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# Kolakoski sequence

Kolakoski sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Kolakoski sequence is an infinite sequence of natural numbers, (excluding zero); with the property that:

if you form a new sequence from the counts of runs of the same number in the first sequence, this new sequence is the same as the first sequence.
Example

This is not a Kolakoski sequence:

`1,1,2,2,2,1,2,2,1,2,...`

Its sequence of run counts, (sometimes called a run length encoding, (RLE); but a true RLE also gives the character that each run encodes), is calculated like this:

Starting from the leftmost number of the sequence we have `2` ones, followed by `3` twos, then `1` ones, `2` twos, `1` one, ...

The above gives the RLE of:

`2, 3, 1, 2, 1, ...`

The original sequence is different from its RLE in this case. It would be the same for a true Kolakoski sequence.

Creating a Kolakoski sequence

Lets start with the two numbers `(1, 2)` that we will cycle through; i.e. they will be used in this order:
`1,2,1,2,1,2,....`

1. We start the sequence `s` with the first item from the cycle `c`:
`1`
2. An index, `k`, into the, (expanding), sequence will step, or index through each item of the sequence `s` from the first, at its own rate.

We will arrange that the `k`'th item of `s` states how many times the last item of `s`should appear at the end of `s`.

We started `s` with `1` and therefore `s[k]` states that it should appear only the `1` time.

1. Increment `k`

2. Get the next item from `c` and append it to the end of sequence `s`. `s` will then become:
`1, 2`

3. `k` was moved to the second item in the list and `s[k]` states that it should appear two times, so append another of the last item to the sequence `s`:
`1, 2,2`

4. Increment `k`

5. Append the next item from the cycle to the list:
`1, 2,2, 1`

6. `k` is now at the third item in the list that states that the last item should appear twice so add another copy of the last item to the sequence `s`:
`1, 2,2, 1,1`

7. increment k

...

Note that the RLE of `1, 2, 2, 1, 1, ...` begins `1, 2, 2` which is the beginning of the original sequence. The generation algorithm ensures that this will always be the case.

1. Create a routine/proceedure/function/... that given an initial ordered list/array/tuple etc of the natural numbers `(1, 2)`, returns the next number from the list when accessed in a cycle.
2. Create another routine that when given the initial ordered list `(1, 2)` and the minimum length of the sequence to generate; uses the first routine and the algorithm above, to generate at least the requested first members of the kolakoski sequence.
3. Create a routine that when given a sequence, creates the run length encoding of that sequence (as defined above) and returns the result of checking if sequence starts with the exact members of its RLE. (But note, due to sampling, do not compare the last member of the RLE).
4. Show, on this page, (compactly), the first 20 members of the sequence generated from `(1, 2)`
5. Check the sequence againt its RLE.
6. Show, on this page, the first 20 members of the sequence generated from `(2, 1)`
7. Check the sequence againt its RLE.
8. Show, on this page, the first 30 members of the Kolakoski sequence generated from `(1, 3, 1, 2)`
9. Check the sequence againt its RLE.
10. Show, on this page, the first 30 members of the Kolakoski sequence generated from `(1, 3, 2, 1)`
11. Check the sequence againt its RLE.

(There are rules on generating Kolakoski sequences from this method that are broken by the last example)

## 11l

Translation of: C++
`F gen_kolakoski(s, n)   [Int] seq   V i = 0   L seq.len < n      V next = s[i % s.len]      seq [+]= [next] * (I i >= seq.len {next} E seq[i])      i++   R seq[0 .< n] F is_possible_kolakoski(s)   [Int] r   V i = 0   L i < s.len      V count = 1      L(j) i + 1 .< s.len         I s[j] != s[i]            L.break         count++      r.append(count)      i += count    L(i) 0 .< r.len      I r[i] != s[i]         R 0B   R 1B L(s) [[1, 2],      [2, 1],      [1, 3, 1, 2],      [1, 3, 2, 1]]   V kol = gen_kolakoski(s, I s.len > 2 {30} E 20)   print(‘Starting with: ’s":\nKolakoski sequence: "kol"\nPossibly kolakoski? "is_possible_kolakoski(kol))`
Output:
```Starting with: [1, 2]:
Kolakoski sequence: [1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possibly kolakoski? 1B
Starting with: [2, 1]:
Kolakoski sequence: [2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possibly kolakoski? 1B
Starting with: [1, 3, 1, 2]:
Kolakoski sequence: [1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possibly kolakoski? 1B
Starting with: [1, 3, 2, 1]:
Kolakoski sequence: [1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possibly kolakoski? 0B
```

