Set

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

Data Structure
This illustrates a data structure, a means of storing data within a program.

You may see other such structures in the Data Structures category.

A   set  is a collection of elements, without duplicates and without order.


Task

Show each of these set operations:

  • Set creation
  • Test m ∈ S -- "m is an element in set S"
  • A ∪ B -- union; a set of all elements either in set A or in set B.
  • A ∩ B -- intersection; a set of all elements in both set A and set B.
  • A ∖ B -- difference; a set of all elements in set A, except those in set B.
  • A ⊆ B -- subset; true if every element in set A is also in set B.
  • A = B -- equality; true if every element of set A is in set B and vice versa.


As an option, show some other set operations.
(If A ⊆ B, but A ≠ B, then A is called a true or proper subset of B, written A ⊂ B or A ⊊ B.)

As another option, show how to modify a mutable set.


One might implement a set using an associative array (with set elements as array keys and some dummy value as the values).

One might also implement a set with a binary search tree, or with a hash table, or with an ordered array of binary bits (operated on with bit-wise binary operators).

The basic test, m ∈ S, is O(n) with a sequential list of elements, O(log n) with a balanced binary search tree, or (O(1) average-case, O(n) worst case) with a hash table.


See also



Ada[edit]

This solution uses the generic Ordered_Sets package from the Ada.Containers standard library, which internally is based on red-black trees. An alternative hash-based solution could use the Hashed_Maps package from Ada.Containers.

with Ada.Text_IO, Ada.Containers.Ordered_Sets;
 
procedure Set_Demo is
 
package CS is new Ada.Containers.Ordered_Sets(Character); use CS;
package IO renames Ada.Text_IO;
 
-- helper functions for string to something conversion, and vice versa
function To_Set(S: String) return Set is
Result: Set;
begin
for I in S'Range loop
begin
Result.Insert(S(I));
-- raises Constraint_Error if S(I) is already in Result
exception
when Constraint_Error => null;
end;
end loop;
return Result;
end To_Set;
 
function Image(S: Set) return String is
C: Character;
T: Set := S;
begin
if T.Is_Empty then
return "";
else
C:= T.First_Element;
T.Delete_First;
return C & Image(T);
end if;
end Image;
 
function Image(C: Ada.Containers.Count_Type) return String renames
Ada.Containers.Count_Type'Image;
 
S1, S2: Set;
begin -- main program
loop
S1 := To_Set(Ada.Text_IO.Get_Line);
exit when S1 = To_Set("quit!");
S2 := To_Set(Ada.Text_IO.Get_Line);
IO.Put_Line("Sets [" & Image(S1) & "], [" & Image(S2) & "] of size"
& Image(S1.Length) & " and" & Image(S2.Length) & ".");
IO.Put_Line("Intersection: [" & Image(Intersection(S1, S2)) & "],");
IO.Put_Line("Union: [" & Image(Union(S1, S2)) & "],");
IO.Put_Line("Difference: [" & Image(Difference(S1, S2)) & "],");
IO.Put_Line("Symmetric Diff: [" & Image(S1 xor S2) & "],");
IO.Put_Line("Subset: " & Boolean'Image(S1.Is_Subset(S2))
& ", Equal: " & Boolean'Image(S1 = S2) & ".");
end loop;
end Set_Demo;
 
Output:
set
demo
Sets [est], [demo] of size 3 and 4.
Intersection:   [e],
Union:          [demost],
Difference:     [st],
Symmetric Diff: [dmost],
Subset: FALSE, Equal: FALSE.
quit!

Apex[edit]

In Apex, Sets are unordered collections of elements. Although elements can be anything including primitives, Ids, Apex classes, or sObjects, typically they are used with primitives and Ids.

 
public class MySetController{
public Set<String> strSet {get; private set; }
public Set<Id> idSet {get; private set; }
 
public MySetController(){
//Initialize to an already known collection. Results in a set of abc,def.
this.strSet = new Set<String>{'abc','abc','def'};
 
//Initialize to empty set and add in entries.
this.strSet = new Set<String>();
this.strSet.add('abc');
this.strSet.add('def');
this.strSet.add('abc');
//Results in {'abc','def'}
 
//You can also get a set from a map in Apex. In this case, the account ids are fetched from a SOQL query.
Map<Id,Account> accountMap = new Map<Id,Account>([Select Id,Name From Account Limit 10]);
Set<Id> accountIds = accountMap.keySet();
 
//If you have a set, you can also use it with the bind variable syntax in SOQL:
List<Account> accounts = [Select Name From Account Where Id in :accountIds];
 
//Like other collections in Apex, you can use a for loop to iterate over sets:
for(Id accountId : accountIds){
Account a = accountMap.get(accountId);
//Do account stuffs here.
}
}
}
 

</pre>

AutoHotkey[edit]

test(Set,element){
for i, val in Set
if (val=element)
return true
return false
}
 
Union(SetA,SetB){
SetC:=[], Temp:=[]
for i, val in SetA
SetC.Insert(val), Temp[val] := true
for i, val in SetB
if !Temp[val]
SetC.Insert(val)
return SetC
}
 
intersection(SetA,SetB){
SetC:=[], Temp:=[]
for i, val in SetA
Temp[val] := true
for i, val in SetB
if Temp[val]
SetC.Insert(val)
return SetC
}
 
difference(SetA,SetB){
SetC:=[], Temp:=[]
for i, val in SetB
Temp[val] := true
for i, val in SetA
if !Temp[val]
SetC.Insert(val)
return SetC
}
 
subset(SetA,SetB){
Temp:=[], A:=B:=0
for i, val in SetA
Temp[val] := true , A++
for i, val in SetB
if Temp[val]{
B++
IfEqual, A, %B%, return 1
} return 0
}
 
equal(SetA,SetB){
return (SetA.MaxIndex() = SetB.MaxIndex() && subset(SetA,SetB)) ? 1: 0
}
Examples:
A:= ["apple", "cherry", "elderberry", "grape"]
B:= ["banana", "cherry", "date", "elderberry", "fig"]
C:= ["apple", "cherry", "elderberry", "grape", "orange"]
D:= ["apple", "cherry", "elderberry", "grape"]
E:= ["apple", "cherry", "elderberry"]
M:= "banana"
 
Res =
(
A:= ["apple", "cherry", "elderberry", "grape"]
B:= ["banana", "cherry", "date", "elderberry", "fig"]
C:= ["apple", "cherry", "elderberry", "grape", "orange"]
D:= ["apple", "cherry", "elderberry", "grape"]
E:= ["apple", "cherry", "elderberry"]
M:= "banana"
 
)
 
Res .= "`nM is " (test(A,M)?"":"not ") "an element of Set A"
Res .= "`nM is " (test(B,M)?"":"not ") "an element of Set B"
 
Res .= "`nUnion(A,B) = "
for i, val in Union(A,B)
Res.= (A_Index=1?"`t":", ") val
 
Res .= "`nintersection(A,B) = "
for i, val in intersection(A,B)
Res.= (A_Index=1?"`t":", ") val
 
Res .= "`ndifference(A,B) = "
for i, val in difference(A,B)
Res.= (A_Index=1?"`t":", ") val
 
Res .= "`n`nA is " (subset(A,C)?"":"not ") "a subset of Set C"
Res .= "`nA is " (subset(A,D)?"":"not ") "a subset of Set D"
Res .= "`nA is " (subset(A,E)?"":"not ") "a subset of Set E"
 
Res .= "`n`nA is " (equal(A,C)?"":"not ") "a equal to Set C"
Res .= "`nA is " (equal(A,D)?"":"not ") "a equal to Set D"
Res .= "`nA is " (equal(A,E)?"":"not ") "a equal to Set E"
 
MsgBox % Res
Output:
A:= ["apple", "cherry", "elderberry", "grape"]
B:= ["banana", "cherry", "date", "elderberry", "fig"]
C:= ["apple", "cherry", "elderberry", "grape", "orange"]
D:= ["apple", "cherry", "elderberry", "grape"]
E:= ["apple", "cherry", "elderberry"]
M:= "banana"

M is not an element of Set A
M is an element of Set B
Union(A,B) = 	apple, cherry, elderberry, grape, banana, date, fig
intersection(A,B) = 	cherry, elderberry
difference(A,B) = 	apple, grape

A is a subset of Set C
A is a subset of Set D
A is not a subset of Set E

A is not a equal to Set C
A is a equal to Set D
A is not a equal to Set E

BBC BASIC[edit]

The sets are represented as 32-bit integers, which means that the maximum number of elements is 32.

      DIM list$(6)
list$() = "apple", "banana", "cherry", "date", "elderberry", "fig", "grape"
 
setA% = %1010101
PRINT "Set A: " FNlistset(list$(), setA%)
setB% = %0111110
PRINT "Set B: " FNlistset(list$(), setB%)
elementM% = %0000010
PRINT "Element M: " FNlistset(list$(), elementM%) '
 
IF elementM% AND setA% THEN
PRINT "M is an element of set A"
ELSE
PRINT "M is not an element of set A"
ENDIF
IF elementM% AND setB% THEN
PRINT "M is an element of set B"
ELSE
PRINT "M is not an element of set B"
ENDIF
 
PRINT '"The union of A and B is " FNlistset(list$(), setA% OR setB%)
PRINT "The intersection of A and B is " FNlistset(list$(), setA% AND setB%)
PRINT "The difference of A and B is " FNlistset(list$(), setA% AND NOT setB%)
 
IF (setA% AND setB%) = setA% THEN
PRINT '"Set A is a subset of set B"
ELSE
PRINT '"Set A is not a subset of set B"
ENDIF
IF setA% = setB% THEN
PRINT "Set A is equal to set B"
ELSE
PRINT "Set A is not equal to set B"
ENDIF
END
 
DEF FNlistset(list$(), set%)
LOCAL i%, o$
FOR i% = 0 TO 31
IF set% AND 1 << i% o$ += list$(i%) + ", "
NEXT
= LEFT$(LEFT$(o$))
Output:
Set A: apple, cherry, elderberry, grape
Set B: banana, cherry, date, elderberry, fig
Element M: banana

M is not an element of set A
M is an element of set B

The union of A and B is apple, banana, cherry, date, elderberry, fig, grape
The intersection of A and B is cherry, elderberry
The difference of A and B is apple, grape

Set A is not a subset of set B
Set A is not equal to set B

C[edit]

Building a set highly depends on the datatype and use case. For example, a set of string could be implemented by hash table, sort tree or trie, but if all sets are known to have very few number of elements, it might be best to use just flat arrays. There isn't, and shouldn't be, an all-purpose set type for C.

A frequent use of set is that of small, non-negative integers, implemented as a bit field as shown below.

#include <stdio.h>
 
typedef unsigned int set_t; /* probably 32 bits; change according to need */
 
void show_set(set_t x, const char *name)
{
int i;
printf("%s is:", name);
for (i = 0; (1U << i) <= x; i++)
if (x & (1U << i))
printf(" %d", i);
putchar('\n');
}
 
int main(void)
{
int i;
set_t a, b, c;
 
a = 0; /* empty set */
for (i = 0; i < 10; i += 3) /* add 0 3 6 9 to set a */
a |= (1U << i);
show_set(a, "a");
 
for (i = 0; i < 5; i++)
printf("\t%d%s in set a\n", i, (a & (1U << i)) ? "":" not");
 
b = a;
b |= (1U << 5); b |= (1U << 10); /* b is a plus 5, 10 */
b &= ~(1U << 0); /* sans 0 */
show_set(b, "b");
 
show_set(a | b, "union(a, b)");
show_set(c = a & b, "c = common(a, b)");
show_set(a & ~b, "a - b"); /* diff, not arithmetic minus */
show_set(b & ~a, "b - a");
printf("b is%s a subset of a\n", !(b & ~a) ? "" : " not");
printf("c is%s a subset of a\n", !(c & ~a) ? "" : " not");
 
printf("union(a, b) - common(a, b) %s union(a - b, b - a)\n",
((a | b) & ~(a & b)) == ((a & ~b) | (b & ~a))
? "equals" : "does not equal");
 
return 0;
}

C#[edit]

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
 
class Program
{
static void PrintCollection(IEnumerable<int> x)
{
Console.WriteLine(string.Join(" ", x));
}
static void Main(string[] args)
{
Console.OutputEncoding = Encoding.UTF8;
Console.WriteLine("Set creation");
var A = new HashSet<int> { 4, 12, 14, 17, 18, 19, 20 };
var B = new HashSet<int> { 2, 5, 8, 11, 12, 13, 17, 18, 20 };
 
PrintCollection(A);
PrintCollection(B);
 
Console.WriteLine("Test m ∈ S -- \"m is an element in set S\"");
Console.WriteLine("14 is an element in set A: {0}", A.Contains(14));
Console.WriteLine("15 is an element in set A: {0}", A.Contains(15));
 
Console.WriteLine("A ∪ B -- union; a set of all elements either in set A or in set B.");
var aUb = A.Union(B);
PrintCollection(aUb);
 
Console.WriteLine("A ∖ B -- difference; a set of all elements in set A, except those in set B.");
var aDb = A.Except(B);
PrintCollection(aDb);
 
Console.WriteLine("A ⊆ B -- subset; true if every element in set A is also in set B.");
Console.WriteLine(A.IsSubsetOf(B));
var C = new HashSet<int> { 14, 17, 18 };
Console.WriteLine(C.IsSubsetOf(A));
 
Console.WriteLine("A = B -- equality; true if every element of set A is in set B and vice versa.");
Console.WriteLine(A.SetEquals(B));
var D = new HashSet<int> { 4, 12, 14, 17, 18, 19, 20 };
Console.WriteLine(A.SetEquals(D));
 
Console.WriteLine("If A ⊆ B, but A ≠ B, then A is called a true or proper subset of B, written A ⊂ B or A ⊊ B");
Console.WriteLine(A.IsProperSubsetOf(B));
Console.WriteLine(C.IsProperSubsetOf(A));
 
Console.WriteLine("Modify a mutable set. (Add 10 to A; remove 12 from B).");
A.Add(10);
B.Remove(12);
PrintCollection(A);
PrintCollection(B);
 
Console.ReadKey();
}
}
Output:
Set creation
4 12 14 17 18 19 20
2 5 8 11 12 13 17 18 20
Test m ∈ S -- "m is an element in set S"
14 is an element in set A: True
15 is an element in set A: False
A ∪ B -- union; a set of all elements either in set A or in set B.
4 12 14 17 18 19 20 2 5 8 11 13
A ∖ B -- difference; a set of all elements in set A, except those in set B.
4 14 19
A ⊆ B -- subset; true if every element in set A is also in set B.
False
True
A = B -- equality; true if every element of set A is in set B and vice versa.
False
True
If A ⊆ B, but A ≠ B, then A is called a true or proper subset of B, written A ⊂ B or A ⊊ B
False
True
Modify a mutable set.  (Add 10 to A; remove 12 from B).
4 12 14 17 18 19 20 10
2 5 8 11 13 17 18 20

C++[edit]

C++ standard library contains a set class, which is a sorted container without duplicates and implemented as a binary tree. Additional set functionality can be implemented in terms of standard library algorithms.

C++11 standard library also contains unordered_set based on a hash table. However, algorithms like std::set_intersection etc take sorted ranges, so set-specific functions should be hand-rolled.

 
#include <set>
#include <iostream>
#include <iterator>
#include <algorithm>
 
namespace set_display {
template <class T>
std::ostream& operator<<(std::ostream& os, const std::set<T>& set)
{
os << '[';
if (!set.empty()) {
std::copy(set.begin(), --set.end(), std::ostream_iterator<T>(os, ", "));
os << *--set.end();
}
return os << ']';
}
}
 
template <class T>
bool contains(const std::set<T>& set, const T& key)
{
return set.count(key) != 0;
}
 
template <class T>
std::set<T> set_union(const std::set<T>& a, const std::set<T>& b)
{
std::set<T> result;
std::set_union(a.begin(), a.end(), b.begin(), b.end(), std::inserter(result, result.end()));
return result;
}
 
template <class T>
std::set<T> set_intersection(const std::set<T>& a, const std::set<T>& b)
{
std::set<T> result;
std::set_intersection(a.begin(), a.end(), b.begin(), b.end(), std::inserter(result, result.end()));
return result;
}
 
template <class T>
std::set<T> set_difference(const std::set<T>& a, const std::set<T>& b)
{
std::set<T> result;
std::set_difference(a.begin(), a.end(), b.begin(), b.end(), std::inserter(result, result.end()));
return result;
}
 
template <class T>
bool is_subset(const std::set<T>& set, const std::set<T>& subset)
{
return std::includes(set.begin(), set.end(), subset.begin(), subset.end());
}
 
int main()
{
using namespace set_display;
std::set<int> a{2, 5, 7, 5, 9, 2}; //C++11 initialization syntax
std::set<int> b{1, 5, 9, 7, 4 };
std::cout << "a = " << a << '\n';
std::cout << "b = " << b << '\n';
 
int value1 = 8, value2 = 5;
std::cout << "Set a " << (contains(a, value1) ? "contains " : "does not contain ") << value1 << '\n';
std::cout << "Set a " << (contains(a, value2) ? "contains " : "does not contain ") << value2 << '\n';
 
std::cout << "Union of a and b: " << set_union(a, b) << '\n';
std::cout << "Intersection of a and b: " << set_intersection(a, b) << '\n';
std::cout << "Difference of a and b: " << set_difference(a, b) << '\n';
 
std::set<int> sub{5, 9};
std::cout << "Set b " << (is_subset(a, b) ? "is" : "is not") << " a subset of a\n";
std::cout << "Set " << sub << ' ' << (is_subset(a, sub) ? "is" : "is not") << " a subset of a\n";
 
std::set<int> copy = a;
std::cout << "a " << (a == copy ? "equals " : "does not equal ") << copy << '\n';
 
return 0;
}
 

Clojure[edit]

(require 'clojure.set)
 
; sets can be created using the set method or set literal syntax
(def a (set [1 2 3 4]))
(def b #{4 5 6 7})
 
(a 10) ; returns the element if it's contained in the set, otherwise nil
 
(clojure.set/union a b)
 
(clojure.set/intersection a b)
 
(clojure.set/difference a b)
 
(clojure.set/subset? a b)

CoffeeScript[edit]

This implements functions from the task, along with an iteration helper called "each".

 
# For ad-hoc set features, it sometimes makes sense to use hashes directly,
# rather than abstract to this level, but I'm showing a somewhat heavy
# solution to show off CoffeeScript class syntax.
class Set
constructor: (elems...) ->
@hash = {}
for elem in elems
@hash[elem] = true
 
add: (elem) ->
@hash[elem] = true
 
remove: (elem) ->
delete @hash[elem]
 
has: (elem) ->
@hash[elem]?
 
union: (set2) ->
set = new Set()
for elem of @hash
set.add elem
for elem in set2.to_array()
set.add elem
set
 
intersection: (set2) ->
set = new Set()
for elem of @hash
set.add elem if set2.has elem
set
 
minus: (set2) ->
set = new Set()
for elem of @hash
set.add elem if !set2.has elem
set
 
is_subset_of: (set2) ->
for elem of @hash
return false if !set2.has elem
true
 
equals: (set2) ->
this.is_subset_of(set2) and set2.is_subset_of this
 
to_array: ->
(elem for elem of @hash)
 
each: (f) ->
for elem of @hash
f(elem)
 
to_string: ->
@to_array()
 
run_tests = ->
set1 = new Set("apple", "banana") # creation
console.log set1.has "apple" # true (membership)
console.log set1.has "worms" # false (membership)
 
set2 = new Set("banana", "carrots")
console.log set1.union(set2).to_string() # [ 'apple', 'banana', 'carrots' ] (union)
console.log set1.intersection(set2).to_string() # [ 'banana' ] (intersection)
console.log set1.minus(set2).to_string() # [ 'apple' ] (difference)
 
set3 = new Set("apple")
console.log set3.is_subset_of set1 # true
console.log set3.is_subset_of set2 # false
 
set4 = new Set("apple", "banana")
console.log set4.equals set1 # true
console.log set4.equals set2 # false
 
set5 = new Set("foo")
set5.add "bar" # add
console.log set5.to_string() # [ 'foo', 'bar' ]
set5.remove "bar" # remove
console.log set5.to_string() # [ 'foo' ]
 
# iteration, prints apple then banana (order not guaranteed)
set1.each (elem) ->
console.log elem
 
run_tests()
 

Common Lisp[edit]

Common Lisp provides some set operations on lists.

