### 11.4 Practical Session

Try the following two programming exercises:

1. Sets can be thought of as lists that don’t contain any repeated elements. For example, [a,4,6] is a set, but [a,4,6,a] is not (as it contains two occurrences of a) . Write a Prolog program subset/2 that is satisfied when the first argument is a subset of the second argument (that is, when every element of the first argument is a member of the second argument). For example:
?-  subset([a,b],[a,b,c])
yes

?-  subset([c,b],[a,b,c])
yes

?-  subset([],[a,b,c])
yes

Your program should be capable of generating all subsets of an input set by backtracking. For example, if you give it as input

?-  subset(X,[a,b,c])

it should successively generate all eight subsets of [a,b,c] .

2. Using the subset predicate you have just written, and findall/3 , write a predicate powerset/2 that takes a set as its first argument, and returns the powerset of this set as the second argument. (The powerset of a set is the set of all its subsets.) For example:
?-  powerset([a,b,c],P)

should return

P  =  [[],[a],[b],[c],[a,b],[a,c],[b,c],[a,b,c]]

It doesn’t matter if the sets are returned in some other order. For example,

P  =  [[a],[b],[c],[a,b,c],[],[a,b],[a,c],[b,c]]

is fine too. © 2006-2012 Patrick Blackburn, Johan Bos, Kristina Striegnitz