# Fairshare between two and more

Fairshare between two and more
You are encouraged to solve this task according to the task description, using any language you may know.

The Thue-Morse sequence is a sequence of ones and zeros that if two people take turns in the given order, the first persons turn for every '0' in the sequence, the second for every '1'; then this is shown to give a fairer, more equitable sharing of resources. (Football penalty shoot-outs for example, might not favour the team that goes first as much if the penalty takers take turns according to the Thue-Morse sequence and took 2^n penalties)

The Thue-Morse sequence of ones-and-zeroes can be generated by:

"When counting in binary, the digit sum modulo 2 is the Thue-Morse sequence"

Sharing fairly between two or more

Use this method:

When counting base b, the digit sum modulo b is the Thue-Morse sequence of fairer sharing between b people.

Counting from zero;   using a function/method/routine to express an integer count in base b,
sum the digits modulo b to produce the next member of the Thue-Morse fairshare series for b people.

Show the first 25 terms of the fairshare sequence:

•   For two people:
•   For three people
•   For five people
•   For eleven people

## 11l

Translation of: Python
```F _basechange_int(=num, b)
‘
Return list of ints representing positive num in base b
’
I num == 0
R [0]
[Int] result
L num != 0
(num, V d) = divmod(num, b)
result.append(d)
R reversed(result)

F fairshare(b, n)
[Int] r
L(i) 0..
r [+]= sum(_basechange_int(i, b)) % b
I r.len == n
L.break
R r

L(b) (2, 3, 5, 11)
print(‘#2’.format(b)‘: ’String(fairshare(b, 25))[1 .< (len)-1])```
Output:
``` 2: 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0
3: 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1
5: 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3
11: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4
```

## ALGOL 68

```BEGIN # Use the generalised Thue-Morse sequence to generate Fairshare        #
# sequences for various numbers of people                              #
# returns the digit sum of n in the specified base                       #
PROC digit sum = ( INT n, base )INT:
IF n = 0 THEN 0
ELSE
INT result := 0;
INT v      := ABS n;
WHILE v > 0 DO
result +:= v MOD base;
v   OVERAB base
OD;
result
FI # digit sum # ;
# show the first n terms of the fairshare sequence in the specified base #
PROC show fairshare = ( INT n, base )VOID:
BEGIN
print( ( whole( base, -2 ), ":" ) );
FOR i FROM 0 TO n - 1 DO
print( ( " ", whole( digit sum( i, base ) MOD base, -2 ) ) )
OD;
print( ( newline ) )
END # show fairshare # ;

show fairshare( 25,  2 );
show fairshare( 25,  3 );
show fairshare( 25,  5 );
show fairshare( 25, 11 )
END```
Output:
``` 2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
```

## APL

Works with: Dyalog APL
```fairshare ← ⊣|(+/⊥⍣¯1)
2 3 5 11 ∘.fairshare 0,⍳24
```
Output:
```0 1 1 0 1 0 0 1 1 0  0 1 0 1 1 0 1 0 0 1  0 1 1 0 0
0 1 2 1 2 0 2 0 1 1  2 0 2 0 1 0 1 2 2 0  1 0 1 2 1
0 1 2 3 4 1 2 3 4 0  2 3 4 0 1 3 4 0 1 2  4 0 1 2 3
0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4```

## AppleScript

```-- thueMorse :: Int -> [Int]
on thueMorse(base)
-- Non-finite sequence of Thue-Morse terms for a given base.
fmapGen(baseDigitsSumModBase(base), enumFrom(0))
end thueMorse

-- baseDigitsSumModBase :: Int -> Int -> Int
on baseDigitsSumModBase(b)
script
on |λ|(n)
script go
on |λ|(x)
if 0 < x then
Just(Tuple(x mod b, x div b))
else
Nothing()
end if
end |λ|
end script
sum(unfoldl(go, n)) mod b
end |λ|
end script
end baseDigitsSumModBase

-------------------------- TEST ---------------------------
on run
script rjust
on |λ|(x)
justifyRight(2, space, str(x))
end |λ|
end script

script test
on |λ|(n)
|λ|(n) of rjust & " -> " & ¬
showList(map(rjust, take(25, thueMorse(n))))
end |λ|
end script

unlines({"First 25 fairshare terms for N players:"} & ¬
map(test, {2, 3, 5, 11}))
end run

-------------------- GENERIC FUNCTIONS --------------------

-- Just :: a -> Maybe a
on Just(x)
-- Constructor for an inhabited Maybe (option type) value.
-- Wrapper containing the result of a computation.
{type:"Maybe", Nothing:false, Just:x}
end Just

-- Nothing :: Maybe a
on Nothing()
-- Constructor for an empty Maybe (option type) value.
-- Empty wrapper returned where a computation is not possible.
{type:"Maybe", Nothing:true}
end Nothing

-- Tuple (,) :: a -> b -> (a, b)
on Tuple(a, b)
-- Constructor for a pair of values, possibly of two different types.
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple

-- enumFrom :: Enum a => a -> [a]
on enumFrom(x)
script
property v : missing value
property blnNum : class of x is not text
on |λ|()
if missing value is not v then
if blnNum then
set v to 1 + v
else
set v to succ(v)
end if
else
set v to x
end if
return v
end |λ|
end script
end enumFrom

-- fmapGen <\$> :: (a -> b) -> Gen [a] -> Gen [b]
on fmapGen(f, gen)
script
property g : mReturn(f)
on |λ|()
set v to gen's |λ|()
if v is missing value then
v
else
g's |λ|(v)
end if
end |λ|
end script
end fmapGen

-- 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

-- intercalateS :: String -> [String] -> String
on intercalate(delim, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, delim}
set s to xs as text
set my text item delimiters to dlm
s
end intercalate

-- justifyRight :: Int -> Char -> String -> String
on justifyRight(n, cFiller, s)
if n > length of s then
text -n thru -1 of ((replicate(n, cFiller) as text) & s)
else
s
end if
end justifyRight

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of 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

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

-- Egyptian multiplication - progressively doubling a list, appending
-- stages of doubling to an accumulator where needed for binary
-- assembly of a target length
-- replicate :: Int -> a -> [a]
on replicate(n, a)
set out to {}
if 1 > n then return out
set dbl to {a}

repeat while (1 < n)
if 0 < (n mod 2) then set out to out & dbl
set n to (n div 2)
set dbl to (dbl & dbl)
end repeat
return out & dbl
end replicate

-- showList :: [a] -> String
on showList(xs)
"[" & intercalate(",", map(my str, xs)) & "]"
end showList

-- str :: a -> String
on str(x)
x as string
end str

-- sum :: [Num] -> Num
on sum(xs)
on |λ|(a, b)
a + b
end |λ|
end script

end sum

-- 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

-- > unfoldl (\b -> if b == 0 then Nothing else Just (b, b-1)) 10
-- > [1,2,3,4,5,6,7,8,9,10]
-- unfoldl :: (b -> Maybe (b, a)) -> b -> [a]
on unfoldl(f, v)
set xr to Tuple(v, v) -- (value, remainder)
set xs to {}
tell mReturn(f)
repeat -- Function applied to remainder.
set mb to |λ|(|2| of xr)
if Nothing of mb then
exit repeat
else -- New (value, remainder) tuple,
set xr to Just of mb
-- and value appended to output list.
set xs to ({|1| of xr} & xs)
end if
end repeat
end tell
return xs
end unfoldl

-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
```
Output:
```First 25 fairshare terms for N players:
2 -> [ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
3 -> [ 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
5 -> [ 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
11 -> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 0, 2, 3, 4]```

## Arturo

```thueMorse: function [base, howmany][
i: 0
result: new []
while [howmany > size result][
'result ++ (sum digits.base:base i) % base
i: i + 1
]

return result
]

loop [2 3 5 11] 'b ->
print [
(pad.right "Base "++(to :string b) 7)++" =>"
join.with:" " map to [:string] thueMorse b 25 'x -> pad x 2
]
```
Output:
```Base 2  =>  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base 3  =>  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base 5  =>  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11 =>  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4```

## BASIC

