Look-and-say sequence
You are encouraged to solve this task according to the task description, using any language you may know.
The Look and say sequence is a recursively defined sequence of numbers studied most notably by John Conway.
The look-and-say sequence is also known as the Morris Number Sequence, after cryptographer Robert Morris, and the puzzle What is the next number in the sequence 1, 11, 21, 1211, 111221? is sometimes referred to as the Cuckoo's Egg, from a description of Morris in Clifford Stoll's book The Cuckoo's Egg.
Sequence Definition
- Take a decimal number
- Look at the number, visually grouping consecutive runs of the same digit.
- Say the number, from left to right, group by group; as how many of that digit there are - followed by the digit grouped.
- This becomes the next number of the sequence.
An example:
- Starting with the number 1, you have one 1 which produces 11
- Starting with 11, you have two 1's. I.E.: 21
- Starting with 21, you have one 2, then one 1. I.E.: (12)(11) which becomes 1211
- Starting with 1211, you have one 1, one 2, then two 1's. I.E.: (11)(12)(21) which becomes 111221
- Task
Write a program to generate successive members of the look-and-say sequence.
- Related tasks
- Fours is the number of letters in the ...
- Number names
- Self-describing numbers
- Self-referential sequence
- Spelling of ordinal numbers
- See also
- Look-and-Say Numbers (feat John Conway), A Numberphile Video.
- This task is related to, and an application of, the Run-length encoding task.
- Sequence A005150 on The On-Line Encyclopedia of Integer Sequences.
8080 Assembly
<lang 8080asm>bdos: equ 5 ; CP/M calls puts: equ 9
nmemb: equ 15 ; Change this to print more or fewer members
org 100h
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Generate and output members of the sequence
mvi b,nmemb ; Counter
outloop: push b ; Preserve counter across calls
mvi c,puts ; Output current member lxi d,memb call bdos ; And newline mvi c,puts lxi d,newline call bdos
lxi h,memb ; Generate next member call looksay
pop b ; Restore counter dcr b ; Done yet? jnz outloop rst 0
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;; Given a $-terminated string under HL, representing ;; a member of the look and say sequence, generate ;; the next one in place (ish). The memory after the ;; string is assumed to be free. looksay: push h ; Save start of string on stack mov d,h ; And in DE mov e,l mvi a,'$' ; Find end of string findend: cmp m inx h jnz findend xchg ; HL=string, DE=destination push d ; Save start of new string on stack
lookchar: mvi b,0 ; Zero counter lookloop: mov a,m ; Get current character inr b ; Compare next character inx h cmp m ; While it is the same, keep going jz lookloop
mov c,a ; Keep character mvi a,'0' ; There are B amount of these characters add b stax d ; Store the amount inx d ; And in the next location mov a,c ; Store the character stax d inx d
mvi a,'$' ; Are we done? cmp m jnz lookchar ; If not, do next character stax d ; If yes, terminate new string
;; Free up memory by copying the new string to where the old ;; string began. pop d ; Start of new string pop h ; Start of old string copyloop: ldax d ; Get char from new string mov m,a ; Store char where old string was cpi '$' ; are we done yet? inx d inx h jnz copyloop ret
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; newline: db 13,10,'$' ;; This is where the string will be stored. memb: db '1$' ; First item ; Due to how CP/M loads programs, the memory after here ; is free until we hit the stack. </lang>
Ada
<lang ada>with Ada.Text_IO, Ada.Strings.Fixed; use Ada.Text_IO, Ada.Strings, Ada.Strings.Fixed;
function "+" (S : String) return String is
Item : constant Character := S (S'First);
begin
for Index in S'First + 1..S'Last loop if Item /= S (Index) then return Trim (Integer'Image (Index - S'First), Both) & Item & (+(S (Index..S'Last))); end if; end loop; return Trim (Integer'Image (S'Length), Both) & Item;
end "+";</lang> This function can be used as follows: <lang ada>Put_Line (+"1"); Put_Line (+(+"1")); Put_Line (+(+(+"1"))); Put_Line (+(+(+(+"1")))); Put_Line (+(+(+(+(+"1"))))); Put_Line (+(+(+(+(+(+"1")))))); Put_Line (+(+(+(+(+(+(+"1"))))))); Put_Line (+(+(+(+(+(+(+(+"1")))))))); Put_Line (+(+(+(+(+(+(+(+(+"1"))))))))); Put_Line (+(+(+(+(+(+(+(+(+(+"1"))))))))));</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
ALGOL 68
<lang algol68>OP + = (STRING s)STRING: BEGIN
CHAR item = s[LWB s]; STRING out; FOR index FROM LWB s + 1 TO UPB s DO IF item /= s [index] THEN out := whole(index - LWB s, 0) + item + (+(s [index:UPB s])); GO TO return out FI OD; out := whole (UPB s, 0) + item; return out: out
END # + #;
OP + = (CHAR s)STRING:
+ STRING(s);
print ((+"1", new line)); print ((+(+"1"), new line)); print ((+(+(+"1")), new line)); print ((+(+(+(+"1"))), new line)); print ((+(+(+(+(+"1")))), new line)); print ((+(+(+(+(+(+"1"))))), new line)); print ((+(+(+(+(+(+(+"1")))))), new line)); print ((+(+(+(+(+(+(+(+"1"))))))), new line)); print ((+(+(+(+(+(+(+(+(+"1")))))))), new line)); print ((+(+(+(+(+(+(+(+(+(+"1"))))))))), new line))</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
APL
<lang apl>
⎕IO←0 d←{(1↓⍵)-¯1↓⍵} f←{m←(0≠d ⍵),1 ⋄ ,(d ¯1,m/⍳⍴⍵),[.5](m/⍵)} {(f⍣⍵) ,1}¨⍳10
</lang>
AutoHotkey
<lang autohotkey>AutoExecute:
Gui, -MinimizeBox Gui, Add, Edit, w500 r20 vInput, 1 Gui, Add, Button, x155 w100 Default, &Calculate Gui, Add, Button, xp+110 yp wp, E&xit Gui, Show,, Look-and-Say sequence
Return
ButtonCalculate:
Gui, Submit, NoHide GuiControl,, Input, % LookAndSay(Input)
Return
GuiClose:
ButtonExit:
ExitApp
Return
- ---------------------------------------------------------------------------
LookAndSay(Input) {
- ---------------------------------------------------------------------------
; credit for this function goes to AutoHotkey forum member Laslo ; http://www.autohotkey.com/forum/topic44657-161.html ;----------------------------------------------------------------------- Loop, Parse, Input ; look at every digit If (A_LoopField = d) ; I've got another one! (of the same value) c += 1 ; Let's count them ... Else { ; No, this one is different! r .= c d ; remember what we've got so far c := 1 ; It is the first one in a row d := A_LoopField ; Which one is it? } Return, r c d
}</lang>
AWK
<lang awk>function lookandsay(a) {
s = "" c = 1 p = substr(a, 1, 1) for(i=2; i <= length(a); i++) { if ( p == substr(a, i, 1) ) { c++ } else { s = s sprintf("%d%s", c, p) p = substr(a, i, 1) c = 1 } } s = s sprintf("%d%s", c, p) return s
}
BEGIN {
b = "1" print b for(k=1; k <= 10; k++) { b = lookandsay(b) print b }
}</lang>
BASIC256
<lang BASIC256>
- look and say
dim a$(2)
i = 0 # input string index
a$[i] = "1"
print a$[i]
for n=1 to 10
j = 1 - i # output string index a$[j] = "" k = 1 while (k <= length(a$[i])) k0 = k + 1 while ((k0 <= length(a$[i])) and (mid(a$[i], k, 1) = mid(a$[i], k0, 1))) k0 = k0 + 1 end while a$[j] += string(k0 - k) + mid(a$[i], k, 1) k = k0 end while i = j print a$[j]
next n </lang>
BBC BASIC
<lang bbcbasic> number$ = "1"
FOR i% = 1 TO 10 number$ = FNlooksay(number$) PRINT number$ NEXT END DEF FNlooksay(n$) LOCAL i%, j%, c$, o$ i% = 1 REPEAT c$ = MID$(n$,i%,1) j% = i% + 1 WHILE MID$(n$,j%,1) = c$ j% += 1 ENDWHILE o$ += STR$(j%-i%) + c$ i% = j% UNTIL i% > LEN(n$) = o$</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Bracmat
In this example we use a non-linear pattern and a negation of a pattern: the end of e sequence of equal digits is (1) the end of the string or (2) the start of a sequence starting with a different digit. <lang bracmat>( 1:?number & 0:?lines & whl
' ( 1+!lines:~>10:?lines & :?say { This will accumulate all that has to be said after one iteration. } & 0:?begin & ( @( !number { Pattern matching. The '@' indicates we're looking in a string rather than a tree structure. } : ? ( [!begin %@?digit ? [?end ( (|(%@:~!digit) ?) { The %@ guarantees we're testing one character - not less (%) and not more (@). The ? takes the rest. } & !say !end+-1*!begin !digit:?say & !end:?begin { When backtracking, 'begin' advances to the begin of the next sequence, or to the end of the string. } ) & ~ { fail! This forces backtracking. Backtracking stops when all begin positions have been tried. } ) ) | out$(str$!say:?number) { After backtracking, output string and set number to string for next iteration. } ) )
);</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
C
This program will not stop until killed or running out of memory. <lang c>#include <stdio.h>
- include <stdlib.h>
int main() { char *a = malloc(2), *b = 0, *x, c; int cnt, len = 1;
for (sprintf(a, "1"); (b = realloc(b, len * 2 + 1)); a = b, b = x) { puts(x = a); for (len = 0, cnt = 1; (c = *a); ) { if (c == *++a) cnt++; else if (c) { len += sprintf(b + len, "%d%c", cnt, c); cnt = 1; } } }
return 0; }</lang>
C#
<lang csharp>using System; using System.Text; using System.Linq;
class Program {
static string lookandsay(string number) { StringBuilder result = new StringBuilder();
char repeat = number[0]; number = number.Substring(1, number.Length-1)+" "; int times = 1; foreach (char actual in number) { if (actual != repeat) { result.Append(Convert.ToString(times)+repeat); times = 1; repeat = actual; } else { times += 1; } } return result.ToString(); }
static void Main(string[] args) { string num = "1";
foreach (int i in Enumerable.Range(1, 10)) { Console.WriteLine(num); num = lookandsay(num); } }
}</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
Alternate version using Regex (C#2 syntax only): <lang csharp> using System; using System.Text.RegularExpressions;
namespace RosettaCode_Cs_LookAndSay {
public class Program { public static int Main(string[] args) { Array.Resize<string>(ref args, 2); string ls = args[0] ?? "1"; int n; if (!int.TryParse(args[1], out n)) n = 10; do { Console.WriteLine(ls); if (--n <= 0) break; ls = say(look(ls)); } while(true);
return 0; }
public static string[] look(string input) { int i = -1; return Array.FindAll(Regex.Split(input, @"((\d)\2*)"), delegate(string p) { ++i; i %= 3; return i == 1; } ); }
public static string say(string[] groups) { return string.Concat( Array.ConvertAll<string, string>(groups, delegate(string p) { return string.Concat(p.Length, p[0]); } ) ); } }
}</lang>
- Output:
(with args
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
C++
<lang cpp>#include <iostream>
- include <sstream>
- include <string>
std::string lookandsay(const std::string& s) {
std::ostringstream r;
for (std::size_t i = 0; i != s.length();) { auto new_i = s.find_first_not_of(s[i], i + 1);
if (new_i == std::string::npos) new_i = s.length();
r << new_i - i << s[i]; i = new_i; } return r.str();
}
int main() {
std::string laf = "1";
std::cout << laf << '\n'; for (int i = 0; i < 10; ++i) { laf = lookandsay(laf); std::cout << laf << '\n'; }
}</lang>
Ceylon
<lang ceylon>shared void run() {
function lookAndSay(Integer|String input) { variable value digits = if (is Integer input) then input.string else input; value builder = StringBuilder(); while (exists currentChar = digits.first) { if (exists index = digits.firstIndexWhere((char) => char != currentChar)) { digits = digits[index...]; builder.append("``index````currentChar``"); } else { builder.append("``digits.size````currentChar``"); break; } } return builder.string; } variable String|Integer result = 1; print(result); for (i in 1..14) { result = lookAndSay(result); print(result); }
}</lang>
Clojure
No ugly int-to-string-and-back conversions.
