Look-and-say sequence: Difference between revisions
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=={{header|Mathematica}}/{{header|Wolfram Language}}== |
=={{header|Mathematica}}/{{header|Wolfram Language}}== |
||
The function: |
The function: |
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<lang Mathematica> LookAndSay[n_Integer?Positive]:=Map[Reverse, Tally/@ Split @ IntegerDigits @ n,2] // Flatten // FromDigits</lang> |
<lang Mathematica> LookAndSay[n_Integer?Positive]:= Map[Reverse, Tally/@ Split @ IntegerDigits @ n, 2] // Flatten // FromDigits</lang> |
||
takes an *arbitrary* positive integer, and generates the next member of the ‘Look and Say’ sequence. |
takes an *arbitrary* positive integer, and generates the next member of the ‘Look and Say’ sequence. |
||
Revision as of 07:23, 25 January 2022
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.
11l
<lang 11l>F lookandsay(=number)
V result = ‘’ V repeat = number[0] number = number[1..]‘ ’ V times = 1
L(actual) number
I actual != repeat
result ‘’= String(times)‘’repeat
times = 1
repeat = actual
E
times++
R result
V num = ‘1’
L 10
print(num) num = lookandsay(num)</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
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>
8086 Assembly
<lang asm> bits 16 cpu 8086 puts: equ 9h ; MS/DOS system call to print a string nmemb: equ 15 ; Change this to print more or fewer members section .text org 100h mov cx,nmemb ; CX = how many members to print outloop: mov dx,memb ; Print current member mov ah,puts int 21h mov dx,newline ; Print newline int 21h mov di,memb ; Generate next member call looksay loop outloop ; Decrease CX, and loop until zero. ret ; Go back to DOS. ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;;; Given a look and say string in ES:DI, generate the next ;;; one in place. Assumption: DS = ES. looksay: push cx ; Keep the counter register mov si,di ; Store pointer to string begin in SI mov bx,di ; And another in BX mov al,'$' ; Find the end of the string xor cx,cx ; Max. 65535 tries dec cx repne scasb ; The 8086 has dedicated string search mov dx,di ; Store copy of start of new str in DX ;;; Process one character .procchar: mov al,'0' ; Set counter to ASCII 0 mov ah,[bx] ; Get current character of string cmp ah,'$' ; Done? je .done .samechar: inc bx ; Increment pointer inc al ; Increment counter cmp ah,[bx] ; Still the same character? je .samechar ; If yes, test next character mov [di],ax ; Store counter and character inc di ; Move ahead two characters inc di jmp .procchar ; Do next character ;;; Copy new string into old location .done: mov byte [di],'$' ; Terminate the string mov cx,di ; Calculate how many bytes to copy sub cx,dx ; end + 1 - start, so one too few here shr cx,1 ; Divide by 2 = words inc cx ; Compensate for the missing +1 mov di,dx ; Pointer to begin of new string xchg si,di ; Set SI = new string and DI = old rep movsw ; Copy 16 bits at a time pop cx ; Restore counter register ret section .data newline: db 13,10,'$' ; Newline to print in between members memb: db '1$' ; This is where the current member is stored</lang>
Action!
