notes for cs3310 artificial intelligence part 5: prolog arithmetic and lists prof. neil c. rowe...
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Notes for CS3310 Artificial Intelligence
Part 5: Prolog arithmetic and lists
Prof. Neil C. Rowe Naval Postgraduate School
Version of July 2009
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Prolog arithmeticComparisons are written in infix form, not the usual
prefix form:A == 3 (A=3 also works), A > 3, X < Y, X
=< 3, Y >= 4 Arithmetic assignment is infix too, with operators
+, -, *, and /.For instance: X is (Y * Z) + 3.This is one-way assignment: Right-side variables
must be bound.The left-side variable usually is unbound. Otherwise
the expression checks whether the value of a variable is equal to that of an arithmetic calculation.
Useful: the built-in predicate var of one argument, which succeeds if its argument is unbound.
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Example: abs (absolute value)
abs(X,X) :- X>0.abs(X,AX) :- X=<0, AX is 0-X.
Behavior:
?-abs(3,A).A=3.?-abs(B,3).B=3.?-abs(-3,C).C=3.?-abs(D,-3).! Error in arithmetic expression: not a number
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Examples: mod (remainder after division) and factorial
Examples:?-mod(3,5,A).A=3.?-mod(7,5,B).B=2.?-mod(-7,5,C).C=3
?- factorial(5,F). F=120
Write definitions for these.
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Example: better_add (more flexible addition)
?-better_add(2,4,A).A=6?-better_add(3,A,5).A=2better_add(X,Y,S) :- \+ var(X), \+ var(Y), var(S),
S is X+Y.better_add(X,Y,S) :- \+ var(X), var(Y), \+ var(S), Y
is S-X.better_add(X,Y,S) :- var(X), \+ var(Y), \+ var(S), X
is S-Y.better_add(X,Y,S) :- \+ var(X), \+ var(Y), \
+ var(S), S2 is X+Y, S=S2.
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Examples of Prolog list notation
[a,b,c]: a list of three constants a, b, and c[a,b,X]: a list of three items a, b, and some
unspecified item represented by a variable X[X | Y]: a list whose first item is represented by
variable X, and whose rest-of-list is represented by variable Y (Y is a list of zero or more items)
[a,b,c | Y]: a list whose first three items are a,b,c, and whose rest is represented by some variable Y
[ ]: the empty list (a list of no items)
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Example queries with lists
Database:employees([tom, dick, harry, mary]).Queries:?- employees(L).L=[tom, dick, harry, mary]?- employees([X | L])X = tom, L = [dick, harry, mary]?- employees([X, Y, Z | L]).X = tom, Y = dick, Z = harry, L = [mary]?- employees([X, dick | L]).X = tom, L = [harry,mary]
?- employees([X, tom | L]).no?- employees([X, Y, Z, W | L])X=tom, Y=dick, Z=harry, W=mary, L=[ ]?- employees([X, Y, Z, W, A | L])no?- employees(L), write([joe, jim | L]), nl.[joe,jim,tom,dick,harry, mary]L=[tom,dick,harry,mary]("nl" does carriage return)
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Classic list-processing programs
first(L,X) Item X is first in list L last(L,X) Item X is last in list L member(X,L) Item X is a member of list L append(L1,L2,L) L is the list consisting of the
items of L1 followed by the items of L2
delete(L,X,NL) NL is the result of deleting all occurrences of X from L
deleteone(L,X,NL) NL is the result of deleting the first occurrence of X from list L
length(L,N) List L has N items reverse(L,R) R is the list with the items
of L in reverse order
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First and last
first([X | L],X).last([X], X).last([X | L], Y) :- last(L, Y).employees([tom,dick,harry,mary]).Examples:?- first([a,b,c], A).A=a?- last([a,b,c], B).B=c?- employees(EL), first(EL, E).EL=[tom,dick,harry,mary], E=tom?- employees(EL), last(EL, E).EL=[tom,dick,harry,mary], E=mary
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The member predicate
member(X, [X | L]).member(X, [Y | L]) :- member(X, L).
?- member(c,[a,b,c,d]). X=c, Y=a, L=[b,c,d] ?- member(c,[b,c,d]). X=c, Y=b, L=[c,d] ?- member(c,[c,d]). X=c, Y=c, L=[d]
? -member(E, [a,b,c]).E=a; [line 1]E=b; [line 2, recursion uses line 1]E=c; [line 2, recursion uses line 2 again, which
then uses line 1]no
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More uses of "member”
?-member(a,L). L=[a | _015]; L=[_016,a | _017]?-member(a,[L]). L=a; no?-member(a,[a,b,a,c,a]). yes; yes; yes; no
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Digression: Box diagrams for procedural analysis
Useful way to trace variables in procedural calls. Draw a box for each procedure call, inside the box for the calling procedure; allocate space for each variable.
a(R,S) :- b(R), c(R,T), c(T,R).c(V,W) :- f(V,W), not(f(W,V)),b(W).b(U) :- d(U), e(U).d(1). e(1). f(2,1). Query: ?- a(1,X).
query: X
a call; R: S: T:
b call; U:
c call; V: W:
b call; U:
b call; U:
c call; V: W:
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Example for box diagram for inheritance recursion
Assume query: ?- owns(W,radio_6391).part_of(radio_6391, car_117654).part_of(car_117654, nps_fleet).owns(nps, nps_fleet).owns(P,X) :- part_of(X,Y), owns(P,Y).
