Programming Languages Chapter 2: Syntax4430s001/logic_programming.pdf · A simple example: Baker, Cooper, Fletcher, Miller, and Smith live in a five-story

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Copyright © 2006 The McGraw-Hill Companies, Inc.

Programming Languages2nd edition

Tucker and Noonan

Chapter 15Logic Programming

Q: How many legs does a dog have if you call its tail a leg?

A: Four. Calling a tail a leg doesn’t make it one.

Abraham Lincoln

Copyright © 2006 The McGraw-Hill Companies, Inc. 16-3

Introduction

Logic programming languages, sometimes called declarative programming languages

Express programs in a form of symbolic logic

Use a logical inferencing process to produce results

Declarative rather that procedural:– Only specification of results are stated (not

detailed procedures for producing them)

Concepts of Programming Languages, 9th ed., by Robert W. Sebesta. Addison Wesley, 2010.

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Example: Sorting a List

Describe the characteristics of a sorted list, not the process of rearranging a list

sort(old_list, new_list)

permute (old_list, new_list) sorted (new_list)

sorted (list) j such that 1 j < n, list(j) list (j+1)

16-4Concepts of Programming Languages, 9th ed., by Robert W. Sebesta. Addison Wesley, 2010.


15.1 Logic and Horn Clauses

A Horn clause has a head h, which is a predicate, and a body, which is a list of predicates p1, p2, …, pn.

It is written as:

h p1, p2, …, pn

This means, “h is true only if p1, p2, …, and pn are simultaneously true.”

E.g., the Horn clause:

snowing(C) precipitation(C), freezing(C)

says, “it is snowing in city C only if there is precipitation in city C and it is freezing in city C.”

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Horn Clauses and Predicates

Any Horn clause

h p1, p2, …, pn

can be written as a predicate:

p1 p2 … pn hor equivalently:

(p1 p2 … pn) h

But not every predicate can be written as a Horn clause.

E.g., literate(x) reads(x) writes(x)

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Resolution and Unification

If h is the head of a Horn clause

h terms

and it matches one of the terms of another Horn clause:

t t1, h, t2

then that term can be replaced by h’s terms to form:

t t1, terms, t2

During resolution, assignment of variables to values is called instantiation.

Unification is a pattern-matching process that determines what particular instantiations can be made to variables during a series of resolutions.

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Example

The two clauses:

speaks(Mary, English)

talkswith(X, Y) speaks(X, L), speaks(Y, L), XY

can resolve to:

talkswith(Mary, Y) speaks(Mary, English),

speaks(Y, English), MaryY

The assignment of values Mary and English to the variables X and L is an instantiation for which this resolution can be made.

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15.2 Logic Programming in Prolog

In logic programming the program declares the goals of the computation, not the method for achieving them.

Logic programming has applications in AI and databases.

– Natural language processing (NLP)

– Automated reasoning and theorem proving

– Expert systems (e.g., MYCIN)

– Database searching, as in SQL (Structured Query Language)

Prolog emerged in the 1970s. Distinguishing features:

– Nondeterminism

– Backtracking

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15.2.1 Prolog Program Elements

Prolog programs are made from terms, which can be:

– Variables

– Constants

– Structures

Variables begin with a capital letter, like Bob.

Constants are either integers, like 24, or atoms, like the, zebra, ‘Bob’, and ‘.’.

Structures are predicates with arguments, like:

n(zebra), speaks(Y, English), and np(X, Y)

– The arity of a structure is its number of arguments (1, 2, and 2 for these examples).

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Facts, Rules, and Programs

A Prolog fact is a Horn clause without a right-hand side. Its form is (note the required period .):

term.

A Prolog rule is a Horn clause with a right-hand side. Its form is (note :- represents and the required period .):

term :- term1, term2, … termn.

A Prolog program is a collection of facts and rules.

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Inferencing Process of Prolog

Queries are called goals

If a goal is a compound proposition, each of the facts is a subgoal

To prove a goal is true, must find a chain of inference rules and/or facts. For goal Q:

A.

B :- A.

C :- B.

…

Q :- P.

