# Combinations with repetitions

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, ${\displaystyle S}$ (of cardinality ${\displaystyle n}$), and a size of resulting selection, ${\displaystyle k}$, by reporting the sets of cardinality ${\displaystyle k}$ where each member of those sets is chosen from ${\displaystyle S}$. 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., ${\displaystyle S}$ is ${\displaystyle \{\mathrm {iced} ,\mathrm {jam} ,\mathrm {plain} \}}$, ${\displaystyle |S|=3}$, and ${\displaystyle k=2}$.)
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.
Also note that doughnut can also be spelled donut.

• Write a function/program/routine/.. to generate all the combinations with repetitions of ${\displaystyle n}$ types of things taken ${\displaystyle k}$ 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

The number of samples of size k from n objects.

With   combinations and permutations   generation tasks.

Order Unimportant Order Important
Without replacement ${\displaystyle {\binom {n}{k}}=^{n}\operatorname {C} _{k}={\frac {n(n-1)\ldots (n-k+1)}{k(k-1)\dots 1}}}$ ${\displaystyle ^{n}\operatorname {P} _{k}=n\cdot (n-1)\cdot (n-2)\cdots (n-k+1)}$
With replacement ${\displaystyle {\binom {n+k-1}{k}}=^{n+k-1}\operatorname {C} _{k}={(n+k-1)! \over (n-1)!k!}}$ ${\displaystyle n^{k}}$

## 11l

Translation of: Nim
F combsReps(lst, k)
T Ty = T(lst[0])
I k == 0
R [[Ty]()]
I lst.empty
R [[Ty]]()
R combsReps(lst, k - 1).map(x -> @lst[0] [+] x) [+] combsReps(lst[1..], k)

print(combsReps([‘iced’, ‘jam’, ‘plain’], 2))
print(combsReps(Array(1..10), 3).len)
Output:
[[iced, iced], [iced, jam], [iced, plain], [jam, jam], [jam, plain], [plain, plain]]
220

## 360 Assembly

*        Combinations with repetitions - 16/04/2019
COMBREP  CSECT
USING  COMBREP,R13        base register
B      72(R15)            skip savearea
DC     17F'0'             savearea
SAVE   (14,12)            save previous context
MVC    FLAVORS(9),SET1    flavors=3,draws=2,tell=1
BAL    R14,COMBINE        call combine
MVC    FLAVORS(9),SET2    flavors=10,draws=3,tell=0
BAL    R14,COMBINE        call combine
L      R13,4(0,R13)       restore previous savearea pointer
RETURN (14,12),RC=0       restore registers from calling sav
COMBINE  SR     R9,R9              n=0
MVI    V,X'00'            ~
MVC    V+1(NN*L'V-1),V    v=0
IF   CLI,TELL,EQ,X'01' THEN if tell then
XPRNT  =C'list:',5          print
ENDIF    ,                  endif
LOOP     LA     R6,1               i=1
DO WHILE=(C,R6,LE,DRAWS)    do i=1 to draws
LR     R1,R6                i
SLA    R1,2                 ~
L      R2,V-4(R1)           v(i)
L      R3,FLAVORS           flavors
BCTR   R3,0                 flavors-1
IF    CR,R2,GT,R3 THEN        if v(i)>flavors-1 then
LR     R1,R6                  i
SLA    R1,2                   ~
LA     R1,V(R1)               @v(i+1)
L      R2,0(R1)               v(i+1)
LA     R2,1(R2)               v(i+1)+1
ST     R2,0(R1)               v(i+1)=v(i+1)+1
LR     R7,R6                  j=i
DO WHILE=(C,R7,GE,=A(1))        do j=i to 1 by -1
LR     R1,R6                    i
SLA    R1,2                     ~
L      R2,V(R1)                 v(i+1)
LR     R1,R7                    j
SLA    R1,2                     ~
ST     R2,V-4(R1)               v(j)=v(i+1)
BCTR   R7,0                     j--
ENDDO    ,                      enddo j
ENDIF    ,                    endif
LA     R6,1(R6)             i++
ENDDO    ,                  enddo i
L      R1,DRAWS           draws
LA     R1,1(R1)           draws+1
SLA    R1,2               ~
L      R2,V-4(R1)         v(draws+1)
LTR    R2,R2              if v(draws+1)>0
BP     EXITLOOP           then exit loop
LA     R9,1(R9)           n=n+1
IF   CLI,TELL,EQ,X'01' THEN if tell then
MVC    BUF,=CL60' '         buf=' '
LA     R10,1                ibuf=1
LA     R6,1                 i=1
DO WHILE=(C,R6,LE,DRAWS)      do i=1 to draws
LR     R1,R6                  i
SLA    R1,2                   ~
L      R1,V-4(R1)             v(i)
MH     R1,=AL2(L'ITEMS)       ~
LA     R4,ITEMS(R1)           @items(v(i)+1)
LA     R5,BUF-1               @buf-1
AR     R5,R10                 +ibuf
MVC    0(6,R5),0(R4)          substr(buf,ibuf,6)=items(v(i)+1)
LA     R10,L'ITEMS(R10)       ibuf=ibuf+length(items)
LA     R6,1(R6)               i++
ENDDO    ,                    enddo i
XPRNT  BUF,L'BUF            print buf
ENDIF    ,                    endif
L      R2,V                 v(1)
LA     R2,1(R2)             v(1)+1
ST     R2,V                 v(1)=v(1)+1
B      LOOP               loop
EXITLOOP L      R1,FLAVORS           flavors
XDECO  R1,XDEC              edit flavors
MVC    PG+4(2),XDEC+10      output flavors
L      R1,DRAWS             draws
XDECO  R1,XDEC              edit draws
MVC    PG+7(2),XDEC+10      output draws
XDECO  R9,PG+11             edit & output n
XPRNT  PG,L'PG              print buffer
BR     R14                return
NN       EQU    16
ITEMS    DC     CL6'iced',CL6'jam',CL6'plain'
FLAVORS  DS     F
DRAWS    DS     F
TELL     DS     X
SET1     DC     F'3',F'2',X'01'    flavors=3,draws=2,tell=1
SET2     DC     F'10',F'3',X'00'   flavors=10,draws=3,tell=0
V        DS     (NN)F              v(nn)
BUF      DS     CL60               buf
PG       DC     CL40'cwr(..,..)=............'
XDEC     DS     CL12               temp for xdeco
REGEQU
END    COMBREP
Output:
list:
iced  iced
jam   iced
plain iced
jam   jam
plain jam
plain plain
cwr( 3, 2)=           6
cwr(10, 3)=         220

## Acornsoft Lisp

Translation of: Scheme
(defun samples (k items)
(cond
((zerop k) '(()))
((null items) '())
(t (append
(mapc '(lambda (c) (cons (car items) c))
(samples (sub1 k) items))
(samples k (cdr items))))))

(defun append (a b)
(cond ((null a) b)
(t (cons (car a) (append (cdr a) b)))))

(defun length (list (len . 0))
(map '(lambda (e) (setq len (add1 len)))
list)
len)
Output:
Evaluate : (samples 2 '(iced jam plain))

Value is : ((iced iced) (iced jam) (iced plain) (jam jam)
(jam plain) (plain plain))

Evaluate : (length (samples 3 '(1 2 3 4 5 6 7 8 9 10)))

Value is : 220

## Action!

PROC PrintComb(BYTE ARRAY c BYTE len)
BYTE i,ind

FOR i=0 TO len-1
DO
IF i>0 THEN Put('+) FI
ind=c(i)
IF ind=0 THEN
Print("iced")
ELSEIF ind=1 THEN
Print("jam")
ELSE
Print("plain")
FI
OD
PutE()
RETURN

BYTE FUNC NotDecreasing(BYTE ARRAY c BYTE len)
BYTE i

IF len<2 THEN RETURN (1) FI

FOR i=0 TO len-2
DO
IF c(i)>c(i+1) THEN
RETURN (0)
FI
OD
RETURN (1)

BYTE FUNC NextComb(BYTE ARRAY c BYTE n,k)
INT pos,i

DO
pos=k-1
DO
c(pos)==+1
IF c(pos)<n THEN
EXIT
ELSE
pos==-1
IF pos<0 THEN RETURN (0) FI
FI
FOR i=pos+1 TO k-1
DO
c(i)=c(pos)
OD
OD
UNTIL NotDecreasing(c,k)
OD
RETURN (1)

PROC Comb(BYTE n,k,show)
BYTE ARRAY c(10)
BYTE i,count

IF k>n THEN
Print("Error! k is greater than n.")
Break()
FI

PrintF("Choices of %B from %B:%E",k,n)
FOR i=0 TO k-1
DO
c(i)=0
OD

count=0
DO
count==+1
IF show THEN
PrintF(" %B. ",count)
PrintComb(c,k)
FI
UNTIL NextComb(c,n,k)=0
OD
PrintF("Total choices %B%E%E",count)
RETURN

PROC Main()
Comb(3,2,1)
Comb(10,3,0)
RETURN
Output:
Choices of 2 from 3:
1. iced+iced
2. iced+jam
3. iced+plain
4. jam+jam
5. jam+plain
6. plain+plain
Total choices 6

Choices of 3 from 10:
Total choices 220

Should work for any discrete type: integer, modular, or enumeration.

