Say What You See

Question

The "Look and say" or "Say what you see" sequence is a series of numbers where each describes the last.

1
11 (one one)
21 (two ones)
1211 (one two, one one)
111221 (one one, one two, two ones)
312211 (three ones, two twos, one one)

and on and on... https://oeis.org/A005150

Anyway, this is a regular code golf challenge (least byte count wins) to make a program that takes two arguments, an initial number and the amount of iterations. For example if you plugged in "1" and "2" the result would be "21". If you plugged in "2" and "4" the result would be "132112". Have fun!

Answer 1

Pyth, 10 8 bytes

-2 bytes by @FryAmTheEggman

ussrG8Qz

Explanation:

            Implicit: z=first line as string, Q=second line
u         the result of reducing lambda G:
  s s rG8   flattened run-length-encoded G
  Q       Q times
  z     starting with z
  

Try it here.

Answer 2

CJam, 8 bytes

q~{se`}*

Input format is the initial number first, iterations second, separated by some whitespace.

Test it here.

Explanation

q~   e# Read and evaluate input, dumping both numbers on the stack.
{    e# Run this block once for each iteration...
  s  e#   Convert to string... in the first iteration this just stringifies the input
     e#   number again. In subsequent iterations it flattens and then stringifies the
     e#   array we get from the run-length encoding.
  e` e#   Run-length encode.
}*

The array is also flattened before being printed so the result is just the required number.

Answer 3

JavaScript, 57 bytes

F=(a,b)=>b?F(a.replace(/(.)\1*/g,c=>c.length+c[0]),b-1):a

Recursion works well for this problem. The first parameter is the initial number as a string, and the second is the number of iterations.

Answer 4

R, 87 bytes

function(a,n){for(i in 1:n){r=rle(el(strsplit(a,"")));a=paste0(r$l,r$v,collapse="")};a}

Ungolfed & explained

f=function(a,n){
    for(i in 1:n){                      # For 1...n
        r=rle(el(strsplit(a,"")))       # Run length encoding
        a=paste0(r$l,r$v,collapse="")   # concatenate length vector and values vector and collapse
    };
    a                                   # print final result
}

Answer 5

Ruby, 63 bytes

A full program, since the question seems to ask for that. Takes input as command line arguments.

i,n=$*
n.to_i.times{i=i.gsub(/(.)\1*/){"#{$&.size}#$1"}}
puts i

^{No, gsub! can't be used, since the strings in $* are frozen :/}

Answer 6

MATL, 9 bytes

:"Y'wvX:!

Inputs are: number of iterations, initial number.

Try it online!

:      % implicit input: number of iterations. Create vector with that size
"      % for loop
  Y'   %   RLE. Pushes two arrays: elements and numbers of repetitions.
       %   First time implicitly asks for input: initial number
  w    %   swap
  v    %   concatenate vertically
  X:   %   linearize to column array
  !    %   transpose to row array
       % implicitly end loop
       % implicitly display

Answer 7

R, 61 57 bytes

-4 thanks to @JayCe, just when I was sure it couldn't be done any simpler!

f=function(a,n)`if`(n,f(t(sapply(rle(c(a)),c)),n-1),c(a))

Try it online!

Answer 8

Perl 6, 63 bytes

say (@*ARGS[0],*.trans(/(.)$0*/=>{$/.chars~$0})…*)[@*ARGS[1]]

This is as short as I could get it for now, there might be some tricky flags that could reduce it, I'm not sure

Answer 9

Retina, 46 45 27 bytes

Martin did lots to help golf this.

+`(\d)(\1?)*(?=.*_)_?
$#2$1

Try it online

Takes input in the format:

<start><count>

<start> is the initial number.

<count> is in unary, all underscores, and is how many iterations are performed.

Single iteration, 20 16 bytes:

(\d)(\1?)*
$#2$1

Answer 10

Haskell, 62 bytes

import Data.List
0%y=y
x%y=do x<-group$(x-1)%y;[length x,x!!0]

Try it online!

Answer 11

JavaScript ES6, 71 bytes

(m,n)=>[...Array(n)].map(_=>m=m.replace(/(.)\1*/g,x=>x.length+x[0]))&&m

Takes input as a string and a number.

Answer 12

Perl 5, 50 bytes

$_=pop;for$i(1..pop){s/(.)\1*/length($&).$1/ge}say

The arguments are in reverse order (number of iterations then seed). Example:

> perl -E'$_=pop;for$i(1..pop){s/(.)\1*/length($&).$1/ge}say' 4 2
132112
> perl -E'$_=pop;for$i(1..pop){s/(.)\1*/length($&).$1/ge}say' 0 2
2
> perl -E'$_=pop;for$i(1..pop){s/(.)\1*/length($&).$1/ge}say' 2 0
1110
> perl -E'$_=pop;for$i(1..pop){s/(.)\1*/length($&).$1/ge}say' 1 10
1110
> perl -E'$_=pop;for$i(1..pop){s/(.)\1*/length($&).$1/ge}say' 11 1
3113112221232112111312211312113211

Answer 13

Python 3.6, 100 98 93 bytes

import re
f=lambda s,n:n and eval("f'"+re.sub(r'((.)\2*)',r'{len("\1")}\2',f(s,n-1))+"'")or s

Try it online!

Note this creates a lambda that takes a string and an integer, and returns a string. Example: f('1', 5) == '312211'

Finds all repeated characters (((.)\2*) regex), makes a f-string out of their length and the character itself (r'{len("\1")}\2'), then evaluates it. Uses recursion on the counter (n and ...f(s,n-1)... or s) to avoid having to define a proper function and a loop.

Answer 14

Jelly, 6 bytes

ŒrUFµ¡

Try it online!