(setf a '(1 2 3 4))
(setf b '(2 3 4 5))
 
(format t "sets: ~a ~a~%" a b)
 
;;; element
(loop for x from 1 to 6 do
(format t (if (member x a)
"~d ∈ A~%"
"~d ∉ A~%") x))
 
(format t "A ∪ B: ~a~%" (union a b))
(format t "A ∩ B: ~a~%" (intersection a b))
(format t "A \\ B: ~a~%" (set-difference a b))
(format t (if (subsetp a b)
"~a ⊆ ~a~%"
"~a ⊈ ~a~%") a b)
 
(format t (if (and (subsetp a b)
(subsetp b a))
"~a = ~a~%"
"~a ≠ ~a~%") a b)

D[edit]

void main() {
import std.stdio, std.algorithm, std.range;
 
// Not true sets, items can be repeated, but must be sorted.
auto s1 = [1, 2, 3, 4, 5, 6].assumeSorted;
auto s2 = [2, 5, 6, 3, 4, 8].sort(); // [2,3,4,5,6,8].
auto s3 = [1, 2, 5].assumeSorted;
 
assert(s1.canFind(4)); // Linear search.
assert(s1.contains(4)); // Binary search.
assert(s1.setUnion(s2).equal([1,2,2,3,3,4,4,5,5,6,6,8]));
assert(s1.setIntersection(s2).equal([2, 3, 4, 5, 6]));
assert(s1.setDifference(s2).equal([1]));
assert(s1.setSymmetricDifference(s2).equal([1, 8]));
assert(s3.setDifference(s1).empty); // It's a subset.
assert(!s1.equal(s2));
 
auto s4 = [[1, 4, 7, 8], [1, 7], [1, 7, 8], [4], [7]];
const s5 = [1, 1, 1, 4, 4, 7, 7, 7, 7, 8, 8];
assert(s4.nWayUnion.equal(s5));
}


D[edit]

 
module set;
import std.typecons : Tuple, tuple;
struct Set(V) { // Limited set of V-type elements // here 'this' is named A, s is B, v V-type item
 
protected V[] array;
 
this(const Set s) { // construct A by copy of B
array = s.array.dup;
}
 
this(V[] arg...){ // construct A with items
foreach(v; arg) if (v.isNotIn(array)) array ~= v;
}
 
enum : Set { empty = Set() } // ∅
 
ref Set opAssign()(const Set s) { // A = B
array = s.array.dup;
return this;
}
 
bool opBinaryRight(string op : "in")(const V v) const { // v ∈ A
return v.isIn(array);
}
 
ref Set opOpAssign(string op)(const V v) if (op == "+" || op == "|") { // A += {v} // + = ∪ = |
if (v.isIn(array)) return this;
array ~= v;
return this;
}
 
ref Set opOpAssign(string op)(const Set s) if (op == "+" || op == "|") { // A += B
foreach(x; s.array) if (x.isNotIn(array)) array ~= x;
return this;
}
 
Set opBinary(string op)(const V v) const if (op == "+" || op == "|"){ // A + {v}
Set result = this;
result += v;
return result;
}
 
Set opBinaryRight(string op)(const V v) const if (op == "+" || op == "|") { // {v} + A
Set result = this;
result += v;
return result;
}
 
Set opBinary(string op)(const Set s) const if (op == "+" || op == "|") { // A + B
Set result = this;
result += s;
return result;
}
 
Set opBinary(string op : "&")(const Set s) const{ // A ∩ B // ∩ = &
Set result;
foreach(x; array) if(x.isIn(s.array)) result += x;
return result;
}
 
ref Set opOpAssign(string op : "&")(const Set s) { // A ∩= B
return this(this & s);
}
 
Set opBinary(string op : "^")(const Set s) const { // (A ∪ B) - (A ∩ B) // = A ^ B
Set result;
foreach(x; array) if (x.isNotIn(s.array)) result += x;
foreach(x; s.array) if(x.isNotIn(array)) result += x;
return result;
}
 
ref opOpAssign(string op : "^")(const Set s) {
return this = this ^ s;
}
 
Set opBinary(string op : "-")(const Set s) const { // A - B
Set r;
foreach(x; array) if(x.isNot(s.array)) r += x;
return r;
}
 
ref Set opOpAssign(string op : "-")(const Set s) { // A -= B
return this = this - s;
}
 
Set!(Tuple!(V,U)) opBinary(U, string op : "*")(const Set!U s) const { // A × B = { (x, y) | ∀x ∈ A ∧ ∀y ∈ B }
Set!(Tuple!(V, U)) r;
foreach(x; array) foreach(y; s.array) r += tuple(x, y);
return r;
}
 
bool isEmpty() const { return !array.length;} // A ≟ ∅
 
bool opBinary(string op : "in")(const Set s) const { // A ⊂ s
foreach(v; array) if(v.isNotIn(s.array)) return false;
return true;
}
 
bool opEquals(const Set s) const { // A ≟ B
if (array.length != s.array.length) return false;
return this in s;
}
 
T[] array() const @property { return array.dup;}
 
}
 
Set!(Tuple!(T, T)) sqr(T)(const Set!T s) { return s * s; } // A²
 
auto pow(T, uint n : 0)(const Set!T s) { // A ^ 0
return Set!T.empty;
}
 
auto pow(T, uint n : 1)(const Set!T s) { // A ^ 1 = A
return s;
}
 
auto pow(T, uint n : 2)(const Set!T s) { // A ^ 2 (=A²)
return sqr!T(s);
}
 
auto pow(T, uint n)(const Set!T s) if(n % 2) { // if n Odd, A^n = A * (A^(n/2))²
return s * sqr!T(pow!(T, n/2)(s));
}
 
auto pow(T, uint n)(const Set!T s) if(!(n % 2)) { // if n Even, A^n = (A^(n/2))²
return sqr!T(pow!(T, n/2)(s));
}
 
size_t Card(T)(const Set!T s) {return s.length; } // Card(A)
 
Set!(Set!T) power(T)(Set!T s) { // ∀B ∈ P(A) ⇒ B ⊂ A
Set!(Set!T) ret;
foreach(e; s.array) {
Set!(Set!T) rs;
foreach(x; ret.array) {
x += e;
rs += x;
}
ret += rs;
}
return ret;
}
 
bool isIn(T)(T x, T[] array){
foreach(a; array) if(a == x) return true;
return false;
}
bool isNotIn(T)(T x, T[] array){
foreachj(a; array) if(a == x) return false;
return true;
}
 

Dart[edit]

void main(){
//Set Creation
Set A = new Set.from([1,2,3]);
Set B = new Set.from([1,2,3,4,5]);
Set C = new Set.from([1,2,4,5]);
 
print('Set A = $A');
print('Set B = $B');
print('Set C = $C');
print('');
//Test if element is in set
int m = 3;
print('m = 5');
print('m in A = ${A.contains(m)}');
print('m in B = ${B.contains(m)}');
print('m in C = ${C.contains(m)}');
print('');
//Union of two sets
Set AC = A.union(C);
print('Set AC = Union of A and C = $AC');
print('');
//Intersection of two sets
Set A_C = A.intersection(C);
print('Set A_C = Intersection of A and C = $A_C');
print('');
//Difference of two sets
Set A_diff_C = A.difference(C);
print('Set A_diff_C = Difference between A and C = $A_diff_C');
print('');
//Test if set is subset of another set
print('A is a subset of B = ${B.containsAll(A)}');
print('C is a subset of B = ${B.containsAll(C)}');
print('A is a subset of C = ${C.containsAll(A)}');
print('');
//Test if two sets are equal
print('A is equal to B = ${B.containsAll(A) && A.containsAll(B)}');
print('B is equal to AC = ${B.containsAll(AC) && AC.containsAll(B)}');
}
Output:
Set A = {1, 2, 3}
Set B = {1, 2, 3, 4, 5}
Set C = {1, 2, 4, 5}

m = 5
m in A = true
m in B = true
m in C = false

Set AC = Union of A and C = {1, 2, 3, 4, 5}

Set A_C = Intersection of A and C = {1, 2}

Set A_diff_C = Difference between A and C = {3}

A is a subset of B = true
C is a subset of B = true
A is a subset of C = false

A is equal to B  = false
B is equal to AC = true

EchoLisp[edit]

EchoLisp sets are lists, i.e the set of all sets is a proper subset of the set of all lists. Sets elements may be any object, including sets.

The set operations are: ∩ ∪ ⊆ / ∈ = ∆ ×

 
; use { } to read a set
(define A { 1 2 3 4 3 5 5}){ 1 2 3 4 5 } ; duplicates are removed from a set
; or use make-set to make a set from a list
(define B (make-set ' ( 3 4 5 6 7 8 8))){ 3 4 5 6 7 8 }
(set-intersect A B){ 3 4 5 }
(set-intersect? A B) → #t ; predicate
(set-union A B){ 1 2 3 4 5 6 7 8 }
(set-substract A B){ 1 2 }
(set-sym-diff A B){ 1 2 6 7 8 } ; ∆ symmetric difference
(set-equal? A B) → #f
(set-equal? { a b c} { c b a}) → #t ; order is unimportant
(set-subset? A B) → #f ; B in A or B = A
(set-subset? A { 3 4 }) → #t
(member 4 A)(4 5) ; same as #t : true
(member 9 A) → #f
 
; check basic equalities
(set-equal? A (set-union (set-intersect A B) (set-substract A B))) → #t
(set-equal? (set-union A B) (set-union (set-sym-diff A B) (set-intersect A B))) → #t
 
; × : cartesian product of two sets : all pairs (a . b) , a in A, b in B
; returns a list (not a set)
(define A { albert simon})
(define B { antoinette ornella marylin})
 
(set-product A B)
((albert . antoinette) (albert . marylin) (albert . ornella) (simon . antoinette) (simon . marylin) (simon . ornella))
 
; sets elements may be sets
{ { a b c} {c b a } { a b d}}{ { a b c } { a b d } } ; duplicate removed
 
; A few functions return sets :
(primes 10){ 2 3 5 7 11 13 17 19 23 29 }
 

Elixir[edit]

Works with: Elixir version 1.1
iex(1)> s = MapSet.new
#MapSet<[]>
iex(2)> sa = MapSet.put(s, :a)
#MapSet<[:a]>
iex(3)> sab = MapSet.put(sa, :b)
#MapSet<[:a, :b]>
iex(4)> sbc = Enum.into([:b, :c], MapSet.new)
#MapSet<[:b, :c]>
iex(5)> MapSet.member?(sab, :a)
true
iex(6)> MapSet.member?(sab, :c)
false
iex(7)> :a in sab
true
iex(8)> MapSet.union(sab, sbc)
#MapSet<[:a, :b, :c]>
iex(9)> MapSet.intersection(sab, sbc)
#MapSet<[:b]>
iex(10)> MapSet.difference(sab, sbc)
#MapSet<[:a]>
iex(11)> MapSet.disjoint?(sab, sbc)
false
iex(12)> MapSet.subset?(sa, sab)
true
iex(13)> MapSet.subset?(sab, sa)
false
iex(14)> sa == sab
false

Erlang[edit]

Built in.

2> S = sets:new().
3> Sa = sets:add_element(a, S).
4> Sab = sets:from_list([a, b]).
5> sets:is_element(a, Sa).
true
6> Union = sets:union(Sa, Sab).
7> sets:to_list(Union).
[a,b]
8> Intersection = sets:intersection(Sa, Sab).
9> sets:to_list(Intersection).
[a]
10> Subtract = sets:subtract(Sab, Sa).
11> sets:to_list(Subtract).
[b]
12> sets:is_subset(Sa, Sab). 
true
13> Sa =:= Sab.
false

F#[edit]

The Collections.Set<'T> class implements "Immutable sets based on binary trees, where comparison is the F# structural comparison function, potentially using implementations of the IComparable interface on key values." (http://msdn.microsoft.com/en-us/library/ee353619.aspx)

[<EntryPoint>]
let main args =
// Create some sets (of int):
let s1 = Set.ofList [1;2;3;4;3]
let s2 = Set.ofArray [|3;4;5;6|]
 
printfn "Some sets (of int):"
printfn "s1 = %A" s1
printfn "s2 = %A" s2
printfn "Set operations:"
printfn "2 ∈ s1? %A" (s1.Contains 2)
printfn "10 ∈ s1? %A" (s1.Contains 10)
printfn "s1 ∪ s2 = %A" (Set.union s1 s2)
printfn "s1 ∩ s2 = %A" (Set.intersect s1 s2)
printfn "s1 ∖ s2 = %A" (Set.difference s1 s2)
printfn "s1 ⊆ s2? %A" (Set.isSubset s1 s1)
printfn "{3, 1} ⊆ s1? %A" (Set.isSubset (Set.ofList [3;1]) s1)
printfn "{3, 2, 4, 1} = s1? %A" ((Set.ofList [3;2;4;1]) = s1)
printfn "s1 = s2? %A" (s1 = s2)
printfn "More set operations:"
printfn "#s1 = %A" s1.Count
printfn "s1 ∪ {99} = %A" (s1.Add 99)
printfn "s1 ∖ {3} = %A" (s1.Remove 3)
printfn "s1 ⊂ s1? %A" (Set.isProperSubset s1 s1)
printfn "s1 ⊂ s2? %A" (Set.isProperSubset s1 s2)
0
Output:
Some sets (of int):
s1 = set [1; 2; 3; 4]
s2 = set [3; 4; 5; 6]
Set operations:
2 ∈ s1? true
10 ∈ s1? false
s1 ∪ s2 = set [1; 2; 3; 4; 5; 6]
s1 ∩ s2 = set [3; 4]
s1 ∖ s2 = set [1; 2]
s1 ⊆ s2? true
{3, 1} ⊆ s1? true
{3, 2, 4, 1} = s1? true
s1 = s2? false
More set operations:
#s1 = 4
s1 ∪ {99} = set [1; 2; 3; 4; 99]
s1 ∖ {3} = set [1; 2; 4]
s1 ⊂ s1? false
s1 ⊂ s2? false

Factor[edit]

We will use Factor's hash-sets for this task. A hash-set is created with HS{ ... }.

( scratchpad ) USE: sets
( scratchpad ) HS{ 2 5 4 3 } HS{ 5 6 7 } union .
HS{ 2 3 4 5 6 7 }
( scratchpad ) HS{ 2 5 4 3 } HS{ 5 6 7 } intersect .
HS{ 5 }
( scratchpad ) HS{ 2 5 4 3 } HS{ 5 6 7 } diff .
HS{ 2 3 4 }
( scratchpad ) HS{ 2 5 4 3 } HS{ 5 6 7 } subset? .
f
( scratchpad ) HS{ 5 6 } HS{ 5 6 7 } subset? .
t
( scratchpad ) HS{ 5 6 } HS{ 5 6 7 } set= .
f
( scratchpad ) HS{ 6 5 7 } HS{ 5 6 7 } set= .
t

Forth[edit]

Works with: Forth

Works with any ANS Forth. Needs the FMS-SI (single inheritance) library code located here: http://soton.mpeforth.com/flag/fms/index.html

include FMS-SI.f
include FMS-SILib.f
 
: union {: a b -- c :}
begin
b each:
while dup
a indexOf: if 2drop else a add: then
repeat b <free a dup sort: ; ok
 
i{ 2 5 4 3 } i{ 5 6 7 } union p: i{ 2 3 4 5 6 7 } ok
 
 
: free2 ( a b -- ) <free <free ;
: intersect {: a b | c -- c :}
heap> 1-array2 to c
begin
b each:
while dup
a indexOf: if drop c add: else drop then
repeat a b free2 c dup sort: ;
 
i{ 2 5 4 3 } i{ 5 6 7 } intersect p: i{ 5 } ok
 
 
: diff {: a b | c -- c :}
heap> 1-array2 to c
begin
a each:
while dup
b indexOf: if 2drop else c add: then
repeat a b free2 c dup sort: ;
 
i{ 2 5 4 3 } i{ 5 6 7 } diff p: i{ 2 3 4 } ok
 
: subset {: a b -- flag :}
begin
a each:
while
b indexOf: if drop else false exit then
repeat a b free2 true ;
 
i{ 2 5 4 3 } i{ 5 6 7 } subset . 0 ok
i{ 5 6 } i{ 5 6 7 } subset . -1 ok
 
 
: set= {: a b -- flag :}
a size: b size: <> if a b free2 false exit then
a sort: b sort:
begin
a each: drop b each:
while
<> if a b free2 false exit then
repeat a b free2 true ;
 
i{ 5 6 } i{ 5 6 7 } set= . 0 ok
i{ 6 5 7 } i{ 5 6 7 } set= . -1 ok
 

Frink[edit]

 
a = new set[1, 2]
b = toSet[[2,3]] // Construct a set from an array
 
a.contains[2] // Element test (returns true)
union[a,b]
intersection[a,b]
setDifference[a,b]
isSubset[a,b] // Returns true if a is a subset of b
a==b // set equality test
 

FunL[edit]

A = {1, 2, 3}
B = {3, 4, 5}
C = {1, 2, 3, 4, 5}
D = {2, 1, 3}
 
println( '2 is in A: ' + (2 in A) )
println( '4 is in A: ' + (4 in A) )
println( 'A union B: ' + A.union(B) )
println( 'A intersect B: ' + A.intersect(B) )
println( 'A difference B: ' + A.diff(B) )
println( 'A subset of B: ' + A.subsetOf(B) )
println( 'A subset of B: ' + A.subsetOf(C) )
println( 'A equal B: ' + (A == B) )
println( 'A equal D: ' + (A == D) )
 
S = set( A )
 
println( 'S (mutable version of A): ' + S )
S.add( 4 )
println( 'S with 4 added: ' + S )
println( 'S subset of C: ' + S.subsetOf(C) )
S.remove( 1 )
println( 'S after 1 removed: ' + S )
Output:
2 is in A: true                                                                                                                                                                             
4 is in A: false                                                                                                                                                                            
A union B: {4, 5, 1, 2, 3}                                                                                                                                                                  
A intersect B: {3}                                                                                                                                                                          
A difference B: {1, 2}                                                                                                                                                                      
A subset of B: false                                                                                                                                                                        
A subset of B: true                                                                                                                                                                         
A equal B: false                                                                                                                                                                            
A equal D: true                                                                                                                                                                             
S (mutable version of A): {1, 2, 3}                                                                                                                                                         
S with 4 added: {1, 2, 3, 4}                                                                                                                                                                
S subset of C: true                                                                                                                                                                         
S after 1 removed: {2, 3, 4}

Go[edit]

A common complaint is that Go has no native set type and so there are a number of third-party libraries offering to fill this perceived gap. Yet Go has good native support for most applications for sets.

