Programming Languages

2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 Introduction Objectives Foundations of Prolog Notations in Prolog for Building Blocks How Prolog System Solves Problems Back Tracking Data Types and Structures in Prolog Operations on Lists in Prolog The Equality Predicate ‘=’ Arithmetic in Prolog The Operator Cut Cut and Fail Summary Solutions/Answers Further Readings

Page Nos.
42 43 43 46 50 54 55 57 61 62 63 65 66 67 70

We mentioned in the previous unit that there are different styles of problem solving with the help of a computer. For a given type of application, some style is more appropriate than others. Further, for each style, some programming languages have been developed to support it. In this context, we have already discussed two styles of solving problems viz imperative style and functional style. Imperative style is supported by a number of languages including C and FORTRAN. Functional style is supported by, among others, the language LISP. The language LISP is more appropriate for A. I. applications. There is another style viz declarative style of problem solving. A declarative style is non-procedural in the sense that a program written according to this style does not state exactly how the computational process is to be carried out. Rather, a program consists of mainly a number of declarations representing relevant facts and rules concerning the problem domain. The solution to be discovered is also expressed as a question to be answered or, to be more precise, a goal to be achieved. This question/goal also forms a part of the PROLOG program that is intended to solve the problem under consideration. The main technique, in this style, based on resolution method suggested by Robinson (1965), is that of matching goals (to be discussed) with facts and rules. The matching process generates new facts and goals. The matching process is repeated for the whole set of goals, facts and rules, including the newly generated ones. The process, terminates when either all the initial goals, alongwith new goals generated later, are satisfied or when it may be judged or proved that the goals in the original question are not satisfiable. Logic programming is a special type of declarative style of programming, in which the various program elements and constructs are expressed in the notations similar to that of predicate logic. Two interesting features of logic programs are nondeterminism and backtracking. A non-deterministic program may find a number of solutions, rather than just one, to a given problem. Backtracking mechanism allows exploration of potential alternative directions for solutions, when some direction, currently being investigated, fails to find an appropriate solution. The language


PROLOG, which sports Logic Programming style, has been found useful in A.I. applications. In this unit, we have introduce the programming language PROLOG,

A.I. Languages-2: PROLOG

After going through this unit, you should be able to: • explain the declarative style and logic programming style of problem solving, which form the basis of PROLOG programming; • explain the difference between declarative style and imperative style, where the later forms the basis of conventional programming languages like, FORTRAN, Pascal, C etc. • tell us the rules of syntax of PROLOG; • to write PROLOG programs, using the syntax of the language; • explain how PROLOG system solves a problem; • explain the concept of backtracking and its use in solving problems; • enumerate the various data types and data structures available in PROLOG, and further be able to use the available types and structures in defining the data for a program; • enumerate and apply various PROLOG operations on lists; • enumerate and use predicates in expressions, and • explain the significance of the operator Cut and operator Fail and further should be able to use the operators ‘cut’ and ‘fail’ in programs to solve problems.

The programming language PROLOG derives its name from PROgramming in LOGic . The fundamentals of PROLOG were developed in the early 1970’s by Alain Colmerauer and Phillipe Roussel of the Artificial Intelligence group at the University of Marseille together with Robert Kowalski of Department of Artificial Intelligence at University of Edinbury, U.K. There are a number of dialects of PROLOG. Two most important characteristics of PROLOG are as given below: (i) The facts and rules are represented using a syntax similar to that of predicate logic. These facts and rules constitute what is called PROLOG database/knowledge base. The process of problem solving through PROLOG is carried out mainly using an in-built inferencing mechanism based on Robinson’s resolutions method. Through the inferencing mechanism, in the process of meeting the initially given goal(s), new facts and goals are generated which also become part of the PROLOG database. The syntax of PROLOG is based on Horn Clause, a special type of clause in predicate logic. We know that a Clause is a predicate logic expression of the form p1 ∨ p2…..∨ pi ∨ …..pr, where each pi is a literal, and ‘∨’ denotes disjunction. Further a literal q is of the form p or ~ p (negation of p) where p is an atomic symbol. A Horn Clause is a clause that has at most one positive literal, i.e., an atom.



Q. it is a sentence to which a truth value TRUE is assigned. 44 . where the names like Mohan etc.I. Aslam is richer than John. john). may be stated as is_richer (aslam. Further the conjunction symbol is replaced by ‘.H. Repeating. Ram and Sita are parents of Kush. as a variable. Mohan is tall.. the symbol ‘:-’ is read as ‘if’ and comma on R. The formula of predicate logic P ∧ Q ∧ R → S is written as S: . the predicate logic clause P ∧ Q ∧ R → S is equivalently represented in PROLOG as S: . in the domain of the problem under consideration. The integer 5 is greater than the integer 4 may be stated as: is_greater (5. is treated. if true. 4. R. Examples of the statements that may be facts are: 1.’ (Comma).S stands for conjunction. denote a constant. then the fact may be stated as is_tall (mohan).e.A. if true. Anuj is father of Gopal. sita). 5. Gold is a precious metal may be stated as: is_precious (gold). R. 2. hence. A fact is a proposition. in PROLOG. Programming Languages There is some minor difference between the predicate logic notation and notation used in PROLOG. gopal). Details are given later on. (note the full stop following the last symbol R) Where the symbol obtained from writing ‘:’ (colon) followed by ‘-’ (hyphen) is read as ‘if ’. (Note the full stop at the end) Problem solving style using PROLOG requires statements of the relevant facts that are true in the problem domain and rules that are valid. 4). denote a particular person and. In this case is_tall is a property and if Mohan is actually tall (by some criteria).P. i. (the only other truth value is FALSE). 3.P. may be stated as: father (anuj. Q. 6. (the reason for starting the name Mohan with lower-case ‘m’ and not with upper-case letter ‘M’ is that any sequence of letters starting with an upper-case letter. again. may be stated as parents (kush. Summarizing. ram. Facts are about various properties of objects of the problem domain and about expected-to-be useful relations between objects of the problem domain.

Numerals like. Gold etc. we may use the general statement involving variables.I. there are facts such as: If x is father of y and y is father of z then x is grand_father of z. . But numerals are also names of course. parents (X. parent. . facts and rules can be stated in terms of (i) (ii) Atoms (which represent object names) like. Also.. parents (Y. viz. .female(X). and Predicates or relation names like. which are better stated in general terms. using conjunct notations of predicate logic. The above statements show that facts are constituted of the following types of entities viz (i) Objects (or rather object names) like. Similarly. though. Mohan is tall and Aslam is richer than John may be stated . slightly incorrectly. it is now clear that knowledge of a problem domain can be stated in terms of facts and rules. as is_tall (mohan) ∧ is_richer (aslam. is_precious. Mohan. in stead of being stated though innumerable number of facts in which x’s are given specific names of persons. In a PROLOG statement.3 etc. F. A. 36. Sabina is sister of Aslam. On the other hand. generally. in stead of stating innumerable number of facts about the relation of sister as Anita is sister of Anuj. Jane is sister of Johan. 14. numerals are called numbers. Gold etc. The above rule may be stated in PROLOG as: is_sister-of (X. a constant is either an atom or a number. the following type of entities will be introduced soon: (iv) Variables 8. john). father. Languages-2: PROLOG (ii) (iii) Predicates are also called Functors in PROLOG Further. F. However. M and F. Y):. . Further. X. M). M). in the form of the following rule: X is sister of Y if X is a female and X and Y have the same parents M and F. that of numbers. From the above discussion. Ram.7. In PROLOG names of the objects are called atoms. Y. y’s specific names of their respective fathers and z’s specific names of respective grand-fathers. numbers 45 . Mohan. Ram. is_ richer and is_greater etc.