<lang Action!>BYTE FUNC GetLength(CHAR ARRAY s BYTE pos)
CHAR c BYTE len
c=s(pos)
len=1
DO
pos==+1
IF pos<=s(0) AND s(pos)=c THEN
len==+1
ELSE
EXIT
FI
OD
RETURN (len)
PROC Append(CHAR ARRAY text,suffix)
BYTE POINTER srcPtr,dstPtr BYTE len
len=suffix(0) IF text(0)+len>255 THEN len=255-text(0) FI IF len THEN srcPtr=suffix+1 dstPtr=text+text(0)+1 MoveBlock(dstPtr,srcPtr,len) text(0)==+suffix(0) FI
RETURN
PROC LookAndSay(CHAR ARRAY in,out)
BYTE pos,len CHAR ARRAY tmp(5)
pos=1 len=0 out(0)=0 WHILE pos<=in(0) DO len=GetLength(in,pos) StrB(len,tmp) Append(out,tmp) out(0)==+1 out(out(0))=in(pos) pos==+len OD
RETURN
PROC Main()
CHAR ARRAY s1(256),s2(256) BYTE i
SCopy(s1,"1")
PrintE(s1)
FOR i=1 TO 11
DO
IF (i&1)=0 THEN
LookAndSay(s2,s1)
PrintE(s1)
ELSE
LookAndSay(s1,s2)
PrintE(s2)
FI
OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
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
ALGOL-M
<lang algolm>begin
string(1) function digit(n);
integer n;
case n of begin
digit := "0"; digit := "1"; digit := "2";
digit := "3"; digit := "4"; digit := "5";
digit := "6"; digit := "7"; digit := "8";
digit := "9";
end;
string(1) array cur[1:128];
string(1) array next[1:128];
integer curlen, i, cnt, j, n;
cur[1] := "1";
curlen := 1;
for n := 1 step 1 until 15 do begin
write("");
for i := 1 step 1 until curlen do
writeon(cur[i]);
i := j := 1;
while i <= curlen do begin
cnt := 1;
while cur[i + cnt] = cur[i] do
cnt := cnt + 1;
next[j] := digit(cnt);
next[j + 1] := cur[i];
j := j + 2;
i := i + cnt;
end;
for i := 1 step 1 until j-1 do
cur[i] := next[i];
curlen := j - 1;
end;
end</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
APL
<lang apl>
⎕IO←0
d←{(1↓⍵)-¯1↓⍵}
f←{m←(0≠d ⍵),1 ⋄ ,(d ¯1,m/⍳⍴⍵),[.5](m/⍵)}
{(f⍣⍵) ,1}¨⍳10
</lang>
This is an ugly little APL2 function that accepts a numeric vector (or scalar) and returns the result.
Apologies for the labeled loop...
<lang apl2>
R←LNS V;T
R←0⍴0 ⍝ initiate empty reply
LOOP:T←↑⍴↑(=\V)⊂V←,V ⍝ t is the length of the 1st digit's run
R←R,T,↑V ⍝ append t and the 1st digit
→(0≠↑⍴V←T↓V)/LOOP ⍝ drop t digits and iterate
</lang>
AppleScript
<lang applescript>on lookAndSay(startNumber, howMany)
if (howMany < 1) then return {}
-- The numbers are handled as lists of digit-value integers for efficiency and output as a list of strings.
script o
property previousNumber : {}
property newNumber : {}
property output : {}
end script
-- "Digitise" the start number.
repeat
set beginning of o's newNumber to startNumber mod 10 as integer
set startNumber to startNumber div 10
if (startNumber is 0) then exit repeat
end repeat
-- Add it to the output as text and successively derive the remaining numbers.
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ""
set end of o's output to o's newNumber as text
repeat (howMany - 1) times
set o's previousNumber to o's newNumber
set o's newNumber to {}
set i to 1
set previousLength to (o's previousNumber's length)
set currentDigit to beginning of o's previousNumber
repeat with j from 2 to previousLength
set thisDigit to item j of o's previousNumber
if (thisDigit is not currentDigit) then
set end of o's newNumber to j - i
set end of o's newNumber to currentDigit
set i to j
set currentDigit to thisDigit
end if
end repeat
set end of o's newNumber to previousLength - i + 1
set end of o's newNumber to currentDigit
set end of o's output to o's newNumber as text
end repeat
set AppleScript's text item delimiters to astid
return o's output
end lookAndSay
-- Test code: return lookAndSay(1, 10)</lang>
- Output:
<lang applescript>{"1", "11", "21", "1211", "111221", "312211", "13112221", "1113213211", "31131211131221", "13211311123113112211"}</lang>
Arturo
<lang rebol>lookAndSay: function [n][
if n=0 -> return "1" previous: lookAndSay n-1
