===================================================
'Clancy Bouvier'=='Jacqueline Gurney'
'Carnival Woman'=='Abraham Simpson' |
| 'Abraham Simpson'=='Penelope Olsen' +------------------+--------------+
| | | | |
'Herbert Powell' 'Homer Simpson'=='Marge Bouvier' 'Patricia Bouvier' 'Selma Bouvier'
| |
+-----------------+----------------+ 'Ling Bouvier'
| | | (adopted)
'Bart Simpson' 'Lisa Simpson' 'Maggie Simpson'
===================================================
(A==B means that A married B; their children are listed underneath the ==. Note that 'Abraham Simpson' married twice. If you are into this Simpsons thing, please accept my apologies that I omitted Amber Simpson from the tree --- it is just too weird to draw that she was married to both Homer and Abraham Simpson!
Part 1a.
Make a file named simpsons.pl to hold the Prolog clauses that model the above tree structure.
Here is part of the coding;
you do the rest:
===================================================
female('Carnival Woman').
male('Abraham Simpson').
female('Penelope Olsen').
male('Herbert Powell').
male('Homer Simpson').
male('Bart Simpson').
/* ...you do the remainder of the gender clauses;
consult the above web link if you are unsure about any person's gender. */
female('Ling Bouvier').
/* predicate parents(CHILD, FATHER, MOTHER) asserts that
CHILD has FATHER as its father and MOTHER as its mother:
*/
parents('Herbert Powell', 'Abraham Simpson', 'Carnival Woman').
parents('Homer Simpson', 'Abraham Simpson', 'Penelope Olsen').
parents('Bart Simpson', 'Homer Simpson', 'Marge Bouvier').
/* ...you do the remainder of the parent clauses up to this last one: */
parents('Ling Bouvier', none, 'Selma Bouvier')
/* sister(X, Y) defines that X has Y as a sister if
Y is female, X's and Y's parents are the same, and X != Y */
sister(X,Y) :- female(Y),
parents(X,Father,Mother), parents(Y,Father,Mother),
X \= Y.
Run this test:
?- sister('Marge Bouvier', Z).
type semicolons till the prover prints False (or No). You should obtain:
Z = 'Patricia Bouvier' ;
Z = 'Selma Bouvier' ;
False
Part 1b.
Code these Prolog predicates and add them to simpsons.pl:
/* aunt(X, A) defines that X has A as an aunt */(Hint: Code your definition of aunt so that it matches exactly what I said above.)
/* uncle(X, U) defines that X has U as an uncle */(Hint: define a predicate, brother, and use it like you used sister to define aunt.)
Test your definitions with at least these test cases:
?- aunt('Bart Simpson', Z).
Z = 'Patricia Bouvier' ;
Z = 'Selma Bouvier' ;
false.
?- aunt(X, 'Patricia Bouvier').
X = 'Bart Simpson' ;
X = 'Lisa Simpson' ;
X = 'Maggie Simpson' ;
X = 'Ling Bouvier'.
?- aunt(X,Y).
(there should be a long list!)
?- brother(A,B).
A = 'Lisa Simpson',
B = 'Bart Simpson' ;
A = 'Maggie Simpson',
B = 'Bart Simpson' ;
false.
?- uncle(X,Y).
false.
Copy and
paste your test cases and their outputs into a txt file
named simpsons.txt.
Copy and paste these clauses into a file named library.pl:
===================================================
/* Functor for defining book and dvd objects:
ITEM ::= book(TITLE, AUTHOR) | dvd(TITLE)
where TITLE and AUTHOR are strings
Predicates that define ownership of an object
and when an object is borrowed from the library:
PRED ::= owns(KEY, ITEM) | borrowed(KEY, PERSON, DUEDATE)
where KEY is an atom that begins with k
ITEM is defined above
PERSON is an atom
DUEDATE is an int
*/
owns(k0, book('David Copperfied', 'Charles Dickens')).
owns(k1, book('Tale of Two Cities', 'Charles Dickens')).
owns(k2, book('Tale of Two Cities', 'Charles Dickens')).
owns(k3, dvd('Tale of Two Cities')).
owns(k4, book('Moby Dick', 'Herman Melville')).
owns(k5, dvd('Brothers Karamazov', 'Fyodor Dostoyevski')).
borrowed(k2, 'Homer', 10).
borrowed(k4, 'Homer', 20).
borrowed(k3, 'Lisa', 40).
borrowed(k0, 'Lisa', 45).
borrowed(k5, 'Homer', 90).
/* overdue holds true when ItemKey is borrowed (by Who) and is due earlier than Today */
overdue(ItemKey, Today) :- borrowed(ItemKey, _, DueDate),
Today > DueDate.
/* fine calculates the int value of HowMuch when ItemKey is overdue as of Today */
fine(ItemKey, Today, HowMuch) :- overdue(ItemKey, Today),
borrowed(ItemKey, _, DueDate),
HowMuch is Today - DueDate.
/* getBorrowed(Who, ItemList) calculates a list of KEYs
of all the items Who has borrowed. */
getBorrowed(Who, ItemList) :- findall(Key, borrowed(Key, Who, _), ItemList).
===================================================
Part 2a.
Write this definition in Prolog and add it to library.pl:
/* getOverdue(Who, Today, ItemList) calculates ItemList, which is a list of KEYs
of all the items Who has borrowed that are overdue as of Today. */
Hint: your answer will look a lot like getBorrowed's definition.
Test your coding on at least these examples:
?- getOverdue('Homer', 25, List).
List = [k2, k4].
?- getOverdue('Lisa', 25, List).
List = [].
?- getOverdue(_, 50, List).
List = [k2, k4, k3, k0].
Copy
your test cases and their outputs into the txt file,
library.txt.
Part 2b.
Write this definition in Prolog and add it to library.pl:
/* totalfine(Who, Today, Amount) calculates Amount, which is an int
that states the total fine that Who owes due to overdue items as of Today */
Hint: use this code to total the fine:
/* sum(L, T) holds true when T is the sum of all the ints in list L. */
sum([], 0).
sum([N|Rest], Total) :- sum(Rest, M), Total is N + M.
Test your coding on at least these examples:
?- totalfine('Homer', 25, Amt).
Amt = 20.
?- totalfine('Lisa', 25, Amt).
Amt = 0.
?- totalfine(_, 100, Total).
Total = 295.
Copy and
paste your test cases and their outputs into library.txt.
You download Prolog at http://www.swi-prolog.org/Download.html. (I am happy using an "old version", 5.4.7, but you are welcome to try a newer release.)
When you load a Prolog program, verify there are no errors! (Often an error is triggered by a missing period or comma on the previous line.) Use the listing and trace predicates to see what is happening inside Prolog when you use it.
Plan for several days to do the exercise, because if you get stuck, it might take a day to understand what to do.