Combinations with repetitions
From Rosetta Code
You are encouraged to solve this task according to the task description, using any language you may know.
The set of combinations with repetitions is computed from a set, S (of cardinality n), and a size of resulting selection, k, by reporting the sets of cardinality k where each member of those sets is chosen from S. In the real world, it is about choosing sets where there is a “large” supply of each type of element and where the order of choice does not matter. For example:
- Q: How many ways can a person choose two doughnuts from a store selling three types of doughnut: iced, jam, and plain? (i.e., S is {iced,jam,plain}, | S | = 3, and k = 2.)
- A: 6: {iced, iced}; {iced, jam}; {iced, plain}; {jam, jam}; {jam, plain}; {plain, plain}.
Note that both the order of items within a pair, and the order of the pairs given in the answer is not significant; the pairs represent multisets.
Task description
- Write a function/program/routine/.. to generate all the combinations with repetitions of n types of things taken k at a time and use it to show an answer to the doughnut example above.
- For extra credit, use the function to compute and show just the number of ways of choosing three doughnuts from a choice of ten types of doughnut. Do not show the individual choices for this part.
References:
Contents |
[edit] Ada
Should work for any discrete type: integer, modular, or enumeration.
combinations.adb:
with Ada.Text_IO;
procedure Combinations is
generic
type Set is (<>);
function Combinations
(Count : Positive;
Output : Boolean := False)
return Natural;
function Combinations
(Count : Positive;
Output : Boolean := False)
return Natural
is
package Set_IO is new Ada.Text_IO.Enumeration_IO (Set);
type Set_Array is array (Positive range <>) of Set;
Empty_Array : Set_Array (1 .. 0);
function Recurse_Combinations
(Number : Positive;
First : Set;
Prefix : Set_Array)
return Natural
is
Combination_Count : Natural := 0;
begin
for Next in First .. Set'Last loop
if Number = 1 then
Combination_Count := Combination_Count + 1;
if Output then
for Element in Prefix'Range loop
Set_IO.Put (Prefix (Element));
Ada.Text_IO.Put ('+');
end loop;
Set_IO.Put (Next);
Ada.Text_IO.New_Line;
end if;
else
Combination_Count := Combination_Count +
Recurse_Combinations
(Number - 1,
Next,
Prefix & (1 => Next));
end if;
end loop;
return Combination_Count;
end Recurse_Combinations;
begin
return Recurse_Combinations (Count, Set'First, Empty_Array);
end Combinations;
type Donuts is (Iced, Jam, Plain);
function Donut_Combinations is new Combinations (Donuts);
subtype Ten is Positive range 1 .. 10;
function Ten_Combinations is new Combinations (Ten);
Donut_Count : constant Natural :=
Donut_Combinations (Count => 2, Output => True);
Ten_Count : constant Natural := Ten_Combinations (Count => 3);
begin
Ada.Text_IO.Put_Line ("Total Donuts:" & Natural'Image (Donut_Count));
Ada.Text_IO.Put_Line ("Total Tens:" & Natural'Image (Ten_Count));
end Combinations;
Output:
ICED+ICED ICED+JAM ICED+PLAIN JAM+JAM JAM+PLAIN PLAIN+PLAIN Total Donuts: 6 Total Tens: 220
[edit] Clojure
(defn combinations [coll k]
(when-let [[x & xs] coll]
(if (= k 1)
(map list coll)
(concat (map (partial cons x) (combinations coll (dec k)))
(combinations xs k)))))
Example output:
user> (combinations '[iced jam plain] 2)
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))
[edit] C
#include <stdio.h>Output:
const char * donuts[] = { "iced", "jam", "plain", "something completely different" };
long choose(int * got, int n_chosen, int len, int at, int max_types)
{
int i;
long count = 0;
if (n_chosen == len) {
if (!got) return 1;
for (i = 0; i < len; i++)
printf("%s\t", donuts[got[i]]);
printf("\n");
return 1;
}
for (i = at; i < max_types; i++) {
if (got) got[n_chosen] = i;
count += choose(got, n_chosen + 1, len, i, max_types);
}
return count;
}
int main()
{
int chosen[3];
choose(chosen, 0, 2, 0, 3);
printf("\nWere there ten donuts, we'd have had %ld choices of three\n",
choose(0, 0, 3, 0, 10));
return 0;
}
iced icediced jam iced plain jam jam jam plain plain plain
Were there ten donuts, we'd have had 220 choices of three
[edit] CoffeeScript
combos = (arr, k) ->
return [ [] ] if k == 0
return [] if arr.length == 0
combos_with_head = ([arr[0]].concat combo for combo in combos arr, k-1)
combos_sans_head = combos arr[1...], k
combos_with_head.concat combos_sans_head
arr = ['iced', 'jam', 'plain']
console.log "valid pairs from #{arr.join ','}:"
console.log combos arr, 2
console.log "#{combos([1..10], 3).length} ways to order 3 donuts given 10 types"
output
jam,plain:
[ [ 'iced', 'iced' ],
[ 'iced', 'jam' ],
[ 'iced', 'plain' ],
[ 'jam', 'jam' ],
[ 'jam', 'plain' ],
[ 'plain', 'plain' ] ]
220 ways to order 3 donuts given 10 types
[edit] D
Using lexicographic next bit permutation to generate combinations with repetitions.