## Arturo

`kolakoski: function [a, length][    result: array.of: length 0    i: new 0    k: new 0     loop.forever a 'x [        result\[i]: x        if result\[k] > 1 [            loop 1..dec result\[k] 'j [                inc 'i                if i = length -> return result                result\[i]: result\[i-1]            ]        ]        inc 'i        if i = length -> return result        inc 'k    ]    return result] possibleKolakoski?: function [seq][    prev: seq\0    count: new 1    rle: new []     loop 1..dec size seq 'i [        if? seq\[i] = prev -> inc 'count        else [            'rle ++ count            count: new 1            prev: seq\[i]        ]    ]     loop.with:'i rle 'val [        if val <> seq\[i] -> return false    ]    return true] Seqs: [[1 2] [2 1] [1 3 1 2] [1 3 2 1]]Lens: [20 20 30 30] loop combine Seqs Lens 'c [    generated: kolakoski c\0 c\1    print ["First" c\1 "members of the sequence generated by" c\0 ":"]    print generated    print ["Possible Kolakoski sequence?" possibleKolakoski? generated]    print ""]`
Output:
```First 20 members of the sequence generated by [1 2] :
1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1
Possible Kolakoski sequence? true

First 20 members of the sequence generated by [2 1] :
2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2
Possible Kolakoski sequence? true

First 30 members of the sequence generated by [1 3 1 2] :
1 3 3 3 1 1 1 2 2 2 1 3 1 2 2 1 1 3 3 1 2 2 2 1 3 3 1 1 2 1
Possible Kolakoski sequence? true

First 30 members of the sequence generated by [1 3 2 1] :
1 3 3 3 2 2 2 1 1 1 1 1 3 3 2 2 1 1 3 2 1 1 1 1 3 3 3 2 2 1
Possible Kolakoski sequence? false```

## C

Translation of: Kotlin
`#include <stdio.h>#include <stdlib.h> #define TRUE 1#define FALSE 0 typedef int bool; int next_in_cycle(int *c, int len, int index) {    return c[index % len];} void kolakoski(int *c, int *s, int clen, int slen) {    int i = 0, j, k = 0;    while (TRUE) {        s[i] = next_in_cycle(c, clen, k);        if (s[k] > 1) {            for (j = 1; j < s[k]; ++j) {                if (++i == slen) return;                s[i] = s[i - 1];            }        }        if (++i == slen) return;        k++;    }} bool possible_kolakoski(int *s, int len) {    int i, j = 0, prev = s[0], count = 1;    int *rle = calloc(len, sizeof(int));    bool result = TRUE;    for (i = 1; i < len; ++i) {        if (s[i] == prev) {            count++;        }        else {            rle[j++] = count;            count = 1;            prev = s[i];        }    }    /* no point adding final 'count' to rle as we're not going to compare it anyway */    for (i = 0; i < j; i++) {        if (rle[i] != s[i]) {           result = FALSE;           break;        }    }    free(rle);    return result;} void print_array(int *a, int len) {    int i;    printf("[");    for (i = 0; i < len; ++i) {       printf("%d", a[i]);       if (i < len - 1) printf(", ");    }    printf("]");} int main() {    int i, clen, slen, *s;    int c0[2] = {1, 2};    int c1[2] = {2, 1};    int c2[4] = {1, 3, 1, 2};    int c3[4] = {1, 3, 2, 1};    int *cs[4] = {c0, c1, c2, c3};    bool p;    int clens[4] = {2, 2, 4, 4};    int slens[4] = {20, 20, 30, 30};    for (i = 0; i < 4; ++i) {        clen = clens[i];        slen = slens[i];        s = calloc(slen, sizeof(int));        kolakoski(cs[i], s, clen, slen);        printf("First %d members of the sequence generated by ", slen);        print_array(cs[i], clen);        printf(":\n");        print_array(s, slen);        printf("\n");        p = possible_kolakoski(s, slen);        printf("Possible Kolakoski sequence? %s\n\n", p ? "True" : "False");        free(s);     }    return 0;}`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? True

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? True

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? True

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? False
```

## C#

Translation of: Java
`using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Threading.Tasks; namespace KolakoskiSequence {    class Crutch {        public readonly int len;        public int[] s;        public int i;         public Crutch(int len) {            this.len = len;            s = new int[len];            i = 0;        }         public void Repeat(int count) {            for (int j = 0; j < count; j++) {                if (++i == len) return;                s[i] = s[i - 1];            }        }    }     static class Extension {        public static int NextInCycle(this int[] self, int index) {            return self[index % self.Length];        }         public static int[] Kolakoski(this int[] self, int len) {            Crutch c = new Crutch(len);             int k = 0;            while (c.i < len) {                c.s[c.i] = self.NextInCycle(k);                if (c.s[k] > 1) {                    c.Repeat(c.s[k] - 1);                }                if (++c.i == len) return c.s;                k++;            }            return c.s;        }         public static bool PossibleKolakoski(this int[] self) {            int[] rle = new int[self.Length];            int prev = self[0];            int count = 1;            int pos = 0;            for (int i = 1; i < self.Length; i++) {                if (self[i] == prev) {                    count++;                }                else {                    rle[pos++] = count;                    count = 1;                    prev = self[i];                }            }            // no point adding final 'count' to rle as we're not going to compare it anyway            for (int i = 0; i < pos; i++) {                if (rle[i] != self[i]) {                    return false;                }            }            return true;        }         public static string AsString(this int[] self) {            StringBuilder sb = new StringBuilder("[");            int count = 0;            foreach (var item in self) {                if (count > 0) {                    sb.Append(", ");                }                sb.Append(item);                count++;            }            return sb.Append("]").ToString();        }    }     class Program {        static void Main(string[] args) {            int[][] ias = {                new int[]{1, 2},                new int[]{2, 1},                new int[]{1, 3, 1, 2},                new int[]{1, 3, 2, 1}            };            int[] lens = { 20, 20, 30, 30 };             for (int i = 0; i < ias.Length; i++) {                int len = lens[i];                int[] kol = ias[i].Kolakoski(len);                 Console.WriteLine("First {0} members of the sequence by {1}: ", len, ias[i].AsString());                Console.WriteLine(kol.AsString());                Console.WriteLine("Possible Kolakoski sequence? {0}", kol.PossibleKolakoski());                Console.WriteLine();            }        }    }}`