<lang clojure>(defn digits-seq
"Returns a seq of the digits of a number (L->R)." [n] (loop [digits (), number n] (if (zero? number) (seq digits) (recur (cons (mod number 10) digits) (quot number 10)))))
(defn join-digits
"Converts a digits-seq back in to a number." [ds] (reduce (fn [n d] (+ (* 10 n) d)) ds))
(defn look-and-say [n]
(->> n digits-seq (partition-by identity) (mapcat (juxt count first)) join-digits))</lang>
- Output:
<lang clojure>user> (take 8 (iterate look-and-say 1)) (1 11 21 1211 111221 312211 13112221 1113213211)</lang>
Common Lisp
<lang lisp>(defun compress (array &key (test 'eql) &aux (l (length array)))
"Compresses array by returning a list of conses each of whose car is
a number of occurrences and whose cdr is the element occurring. For instance, (compress \"abb\") produces ((1 . #\a) (2 . #\b))."
(if (zerop l) nil (do* ((i 1 (1+ i)) (segments (acons 1 (aref array 0) '()))) ((eql i l) (nreverse segments)) (if (funcall test (aref array i) (cdar segments)) (incf (caar segments)) (setf segments (acons 1 (aref array i) segments))))))
(defun next-look-and-say (number)
(reduce #'(lambda (n pair) (+ (* 100 n) (* 10 (car pair)) (parse-integer (string (cdr pair))))) (compress (princ-to-string number)) :initial-value 0))</lang>
Example use:
<lang lisp>(next-look-and-say 9887776666) ;=> 19283746</lang>
Straight character counting: <lang lisp>(defun look-and-say (s)
(let ((out (list (char s 0) 0))) (loop for x across s do
(if (char= x (first out)) (incf (second out)) (setf out (list* x 1 out))))
(format nil "~{~a~^~}" (nreverse out))))
(loop for s = "1" then (look-and-say s)
repeat 10 do (write-line s))</lang>
Crystal
The simplest one: <lang ruby>class String
def look_and_say gsub(/(.)\1*/){ |s| s.size.to_s + s[0] } end
end
ss = '1' 12.times { puts ss; ss = ss.to_s.look_and_say }</lang>
<lang ruby>def lookandsay(str)
str.gsub(/(.)\1*/) { |s| s.size.to_s + $1 }
end
num = "1" 12.times { puts num; num = lookandsay(num) }</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
D
Short Functional Version
<lang d>import std.stdio, std.algorithm, std.range;
enum say = (in string s) pure => s.group.map!q{ text(a[1],a[0]) }.join;
void main() {
"1".recurrence!((t, n) => t[n - 1].say).take(8).writeln;
}</lang>
- Output:
["1", "11", "21", "1211", "111221", "312211", "13112221", "1113213211"]
Fast Imperative Version
Same output. <lang d>import core.stdc.stdio, std.math, std.conv, std.algorithm, std.array;
void showLookAndSay(bool showArrays)(in uint n) nothrow {
if (n == 0) // No sequences to generate and show. return;
enum Digit : char { nil = '\0', one = '1', two = '2', thr = '3' }
// Allocate an approximate upper bound size for the array. static Digit* allocBuffer(in uint m) nothrow { immutable len = cast(size_t)(100 + 1.05 * exp(0.269 * m + 0.2686)) + 1; auto a = len.uninitializedArray!(Digit[]); printf("Allocated %d bytes.\n", a.length * Digit.sizeof); return a.ptr; }
// Can't be expressed in the D type system: // a1 and a2 are immutable pointers to mutable data. auto a1 = allocBuffer(n % 2 ? n : n - 1); auto a2 = allocBuffer(n % 2 ? n - 1 : n); printf("\n");
a1[0] = Digit.one; size_t len1 = 1; a1[len1] = Digit.nil;
foreach (immutable i; 0 .. n - 1) { static if (showArrays) printf("%2u: %s\n", i + 1, a1); else printf("%2u: n. digits: %u\n", i + 1, len1); auto p1 = a1, p2 = a2;
S0: final switch (*p1++) with (Digit) { // Initial state. case nil: goto END; case one: goto S1; case two: goto S2; case thr: goto S3; } S1: final switch (*p1++) with (Digit) { case nil: *p2++ = one; *p2++ = one; goto END; case one: goto S11; case two: *p2++ = one; *p2++ = one; goto S2; case thr: *p2++ = one; *p2++ = one; goto S3; } S2: final switch (*p1++) with (Digit) { case nil: *p2++ = one; *p2++ = two; goto END; case one: *p2++ = one; *p2++ = two; goto S1; case two: goto S22; case thr: *p2++ = one; *p2++ = two; goto S3; } S3: final switch (*p1++) with (Digit) { case nil: *p2++ = one; *p2++ = thr; goto END; case one: *p2++ = one; *p2++ = thr; goto S1; case two: *p2++ = one; *p2++ = thr; goto S2; case thr: goto S33; } S11: final switch (*p1++) with (Digit) { case nil: *p2++ = two; *p2++ = one; goto END; case one: *p2++ = thr; *p2++ = one; goto S0; case two: *p2++ = two; *p2++ = one; goto S2; case thr: *p2++ = two; *p2++ = one; goto S3; } S22: final switch (*p1++) with (Digit) { case nil: *p2++ = two; *p2++ = two; goto END; case one: *p2++ = two; *p2++ = two; goto S1; case two: *p2++ = thr; *p2++ = two; goto S0; case thr: *p2++ = two; *p2++ = two; goto S3; } S33: final switch (*p1++) with (Digit) { case nil: *p2++ = two; *p2++ = thr; goto END; case one: *p2++ = two; *p2++ = thr; goto S1; case two: *p2++ = two; *p2++ = thr; goto S2; case thr: *p2++ = thr; *p2++ = thr; goto S0; } END: immutable len2 = p2 - a2; a2[len2] = Digit.nil; a1.swap(a2); len1 = len2; }
static if (showArrays) printf("%2u: %s\n", n, a1); else printf("%2u: n. digits: %u\n", n, len1);
}
void main(in string[] args) {
immutable n = (args.length == 2) ? args[1].to!uint : 10; n.showLookAndSay!true;
}</lang>
- Output:
Allocated 116 bytes. Allocated 121 bytes. 1: 1 2: 11 3: 21 4: 1211 5: 111221 6: 312211 7: 13112221 8: 1113213211 9: 31131211131221 10: 13211311123113112211
With: <lang d>70.showLookAndSay!false;</lang>
- Output:
Allocated 158045069 bytes. Allocated 206826462 bytes. 1: n. digits: 1 2: n. digits: 2 3: n. digits: 2 4: n. digits: 4 5: n. digits: 6 ... 60: n. digits: 12680852 61: n. digits: 16530884 62: n. digits: 21549544 63: n. digits: 28091184 64: n. digits: 36619162 65: n. digits: 47736936 66: n. digits: 62226614 67: n. digits: 81117366 68: n. digits: 105745224 69: n. digits: 137842560 70: n. digits: 179691598
Using the LDC2 compiler with n=70 the run-time is about 3.74 seconds.
Intermediate Version
This mostly imperative version is intermediate in both speed and code size: <lang d>void main(in string[] args) {
import std.stdio, std.conv, std.algorithm, std.array, std.string;
immutable n = (args.length == 2) ? args[1].to!uint : 10; if (n == 0) return;
auto seq = ['1']; writefln("%2d: n. digits: %d", 1, seq.length); foreach (immutable i; 2 .. n + 1) { Appender!(typeof(seq)) result; foreach (const digit, const count; seq.representation.group) { result ~= "123"[count - 1]; result ~= digit; } seq = result.data; writefln("%2d: n. digits: %d", i, seq.length); }
}</lang> The output is the same as the second version.
If you modify the first program to print only the lengths of the strings
(with a .map!(s => s.length)
),
compiling with LDC2 the run-times of the three versions
with n=55 are about 31.1, 0.10 and 0.23 seconds.
More Direct Version
Translated and modified from C code by Reddit user "skeeto": http://www.reddit.com/r/dailyprogrammer/comments/2ggy30/9152014_challenge180_easy_looknsay/
Using ideas from: http://www.njohnston.ca/2010/10/a-derivation-of-conways-degree-71-look-and-say-polynomial/
This recursive version is able to generate very large sequences in a short time without memory for the intermediate sequence (and with stack space proportional to the sequence order).