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));
end loop;
Set_IO.Put (Next);
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

## AppleScript

Translation of: Python
--------------- COMBINATIONS WITH REPETITION -------------

-- combinationsWithRepetition :: Int -> [a] -> [kTuple a]
on combinationsWithRepetition(k, xs)
-- A list of lists, representing
-- sets of cardinality k, with
-- members drawn from xs.

script combinationsBySize
script f
on |λ|(a, x)
script prefix
on |λ|(z)
{x} & z
end |λ|
end script

script go
on |λ|(ys, xs)
xs & map(prefix, ys)
end |λ|
end script

scanl1(go, a)
end |λ|
end script

on |λ|(xs)
foldl(f, {{{}}} & take(k, |repeat|({})), xs)
end |λ|
end script

|Just| of |index|(|λ|(xs) of combinationsBySize, 1 + k)
end combinationsWithRepetition

--------------------------- TEST -------------------------
on run
{length of combinationsWithRepetition(3, enumFromTo(0, 9)), ¬
combinationsWithRepetition(2, {"iced", "jam", "plain"})}
end run

------------------------- GENERIC ------------------------

-- Just :: a -> Maybe a
on Just(x)
{type:"Maybe", Nothing:false, Just:x}
end Just

-- Nothing :: Maybe a
on Nothing()
{type:"Maybe", Nothing:true}
end Nothing

-- enumFromTo :: (Int, Int) -> [Int]
on enumFromTo(m, n)
if m  n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
return lst
else
return {}
end if
end enumFromTo

-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl

-- index (!!) :: [a] -> Int -> Maybe a
-- index (!!) :: Gen [a] -> Int -> Maybe a
-- index (!!) :: String -> Int -> Maybe Char
on |index|(xs, i)
if script is class of xs then
repeat with j from 1 to i
set v to |λ|() of xs
end repeat
if missing value is not v then
Just(v)
else
Nothing()
end if
else
if length of xs < i then
Nothing()
else
Just(item i of xs)
end if
end if
end |index|

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn

-- repeat :: a -> Generator [a]
on |repeat|(x)
script
on |λ|()
return x
end |λ|
end script
end |repeat|

-- scanl :: (b -> a -> b) -> b -> [a] -> [b]
on scanl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return lst
end tell
end scanl

-- scanl1 :: (a -> a -> a) -> [a] -> [a]
on scanl1(f, xs)
if 0 < length of xs then
scanl(f, item 1 of xs, rest of xs)
else
{}
end if
end scanl1

-- take :: Int -> [a] -> [a]
-- take :: Int -> String -> String
on take(n, xs)
set c to class of xs
if list is c then
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
else if string is c then
if 0 < n then
text 1 thru min(n, length of xs) of xs
else
""
end if
else if script is c then
set ys to {}
repeat with i from 1 to n
set v to |λ|() of xs
if missing value is v then
return ys
else
set end of ys to v
end if
end repeat
return ys
else
missing value
end if
end take
Output:
{220, {{"iced", "iced"}, {"jam", "iced"}, {"jam", "jam"}, {"plain", "iced"}, {"plain", "jam"}, {"plain", "plain"}}}

## Arturo

print combine.repeated.by:2 ["iced" "jam" "plain"]

print combine.count.repeated.by:3 @1..10
Output:
[iced iced] [iced jam] [iced plain] [jam jam] [jam plain] [plain plain]
220

## AutoHotkey

;===========================================================
; based on "https://www.geeksforgeeks.org/combinations-with-repetitions/"
;===========================================================
CombinationRepetition(arr, k:=0, Delim:="") {
CombinationRepetitionUtil(arr, k?k:str.count(), Delim, [k+1], result:=[])
return result
} ;===========================================================
CombinationRepetitionUtil(arr, k, Delim, chosen, result , index:=1, start:=1){
line := [], i:=0, 	res := ""
if (index = k+1){
while (++i <= k)
res .= arr[chosen[i]] Delim,	line.push(arr[chosen[i]])
return result.Push(Trim(res, Delim))
}
i:=start
while (i <= arr.count())
chosen[Index]:=i, 	CombinationRepetitionUtil(arr, k, Delim, chosen, result, index+1, i++)
} ;===========================================================

Examples:

result := CombinationRepetition(["iced","jam","plain"], 2, " + ")
for k, v in result
res .=  v "n"
res := trim(res, ",") "n"
MsgBox % result.count() " Combinations with Repetition found:n" res
MsgBox % CombinationRepetition([0,1,2,3,4,5,6,7,8,9], 3).Count()

Outputs:

---------------------------
6 Combinations with Repetition found:
iced + iced
iced + jam
iced + plain
jam + jam
jam + plain
plain + plain
---------------------------
220
---------------------------

## AWK

# syntax: GAWK -f COMBINATIONS_WITH_REPETITIONS.AWK
BEGIN {
n = split("iced,jam,plain",donuts,",")
for (i=1; i<=n; i++) {
for (j=1; j<=n; j++) {
if (donuts[i] < donuts[j]) {
key = sprintf("%s %s",donuts[i],donuts[j])
}
else {
key = sprintf("%s %s",donuts[j],donuts[i])
}
arr[key]++
}
}
cmd = "SORT"
for (i in arr) {
printf("%s\n",i) | cmd
choices++
}
close(cmd)
printf("choices = %d\n",choices)
exit(0)
}

output:

iced iced
iced jam
iced plain
jam jam
jam plain
plain plain
choices = 6

## BASIC

### BASIC256

arraybase 0
print "Enter n comb m.  "
input integer "n:  ", n
input integer "m:  ", m
outstr$= "" dim names$(m)

for i = 0 to m - 1
print "Name for item "; i; ":  ";
input string names$[i] next i call iterate (outstr$, 0, m-1, n-1, names$) end subroutine iterate(curr$, start, stp, depth, names$) for i = start to stp if depth = 0 then print curr$; " "; names$[i] else call iterate (curr$ & " " & names$[i], i, stp, depth-1, names$)
end if
next i
return
end subroutine

### BBC BASIC

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%
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

### IS-BASIC

100 PROGRAM "Combinat.bas"
120 STRING D$(1 TO N)*5 130 FOR I=1 TO N 140 READ D$(I)
150 NEXT
160 FOR I=1 TO N
170   FOR J=I TO N
180     PRINT D$(I);" ";D$(J)
190   NEXT
200 NEXT
210 DATA 3,iced,jam,plain

## Bracmat

This minimalist solution expresses the answer as a sum of products. Bracmat automatically normalises such expressions: terms and factors are sorted alphabetically, products containing a sum as a factor are decomposed in a sum of factors (unless the product is not itself term in a multiterm expression). Like factors are converted to a single factor with an appropriate exponent, so ice^2 is to be understood as ice twice.