Constants in PROLOG represent specific objects and specific relationships. Atomic formula or structure (atomic formula is different from atom) (vi) Out of the eight examples of statements which were considered a while ago. Programming Languages (iii) (iv) (v) Variables like. atoms. and Comments: A string of characters strings generally enclosed between the pair of signs. predicates. variables and structures. The comments are ignored by PROLOG system. numbers. atomic formulae are called structures. numbers. And.I. whether natural or formal. statements etc. in logic programming style of problem solving. . Terms: functors. 2. viz. We will discuss structure again under data structures of PROLOG. X. numbers variables and even other structures. are formed from the elements of a set of characters. In a logical language. ∨ (disjunction) → (implication) and ↔ (biimplication). viz. 46 . this knowledge plays very important role in solving problems. viz.A. The alphabet set of PROLOG is: ABCDEFGHIJKLMNOPQRSTUVWXYZ a b c d ef gh ij kl m n o p q r s t u v w x yz 0 1 2 345 6 789 + – * / \ ^< >= ‘ ~ : . parent. Predicates or relation names like. structures and constants including numbers are called terms. the first six do not involve any of the logical operators. a structure is of the form: functor (parameter list) where functor is a predicate and parameter list is list of atoms.. as mentioned earlier. In general. This set is called alphabet set of the language. ∧ (conjunction).3 NOTATIONS IN PROLOG FOR BUILDING BLOCKS Alphabet Set of a Language: In any written language. Next we discuss these rules. the atoms. is-precious. viz. father. In PROLOG. the various linguistic constructs like words expressions. Such statements which do not contain any logical operators. denotes a comment in PROLOG. there are different rules to build the type of terms.. A variable may be thought of as something that stands for some object from a set but. numbers and atoms. variables and atomic formulas are basic building blocks for expression of facts and rules constituting knowledge of the domain under consideration. Y and Z. (upper-case letters) (lower-case letters) (numbers) For each type of terms. it is not known for which particular object. is_richer and is_greater etc. Constants are of two types. ‘/*’ and ‘*/’. ~ (negation). ? @ # $ & ! ” %( )’ | { } [ ] . are called atomic formulae.

For example. then we may use the following statement: ?-likes (_ . represents a single atom. digits and underscores which begins with an underscore. digits and underscores that begins with an uppercase letter. but within single quotes all character of the alphabet are allowed) (xi) ‘Anand Prakash’ (the blank symbol is not allowed within a single atom.03e –12 In the above. 87. different occurrences of “-” may be associated with different constants: ?.6 e 2 denotes the number 87. if we want to know if there is anyone who likes Mohan. X). For example. Underscore denotes a special type of variable.’ is associated with Ankit and the second occurrence of ‘-’ is associated with ‘Suresh’. (iii) all special symbols like “?-” and “:-” are also atoms in PROLOG Examples of Atoms (i) circle (ii) b (iii) =(equal to sign) (iv) _(underscore) (v) ‘→’ (vi) _beta (vii) mohan (viii) 3 (ix) abdul_kalam(uses underscore) (x)‘abdul-kalam’ (uses hyphen. the second someone may be different from the first someone. (ii) any sequence of letters. However.I. digits and underscores (but no sign character) (ii) a string or a sequence of characters from the alphabet set (including sign characters) enclosed between apostrophes. The term 35.) But the whole sequence of characters in ‘Anand Prakash’ including single quotes. The PROLOG system. Thus. in the following query.58 0 87. the above statement in this case says ‘Ankit like himself’. Hyphen is not allowed to be a part of an atom. the two occurrences of X denote the same person. Thus. _result (iii) The symbol “_” (underscore) alone. called anonimous variable.Numbers: How numbers are represented in PROLOG is illustrated through the following examples of representation of numbers: 8 7 – 3. -).03e–12 denotes 35. e. the “e” notation is used to denote a power of 10. without being interested in who likes Mohan.likes (-. The following are not atoms (i) 3mohan (starts with a number) (ii) abdul-kalam (has hyphen in-between) (iii) Abdul_kalam (starts with a capital letter) Predicate: Predicate have the same notations as atoms Variables: The following notations are used for variables in PROLOG: (i) A string of letters.likes (-.g. The difference between the behaviours of other variables like ‘X’ and ‘_’ is explained by the following two queries. if first occurrence of X is associated with Ankit then second occurrence of X is automatically associated with Ankit. may respond as ‘Ankit likes Suresh’. where first occurrence of ‘. Languages-2: PROLOG 47 . The query above asks to find someone who likes someone. for ?. In the query ?-likes (X. A.-). hence. mohan).6 x 102 or 8760.03 x 10–12 Atom: an atom is represented by (i) either a string in which the first symbol is a lower-case letter and other characters in the string are letters.6e2 35.

. is_richer (aslam. ankur). owns (mohan. a2. we know a Horn Clause can have at most one positive literal.Body. Rule Statements: In general. we have also already discussed representations of constants and structures. And. the syntax of PROLOG does not allow arbitrary clauses but only Horn Clauses. Query/Question Statements: We mentioned earlier. the following single (non-atomic) statement of predicate calculus is_tall (mohan) ∧ is_richer (aslam. However.I. about objects and their relationships 48 . (states the fact that Anita is a sister of Ankur). Term: We have already mentioned that a term is either a constant or a structure. a2. a3. john) has to be expressed as two Prolog statements as follows: is_tall (mohan). a4 is the Head of the rule and the R. Programming Languages Structure: As mentioned earlier. ‘a1. (states the fact that mohan owns a book) Fact with Compound Statement: For expressing facts and relations. a5’ is the body of the rule. Therefore. a rule in PROLOG is of the form: Head:. The Head of a rule represents a goal to be achieved and the Body represents one or more of the subgoals (each represented by single (atomic) structure) each of which must be achieved if the goal represented by the Head can be said to have been achieved. i. (note the dot at the end of each statement) Head & Body of a Rule In the rule a4:.e. Fact: A simple statement (i. statement not involving logical operators: ~ ∧ ∨ → and ↔) is represented by a structure followed by a full-stop.H. a5. book). computer programming in PROLOG consists of: • specifying facts.. Repeating what has already been said. in PROLOG goals may be conjuncted.S of ‘:-’ . • defining rules. where Head is a (single) structure and Body is a finite sequence of structures separated by commas. And. Example is_sister(anita. john). a3. a structure is of the form: Predicate (parameter list) where parameter list consists of atoms and variables separated by commas .e.A. and • asking questions.a1.