import std.stdio, std.range;
/*immutable*/ const struct CombRep {
immutable uint nt, nc;
private /*immutable*/ ulong[] combVal;
this(in uint numType, in uint numChoice) pure nothrow {
assert(0 < numType && numType + numChoice <= 64,
"valid only for nt + nc <= 64 (ulong bit size)");
nt = numType;
nc = numChoice;
if (nc == 0)
return;
auto v = (1UL << (nt - 1)) - 1;
// init to smallest number that has nt-1 bit set
// a set bit is metaphored as a _type_ seperator
immutable limit = v << nc;
// limit is the largest nt-1 bit set number that has nc
// zero-bit a zero-bit means a _choice_ between _type_
// seperators
while (v <= limit) {
combVal ~= v;
if (v == 0)
break;
// get next nt-1 bit number
immutable t = (v | (v - 1)) + 1;
v = t | ((((t & -t) / (v & -v)) >> 1) - 1);
}
}
uint length() @property const pure nothrow {
return combVal.length;
}
uint[] opIndex(in uint idx) const pure nothrow {
return val2set(combVal[idx]);
}
int opApply(int delegate(ref uint[]) dg) {
foreach (v; combVal) {
auto set = val2set(v);
if (dg(set))
break;
}
return 1;
}
private uint[] val2set(in ulong v) const pure nothrow {
// convert bit pattern to selection set
immutable uint bitLimit = nt + nc - 1;
uint typeIdx = 0;
uint[] set;
foreach (bitNum; 0 .. bitLimit)
if (v & (1 << (bitLimit - bitNum - 1)))
typeIdx++;
else
set ~= typeIdx;
return set;
}
}
// For finite Random Access Range
auto combRep(R)(R types, in uint numChoice) /*pure nothrow*/
if (hasLength!R && isRandomAccessRange!R) {
ElementType!R[][] result;
foreach (s; CombRep(types.length, numChoice)) {
ElementType!R[] r;
foreach (i; s)
r ~= types[i];
result ~= r;
}
return result;
}
void main() {
foreach (e; combRep(["iced", "jam", "plain"], 2))
// writefln("%5s", e);
writeln(e);
writeln("Ways to select 3 from 10 types is ",
CombRep(10, 3).length);
}
output:
["iced", "iced"] ["iced", "jam"] ["iced", "plain"] ["jam", "jam"] ["jam", "plain"] ["plain", "plain"] Ways to select 3 from 10 types is 220
[edit] Go
[edit] Concise recursive
package main
import "fmt"
func combrep(n int, lst []string) [][]string {
if n == 0 {
return [][]string{nil}
}
if len(lst) == 0 {
return nil
}
r := combrep(n, lst[1:])
for _, x := range combrep(n-1, lst) {
r = append(r, append(x, lst[0]))
}
return r
}
func main() {
fmt.Println(combrep(2, []string{"iced", "jam", "plain"}))
fmt.Println(len(combrep(3,
[]string{"1", "2", "3", "4", "5", "6", "7", "8", "9", "10"})))
}
Output:
[[plain plain] [plain jam] [jam jam] [plain iced] [jam iced] [iced iced]] 220
[edit] Channel
Using channel and goroutine, showing how to use synced or unsynced communication.
