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[prolog/en] Corrected statement about unifying two free terms (#3033)
* Corrected statement about unifying two free terms While the intricacies of unification would bring us too far, stating that assigning two free 'sides' is wrong. I tried to give a small description about how this works (without going into the details of occurrence checks or unification of more complex structures). * Fixed indentation * Replaced old style of structured comments
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Evert Heylen
Pratik Karki
parent
7d303e5042
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eefc0a9c92
@ -38,9 +38,9 @@ magicNumber(42).
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% predicate names must start with lower case letters. We can now use
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% interactive mode to ask if it is true for different values:
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?- magicNumber(7). % True
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?- magicNumber(8). % False
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?- magicNumber(9). % True
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?- magicNumber(7). % True
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?- magicNumber(8). % False
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?- magicNumber(9). % True
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% Some older Prologs may display "Yes" and "No" instead of True and
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% False.
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@ -50,7 +50,7 @@ magicNumber(42).
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% starting with a capital letter is a variable in Prolog.
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?- magicNumber(Presto). % Presto = 7 ;
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% Presto = 9 ;
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% Presto = 9 ;
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% Presto = 42.
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% Prolog makes magicNumber true by assigning one of the valid numbers to
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@ -66,26 +66,33 @@ magicNumber(42).
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% follows:
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% If both sides are bound (ie, defined), check equality.
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% If one side is free (ie, undefined), assign to match the other side.
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% If both sides are free, abort because this can't be resolved.
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% If both sides are free, the assignment is remembered. With some luck,
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% one of the two sides will eventually be bound, but this isn't
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% necessary.
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%
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% The = sign in Prolog represents unification, so:
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?- 2 = 3. % False - equality test
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?- X = 3. % X = 3 - assignment
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?- X = 2, X = Y. % X = Y = 2 - two assignments
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?- X = 3. % X = 3 - assignment
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?- X = 2, X = Y. % X = Y = 2 - two assignments
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% Note Y is assigned to, even though it is
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% on the right hand side, because it is free
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?- X = 3, X = 2. % False
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% First acts as assignment and binds X=3
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% Second acts as equality because X is bound
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% First acts as assignment and binds X=3
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% Second acts as equality because X is bound
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% Since 3 does not equal 2, gives False
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% Thus in Prolog variables are immutable
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?- X = 3+2. % X = 3+2 - unification can't do arithmetic
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?- X is 3+2. % X = 5 - "is" does arithmetic.
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?- 5 = X+2. % This is why = can't do arithmetic -
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?- 5 = X+2. % This is why = can't do arithmetic -
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% because Prolog can't solve equations
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?- 5 is X+2. % Error. Unlike =, the right hand side of IS
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% must always be bound, thus guaranteeing
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% no attempt to solve an equation.
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?- X = Y, X = 2, Z is Y + 3. % X = Y, Y = 2, Z = 5.
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% X = Y are both free, so Prolog remembers
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% it. Therefore assigning X will also
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% assign Y.
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% Any unification, and thus any predicate in Prolog, can either:
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% Succeed (return True) without changing anything,
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@ -101,11 +108,11 @@ magicNumber(42).
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% example, Prolog has a built in predicate plus which represents
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% arithmetic addition but can reverse simple additions.
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?- plus(1, 2, 3). % True
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?- plus(1, 2, 3). % True
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?- plus(1, 2, X). % X = 3 because 1+2 = X.
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?- plus(1, X, 3). % X = 2 because 1+X = 3.
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?- plus(X, 2, 3). % X = 1 because X+2 = 3.
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?- plus(X, 5, Y). % Error - although this could be solved,
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?- plus(1, X, 3). % X = 2 because 1+X = 3.
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?- plus(X, 2, 3). % X = 1 because X+2 = 3.
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?- plus(X, 5, Y). % Error - although this could be solved,
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% the number of solutions is infinite,
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% which most predicates try to avoid.
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@ -129,9 +136,9 @@ magicNumber(42).
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?- print("Hello"). % "Hello" true.
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?- X = 2, print(X). % 2 true.
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?- X = 2, print(X), X = 3. % 2 false - print happens immediately when
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% it is encountered, even though the overall
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% compound goal fails (because 2 != 3,
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?- X = 2, print(X), X = 3. % 2 false - print happens immediately when
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% it is encountered, even though the overall
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% compound goal fails (because 2 != 3,
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% see the example above).
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% By using Print we can see what actually happens when we give a
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@ -156,7 +163,7 @@ magicNumber(42).
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% the interactive prompt by pressing ;, for example:
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?- magicNumber(X), print(X), X > 8. % 7 9 X = 9 ;
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% 42 X = 42.
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% 42 X = 42.
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% As you saw above we can define our own simple predicates as facts.
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% More complex predicates are defined as rules, like this:
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@ -168,7 +175,7 @@ nearby(X,Y) :- Y is X-1.
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% nearby(X,Y) is true if Y is X plus or minus 1.
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% However this predicate could be improved. Here's why:
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?- nearby(2,3). % True ; False.
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?- nearby(2,3). % True ; False.
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% Because we have three possible definitions, Prolog sees this as 3
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% possibilities. X = Y fails, so Y is X+1 is then tried and succeeds,
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% giving the True answer. But Prolog still remembers there are more
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@ -177,11 +184,11 @@ nearby(X,Y) :- Y is X-1.
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% the option of rejecting the True answer, which doesn't make a whole
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% lot of sense.
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?- nearby(4, X). % X = 4 ;
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% X = 5 ;
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% X = 3. Great, this works
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?- nearby(X, 4). % X = 4 ;
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% error
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?- nearby(4, X). % X = 4 ;
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% X = 5 ;
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% X = 3. Great, this works
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?- nearby(X, 4). % X = 4 ;
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% error
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% After rejecting X = 4 prolog backtracks and tries "Y is X+1" which is
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% "4 is X+1" after substitution of parameters. But as we know from above
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% "is" requires its argument to be fully instantiated and it is not, so
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@ -195,10 +202,10 @@ nearbychk(X,Y) :- Y is X+1, !.