```10 DEFINT A-Z
20 DIM B(4): B(1)=2: B(2)=3: B(3)=5: B(4)=11
30 FOR I=1 TO 4
40 B=B(I)
50 PRINT USING "###:";I;
60 FOR N=0 TO 24: GOSUB 100: PRINT USING "###";S;: NEXT N
70 PRINT
80 NEXT I
90 END
100 REM
101 REM S = N'th item of fairshare sequence for B people
102 REM
110 S=0: Z=N
120 S=S+Z MOD B: Z=Z\B: IF Z>0 THEN 120
130 S=S MOD B
140 RETURN```
Output:
```  1:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
2:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
3:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
4:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4```

## BCPL

```get "libhdr"

let digitsum(n,b) =
n < b -> n,
n rem b + digitsum(n/b, b)

let fairshare(n,b) =
digitsum(n,b) rem b

let start() be
\$(  let bs = table 2, 3, 5, 11
for bi = 0 to 3
\$(  writef("%I2:", bs!bi)
for n = 0 to 24 do writef("%I2 ", fairshare(n, bs!bi))
wrch('*N')
\$)
\$)```
Output:
``` 2: 0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3: 0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5: 0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11: 0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4```

## C

```#include <stdio.h>
#include <stdlib.h>

int turn(int base, int n) {
int sum = 0;
while (n != 0) {
int rem = n % base;
n = n / base;
sum += rem;
}
return sum % base;
}

void fairshare(int base, int count) {
int i;

printf("Base %2d:", base);
for (i = 0; i < count; i++) {
int t = turn(base, i);
printf(" %2d", t);
}
printf("\n");
}

void turnCount(int base, int count) {
int *cnt = calloc(base, sizeof(int));
int i, minTurn, maxTurn, portion;

if (NULL == cnt) {
printf("Failed to allocate space to determine the spread of turns.\n");
return;
}

for (i = 0; i < count; i++) {
int t = turn(base, i);
cnt[t]++;
}

minTurn = INT_MAX;
maxTurn = INT_MIN;
portion = 0;
for (i = 0; i < base; i++) {
if (cnt[i] > 0) {
portion++;
}
if (cnt[i] < minTurn) {
minTurn = cnt[i];
}
if (cnt[i] > maxTurn) {
maxTurn = cnt[i];
}
}

printf("  With %d people: ", base);
if (0 == minTurn) {
printf("Only %d have a turn\n", portion);
} else if (minTurn == maxTurn) {
printf("%d\n", minTurn);
} else {
printf("%d or %d\n", minTurn, maxTurn);
}

free(cnt);
}

int main() {
fairshare(2, 25);
fairshare(3, 25);
fairshare(5, 25);
fairshare(11, 25);

printf("How many times does each get a turn in 50000 iterations?\n");
turnCount(191, 50000);
turnCount(1377, 50000);
turnCount(49999, 50000);
turnCount(50000, 50000);
turnCount(50001, 50000);

return 0;
}
```
Output:
```Base  2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```

## C#

Translation of: Java
```using System;
using System.Collections.Generic;

class FairshareBetweenTwoAndMore
{
static void Main(string[] args)
{
foreach (int baseValue in new List<int> { 2, 3, 5, 11 })
{
Console.WriteLine(\$"Base {baseValue} = {string.Join(", ", ThueMorseSequence(25, baseValue))}");
}
}

private static List<int> ThueMorseSequence(int terms, int baseValue)
{
List<int> sequence = new List<int>();
for (int i = 0; i < terms; i++)
{
int sum = 0;
int n = i;
while (n > 0)
{
// Compute the digit sum
sum += n % baseValue;
n /= baseValue;
}
// Compute the digit sum modulo baseValue.
}
return sequence;
}
}
```
Output:
```Base 2 = 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0
Base 3 = 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1
Base 5 = 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3
Base 11 = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4

```

## C++

Translation of: C
```#include <iostream>
#include <vector>

int turn(int base, int n) {
int sum = 0;
while (n != 0) {
int rem = n % base;
n = n / base;
sum += rem;
}
return sum % base;
}

void fairshare(int base, int count) {
printf("Base %2d:", base);
for (int i = 0; i < count; i++) {
int t = turn(base, i);
printf(" %2d", t);
}
printf("\n");
}

void turnCount(int base, int count) {
std::vector<int> cnt(base, 0);

for (int i = 0; i < count; i++) {
int t = turn(base, i);
cnt[t]++;
}

int minTurn = INT_MAX;
int maxTurn = INT_MIN;
int portion = 0;
for (int i = 0; i < base; i++) {
if (cnt[i] > 0) {
portion++;
}
if (cnt[i] < minTurn) {
minTurn = cnt[i];
}
if (cnt[i] > maxTurn) {
maxTurn = cnt[i];
}
}

printf("  With %d people: ", base);
if (0 == minTurn) {
printf("Only %d have a turn\n", portion);
} else if (minTurn == maxTurn) {
printf("%d\n", minTurn);
} else {
printf("%d or %d\n", minTurn, maxTurn);
}
}

int main() {
fairshare(2, 25);
fairshare(3, 25);
fairshare(5, 25);
fairshare(11, 25);

printf("How many times does each get a turn in 50000 iterations?\n");
turnCount(191, 50000);
turnCount(1377, 50000);
turnCount(49999, 50000);
turnCount(50000, 50000);
turnCount(50001, 50000);

return 0;
}
```
Output:
```Base  2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```

## Cowgol

```include "cowgol.coh";

sub fairshare(index: uint16, base: uint16): (result: uint16) is
result := 0;
while index > 0 loop
result := result + index % base;
index := index / base;
end loop;
result := result % base;
end sub;

sub printSequence(amount: uint16, base: uint16) is
print_i16(base);
print(": ");
var index: uint16 := 0;
while index < amount loop
print_i16(fairshare(index, base));
print(" ");
index := index + 1;
end loop;
print_nl();
end sub;

printSequence(25, 2);
printSequence(25, 3);
printSequence(25, 5);
printSequence(25, 11);```
Output:
```2: 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0
3: 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1
5: 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
11: 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4```

## EasyLang

Translation of: Cowgol
```func fairshare ind base .
while ind > 0
r += ind mod base
ind = ind div base
.
r = r mod base
return r
.
proc sequence n base . .
write base & ": "
for ind range0 n
write (fairshare ind base) & " "
.
print ""
.
sequence 25 2
sequence 25 3
sequence 25 5
sequence 25 11```

## D

Translation of: Kotlin
```import std.array;
import std.stdio;

int turn(int base, int n) {
int sum = 0;
while (n != 0) {
int re = n % base;
n /= base;
sum += re;
}
return sum % base;
}

void fairShare(int base, int count) {
writef("Base %2d:", base);
foreach (i; 0..count) {
auto t = turn(base, i);
writef(" %2d", t);
}
writeln;
}

void turnCount(int base, int count) {
auto cnt = uninitializedArray!(int[])(base);
cnt[] = 0;

foreach (i; 0..count) {
auto t = turn(base, i);
cnt[t]++;
}

auto minTurn = int.max;
auto maxTurn = int.min;
int portion = 0;
foreach (num; cnt) {
if (num > 0) {
portion++;
}
if (num < minTurn) {
minTurn = num;
}
if (maxTurn < num) {
maxTurn = num;
}
}

writef("  With %d people: ", base);
if (minTurn == 0) {
writefln("Only %d have a turn", portion);
} else if (minTurn == maxTurn) {
writeln(minTurn);
} else {
writeln(minTurn," or ", maxTurn);
}
}

void main() {
fairShare(2, 25);
fairShare(3, 25);
fairShare(5, 25);
fairShare(11, 25);

writeln("How many times does each get a turn in 50000 iterations?");
turnCount(191, 50000);
turnCount(1377, 50000);
turnCount(49999, 50000);
turnCount(50000, 50000);
turnCount(50001, 50000);
}
```
Output:
```Base  2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```

## Delphi

Works with: Delphi version 6.0

```procedure DoFairshare(Memo: TMemo; Base: integer);
{Display 25 fairshare sequence items}
var I, N, Sum: integer;
var S: string;
begin
S:=Format('Base - %2d: ',[Base]);
for I:= 0 to 25-1 do
begin
N:= I; Sum:= 0;
while N>0 do
begin
Sum:= Sum + (N mod Base);
N:= N div Base;
end;
S:=S+' '+IntToStr(Sum mod Base);
end;
end;

procedure ShowFairshare(Memo: TMemo);
begin
DoFairshare(Memo,2);
DoFairshare(Memo,3);
DoFairshare(Memo,5);
DoFairshare(Memo,11);
end;
```
Output:
```Base -  2:  0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0
Base -  3:  0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1
Base -  5:  0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
Base - 11:  0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4