Case (iv) Still further. represents the fact which when expressed in English becomes ‘Mohan is tall’. The above rule may be stated in PROLOG as: is_sister of (X. F. The only difference in the representations is that representation of a fact when preceded by the symbol ‘?-’ (question mark followed by hyphen) becomes the representation of a question. X). Y) of. parents (X. M).is_tall (mohan). However. the conjunct of atomic goals representing the (composite) question is represented in Prolog by writing the atomic goals separated by commas. john). Y). A PROLOG program consists of a finite sequence of facts. represents the command: Find all pairs (represented by pair (X.female(X). denotes the question which when expressed in English becomes: ‘Is Mohan tall?’ We mentioned earlier that for representing facts and rules in PROLOG. we discuss PROLOG representation of questions. the above PROLOG statement is a sort of the following command: Find the names (represents by upper-case letter X) of (all) sisters of John. the solving of a problem using a PROLOG system. rules and a query or goal statement. requires asking questions. However. For example. A. Case (iii) Further. persons in which the first person is a sister of the second person in the pair. Case (i) Then the question in PROLOG ?. the following PROLOG notation ?. we are restricted to using only Horn Clauses. M). the PROLOG question: ?-is_sister (X.I. richer (aslam. The PROLOG representation of questions is almost the same as that for facts. where second person may be a male or a female. Y):. parents (Y. john). the statement is_tall (mohan). F. john). any one of which may be an atomic goal or may be a conjunct of more than one atomic goals. For example.is_tal (mohan). In the later case. ?-is_sister (X. represents the question (in English): Is Jane sister of John? Case (ii) Then the question in PROLOG. Next. represents the question: Who (represent by upper-case letter X) are John’s sisters. if any? Rather. if we want to know whether ‘Mohan is tall and Aslam is richer than John?’ then the question may be stated in PROLOG as ?. the PROLOG question: ?-is_sister (jane.is_sister (jane. represents the command: Find the names (represented by the upper-case letter X) of all siblings (brothers and sisters) of jane. Languages-2: PROLOG 49 . Some interpretations of question forms in PROLOG Let us consider the rule: X is sister of Y if X is a female and X and Y have the same parents M and F.We have already discussed PROLOG notation for specifying facts and defining rules.

These topics will be taken up later on. in addition to what we have discussed so far we need to discuss in some detail representation of arithmetic facts. f) and father (mohan.is_sister (sita. m). parents (Y. moha). is not given in the PROLOG base then also. even if the facts in the query. which mainly consists of three mechanisms viz (i) Backtracking. m. f). we are not given any of the following: (i) is_sister (sita. then PROLOG system would answer the query: ?. but some statement equivalent to the fact female (sita). For example: Suppose that ‘Sita is a sister of Mohan’ is a fact in the real world. (v) father (sita. PROLOG inferencing system.female (X). Further. (iii) Resolution. M. if in the database. Programming Languages In order to discuss solutions of problems with a PROLOG system.A. m) and mother (mohan. Also. PROLOG database consists of facts and rules. but the following rule (vi) is_sister (X.4 HOW PROLOG SYSTEM SOLVES PROBLEMS? We have mentioned earlier also that solution of a problem through PROLOG system depends on (i) (contents of) PROLOG database or knowledge base. as something like: “Sita is a sister of Mohan” is not true. f). M. 2. More generally. (iv) mother (sita. parents (X. Next. the system would answer as something like: “Sita is a sister of Mohan” is not true. even if all the statements given under (ii) and (iii) and even rule (vi) are in the database. even if all the statements given under (i) to (v) above are in the database. be true in everyday life. 50 . f). if we are not given any set of statements or relations from which we can conclude that Sita is sister of Mohan. m. mohan). F). we discuss how a PROLOG system solves a problem. Y):. (ii) If a PROLOG database does not contain sufficient relevant facts and rules in respect of a particular query. However.I. then the PROLOG system will say ‘fail’ or ‘no’. (ii) Unification. (iii) parents (mohan. is not given. F). rules and goals. we need to discuss in some details data structures in PROLOG. Further. (ii) parents (sita. then also the system would answer as something like: “Sita is a sister of Mohan” is not true.

(jane is half sister of Phillips and according to the definition of the PROLOG database. (xiii) father (jane. Y. Z):. Then PROLOG system passes to the next statement and attempts to match is_sister with the functor also female of the second statement in the database.) (xix) is_sister (X. roshnara. In order to explain the in-built mechanism of PROLOG system. 1 For the database given above. the corresponding functors do A. (ii) female (zarina). (xv) father (Phillips. mother (Y. luxmi. (vi) is_sister (anita. (x) parents (isaac. Y):. (xvii) mother (john. roshnara. father (Y. Y):. Languages-2: PROLOG 51 . albert). M.mother (X.female (X). (xi) parents (aslam. Now. In this case also. F). The last rule could have been replaced by the following rule: (xxi) parents (X. raj). mother (X. we discuss how the PROLOG system solves a problem under consideration. parents (Y. we illustrate the way in which a PROLOG system solves problems. (iii) female (sabina) (iv) female (jane). father (X.albert). ann). mary). (ix) parents (sabina. we start with the particular database given below and some queries and then generalize the results of our discussions. The PROLOG system starts from the top of the database and matches the functor is_sister of the query with the functor female of the statement which do not match (both being constants must be identical for matching). raj). Given Database (i) female (sita). Y). parents (X. Then. father (X.I. anil). roshnara. (viii) parents (sarita. F). through a number of query examples. M). (xii) parents (zarina. F). mary. (xviii) mother (phillps. albert) (xvi) mother (jane. mary). Query No. F). (v) is_sister (sita. (xx) is_sister (X.Let us Assume the PROLOG database is complete in the sense that the database contains all the required facts and rules about the real world that are sufficient to answer any relevant query. (xiv) father (john.female (X). M). Kasim). Z) But to explain some important points we consider rule (xx) and do not consider the last rule. consider the query: ?-is_sister (anita. Kasim). jane is not a sister of Phillips. (vii) parents (sita. albert). M. sarita). anil). Kasim). luxmi.

out of which at least one is a variable (i. of the facts and rules of the database. F). M. In this case. M.. the corresponding argument. the PROLOG system attempts to match with Fact (vi) which also does not match the query. 52 . of the Rule (xix). Hence. (ii) If both the terms are variables. Thus. one by one. The first possibility of match is in Fact (v) which also has functor is_sister. is defined as follows: (i) If the term other than the variable term is a constant. the functor is_sister in the query matches the functor is_sister of the L. aslam). then the constant value is temporarily associated with the variable. PROLOG system attempts to match second arguments of the functor is_sister of the query and of L.female (sabina). Unification of two terms. The functors in the query and the third statement are identical. H. then both are temporarily made synonym in the sense any value associated with one variable will be assumed to be associated temporarily with the other variable. rule (xix) takes the following form: is_sister (sabina. S. Here matching takes a different meaning called unification.is_sister (Sabina. However. Programming Languages not match. parents (aslam. The various components of the query match (in this case are actually identical) to the corresponding components of the sixth statement. And further throughout the statement or rule. Next. the query is answered as ‘yes’. S. consider the query ?. fifth) statement.e. the PROLOG system passes to sixth statement in the database. aslam):. the first argument of the functor in Fact (v) is sita which does not match sabina. the rest of the parts of the query with the corresponding parts of the fifth statement. The PROLOG system attempts to match the functor is_sister of the query with funtors. In other words. 2 With the same database. of rule (xix).. Similarly. The PROLOG system is not able to find any appropriate matching upto fact (xviii).e. Hence. Hence. Hence. The PROLOG system passes to next (i. The first argument viz anita of the query does not match (being constants. F). the first letter of the term is an upper case letter). the query system attempts to match. parents (sabina. are required to be identical for matching) to the first argument viz sita of the fifth statement. functor in the query does not match the functor in each of the next two statements. H. Again this matching is a case of unification of Y being temporarily associated with the constant aslam through out the rule (xix). in this case. the variable is temporarily replaced by the constant. the variable temporarily associated with the constant sabina for all occurrence of X in rule (xix). one by one.I. However. Query No.A. PROLOG system attempts to match the constant sabina given in the query with the variable X given in the rule (xix).