( ( choices
=   n things thing result
.   !arg:(?n.?things)
& ( !n:0&1
|   0:?result
& (   !things
:   ?
( %?thing ?:?things
&   !thing*choices$(!n+-1.!things)+!result : ?result & ~ ) | !result ) ) ) & out$(choices$(2.iced jam plain)) & out$(choices$(3.iced jam plain butter marmite tahin fish salad onion grass):?+[?N&!N) ); Output: iced^2+jam^2+plain^2+iced*jam+iced*plain+jam*plain 220 ## C Non recursive solution #include <stdio.h> const char *donuts[] = {"iced", "jam", "plain", "something completely different"}; int pos[] = {0, 0, 0, 0}; void printDonuts(int k) { for (size_t i = 1; i < k + 1; i += 1) // offset: i:1..N, N=k+1 printf("%s\t", donuts[pos[i]]); // str:0..N-1 printf("\n"); } // idea: custom number system with 2s complement like 0b10...0==MIN stop case void combination_with_repetiton(int n, int k) { while (1) { for (int i = k; i > 0; i -= 1) { if (pos[i] > n - 1) // if number spilled over: xx0(n-1)xx { pos[i - 1] += 1; // set xx1(n-1)xx for (int j = i; j <= k; j += 1) pos[j] = pos[j - 1]; // set xx11..1 } } if (pos[0] > 0) // stop condition: 1xxxx break; printDonuts(k); pos[k] += 1; // xxxxN -> xxxxN+1 } } int main() { combination_with_repetiton(3, 2); return 0; } #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; } 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 ## C# Translation of: PHP using System; using System.Collections.Generic; using System.Linq; public static class MultiCombinations { private static void Main() { var set = new List<string> { "iced", "jam", "plain" }; var combinations = GenerateCombinations(set, 2); foreach (var combination in combinations) { string combinationStr = string.Join(" ", combination); Console.WriteLine(combinationStr); } var donuts = Enumerable.Range(1, 10).ToList(); int donutsCombinationsNumber = GenerateCombinations(donuts, 3).Count; Console.WriteLine("{0} ways to order 3 donuts given 10 types", donutsCombinationsNumber); } private static List<List<T>> GenerateCombinations<T>(List<T> combinationList, int k) { var combinations = new List<List<T>>(); if (k == 0) { var emptyCombination = new List<T>(); combinations.Add(emptyCombination); return combinations; } if (combinationList.Count == 0) { return combinations; } T head = combinationList[0]; var copiedCombinationList = new List<T>(combinationList); List<List<T>> subcombinations = GenerateCombinations(copiedCombinationList, k - 1); foreach (var subcombination in subcombinations) { subcombination.Insert(0, head); combinations.Add(subcombination); } combinationList.RemoveAt(0); combinations.AddRange(GenerateCombinations(combinationList, k)); return combinations; } } Output: iced iced iced jam iced plain jam jam jam plain plain plain 220 ways to order 3 donuts given 10 types Recursive version using System; class MultiCombination { static string [] set = { "iced", "jam", "plain" }; static int k = 2, n = set.Length; static string [] buf = new string [k]; static void Main() { rec(0, 0); } static void rec(int ind, int begin) { for (int i = begin; i < n; i++) { buf [ind] = set[i]; if (ind + 1 < k) rec(ind + 1, i); else Console.WriteLine(string.Join(",", buf)); } } } ## C++ Non recursive version. #include <iostream> #include <string> #include <vector> void print_vector(const std::vector<int> &pos, const std::vector<std::string> &str) { for (size_t i = 1; i < pos.size(); ++i) // offset: i:1..N printf("%s\t", str[pos[i]].c_str()); // str: 0..N-1 printf("\n"); } // idea: custom number system with 2s complement like 0b10...0==MIN stop case void combination_with_repetiton(int n, int k, const std::vector<std::string> &str) { std::vector<int> pos(k + 1, 0); while (true) { for (int i = k; i > 0; i -= 1) { if (pos[i] > n - 1) // if number spilled over: xx0(n-1)xx { pos[i - 1] += 1; // set xx1(n-1)xx for (int j = i; j <= k; j += 1) pos[j] = pos[j - 1]; // set xx11..1 } } if (pos[0] > 0) // stop condition: 1xxxx break; print_vector(pos, str); pos[k] += 1; // xxxxN -> xxxxN+1 } } int main() { std::vector<std::string> str{"iced", "jam", "plain"}; combination_with_repetiton(3, 2, str); return 0; } Output: iced iced iced jam iced plain jam jam jam plain plain plain ## Clojure Translation of: Scheme (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))))) Output: user> (combinations '[iced jam plain] 2) ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) ## 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 ## Common Lisp The code below is a modified version of the Clojure solution. (defun combinations (xs k) (let ((x (car xs))) (cond ((null xs) nil) ((= k 1) (mapcar #'list xs)) (t (append (mapcar (lambda (ys) (cons x ys)) (combinations xs (1- k))) (combinations (cdr xs) k)))))) Output: ((:ICED :ICED) (:ICED :JAM) (:ICED :PLAIN) (:JAM :JAM) (:JAM :PLAIN) (:PLAIN :PLAIN)) ## Crystal Translation of: Ruby possible_doughnuts = ["iced", "jam", "plain"].repeated_combinations(2) puts "There are #{possible_doughnuts.size} possible doughnuts:" possible_doughnuts.each{|doughnut_combi| puts doughnut_combi.join(" and ")} # Extra credit possible_doughnuts = (1..10).to_a.repeated_combinations(3) # size returns the size of the enumerator, or nil if it can’t be calculated lazily. puts "", "#{possible_doughnuts.size} ways to order 3 donuts given 10 types." Output: There are 6 possible doughnuts: iced and iced iced and jam iced and plain jam and jam jam and plain plain and plain 220 ways to order 3 donuts given 10 types. ## D Using lexicographic next bit permutation to generate combinations with repetitions. 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 @safe 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 @safe { return combVal.length; } uint[] opIndex(in uint idx) const pure nothrow @safe { return val2set(combVal[idx]); } int opApply(immutable int delegate(in ref uint[]) pure nothrow @safe dg) pure nothrow @safe { foreach (immutable v; combVal) { auto set = val2set(v); if (dg(set)) break; } return 1; } private uint[] val2set(in ulong v) const pure nothrow @safe { // 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 @safe 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() @safe { foreach (const e; combRep(["iced", "jam", "plain"], 2)) writefln("%-(%5s %)", 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 ### Short Recursive Version 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;
}
Output:
[["iced", "iced"], ["iced", "jam"], ["iced", "plain"], ["jam", "jam"], ["jam", "plain"], ["plain", "plain"]]
220

## EasyLang

items$[] = [ "iced" "jam" "plain" ] n = len items$[]
k = 2
len result[] k
n_results = 0
#
proc output . .
n_results += 1
if len items$[] > 0 s$ = ""
for i = 1 to k
s$&= items$[result[i]] & " "
.
print s$. . proc combine pos val . . if pos > k output else for i = val to n result[pos] = i combine pos + 1 i . . . combine 1 1 # n = 10 k = 3 len result[] k items$[] = [ ]
n_results = 0
combine 1 1
print ""
print n_results & " results with 10 donuts"
Output:
iced iced
iced jam
iced plain
jam jam
jam plain
plain plain

220 results with 10 donuts

## EchoLisp

We can use the native combinations/rep function, or use a combinator iterator, or implement the function.

;;
;; native function : combinations/rep in list.lib
;;
(lib 'list)

(combinations/rep '(iced jam plain) 2)
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

;;
;; using a combinator iterator
;;
(lib 'sequences)

(take (combinator/rep '(iced jam plain) 2) 8)
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

;;
;; or, implementing the function
;;

(define (comb/rep nums k)
(cond
[(null? nums) null]
[(<= k 0) null]
[(= k 1) (map list nums)]
[else
(for/fold (acc null) ((anum nums))
(append acc
(for/list ((xs (comb/rep nums (1- k))))
#:continue (< (first xs) anum)
(cons anum xs))))]))

(map (curry list-permute '(iced jam plain)) (comb/rep (iota 3) 2))
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

;;
;; extra credit
;;

(length (combinator/rep (iota 10) 3))
220

## Egison

(define $comb/rep (lambda [$n $xs] (match-all xs (list something) [(loop$i [1 ,n] <join _ (& <cons $a_i _> ...)> _) a]))) (test (comb/rep 2 {"iced" "jam" "plain"})) Output: {[|"iced" "iced"|] [|"iced" "jam"|] [|"jam" "jam"|] [|"iced" "plain"|] [|"jam" "plain"|] [|"plain" "plain"|]} ## Elixir Translation of: Erlang defmodule RC do def comb_rep(0, _), do: [[]] def comb_rep(_, []), do: [] def comb_rep(n, [h|t]=s) do (for l <- comb_rep(n-1, s), do: [h|l]) ++ comb_rep(n, t) end end s = [:iced, :jam, :plain] Enum.each(RC.comb_rep(2, s), fn x -> IO.inspect x end) IO.puts "\nExtra credit: #{length(RC.comb_rep(3, Enum.to_list(1..10)))}" Output: [:iced, :iced] [:iced, :jam] [:iced, :plain] [:jam, :jam] [:jam, :plain] [:plain, :plain] Extra credit: 220 ## 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). Output: 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 ## Factor See the implementation of all-combinations-with-replacement here. Works with: Factor version 0.99 2022-04-03 USING: math.combinatorics prettyprint qw ; qw{ iced jam plain } 2 all-combinations-with-replacement . Output: { { "iced" "iced" } { "iced" "jam" } { "iced" "plain" } { "jam" "jam" } { "jam" "plain" } { "plain" "plain" } } ## 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 Output: iced iced iced jam iced plain jam jam jam plain plain plain Total = 6 ## FreeBASIC sub iterate( byval curr as string, byval start as uinteger,_ byval stp as uinteger, byval depth as uinteger,_ names() as string ) dim as uinteger i for i = start to stp if depth = 0 then print curr + " " + names(i) else iterate curr+" "+names(i), i, stp, depth-1, names() end if next i return end sub dim as uinteger m, n, o, i input "Enter n comb m. ", n, m dim as string outstr = "" dim as string names(0 to m-1) for i = 0 to m - 1 print "Name for item ",+i input names(i) next i iterate outstr, 0, m-1, n-1, names() Output: Enter n comb m. 2,3 Name for item 0 ? Iced Name for item 1 ? Jam Name for item 2 ? Plain Iced Iced Iced Jam Iced Plain Jam Jam Jam Plain Plain Plain ## GAP # Built-in UnorderedTuples(["iced", "jam", "plain"], 2); ## Go ### 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 ### 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 ### 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 ## 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 :: Int -> [a] -> [[a]] combsWithRep 0 _ = [[]] combsWithRep _ [] = [] combsWithRep k xxs@(x:xs) = (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 :: Int -> [a] -> Int
countCombsWithRep k lst = binomial (k - 1 + length lst) k