<lang d>import core.stdc.stdio, std.conv;
// On Windows this uses the printf from the Microsoft C runtime, // that doesn't handle real type and some of the C99 format // specifiers, but it's faster for bulk printing. version (LDC) version (Windows) extern(C) nothrow @nogc int printf(const char*, ...);
// http://www.njohnston.ca/2010/10/a-derivation-of-conways-degree-71-look-and-say-polynomial/ struct Sequence {
string seq; uint[6] next;
}
immutable Sequence[93] sequences = [
{"", []}, {"1112", [63]}, {"1112133", [64, 62]}, {"111213322112", [65]}, {"111213322113", [66]}, {"1113", [68]}, {"11131", [69]}, {"111311222112", [84, 55]}, {"111312", [70]}, {"11131221", [71]}, {"1113122112", [76]}, {"1113122113", [77]}, {"11131221131112", [82]}, {"111312211312", [78]}, {"11131221131211", [79]}, {"111312211312113211", [80]}, {"111312211312113221133211322112211213322112", [81, 29, 91]}, {"111312211312113221133211322112211213322113", [81, 29, 90]}, {"11131221131211322113322112", [81, 30]}, {"11131221133112", [75, 29, 92]}, {"1113122113322113111221131221", [75, 32]}, {"11131221222112", [72]}, {"111312212221121123222112", [73]}, {"111312212221121123222113", [74]}, {"11132", [83]}, {"1113222", [86]}, {"1113222112", [87]}, {"1113222113", [88]}, {"11133112", [89, 92]}, {"12", [1]}, {"123222112", [3]}, {"123222113", [4]}, {"12322211331222113112211", [2, 61, 29, 85]}, {"13", [5]}, {"131112", [28]}, {"13112221133211322112211213322112", [24, 33, 61, 29, 91]}, {"13112221133211322112211213322113", [24, 33, 61, 29, 90]}, {"13122112", [7]}, {"132", [8]}, {"13211", [9]}, {"132112", [10]}, {"1321122112", [21]}, {"132112211213322112", [22]}, {"132112211213322113", [23]}, {"132113", [11]}, {"1321131112", [19]}, {"13211312", [12]}, {"1321132", [13]}, {"13211321", [14]}, {"132113212221", [15]}, {"13211321222113222112", [18]}, {"1321132122211322212221121123222112", [16]}, {"1321132122211322212221121123222113", [17]}, {"13211322211312113211", [20]}, {"1321133112", [6, 61, 29, 92]}, {"1322112", [26]}, {"1322113", [27]}, {"13221133112", [25, 29, 92]}, {"1322113312211", [25, 29, 67]}, {"132211331222113112211", [25, 29, 85]}, {"13221133122211332", [25, 29, 68, 61, 29, 89]}, {"22", [61]}, {"3", [33]}, {"3112", [40]}, {"3112112", [41]}, {"31121123222112", [42]}, {"31121123222113", [43]}, {"3112221", [38, 39]}, {"3113", [44]}, {"311311", [48]}, {"31131112", [54]}, {"3113112211", [49]}, {"3113112211322112", [50]}, {"3113112211322112211213322112", [51]}, {"3113112211322112211213322113", [52]}, {"311311222", [47, 38]}, {"311311222112", [47, 55]}, {"311311222113", [47, 56]}, {"3113112221131112", [47, 57]}, {"311311222113111221", [47, 58]}, {"311311222113111221131221", [47, 59]}, {"31131122211311122113222", [47, 60]}, {"3113112221133112", [47, 33, 61, 29, 92]}, {"311312", [45]}, {"31132", [46]}, {"311322113212221", [53]}, {"311332", [38, 29, 89]}, {"3113322112", [38, 30]}, {"3113322113", [38, 31]}, {"312", [34]}, {"312211322212221121123222113", [36]}, {"312211322212221121123222112", [35]}, {"32112", [37]}
];
void evolve(in uint seq, in uint n) nothrow @nogc {
if (n <= 0) { printf(sequences[seq].seq.ptr); } else { foreach (immutable next; sequences[seq].next) { if (next == 0) break; evolve(next, n - 1); } }
}
void main(in string[] args) {
immutable uint n = (args.length != 2) ? 10 : args[1].to!uint;
immutable base = 8; immutable string[base] results = ["", "1", "11", "21", "1211", "111221", "312211", "13112221"]; if (n < base) { printf("%s\n", results[n].ptr); return; }
evolve(24, n - base); evolve(39, n - base); '\n'.putchar;
}</lang>
E
<lang e>def lookAndSayNext(number :int) {
var seen := null var count := 0 var result := "" def put() { if (seen != null) { result += count.toString(10) + E.toString(seen) } } for ch in number.toString(10) { if (ch != seen) { put() seen := ch count := 0 } count += 1 } put() return __makeInt(result, 10)
}
var number := 1 for _ in 1..20 {
println(number) number := lookAndSayNext(number)
}</lang>
EchoLisp
<lang scheme> (lib 'math) ;; for (number->list) = explode function (lib 'list) ;; (group)
(define (next L) (for/fold (acc null) ((g (group L))) (append acc (list (length g) (first g)))))
(define (task n starter)
(for/fold (L (number->list starter)) ((i n))
(writeln (list->string L))
(next L)))
</lang>
- Output:
<lang scheme> (task 10 1) 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 </lang>
Elixir
<lang elixir>defmodule LookAndSay do
def next(n) do Enum.chunk_by(to_char_list(n), &(&1)) |> Enum.map(fn cl=[h|_] -> Enum.concat(to_char_list(length cl), [h]) end) |> Enum.concat |> List.to_integer end def sequence_from(n) do Stream.iterate n, &(next/1) end def main([start_str|_]) do {start_val,_} = Integer.parse(start_str) IO.inspect sequence_from(start_val) |> Enum.take 9 end def main([]) do main(["1"]) end
end
LookAndSay.main(System.argv)</lang>
- Output:
[1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221]
Regex version: <lang elixir>defmodule RC do
def look_and_say(n) do Regex.replace(~r/(.)\1*/, to_string(n), fn x,y -> [to_string(String.length(x)),y] end) |> String.to_integer end
end
IO.inspect Enum.reduce(1..9, [1], fn _,acc -> [RC.look_and_say(hd(acc)) | acc] end) |> Enum.reverse</lang>
- Output:
[1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, 13211311123113112211]
Erlang
<lang erlang>-module(str). -export([look_and_say/1, look_and_say/2]).
%% converts a single number look_and_say([H|T]) -> lists:reverse(look_and_say(T,H,1,"")).
%% converts and accumulates as a loop look_and_say(_, 0) -> []; look_and_say(Start, Times) when Times > 0 ->
[Start | look_and_say(look_and_say(Start), Times-1)].
%% does the actual conversion for a number look_and_say([], Current, N, Acc) ->
[Current, $0+N | Acc];
look_and_say([H|T], H, N, Acc) ->
look_and_say(T, H, N+1, Acc);
look_and_say([H|T], Current, N, Acc) ->
look_and_say(T, H, 1, [Current, $0+N | Acc]).</lang>
- Output:
1> c(str). {ok,str} 2> str:look_and_say("1"). "11" 3> str:look_and_say("111221"). "312211" 4> str:look_and_say("1",10). ["1","11","21","1211","111221","312211","13112221", "1113213211","31131211131221","13211311123113112211"]
ERRE
<lang> PROGRAM LOOK
PROCEDURE LOOK_AND_SAY(N$->N$)
LOCAL I%,J%,C$,O$ I%=1 REPEAT C$=MID$(N$,I%,1) J%=I%+1 WHILE MID$(N$,J%,1)=C$ DO J%+=1 END WHILE O$+=MID$(STR$(J%-I%),2)+C$ I%=J% UNTIL I%>LEN(N$) N$=O$
END PROCEDURE
BEGIN
NUMBER$="1" FOR I%=1 TO 10 DO LOOK_AND_SAY(NUMBER$->NUMBER$) PRINT(NUMBER$) END FOR
END PROGRAM </lang>
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
F#
Library functions somehow missing in F# out of the box (but present in haskell) <lang fsharp> let rec brk p lst =
match lst with | [] -> (lst, lst) | x::xs -> if p x then ([], lst) else let (ys, zs) = brk p xs (x::ys, zs)
let span p lst = brk (not << p) lst
let rec groupBy eq lst =
match lst with | [] -> [] | x::xs -> let (ys,zs) = span (eq x) xs
(x::ys)::groupBy eq zs
let group lst : list<list<'a>> when 'a : equality = groupBy (=) lst </lang>
Implementation <lang fsharp> let lookAndSay =
let describe (xs: char list) = List.append (List.ofSeq <| (List.length xs).ToString()) [List.head xs] let next xs = List.collect describe (group xs) let toStr xs = String (Array.ofList xs) Seq.map toStr <| Seq.unfold (fun xs -> Some (xs, next xs)) ['1']
let getNthLookAndSay n = Seq.nth n lookAndSay
Seq.take 10 lookAndSay </lang>
Factor
<lang factor>: (look-and-say) ( str -- )
unclip-slice swap [ 1 ] 2dip [ 2dup = [ drop [ 1 + ] dip ] [ [ [ number>string % ] dip , 1 ] dip ] if ] each [ number>string % ] [ , ] bi* ;
- look-and-say ( str -- str' ) [ (look-and-say) ] "" make ;
"1" 10 [ dup print look-and-say ] times print</lang>
Forth
<lang forth>create buf1 256 allot create buf2 256 allot buf1 value src buf2 value dest
s" 1" src place
- append-run ( digit run -- )
dest count + tuck c! 1+ c! dest c@ 2 + dest c! ;
- next-look-and-say
0 dest c! src 1+ c@ [char] 0 ( digit run ) src count bounds do over i c@ = if 1+ else append-run i c@ [char] 1 then loop append-run src dest to src to dest ;
- look-and-say ( n -- )
0 do next-look-and-say cr src count type loop ;
10 look-and-say</lang>
Fortran
<lang fortran>module LookAndSay
implicit none
contains
subroutine look_and_say(in, out) character(len=*), intent(in) :: in character(len=*), intent(out) :: out
integer :: i, c character(len=1) :: x character(len=2) :: d
out = "" c = 1 x = in(1:1) do i = 2, len(trim(in)) if ( x == in(i:i) ) then c = c + 1 else write(d, "(I2)") c out = trim(out) // trim(adjustl(d)) // trim(x) c = 1 x = in(i:i) end if end do write(d, "(I2)") c out = trim(out) // trim(adjustl(d)) // trim(x) end subroutine look_and_say
end module LookAndSay</lang>
<lang fortran>program LookAndSayTest
use LookAndSay implicit none integer :: i character(len=200) :: t, r t = "1" print *,trim(t) call look_and_say(t, r) print *, trim(r) do i = 1, 10 call look_and_say(r, t) r = t print *, trim(r) end do
end program LookAndSayTest</lang>
FreeBASIC
<lang freebasic> Dim As Integer n, j, k, k0, r Dim As String X(2) Dim As Integer i = 0 ' índice de cadena de entrada X(0) = "1"
Input "Indica cuantas repeticiones: ", r Print Chr(10) & "Secuencia:"
Print X(i) For n = 1 To r-1
j = 1 - i ' índice de cadena de salida X(j) = "" k = 1 While k <= Len(X(i)) k0 = k + 1 While ((k0 <= Len(X(i))) And (Mid(X(i), k, 1) = Mid(X(i), k0, 1))) k0 += 1 Wend X(j) += Str(k0 - k) + Mid(X(i), k, 1) k = k0 Wend i = j Print X(j)
Next n End </lang>
- Output:
Indica cuantas repeticiones: 10 Secuencia: 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
Frink
<lang frink> LookAndSay = "12211123" println["Starting Value: " + LookAndSay]
/* Anonymous function
There was an issue processing a standard function with MapList. However, the anonymous function formatting works without issue with the MapList function.
