Check Admin
sakai to make sure your grades are what you expect Homework 4, due monday before lecture Midterm Programming Problems
#1, #2, #3, #6 can earn back up to 2 points per problem figure out and type up correct solution in electronic form email mcore AT rci.rutgers.edu and elqursh AT cs.rutgers.edu DUE MONDAY BEFORE LECTURE
Logic Programming, MGCore, BG Ryder 1
Admin
either download prolog from http://www.swi-prolog.org or use it from spanky.rutgers.edu (command is called pl)
Logic Programming, MGCore, BG Ryder
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Course Progress
Python
imperative programming paradigm use of iteration / polymorphism
Regular Expressions / Context Free Grammars Scheme and Python
functional programming paradigm
Semantics Prolog
logic programming paradigm
Semantics (cont) parameter passing, data types
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Logic Programming Paradigm
formulate your problem as a fact to be proven given other facts and rules facts and rules encoded in logic
Prolog: uses horn clause syntax subset of first order predicate logic
query: fact to be proven solution: can proof be found? what are the values of variables used in the proof?
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Logic Programming
solution comprises:
result: proof succeeds or not variables set during proof
Separate problem description from how to solve the problem Comerical interpreters available: http://www.sics.se/isl/sicstuswww/site/customers. html Wikipedia: http://en.wikipedia.org/wiki/Prolog a tool that can be used in conjunction with other languages/paradigms
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Horn Clauses
raining. cold.
body . acts to end Horn clauses in Prolog
body1 , ... , bodyn . asserts body1...bodyn are true
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Horn Clauses
rainy(rochester). cold(rochester).
body . acts to end Horn clauses in Prolog
body1 , ... , bodyn . asserts body1...bodyn are true
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Horn Clauses
snowy(X) :- rainy(X), cold(X).
head implies and . acts to end Horn clauses in Prolog
body
head :- body1 , ... , bodyn . if body1...bodyn are true then head is true if body then head is asserted to be true
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Prolog
Based on predicate calculus Main language constructs
variables / constants youve seen them before but remember the Prolog convention: variables start with a CAPITAL letter predicates define relations:
rainy(rochester). means rochester belongs to set of rainy things lecturer(core,cs314). means pair: core, cs314 belongs to set of lecturers and their classes
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Prolog Problem Solving
Programmer
specifies facts and rules specifics query to be proven
Prolog
backward chaining (i.e., start with query and match against facts and rules) if query matches fact then query is proven if query matches left hand side of rule, then try to prove right hand side elements (recursively)
work left to right if fail then backtrack and try again using different rules
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Resolution
Resolution principle:
Given Horn clauses C1 and C2 If head of C1 matches one of the terms of C2s body then this term can be replaced with the body of C1
Ex.
snowy(X) :- rainy(X), cold(X). has_snowplows(X) :- snowy(X). has_snowplows(X) :- rainy(X), cold(X)
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variables
snowy(X) :- rainy(X), cold(X).
head implies and . acts to end Horn clauses in Prolog
body
all three Xs here must have same value
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Unification
Variables appear in Horn clauses with no definition If we wrote these clauses in first order predicate calculus we would explicit define the variables as follows:
variables that appear in the head are universally quantified: e.g., forall(X) snowy(X) :- rainy(X) , cold(X) variables that only appear in the body are existentially quantified: e.g., forall(X) forall(Y) sister_of(X,Y) :exists(M) exists(F) female(X) , parents(X,M,F) , parents(Y,M,F)
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Unification
Prolog binds variables during proof We can use the = operator to check whether two atomic formula unify e.g., a =a. A=a. likes(eve,tom) = likes(eve,tom). see page 566 of text
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Unification Rules
Constants must match Predicates with the same name match if their arguments match recursively an uninstantiated variable unifies with anything
if it unifies with a constant or instantiated variable, it takes the value of that term if two uninstantiated variables unify they are linked to always share the same value
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Unification Examples
a = a. a = b. X = Y. X = Y, X = a. X = Y, X = a, Y = b. mortal(socrates) = mortal(X).
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Experimenting with Prolog Queries
Type facts Query to get results use examples from text and/or make up your own: rainy(X). takes(jane_doe, cs254). takes(X,cs254).
