30
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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!

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  • 2
    \$\begingroup\$ Can we receive/return a list of digits? \$\endgroup\$ – LegionMammal978 Feb 2 '16 at 23:38
  • 5
    \$\begingroup\$ I'd close the older questions as dupes if necessary; this doesn't have the restrictions. \$\endgroup\$ – lirtosiast Feb 2 '16 at 23:43
  • 4
    \$\begingroup\$ 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. \$\endgroup\$ – trichoplax Feb 2 '16 at 23:56
  • 3
    \$\begingroup\$ Are we allowed to output as a list of numbers? \$\endgroup\$ – lirtosiast Feb 3 '16 at 0:23
  • 3
    \$\begingroup\$ Discussion on meta about duplicates without restrictions \$\endgroup\$ – trichoplax Feb 3 '16 at 15:06

24 Answers 24

9
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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.

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  • \$\begingroup\$ But at least I don't output brackets and commas; only spaces between the numbers :-P \$\endgroup\$ – Luis Mendo Feb 3 '16 at 0:43
  • 2
    \$\begingroup\$ In Soviet Russia, ussrG8Qz \$\endgroup\$ – mbomb007 Feb 1 '17 at 17:27
8
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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.

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6
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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.

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  • \$\begingroup\$ 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 ;) \$\endgroup\$ – ETHproductions Feb 1 '17 at 12:32
4
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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
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  • \$\begingroup\$ If you can output as an array then Pyth has 8. \$\endgroup\$ – lirtosiast Feb 3 '16 at 0:24
  • \$\begingroup\$ @ThomasKwa Good point. I assumed it was possible \$\endgroup\$ – Luis Mendo Feb 3 '16 at 0:35
4
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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
}
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3
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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

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3
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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 :/

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  • \$\begingroup\$ 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. \$\endgroup\$ – Value Ink Dec 24 '16 at 5:18
3
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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
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3
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Haskell, 62 bytes

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

Try it online!

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2
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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.

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  • \$\begingroup\$ ('1',2) gives me 12, when it should be 21. Your length should come before the character in the replace. \$\endgroup\$ – Mwr247 Feb 3 '16 at 18:52
  • \$\begingroup\$ @Mwr247 Whoops, sorry. \$\endgroup\$ – ETHproductions Feb 3 '16 at 20:58
2
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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
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  • \$\begingroup\$ 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. \$\endgroup\$ – msh210 Feb 3 '16 at 21:10
2
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05AB1E, 9 bytes (Non-competing)

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

F.¡vygyÙJ

Try it online!

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  • 1
    \$\begingroup\$ 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 \$\endgroup\$ – Emigna Feb 1 '17 at 15:36
  • 1
    \$\begingroup\$ And the byte lost by that could be regained by replacing Dgs with gy. \$\endgroup\$ – Emigna Feb 1 '17 at 15:39
  • \$\begingroup\$ @Emigna what's y do in that context? \$\endgroup\$ – Magic Octopus Urn Feb 1 '17 at 17:21
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Emigna Feb 1 '17 at 21:19
  • \$\begingroup\$ @Emigna it should seem I still have a lot to learn haha. \$\endgroup\$ – Magic Octopus Urn Feb 7 '17 at 21:00
2
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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!

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  • 1
    \$\begingroup\$ Slightly golfed: TIO \$\endgroup\$ – JayCe Sep 6 '18 at 15:14
  • \$\begingroup\$ That t(sapply(z,c)) call is clever. \$\endgroup\$ – J.Doe Sep 6 '18 at 17:58
1
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Mathematica, 81 73 bytes

FromDigits@Nest[Flatten[(Tally/@Split@#)~Reverse~3]&,IntegerDigits@#,#2]&
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  • \$\begingroup\$ Prepend your code with four spaces to get it show up as code :) \$\endgroup\$ – Ogaday Feb 4 '16 at 11:50
1
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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
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1
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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!

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0
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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.

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0
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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
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0
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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. 
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0
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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.

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0
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APL (Dyalog Classic), 22 bytes

{∊(≢,⊃)¨⍵⊂⍨1,2≠/⍵}⍣⎕,⎕

Try it online!

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0
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Retina, 27 bytes

~(`^
K`
$
+`(.)\1$*¶$$.&$$1

Try it online!

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0
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K (ngn/k), 30 bytes

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

Try it online!

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0
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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.

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