Elapsed Time: 4.753 ms.

```

## Draco

```proc fairshare(word n, base) word:
word result;
result := 0;
while n>0 do
result := result + n % base;
n := n / base
od;
result % base
corp

proc main() void:
[4]word bases = (2,3,5,11);
word b, n;
for b from 0 upto 3 do
write(bases[b]:2, ':');
for n from 0 upto 24 do
write(fairshare(n, bases[b]):3)
od;
writeln()
od
corp```
Output:
``` 2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4```

## Factor

Works with: Factor version 0.99 2020-01-23
```USING: formatting kernel math math.parser sequences ;

: nth-fairshare ( n base -- m )
[ >base string>digits sum ] [ mod ] bi ;

: <fairshare> ( n base -- seq )
[ nth-fairshare ] curry { } map-integers ;

{ 2 3 5 11 }
[ 25 over <fairshare> "%2d -> %u\n" printf ] each
```
Output:
``` 2 -> { 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 }
3 -> { 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1 }
5 -> { 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3 }
11 -> { 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4 }
```

## FreeBASIC

Translation of: Visual Basic .NET
```Function Turn(mibase As Integer, n As Integer) As Integer
Dim As Integer sum = 0
While n <> 0
Dim As Integer re = n Mod mibase
n \= mibase
sum += re
Wend
Return sum Mod mibase
End Function

Sub Fairshare(mibase As Integer, count As Integer)
Print Using "mibase ##:"; mibase;
For i As Integer = 1 To count
Dim As Integer t = Turn(mibase, i - 1)
Print Using " ##"; t;
Next i
Print
End Sub

Sub TurnCount(mibase As Integer, count As Integer)
Dim As Integer cnt(mibase), i
For i = 1 To mibase
cnt(i - 1) = 0
Next i

For i = 1 To count
Dim As Integer t = Turn(mibase, i - 1)
cnt(t) += 1
Next i

Dim As Integer minTurn = 4294967295 'MaxValue of uLong
Dim As Integer maxTurn = 0 'MinValue of uLong
Dim As Integer portion = 0
For i As Integer = 1 To mibase
Dim As Integer num = cnt(i - 1)
If num > 0 Then portion += 1
If num < minTurn Then minTurn = num
If num > maxTurn Then maxTurn = num
Next i

Print Using "  With ##### people: "; mibase;
If 0 = minTurn Then
Print Using "Only & have a turn"; portion
Elseif minTurn = maxTurn Then
Print minTurn
Else
Print Using "& or &"; minTurn; maxTurn
End If
End Sub

Fairshare(2, 25)
Fairshare(3, 25)
Fairshare(5, 25)
Fairshare(11, 25)

Print "How many times does each get a turn in 50000 iterations?"
TurnCount(191, 50000)
TurnCount(1377, 50000)
TurnCount(49999, 50000)
TurnCount(50000, 50000)
TurnCount(50001, 50000)
Sleep```
Output:
```Igual que la entrada de Visual Basic .NET.
```

## Go

```package main

import (
"fmt"
"sort"
"strconv"
"strings"
)

func fairshare(n, base int) []int {
res := make([]int, n)
for i := 0; i < n; i++ {
j := i
sum := 0
for j > 0 {
sum += j % base
j /= base
}
res[i] = sum % base
}
return res
}

func turns(n int, fss []int) string {
m := make(map[int]int)
for _, fs := range fss {
m[fs]++
}
m2 := make(map[int]int)
for _, v := range m {
m2[v]++
}
res := []int{}
sum := 0
for k, v := range m2 {
sum += v
res = append(res, k)
}
if sum != n {
return fmt.Sprintf("only %d have a turn", sum)
}
sort.Ints(res)
res2 := make([]string, len(res))
for i := range res {
res2[i] = strconv.Itoa(res[i])
}
return strings.Join(res2, " or ")
}

func main() {
for _, base := range []int{2, 3, 5, 11} {
fmt.Printf("%2d : %2d\n", base, fairshare(25, base))
}
fmt.Println("\nHow many times does each get a turn in 50000 iterations?")
for _, base := range []int{191, 1377, 49999, 50000, 50001} {
t := turns(base, fairshare(50000, base))
fmt.Printf("  With %d people: %s\n", base, t)
}
}
```
Output:
``` 2 : [ 0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0]
3 : [ 0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1]
5 : [ 0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3]
11 : [ 0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4]

How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: only 50000 have a turn
```

```import Data.Bool (bool)
import Data.List (intercalate, unfoldr)
import Data.Tuple (swap)

------------- FAIR SHARE BETWEEN TWO AND MORE ------------

thueMorse :: Int -> [Int]
thueMorse base = baseDigitsSumModBase base <\$> [0 ..]

baseDigitsSumModBase :: Int -> Int -> Int
baseDigitsSumModBase base n =
mod
( sum \$
unfoldr
( bool Nothing
. Just
. swap
. flip quotRem base
<*> (0 <)
)
n
)
base

--------------------------- TEST -------------------------
main :: IO ()
main =
putStrLn \$
fTable
( "First 25 fairshare terms "
<> "for a given number of players:\n"
)
show
( ('[' :) . (<> "]") . intercalate ","
. fmap show
)
(take 25 . thueMorse)
[2, 3, 5, 11]

------------------------- DISPLAY ------------------------
fTable ::
String ->
(a -> String) ->
(b -> String) ->
(a -> b) ->
[a] ->
String
fTable s xShow fxShow f xs =
unlines \$
s :
fmap
( ((<>) . justifyRight w ' ' . xShow)
<*> ((" -> " <>) . fxShow . f)
)
xs
where
w = maximum (length . xShow <\$> xs)

justifyRight :: Int -> Char -> String -> String
justifyRight n c = (drop . length) <*> (replicate n c <>)
```
Output:
```First 25 fairshare terms for a given number of players:

2 -> [0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0]
3 -> [0,1,2,1,2,0,2,0,1,1,2,0,2,0,1,0,1,2,2,0,1,0,1,2,1]
5 -> [0,1,2,3,4,1,2,3,4,0,2,3,4,0,1,3,4,0,1,2,4,0,1,2,3]
11 -> [0,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,0,2,3,4]```

## J

```   fairshare=: [ | [: +/"1 #.inv

2 3 5 11 fairshare"0 1/i.25
0 1 1 0 1 0 0 1 1 0  0 1 0 1 1 0 1 0 0 1  0 1 1 0 0
0 1 2 1 2 0 2 0 1 1  2 0 2 0 1 0 1 2 2 0  1 0 1 2 1
0 1 2 3 4 1 2 3 4 0  2 3 4 0 1 3 4 0 1 2  4 0 1 2 3
0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4

NB. In the 1st 50000 how many turns does each get?  answered:
<@({.,#)/.~"1 ] 2 3 5 11 fairshare"0 1/i.50000
+-------+-------+-------+-------+-------+------+------+------+------+------+-------+
|0 25000|1 25000|       |       |       |      |      |      |      |      |       |
+-------+-------+-------+-------+-------+------+------+------+------+------+-------+
|0 16667|1 16667|2 16666|       |       |      |      |      |      |      |       |
+-------+-------+-------+-------+-------+------+------+------+------+------+-------+
|0 10000|1 10000|2 10000|3 10000|4 10000|      |      |      |      |      |       |
+-------+-------+-------+-------+-------+------+------+------+------+------+-------+
|0 4545 |1 4545 |2 4545 |3 4545 |4 4546 |5 4546|6 4546|7 4546|8 4546|9 4545|10 4545|
+-------+-------+-------+-------+-------+------+------+------+------+------+-------+
```

## Java

```import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class FairshareBetweenTwoAndMore {

public static void main(String[] args) {
for ( int base : Arrays.asList(2, 3, 5, 11) ) {
System.out.printf("Base %d = %s%n", base, thueMorseSequence(25, base));
}
}

private static List<Integer> thueMorseSequence(int terms, int base) {
List<Integer> sequence = new ArrayList<Integer>();
for ( int i = 0 ; i < terms ; i++ ) {
int sum = 0;
int n = i;
while ( n > 0 ) {
//  Compute the digit sum
sum += n % base;
n /= base;
}
//  Compute the digit sum module base.
}
return sequence;
}

}
```
Output:
```Base 2 = [0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
Base 3 = [0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
Base 5 = [0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
Base 11 = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4]
```

## JavaScript