Process of proving a subgoal called matching, satisfying, or resolution

Concepts of Programming Languages, 9th ed., by Robert W. Sebesta. Addison Wesley, 2010. 16-12


Copyright © 2009 Addison-Wesley. All rights reserved. 16-13

Approaches

Bottom-up resolution, forward chaining– Begin with facts and rules of database and

attempt to find sequence that leads to goal– Works well with a large set of possibly correct

answersA. {A}

B :- A. {A, B}

C :- B. {A, B, C}

…

Q :- P. {A, B, C, …, P, Q}

? Q. true

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Approaches

Top-down resolution, backward chaining

– Begin with goal and attempt to find sequence that leads to set of facts in database

– Works well with a small set of possibly correct answers

A. {A, B, C, …, P, Q}

B :- A. {A?, B?, C?, …, P?, Q?}

C :- B. {B?, C?, …, P?, Q?}

…

Q :- P. {P?, Q?}

? Q. {Q?}

Prolog implementations use backward chaining

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Subgoal Strategies

When goal has more than one subgoal, can use either

– Depth-first search: find a complete proof for the first subgoal before working on others

– Breadth-first search: work on all subgoals in parallel

Prolog uses depth-first search

– Can be done with fewer computer resources

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Backtracking

With a goal with multiple subgoals, if fail to show truth of one of subgoals, reconsider previous subgoal to find an alternative solution: backtracking

Begin search where previous search left off

Can take lots of time and space because may find all possible proofs to every subgoal

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Simple Arithmetic

Prolog supports integer variables and integer arithmetic

is operator: takes an arithmetic expression as right operand and variable as left operand

A is B / 17 + C

Not the same as an assignment statement!

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Prolog Example

brb0164@faculty% cat min.pro

min(X, Y, X) :- X < Y.

min(X, Y, Y) :- X >= Y.

brb0164@faculty% gprolog

GNU Prolog 1.2.19

| ?- consult(‘min’).

compiling min.pro for byte code...

min.pro compiled, 2 lines read - 455 bytes written, 17 ms

yes

| ?- min(0, 1, 0).

true ?

yes

| ?- min(17, 45, Minimum).

Minimum = 17 ?

yes

| ?- min(65, 3, Minimum).

Minimum = 3

yes

| ?- ^D

Copyright © 2014 Barrett R. Bryant. 16-18

Copyright © 2006 The McGraw-Hill Companies, Inc. 16-19

Prolog Example

brb0164@faculty% cat factorial.pro

factorial(0, 1).

factorial(N, Factorial) :-

M is N – 1, factorial(M, M_Factorial),

Factorial is N * M_Factorial.


…

| ?- consult(‘factorial’).

compiling factorial.pro for byte code...

factorial.pro compiled, 3 lines read - 847 bytes written, 58 ms

yes

| ?- factorial(5, Factorial).

Factorial = 120 ?

yes

| ?- factorial(10, Factorial).

Factorial = 3628800 ?

yes

| ?- ^D

Copyright © 2014 Barrett R. Bryant.

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Example Program

speaks(allen, russian).

speaks(bob, english).

speaks(mary, russian).

speaks(mary, english).

talkswith(X, Y) :- speaks(X, L), speaks(Y, L), X \= Y.

This program has four facts and one rule.

The rule succeeds for any instantiation of its variables in which all the terms on the right of := are simultaneously true. E.g., this rule succeeds for the instantiation X=allen, Y=mary, and L=russian.

For other instantiations, like X=allen and Y=bob, the rule fails.

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Lists

A list is a series of terms separated by commas and enclosed in brackets.

– The empty list is written [].

– The sentence “The giraffe dreams” can be written as a list: [the, giraffe, dreams]

– A “don’t care” entry is signified by _, as in

[_, X, Y]

– A list can also be written in the form:

[Head | Tail]

– The functions

append joins two lists, and

member tests for list membership.

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append Function

append([], X, X).

append([Head | Tail], Y, [Head | Z]) :-

append(Tail, Y, Z).

This definition says:

1. Appending the empty list to any list (X) returns an unchanged list (X again).

2. If Tail is appended to Y to get Z, then a list one element larger [Head | Tail] can be appended to Y to get [Head | Z].

Note: The last parameter designates the result of the function. So a variable must be passed as an argument.

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member Function

member(X, [X | _]).member(X, [_ | Y]) :- member(X, Y).

The test for membership succeeds if either:

1. X is the head of the list [X | _]

2. X is not the head of the list [_ | Y] , but X is a member of the list Y.

Notes: pattern matching governs tests for equality.

Don’t care entries (_) mark parts of a list that aren’t important to the rule.

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More List Functions

X is a prefix of Z if there is a list Y that can be appended to X to make Z.

That is:

prefix(X, Z) :- append(X, Y, Z).

Similarly, Y is a suffix of Z if there is a list X to which Y can be appended to

make Z. That is:

suffix(Y, Z) :- append(X, Y, Z).