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The “append” predicate
The definition of “append” is built-in in most Prolog dialects -- but here’s how you could define it otherwise. Examining this definition should make it clear how it works.
append([],L,L).append([X|L1],L2,[X|L]) :-
append(L1,L2,L).
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Demo of the built-in append predicate
?- append([a,b], [c,d,e], L).L = [a,b,c,d,e];no?- append(a, [c,d,e], L).no?- append([a,b,c], [d,e], [X,Y|L3]).X = a Y = b L3 = [c,d,e];no?- append(L, [r,s], [a,b,r,s]).L = [a,b];no
?- append(L1,L2,[a,b,c,d]).L1 = [] L2 = [a,b,c,d];L1 = [a] L2 = [b,c,d];L1 = [a,b] L2 = [c,d];L1 = [a,b,c] L2 = [d];L1 = [a,b,c,d] L2 = [];no?- append([I1|L1],[I2|L2], [a,b,c,d]).I1 = a L1 = [] I2 = b L2 = [c,d];I1 = a L1 = [b] I2 = c L2 = [d];I1 = a L1 = [b,c] I2 = d L2 = [];no
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Using "member", "append", and "delete" with a databaseDatabase:engineering_depts([ec,me,aa,cs])science_depts([ph,or,oc,mr])(a) Ask whether CS is an engineering department.(b) Ask whether CS is an engineering or science
department.(c) Make a list of all the engineering and science
departments besides CS.
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Exercises with "append”
Using only append on the right side, define:(a) front(List1, List2) (whether List1 items are the
front of List2)(b) member(Item, List)(c) deleteone(List, Item, Newlist) (deletes first
occurrence of an item in a list)(d) substitute(List, Olditem, Newitem, Newlist)
(replaces the first occurrence of Olditem by Newitem in List)
(e) twomembers(Item1, Item2, List) (gives two items in a list such that the first item occurs somewhere before the second item)
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You can't rebind a bound variable in Prolog
?- p(X,Y), p(Y,Z). Here you bind Y in the first predicate expression. The second Y refers to earlier value. You can't force Y to a new value. That's how specifications are -- a variable means the same thing everywhere.
?- Y is 10, Y is Y+1. Similarly, this always fails because Y can't be rebound. The second "is” becomes a check whether Y is equal to Y+1, which is impossible.
?- p(L), append(L1,L2,L), append(L3,L,L1). Similarly, this try to split a list into three pieces always fails. Once L is bound in p, you can't set it to something new (a sublist of L1).
Hence you need additional variables in Prolog to handle new values.
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Defining delete, length, and reverse
This deletes all occurrences from a list.delete([],X,[]).delete([X|L],X,NL) :- delete (L,X,NL).delete([Y|L],X,[Y|NL]) :- \+ X=Y,
delete(L,X,NL).This counts the number of items in a list.length([],0).length([X|L],N) :- length(L,N2), N is N2+1.This reverses a list.reverse(L,R) :- reverse2(L,[],R).reverse2([],R,R).reverse2([X|L],M,R) :- reverse2(L,[X|M],R).
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Recursive-rule exercise
Define a "sort(List,Slist)" that uses insertion sort to sort List into Slist.
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Comparing speed of Prolog and C
Same algorithms were used in both languages, but linked lists were used in Prolog for C arrays. Prolog is better than C for several programs, and not much worse otherwise.
Benchmark CompiledQuintusProlog
UnoptimizedAquarius
Prolog
MIPS C OptimizedMIPS C
recursiveinteger
arithmetic
22.0 1.2 2.1 1.6
Fibonacciseries
- 1.5 2.0 1.6
Towers ofHanoi
- 1.3 1.6 1.5
Quicksort 8.4 2.8 3.3 1.4
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Three new terms
Temporal logic: Rules for reasoning about time. Typically they explain time phenomenon like "before", "until", and "periodically".
Constraints: Predicate expressions that restrict solutions to some problem.
Constraint programming: Programming that focuses on satisfying constraints on a solution. Examples are scheduling, resource allocation, and certain aspects of computer vision.
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Example: reasoning about time
Assume we have a database of facts of the form: event(<name>,<start-time>,<end-time>) where the last two argument are floating-point numbers representing the number of days since January 1, 1950.
We want to define the following:before(X,Y): X is before Yafter(X,Y): X is after Y --Use "before" to define itduring(X,Y): X is during Ybetween(X,Y,Z): Y is between X and Zfirstevent(X): X is the first known eventAlso, why can't we handle because(X,Y)?