```(() => {
'use strict';

// thueMorse :: Int -> [Int]
const thueMorse = base =>
// Thue-Morse sequence for a given base
fmapGen(baseDigitsSumModBase(base))(
enumFrom(0)
)

// baseDigitsSumModBase :: Int -> Int -> Int
const baseDigitsSumModBase = base =>
// For any integer n, the sum of its digits
// in a given base, modulo that base.
n => sum(unfoldl(
x => 0 < x ? (
Just(quotRem(x)(base))
) : Nothing()
)(n)) % base

// ------------------------TEST------------------------
const main = () =>
console.log(
fTable(
'First 25 fairshare terms for a given number of players:'
)(str)(
xs => '[' + map(
compose(justifyRight(2)(' '), str)
)(xs) + ' ]'
)(
compose(take(25), thueMorse)
)([2, 3, 5, 11])
);

// -----------------GENERIC FUNCTIONS------------------

// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});

// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});

// Tuple (,) :: a -> b -> (a, b)
const Tuple = a => b => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});

// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
x => fs.reduceRight((a, f) => f(a), x);

// enumFrom :: Enum a => a -> [a]
function* enumFrom(x) {
// A non-finite succession of enumerable
// values, starting with the value x.
let v = x;
while (true) {
yield v;
v = 1 + v;
}
}

// fTable :: String -> (a -> String) -> (b -> String)
//                      -> (a -> b) -> [a] -> String
const fTable = s => xShow => fxShow => f => xs => {
// Heading -> x display function ->
//           fx display function ->
//    f -> values -> tabular string
const
ys = xs.map(xShow),
w = Math.max(...ys.map(length));
return s + '\n' + zipWith(
a => b => a.padStart(w, ' ') + ' -> ' + b
)(ys)(
xs.map(x => fxShow(f(x)))
).join('\n');
};

// fmapGen <\$> :: (a -> b) -> Gen [a] -> Gen [b]
const fmapGen = f =>
function*(gen) {
let v = take(1)(gen);
while (0 < v.length) {
yield(f(v[0]))
v = take(1)(gen)
}
};

// justifyRight :: Int -> Char -> String -> String
const justifyRight = n =>
// The string s, preceded by enough padding (with
// the character c) to reach the string length n.
c => s => n > s.length ? (
) : s;

// length :: [a] -> Int
const length = xs =>
// Returns Infinity over objects without finite
// length. This enables zip and zipWith to choose
// the shorter argument when one is non-finite,
// like cycle, repeat etc
(Array.isArray(xs) || 'string' === typeof xs) ? (
xs.length
) : Infinity;

// map :: (a -> b) -> [a] -> [b]
const map = f =>
// The list obtained by applying f to each element of xs.
// (The image of xs under f).
xs => (Array.isArray(xs) ? (
xs
) : xs.split('')).map(f);

// quotRem :: Int -> Int -> (Int, Int)
const quotRem = m => n =>
Tuple(Math.floor(m / n))(
m % n
);

// str :: a -> String
const str = x => x.toString();

// sum :: [Num] -> Num
const sum = xs =>
// The numeric sum of all values in xs.
xs.reduce((a, x) => a + x, 0);

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => 'GeneratorFunction' !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));

// unfoldl :: (b -> Maybe (b, a)) -> b -> [a]
const unfoldl = f => v => {
// Dual to reduce or foldl.
// Where these reduce a list to a summary value, unfoldl
// builds a list from a seed value.
// Where f returns Just(a, b), a is appended to the list,
// and the residual b is used as the argument for the next
// application of f.
// Where f returns Nothing, the completed list is returned.
let
xr = [v, v],
xs = [];
while (true) {
const mb = f(xr[0]);
if (mb.Nothing) {
return xs
} else {
xr = mb.Just;
xs = [xr[1]].concat(xs);
}
}
};

// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
xs => ys => {
const
lng = Math.min(length(xs), length(ys)),
vs = take(lng)(ys);
return take(lng)(xs)
.map((x, i) => f(x)(vs[i]));
};

// MAIN ---
return main();
})();
```
Output:
```First 25 fairshare terms for a given number of players:
2 -> [ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0 ]
3 -> [ 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1 ]
5 -> [ 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3 ]
11 -> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 0, 2, 3, 4 ]```

## jq

jq has a built-in for limiting a generator, but does not have a built-in for generating the integers in an arbitrary base, so we begin by defining `integers(\$base)` for generating integers as arrays beginning with the least-significant digit.

```# Using a "reverse array" representations of the integers base b (b>=2),
# generate an unbounded stream of the integers from [0] onwards.
# E.g. for binary: [0], [1], [0,1], [1,1] ...

def integers(\$base):
[foreach (.[], null) as \$d ({carry: 1};
if \$d then (\$d + .carry ) as \$r
| if \$r >= \$base
then {carry: 1, emit: (\$r - \$base)}
else {carry: 0, emit: \$r }
end
elif (.carry == 0) then .emit = null
else .emit = .carry
end;
select(.emit).emit)];

def fairshare(\$base; \$numberOfTerms):
limit(\$numberOfTerms;
integers(\$base) | add | . % \$base);

(2,3,5,11)
| "Fairshare \((select(.>2)|"among") // "between") \(.) people: \([fairshare(.; 25)])"```
Output:

As for Julia.

## Julia

```fairshare(nplayers,len) = [sum(digits(n, base=nplayers)) % nplayers for n in 0:len-1]

for n in [2, 3, 5, 11]
println("Fairshare ", n > 2 ? "among" : "between", " \$n people: ", fairshare(n, 25))
end
```
Output:
```Fairshare between 2 people: [0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
Fairshare among 3 people: [0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
Fairshare among 5 people: [0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
Fairshare among 11 people: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4]
```

## Kotlin

Translation of: Visual Basic .NET
```fun turn(base: Int, n: Int): Int {
var sum = 0
var n2 = n
while (n2 != 0) {
val re = n2 % base
n2 /= base
sum += re
}
return sum % base
}

fun fairShare(base: Int, count: Int) {
print(String.format("Base %2d:", base))
for (i in 0 until count) {
val t = turn(base, i)
print(String.format(" %2d", t))
}
println()
}

fun turnCount(base: Int, count: Int) {
val cnt = IntArray(base) { 0 }
for (i in 0 until count) {
val t = turn(base, i)
cnt[t]++
}

var minTurn = Int.MAX_VALUE
var maxTurn = Int.MIN_VALUE
var portion = 0
for (i in 0 until base) {
val num = cnt[i]
if (num > 0) {
portion++
}
if (num < minTurn) {
minTurn = num
}
if (num > maxTurn) {
maxTurn = num
}
}

print("  With \$base people: ")
when (minTurn) {
0 -> {
println("Only \$portion have a turn")
}
maxTurn -> {
println(minTurn)
}
else -> {
println("\$minTurn or \$maxTurn")
}
}
}

fun main() {
fairShare(2, 25)
fairShare(3, 25)
fairShare(5, 25)
fairShare(11, 25)

println("How many times does each get a turn in 50000 iterations?")
turnCount(191, 50000)
turnCount(1377, 50000)
turnCount(49999, 50000)
turnCount(50000, 50000)
turnCount(50001, 50000)
}
```
Output:
```Base  2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```

## Lua

Translation of: D
```function turn(base, n)
local sum = 0
while n ~= 0 do
local re = n % base
n = math.floor(n / base)
sum = sum + re
end
return sum % base
end

function fairShare(base, count)
io.write(string.format("Base %2d:", base))
for i=1,count do
local t = turn(base, i - 1)
io.write(string.format(" %2d", t))
end
print()
end

function turnCount(base, count)
local cnt = {}

for i=1,base do
cnt[i - 1] = 0
end

for i=1,count do
local t = turn(base, i - 1)
if cnt[t] ~= nil then
cnt[t] = cnt[t] + 1
else
cnt[t] = 1
end
end

local minTurn = count
local maxTurn = -count
local portion = 0
for _,num in pairs(cnt) do
if num > 0 then
portion = portion + 1
end
if num < minTurn then
minTurn = num
end
if maxTurn < num then
maxTurn = num
end
end

io.write(string.format("  With %d people: ", base))
if minTurn == 0 then
print(string.format("Only %d have a turn", portion))
elseif minTurn == maxTurn then
print(minTurn)
else
print(minTurn .. " or " .. maxTurn)
end
end

function main()
fairShare(2, 25)
fairShare(3, 25)
fairShare(5, 25)
fairShare(11, 25)

print("How many times does each get a turn in 50000 iterations?")
turnCount(191, 50000)
turnCount(1377, 50000)
turnCount(49999, 50000)
turnCount(50000, 50000)
turnCount(50001, 50000)
end

main()
```
Output:
```Base  2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```

## Mathematica / Wolfram Language

