Set consolidation: Difference between revisions

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elif i == j then del(.[i])
elif i == j then del(.[i])
else
else
.[i] as $a | .[j] as $b | union($a;$b) as $c
union(.[i]; .[j]) as $c
| union(del(.[j]) | del(.[i]); [$c])
| union(del(.[j]) | del(.[i]); [$c])
end;
end;

Revision as of 06:22, 5 October 2014

Task
Set consolidation
You are encouraged to solve this task according to the task description, using any language you may know.

Given two sets of items then if any item is common to any set then the result of applying consolidation to those sets is a set of sets whose contents is:

  • The two input sets if no common item exists between the two input sets of items.
  • The single set that is the union of the two input sets if they share a common item.

Given N sets of items where N>2 then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible. If N<2 then consolidation has no strict meaning and the input can be returned.

Example 1:
Given the two sets {A,B} and {C,D} then there is no common element between the sets and the result is the same as the input.
Example 2:
Given the two sets {A,B} and {B,D} then there is a common element B between the sets and the result is the single set {B,D,A}. (Note that order of items in a set is immaterial: {A,B,D} is the same as {B,D,A} and {D,A,B}, etc).
Example 3:
Given the three sets {A,B} and {C,D} and {D,B} then there is no common element between the sets {A,B} and {C,D} but the sets {A,B} and {D,B} do share a common element that consolidates to produce the result {B,D,A}. On examining this result with the remaining set, {C,D}, they share a common element and so consolidate to the final output of the single set {A,B,C,D}
Example 4:
The consolidation of the five sets:
{H,I,K}, {A,B}, {C,D}, {D,B}, and {F,G,H}
Is the two sets:
{A, C, B, D}, and {G, F, I, H, K}

See also:

Ada

We start with specifying a generic package Set_Cons that provides the neccessary tools, such as contructing and manipulating sets, truning them, ect.:

<lang Ada>generic 

  type Element is (<>);
  with function Image(E: Element) return String;

package Set_Cons is 

  type Set is private;

  -- constructor and manipulation functions for type Set
  function "+"(E: Element) return Set;
  function "+"(Left, Right: Element) return Set;
  function "+"(Left: Set; Right: Element) return Set;
  function "-"(Left: Set; Right: Element) return Set;

  -- compare, unite or output a Set
  function Nonempty_Intersection(Left, Right: Set) return Boolean;
  function Union(Left, Right: Set) return Set;
  function Image(S: Set) return String;

  type Set_Vec is array(Positive range <>) of Set;

  -- output a Set_Vec
  function Image(V: Set_Vec) return String;

private 

  type Set is array(Element) of Boolean;

end Set_Cons;</lang>

Here is the implementation of Set_Cons:

<lang Ada>package body Set_Cons is 

  function "+"(E: Element) return Set is
     S: Set := (others => False);
  begin
     S(E) := True;
     return S;
  end "+";

  function "+"(Left, Right: Element) return Set is
  begin
     return (+Left) + Right;
  end "+";

  function "+"(Left: Set; Right: Element) return Set is
     S: Set := Left;
  begin
     S(Right) := True;
     return S;
  end "+";

  function "-"(Left: Set; Right: Element) return Set is
     S: Set := Left;
  begin
     S(Right) := False;
     return S;
  end "-";

  function Nonempty_Intersection(Left, Right: Set) return Boolean is
  begin
     for E in Element'Range loop
        if Left(E) and then Right(E) then return True;
        end if;
     end loop;
     return False;
  end Nonempty_Intersection;

  function Union(Left, Right: Set) return Set is
     S: Set := Left;
  begin
     for E in Right'Range loop
        if Right(E) then S(E) := True;
        end if;
     end loop;
     return S;
  end Union;

  function Image(S: Set) return String is

     function Image(S: Set; Found: Natural) return String is
     begin
        for E in S'Range loop
           if S(E) then
              if Found = 0 then
                 return Image(E) & Image((S-E), Found+1);
              else
                 return "," & Image(E) & Image((S-E), Found+1);
              end if;
           end if;
        end loop;
        return "";
     end Image;

  begin
     return "{" & Image(S, 0) & "}";
  end Image;

  function Image(V: Set_Vec) return String is
   begin
     if V'Length = 0 then
        return "";
     else
        return Image(V(V'First)) & Image(V(V'First+1 .. V'Last));
     end if;
  end Image;

end Set_Cons;</lang>

Given that package, the task is easy:

<lang Ada>with Ada.Text_IO, Set_Cons;

procedure Set_Consolidation is 

  type El_Type is (A, B, C, D, E, F, G, H, I, K);

  function Image(El: El_Type) return String is
  begin
     return El_Type'Image(El);
  end Image;

  package Helper is new Set_Cons(Element => El_Type, Image => Image);
  use Helper;

  function Consolidate(List: Set_Vec) return Set_Vec is
  begin
     for I in List'First .. List'Last - 1 loop
        for J in I+1 .. List'Last loop
           -- if List(I) and List(J) share an element
           -- then recursively consolidate
           --   (List(I) union List(J)) followed by List(K), K not in {I, J}
           if Nonempty_Intersection(List(I), List(J)) then
              return Consolidate
                (Union(List(I), List(J)) 
                   & List(List'First .. I-1) 
                   & List(I+1        .. J-1) 
                   & List(J+1        .. List'Last));
           end if;
        end loop;
     end loop;
     return List;
  end Consolidate;

begin 

  Ada.Text_IO.Put_Line(Image(Consolidate((A+B) & (C+D))));
  Ada.Text_IO.Put_Line(Image(Consolidate((A+B) & (B+D))));
  Ada.Text_IO.Put_Line(Image(Consolidate((A+B) & (C+D) & (D+B))));
  Ada.Text_IO.Put_Line
    (Image(Consolidate((H+I+K) & (A+B) & (C+D) & (D+B) & (F+G+H))));

end Set_Consolidation;</lang>

This generates the following output: 

