Combinations with repetitions
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, (of cardinality ), and a size of resulting selection, , by reporting the sets of cardinality where each member of those sets is chosen from . 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., is , , and .)
- 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.
Also note that doughnut can also be spelled donut.
Task description
- Write a function/program/routine/.. to generate all the combinations with repetitions of types of things taken 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:
See Also:
The number of samples of size k from n objects.
With combinations and permutations generation tasks.
Order Unimportant Order Important Without replacement Task: Combinations Task: Permutations With replacement Task: Combinations with repetitions Task: Permutations with repetitions
Ada
Should work for any discrete type: integer, modular, or enumeration.
combinations.adb: <lang Ada>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;</lang>
Output:
ICED+ICED ICED+JAM ICED+PLAIN JAM+JAM JAM+PLAIN PLAIN+PLAIN Total Donuts: 6 Total Tens: 220
BBC BASIC
<lang bbcbasic> DIM list$(2), chosen%(2)
list$() = "iced", "jam", "plain" PRINT "Choices of 2 from 3:" choices% = FNchoose(0, 2, 0, 3, chosen%(), list$()) PRINT "Total choices = " ; choices% PRINT '"Choices of 3 from 10:" choices% = FNchoose(0, 3, 0, 10, chosen%(), nul$()) PRINT "Total choices = " ; choices% END DEF FNchoose(n%, l%, p%, m%, g%(), RETURN n$()) LOCAL i%, c% IF n% = l% THEN IF !^n$() THEN FOR i% = 0 TO n%-1 PRINT " " n$(g%(i%)) ; NEXT PRINT ENDIF = 1 ENDIF FOR i% = p% TO m%-1 g%(n%) = i% c% += FNchoose(n% + 1, l%, i%, m%, g%(), n$()) NEXT = c%</lang>
Output:
Choices of 2 from 3: iced iced iced jam iced plain jam jam jam plain plain plain Total choices = 6 Choices of 3 from 10: Total choices = 220
Clojure
<lang 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)))))
</lang>
Example output:
<lang clojure> user> (combinations '[iced jam plain] 2) ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) </lang>
C
<lang C>#include <stdio.h>
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;
}
</lang>Output:
iced iced iced jam iced plain jam jam jam plain plain plain Were there ten donuts, we'd have had 220 choices of three
Fortran
<lang Fortran> program main integer :: chosen(4) integer :: ssize
character(len=50) :: donuts(4) = [ "iced", "jam", "plain", "something completely different" ]
ssize = choose( chosen, 2, 3 ) write(*,*) "Total = ", ssize
contains
recursive function choose( got, len, maxTypes, nChosen, at ) result ( output ) integer :: got(:) integer :: len integer :: maxTypes integer :: output integer, optional :: nChosen integer, optional :: at
integer :: effNChosen integer :: effAt
integer :: i integer :: counter
effNChosen = 1 if( present(nChosen) ) effNChosen = nChosen
effAt = 1 if( present(at) ) effAt = at
if ( effNChosen == len+1 ) then do i=1,len write(*,"(A10,5X)", advance='no') donuts( got(i)+1 ) end do
write(*,*) ""
output = 1 return end if
counter = 0 do i=effAt,maxTypes got(effNChosen) = i-1 counter = counter + choose( got, len, maxTypes, effNChosen + 1, i ) end do
output = counter return end function choose
end program main </lang> Output:
iced iced iced jam iced plain jam jam jam plain plain plain Total = 6
CoffeeScript
<lang 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" </lang>
output <lang> jam,plain: [ [ 'iced', 'iced' ],
[ 'iced', 'jam' ], [ 'iced', 'plain' ], [ 'jam', 'jam' ], [ 'jam', 'plain' ], [ 'plain', 'plain' ] ]
220 ways to order 3 donuts given 10 types </lang>
D
Using lexicographic next bit permutation to generate combinations with repetitions. <lang d>import std.stdio, std.range;
const struct CombRep {
immutable uint nt, nc; private const ulong[] combVal;
this(in uint numType, in uint numChoice) pure nothrow in { assert(0 < numType && numType + numChoice <= 64, "Valid only for nt + nc <= 64 (ulong bit size)"); } body { nt = numType; nc = numChoice; if (nc == 0) return; ulong 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;
ulong[] localCombVal; // 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) { localCombVal ~= 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); } this.combVal = localCombVal; }
uint length() @property const pure nothrow { return combVal.length; }
uint[] opIndex(in uint idx) const pure nothrow { return val2set(combVal[idx]); }
int opApply(immutable int delegate(in ref uint[]) dg) { foreach (immutable 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 (immutable 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 (const s; CombRep(types.length, numChoice)) { ElementType!R[] r; foreach (immutable i; s) r ~= types[i]; result ~= r; }
return result;
}
void main() {
foreach (const e; combRep(["iced", "jam", "plain"], 2)) writefln("%-(%5s %)", e); writeln("Ways to select 3 from 10 types is ", CombRep(10, 3).length);
}</lang>
- Output:
iced iced iced jam iced plain jam jam jam plain plain plain Ways to select 3 from 10 types is 220
Short Recursive Version
<lang d>import std.stdio, std.range, std.algorithm;
T[][] combsRep(T)(T[] lst, in int k) {
if (k == 0) return [[]]; if (lst.empty) return null;
return combsRep(lst, k - 1).map!(L => lst[0] ~ L).array ~ combsRep(lst[1 .. $], k);
}
void main() {
["iced", "jam", "plain"].combsRep(2).writeln; 10.iota.array.combsRep(3).length.writeln;
}</lang>
- Output:
[["iced", "iced"], ["iced", "jam"], ["iced", "plain"], ["jam", "jam"], ["jam", "plain"], ["plain", "plain"]] 220
Egison
<lang egison> > (match-all {"iced" "jam" "plain"} (list something) [<join _ (& <cons $x _> <join _ <cons $y _>>)> [x y]]) {["iced" "iced"] ["iced" "jam"] ["jam" "jam"] ["iced" "plain"] ["jam" "plain"] ["plain" "plain"]} </lang>
Erlang
<lang erlang> -module(comb). -compile(export_all).