## C++

`#include <iostream>#include <vector> using Sequence = std::vector<int>; std::ostream& operator<<(std::ostream& os, const Sequence& v) {  os << "[ ";  for (const auto& e : v) {    std::cout << e << ", ";  }  os << "]";  return os;} int next_in_cycle(const Sequence& s, size_t i) {  return s[i % s.size()];} Sequence gen_kolakoski(const Sequence& s, int n) {  Sequence seq;  for (size_t i = 0; seq.size() < n; ++i) {    const int next = next_in_cycle(s, i);    Sequence nv(i >= seq.size() ? next : seq[i], next);    seq.insert(std::end(seq), std::begin(nv), std::end(nv));  }  return { std::begin(seq), std::begin(seq) + n };} bool is_possible_kolakoski(const Sequence& s) {  Sequence r;  for (size_t i = 0; i < s.size();) {    int count = 1;    for (size_t j = i + 1; j < s.size(); ++j) {      if (s[j] != s[i]) break;      ++count;    }    r.push_back(count);    i += count;  }  for (size_t i = 0; i < r.size(); ++i) if (r[i] != s[i]) return false;  return true;} int main() {  std::vector<Sequence> seqs = {    { 1, 2 },    { 2, 1 },    { 1, 3, 1, 2 },    { 1, 3, 2, 1 }  };  for (const auto& s : seqs) {    auto kol = gen_kolakoski(s, s.size() > 2 ? 30 : 20);    std::cout << "Starting with: " << s << ": " << std::endl << "Kolakoski sequence: "       << kol << std::endl << "Possibly kolakoski? " << is_possible_kolakoski(kol) << std::endl;		  }  return 0;}`
Output:
```Starting with: [ 1, 2, ]:
Kolakoski sequence: [ 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, ]
Possibly kolakoski? 1
Starting with: [ 2, 1, ]:
Kolakoski sequence: [ 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, ]
Possibly kolakoski? 1
Starting with: [ 1, 3, 1, 2, ]:
Kolakoski sequence: [ 1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1, ]
Possibly kolakoski? 1
Starting with: [ 1, 3, 2, 1, ]:
Kolakoski sequence: [ 1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1, ]
Possibly kolakoski? 0```

## D

Translation of: Kotlin
`import std.stdio; void repeat(int count, void delegate(int) action) {    for (int i=0; i<count; i++) {        action(i);    }} T nextInCycle(T)(T[] self, int index) {    return self[index % self.length];} T[] kolakoski(T)(T[] self, int len) {    T[] s;    s.length = len;    int i;    int k;    while (i<len) {        s[i] = self.nextInCycle(k);        if (s[k] > 1) {            repeat(s[k] - 1,                (int j) {                    if (++i == len) return;                    s[i] = s[i-1];                }            );        }        if (++i == len) return s;        k++;    }    return s;} bool possibleKolakoski(T)(T[] self) {    auto len = self.length;    T[] rle;    auto prev = self[0];    int count = 1;    foreach (i; 1..len) {        if (self[i] == prev) {            count++;        } else {            rle ~= count;            count = 1;            prev = self[i];        }    }    // no point adding final 'count' to rle as we're not going to compare it anyway    foreach (i; 0..rle.length) {        if (rle[i] != self[i]) {            return false;        }    }    return true;} void main() {    auto ias = [[1,2],[2,1],[1,3,1,2],[1,3,2,1]];    auto lens = [20,20,30,30];     foreach (i,ia; ias) {        auto len = lens[i];        auto kol = ia.kolakoski(len);        writeln("First ", len, " members of the sequence generated by ", ia, ":");        writeln(kol);        write("Possible Kolakoski sequence? ");        if (kol.possibleKolakoski) {            writeln("Yes");        } else {            writeln("no");        }        writeln;    }}`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? Yes

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? no```

## Go

Translation of: Kotlin
`package main import "fmt" func nextInCycle(c []int, index int) int {    return c[index % len(c)]} func kolakoski(c []int, slen int) []int {    s := make([]int, slen)    i, k := 0, 0    for {        s[i] = nextInCycle(c, k)        if s[k] > 1 {            for j := 1; j < s[k]; j++ {                i++                if i == slen {                    return s                }                s[i] = s[i - 1]            }        }        i++        if i == slen {            return s        }        k++    }} func possibleKolakoski(s []int) bool {    slen := len(s)    rle := make([]int, 0, slen)    prev := s[0]    count := 1    for i := 1; i < slen; i++ {        if s[i] == prev {            count++        } else {            rle = append(rle, count)            count = 1            prev = s[i]        }    }    // no point adding final 'count' to rle as we're not going to compare it anyway    for i := 0; i < len(rle); i++ {        if rle[i] != s[i] {            return false        }    }    return true} func printInts(ia []int, suffix string) {    fmt.Print("[")    alen := len(ia)    for i := 0; i < alen; i++ {        fmt.Print(ia[i])        if i < alen - 1 {            fmt.Print(", ")        }    }    fmt.Printf("]%s\n", suffix)} func main() {    ias := make([][]int, 4)    ias[0] = []int{1, 2}    ias[1] = []int{2, 1}    ias[2] = []int{1, 3, 1, 2}    ias[3] = []int{1, 3, 2, 1}    slens := []int{20, 20, 30, 30}    for i, ia := range ias {        slen := slens[i]        kol := kolakoski(ia, slen)        fmt.Printf("First %d members of the sequence generated by ", slen)        printInts(ia, ":")        printInts(kol, "")        p := possibleKolakoski(kol)        poss := "Yes"        if !p {            poss = "No"        }        fmt.Println("Possible Kolakoski sequence?", poss, "\n")    }}`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? Yes

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? No
```

`import Data.List (group)import Control.Monad (forM_) replicateAtLeastOne :: Int -> a -> [a]replicateAtLeastOne n x = x : replicate (n-1) x zipWithLazy :: (a -> b -> c) -> [a] -> [b] -> [c]zipWithLazy f ~(x:xs) ~(y:ys) = f x y : zipWithLazy f xs ys kolakoski :: [Int] -> [Int]kolakoski items = s  where s = concat \$ zipWithLazy replicateAtLeastOne s \$ cycle items rle :: Eq a => [a] -> [Int]rle = map length . group sameAsRleUpTo :: Int -> [Int] -> BoolsameAsRleUpTo n s = r == take (length r) prefix  where prefix = take n s        r = init \$ rle prefix main :: IO ()main = forM_ [([1, 2], 20),              ([2, 1], 20),               ([1, 3, 1, 2], 30),              ([1, 3, 2, 1], 30)]        \$ \(items, n) -> do          putStrLn \$ "First " ++ show n ++ " members of the sequence generated by " ++ show items ++ ":"          let s = kolakoski items          print \$ take n s          putStrLn \$ "Possible Kolakoski sequence? " ++ show (sameAsRleUpTo n s)          putStrLn ""`
Output:
```First 20 members of the sequence generated by [1,2]:
[1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1]
Possible Kolakoski sequence? True

First 20 members of the sequence generated by [2,1]:
[2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2]
Possible Kolakoski sequence? True

First 30 members of the sequence generated by [1,3,1,2]:
[1,3,3,3,1,1,1,2,2,2,1,3,1,2,2,1,1,3,3,1,2,2,2,1,3,3,1,1,2,1]
Possible Kolakoski sequence? True

First 30 members of the sequence generated by [1,3,2,1]:
[1,3,3,3,2,2,2,1,1,1,1,1,3,3,2,2,1,1,3,2,1,1,1,1,3,3,3,2,2,1]
Possible Kolakoski sequence? False