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Answer to reasoning about time
(Quintus Prolog "less than or equals" is "=<".)before(X,Y) :- event(X,SX,EX), event(Y,SY,EY),
EX =< SY.after(X,Y) :- before(Y,X).during(X,Y) :- event(X,SX,EX), event(Y,SY,EY), SY =< SX, EX =< EY.between(X,Y,Z) :- before(X,Y), before(Y,Z).between(X,Y,Z) :- before(Z,Y), before(Y,X).firstevent(X) :- event(X,SX,EX), \+ before(W,X).
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Example: scheduling programThis picks 5 meeting times during a week.schedule(Times) :- Times=[T1,T2,T3,T4,T5], classtime(T1), \+ occupied(T1), classtime(T2), \+ occupied(T2), classtime(T3), \+ occupied(T3), classtime(T4), \+ occupied(T4), classtime(T5), \+ occupied(T5), \+ duplication(Times), \+ two_consecutive_hours(Times), \+ three_classes_same_day(Times).classtime([Day,Hour]) :- member(Day, [monday,
tuesday, wednesday, thursday, friday]), member(Hour, [800, 900, 1000, 1100, 1200, 1300, 1400, 1500]).
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The scheduling program, cont.
two_consecutive_hours(TL) :- member([Day,Hour1],TL), member([Day,Hour2],TL), H2 is Hour1+100, H2=Hour2.
three_classes_same_day(TL) :- member([Day,Hour1],TL), member([Day,Hour2],TL), member([Day,Hour3],TL), \+ duplication([Hour1,Hour2,Hour3]).
duplication([X | L]) :- member(X,L).duplication([X | L]) :- duplication(L).member(X,[X | L]).member(X,[Y | L]) :- member(X,L).
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The scheduling program, continued
Sample facts about times not available:occupied([X,900]).occupied([X,1000]).occupied([X,1200]).occupied([X,1500]).occupied([wednesday,Y]).occupied([friday,Y]).First three answers of the program with above facts:| ?- schedule(S).S = [[monday,800],[monday,1100],[tuesday,800],
[tuesday,1100],[thursday,800]] ;S = [[monday,800],[monday,1100],[tuesday,800],
[tuesday,1100],[thursday,1100]] ;S = [[monday,800],[monday,1100],[tuesday,800],
[tuesday,1100],[thursday,1300]] ;
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Exercises with the scheduling program (mutually exclusive)
(a) Modify it so all schedules avoid Friday for any database.
(b) Modify it so all classes are on different days.(c) Modify it so all class hours are within 2 hours apart.(d) Paraphrase in 20 words or less what the following
routine does. Assume it is called last in schedule with the calling expression \+ test([T1,T2,T3,T4,T5]).
test(TL) :- member([D1,U1],TL), member([D1,U2],TL), member([D2,U3],TL), member([D2,U4],TL), \+ U1=U2, \+ U3=U4, \+ D1=D2.
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Improving the speed of scheduling (1)
betterschedule([T1,T2,T3,T4,T5]) :- classtime(T1), \+ occupied(T1), classtime(T2), \+ occupied(T2), \+ duplication([T1,T2]), classtime(T3), \+ occupied(T3), \+ duplication([T1,T2,T3]), classtime(T4), \+ occupied(T4), \+ duplication([T1,T2,T3,T4]), classtime(T5), \+ occupied(T5), \+ duplication([T1,T2,T3,T4,T5]), \+ two_consecutive_hours([T1,T2,T3,T4,T5]), \+ three_classes_same_day([T1,T2,T3,T4,T5]).Note this is faster than the original (and shorter)
program.
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Improving the speed of scheduling (2)
betterschedule2([T1,T2,T3,T4,T5]) :- classtime(T1), \+ occupied(T1), classtime(T2), \+ occupied(T2), before(T1,T2), classtime(T3), \+ occupied(T3), before(T2,T3), classtime(T4), \+ occupied(T4), before(T3,T4), classtime(T5), \+ occupied(T5), before(T4,T5), \+ two_consecutive_hours([T1,T2,T3,T4,T5]), \+ three_classes_same_day([T1,T2,T3,T4,T5]).before([Day1,_],[Day2,_]) :- append(_,[Day1|L2],
[monday,tuesday,wednesday,thursday,friday]), append(_,[Day2|_],L2).before([Day,Hour1],[Day,Hour2]) :- Hour1 < Hour2.This program is faster still.
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NPS class scheduling (2009)• Automated class scheduling at NPS uses several
sophisticated algorithms.• Students and professors specify preferences (days
and times, rooms, consecutive classes).• The program first tries to assign class sections to
rooms ignoring students and professors.• If this works, students then assigned to class
sections.• If this works, professors are assigned to sections.• Any failures are flagged and options can be
selected: Ignoring preferences, reassigning students to sections, introducing new sections, etc. Then scheduling is run again.