Maps[edit]

Go maps meet the task description in that they do not require orderable elements. To demonstrate that, a set of complex numbers is shown here. Complex numbers can be compared for equality but are not ordered.

package main
 
import "fmt"
 
// Define set as a type to hold a set of complex numbers. A type
// could be defined similarly to hold other types of elements. A common
// variation is to make a map of interface{} to represent a set of
// mixed types. Also here the map value is a bool. By always storing
// true, the code is nicely readable. A variation to use less memory
// is to make the map value an empty struct. The relative advantages
// can be debated.
type set map[complex128]bool
 
func main() {
// task: set creation
s0 := make(set) // create empty set
s1 := set{3: true} // create set with one element
s2 := set{3: true, 1: true} // create set with two elements
 
// option: another way to create a set
s3 := newSet(3, 1, 4, 1, 5, 9)
 
// option: output!
fmt.Println("s0:", s0)
fmt.Println("s1:", s1)
fmt.Println("s2:", s2)
fmt.Println("s3:", s3)
 
// task: element predicate
fmt.Printf("%v ∈ s0: %t\n", 3, s0.hasElement(3))
fmt.Printf("%v ∈ s3: %t\n", 3, s3.hasElement(3))
fmt.Printf("%v ∈ s3: %t\n", 2, s3.hasElement(2))
 
// task: union
b := set{4: true, 2: true}
fmt.Printf("s3 ∪ %v: %v\n", b, union(s3, b))
 
// task: intersection
fmt.Printf("s3 ∩ %v: %v\n", b, intersection(s3, b))
 
// task: difference
fmt.Printf("s3 \\ %v: %v\n", b, difference(s3, b))
 
// task: subset predicate
fmt.Printf("%v ⊆ s3: %t\n", b, subset(b, s3))
fmt.Printf("%v ⊆ s3: %t\n", s2, subset(s2, s3))
fmt.Printf("%v ⊆ s3: %t\n", s0, subset(s0, s3))
 
// task: equality
s2Same := set{1: true, 3: true}
fmt.Printf("%v = s2: %t\n", s2Same, equal(s2Same, s2))
 
// option: proper subset
fmt.Printf("%v ⊂ s2: %t\n", s2Same, properSubset(s2Same, s2))
fmt.Printf("%v ⊂ s3: %t\n", s2Same, properSubset(s2Same, s3))
 
// option: delete. it's built in.
delete(s3, 3)
fmt.Println("s3, 3 deleted:", s3)
}
 
func newSet(ms ...complex128) set {
s := make(set)
for _, m := range ms {
s[m] = true
}
return s
}
 
func (s set) String() string {
if len(s) == 0 {
return "∅"
}
r := "{"
for e := range s {
r = fmt.Sprintf("%s%v, ", r, e)
}
return r[:len(r)-2] + "}"
}
 
func (s set) hasElement(m complex128) bool {
return s[m]
}
 
func union(a, b set) set {
s := make(set)
for e := range a {
s[e] = true
}
for e := range b {
s[e] = true
}
return s
}
 
func intersection(a, b set) set {
s := make(set)
for e := range a {
if b[e] {
s[e] = true
}
}
return s
}
 
func difference(a, b set) set {
s := make(set)
for e := range a {
if !b[e] {
s[e] = true
}
}
return s
}
 
func subset(a, b set) bool {
for e := range a {
if !b[e] {
return false
}
}
return true
}
 
func equal(a, b set) bool {
return len(a) == len(b) && subset(a, b)
}
 
func properSubset(a, b set) bool {
return len(a) < len(b) && subset(a, b)
}
Output:
s0: ∅
s1: {(3+0i)}
s2: {(3+0i), (1+0i)}
s3: {(3+0i), (1+0i), (4+0i), (5+0i), (9+0i)}
3 ∈ s0: false
3 ∈ s3: true
2 ∈ s3: false
s3 ∪ {(4+0i), (2+0i)}: {(5+0i), (9+0i), (2+0i), (3+0i), (1+0i), (4+0i)}
s3 ∩ {(2+0i), (4+0i)}: {(4+0i)}
s3 \ {(4+0i), (2+0i)}: {(5+0i), (9+0i), (3+0i), (1+0i)}
{(2+0i), (4+0i)} ⊆ s3: false
{(3+0i), (1+0i)} ⊆ s3: true
∅ ⊆ s3: true
{(1+0i), (3+0i)} = s2: true
{(1+0i), (3+0i)} ⊂ s2: false
{(1+0i), (3+0i)} ⊂ s3: true
s3, 3 deleted: {(5+0i), (9+0i), (1+0i), (4+0i)}

Big.Int[edit]

If elements of your set are integers or can be indexed by integers, are zero based and relatively "dense", then the big.Int type in the standard library can serve efficiently as a set. The solution here doesn't even bother to define a set type, it just defines functions that use big.Ints directly as sets.

Note that the elements here, being integers, are of course ordered and so might not meet a strict reading of the task requirements.

package main
 
import (
"fmt"
"math/big"
)
 
func main() {
// create an empty set
var s0 big.Int
 
// create sets with elements
s1 := newSet(3)
s2 := newSet(3, 1)
s3 := newSet(3, 1, 4, 1, 5, 9)
 
// output
fmt.Println("s0:", format(s0))
fmt.Println("s1:", format(s1))
fmt.Println("s2:", format(s2))
fmt.Println("s3:", format(s3))
 
// element predicate
fmt.Printf("%v ∈ s0: %t\n", 3, hasElement(s0, 3))
fmt.Printf("%v ∈ s3: %t\n", 3, hasElement(s3, 3))
fmt.Printf("%v ∈ s3: %t\n", 2, hasElement(s3, 2))
 
// union
b := newSet(4, 2)
fmt.Printf("s3 ∪ %v: %v\n", format(b), format(union(s3, b)))
 
// intersection
fmt.Printf("s3 ∩ %v: %v\n", format(b), format(intersection(s3, b)))
 
// difference
fmt.Printf("s3 \\ %v: %v\n", format(b), format(difference(s3, b)))
 
// subset predicate
fmt.Printf("%v ⊆ s3: %t\n", format(b), subset(b, s3))
fmt.Printf("%v ⊆ s3: %t\n", format(s2), subset(s2, s3))
fmt.Printf("%v ⊆ s3: %t\n", format(s0), subset(s0, s3))
 
// equality
s2Same := newSet(1, 3)
fmt.Printf("%v = s2: %t\n", format(s2Same), equal(s2Same, s2))
 
// proper subset
fmt.Printf("%v ⊂ s2: %t\n", format(s2Same), properSubset(s2Same, s2))
fmt.Printf("%v ⊂ s3: %t\n", format(s2Same), properSubset(s2Same, s3))
 
// delete
remove(&s3, 3)
fmt.Println("s3, 3 removed:", format(s3))
}
 
func newSet(ms ...int) (set big.Int) {
for _, m := range ms {
set.SetBit(&set, m, 1)
}
return
}
 
func remove(set *big.Int, m int) {
set.SetBit(set, m, 0)
}
 
func format(set big.Int) string {
if len(set.Bits()) == 0 {
return "∅"
}
r := "{"
for e, l := 0, set.BitLen(); e < l; e++ {
if set.Bit(e) == 1 {
r = fmt.Sprintf("%s%v, ", r, e)
}
}
return r[:len(r)-2] + "}"
}
 
func hasElement(set big.Int, m int) bool {
return set.Bit(m) == 1
}
 
func union(a, b big.Int) (set big.Int) {
set.Or(&a, &b)
return
}
 
func intersection(a, b big.Int) (set big.Int) {
set.And(&a, &b)
return
}
 
func difference(a, b big.Int) (set big.Int) {
set.AndNot(&a, &b)
return
}
 
func subset(a, b big.Int) bool {
ab := a.Bits()
bb := b.Bits()
if len(ab) > len(bb) {
return false
}
for i, aw := range ab {
if aw&^bb[i] != 0 {
return false
}
}
return true
}
 
func equal(a, b big.Int) bool {
return a.Cmp(&b) == 0
}
 
func properSubset(a, b big.Int) (p bool) {
ab := a.Bits()
bb := b.Bits()
if len(ab) > len(bb) {
return false
}
for i, aw := range ab {
bw := bb[i]
if aw&^bw != 0 {
return false
}
if aw != bw {
p = true
}
}
return
}
Output:
s0: ∅
s1: {3}
s2: {1, 3}
s3: {1, 3, 4, 5, 9}
3 ∈ s0: false
3 ∈ s3: true
2 ∈ s3: false
s3 ∪ {2, 4}: {1, 2, 3, 4, 5, 9}
s3 ∩ {2, 4}: {4}
s3 \ {2, 4}: {1, 3, 5, 9}
{2, 4} ⊆ s3: false
{1, 3} ⊆ s3: true
∅ ⊆ s3: true
{1, 3} = s2: true
{1, 3} ⊂ s2: false
{1, 3} ⊂ s3: true
s3, 3 removed: {1, 4, 5, 9}

Intsets[edit]

Not quite in the stanard library but still in the official "sub repository", intsets are a sparse bit set. Like big.Int they use a single bit to represent a possible element, but the sparse representation efficiently allows for large "holes" in the element sequence. Also the intsets API provides a more set-like terminology so the RC task can be coded more directly.

package main
 
import (
"fmt"
 
"golang.org/x/tools/container/intsets"
)
 
func main() {
var s0, s1 intsets.Sparse // create some empty sets
s1.Insert(3) // insert an element
s2 := newSet(3, 1) // create sets with elements
s3 := newSet(3, 1, 4, 1, 5, 9)
 
// output
fmt.Println("s0:", &s0)
fmt.Println("s1:", &s1)
fmt.Println("s2:", s2)
fmt.Println("s3:", s3)
 
// element predicate
fmt.Printf("%v ∈ s0: %t\n", 3, s0.Has(3))
fmt.Printf("%v ∈ s3: %t\n", 3, s3.Has(3))
fmt.Printf("%v ∈ s3: %t\n", 2, s3.Has(2))
 
// union
b := newSet(4, 2)
var s intsets.Sparse
s.Union(s3, b)
fmt.Printf("s3 ∪ %v: %v\n", b, &s)
 
// intersection
s.Intersection(s3, b)
fmt.Printf("s3 ∩ %v: %v\n", b, &s)
 
// difference
s.Difference(s3, b)
fmt.Printf("s3 \\ %v: %v\n", b, &s)
 
// subset predicate
fmt.Printf("%v ⊆ s3: %t\n", b, b.SubsetOf(s3))
fmt.Printf("%v ⊆ s3: %t\n", s2, s2.SubsetOf(s3))
fmt.Printf("%v ⊆ s3: %t\n", &s0, s0.SubsetOf(s3))
 
// equality
s2Same := newSet(1, 3)
fmt.Printf("%v = s2: %t\n", s2Same, s2Same.Equals(s2))
 
// delete
s3.Remove(3)
fmt.Println("s3, 3 removed:", s3)
}
 
func newSet(ms ...int) *intsets.Sparse {
var set intsets.Sparse
for _, m := range ms {
set.Insert(m)
}
return &set
}
Output:
s0: {}
s1: {3}
s2: {1 3}
s3: {1 3 4 5 9}
3 ∈ s0: false
3 ∈ s3: true
2 ∈ s3: false
s3 ∪ {2 4}: {1 2 3 4 5 9}
s3 ∩ {2 4}: {4}
s3 \ {2 4}: {1 3 5 9}
{2 4} ⊆ s3: false
{1 3} ⊆ s3: true
{} ⊆ s3: true
{1 3} = s2: true
s3, 3 removed: {1 4 5 9}

Groovy[edit]

def s1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] as Set
def m1 = 6
def m2 = 7
def s2 = [0, 2, 4, 6, 8] as Set
assert m1 in s1  : 'member'
assert ! (m2 in s2)  : 'not a member'
def su = s1 + s2
assert su == [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] as Set : 'union'
def si = s1.intersect(s2)
assert si == [8, 6, 4, 2] as Set  : 'intersection'
def sd = s1 - s2
assert sd == [1, 3, 5, 7, 9, 10] as Set  : 'difference'
assert s1.containsAll(si)  : 'subset'
assert ! s1.containsAll(s2)  : 'not a subset'
assert (si + sd) == s1  : 'equality'
assert (s2 + sd) != s1  : 'inequality'
assert s1 != su && su.containsAll(s1)  : 'proper subset'
s1 << 0
assert s1 == su  : 'added element 0 to s1'

Haskell[edit]

Works with: GHC

GHC offers a functional, persistent set data structure in its Data.Set module. It is implemented using a binary search tree. Elements must be of an orderable type (instance of Ord).

Prelude> import Data.Set
Prelude Data.Set> empty :: Set Integer -- Empty set
fromList []
Prelude Data.Set> let s1 = fromList [1,2,3,4,3] -- Convert list into set
Prelude Data.Set> s1
fromList [1,2,3,4]
Prelude Data.Set> let s2 = fromList [3,4,5,6]
Prelude Data.Set> union s1 s2 -- Union
fromList [1,2,3,4,5,6]
Prelude Data.Set> intersection s1 s2 -- Intersection
fromList [3,4]
Prelude Data.Set> s1 \\ s2 -- Difference
fromList [1,2]
Prelude Data.Set> s1 `isSubsetOf` s1 -- Subset
True
Prelude Data.Set> fromList [3,1] `isSubsetOf` s1
True
Prelude Data.Set> s1 `isProperSubsetOf` s1 -- Proper subset
False
Prelude Data.Set> fromList [3,1] `isProperSubsetOf` s1
True
Prelude Data.Set> fromList [3,2,4,1] == s1 -- Equality
True
Prelude Data.Set> s1 == s2
False
Prelude Data.Set> 2 `member` s1 -- Membership
True
Prelude Data.Set> 10 `notMember` s1
True
Prelude Data.Set> size s1 -- Cardinality
4
Prelude Data.Set> insert 99 s1 -- Create a new set by inserting
fromList [1,2,3,4,99]
Prelude Data.Set> delete 3 s1 -- Create a new set by deleting
fromList [1,2,4]

Regular lists can also be used as sets. Haskell has some functions to help with using lists as sets. No requirement is made of element type. However, these are not very efficient because they require linear time to find an element.

Prelude> import Data.List
Prelude Data.List> let s3 = nub [1,2,3,4,3] -- Remove duplicates from list
Prelude Data.List> s3
[1,2,3,4]
Prelude Data.List> let s4 = [3,4,5,6]
Prelude Data.List> union s3 s4 -- Union
[1,2,3,4,5,6]
Prelude Data.List> intersect s3 s4 -- Intersection
[3,4]
Prelude Data.List> s3 \\ s4 -- Difference
[1,2]
Prelude Data.List> 42 : s3 -- Return new list with element inserted at the beginning
[42,1,2,3,4]
Prelude Data.List> delete 3 s3 -- Return new list with first occurrence of element removed
[1,2,4]

Icon and Unicon[edit]

The set is a basic datatype (structure) in Icon and Unicon, which supports 'member', 'union', 'intersection' and 'difference' operations. Subset and equality must be implemented separately, or use the routines in the 'sets' library.

Implemented directly:

 
procedure display_set (s)
writes ("[")
every writes (!s || " ")
write ("]")
end
 
# fail unless s1 and s2 contain the same elements
procedure set_equals (s1, s2)
return subset(s1, s2) & subset(s2, s1)
end
 
# fail if every element in s2 is not contained in s1
procedure subset (s1, s2)
every (a := !s2) do {
if not(member(s1,a)) then fail
}
return s2
end
 
procedure main ()
a := set(1, 1, 2, 3, 4)
b := set(2, 3, 5)
writes ("a: ")
display_set (a)
writes ("b: ")
display_set (b)
# basic set operations
writes ("Intersection: ")
display_set (a ** b)
writes ("Union: ")
display_set (a ++ b)
writes ("Difference: ")
display_set (a -- b)
# membership
if member(a, 2) then
write ("2 is a member of a")
else
write ("2 is not a member of a")
if member(a, 5) then
write ("5 is a member of a")
else
write ("5 is not a member of a")
# equality
if set_equals(a, set(1,2,3,4,4)) then
write ("a equals set(1,2,3,4,4)")
else
write ("a does not equal set(1,2,3,4,4)")
if set_equals(a, b) then
write ("a equals b")
else
write ("a does not equal b")
# subset
if subset(a, set(1,2)) then
write ("(1,2) is included in a")
else
write ("(1,2) is not included in a")
if subset(a, set(1,2,5)) then
write ("(1,2,5) is included in a")
else
write ("(1,2,5) is not included in a")
end
 
Output:
a: [2 4 1 3 ]
b: [5 2 3 ]
Intersection: [2 3 ]
Union: [5 2 4 1 3 ]
Difference: [4 1 ]
2 is a member of a
5 is not a member of a
a equals set(1,2,3,4,4)
a does not equal b
(1,2) is included in a
(1,2,5) is not included in a

Using library:

 
link sets
 
procedure main ()
a := set(1, 1, 2, 3, 4)
b := set(2, 3, 5)
write ("a: ", simage(a))
write ("b: ", simage(b))
# basic set operations
write ("Intersection: ", simage (a**b))
write ("Union: ", simage (a++b))
write ("Difference: ", simage (a--b))
# membership
if member(a, 2) then
write ("2 is a member of a")
else
write ("2 is not a member of a")
if member(a, 5) then
write ("5 is a member of a")
else
write ("5 is not a member of a")
# equality
if seteq(a, set(1,2,3,4,4)) then
write ("a equals set(1,2,3,4,4)")
else
write ("a does not equal set(1,2,3,4,4)")
if seteq(a, b) then
write ("a equals b")
else
write ("a does not equal b")
# check subset
if setlt(set(1,2), a) then
write ("(1,2) is included in a")
else
write ("(1,2) is not included in a")
if setlt(a, set(1,2,5), a) then
write ("(1,2,5) is included in a")
else
write ("(1,2,5) is not included in a")
end
 
Output:
a: { 2, 4, 1, 3 }
b: { 5, 2, 3 }
Intersection: { 2, 3 }
Union: { 5, 2, 4, 1, 3 }
Difference: { 4, 1 }
2 is a member of a
5 is not a member of a
a equals set(1,2,3,4,4)
a does not equal b
(1,2) is included in a
(1,2,5) is not included in a


J[edit]

In J, we use a sequence to represent a set. This actually winds up being a family of set implementations. In this example, we chose to ignore order and specify that duplicate elements are not allowed.

Here are definitions for the required operations:

union=: ~.@,
intersection=: [ -. -.
difference=: -.
subset=: [email protected]
equality=: -:&(/:~)

Examples:

  2 4 6 8 ~.@, 2 3 5 7
2 4 6 8 3 5 7
2 4 6 8 ([ -. -.) 2 3 5 7
2
2 4 6 8 -. 2 3 5 7
4 6 8
2 4 6 8 [email protected] 2 3 5 7
0
'' [email protected] 2 3 5 7
1
2 4 6 8 3 5 7 -:&(/:~) 8 7 6 5 4 3 2
1

Examples again, using names rather than code:

   2 4 6 8 union 2 3 5 7
2 4 6 8 3 5 7
2 4 6 8 intersection 2 3 5 7
2
2 4 6 8 difference 2 3 5 7
4 6 8
2 4 6 8 subset 2 3 5 7
0
'' subset 2 3 5 7
1
2 4 6 8 3 5 7 equality 8 7 6 5 4 3 2
1

Note that J uses 1 for true and 0 for false. Mathematical revisionists object to this, but this is consistent with the original (and revised) formulations of boolean algebra. (And there are deep ties to Bayes' rule.)

Note that these operations can be combined in sentences with other operations. For example we could define

properSubset=: subset * 1 - equality

Java[edit]

Works with: Java version 7+

To use this in Java 5 replace all "<>" with "<Integer>".

import java.util.Arrays;
import java.util.Collections;
import java.util.Set;
import java.util.TreeSet;
 
public class Sets {
public static void main(String[] args){
Set<Integer> a = new TreeSet<>();
//TreeSet sorts on natural ordering (or an optional comparator)
//other options: HashSet (hashcode)
// LinkedHashSet (insertion order)
// EnumSet (optimized for enum values)
//others at: http://download.oracle.com/javase/7/docs/api/java/util/Set.html
Set<Integer> b = new TreeSet<>();
Set<Integer> c = new TreeSet<>();
Set<Integer> d = new TreeSet<>();
 
a.addAll(Arrays.asList(1, 2, 3, 4, 5));
b.addAll(Arrays.asList(2, 3, 4, 5, 6, 8));
c.addAll(Arrays.asList(2, 3, 4));
d.addAll(Arrays.asList(2, 3, 4));
System.out.println("a: " + a);
System.out.println("b: " + b);
System.out.println("c: " + c);
System.out.println("d: " + d);
 
System.out.println("2 in a: " + a.contains(2));
System.out.println("6 in a: " + a.contains(6));
 
Set<Integer> ab = new TreeSet<>();
ab.addAll(a);
ab.addAll(b);
System.out.println("a union b: " + ab);
 
Set<Integer> a_b = new TreeSet<>();
a_b.addAll(a);
a_b.removeAll(b);
System.out.println("a - b: " + a_b);
 
System.out.println("c subset of a: " + a.containsAll(c));
//use a.conatins() for single elements
 
System.out.println("c = d: " + c.equals(d));
System.out.println("d = c: " + d.equals(c));
 
Set<Integer> aib = new TreeSet<>();
aib.addAll(a);
aib.retainAll(b);
System.out.println("a intersect b: " + aib);
 
System.out.println("add 7 to a: " + a.add(7));
System.out.println("add 2 to a again: " + a.add(2));
 
//other noteworthy things related to sets:
Set<Integer> empty = Collections.EMPTY_SET; //immutable empty set
//empty.add(2); would fail
empty.isEmpty(); //test if a set is empty
empty.size();
Collections.disjoint(a, b); //returns true if the sets have no common elems (based on their .equals() methods)
Collections.unmodifiableSet(a); //returns an immutable copy of a
}
}
Output:
a: [1, 2, 3, 4, 5]
b: [2, 3, 4, 5, 6, 8]
c: [2, 3, 4]
d: [2, 3, 4]
2 in a: true
6 in a: false
a union b: [1, 2, 3, 4, 5, 6, 8]
a - b: [1]
c subset of a: true
c = d: true
d = c: true
a intersect b: [2, 3, 4, 5]
add 7 to a: true
add 2 to a again: false

JavaScript[edit]

JavaScript does not support native sets before ECMAScript 6.

 
var set = new Set();
 
set.add(0);
set.add(1);
set.add('two');
set.add('three');
 
set.has(0); //=> true
set.has(3); //=> false
set.has('two'); // true
set.has(Math.sqrt(4)); //=> false
set.has('TWO'.toLowerCase()); //=> true
 
set.size; //=> 4
 
set.delete('two');
set.has('two'); //==> false
set.size; //=> 3
 
//iterating set using ES6 for..of
//Set order is preserved in order items are added.
for (var item of set) {
console.log('item is ' + item);
}

jq[edit]

Works with: jq version 1.4

Neither JSON nor jq has a "set" type, but as explained below in the first part of this entry, finite sets of Unicode strings can be directly represented in JSON and thus jq.