comb_rep(0,_) ->
[[]];
comb_rep(_,[]) ->
[];
comb_rep(N,[H|T]=S) ->
[[H|L] || L <- comb_rep(N-1,S)]++comb_rep(N,T).
</lang> Output: <lang erlang> 94> comb:comb_rep(2,[iced,jam,plain]). [[iced,iced],
[iced,jam], [iced,plain], [jam,jam], [jam,plain], [plain,plain]]
95> length(comb:comb_rep(3,lists:seq(1,10))). 220 </lang>
GAP
<lang gap># Built-in UnorderedTuples(["iced", "jam", "plain"], 2);</lang>
Go
Concise recursive
<lang go>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"})))
}</lang> Output:
[[plain plain] [plain jam] [jam jam] [plain iced] [jam iced] [iced iced]] 220
Channel
Using channel and goroutine, showing how to use synced or unsynced communication. <lang go>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) }</lang> Output:
plain plain plain jam plain iced jam jam jam iced iced iced picking 3 of 10: 220
Multiset
This version has proper representation of sets and multisets. <lang go>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)))
}</lang> 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
Haskell
<lang haskell>-- 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
binomial n m = (f n) `div` (f (n - m)) `div` (f m) where f n = if n == 0 then 1 else n * f (n - 1)
countCombsWithRep k lst = binomial (k - 1 + length lst) k -- countCombsWithRep k = length . combsWithRep k
main = do
print $ combsWithRep 2 ["iced","jam","plain"] print $ countCombsWithRep 3 [1..10]</lang>
Example output:
<lang haskell> [["iced","iced"],["iced","jam"],["iced","plain"],["jam","jam"],["jam","plain"],["plain","plain"]] 220 </lang>
Icon and Unicon
Following procedure is a generator, which generates each combination of length n in turn: <lang Icon>
- 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 </lang>
Test procedure:
<lang Icon>
- 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 </lang>
Output:
plain plain jam plain jam jam iced plain iced jam iced iced There are 220 possible combinations of 3 from 10
J
<lang j>rcomb=: >@~.@:(/:~&.>)@,@{@# <</lang>
Example use:
<lang j> 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</lang>
Java
MultiCombinationsTester.java <lang 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 </lang>
MultiCombinations.java <lang 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 </lang>
Output:
iced iced iced jam iced plain jam jam jam plain plain plain ---------- The ways to choose 3 items from 10 with replacement = 220
JavaScript
<lang javascript><html><head><title>Donuts</title></head>
<body>
<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></lang>output<lang>iced iced iced jam iced plain jam jam jam plain plain plain 6 combos pick 3 out of 10: 220 combos</lang>
Mathematica
This method will only work for small set and sample sizes (as it generates all Tuples then filters duplicates - Length[Tuples[Range[10],10]] is already bigger than Mathematica can handle). <lang 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 </lang>
A better method therefore:
<lang Mathematica>CombinWithRep[S_List, k_] := Module[{occupation, assignment},
occupation = Flatten[Permutations /@ IntegerPartitions[k, {Length[S]}, Range[0, k]], 1]; assignment = Flatten[Table[ConstantArray[z, {#z}], {z, Length[#]}]] & /@ occupation; Thread[S#] & /@ assignment ]
In[2]:= CombinWithRep[{"iced", "jam", "plain"}, 2]
Out[2]= {{"iced", "iced"}, {"jam", "jam"}, {"plain",
"plain"}, {"iced", "jam"}, {"iced", "plain"}, {"jam", "plain"}}
</lang>
Which can handle the Length[S] = 10, k=10 situation in still only seconds.
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.
<lang Mercury>:- 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.</lang>
Usage:
<lang Mercury>:- 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).</lang>
Output:
[[iced, iced], [jam, jam], [plain, plain], [iced, jam], [iced, plain], [jam, plain]] 220 choices.