-- countCombsWithRep k = length . combsWithRep k
main :: IO ()
main = do
print $combsWithRep 2 ["iced", "jam", "plain"] print$ countCombsWithRep 3 [1 .. 10]
Output:
[["iced","iced"],["iced","jam"],["iced","plain"],["jam","jam"],["jam","plain"],["plain","plain"]]
220

### Dynamic Programming

The first solution is inefficient because it repeatedly calculates the same subproblem in different branches of recursion. For example, combsWithRep k (x:xs) involves computing combsWithRep (k-1) (x:xs) and combsWithRep k xs, both of which (separately) compute combsWithRep (k-1) xs. To avoid repeated computation, we can use dynamic programming:

combsWithRep :: Int -> [a] -> [[a]]
combsWithRep k xs = combsBySize xs !! k
where
combsBySize = foldr f ([[]] : repeat [])
f x = scanl1 $(++) . map (x :) main :: IO () main = print$ combsWithRep 2 ["iced", "jam", "plain"]

and another approach, using manual recursion:

--------------- COMBINATIONS WITH REPETITION -------------

combinationsWithRepetition ::
(Eq a) =>
Int ->
[a] ->
[[a]]
combinationsWithRepetition k xs = cmb k []
where
cmb 0 ys = ys
cmb n [] = cmb (pred n) (pure <$> xs) cmb n peers = cmb (pred n) (peers >>= nextLayer) where nextLayer [] = [] nextLayer ys@(h : _) = (: ys) <$> dropWhile (/= h) xs

-------------------------- TESTS -------------------------
main :: IO ()
main = do
print $combinationsWithRepetition 2 ["iced", "jam", "plain"] print$ length $combinationsWithRepetition 3 [0 .. 9] Output: [["iced","iced"],["jam","iced"],["plain","iced"],["jam","jam"],["plain","jam"],["plain","plain"]] 220 ## 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 ## J Cartesian product, the monadic j verb { solves the problem. The rest of the code handles the various data types, order, and quantity to choose, and makes a set from the result. rcomb=: >@~.@:(/:~&.>)@,@{@# < Example use: 2 rcomb ;:'iced jam plain' ┌─────┬─────┐ iced iced ├─────┼─────┤ iced jam ├─────┼─────┤ iced plain ├─────┼─────┤ jam jam ├─────┼─────┤ jam plain ├─────┼─────┤ plainplain └─────┴─────┘ #3 rcomb i.10 NB. # ways to choose 3 items from 10 with repetitions 220 ### J Alternate implementation Considerably faster: require 'stats' rcomb=: (combrep #) { ] This definition of rcomb behaves identically to the previous one, and combrep calculates indices: 2 combrep 3 0 0 0 1 0 2 1 1 1 2 2 2 combrep computes 2 comb 4 (note that 4 = (2 + 3)-1) and then subtracts from each column the minimum value in each column (i. 2). ## 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 ## JavaScript ### ES5 #### Imperative <html><head><title>Donuts</title></head> <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> Output: iced iced iced jam iced plain jam jam jam plain plain plain 6 combos pick 3 out of 10: 220 combos #### Functional (function () { // n -> [a] -> [[a]] function combsWithRep(n, lst) { return n ? ( lst.length ? combsWithRep(n - 1, lst).map(function (t) { return [lst[0]].concat(t); }).concat(combsWithRep(n, lst.slice(1))) : [] ) : [[]]; }; // If needed, we can derive a significantly faster version of // the simple recursive function above by memoizing it // f -> f function memoized(fn) { m = {}; return function (x) { var args = [].slice.call(arguments), strKey = args.join('-'); v = m[strKey]; if ('u' === (typeof v)[0]) m[strKey] = v = fn.apply(null, args); return v; } } // [m..n] function range(m, n) { return Array.apply(null, Array(n - m + 1)).map(function (x, i) { return m + i; }); } return [ combsWithRep(2, ["iced", "jam", "plain"]), // obtaining and applying a memoized version of the function memoized(combsWithRep)(3, range(1, 10)).length ]; })(); Output: [ [["iced", "iced"], ["iced", "jam"], ["iced", "plain"], ["jam", "jam"], ["jam", "plain"], ["plain", "plain"]], 220 ] ### ES6 Translation of: Haskell (() => { 'use strict'; // COMBINATIONS WITH REPETITIONS ------------------------------------------- // combsWithRep :: Int -> [a] -> [[a]] const combsWithRep = (k, xs) => { const comb = (n, ys) => { if (0 === n) return ys; if (isNull(ys)) return comb(n - 1, map(pure, xs)); return comb(n - 1, concatMap(zs => { const h = head(zs); return map(x => [x].concat(zs), dropWhile(x => x !== h, xs)); }, ys)); }; return comb(k, []); }; // GENERIC FUNCTIONS ------------------------------------------------------ // concatMap :: (a -> [b]) -> [a] -> [b] const concatMap = (f, xs) => [].concat.apply([], xs.map(f)); // dropWhile :: (a -> Bool) -> [a] -> [a] const dropWhile = (p, xs) => { let i = 0; for (let lng = xs.length; (i < lng) && p(xs[i]); i++) {} return xs.slice(i); }; // enumFromTo :: Int -> Int -> [Int] const enumFromTo = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i); // head :: [a] -> Maybe a const head = xs => xs.length ? xs[0] : undefined; // isNull :: [a] -> Bool const isNull = xs => (xs instanceof Array) ? xs.length < 1 : undefined; // length :: [a] -> Int const length = xs => xs.length; // map :: (a -> b) -> [a] -> [b] const map = (f, xs) => xs.map(f); // pure :: a -> [a] const pure = x => [x]; // show :: a -> String const show = x => JSON.stringify(x, null, 2); // TEST ------------------------------------------------------------------- return show({ twoFromThree: combsWithRep(2, ['iced', 'jam', 'plain']), threeFromTen: length(combsWithRep(3, enumFromTo(0, 9))) }); })(); Output: { "twoFromThree": [ [ "iced", "iced" ], [ "jam", "iced" ], [ "plain", "iced" ], [ "jam", "jam" ], [ "plain", "jam" ], [ "plain", "plain" ] ], "threeFromTen": 220 } ## jq def pick(n): def pick(n; m): # pick n, from m onwards if n == 0 then [] elif m == length then empty elif n == 1 then (.[m:][] | [.]) else ([.[m]] + pick(n-1; m)), pick(n; m+1) end; pick(n;0) ; The task: "Pick 2:", (["iced", "jam", "plain"] | pick(2)), ([[range(0;10)] | pick(3)] | length) as$n
| "There are \($n) ways to pick 3 objects with replacement from 10." Output:$ jq -n -r -c -f pick.jq
Pick 2:
["iced","iced"]
["iced","jam"]
["iced","plain"]
["jam","jam"]
["jam","plain"]
["plain","plain"]
There are 220 ways to pick 3 objects with replacement from 10.