The second part of this entry focuses on a jq library of set-theoretic functions that support finite sets of arbitrary JSON entities.

Finite Sets of Unicode Strings[edit]

There is an obvious 1-1 mapping between the collection of finite sets of Unicode strings and the collection of JSON objects with distinct keys the values of which all have the boolean value "true". For example, the set of strings {"a", "b"} corresponds to the JSON object {"a": true, "b": true }.

When restricted to such JSON objects, jq's equality operator ("==") yields set-theoretic semantics, and similarly, jq's + operator yields set-theoretic union.

For example:

{"a":true, "b":true } == {"b":true, "a":true}.
{"a":true} + {"b":true } == { "a":true, "b":true}

Thus, it can be seen that jq has built-in support for sets of finite-length Unicode strings.

For simplicity and to avoid confusion, we shall refer to JSON objects all of whose keys are distinct and all values of which have the boolean value "true" as "string sets".

String-set test

Here is a jq filter for determining whether a JSON object is a "string set":

def is_stringset:
. as $in | type == "object" and reduce keys[] as $key (true; . and $in[$key] == true);

String-set membership:

The test for set membership, m ∈ S, where m is a string and S is a set of strings, corresponds exactly to the jq test:

T | has(m)

where T is the JSON object corresponding to S. This test is also efficient.

String-set intersection

# Set-intersection: A ∩ B
def stringset_intersection(A;B):
reduce (A|keys)[] as $k
({}; if (B|has($k)) then . + {($k):true} else . end);

String-set difference

# stringset_difference: A \ B
def stringset_difference(A;B):
reduce (A|keys)[] as $k
({}; if (B|has($k)) then . else . + {($k):true} end);

Subset

# A ⊆ B iff string_subset(A;B)
def stringset_subset(A;B):
reduce (A|keys)[] as $k
(true; . and (B|has($k)));

Finite Sets of JSON Entities[edit]

Finite sets of arbitrary JSON entities can be represented by sets of strings using an invertible serialization of JSON entities, but in the remainder of this entry, we provide a more straightforward and probably more efficient implementation of finite sets using JSON arrays.

Specifically, the empty set is represented by [] and a non-empty set of JSON entities with distinct members m1, m2, ... mN is represented by the JSON array [s1, s2, ... sN] where:

[s1, s2, ... sN] is the result of ([m1, m2, mN] | sort)

When confined to sorted arrays, jq's equality operator (==) yields set-theoretic semantics, and therefore, for the remainder of this entry, we shall refer to sorted arrays simply as sets.

To convert an arbitrary jq or JSON array to a set, we can simply use the built-in jq operator "unique". To test whether an arbitrary JSON entity is a set without sorting:

def is_set:
. as $in
| type == "array" and
reduce range(0;length-1) as $i
(true; if . then $in[$i] < $in[$i+1] else false end);

The following library of set-theoretic functions is intended for use with jq version 1.4 or later. However, as noted below, if used with a version of jq that does not have bsearch, then it is assumed that a definition of bsearch equivalent to that given in Binary search is available.

Set creation

  • [] is the empty set;
  • if m1 <= m2 <= ... mN then [m1, m2, ... mN] is the set containing the listed elements;
  • The set of elements in an array, a, can be constructed by writing: a | unique
  • The set of strings in the string-set SS is: SS|keys

m ∈ S

If m is a JSON entity and S a set, then the jq expression S[m] can be used to test whether m is an element of S, but for large sets, this is inefficient. A generally more efficient test membership of m in S would use bsearch as defined at Binary search or as provided in recent versions of jq:

def is_member(m):  bsearch(m) > -1;

Intersection

# If A and B are sets, then intersection(A;B) emits their intersection:
def intersection(A;B):
(A|length) as $al
| (B|length) as $bl
| if $al == 0 or $bl == 0 then []
else
reduce range(0; $al + $bl) as $k
( [0, 0, []];
.[0] as $i | .[1] as $j
| if $i < $al and $j < $bl then
if A[$i] == B[$j] then [ $i+1 , $j+1, .[2] + [A[$i]]]
elif A[$i] < B[$j] then [ $i+1 , $j, .[2] ]
else [ $i , $j+1, .[2] ]
end
else .
end
) | .[2]
end ;

Difference

# If A and B are sets, then A-B is emitted
def difference(A;B):
(A|length) as $al
| (B|length) as $bl
| if $al == 0 then [] elif $bl == 0 then A
else
reduce range(0; $al + $bl) as $k
( [0, 0, []];
.[0] as $i | .[1] as $j
| if $i < $al and $j < $bl then
if A[$i] == B[$j] then [ $i+1, $j+1, .[2] ]
elif A[$i] < B[$j] then [ $i+1, $j, .[2] + [A[$i]] ]
else [ $i , $j+1, .[2] ]
end
elif $i < $al then [ $i+1, $j, .[2] + [A[$i]] ]
else .
end
) | .[2]
end ;

Union

A simple but inefficient implementation would use: (A + B) | unique

To compute the union of two sets efficiently, it is helpful to define a function for merging sorted arrays.

# merge input array with array x by comparing the heads of the arrays in turn;
# if both arrays are sorted, the result will be sorted:
def merge(x):
length as $length
| (x|length) as $xl
| if $length == 0 then x
elif $xl == 0 then .
else
. as $in
| reduce range(0; $xl + $length) as $z
# state [ix, xix, ans]
( [0, 0, []];
if .[0] < $length and ((.[1] < $xl and $in[.[0]] <= x[.[1]]) or .[1] == $xl)
then [(.[0] + 1), .[1], (.[2] + [$in[.[0]]]) ]
else [.[0], (.[1] + 1), (.[2] + [x[.[1]]]) ]
end
) | .[2]
end ;
 
def union(A;B):
A|merge(B)
| reduce .[] as $m ([]; if length == 0 or .[length-1] != $m then . + [$m] else . end);

A ⊆ B

def subset(A;B):
# TCO
def _subset:
if .[0]|length == 0 then true
elif .[1]|length == 0 then false
elif .[0][0] == .[1][0] then [.[0][1:], .[1][1:]] | _subset
elif .[0][0] < .[1][0] then false
else [ .[0], .[1][1:] ] | _subset
end;
[A,B] | _subset;

Test whether two sets intersect

The following implementation assumes a version of jq with bsearch/1.

If A and B are sets (i.e. A == (A|unique) and B == (B|unique)), then [A,B] | intersect emits true if A and B have at least one element in common:

def intersect:
.[0] as $A | .[1] as $B
| ($A|length) as $al
| ($B|length) as $bl
| if $al == 0 or $bl == 0 then false
else
($B | bsearch($A[0])) as $b
| if $b >= 0 then true
else [$A[1:], $B[- (1 + $b) :]] | intersect
end
end;
 

Julia[edit]

julia> S1 = Set(1:4) ; S2 = Set(3:6) ; println(S1,"\n",S2)
Set{Int64}({4,2,3,1})
Set{Int64}({5,4,6,3})

julia> 5 in S1 , 5 in S2
(false,true)

julia> intersect(S1,S2)
Set{Int64}({4,3})

julia> union(S1,S2)
Set{Int64}({5,4,6,2,3,1})

julia> setdiff(S1,S2)
Set{Int64}({2,1})

julia> issubset(S1,S2)
false

julia> isequal(S1,S2)
false

julia> symdiff(S1,S2)
Set{Int64}({5,6,2,1})

Kotlin[edit]

// version 1.0.6
 
fun main(args: Array<String>) {
val fruits = setOf("apple", "pear", "orange", "banana")
println("fruits  : $fruits")
val fruits2 = setOf("melon", "orange", "lemon", "gooseberry")
println("fruits2 : $fruits2\n")
 
println("fruits contains 'banana'  : ${"banana" in fruits}")
println("fruits2 contains 'elderberry' : ${"elderbury" in fruits2}\n")
 
println("Union  : ${fruits.union(fruits2)}")
println("Intersection : ${fruits.intersect(fruits2)}")
println("Difference  : ${fruits.minus(fruits2)}\n")
 
println("fruits2 is a subset of fruits : ${fruits.containsAll(fruits2)}\n")
val fruits3 = fruits
println("fruits3 : $fruits3\n")
var areEqual = fruits.containsAll(fruits2) && fruits3.containsAll(fruits)
println("fruits2 and fruits are equal  : $areEqual")
areEqual = fruits.containsAll(fruits3) && fruits3.containsAll(fruits)
println("fruits3 and fruits are equal  : $areEqual\n")
 
val fruits4 = setOf("apple", "orange")
println("fruits4 : $fruits4\n")
var isProperSubset = fruits.containsAll(fruits3) && !fruits3.containsAll(fruits)
println("fruits3 is a proper subset of fruits : $isProperSubset")
isProperSubset = fruits.containsAll(fruits4) && !fruits4.containsAll(fruits)
println("fruits4 is a proper subset of fruits : $isProperSubset\n")
 
val fruits5 = mutableSetOf("cherry", "blueberry", "raspberry")
println("fruits5 : $fruits5\n")
fruits5 += "guava"
println("fruits5 + 'guava'  : $fruits5")
println("fruits5 - 'cherry' : ${fruits5 - "cherry"}")
}
Output:
fruits  : [apple, pear, orange, banana]
fruits2 : [melon, orange, lemon, gooseberry]

fruits  contains 'banana'     : true
fruits2 contains 'elderberry' : false

Union        : [apple, pear, orange, banana, melon, lemon, gooseberry]
Intersection : [orange]
Difference   : [apple, pear, banana]

fruits2 is a subset of fruits : false

fruits3 : [apple, pear, orange, banana]

fruits2 and fruits are equal  : false
fruits3 and fruits are equal  : true

fruits4 : [apple, orange]

fruits3 is a proper subset of fruits : false
fruits4 is a proper subset of fruits : true

fruits5 : [cherry, blueberry, raspberry]

fruits5 + 'guava'  : [cherry, blueberry, raspberry, guava]
fruits5 - 'cherry' : [blueberry, raspberry, guava]

Lasso[edit]

// Extend set type
define set->issubsetof(p::set) => .intersection(#p)->size == .size
define set->oncompare(p::set) => .intersection(#p)->size - .size
 
// Set creation
local(set1) = set('j','k','l','m','n')
local(set2) = set('m','n','o','p','q')
 
//Test m ∈ S -- "m is an element in set S"
#set1 >> 'm'
 
// A ∪ B -- union; a set of all elements either in set A or in set B.
#set1->union(#set2)
 
//A ∩ B -- intersection; a set of all elements in both set A and set B.
#set1->intersection(#set2)
 
//A ∖ B -- difference; a set of all elements in set A, except those in set B.
#set1->difference(#set2)
 
//A ⊆ B -- subset; true if every element in set A is also in set B.
#set1->issubsetof(#set2)
 
//A = B -- equality; true if every element of set A is in set B and vice-versa.
#set1 == #set2
Output:
true
set(j, k, l, m, n, o, p, q)
set(m, n)
set(j, k, l)
false
false

LFE[edit]

Translation of: Erlang
 
> (set set-1 (sets:new))
#(set 0 16 16 8 80 48 ...)
> (set set-2 (sets:add_element 'a set-1))
#(set 1 16 16 8 80 48 ...)
> (set set-3 (sets:from_list '(a b)))
#(set 2 16 16 8 80 48 ...)
> (sets:is_element 'a set-2)
true
> (set union (sets:union set-2 set-3))
#(set 2 16 16 8 80 48 ...)
> (sets:to_list union)
(a b)
> (set intersect (sets:intersection set-2 set-3))
#(set 1 16 16 8 80 48 ...)
> (sets:to_list intersect)
(a)
> (set subtr (sets:subtract set-3 set-2))
#(set 1 16 16 8 80 48 ...)
> (sets:to_list subtr)
(b)
> (sets:is_subset set-2 set-3)
true
> (=:= set-2 set-3)
false
> (set set-4 (sets:add_element 'b set-2))
#(set 2 16 16 8 80 48 ...)
> (=:= set-3 set-4)
true
 

Liberty BASIC[edit]

Sets are not natively available- implemented here in string form so no need to dim/redim or pass number of elements.

 
A$ ="red hot chili peppers rule OK"
B$ ="lady in red"
 
print " New set, in space-separated form. Extra spaces and duplicates will be removed. "
input newSet$
newSet$ =trim$( newSet$)
newSet$ =stripBigSpaces$( newSet$)
newSet$ =removeDupes$( newSet$)
print " Set stored as the string '"; newSet$; "'"
 
print
print " 'red' is an element of '"; A$; "' is "; isAnElementOf$( "red", A$)
print " 'blue' is an element of '"; A$; "' is "; isAnElementOf$( "blue", A$)
print " 'red' is an element of '"; B$; "' is "; isAnElementOf$( "red", B$)
print
print " Union of '"; A$; "' & '"; B$; "' is '"; unionOf$( A$, B$); "'."
print
print " Intersection of '"; A$; "' & '"; B$; "' is '"; intersectionOf$( A$, B$); "'."
print
print " Difference of '"; A$; "' & '"; B$; "' is '"; differenceOf$( A$, B$); "'."
print
print " '"; A$; "' equals '"; A$; "' is "; equalSets$( A$, A$)
print " '"; A$; "' equals '"; B$; "' is "; equalSets$( A$, B$)
print
print " '"; A$; "' is a subset of '"; B$; "' is "; isSubsetOf$( A$, B$)
print " 'red peppers' is a subset of 'red hot chili peppers rule OK' is "; isSubsetOf$( "red peppers", "red hot chili peppers rule OK")
 
end
 
function removeDupes$( a$)
numElements =countElements( a$)
redim elArray$( numElements) ' ie 4 elements are array entries 1 to 4 and 0 is spare =""
for m =0 to numElements
el$ =word$( a$, m, " ")
elArray$( m) =el$
next m
sort elArray$(), 0, numElements
b$ =""
penultimate$ ="999"
for jk =0 to numElements ' do not use "" ( nuls) or elementsalready seen
if elArray$( jk) ="" then [on]
if elArray$( jk) <>penultimate$ then b$ =b$ +elArray$( jk) +" ": penultimate$ =elArray$( jk)
[on]
next jk
b$ =trim$( b$)
removeDupes$ =b$
end function
 
function stripBigSpaces$( a$) ' copy byte by byte, but id=f a space had a preceding space, ignore it.
lenA =len( a$)
penul$ =""
for i =1 to len( a$)
c$ =mid$( a$, i, 1)
if c$ <>" " then
if penul$ <>" " then
b$ =b$ +c$
else
b$ =b$ +" " +c$
end if
end if
penul$ =c$
next i
stripBigSpaces$ =b$
end function
 
function countElements( a$) ' count elements repr'd by space-separated words in string rep'n.
if isNul$( a$) ="True" then countElements =0: exit function
i =0
do
el$ =word$( a$, i +1, " ")
i =i +1
loop until el$ =""
countElements =i -1
end function
 
function isNul$( a$) ' a nul set implies its string rep'n is length zero.
if a$ ="" then isNul$ ="True" else isNul$ ="False"
end function
 
function isAnElementOf$( a$, b$) ' check element a$ exists in set b$.
isAnElementOf$ ="False"
i =0
do
el$ =word$( b$, i +1, " ")
if a$ =el$ then isAnElementOf$ ="True"
i =i +1
loop until el$ =""
end function
 
function unionOf$( a$, b$)
i =1
o$ =a$
do
w$ =word$( b$, i, " ")
if w$ ="" then exit do
if isAnElementOf$( w$, a$) ="False" then o$ =o$ +" " +w$
i =i +1
loop until w$ =""
unionOf$ =o$
end function
 
function intersectionOf$( a$, b$)
i =1
o$ =""
do
el$ =word$( a$, i, " ")
if el$ ="" then exit do
if ( isAnElementOf$( el$, b$) ="True") and ( o$ ="") then o$ =el$
if ( isAnElementOf$( el$, b$) ="True") and ( o$ <>el$) then o$ =o$ +" " +el$
i =i +1
loop until el$ =""
intersectionOf$ =o$
end function
 
function equalSets$( a$, b$)
if len( a$) <>len( b$) then equalSets$ ="False": exit function
i =1
do
el$ =word$( a$, i, " ")
if isAnElementOf$( el$, b$) ="False" then equalSets$ ="False": exit function
i =i +1
loop until w$ =""
equalSets$ ="True"
end function
 
function differenceOf$( a$, b$)
i =1
o$ =""
do
el$ =word$( a$, i, " ")
if el$ ="" then exit do
if ( isAnElementOf$( el$, b$) ="False") and ( o$ ="") then o$ =el$
if ( isAnElementOf$( el$, b$) ="False") and ( o$ <>el$) then o$ =o$ +" " +el$
i =i +1
loop until el$ =""
differenceOf$ =o$
end function
 
function isSubsetOf$( a$, b$)
isSubsetOf$ ="True"
i =1
do
el$ =word$( a$, i, " ")
if el$ ="" then exit do
if ( isAnElementOf$( el$, b$) ="False") then isSubsetOf$ ="False": exit function
i =i +1
loop until el$ =""
end function
 
 


New set, in space-separated form. Extra spaces and duplicates will be removed.
? now is the the time for all good all men
Set stored as the string 'all for good is men now the time'
'red' is an element of 'red hot chili peppers rule OK' is True
'blue' is an element of 'red hot chili peppers rule OK' is False
'red' is an element of 'lady in red' is True
Union of 'red hot chili peppers rule OK' & 'lady in red' is 'red hot chili peppers rule OK lady in'.
Intersection of 'red hot chili peppers rule OK' & 'lady in red' is 'red'.
Difference of 'red hot chili peppers rule OK' & 'lady in red' is 'hot chili peppers rule OK'.
'red hot chili peppers rule OK' equals 'red hot chili peppers rule OK' is True
'red hot chili peppers rule OK' equals 'lady in red' is False
'red hot chili peppers rule OK' is a subset of 'lady in red' is False
'red peppers' is a subset of 'red hot chili peppers rule OK' is True

Lua[edit]

Works with: lua version 5.1
function emptySet()         return { }  end
function insert(set, item) set[item] = true end
function remove(set, item) set[item] = nil end
function member(set, item) return set[item] end
function size(set)
local result = 0
for _ in pairs(set) do result = result + 1 end
return result
end
function fromTable(tbl) -- ignore the keys of tbl
local result = { }
for _, val in pairs(tbl) do
result[val] = true
end
return result
end
function toArray(set)
local result = { }
for key in pairs(set) do
table.insert(result, key)
end
return result
end
function printSet(set)
print(table.concat(toArray(set), ", "))
end
function union(setA, setB)
local result = { }
for key, _ in pairs(setA) do
result[key] = true
end
for key, _ in pairs(setB) do
result[key] = true
end
return result
end
function intersection(setA, setB)
local result = { }
for key, _ in pairs(setA) do
if setB[key] then
result[key] = true
end
end
return result
end
function difference(setA, setB)
local result = { }
for key, _ in pairs(setA) do
if not setB[key] then
result[key] = true
end
end
return result
end
function subset(setA, setB)
for key, _ in pairs(setA) do
if not setB[key] then
return false
end
end
return true
end
function properSubset(setA, setB)
return subset(setA, setB) and (size(setA) ~= size(setB))
end
function equals(setA, setB)
return subset(setA, setB) and (size(setA) == size(setB))
end
 

Maple[edit]

Sets in Maple are built-in, native data structures, and are immutable. Sets are formed by enclosing a sequence of objects between braces. You can get something essentially equivalent to set comprehensions by using {seq}, which applies the set constructor ("{}") to the sequencing operation "seq".