package main
import "fmt"
func picks(picked []int, pos, need int, c chan[]int, do_wait bool) {
if need == 0 {
if do_wait {
c <- picked
<-c
} else { // if we want only the count, there's no need to
// sync between coroutines; let it clobber the array
c <- []int {}
}
return
}
if pos <= 0 {
if need == len(picked) { c <- nil }
return
}
picked[len(picked) - need] = pos - 1
picks(picked, pos, need - 1, c, do_wait) // choose the current donut
picks(picked, pos - 1, need, c, do_wait) // or don't
}
func main() {
donuts := []string {"iced", "jam", "plain" }
picked := make([]int, 2)
ch := make(chan []int)
// true: tell the channel to wait for each sending, because
// otherwise the picked array may get clobbered before we can do
// anything to it
go picks(picked, len(donuts), len(picked), ch, true)
var cc []int
for {
if cc = <-ch; cc == nil { break }
for _, i := range cc {
fmt.Printf("%s ", donuts[i])
}
fmt.Println()
ch <- nil // sync
}
picked = make([]int, 3)
// this time we only want the count, so tell goroutine to keep going
// and work the channel buffer
go picks(picked, 10, len(picked), ch, false)
count := 0
for {
if cc = <-ch; cc == nil { break }
count++
}
fmt.Printf("\npicking 3 of 10: %d\n", count)
}
Output:
plain plain plain jam plain iced jam jam jam iced iced iced picking 3 of 10: 220
[edit] Multiset
This version has proper representation of sets and multisets.
package main
import (
"fmt"
"sort"
"strconv"
)
// Go maps are an easy representation for sets as long as the element type
// of the set is valid as a key type for maps. Strings are easy. We follow
// the convention of always storing true for the value.
type set map[string]bool
// Multisets of strings are easy in the same way. We store the multiplicity
// of the element as the value.
type multiset map[string]int
// But multisets are not valid as a map key type so we must do something
// more involved to make a set of multisets, which is the desired return
// type for the combrep function required by the task. We can store the
// multiset as the value, but we derive a unique string to use as a key.
type msSet map[string]multiset
// The key method returns this string. The string will simply be a text
// representation of the contents of the multiset. The standard
// printable representation of the multiset cannot be used however, because
// Go maps are not ordered. Instead, the contents are copied to a slice,
// which is sorted to produce something with a printable representation
// that will compare == for mathematically equal multisets.
//
// Of course there is overhead for this and if performance were important,
// a different representation would be used for multisets, one that didn’t
// require sorting to produce a key...
func (m multiset) key() string {
pl := make(pairList, len(m))
i := 0
for k, v := range m {
pl[i] = msPair{k, v}
i++
}
sort.Sort(pl)
return fmt.Sprintf("%v", pl)
}
// Types and methods needed for sorting inside of mulitset.key()
type msPair struct {
string
int
}
type pairList []msPair
func (p pairList) Len() int { return len(p) }
func (p pairList) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
func (p pairList) Less(i, j int) bool { return p[i].string < p[j].string }
// Function required by task.
func combrep(n int, lst set) msSet {
if n == 0 {
var ms multiset
return msSet{ms.key(): ms}
}
if len(lst) == 0 {
return msSet{}
}
var car string
var cdr set
for ele := range lst {
if cdr == nil {
car = ele
cdr = make(set)
} else {
cdr[ele] = true
}
}
r := combrep(n, cdr)
for _, x := range combrep(n-1, lst) {
c := multiset{car: 1}
for k, v := range x {
c[k] += v
}
r[c.key()] = c
}
return r
}
// Driver for examples required by task.
func main() {
// Input is a set.
three := set{"iced": true, "jam": true, "plain": true}
// Output is a set of multisets. The set is a Go map:
// The key is a string representation that compares equal
// for equal multisets. We ignore this here. The value
// is the multiset. We print this.
for _, ms := range combrep(2, three) {
fmt.Println(ms)
}
ten := make(set)
for i := 1; i <= 10; i++ {
ten[strconv.Itoa(i)] = true
}
fmt.Println(len(combrep(3, ten)))
}
Output:
map[plain:1 jam:1] map[plain:2] map[iced:1 jam:1] map[jam:2] map[iced:1 plain:1] map[iced:2] 220
[edit] Haskell
import Data.List
-- Return the combinations, with replacement, of k items from the
-- list. We ignore the case where k is greater than the length of
-- the list.