- /
LASStr = { |LaS|
length[LaS@0] + length[LaS@1] + LaS@0 /* The returned results from the Regex are divided between a distinct matching character and any following identical characters. For example, a string of 2222 would be returned from this function as [2,222]. The function adds the length of both elements (1 and 3 in the example above) and returns that value with the matching character. i.e. Length of 1st element = 1, Length of 2nd element = 3, value of 1st element = 2 1 + 3 = 4 & value 2. Returned result is "42" i.e. "Four 2s." */
}
// Calculate the next 10 Look and Say Sequence Values
for i = 1 to 10 {
LookAndSayReg = LookAndSay =~ %r/(\d)(\1{0,})/g LookAndSay = join["",mapList[LASStr,LookAndSayReg]] println["$i - $LookAndSay"]
} </lang>
- Output:
Starting Value: 1 1 - 11 2 - 21 3 - 1211 4 - 111221 5 - 312211 6 - 13112221 7 - 1113213211 8 - 31131211131221 9 - 13211311123113112211 10 - 11131221133112132113212221
Gambas
Code is modified from the [PureBasic] example
Click this link to run this code <lang gambas>Public Sub Main() Dim i, j, cnt As Integer Dim txt$, curr$, result$ As String
txt$ = "1211" i = 1
Print "Sequence: " & txt$ & " = ";
Repeat j = 1 result$ = "" Repeat curr$ = Mid(txt$, j, 1) cnt = 0 Repeat Inc cnt Inc j Until Mid(txt$, j, 1) <> curr$ result$ &= Str(cnt) & curr$ Until j > Len(txt$) Print result$ txt$ = result$ Dec i Until i <= 0
End</lang> Output:
Sequence: 1211 = 111221
GAP
<lang gap>LookAndSay := function(s)
local c, r, cur, ncur, v; v := "123"; r := ""; cur := 0; ncur := 0; for c in s do if c = cur then ncur := ncur + 1; else if ncur > 0 then Add(r, v[ncur]); Add(r, cur); fi; cur := c; ncur := 1; fi; od; Add(r, v[ncur]); Add(r, cur); return r;
end;
LookAndSay("1"); # "11" LookAndSay(last); # "21" LookAndSay(last); # "1211" LookAndSay(last); # "111221" LookAndSay(last); # "312211" LookAndSay(last); # "13112221" LookAndSay(last); # "1113213211" LookAndSay(last); # "31131211131221" LookAndSay(last); # "13211311123113112211" LookAndSay(last); # "11131221133112132113212221" LookAndSay(last); # "3113112221232112111312211312113211" LookAndSay(last); # "1321132132111213122112311311222113111221131221" LookAndSay(last); # "11131221131211131231121113112221121321132132211331222113112211" LookAndSay(last); # "311311222113111231131112132112311321322112111312211312111322212311322113212221"</lang>
Go
<lang go>package main
import (
"fmt" "strconv"
)
func lss(s string) (r string) {
c := s[0] nc := 1 for i := 1; i < len(s); i++ { d := s[i] if d == c { nc++ continue } r += strconv.Itoa(nc) + string(c) c = d nc = 1 } return r + strconv.Itoa(nc) + string(c)
}
func main() {
s := "1" fmt.Println(s) for i := 0; i < 8; i++ { s = lss(s) fmt.Println(s) }
}</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221
Groovy
<lang groovy>def lookAndSay(sequence) {
def encoded = new StringBuilder() (sequence.toString() =~ /(([0-9])\2*)/).each { matcher -> encoded.append(matcher[1].size()).append(matcher[2]) } encoded.toString()
}</lang> Test Code <lang groovy>def sequence = "1" (1..12).each {
println sequence sequence = lookAndSay(sequence)
}</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
Haskell
<lang haskell>import Control.Monad (liftM2) import Data.List (group)
-- this function is composed out of many functions; data flows from the bottom up lookAndSay :: Integer -> Integer lookAndSay = read -- convert digits to integer
. concatMap -- concatenate for each run, (liftM2 (++) (show . length) -- the length of it (take 1)) -- and an example member . group -- collect runs of the same digit . show -- convert integer to digits
-- less comments lookAndSay2 :: Integer -> Integer lookAndSay2 = read . concatMap (liftM2 (++) (show . length)
(take 1)) . group . show
-- same thing with more variable names
lookAndSay3 :: Integer -> Integer
lookAndSay3 n = read (concatMap describe (group (show n)))
where describe run = show (length run) ++ take 1 run
main = mapM_ print (iterate lookAndSay 1) -- display sequence until interrupted</lang>
Haxe
<lang haxe>using Std;
class Main {
static function main() { var test = "1"; for (i in 0...11) { Sys.println(test); test = lookAndSay(test); } }
static function lookAndSay(s:String) { if (s == null || s == "") return "";
var results = ""; var repeat = s.charAt(0); var amount = 1; for (i in 1...s.length) { var actual = s.charAt(i); if (actual != repeat) { results += amount.string(); results += repeat; repeat = actual; amount = 0; } amount++; } results += amount.string(); results += repeat;
return results; } }</lang>
Icon and Unicon
<lang Icon>procedure main() every 1 to 10 do
write(n := nextlooknsayseq(\n | 1))
end
procedure nextlooknsayseq(n) #: return next element in look and say sequence n2 := "" n ? until pos(0) do {
i := tab(any(&digits)) | fail # or fail if not digits move(-1) n2 ||:= *tab(many(i)) || i # accumulate count+digit }
return n2 end</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
J
Solution:
<lang j>las=: ,@((# , {.);.1~ 1 , 2 ~:/\ ])&.(10x&#.inv)@]^:(1+i.@[)</lang>
Example: <lang j> 10 las 1 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221</lang>
Note the result is an actual numeric sequence (cf. the textual solutions given in other languages).
Java
<lang java5>public static String lookandsay(String number){ StringBuilder result= new StringBuilder();
char repeat= number.charAt(0); number= number.substring(1) + " "; int times= 1;
for(char actual: number.toCharArray()){ if(actual != repeat){ result.append(times + "" + repeat); times= 1; repeat= actual; }else{ times+= 1; } } return result.toString(); }</lang> Testing: <lang java5>public static void main(String[] args){ String num = "1";
for (int i=1;i<=10;i++) { System.out.println(num); num = lookandsay(num); } }</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
JavaScript
<lang javascript>function lookandsay(str) {
return str.replace(/(.)\1*/g, function(seq, p1){return seq.length.toString() + p1})
}
var num = "1"; for (var i = 10; i > 0; i--) {
alert(num); num = lookandsay(num);
}</lang>
Without RegExp
<lang javascript>function lookSay(digits) {
var result = , chars = (digits + ' ').split(), lastChar = chars[0], times = 0; chars.forEach(function(nextChar) { if (nextChar === lastChar) { times++; } else { result += (times + ) + lastChar; lastChar = nextChar; times = 1; } }); return result;
}
(function output(seed, iterations) {
for (var i = 0; i < iterations; i++) { console.log(seed); seed = lookSay(seed); }
})("1", 10);</lang>
jq
<lang jq>def look_and_say:
def head(c; n): if .[n:n+1] == c then head(c; n+1) else n end; tostring | if length == 0 then "" else head(.[0:1]; 1) as $len | .[0:$len] as $head | ($len | tostring) + $head[0:1] + (.[$len:] | look_and_say) end ;
- look and say n times
def look_and_say(n):
if n == 0 then empty else look_and_say as $lns | $lns, ($lns|look_and_say(n-1)) end ;</lang>
Example
1 | look_and_say(10)
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Julia
<lang julia>function lookandsay(s::String)
rst = IOBuffer() c = 1 for i in 1:length(s) if i != length(s) && s[i] == s[i+1] c += 1 else print(rst, c, s[i]) c = 1 end end String(take!(rst))
end
function lookandsayseq(n::Integer)
rst = Vector{String}(undef, n) rst[1] = "1" for i in 2:n rst[i] = lookandsay(rst[i-1]) end rst
end
println(lookandsayseq(10))</lang>
- Output:
String["1", "11", "21", "1211", "111221", "312211", "13112221", "1113213211", "31131211131221", "13211311123113112211"]
K
<lang k> las: {x{0$,//$(#:'n),'*:'n:(&1,~=':x)_ x:0$'$x}\1}
las 8
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221</lang>
Kotlin
<lang scala>// version 1.0.6
fun lookAndSay(s: String): String {
val sb = StringBuilder() var digit = s[0] var count = 1 for (i in 1 until s.length) { if (s[i] == digit) count++ else { sb.append("$count$digit") digit = s[i] count = 1 } } return sb.append("$count$digit").toString()
}
fun main(args: Array<String>) {
var las = "1" for (i in 1..15) { println(las) las = lookAndSay(las) }
}</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
Lasso
The Look-and-say sequence is a recursive RLE, so the solution can leverage the same method as used for RLE. <lang Lasso>define rle(str::string)::string => { local(orig = #str->values->asCopy,newi=array, newc=array, compiled=string) while(#orig->size) => { if(not #newi->size) => { #newi->insert(1) #newc->insert(#orig->first) #orig->remove(1) else if(#orig->first == #newc->last) => { #newi->get(#newi->size) += 1 else #newi->insert(1) #newc->insert(#orig->first) } #orig->remove(1) } } loop(#newi->size) => { #compiled->append(#newi->get(loop_count)+#newc->get(loop_count)) } return #compiled } define las(n::integer,run::integer) => { local(str = #n->asString) loop(#run) => { #str = rle(#str) } return #str } loop(15) => {^ las(1,loop_count) + '\r' ^}</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221 132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211
LiveCode
This function takes a string and returns the next Look-And-Say iteration of it: <lang Lua>function lookAndSay S
put 0 into C put char 1 of S into lastChar repeat with i = 2 to length(S) add 1 to C if char i of S is lastChar then next repeat put C & lastChar after R put 0 into C put char i of S into lastChar end repeat return R & C + 1 & lastChar
end lookAndSay
on demoLookAndSay
put 1 into x repeat 10 put x & cr after message put lookAndSay(x) into x end repeat put x after message
end demoLookAndSay</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Logo
<lang logo>to look.and.say.loop :in :run :c :out
if empty? :in [output (word :out :run :c)] if equal? first :in :c [output look.and.say.loop bf :in :run+1 :c :out] output look.and.say.loop bf :in 1 first :in (word :out :run :c)
end to look.and.say :in
if empty? :in [output :in] output look.and.say.loop bf :in 1 first :in "||
end
show cascade 10 [print ? look.and.say ?] 1</lang>
Lua
<lang lua>--returns an iterator over the first n copies of the look-and-say sequence function lookandsayseq(n)
local t = {1} return function() local ret = {} for i, v in ipairs(t) do if t[i-1] and v == t[i-1] then ret[#ret - 1] = ret[#ret - 1] + 1 else ret[#ret + 1] = 1 ret[#ret + 1] = v end end t = ret n = n - 1 if n > 0 then return table.concat(ret) end end
end for i in lookandsayseq(10) do print(i) end</lang>
Alternative solution, using LPeg: <lang lua>require "lpeg" local P, C, Cf, Cc = lpeg.P, lpeg.C, lpeg.Cf, lpeg.Cc lookandsay = Cf(Cc"" * C(P"1"^1 + P"2"^1 + P"3"^1)^1, function (a, b) return a .. #b .. string.sub(b,1,1) end) t = "1" for i = 1, 10 do
print(t) t = lookandsay:match(t)
end</lang>
M4
Using regular expressions:
<lang M4>divert(-1) define(`for',
`ifelse($#,0,``$0, `ifelse(eval($2<=$3),1, `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')
define(`las',
`patsubst(`$1',`\(\(.\)\2*\)',`len(\1)`'\2')')
define(`v',1)
divert
for(`x',1,10,
`v
define(`v',las(v))')dnl v</lang>
Mathematica
Custom Functions: <lang Mathematica>RunLengthEncode[x_List]:=(Through[{First,Length}[#]]&)/@Split[x]
LookAndSay[n_,d_:1]:=NestList[Flatten[Reverse/@RunLengthEncode[#]]&,{d},n-1]</lang>