result: new ""
currentCounter: 0
currentCh: first previous
loop previous 'ch [
if? currentCh <> ch [
if not? zero? currentCounter ->
'result ++ (to :string currentCounter) ++ currentCh
currentCounter: 1
currentCh: ch
]
else ->
currentCounter: currentCounter + 1
]
'result ++ (to :string currentCounter) ++ currentCh
return result
]
loop 0..10 'x [
print [x "->" lookAndSay x]
]</lang>
- Output:
0 -> 1 1 -> 11 2 -> 21 3 -> 1211 4 -> 111221 5 -> 312211 6 -> 13112221 7 -> 1113213211 8 -> 31131211131221 9 -> 13211311123113112211 10 -> 11131221133112132113212221
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>
BASIC
<lang BASIC>10 DEFINT A-Z: I$="1" 20 FOR Z=1 TO 15 30 PRINT I$ 40 O$="" 50 FOR I=1 TO LEN(I$) 60 C=1 70 IF MID$(I$,I,1)=MID$(I$,I+C,1) THEN C=C+1: GOTO 70 80 O$=O$+CHR$(C+48)+MID$(I$,I,1) 90 I=I+C-1 100 NEXT I 110 I$=O$ 120 NEXT Z</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
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
BCPL
<lang bcpl>get "libhdr"
manifest $(
amount = 15 bufsize = 128
$)
let move(dest,src) be $( until !src = 0 do
$( !dest := !src
dest := dest + 1
src := src + 1
$)
!dest := 0
$)
let count(v) = valof $( let i=1
while v!i = !v do i := i + 1 resultis i
$)
let looksay(in,out) be $( until !in = 0 do
$( let n = count(in)
out!0 := n
out!1 := !in
out := out + 2
in := in + n
$)
!out := 0
$)
let show(v) be $( until !v = 0 do
$( writen(!v)
v := v + 1
$)
wrch('*N')
$)
let start() be $( let buf1 = vec bufsize and buf2 = vec bufsize
buf1!0 := 1
buf1!1 := 0
for n = 1 to amount do
$( show(buf1)
looksay(buf1,buf2)
move(buf1,buf2)
$)
$)</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
BQN
<lang bqn>LookSay ← ∾´((⊑∾˜ ≠+'0'˙)¨1↓((+`»≠⊢)⊸⊔))
>((⌈´≠¨)↑¨⊢) LookSay⍟(↕15)"1"</lang>
- Output:
┌─
╵"1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221"
┘
Bracmat
In this example we use a non-linear pattern and a negation of a pattern: the end of e sequence of equal digits is (1) the end of the string or (2) the start of a sequence starting with a different digit. <lang bracmat>( 1:?number & 0:?lines & whl
' ( 1+!lines:~>10:?lines
& :?say { This will accumulate all that has to be said after one iteration. }
& 0:?begin
& ( @( !number { Pattern matching. The '@' indicates we're looking in a string rather than a tree structure. }
: ?
( [!begin
%@?digit
?
[?end
( (|(%@:~!digit) ?) { The %@ guarantees we're testing one character - not less (%) and not more (@). The ? takes the rest. }
& !say !end+-1*!begin !digit:?say
& !end:?begin { When backtracking, 'begin' advances to the begin of the next sequence, or to the end of the string. }
)
& ~ { fail! This forces backtracking. Backtracking stops when all begin positions have been tried. }
)
)
| out$(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>
CLU
<lang clu>look_and_say = proc (s: string) returns (string)
out: array[char] := array[char]$[]
count: int := 0
last: char := '\000'
for c: char in string$chars(s) do
if c ~= last then
if count ~= 0 then
array[char]$addh(out, char$i2c(count + 48))
array[char]$addh(out, last)
end
last := c
count := 1
else
count := count + 1
end
end
array[char]$addh(out, char$i2c(count + 48))
array[char]$addh(out, last)
return (string$ac2s(out))
end look_and_say
start_up = proc ()
lines = 15
po: stream := stream$primary_output()
cur: string := "1"
for i: int in int$from_to(1, lines) do
stream$putl(po, cur)
cur := look_and_say(cur)
end
end start_up</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
COBOL
<lang cobol> IDENTIFICATION DIVISION.
PROGRAM-ID. LOOK-AND-SAY-SEQ.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 SEQUENCES.
02 CUR-SEQ PIC X(80) VALUE "1".
02 CUR-CHARS REDEFINES CUR-SEQ
PIC X OCCURS 80 TIMES INDEXED BY CI.