combsWithRep 0 _ = [[]]
combsWithRep _ [] = []
combsWithRep k xxs@(x:xs) = map (x:) (combsWithRep (k-1) xxs) ++ combsWithRep k xs
countCombsWithRep k = length . combsWithRep k
main = do
print $ combsWithRep 2 ["iced","jam","plain"]
print $ countCombsWithRep 3 [1..10]
Example output:
[["iced","iced"],["iced","jam"],["iced","plain"],["jam","jam"],["jam","plain"],["plain","plain"]]
220
[edit] Icon and Unicon
Following procedure is a generator, which generates each combination of length n in turn:
# generate all combinations of length n from list L,
# including repetitions
procedure combinations_repetitions (L, n)
if n = 0
then suspend [] # if reach 0, then return an empty list
else if *L > 0
then {
# keep the first element
item := L[1]
# get all of length n in remaining list
every suspend (combinations_repetitions (L[2:0], n))
# get all of length n-1 in remaining list
# and add kept element to make list of size n
every i := combinations_repetitions (L, n-1) do
suspend [item] ||| i
}
end
Test procedure:
# convenience function
procedure write_list (l)
every (writes (!l || " "))
write ()
end
# testing routine
procedure main ()
# display all combinations for 2 of iced/jam/plain
every write_list (combinations_repetitions(["iced", "jam", "plain"], 2))
# get a count for number of ways to select 3 items from 10
every push(num_list := [], 1 to 10)
count := 0
every combinations_repetitions(num_list, 3) do count +:= 1
write ("There are " || count || " possible combinations of 3 from 10")
end
Output:
plain plain jam plain jam jam iced plain iced jam iced iced There are 220 possible combinations of 3 from 10
[edit] J
rcomb=: >@~.@:(/:~&.>)@,@{@# <
Example use:
2 rcomb ;:'iced jam plain'
┌─────┬─────┐
│iced │iced │
├─────┼─────┤
│iced │jam │
├─────┼─────┤
│iced │plain│
├─────┼─────┤
│jam │jam │
├─────┼─────┤
│jam │plain│
├─────┼─────┤
│plain│plain│
└─────┴─────┘
#3 rcomb i.10 NB. ways to choose 3 items from 10 with replacement
220
[edit] Java
MultiCombinationsTester.java
import com.objectwave.utility.*;
public class MultiCombinationsTester {
public MultiCombinationsTester() throws CombinatoricException {
Object[] objects = {"iced", "jam", "plain"};
//Object[] objects = {"abba", "baba", "ab"};
//Object[] objects = {"aaa", "aa", "a"};
//Object[] objects = {(Integer)1, (Integer)2, (Integer)3, (Integer)4};
MultiCombinations mc = new MultiCombinations(objects, 2);
while (mc.hasMoreElements()) {
for (int i = 0; i < mc.nextElement().length; i++) {
System.out.print(mc.nextElement()[i].toString() + " ");
}
System.out.println();
}
// Extra credit:
System.out.println("----------");
System.out.println("The ways to choose 3 items from 10 with replacement = " + MultiCombinations.c(10, 3));
} // constructor
public static void main(String[] args) throws CombinatoricException {
new MultiCombinationsTester();
}
} // class
MultiCombinations.java
import com.objectwave.utility.*;
import java.util.*;
public class MultiCombinations {
private HashSet<String> set = new HashSet<String>();
private Combinations comb = null;
private Object[] nextElem = null;
public MultiCombinations(Object[] objects, int k) throws CombinatoricException {
k = Math.max(0, k);
Object[] myObjects = new Object[objects.length * k];
for (int i = 0; i < objects.length; i++) {
for (int j = 0; j < k; j++) {
myObjects[i * k + j] = objects[i];
}
}
comb = new Combinations(myObjects, k);
} // constructor
boolean hasMoreElements() {
boolean ret = false;
nextElem = null;