```

## J

` NB. cyclic create_cycle_=: 3 :0 I=: 0 A=: y N=: # A) next_cycle_=: 3 :0 r=. A {~ N | I I=: >: I r) NB. kolakoski kolakoski =: 30&\$: :(dyad define) NB. TERMS kolakoski ALPHABET c=. y conew'cycle' s=. i. 0 term=. 0 while. x > # s do.   s=. (, ([: #~ next__c)`(term&{ # next__c)@.(term < #)) s  term=. >: term end. s)  test=: (({.~ #) -: ]) }:@:(#;.1~ (1 , 2&(~:/\))) `

test cuts the data at a vector of frets where successive pairs are unequal. The groups are tallied, giving run length.

```   f=: (;~ test)@:kolakoski

(; f)&> 1 2 ; 2 1 ; 1 3 1 2 ; 1 3 2 1
┌───────┬─┬─────────────────────────────────────────────────────────────┐
│1 2    │1│1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2 1 1 2 1 2 2 1 1 2  │
├───────┼─┼─────────────────────────────────────────────────────────────┤
│2 1    │1│2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2 1 1 2 1 2 2 1 1 2 1 1│
├───────┼─┼─────────────────────────────────────────────────────────────┤
│1 3 1 2│1│1 3 3 3 1 1 1 2 2 2 1 3 1 2 2 1 1 3 3 1 2 2 2 1 3 3 1 1 2 1  │
├───────┼─┼─────────────────────────────────────────────────────────────┤
│1 3 2 1│0│1 3 3 3 2 2 2 1 1 1 1 1 3 3 2 2 1 1 3 2 1 1 1 1 3 3 3 2 2 1 1│
└───────┴─┴─────────────────────────────────────────────────────────────┘
```

## Java

Translation of: Kotlin
`import java.util.Arrays; public class Kolakoski {    private static class Crutch {        final int len;        int[] s;        int i;         Crutch(int len) {            this.len = len;            s = new int[len];            i = 0;        }         void repeat(int count) {            for (int j = 0; j < count; j++) {                if (++i == len) return;                s[i] = s[i - 1];            }        }    }     private static int nextInCycle(final int[] self, int index) {        return self[index % self.length];    }     private static int[] kolakoski(final int[] self, int len) {        Crutch c = new Crutch(len);         int k = 0;        while (c.i < len) {            c.s[c.i] = nextInCycle(self, k);            if (c.s[k] > 1) {                c.repeat(c.s[k] - 1);            }            if (++c.i == len) return c.s;            k++;        }        return c.s;    }     private static boolean possibleKolakoski(final int[] self) {        int[] rle = new int[self.length];        int prev = self[0];        int count = 1;        int pos = 0;        for (int i = 1; i < self.length; i++) {            if (self[i] == prev) {                count++;            } else {                rle[pos++] = count;                count = 1;                prev = self[i];            }        }        // no point adding final 'count' to rle as we're not going to compare it anyway        for (int i = 0; i < pos; i++) {            if (rle[i] != self[i]) {                return false;            }        }        return true;    }     public static void main(String[] args) {        int[][] ias = new int[][]{            new int[]{1, 2},            new int[]{2, 1},            new int[]{1, 3, 1, 2},            new int[]{1, 3, 2, 1}        };        int[] lens = new int[]{20, 20, 30, 30};         for (int i=0; i<ias.length; i++) {            int len = lens[i];            int[] kol = kolakoski(ias[i], len);             System.out.printf("First %d members of the sequence generated by %s: \n", len, Arrays.toString(ias[i]));            System.out.printf("%s\n", Arrays.toString(kol));            System.out.printf("Possible Kolakoski sequence? %s\n\n", possibleKolakoski(kol));        }    }}`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? true

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? true

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? true

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? false```

## jq

This section is based on a kolakoski generator that continues indefinitely.

Thanks to jq's backtracking semantics, we only need a "cycle" generator that cycles indefinitely often:

`def cycle:  def c: .[], c;  c;`
` # Input: an array# Output: the corresponding kolakoski sequence.# This version of the kolakoski generator is optimized to the extent# that it avoids storing the full sequence by removing the first item# in the .s array at each iteration.def kolakoski:  foreach cycle as \$next ( {s: []};      # ensure the next element occurs .s[0] times      .s += [\$next]      | .extra = [range(0; .s[0]-1) as \$i | \$next]      | .s = .s[1:] + .extra      ;      \$next, .extra[] ) ; def kolakoski(\$len): limit(\$len; kolakoski); def iskolakoski:  def rle:    . as \$seq    | reduce range(1;length) as \$i ({rle:[], count:1};        if \$seq[\$i] == \$seq[\$i - 1]        then .count += 1        else .rle = .rle + [.count]        | .count = 1        end)    | .rle;  rle | . == .[0 : length] ; `

Testing

` def tests: [[[1, 2], 20], [[2, 1] ,20], [[1, 3, 1, 2], 30], [[1, 3, 2, 1], 30]]; tests[] as [\$a, \$n]| \$a| [kolakoski(\$n)] as \$k| "First \(\$n) of kolakoski sequence for \(\$a):", \$k, "check: \(\$k | if iskolakoski then "✓" else "❌" end )", "" `
Output:

Invocation: jq -nr -f kolakoski.jq

```First 20 of kolakoski sequence for [1,2]:
[1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1]
check: ✓

First 20 of kolakoski sequence for [2,1]:
[2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2]
check: ✓

First 30 of kolakoski sequence for [1,3,1,2]:
[1,3,3,3,1,1,1,2,2,2,1,3,1,2,2,1,1,3,3,1,2,2,2,1,3,3,1,1,2,1]
check: ✓

First 30 of kolakoski sequence for [1,3,2,1]:
[1,3,3,3,2,2,2,1,1,1,1,1,3,3,2,2,1,1,3,2,1,1,1,1,3,3,3,2,2,1]
check: ✓
```

## Julia

Translation of: C
`function kolakoski(vec, len)    seq = Vector{Int}()    k = 0    denom = length(vec)    while length(seq) < len        n = vec[k % denom + 1]        k += 1        seq = vcat(seq, repeat([n], k > length(seq) ? n : seq[k]))    end    seq[1:len]end function iskolakoski(seq)    count = 1    rle = Vector{Int}()    for i in 2:length(seq)        if seq[i] == seq[i - 1]            count += 1        else            push!(rle, count)            count = 1        end    end    rle == seq[1:length(rle)]end const tests = [[[1, 2], 20],[[2, 1] ,20], [[1, 3, 1, 2], 30], [[1, 3, 2, 1], 30]] for t in tests    vec, n = t[1], t[2]    seq = kolakoski(vec, n)    println("Kolakoski from \$(vec): first \$n numbers are \$seq.")    println("\t\tDoes this look like a Kolakoski sequence? ", iskolakoski(seq) ? "Yes" : "No")end `
Output:
```
Kolakoski from [1, 2]: first 20 numbers are [1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1].
Does this look like a Kolakoski sequence? Yes
Kolakoski from [2, 1]: first 20 numbers are [2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2].
Does this look like a Kolakoski sequence? Yes
Kolakoski from [1, 3, 1, 2]: first 30 numbers are [1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1].
Does this look like a Kolakoski sequence? Yes
Kolakoski from [1, 3, 2, 1]: first 30 numbers are [1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1].
Does this look like a Kolakoski sequence? No