02 CUR-LENGTH PIC 99 COMP VALUE 1.
02 NEXT-SEQ PIC X(80).
02 NEXT-CHARS REDEFINES NEXT-SEQ
PIC X OCCURS 80 TIMES INDEXED BY NI.
01 ALG-STATE.
02 STEP-AMOUNT PIC 99 VALUE 14.
02 ITEM-COUNT PIC 9.
PROCEDURE DIVISION.
LOOK-AND-SAY.
DISPLAY CUR-SEQ.
SET CI TO 1.
SET NI TO 1.
MAKE-NEXT-ENTRY.
MOVE 0 TO ITEM-COUNT.
IF CI IS GREATER THAN CUR-LENGTH GO TO STEP-DONE.
TALLY-ITEM.
ADD 1 TO ITEM-COUNT.
SET CI UP BY 1.
IF CI IS NOT GREATER THAN CUR-LENGTH
AND CUR-CHARS(CI) IS EQUAL TO CUR-CHARS(CI - 1)
GO TO TALLY-ITEM.
INSERT-ENTRY.
MOVE ITEM-COUNT TO NEXT-CHARS(NI).
MOVE CUR-CHARS(CI - 1) TO NEXT-CHARS(NI + 1).
SET NI UP BY 2.
GO TO MAKE-NEXT-ENTRY.
STEP-DONE.
MOVE NEXT-SEQ TO CUR-SEQ.
SET NI DOWN BY 1.
SET CUR-LENGTH TO NI.
SUBTRACT 1 FROM STEP-AMOUNT.
IF STEP-AMOUNT IS NOT EQUAL TO ZERO GO TO LOOK-AND-SAY.
STOP RUN.</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211
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>
Cowgol
<lang cowgol>include "cowgol.coh"; include "strings.coh";
- Given a string with a member of the look-and-say sequence,
- generate the next member of the sequence.
sub LookSay(cur: [uint8], next: [uint8]) is
while [cur] != 0 loop
var count: uint8 := 1;
var curch: uint8 := [cur];
# count how many of this character we have
loop
cur := @next cur;
if [cur] != curch then break; end if;
count := count + 1;
end loop;
# add it and its count to the next sequence
[next] := '0' + count;
next := @next next;
[next] := curch;
next := @next next;
end loop;
[next] := 0;
end sub;
- amount of members to print
- (don't forget to enlarge the buffers if you make this bigger)
var members: uint8 := 15;
- define two buffers
var curbuf: uint8[255]; var nextbuf: uint8[255];
- start with "1"
CopyString("1", &curbuf as [uint8]);
- generate and print successive members
while members > 0 loop
print(&curbuf as [uint8]); print_nl(); LookSay(&curbuf as [uint8], &nextbuf as [uint8]); CopyString(&nextbuf as [uint8], &curbuf as [uint8]); members := members - 1;
end loop;</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
Crystal
The simplest one: <lang ruby>class String
def lookandsay
gsub(/(.)\1*/){ |s| s.size.to_s + s[0] }
end
end
ss = '1' 12.times { puts ss; ss = ss.to_s.lookandsay }</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>
Using Enumerable#chunks <lang ruby>def lookandsay(str)
str.chars.chunks(&.itself).map{ |(c, x)| x.size.to_s + c }.join
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"]
Beginner Imperative Version
<lang d>import std.stdio, std.conv, std.array;
pure string lookAndSay(string s){
auto result = appender!string;
auto i=0, j=i+1;
while(i<s.length){
while(j<s.length && s[i]==s[j]) j++;
result.put( text(j-i) ~ s[i] );
i = j++;
}
return result.data;
} void main(){
auto s="1"; for(auto i=0; i<10; i++) (s = s.lookAndSay).writeln;
}</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Fast Imperative Version
<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>
Delphi
See Pascal.