{A,B}{C,D}
{A,B,D}
{A,B,C,D}
{A,B,C,D}{F,G,H,I,K}

Aime

<lang aime>void display(list l) { 

   integer i;
   text s;

   i = -l_length(l);
   while (i) {
       record r;
       text u, v;

       o_text(s);
       s = ", ";
       o_text("{");
       r = l_q_record(l, i);
       if (r_first(r, u)) {
           do {
               o_text(v);
               v = ", ";
               o_text(u);
           } while (r_greater(r, u, u));
       }
       o_text("}");
       i += 1;
   }

   o_text("\n");

}

integer intersect(record r, record u) { 

   integer a;
   text s;

   a = 0;
   if (r_first(r, s)) {
       do {
           if (r_key(u, s)) {
               a = 1;
               break;
           }
       } while (r_greater(r, s, s));
   }

   return a;

}

void merge(record u, record r) { 

   text s;

   if (r_first(r, s)) {
       do {
           r_add(u, s, r_query(r, s));
       } while (r_greater(r, s, s));
   }

}

list consolidate(list l) { 

   integer i;

   i = -l_length(l);
   while (i) {
       integer j;
       record r;

       r = l_q_record(l, i);
       i += 1;
       j = i;
       while (j) {
           record u;

           u = l_q_record(l, j);
           j += 1;
           if (intersect(r, u)) {
               merge(u, r);
               l_delete(l, i - 1);
               break;
           }
       }
   }

   return l;

}

list L(...) { 

   integer i;
   list l;

   i = -count();
   while (i) {
       l_link(l, -1, $i);
       i += 1;
   }

   return l;

}

record R(...) { 

   integer i;
   record r;

   i = -count();
   while (i) {
       r_p_integer(r, $i, 0);
       i += 1;
   }

   return r;

}

integer main(void) { 

   display(consolidate(L(R("A", "B"), R("C", "D"))));
   display(consolidate(L(R("A", "B"), R("B", "D"))));
   display(consolidate(L(R("A", "B"), R("C", "D"), R("D", "B"))));
   display(consolidate(L(R("H", "I", "K"), R("A", "B"), R("C", "D"),
                         R("D", "B"), R("F", "G", "K"))));

   return 0;

}</lang>

Output:
{A, B}, {C, D}
{A, B, D}
{A, B, C, D}
{A, B, C, D}, {F, G, H, I, K}

Bracmat

<lang bracmat>( ( consolidate 

 =   a m z mm za zm zz
   .     ( removeNumFactors
         =   a m z
           .     !arg:?a+#%*?m+?z
               & !a+!m+removeNumFactors$!z
             | !arg
         )
       &   !arg
         :   ?a
             %?`m
             ( %?z
             &   !m
               :   ?
                 + ( %@?mm
                   & !z:?za (?+!mm+?:?zm) ?zz
                   )
                 + ?
             )
       & consolidate$(!a removeNumFactors$(!m+!zm) !za !zz)
     | !arg
 )

& (test=.out$(!arg "==>" consolidate$!arg)) & test$(A+B C+D) & test$(A+B B+D) & test$(A+B C+D D+B) & test$(H+I+K A+B C+D D+B F+G+H) );</lang> Output: 

A+B C+D ==> A+B C+D
A+B B+D ==> A+B+D
A+B C+D B+D ==> A+B+C+D
  H+I+K
  A+B
  C+D
  B+D
  F+G+H
  ==>
  F+G+H+I+K
  A+B+C+D

C

<lang c>#include <stdio.h>

  1. define s(x) (1U << ((x) - 'A'))

typedef unsigned int bitset;

int consolidate(bitset *x, int len) { int i, j; for (i = len - 2; i >= 0; i--) for (j = len - 1; j > i; j--) if (x[i] & x[j]) x[i] |= x[j], x[j] = x[--len]; return len; }

void show_sets(bitset *x, int len) { bitset b; while(len--) { for (b = 'A'; b <= 'Z'; b++) if (x[len] & s(b)) printf("%c ", b); putchar('\n'); } }

int main(void) { bitset x[] = { s('A') | s('B'), s('C') | s('D'), s('B') | s('D'), s('F') | s('G') | s('H'), s('H') | s('I') | s('K') };

int len = sizeof(x) / sizeof(x[0]);

puts("Before:"); show_sets(x, len); puts("\nAfter:"); show_sets(x, consolidate(x, len)); return 0; }</lang>

The above is O(N2) in terms of number of input sets. If input is large (many sets or huge number of elements), here's an O(N) method, where N is the sum of the sizes of all input sets: <lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <string.h>

struct edge { int to; struct edge *next; }; struct node { int group; struct edge *e; };

int **consolidate(int **x) {

  1. define alloc(v, size) v = calloc(size, sizeof(v[0]));

int group, n_groups, n_nodes; int n_edges = 0; struct edge *edges, *ep; struct node *nodes; int pos, *stack, **ret;

void add_edge(int a, int b) { ep->to = b; ep->next = nodes[a].e; nodes[a].e = ep; ep++; }

void traverse(int a) { if (nodes[a].group) return;

nodes[a].group = group; stack[pos++] = a;

for (struct edge *e = nodes[a].e; e; e = e->next) traverse(e->to); }

n_groups = n_nodes = 0; for (int i = 0; x[i]; i++, n_groups++) for (int j = 0; x[i][j]; j++) { n_edges ++; if (x[i][j] >= n_nodes) n_nodes = x[i][j] + 1; }

alloc(ret, n_nodes); alloc(nodes, n_nodes); alloc(stack, n_nodes); ep = alloc(edges, n_edges);

for (int i = 0; x[i]; i++) for (int *s = x[i], j = 0; s[j]; j++) add_edge(s[j], s[j + 1] ? s[j + 1] : s[0]);

group = 0; for (int i = 1; i < n_nodes; i++) { if (nodes[i].group) continue;

group++, pos = 0; traverse(i);

stack[pos++] = 0; ret[group - 1] = malloc(sizeof(int) * pos); memcpy(ret[group - 1], stack, sizeof(int) * pos); }

free(edges); free(stack); free(nodes);

// caller is responsible for freeing ret return realloc(ret, sizeof(ret[0]) * (1 + group));