```

## Kotlin

`// Version 1.2.41 fun IntArray.nextInCycle(index: Int) = this[index % this.size] fun IntArray.kolakoski(len: Int): IntArray {    val s = IntArray(len)    var i = 0    var k = 0    while (true) {        s[i] = this.nextInCycle(k)        if (s[k] > 1) {            repeat(s[k] - 1) {                if (++i == len) return s                s[i] = s[i - 1]            }        }        if (++i == len) return s        k++    }} fun IntArray.possibleKolakoski(): Boolean {    val len = this.size    val rle = mutableListOf<Int>()    var prev = this[0]    var count = 1    for (i in 1 until len) {        if (this[i] == prev) {            count++        }        else {            rle.add(count)            count = 1            prev = this[i]        }          }    // no point adding final 'count' to rle as we're not going to compare it anyway    for (i in 0 until rle.size) {        if (rle[i] != this[i]) return false    }    return true} fun main(args: Array<String>) {    val ias = listOf(        intArrayOf(1, 2), intArrayOf(2, 1),        intArrayOf(1, 3, 1, 2), intArrayOf(1, 3, 2, 1)    )    val lens = intArrayOf(20, 20, 30, 30)    for ((i, ia) in ias.withIndex()) {        val len = lens[i]        val kol = ia.kolakoski(len)        println("First \$len members of the sequence generated by \${ia.asList()}:")        println(kol.asList())        val p = kol.possibleKolakoski()        println("Possible Kolakoski sequence? \${if (p) "Yes" else "No"}\n")    }}`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? Yes

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? No

```

## Lua

Translation of: C
`function next_in_cycle(c,length,index)    local pos = index % length    return c[pos]end function kolakoski(c,s,clen,slen)    local i = 0    local k = 0     while true do        s[i] = next_in_cycle(c,clen,k)        if s[k] > 1 then            for j=1,s[k]-1 do                i = i + 1                if i == slen then                    return nil                end                s[i] = s[i - 1]            end        end        i = i + 1        if i == slen then            return nil        end        k = k + 1    end    return nilend function possible_kolakoski(s,length)    local j = 0    local prev = s[0]    local count = 1    local rle = {}    local result = "True"     for i=0,length do        rle[i] = 0    end     for i=1,length-1 do        if s[i] == prev then            count = count + 1        else            rle[j] = count            j = j + 1            count = 1            prev = s[i]        end    end     -- no point adding the final 'count' to rle as we're not going to compare it anyway    for i=0,j-1 do        if rle[i] ~= s[i] then            result = "False"            break        end    end     return resultend function print_array(a)    io.write("[")    for i=0,#a do        if i>0 then            io.write(", ")        end        io.write(a[i])    end    io.write("]")end -- mainlocal c0 =    {[0]=1,  [1]=2}local c1 =    {[0]=2,  [1]=1}local c2 =    {[0]=1,  [1]=3,  [2]=1,  [3]=2}local c3 =    {[0]=1,  [1]=3,  [2]=2,  [3]=1} local cs =    {[0]=c0, [1]=c1, [2]=c2, [3]=c3}local clens = {[0]=2,  [1]=2,  [2]=4,  [3]=4}local slens = {[0]=20, [1]=20, [2]=30, [3]=30} for i=0,3 do    local clen = clens[i]    local slen = slens[i]    local s = {}     for j=0,slen-1 do        s[j] = 0    end     kolakoski(cs[i],s,clen,slen)    io.write(string.format("First %d members of the sequence generated by ", slen))    print_array(cs[i])    print(":")    print_array(s)    print()     local p = possible_kolakoski(s,slen)    print(string.format("Possible Kolakoski sequence? %s", p))     print()end`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? True

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? True

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? True

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? False```

## Mathematica/Wolfram Language

`ClearAll[KolakoskiGen]KolakoskiGen[start_List, its_Integer] := Module[{c, s, k, cnext, sk},  s = {};  k = 1;  c = start;  Do[   cnext = First[c];   c = RotateLeft[c];   AppendTo[s, cnext];   sk = s[[k]];   If[sk > 1,    s = Join[s, ConstantArray[cnext, sk - 1]]    ];   k += 1;   ,   {its}   ];  s  ] run = Take[KolakoskiGen[{1, 2}, 20], 20]check = Length /@ Split[%];check === Take[run, Length[check]] run = Take[KolakoskiGen[{2, 1}, 20], 20]check = Length /@ Split[%];check === Take[run, Length[check]] run = Take[KolakoskiGen[{1, 3, 1, 2}, 30], 30]check = Length /@ Split[%];check === Take[run, Length[check]]`
Output:
```{1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1}
True
{2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2}
True
{1,3,3,3,1,1,1,2,2,2,1,3,1,2,2,1,1,3,3,1,2,2,2,1,3,3,1,1,2,1}
True
{1,3,3,3,2,2,2,1,1,1,1,1,3,3,2,2,1,1,3,2,1,1,1,1,3,3,3,2,2,1}
False```