If second argument is omitted the sequence is started with 1. Second argument is supposed to be a digits from 0 to 9. If however a larger number is supplied it will be seen as 1 number, not multiple digits. However if one wants to start with a 2 or more digit number, one could reverse the sequence to go back to a single digit start. First example will create the first 13 numbers of the sequence starting with 1, the next example starts with 7:
<lang Mathematica>FromDigits /@ LookAndSay[13] // Column FromDigits /@ LookAndSay[13, 7] // Column</lang>
gives back:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 7 17 1117 3117 132117 1113122117 311311222117 13211321322117 1113122113121113222117 31131122211311123113322117 132113213221133112132123222117 11131221131211132221232112111312111213322117 31131122211311123113321112131221123113111231121123222117
Maxima
<lang maxima>collect(a) := block(
[n: length(a), b: [ ], x: a[1], m: 1], for i from 2 thru n do (if a[i] = x then m: m + 1 else (b: endcons([x, m], b), x: a[i], m: 1)), b: endcons([x, m], b)
)$
look_and_say(s) := apply(sconcat, map(lambda([p], sconcat(string(p[2]), p[1])), collect(charlist(s))))$
block([s: "1"], for i from 1 thru 10 do (disp(s), s: look_and_say(s))); /* "1"
"11" "21" "1211" "111221" "312211" "13112221" "1113213211" "31131211131221" "13211311123113112211" */</lang>
MAXScript
<lang maxscript>fn lookAndSay num = (
local result = "" num += " " local current = num[1] local numReps = 1
for digit in 2 to num.count do ( if num[digit] != current then ( result += (numReps as string) + current numReps = 1 current = num[digit] ) else ( numReps += 1 ) ) result
)
local num = "1"
for i in 1 to 10 do (
print num num = lookAndSay num
)</lang>
Metafont
<lang metafont>vardef lookandsay(expr s) = string r; r := ""; if string s:
i := 0; forever: exitif not (i < length(s)); c := i+1; forever: exitif ( (substring(c,c+1) of s) <> (substring(i,i+1) of s) ); c := c + 1; endfor r := r & decimal (c-i) & substring(i,i+1) of s; i := c; endfor
fi r enddef;
string p; p := "1"; for el := 1 upto 10:
message p; p := lookandsay(p);
endfor
end</lang>
MiniScript
<lang MiniScript>// Look and Say Sequence repeats = function(digit, string) count = 0 for c in string if c != digit then break count = count + 1 end for return str(count) end function
numbers = "1" print numbers for i in range(1,10) // warning, loop size > 15 gets long numbers very quickly number = "" position = 0 while position < numbers.len repeatCount = repeats(numbers[position], numbers[position:]) number = number + repeatCount + numbers[position] position = position + repeatCount.val end while print number numbers = number end for</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Nim
<lang nim>proc NextInLookAndSaySequence (current: string): string =
assert(len(current) > 0) Result = "" var ch = current[0] var count = 1 for i in countup(1, len(current)-1): if current[i] != ch: Result &= $count & ch ch = current[i] count = 1 else: count += 1 Result &= $count & ch
proc LookAndSay (n = 10) =
var next = "1" for i in countup(1, n): next = NextInLookAndSaySequence(next) echo next
LookAndSay() </lang>
Objective-C
<lang objc>#import <Foundation/Foundation.h>
-(NSString*)lookAndSay:(NSString *)word{
if (!word) { return nil; } NSMutableString *result = [NSMutableString new]; char repeat = [word characterAtIndex:0]; int times = 1; word = [NSString stringWithFormat:@"%@ ",[word substringFromIndex:1] ]; for (NSInteger index = 0; index < word.length; index++) { char actual = [word characterAtIndex:index]; if (actual != repeat) { [result appendFormat:@"%d%c", times, repeat]; times = 1; repeat = actual; } else { times ++; } }
return [result copy];
}
- (void)applicationDidFinishLaunching:(NSNotification *)aNotification {
NSString *num = @"1"; for (int i=1;i<=10;i++) { NSLog(@"%@", num);
num = [self lookAndSay:num]; }
} </lang>
OCaml
Functional
This function computes a see-and-say sequence from the previous one: <lang ocaml>let rec seeAndSay = function
| [], nys -> List.rev nys | x::xs, [] -> seeAndSay(xs, [x; 1]) | x::xs, y::n::nys when x=y -> seeAndSay(xs, y::1+n::nys) | x::xs, nys -> seeAndSay(xs, x::1::nys)</lang>
It can be used like this: <lang ocaml>> let gen n =
let xs = Array.create n [1] in for i=1 to n-1 do xs.(i) <- seeAndSay(xs.(i-1), []) done; xs;;
val gen : int -> int list array = <fun>
> gen 10;; - : int list array =
[|[1]; [1; 1]; [2; 1]; [1; 2; 1; 1]; [1; 1; 1; 2; 2; 1]; [3; 1; 2; 2; 1; 1]; [1; 3; 1; 1; 2; 2; 2; 1]; [1; 1; 1; 3; 2; 1; 3; 2; 1; 1]; [3; 1; 1; 3; 1; 2; 1; 1; 1; 3; 1; 2; 2; 1]; [1; 3; 2; 1; 1; 3; 1; 1; 1; 2; 3; 1; 1; 3; 1; 1; 2; 2; 1; 1]|]</lang>
With regular expressions in the Str library
<lang ocaml>#load "str.cma";;
let lookandsay =
Str.global_substitute (Str.regexp "\\(.\\)\\1*") (fun s -> string_of_int (String.length (Str.matched_string s)) ^ Str.matched_group 1 s)
let () =
let num = ref "1" in print_endline !num; for i = 1 to 10 do num := lookandsay !num; print_endline !num; done</lang>
With regular expressions in the Pcre library
<lang ocaml>open Pcre
let lookandsay str =
let rex = regexp "(.)\\1*" in let subs = exec_all ~rex str in let ar = Array.map (fun sub -> get_substring sub 0) subs in let ar = Array.map (fun s -> String.length s, s.[0]) ar in let ar = Array.map (fun (n,c) -> (string_of_int n) ^ (String.make 1 c)) ar in let res = String.concat "" (Array.to_list ar) in (res)
let () =
let num = ref(string_of_int 1) in for i = 1 to 10 do num := lookandsay !num; print_endline !num; done</lang>
run this example with 'ocaml -I +pcre pcre.cma script.ml'
Imperative
<lang ocaml>(* see http://oeis.org/A005150 *)
let look_and_say s = let n = String.length s and buf = Buffer.create 0 and prev = ref s.[0] and count = ref 0 in let append () = Buffer.add_char buf (char_of_int (48 + !count));
Buffer.add_char buf !prev in
String.iter (fun c ->
if c = !prev then incr count else begin append (); prev := c; count := 1 end
) s; append (); Buffer.contents buf;;
(* what about length of successive strings ? *) let iter f a n = let rec aux r n v = if n = 0
then List.rev(r::v) else aux (f r) (n - 1) (r::v) in
aux a n [];;
let las = iter look_and_say "1";;
(* the first sixty terms *)
List.map (String.length) (las 59);; (*
[1; 2; 2; 4; 6; 6; 8; 10; 14; 20; 26; 34; 46; 62; 78; 102; 134; 176; 226; 302; 408; 528; 678; 904; 1182; 1540; 2012; 2606; 3410; 4462; 5808; 7586; 9898; 12884; 16774; 21890; 28528; 37158; 48410; 63138; 82350; 107312; 139984; 182376; 237746; 310036; 403966; 526646; 686646; 894810; 1166642; 1520986; 1982710; 2584304; 3369156; 4391702; 5724486; 7462860; 9727930; 12680852]
- )
(* see http://oeis.org/A005341 *)</lang>
Oforth
<lang Oforth>import: mapping
- lookAndSay ( n -- )
[ 1 ] #[ dup .cr group map( [#size, #first] ) expand ] times( n ) ;</lang>
- Output:
for n = 10
[1] [1, 1] [2, 1] [1, 2, 1, 1] [1, 1, 1, 2, 2, 1] [3, 1, 2, 2, 1, 1] [1, 3, 1, 1, 2, 2, 2, 1] [1, 1, 1, 3, 2, 1, 3, 2, 1, 1] [3, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1] [1, 3, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1]
Oz
<lang oz>declare
%% e.g. "21" -> "1211" fun {LookAndSayString S} for DigitGroup in {Group S} append:Add do {Add {Int.toString {Length DigitGroup}}} {Add [DigitGroup.1]} end end
%% lazy sequence of integers starting with N fun {LookAndSay N} fun lazy {Loop S} {String.toInt S}|{Loop {LookAndSayString S}} end in {Loop {Int.toString N}} end
%% like Haskell's "group" fun {Group Xs} case Xs of nil then nil [] X|Xr then
Ys Zs
{List.takeDropWhile Xr fun {$ W} W==X end ?Ys ?Zs} in (X|Ys) | {Group Zs} end end
in
{ForAll {List.take {LookAndSay 1} 10} Show}</lang>
PARI/GP
<lang parigp>step(n)={
my(v=eval(Vec(Str(n))),cur=v[1],ct=1,out=""); v=concat(v,99); for(i=2,#v, if(v[i]==cur, ct++ , out=Str(out,ct,cur); cur=v[i]; ct=1 ) ); eval(out)
}; n=1;for(i=1,20,print(n);n=step(n))</lang>
Pascal
<lang pascal>program LookAndSayDemo(input, output);
uses
SysUtils;
function LookAndSay (s: string): string;
var item: char; index: integer; count: integer; begin LookAndSay := ; item := s[1]; count := 1; for index:= 2 to length(s) do if item = s[index] then inc(count) else begin
LookAndSay := LookAndSay + intTostr(count) + item;
item := s[index];
count := 1;
end; LookAndSay := LookAndSay + intTostr(count) + item; end;
var
number: string;
begin
writeln('Press RETURN to continue and ^C to stop.'); number := '1'; while not eof(input) do begin write(number); readln; number := LookAndSay(number); end;
end.</lang>
- Output:
% ./LookAndSay Press RETURN to continue and ^C to stop. 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211^C
Even faster imperative Version Improvement: setlength of result,no inttoStr and using pChar But the Code Alignment is very important.
<lang pascal> program LookAndSayDemo(input, output); {$IFDEF FPC}
{$Mode Delphi} // using result {$optimization ON}
// i3-4330 3.5 Ghz // {$CodeAlign proc=16,loop=8} //2,6 secs
{$CodeAlign proc=16,loop=1} //1,6 secs so much faster ???
{$ENDIF}
uses
SysUtils;
const
cntChar : array[1..9] of char = ('1','2','3','4','5','6','7','8','9');
function LookAndSay2 (const s: string): string; //using pChar for result var
source, destin : pChar; len, idxFrom, idxTo : integer; cnt: integer;
item: char;
begin
idxFrom := length(s); source := @s[1];
//adjust length of result len := round(length(s)* 1.306+10); setlength(result,len); destin := @result[1]; dec(destin);
idxto := 1; item := source^; inc(source); cnt := 1; for idxFrom := idxFrom downto 2 do begin if item <> source^ then begin destin[idxTo] := cntChar[cnt]; destin[idxTo+1]:= item; item := source^; cnt := 1; inc(idxto,2); end else inc(cnt); inc(source); end; destin[idxTo] := cntChar[cnt]; destin[idxTo+1]:= item; setlength(result,idxto+1);
end;
var
number: string; l1,l2, i : integer;
begin
number := '1'; writeln(number); writeln(1:4,length(number):16,1/1:10:6);
For i := 2 to 70 do begin l1 := length(number); number := LookAndSay2(number); l2 := length(number); IF i <10 then writeln(number); writeln(i:4,length(number):16,l2/l1:10:6); end;
end.</lang>
- Output:
1 1 1 1.000000 11 2 2 2.000000 21 3 2 1.000000 1211 4 4 2.000000 111221 5 6 1.500000 312211 6 6 1.000000 13112221 7 8 1.333333 1113213211 8 10 1.250000 31131211131221 9 14 1.400000 10 20 1.428571 11 26 1.300000 12 34 1.307692 13 46 1.352941 14 62 1.347826 15 78 1.258065 16 102 1.307692 ........ 67 81117366 1.303580 68 105745224 1.303608 69 137842560 1.303535 70 179691598 1.303600 real 0m1.639s user 0m1.593s sys 0m0.043s
Perl
<lang perl>sub lookandsay {
my $str = shift; $str =~ s/((.)\2*)/length($1) . $2/ge; return $str;
}
my $num = "1"; foreach (1..10) {
print "$num\n"; $num = lookandsay($num);
}</lang>
Using string as a cyclic buffer: <lang perl>for (local $_ = "1\n"; s/((.)\2*)//s;) { print $1; $_ .= ($1 ne "\n" and length($1)).$2 }</lang>
Phix
<lang Phix>function lookandsay(string s) string res = "" integer p = s[1], c = 1
for i=2 to length(s) do if p=s[i] then c += 1 else res &= sprintf("%d%s",{c,p}) p = s[i] c = 1 end if end for res &= sprintf("%d%s",{c,p}) return res
end function
string s = "1" ?s for i=1 to 10 do
s = lookandsay(s) ?s
end for</lang>
- Output:
"1" "11" "21" "1211" "111221" "312211" "13112221" "1113213211" "31131211131221" "13211311123113112211" "11131221133112132113212221"
PHP
<lang php><?php
function lookAndSay($str) {
return preg_replace_callback('#(.)\1*#', function($matches) {
return strlen($matches[0]).$matches[1]; }, $str); }
$num = "1";
foreach(range(1,10) as $i) {
echo $num."