int oldCount = set.size();
while (comb.hasMoreElements()) {
Object[] elem = (Object[]) comb.nextElement();
String str = "";
for (int i = 0; i < elem.length; i++) {
str += ("%" + elem[i].toString() + "~");
}
set.add(str);
if (set.size() > oldCount) {
nextElem = elem;
ret = true;
break;
}
}
return ret;
} // hasMoreElements()
Object[] nextElement() {
return nextElem;
}
static java.math.BigInteger c(int n, int k) throws CombinatoricException {
return Combinatoric.c(n + k - 1, k);
}
} // class
Output:
iced iced iced jam iced plain jam jam jam plain plain plain ---------- The ways to choose 3 items from 10 with replacement = 220
[edit] JavaScript
<html><head><title>Donuts</title></head>output
<body><pre id='x'></pre><script type="application/javascript">
function disp(x) {
var e = document.createTextNode(x + '\n');
document.getElementById('x').appendChild(e);
}
function pick(n, got, pos, from, show) {
var cnt = 0;
if (got.length == n) {
if (show) disp(got.join(' '));
return 1;
}
for (var i = pos; i < from.length; i++) {
got.push(from[i]);
cnt += pick(n, got, i, from, show);
got.pop();
}
return cnt;
}
disp(pick(2, [], 0, ["iced", "jam", "plain"], true) + " combos");
disp("pick 3 out of 10: " + pick(3, [], 0, "a123456789".split(''), false) + " combos");
</script></body></html>
iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
6 combos
pick 3 out of 10: 220 combos
[edit] Mathematica
DeleteDuplicates[Tuples[{"iced", "jam", "plain"}, 2],Sort[#1] == Sort[#2] &]
->{{"iced", "iced"}, {"iced", "jam"}, {"iced", "plain"}, {"jam", "jam"}, {"jam", "plain"}, {"plain", "plain"}}
Combi[x_, y_] := Binomial[(x + y) - 1, y]
Combi[3, 2]
-> 6
Combi[10, 3]
->220
[edit] Mercury
comb.choose uses a nondeterministic list.member/2 to pick values from the list, and just puts them into a bag (a multiset). comb.choose_all gathers all of the possible bags that comb.choose would produce for a given list and number of picked values, and turns them into lists (for readability).
comb.count_choices shows off solutions.aggregate (which allows you to fold over solutions as they're found) rather than list.length and the factorial function.
:- module comb.
:- interface.
:- import_module list, int, bag.
:- pred choose(list(T)::in, int::in, bag(T)::out) is nondet.
:- pred choose_all(list(T)::in, int::in, list(list(T))::out) is det.
:- pred count_choices(list(T)::in, int::in, int::out) is det.
:- implementation.
:- import_module solutions.
choose(L, N, R) :- choose(L, N, bag.init, R).
:- pred choose(list(T)::in, int::in, bag(T)::in, bag(T)::out) is nondet.
choose(L, N, !R) :-
( N = 0 ->
true
;
member(X, L),
bag.insert(!.R, X, !:R),
choose(L, N - 1, !R)
).
choose_all(L, N, R) :-
solutions(choose(L, N), R0),
list.map(bag.to_list, R0, R).
count_choices(L, N, Count) :-
aggregate(choose(L, N), count, 0, Count).
:- pred count(T::in, int::in, int::out) is det.
count(_, N0, N) :- N0 + 1 = N.
Usage:
:- module comb_ex.
:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module comb, list, string.
:- type doughtnuts
---> iced ; jam ; plain
; glazed ; chocolate ; cream_filled ; mystery
; cubed ; cream_covered ; explosive.
main(!IO) :-
choose_all([iced, jam, plain], 2, L),
count_choices([iced, jam, plain, glazed, chocolate, cream_filled,
mystery, cubed, cream_covered, explosive], 3, N),
io.write(L, !IO), io.nl(!IO),
io.write_string(from_int(N) ++ " choices.\n", !IO).
Output:
[[iced, iced], [jam, jam], [plain, plain], [iced, jam], [iced, plain], [jam, plain]] 220 choices.