## Nim

Translation of: Kotlin
`template nextInCycle(a: openArray[int]; index: Natural): int =  a[index mod a.len] #--------------------------------------------------------------------------------------------------- func kolakoski(a: openArray[int]; length: Positive): seq[int] =   result.setLen(length)  var i, k = 0   while true:    result[i] = a.nextInCycle(k)    if result[k] > 1:      for j in 1..<result[k]:        inc i        if i == length: return        result[i] = result[i - 1]    inc i    if i == length: return    inc k #--------------------------------------------------------------------------------------------------- func possibleKolakoski(a: openArray[int]): bool =   var    rle: seq[int]    prev = a[0]    count = 1   for i in 1..a.high:    if a[i] == prev:      inc count    else:      rle.add count      count = 1      prev = a[i]   # No point adding final 'count' to rle as we're not going to compare it anyway.  for i, val in rle:    if val != a[i]: return false   result = true #——————————————————————————————————————————————————————————————————————————————————————————————————— when isMainModule:   import sequtils, strformat   const    Ias = [@[1, 2], @[2, 1], @[1, 3, 1, 2], @[1, 3, 2, 1]]    Lengths = [20, 20, 30, 30]   for (length, ia) in zip(Lengths, Ias):    let kol = ia.kolakoski(length)    echo &"First {length} members of the sequence generated by {(\$ia)[1..^1]}:"    echo (\$kol)[1..^1]    let s = if kol.possibleKolakoski(): "Yes" else: "No"    echo "Possible Kolakoski sequence? " & s & '\n'`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? Yes

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? No```

## Perl

Translation of: Raku
`sub kolakoski {    my(\$terms,@seed) = @_;    my @k;    my \$k = \$seed[0] == 1 ? 1 : 0;    if (\$k == 1) { @k = (1, split //, ((\$seed[1]) x \$seed[1])) }    else         { @k = (\$seed[0]) x \$seed[0] }    do {        \$k++;        push @k, (\$seed[\$k % @seed]) x \$k[\$k];    } until \$terms <= @k;    @k[0..\$terms-1]} sub rle {    (my \$string = join '', @_) =~ s/((.)\2*)/length \$1/eg;    split '', \$string} for ([20,1,2], [20,2,1], [30,1,3,1,2], [30,1,3,2,1]) {    \$terms = shift @\$_;    print "\n\$terms members of the series generated from [@\$_] is:\n";    print join(' ', @kolakoski = kolakoski(\$terms, @\$_)) . "\n";    \$status = join('', @rle = rle(@kolakoski)) eq join('', @kolakoski[0..\$#rle]) ? 'True' : 'False';    print "Looks like a Kolakoski sequence?: \$status\n";}`
Output:
```20 members of the series generated from [1 2] is:
1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1
Looks like a Kolakoski sequence?: True

20 members of the series generated from [2 1] is:
2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2
Looks like a Kolakoski sequence?: True

30 members of the series generated from [1 3 1 2] is:
1 3 3 3 1 1 1 2 2 2 1 3 1 2 2 1 1 3 3 1 2 2 2 1 3 3 1 1 2 1
Looks like a Kolakoski sequence?: True

30 members of the series generated from [1 3 2 1] is:
1 3 3 3 2 2 2 1 1 1 1 1 3 3 2 2 1 1 3 2 1 1 1 1 3 3 3 2 2 1
Looks like a Kolakoski sequence?: False```

## Phix

Translation of: C
`function kolakoski(sequence cycle, integer n)    sequence s = {}    integer k = 1    while length(s)<n do        integer c = cycle[mod(k-1,length(cycle))+1]        s &= repeat(c,iff(k>length(s)?c:s[k]))        k += 1    end while    s = s[1..n]    return send function function possible_kolakoski(sequence s)    integer count = 1    sequence rle = {}    for i=2 to length(s) do        if s[i]==s[i-1] then            count += 1        else            rle &= count            count = 1        end if    end for    -- (final count probably incomplete, so ignore it)    return rle = s[1..length(rle)]end function constant cycles = {{1,2},20,                   {2,1},20,                   {1,3,1,2},30,                   {1,3,2,1},30} for i=1 to length(cycles) by 2 do    {sequence c, integer n} = cycles[i..i+1]    sequence s = kolakoski(c,n)    printf(1,"First %d members of the sequence generated by %s\n", {n,sprint(c)})    ?s    bool p = possible_kolakoski(s)    printf(1,"Possible Kolakoski sequence? %s\n\n", {iff(p ? "Yes" : "No")})end for`
Output:
```First 20 members of the sequence generated by {1,2}
{1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1}
Possible Kolakoski sequence? Yes

First 20 members of the sequence generated by {2,1}
{2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2}
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by {1,3,1,2}
{1,3,3,3,1,1,1,2,2,2,1,3,1,2,2,1,1,3,3,1,2,2,2,1,3,3,1,1,2,1}
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by {1,3,2,1}
{1,3,3,3,2,2,2,1,1,1,1,1,3,3,2,2,1,1,3,2,1,1,1,1,3,3,3,2,2,1}
Possible Kolakoski sequence? No
```

## Python

Python 3.6+

`import itertools def cycler(start_items):	return itertools.cycle(start_items).__next__ def _kolakoski_gen(start_items):    s, k = [], 0    c = cycler(start_items)    while True:        c_next = c()        s.append(c_next)        sk = s[k]        yield sk        if sk > 1:            s += [c_next] * (sk - 1)        k += 1 def kolakoski(start_items=(1, 2), length=20):    return list(itertools.islice(_kolakoski_gen(start_items), length)) def _run_len_encoding(truncated_series):    return [len(list(group)) for grouper, group in itertools.groupby(truncated_series)][:-1] def is_series_eq_its_rle(series):    rle = _run_len_encoding(series)    return (series[:len(rle)] == rle) if rle else not series if __name__ == '__main__':    for start_items, length in [((1, 2), 20), ((2, 1), 20),                                 ((1, 3, 1, 2), 30), ((1, 3, 2, 1), 30)]:        print(f'\n## {length} members of the series generated from {start_items} is:')        s = kolakoski(start_items, length)        print(f'  {s}')        ans = 'YES' if is_series_eq_its_rle(s) else 'NO'        print(f'  Does it look like a Kolakoski sequence: {ans}')`
Output:
```## 20 members of the series generated from (1, 2) is:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Does it look like a Kolakoski sequence: YES

## 20 members of the series generated from (2, 1) is:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Does it look like a Kolakoski sequence: YES

## 30 members of the series generated from (1, 3, 1, 2) is:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Does it look like a Kolakoski sequence: YES

