# Look-and-say: Conway revisited

You should all be familiar with the Conway sequence (a.k.a. 'look-and-say'-sequence) by now:

     1
11
21
1211
111221
312211
etc


You can also start by any arbitrary number as starting point. Let f(s) be the next element of the sequence. Now for every given swe can find f(s). The reverse is not as trivial: it is not for every y possible to find the predecessor s such that f(s) = y. E.g. for y = 1 we cannot find a predecessor. But if y has an even length you can divide it into pairs of digits which describe each one part of a predecessor:

513211 divides in 51,32,11
so: 51 comes from 11111
32 comes from 222
11 comes from 1
put together: 111112221


So this way there we can define an unique predecessor for every y of even length.

Note: The 'predecessor' s defined this way does generally NOT satisfy f(s) = y.

## Goal

Write a function / program snippet that accepts a string of digits as input that

• calculates the next element of the Conway sequence if the length of the input string is odd
• calculates the predecessor of the input string as defined above if the input string length is even.

Shortest code in bytes wins.

Recent Questions based on the look-and-say sequences:

• I'm confused. Can you explain how 513111 divides into 51, 32 and 11? – squeamish ossifrage Nov 27 '14 at 22:44
• I feel like this is a combined duplicate of some look and say sequence challenge and a run-length decoding challenge (I'm sure we had those). – Martin Ender Nov 27 '14 at 22:57
• What would the predecessor of 11111111111111 be? According to your spec, it would be 1111111. You should modify your specification to define a reasonable answer for this. – TheNumberOne Nov 27 '14 at 23:44
• @TheBestOne 11111111111111 simply has no predecessor. It's an illegal input. – kay Nov 28 '14 at 0:37
• @TheBestOne Yes this is correct, I defined an arbitrary rule for the predecessor that does not always match a 'real' predecessor. – flawr Nov 28 '14 at 9:19

