Lists in Prolog

Symbols in Prolog:

atom

variable

number

list (how to assembly and take them apart)

 

Lists are very powerful in prolog because we do not need to do similar but repeatedtasks again and again. Instead, we can put all elements in a list and deal with them in a more general way.

 

If we have a look at the difference between lists in prolog and arrays in C, it is easy to notice that we do not need to define the length of a list in prolog. Instead, we use arecursive way to deal with them one by one until the remaining is empty.

 

How to write a list:

use commas to separate every element: [A, B]

use vertical bars to divide head and tail: [A | B]

 

The two definitions are quite different. 

When using [A, B], the list has only 2 element A and B.

When using [A | B], the list can have at least 1 elements because B is a list and it can be empty or very large.

 

How to operate on a list:

Define a stop condition for the recursion;

Deal with lists recursively (always divide the list by head and tail) decreasing the length of it.

 

Examples:

 

example 1: members in a list

 

elem(A, [A | _]).

elem(A, [_ | B]) :- elem(A, B).

 

If we use the following instead, it will only hold when A is the tail of the list.

elem(A, [A]).

elem(A, [_ | B]) :- elem(A, B).

 

 

example 2: list of unique people

 

uniq([ ]).

uniq([H | T]) :- people(H), \+ member(H, T), uniq(T).

 

example 3: joining two lists

 

join([ ], T, T).

join([H | T], L, [H | W]) :- join(T, L, W).

 

We can learn from these examples that we use [H | T] to take the lists into head and tail decreasing the length of it to solve our problem.

 

Using append predicate

append predicate is powerful is because it can analyze the structure of a list.

Notice that the parameter of append predicate must be lists or variables not atoms.

 

append([a, b], [c, d], L).

L=[a, b, c, d].

 

Examples:

 

L1 is start of L2:

front(L1, L2) :- append(L1, _, L2).

 

E is last element of L:

last(E, L) :- append(_, [E], L).

 

another version of member predicate:

mem(A, B) :- append(_, [A | _], B).

 

X is before Y in L:

before(X, Y, L) :- append(Z, [Y | _], L), append(_, [X | _], Z).