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Experimenting with SWI-Prolog
http://www.swi-prolog.org (works for Windows XP maybe others) type facts and rules into text file
default working directory is: My Documents/Prolog try File menu option Navigator to change
[test].
syntax for load file test.pl from working directory dont forget period at end
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Experimenting with SWI-Prolog
If multiple matches are available, typing ; will list them When no more matches are available, interpreter notifies you with ?- prompt (you cant type ; at the ?- prompt) Try out: A = B. (i.e., is it true that A equals B?). SWI-Prolog gives a smart answer
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Queries (Asking Questions)
likes(eve, pie). food(pie). likes(al, eve).food(apple). likes(eve, tom). person(tom). likes(eve, eve). ?-likes(al,eve). ?-likes(al, pie) ?-likes(eve,al). ?-likes(al,Who). ?-likes(eve,W).
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Queries (Asking Questions)
likes(eve, pie). food(pie). likes(al, eve).food(apple). likes(eve, tom). person(tom). likes(eve, eve). ?-likes(al,eve). yes query ?-likes(al, pie) no answer ?-likes(eve,al). no
variable
?-likes(al,Who). Who=eve ?-likes(eve,W). W=pie ; answer with W=tom ; variable binding W=eve ; force search for MGCore, BG Ryder more answers 21
Logic Programming,
Basic Prolog Queries
List all cases where someone is a person List all pairs of some entity liking another entity List all pairs of some person liking another entity List all entities liked by both eve and al e.g., linked_to(homepage314,Y), linked_to(Y,Z)
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Basic Prolog Queries
List cases where predicate is true
e.g., person(X). likes(X,Y).
In case of multiple matches, type either ; or c
Use variable unification
e.g., likes(X,X). likes(X,Y), person(X), food(Y). likes(eve,Y) , likes(al,Y). e.g., linked_to(homepage314,Y), linked_to(Y,Z)
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Prolog Problem Solving
Prolog Interpreter ?- [royalty]. % select graphical debugger from debug menu ?- spy(sister_of). ?- sister_of(alice,Z). Y = edward ; Y = alice DATABASE: royalty.pl 2. male(albert). 3. female(alice). 4. male(edward). 5. female(victoria). 6. parents(edward,victoria,albert). 7. parents(alice,victoria,albert). 8. sister_of(X,Y) :- female(X), parents(X,M,F), parents(Y,M,F).
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Prolog Problem Solving
Prolog Interpreter ?- [royalty]. ?- sister_of(alice,Z). Y = edward ; Y = alice 1. match query to rule in #7 2. bind X=alice, Y=Z 3. prove female(alice) 4. prove parents(alice,M,F) bind M=victoria, F=albert 5. prove parents (Z,victoria,albert) first match Z=edward Note, each use of the rule involves local variables, so each use is independent by default. DATABASE: royalty.pl 2. male(albert). 3. female(alice). 4. male(edward). 5. female(victoria). 6. parents(edward,victoria,albert). 7. parents(alice,victoria,albert). 8. sister_of(X,Y) :- female(X), parents(X,M,F), parents(Y,M,F).
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Prolog Problem Solving
Prolog Interpreter ?- [royalty]. ?- sister_of(alice,Z). Y = edward ; Y = alice 1. match query to rule in #7 2. bind X=alice, Y=Z 3. prove female(alice) 4. prove parents(alice,M,F) bind M=victoria, F=albert 5. prove parents (Z,victoria,albert) first match Z=edward, second match Z=alice Each recursive prove has its own copy of rules, facts and variable bindings
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DATABASE: royalty.pro 2. male(albert). 3. female(alice). 4. male(edward). 5. female(victoria). 6. parents(edward,victoria,albert). 7. parents(alice,victoria,albert). 8. sister_of(X,Y) :- female(X), parents(X,M,F), parents(Y,M,F).