## Julia

Works with: Julia version 0.6
using Combinatorics

l = ["iced", "jam", "plain"]
println("List: ", l, "\nCombinations:")
for c in with_replacement_combinations(l, 2)
println(c)
end

@show length(with_replacement_combinations(1:10, 3))
Output:
List: String["iced", "jam", "plain"]
Combinations:
String["iced", "iced"]
String["iced", "jam"]
String["iced", "plain"]
String["jam", "jam"]
String["jam", "plain"]
String["plain", "plain"]
length(with_replacement_combinations(1:10, 3)) = 220

## Kotlin

// version 1.0.6

class CombsWithReps<T>(val m: Int, val n: Int, val items: List<T>, val countOnly: Boolean = false) {
private val combination = IntArray(m)
private var count = 0

init {
generate(0)
if (!countOnly) println()
println("There are $count combinations of$n things taken $m at a time, with repetitions") } private fun generate(k: Int) { if (k >= m) { if (!countOnly) { for (i in 0 until m) print("${items[combination[i]]}\t")
println()
}
count++
}
else {
for (j in 0 until n)
if (k == 0 || j >= combination[k - 1]) {
combination[k] = j
generate(k + 1)
}
}
}
}

fun main(args: Array<String>) {
val doughnuts = listOf("iced", "jam", "plain")
CombsWithReps(2, 3, doughnuts)
println()
val generic10 = "0123456789".chunked(1)
CombsWithReps(3, 10, generic10, true)
}
Output:
iced    iced
iced    jam
iced    plain
jam     jam
jam     plain
plain   plain

There are 6 combinations of 3 things taken 2 at a time, with repetitions

There are 220 combinations of 10 things taken 3 at a time, with repetitions

## LFE

Translation of: Erlang

and

Translation of: Clojure

### With List Comprehension

(defun combinations
(('() _)
'())
((coll 1)
(lists:map #'list/1 coll))
(((= (cons head tail) coll) n)
(++ (lc ((<- subcoll (combinations coll (- n 1))))
(combinations tail n))))

### With Map

(defun combinations
(('() _)
'())
((coll 1)
(lists:map #'list/1 coll))
(((= (cons head tail) coll) n)
(++ (lists:map (lambda (subcoll) (cons head subcoll))
(combinations coll (- n 1)))
(combinations tail n))))

Output is the same for both:

> (combinations '(iced jam plain) 2)
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain))

## Lobster

Translation of: C
import std

// set S of length n, choose k

def choose(s, k, f):
let got = map(k): s[0]
let n = s.length

def choosi(n_chosen, at):
var count = 0
if n_chosen == k:
f(got)
return 1
var i = at
while i < n:
got[n_chosen] = s[i]
count += choosi(n_chosen + 1, i)
i += 1
return count
return choosi(0, 0)

let count = choose(["iced", "jam", "plain"], 2): print(_)
print count
let extra = choose(map(10):_, 3): _
print extra
Output:
["iced", "iced"]
["iced", "jam"]
["iced", "plain"]
["jam", "jam"]
["jam", "plain"]
["plain", "plain"]
6
220

## Lua

function GenerateCombinations(tList, nMaxElements, tOutput, nStartIndex, nChosen, tCurrentCombination)
if not nStartIndex then
nStartIndex = 1
end
if not nChosen then
nChosen = 0
end
if not tOutput then
tOutput = {}
end
if not tCurrentCombination then
tCurrentCombination = {}
end

if nChosen == nMaxElements then
-- Must copy the table to avoid all elements referring to a single reference
local tCombination = {}
for k,v in pairs(tCurrentCombination) do
tCombination[k] = v
end

table.insert(tOutput, tCombination)
return
end

local nIndex = 1
for k,v in pairs(tList) do
if nIndex >= nStartIndex then
tCurrentCombination[nChosen + 1] = tList[nIndex]
GenerateCombinations(tList, nMaxElements, tOutput, nIndex, nChosen + 1, tCurrentCombination)
end

nIndex = nIndex + 1
end

end

tDonuts = {"iced", "jam", "plain"}
tCombinations = GenerateCombinations(tDonuts, #tDonuts)
for nCombination,tCombination in ipairs(tCombinations) do
print("Combination " .. tostring(nCombination))
for nIndex,strFlavor in ipairs(tCombination) do
print("+" .. strFlavor)
end
end

## Maple

with(combinat):
chooserep:=(s,k)->choose([seq(op(s),i=1..k)],k):
chooserep({iced,jam,plain},2);
# [[iced, iced], [iced, jam], [iced, plain], [jam, jam], [jam, plain], [plain, plain]]
numbchooserep:=(s,k)->binomial(nops(s)+k-1,k);
numbchooserep({iced,jam,plain},2);
# 6

## Mathematica / Wolfram Language

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).

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

A better method therefore:

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;
]

In[2]:= CombinWithRep[{"iced", "jam", "plain"}, 2]

Out[2]= {{"iced", "iced"}, {"jam", "jam"}, {"plain",
"plain"}, {"iced", "jam"}, {"iced", "plain"}, {"jam", "plain"}}

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.

:- 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.

## Nim

Translation of: D
import sugar, sequtils

proc combsReps[T](lst: seq[T], k: int): seq[seq[T]] =
if k == 0:
@[newSeq[T]()]
elif lst.len == 0:
@[]
else:
lst.combsReps(k - 1).map((x: seq[T]) => lst[0] & x) &
lst[1 .. ^1].combsReps(k)

echo(@["iced", "jam", "plain"].combsReps(2))
echo toSeq(1..10).combsReps(3).len
Output:
@[@[iced, iced], @[iced, jam], @[iced, plain], @[jam, jam], @[jam, plain], @[plain, plain]]
220

## 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

### Dynamic programming

let combs_with_rep m xs =
let arr = Array.make (m+1) [] in
arr.(0) <- [[]];
List.iter (fun x ->
for i = 1 to m do
arr.(i) <- arr.(i) @ List.map (fun xs -> x::xs) arr.(i-1)
done
) xs;
arr.(m)

in the interactive loop:

# combs_with_rep 2 ["iced"; "jam"; "plain"] ;;
- : string list list =
[["iced"; "iced"]; ["jam"; "iced"]; ["jam"; "jam"]; ["plain"; "iced"];
["plain"; "jam"]; ["plain"; "plain"]]

# List.length (combs_with_rep 3 [1;2;3;4;5;6;7;8;9;10]) ;;
- : int = 220

## 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"])

## Pascal

program CombWithRep;
//combinations with repetitions
//Limit = count of elements
//Maxval = value of top , lowest is always 0
//so 0..Maxvalue => maxValue+1 elements
{$IFDEF FPC} // {$R+,O+}
{$MODE DELPHI}{$OPTIMIZATION ON,ALL}{$COPERATORS ON} {$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
SysUtils;//GetTickCount64

var
CombIdx: array of byte;

function InitCombIdx(ElemCount: Int32): pbyte;
begin
setlength(CombIdx, ElemCount + 1);
Fillchar(CombIdx[0], sizeOf(CombIdx[0]) * (ElemCount + 1), #0);
Result := @CombIdx[0];
end;

function NextCombWithRep(pComb: pByte; MaxVal, ElemCount: UInt32): boolean;
var
i, dgt: NativeInt;
begin
i := -1;
repeat
i += 1;
dgt := pComb[i];
if dgt < MaxVal then
break;
until i > ElemCount;
Result := i >= ElemCount;
dgt +=1;
repeat
pComb[i] := dgt;
i -= 1;
until i < 0;
end;

procedure GetDoughnuts(ElemCount: NativeInt);
const
doughnuts: array[0..2] of string = ('iced', 'jam', 'plain');
var
pComb: pByte;
MaxVal, i: Uint32;
begin
if ElemCount < 1 then
EXIT;
MaxVal := High(doughnuts);
writeln('Getting ', ElemCount, ' elements of ', MaxVal + 1, ' different');
pComb := InitCombIdx(ElemCount);
repeat
Write('[');
for i := 0 to ElemCount - 2 do
Write(pComb[i], ',');
Write(pComb[ElemCount - 1], ']');
Write('{');
for i := 0 to ElemCount - 2 do
Write(doughnuts[pComb[i]], ',');
Write(doughnuts[pComb[ElemCount - 1]], '}');
until NextCombWithRep(pComb, MaxVal, ElemCount);
writeln(#10);
end;

procedure Check(ElemCount: Int32; ElemRange: Uint32);
var
pComb: pByte;
T0: int64;
rec_cnt: NativeInt;
begin
T0 := GetTickCount64;
rec_cnt := 0;
ElemRange -= 1;
pComb := InitCombIdx(ElemCount);
repeat
Inc(rec_cnt);
until NextCombWithRep(pComb, ElemRange, ElemCount);
T0 := GetTickCount64 - T0;
writeln('Getting ', ElemCount, ' elements of ', ElemRange + 1, ' different');
writeln(rec_cnt: 10, t0: 10,' ms');
end;

begin
GetDoughnuts(2);
GetDoughnuts(3);
Check(3, 10);
Check(9, 10);
Check(15, 16);
{$IFDEF WINDOWS} readln; {$ENDIF}
end.
Output:
TIO.RUN
Getting 2 elements of 3 different
[0,0]{iced,iced}[1,0]{jam,iced}[2,0]{plain,iced}[1,1]{jam,jam}[2,1]{plain,jam}[2,2]{plain,plain}

Getting 3 elements of 3 different
[0,0,0]{iced,iced,iced}[1,0,0]{jam,iced,iced}[2,0,0]{plain,iced,iced}[1,1,0]{jam,jam,iced}[2,1,0]{plain,jam,iced}[2,2,0]{plain,plain,iced}[1,1,1]{jam,jam,jam}[2,1,1]{plain,jam,jam}[2,2,1]{plain,plain,jam}[2,2,2]{plain,plain,plain}

Getting 3 elements of 10 different
220         0 ms
Getting 9 elements of 10 different
48620         0 ms
Getting 15 elements of 16 different
155117520       735 ms