           Implicit input: first argument.
     µ¡    Do this to it <second argument> times:
Œr            Run-length encode into [value, times] pairs
  U           Flip them
   F          Flatten list
    

Answer 15

Perl, 38 + 2 bytes

for$i(1..<>){s/(.)\1*/(length$&).$1/ge}

Requires the -p flag:

$ perl -pe'for$i(1..<>){s/(.)\1*/(length$&).$1/ge}' <<< $'1\n5'
312211

Input is a multi line string:

input number
numbers of iterations

If all the steps are required as well then we can change it to the following, which is 44 + 2 bytes:

$ perl -nE'for$i(1..<>){s/(.)\1*/(length$&).$1/ge,print}' <<< $'1\n5'
11
21
1211
111221
312211

Answer 16

Mathematica, 81 73 bytes

FromDigits@Nest[Flatten[(Tally/@Split@#)~Reverse~3]&,IntegerDigits@#,#2]&

Answer 17

05AB1E, 9 bytes (Non-competing)

Corrected due to Emigna's comments, see below/edits.

F.¡vygyÙJ

Try it online!

Answer 18

Stax, 10 bytes

Çα▲ì4↔┌j█♀

Run and debug online!

Spent too many bytes on proper IO format ...

Explanation

Uses the unpacked version to explain.

DE|R{rm:f$e
D              Do `2nd parameter` times
 E             Convert number to digits
                   Starting from the `1st parmeter`
  |R           Convert to [element, count] pairs for each run
    {rm        Revert each pair
       :f      Flatten the array
         $     Convert array to string of digits
          e    Convert string of digits to integer

The essential part is D|R{rm:f(8 bytes).

If the first input can be taken as an array of digits, the whole program can be written in 9 bytes: Run and debug online!

Answer 19

K (ngn/k), 30 bytes

{y{,/{(#x;*x)}'(&~=':x)_x}/,x}

Try it online!

Answer 20

J, 26 bytes

2&([:;](#<@,{.);.1~1,~:/\)

Try it online!

Answer 21

05AB1E, 6 bytes

FÅγøí˜

Try it online!

FÅγøí˜  # full program
F       # for N in [0, 1, ...,
        # ..., implicit input...
F       # ... minus 1]...
     ˜  # flatten...
   ø    # zipped...
 Åγ     # list of chars used in runs of the same char in...
        # implicit input...
 Åγ     # or top of stack if not first iteration...
   ø    # with...
 Åγ     # list of lengths of runs of the same char in...
        # implicit input...
 Åγ     # or top of stack if not first iteration...
    í   # with each element of the list reversed
        # (implicit) exit loop
        # implicit output

øí can also be sø with no change in functionality. Try it online!

Answer 22

Husk, 11 bytes

!¡(dṁ§eL←gd

Try it online!

!              # get the arg2-th element of
 ¡             # the infinite list by repeatedly applying
               # (starting with arg1):
  (d           # get the digits of
    ṁ          # applying to each of 
         gd    # the groups of identical neighbouring digits:
     §eL←      # combine length + first element

Answer 23

Go, 175 bytes

import."fmt"
func g(n int,s string)string{for N:=0;N<n;N++{c,o,f,k:=1,"",rune(s[0]),s[1:]+"_"
for _,r:=range k{if r==f{c++}else{o+=Sprintf("%d%c",c,f);f,c=r,1}}
s=o}
return s}

Attempt This Online!

A non-recursive port of Ogaday's answer.

Answer 24

J-uby, 41 bytes

Takes curried arguments with number of iterations first.

:**&(~(:gsub+:+%[:+@|S,~:[]&0])&/(.)\1*/)

Attempt This Online!

Non-regex solution, 47 bytes

:**&(A|:slice_when+:!=|:*&-[:+@,~:[]&0]|~:*&"")

Attempt This Online!

Answer 25

Vyxal, 31 bits^v2, 3.875 bytes

(øeRf

Try it Online!

Bitstring:

0011010110011001011100101001100

inputs are reversed

Answer 26

Japt -h, 14 bytes

ÆV=ìÈòÎcÈâ iXÊ

Try it

Answer 27

YASEPL, 132 bytes

=p'=n'(+`2£s©1`1=q)""=x=l®"p"`3=t$x!h¥t,"p"`4!x+}1,l,5!y¥x,"p"}3,h,4`5!f$x-tſ""!q+f+h!x}2,l,3!n-!p$q<!o$p(!s©o!pſ""!n}!s<!i+}2,n,2

prompts you twice. enter starting number and amount of iterations

Answer 28

Python 3, 138 bytes

I used a recursive approach.

def g(a,b):
 if b<1:return a
 else:
  c,n=1,'';f,*a=str(a)+'_'
  for i in a:
   if i==f:c+=1
   else:n+=str(c)+f;f,c=i,1
  return g(n,b-1)

The function accepts two ints, a and b as described.

I'm amazed at how terse the entries here are! Maybe someone will come along with a better Python method too.

Answer 29

Pylons, 11

i:At,{n,A}j

How it works:

i      # Get input from command line.
:A     # Initialize A
  t    # Set A to the top of the stack.
,      # Pop the top of the stack.
{      # Start a for loop.
 n     # Run length encode the stack.
  ,    # Seperate command and iteration
   A   # Repeat A times.
    }  # End for loop.
j      # Join the stack with '' and print it and then exit. 

Answer 30

SmileBASIC, 100 98 bytes

DEF S N,T?N
WHILE""<N
C=C+1C$=SHIFT(N)IF C$!=(N+@L)[0]THEN O$=O$+STR$(C)+C$C=0
WEND
S O$,T-T/T
END

Prints out all the steps. T/T is there to end the program when T is 0.