  1. undef alloc

}

void show_sets(int **x) { for (int i = 0; x[i]; i++) { printf("%d: ", i); for (int j = 0; x[i][j]; j++) printf(" %d", x[i][j]); putchar('\n'); } }

int main(void) { int *x[] = { (int[]) {1, 2, 0}, // 0: end of set (int[]) {3, 4, 0}, (int[]) {3, 1, 0}, (int[]) {0}, // empty set (int[]) {5, 6, 0}, (int[]) {7, 6, 0}, (int[]) {3, 9, 10, 0}, 0 // 0: end of sets };

puts("input:"); show_sets(x);

puts("components:"); show_sets(consolidate(x));

return 0; }</lang>

Common Lisp

Translation of: Racket

<lang lisp>(defun consolidate (ss) 

 (labels ((comb (cs s)
            (cond ((null s) cs)
                  ((null cs) (list s))
                  ((null (intersection s (first cs)))
                   (cons (first cs) (comb (rest cs) s)))
                  ((consolidate (cons (union s (first cs)) (rest cs)))))))
   (reduce #'comb ss :initial-value nil)))</lang>
Output:
> (consolidate '((a b) (c d)))
((A B) (C D))
> (consolidate '((a b) (b c)))
((C A B))
> (consolidate '((a b) (c d) (d b)))
((C D A B))
> (consolidate '((h i k) (a b) (c d) (d b) (f g h)))
((F G H I K) (C D A B))

D

Translation of: Go

<lang d>import std.stdio, std.algorithm, std.array;

dchar[][] consolidate(dchar[][] sets) { 

   foreach (set; sets)
       set.sort;

   foreach (i, ref si; sets[0 .. $ - 1]) {
       if (si.empty)
           continue;
       foreach (ref sj; sets[i + 1 .. $])
           if (!sj.empty && !si.setIntersection(sj).empty) {
               sj = si.setUnion(sj).uniq.array;
               si = null;
           }
   }

   return sets.filter!"!a.empty".array;

}

void main() { 

   [['A', 'B'], ['C','D']].consolidate.writeln;

   [['A','B'], ['B','D']].consolidate.writeln;

   [['A','B'], ['C','D'], ['D','B']].consolidate.writeln;

   [['H','I','K'], ['A','B'], ['C','D'],
    ['D','B'], ['F','G','H']].consolidate.writeln;

}</lang>

Output:
["AB", "CD"]
["ABD"]
["ABCD"]
["ABCD", "FGHIK"]

Recursive version, as described on talk page. <lang d>import std.stdio, std.algorithm, std.array;

dchar[][] consolidate(dchar[][] sets) { 

   foreach (set; sets)
       set.sort;

   dchar[][] consolidateR(dchar[][] s) {
       if (s.length < 2)
           return s;
       auto r = [s[0]];
       foreach (x; consolidateR(s[1 .. $])) {
           if (!r[0].setIntersection(x).empty) {
               r[0] = r[0].setUnion(x).uniq.array;
           } else
               r ~= x;
       }
       return r;
   }

   return consolidateR(sets);

}

void main() { 

   [['A', 'B'], ['C','D']].consolidate.writeln;

   [['A','B'], ['B','D']].consolidate.writeln;

   [['A','B'], ['C','D'], ['D','B']].consolidate.writeln;

   [['H','I','K'], ['A','B'], ['C','D'],
    ['D','B'], ['F','G','H']].consolidate.writeln;

}</lang> 

["AB", "CD"]
["ABD"]
["ABCD"]
["FGHIK", "ABCD"]

Egison

<lang egison> (define $consolidate 

 (lambda [$xss]
   (match xss (multiset (set char))
     {[<cons <cons $m $xs>
             <cons <cons ,m $ys>
                   $rss>>
       (consolidate {(unique/m char {m @xs @ys}) @rss})]
      [_ xss]})))

(test (consolidate {{'H' 'I' 'K'} {'A' 'B'} {'C' 'D'} {'D' 'B'} {'F' 'G' 'H'}})) </lang> Output: <lang egison> {"DBAC" "HIKFG"} </lang>

Ela

This solution emulate sets using linked lists: <lang ela>open list

merge [] ys = ys merge (x::xs) ys | x `elem` ys = merge xs ys 

                | else = merge xs (x::ys)

consolidate (_::[])@xs = xs consolidate (x::xs) = conso [x] (consolidate xs) 

               where conso xs [] = xs
                     conso (x::xs)@r (y::ys) | intersect x y <> [] = conso ((merge x y)::xs) ys
                                             | else = conso (r ++ [y]) ys</lang>

Usage: <lang ela>open console

consolidate [['H','I','K'], ['A','B'], ['C','D'], ['D','B'], ['F','G','H']] |> writen $ 

 consolidate [['A','B'], ['B','D']] |> writen</lang>
Output:
[[K,I,F,G,H],[A,C,D,B]]
[[A,B,D]]

F#

<lang fsharp>let (|SeqNode|SeqEmpty|) s = 

   if Seq.isEmpty s then SeqEmpty
   else SeqNode ((Seq.head s), Seq.skip 1 s)

let SetDisjunct x y = Set.isEmpty (Set.intersect x y)

let rec consolidate s = seq { 

   match s with
   | SeqEmpty -> ()
   | SeqNode (this, rest) ->
       let consolidatedRest = consolidate rest
       for that in consolidatedRest do
           if (SetDisjunct this that) then yield that
       yield Seq.fold (fun x y -> if not (SetDisjunct x y) then Set.union x y else x) this consolidatedRest

}

[<EntryPoint>] let main args = 

   let makeSeqOfSet listOfList = List.map (fun x -> Set.ofList x) listOfList |> Seq.ofList
   List.iter (fun x -> printfn "%A" (consolidate (makeSeqOfSet x))) [
       [["A";"B"]; ["C";"D"]];
       [["A";"B"]; ["B";"C"]];
       [["A";"B"]; ["C";"D"]; ["D";"B"]];
       [["H";"I";"K"]; ["A";"B"]; ["C";"D"]; ["D";"B"]; ["F";"G";"H"]]
   ]
   0</lang>

