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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 arithmetic Comparisons are written in infix form, not the usual prefix form: A == 3 (A=3 also works), A > 3, X = 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=2 better_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

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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; 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: 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 database Database: 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.

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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(,, ) 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 Y after(X,Y): X is after Y --Use "before" to define it during(X,Y): X is during Y between(X,Y,Z): Y is between X and Z firstevent(X): X is the first known event Also, 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 program This 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.

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Section 4.5 Exp. & Log Equations

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