0, succ , and the left and right parentheses.
The following is the knowledge base for this representation of a numeral.
The predicate 'add' is the definition of adding this representation of numbers.
Use this base you should use to answer the question.
(Reference 3.1 example 3 Learn Prolog Now!
http://www.learnprolognow.org/lpnpage.php?pagetype=html&pageid=lpn-htmlse9)
*/
numeral(0).
numeral(succ(X)) :- numeral(X).
add(0,Y,Y).
add(succ(X),Y,succ(Z)) :-
add(X,Y,Z).
/* 1. Name the clauses, predicates, rules, and facts.
2. Name the atoms, variables, and data structure constructors. */
% There are four clauses, two are facts and two are rules. The facts
% are 'numeral(0).' and 'add(0,Y,Y).' The rules are 'numeral(succ(X)) :- numeral(X).' and
% 'add(succ(X),Y,succ(Z)) :- add(X,Y,Z).'
% The predicates are numeral/1 and add/3.
%
% The atom is '0'. The variables are X, Y and Z. The constructor succ takes
% 1 number.
/* Problem 2:
The following are two basic predicates for list manipulation:
my_first/2 and my_last/2. We may refer to a predicate by writings it
as name/arity; hence, my_first/2 means a predicate named my_first with
two arguments.
my_first(X,Y) succeeds if X is the first element of list Y.
my_last(X,Y) succeeds if X is the last element of list Y.
Write definitions for my_first and my_last.
*/
% No recursion, see if head equals X
my_first(X, [X|_]).
% Tail should be empty if X is the last element
my_last(X, [X|[]]).
my_last(X, [_|T]) :- my_last(X, T).
/* Problem 2 Test: */
:- my_first(a, [a]). % SUCCEED
:- my_first(3, [3, 2, 1]). % SUCCEED
:- my_first(3, [4, 3, 2, 1]) -> fail ; true. % FAIL
:- my_last(a, [a]). % SUCCEED
:- my_last(1, [3, 2, 1]). % SUCCEED
:- my_last(1, [3, 2, 1, 0]) -> fail ; true. % FAIL
/* Problem 3:
Write a predicate init(All, BLst) that succeeds if Blst has all the items of ALL
except the last item. The items in BLst are in the same order as ALL.
*/
% Base case : On the last element of All, BLst should be empty
init([_|[]],[]).
% Compare the nth elements then recurse
init([X|TA],[X|TB]) :- init(TA,TB).
/* Problem 3 Test: */
:- init([1], []). % SUCCEED
:- init([1,2,3], [1,2]).% SUCCEED
:- init([1,2], [1,2]) -> fail ; true. % FAIL
:- init([1,2], [2]) -> fail ; true. % FAIL
/* Problem 4:
Write a predicate is_decreasing(X) that succeeds if X is a list of decreasing numbers -- Each number is either the same or lower than the preceding number.
NOTE: You may match two elements at a time against a list: [X,Y|Xs] = List. It's preferable to do it in the rule head however...
some_rule([X,Y|Xs]) :- ... */
% Base case if there's 0 elements
is_decreasing([]).
% Base case if there's only 1 element
is_decreasing([_|[]]).
% Make sure X is more or equal to Y and make Y the new head
is_decreasing([X,Y|Xs]) :- X >= Y, is_decreasing([Y|Xs]).
/* Problem 5 Test: */
:- is_decreasing([]). % SUCCEED
:- is_decreasing([10]). % SUCCEED
:- is_decreasing([10,9]). % SUCCEED
:- is_decreasing([10,9,7]). % SUCCEED
:- is_decreasing([10,9,7,7,2]). % SUCCEED
:- is_decreasing([1,1,1,1,1]). % SUCCEED
:- is_decreasing([10,9,7,9]) -> fail ; true. % FAIL
:- is_decreasing([2,3,1]) -> fail ; true. % FAIL
:- is_decreasing([1,2,3]) -> fail ; true. % FAIL
:- is_decreasing([7,19])-> fail ; true. % FAIL
/* Problem 5:
Write a predicate element_at(X,Y,N) that succeeds if X is the Nth element of list Y. Y is 0-indexed.
NOTE: Don't worry about the error cases: i.e, N greater than the length of Y. */
% Base Case: 0th element will be at head
element_at(X, [X|_], 0).
% shift head to the right, decrease index by 1 to keep it in same spot relative to head
element_at(X, [_|T], N) :- D is N-1, element_at(X, T, D).
/* Problem 5 Test: */
:- element_at(3,[1,2,3],2). % SUCCEED
:- element_at(1,[1,2,3],0). % SUCCEED
:- element_at(1,[1,2,3],1) -> fail ; true. % FAIL
/* Problem 6:
Write a predicate insert_at(E,Y,N,Z) that succeeds if Z is the list Y with E inserted at index N -- Insert X at index N in Y.