[edit] OCaml
let rec combs_with_rep k xxs =
match k, xxs with
| 0, _ -> [[]]
| _, [] -> []
| k, x::xs ->
List.map (fun ys -> x::ys) (combs_with_rep (k-1) xxs)
@ combs_with_rep k xs
in the interactive loop:
# combs_with_rep 2 ["iced"; "jam"; "plain"] ;;
- : string list list =
[["iced"; "iced"]; ["iced"; "jam"]; ["iced"; "plain"]; ["jam"; "jam"];
["jam"; "plain"]; ["plain"; "plain"]]
# List.length (combs_with_rep 3 [1;2;3;4;5;6;7;8;9;10]) ;;
- : int = 220
[edit] PARI/GP
ways(k,v,s=[])={
if(k==0,return([]));
if(k==1,return(vector(#v,i,concat(s,[v[i]]))));
if(#v==1,return(ways(k-1,v,concat(s,v))));
my(u=vecextract(v,2^#v-2));
concat(ways(k-1,v,concat(s,[v[1]])),ways(k,u,s))
};
xc(k,v)=binomial(#v+k-1,k);
ways(2, ["iced","jam","plain"])
[edit] Perl
The highly readable version:
sub p { $_[0] ? map p($_[0] - 1, [@{$_[1]}, $_[$_]], @_[$_ .. $#_]), 2 .. $#_ : $_[1] }
sub f { $_[0] ? $_[0] * f($_[0] - 1) : 1 }
sub pn{ f($_[0] + $_[1] - 1) / f($_[0]) / f($_[1] - 1) }
for (p(2, [], qw(iced jam plain))) {
print "@$_\n";
}
printf "\nThere are %d ways to pick 7 out of 10\n", pn(7,10);
Prints:
iced iced iced jam iced plain jam jam jam plain plain plain There are 11440 ways to pick 7 out of 10
[edit] Perl 6
proto combs_with_rep (Int, @) {*}Output:
multi combs_with_rep (0, @) { [] }
multi combs_with_rep ($, []) { () }
multi combs_with_rep ($n, [$head, *@tail]) {
map( { [$head, @^others] },
combs_with_rep($n - 1, [$head, @tail]) ),
combs_with_rep($n, @tail);
}
.perl.say for combs_with_rep( 2, [< iced jam plain >] );
# Extra credit:
sub postfix:<!> { [*] 1..$^n }
sub combs_with_rep_count ($k, $n) { ($n + $k - 1)! / $k! / ($n - 1)! }
say combs_with_rep_count( 3, 10 );
["iced", "iced"] ["iced", "jam"] ["iced", "plain"] ["jam", "jam"] ["jam", "plain"] ["plain", "plain"] 220
[edit] PHP
<?php
function combos($arr, $k) {
if ($k == 0) {
return array(array());
}
if (count($arr) == 0) {
return array();
}
$head = $arr[0];
$combos = array();
$subcombos = combos($arr, $k-1);
foreach ($subcombos as $subcombo) {
array_unshift($subcombo, $head);
$combos[] = $subcombo;
}
array_shift($arr);
$combos = array_merge($combos, combos($arr, $k));
return $combos;
}
$arr = array("iced", "jam", "plain");
$result = combos($arr, 2);
foreach($result as $combo) {
echo implode(' ', $combo), "<br>";
}
$donuts = range(1, 10);
$num_donut_combos = count(combos($donuts, 3));
echo "$num_donut_combos ways to order 3 donuts given 10 types";
?>
output in the browser:
iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
220 ways to order 3 donuts given 10 types
[edit] PicoLisp
(de combrep (N Lst)
(cond
((=0 N) '(NIL))
((not Lst))
(T
(conc
(mapcar
'((X) (cons (car Lst) X))
(combrep (dec N) Lst) )
(combrep N (cdr Lst)) ) ) ) )
Output:
: (combrep 2 '(iced jam plain)) -> ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) : (length (combrep 3 (range 1 10))) -> 220
[edit] PureBasic
Procedure nextCombination(Array combIndex(1), elementCount)The nextCombination() procedure operates on an array of indexes to produce the next combination. This generalization allows producing a combination from any collection of elements. nextCombination() returns the value #False when the indexes have reach their maximum values and are then reset.