# CJam, 46454443 42 bytes

l_,2%{0\{@)@@_2$=!{0\0}*;}*\)\}{2/{(~*}%}?  Test it here. It takes the number on STDIN and prints the result to STDOUT. • si -> ~ = 45 – Optimizer Nov 28 '14 at 14:19 • @Optimizer Thanks, I forget you can eval a character. – Martin Ender Nov 28 '14 at 14:21 # Ruby, 125 120 119 101 bytes f=->n{c=n.chars;(n.size%2>0?c.chunk{|x|x}.map{|a,b|[b.size,a]}:c.each_slice(2).map{|a,b|b*a.hex})*''}  String input taken through function f: f['111221'] # => "1211" f['513211'] # => "111112221" f['111112221'] # => "513211"  Expanded with notes: # define lambda that takes one arg f = -> (n) { # store digits in c c = n.chars # n is of odd length if n.size % 2 > 0 # group identical numbers c.chunk{ |x| x }.map do |a, b| # array of [digit count, digit value] [b.size, a] end else # slice array into groups of two elements c.each_slice(2).map do |a, b| # repeat the second digit in a pair # the first digit-times. b * a.hex end end * '' # join array }  ## Prolog - 170 bytes []/[]. T/R:-0*_*T*R. C*X*[X|T]*R:-(C+1)*X*T*R. C*X*T*Y:-10*C+X+Y+R,T/R. N+R+A:-N=:=0,A=R;D is N mod 10,N//10+R+[D|A]. F-C-R:-C/N,(N=F,R=C;F-N-R). [1]-[1,1]. S-T:-S-[1]-T.  This snipped defines the function (-)/2. You can invoke it like ?- [1,1,1,3,2,1,3,2,1,1]-T. T = [1, 3, 1, 1, 2, 2, 2, 1] . ?- [1]-T. T = [1, 1] .  There seems to be only one lengths in this sequence with an odd parity: the initial [1]. wr_len :- wr_len(1, [1]). wr_len(N, Cur) :- length(Cur, Len), TrailingZeroes is lsb(Len), (TrailingZeroes > 0 -> Par = 'even'; Par = 'odd'), writef('%t\t%t\t%t\t%t\n', [N, Len, Par, TrailingZeroes]), get_next(Cur, Next), succ(N, O), !, wr_len(O, Next).  % index, length, parity of length, num of trailing 0 in bin presentation of length ?- wr_len. 1 1 odd 0 2 2 even 1 3 2 even 1 4 4 even 2 5 6 even 1 6 6 even 1 7 8 even 3 8 10 even 1 9 14 even 1 10 20 even 2 11 26 even 1 12 34 even 1 13 46 even 1 14 62 even 1 15 78 even 1 16 102 even 1 17 134 even 1 18 176 even 4 19 226 even 1 20 302 even 1 21 408 even 3 22 528 even 4 23 678 even 1 24 904 even 3 25 1182 even 1 26 1540 even 2 27 2012 even 2 28 2606 even 1 29 3410 even 1 30 4462 even 1 31 5808 even 4 32 7586 even 1 33 9898 even 1 34 12884 even 2 35 16774 even 1 36 21890 even 1 37 28528 even 4 38 37158 even 1 39 48410 even 1 40 63138 even 1 41 82350 even 1 42 107312 even 4 43 139984 even 4 44 182376 even 3 45 237746 even 1 46 310036 even 2 47 403966 even 1 48 526646 even 1 49 686646 even 1 50 894810 even 1 51 1166642 even 1 52 1520986 even 1 53 1982710 even 1 54 2584304 even 4 55 3369156 even 2 56 4391702 even 1 57 5724486 even 1 58 7462860 even 2 59 9727930 even 1 ERROR: Out of global stack % I added a few "strategic" cuts (!) to get so far.  Readable: get_next([], []). get_next(Current, Next) :- get_next_sub(0, _, Current, Next). get_next_sub(Length, Digit, [Digit|Tail], Result) :- get_next_sub(Length+1, Digit, Tail, Result). get_next_sub(Length, Digit, Further, Result) :- number_to_list(10*Length+Digit, Result, ResultTail), get_next(Further, ResultTail). number_to_list(Number, Result, Accumulator) :- 0 is Number -> Result = Accumulator; Digit is Number mod 10, number_to_list(Number // 10, Result, [Digit|Accumulator]). get_previous(Stop, Current, Result) :- get_next(Current, Next), ( Next = Stop -> Result = Current ; get_previous(Stop, Next, Result) ). get_prev_or_next(Input, Result) :- length(Input, Length), ( 1 is Length mod 2 -> get_next(Input, Result) ; get_previous(Input, [1], Result) ).  # Python: 139 chars import re f=lambda s:''.join(['len(b)'+a for a,b in re.findall(r'((\d)\2*)',s)] if len(s)%2 else map(lambda a,b:b*int(a),s[::2],s[1::2]))  single test case print f('1111122221111222112') >>> 514241322112 print f('514241322112') >>> 1111122221111222112  • You can remove the space from s)] if to s)]if. – Bakuriu Nov 30 '14 at 8:38 • You can also remove the space in between 2 else – Beta Decay Nov 30 '14 at 10:20 # Haskell, 134 128 115 n=length l x=x#mod(n x)2 (a:b:c)#0=replicate(read[a])b++c#0 (a:b)#1=(\(c,d)->show(1+n c)++a:d#1)$span(==a)b
_#_=[]


If it needs to be from stdin/stdout, add main=interact l for 150 144 131 total chars. The function is called l.

*Main> putStr . unlines $map (\x->x++":\t"++l x) ([replicate n '1'|n<-[5..10]]++map show [0,6..30]++map show [n*n+n|n<-[2..10]]) 11111: 51 111111: 111 1111111: 71 11111111: 1111 111111111: 91 1111111111: 11111 0: 10 6: 16 12: 2 18: 8 24: 44 30: 000 6: 16 12: 2 20: 00 30: 000 42: 2222 56: 66666 72: 2222222 90: 000000000 110: 2110  • Could you please provide usage example? While I got l "11" to work, I get an exception with l "111" or l "1111111111111" – Paul Guyot Nov 28 '14 at 14:35 • @PaulGuyot, Seems like fixing that cut a few chars off my score. Thanks :-) – Zaq Nov 28 '14 at 16:00 # Perl - 98 bytes The size of all these control statements bugs me, but I'm pretty happy with how the regexes worked out. ($_)=@ARGV;if(length()%2){$\=$&,s/^$&+//,print length$&while/^./}else{print$2x$1while s/^(.)(.)//}