the PROLOG system need to check three subgoals viz. Thus. the relation is reflexive. X) must be true). we know the symbol ‘:-’ stands for ‘if ’. discuss how the following query is answered by the Prolog system: ?.. M. As a consequence. At this stage the subgoal is parents(sabina. in order to satisfy the fact (or to answer whether) sabina is sister of aslam. Ex 2: Query No 4. Z) must be true) Ex 1: Query No.is_sister (aslam. Z) is true. Languages-2: PROLOG 53 . F) are variables.e. Y) is true then it is not necessary that is_sister (Y. F). roshnara.e. The PROLOG system answers the query with ‘yes’. the PROLOG system marks the rule (xix) for future reference or for backtracking (to be explained) to some earlier fact or rule. kasim).. (xii). the PROLOG system again starts the exercise of matching/unification from the top of the database. from the first statement in the database.I..….. the constant roshnara gets temporarily (for the discussion of rule (xix) only) gets associated with the variable M and the constant kasim gets associated with the variable F. discuss how the following query is answered by the Prolog system: ?. parents (sabina. Remarks: In respect of the above database. To satisfy this subgoal. Y) is true and if is_sister (Y. However. F) is actually a sort of question of the form: Who are the parents of sabina? For the satisfaction of this subgoal. female (sabina). represent variables. However. (viii). F) and parents (aslam. and the fact (ix) is parents (sabina. the PROLOG system again starts the process of matching from the first statement in the PROLOG database.e. Ultimately.…. because the condition female (Y) will not be satisfied. the relation of is_sister is also not symmetric (i. subgoal is satisfied. if is_sister (X. the first arguments of the functors in (vii) and (viii) do not match the first argument of parents in the subgoal.e.. Hence the PROLOG system proceeds further to Fact (ix) for matching. no male is his own sister. M and F. because of matching with the fact (ix) in the database. M. i. (viii). Remarks: In respect of the above database. F). sabina). if is_sister (X. The first subgoal: female (sabina) is trivially matched with fact (iii). Remarks: However. roshnara. after failure of matching with the first eight facts in the database. In view of the fact that except sabina the other two arguments in the next (sub)goal viz parents (sabina. F) becomes the (sub) goal parents (aslam. i. The possibilities are with facts (vii). From the way we have explained earlier. then is_sister (X. Any female X may be a sister of a male Y. Kasim). but the reverse is not true.. As explained earlier. Hence. the relation of is_sister is (fully) transitive (i. in which the functor parents of the subgoal occurs as the functor of the facts (vii). the relation of is_sister is not reflexive (i. the subgoal parents (sabina. But the relation is symmetric within the set of all females. 3 With the same database. john) A. M. the next goal parents (aslam. M.is_sister (jane. X) is not true for all numbers of the database). M. being upper case letters. With the same database. (xii). M.. Before starting to work on these three subgoals.e. for the set of all females. the exercise of matching becomes exercise of unification.And. is_sister (X.

At this stage.H.5 BACKTRACKING The concept of backtracking is quite significant in PROLOG. Y). In other words. If we are interested in more than one answers. PROLOG system attempts to satisfy the second subgoal which.S. Phillips) 2. is of the form: is_sister (aslam. the PROLOG system attempts to associate some other value to Y. discuss how the following query is answered by the Prolog system: ?. i. In order to prevent the PROLOG system from attempting to find the same answer: sabina again. Programming Languages Ex 3: Query No. For satisfying this goal. further from the point where Y was associated with aslam. S. the system proceeds to satisfy Rule (xix) from Fact (x) (including) onwards. the PROLOG system goes back to the association i. Typing a semi-colon followed by return’ serves as a direction to PROLOG system to search the database from the beginning once again for an alternative solution.’ (i. then. zarina). The first subgoal on R. which.e.is_sister (jane.e. the first solution which the PROLOG system comes with after searching the database is : ‘Associate sabina with Y’. the system puts markers – one on Rule (xix) and after that on Fact (ix) (in that order). H. For satisfying L.A. leads to a failure. Then through Fact (xi). when one of the directions being pursued for finding a solution. Then aslam replaces Y in the rule (xix). specially in view of the fact that it provides a powerful tool in searching alternative directions. Bactracking is also useful when to a query.S. Again rule (xix) is to be used. of (xix) is to satisfy the fact: female (aslam). sabina is one of the possible answers to the query (which is a conjunct of two propositions). Next. is_sister (Y. The query when translated in English becomes: ‘Find the names (denoted by Y) of all those persons for whom sabina is a sister and further who (denoted by Y) is a sister of zarina. The variable X is already associated with constant sabina. and removes this association of Y with aslam. aslam is associated. there are more than one possible solutions. the PROLOG system searches the database from the top again. 5 With the same database. according to the facts and rules given in the database.. zarina). which is not satisfied. for the first occurrence of Y. sabina) and is_sister (sabina. which are possible in this case.H.’ In order to answer the above query. aslam to Y. of (xix) need to be satisfied. we illustrate the concept through an example and then explain the general idea of backtracking.is_sister (sabina.. associate Y with sabina. zarina). of (xix) the subgoal on R.I. Once the instruction from user through semi-colon is received to find another solution. Example 1: For the database given earlier consider the query ?. after the answer ‘sabina’ the user should type the symbol ‘.e. at present. This is what is meant by Backtracking. satisfies is_sister (sabina. type semi-colon). 54 . Next. First.

designation. viz. the subgoal to be searched becomes is_sister (zarina. parents (zarina. for another answer.’. in context of representing facts. A structures in PROLOG is of the form: (predicate (list-of-terms). particularly. that of representing complex data. F) (where zarina is associated with X) and parents (zarina.6 DATA TYPES AND STRUCTURES IN PROLOG We have already discussed the concepts of atom and number. rules and queries. The system has already marked the fact (xii) in the database while finding the answer: zarina. gross-pay.e. zarina). PROLOG has mainly two data structures: (i) List (ii) Structure. For example.. i. Languages-2: PROLOG 2.I. the PROLOG system starts from the next statement. a term may also be a constant or a variable. a rule or a query can be obtained from appropriate structures. it returns ‘fail’ or ‘NO’. the second answer which the system gives is zarina. Hence. a term itself may be a structure again. Structure is a recursive concept. atoms and numbers. Therefore. Structure in this respect is like record structure of other programming languages. A. female (zarina). It can be easily seen that the PROLOG system will not find any more answers. M. M. operations on numbers and statements about numbers later.e. Of course. Again. We also mentioned how atoms and numbers are represented in PROLOG. the user may seek for another answer to the query by typing ‘. Fact (xiii) to search. middle-name. F) (where zarina is associated with Y) can be easily seen to be satisfiable from the database. as 55 . address) but then name may be further structured as Name (first-name. We will discuss briefly: relations between numbers. i. Here we discuss briefly structure from another perspective. Thus. because. These are the only two elementary data types in PROLOG. A fact is represented in PROLOG by simply putting a full stop at the end of appropriate structure. in the representation predicate (list-of-terms). where predicate is an atom and list-of-terms is a list of terms. three subgoals for this goal. we consider the data structure: Structure Structure We have already discussed the concept of structure. Similarly. last-name) further first-name may be written. age. Hence.Next. through Fact (xii). for example. We next discuss how symbolic data is structured out of the two elementary data types viz. viz. zarina is associated with Y and sabina is already associated with with X. To begin with. The discussion is being postponed in view of the fact that main goal of PROLOG is symbolic processing rather than numeric processing. an employee may be defined as employee (name. This goal can be satisfied.