So finding all the prefixes (suffixes) of a list is easy. E.g.:

?- prefix(X, [my, dog, has, fleas]).

X = [];

X = [my];

X = [my, dog];

… 16-24


Prolog Example

brb0164@faculty% cat length.pro

length([], 0).

length([X | Xs], Length) :-

length(Xs, Xs_Length), Length is Xs_Length + 1.


…

| ?- consult(‘length’).

…

yes

| ?- length([a, b, c, d], Length).

Length = 4

yes

| ?- length([a, [b, c]], Length).

Length = 2

Yes

| ?- length([], Length).

Length = 0

yes

| ?- length([a, [b, c], d, [e, f, [g, h]]], Length).

Length = 4

yesCopyright © 2014 Barrett R. Bryant. 16-25

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Lisp Examples in Prolog

brb0164@faculty% cat lispfunctions.pro

car([X | Xs], X).

cdr([X | Xs], Xs).

cons(X, Xs, [X | Xs]).

append([], Ys, Ys).

append([X | Xs], Ys, [X | Zs]) :-

append(Xs, Ys, Zs).





…

| ?- consult(‘lispfunctions’).

…

| ?- cons(x, [], A).

A = [x]

yes

| ?- cons(y, [x], B).

B = [y,x]

yes


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| ?- append([y, x], [z], D).

D = [y,x,z]

yes

| ?- append([y, x], [z], D),

cdr(D, Cdr_D), car(Cdr_D, Cadr_D).

Cadr_D = x

Cdr_D = [x,z]

D = [y,x,z]

yes



Database Example in Prolog

brb0164@faculty% cat ancestors.pro

female(shelley).

female(mary).

female(lisa).

female(joan).

male(bill).

male(jake).

male(bob).

male(frank).

mother(mary, jake).

mother(mary, shelley).

mother(lisa, mary).

mother(joan, bill).

father(bill, jake).

father(bill, shelley).

father(bob, mary).

father(frank, bill).


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parent(Father, Child) :-

father(Father, Child).

parent(Mother, Child) :-

mother(Mother, Child).

parents(Father, Mother, Child) :-

father(Father, Child),

mother (Mother, Child).





…

| ?- consult(‘ancestors’).

…

| ?- parents(Father, Mother, jake).

Father = bill

Mother = mary ?

yes

| ?- parents(bill, mary, Child).

Child = jake ? ;

Child = shelley

yes


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| ?- parents(Father, Mother, Child).

Child = jake

Father = bill

Mother = mary ? ;

Child = shelley

Father = bill

Mother = mary ? ;

Child = mary

Father = bob

Mother = lisa? ;

Child = bill

Father = frank

Mother = joan

yes




sibling(Child1, Child2) :-

father(Father, Child1),

father(Father, Child2),

mother(Mother, Child1),

mother(Mother, Child2).


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| ?- sibling(Child1, Child2).

Child1 = jake

Child2 = jake ? ;

Child1 = jake

Child2 = shelley ? ;

Child1 = shelley

Child2 = jake ? ;

Child1 = shelley


Child1 = mary

Child2 = mary ? ;

Child1 = bill

Child2 = bill

yes




true_sibling(Child1, Child2) :-

sibling(Child1, Child2),

\+ Child1 = Child2. /* \+ is negation */

| ?- true_sibling(Child1, Child2).

Child1 = jake


Child1 = shelley

Child2 = jake? ;

no


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ancestor(Ancestor, Descendant) :-

parent(Ancestor, Descendant).

ancestor(Ancestor, Descendant) :-

parent(Descendants_Parent, Descendant),

ancestor(Ancestor, Descendants_Parent).

| ?- ancestor(Ancestor, jake).

Ancestor = bill ? ;

Ancestor = mary ? ;

Ancestor = frank ? ;

Ancestor = joan ? ;

Ancestor = bob ? ;

Ancestor = lisa ? ;

no



Postfix Example in Prolog

brb0164@faculty% cat postfix.pro

/*

This program converts a postfix expression,

represented as a list of operands and operators,

into a syntax tree.

*/

postfix(Exp, Tree) :- postfix_stack(Exp, [], [Tree]).


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postfix_stack([], Tree, Tree).

postfix_stack([Operator | Rest_exp],

[Top, Next_top | Rest_stack], Tree) :-

operator(Operator),

postfix_stack(Rest_exp,

[tree(Operator, Next_top, Top) | Rest_stack], Tree).

postfix_stack([Operand | Rest_exp], Stack, Tree) :-

postfix_stack(Rest_exp, [Operand | Stack], Tree).

operator(+).

operator(-).

operator(*).

operator(/).