Output 

seq [set ["C"; "D"]; set ["A"; "B"]]
seq [set ["A"; "B"; "C"]]
seq [set ["A"; "B"; "C"; "D"]]
seq [set ["A"; "B"; "C"; "D"]; set ["F"; "G"; "H"; "I"; "K"]]

Go

Translation of: Python

<lang go>package main

import "fmt"

type set map[string]bool

var testCase = []set{ 

   set{"H": true, "I": true, "K": true},
   set{"A": true, "B": true},
   set{"C": true, "D": true},
   set{"D": true, "B": true},
   set{"F": true, "G": true, "H": true},

}

func main() { 

   fmt.Println(consolidate(testCase))

}

func consolidate(sets []set) []set { 

   setlist := []set{}
   for _, s := range sets {
       if s != nil && len(s) > 0 {
           setlist = append(setlist, s)
       }
   }
   for i, s1 := range setlist {
       if len(s1) > 0 {
           for _, s2 := range setlist[i+1:] {
               if s1.disjoint(s2) {
                   continue
               }
               for e := range s1 {
                   s2[e] = true
                   delete(s1, e)
               }
               s1 = s2
           }
       }
   }
   r := []set{}
   for _, s := range setlist {
       if len(s) > 0 {
           r = append(r, s)
       }
   }
   return r

}

func (s1 set) disjoint(s2 set) bool { 

   for e := range s2 {
       if s1[e] {
           return false
       }
   }
   return true

}</lang>

Output:
[map[A:true C:true B:true D:true] map[G:true F:true I:true H:true K:true]]

Haskell

<lang Haskell>import qualified Data.Set as S

consolidate :: Ord a => [S.Set a] -> [S.Set a] consolidate = foldl comb [] 

 where comb [] s' = [s']
       comb (s:ss) s'
         | S.null (s `S.intersection` s') = s : comb ss s'
         | otherwise = comb ss (s `S.union` s')

</lang>

Output:
*Main> consolidate [S.fromList "ab", S.fromList "dc"]
[fromList "ab",fromList "cd"]
*Main> consolidate [S.fromList "ab", S.fromList "bc"]
[fromList "abc"]
*Main> consolidate [S.fromList "ab", S.fromList "cd", S.fromList "db"]
[fromList "abcd"]
*Main> consolidate [S.fromList "hik", S.fromList "ab", S.fromList "cd", S.fromList "db", S.fromList "fgh"]
[fromList "abcd",fromList "fghik"]

J

<lang J>consolidate=:4 :0/ 

 b=. y 1&e.@e.&> x
 (1,-.b)#(~.;x,b#y);y

)</lang>

Examples:

<lang J> consolidate 'ab';'cd' ┌──┬──┐ │ab│cd│ └──┴──┘ 

  consolidate 'ab';'bd'

┌───┐ │abd│ └───┘ 

  consolidate 'ab';'cd';'db'

┌────┐ │abcd│ └────┘ 

  consolidate 'hij';'ab';'cd';'db';'fgh'

┌─────┬────┐ │hijfg│abcd│ └─────┴────┘</lang>

Java

Translation of: D
Works with: Java version 7

<lang java>import java.util.*;

public class SetConsolidation { 

   public static void main(String[] args) {
       List<Set<Character>> h1 = hashSetList("AB", "CD");
       System.out.println(consolidate(h1));

       List<Set<Character>> h2 = hashSetList("AB", "BD");
       System.out.println(consolidateR(h2));

       List<Set<Character>> h3 = hashSetList("AB", "CD", "DB");
       System.out.println(consolidate(h3));

       List<Set<Character>> h4 = hashSetList("HIK", "AB", "CD", "DB", "FGH");
       System.out.println(consolidateR(h4));
   }

   // iterative
   private static <E> List<Set<E>>
               consolidate(Collection<? extends Set<E>> sets) {

List<Set<E>> r = new ArrayList<>(); for (Set<E> s : sets) { List<Set<E>> new_r = new ArrayList<>(); new_r.add(s); for (Set<E> x : r) { if (!Collections.disjoint(s, x)) { s.addAll(x); } else { new_r.add(x); } } r = new_r; } return r; 

   }

   // recursive
   private static <E> List<Set<E>> consolidateR(List<Set<E>> sets) {
       if (sets.size() < 2)
           return sets;
       List<Set<E>> r = new ArrayList<>();
       r.add(sets.get(0));
       for (Set<E> x : consolidateR(sets.subList(1, sets.size()))) {
           if (!Collections.disjoint(r.get(0), x)) {
               r.get(0).addAll(x);
           } else {
               r.add(x);
           }
       }
       return r;
   }

   private static List<Set<Character>> hashSetList(String... set) {
       List<Set<Character>> r = new ArrayList<>();
       for (int i = 0; i < set.length; i++) {
           r.add(new HashSet<Character>());
           for (int j = 0; j < set[i].length(); j++)
               r.get(i).add(set[i].charAt(j));
       }
       return r;
   }

}</lang> 

[A, B] [D, C] 
[D, A, B] 
[D, A, B, C] 
[F, G, H, I, K] [D, A, B, C]

jq

Infrastructure:

Currently, jq does not have a "Set" library, so to save space here, we will use simple but inefficient implementations of set-oriented functions as they are fast for sets of moderate size. Nevertheless, we will represent sets as sorted arrays. <lang jq>def to_set: unique;

def union(A; B): (A + B) | unique;

  1. boolean

def intersect(A;B): 

 reduce A[] as $x (false; if . then . else (B|index($x)) end) | not | not;</lang>

Consolidation:

For clarity, the helper functions are presented as top-level functions, but they could be defined as inner functions of the main function, consolidate/0.