. (noun N2) ). (designation assistant_registrar). (age 43). _ ) (gross_pay X)). Further. (verbphrase ( verb v). A query of the form ?_sentence ((noun mohan). Tanenbaum published by Prentice Hall PTR in the year 2003 as a structure in PROLOG. (verb_phrase (verb eats). In general. Programming Languages (first-name mohan) Thus. assuming the author’s name is not given Example 2: The following simple sentence: Mohan eats banana may be represented in PROLOG as sentence ((noun mohan). _ . the information about an employee Mohan may be written in PROLOG as a fact using a (complex) structure: employee (name (first_name mohan). Informally. array has a fixed length). Formally.e. list is a recursive concept.A. Ex 4: Give the information about the book: Computer Networks. The elements of a list may be any terms (i. the underscores may represent many different variables. list is an ordered sequence of elements and can have any length (this is how a list differs from the data structure array. (noun X) ) ) tells us about what Mohan eats as follows: (noun banana) Next we consider the data structure: List List: The list is a common data structure which is built-in in almost all programming languages. (noun banana))). Fourth Edition by Andrew S. variables and structures) and even other lists. write a query in PROLOG to know the name of the author. (gross_pay 35000 (address delhi). Similarly address may be further structured. the structure of a sentence of the form given above may be expressed in PROLOG as Sentence ((noun N1). _ ). _ . It will return 3500. (last_name sharma)). (verbphrase (verb eats). as address ( (house-number 139).I. (middle-name lal). Thus. Further. In the above query. a list in PROLOG may be defined recursively as follows: 56 . constants. (street khari-baoli) (area chandni-chowk) (city delhi) ) A query in order to know the gross-pay of Mohan may be given as: ?_ employee (name (mohan). for example. specially programming languages meant for non-numeric processing.

is a list. [3. c] then X is associated with the element a. c] and Y = [ ] (iv) [X |Y] = [ a. [c]] (vi) The operator ‘|’ is not defined for the list [ ]. [2. if X. [1. the first element of the list and Y is associated with the list [ b. However. and Y is associated with [c].3] Also. For example. b]. (iii) [X |Y] = [ [a. Also [_ . c] ] then X = [a.H.e. c] ] then X is associated with the list [a. (ii) [X |Y] = [ [a. X. b] . Expression is a sequence of terms and operators formed according to syntactic rules of the language. X + Y * Z may be an expression. if any. X and Y are numbers.(i) [ ]. ‘|’ is used to associate Head and Tail of a list with two variables as described below: (i) [X |Y] = [a. out of which. X + Y] ] ] is a list provided X + Y is defined. Examples of Lists: [ ]. obtained from the given list by removing its first element. c] ] then X is associated with a and Y is associated with [ [ b.2. [b. if prefix notation is used. e2 . c] ] (v) [X|Y] = [ a. b. [1.3].I.S. then as we shall see later that X + Y is a valid expression. Languages-2: PROLOG 2. [c] ] then then X is associated with a and Y is associated with [ b.…… en ] is a list if each of ei is a term. 3]] (note the last two are different lists and also note that the last list has two elements viz 1 and [2. representing the empty list. (ii) [ e1 . the first element of the list on R. having no elements 57 . b. Y] is a list of three elements. a list or an expression (to be defined).7 OPERATIONS ON LISTS IN PROLOG I. first two are ‘don’t care’ or anonymous variables. For example. Y and Z are numbers and if infix notation for operations is used. but +X∗YZ be may an expression. b. [b. A. c ]. The operation denoted by the vertical bar. [a. b. then X + Y * Z is not an expression. i. _.

pascal is a member of the list [fortran. Again fact (i) of the definition is applied to the new list. pascal. pascal. i. c + d] = [X. [ ]) is not a part of the definition. (i) fails. Programming Languages [a + b .e. Member Function the function is used to determine whether a given argument X is a member of a given list L. i. i. Hence. cobol] to check whether pascal belongs to it. in order to find out whether first argument X is a member of second argument then find out whether X is a member of the list obtained from the second argument by deleting the first element of the second argument. system matches pascal with the Head of the given list [prolog. [fortran. yet the following simple (recursive) program in PROLOG for member achieves the required effect: member(X. i. fortran. Though the member function is a built-in function in almost every implementation of PROLOG. cobol] [fortran|pascal. 3.e. cobol]) The PROLOG system first attempts to verify(i) in the definition of ‘member’. cobol]. hence.e. cobol]. [_ |Y]):.. the PROLOG system returns ‘yes’. Hence. go to definition under (ii). pascal. pascal pascal [fortran. then. X pascal [– | Y] [prolog|fortran. it is clear that the case member (X. The above procedure may be concisely expressed as follows: Goal 1. Hence. then predicate member is true. Again fact (i) is applied to check whether pascal is a member of the current list [pascal. (i) (ii) (vii) Definition of member under (i) above says that if first argument of member is Head of the second argument. member (X. which is actually true. The two constants are not identical.A. From the above definition. pascal. rule (ii) is applied to the new list. if pascal is a member of the tail [pascal. the PROLOG system uses the rule (ii) of the definition to solve the problem. Y] then X is associated with a + b and Y is associated with c + d (Note the difference in the response because vertical bar is replaced by comma) II. Next. cobol]. pascal. Example 3: ?-member (pascal. cobol] of the given list. pascal.e. pascal. According to rule (ii) the system attempts to check whether pascal is a member of the tail [fortran. According to fact (i) for pascal to be a member of the current list pascal should be Head of the current list. cobol] Comment pascal does not match prolog hence apply rule (ii) by rule (ii) using fact (i) 2. Therefore. [prolog. fortran of the new list. As pascal does not match the Head. [Y]). cobol] 58 . If not so.. we discuss example to explain how the PROLOG system responds to the queries involving the member function. with prolog. [X| _ ].I. By rule (ii). The definition of member under (ii) above says that. cobol] of the current list.member (X. the system returns FALSE. fortran..

each of which should be a list. cobol] [pascal|cobol] A. X]. [b. c]. Executing append according to the above definition is a recursive process. (i) (ii) Explanation: The above PROLOG program states that append takes three arguments.append ([Tail]. X]. c. b] [ c . a built-in function in PROLOG. we can see how a simple program in PROLOG can be written for the function. c]. However. The append function takes two lists and returns a single list such that the elements of the first list followed by elements of the second list constitute the elements of the returned list. [b. c].. Languages-2: PROLOG (again apply fact (i)) Head of the current list is identical with pascal. [prolog.e. [a. [ ]) also returns the list [a. In each successive step. The append function in PROLOG may be written as: append ([ ]. Further. append ( [a. b. the result of application of (i) of the definition of append to an empty list in first argument with a list X in second argument is the same list X as given in second argument. APPEND function is also generally. The result of appending the elements of the lists in the first and second arguments. According to rule (ii) above. [tail] of the first argument to the list given as second argument of append. of the third argument) and the tail of the resultant list is obtained by appending tail i. Y]) returns the list [a. c] ]. pascal pascal [pascal. cobol]). 5. c]. b. However.e.pascal is not identical to head of the list Hence apply rule (ii) 4. III. b. [ [ c. X. b. c]) returns the list [a. the system responds with a ‘yes’. X). X. Ex 5: Explain how the PROLOG system responds to the following query: ?-member (pascal. Y]) returns the list [a. append ([Head|Tail]. Y].. append ([a. In the last case. Head of the list in first argument becomes Head of the resultant list (i. b. Y] append ([ ]. Y. the size of the list in the first argument is reduced by one and finally the list in the first argument becomes [ ]. fact (i) is applicable and the process terminates. is stored in the list represented by third argument. X. 59 . Z). if the first argument is not [ ] then apply rule (ii) above. Y. but written in third argument.I. [Head|Z]:. [c. hence. For example (i) (ii) (iii) (iv) append ([a. fortran.