 
> S := { 2, 3, 5, 7, 11, Pi, "foo", { 2/3, 3/4, 4/5 } };
S := {2, 3, 5, 7, 11, "foo", Pi, {2/3, 3/4, 4/5}}
 
> type( S, set );
true
 
> Pi in S;
Pi in {2, 3, 5, 7, 11, "foo", Pi, {2/3, 3/4, 4/5}}
 
> if Pi in S then print( yes ) else print( no ) end:
yes
 
> member( Pi, S );
true
 
> if 4 in S then print( yes ) else print( no ) end:
no
 
> evalb( { 2/3, 3/4, 4/5 } in S );
true
 
> { a, b, c } union { 1, 2, 3 };
{1, 2, 3, a, b, c}
 
> { a, b, c } intersect { b, c, d };
{b, c}
 
> { a, b, c } minus { b, c, d };
{a}
 
> { a, b } subset { a, b, c };
true
 
> { a, d } subset { a, b, c };
false
 
> evalb( { 1, 2, 3 } = { 1, 2, 3 } );
true
 
> evalb( { 1, 2, 3 } = { 1, 2, 4 } );
false
 


Mathematica[edit]

set1 = {"a", "b", "c", "d", "e"}; set2 = {"a", "b", "c", "d", "e", "f", "g"};
MemberQ[set1, "a"]
Union[set1 , set2]
Intersection[set1 , set2]
Complement[set2, set1](*Set Difference*)
MemberQ[Subsets[set2], set1](*Subset*)
set1 == set2(*Equality*)
set1 == set1(*Equality*)
Output:
True
{"a", "b", "c", "d", "e", "f", "g"}
{"a", "b", "c", "d", "e"}
{"f", "g"}
True
False
True

MATLAB / Octave[edit]

There are two types of sets supported, sets with numeric values are stored in a vector, sets with string elements are stored in a cell-array.

 
% Set creation
s = [1, 2, 4]; % numeric values
t = {'a','bb','ccc'}; % cell array of strings
u = unique([1,2,3,3,2,3,2,4,1]); % set consists only of unique elements
% Test m ∈ S -- "m is an element in set S"
ismember(m, S)
% A ∪ B -- union; a set of all elements either in set A or in set B.
union(A, B)
% A ∩ B -- intersection; a set of all elements in both set A and set B.
intersect(A, B)
% A ∖ B -- difference; a set of all elements in set A, except those in set B.
setdiff(A, B)
% A ⊆ B -- subset; true if every element in set A is also in set B.
all(ismember(A, B))
% A = B -- equality; true if every element of set A is in set B and vice-versa.
isempty(setxor(A, B))
 

Maxima[edit]

/* illustrating some functions on sets; names are self-explanatory */
 
a: {1, 2, 3, 4};
{1, 2, 3, 4}
 
b: {2, 4, 6, 8};
{2, 4, 6, 8}
 
intersection(a, b);
{2, 4}
 
union(a, b);
{1, 2, 3, 4, 6, 8}
 
powerset(a);
{{}, {1}, {1, 2}, {1, 2, 3}, {1, 2, 3, 4}, {1, 2, 4}, {1, 3}, {1, 3, 4}, {1, 4}, {2}, {2, 3}, {2, 3, 4}, {2, 4}, {3}, {3, 4}, {4}}
 
set_partitions(a);
{{{1}, {2}, {3}, {4}}, {{1}, {2}, {3, 4}}, {{1}, {2, 3}, {4}}, {{1}, {2, 3, 4}}, {{1}, {2, 4}, {3}}, {{1, 2}, {3}, {4}},
{{1, 2}, {3, 4}}, {{1, 2, 3}, {4}}, {{1, 2, 3, 4}}, {{1, 2, 4}, {3}}, {{1, 3}, {2}, {4}}, {{1, 3}, {2, 4}}, {{1, 3, 4}, {2}},
{{1, 4}, {2}, {3}}, {{1, 4}, {2, 3}}}
 
setdifference(a, b);
{1, 3}
 
emptyp(a);
false
 
elementp(2, a);
true
 
cardinality(a);
4
 
cartesian_product(a, b);
{[1, 2], [1, 4], [1, 6], [1, 8], [2, 2], [2, 4], [2, 6], [2, 8], [3, 2], [3, 4], [3, 6], [3, 8], [4, 2], [4, 4], [4, 6], [4, 8]}
 
subsetp(a, b);
false
 
symmdifference(a, b);
{1, 3, 6, 8}
 
partition_set(union(a, b), evenp);
[{1, 3}, {2, 4, 6, 8}]
 
c: setify(makelist(fib(n), n, 1, 20));
{1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765}
 
equiv_classes(c, lambda([m, n], mod(m - n, 3) = 0));
{{1, 13, 34, 55, 610, 1597, 2584}, {2, 5, 8, 89, 233, 377, 4181}, {3, 21, 144, 987, 6765}}
 
disjointp(a, b);
false
 
adjoin(7, a);
{1, 2, 3, 4, 7}
 
a;
{1, 2, 3, 4}
 
disjoin(1, a);
{2, 3, 4}
 
a;
{1, 2, 3, 4}
 
subset(c, primep);
{2, 3, 5, 13, 89, 233, 1597}
 
permutations(a);
{[1, 2, 3, 4], [1, 2, 4, 3], [1, 3, 2, 4], [1, 3, 4, 2], [1, 4, 2, 3], [1, 4, 3, 2],
[2, 1, 3, 4], [2, 1, 4, 3], [2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1],
[3, 1, 2, 4], [3, 1, 4, 2], [3, 2, 1, 4], [3, 2, 4, 1], [3, 4, 1, 2], [3, 4, 2, 1],
[4, 1, 2, 3], [4, 1, 3, 2], [4, 2, 1, 3], [4, 2, 3, 1], [4, 3, 1, 2], [4, 3, 2, 1]}
 
setequalp(a, b);
false

Nemerle[edit]

The Nemerle.Collections namespace provides an implementation of a Set.

using System.Console;
using Nemerle.Collections;
 
module RCSet
{
HasSubset[T](this super : Set[T], sub : Set[T]) : bool
{
super.ForAll(x => sub.Contains(x))
}
 
Main() : void
{
def names1 = Set(["Bob", "Billy", "Tom", "Dick", "Harry"]);
def names2 = Set(["Bob", "Mary", "Alice", "Louisa"]);
//def names3 = Set(["Bob", "Bob"]); // unfortunately, duplicated elements are not well handled by the stock
// implementation, this statement would throw an ArgumentException
def elem = names1.Contains("Bob"); // element test
def names1u2 = names1.Sum(names2); // union
def names1d2 = names1.Subtract(names2); // difference
def names1i2 = names1.Intersect(names2); // intersection
def same = names1.Equals(names2); // equality
def sub12 = names1.HasSubset(names2); // subset
 
WriteLine($"$names1u2\n$names1d2\n$names1i2");
WriteLine($"$same\t$sub12");
}
}

Nim[edit]

var # creation
s = {0,3,5,10}
t = {3..20, 50..55}
 
if 5 in s: echo "5 is in!" # element test
 
var
c = s + t # union
d = s * t # intersection
e = s - t # difference
 
if s <= t: echo "s ⊆ t" # subset
 
if s <= t: echo "s ⊂ t" # strong subset
 
if s == t: echo "s = s" # equality
 
s.incl(4) # add 4 to set
s.excl(5) # remove 5 from set

Objective-C[edit]

#import <Foundation/Foundation.h>
 
int main (int argc, const char *argv[]) {
@autoreleasepool {
 
NSSet *s1 = [NSSet setWithObjects:@"a", @"b", @"c", @"d", @"e", nil];
NSSet *s2 = [NSSet setWithObjects:@"b", @"c", @"d", @"e", @"f", @"h", nil];
NSSet *s3 = [NSSet setWithObjects:@"b", @"c", @"d", nil];
NSSet *s4 = [NSSet setWithObjects:@"b", @"c", @"d", nil];
NSLog(@"s1: %@", s1);
NSLog(@"s2: %@", s2);
NSLog(@"s3: %@", s3);
NSLog(@"s4: %@", s4);
 
// Membership
NSLog(@"b in s1: %d", [s1 containsObject:@"b"]);
NSLog(@"f in s1: %d", [s1 containsObject:@"f"]);
 
// Union
NSMutableSet *s12 = [NSMutableSet setWithSet:s1];
[s12 unionSet:s2];
NSLog(@"s1 union s2: %@", s12);
 
// Intersection
NSMutableSet *s1i2 = [NSMutableSet setWithSet:s1];
[s1i2 intersectSet:s2];
NSLog(@"s1 intersect s2: %@", s1i2);
 
// Difference
NSMutableSet *s1_2 = [NSMutableSet setWithSet:s1];
[s1_2 minusSet:s2];
NSLog(@"s1 - s2: %@", s1_2);
 
// Subset of
NSLog(@"s3 subset of s1: %d", [s3 isSubsetOfSet:s1]);
 
// Equality
NSLog(@"s3 = s4: %d", [s3 isEqualToSet:s4]);
 
// Cardinality
NSLog(@"size of s1: %lu", [s1 count]);
 
// Has intersection (not disjoint)
NSLog(@"does s1 intersect s2? %d", [s1 intersectsSet:s2]);
 
// Adding and removing elements from a mutable set
NSMutableSet *mut_s1 = [NSMutableSet setWithSet:s1];
[mut_s1 addObject:@"g"];
NSLog(@"mut_s1 after adding g: %@", mut_s1);
[mut_s1 addObject:@"b"];
NSLog(@"mut_s1 after adding b again: %@", mut_s1);
[mut_s1 removeObject:@"c"];
NSLog(@"mut_s1 after removing c: %@", mut_s1);
 
}
return 0;
}
Output:
s1: {(
    d,
    b,
    e,
    c,
    a
)}
s2: {(
    d,
    b,
    e,
    c,
    h,
    f
)}
s3: {(
    b,
    c,
    d
)}
s4: {(
    b,
    c,
    d
)}
b in s1: 1
f in s1: 0
s1 union s2: {(
    c,
    h,
    d,
    e,
    a,
    f,
    b
)}
s1 intersect s2: {(
    d,
    b,
    e,
    c
)}
s1 - s2: {(
    a
)}
s3 subset of s1: 1
s3 = s4: 1
size of s1: 5
does s1 intersect s2? 1
mut_s1 after adding g: {(
    d,
    b,
    g,
    e,
    c,
    a
)}
mut_s1 after adding b again: {(
    d,
    b,
    g,
    e,
    c,
    a
)}
mut_s1 after removing c: {(
    d,
    b,
    g,
    e,
    a
)}

OCaml[edit]

OCaml offers a functional, persistent set data structure in its Set module. It is implemented using a binary search tree. Set works in the functor model, which means you need to first use the Set.Make functor to create a module for your kind of set that you can use. You must give the functor an argument that is a module containing the type and ordering function.

In the interactive toplevel:

Works with: OCaml version 4.02+
# module IntSet = Set.Make(struct type t = int let compare = compare end);; (* Create a module for our type of set *)
module IntSet :
sig
type elt = int
type t
val empty : t
val is_empty : t -> bool
val mem : elt -> t -> bool
val add : elt -> t -> t
val singleton : elt -> t
val remove : elt -> t -> t
val union : t -> t -> t
val inter : t -> t -> t
val diff : t -> t -> t
val compare : t -> t -> int
val equal : t -> t -> bool
val subset : t -> t -> bool
val iter : (elt -> unit) -> t -> unit
val fold : (elt -> 'a -> 'a) -> t -> 'a -> 'a
val for_all : (elt -> bool) -> t -> bool
val exists : (elt -> bool) -> t -> bool
val filter : (elt -> bool) -> t -> t
val partition : (elt -> bool) -> t -> t * t
val cardinal : t -> int
val elements : t -> elt list
val min_elt : t -> elt
val max_elt : t -> elt
val choose : t -> elt
val split : elt -> t -> t * bool * t
val find : elt -> t -> elt
val of_list : elt list -> t
end
# IntSet.empty;; (* Empty set. A set is an abstract type that will not display in the interpreter *)
- : IntSet.t = <abstr>
# IntSet.elements (IntSet.empty);; (* Get the previous set into a list *)
- : IntSet.elt list = []
# let s1 = IntSet.of_list [1;2;3;4;3];;
val s1 : IntSet.t = <abstr>
# IntSet.elements s1;;
- : IntSet.elt list = [1; 2; 3; 4]
# let s2 = IntSet.of_list [3;4;5;6];;
val s2 : IntSet.t = <abstr>
# IntSet.elements s2;;
- : IntSet.elt list = [3; 4; 5; 6]
# IntSet.elements (IntSet.union s1 s2);; (* Union *)
- : IntSet.elt list = [1; 2; 3; 4; 5; 6]
# IntSet.elements (IntSet.inter s1 s2);; (* Intersection *)
- : IntSet.elt list = [3; 4]
# IntSet.elements (IntSet.diff s1 s2);; (* Difference *)
- : IntSet.elt list = [1; 2]
# IntSet.subset s1 s1;; (* Subset *)
- : bool = true
# IntSet.subset (IntSet.of_list [3;1]) s1;;
- : bool = true
# IntSet.equal (IntSet.of_list [3;2;4;1]) s1;; (* Equality *)
- : bool = true
# IntSet.equal s1 s2;;
- : bool = false
# IntSet.mem 2 s1;; (* Membership *)
- : bool = true
# IntSet.mem 10 s1;;
- : bool = false
# IntSet.cardinal s1;; (* Cardinality *)
- : int = 4
# IntSet.elements (IntSet.add 99 s1);; (* Create a new set by inserting *)
- : IntSet.elt list = [1; 2; 3; 4; 99]
# IntSet.elements (IntSet.remove 3 s1);; (* Create a new set by deleting *)
- : IntSet.elt list = [1; 2; 4]

(Note: of_list is only available in OCaml 4.02+. In earlier versions, you can implement one yourself like let of_list lst = List.fold_right IntSet.add lst IntSet.empty;;)

Regular lists can also be used as sets.

In addition, you can use imperative hash tables from the Hashtbl module as a hash table-based set, using the unit type as the "value" for each key.

ooRexx[edit]

-- Set creation
-- Using the OF method
s1 = .set~of(1, 2, 3, 4, 5, 6)
-- Explicit addition of individual items
s2 = .set~new
s2~put(2)
s2~put(4)
s2~put(6)
-- group addition
s3 = .set~new
s3~putall(.array~of(1, 3, 5))
-- Test m ? S -- "m is an element in set S"
say s1~hasindex(1) s3~hasindex(2) -- "1 0", which is "true" and "false"
-- A ? B -- union; a set of all elements either in set A or in set B.
s4 = s2~union(s3) -- {1, 2, 3, 4, 5, 6}
Call show 's4',s4
-- A ? B -- intersection; a set of all elements in both set A and set B.
s5 = s1~intersection(s2) -- {2, 4, 6}
Call show 's5',s5
-- A ? B -- difference; a set of all elements in set A, except those in set B.
s6 = s1~difference(s2) -- {1, 3, 5}
Call show 's6',s6
-- A ? B -- subset; true if every element in set A is also in set B.
say s1~subset(s2) s2~subset(s1) -- "0 1"
-- A = B -- equality; true if every element of set A is in set B and vice-versa.
-- No direct equivalence method, but the XOR method can be used to determine this
say s1~xor(s4)~isempty -- true
Exit
show: Procedure
Use Arg set_name,set
Say set_name':' set~makearray~makestring((LINE),',')
return

The set operators don't come out too well :-(

Output:
1 0
s4: 1,2,3,4,5,6
s5: 2,4,6
s6: 1,3,5
0 1
1

PARI/GP[edit]

Aside from ⊆, all operations are already a part of GP.

setsubset(s,t)={
for(i=1,#s,
if(!setsearch(t,s[i]), return(0))
);
1
};
s=Set([1,2,2])
t=Set([4,2,4])
setsearch(s,1)
setunion(s,t)
setintersect(s,t)
setminus(s,t)
setsubset(s,t)
s==t
Output:
%1 = [1, 2]
%2 = [2, 4]
%3 = 1
%4 = [1, 2, 4]
%5 = [2]
%6 = [1]
%7 = 0
%8 = 0

Pascal[edit]

Works with: Free Pascal

Freepascal/object pascal handles sets very well. --Guionardo 22:03, 7 January 2012 (UTC)

program Rosetta_Set;
 
{$mode objfpc}{$H+}
 
uses {$IFDEF UNIX} {$IFDEF UseCThreads}
cthreads, {$ENDIF} {$ENDIF}
Classes;
 
{$R *.res}
type
CharSet = set of char;
 
var
A, B, C, S: CharSet;
M: char;
 
function SetToString(const ASet: CharSet): string;
var
J: char;
begin
Result := '';
// Test all chars
for J in char do
// If the char is in set, add to result
if J in ASet then
Result := Result + J + ', ';
// Clear the result
if Result > '' then
Delete(Result, Length(Result) - 1, 2);
end;
 
procedure PrintSet(const ASet: CharSet; const ASetName: string;
const ATitle: string = '');
begin
if ATitle > '' then
WriteLn(ATitle);
WriteLn(ASetName, ' = [', SetToString(ASet), ']', #10);
end;
 
procedure ShowEqual(const ASetA, ASetB: CharSet; const ASetNameA, ASetNameB: string);
begin
WriteLn(ASetNameA, ' = [', SetToString(ASetA), ']');
WriteLn(ASetNameB, ' = [', SetToString(ASetB), ']');
if ASetA = ASetB then
WriteLn(ASetNameA, ' = ', ASetNameB)
else
WriteLn(ASetNameA, ' <> ', ASetNameB);
end;
 
 
begin
// Set Creation
A := ['A', 'B', 'C', 'D', 'E', 'F'];
B := ['E', 'F', 'G', 'H', 'I', 'J'];
PrintSet(A, 'A', 'Set Creation');
PrintSet(B, 'B');
 
// Test m ∈ S -- "m is an element in set S"
M := 'A';
if M in A then
WriteLn('"A" is in set A');
 
// A ∪ B -- union; a set of all elements either in set A or in set B.
S := A + B;
PrintSet(S, 'S', 'S = A U B -- union; a set of all elements either in set A or in set B.');
 
// A ∩ B -- intersection; a set of all elements in both set A and set B.
S := A * B;
PrintSet(S, 'S',
'S = A ∩ B -- intersection; a set of all elements in both set A and set B.');
 
// A \ B -- difference; a set of all elements in set A, except those in set B.
S := A - B;
PrintSet(S, 'S',
'S = A \ B -- difference; a set of all elements in set A, except those in set B.');
 
// A ⊆ B -- subset; true if every element in set A is also in set B.
Writeln('A ⊆ B -- subset; true if every element in set A is also in set B.');
if A <= B then
WriteLn('A in B')
else
Writeln('A is not in B');
Writeln;
//A = B -- equality; true if every element of set A is in set B and vice-versa.
Writeln('A = B -- equality; true if every element of set A is in set B and vice-versa.');
 
ShowEqual(A, B, 'A', 'B');
S := A * B;
C := ['E', 'F'];
ShowEqual(S, C, 'S', 'C');
 
readln;
 
end.