";
$num = lookAndSay($num);
}
?></lang>
PicoLisp
<lang PicoLisp>(de las (Lst)
(make (while Lst (let (N 1 C) (while (= (setq C (pop 'Lst)) (car Lst)) (inc 'N) ) (link N C) ) ) ) )</lang>
Usage: <lang PicoLisp>: (las (1)) -> (1 1)
- (las @)
-> (2 1)
- (las @)
-> (1 2 1 1)
- (las @)
-> (1 1 1 2 2 1)
- (las @)
-> (3 1 2 2 1 1)
- (las @)
-> (1 3 1 1 2 2 2 1)
- (las @)
-> (1 1 1 3 2 1 3 2 1 1)
- (las @)
-> (3 1 1 3 1 2 1 1 1 3 1 2 2 1)</lang>
PowerBASIC
This uses the RLEncode
function from the PowerBASIC Run-length encoding entry.
<lang powerbasic>FUNCTION RLEncode (i AS STRING) AS STRING
DIM tmp1 AS STRING, tmp2 AS STRING, outP AS STRING DIM Loop0 AS LONG, count AS LONG
FOR Loop0 = 1 TO LEN(i) tmp1 = MID$(i, Loop0, 1) IF tmp1 <> tmp2 THEN IF count > 1 THEN outP = outP & TRIM$(STR$(count)) & tmp2 tmp2 = tmp1 count = 1 ELSEIF 0 = count THEN tmp2 = tmp1 count = 1 ELSE outP = outP & "1" & tmp2 tmp2 = tmp1 END IF ELSE INCR count END IF NEXT
outP = outP & TRIM$(STR$(count)) & tmp2 FUNCTION = outP
END FUNCTION
FUNCTION lookAndSay(BYVAL count AS LONG) AS STRING
DIM iii AS STRING, tmp AS STRING
IF count > 1 THEN iii = lookAndSay(count - 1) ELSEIF count < 2 THEN iii = "1" END IF
tmp = RLEncode(iii) lookAndSay = tmp
END FUNCTION
FUNCTION PBMAIN () AS LONG
DIM v AS LONG v = VAL(INPUTBOX$("Enter a number.")) MSGBOX lookAndSay(v)
END FUNCTION</lang>
PowerShell
<lang powershell>function Get-LookAndSay ($n = 1) {
$re = [regex] '(.)\1*' $ret = "" foreach ($m in $re.Matches($n)) { $ret += [string] $m.Length + $m.Value[0] } return $ret
}
function Get-MultipleLookAndSay ($n) {
if ($n -eq 0) { return @() } else { $a = 1 $a for ($i = 1; $i -lt $n; $i++) { $a = Get-LookAndSay $a $a } }
}</lang>
- Output:
PS> Get-MultipleLookAndSay 8 1 11 21 1211 111221 312211 13112221 1113213211
Prolog
Works with SWI-Prolog.
<lang Prolog>look_and_say(L) :- maplist(write, L), nl, encode(L, L1), look_and_say(L1).
% This code is almost identical to the code of "run-length-encoding" encode(In, Out) :- packList(In, R1), append(R1,Out).
% use of library clpfd allows packList(?In, ?Out) to works
% in both ways In --> Out and In <-- Out.
- - use_module(library(clpfd)).
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % ?- packList([a,a,a,b,c,c,c,d,d,e], L). % L = [[3,a],[1,b],[3,c],[2,d],[1,e]] . % ?- packList(R, [[3,a],[1,b],[3,c],[2,d],[1,e]]). % R = [a,a,a,b,c,c,c,d,d,e] . % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% packList([],[]).
packList([X],1,X) :- !.
packList([X|Rest],[XRun|Packed]):-
run(X,Rest, XRun,RRest), packList(RRest,Packed).
run(Var,[],[1,Var],[]).
run(Var,[Var|LRest],[N1, Var],RRest):-
N #> 0, N1 #= N + 1, run(Var,LRest,[N, Var],RRest).
run(Var,[Other|RRest], [1,Var],[Other|RRest]):-
dif(Var,Other).
</lang>
- Output:
?- look_and_say([1]). 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 ..........................
Pure
<lang pure>using system;
// Remove the trailing "L" from the string representation of bigints. __show__ x::bigint = init (str x);
say x = val $ strcat $ map (sprintf "%d%s") $ look $ chars $ str x with
look [] = []; look xs@(x:_) = (#takewhile (==x) xs,x) : look (dropwhile (==x) xs);
end;
iteraten 5 say 1; // [1,11,21,1211,111221]
// This prints the entire sequence, press Ctrl-C to abort. do (puts.str) (iterate say 1);</lang>
PureBasic
<lang PureBasic>If OpenConsole()
Define i, j, cnt, txt$, curr$, result$ Print("Enter start sequence: "): txt$=Input() Print("How many repetitions: "): i=Val(Input()) ; PrintN(#CRLF$+"Sequence:"+#CRLF$+txt$) Repeat j=1 result$="" Repeat curr$=Mid(txt$,j,1) cnt=0 Repeat cnt+1 j+1 Until Mid(txt$,j,1)<>curr$ result$+Str(cnt)+curr$ Until j>Len(txt$) PrintN(result$) txt$=result$ i-1 Until i<=0 ; PrintN(#CRLF$+"Press ENTER to exit."): Input() CloseConsole()
EndIf</lang>
- Output:
Enter start sequence: 1 How many repetitions: 7 Sequence: 1 11 21 1211 111221 312211 13112221 1113213211
Python
<lang python>def lookandsay(number):
result = ""
repeat = number[0] number = number[1:]+" " times = 1
for actual in number: if actual != repeat: result += str(times)+repeat times = 1 repeat = actual else: times += 1
return result
num = "1"
for i in range(10):
print num num = lookandsay(num)</lang>
Functional
<lang python>>>> from itertools import groupby >>> def lookandsay(number): return .join( str(len(list(g))) + k for k,g in groupby(number) )
>>> numberstring='1' >>> for i in range(10): print numberstring numberstring = lookandsay(numberstring)</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
As a generator
<lang python>>>> from itertools import groupby, islice
>>>
>>> def lookandsay(number='1'):
while True:
yield number
number = .join( str(len(list(g))) + k
for k,g in groupby(number) )
>>> print('\n'.join(islice(lookandsay(), 10)))
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211</lang>
Using regular expressions
<lang python>import re
def lookandsay(str):
return re.sub(r'(.)\1*', lambda m: str(len(m.group(0))) + m.group(1), str)
num = "1" for i in range(10):
print num num = lookandsay(num)</lang>
Q
<lang q>las:{{raze string[count@'x],'@'[;0]x:where[differ x]_x}\[x;1#"1"]} las 8</lang>
- Output:
,"1" "11" "21" "1211" "111221" "312211" "13112221" "1113213211" "31131211131221"
R
Returning the value as an integer limits how long the sequence can get, so the option for integer or character return values are provided. <lang R>look.and.say <- function(x, return.an.int=FALSE) {
#convert number to character vector xstr <- unlist(strsplit(as.character(x), "")) #get run length encoding rlex <- rle(xstr) #form new string odds <- as.character(rlex$lengths) evens <- rlex$values newstr <- as.vector(rbind(odds, evens)) #collapse to scalar newstr <- paste(newstr, collapse="") #convert to number, if desired if(return.an.int) as.integer(newstr) else newstr
}</lang> Example usage: <lang R>x <- 1 for(i in 1:10) {
x <- look.and.say(x) print(x)
}</lang>
Racket
<lang Racket>
- lang racket
(define (encode str)
(regexp-replace* #px"(.)\\1*" str (lambda (m c) (~a (string-length m) c))))
(define (look-and-say-sequence n)
(reverse (for/fold ([r '("1")]) ([n n]) (cons (encode (car r)) r))))
(for-each displayln (look-and-say-sequence 10)) </lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Raku
(formerly Perl 6)
In Raku it is natural to avoid explicit loops; rather we use the sequence operator to define a lazy infinite sequence. We'll print the first 15 values here.
<lang perl6>.say for ('1', *.subst(/(.)$0*/, { .chars ~ .[0] }, :g) ... *)[^15];</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
REXX
Programming note: this version works with any string (a null is assumed, which causes 1 to be used).
If a negative number is specified (the number of iterations
to be used for the calculations), only the length of
the number (or character string) is shown.
simple version
<lang rexx>/*REXX program displays the sequence (and/or lengths) for the look and say series.*/ parse arg N ! . /*obtain optional arguments from the CL*/ if N== | N=="," then N=20 /*Not specified? Then use the deault. */ if !== | !=="," then !=1 /* " " " " " " */
do j=1 for abs(N) /*repeat a number of times to show NUMS*/ if j\==1 then != $lookAndSay(!) /*invoke function to calculate next #. */ if N<0 then say 'length['j"]:" length(!) /*Also, display the sequence's length.*/ else say '['j"]:" ! /*display the number to the terminal. */ end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ $lookAndSay: procedure; parse arg x,,$ /*obtain the (passed) argument {X}. */ fin = '0'x /*use unique character to end scanning.*/ x=x || fin /*append the FIN character to string.*/
do k=1 by 0 /*now, process the given sequence. */ y= substr(x, k, 1) /*pick off one character to examine. */ if y== fin then return $ /*if we're at the end, then we're done.*/ _= verify(x, y, , k) - k /*see how many characters we have of Y.*/ $= $ || _ || y /*build the "look and say" sequence. */ k= k + _ /*now, point to the next character. */ end /*k*/</lang>
- output when using the default input values of: 20 1
[1]: 1 [2]: 11 [3]: 21 [4]: 1211 [5]: 111221 [6]: 312211 [7]: 13112221 [8]: 1113213211 [9]: 31131211131221 [10]: 13211311123113112211 [11]: 11131221133112132113212221 [12]: 3113112221232112111312211312113211 [13]: 1321132132111213122112311311222113111221131221 [14]: 11131221131211131231121113112221121321132132211331222113112211 [15]: 311311222113111231131112132112311321322112111312211312111322212311322113212221 [16]: 132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211 [17]: 11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221 [18]: 31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211 [19]: 1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221 [20]: 11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211
- output when using the input values of: 17 ggg
[1]: ggg [2]: 3g [3]: 131g [4]: 1113111g [5]: 3113311g [6]: 132123211g [7]: 11131211121312211g [8]: 31131112311211131122211g [9]: 132113311213211231132132211g [10]: 11131221232112111312211213211312111322211g [11]: 3113112211121312211231131122211211131221131112311332211g [12]: 1321132122311211131122211213211321322112311311222113311213212322211g [13]: 1113122113121122132112311321322112111312211312111322211213211321322123211211131211121332211g [14]: 31131122211311122122111312211213211312111322211231131122211311123113322112111312211312111322111213122112311311123112112322211g [15]: 132113213221133122112231131122211211131221131112311332211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211g [16]: 11131221131211132221231122212213211321322112311311222113311213212322211211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211g [17]: 31131122211311123113321112132132112211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211g
- output when using the input value of: -60
length[1]: 1 length[2]: 2 length[3]: 2 length[4]: 4 length[5]: 6 length[6]: 6 length[7]: 8 length[8]: 10 length[9]: 14 length[10]: 20 length[11]: 26 length[12]: 34 length[13]: 46 length[14]: 62 length[15]: 78 length[16]: 102 length[17]: 134 length[18]: 176 length[19]: 226 length[20]: 302 length[21]: 408 length[22]: 528 length[23]: 678 length[24]: 904 length[25]: 1182 length[26]: 1540 length[27]: 2012 length[28]: 2606 length[29]: 3410 length[30]: 4462 length[31]: 5808 length[32]: 7586 length[33]: 9898 length[34]: 12884 length[35]: 16774 length[36]: 21890 length[37]: 28528 length[38]: 37158 length[39]: 48410 length[40]: 63138 length[41]: 82350 length[42]: 107312 length[43]: 139984 length[44]: 182376 length[45]: 237746 length[46]: 310036 length[47]: 403966 length[48]: 526646 length[49]: 686646 length[50]: 894810 length[51]: 1166642 length[52]: 1520986 length[53]: 1982710 length[54]: 2584304 length[55]: 3369156 length[56]: 4391702 length[57]: 5724486 length[58]: 7462860 length[59]: 9727930 length[60]: 12680852
faster version
This version appends the generated parts of the sequence, and after it gets to a certain size (chunkSize),
it appends the sequence generated (so far) to the primary sequence, and starts with a null sequence.