<lang jq># Input: [i, j, sets] with i < j

  1. Return [i,j] for a pair that can be combined, else null

def combinable: 

  .[0] as $i | .[1] as $j | .[2] as $sets
  | ($sets|length) as $length
  | if intersect($sets[$i]; $sets[$j]) then [$i, $j]
    elif $i < $j - 1      then (.[0] += 1 | combinable)
    elif $j < $length - 1 then [0, $j+1, $sets] | combinable
    else null
    end;
  1. Given an array of arrays, remove the i-th and j-th elements,
  2. and add their union:

def update(i;j): 

 if i > j then update(j;i)
 elif i == j then del(.[i])
 else 
   union(.[i]; .[j]) as $c
   | union(del(.[j]) | del(.[i]); [$c])
 end;
  1. Input: a set of sets

def consolidate: 

  if length <= 1 then .
  else
    ([0, 1, .] | combinable) as $c
    | if $c then update($c[0]; $c[1]) | consolidate
      else .
      end
  end;

</lang> Examples: <lang jq>def tests: 

 [["A", "B"], ["C","D"]],
 [["A","B"], ["B","D"]],
 [["A","B"], ["C","D"], ["D","B"]],
 [["H","I","K"], ["A","B"], ["C","D"], ["D","B"], ["F","G","H"]]

def test: 

 tests | to_set | consolidate;

test</lang>

Output:

<lang sh>$ jq -c -n -f Set_consolidation.rc [["A","B"],["C","D"]] "A","B","D" "A","B","C","D" [["A","B","C","D"],["F","G","H","I","K"]]</lang>

Mathematica

<lang Mathematica>reduce[x_] := 

Block[{pairs, unique},
 pairs = 
  DeleteCases[
   Subsets[Range@
     Length@x, {2}], _?(Intersection @@ x# == {} &)];
 unique = Complement[Range@Length@x, Flatten@pairs];
 Join[Union[Flatten[x#]] & /@ pairs, xunique]]

consolidate[x__] := FixedPoint[reduce, {x}]</lang> 

consolidate[{a, b}, {c, d}]
-> {{a, b}, {c, d}}

consolidate[{a, b}, {b, d}]
-> {{a, b, d}}

consolidate[{a, b}, {c, d}, {d, b}]
-> {{a, b, c, d}}

consolidate[{h, i, k}, {a, b}, {c, d}, {d, b}, {f, g, h}]
-> {{a,b,c,d},{f,g,h,i,k}}

OCaml

<lang ocaml>let join a b = 

 List.fold_left (fun acc v ->
   if List.mem v acc then acc else v::acc
 ) b a

let share a b = List.exists (fun x -> List.mem x b) a

let extract p lst = 

 let rec aux acc = function
 | x::xs -> if p x then Some (x, List.rev_append acc xs) else aux (x::acc) xs
 | [] -> None
 in
 aux [] lst

let consolidate sets = 

 let rec aux acc = function
 | [] -> List.rev acc
 | x::xs ->
     match extract (share x) xs with
     | Some (y, ys) -> aux acc ((join x y) :: ys)
     | None -> aux (x::acc) xs
 in
 aux [] sets

let print_sets sets = 

 print_string "{ ";
 List.iter (fun set ->
   print_string "{";
   print_string (String.concat " " set);
   print_string "} "
 ) sets;
 print_endline "}"

let () = 

 print_sets (consolidate [["A";"B"]; ["C";"D"]]);
 print_sets (consolidate [["A";"B"]; ["B";"C"]]);
 print_sets (consolidate [["A";"B"]; ["C";"D"]; ["D";"B"]]);
 print_sets (consolidate [["H";"I";"K"]; ["A";"B"]; ["C";"D"]; ["D";"B"];
                          ["F";"G";"H"]]);
</lang>

Output: 

{ {A B} {C D} }
{ {A B C} }
{ {B A C D} }
{ {K I F G H} {B A C D} }

ooRexx

<lang oorexx>/* REXX ***************************************************************

  • 04.08.2013 Walter Pachl using ooRexx features
  • (maybe not in the best way -improvements welcome!)
  • but trying to demonstrate the algorithm
                                                                                                                                            • /

s.1=.array~of(.set~of('A','B'),.set~of('C','D')) s.2=.array~of(.set~of('A','B'),.set~of('B','D')) s.3=.array~of(.set~of('A','B'),.set~of('C','D'),.set~of('D','B')) s.4=.array~of(.set~of('H','I','K'),.set~of('A','B'),.set~of('C','D'),, 

             .set~of('B','D'),.set~of('F','G','H'))

s.5=.array~of(.set~of('snow','ice','slush','frost','fog'),, 

             .set~of('iceburgs','icecubes'),,
             .set~of('rain','fog','sleet'))

s.6=.array~of('one') s.7=.array~new s.8=.array~of() Do si=1 To 8 /* loop through the test data */ 

 na=s.si                              /* an array of sets           */
 head='Output(s):'
 Say left('Input' si,10) list_as(na)  /* show the input             */
 Do While na~items()>0                /* while the array ain't empty*/
   na=cons(na)                        /* consolidate and get back   */
                                      /*  array of remaining sets   */
   head='          '
   End
 Say '===='                           /* separator line             */
 End

Exit

cons: Procedure Expose head /**********************************************************************

  • consolidate the sets in the given array
                                                                                                                                            • /
 Use Arg a
 w=a                                  /* work on a copy             */
 n=w~items()                          /* number of sets in the array*/
 Select
   When n=0 Then                      /* no set in array            */
     Return .array~new                /* retuns an empty array      */
   When n=1 Then Do                   /* one set in array           */
     Say head list(w[1])              /* show its contents          */
     Return .array~new                /* retuns an empty array      */
     End
   Otherwise Do                       /* at least two sets are there*/
     b=.array~new                     /* use for remaining sets     */
     r=w[n]                           /* start with last set        */
     try=1
     Do until changed=0               /* loop until result is stable*/
       changed=0
       new=0
       n=w~items()                    /* number of sets             */
       Do i=1 To n-try                /* loop first through n-1 sets*/
         try=0                        /* then through all of them   */
         is=r~intersection(w[i])
         If is~items>0 Then Do        /* any elements in common     */
           r=r~union(w[i])            /* result is the union        */
           Changed=1                  /* and result is now larger   */
           End
         Else Do                      /* no elemen in common        */
           new=new+1                  /* add the set to the array   */
           b[new]=w[i]                /* of remaining sets          */
           End
         End
       If b~items()=0 Then Do         /* no remaining sets          */
         w=.array~new
         Leave                        /* we are done                */
         End
       w=b                            /* repeat with remaining sets */
       b=.array~new                   /* prepare for next iteration */
       End
     End
   Say head list(r)                   /* show one consolidated set  */
   End
 Return w                             /* return array of remaining  */

list: Procedure /**********************************************************************