Logic Programming, MGCore, BG Ryder
Solution: Additional Notation (pt 1)
Prolog can either prove or fail to prove a query (cant say the query is actually false) The not operator \+ applied to a predicate creates an expression that is proven when Prolog fails to prove the predicate \+likes(eve,al) can be proven since this fact is not in the database
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Solution: Additional Notation (pt 2)
== (same value) operator is similar to = (unifies) but simpler (no recursive checking) add constraint that you cant be your own sister:
sister_of(X,Y) :- female(X), parents(X,M,F), parents(Y,M,F) , \+(X == Y).
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Additional Notation
NOT = \+
likes(eve,Y) , \+food(Y) Note, order is important. Dont type: \+food(Y), likes(eve,Y) Prolog evaluates left to right NOT acts as a constraint, Y must be bound first Prolog attempts to prove whatever is inside the not, if it fails, the not succeeds but no variable bindings are returned (e.g., it doesnt return all the possible values that Y could take)
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Additional Notation
; = OR
likes(eve,Y) , (food(Y);person(Y)).
_ = temporary variable, dont care, arent reusing
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Transitive Relations
parents(jane,bob,sally). parents(john,bob,sally). parents(sally,al,mary). sally bob parents(bob,mike,ann). parents(mary,joe,lee). jane john ancestor(X,Y) :- parents(X,Y,_). ancestor(X,Y) :- parents(X,_,Y). ancestor(X,Y) :- parents(X,W,_),ancestor(W,Y). ancestor(X,Y) :- ancestor(W,Y). parents(X,_,W), X= al ; ?- ancestor(jane,X). X = mary ; X = bob ; X = joe ; X = sally ; X = lee X = mike ; Logic Programming, MGCore, BG Ryder X = ann ;
Family tree lee joe ann mike mary al
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Prolog v. Logical Programming
NOT operator
Prolog: \+(predicate(...)) can be proven if predicate(...) cannot be proven Logic: NOT(predicate(...)) is true if predicate is false
Order of evaluation
Prolog (deterministic)
prove subgoals left to right follow order of database file
Logic (nondeterministic)
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Horn Clauses
Basis for rules, facts, and queries Built from:
predicates, constants, variables operators: not (\+), and (,), unifies (=), samevalue (==) parentheses
New element: Lists
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Lists
Syntax: [a, b, c]
sequence of atomic formula and lists separated by commas
[] = empty list Conceptually this list is represented as:
a
b
c
[]
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Logic Programming, MGCore, BG Ryder
Lists
Additional syntax for [a, b, c], | = tail [a | [b , c] ] Unify [ H | T] w/ [a b c] and H =a and T = [b,c]
a
b
c
[]
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Logic Programming, MGCore, BG Ryder
Example: Unification of Lists
[H | T]
a
b
c
[]
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Logic Programming, MGCore, BG Ryder
Unifying Lists
[a] = [ H | T].
H=a T = []
[a, b] = [ H | T].
H=a T = [b]
[a, b, c] = [ H | T].
H=a T = [b, c]
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member predicate
member(A, [A|B]). member(A, [B|C]) :- member(A,C). although similar to functions, Prolog rules do not generally have defined inputs and outputs We can use member like a boolean function:
?- member(a,[b,k,z,a,t]). more? c Yes Logic Programming, MGCore, BG Ryder
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member predicate
We can also use member to return the values of a list:
?- member(X,[b,k,z,a,t]).
Or define a variable in a list given that membership holds
?- member(a,[b,k,z,X,t]).
Given these simple examples, what does Prologs search process look like?
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Prolog Search Tree
member(X,[a,b,c]) X=A=a,B=[b,c] X=A,B=a,C=[b,c] X=a success X=A=b,B=[c] X=b success member(X,[b,c]) X=A,B=b,C=[c] member(X,[c]) X=A X=A=c,B=[ ] B=c, C=[ ] X=c member(A,[ ]) success fail fail
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1. member(A, [A | B] ). 2. member(A, [B | C]) :- member (A, C).