```ClearAll[Fairshare]
Fairshare[n_List, b_Integer : 2] := Fairshare[#, b] & /@ n
Fairshare[n_Integer, b_Integer : 2] := Mod[Total[IntegerDigits[n, b]], b]
Fairshare[Range[0, 24], 2]
Fairshare[Range[0, 24], 3]
Fairshare[Range[0, 24], 5]
Fairshare[Range[0, 24], 11]
```
Output:
```{0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0}
{0,1,2,1,2,0,2,0,1,1,2,0,2,0,1,0,1,2,2,0,1,0,1,2,1}
{0,1,2,3,4,1,2,3,4,0,2,3,4,0,1,3,4,0,1,2,4,0,1,2,3}
{0,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,0,2,3,4}```

## Nim

```import math, strutils

#---------------------------------------------------------------------------------------------------

iterator countInBase(b: Positive): seq[Natural] =
## Yield the successive integers in base "b" as a sequence of digits.

var value = @[Natural 0]
yield value

while true:

# Add one to current value.
var c = 1
for i in countdown(value.high, 0):
let val = value[i] + c
if val < b:
value[i] = val
c = 0
else:
value[i] = val - b
c = 1

if c == 1:
# Add a new digit at the beginning.
# In this case, for better performances, we could have add it at the end.
value.insert(c, 0)

yield value

#---------------------------------------------------------------------------------------------------

func thueMorse(b: Positive; count: Natural): seq[Natural] =
## Return the "count" first elements of Thue-Morse sequence for base "b".

var count = count
for n in countInBase(b):
dec count
if count == 0: break

#———————————————————————————————————————————————————————————————————————————————————————————————————

for base in [2, 3, 5, 11]:
echo "Base ", (\$base & ": ").alignLeft(4), thueMorse(base, 25).join(" ")
```
Output:
```Base 2:  0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0
Base 3:  0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1
Base 5:  0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
Base 11: 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4```

## Pascal

### Free Pascal

```program Fairshare;
{\$IFDEF FPC}{\$MODE Delphi}{\$Optimization ON,ALL}{\$ENDIF}
{\$IFDEF WINDOWS}{\$APPTYPE CONSOLE}{\$ENDIF}
uses
sysutils;
const
maxDigitCnt = 32;
type
tdigit  = Uint32;
tDgtSum = record
dgts   : array[0..maxDigitCnt-1] of tdigit;
dgtNum : Uint64;
dgtsum : Uint64;//maxValue maxDigitCnt*(dgtBase-1)
dgtBase,
dgtThue : tDigit;
end;
procedure OutDgtSum(const ds:tDgtSum);
var
i : NativeInt;
Begin
with ds do
Begin
writeln(' base ',dgtBase,' sum of digits :  ',dgtSum,' dec number ',dgtNum);
i := Low(dgts);
repeat
write(dgts[i],'|');
inc(i);
until i > High(dgts);
writeln;
end;
end;

function IncDgtSum(var ds:tDgtSum):boolean;
//add 1 to dgts and corrects sum of Digits
//return false if overflow happens
var
i,base_1 : NativeInt;
Begin
with ds do
begin
i := High(dgts);
base_1 := dgtBase-1;
inc(dgtNum);
repeat
IF dgts[i] < base_1 then
Begin
inc(dgts[i]);
inc(dgtSum);
BREAK;
end
else
Begin
dgts[i] := 0;
dec(dgtSum,base_1);
end;
dec(i);
until i < Low(dgts);
dgtThue := dgtSum MOD (base_1+1);
result := i < Low(dgts)
end;
end;

procedure CheckBase_N_Turns( base:tDigit;turns:NativeUInt);
var
actualNo :tDgtSum;
slots : array of Uint32;
pSlots : pUint32;
i : NativeUInt;
Begin
fillchar(actualNo,SizeOf(actualNo),#0);
setlength(slots,base);
pSlots := @slots[0];
actualNo.dgtBase := Base;
Write(base:3,' [');
while turns>0 do
Begin
inc(pSlots[actualNo.dgtThue],turns);
IncDgtSum(actualNo);
dec(turns);
end;
For i := 0 to Base-1 do
write(pSlots[i],' ');
writeln(']');
end;

procedure SumBase_N_Turns( base:tDigit;turns:NativeUInt);
var
actualNo :tDgtSum;
Begin
fillchar(actualNo,SizeOf(actualNo),#0);
actualNo.dgtBase := Base;
Write(base:3,' [');
while turns>1 do
Begin
write(actualNo.dgtThue,',');
IncDgtSum(actualNo);
dec(turns);
end;
writeln(actualNo.dgtThue,']');
end;

var
turns : NativeInt;
Begin
turns := 25;
SumBase_N_Turns(2,turns);  SumBase_N_Turns(3,turns);
SumBase_N_Turns(5,turns);  SumBase_N_Turns(11,turns);

Writeln;
writeln('Summing up descending numbers from turns downto 0');;
turns := 2*3*5*11;
Writeln(turns,' turns = 2*3*5*11');
CheckBase_N_Turns(2,turns); CheckBase_N_Turns(3,turns);
CheckBase_N_Turns(5,turns); CheckBase_N_Turns(11,turns);

turns := sqr(2)*sqr(3)*sqr(5)*sqr(11);
Writeln(turns,' turns = sqr(2)*sqr(3)*sqr(5)*sqr(11)');
CheckBase_N_Turns(2,turns); CheckBase_N_Turns(3,turns);
CheckBase_N_Turns(5,turns); CheckBase_N_Turns(11,turns);
end.
```
Output:
```  2 [0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0]
3 [0,1,2,1,2,0,2,0,1,1,2,0,2,0,1,0,1,2,2,0,1,0,1,2,1]
5 [0,1,2,3,4,1,2,3,4,0,2,3,4,0,1,3,4,0,1,2,4,0,1,2,3]
11 [0,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,0,2,3,4]

Summing up descending numbers from turns downto 0
330 turns = 2*3*5*11
2 [27307 27308 ]
3 [18206 18204 18205 ]
5 [10925 10924 10923 10922 10921 ]
11 [4961 4953 4956 4959 4962 4965 4968 4971 4974 4977 4969 ]
108900 turns = sqr(2)*sqr(3)*sqr(5)*sqr(11)
2 [2964829725 2964829725]
3 [1976553150 1976553150 1976553150]
5 [1185931890 1185931890 1185931890 1185931890 1185931890]
11 [539059950 539059950 539059950 539059950 539059950 539059950 539059950 539059950 539059950 539059950 539059950]

```

## Perl

Translation of: Raku
```use strict;
use warnings;
use Math::AnyNum qw(sum polymod);

sub fairshare {
my(\$b, \$n) = @_;
sprintf '%3d'x\$n, map { sum ( polymod(\$_, \$b, \$b )) % \$b } 0 .. \$n-1;
}

for (<2 3 5 11>) {
printf "%3s:%s\n", \$_, fairshare(\$_, 25);
}
```
Output:
```  2:  0  1  1  0  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  1  0  0  1  0
3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4```

## Phix

Translation of: Go
```function fairshare(integer n, base)
sequence res = repeat(0,n)
for i=1 to n do
integer j = i-1,
t = 0
while j>0 do
t += remainder(j,base)
j = floor(j/base)
end while
res[i] = remainder(t,base)
end for
return {base,res}
end function

constant tests = {2,3,5,11}
for i=1 to length(tests) do
printf(1,"%2d : %v\n", fairshare(25, tests[i]))
end for
```
Output:
``` 2 : {0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0}
3 : {0,1,2,1,2,0,2,0,1,1,2,0,2,0,1,0,1,2,2,0,1,0,1,2,1}
5 : {0,1,2,3,4,1,2,3,4,0,2,3,4,0,1,3,4,0,1,2,4,0,1,2,3}
11 : {0,1,2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,0,2,3,4}
```