This avoids appending a small character string to a growing larger and larger character string.
<lang rexx>/*REXX program displays the sequence (and/or lengths) for the look and say series.*/
parse arg N ! . /*obtain optional arguments from the CL*/
if N== | N=="," then N=20 /*Not specified? Then use the default.*/
if !== | !=="," then !=1 /* " " " " " " */
/* [↑] !: starting char for the seq.*/ do j=1 for abs(N) /*repeat a number of times to show NUMS*/ if j\==1 then != $lookAndSay(!) /*invoke function to calculate next #. */ if N<0 then say 'length['j"]:" length(!) /*Also, display the sequence's length.*/ else say '['j"]:" ! /*display the number to the terminal. */ end /*j*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ $lookAndSay: procedure; parse arg x,,$ ! /*obtain the (passed) argument {X}. */
chSize= 1000 /*define a sensible chunk size. */ fin = '0'x /*use unique character to end scanning.*/ x=x || fin /*append the FIN character to string.*/ do k=1 by 0 /*now, process the given sequence. */ y= substr(x, k, 1) /*pick off one character to examine. */ if y== fin then return $ /*if we're at the end, then we're done.*/ _= verify(x, y, , k) - k /*see how many characters we have of Y.*/ $= $ || _ || y /*build the "look and say" sequence. */ k= k + _ /*now, point to the next character. */ if length($)<chSize then iterate /*Less than chunkSize? Then keep going*/ != ! || $ /*append $ to the ! string. */ $= /*now, start $ from scratch. */ chSize= chSize + 100 /*bump the chunkSize (length) counter.*/ end /*k*/ return ! || $ /*return the ! string plus the $ string*/</lang>
- output is identical to the 1st REXX version (the simple version).
Ring
<lang ring> number = "1" for nr = 1 to 10
number = lookSay(number) see number + nl
next
func lookSay n
i = 0 j = 0 c="" o="" i = 1 while i <= len(n) c = substr(n,i,1) j = i + 1 while substr(n,j,1) = c j += 1 end o += string(j-i) + c i = j end return o
</lang>
Ruby
The simplest one: <lang ruby> class String
def look_and_say gsub(/(.)\1*/){|s| s.size.to_s + s[0]} end
end
ss = '1' 12.times {puts ss; ss = ss.look_and_say} </lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
<lang ruby>def lookandsay(str)
str.gsub(/(.)\1*/) {$&.length.to_s + $1}
end
num = "1" 10.times do
puts num num = lookandsay(num)
end</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
Using Enumerable#chunk <lang ruby>def lookandsay(str)
str.chars.chunk{|c| c}.map{|c,x| [x.size, c]}.join
end
puts num = "1" 9.times do
puts num = lookandsay(num)
end</lang> The output is the same above.
Without regular expression:
<lang ruby># Adding clusterization (http://apidock.com/rails/Enumerable/group_by) module Enumerable
# clumps adjacent elements together # >> [2,2,2,3,3,4,2,2,1].cluster # => [[2, 2, 2], [3, 3], [4], [2, 2], [1]] def cluster cluster = [] each do |element| if cluster.last && cluster.last.last == element cluster.last << element else cluster << [element] end end cluster end
end</lang>
Using Array#cluster defined above:
<lang ruby>def print_sequence(input_sequence, seq=10)
return unless seq > 0 puts input_sequence.join result_array = input_sequence.cluster.map do |cluster| [cluster.count, cluster.first] end print_sequence(result_array.flatten, seq-1)
end
print_sequence([1])</lang> The output is the same above.
Rust
<lang rust>fn next_sequence(in_seq: &[i8]) -> Vec<i8> {
assert!(!in_seq.is_empty());
let mut result = Vec::new(); let mut current_number = in_seq[0]; let mut current_runlength = 1;
for i in &in_seq[1..] { if current_number == *i { current_runlength += 1; } else { result.push(current_runlength); result.push(current_number); current_runlength = 1; current_number = *i; } } result.push(current_runlength); result.push(current_number); result
}
fn main() {
let mut seq = vec![1];
for i in 0..10 { println!("Sequence {}: {:?}", i, seq); seq = next_sequence(&seq); }
}</lang>
- Output:
Sequence 0: [1] Sequence 1: [1, 1] Sequence 2: [2, 1] Sequence 3: [1, 2, 1, 1] Sequence 4: [1, 1, 1, 2, 2, 1] Sequence 5: [3, 1, 2, 2, 1, 1] Sequence 6: [1, 3, 1, 1, 2, 2, 2, 1] Sequence 7: [1, 1, 1, 3, 2, 1, 3, 2, 1, 1] Sequence 8: [3, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1] Sequence 9: [1, 3, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1]
Scala
Recursive
<lang Scala>import scala.annotation.tailrec
object LookAndSay extends App {
loop(10, "1")
@tailrec private def loop(n: Int, num: String): Unit = { println(num) if (n <= 0) () else loop(n - 1, lookandsay(num)) }
private def lookandsay(number: String): String = { val result = new StringBuilder
@tailrec def loop(numberString: String, repeat: Char, times: Int): String = if (numberString.isEmpty) result.toString() else if (numberString.head != repeat) { result.append(times).append(repeat) loop(numberString.tail, numberString.head, 1) } else loop(numberString.tail, numberString.head, times + 1)
loop(number.tail + " ", number.head, 1) }
}</lang>
- Output:
See it running in your browser by (JavaScript, non JVM) or by Scastie (JVM).
using Iterator
<lang scala>def lookAndSay(seed: BigInt) = {
val s = seed.toString ( 1 until s.size).foldLeft((1, s(0), new StringBuilder)) { case ((len, c, sb), index) if c != s(index) => sb.append(len); sb.append(c); (1, s(index), sb) case ((len, c, sb), _) => (len + 1, c, sb) } match { case (len, c, sb) => sb.append(len); sb.append(c); BigInt(sb.toString) }
}
def lookAndSayIterator(seed: BigInt) = Iterator.iterate(seed)(lookAndSay)</lang>
using Stream
<lang Scala>object Main extends App {
def lookAndSay(previous: List[BigInt]): Stream[List[BigInt]] = {
def next(num: List[BigInt]): List[BigInt] = num match { case Nil => Nil case head :: Nil => 1 :: head :: Nil case head :: tail => val size = (num takeWhile (_ == head)).size List(BigInt(size), head) ::: next(num.drop(size)) } val x = next(previous) x #:: lookAndSay(x) }
(lookAndSay(1 :: Nil) take 10).foreach(s => println(s.mkString("")))
}</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func string: lookAndSay (in integer: level, in string: stri) is func
result var string: lookAndSay is ""; local var integer: index is 2; begin if level = 1 then if stri <> "" then while index <= length(stri) and stri[index] = stri[1] do incr(index); end while; lookAndSay := str(pred(index)) & stri[1 len 1] & lookAndSay(level, stri[index ..]); end if; else lookAndSay := lookAndSay(1, lookAndSay(pred(level), stri)); end if; end func;
const proc: main is func
local var integer: level is 0; begin for level range 1 to 14 do writeln(lookAndSay(level, "1")); end for; end func;</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
Sidef
<lang ruby>func lookandsay(str) {
str.gsub(/((.)\2*)/, {|a,b| a.len.to_s + b });
}
var num = "1"; {
say num; num = lookandsay(num);
} * 10;</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
Smalltalk
<lang smalltalk>String extend [
lookAndSay [ |anElement nextElement counter coll newColl| coll := (self asOrderedCollection). newColl := OrderedCollection new. counter := 0. anElement := (coll first). [ coll size > 0 ] whileTrue: [ nextElement := coll removeFirst.
( anElement == nextElement ) ifTrue: [
counter := counter + 1. ] ifFalse: [
newColl add: (counter displayString). newColl add: (anElement asString). anElement := nextElement. counter := 1.
] ]. newColl add: (counter displayString). newColl add: (anElement asString). ^(newColl join) ]
].
|r| r := '1'. 10 timesRepeat: [
r displayNl. r := r lookAndSay.
]</lang>
<lang smalltalk>String compile:
'lookAndSay |anElement nextElement counter coll newColl| coll := (self asOrderedCollection). newColl := OrderedCollection new. counter := 0. anElement := (coll first). [ coll size > 0 ] whileTrue: [ nextElement := coll removeFirst.
( anElement == nextElement ) ifTrue: [
counter := counter + 1. ] ifFalse: [
newColl add: (counter displayString). newColl add: (anElement asString). anElement := nextElement. counter := 1.
] ]. newColl add: (counter displayString). newColl add: (anElement asString). ^(' join: newColl)' classified: 'toys'.
result := OrderedCollection new. r := '1'. result add: r. result addAll: ((1 to: 10) collect: [ :i |
r := r lookAndSay.
]). result. </lang>
Output:
an OrderedCollection('1' '11' '21' '1211' '111221' '312211' '13112221' '1113213211' '31131211131221' '13211311123113112211' '11131221133112132113212221')
SNOBOL4
The look-and-say sequence is an iterative run-length string encoding. So looksay( ) is just a wrapper around the Run-length Encoding task. This is by far the easiest solution.