NOTE: Don't worry about the error cases: i.e, N greater than the length of Y. */
% If N = 0, Z will have a head of E and tail of list Y
insert_at(E,Y,0,[E|Y]).
% Similar to recursive step in problem 5
insert_at(E, [H|YT], N, [H|ZT]) :- N1 is N-1, insert_at(E, YT, N1, ZT ).
/* Problem 6 Test: */
:- insert_at(3,[1,2,3],2,[1,2,3,3]). % SUCCEED
:- insert_at(1,[1,2,3],0,[1,1,2,3]). % SUCCEED
:- insert_at(a,[1,2,3],1,[1,a,2,3]). % SUCCEED
:- insert_at(1,[1,2,3],0,[1,2,3]) -> fail ; true. % FAIL
/* Problem 7 :
Write a predicate delete_at(E,Y,N,Z) that succeeds if Z is the list Y with E delete at index N -- Delete E at index N in Y.
YOU MUST USE the predicate defined in the above problem to solve this problem.
NOTE: Don't worry about the error cases: i.e, N greater than the length of Z. */
% The insert algorithm ran with swapped list arguments gives us the delete behavior
delete_at(E, Y, N, Z) :- insert_at(E, Z, N, Y).
/* Problem 7 Test: */
:- delete_at(3,[1,2,3,3],2,[1,2,3]). % SUCCEED
:- delete_at(1,[1,1,2,3],0,[1,2,3]). % SUCCEED
:- delete_at(a,[1,a,2,3],1,[1,2,3]). % SUCCEED
:- delete_at(1,[1,2,3],0,[1,2,3]) -> fail ; true. % FAIL
/* Problem 8:
Write a predicate zip(Xs,Ys,Zs) that succeeds if Zs is a list where each element is a tuple, (X,Y), with Xs and Ys paired together.
For example...
zip([1,2,3],[a,b,c],Zs) should give Zs = [(1,a),(2,b),(3,c)]
zip([1],[a],Zs) should give Zs = [(1,a)]
NOTE: You may assume X and Y have the same length. */
% Base Cases
% 0 items in Xs and Ys
zip([],[],[]).
% See if X and Y are in a tuple in the Zs at each iteration
zip([X|XT],[Y|YT],[(X,Y)|ZT]) :- zip(XT, YT, ZT).
/* Problem 8 Test: */
:- zip([1,2,3],[a,b,c],[(1,a),(2,b),(3,c)]). % SUCCEED
:- zip([],[],[]). % SUCCEED
:- zip([1],[2],[(1,2)]). % SUCCEED
:- zip([1],[2],[(2,3)]) -> fail ; true. % FAIL
:- zip([1],[2,3],[(1,2)]) -> fail ; true. % FAIL
/* Problem 9:
Write a predicate zip2(Xs,Ys,Zs) that succeeds if Zs is a list where each element is a tuple, (X,Y), with Xs and Ys paired together. However, the length of Zs will be equal to the length of Xs or Ys which ever is less.
For example...
zip2([1,2,3,4],[a,b,c],Zs) should give Zs = [(1,a),(2,b),(3,c)]
zip2([1],[a,b],Zs) should give Zs = [(1,a)] */
% Base Cases
% If either Xs or Ys are empty, Zs are empty
zip2([],[],[]).
zip2(_,[],[]).
zip2([],_,[]).
% Less Ys than Xs
zip2([X|_],[Y|[]],[(X,Y)]).
% Less Xs than Ys
zip2([X|[]],[Y|_],[(X,Y)]).
% Same as recursive step in problem 8
zip2([X|XT],[Y|YT],[(X,Y)|ZT]) :- zip2(XT, YT, ZT).
/* Problem 9 Test: */
:- zip2([1,2,3],[a,b,c],[(1,a),(2,b),(3,c)]). % SUCCEED
:- zip2([],[a,b,c],[]). % SUCCEED
:- zip2([1,3],[],[]). % SUCCEED
:- zip2([1,3],[2],[(1,2)]). % SUCCEED
:- zip2([1],[2],[(2,3)]) -> fail ; true. % FAIL
:- zip2([1],[a,b],[(1,a),(1,b)]) -> fail ; true. % FAIL
/* Problem 10:
See Problem 1b above for the knowledge base used for defining greater_than/2 .
Exercise 3.4 (http://www.learnprolognow.org/lpnpage.php?pagetype=html&pageid=lpn-htmlse11)
Define a predicate greater_than/2 that takes two numerals in the notation
that we introduced in the text (that is, 0, succ(0), succ(succ(0)), and so on)
as arguments and decides whether the first one is greater than the second one.