## 30 members of the series generated from (1, 3, 2, 1) is:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Does it look like a Kolakoski sequence: NO```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2018.04.01
`sub kolakoski (*@seed) {    my \$k = @seed[0] == 1 ?? 1 !! 0;    my @k = flat @seed[0] == 1 ?? (1, @seed[1] xx @seed[1]) !! @seed[0] xx @seed[0],      { \$k++; @seed[\$k % @seed] xx @k[\$k] } … *} sub rle (*@series) { @series.join.subst(/((.)\$0*)/, -> { \$0.chars }, :g).comb».Int } # Testingfor [1, 2], 20,    [2, 1], 20,    [1, 3, 1, 2], 30,    [1, 3, 2, 1], 30  -> @seed, \$terms {    say "\n## \$terms members of the series generated from { @seed.perl } is:\n   ",    my @kolakoski = kolakoski(@seed)[^\$terms];    my @rle = rle @kolakoski;    say "   Looks like a Kolakoski sequence?: ", @rle[*] eqv @kolakoski[^@rle];}`
Output:
```## 20 members of the series generated from [1, 2] is:
[1 2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1]
Looks like a Kolakoski sequence?: True

## 20 members of the series generated from [2, 1] is:
[2 2 1 1 2 1 2 2 1 2 2 1 1 2 1 1 2 2 1 2]
Looks like a Kolakoski sequence?: True

## 30 members of the series generated from [1, 3, 1, 2] is:
[1 3 3 3 1 1 1 2 2 2 1 3 1 2 2 1 1 3 3 1 2 2 2 1 3 3 1 1 2 1]
Looks like a Kolakoski sequence?: True

## 30 members of the series generated from [1, 3, 2, 1] is:
[1 3 3 3 2 2 2 1 1 1 1 1 3 3 2 2 1 1 3 2 1 1 1 1 3 3 3 2 2 1]
Looks like a Kolakoski sequence?: False```

## Ruby

`def create_generator(ar)  Enumerator.new do |y|    cycle = ar.cycle    s = []    loop do      t = cycle.next      s.push(t)      v = s.shift      y << v      (v-1).times{s.push(t)}    end  endend def rle(ar)  ar.slice_when{|a,b| a != b}.map(&:size)end [[20, [1,2]],  [20, [2,1]],  [30, [1,3,1,2]], [30, [1,3,2,1]]].each do |num,ar|  puts "\nFirst #{num} of the sequence generated by #{ar.inspect}:"  p res = create_generator(ar).take(num)  puts "Possible Kolakoski sequence? #{res.join.start_with?(rle(res).join)}"end`
Output:
```First 20 of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? true

First 20 of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? true

First 30 of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? true

First 30 of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? false
```

## Rust

` use itertools::Itertools; fn get_kolakoski_sequence(iseq: &[usize], size: &usize) -> Vec<usize> {    assert!(*size > 0);    assert!(!iseq.is_empty());     let mut kseq: Vec<usize> = Vec::default();     // create an itertor which keeps repeating the initial sequence infinitely    let repeater = iseq.iter().cloned().cycle();     // push the very first element, repeated as many times as the number    kseq.extend_from_slice(&vec![*iseq.get(0).unwrap()].repeat(*iseq.get(0).unwrap()));     //start cycling throught the initial sequence, but skip the very first one    for (k_counter, elem) in repeater.enumerate().skip(1) {        // push the given element        kseq.push(elem);         // and repeat the current element as many times        // as it's needed based on the previous elements        kseq.extend_from_slice(&vec![elem].repeat(*kseq.get(k_counter).unwrap() - 1));         // finish generation when the Kolakoski sequence has reached the given length        if kseq.len() >= *size {            break;        }    }     // truncate it as it might have more elements than needed    kseq[0..*size].to_vec()} fn is_kolakoski(kseq: &[usize]) -> bool {    assert!(!kseq.is_empty());     // calculate the RLE    let rle: Vec<usize> = kseq        .iter()        .batching(|it| {            it.next()                .map(|v| it.take_while_ref(|&v2| v2 == v).count() + 1)        })        .collect();     rle.iter().zip(kseq).filter(|&(a, b)| a == b).count() == rle.len()} fn main() {    let lengths = vec![20, 20, 30, 30];    let seqs = vec![vec![1, 2], vec![2, 1], vec![1, 3, 1, 2], vec![1, 3, 2, 1]];     for (seq, length) in seqs.iter().zip(&lengths) {        let kseq = get_kolakoski_sequence(&seq, length);         println!("Starting sequence: {:?}", seq);        println!("Kolakoski sequence: {:?}", kseq);        println!("Possible Kolakoski sequence? {:?}", is_kolakoski(&kseq));    }} `
Output:
```Starting sequence: [1, 2]
Kolakoski sequence: [1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? true
Starting sequence: [2, 1]
Kolakoski sequence: [2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? true
Starting sequence: [1, 3, 1, 2]
Kolakoski sequence: [1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? true
Starting sequence: [1, 3, 2, 1]
Kolakoski sequence: [1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? false
```

## Sidef

Translation of: Ruby
`func create_generator(arr) {    Enumerator({|f|        var s = []        var i = 0        loop {            var t = arr[i++ % arr.len]            s << t            f(var v = s.shift)            s << (v-1).of(t)...        }    })} var tests = [    [20, [1,2]],    [20, [2,1]],    [30, [1,3,1,2]],    [30, [1,3,2,1]]] for num,arr in (tests) {    say "\nFirst #{num} of the sequence generated by #{arr}:"    var res = create_generator(arr).first(num)    var rle = res.run_length.map{.tail}    say "#{res}\nPossible Kolakoski sequence? #{res.first(rle.len) == rle}"}`
Output:
```First 20 of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? true

