# Say What You See

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!

• Can we receive/return a list of digits? Feb 2 '16 at 23:38
• I'd close the older questions as dupes if necessary; this doesn't have the restrictions. Feb 2 '16 at 23:43
• I don't see this as a duplicate. Both the previous look and say challenges were very restrictive (one with no numbers in the source code, the other with no named variables, named functions or named arguments). Very few languages will allow answers to the previous challenges that are also competitive here. Feb 2 '16 at 23:56
• Are we allowed to output as a list of numbers? Feb 3 '16 at 0:23
• Discussion on meta about duplicates without restrictions Feb 3 '16 at 15:06

## 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.

• But at least I don't output brackets and commas; only spaces between the numbers :-P Feb 3 '16 at 0:43
• In Soviet Russia, ussrG8Qz Feb 1 '17 at 17:27

## 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.

# 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.

• You can save three bytes with a weird recursive curry: b=>F=a=>b--?F(a.replace(/(.)\1*/g,c=>c.length+c[0])):a Found that while golfing my answer before I realized it was pretty much identical to yours ;) Feb 1 '17 at 12:32

## 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
}


# 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

• If you can output as an array then Pyth has 8. Feb 3 '16 at 0:24
• @ThomasKwa Good point. I assumed it was possible Feb 3 '16 at 0:35

# 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 :/

• Could you perhaps use the -p flag to save bytes? If you use it, gsub operates on a line of STDIN as if it were $_.gsub!. Then the command line argument is the iterations, so n,=$*, and the other input is read from STDIN. Dec 24 '16 at 5:18

# Perl 6, 63 bytes

say (@*ARGS[0],*.trans(/(.)$0*/=>{$/.chars~$0})…*)[@*ARGS[1]]  This is as short as I could get it for now, there might be some tricky flags that could reduce it, I'm not sure # Retina, 4645 27 bytes Martin did lots to help golf this. +(\d)(\1?)*(?=.*_)_?$#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$1  # R, 61 57 bytes -4 thanks to @JayCe, just when I was sure it couldn't be done any simpler! f=function(a,n)if(n,f(t(sapply(rle(c(a)),c)),n-1),c(a))  Try it online! • Slightly golfed: TIO Sep 6 '18 at 15:14 • That t(sapply(z,c)) call is clever. Sep 6 '18 at 17:58 # Haskell, 62 bytes import Data.List 0%y=y x%y=do x<-group$(x-1)%y;[length x,x!!0]


Try it online!

# 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.

• ('1',2) gives me 12, when it should be 21. Your length should come before the character in the replace. Feb 3 '16 at 18:52
• @Mwr247 Whoops, sorry. Feb 3 '16 at 20:58

# 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

• As a subroutine, I can shave a byte by ending it with $_ instead of say, I suppose, but I haven't tested it. The current solution is a program. Feb 3 '16 at 21:10 ## Perl, 38 + 2 bytes for$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


# Mathematica, 81 73 bytes

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

• Prepend your code with four spaces to get it show up as code :) Feb 4 '16 at 11:50

# Jelly, 6 bytes (non-competing)

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


# 05AB1E, 9 bytes (Non-competing)

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

F.¡vygyÙJ


Try it online!

• I think you missed the part about taking 2 arguments (initial number and number of iterations). Luckily you can just add F at the beginning and take the arguments as iterations,initialNo Feb 1 '17 at 15:36
• And the byte lost by that could be regained by replacing Dgs with gy. Feb 1 '17 at 15:39
• @Emigna what's y do in that context? Feb 1 '17 at 17:21
• Same as the first y, push the current value in the loop. So instead of duplicating y and swapping it to the top, you just push it again when you need it. Feb 1 '17 at 21:19
• @Emigna it should seem I still have a lot to learn haha. Feb 7 '17 at 21:00

# 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!

• -1 if you use $ instead of :f Feb 27 '21 at 2:25 # K (ngn/k), 30 bytes {y{,/{(#x;*x)}'(&~=':x)_x}/,x}  Try it online! # Python 3.6, 10098 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. # 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  # 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. # 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.  # 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.

# APL (Dyalog Classic), 22 bytes

{∊(≢,⊃)¨⍵⊂⍨1,2≠/⍵}⍣⎕,⎕


Try it online!

# Retina, 27 bytes

~(^
K
$+(.)\1$*¶$$.&$$1


Try it online!

# 05AB1E, 7 bytes

FÅγs.ιJ


Try it online!

# J, 26 bytes

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


Try it online!

# 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!

# GolfScript, 32 bytes

~{-1%1/(1@{.3$={;)}1if}/]{\+}*}*  Try it online! Input with first arg as string, +2 bytes for first arg as num: ~{-1%1/(1@{.3$={;)}1if}/]{\+}*~}*
`