…

| ?- consult(‘postfix’).

…

| ?- postfix([a,b,c,*,+],Tree).

Tree = tree(+,a,tree(*,b,c)) ?

yes

| ?- postfix([a,b,+,c,*,d,/,e,-],Tree).

Tree = tree(-,tree(/,tree(*,tree(+,a,b),c),d),e) ?

yes


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15.2.2 Practical Aspects of Prolog

• Tracing

• The Cut

• Negation

• The is, not, and Other Operators

• The Assert Function

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Tracing

To see the dynamics of a function call, the trace function can be used. E.g., if we want to trace a call to the following function:

factorial(0, 1).

factorial(N, Result) :- N > 0, M is N - 1,

factorial(M, SubRes), Result is N * SubRes.

we can activate trace and then call the function:

?- trace(factorial/2).

?- factorial(4, X).

Note: the argument to trace may include the function’s arity.

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Tracing Output

?- factorial(4, X).

Call: ( 7) factorial(4, _G173)

Call: ( 8) factorial(3, _L131)




Exit: ( 11) factorial(0, 1)





X = 24

These are

temporary

variables

These are

levels in the

search tree

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The Cut

The cut is an operator (!) inserted on the right-hand side of a rule.

semantics: the cut forces those subgoals not to be retried if the right-hand side succeeds once.

E.g (bubble sort):

bsort(L, S) :- append(U, [A, B | V], L),

B < A, !,

append(U, [B, A | V], M),

bsort(M, S).

bsort(L, L).

So this code gives one answer rather than many.

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Bubble Sort Trace

?- bsort([5,2,3,1], Ans).

Call: ( 7) bsort([5, 2, 3, 1], _G221)

Call: ( 8) bsort([2, 5, 3, 1], _G221)

…

Call: ( 12) bsort([1, 2, 3, 5], _G221)

Redo: ( 12) bsort([1, 2, 3, 5], _G221)

…

Exit: ( 7) bsort([5, 2, 3, 1], [1, 2, 3, 5])

Ans = [1, 2, 3, 5] ;

No

Without the cut, this

would have given some

wrong answers.

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The assert Function

The assert function can update the facts and rules of a program dynamically. E.g., if we add the following to the foregoing database program:

?- assert(mother(jane, joe)).

Then the query:

?- mother(jane, X).

gives:

X = ron ;

X = joe;

No

F

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15.3.2 Solving Word Problems

A simple example:

Baker, Cooper, Fletcher, Miller, and Smith live in a five-story building. Baker doesn't live on the 5th floor and Cooper doesn't live on the first. Fletcher doesn't live on the top or the bottom floor, and he is not on a floor adjacent to Smith or Cooper. Miller lives on some floor above Cooper. Who lives on what floors?

We can set up the solution as a list of five entries:

[floor(_,5), floor(,4), floor(_,3), floor(_,2), floor(_,1)]

The don’t care entries are placeholders for the five names.

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Modeling the solution

We can identify the variables B, C, F, M, and S with the five persons,

and the structure floors(Floors) as a function whose argument is the list to be solved.

Here’s the first constraint:

member(floor(baker, B), Floors), B\=5

which says that Baker doesn't live on the 5th floor.

The other four constraints are coded similarly, leading to the following program:

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Prolog solution

floors([floor(_,5),floor(_,4),floor(_,3),floor(_,2),

floor(_,1)]).

building(Floors) :- floors(Floors),

member(floor(baker, B), Floors), B \= 5,

member(floor(cooper, C), Floors), C \= 1,

member(floor(fletcher, F), Floors), F \= 1, F \= 5,

member(floor(miller, M), Floors), M > C,

member(floor(smith, S), Floors), not(adjacent(S, F)),

not(adjacent(F, C)),

print_floors(Floors).

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Auxiliary functions

Floor adjacency:

adjacent(X, Y) :- X =:= Y+1.adjacent(X, Y) :- X =:= Y-1.

Note: =:= tests for numerical equality.

Displaying the results:

print_floors([A | B]) :- write(A), nl, print_floors(B).print_floors([]).

Note: write is a Prolog function and nl stands for “new line.”

Solving the puzzle is done with the query:

?- building(X).which finds an instantiation for X that satisfies all the constraints.

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Documents

Programming Languages Chapter 2: Syntax4430s001/logic_programming.pdf · A simple example: Baker, Cooper, Fletcher, Miller, and Smith live in a five-story