Prolog how does append work

There are multiple answers! On the other hand, flexibility in all things is an asset! Here, we append atom foo to a list [a,b,c,d] resulting in something that is not a list, as it starts with a prefix but does not end in [] :?

Suffix extraction Get me a suffix of "anything beyond and including c ":? Is it efficient? I don't know, but it works. Prefix extraction Get me a prefix of "anything up to and including c ". There is no way to express "a list where the last element is c " as a syntactic literal in Prolog, so we have to resort to more extensive code:?

This predicate successfully finds prefixes of lists, and moreover, via backtracking, finds them all:. In a similar fashion, we can define a program which finds suffixes of lists. For example, the suffixes of [a,b,c,d] are [] , [d] , [c,d] , [b,c,d] , and [a,b,c,d]. That is, list S is a suffix of list L if there is some list such that L is the result of concatenating that list with S. This predicate successfully finds suffixes of lists, and moreover, via backtracking, finds them all:. And now it's very easy to define a program that finds sublists of lists.

If a list contains repeated elements they should be placed in separate sublists. Use the result of problem P09 to implement the so-called run-length encoding data compression method. Consecutive duplicates of elements are encoded as terms [N,E] where N is the number of duplicates of the element E. Modify the result of problem P10 in such a way that if an element has no duplicates it is simply copied into the result list.

Only elements with duplicates are transferred as [N,E] terms. Given a run-length code list generated as specified in problem P Construct its uncompressed version. Implement the so-called run-length encoding data compression method directly. As in problem P11, simplify the result list by replacing the singleton terms [1,X] by X.

Do not use any predefined predicates. Given two indices, I and K, the slice is the list containing the elements between the I'th and K'th element of the original list both limits included.

Start counting the elements with 1. The selected items shall be put into a result list. The selected numbers shall be put into a result list. For pure mathematicians, this result may be great. But we want to really generate all the possibilities via backtracking. Write a predicate that generates all the possibilities via backtracking. Note that we do not want permutations of the group members; i.

However, we make a difference between [[aldo,beat],[carla,david], You may find more about this combinatorial problem in a good book on discrete mathematics under the term "multinomial coefficients". The objective is to sort the elements of InList according to their length. But this time the objective is to sort the elements of InList according to their length frequency ; i. The third and forth list have length 3 which appears, there are two list of this length.

And finally, the last three lists have length 2. This is the most frequent length. Use Euclid's algorithm. Two numbers are coprime if their greatest common divisor equals 1.

Euler's totient function plays an important role in one of the most widely used public key cryptography methods RSA. Asked 9 years, 4 months ago. Active 1 month ago. Viewed 99k times. Example, appendlist [1,2],[3,4,5],X. Also I don't know what happening in the recursion. I traced it but didn't understand EDIT: What I want to know is how it should be coded to function like the predefined append in Prolog. Improve this question.

Mateusz Piotrowski 6, 9 9 gold badges 46 46 silver badges 73 73 bronze badges. Zik Zik 1, 6 6 gold badges 21 21 silver badges 31 31 bronze badges. Add a comment. Active Oldest Votes. The order of clauses just needs to be swapped in order to make this predicate definition productive, when used in a generative fashion : append [], X, X. Or from left to right: a b c. Improve this answer.