Perl[edit]

For real code, try Set::Object from CPAN. Here we provide a primitive implementation using hashes.

use strict;
 
package Set; # likely will conflict with stuff on CPAN
use overload
'""' => \&str,
'bool' => \&count,
'+=' => \&add,
'-=' => \&del,
'-' => \&diff,
'==' => \&eq,
'&' => \&intersection,
'|' => \&union,
'^' => \&xdiff;
 
sub str {
my $set = shift;
# This has drawbacks: stringification is used as set key
# if the set is added to another set as an element, which
# may cause inconsistencies if the element set is modified
# later. In general, a hash key loses its object identity
# anyway, so it's not unique to us.
"Set{ ". join(", " => sort map("$_", values %$set)) . " }"
}
 
sub new {
my $pkg = shift;
my $h = bless {};
$h->add($_) for @_;
$h
}
 
sub add {
my ($set, $elem) = @_;
$set->{$elem} = $elem;
$set
}
 
sub del {
my ($set, $elem) = @_;
delete $set->{$elem};
$set
}
 
sub has { # set has element
my ($set, $elem) = @_;
exists $set->{$elem}
}
 
sub union {
my ($this, $that) = @_;
bless { %$this, %$that }
}
 
sub intersection {
my ($this, $that) = @_;
my $s = new Set;
for (keys %$this) {
$s->{$_} = $this->{$_} if exists $that->{$_}
}
$s
}
 
sub diff {
my ($this, $that) = @_;
my $s = Set->new;
for (keys %$this) {
$s += $this->{$_} unless exists $that->{$_}
}
$s
}
 
sub xdiff { # xor, symmetric diff
my ($this, $that) = @_;
my $s = new Set;
bless { %{ ($this - $that) | ($that - $this) } }
}
 
sub count { scalar(keys %{+shift}) }
 
sub eq {
my ($this, $that) = @_;
!($this - $that) && !($that - $this);
}
 
sub contains { # this is a superset of that
my ($this, $that) = @_;
for (keys %$that) {
return 0 unless $this->has($_)
}
return 1
}
 
package main;
my ($x, $y, $z, $w);
 
$x = Set->new(1, 2, 3);
$x += $_ for (5 .. 7);
$y = Set->new(1, 2, 4, $x); # not the brightest idea
 
print "set x is: $x\nset y is: $y\n";
for (1 .. 4, $x) {
print "$_ is", $y->has($_) ? "" : " not", " in y\n";
}
 
print "union: ", $x | $y, "\n";
print "intersect: ", $x & $y, "\n";
print "z = x - y = ", $z = $x - $y, "\n";
print "y is ", $x->contains($y) ? "" : "not ", "a subset of x\n";
print "z is ", $x->contains($z) ? "" : "not ", "a subset of x\n";
print "z = (x | y) - (x & y) = ", $z = ($x | $y) - ($x & $y), "\n";
print "w = x ^ y = ", $w = ($x ^ $y), "\n";
print "w is ", ($w == $z) ? "" : "not ", "equal to z\n";
print "w is ", ($w == $x) ? "" : "not ", "equal to x\n";

Perl 6[edit]

use Test;
 
my $a = set <a b c>;
my $b = set <b c d>;
my $c = set <a b c d e>;
 
ok 'c'$a, "c is an element in set A";
nok 'd'$a, "d is not an element in set A";
 
is $a$b, set(<a b c d>), "union; a set of all elements either in set A or in set B";
is $a$b, set(<b c>), "intersection; a set of all elements in both set A and set B";
is $a (-) $b, set(<a>), "difference; a set of all elements in set A, except those in set B";
 
ok $a$c, "subset; true if every element in set A is also in set B";
nok $c$a, "subset; false if every element in set A is not also in set B";
ok $a$c, "strict subset; true if every element in set A is also in set B";
nok $a$a, "strict subset; false for equal sets";
ok $a === set(<a b c>), "equality; true if every element of set A is in set B and vice-versa";
nok $a === $b, "equality; false for differing sets";
Output:
ok 1 - c is an element in set A
ok 2 - d is not an element in set A
ok 3 - union; a set of all elements either in set A or in set B
ok 4 - intersection; a set of all elements in both set A and set B
ok 5 - difference; a set of all elements in set A, except those in set B
ok 6 - subset; true if every element in set A is also in set B
ok 7 - subset; false if every element in set A is not also in set B
ok 8 - strict subset; true if every element in set A is also in set B
ok 9 - strict subset; false for equal sets
ok 10 - equality; true if every element of set A is in set B and vice-versa
ok 11 - equality; false for differing sets

Phix[edit]

First, a simple implementaion using native sequences:

sequence set1 = {1,2,3},
set2 = {3,4,5}
 
function element(object x, sequence set)
return find(x,set)!=0
end function
 
function union(sequence set1, sequence set2)
for i=1 to length(set2) do
if not element(set2[i],set1) then
set1 = append(set1,set2[i])
end if
end for
return set1
end function
 
function intersection(sequence set1, sequence set2)
sequence res = {}
for i=1 to length(set1) do
if element(set1[i],set2) then
res = append(res,set1[i])
end if
end for
return res
end function
 
function difference(sequence set1, sequence set2)
sequence res = {}
for i=1 to length(set1) do
if not element(set1[i],set2) then
res = append(res,set1[i])
end if
end for
return res
end function
 
function subset(sequence set1, sequence set2)
for i=1 to length(set1) do
if not element(set1[i],set2) then
return false
end if
end for
return true
end function
 
function equality(sequence set1, sequence set2)
if length(set1)!=length(set2) then
return false
end if
return subset(set1,set2)
end function
 
--test code:
?element(3,set1) -- 1
?element(4,set1) -- 0
?union(set1,set2) -- {1,2,3,4,5}
?intersection(set1,set2) -- {3}
?difference(set1,set2) -- {1,2}
?subset(set1,set2) -- 0
?subset({1,2},set1) -- 1
?equality(set1,set2) -- 0
?equality(set1,{3,1,2}) -- 1
Output:
1
0
{1,2,3,4,5}
{3}
{1,2}
0
1
0
1

Alternative using dictionaries, which needs several additional visitor routines (at a pinch they could be merged), but performance is better on larger sets:

integer set1 = new_dict(),
set2 = new_dict()
setd(3,0,set1)
setd(1,0,set1)
setd(2,0,set1)
setd(5,0,set2)
setd(3,0,set2)
setd(4,0,set2)
 
function element(object x, integer set)
return getd_index(x,set)!=0
end function
 
function u_visitor(object key, object data, object user_data)
integer {union_set,set2} = user_data
if set2=0
or not element(key,union_set) then
setd(key,data,union_set)
end if
return 1
end function
 
function union(integer set1, integer set2)
integer union_set = new_dict()
traverse_dict(routine_id("u_visitor"),{union_set,0},set1)
traverse_dict(routine_id("u_visitor"),{union_set,set2},set2)
return union_set
end function
 
function i_visitor(object key, object data, object user_data)
integer {union_set,set2} = user_data
if element(key,set2) then
setd(key,data,union_set)
end if
return 1
end function
 
function intersection(integer set1, integer set2)
integer union_set = new_dict()
traverse_dict(routine_id("i_visitor"),{union_set,set2},set1)
return union_set
end function
 
function d_visitor(object key, object data, object user_data)
integer {union_set,set2} = user_data
if not element(key,set2) then
setd(key,data,union_set)
end if
return 1
end function
 
function difference(integer set1, integer set2)
integer union_set = new_dict()
traverse_dict(routine_id("d_visitor"),{union_set,set2},set1)
return union_set
end function
 
integer res
function s_visitor(object key, object data, object user_data)
integer set2 = user_data
if not element(key,set2) then
res = 0
return 0 -- cease traversal
end if
return 1
end function
 
function subset(integer set1, integer set2)
res = 1
traverse_dict(routine_id("s_visitor"),set2,set1)
return res
end function
 
function equality(integer set1, integer set2)
if dict_size(set1)!=dict_size(set2) then
return false
end if
return subset(set1,set2)
end function
 
include builtins/map.e -- for keys()
 
-- matching test code:
?element(3,set1) -- 1
?element(4,set1) -- 0
?keys(union(set1,set2)) -- {1,2,3,4,5}
?keys(intersection(set1,set2)) -- {3}
?keys(difference(set1,set2)) -- {1,2}
?subset(set1,set2) -- 0
integer set3 = new_dict()
setd(2,0,set3)
setd(1,0,set3)
?subset(set3,set1) -- 1
?equality(set1,set2) -- 0
setd(3,0,set3)
?equality(set1,set3) -- 1

same output as above

PicoLisp[edit]

We may use plain lists, or 'idx' structures for sets. A set may contain any type of data.

Using lists[edit]

(setq
Set1 (1 2 3 7 abc "def" (u v w))
Set2 (2 3 5 hello (x y z))
Set3 (3 hello (x y z)) )
 
 
# Element tests (any non-NIL value means "yes")
: (member "def" Set1)
-> ("def" (u v w))
 
: (member "def" Set2)
-> NIL
 
: (member '(x y z) Set2)
-> ((x y z))
 
 
# Union
: (uniq (append Set1 Set2))
-> (1 2 3 7 abc "def" (u v w) 5 hello (x y z))
 
 
# Intersection
: (sect Set1 Set2)
-> (2 3)
 
 
# Difference
: (diff Set1 Set2)
-> (1 7 abc "def" (u v w))
 
 
# Test for subset
: (not (diff Set1 Set2))
-> NIL # Set1 is not a subset of Set2
 
: (not (diff Set3 Set2))
-> T # Set3 is a subset of Set2
 
 
# Test for equality
: (= (sort (copy Set1)) (sort (copy Set2)))
-> NIL
 
: (= (sort (copy Set2)) (sort (copy Set2)))
-> T

Using 'idx' structures[edit]

# Create three test-sets
(balance 'Set1 (1 2 3 7 abc "def" (u v w)))
(balance 'Set2 (2 3 5 hello (x y z)))
(balance 'Set3 (3 hello (x y z)))
 
 
# Get contents
: (idx 'Set1)
-> (1 2 3 7 abc "def" (u v w))
 
: (idx 'Set2)
-> (2 3 5 hello (x y z))
 
 
# Element tests (any non-NIL value means "yes")
: (idx 'Set1 "def")
-> ("def" (abc) (u v w))
 
: (idx 'Set2 "def")
-> NIL
 
: (idx 'Set2 '(x y z))
-> ((x y z))
 
 
# Union
: (use S
(balance 'S (idx 'Set1))
(balance 'S (idx 'Set2) T)
(idx 'S) )
-> (1 2 3 5 7 abc "def" hello (u v w) (x y z))
 
 
# Intersection
: (sect (idx 'Set1) (idx 'Set2))
-> (2 3)
 
 
# Difference
: (diff (idx 'Set1) (idx 'Set2))
-> (1 7 abc "def" (u v w))
 
 
# Test for subset
: (not (diff (idx 'Set1) (idx 'Set2)))
-> NIL # Set1 is not a subset of Set2
 
: (not (diff (idx 'Set3) (idx 'Set2)))
-> T # Set3 is a subset of Set2
 
 
# Test for equality
: (= (idx 'Set1) (idx 'Set2))
-> NIL
 
: (= (idx 'Set2) (idx 'Set2))
-> T

PowerShell[edit]

.NET offers the HashSet type which seems to act in most ways like a set.

When used in PowerShell, the syntax is clumsy. In addition, the "reference" set ($set1) is modified in place to become the result. (All examples assume the variable $set1 contains the value @(1,2,3,4))

 
[System.Collections.Generic.HashSet[object]]$set1 = 1..4
[System.Collections.Generic.HashSet[object]]$set2 = 3..6
 
# Operation + Definition + Result
#--------------------------------+---------------------+-------------------------
$set1.UnionWith($set2) # Union $set1 = 1, 2, 3, 4, 5, 6
$set1.IntersectWith($set2) # Intersection $set1 = 3, 4
$set1.ExceptWith($set2) # Difference $set1 = 1, 2
$set1.SymmetricExceptWith($set2) # Symmetric difference $set1 = 1, 2, 6, 5
$set1.IsSupersetOf($set2) # Test superset False
$set1.IsSubsetOf($set2) # Test subset False
$set1.Equals($set2) # Test equality False
$set1.IsProperSupersetOf($set2) # Test proper superset False
$set1.IsProperSubsetOf($set2) # Test proper subset False
 
5 -in $set1 # Test membership False
7 -notin $set1 # Test non-membership True
 

Prolog[edit]

Works with SWI-Prolog, library(lists).

:- use_module(library(lists)).
 
set :-
A = [2, 4, 1, 3],
B = [5, 2, 3, 2],
( is_set(A) -> format('~w is a set~n', [A])
; format('~w is not a set~n', [A])),
( is_set(B) -> format('~w is a set~n', [B])
; format('~w is not a set~n', [B])),
 
% create a set from a list
 
list_to_set(B, BS),
( is_set(BS) -> format('~nCreate a set from a list~n~w is a set~n', [BS])
; format('~w is not a set~n', [BS])),
 
intersection(A, BS, I),
format('~n~w intersection ~w => ~w~n', [A, BS, I]),
union(A, BS, U),
format('~w union ~w => ~w~n', [A, BS, U]),
difference(A, BS, D),
format('~w difference ~w => ~w~n', [A, BS, D]),
 
X = [1,2],
( subset(X, A) -> format('~n~w is a subset of ~w~n', [X, A])
; format('~w is not a subset of ~w~n', [X, A])),
Y = [1,5],
( subset(Y, A) -> format('~w is a subset of ~w~n', [Y, A])
; format('~w is not a subset of ~w~n', [Y, A])),
Z = [1, 2, 3, 4],
( equal(Z, A) -> format('~n~w is equal to ~w~n', [Z, A])
; format('~w is not equal to ~w~n', [Z, A])),
T = [1, 2, 3],
( equal(T, A) -> format('~w is equal to ~w~n', [T, A])
; format('~w is not equal to ~w~n', [T, A])).
 
 
 
% compute difference of sets
difference(A, B, D) :-
exclude(member_(B), A, D).
 
member_(L, X) :-
member(X, L).
 
equal([], []).
equal([H1 | T1], B) :-
select(H1, B, B1),
equal(T1, B1).
 
Output:
 ?- set.
[2,4,1,3] is a set
[5,2,3,2] is not a set

Create a set from a list
[5,2,3] is a set

[2,4,1,3] intersection [5,2,3] => [2,3]
[2,4,1,3] union [5,2,3] => [4,1,5,2,3]
[2,4,1,3] difference [5,2,3] => [4,1]

[1,2] is a subset of [2,4,1,3]
[1,5] is not a subset of [2,4,1,3]

[1,2,3,4] is equal to [2,4,1,3]
[1,2,3] is not equal to [2,4,1,3]
true.


Works with: SWI-Prolog version library(ordset)

SWI-Prolog provides a standard library(ordsets). It is loaded by default. I demonstrate almost all of these predicates in the interactive top-level (`?-` is the prompt). Variables prefixed with `$` refer back to the value of the last instantiation. It treats sets as ordered lists of unique elements:

 
%% Set creation
 
?- list_to_ord_set([1,2,3,4], A), list_to_ord_set([2,4,6,8], B).
A = [1, 2, 3, 4],
B = [2, 4, 6, 8].
 
%% Test m ∈ S -- "m is an element in set S"
 
?- ord_memberchk(2, $A).
true.
 
%% A ∪ B -- union; a set of all elements either in set A or in set B.
 
?- ord_union($A, $B, Union).
Union = [1, 2, 3, 4, 6, 8].
 
%% A ∩ B -- intersection; a set of all elements in both set A and set B.
 
?- ord_intersection($A, $B, Intersection).
Intersection = [2, 4].
 
%% A ∖ B -- difference; a set of all elements in set A, except those in set B.
 
?- ord_subtract($A, $B, Diff).
Diff = [1, 3].
 
%% A ⊆ B -- subset; true if every element in set A is also in set B.
 
?- ord_subset($A, $B).
false.
 
?- ord_subset([2,4], $B).
true.
 
%% A = B -- equality; true if every element of set A is in set B and vice-versa.
 
?- $A == $B.
false.
 
?- $A == [1,2,3,4].
true.
 
%% Definition of a proper subset:
 
ord_propsubset(A, B) :-
ord_subset(A, B),
\+(A == B).
 
%% add/remove elements
 
?- ord_add_element($A, 19, NewA).
NewA = [1, 2, 3, 4, 19].
 
?- ord_del_element($NewA, 3, NewerA).
NewerA = [1, 2, 4, 19].
 

PureBasic[edit]

This solution uses PureBasic's maps (hash tables).

Procedure.s booleanText(b) ;returns 'True' or 'False' for a boolean input
If b: ProcedureReturn "True": EndIf
ProcedureReturn "False"
EndProcedure
 
Procedure.s listSetElements(Map a(), delimeter.s = " ") ;format elements for display
Protected output$
 
ForEach a()
output$ + MapKey(a()) + delimeter
Next
 
ProcedureReturn "(" + RTrim(output$, delimeter) + ")"
EndProcedure
 
Procedure.s listSortedSetElements(Map a(), delimeter.s = " ") ;format elements for display as sorted for easy comparison
Protected output$
NewList b.s()
 
ForEach a()
AddElement(b()): b() = MapKey(a())
Next
SortList(b(), #PB_Sort_Ascending | #PB_Sort_NoCase)
ForEach b()
output$ + b() + delimeter
Next
 
ProcedureReturn "(" + RTrim(output$, delimeter) + ")"
EndProcedure
 
Procedure cardinalityOf(Map a())
ProcedureReturn MapSize(a())
EndProcedure
 
Procedure createSet(elements.s, Map o(), delimeter.s = " ", clearSet = 1)
Protected i, elementCount
 
If clearSet: ClearMap(o()): EndIf
elementCount = CountString(elements, delimeter) + 1 ;add one for the last element which won't have a delimeter
For i = 1 To elementCount
AddMapElement(o(), StringField(elements, i, delimeter))
Next
 
ProcedureReturn MapSize(o())
EndProcedure
 
Procedure adjoinTo(elements.s, Map o(), delimeter.s = " ")
ProcedureReturn createSet(elements, o(), delimeter, 0)
EndProcedure
 
Procedure disjoinFrom(elements.s, Map o(), delimeter.s = " ")
Protected i, elementCount
 
elementCount = CountString(elements, delimeter) + 1 ;add one for the last element which won't have a delimeter
For i = 1 To elementCount
DeleteMapElement(o(), StringField(elements, i, delimeter))
Next
 
ProcedureReturn MapSize(o())
EndProcedure
 
Procedure isElementOf(element.s, Map a())
ProcedureReturn FindMapElement(a(), element)
EndProcedure
 
 
 
Procedure unionOf(Map a(), Map b(), Map o())
CopyMap(a(), o())
ForEach b()
AddMapElement(o(), MapKey(b()))
Next
 
ProcedureReturn MapSize(o())
EndProcedure
 
Procedure intersectionOf(Map a(), Map b(), Map o())
ClearMap(o())
ForEach a()
If FindMapElement(b(), MapKey(a()))
AddMapElement(o(), MapKey(a()))
EndIf
Next
 
ProcedureReturn MapSize(o())
EndProcedure
 
Procedure differenceOf(Map a(), Map b(), Map o())
CopyMap(a(), o())
ForEach b()
If FindMapElement(o(), MapKey(b()))
DeleteMapElement(o())
Else
AddMapElement(o(), MapKey(b()))
EndIf
Next
 
ProcedureReturn MapSize(o())
EndProcedure
 
Procedure isSubsetOf(Map a(), Map b()) ;boolean
ForEach a()
If Not FindMapElement(b(), MapKey(a()))
ProcedureReturn 0
EndIf
Next
ProcedureReturn 1
EndProcedure
 
Procedure isProperSubsetOf(Map a(), Map b()) ;boolean
If MapSize(a()) = MapSize(b())
ProcedureReturn 0
EndIf
ProcedureReturn isSubsetOf(a(), b())
EndProcedure
 
Procedure isEqualTo(Map a(), Map b())
If MapSize(a()) = MapSize(b())
ProcedureReturn isSubsetOf(a(), b())
EndIf
ProcedureReturn 0
EndProcedure
 
Procedure isEmpty(Map a()) ;boolean
If MapSize(a())
ProcedureReturn 0
EndIf
ProcedureReturn 1
EndProcedure
 
If OpenConsole()
NewMap a()
NewMap b()
NewMap o() ;for output sets
NewMap c()
 
createSet("red blue green orange yellow", a())
PrintN("Set A = " + listSortedSetElements(a()) + " of cardinality " + Str(cardinalityOf(a())) + ".")
createSet("lady green red", b())
PrintN("Set B = " + listSortedSetElements(b()) + " of cardinality " + Str(cardinalityOf(b())) + ".")
PrintN("'red' is an element of A is " + booleanText(isElementOf("red", a())) + ".")
PrintN("'red' is an element of B is " + booleanText(isElementOf("red", b())) + ".")
PrintN("'blue' is an element of B is " + booleanText(isElementOf("blue", b())) + ".")
 
unionOf(a(), b(), o())
PrintN(#crlf$ + "Union of A & B is " + listSortedSetElements(o()) + ".")
intersectionOf(a(), b(), o())
PrintN("Intersection of A & B is " + listSortedSetElements(o()) + ".")
differenceOf(a(), b(), o())
PrintN("Difference of A & B is " + listSortedSetElements(o()) + ".")
 