Uncompressed:

($_)=@ARGV; if(length()%2) {$\=$&, #Assigning the character to$\ causes it to be appended to the print (thanks, tips thread!)
s/^$&+//, #Extract the string of characters print length$& #Print its length
while/^./ #Get next character into $& }else{ print$2x$1 while s/^(.)(.)// }  # Erlang, 205 f(L)->g(L,L,[]). g([A,B|T],L,C)->g(T,L,lists:duplicate(A-$0,B)++C);g([],_,R)->R;g(_,L,_)->i(L,[]).
h([A|T],A,N,B)->h(T,A,N+1,B);h(L,B,N,C)->i(L,integer_to_list(N)++[B|C]).
i([H|T],A)->h(T,H,1,A);i(_,R)->R.


Main function is f, taking the input as an Erlang string and returning the output as a string as well.

f("11"). % returns "1"
f("111"). % returns "31"
f("1111111111111"). % returns "131"


The function can be made 15 bytes shorter (190) by dropping the more than 9 identical characters case requirement. f calls g which computes the predecessor recursively, and if the number of characters is odd (found at when computation ends), it calls function i which, paired with h, computes the next element.

import Data.List
l=length
c r|odd\$l r=group r>>=(l>>=(.take 1).shows)|x:y:z<-r=[y|_<-['1'..x]]++c z|0<1=r


I think it's nice it turned out not using any helper functions :-).

• |x:y:z<-r - I totally did not know you could do that. That's so cool! – Flonk Nov 28 '14 at 17:35
• @Flonk Its called "pattern guards". It used to be an extension and in one of the versions of Haskell it was added :-) – proud haskeller Nov 28 '14 at 17:40
• This makes a lot of things so much easier. You learn something new every day! :) – Flonk Nov 28 '14 at 17:48

## APL (45)

∊{2|⍴⍵:(⊃,⍨⍕∘⍴)¨⍵⊂⍨1,2≠/⍵⋄/⍨∘⍎/¨⌽¨⍵⊂⍨1 0⍴⍨⍴⍵}


Yes, that's a valid function definition, even with the ∊ on the outside.

      ∊{2|⍴⍵:(⊃,⍨⍕∘⍴)¨⍵⊂⍨1,2≠/⍵⋄/⍨∘⍎/¨⌽¨⍵⊂⍨1 0⍴⍨⍴⍵}'513211'
111112221
∊{2|⍴⍵:(⊃,⍨⍕∘⍴)¨⍵⊂⍨1,2≠/⍵⋄/⍨∘⍎/¨⌽¨⍵⊂⍨1 0⍴⍨⍴⍵}'111112221'
513211
f←∊{2|⍴⍵:(⊃,⍨⍕∘⍴)¨⍵⊂⍨1,2≠/⍵⋄/⍨∘⍎/¨⌽¨⍵⊂⍨1 0⍴⍨⍴⍵}
f ¨ '513211' '111112221'
┌─────────┬──────┐
│111112221│513211│
└─────────┴──────┘


# Java 7, Score = 252 235 bytes

Yep, it's java again; the worst golfing language in the world. This approach uses strings. Arbitrarily large integers are supported in java but would take much more room to code in.

Call with f(intputString). Returns the corresponding string.

Golfed:

String f(String a){char[]b=a.toCharArray();a="";int i=0,j=0,d=0,e=b.length;if(e%2==0)for(;i<e;i+=2)for(j=0;j<b[i]-48;j++)a+=b[i+1];else{for(;i<e;i++)d=d==0?(j=b[i]-48)*0+1:b[i]-48!=j?(a+=d+""+j).length()*--i*0:d+1;a+=d;a+=j;}return a;}


Golfed Expanded with structure code:

public class LookAndSayExpandedGolfed{

public static void main(String[] args){
System.out.println(new LookAndSayExpandedGolfed().f(args[0]));
}