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nearbychk(X,Y) :- Y is X-1.
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% This solves the first problem:
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?- nearbychk(2,3). % True.
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?- nearbychk(2,3). % True.
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% But unfortunately it has consequences:
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?- nearbychk(2,X). % X = 2.
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?- nearbychk(2,X). % X = 2.
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% Because Prolog cannot backtrack past the cut after X = Y, it cannot
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% try the possibilities "Y is X+1" and "Y is X-1", so it only generates
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% one solution when there should be 3.
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@ -230,9 +237,9 @@ nearby3(X,Y) :- nearby2(X,Y).
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% Here is the structured comment declaration for nearby3:
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%% nearby3(+X:Int, +Y:Int) is semideterministic.
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%% nearby3(+X:Int, -Y:Int) is multi.
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%% nearby3(-X:Int, +Y:Int) is multi.
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%! nearby3(+X:Int, +Y:Int) is semideterministic.
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%! nearby3(+X:Int, -Y:Int) is multi.
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%! nearby3(-X:Int, +Y:Int) is multi.
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% For each variable we list a type. The + or - before the variable name
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% indicates if the parameter is bound (+) or free (-). The word after
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@ -250,13 +257,13 @@ nearby3(X,Y) :- nearby2(X,Y).
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% An unusual feature of Prolog is its support for atoms. Atoms are
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% essentially members of an enumerated type that are created on demand
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% whenever an unquoted non variable value is used. For example:
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character(batman). % Creates atom value batman
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character(robin). % Creates atom value robin
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character(joker). % Creates atom value joker
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character(darthVader). % Creates atom value darthVader
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?- batman = batman. % True - Once created value is reused
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?- batman = batMan. % False - atoms are case sensitive
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?- batman = darthVader. % False - atoms are distinct
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character(batman). % Creates atom value batman
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character(robin). % Creates atom value robin
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character(joker). % Creates atom value joker
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character(darthVader). % Creates atom value darthVader
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?- batman = batman. % True - Once created value is reused
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?- batman = batMan. % False - atoms are case sensitive
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?- batman = darthVader. % False - atoms are distinct
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% Atoms are popular in examples but were created on the assumption that
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% Prolog would be used interactively by end users - they are less
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@ -267,54 +274,55 @@ character(darthVader). % Creates atom value darthVader
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% Note that below, writeln is used instead of print because print is
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% intended for debugging.
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%% countTo(+X:Int) is deterministic.
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%% countUpTo(+Value:Int, +Limit:Int) is deterministic.
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%! countTo(+X:Int) is deterministic.
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%! countUpTo(+Value:Int, +Limit:Int) is deterministic.
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countTo(X) :- countUpTo(1,X).
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countUpTo(Value, Limit) :- Value = Limit, writeln(Value), !.
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countUpTo(Value, Limit) :- Value \= Limit, writeln(Value),
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NextValue is Value+1,
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countUpTo(NextValue, Limit).
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NextValue is Value+1,
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countUpTo(NextValue, Limit).
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?- countTo(10). % Outputs 1 to 10
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?- countTo(10). % Outputs 1 to 10
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% Note the use of multiple declarations in countUpTo to create an
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% IF test. If Value = Limit fails the second declaration is run.
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% There is also a more elegant syntax.
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%% countUpTo2(+Value:Int, +Limit:Int) is deterministic.
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%! countUpTo2(+Value:Int, +Limit:Int) is deterministic.
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countUpTo2(Value, Limit) :- writeln(Value),
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Value = Limit -> true ; (
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NextValue is Value+1,
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countUpTo2(NextValue, Limit)).
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Value = Limit -> true ; (
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NextValue is Value+1,
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countUpTo2(NextValue, Limit)).
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?- countUpTo2(1,10). % Outputs 1 to 10
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?- countUpTo2(1,10). % Outputs 1 to 10
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% If a predicate returns multiple times it is often useful to loop
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% through all the values it returns. Older Prologs used a hideous syntax
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% called a "failure-driven loop" to do this, but newer ones use a higher
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% order function.
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%% countTo2(+X:Int) is deterministic.
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%! countTo2(+X:Int) is deterministic.
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countTo2(X) :- forall(between(1,X,Y),writeln(Y)).
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?- countTo2(10). % Outputs 1 to 10
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?- countTo2(10). % Outputs 1 to 10
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% Lists are given in square brackets. Use memberchk to check membership.
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% A group is safe if it doesn't include Joker or does include Batman.
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%% safe(Group:list(atom)) is deterministic.
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%! safe(Group:list(atom)) is deterministic.
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safe(Group) :- memberchk(joker, Group) -> memberchk(batman, Group) ; true.
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?- safe([robin]). % True
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?- safe([joker]). % False
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?- safe([joker, batman]). % True
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?- safe([robin]). % True
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?- safe([joker]). % False
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?- safe([joker, batman]). % True
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% The member predicate works like memberchk if both arguments are bound,
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% but can accept free variables and thus can be used to loop through
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% lists.
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?- member(X, [1,2,3]). % X = 1 ; X = 2 ; X = 3 .
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?- member(X, [1,2,3]). % X = 1 ; X = 2 ; X = 3 .
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?- forall(member(X,[1,2,3]),
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(Y is X+1, writeln(Y))). % 2 3 4
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(Y is X+1, writeln(Y))). % 2 3 4
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% The maplist function can be used to generate lists based on other
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% lists. Note that the output list is a free variable, causing an
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