  • list elements of given set
                                                                                                                                            • /
 Call trace ?O
 Use Arg set
 arr=set~makeArray
 arr~sort()
 ol='('
 Do i=1 To arr~items()
   If i=1 Then
     ol=ol||arr[i]
   Else
     ol=ol||','arr[i]
   End
 Return ol')'

list_as: Procedure /**********************************************************************

  • List an array of sets
                                                                                                                                            • /
 Call trace ?O
 Use Arg a
 n=a~items()
 If n=0 Then
   ol='no element in array'
 Else Do
   ol=
   Do i=1 To n
     ol=ol '('
     arr=a[i]~makeArray
     Do j=1 To arr~items()
       If j=1 Then
         ol=ol||arr[j]
       Else
         ol=ol','arr[j]
       End
     ol=ol') '
     End
   End
 Return strip(ol)</lang>

Output: 

Input 1    (B,A)  (C,D)
Output(s): (C,D)
           (A,B)
====
Input 2    (B,A)  (B,D)
Output(s): (A,B,D)
====
Input 3    (B,A)  (C,D)  (B,D)
Output(s): (A,B,C,D)
====
Input 4    (H,I,K)  (B,A)  (C,D)  (F,G,H)
Output(s): (F,G,H,I,K)
           (A,B,C,D)
====
Input 5    (snow,fog,ice,frost,slush)  (icecubes,iceburgs)  (fog,sleet,rain)
Output(s): (fog,frost,ice,rain,sleet,slush,snow)
           (iceburgs,icecubes)
====
Input 6    (one)
Output(s): (one)
====
Input 7    no element in array
====
Input 8    ()
Output(s): ()
====

PARI/GP

<lang parigp>cons(V)={ 

 my(v,u,s);
 for(i=1,#V,
   v=V[i];
   for(j=i+1,#V,
     u=V[j];
     if(#setintersect(u,v),V[i]=v=vecsort(setunion(u,v));V[j]=[];s++)
   )
 );
 V=select(v->#v,V);
 if(s,cons(V),V)

};</lang>

Perl 6

<lang perl6>multi consolidate() { () } multi consolidate(Set \this is copy, *@those) { 

   gather {
       for consolidate |@those -> \that {
           if this ∩ that { this = this ∪ that }
           else           { take that }
       }
       take this;
   }

}

enum Elems <A B C D E F G H I J K>; say $_, "\n ==> ", consolidate |$_ 

   for [set(A,B), set(C,D)],
       [set(A,B), set(B,D)],
       [set(A,B), set(C,D), set(D,B)],
       [set(H,I,K), set(A,B), set(C,D), set(D,B), set(F,G,H)];</lang>
Output:
set(A, B) set(C, D)
    ==> set(C, D) set(A, B)
set(A, B) set(B, D)
    ==> set(A, B, D)
set(A, B) set(C, D) set(D, B)
    ==> set(A, B, C, D)
set(H, I, K) set(A, B) set(C, D) set(D, B) set(F, G, H)
    ==> set(A, B, C, D) set(H, I, K, F, G)

PicoLisp

Translation of: Python

<lang PicoLisp>(de consolidate (S) 

  (when S
     (let R (cons (car S))
        (for X (consolidate (cdr S))
           (if (mmeq X (car R))
              (set R (uniq (conc X (car R))))
              (conc R (cons X)) ) )
        R ) ) )</lang>

Test: <lang PicoLisp>: (consolidate '((A B) (C D))) -> ((A B) (C D))

(consolidate '((A B) (B D)))

-> ((B D A))

(consolidate '((A B) (C D) (D B)))

-> ((D B C A))

(consolidate '((H I K) (A B) (C D) (D B) (F G H)))

-> ((F G H I K) (D B C A))</lang>

PL/I

<lang PL/I>Set: procedure options (main); /* 13 November 2013 */ 

  declare set(20) character (200) varying;
  declare e character (1);
  declare (i, n) fixed binary;

  set = ;
  n = 1;
  do until (e = ']');
     get edit (e) (a(1)); put edit (e) (a(1));
     if e = '}' then n = n + 1; /* end of set. */
     if e ^= '{' & e ^= ',' & e ^= '}' & e ^= ' ' then
        set(n) = set(n) || e;   /* Build set */
  end;
  /* We have read in all sets. */
  n = n - 1; /* we have n sets */
  /* Display the sets: */
  put skip list ('The original sets:');
  do i = 1 to n;
     call print(i);
  end;
  /* Look for sets to combine: */
  do i = 2 to n;
     if length(set(i)) > 0 then
        if search(set(1), set(i)) > 0 then
           /* there's at least one common element */
           do; call combine (1, i); set(i) = ;  end;
  end;

  put skip (2) list ('Results:');
  do i = 1 to n;
     if length(set(i)) > 0 then call print (i);
  end;

combine: procedure (p, q); 

  declare (p, q) fixed binary;
  declare e character (1);
  declare i fixed binary;

  do i = 1 to length(set(q));
     e = substr(set(q), i, 1);
     if index(set(p), e) = 0 then set(p) = set(p) || e;
  end;

end combine;

print: procedure(k); 

  declare k fixed binary;
  declare i fixed binary;

  put edit ('{') (a);
  do i = 1 to length(set(k));
     put edit (substr(set(k), i, 1)) (a);
     if i < length(set(k)) then put edit (',') (a);
  end;
  put edit ('} ') (a);

end print;

end Set;</lang> 

The original sets: {A,B} 

Results: {A,B}

The original sets: {A,B} {C,D} 

Results: {A,B} {C,D}

The original sets: {A,B} {B,C} 

Results: {A,B,C}

The original sets: {A,B} {C,D} {E,B,F,G,H} 

Results: {A,B,E,F,G,H} {C,D}

Python

Python: Iterative

<lang python>def consolidate(sets): 

   setlist = [s for s in sets if s]
   for i, s1 in enumerate(setlist):
       if s1:
           for s2 in setlist[i+1:]:
               intersection = s1.intersection(s2)
               if intersection:
                   s2.update(s1)
                   s1.clear()
                   s1 = s2
   return [s for s in setlist if s]</lang>

