The examples used here are from the weekly challenge problem statement and demonstrate the working solution.
You are given a list of strings S. Write a script to group Anagrams together in any random order.
:-initialization(main). letter_factor(e, 2). letter_factor(t, 3). letter_factor(a, 5). letter_factor(o, 7). letter_factor(i, 11). letter_factor(n, 13). letter_factor(s, 17). letter_factor(h, 19). letter_factor(r, 23). letter_factor(d, 29). letter_factor(l, 31). letter_factor(c, 37). letter_factor(u, 41). letter_factor(m, 43). letter_factor(w, 47). letter_factor(f, 53). letter_factor(g, 59). letter_factor(y, 61). letter_factor(p, 67). letter_factor(b, 71). letter_factor(v, 73). letter_factor(k, 79). letter_factor(j, 83). letter_factor(x, 89). letter_factor(q, 97). letter_factor(z, 101). chars_product(, 1). chars_product([H|T], Product):- letter_factor(H, Factor), chars_product(T, Product0), Product is Factor * Product0. word_product(Word, Product):- atom_chars(Word, Chars), chars_product(Chars, Product). organize():- findall(Words, bagof(Word, word_product(Word-_), Words), WordsList), write(WordsList). organize([H|T]):- word_product(H, P), assertz(word_product(H-P)), organize(T). main:- Anagrams = [opt, bat, saw, tab, pot, top, was], organize(Anagrams), nl, halt.
$ gplc ch-1.p $ ./ch-1 [[bat,tab],[opt,pot,top],[saw,was]]
This is a Prolog version of Perl code I wrote for the same problem. In translating the approach of using the Fundamental Theorem of Arithmetic I tried to make sure the result was idiomatic Prolog as much as I could. I think I pulled that off!?!? In Prolog there is sometimes an over temptation to end up writing a bunch of mostly functional code. That is, code that could just as easily be written in, say, Haskell or ML. The
organize/1 predicate is pretty much that sort of functional code but it is very short and wraps Prolog concepts such as
I am using the same mathematical trick that I have used for anagrams in the past, starting with Challenge 005. The By the Fundamental Theorem of Arithmetic every integer greater than 1 is either a prime number itself or can be represented as the unique product of prime numbers. We use that to our advantage by having a prime number associated with each letter. Each word is a product of these numbers and words with the same product are anagrams.
In this way we
assert/1Word-Product pairs for all the given anagrams and once done use
bagof/3to collect the solutions for printing as desired. This method of collecting solutions is given a nice description here.
The choice of letters and prime numbers is based on the Lewand Ordering and it isn’t at all necessary (it stems from an early, unnecessary, design decision) but it does little harm so I left it in anyway.
You are given a binary tree. Write a script to represent the given binary tree as an object and flatten it to a linked list object. Finally, print the linked list object.
:-initialization(main). dfs(Node, Graph, [Node|Path]):- dfs(Node, Graph, Path, ). dfs(_, _, , _). dfs(Node, Graph, [AdjNode|Path], Seen) :- member(r(Node, Adjacent), Graph), member(AdjNode, Adjacent), \+ memberchk(AdjNode, Seen), dfs(AdjNode, Graph, Path, [Node|Seen]). unseen_nodes(Nodes, NodeList, Unseen):- unseen_nodes(Nodes, NodeList, , Unseen). unseen_nodes(, _, Unseen, Unseen). unseen_nodes([H|T], NodeList, UnseenAccum, Unseen):- \+ memberchk(H, NodeList), unseen_nodes(T, NodeList, [H|UnseenAccum], Unseen). unseen_nodes([H|T], NodeList, UnseenAccum, Unseen):- memberchk(H, NodeList), unseen_nodes(T, NodeList, UnseenAccum, Unseen). paths_list(Paths, List):- paths_list(Paths, , List). paths_list(, List, List). paths_list([H|T], ListAccum, List):- unseen_nodes(H, ListAccum, Unseen), append(ListAccum, Unseen, ListAccum0), paths_list(T, ListAccum0, List). print_list([H|]):- format("~d~n", [H]). print_list([H|T]):- format("~d -> ", [H]), print_list(T). main:- findall(Path, dfs(1,[r(1,),r(1,),r(2,[4,5]),r(5,[6,7])], Path), Paths), paths_list(Paths, List), print_list(List), halt.
$ gplc ch-2.p -bash-5.0$ ./ch-2 1 -> 2 -> 4 -> 5 -> 6 -> 7 -> 3
Some of this code is re-used from last week
The idea of a (singly) Linked List in Prolog is fundamental. We might consider any ordinary list in this way since Prolog allows direct access to the head and tail of
lists. This solution, therefore ends up being simpler than even the Perl solution to the same problem.
The code above simply does a DFS of the given tree and then adds new nodes to a list in the order in which they are seen.