## 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 "@$_\n"; } printf "\nThere are %d ways to pick 7 out of 10\n", pn(7,10); Prints: iced iced iced jam iced plain jam jam jam plain plain plain There are 11440 ways to pick 7 out of 10 With a module: use Algorithm::Combinatorics qw/combinations_with_repetition/; my$iter = combinations_with_repetition([qw/iced jam plain/], 2);
while (my $p =$iter->next) {
print "@$p\n"; } # Not an efficient way: generates them all in an array! my$count =()= combinations_with_repetition([1..10],7);
print "There are $count ways to pick 7 out of 10\n"; ## Phix with javascript_semantics procedure show_choices(sequence set, integer n, at=1, sequence res={}) if length(res)=n then ?res else for i=at to length(set) do show_choices(set,n,i,append(deep_copy(res),set[i])) end for end if end procedure show_choices({"iced","jam","plain"},2) Output: {"iced","iced"} {"iced","jam"} {"iced","plain"} {"jam","jam"} {"jam","plain"} {"plain","plain"} The second part suggests enough differences (collecting and showing vs only counting) to strike me as ugly and confusing. While I could easily, say, translate the C version, I'd rather forego the extra credit and use a completely different routine: with javascript_semantics function count_choices(integer set_size, n, at=1, taken=0) integer count = 0 if taken=n then return 1 end if taken += 1 for i=at to set_size do count += count_choices(set_size,n,i,taken) end for return count end function ?count_choices(10,3) Output: 220 As of 1.0.2 there is a builtin combinations_with_repetitions() function. Using a string here for simplicity and neater output, but it works with any sequence: ?join(combinations_with_repetitions("ijp",2),',') ?length(combinations_with_repetitions(tagset(10),3)) Output: "ii,ij,ip,jj,jp,pp" 220 ## PHP <?php function combos($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 "$num_donut_combos ways to order 3 donuts given 10 types"; ?> Output: in the browser iced iced iced jam iced plain jam jam jam plain plain plain 220 ways to order 3 donuts given 10 types Non-recursive algorithm to generate all combinations with repetitons. Taken from here: [1] You must set k n variables and fill arrays b and c. <?php //Author Ivan Gavryushin @dcc0$k=3;
$n=5; //set amount of elements as in$n var
$c=array(1,2,3,4,5); //set amount of 1 as in$k var
$b=array(1,1,1);$j=$k-1; //Вывод function printt($b,$k) {$z=0;

while ($z <$k) print $b[$z++].' ';
print '<br>';
}
printt ($b,$k);

while (1) {
//Увеличение на позиции K до N

if (array_search($b[$j]+1,$c)!==false ) {$b[$j]=$b[$j]+1; printt ($b,$k); } if ($b[$k-1]==$n) {
$i=$k-1;
//Просмотр массива справа налево
while ($i >= 0) { //Условие выхода if ($i == 0 && $b[$i] == $n) break 2; //Поиск элемента для увеличения$m=array_search($b[$i]+1,$c); if ($m!==false) {
$c[$m]=$c[$m]-1;
$b[$i]=$b[$i]+1;

$g=$i;
//Сортировка массива B
while ($g !=$k-1) {
array_unshift ($c,$b[$g+1]);$b[$g+1]=$b[$i];$g++;
}
//Удаление повторяющихся значений из C
$c=array_diff($c,$b); printt ($b,$k); array_unshift ($c, $n); break; }$i--;

}

}
}

?>

## 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

## Prolog

combinations_of_length(_,[]).
combinations_of_length([X|T],[X|Combinations]):-
combinations_of_length([X|T],Combinations).
combinations_of_length([_|T],[X|Combinations]):-
combinations_of_length(T,[X|Combinations]).
?- [user].
|: combinations_of_length(_,[]).
|: combinations_of_length([X|T],[X|Combinations]):-
|:     combinations_of_length([X|T],Combinations).
|: combinations_of_length([_|T],[X|Combinations]):-
|:     combinations_of_length(T,[X|Combinations]).
|: ^D% user://1 compiled 0.01 sec, 3 clauses
true.

?- combinations_of_length([0,1,2,4],[L0, L1, L2, L3, L4, L5]).
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = L5, L5 = 0 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 1 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = L4, L4 = 0,
L5 = 4 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = L5, L5 = 1 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 1,
L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 1,
L5 = 4 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = L5, L5 = 2 ;
L0 = L1, L1 = L2, L2 = L3, L3 = 0,
L4 = 2,
.
.
.

## 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(n) + " dougnut types taken " + Str(k) + " 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 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. Output: Combinations of 3 dougnut types taken 2 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 >>> 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: Or, assembling our own combsWithRep, by composition of functional primitives: Translation of: Haskell Works with: Python version 3.7 '''Combinations with repetitions''' from itertools import (accumulate, chain, islice, repeat) from functools import (reduce) # combsWithRep :: Int -> [a] -> [kTuple a] def combsWithRep(k): '''A list of tuples, representing sets of cardinality k, with elements drawn from xs. ''' def f(a, x): def go(ys, xs): return xs + [[x] + y for y in ys] return accumulate(a, go) def combsBySize(xs): return reduce( f, xs, chain( [[[]]], islice(repeat([]), k) ) ) return lambda xs: [ tuple(x) for x in next(islice( combsBySize(xs), k, None )) ] # TEST ---------------------------------------------------- def main(): '''Test the generation of sets of cardinality k with elements drawn from xs. ''' print( combsWithRep(2)(['iced', 'jam', 'plain']) ) print( len(combsWithRep(3)(enumFromTo(0)(9))) ) # GENERIC ------------------------------------------------- # enumFromTo :: (Int, Int) -> [Int] def enumFromTo(m): '''Integer enumeration from m to n.''' return lambda n: list(range(m, 1 + n)) # showLog :: a -> IO String def showLog(*s): '''Arguments printed with intercalated arrows.''' print( ' -> '.join(map(str, s)) ) # MAIN --- if __name__ == '__main__': main() Output: [('iced', 'iced'), ('jam', 'iced'), ('jam', 'jam'), ('plain', 'iced'), ('plain', 'jam'), ('plain', 'plain')] 220 ## Quackery Regarding the extra credit portion of the task, while this is clearly not the most efficient way of computing the number, it does serve to illustrate that, at the very least, the algorithm generates the correct number of choices, so I am content to comply with the specification rather than demonstrate a more efficient method. "plaindrome" is not a typo. It is however, a neologism that appears to have little to no currency outside of The On-Line Encyclopedia of Integer Sequences. ( nextplain generates the next plaindrome in the current base by adding one to a given plaindrome, then replacing each trailing zero with the least significant non-zero digit of the number See: https://oeis.org/search?q=plaindromes 4 base put -1 10 times [ nextplain dup echo sp ] drop base release prints "0 1 2 3 11 12 13 22 23 33" i.e. decimal "0 1 2 3 5 6 7 10 11 15" Right padding the base 4 representations with zeros gives all the combinations with repetitions for selecting two doughnuts in a store selling four types of doughnut, numbered 0, 1, 2, and 3. 00 01 02 03 11 12 13 22 23 33 ) [ 1+ dup 0 = if done 0 swap [ base share /mod dup 0 = while drop dip 1+ again ] swap rot 1+ times [ base share * over + ] nip ] is nextplain ( n --> n ) [ dup base put swap ** 1 - [] swap -1 [ 2dup > while nextplain rot over join unrot again ] base release 2drop ] is kcombnums ( n n --> [ ) [ [] unrot times [ base share /mod rot join swap ] drop ] is ndigits ( n n --> [ ) [ [] unrot witheach [ dip dup peek nested rot swap join swap ] drop ] is [peek] ( [ [ --> [ ) [ dup temp put size dup base put dip dup kcombnums [] unrot witheach [ over ndigits temp share swap [peek] nested rot swap join swap ] temp release base release drop ] is kcombs ( n [ --> [ ) 2$ "jam iced plain" nest$kcombs witheach [ witheach [ echo$ sp ] cr ]
cr
3 10 kcombnums size echo
Output:
jam jam
iced jam
plain jam
iced iced
plain iced
plain plain

220

## R

The idiomatic solution is to just use a library.

library(gtools)
combinations(3, 2, c("iced", "jam", "plain"), set = FALSE, repeats.allowed = TRUE)
nrow(combinations(10, 3, repeats.allowed = TRUE))
Output:
> combinations(3, 2, c("iced", "jam", "plain"), set = FALSE, repeats.allowed = TRUE)
[,1]    [,2]
[1,] "iced"  "iced"
[2,] "iced"  "jam"
[3,] "iced"  "plain"
[4,] "jam"   "jam"
[5,] "jam"   "plain"
[6,] "plain" "plain"
> nrow(combinations(10, 3, repeats.allowed = TRUE))
[1] 220

## 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))))]))

## Raku

(formerly Perl 6)