Programming Languages Next. Further. [C. [b. Z).A. In this case however rule (i) is applicable which returns [C.append (X. b] is not [ ]. Y. Z). list is the Head of new first argument and [ ] is the tail of the new first argument.append ([a. returns yes. prefix (X. C. lisp. such that X is either [ ] or X contains any number of consecutive elements of the list Z starting from the first element of Z onwards. by the PROLOG system may be described as follows: For satisfying the query. fortran] Next. the rule becomes prefix ([a. b]. Y. lisp]. fortran] and the result will form the tail of the originally required resultant list. Which on another application of rule (ii) takes the form append ([a |]. Y. As first list is not [ ]. The last element of X when placed in Z is followed by elements of Y. [Head | Z]):. second element is lisp and the rest of the elements of the resultant list will be obtained by appending [ ] to [C. we have already defined append as append ( [ ]. [a. fortran]. Y . Z returns Z as a list in which first. X. b]. [c]). However. Z). Y.append ([Tail]. and which will form the tail of the resultant list. we consider another list function. c]. [ c]):. [a. b]. we explain how PROLOG system responds to a query involving append. next two-statement program append says that append. Z) function which returns true if X is a sublist of Z. the system need to satisfy the goal: append ([a. Z) returns ‘No’ or ‘False’. Further explanation may be given with the following example query: ?-prefix ([a. again preserving their order in Y. append ([Head | Tail]. Y. b. fortran]. Otherwise. therefore the system associates prolog with Head and [lisp] with Tail. c]). The PROLOG system responds with Y = [c] 60 . fortran]. Then by rule (ii) prolog becomes the Head of the resultant list and after that PROLOG system attempts to append the list [lisp] with [C. The result of the second append is a list whose first element is prolog. b. Further explain. First fact of append is not applicable as [a. returns true. justify and trace your program with an example. [a | b.append ([ ]. c]) Next. Y. if by appending any list Y to X we get Z. Finally. The PROLOG program prefix is just one statement program: prefix (X. b. Z): . The rule (ii) takes the form append ([a | b]. c]). prefix (X. b]. IV. Y. all the elements of X occur preserving their original order in X. Y is associated with [c]. (i) (ii) Explanation: The one-line program. c]):. The processing for the response. c]):-append ([a]. second argument remains unchanged. X]).I. where. [a. the resultant list is [prolog. Next. Example 4: Let us consider the query ?-append ([prolog. fortran] as the result of append [ ] to [C. a function of three arguments X. [a. b. Y. Y.

b]. b. c] Explain the sequence of steps of processing by PROLOG system.1287 =1289 fails. is discussed below: Case I: Both Term1 and Term2 are constants and. and. if still another response is required by the user. if X is already instantiated to any other constant say flower. [a. Then the success or failure of the goal ?-Term1 = Term2. Z) which returns true if ‘Y is a sublist of Z such that either Y is [ ] or Y contains any number of consecutive elements of Z starting anywhere in the list but the last element of Y must be the last element of Z. Finally.Let us consider another query: ?-prefix (X. replace X by tree in the given query. Applying Case I above. explain. c]). PROLOG system returns success. ) to find another answer. Case II: One of the terms is a variable and if the variable is not instantiated then the goal always succeeds For example ?-brother (mohan. c] is returned. which takes the new form: ?tree = tree On encountering this new query. For example. sita) = X. Ex 7: Write a PROLOG program suffix (Y. if both are identical then the goal succeeds. If. where a term may be a constant. Languages-2: PROLOG Ex 6: Another Example query ?-prefix ([a. Further. then PROLOG system returns X as the list [a]. consider the query ?-X = tree. sita) Case III: One of the terms is a variable and the variable is instantiated: Then apply ‘=’ recursively to the case which is obtained by replacing the variable by what is its instantiation. If still another answer is required. b. if not identical. Examples: The query ?-mohan = mohan succeeds ?-1287 = 1287 succeeds ?-program = programme fails ?. then the new query becomes: ?-flower = tree. then the system responds with a ‘No’. the answer [a. then the goal fails. say. c]. And X is already instantiated to tree. further. 2. first returns X as [ ]. further another answer is required then PROLOG system returns X as the list [a.8 THE EQUALITY PREDICATE ‘=’ Let Term1 and Term2 be two terms of PROLOG. In response to the query the PROLOG system. Then the goal succeeds and the variable X is instantiated to the term brother (mohan. a variable or a structure. b. A. [a. we get Fail. justify and trace your program with an example. However. then if the user gives ( . the query is of the form: 61 . then. Case IV: If both terms are variables.I.

2. Case V: Both terms are structures: Two structures satisfy ‘=’ if they have the same functor and the same number of components and. X) = brother (Y. b). U to u and the variable V to v. For this purpose. (b) If one or both are instantiated. because. Programming Languages ?-X = Y then if (a) both are uninstantiated the query succeeds. but. The symbol for ‘not equal to’ is /= Please note the notation for ‘less than or equal to’ and ‘greater than or equal to’ 62 .noun (alpha) = Noun. > (greater than).I applications. However. sita).e. the functors. p on the two sides of ‘=’ are identical. yet they become coreferred variables in the sense that if one of these is instantiated to some constant then the other also gets instantiated to the same constant. the variable S to s. c)) (ii) ?-letter(H) = word (letter) (iii) ?-g(X. X) = g(a. if during further processing one of X or Y gets instantiated to some term then other variable also gets instantiated to the same term. i. r. U. (v) ?-noun (alpha) = noun.. Exercise 8: Discuss success/failure of each of the following queries: (i) ?-g(X. Further relational operators viz < (less than). / (division) and ∧ (exponentiation) are built-in and are used in infix notation. Next. Further. if the definition of ‘=’is applied recursively to the corresponding components. > = (less than or equal to) have their usual meanings and infix notation is used in expressions using these relational operators. then replace X and/or Y by their respective instantiations to generate a new query and apply ‘=’ to the new query. For example: ?-brother (mohan.9 ARITHMETICS IN PROLOG The language PROLOG has been designed mainly for symbolic processing for A. success of ‘=’ for the terms at corresponding positions is discussed below: variable Q is instantiated to q. first of all. then each pair of corresponding components satisfies ‘=’. a (Z. = < (greater than or equal to). – (substraction). s. r. v) = p(q. (iv) ?. = (equal to). T. the variables T and W are. a(b. further. W. V). c) ) = g (Z. some facilities for numeric processing are also included in PROLOG. though. u. the arithmetic operations viz + (addition). not instantiated.I. S. Then the above goal succeeds and X is instantiated to sita and Y is instantiated to mohan Similarly the following goal also succeeds ?-p(Q.A. * (multiplication).

in response to the query ?. through the operator ‘is’. the expression ‘3 + 7’ is not the same as the number 10. Example: It is required to check whether an element is a member of a list X.. Z):. The operation of ‘+’ does not execute automatically. b]). result of computation for factorial of any number Number will be associated with the variable Result and the goal can be achieved through the following four subgoals: (i) (ii) Number > 0 should be true.Number > 0. the problem is taken as solved without need for iterations any more.e. of the number New. Languages-2: PROLOG 2. each of Number.. Partial). for the following one statement PROLOG program prefix (X. as discussed by the following example. Z). Further. if we test the elements of X from left to right for equality with a. [a. 1).append (X. i. factorial (Number. factorial (New. For example. A. Partial and Result. (the result of the factorial of New (i. New. Result is Number * Partial. the variable X gets instantiated to 10 and the prolog system responds as X = 10 Yes In order to explain numerical programming in PROLOG.10 THE OPERATOR CUT In some situations. c}. Explanation of PROLOG program for factorial: If the given number is 0. if the goal is met once. Result):. Note: In the above definition. even after satisfaction of the goal once. For the purpose of execution of an operation. then first X is associates with [ ] as X = [ ] as an answer to the query. 63 . However. (iii) Factorial of the number (Number – 1) through the operator ‘is’. Thus.X is 3 + 7.The ‘is’ operator It may be noted that in PROLOG. of Number – 1)) is calculated and associated with Result. PROLOG provides the operator ‘is’. d. a. Y. Thus in such a situation. where. say X = {c. then its factorial is 1. we know that PROLOG system attempts to re-satisfy a goal. The number Number – 1 is calculated through the operator ‘is’ and is associated with variable New. is recursively calculated and the result of the calculation is associated with the variable Partial (iv) Finally. we are not interested in any more occurrences of a in the list X. let us consider the PROLOG program for factorial: factorial (0. a. New is Number – 1.I. then.e. if the query is ?-prefix (X. then the third element in the list matches a and once that happens. the product of the Number with Partial. b. denotes a variable. it is required that the PROLOG system to stop any further computation.