## PL/M

```100H:

BDOS: PROCEDURE (F,A); DECLARE F BYTE, A ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; GO TO 0; END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;

PRINT\$NUMBER: PROCEDURE (N);
DECLARE S (7) BYTE INITIAL ('..... \$');
DECLARE (N, P) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1;
C = N MOD 10 + '0';
N = N / 10;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT\$NUMBER;

SUM = 0;
DO WHILE N>0;
SUM = SUM + N MOD BASE;
N = N / BASE;
END;
RETURN SUM MOD BASE;
END FAIR\$SHARE;

DECLARE BASES (4) BYTE INITIAL (2, 3, 5, 11);
DECLARE (I, N) BYTE;

DO I=0 TO LAST(BASES);
CALL PRINT\$NUMBER(BASES(I));
CALL PRINT(.': \$');
DO N=0 TO 24;
CALL PRINT\$NUMBER(FAIR\$SHARE(N, BASES(I)));
END;
CALL PRINT(.(13,10,'\$'));
END;
CALL EXIT;
EOF```
Output:
```2 : 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0
3 : 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1
5 : 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
11 : 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4```

## Python

### Procedural

```from itertools import count, islice

def _basechange_int(num, b):
"""
Return list of ints representing positive num in base b

>>> b = 3
>>> print(b, [_basechange_int(num, b) for num in range(11)])
3 [[0], [1], [2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [1, 0, 0], [1, 0, 1]]
>>>
"""
if num == 0:
return [0]
result = []
while num != 0:
num, d = divmod(num, b)
result.append(d)
return result[::-1]

def fairshare(b=2):
for i in count():
yield sum(_basechange_int(i, b)) % b

if __name__ == '__main__':
for b in (2, 3, 5, 11):
print(f"{b:>2}: {str(list(islice(fairshare(b), 25)))[1:-1]}")
```
Output:
``` 2: 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0
3: 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1
5: 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3
11: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4```

### Functional

```'''Fairshare between two and more'''

from itertools import count, islice
from functools import reduce

# thueMorse :: Int -> [Int] -> [Int]
def thueMorse(base):
'''Thue-Morse sequence for a given base.'''
return fmapNext(baseDigitsSumModBase(base))(
count(0)
)

# baseDigitsSumModBase :: Int -> Int -> Int
def baseDigitsSumModBase(base):
'''For any integer n, the sum of its digits
in a given base, modulo that base.
'''
def go(n):
return sum(unfoldl(
lambda x: Just(divmod(x, base)) if 0 < x else Nothing()
)(n)) % base

return go

# -------------------------- TEST --------------------------
# main :: IO ()
def main():
'''First 25 fairshare terms for a given number of players'''

print(
fTable(
main.__doc__ + ':\n'
)(repr)(
lambda xs: '[' + ','.join(
[str(x).rjust(2, ' ') for x in xs]
) + ' ]'
)(
compose(take(25), thueMorse)
)([2, 3, 5, 11])
)

# ------------------------ GENERIC -------------------------

# Just :: a -> Maybe a
def Just(x):
'''Constructor for an inhabited Maybe (option type) value.
Wrapper containing the result of a computation.
'''
return {'type': 'Maybe', 'Nothing': False, 'Just': x}

# Nothing :: Maybe a
def Nothing():
'''Constructor for an empty Maybe (option type) value.
Empty wrapper returned where a computation is not possible.
'''
return {'type': 'Maybe', 'Nothing': True}

# compose :: ((a -> a), ...) -> (a -> a)
def compose(*fs):
'''Composition, from right to left,
of a series of functions.
'''
def go(f, g):
def fg(x):
return f(g(x))
return fg
return reduce(go, fs, identity)

# fmapNext <\$> :: (a -> b) -> Iter [a] -> Iter [b]
def fmapNext(f):
'''A function f mapped over a
possibly non-finite iterator.
'''
def go(g):
v = next(g, None)
while None is not v:
yield f(v)
v = next(g, None)
return go

# fTable :: String -> (a -> String) ->
# (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)

# identity :: a -> a
def identity(x):
'''The identity function.'''
return x

# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''
return lambda xs: (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)

# unfoldl(lambda x: Just(((x - 1), x)) if 0 != x else Nothing())(10)
# -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# unfoldl :: (b -> Maybe (b, a)) -> b -> [a]
def unfoldl(f):
'''Dual to reduce or foldl.
Where these reduce a list to a summary value, unfoldl
builds a list from a seed value.
Where f returns Just(a, b), a is appended to the list,
and the residual b is used as the argument for the next
application of f.
When f returns Nothing, the completed list is returned.
'''
def go(v):
x, r = v, v
xs = []
while True:
mb = f(x)
if mb['Nothing']:
return xs
else:
x, r = mb['Just']
xs.insert(0, r)
return xs
return go

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
```First 25 fairshare terms for a given number of players:

2 -> [ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0 ]
3 -> [ 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1 ]
5 -> [ 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3 ]
11 -> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 0, 2, 3, 4 ]```

## Quackery

`digitsum` is defined at Sum digits of an integer.

```  [ dup dip digitsum mod ] is fairshare ( n n --> n )

' [ 2 3 5 11 ]
witheach
[ dup echo say ": "
25 times
[ i^ over fairshare echo sp ]
drop cr ]```
Output:
```2: 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0
3: 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1
5: 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3
11: 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4 ```

## Racket

```#lang racket

(define (Thue-Morse base)
(letrec ((q/r (curryr quotient/remainder base))
(inner (λ (n (s 0))
(match n
[0 (modulo s base)]
[(app q/r q r) (inner q (+ s r))]))))
inner))

(define (report-turns B n)
(printf "Base:\t~a\t~a~%" B (map (Thue-Morse B) (range n))))

(define (report-stats B n)
(define TM (Thue-Morse B))
(define h0 (for/hash ((b B)) (values b 0)))
(define d (for/fold ((h h0)) ((i n)) (hash-update h (TM i) add1 0)))
(define d′ (for/fold ((h (hash))) (([k v] (in-hash d))) (hash-update h v add1 0)))
(define d′′ (hash-map d′ (λ (k v) (format "~a people have ~a turn(s)" v k))))
(printf "Over ~a turns for ~a people:~a~%" n B (string-join d′′ ", ")))

(define (Fairshare-between-two-and-more)
(report-turns 2 25)
(report-turns 3 25)
(report-turns 5 25)
(report-turns 11 25)
(newline)
(report-stats 191 50000)
(report-stats 1377 50000)
(report-stats 49999 50000)
(report-stats 50000 50000)
(report-stats 50001 50000))

(module+ main
(Fairshare-between-two-and-more))
```
Output:
```Base:	2	(0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0)
Base:	3	(0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1)
Base:	5	(0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3)
Base:	11	(0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4)

Over 50000 turns for 191 people:42 people have 261 turn(s), 149 people have 262 turn(s)
Over 50000 turns for 1377 people:428 people have 37 turn(s), 949 people have 36 turn(s)
Over 50000 turns for 49999 people:49998 people have 1 turn(s), 1 people have 2 turn(s)
Over 50000 turns for 50000 people:50000 people have 1 turn(s)
Over 50000 turns for 50001 people:50000 people have 1 turn(s), 1 people have 0 turn(s)```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2020.01

Add an extension showing the relative fairness correlation of this selection algorithm. An absolutely fair algorithm would have a correlation of 1 for each person (no person has an advantage or disadvantage due to an algorithmic artefact.) This algorithm is fair, and gets better the more iterations are done.