Python: Recursive

<lang python>def conso(s): if len(s) < 2: return s

r, b = [s[0]], conso(s[1:]) for x in b: if r[0].intersection(x): r[0].update(x) else: r.append(x) return r</lang>

Python: Testing

The _test function contains solutions to all the examples as well as a check to show the order-independence of the sets given to the consolidate function. <lang python>def _test(consolidate=consolidate): 

   def freze(list_of_sets):
       'return a set of frozensets from the list of sets to allow comparison'
       return set(frozenset(s) for s in list_of_sets)
       
   # Define some variables
   A,B,C,D,E,F,G,H,I,J,K = 'A,B,C,D,E,F,G,H,I,J,K'.split(',')
   # Consolidate some lists of sets
   assert (freze(consolidate([{A,B}, {C,D}])) == freze([{'A', 'B'}, {'C', 'D'}]))
   assert (freze(consolidate([{A,B}, {B,D}])) == freze([{'A', 'B', 'D'}]))
   assert (freze(consolidate([{A,B}, {C,D}, {D,B}])) == freze([{'A', 'C', 'B', 'D'}]))
   assert (freze(consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}])) ==
            freze([{'A', 'C', 'B', 'D'}, {'G', 'F', 'I', 'H', 'K'}]))
   assert (freze(consolidate([{A,H}, {H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}])) ==
            freze([{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}]))
   assert (freze(consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}])) ==
            freze([{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}]))
   # Confirm order-independence
   from copy import deepcopy
   import itertools
   sets = [{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}]
   answer = consolidate(deepcopy(sets))
   for perm in itertools.permutations(sets):
           assert consolidate(deepcopy(perm)) == answer

   assert (answer == [{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}])
   assert (len(list(itertools.permutations(sets))) == 720)
   
   print('_test(%s) complete' % consolidate.__name__)

if __name__ == '__main__': 

   _test(consolidate)
   _test(conso)</lang>
Output:
_test(consolidate) complete
_test(conso) complete

Racket

<lang racket>

  1. lang racket

(define (consolidate ss) 

 (define (comb s cs)
   (cond [(set-empty? s) cs]
         [(empty? cs) (list s)]
         [(set-empty? (set-intersect s (first cs)))
          (cons (first cs) (comb s (rest cs)))]
         [(consolidate (cons (set-union s (first cs)) (rest cs)))]))
 (foldl comb '() ss))

(consolidate (list (set 'a 'b) (set 'c 'd))) (consolidate (list (set 'a 'b) (set 'b 'c))) (consolidate (list (set 'a 'b) (set 'c 'd) (set 'd 'b))) (consolidate (list (set 'h 'i 'k) (set 'a 'b) (set 'c 'd) (set 'd 'b) (set 'f 'g 'h))) </lang> Output: <lang racket> (list (set 'b 'a) (set 'd 'c)) (list (set 'a 'b 'c)) (list (set 'a 'b 'd 'c)) (list (set 'g 'h 'k 'i 'f) (set 'a 'b 'd 'c)) </lang>


REXX

<lang rexx>/*REXX program shows how to consolidate a sample bunch of sets. */ sets.= /*assign all SETS. to null. */ sets.1 = '{A,B} {C,D}' sets.2 = "{A,B} {B,D}" sets.3 = '{A,B} {C,D} {D,B}' sets.4 = '{H,I,K} {A,B} {C,D} {D,B} {F,G,H}' sets.5 = '{snow,ice,slush,frost,fog} {iceburgs,icecubes} {rain,fog,sleet}' 

       do j=1 while sets.j\==       /*traipse through the sample sets*/
       call SETcombo sets.j           /*have the other guy do the work.*/
       end   /*j*/

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────SETcombo subroutine─────────────────*/ SETcombo: procedure; parse arg bunch; n=words(bunch); newBunch= say ' the old sets=' space(bunch) 

 do k=1 for n                         /*change all commas to a blank.  */
 @.k=translate(word(bunch,k),,'},{')  /*create a list of words (=a set)*/
 end    /*k*/                              /*... and also remove the braces.*/

 do until \changed;   changed=0       /*consolidate some sets  (maybe).*/
      do set=1 for n-1
          do item=1 for words(@.set);   x=word(@.set,item)
              do other=set+1 to n
              if isIn(x,@.other) then do;   changed=1
                                      @.set=@.set @.other;   @.other=
                                      iterate set
                                      end
              end   /*other*/
          end       /*item*/
      end           /*set*/
 end                /*until ¬changed*/

    do set=1 for n;   new=            /*remove duplicates in a set.    */
      do items=1 for words(@.set)
      x=word(@.set,items);   if x==',' then iterate;  if x== then leave
      new=new x                       /*start building the new set.    */
        do forever;  if \isIn(x,@.set) then leave
        _=wordpos(x,@.set)
        @.set=subword(@.set,1,_-1) ',' subword(@.set,_+1) /*purify set.*/
        end    /*forever*/
      end      /*items*/
    @.set=translate(strip(new),','," ")
    end        /*set*/

 do new=1 for n;   if @.new== then iterate
 newBunch=space(newbunch '{'@.new"}")
 end   /*new*/

say ' the new sets=' newBunch; say return /*──────────────────────────────────isIn subroutine─────────────────────*/ isIn: return wordpos(arg(1),arg(2))\==0 /*is (word) arg1 in set arg2? */</lang> output when using the default supplied sample sets 

 the old sets= {A,B} {C,D}
 the new sets= {A,B} {C,D}

 the old sets= {A,B} {B,D}
 the new sets= {A,B,D}

 the old sets= {A,B} {C,D} {D,B}
 the new sets= {A,B,D,C}

 the old sets= {H,I,K} {A,B} {C,D} {D,B} {F,G,H}
 the new sets= {H,I,K,F,G} {A,B,D,C}

 the old sets= {snow,ice,slush,frost,fog} {iceburgs,icecubes} {rain,fog,sleet}
 the new sets= {snow,ice,slush,frost,fog,rain,sleet} {iceburgs,icecubes}