String f(String a){
char[]b=a.toCharArray();
a="";
int i=0,j=0,d=0,e=b.length;
if(e%2==0)
for(;i<e;i+=2)
for(j=0;j<b[i]-48;j++)
a+=b[i+1];
else{
for(;i<e;i++)
d=d==0?(j=b[i]-48)*0+1:b[i]-48!=j?(a+=d+""+j).length()*--i*0:d+1;
a+=d;
a+=j;
}
return a;
}

}


Partially Golfed:

public class LookAndSayPartiallyGolfed{

public static void main(String[] args){
System.out.println(new LookAndSayPartiallyGolfed().f(args[0]));
}

String f(String a){
char[] number = a.toCharArray();
int i, j, longestStreakLength = 0;
if (number.length % 2 == 0){
for (i = 0; i < number.length; i += 2)
for (j = 0; j < number[i] - 48; j++)
} else{
j = 0;
for (i = 0; i < number.length; i++)
longestStreakLength = longestStreakLength == 0 ? (j = number[i] - 48) * 0 + 1 : number[i] - 48 != j ? (answer += longestStreakLength + "" + j).length()*--i*0 : longestStreakLength + 1;
}
}

}


Completely expanded:

public class LookAndSay{

public static void main(String[] args){
System.out.println(new LookAndSay().function(args[0]));
}

String function(String a){
char[] number = a.toCharArray();
if (number.length % 2 == 0){
for (int i = 0; i < number.length; i += 2){
for (int j = 0; j < number[i] - '0'; j++){
}
}
}
int longestStreakLength = 0;
int longestStreakNum = 0;
for (int i = 0; i < number.length; i++){
if (longestStreakLength == 0){
longestStreakNum = number[i] - '0';
longestStreakLength = 1;
continue;
}
if (number[i] - '0' != longestStreakNum){
longestStreakLength = 0;
i--;
} else {
longestStreakLength++;
}
}
}

}


To run, first compile the second entry with: javac LookAndSayExpandedGolfed.java

Then run with: java LookAndSayExpandedGolfed

Edit: Fixed error.

• Doesn't work for me, Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 4 at java.lang.String.charAt(String.java:658) – Octavia Togami Nov 30 '14 at 4:18
• @KenzieTogami Fixed. – TheNumberOne Nov 30 '14 at 23:51
• I don't think so, is --1 supposed to be --i? – Octavia Togami Dec 1 '14 at 0:03
• @KenzieTogamie Yes, it is. It's fixed now. – TheNumberOne Dec 1 '14 at 0:33
• Cool, that works. The reverser fails, however. 513211 -> 11111. – Octavia Togami Dec 1 '14 at 0:59

# Javascript (in-browser, ES5, IE8+), 152

var s=prompt(),r='',n=r,c=r,i=0;do{c!=s[i]?(r+=n+c,n=1,c=s[i]):n++;i++}while(c)n='';i=0;while(s[i]){n+=Array(1*s[i]+1).join(c=s[i+1]);i+=2}alert(c?n:r)


Can be shortened by 4 chars if you skip var, or a few more chars with other intermediate unvarred globals too, but let's pretend we are not bad programmers for a minute.

Switching to ES6 short-syntax function with argument and return value instead of using prompt, alert for IO can save more.

JSFiddle here: http://jsfiddle.net/86L1w6Lk/

• Don't worry about those vars... we're all "bad programmers" here. ;) – Martin Ender Nov 28 '14 at 12:30

# Python 3 - 159 bytes

import re;l=input();x=len;print(''.join([str(x(i))+j for i,j in re.findall('((.)\2*)',l)]if x(l)%2 else[j*int(i)for i,j in[l[i:i+2]for i in range(0,x(l),2)]]))


# Cobra - 217

(186 if I can assume a use statement for System.Text.RegularExpressions exists elsewhere)

do(s='')=if((l=s.length)%2,System.Text.RegularExpressions.Regex.replace(s,(for x in 10 get'[x]+|').join(''),do(m=Match())="['[m]'.length]"+'[m]'[:1]),(for x in 0:l:2get'[s[x+1]]'.repeat(int.parse('[s[x]]'))).join(''))


# JavaScript (ES6) 85

Using regular expression replace with function. Different regexp and different function depending on input length being even or odd.

F=s=>s.replace((i=s.length&1)?/(.)(\1*)/g:/../g,x=>i?x.length+x[0]:x[1].repeat(x[0]))