Logic Programming, MGCore, BG Ryder
Prolog Search Tree
prove: member(a,[b,c,X]).
apply member(A,[A|B])
a not equal b unification fail
apply member(A,[B|C]) :- member(A,C).
bind A=a, B=b, C=[c,X] prove: member(a,[c,X]) apply member(A,[A|B])
a not equal c fail
apply member(A,[B|C]) :- member(A,C).
bind A=a, B=c, C=[X] prove: member(a,[X]) apply member(A,[A|B]) bind A=a, A=X, B = [] X=a (; backtrack) apply member(A,[B|C]) :- member(A,C). bind A=a, B=X C=[], prove: member(a,[ ])
Prolog Search Tree
prove: member(X,[a,b,c]).
apply member(A,[A|B])
bind X=A, A=a, B=[b,c], X=a (; backtrack)
apply member(A,[B|C]) :- member(A,C).
bind A=X, B=a, C=[b,c] prove: member(X,[b,c]) apply member(A,[A|B])
bind X=A, A=b, B = [c], X=b (; backtrack)
apply member(A,[B|C]) :- member(A,C).
bind A=X, B=b, C=[c] prove: member(X,[c]) apply member(A,[A|B]) bind X=A, A=c, B = [] X=c (; backtrack) apply member(A,[B|C]) :- member(A,C). bind X=A, B=c C=[], prove: member(X,[ ])
member(X,Y)
prove: member(X,Y).
apply member(A,[A|B])
bind X=A, Y=[A|B], Y=[X|B] (; backtrack)
INFINITE LOOP: You can type ; forever Lazy evaluation: extends search as needed (e.g., when we type ; )
apply member(A,[B|C]) :- member(A,C).
bind X=A, Y=[B|C] prove: member(X,C) apply member(A,[A|B]) apply member(A,[B|C]) :- member(A,C).
bind X=A, C=[A|B], Y=[B|[X|B]]=[B, X | B] (; backtrack) bind X=A, C=[B|C] prove: member(X,C) apply member(A,[A|B]) bind X=A, C=[A|B], Y= [B|[B|[X|B]]]=[B, B, X | B]
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Alternate member predicate
Note, we didnt really use the B variable:
member(A, [A|B]). member(A, [B|C]) :- member(A,C).
So we can write
member(A, [A| _ ]). member(A, [ _ |C]) :- member(A,C).
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minimum
basic idea: define minimum predicate ex: minimum([1,2,4,-1,7],Min). Min = -1 second argument stores return value of function
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minimum
basic idea: define minimum predicate ex: minimum([1,2,4,-1,7],Min). Min = -1 insight 1: first need to define predicate: compute_minimum(1,[2,4,-1,7],Min) which has an extra argument for the current minimum
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minimum
define compute_minimum predicate ex: compute_minimum(1,[1,2,4,-1,7],Min). Min = -1 insight 2: base case compute_minimum(CurrMin,[],CurrMin).
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minimum
define compute_minimum predicate ex: compute_minimum(1,[1,2,4,-1,7],Min). Min = -1 insight 3: recursive rule 1 (head > CurrMin) compute_minimum(CurrMin,[H|T],Ret) :H > CurrMin, compute_minimum(CurrMin,T,Ret)
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minimum
define compute_minimum predicate ex: compute_minimum(1,[1,2,4,-1,7],Min). Min = -1 insight 4: recursive rule 2 not(head > CurrMin) compute_minimum(CurrMin,[H|T],Ret) :not(H > CurrMin), compute_minimum(H,T,Ret)
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minimum
define minimum using compute_minimum predicate
minimum([H|T],Ret) :compute_minimum(H,T,Ret). assumes list has one element
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minimum in action
minimum([1,2,4,-1,7],1) = No minimum([1,2,4,X,7], compute_minimum(CurrMin,[H|T],Ret) :not(H > CurrMin), compute_minimum(H,T,Ret) % mathematical comparison doesnt work % with variables
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append
idea: append([a],[b],Ret) Ret = [a,b] base case:
append([ ], A, A).
recursive case:
append([H|T],List2,%fill in ret val) :append(T,List2,Ret).
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append
idea: append([a],[b],Ret) Ret = [a,b] base case:
append([ ], A, A).
recursive case:
append([H|T],List2,[H|Ret]) :append(T,List2,Ret).
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append predicate
1. append([ ], A, A). (see page 565 of text) 2. append([H|T], A, [H|Ret]) :- append(T,A,Ret). multiple functions, main o...