OCaml
<lang 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</lang>
in the interactive loop:
<lang ocaml># 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</lang>
PARI/GP
<lang parigp>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"])</lang>
Perl
The highly readable version: <lang perl>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); </lang>
Prints:
iced iced iced jam iced plain jam jam jam plain plain plain There are 11440 ways to pick 7 out of 10
Perl 6
<lang perl6>proto combs_with_rep (Int, @) {*}
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 );</lang>
Output:
["iced", "iced"] ["iced", "jam"] ["iced", "plain"] ["jam", "jam"] ["jam", "plain"] ["plain", "plain"] 220
PHP
<lang 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), "
"; } $donuts = range(1, 10); $num_donut_combos = count(combos($donuts, 3)); echo "$num_donut_combos ways to order 3 donuts given 10 types";
?></lang> output in the browser: <lang> iced iced iced jam iced plain jam jam jam plain plain plain 220 ways to order 3 donuts given 10 types </lang>
PicoLisp
<lang 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)) ) ) ) )</lang>
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
PureBasic
<lang PureBasic>Procedure nextCombination(Array combIndex(1), elementCount)
;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 </lang>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.
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
Python
<lang 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 >>> </lang>
References:
Racket
<lang racket>
- lang racket
(define (combinations xs k)
(cond [(= k 0) '(())] [(empty? xs) '()] [(append (combinations (rest xs) k) (map (λ(x) (cons (first xs) x)) (combinations xs (- k 1))))]))
</lang>
REXX
<lang rexx>/*REXX program shows combination sets for X things taken Y at a time*/ parse arg x y symbols .; if x== | x==',' then x=3
if y== | y==',' then y=2
if symbols== then symbols='iced jam plain' /*symbol table words.*/
say "────────────" abs(x) 'doughnut selection taken' y "at a time:" say "────────────" RcombN(x,y) 'combinations.'; if x\== then exit # say
x= -10 /*indicate that the combinations aren't to be shown.*/ y= 3
say "────────────" abs(x) 'doughnut selection choose' y "at a time:" say "────────────" RcombN(x,y) 'combinations.' exit /*stick a fork in it, we're done.*/ /*─────────────────────────────────────RCOMBN subroutine────────────────*/ RcombN: procedure expose # symbols; parse arg x 1 ox,y; x=abs(x); base=x !.=1
do #=1; if ox>0 then do; L=; do d=1 for y while ox>0 L=L word(symbols,!.d) end /*d*/ ; say L end !.y=!.y+1; if !.y==base then if .RcombN(y-1) then leave end /*#*/
return # /*─────────────────────────────────────.RCOMBN subroutine───────────────*/ .RcombN: procedure expose !. y base; parse arg d; if d==0 then return 1 p=!.d+1; if p==base then return .RcombN(d-1)
do u=d to y !.u=p end /*u*/
return 0</lang> output using the defaults for input:
──────────── 3 doughnut selection taken 2 at a time: iced iced iced jam iced plain jam jam jam plain plain plain ──────────── 6 combinations. ──────────── 10 doughnut selection choose 3 at a time: ──────────── 220 combinations.
Ruby
Ruby 1.9.2 <lang ruby> 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 ')}
</lang> Output:
There are 6 possible doughnuts: iced and iced iced and jam iced and plain jam and jam jam and plain plain and plain
Scala
Scala has a combinations method in the standard library. <lang scala> 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") }
} </lang>
Output
iced,iced iced,jam iced,plain jam,jam jam,plain plain,plain There are 220 ways to choose 3 items from 10 choices
Scheme
<lang 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))))))
</lang>
Tcl
<lang 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]]</lang> Output:
{iced iced} {iced jam} {iced plain} {jam jam} {jam plain} {plain plain} 220
Ursala
<lang 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)</lang> output:
( { <'iced','iced'>, <'iced','jam'>, <'iced','plain'>, <'jam','jam'>, <'jam','plain'>, <'plain','plain'>}, 220)
XPL0
<lang XPL0>code ChOut=8, CrLf=9, IntOut=11, Text=12; int Count, Array(10);
proc Combos(D, S, K, N, Names); \Generate all size K combinations of N objects int D, S, K, N, Names; \depth of recursion, starting value of N, etc. int I; [if D<K then \depth < size
[for I:= S to N-1 do [Array(D):= I; Combos(D+1, I, K, N, Names); ]; ]
else [Count:= Count+1;
if Names(0) then [for I:= 0 to K-1 do [Text(0, Names(Array(I))); ChOut(0, ^ )]; CrLf(0); ]; ];
];
[Count:= 0; Combos(0, 0, 2, 3, ["iced", "jam", "plain"]); Text(0, "Combos = "); IntOut(0, Count); CrLf(0); Count:= 0; Combos(0, 0, 3, 10, [0]); Text(0, "Combos = "); IntOut(0, Count); CrLf(0); ]</lang>
Output:
iced iced iced jam iced plain jam jam jam plain plain plain Combos = 6 Combos = 220