First 20 of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? true

First 30 of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? true

First 30 of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? false
```

## Visual Basic .NET

Translation of: C#
`Imports System.Runtime.CompilerServicesImports System.Text Module Module1     Class Crutch        Public ReadOnly len As Integer        Public s() As Integer        Public i As Integer         Public Sub New(len As Integer)            Me.len = len            s = New Integer(len - 1) {}            i = 0        End Sub         Public Sub Repeat(count As Integer)            For j = 1 To count                i += 1                If i = len Then                    Return                End If                s(i) = s(i - 1)            Next        End Sub    End Class     <Extension()>    Public Function NextInCycle(self As Integer(), index As Integer) As Integer        Return self(index Mod self.Length)    End Function     <Extension()>    Public Function Kolakoski(self As Integer(), len As Integer) As Integer()        Dim c As New Crutch(len)         Dim k = 0        While c.i < len            c.s(c.i) = self.NextInCycle(k)            If c.s(k) > 1 Then                c.Repeat(c.s(k) - 1)            End If            c.i += 1            If c.i = len Then                Return c.s            End If            k += 1        End While         Return c.s    End Function     <Extension()>    Public Function PossibleKolakoski(self As Integer()) As Boolean        Dim rle(self.Length) As Integer        Dim prev = self(0)        Dim count = 1        Dim pos = 0        For i = 2 To self.Length            If self(i - 1) = prev Then                count += 1            Else                rle(pos) = count                pos += 1                 count = 1                prev = self(i - 1)            End If        Next        REM no point adding final 'count' to rle as we're not going to compare it anyway        For i = 1 To pos            If rle(i - 1) <> self(i - 1) Then                Return False            End If        Next        Return True    End Function     <Extension()>    Public Function AsString(self As Integer()) As String        Dim sb As New StringBuilder("[")        Dim it = self.GetEnumerator()        If it.MoveNext Then            sb.Append(it.Current)        End If        While it.MoveNext            sb.Append(", ")            sb.Append(it.Current)        End While        Return sb.Append("]").ToString    End Function     Sub Main()        Dim ias()() As Integer = {New Integer() {1, 2}, New Integer() {2, 1}, New Integer() {1, 3, 1, 2}, New Integer() {1, 3, 2, 1}}        Dim lens() As Integer = {20, 20, 30, 30}         For i = 1 To ias.Length            Dim len = lens(i - 1)            Dim kol = ias(i - 1).Kolakoski(len)             Console.WriteLine("First {0} members of the sequence by {1}: ", len, ias(i - 1).AsString)            Console.WriteLine(kol.AsString)            Console.WriteLine("Possible Kolakoski sequence? {0}", kol.PossibleKolakoski)            Console.WriteLine()        Next    End Sub End Module`
Output:
```First 20 members of the sequence by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? True

First 20 members of the sequence by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? True

First 30 members of the sequence by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? True

First 30 members of the sequence by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? False```

## Wren

Translation of: Go
`var kolakoski = Fn.new { |c, slen|    var s = List.filled(slen, 0)    var i = 0    var k = 0    while (true) {        s[i] = c[k % c.count]        if (s[k] > 1) {            for (j in 1...s[k]) {                i = i + 1                if (i == slen) return s                s[i] = s[i-1]            }        }        i = i + 1        if (i == slen) return s        k = k + 1    }} var possibleKolakoski = Fn.new { |s|    var slen = s.count    var rle = []    var prev = s[0]    var count = 1    for (i in 1...slen) {        if (s[i] == prev) {            count = count + 1        } else {            rle.add(count)            count = 1            prev = s[i]        }    }    // no point adding final 'count' to rle as we're not going to compare it anyway    for (i in 0...rle.count) {        if (rle[i] != s[i]) return false    }    return true} var ias = [    [1, 2],    [2, 1],    [1, 3, 1, 2],    [1 ,3, 2, 1]]var slens = [20, 20, 30, 30]var i = 0for (ia in ias) {    var slen = slens[i]    var kol = kolakoski.call(ia, slen)    System.write("First %(slen) members of the sequence generated by ")    System.print("%(ia):")    System.print("%(kol)")    var p = possibleKolakoski.call(kol)    var poss = p ? "Yes" : "No"    System.print("Possible Kolakoski sequence? %(poss)\n")    i = i + 1}`
Output:
```First 20 members of the sequence generated by [1, 2]:
[1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1]
Possible Kolakoski sequence? Yes

First 20 members of the sequence generated by [2, 1]:
[2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 1, 2]:
[1, 3, 3, 3, 1, 1, 1, 2, 2, 2, 1, 3, 1, 2, 2, 1, 1, 3, 3, 1, 2, 2, 2, 1, 3, 3, 1, 1, 2, 1]
Possible Kolakoski sequence? Yes

First 30 members of the sequence generated by [1, 3, 2, 1]:
[1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 3, 3, 2, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 3, 3, 2, 2, 1]
Possible Kolakoski sequence? No
```

## zkl

Translation of: Python
`fcn kolakoski(start_items=List(1,2), length=20){  //-->List   Walker.tweak(fcn(s,rk,cw){	// infinite iterator      s.append( c_next:=cw() );      sk:=s[rk.inc()];		// inc returns previous value, ie k++      if(sk>1) s.extend((List.createLong(sk - 1,c_next)));  // list of sk cn's      sk		// where we are in s, not end of s   }.fp(List(), Ref(0), Walker.cycle(start_items).next) )   .walk(length);	// iterate length times, return list}`
`fcn _run_len_encoding(truncated_series){  //List-->List   truncated_series.reduce(fcn(a,b,rm,s){ # if trailing singleton, it is ignored      if(a==b){ rm.inc(); return(b); }      s.append(rm.value);      rm.set(1);      b   }.fp2(Ref(1),s:=List()) );   s} fcn is_series_eq_its_rle(series){	//-->Bool   rle:=_run_len_encoding(series);   series[0,rle.len()]==rle}`
`foreach sl in (List( L( L(1,2), 20), L( L(2, 1), 20),                     L( L(1,3,1,2), 30), L( L(1,3,2,1), 30) )){   start_items, length := sl;   println("First %d members of the series generated from (%s) are:"           .fmt(length,start_items.concat(",")));   println("   (%s)".fmt(( s:=kolakoski(start_items, length) ).concat(",") ));   println("   Does it look like a Kolakoski sequence: ",is_series_eq_its_rle(s) )}`
Output:
```First 20 members of the series generated from (1,2) are:
(1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1)
Does it look like a Kolakoski sequence: True
First 20 members of the series generated from (2,1) are:
(2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2)
Does it look like a Kolakoski sequence: True
First 30 members of the series generated from (1,3,1,2) are:
(1,3,3,3,1,1,1,2,2,2,1,3,1,2,2,1,1,3,3,1,2,2,2,1,3,3,1,1,2,1)
Does it look like a Kolakoski sequence: True
First 30 members of the series generated from (1,3,2,1) are:
(1,3,3,3,2,2,2,1,1,1,1,1,3,3,2,2,1,1,3,2,1,1,1,1,3,3,3,2,2,1)
Does it look like a Kolakoski sequence: False
```