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 sshould appear at the end of s.

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

3. Increment k

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

5. 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

6. Increment k

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

8. 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

9. 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.

Task

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 couple 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 

Haskell

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] -> Bool
sameAsRleUpTo 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 nil
end

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 result
end

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

-- main
local 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

with javascript_semantics
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 s
end 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 }

# Testing
for [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
  end
end

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.CompilerServices
Imports 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

V (Vlang)

Translation of: Go

fn next_in_cycle(c []int, index int) int {
    return c[index % c.len]
}
 
fn kolakoski(c []int, slen int) []int {
    mut s := []int{len: slen}
    mut i, mut k := 0, 0
    for {
        s[i] = next_in_cycle(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++
    }
    return s
}
 
fn possible_kolakoski(s []int) bool {
    slen := s.len
    mut rle := []int{len: 0, cap:slen}
    mut prev := s[0]
    mut count := 1
    for i in 1..slen {
        if s[i] == prev {
            count++
        } else {
            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 in 0..rle.len {
        if rle[i] != s[i] {
            return false
        }
    }
    return true
}
 
fn print_ints(ia []int, suffix string) {
    print("[")
    alen := ia.len
    for i in 0.. alen {
        print(ia[i])
        if i < alen - 1 {
            print(", ")
        }
    }
    println("]$suffix")
}
 
fn main() {
    mut ias := [][]int{len: 4}
    ias[0] = [1, 2]
    ias[1] = [2, 1]
    ias[2] = [1, 3, 1, 2]
    ias[3] = [1, 3, 2, 1]
    slens := [20, 20, 30, 30]
    for i, ia in ias {
        slen := slens[i]
        kol := kolakoski(ia, slen)
        print("First $slen members of the sequence generated by ")
        print_ints(ia, ":")
        print_ints(kol, "")
        p := possible_kolakoski(kol)
        mut poss := "Yes"
        if !p {
            poss = "No"
        }
        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 

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 = 0
for (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