;combIndex() must be dimensioned to 'k' - 1, elementCount equals 'n' - 1
;combination produced includes repetition of elements and is represented by the array combIndex()
Protected i, indexValue, combSize = ArraySize(combIndex()), curIndex
;update indexes
curIndex = combSize
Repeat
combIndex(curIndex) + 1
If combIndex(curIndex) > elementCount
curIndex - 1
If curIndex < 0
For i = 0 To combSize
combIndex(i) = 0
Next
ProcedureReturn #False ;array reset to first combination
EndIf
ElseIf curIndex < combSize
indexValue = combIndex(curIndex)
Repeat
curIndex + 1
combIndex(curIndex) = indexValue
Until curIndex = combSize
EndIf
Until curIndex = combSize
ProcedureReturn #True ;array contains next combination
EndProcedure
Procedure.s display(Array combIndex(1), Array dougnut.s(1))
Protected i, elementCount = ArraySize(combIndex()), output.s = " "
For i = 0 To elementCount
output + dougnut(combIndex(i)) + " + "
Next
ProcedureReturn Left(output, Len(output) - 3)
EndProcedure
DataSection
Data.s "iced", "jam", "plain"
EndDataSection
If OpenConsole()
Define n = 3, k = 2, i, combinationCount
Dim combIndex(k - 1)
Dim dougnut.s(n - 1)
For i = 0 To n - 1: Read.s dougnut(i): Next
PrintN("Combinations of " + Str(k) + " dougnuts taken " + Str(n) + " at a time with repetitions.")
combinationCount = 0
Repeat
PrintN(display(combIndex(), dougnut()))
combinationCount + 1
Until Not nextCombination(combIndex(), n - 1)
PrintN("Total combination count: " + Str(combinationCount))
;extra credit
n = 10: k = 3
Dim combIndex(k - 1)
combinationCount = 0
Repeat: combinationCount + 1: Until Not nextCombination(combIndex(), n - 1)
PrintN(#CRLF$ + "Ways to select " + Str(k) + " items from " + Str(n) + " types: " + Str(combinationCount))
Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
CloseConsole()
EndIf
Sample output:
Combinations of 2 dougnuts taken 3 at a time with repetitions. iced + iced iced + jam iced + plain jam + jam jam + plain plain + plain Total combination count: 6 Ways to select 3 items from 10 types: 220
[edit] Python
>>> from itertools import combinations_with_replacement
>>> n, k = 'iced jam plain'.split(), 2
>>> list(combinations_with_replacement(n,k))
[('iced', 'iced'), ('iced', 'jam'), ('iced', 'plain'), ('jam', 'jam'), ('jam', 'plain'), ('plain', 'plain')]
>>> # Extra credit
>>> len(list(combinations_with_replacement(range(10), 3)))
220
>>>
References:
[edit] REXX
/*REXX program shows a set of combinations with repetitions. */
stores=3
doughnuts=2
/*the following uses a subroutine ($PERMSETS) that */
/*can calculat PermSets, CombSets, or rCombSets,*/
/*depending on the setting of the F variable. */
f='RCOMBSETS'
sep=','
list=$permsets(stores,doughnuts,sep,"iced jam plain")
do j=1 for words(list)
say center(word(list,j),30)
end
/*The task was to use the same function to calculate */
/* the number of ways of choosing three doughnuts */
/* from a choice of ten types of doughnuts. */
/*It would be simplier to use the RCOMB function. */
types=10
doughnuts=3
list=$permsets(types,doughnuts)
say
say 'Number of ways to choose three doughnuts from ten types: ' words(list)
say 'Number of ways to choose 3 doughnuts from a choice of 10:' rcomb(10,3)
exit
/*────────────────────────────────$PERMSETS subroutine──────────────────*/
$permsets: procedure expose f /*gen COMB or PERM sets. */
@abc='abcdefghijklmnopqrstuvwxyz'; @abcu=@abc; upper @abcu
@abcs=@abcu||@abc; @0abcs=123456789||@abcs
parse arg x,y,between,usyms /*take X things Y at a time.*/
/*X can't be > length(@0abcs). */
@.=
$.=
sep=
k=0
@0abcs_=@0abcs||xrange(' ','fe'x)
!=' '
do j=1 for words(usyms) /*build list of symbols. */
_=word(usyms,j) /*get a user symbol. */
if wordpos(_,!)\==0 then iterate
k=k+1
$.k=_ /*append to the sumbol list.*/
!=! _
if length(_)\==1 then sep='-'
end
do j=1 for x while k<x
_=substr(@0abcs_,j,1) /*get a character for symbol*/
if wordpos(_,!)\==0 then iterate
k=k+1
$.k=_ /*append to the sumbol list.*/
!=! _
end
!=
if between=='' then between=sep /*use appropriate seperator.*/
return $permgen(1)
/*────────────────────────────────$PERMGEN subroutine───────────────────*/
$permgen: procedure expose @. $. ! between x y f; parse arg ?