And hence PROLOG system attempts to satisfy the subgoal d. the comparison X < Y is not required if the first rule fails. the PROLOG system responds with X = [a. b]. the PROLOG system associates the list [a] to X. Example 6: We define below the membership for a set (in a set each element occurs only once). once d fails and crosses on the left to the operator !.S. 64 . f and g. Y. However. otherwise Y is the maximum. And finally. e. may be defined as member (X. This wastage of effort may be saved by using the following program for maximum. in the case when the operator is the cut symbol then PROLOG system is not allowed to go to its Left. must succeed.H. f and g. b and c as long as it is required by the system answer the query. [_| Y]: . (i) /* third argument is for the variable supposed to be associated with the result */ max (X. Further. consists of only the cut symbol ‘!’ which is always true. The above rule says predicate trial succeeds if the subgoals on R.H. The operator cut is denoted as ! (the exclamation mark) and inserted at a place where the interruption for not retrying is to be inserted. (i) member (X. in stead of the one given through (i) and (ii) above. Hence. d. !. for the occurrence of the element in a set.S. [X | – ]):. Y. if required. then no more attempts will be made to re-satisfy c on the left of the operator !. R. g. (ii) The statement (i) above says that if the element X is the Head of the list (i. In the process.X < Y. the system responds with a ‘no’. S of a rule. provides a special mechanism or function called ‘cut’ by which..S. Y): . c. The PROLOG system may backtrack between subgoals a.X = [a]. b. Programming Languages In the next iteration. once a particular element is found to occur then there is no need for further testing. in this sense.member (X. X is the first element of the list) then the R. f. However.. if d succeeds then backtracking may occur between d.H. The predicate member. i. H. In the still next iteration. succeed.e. Y). it is possible that several backtrackings may occur between a. Further. Example 5: Through the following general example we explain how PROLOG system behaves in the presence of cut denoted by ! when inserted on the R. PROLOG system always succeeds on the cut operator represented by ‘!’. The change in the behaviour of PROLOG system due to the presence of ‘!’ occurs only after the subgoal c succeeds and PROLOG system encounters the cut symbol ‘!’. if all subgoals succeed once.A.a. The PROLOG system.!. b and c. for the predicates a.e.X > = Y. X):. e. the subgoals on the right-hand side of a rule may be prevented from being tried for more than once. e. . b and c and several (independent) backtrackings may occur between d. Consider the rule trial: . Hence. the PROLOG system after succeeding on ‘!’ exits program execution because of the restriction that the system is not allowed to backtrack from ‘!’. Example 7: One of the possible PROLOG programs for finding maximum of two given numbers is the following two-statement program: max (X.I. In other words. (ii) But we know that if X > = Y then X is the maximum of the two.

L) /* the system after executing only the first statement returns */ L = [d. add (X. we want to write a program to calculate tax payable by a person for a given income. [d. Z):. suppose sum (X. Y. Y). we see how the procedure add behaves on various arguments.fail. 65 . 1) :. A. L): . tax (Name. Y. Then add can be written as add (X. number (Y).number (X). Y. L does not contain any duplicate elements). F is F1 * N. f]. Tax):-………. !. We define a function add (X. F1). [d.member (X. d. Languages-2: PROLOG 2. sum (X. Then one of the possible (incorrect) programs may be written as: tax (Name.M is N – 1. Thus the program may be written as add (X.fail. ?-add (c. else it fails.X > = Y. The predicate fail wherever occurs always fails and initiates backtracking. . . Y. tax (Name.. max (X. Z):. [X | L]).e. then nothing needs to be done. ! fact (M. Therefore. Y. Further. X): . However. L). a foreigner is not required to pay tax.The more efficient program for the required solution is max (X.I. L) /* the system responds with */ L = [c. . Tax):-………. f]. a new list L1 may be returned in which X is the Head of L1 and additionally contains all the elements of L and no other elements. Let number denote a predicate such that number (X) is true whenever X is a number. L. except that the function add first verifies that each of X and Y is a number. Next. For example. according to the law of a country. L. Income. Y. !. Z) associates with Z the sum of the numbers X and Y. Cut & fail Example: Suppose. Y. else. we write a program to find the factorial(N) using cut as follows: fact (N. Income. let us first discuss the predicate fail. f] X=d Example: In order to explain the use of cut. Tax):. Income. Z). add (X.11 CUT AND FAIL Before discussion of the cut and fail combination.foreigner(Name). Z). Y. Example 8: To write a PROLOG program that adds an element X to a given set L (i. Z) which is similar to sum (X. f] ?-add (X. ! fact (N.N < = 1. F):. if X ∈ L.

are discussed and defined in English. if foreigner (name) succeeds. concession/rebate etc. in Section 2. despite the fact that the tax for albert should not be calculated.5. i. However. some of the later goals may succeed. . important concepts like matching and unification are introduced and defined. If foreigner (name) succeeds. subgoal and program etc are (formally) defined in Section 2. Hence. these other rules do not involve the predicate foreigner but may depend upon income. The syntax for various elements and constructs of PROLOG including for the abovementioned concepts and also for the concepts of structure. Tax):-………. goal. Tax):-………. fail. atomic formula. . Programming Languages Except for the first one. However. 2.e.A. Income. !. The concept of backtracking is explained in Section 2.H. tax (Name.I. The issues about these paradigms are discuss in the Introduction to the unit. Now suppose Alberts is a foreigner. Hence. all others rules are left unspecified.4 is mainly devoted to explaining how a PROLOG system solves a problem where the problem is defined to the system in the way discussed above. Tax):. rule. including those of atom. the program execution stops after returning back to ‘!’. no backtracking takes place after returning to cut from the execution of fail. tax (Name. variable. Because of the presence of cut in the first rule before fail. goal and fact etc. These paradigms are quite different from the normally used paradigm viz imperative paradigm of solving problems. term. if foreigner (Name) fails.. However. Alongwith matching and unification.4. then processing by PROLOG system continues from the second rule onwards. predicate. Income.foreigner (Name).S not to succeed. 66 . another mechanism viz backtrakcing play a significant role in problem solving by PROLOG system. The concept of database is defined in Section 2. Income. The following programme outline gives the correct result in the case of foreigners: tax (Name.12 SUMMARY The programming language PROLOG is based on Declarative paradigm and Logic Programming paradigm of solving problems. However. A problem in PROLOG is defined by (i) defining a database and (ii) a query representing the goal to be achieved. Then According to the first rule foreigner (alberts) succeeds and then the predicate fail makes the goal on L. according to normal rules of PROLOG execution. the error can be rectified by using cut. Hence PROLOG system backtracks and attempts the second and later rules which do not involve foreigner. In the process of explanation. . Section 2.0. query. In Section 2.2 the basic concepts of PROLOG.3.