One could simply generate all permutations, and then remove "duplicates":

Works with: Rakudo version 2016.07
my @S = <iced jam plain>;
my $k = 2; .put for [X](@S xx$k).unique(as => *.sort.cache, with => &[eqv])
Output:
iced iced
iced jam
iced plain
jam jam
jam plain
plain plain

Alternatively, a recursive solution:

proto combs_with_rep (UInt, @) {*}

multi combs_with_rep (0,  @)  { () }
multi combs_with_rep (1,  @a)  { map { $_, }, @a } multi combs_with_rep ($,  []) { () }
multi combs_with_rep ($n, [$head, *@tail]) {
|combs_with_rep($n - 1, ($head, |@tail)).map({ $head, |@_ }), |combs_with_rep($n, @tail);
}

.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! / ($n - 1)! } say combs_with_rep_count( 3, 10 ); Output: (iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain) 220 ## REXX ### version 1 This REXX version uses a type of anonymous recursion. /*REXX pgm displays combination sets with repetitions for X things taken Y at a time*/ call RcombN 3, 2, 'iced jam plain' /*The 1st part of Rosetta Code task. */ call RcombN -10, 3, 'Iced jam plain' /* " 2nd " " " " " */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ RcombN: procedure; parse arg x,y,syms; tell= x>0; x=abs(x); z=x+1 /*X>0? Show combo*/ say copies('─',15) x "doughnut selection taken" y 'at a time:' /*display title. */ do i=1 for words(syms);$.i=word(syms, i)    /*assign symbols.*/
end   /*i*/
@.=1                                                           /*assign default.*/
do #=1;        if tell    then  call show               /*display combos?*/
@.y=@.y + 1;   if @.y==z  then  if .(y-1)  then leave   /* ◄─── recursive*/
end   /*#*/
say copies('═',15)  #  "combinations.";    say;   say          /*display answer.*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
.: procedure expose @. y z;   parse arg ?;     if ?==0  then return 1;            p=@.? +1
if p==z  then return .(? -1);      do j=?  to y;   @.j=p;   end  /*j*/;   return 0
/*──────────────────────────────────────────────────────────────────────────────────────*/
show:   L=;      do c=1  for y;   _=@.c;   L=L $._; end /*c*/; say L; return output: ─────────────── 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 taken 3 at a time: ═══════════════ 220 combinations. ### version 2 recursive (taken from version 1) Reformatted and variable names suitable for OoRexx. /*REXX compute (and show) combination sets for nt things in ns places*/ debug=0 Call time 'R' Call RcombN 3,2,'iced,jam,plain' /* The 1st part of the task */ Call RcombN -10,3,'iced,jam,plain,d,e,f,g,h,i,j' /* 2nd part */ Call RcombN -10,9,'iced,jam,plain,d,e,f,g,h,i,j' /* extra part */ Say time('E') 'seconds' Exit /*-------------------------------------------------------------------*/ Rcombn: Procedure Expose thing. debug Parse Arg nt,ns,thinglist tell=nt>0 nt=abs(nt) Say '------------' nt 'doughnut selection taken' ns 'at a time:' If tell=0 Then Say ' list output suppressed' Do i=1 By 1 While thinglist>'' Parse Var thinglist thing.i ',' thinglist /* assign things. */ End index.=1 Do cmb=1 By 1 If tell Then /* display combinations */ Call show /* show this one */ index.ns=index.ns+1 Call show_index 'A' If index.ns==nt+1 Then If proc(ns-1) Then Leave End Say '------------' cmb 'combinations.' Say Return /*-------------------------------------------------------------------*/ proc: Procedure Expose nt ns thing. index. debug Parse Arg recnt If recnt>0 Then Do p=index.recnt+1 If p=nt+1 Then Return proc(recnt-1) Do i=recnt To ns index.i=p End Call show_index 'C' End Return recnt=0 /*-------------------------------------------------------------------*/ show: Procedure Expose index. thing. ns debug l='' Call show_index 'B----------------------->' Do i=1 To ns j=index.i l=l thing.j End Say l Return show_index: Procedure Expose index. ns debug If debug Then Do Parse Arg tag l=tag Do i=1 To ns l=l index.i End Say l End Return Output: ----------- 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 taken 3 at a time: list output suppressed ------------ 220 combinations. ------------ 10 doughnut selection taken 9 at a time: list output suppressed ------------ 48620 combinations. 0.125000 seconds ### version 3 iterative (transformed from version 1) /*REXX compute (and show) combination sets for nt things in ns places*/ Numeric Digits 20 debug=0 Call time 'R' Call IcombN 3,2,'iced,jam,plain' /* The 1st part of the task */ Call IcombN -10,3,'iced,jam,plain,d,e,f,g,h,i,j' /* 2nd part */ Call IcombN -10,9,'iced,jam,plain,d,e,f,g,h,i,j' /* extra part */ Say time('E') 'seconds' Exit IcombN: Procedure Expose thing. debug Parse Arg nt,ns,thinglist tell=nt>0 nt=abs(nt) Say '------------' nt 'doughnut selection taken' ns 'at a time:' If tell=0 Then Say ' list output suppressed' Do i=1 By 1 While thinglist>'' Parse Var thinglist thing.i ',' thinglist /* assign things. */ End index.=1 cmb=0 Call show i=ns+1 Do While i>1 i=i-1 Do j=1 By 1 While index.i<nt index.i=index.i+1 Call show End i1=i-1 If index.i1<nt Then Do index.i1=index.i1+1 Do ii=i To ns index.ii=index.i1 End Call show i=ns+1 End If index.1=nt Then Leave End Say cmb Return show: Procedure Expose ns index. thing. tell cmb cmb=cmb+1 If tell Then Do l='' Do i=1 To ns j=index.i l=l thing.j End Say l End Return Output: ------------ 3 doughnut selection taken 2 at a time: iced iced iced jam iced plain jam jam jam plain plain plain 6 ------------ 10 doughnut selection taken 3 at a time: list output suppressed 220 ------------ 10 doughnut selection taken 9 at a time: list output suppressed 48620 0.109000 seconds Slightly faster ## Ring # Project : Combinations with repetitions n = 2 k = 3 temp = [] comb = [] num = com(n, k) combinations(n, k) comb = sortfirst(comb, 1) see showarray(comb) + nl func combinations(n, k) while true temp = [] for nr = 1 to k tm = random(n-1) + 1 add(temp, tm) next add(comb, temp) for p = 1 to len(comb) - 1 for q = p + 1 to len(comb) if (comb[p][1] = comb[q][1]) and (comb[p][2] = comb[q][2]) and (comb[p][3] = comb[q][3]) del(comb, p) ok next next if len(comb) = num exit ok end func com(n, k) res = pow(n, k) return res func showarray(vect) svect = "" for nrs = 1 to len(vect) svect = "[" + vect[nrs][1] + " " + vect[nrs][2] + " " + vect[nrs][3] + "]" + nl see svect next Func sortfirst(alist, ind) aList = sort(aList,ind) for n = 1 to len(alist)-1 for m= n + 1 to len(aList) if alist[n][1] = alist[m][1] and alist[m][2] < alist[n][2] temp = alist[n] alist[n] = alist[m] alist[m] = temp ok next next for n = 1 to len(alist)-1 for m= n + 1 to len(aList) if alist[n][1] = alist[m][1] and alist[n][2] = alist[m][2] and alist[m][3] < alist[n][3] temp = alist[n] alist[n] = alist[m] alist[m] = temp ok next next return aList Output: [1 1 1] [1 1 2] [1 2 1] [1 2 2] [2 1 1] [2 1 2] [2 2 1] [2 2 2] ## Ruby Works with: Ruby version 2.0 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 ')} # Extra credit possible_doughnuts = [*1..1000].repeated_combination(30) # size returns the size of the enumerator, or nil if it can’t be calculated lazily. puts "", "#{possible_doughnuts.size} ways to order 30 donuts given 1000 types." Output: There are 6 possible doughnuts: iced and iced iced and jam iced and plain jam and jam jam and plain plain and plain 5799990040867421088231767567302459508715964845043404147200 ways to order 30 donuts given 1000 types. ## Rust // Iterator for the combinations of arr with k elements with repetitions. // Yields the combinations in lexicographical order. struct CombinationsWithRepetitions<'a, T: 'a> { // source array to get combinations from arr: &'a [T], // length of the combinations k: u32, // current counts of each object that represent the next combination counts: Vec<u32>, // whether there are any combinations left remaining: bool, } impl<'a, T> CombinationsWithRepetitions<'a, T> { fn new(arr: &[T], k: u32) -> CombinationsWithRepetitions<T> { let mut counts = vec![0; arr.len()]; counts[arr.len() - 1] = k; CombinationsWithRepetitions { arr, k, counts, remaining: true, } } } impl<'a, T> Iterator for CombinationsWithRepetitions<'a, T> { type Item = Vec<&'a T>; fn next(&mut self) -> Option<Vec<&'a T>> { if !self.remaining { return None; } let mut comb = Vec::new(); for (count, item) in self.counts.iter().zip(self.arr.iter()) { for _ in 0..