PrintN(listSortedSetElements(a()) + " equals " + listSortedSetElements(a()) + " is " + booleanText(isEqualTo(a(), a())) + ".")
PrintN(listSortedSetElements(a()) + " equals " + listSortedSetElements(b()) + " is " + booleanText(isEqualTo(a(), b())) + ".")
 
createSet("red green", c())
PrintN(#crlf$ + listSortedSetElements(c()) + " is a subset of " + listSortedSetElements(a()) + " is "+ booleanText(isSubsetOf(c(), a())) + ".")
PrintN(listSortedSetElements(c()) + " is a proper subset of " + listSortedSetElements(b()) + " is "+ booleanText(isProperSubsetOf(c(), b())) + ".")
PrintN(listSortedSetElements(c()) + " is a proper subset of " + listSortedSetElements(a()) + " is "+ booleanText(isProperSubsetOf(c(), a())) + ".")
PrintN(listSortedSetElements(b()) + " is a proper subset of " + listSortedSetElements(b()) + " is "+ booleanText(isProperSubsetOf(b(), b())) + ".")
 
PrintN(#crlf$ + "Set C = " + listSortedSetElements(c()) + " of cardinality " + Str(cardinalityOf(c())) + ".")
adjoinTo("dog cat mouse", c())
PrintN("Add 'dog cat mouse' to C to get " + listSortedSetElements(c()) + " of cardinality " + Str(cardinalityOf(c())) + ".")
disjoinFrom("red green dog", c())
PrintN("Take away 'red green dog' from C to get " + listSortedSetElements(c()) + " of cardinality " + Str(cardinalityOf(c())) + ".")
 
 
Print(#crlf$ + #crlf$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
Output:
Set A = (blue green orange red yellow) of cardinality 5.
Set B = (green lady red) of cardinality 3.
'red' is an element of A is True.
'red' is an element of B is True.
'blue' is an element of B is False.

Union of A & B is (blue green lady orange red yellow).
Intersection of  A & B is (green red).
Difference of  A & B is (blue lady orange yellow).
(blue green orange red yellow) equals (blue green orange red yellow) is True.
(blue green orange red yellow) equals (green lady red) is False.

(green red) is a subset of (blue green orange red yellow) is True.
(green red) is a proper subset of (green lady red) is True.
(green red) is a proper subset of (blue green orange red yellow) is True.
(green lady red) is a proper subset of (green lady red) is False.

Set C = (green red) of cardinality 2.
Add 'dog cat mouse' to C to get (cat dog green mouse red) of cardinality 5.
Take away 'red green dog' from C to get (cat mouse) of cardinality 2.

Python[edit]

In Python, set is a standard type since Python 2.4. There is also frozenset which is an immutable version of set. (In Python 2.3, they were provided as Set and ImmutableSet types in the module sets.)

Language syntax for set literals is supported starting in Python 3.0 and 2.7. (For versions prior to 2.7, use set([1, 2, 3, 4]) instead of {1, 2, 3, 4}. Even in Python 2.7+ and 3.0+, it is necessary to write set() to express the empty set.)

Works with: Python version 2.7+ and 3.0+
>>> s1, s2 = {1, 2, 3, 4}, {3, 4, 5, 6}
>>> s1 | s2 # Union
{1, 2, 3, 4, 5, 6}
>>> s1 & s2 # Intersection
{3, 4}
>>> s1 - s2 # Difference
{1, 2}
>>> s1 < s1 # True subset
False
>>> {3, 1} < s1 # True subset
True
>>> s1 <= s1 # Subset
True
>>> {3, 1} <= s1 # Subset
True
>>> {3, 2, 4, 1} == s1 # Equality
True
>>> s1 == s2 # Equality
False
>>> 2 in s1 # Membership
True
>>> 10 not in s1 # Non-membership
True
>>> {1, 2, 3, 4, 5} > s1 # True superset
True
>>> {1, 2, 3, 4} > s1 # True superset
False
>>> {1, 2, 3, 4} >= s1 # Superset
True
>>> s1 ^ s2 # Symmetric difference
{1, 2, 5, 6}
>>> len(s1) # Cardinality
4
>>> s1.add(99) # Mutability
>>> s1
{99, 1, 2, 3, 4}
>>> s1.discard(99) # Mutability
>>> s1
{1, 2, 3, 4}
>>> s1 |= s2 # Mutability
>>> s1
{1, 2, 3, 4, 5, 6}
>>> s1 -= s2 # Mutability
>>> s1
{1, 2}
>>> s1 ^= s2 # Mutability
>>> s1
{1, 2, 3, 4, 5, 6}
>>>

Racket[edit]

 
#lang racket
 
(define A (set 1 2 3 4))
(define B (set 3 4 5 6))
(define C (set 4 5))
 
(set-union A B)  ; gives (set 1 2 3 4 5 6)
(set-intersect A B) ; gives (set 3 4)
(set-subtract A B)  ; gives (set 1 2)
(set=? A B)  ; gives #f
(subset? C A)  ; gives #f
(subset? C B)  ; gives #t
 

REXX[edit]

REXX doesn't have native set support, but can be easily coded to handle lists as sets.

/*REXX program  demonstrates some  common   SET   functions.                            */
truth.0= 'false'; truth.1= "true" /*two common names for a truth table. */
set.= /*the order of sets isn't important. */
 
call setAdd 'prime',2 3 2 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
call setSay 'prime' /*a small set of some prime numbers. */
 
call setAdd 'emirp',97 97 89 83 79 73 71 67 61 59 53 47 43 41 37 31 29 23 19 17 13 11 7 5 3 2
call setSay 'emirp' /*a small set of backward primes. */
 
call setAdd 'happy',1 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 100 94 97 97 97 97 97
call setSay 'happy' /*a small set of some happy numbers. */
 
do j=11 to 100 by 10 /*see if PRIME contains some numbers. */
call setHas 'prime', j
say ' prime contains' j":" truth.result
end /*j*/
 
call setUnion 'prime','happy','eweion'; call setSay 'eweion' /* (sic). */
call setCommon 'prime','happy','common'; call setSay 'common'
call setDiff 'prime','happy','diff'  ; call setSay 'diff'; _=left('', 12)
call setSubset 'prime','happy'  ; say _ 'prime is a subset of happy:' truth.result
call setEqual 'prime','emirp'  ; say _ 'prime is equal to emirp:' truth.result
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
setHas: procedure expose set.; arg _ .,! .; return wordpos(!, set._)\==0
setAdd: return set$('add' , arg(1), arg(2))
setDiff: return set$('diff' , arg(1), arg(2), arg(3))
setSay: return set$('say' , arg(1), arg(2))
setUnion: return set$('union' , arg(1), arg(2), arg(3))
setCommon: return set$('common' , arg(1), arg(2), arg(3))
setEqual: return set$('equal' , arg(1), arg(2))
setSubset: return set$('subSet' , arg(1), arg(2))
/*──────────────────────────────────────────────────────────────────────────────────────*/
set$: procedure expose set.; arg $,_1,_2,_3; set_=set._1; t=_3; s=t;  !=1
if $=='SAY' then do; say "[set."_1']= 'set._1; return set._1; end
if $=='UNION' then do
call set$ 'add', _3, set._1
call set$ 'add', _3, set._2
return set._3
end
add=$=='ADD'; common=$=='COMMON'; diff=$=='DIFF'; eq=$=='EQUAL'; subset=$=='SUBSET'
if common | diff | eq | subset then s=_2
if add then do; set_=_2; t=_1; s=_1; end
 
do j=1 for words(set_); _=word(set_, j); has=wordpos(_, set.s)\==0
if (add & \has) |,
(common & has) |,
(diff & \has) then set.t=space(set.t _)
if (eq | subset) & \has then return 0
end /*j*/
 
if subset then return 1
if eq then if arg()>3 then return 1
else return set$('equal', _2, _1, 1)
return set.t

output

[set.PRIME]=2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
[set.EMIRP]=97 89 83 79 73 71 67 61 59 53 47 43 41 37 31 29 23 19 17 13 11 7 5 3 2
[set.HAPPY]=1 7 10 13 19 23 28 31 32 44 49 68 70 79 82 86 91 100 94 97
             prime contains 11: true
             prime contains 21: false
             prime contains 31: true
             prime contains 41: true
             prime contains 51: false
             prime contains 61: true
             prime contains 71: true
             prime contains 81: false
             prime contains 91: false
[set.EWEION]=2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 1 10 28 32 44 49 68 70 82 86 91 100 94
[set.COMMON]=7 13 19 23 31 79 97
[set.DIFF]=2 3 5 11 17 29 37 41 43 47 53 59 61 67 71 73 83 89
             prime is a subset of happy: false
             prime is  equal   to emirp: true

Ruby[edit]

Ruby's standard library contains a "set" package, which provides Set and SortedSet classes.

>> require 'set'
=> true
>> s1, s2 = Set[1, 2, 3, 4], [3, 4, 5, 6].to_set # different ways of creating a set
=> [#<Set: {1, 2, 3, 4}>, #<Set: {5, 6, 3, 4}>]
>> s1 | s2 # Union
=> #<Set: {5, 6, 1, 2, 3, 4}>
>> s1 & s2 # Intersection
=> #<Set: {3, 4}>
>> s1 - s2 # Difference
=> #<Set: {1, 2}>
>> s1.proper_subset?(s1) # Proper subset
=> false
>> Set[3, 1].proper_subset?(s1) # Proper subset
=> true
>> s1.subset?(s1) # Subset
=> true
>> Set[3, 1].subset?(s1) # Subset
=> true
>> Set[3, 2, 4, 1] == s1 # Equality
=> true
>> s1 == s2 # Equality
=> false
>> s1.include?(2) # Membership
=> true
>> Set[1, 2, 3, 4, 5].proper_superset?(s1) # Proper superset
=> true
>> Set[1, 2, 3, 4].proper_superset?(s1) # Proper superset
=> false
>> Set[1, 2, 3, 4].superset?(s1) # Superset
=> true
>> s1 ^ s2 # Symmetric difference
=> #<Set: {5, 6, 1, 2}>
>> s1.size # Cardinality
=> 4
>> s1 << 99 # Mutability (or s1.add(99) )
=> #<Set: {99, 1, 2, 3, 4}>
>> s1.delete(99) # Mutability
=> #<Set: {1, 2, 3, 4}>
>> s1.merge(s2) # Mutability
=> #<Set: {5, 6, 1, 2, 3, 4}>
>> s1.subtract(s2) # Mutability
=> #<Set: {1, 2}>
>>

Run BASIC[edit]

 
A$ = "apple cherry elderberry grape"
B$ = "banana cherry date elderberry fig"
C$ = "apple cherry elderberry grape orange"
D$ = "apple cherry elderberry grape"
E$ = "apple cherry elderberry"
M$ = "banana"
 
print "A = ";A$
print "B = ";B$
print "C = ";C$
print "D = ";D$
print "E = ";E$
print "M = ";M$
 
if instr(A$,M$) = 0 then a$ = "not "
print "M is ";a$; "an element of Set A"
a$ = ""
if instr(B$,M$) = 0 then a$ = "not "
print "M is ";a$; "an element of Set B"
 
un$ = A$ + " "
for i = 1 to 5
if instr(un$,word$(B$,i)) = 0 then un$ = un$ + word$(B$,i) + " "
next i
print "union(A,B) = ";un$
 
for i = 1 to 5
if instr(A$,word$(B$,i)) <> 0 then ins$ = ins$ + word$(B$,i) + " "
next i
print "Intersection(A,B) = ";ins$
 
for i = 1 to 5
if instr(B$,word$(A$,i)) = 0 then dif$ = dif$ + word$(A$,i) + " "
next i
print "Difference(A,B) = ";dif$
 
a = subs(A$,B$,"AB")
a = subs(A$,C$,"AC")
a = subs(A$,D$,"AD")
a = subs(A$,E$,"AE")
 
a = eqs(A$,B$,"AB")
a = eqs(A$,C$,"AC")
a = eqs(A$,D$,"AD")
a = eqs(A$,E$,"AE")
end
 
function subs(a$,b$,sets$)
for i = 1 to 5
if instr(b$,word$(a$,i)) <> 0 then subs = subs + 1
next i
if subs = 4 then
print left$(sets$,1);" is a subset of ";right$(sets$,1)
else
print left$(sets$,1);" is not a subset of ";right$(sets$,1)
end if
end function
 
function eqs(a$,b$,sets$)
for i = 1 to 5
if word$(a$,i) <> "" then a = a + 1
if word$(b$,i) <> "" then b = b + 1
if instr(b$,word$(a$,i)) <> 0 then c = c + 1
next i
if (a = b) and (a = c) then
print left$(sets$,1);" is equal ";right$(sets$,1)
else
print left$(sets$,1);" is not equal ";right$(sets$,1)
end if
end function
Output:
A = apple cherry elderberry grape
B = banana cherry date elderberry fig
C = apple cherry elderberry grape orange
D = apple cherry elderberry grape
E = apple cherry elderberry
M = banana
M is not an element of Set A
M is an element of Set B
union(A,B) = apple cherry elderberry grape banana date fig 
Intersection(A,B) = cherry elderberry 
Difference(A,B) = apple grape  
A is not a subset of B
A is a subset of C
A is a subset of D
A is not a subset of E
A is not equal B
A is not equal C
A is equal D
A is not equal E

Rust[edit]

use std::collections::HashSet;
 
fn main() {
let a = vec![1, 3, 4].into_iter().collect::<HashSet<i32>>();
let b = vec![3, 5, 6].into_iter().collect::<HashSet<i32>>();
 
println!("Set A: {:?}", a.iter().collect::<Vec<_>>());
println!("Set B: {:?}", b.iter().collect::<Vec<_>>());
println!("Does A contain 4? {}", a.contains(&4));
println!("Union: {:?}", a.union(&b).collect::<Vec<_>>());
println!("Intersection: {:?}", a.intersection(&b).collect::<Vec<_>>());
println!("Difference: {:?}", a.difference(&b).collect::<Vec<_>>());
println!("Is A a subset of B? {}", a.is_subset(&b));
println!("Is A equal to B? {}", a == b);
}

Scala[edit]

 
object sets {
val set1 = Set(1,2,3,4,5)
val set2 = Set(3,5,7,9)
println(set1 contains 3)
println(set1 | set2)
println(set1 & set2)
println(set1 diff set2)
println(set1 subsetOf set2)
println(set1 == set2)
}
 

Scheme[edit]

Implemented based on lists. Not efficient on large sets.

(define (element? a lst)
(and (not (null? lst))
(or (eq? a (car lst))
(element? a (cdr lst)))))
 
; util, not strictly needed
(define (uniq lst)
(if (null? lst) lst
(let ((a (car lst)) (b (cdr lst)))
(if (element? a b)
(uniq b)
(cons a (uniq b))))))
 
(define (intersection a b)
(cond ((null? a) '())
((null? b) '())
(else
(append (intersection (cdr a) b)
(if (element? (car a) b)
(list (car a))
'())))))
 
(define (union a b)
(if (null? a) b
(union (cdr a)
(if (element? (car a) b)
b
(cons (car a) b)))))
 
(define (diff a b) ; a - b
(if (null? a) '()
(if (element? (car a) b)
(diff (cdr a) b)
(cons (car a) (diff (cdr a) b)))))
 
(define (subset? a b) ; A ⊆ B
(if (null? a) #t
(and (element? (car a) b)
(subset? (cdr a) b))))
 
(define (set-eq? a b)
(and (subset? a b)
(subset? b a)))

Seed7[edit]

$ include "seed7_05.s7i";
 
const type: charSet is set of char;
enable_output(charSet);
 
const proc: main is func
local
const charSet: A is {'A', 'B', 'C', 'D', 'E', 'F'};
var charSet: B is charSet.value;
var char: m is 'A';
begin
B := {'E', 'F', 'G', 'H', 'I', 'K'};
incl(B, 'J'); # Add 'J' to set B
excl(B, 'K'); # Remove 'K' from set B
writeln("A: " <& A);
writeln("B: " <& B);
writeln("m: " <& m);
writeln("m in A -- m is an element in A: " <& m in A);
writeln("A | B -- union: " <& A | B);
writeln("A & B -- intersection: " <& A & B);
writeln("A - B -- difference: " <& A - B);
writeln("A >< B -- symmetric difference: " <& A >< B);
writeln("A <= A -- subset: " <& A <= A);
writeln("A < A -- proper subset: " <& A < A);
writeln("A = B -- equality: " <& A = B);
end func;
Output:
A: {A, B, C, D, E, F}
B: {E, F, G, H, I, J}
m: A
m in A -- m is an element in A: TRUE
A | B  -- union:                {A, B, C, D, E, F, G, H, I, J}
A & B  -- intersection:         {E, F}
A - B  -- difference:           {A, B, C, D}
A >< B -- symmetric difference: {A, B, C, D, G, H, I, J}
A <= A -- subset:               TRUE
A < A  -- proper subset:        FALSE
A = B  -- equality:             FALSE

SETL[edit]

 
A := {1, 2, 3, 4};
B := {3, 4, 5, 6};
C := {4, 5};
 
-- Union, Intersection, Difference, Subset, Equality
print(A + B); -- {1, 2, 3, 4, 5, 6}
print(A * B); -- {3, 4}
print(A - B); -- {1, 2}
print(C subset B); -- #T
print(C = B); -- #F
 

Sidef[edit]

Translation of: Perl
class Set(*set) {
 
method init {
var elems = set;
set = Hash.new;
elems.each { |e| self += e }
}
 
method +(elem) {
set{elem} = elem;
self;
}
 
method del(elem) {
set.delete(elem);
}
 
method has(elem) {
set.has_key(elem);
}
 
method ∪(Set that) {
Set(set.values..., that.values...);
}
 
method ∩(Set that) {
Set(set.keys.grep{ |k| k ∈ that } \
.map { |k| set{k} }...);
}
 
method ∖(Set that) {
Set(set.keys.grep{|k| !(k ∈ that) } \
.map {|k| set{k} }...);
}
 
method ^(Set that) {
var d = ((self ∖ that)(that ∖ self));
Set(d.values...);
}
 
method count { set.len }
 
method ≡(Set that) {
(self ∖ that -> count.is_zero) && (that ∖ self -> count.is_zero);
}
 
method values { set.values }
 
method ⊆(Set that) {
that.set.keys.each { |k|
k ∈ self || return false;
}
return true;
}
 
method to_s {
"Set{" + set.values.map{|e| "#{e}"}.sort.join(', ') + "}"
}
}
 
class Object {
method ∈(Set set) {
set.has(self);
}
}

Usage example:

var x = Set(1, 2, 3);
5..7 -> each { |i| x += i };
 
var y = Set(1, 2, 4, x);
 
say "set x is: #{x}";
say "set y is: #{y}";
 
[1,2,3,4,x].each { |elem|
say ("#{elem} is ", elem ∈ y ? '' : 'not', " in y");
}
 
var (w, z);
say ("union: ", x ∪ y);
say ("intersect: ", x ∩ y);
say ("z = x ∖ y = ", z = (x ∖ y) );
say ("y is ", x ⊆ y ? "" : "not ", "a subset of x");
say ("z is ", x ⊆ z ? "" : "not ", "a subset of x");
say ("z = (x ∪ y) ∖ (x ∩ y) = ", z = ((x ∪ y)(x ∩ y)));
say ("w = x ^ y = ", w = (x ^ y));
say ("w is ", w ≡ z ? "" : "not ", "equal to z");
say ("w is ", w ≡ x ? "" : "not ", "equal to x");
Output:
set x is: Set{1, 2, 3, 5, 6, 7}
set y is: Set{1, 2, 4, Set{1, 2, 3, 5, 6, 7}}
1 is  in y
2 is  in y
3 is not in y
4 is  in y
Set{1, 2, 3, 5, 6, 7} is  in y
union: Set{1, 2, 3, 4, 5, 6, 7, Set{1, 2, 3, 5, 6, 7}}
intersect: Set{1, 2}
z = x ∖ y = Set{3, 5, 6, 7}
y is not a subset of x
z is a subset of x
z = (x ∪ y) ∖ (x ∩ y) = Set{3, 4, 5, 6, 7, Set{1, 2, 3, 5, 6, 7}}
w = x ^ y = Set{3, 4, 5, 6, 7, Set{1, 2, 3, 5, 6, 7}}
w is equal to z
w is not equal to x