*/
greater_than(succ(_), 0).
greater_than(succ(X), succ(Y)) :- greater_than(X, Y).
/* Problem 10 Test: */
:- greater_than(succ(succ(succ(0))),succ(0)). % SUCCEED
:- greater_than(succ(succ(0)),succ(succ(succ(0)))) -> fail ; true. % FAIL
/* Problem 11:
See Problem 1b above for the knowledge base used for defining subtract/3 .
Define substract(Num1,Num2,Result) to succeed if Result is the result of
Num1 - Num2. Num1, Num2 and Result use four symbols: 0, succ , and the left and right parentheses
to represent numbers.
Use the add/3, from problem 1b, definition to define subtract/3. Do not
write a recursive definition for subtract/3.
*/
subtract(X,0,X).
% X - Y = R -> R + Y = X
subtract(X,Y,R) :- add(R,Y,X).
/* Problem 11 Test: */
:- subtract(succ(succ(0)), succ(0), succ(0)). % SUCCEED
:- subtract(succ(succ(0)), 0, succ(succ(0))). % SUCCEED
:- subtract(succ(succ(0)), succ(succ(0)), 0). % SUCCEED
:- subtract(succ(succ(0)), 0, 0) -> fail ; true. % FAIL
:- subtract(succ(succ(0)), succ(0), succ(succ(0))) -> fail ; true. % FAIL
/* Problem 12:
Write a predicate has_subseq(X,Y) that succeeds if Y is a list that is a subsequence of a list X.
For example...
has_subseq([a,b,c,d],[b,d]) should succeed, but has_subseq([a,b,c,d],[b,e]) should fail. */
has_subseq(_,[]).
% If we find a common element move heads of both lists to the next element
has_subseq([X|XT], [Y|YT]) :- X == Y, has_subseq(XT,YT).
% If not equal traverse Xs but stay at Ys head
has_subseq([_|XT], [Y|YT]) :- has_subseq(XT, [Y|YT]).
/* Problem 12 Test: */
:- has_subseq([a,g,b,d],[g,b]). % SUCCEED
:- has_subseq([1,2,3,4],[2,4]). % SUCCEED
:- has_subseq([1,2,3,4],[2,3]). % SUCCEED
:- has_subseq([1,2,3,4],[]). % SUCCEED
:- has_subseq([1,2,3,4],[2,5]) -> fail ; true. % FAIL
:- has_subseq([1,2,3,4],[4,3]) -> fail ; true. % FAIL
/* Problem 13:
Write a predicate bubblesort(X,Y) that succeeds if Y is X, sorted from least to greatest. Items do not need to be unique.
The implementation of bubblesort should be based on the bubble sort algorithm
Hint: It might be helpful to define a helper relation bubble, in addition to bubblesort itself */
% Base Case:
bubblesort(Xs, Sorted) :-
% The outer loop, do the next pass if Xs aren't Sorted
( bubble(Xs, L2) -> bubblesort(L2, Sorted)
% else Xs are sorted, the two lists are equal
; Xs = Sorted ).
bubble([A,B | T], L) :-
% If B is less than A, swap them.
( B < A -> L = [B, A|T] ;
% Else continue bubbling with B as new head
L = [A | L1], bubble([B|T], L1)).
/* Problem 13 Test: */
:- bubblesort([],[]). % SUCCEED
:- bubblesort([4, 3, 2, 1],[1, 2, 3, 4]). % SUCCEED
:- bubblesort([4, 3, 2, 1, 4],[1, 2, 3, 4, 4]). % SUCCEED
:- bubblesort([4, 3, 2, 1],[1, 2, 4, 3]) -> fail ; true. % FAIL
/* Problem 14:
Write a predicate merge(A,B,M) that succeed if the list M has all the items from lists A and B in increasing order. Items do not need to be unique.
Hint: You may use predicates defined in previous problems or write helper predicates to aid in solving this problem.
*/
append([], Xs, Xs).
append([X|Xs], Ys, [X|Zs]) :- append(Xs, Ys, Zs).
% Base Cases. Merging an empty list will be the same list
merge([],[],[]).
merge([A],[],[A]).
merge([],[B],[B]).
% Append the lists and sort
merge(A, B, M) :- append(A, B, X), bubblesort(X, S), M = S.
/* Problem 14 Test: */
:- merge([10,3,2],[11,5,2],[2,2,3,5,10,11]) . % SUCCEED
:- merge([0],[],[0]). % SUCCEED
:- merge([],[3],[3]). % SUCCEED
:- merge([3,4],[3],[3]) -> fail ; true. % FAIL
```