<lang SNOBOL4>* # Encode RLE
define('rle(str)c,n') :(rle_end)
rle str len(1) . c :f(return)
str span(c) @n = rle = rle n c :(rle)
rle_end
- # First m members of sequence with seed n
define('looksay(n,m)') :(looksay_end)
looksay output = n; m = gt(m,1) m - 1 :f(return)
n = rle(n) :(looksay)
looksay_end
- Test and display
looksay(1,10)
end</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
SQL
<lang sql>DROP VIEW delta; CREATE VIEW delta AS
SELECT sequence1.v AS x, (sequence1.v<>sequence2.v)*sequence1.c AS v, sequence1.c AS c FROM sequence AS sequence1, sequence AS sequence2 WHERE sequence1.c = sequence2.c+1;
DROP VIEW rle0; CREATE VIEW rle0 AS
SELECT delta2.x AS x, SUM(delta2.v) AS v, delta2.c AS c FROM delta AS delta1, delta as delta2 WHERE delta1.c >= delta2.c GROUP BY delta1.c;
DROP VIEW rle1; CREATE VIEW rle1 AS
SELECT sum(x)/x AS a, x AS b, c AS c FROM rle0 GROUP BY v;
DROP VIEW rle2; CREATE VIEW rle2 AS
SELECT a as v, 1 as o, 2*c+0 as c FROM rle1 UNION SELECT b as v, 1 as o, 2*c+1 as c FROM rle1;
DROP VIEW normed; CREATE VIEW normed AS
SELECT r1.v as v, SUM(r2.o) as c FROM rle2 AS r1, rle2 AS r2 WHERE r1.c >= r2.c GROUP BY r1.c;
DROP TABLE rle; CREATE TABLE rle(v int, c int); INSERT INTO rle SELECT * FROM normed ORDER BY c;
DELETE FROM sequence; INSERT INTO sequence VALUES(-1,0); INSERT INTO sequence SELECT * FROM rle;</lang>
Usage:
% sqlite3 SQLite version 3.4.0 Enter ".help" for instructions sqlite> CREATE TABLE sequence(v int, c int); sqlite> INSERT INTO sequence VALUES(-1,0); sqlite> INSERT INTO sequence VALUES(1,1); sqlite> SELECT * FROM sequence; -1|0 1|1 sqlite> .read look.sql sqlite> SELECT * FROM sequence; -1|0 1|1 1|2 sqlite> .read look.sql sqlite> SELECT * FROM sequence; -1|0 2|1 1|2 sqlite> .read look.sql sqlite> SELECT * FROM sequence; -1|0 1|1 2|2 1|3 1|4 sqlite> .read look.sql sqlite> SELECT * FROM sequence; -1|0 1|1 1|2 1|3 2|4 2|5 1|6
SQL PL
version 9.7 or higher.
With SQL PL: <lang sql pl> SET SERVEROUTPUT ON @
BEGIN
DECLARE NMBR VARCHAR(100) DEFAULT '1'; DECLARE J SMALLINT DEFAULT 1;
CALL DBMS_OUTPUT.PUT_LINE(NMBR); WHILE (J < 10) DO BEGIN DECLARE I SMALLINT; DECLARE SIZE SMALLINT; DECLARE ACTUAL CHAR(1); DECLARE REPEAT CHAR(1); DECLARE RESULT VARCHAR(100); DECLARE TIMES SMALLINT;
SET REPEAT = SUBSTR(NMBR, 1, 1); SET NMBR = SUBSTR(NMBR, 2) || ' '; SET TIMES = 1; SET I = 1; SET SIZE = LENGTH(NMBR);
WHILE (I <= SIZE) DO SET ACTUAL = SUBSTR(NMBR, I, 1); IF (ACTUAL <> REPEAT) THEN SET RESULT = COALESCE(RESULT, ) || TIMES || || REPEAT; SET TIMES = 1; SET REPEAT = ACTUAL; ELSE SET TIMES = TIMES + 1; END IF; SET I = I + 1; END WHILE;
CALL DBMS_OUTPUT.PUT_LINE(RESULT); SET NMBR = RESULT; END ; SET J = J + 1; END WHILE;
END @ </lang> Output:
db2 => BEGIN ... db2 (cont.) => END @ DB20000I The SQL command completed successfully. 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
Swift
<lang swift>func lookAndSay(_ seq: [Int]) -> [Int] {
var result = [Int]() var cur = seq[0] var curRunLength = 1
for i in seq.dropFirst() { if cur == i { curRunLength += 1 } else { result.append(curRunLength) result.append(cur) curRunLength = 1 cur = i } }
result.append(curRunLength) result.append(cur)
return result
}
var seq = [1]
for i in 0..<10 {
print("Seq \(i): \(seq)") seq = lookAndSay(seq)
}</lang>
- Output:
Seq 0: [1] Seq 1: [1, 1] Seq 2: [2, 1] Seq 3: [1, 2, 1, 1] Seq 4: [1, 1, 1, 2, 2, 1] Seq 5: [3, 1, 2, 2, 1, 1] Seq 6: [1, 3, 1, 1, 2, 2, 2, 1] Seq 7: [1, 1, 1, 3, 2, 1, 3, 2, 1, 1] Seq 8: [3, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1] Seq 9: [1, 3, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1]
Tcl
<lang tcl>proc lookandsay n {
set new "" while {[string length $n] > 0} { set char [string index $n 0] for {set count 1} {[string index $n $count] eq $char} {incr count} {} append new $count $char set n [string range $n $count end] } interp alias {} next_lookandsay {} lookandsay $new return $new
}
puts 1 ;# ==> 1 puts [lookandsay 1] ;# ==> 11 puts [next_lookandsay] ;# ==> 21 puts [next_lookandsay] ;# ==> 1211 puts [next_lookandsay] ;# ==> 111221 puts [next_lookandsay] ;# ==> 312211</lang>
Alternatively, with coroutines:
<lang tcl>proc seq_lookandsay {n {coroName next_lookandsay}} {
coroutine $coroName apply {n { for {} {[yield $n] ne "stop"} {set n $new} { set new "" foreach subseq [regexp -all -inline {0+|1+|2+|3+|4+|5+|6+|7+|8+|9+} $n] { append new [string length $subseq] [string index $subseq 0] } } }} $n
}
puts [seq_lookandsay 1] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay]</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT,{} num=1,say=""
LOOP look digits=STRINGS (num," ? ") digitgrouped=ACCUMULATE (digits,howmany) LOOP/CLEAR h=howmany,digit=digitgrouped say=JOIN (say,"",h,digit) ENDLOOP PRINT say num=VALUE(say),say="" IF (look==14) EXIT ENDLOOP
</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
UNIX Shell
<lang bash>lookandsay() {
local num=$1 char seq i for ((i=0; i<=${#num}; i++)); do char=${num:i:1} if [[ $char == ${seq:0:1} ]]; then seq+=$char else -n $seq && printf "%d%s" ${#seq} ${seq:0:1} seq=$char fi done
}
for ((num=1, i=1; i<=10; i++)); do
echo $num num=$( lookandsay $num )
done</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
Ursala
The look_and_say function returns the first n results by iterating the function that maps a given sequence to its successor. <lang Ursala>#import std
- import nat
look_and_say "n" = ~&H\'1' next"n" rlc~&E; *= ^lhPrT\~&hNC %nP+ length
- show+
main = look_and_say 10</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
VBA
<lang VBA> Public Sub LookAndSay(Optional Niter As Integer = 10) 'generate "Niter" members of the look-and-say sequence '(argument is optional; default is 10)
Dim s As String 'look-and-say number Dim news As String 'next number in sequence Dim curdigit As String 'current digit in s Dim newdigit As String 'next digit in s Dim curlength As Integer 'length of current run Dim p As Integer 'position in s Dim L As Integer 'length of s
On Error GoTo Oops 'to catch overflow, i.e. number too long
'start with "1" s = "1" For i = 1 To Niter
'initialise L = Len(s) p = 1 curdigit = Left$(s, 1) curlength = 1 news = "" For p = 2 To L 'check next digit in s newdigit = Mid$(s, p, 1) If curdigit = newdigit Then 'extend current run curlength = curlength + 1 Else ' "output" run and start new run news = news & CStr(curlength) & curdigit curdigit = newdigit curlength = 1 End If Next p ' "output" last run news = news & CStr(curlength) & curdigit Debug.Print news s = news
Next i Exit Sub
Oops:
Debug.Print If Err.Number = 6 Then 'overflow Debug.Print "Oops - number too long!" Else Debug.Print "Error: "; Err.Number, Err.Description End If
End Sub </lang>
- Output:
LookAndSay 7 11 21 1211 111221 312211 13112221 1113213211
(Note: overflow occurs at 38th iteration!)
VBScript
Implementation
<lang vb>function looksay( n ) dim i dim accum dim res dim c res = vbnullstring do if n = vbnullstring then exit do accum = 0 c = left( n,1 ) do while left( n, 1 ) = c accum = accum + 1 n = mid(n,2) loop if accum > 0 then res = res & accum & c end if loop looksay = res end function</lang>
Invocation
<lang vb>dim m m = 1 for i = 0 to 13 m = looksay(m) wscript.echo m next</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
Vedit macro language
This implementation generates look-and-say sequence starting from the sequence on cursor line in edit buffer. Each new sequence is inserted as a new line. 10 sequences are created in this example.
<lang vedit>Repeat(10) {
BOL Reg_Empty(20) While (!At_EOL) { Match("(.)\1*", REGEXP+ADVANCE) Num_Str(Chars_Matched, 20, LEFT+APPEND) Reg_Copy_Block(20, CP-1, CP, APPEND) } Ins_Newline Reg_Ins(20)
}</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Yabasic
<lang Yabasic> dim X$(2) i = 0 // índice de cadena de entrada X$(i) = "1"
input "Indica cuantas repeticiones: " r print "\nSecuencia:"
print X$(i) for n = 1 to r-1
j = 1 - i // índice de cadena de salida X$(j) = "" k = 1 while k <= len(X$(i)) k0 = k + 1 while ((k0 <= len(X$(i))) and (mid$(X$(i), k, 1) = mid$(X$(i), k0, 1))) k0 = k0 + 1 wend X$(j) = X$(j) + str$(k0 - k) + mid$(X$(i), k, 1) k = k0 wend i = j print X$(j)
next n print </lang>
- Output:
La salida es similar a la de FreeBASIC, mostrada arriba.
Yorick
<lang yorick>func looksay(input) {
// Special case: one digit if(strlen(input) == 1) return "1" + input; // Convert string into an array of digits digits = strchar(input)(:-1); // Find indices where each run starts w = where(digits(dif)); start = numberof(w) ? grow(1, w+1) : [1]; // Find length of each run len = grow(start, numberof(digits)+1)(dif); // Find digits for each run run = digits(start); // Construct output array result = array(string, numberof(start)*2); // Fill in lengths result(1::2) = swrite(format="%d", len); // Fill in digits; first must add trailing nulls to coerce single string // into an array of strings. run = transpose([run, array(char(0), numberof(run))])(*); result(2::2) = strchar(run); // Merge string array into single string return result(sum);
}
val = "1"; do {
write, val; val = looksay(val);
} while(strlen(val) < 80);</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
zkl
Treating the task as a string manipulation problem.
<lang zkl>fcn lookAndSay(seed){ // numeric String --> numeric String
len,c:=[1..seed.len()-1].reduce(fcn([(len,c)]lc,index,s,sb){ if(c!=s[index]) { sb.write(len); sb.write(c); lc.clear(1,s[index]) } else lc.clear(len+1,c); },L(1,seed[0]), seed,sb:=Sink(String)); sb.write(len); sb.write(c); sb.close();
}</lang>
- Output:
(0).reduce(10,fcn(seed,_){ lookAndSay(seed).println() },"1"); 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
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