A lower correlation factor corresponds to an advantage, higher to a disadvantage, the closer to 1 it is, the fairer the algorithm. Absolute best possible advantage correlation is 0. Absolute worst is 2.

```sub fairshare (\b) { ^∞ .hyper.map: { .polymod( b xx * ).sum % b } }

.say for <2 3 5 11>.map: { .fmt('%2d:') ~ .&fairshare[^25]».fmt('%2d').join: ', ' }

say "\nRelative fairness of this method. Scaled fairness correlation. The closer to 1.0 each person
is, the more fair the selection algorithm is. Gets better with more iterations.";

for <2 3 5 11 39> -> \$people {
print "\n\$people people: \n";
for \$people * 1, \$people * 10, \$people * 1000 -> \$iterations {
my @fairness;
fairshare(\$people)[^\$iterations].kv.map: { @fairness[\$^v % \$people] += \$^k }
my \$scale = @fairness.sum / @fairness;
my @range = @fairness.map( { \$_ / \$scale } );
printf "After round %4d: Best advantage: %-10.8g - Worst disadvantage: %-10.8g - Spread between best and worst: %-10.8g\n",
\$iterations/\$people, @range.min, @range.max, @range.max - @range.min;
}
}
```
``` 2: 0,  1,  1,  0,  1,  0,  0,  1,  1,  0,  0,  1,  0,  1,  1,  0,  1,  0,  0,  1,  0,  1,  1,  0,  0
3: 0,  1,  2,  1,  2,  0,  2,  0,  1,  1,  2,  0,  2,  0,  1,  0,  1,  2,  2,  0,  1,  0,  1,  2,  1
5: 0,  1,  2,  3,  4,  1,  2,  3,  4,  0,  2,  3,  4,  0,  1,  3,  4,  0,  1,  2,  4,  0,  1,  2,  3
11: 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,  0,  2,  3,  4

Relative fairness of this method. Scaled fairness correlation. The closer to 1.0 each person
is, the more fair the selection algorithm is. Gets better with more iterations.

2 people:
After round    1: Best advantage: 0          - Worst disadvantage: 2          - Spread between best and worst: 2
After round   10: Best advantage: 1          - Worst disadvantage: 1          - Spread between best and worst: 0
After round 1000: Best advantage: 1          - Worst disadvantage: 1          - Spread between best and worst: 0

3 people:
After round    1: Best advantage: 0          - Worst disadvantage: 2          - Spread between best and worst: 2
After round   10: Best advantage: 0.99310345 - Worst disadvantage: 1.0068966  - Spread between best and worst: 0.013793103
After round 1000: Best advantage: 0.99999933 - Worst disadvantage: 1.0000007  - Spread between best and worst: 1.3337779e-06

5 people:
After round    1: Best advantage: 0          - Worst disadvantage: 2          - Spread between best and worst: 2
After round   10: Best advantage: 1          - Worst disadvantage: 1          - Spread between best and worst: 0
After round 1000: Best advantage: 1          - Worst disadvantage: 1          - Spread between best and worst: 0

11 people:
After round    1: Best advantage: 0          - Worst disadvantage: 2          - Spread between best and worst: 2
After round   10: Best advantage: 0.99082569 - Worst disadvantage: 1.0091743  - Spread between best and worst: 0.018348624
After round 1000: Best advantage: 0.99999909 - Worst disadvantage: 1.0000009  - Spread between best and worst: 1.8183471e-06

39 people:
After round    1: Best advantage: 0          - Worst disadvantage: 2          - Spread between best and worst: 2
After round   10: Best advantage: 0.92544987 - Worst disadvantage: 1.0745501  - Spread between best and worst: 0.14910026
After round 1000: Best advantage: 0.99999103 - Worst disadvantage: 1.000009   - Spread between best and worst: 1.7949178e-05```

## REXX

Programming note:   the   base   function (as coded herein) handles bases from base ten up to   36.

```/*REXX program calculates  N  terms of the fairshare sequence for some group of peoples.*/
parse arg n g                                    /*obtain optional arguments from the CL*/
if n=='' | n==","  then n= 25                    /*Not specified?  Then use the default.*/
if g=''  | g=","   then g=  2 3 5 11             /* "      "         "   "   "     "    */
/* [↑]  a list of a number of peoples. */
do p=1  for words(g);        r= word(g, p)  /*traipse through the bases specfiied. */
\$= 'base' right(r, 2)': '                   /*construct start of the 1─line output.*/
do j=0  for n;   \$= \$ right( sumDigs( base(j, r))  //  r, 2)','
end   /*j*/                              /* [↑] append # (base R) mod R──►\$ list*/
say strip(\$, , ",")                         /*elide trailing comma from the \$ list.*/
end      /*p*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
base: parse arg #,b,,y;          @= 0123456789abcdefghijklmnopqrstuvwxyz;  @@= substr(@,2)
do while #>=b; y= substr(@, #//b + 1, 1)y; #= #%b; end;  return substr(@, #+1, 1)y
/*──────────────────────────────────────────────────────────────────────────────────────*/
sumDigs: parse arg x; !=0; do i=1 for length(x); != !+pos(substr(x,i,1),@@); end; return !
```
output   when using the default inputs:
```base  2:   0,  1,  1,  0,  1,  0,  0,  1,  1,  0,  0,  1,  0,  1,  1,  0,  1,  0,  0,  1,  0,  1,  1,  0,  0
base  3:   0,  1,  2,  1,  2,  0,  2,  0,  1,  1,  2,  0,  2,  0,  1,  0,  1,  2,  2,  0,  1,  0,  1,  2,  1
base  5:   0,  1,  2,  3,  4,  1,  2,  3,  4,  0,  2,  3,  4,  0,  1,  3,  4,  0,  1,  2,  4,  0,  1,  2,  3
base 11:   0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10,  0,  2,  3,  4
```

## Ring

```str    = []
people = [2,3,5,11]

result = people
for i in people
str = []
see "" + i + ": "
fair(25, i)
for n in result
next
showarray(str)
next

func fair n,base

result = list(n)
for i=1 to n
j = i-1
t = 0
while j>0
t = t + j % base
j = floor(j/base)
end
result[i] = t % base
next

func showarray vect
svect = ""
for n in vect
svect += " " + n + ","
next
svect = left(svect, len(svect) - 1)
? "[" + svect + "]"```
Output:
```2: [ 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
3: [ 0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
5: [ 0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
11: [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4]
```

## RPL

`DIGITS` is defined at Sum digits of an integer.

`D→B` is a slightly modified version of the program defined at Non-decimal radices/Convert, using uppercase characters to be compatible with `DIGITS`

```≪ → b
≪ "" SWAP
WHILE DUP REPEAT
b MOD LAST / IP
SWAP DUP 9 > 55 48 IFTE + CHR
ROT + SWAP
END DROP
≫ ≫ 'D→B' STO

≪ → b
≪ { 0 } 1 24 FOR j
j b D→B DIGITS b MOD + NEXT
```
```2 TASK 3 TASK 5 TASK 11 TASK
```
Output:
```4: { 0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 0 1 1 0 0 }
3: { 0 1 2 1 2 0 2 0 1 1 2 0 2 0 1 0 1 2 2 0 1 0 1 2 1 }
2: { 0 1 2 3 4 1 2 3 4 0 2 3 4 0 1 3 4 0 1 2 4 0 1 2 3 }
1: { 0 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 2 3 4 }
```

## Ruby

Translation of: C
```def turn(base, n)
sum = 0
while n != 0 do
rem = n % base
n = n / base
sum = sum + rem
end
return sum % base
end

def fairshare(base, count)
print "Base %2d: " % [base]
for i in 0 .. count - 1 do
t = turn(base, i)
print " %2d" % [t]
end
print "\n"
end

def turnCount(base, count)
cnt = Array.new(base, 0)

for i in 0 .. count - 1 do
t = turn(base, i)
cnt[t] = cnt[t] + 1
end

minTurn = base * count
maxTurn = -1
portion = 0
for i in 0 .. base - 1 do
if cnt[i] > 0 then
portion = portion + 1
end
if cnt[i] < minTurn then
minTurn = cnt[i]
end
if cnt[i] > maxTurn then
maxTurn = cnt[i]
end
end

print "  With %d people: " % [base]
if 0 == minTurn then
print "Only %d have a turn\n" % portion
elsif minTurn == maxTurn then
print "%d\n" % [minTurn]
else
print "%d or %d\n" % [minTurn, maxTurn]
end
end

def main
fairshare(2, 25)
fairshare(3, 25)
fairshare(5, 25)
fairshare(11, 25)

puts "How many times does each get a turn in 50000 iterations?"
turnCount(191, 50000)
turnCount(1377, 50000)
turnCount(49999, 50000)
turnCount(50000, 50000)
turnCount(50001, 50000)
end

main()
```
Output:
```Base  2:   0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:   0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:   0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:   0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```
Translation of: Sidef
```def fairshare(base, upto) = (0...upto).map{|n| n.digits(base).sum % base}

upto = 25
[2, 3, 5, 11].each{|b| puts"#{'%2d' % b}: " + " %2d"*upto % fairshare(b, upto)}
```
Output:
``` 2:   0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3:   0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5:   0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11:   0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
```

## Rust

Translation of: Python
```struct Digits {
rest: usize,
base: usize,
}

impl Iterator for Digits {
type Item = usize;

fn next(&mut self) -> Option<usize> {
if self.rest == 0 {
return None;
}
let (digit, rest) = (self.rest % self.base, self.rest / self.base);
self.rest = rest;
Some(digit)
}
}

fn digits(num: usize, base: usize) -> Digits {
Digits { rest: num, base: base }
}

struct FairSharing {
participants: usize,
index: usize,
}

impl Iterator for FairSharing {
type Item = usize;
fn next(&mut self) -> Option<usize> {
let digit_sum: usize = digits(self.index, self.participants).sum();
let selected = digit_sum % self.participants;
self.index += 1;
Some(selected)
}
}

fn fair_sharing(participants: usize) -> FairSharing {
FairSharing { participants: participants, index: 0 }
}

fn main() {
for i in vec![2, 3, 5, 7] {
println!("{}: {:?}", i, fair_sharing(i).take(25).collect::<Vec<usize>>());
}
}
```
Output:
```2: [0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
3: [0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
5: [0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
7: [0, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 0, 2, 3, 4, 5, 6, 0, 1, 3, 4, 5, 6]```