Simula[edit]

SIMSET
BEGIN
 
 ! WE DON'T SUBCLASS HEAD BUT USE COMPOSITION FOR CLASS SET ;
CLASS SET;
BEGIN
PROCEDURE ADD(E); REF(ELEMENT) E;
BEGIN
IF NOT ISIN(E, THIS SET) THEN E.CLONE.INTO(H);
END**OF**ADD;
 
BOOLEAN PROCEDURE EMPTY; EMPTY := H.EMPTY;
REF(LINK) PROCEDURE FIRST; FIRST :- H.FIRST;
 
REF(HEAD) H;
H :- NEW HEAD;
END**OF**SET;
 
 ! WE SUBCLASS LINK FOR THE ELEMENTS CONTAINED IN THE SET ;
LINK CLASS ELEMENT;
VIRTUAL:
PROCEDURE ISEQUAL IS
BOOLEAN PROCEDURE ISEQUAL(OTHER); REF(ELEMENT) OTHER;;
PROCEDURE REPR IS
TEXT PROCEDURE REPR;;
PROCEDURE REPR IS
REF(ELEMENT) PROCEDURE CLONE;;
BEGIN
END**OF**ELEMENT;
 
REF(SET) PROCEDURE UNION(S1, S2); REF(SET) S1, S2;
BEGIN REF(SET) SU, S;
SU :- NEW SET;
FOR S :- S1, S2 DO
BEGIN
IF NOT S.EMPTY THEN
BEGIN REF(ELEMENT) E;
E :- S.FIRST;
WHILE E =/= NONE DO
BEGIN SU.ADD(E); E :- E.SUC;
END;
END;
END;
UNION :- SU;
END**OF**UNION;
 
REF(SET) PROCEDURE INTERSECTION(S1, S2); REF(SET) S1, S2;
BEGIN REF(SET) SI;
SI :- NEW SET;
IF NOT S1.EMPTY THEN
BEGIN REF(ELEMENT) E;
E :- S1.FIRST;
WHILE E =/= NONE DO
BEGIN IF ISIN(E, S2) THEN SI.ADD(E); E :- E.SUC;
END;
END;
INTERSECTION :- SI;
END**OF**INTERSECTION;
 
REF(SET) PROCEDURE MINUS(S1, S2); REF(SET) S1, S2;
BEGIN REF(SET) SM;
SM :- NEW SET;
IF NOT S1.EMPTY THEN
BEGIN REF(ELEMENT) E;
E :- S1.FIRST;
WHILE E =/= NONE DO
BEGIN IF NOT ISIN(E, S2) THEN SM.ADD(E); E :- E.SUC;
END;
END;
MINUS :- SM;
END**OF**MINUS;
 
BOOLEAN PROCEDURE ISSUBSET(S1, S2); REF(SET) S1, S2;
BEGIN BOOLEAN B;
B := TRUE;
IF NOT S1.EMPTY THEN
BEGIN REF(ELEMENT) E;
E :- S1.FIRST;
WHILE B AND E =/= NONE DO
BEGIN
B := ISIN(E, S2);
E :- E.SUC;
END;
END;
ISSUBSET := B;
END**OF**ISSUBSET;
 
BOOLEAN PROCEDURE ISEQUAL(S1, S2); REF(SET) S1, S2;
BEGIN
ISEQUAL := ISSUBSET(S1, S2) AND THEN ISSUBSET(S2, S1)
END**OF**ISEQUAL;
 
BOOLEAN PROCEDURE ISIN(ELE,S); REF(ELEMENT) ELE; REF(SET) S;
BEGIN
REF(ELEMENT) E; BOOLEAN FOUND;
IF NOT S.EMPTY THEN
BEGIN
E :- S.FIRST;
FOUND := E.ISEQUAL(ELE);
WHILE NOT FOUND AND E =/= NONE DO
BEGIN FOUND := E.ISEQUAL(ELE); E :- E.SUC;
END;
END;
ISIN := FOUND
END**OF**ISIN;
 
PROCEDURE OUTSET(S); REF(SET) S;
BEGIN
REF(ELEMENT) E;
OUTCHAR('{');
IF NOT S.EMPTY THEN
BEGIN
E :- S.FIRST; OUTTEXT(E.REPR);
FOR E :- E.SUC WHILE E =/= NONE DO
BEGIN OUTTEXT(", "); OUTTEXT(E.REPR);
END;
END;
OUTCHAR('}');
END**OF**OUTSET;
 
 
COMMENT ============== EXAMPLE USING SETS OF NUMBERS ============== ;
 
 
ELEMENT CLASS NUMBER(N); INTEGER N;
BEGIN
BOOLEAN PROCEDURE ISEQUAL(OTHER); REF(ELEMENT) OTHER;
ISEQUAL := N = OTHER QUA NUMBER.N;
TEXT PROCEDURE REPR;
BEGIN TEXT T; INTEGER I;
T :- BLANKS(20); T.PUTINT(N);
T.SETPOS(1);
WHILE T.GETCHAR = ' ' DO;
REPR :- T.SUB(T.POS - 1, T.LENGTH - T.POS + 2);
END;
REF(ELEMENT) PROCEDURE CLONE;
CLONE :- NEW NUMBER(N);
END**OF**NUMBER;
 
PROCEDURE REPORT(S1, MSG1, S2, MSG2, S3); REF(SET) S1, S2, S3; TEXT MSG1, MSG2;
BEGIN
OUTSET(S1); OUTCHAR(' ');
OUTTEXT(MSG1); OUTCHAR(' ');
OUTSET(S2); OUTCHAR(' ');
OUTTEXT(MSG2); OUTCHAR(' ');
OUTSET(S3);
OUTIMAGE;
END**OF**REPORT;
 
PROCEDURE REPORTBOOL(S1, MSG1, S2, MSG2, B); REF(SET) S1, S2; TEXT MSG1, MSG2; BOOLEAN B;
BEGIN
OUTSET(S1); OUTCHAR(' ');
OUTTEXT(MSG1); OUTCHAR(' ');
OUTSET(S2); OUTCHAR(' ');
OUTTEXT(MSG2); OUTCHAR(' ');
OUTTEXT(IF B THEN "T" ELSE "F");
OUTIMAGE;
END**OF**REPORTBOOL;
 
PROCEDURE REPORTNUMBOOL(N1, MSG1, S1, MSG2, B); REF(ELEMENT) N1; REF(SET) S1; TEXT MSG1, MSG2; BOOLEAN B;
BEGIN
OUTTEXT(N1.REPR); OUTCHAR(' ');
OUTTEXT(MSG1); OUTCHAR(' ');
OUTSET(S1); OUTCHAR(' ');
OUTTEXT(MSG2); OUTCHAR(' ');
OUTTEXT(IF B THEN "T" ELSE "F");
OUTIMAGE;
END**OF**REPORTNUMBOOL;
 
REF(SET) S1, S2, S3, S4, S5;
REF(ELEMENT) E;
INTEGER I;
 
S1 :- NEW SET; FOR I := 1, 2, 3, 4 DO S1.ADD(NEW NUMBER(I));
S2 :- NEW SET; FOR I := 3, 4, 5, 6 DO S2.ADD(NEW NUMBER(I));
S3 :- NEW SET; FOR I := 3, 1 DO S3.ADD(NEW NUMBER(I));
S4 :- NEW SET; FOR I := 1, 2, 3, 4, 5 DO S4.ADD(NEW NUMBER(I));
S5 :- NEW SET; FOR I := 4, 3, 2, 1 DO S5.ADD(NEW NUMBER(I));
 
REPORT(S1, "UNION", S2, " = ", UNION(S1, S2));
 
REPORT(S1, "INTERSECTION", S2, " = ", INTERSECTION(S1, S2));
 
REPORT(S1, "MINUS", S2, " = ", MINUS(S1, S2));
 
REPORT(S2, "MINUS", S1, " = ", MINUS(S2, S1));
 
E :- NEW NUMBER(2);
REPORTNUMBOOL(E, "IN", S1, " = ", ISIN(E, S1));
 
E :- NEW NUMBER(10);
REPORTNUMBOOL(E, "NOT IN", S1, " = ", NOT ISIN(E, S1));
 
REPORTBOOL(S1, "IS SUBSET OF", S1, " = ", ISSUBSET(S1, S1));
REPORTBOOL(S3, "IS SUBSET OF", S1, " = ", ISSUBSET(S3, S1));
REPORTBOOL(S4, "IS SUPERSET OF", S1, " = ", ISSUBSET(S1, S4));
 
REPORTBOOL(S1, "IS EQUAL TO", S2, " = ", ISEQUAL(S1, S2));
REPORTBOOL(S2, "IS EQUAL TO", S2, " = ", ISEQUAL(S2, S2));
REPORTBOOL(S1, "IS EQUAL TO", S5, " = ", ISEQUAL(S1, S5));
 
END.
 
Output:
{1, 2, 3, 4} UNION {3, 4, 5, 6}  =  {1, 2, 3, 4, 5, 6}
{1, 2, 3, 4} INTERSECTION {3, 4, 5, 6}  =  {3, 4}
{1, 2, 3, 4} MINUS {3, 4, 5, 6}  =  {1, 2}
{3, 4, 5, 6} MINUS {1, 2, 3, 4}  =  {5, 6}
2 IN {1, 2, 3, 4}  =  T
10 NOT IN {1, 2, 3, 4}  =  T
{1, 2, 3, 4} IS SUBSET OF {1, 2, 3, 4}  =  T
{3, 1} IS SUBSET OF {1, 2, 3, 4}  =  T
{1, 2, 3, 4, 5} IS SUPERSET OF {1, 2, 3, 4}  =  T
{1, 2, 3, 4} IS EQUAL TO {3, 4, 5, 6}  =  F
{3, 4, 5, 6} IS EQUAL TO {3, 4, 5, 6}  =  T
{1, 2, 3, 4} IS EQUAL TO {4, 3, 2, 1}  =  T

Smalltalk[edit]

Works with: Pharo version 1.3-13315
 
#(1 2 3) asSet union: #(2 3 4) asSet.
"a Set(1 2 3 4)"
 
#(1 2 3) asSet intersection: #(2 3 4) asSet.
"a Set(2 3)"
 
#(1 2 3) asSet difference: #(2 3 4) asSet.
"a Set(1)"
 
#(1 2 3) asSet includesAllOf: #(1 3) asSet.
"true"
 
#(1 2 3) asSet includesAllOf: #(1 3 4) asSet.
"false"
 
#(1 2 3) asSet = #(2 1 3) asSet.
"true"
 
#(1 2 3) asSet = #(1 2 4) asSet.
"false"
 

SQL[edit]

Works with: Oracle
 
-- set of numbers is a table
-- create one set with 3 elements
 
CREATE TABLE myset1 (element NUMBER);
 
INSERT INTO myset1 VALUES (1);
INSERT INTO myset1 VALUES (2);
INSERT INTO myset1 VALUES (3);
 
commit;
 
-- check if 1 is an element
 
SELECT 'TRUE' BOOL FROM dual
WHERE 1 IN
(SELECT element FROM myset1);
 
-- create second set with 3 elements
 
CREATE TABLE myset2 (element NUMBER);
 
INSERT INTO myset2 VALUES (1);
INSERT INTO myset2 VALUES (5);
INSERT INTO myset2 VALUES (6);
 
commit;
 
-- union sets
 
SELECT element FROM myset1
UNION
SELECT element FROM myset2;
 
-- intersection
 
SELECT element FROM myset1
INTERSECT
SELECT element FROM myset2;
 
-- difference
 
SELECT element FROM myset1
minus
SELECT element FROM myset2;
 
-- subset
 
-- change myset2 to only have 1 as element
 
DELETE FROM myset2 WHERE NOT element = 1;
 
commit;
 
-- check if myset2 subset of myset1
 
SELECT 'TRUE' BOOL FROM dual
WHERE 0 = (SELECT COUNT(*) FROM
(SELECT element FROM myset2
minus
SELECT element FROM myset1));
 
-- equality
 
-- change myset1 to only have 1 as element
 
DELETE FROM myset1 WHERE NOT element = 1;
 
commit;
 
-- check if myset2 subset of myset1 and
-- check if myset1 subset of myset2 and
 
SELECT 'TRUE' BOOL FROM dual
WHERE
0 = (SELECT COUNT(*) FROM
(SELECT element FROM myset2
minus
SELECT element FROM myset1)) AND
0 =
(SELECT COUNT(*) FROM
(SELECT element FROM myset1
minus
SELECT element FROM myset2));
 
SQL> 
SQL> -- set of numbers is a table
SQL> -- create one set with 3 elements
SQL> 
SQL> create table myset1 (element number);

Table created.

SQL> 
SQL> insert into myset1 values (1);

1 row created.

SQL> insert into myset1 values (2);

1 row created.

SQL> insert into myset1 values (3);

1 row created.

SQL> 
SQL> commit;

Commit complete.

SQL> 
SQL> -- check if 1 is an element
SQL> 
SQL> select 'TRUE' BOOL from dual
  2  where 1 in
  3  (select element from myset1);

BOOL                                                                            
----                                                                            
TRUE                                                                            

SQL> 
SQL> -- create second set with 3 elements
SQL> 
SQL> create table myset2 (element number);

Table created.

SQL> 
SQL> insert into myset2 values (1);

1 row created.

SQL> insert into myset2 values (5);

1 row created.

SQL> insert into myset2 values (6);

1 row created.

SQL> 
SQL> commit;

Commit complete.

SQL> 
SQL> -- union sets
SQL> 
SQL> select element from myset1
  2  union
  3  select element from myset2;

   ELEMENT                                                                      
----------                                                                      
         1                                                                      
         2                                                                      
         3                                                                      
         5                                                                      
         6                                                                      

SQL> 
SQL> -- intersection
SQL> 
SQL> select element from myset1
  2  intersect
  3  select element from myset2;

   ELEMENT                                                                      
----------                                                                      
         1                                                                      

SQL> 
SQL> -- difference
SQL> 
SQL> select element from myset1
  2  minus
  3  select element from myset2;

   ELEMENT                                                                      
----------                                                                      
         2                                                                      
         3                                                                      

SQL> 
SQL> -- subset
SQL> 
SQL> -- change myset2 to only have 1 as element
SQL> 
SQL> delete from myset2 where not element = 1;

2 rows deleted.

SQL> 
SQL> commit;

Commit complete.

SQL> 
SQL> -- check if myset2 subset of myset1
SQL> 
SQL> select 'TRUE' BOOL from dual
  2  where 0 =  (select count(*) from
  3  (select element from myset2
  4  minus
  5  select element from myset1));

BOOL                                                                            
----                                                                            
TRUE                                                                            

SQL> 
SQL> -- equality
SQL> 
SQL> -- change myset1 to only have 1 as element
SQL> 
SQL> delete from myset1 where not element = 1;

2 rows deleted.

SQL> 
SQL> commit;

Commit complete.

SQL> 
SQL>  -- check if myset2 subset of myset1 and
SQL>  -- check if myset1 subset of myset2 and
SQL> 
SQL> select 'TRUE' BOOL from dual
  2  where
  3  0 =  (select count(*) from
  4  (select element from myset2
  5  minus
  6  select element from myset1)) and
  7  0 =
  8  (select count(*) from
  9  (select element from myset1
 10  minus
 11  select element from myset2));

BOOL                                                                            
----                                                                            
TRUE            

Swift[edit]

Works with: Swift version 1.2+
var s1 : Set<Int> = [1, 2, 3, 4]
let s2 : Set<Int> = [3, 4, 5, 6]
println(s1.union(s2)) // union; prints "[5, 6, 2, 3, 1, 4]"
println(s1.intersect(s2)) // intersection; prints "[3, 4]"
println(s1.subtract(s2)) // difference; prints "[2, 1]"
println(s1.isSubsetOf(s1)) // subset; prints "true"
println(Set<Int>([3, 1]).isSubsetOf(s1)) // subset; prints "true"
println(s1.isStrictSubsetOf(s1)) // proper subset; prints "false"
println(Set<Int>([3, 1]).isStrictSubsetOf(s1)) // proper subset; prints "true"
println(Set<Int>([3, 2, 4, 1]) == s1) // equality; prints "true"
println(s1 == s2) // equality; prints "false"
println(s1.contains(2)) // membership; prints "true"
println(Set<Int>([1, 2, 3, 4]).isSupersetOf(s1)) // superset; prints "true"
println(Set<Int>([1, 2, 3, 4]).isStrictSupersetOf(s1)) // proper superset; prints "false"
println(Set<Int>([1, 2, 3, 4, 5]).isStrictSupersetOf(s1)) // proper superset; prints "true"
println(s1.exclusiveOr(s2)) // symmetric difference; prints "[5, 6, 2, 1]"
println(s1.count) // cardinality; prints "4"
s1.insert(99) // mutability
println(s1) // prints "[99, 2, 3, 1, 4]"
s1.remove(99) // mutability
println(s1) // prints "[2, 3, 1, 4]"
s1.unionInPlace(s2) // mutability
println(s1) // prints "[5, 6, 2, 3, 1, 4]"
s1.subtractInPlace(s2) // mutability
println(s1) // prints "[2, 1]"
s1.exclusiveOrInPlace(s2) // mutability
println(s1) // prints "[5, 6, 2, 3, 1, 4]"

Tcl[edit]

Sets in Tcl are modeled as lists of items, with operations that preserve uniqueness of membership.

Library: Tcllib (Package: struct::set)
package require struct::set
 
# Many ways to build sets
set s1 [list 1 2 3 4]
set s2 {3 4 5 6}
struct::set add s3 {2 3 4 3 2}; # $s3 will be proper set...
set item 5
 
puts "union: [struct::set union $s1 $s2]"
puts "intersection: [struct::set intersect $s1 $s2]"
puts "difference: [struct::set difference $s1 $s2]"
puts "membership predicate: [struct::set contains $s1 $item]"
puts "subset predicate: [struct::set subsetof $s1 $s2]"; # NB: not strict subset test!
puts "equality predicate: [struct::set equal $s1 $s2]"
 
# Adding an element to a set (note that we pass in the name of the variable holding the set):
struct::set include s3 $item
# Removing an element from a set:
struct::set exclude s3 $item
# Getting the cardinality:
puts "cardinality: [struct::set size $s3]

zkl[edit]

A simplistic implementation that is fine for smallish sets

var [const] unique = Utils.Helpers.listUnique;
class Set {
fcn init { var [const] set = (vm.arglist.copy() : unique(_)) }
fcn holds(x) { set.holds(x) }
fcn union(setB) { self(set.xplode(),setB.set.xplode()) }
fcn intersection(setB){ sb:=setB.set;
C:=self(); sc:=C.set;
foreach x in (set){ if (sb.holds(x)) sc.append(x) }
C
}
fcn diff(setB){ C:=self(); C.set.extend(set);
setB.set.pump(Void,C.set.remove);
C
}
fcn isSubset(setB){ sb:=setB.set;
set.pump(Void,'wrap(x){
if (not sb.holds(x)) return(Void.Stop,False); True
})
}
fcn __opEQ(setB) { ((set.len() == setB.set.len()) and self.isSubset(setB)) }
}
A := Set(1,2,3,3,3,4);
A.set.println();    //--> L(1,2,3,4)
A.holds(3).println();  //--> True
A.holds(9).println();  //--> False

B:=Set("one",2,"three");
A.union(B).set.println(); // -->L(1,2,3,4,"one","three")
B.union(A).set.println(); // -->L("one",2,"three",1,3,4)
A.union(B).diff(B.union(A)).set.println(); // -->L()

A.intersection(B).set.println(); //-->L(2)
B.intersection(A).set.println(); //-->L(2)

A.diff(B).set.println();  //-->L(1,3,4)
B.diff(A).set.println();  //-->L("one","three")

A.isSubset(B).println();  //-->False
A.isSubset(A).println();  //-->True
Set("three",2,2,2,2,2).isSubset(B).println();  //-->True

(A==B).println();  //-->False
(A==A).println();  //-->True
(A==Set(2,3,1,4)).println(); //-->True