Ruby

<lang ruby>require 'set'

tests = [[['A', 'B'], ['C','D']], [['A','B'], ['B','D']], [['A','B'], ['C','D'], ['D','B']], [['H','I','K'], ['A','B'], ['C','D'], ['D','B'], ['F','G','H']]] tests = tests.map{|sets| sets.map(&:to_set)}

tests.map do |sets| 

 loop until sets.combination(2).none? do |a,b|
   if a.intersect?(b) then
     a.merge(b)
     sets.delete(b)
   end
 end
 p sets

end </lang>

Output:
[#<Set: {"A", "B"}>, #<Set: {"C", "D"}>]
[#<Set: {"A", "B", "D"}>]
[#<Set: {"A", "B", "D", "C"}>]
[#<Set: {"H", "I", "K", "F", "G"}>, #<Set: {"A", "B", "D", "C"}>]

Scala

<lang Scala>object SetConsolidation extends App { 

   def consolidate[Type](sets: Set[Set[Type]]): Set[Set[Type]] = {
       var result = sets // each iteration combines two sets and reiterates, else returns
       for (i <- sets; j <- sets - i; k = i.intersect(j);
           if result == sets && k.nonEmpty) result = result - i - j + i.union(j)
       if (result == sets) sets else consolidate(result)
   }

   // Tests:
   def parse(s: String) =
       s.split(",").map(_.split("").toSet).toSet
   def pretty[Type](sets: Set[Set[Type]]) =
       sets.map(_.mkString("{",",","}")).mkString(" ")
   val tests = List(
       parse("AB,CD") -> Set(Set("A", "B"), Set("C", "D")),
       parse("AB,BD") -> Set(Set("A", "B", "D")),
       parse("AB,CD,DB") -> Set(Set("A", "B", "C", "D")),
       parse("HIK,AB,CD,DB,FGH") -> Set(Set("A", "B", "C", "D"), Set("F", "G", "H", "I", "K"))
   )
   require(Set("A", "B", "C", "D") == Set("B", "C", "A", "D"))
   assert(tests.forall{case (test, expect) =>
       val result = consolidate(test)
       println(s"${pretty(test)} -> ${pretty(result)}")
       expect == result
   })

}</lang>

Output:
{A,B} {C,D} -> {A,B} {C,D}
{A,B} {B,D} -> {A,B,D}
{A,B} {C,D} {D,B} -> {C,D,A,B}
{D,B} {F,G,H} {A,B} {C,D} {H,I,K} -> {F,I,G,H,K} {A,B,C,D}

Tcl

Translation of: Python
Library: Tcllib (Package: struct::set)

This uses just the recursive version, as this is sufficient to handle substantial merges. <lang tcl>package require struct::set

proc consolidate {sets} { 

   if {[llength $sets] < 2} {

return $sets 

   }

   set r [list {}]
   set r0 [lindex $sets 0]
   foreach x [consolidate [lrange $sets 1 end]] {

if {[struct::set size [struct::set intersect $x $r0]]} { struct::set add r0 $x } else { lappend r $x } 

   }
   return [lset r 0 $r0]

}</lang> Demonstrating: <lang tcl>puts 1:[consolidate {{A B} {C D}}] puts 2:[consolidate {{A B} {B D}}] puts 3:[consolidate {{A B} {C D} {D B}}] puts 4:[consolidate {{H I K} {A B} {C D} {D B} {F G H}}]</lang>

Output:
1:{A B} {C D}
2:{D A B}
3:{D A B C}
4:{H I F G K} {D A B C}

TXR

Original solution:

<lang txr>@(do 

  (defun mkset (p x) (set [p x] (or [p x] x)))

  (defun fnd (p x) (if (eq [p x] x) x (fnd p [p x])))

  (defun uni (p x y)
    (let ((xr (fnd p x)) (yr (fnd p y)))
      (set [p xr] yr)))

  (defun consoli (sets)
    (let ((p (hash)))
      (each ((s sets))
        (each ((e s))
          (mkset p e)
          (uni p e (car s))))
      (hash-values 
        [group-by (op fnd p) (hash-keys 
                               [group-by identity (flatten sets)])])))

  ;; tests

  (each ((test '(((a b) (c d))
                 ((a b) (b d))
                 ((a b) (c d) (d b))
                 ((h i k) (a b) (c d) (d b) (f g h)))))
    (format t "~s -> ~s\n" test (consoli test))))</lang>

Output: 

((a b) (c d)) -> ((d c) (b a))
((a b) (b d)) -> ((d b a))
((a b) (c d) (d b)) -> ((d c b a))
((h i k) (a b) (c d) (d b) (f g h)) -> ((d c b a) (g f k i h))
Translation of: Racket

<lang txr>@(do 

  (defun mkset (items) [group-by identity items])

  (defun empty-p (set) (zerop (hash-count set)))

  (defun consoli (ss)
    (defun comb (cs s)
      (cond ((empty-p s) cs)
            ((null cs) (list s))
            ((empty-p (hash-isec s (first cs)))
             (cons (first cs) (comb (rest cs) s)))
            (t (consoli (cons (hash-uni s (first cs)) (rest cs))))))
    [reduce-left comb ss nil])

  ;; tests
  (each ((test '(((a b) (c d))
                 ((a b) (b d))
                 ((a b) (c d) (d b))
                 ((h i k) (a b) (c d) (d b) (f g h)))))
    (format t "~s -> ~s\n" test
            [mapcar hash-keys (consoli [mapcar mkset test])])))</lang>

Output: 

((a b) (c d)) -> ((b a) (d c))
((a b) (b d)) -> ((b a d))
((a b) (c d) (d b)) -> ((b a d c))
((h i k) (a b) (c d) (d b) (f g h)) -> ((g f k i h) (b a d c))
Retrieved from "https://rosettacode.org/wiki/Set_consolidation?oldid=214594"