## Sidef

```for b in (2,3,5,11) {
say ("#{'%2d' % b}: ", 25.of { .sumdigits(b) % b })
}
```
Output:
``` 2: [0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
3: [0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
5: [0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
11: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4]
```

## Visual Basic .NET

Translation of: C
```Module Module1

Function Turn(base As Integer, n As Integer) As Integer
Dim sum = 0
While n <> 0
Dim re = n Mod base
n \= base
sum += re
End While
Return sum Mod base
End Function

Sub Fairshare(base As Integer, count As Integer)
Console.Write("Base {0,2}:", base)
For i = 1 To count
Dim t = Turn(base, i - 1)
Console.Write(" {0,2}", t)
Next
Console.WriteLine()
End Sub

Sub TurnCount(base As Integer, count As Integer)
Dim cnt(base) As Integer
For i = 1 To base
cnt(i - 1) = 0
Next

For i = 1 To count
Dim t = Turn(base, i - 1)
cnt(t) += 1
Next

Dim minTurn = Integer.MaxValue
Dim maxTurn = Integer.MinValue
Dim portion = 0
For i = 1 To base
Dim num = cnt(i - 1)
If num > 0 Then
portion += 1
End If
If num < minTurn Then
minTurn = num
End If
If num > maxTurn Then
maxTurn = num
End If
Next

Console.Write("  With {0} people: ", base)
If 0 = minTurn Then
Console.WriteLine("Only {0} have a turn", portion)
ElseIf minTurn = maxTurn Then
Console.WriteLine(minTurn)
Else
Console.WriteLine("{0} or {1}", minTurn, maxTurn)
End If
End Sub

Sub Main()
Fairshare(2, 25)
Fairshare(3, 25)
Fairshare(5, 25)
Fairshare(11, 25)

Console.WriteLine("How many times does each get a turn in 50000 iterations?")
TurnCount(191, 50000)
TurnCount(1377, 50000)
TurnCount(49999, 50000)
TurnCount(50000, 50000)
TurnCount(50001, 50000)
End Sub

End Module
```
Output:
```Base  2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
Base  3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
Base  5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
Base 11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: Only 50000 have a turn```

## V (Vlang)

Translation of: Go
```fn fairshare(n int, base int) []int {
mut res := []int{len: n}
for i in 0..n {
mut j := i
mut sum := 0
for j > 0 {
sum += j % base
j /= base
}
res[i] = sum % base
}
return res
}

fn turns(n int, fss []int) string {
mut m := map[int]int{}
for fs in fss {
m[fs]++
}
mut m2 := map[int]int{}
for _,v in m {
m2[v]++
}
mut res := []int{}
mut sum := 0
for k, v in m2 {
sum += v
res << k
}
if sum != n {
return "only \$sum have a turn"
}
res.sort()
mut res2 := []string{len: res.len}
for i,_ in res {
res2[i] = '\${res[i]}'
}
return res2.join(" or ")
}

fn main() {
for base in [2, 3, 5, 11] {
println("\${base:2} : \${fairshare(25, base):2}")
}
println("\nHow many times does each get a turn in 50000 iterations?")
for base in [191, 1377, 49999, 50000, 50001] {
t := turns(base, fairshare(50000, base))
println("  With \$base people: \$t")
}
}```
Output:
``` 2 : [0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0]
3 : [0, 1, 2, 1, 2, 0, 2, 0, 1, 1, 2, 0, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 1, 2, 1]
5 : [0, 1, 2, 3, 4, 1, 2, 3, 4, 0, 2, 3, 4, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 2, 3]
11 : [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 2, 3, 4]

How many times does each get a turn in 50000 iterations?
With 191 people: 261 or 262
With 1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: only 50000 have a turn
```

## Wren

Translation of: Go
Library: Wren-fmt
Library: Wren-sort
```import "./fmt" for Fmt
import "./sort" for Sort

var fairshare = Fn.new { |n, base|
var res = List.filled(n, 0)
for (i in 0...n) {
var j = i
var sum = 0
while (j > 0) {
sum = sum + (j%base)
j = (j/base).floor
}
res[i] = sum % base
}
return res
}

var turns = Fn.new { |n, fss|
var m = {}
for (fs in fss) {
var k = m[fs]
if (!k) {
m[fs] = 1
} else {
m[fs] = k + 1
}
}
var m2 = {}
for (k in m.keys) {
var v = m[k]
var k2 = m2[v]
if (!k2) {
m2[v] = 1
} else {
m2[v] = k2 + 1
}
}
var res = []
var sum = 0
for (k in m2.keys) {
var v = m2[k]
sum = sum + v
}
if (sum != n) return "only %(sum) have a turn"
Sort.quick(res)
var res2 = List.filled(res.count, "")
for (i in 0...res.count) res2[i] = res[i].toString
return res2.join(" or ")
}

for (base in [2, 3, 5, 11]) {
Fmt.print("\$2d : \$2d", base, fairshare.call(25, base))
}
System.print("\nHow many times does each get a turn in 50,000 iterations?")
for (base in [191, 1377, 49999, 50000, 50001]) {
var t = turns.call(base, fairshare.call(50000, base))
Fmt.print("  With \$5d people: \$s", base, t)
}
```
Output:
``` 2 :  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3 :  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5 :  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11 :  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4

How many times does each get a turn in 50,000 iterations?
With   191 people: 261 or 262
With  1377 people: 36 or 37
With 49999 people: 1 or 2
With 50000 people: 1
With 50001 people: only 50000 have a turn
```

## XPL0

```proc Fair(Base); \Show first 25 terms of fairshare sequence
int  Base, Count, Sum, N, Q;
[RlOut(0, float(Base));  Text(0, ": ");
for Count:= 0 to 25-1 do
[Sum:= 0;  N:= Count;
while N do
[Q:= N/Base;
Sum:= Sum + rem(0);
N:= Q;
];
RlOut(0, float(rem(Sum/Base)));
];
CrLf(0);
];

[Format(3,0);
Fair(2);  Fair(3);  Fair(5);  Fair(11);
]```
Output:
```  2:   0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3:   0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5:   0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11:   0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
```

## zkl

```fcn fairShare(n,b){    // b<=36
n.pump(List,'wrap(n){ n.toString(b).split("").apply("toInt",b).sum(0)%b })
}
foreach b in (T(2,3,5,11)){
println("%2d: %s".fmt(b,fairShare(25,b).pump(String,"%2d ".fmt)));
}```
Output:
``` 2:  0  1  1  0  1  0  0  1  1  0  0  1  0  1  1  0  1  0  0  1  0  1  1  0  0
3:  0  1  2  1  2  0  2  0  1  1  2  0  2  0  1  0  1  2  2  0  1  0  1  2  1
5:  0  1  2  3  4  1  2  3  4  0  2  3  4  0  1  3  4  0  1  2  4  0  1  2  3
11:  0  1  2  3  4  5  6  7  8  9 10  1  2  3  4  5  6  7  8  9 10  0  2  3  4
```

For any base > 1

```fcn fairShare(n,base){
sum,r := 0,0; while(n){ n,r = n.divr(base); sum+=r }
sum%base
}
foreach b in (T(2,3,5,11)){
println("%2d: %s".fmt(b,[0..24].pump(String,fairShare.fp1(b),"%2d ".fmt)));
}```