mother (X. M). some numeric processing may also be required. F) mother (Y. father (Y.. of rule (xx) generates the following five (new) subgoals viz female (jane). PROLOG is mainly a symbol processing language. matching becomes the exercise of unification of the goal. In the process. and variable Y with constant john throughout rule (xx).10 and 2. But.female (X). father (jane.11. Prolog has basically two (elementary) data types viz atom and number. requires to satisfy two subgoals. The operations available in PROLOG. The predicate equal (denoted as ‘=’) which plays an important role in PROLOG. father (john.e. because. the arithmetic facilities built in PROLOG are discussed and defined in Section 2. Data types and data structures play a significant role in problem solving in PROLOG environment also and are discussed in Section 2.We know built-in data types and data structures play an important role in any programming environment which is generally based on some problem solving style/paradigm including the imperative style which is the dominant style of solving problems using computer. But. one of the subgoals generated is female (aslam). has functor is_sister in the goal and hence matches the functor of the goal. And. is discussed in Section 2. mother (john. PROLOG has mainly two data structures viz List and Structure. These predicate are discussed in Sections 2. of is_sister (jane. i. M).9. parents (john. Y):. john) with is_sister (X. Recursion and iteration are among major mechanism of PROLOG problem solving process. However. M.8. which after appropriate unification. R. become parents(jane. M). which is the goal of the rule (xx).S.6. F). for data structure List are discussed in Section 2. 67 . A. The next rule (xx) is_sister (X. the PROLOG system while attempting to match and then rejecting the previous facts and rules. the first of these fails and hence the whole of rule (xix) itself is taken as not matching and is abandoned for the next rule/fact in the database. mother (jane. Through unification. after matching with either rule (xix) or rule (xx) of the query. variable X is temporarily associated throughout rule (xx) with constant jane. As explained earlier. Also. F).13 SOLUTIONS/ANSWERS Ex 1: The goal fails. in order to check undesirable repetitions (through recursion or iteration) PROLOG system provides for two very important predicates viz Cut and Fail. Y).I. as in the previous query.H. Languages-2: PROLOG 2.7. For this purpose. F). father (X. Ex 2: As in the previous query. F). reaches rule (xix). But this sub-goal can not be satisfied by the facts and other rules in the database. F). M). in view of the fact that even for the problems that require for their solutions dominantly symbol processing. M. specially in matching and unification.

Thus. the PROLOG system takes similar steps upto Step 4 above. except the constant prolog does not occur in the list under [– | Y]. the last subgoal becomes father (john. (middle_initial. mary). the variable M temporarily (for the whole of rule (xx)) gets associated with the constant mary. But there is no fact that matches the last subgoal. X pascal pascal [– | Y] [cobol |] [ ] Comment fact (i) not satisfied. Hence apply rule (ii) cannot be satisfied. From step 5 onwards the procedure adopted by PROLOG system is as follows Goal 5. while satisfying subgoal mother (jane. M) is a sort of question of the form: Who is jane’s mother?. F). Second of these subgoals viz mother (jane. (middle_initial S). M) becomes mother (phillips. Fourth subgoal father (jane. viz. Programming Languages First of these subgoals is easily satisfied by the Fact (iv) of the database. The system returns the name of the author as (author ((first_name Andrew). Ex 6: The PROLOG system processes the query as follows: Using the only rule for prefix. The system says the goal cannot be satisfied. 6. mother (jane. (edition fourth).A. For the following query: ?_book ((tittle computer_networks).’ becomes the goal ‘mother (john. (author (first_name Andrew). Ex 5: As discussed in the preceding Example 3. But this goal is given as Fact (xiv) in the database and hence is satisfied. (last_name tanenbaum) ) (year 2003). Hence the query fails.I. M). Fact (xvi) in the database tells us that mary is mother of jane. mary). albert). F) is equivalent to the question: ‘Who is janes Father?’ The Fact (xiii) in the database associates to variable F the constant albert for the whole of rule (xx).’ which is given as Fact (xvii) and hence is satisfied. (year 2003). (edition fourth). the system instantiates variables to get 68 . Hence the subgoal mother (phillips. The PROLOG system answers the query as ‘yes’. X. female (jane). As. F). mother (phillips. M). Ex 3: The PROLOG system proceeds as in previous query and generates five subgoals. In the light of this association. the next subgoal ‘mother (john. and hence. father (phillips. M) father (jane. (publisher prentice_hall_ptr)). the variable M is associated with mary for all the subgoals. The PROLOG system respeonds with ‘No’. Ex 4: book ((title computer_networks). in the case of previous query. M). (publisher prentice_hall_ptr) ). (last_name tanenbaum)).

for which the value of Y. Y. c]):-append (X. c]).H. Y = [ ] is returned. first of all the two variables X and Z become co-referred variables. c]):-append (Tail. c]. then ‘No’ is returned. X . in order to satisfy the query. Again to satisfy the new goal given on R. [b. X]. If no such value exists. b. b. Y = [c] is returned. b. c]. Y. 69 . b. Z) A. Y. (iii) also fails.e. Y. Y. if another iteration is desired by the user. c] to Y and returns Y = [b. and [Head|Z] is not able to associate any value to Head. [a. c]). b. c]):. [a. c] If the user desires another answer by a typing ‘. we get the subgoal given on R. we get through the application of rule (ii) of append. Ex 8: (i) succeeds. [a. is to be found. the system associates [b. above is satisfied by associating X to [ ]. [a. append ([Head|Tail]. Y. then the system returns: No Next the rule in the recursive definition of append gives Append ([a|c].’ then the system applies rule (ii) of append to Append [(Head|Tail). Hence PROLOG system returns: No Ex 7: The suffix PROLOG program is just one statement program:suffix (Y. to the constant b. Therefore the system returns yes with Y = [a. because the variable X gets instantiated to the constant a but then system can not instantiate the second occurrence of X to some other constant. c]). [a. the two predicates/functors do not equal. if it exists. Then Z gets instantiated to c and hence X also gets instantiated to c (ii) fails. c]).append (Tail.S above.i. [a|b. Z).append (X. c]) which when written as ([c|]. the system has a new goal viz append ([a. The process generated by the program is explained through the following examples: Example query ?-suffix (Y. in this particular case.H. Y. because.append ([c]. Z):. where we have already defined append as append ([ ]. [b|c]) and when matched with append ([Head|Tail]. c]):. through another iteration.append ([a. finally. Through still another iteration.. Y. b. b. c] Similarly. we get suffix (Y.I. [a. c]). . Y. Y. And. b. c]. Then through the rule in the definition. Applying rule (ii) of append. Y. because. and ‘a’ is associated to Head.S. [Head|Z]):. [a|b. Y to [a. [b. Y.prefix ([a. Languages-2: PROLOG (i) (ii) Explanation The only statement in the program. c]. states that the statement: ‘The list Y is a suffix of list Z’ is true if there is a list X to which if Y is appended then the list Z is obtained.

& Mellish. Programming Languages (iv) succeeds because the R. : Programming Languages: Principles and Paradigms. R. W. C. : Programming in Prolog (Fifth Edition).14 FURTHER READINGS 1. 70 . Clocksin. Sebesta.S. Springer (2003). 3.H.F. W. W.S. Tucker. Pearson Education Asia (2002). & Noonan. Clocksin. is a variable. Narosa Publishing House (1981). 2.S.I. 4. & Mellish. C. : Programming in Prolog (Third Edition). The variable Noun is associated to the constant noun (alpha) (v) fails. Tata McGraw-Hill Publishing Company Limited (2002). A.F.A. : Concepts of Programming Languages. R. both sides of ‘=’ are constants but not identical: 2.

Sign up to vote on this title
UsefulNot useful