_=left(f,1)
fr=_=='R'
fc=_=='C'|fr
if ?>y then do
c=@.1; _=$.c
do j=2 to y
c=@.j; _=_||between||$.c
end
!=! _
end
else do
e=x
if fc then if \fr & ?==1 then e=x-1
do q=1 for e
do k=1 for ?-1
if \fr then if @.k==q then iterate q
if fc then if q<@.k then iterate q
end
@.?=q
call $permgen(?+1) /*construction permutation recursively*/
end
end
return strip(!)
/*────────────────────────────────RCOMB subroutine──────────────────────*/
rcomb: procedure; arg x,y; return $comb(x+y-1,y)
/*────────────────────────────────$COMB subroutine──────────────────────*/
$comb: procedure; arg x,y; if x=y then return 1; if y>x then return 0;
if x-y<y then y=x-y; _=1; do j=x-y+1 to x; _=_*j; end; return _/$fact(y)
/*────────────────────────────────$FACT subroutine──────────────────────*/
$fact: procedure; parse arg x; !=1; do j=2 to x; !=!*j; end; return !
Output:
iced,iced
iced,jam
iced,plain
jam,jam
jam,plain
plain,plain
Number of ways to choose three doughnuts from ten types: 220
Number of ways to choose 3 doughnuts from a choice of 10: 220
[edit] Ruby
Ruby 1.9.2
possible_doughnuts = ['iced', 'jam', 'plain'].repeated_combination(2)
puts "There are #{possible_doughnuts.count} possible doughnuts:"
possible_doughnuts.each{|doughnut_combi| puts doughnut_combi.join(' and ')}
Output:
There are 6 possible doughnuts: iced and iced iced and jam iced and plain jam and jam jam and plain plain and plain
[edit] Scala
Scala has a combinations method in the standard library.
object CombinationsWithRepetition {
def multi[A](as: List[A], k: Int): List[List[A]] =
(List.fill(k)(as)).flatten.combinations(k).toList
def main(args: Array[String]): Unit = {
val doughnuts = multi(List("iced", "jam", "plain"), 2)
for (combo <- doughnuts) println(combo.mkString(","))
val bonus = multi(List(0,1,2,3,4,5,6,7,8,9), 3).size
println("There are "+bonus+" ways to choose 3 items from 10 choices")
}
}
Output
iced,iced iced,jam iced,plain jam,jam jam,plain plain,plain There are 220 ways to choose 3 items from 10 choices
[edit] Scheme
(define combinations
(lambda (lst k)
(cond ((= k 0) '(()))
((null? lst) '())
(else
(append
(map
(lambda (x)
(cons (car lst) x))
(combinations lst (- k 1)))
(combinations (cdr lst) k))))))
[edit] Tcl
package require Tcl 8.5
proc combrepl {set n {presorted no}} {
if {!$presorted} {
set set [lsort $set]
}
if {[incr n 0] < 1} {
return {}
} elseif {$n < 2} {
return $set
}
# Recursive call
set res [combrepl $set [incr n -1] yes]
set result {}
foreach item $set {
foreach inner $res {
dict set result [lsort [list $item {*}$inner]] {}
}
}
return [dict keys $result]
}
puts [combrepl {iced jam plain} 2]
puts [llength [combrepl {1 2 3 4 5 6 7 8 9 10} 3]]
Output:
{iced iced} {iced jam} {iced plain} {jam jam} {jam plain} {plain plain}
220
[edit] UNIX Shell
$ echo {iced,jam,plain}-{iced,jam,plain} # trick question???
iced-iced iced-jam iced-plain jam-iced jam-jam jam-plain plain-iced plain-jam plain-plain
[edit] Ursala
#import std
#import nat
cwr = ~&s+ -<&*+ ~&K0=>&^|DlS/~& iota # takes a set and a selection size
#cast %gLSnX
main = ^|(~&,length) cwr~~/(<'iced','jam','plain'>,2) ('1234567890',3)
output:
(
{
<'iced','iced'>,
<'iced','jam'>,
<'iced','plain'>,
<'jam','jam'>,
<'jam','plain'>,
<'plain','plain'>},
220)