*count { comb.push(item); } } // this is lexicographically largest, and thus the last combination if self.counts[0] == self.k { self.remaining = false; } else { let n = self.counts.len(); for i in (1..n).rev() { if self.counts[i] > 0 { let original_value = self.counts[i]; self.counts[i - 1] += 1; for j in i..(n - 1) { self.counts[j] = 0; } self.counts[n - 1] = original_value - 1; break; } } } Some(comb) } } fn main() { let collection = vec!["iced", "jam", "plain"]; for comb in CombinationsWithRepetitions::new(&collection, 2) { for item in &comb { print!("{} ", item) } println!() } } Output: plain plain jam plain jam jam iced plain iced jam iced iced ## 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 ## Scheme Translation of: PicoLisp (define (combs-with-rep k lst) (cond ((= k 0) '(())) ((null? lst) '()) (else (append (map (lambda (x) (cons (car lst) x)) (combs-with-rep (- k 1) lst)) (combs-with-rep k (cdr lst)))))) (display (combs-with-rep 2 (list "iced" "jam" "plain"))) (newline) (display (length (combs-with-rep 3 '(1 2 3 4 5 6 7 8 9 10)))) (newline) Output: ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) 220 ### Dynamic programming (define (combs-with-rep m lst) (define arr (make-vector (+ m 1) '())) (vector-set! arr 0 '(())) (for-each (lambda (x) (do ((i 1 (+ i 1))) ((> i m)) (vector-set! arr i (append (vector-ref arr i) (map (lambda (xs) (cons x xs)) (vector-ref arr (- i 1))))) ) ) lst) (vector-ref arr m)) (display (combs-with-rep 2 (list "iced" "jam" "plain"))) (newline) (display (length (combs-with-rep 3 '(1 2 3 4 5 6 7 8 9 10)))) (newline) Output: ((iced iced) (jam iced) (jam jam) (plain iced) (plain jam) (plain plain)) 220 ## Sidef Translation of: Perl func cwr (n, l, a = []) { n>0 ? (^l -> map {|k| __FUNC__(n-1, l.slice(k), [a..., l[k]]) }) : a } cwr(2, %w(iced jam plain)).each {|a| say a.map{ .join(' ') }.join("\n") } Also built-in: %w(iced jam plain).combinations_with_repetition(2, {|*a| say a.join(' ') }) Output: iced iced iced jam iced plain jam jam jam plain plain plain Efficient count of the total number of combinations with repetition: func cwr_count (n, m) { binomial(n + m - 1, m) } printf("\nThere are %s ways to pick 7 out of 10 with repetition\n", cwr_count(10, 7)) Output: There are 11440 ways to pick 7 out of 10 with repetition ## Standard ML Translation of: Haskell 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"]) ; val it = [["iced","iced"],["iced","jam"],["iced","plain"],["jam","jam"], ["jam","plain"],["plain","plain"]] : string list list - length (combs_with_rep (3, [1,2,3,4,5,6,7,8,9,10])) ; val it = 220 : int ### Dynamic programming fun combs_with_rep (m, xs) = let val arr = Array.array (m+1, []) in Array.update (arr, 0, [[]]); app (fn x => Array.modifyi (fn (i, y) => if i = 0 then y else y @ map (fn xs => x::xs) (Array.sub (arr, i-1)) ) arr ) xs; Array.sub (arr, m) end in the interactive loop: - combs_with_rep (2, ["iced", "jam", "plain"]) ; val it = [["iced","iced"],["jam","iced"],["jam","jam"],["plain","iced"], ["plain","jam"],["plain","plain"]] : string list list - length (combs_with_rep (3, [1,2,3,4,5,6,7,8,9,10])) ; val it = 220 : int ## Stata function combrep(v,k) { n = cols(v) a = J(comb(n+k-1,k),k,v[1]) u = J(1,k,1) for (i=2; 1; i++) { for (j=k; j>0; j--) { if (u[j]<n) break } if (j<1) return(a) m = u[j]+1 for (; j<=k; j++) u[j] = m a[i,.] = v[u] } } combrep(("iced","jam","plain"),2) a = combrep(1..10,3) rows(a) Output 1 2 +-----------------+ 1 | iced iced | 2 | iced jam | 3 | iced plain | 4 | jam jam | 5 | jam plain | 6 | plain plain | +-----------------+ 220 ## Swift func combosWithRep<T>(var objects: [T], n: Int) -> [[T]] { if n == 0 { return [[]] } else { var combos = [[T]]() while let element = objects.last { combos.appendContentsOf(combosWithRep(objects, n: n - 1).map{$0 + [element] })
objects.removeLast()
}
return combos
}
}
print(combosWithRep(["iced", "jam", "plain"], n: 2).map {$0.joinWithSeparator(" and ")}.joinWithSeparator("\n")) Output: plain and plain jam and plain iced and plain jam and jam iced and jam iced and iced ## Tcl package require Tcl 8.5 proc combrepl {set n {presorted no}} { if {!$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 ## TXR txr -p "(rcomb '(iced jam plain) 2)" Output: ((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) txr -p "(length-list (rcomb '(0 1 2 3 4 5 6 7 8 9) 3))" Output: 220 ## 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) ## VBScript ' Combinations with repetitions - iterative - VBScript Sub printc(vi,n,vs) Dim i, w For i=0 To n-1 w=w &" "& vs(vi(i)) Next 'i Wscript.Echo w End Sub Sub combine(flavors, draws, xitem, tell) Dim n, i, j ReDim v(draws) If tell Then Wscript.Echo "list of cwr("& flavors &","& draws &") :" Do While True For i=0 To draws-1 If v(i)>flavors-1 Then v(i+1)=v(i+1)+1 For j=i To 0 Step -1 v(j)=v(i+1) Next 'j End If Next 'i If v(draws)>0 Then Exit Do n=n+1 If tell Then Call printc(v, draws, xitem) v(0)=v(0)+1 Loop Wscript.Echo "cwr("& flavors &","& draws &")="&n End Sub items=Array( "iced", "jam", "plain" ) combine 3, 2, items, True combine 10, 3, , False combine 10, 7, , False combine 10, 9, , False Output: list of cwr(3,2) : iced iced jam iced plain iced jam jam plain jam plain plain cwr(3,2)=6 cwr(10,3)=220 cwr(10,7)=11440 cwr(10,9)=48620 ## Wren ### Concise recursive Translation of: Go Produces results in no particular order. var Combrep = Fn.new { |n, lst| if (n == 0 ) return [[]] if (lst.count == 0) return [] var r = Combrep.call(n, lst[1..-1]) for (x in Combrep.call(n-1, lst)) { var y = x.toList y.add(lst[0]) r.add(y) } return r } System.print(Combrep.call(2, ["iced", "jam", "plain"])) System.print(Combrep.call(3, (1..10).toList).count) Output: [[plain, plain], [plain, jam], [jam, jam], [plain, iced], [jam, iced], [iced, iced]] 220 ### Library based Library: Wren-seq Library: Wren-perm Produces results in lexicographic order. import "./seq" for Lst import "./perm" for Comb var a = ["iced", "jam", "plain"] a = Lst.flatten(Lst.zip(a, a)) System.print(Comb.listDistinct(a, 2)) a = (1..10).toList a = Lst.flatten(Lst.zip(Lst.zip(a, a), a)) System.print(Comb.listDistinct(a, 3).count) Output: [[iced, iced], [iced, jam], [iced, plain], [jam, jam], [jam, plain], [plain, plain]] 220 ## 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); ] Output: iced iced iced jam iced plain jam jam jam plain plain plain Combos = 6 Combos = 220 ## zkl Translation of: Clojure fcn combosK(k,seq){ // repeats, seq is finite if (k==1) return(seq); if (not seq) return(T); self.fcn(k-1,seq).apply(T.extend.fp(seq[0])).extend(self.fcn(k,seq[1,*])); } combosK(2,T("iced","jam","plain")).apply("concat",","); combosK(3,T(0,1,2,3,4,5,6,7,8,9)).len(); Output: L("iced,iced","iced,jam","iced,plain","jam,jam","jam,plain","plain,plain") 220 ## ZX Spectrum Basic 10 READ n 20 DIM d$(n,5)
30 FOR i=1 TO n
40 READ d$(i) 50 NEXT i 60 DATA 3,"iced","jam","plain" 70 FOR i=1 TO n 80 FOR j=i TO n 90 PRINT d$(i);" ";d\$(j)
100 NEXT j
110 NEXT i