Draco
<lang draco>\util.g
proc nonrec look_and_say(*char inp, outp) void:
char cur;
byte count;
channel output text outch;
open(outch, outp);
while cur := inp*; cur ~= '\e' do
count := 1;
while
inp := inp + 1;
inp* = cur
do
count := count + 1
od;
write(outch; count, cur)
od;
close(outch)
corp
proc nonrec main() void:
[256] char buf1, buf2;
byte i;
byte LINES = 14;
CharsCopy(&buf1[0], "1");
for i from 1 upto LINES do
writeln(&buf1[0]);
look_and_say(&buf1[0], &buf2[0]);
CharsCopy(&buf1[0], &buf2[0])
od
corp</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211
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>
Fennel
<lang fennel>(fn look-and-say [t]
(let [lst t
ret []]
(while (> (length lst) 0)
(var (num cnt) (values (table.remove lst 1) 1))
(while (= num (. lst 1))
(set cnt (+ cnt 1))
(when (> (length lst) 0)
(set num (table.remove lst 1))))
(tset ret (+ (length ret) 1) cnt)
(tset ret (+ (length ret) 1) num))
ret))
(var lst [1]) (for [i 1 10]
(print (table.concat lst)) (set lst (look-and-say lst)))</lang>
Alternative solution <lang fennel>(fn look-and-say [s]
(var ret [])
(var (num cnt) (values (s:sub 1 1) 1))
(for [i 2 (length s)]
(var cur-num (s:sub i i))
(if (= num cur-num)
(set cnt (+ cnt 1))
(do
(table.insert ret (.. cnt num))
(set cnt 1)
(set num cur-num))))
(table.insert ret (.. cnt num))
(table.concat ret))
(var str "1") (for [i 1 10]
(print str) (set str (look-and-say str)))</lang>
FOCAL
<lang focal>01.10 A "HOW MANY",M 01.20 S B(0)=1;S B(1)=0 01.30 F Z=1,M;D 4;D 2 01.40 Q
02.10 S X=0;S Y=0 02.15 I (B(X)),2.5 02.17 S CN=0 02.20 S CN=CN+1 02.25 S X=X+1 02.30 I (FABS(B(X)-B(X-1))),2.2 02.35 S C(Y)=CN;S C(Y+1)=B(X-1) 02.40 S Y=Y+2 02.45 G 2.15 02.50 S C(Y)=0 02.55 F X=0,Y;S B(X)=C(X)
03.10 I (A-9)3.2;T "9";R 03.20 I (A-8)3.3;T "8";R 03.30 I (A-7)3.4;T "7";R 03.40 I (A-6)3.5;T "6";R 03.50 I (A-5)3.6;T "5";R 03.60 I (A-4)3.7;T "4";R 03.70 I (A-3)3.8;T "3";R 03.80 I (A-2)3.9;T "2";R 03.90 T "1"
04.10 S X=0 04.20 S A=B(X) 04.30 I (-A)4.4;T !;R 04.40 D 3 04.50 S X=X+1 04.60 G 4.2</lang>
- Output:
HOW MANY:12 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
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]
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
With RegExp
<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>
Imperative version
<lang javascript>function lookAndSay( s="" ){
var tokens=[]
var i=0, j=1
while( i<s.length ) {
while( j<s.length && s[j]===s[i] ) j++
tokens.push( `${j-i}${s[i]}` )
i=j++
}
return tokens.join("")
} var phrase="1" for(var n=0; n<10; n++ )
console.log( phrase = lookAndSay( phrase ) )</lang>
- Output:
11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
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>
Alternative solution, using Lua Pattern: <lang lua>function lookandsay(t)
return t:gsub("(1*)(2*)(3*)", function (...)
local ret = {}
for i = 1, select("#", ...) do
local v = select(i, ...)
if #v > 0 then
ret[#ret + 1] = #v
ret[#ret + 1] = v:sub(1,1)
end
end
return table.concat(ret)
end)
end
local t = "1" for i = 1, 10 do
print(t) t = lookandsay(t)
end</lang>
<lang lua>function lookandsay2(t)
return t:gsub("(1*)(2*)(3*)", function (x, y, z)
return (x == "" and x or (#x .. x:sub(1, 1))) ..
(y == "" and y or (#y .. y:sub(1, 1))) ..
(z == "" and z or (#z .. z:sub(1, 1)))
end)
end
local t = "1" for i = 1, 10 do
print(t) t = lookandsay2(t)
end</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211
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>
Maple
<lang Maple>generate_seq := proc(s) local times, output, i; times := 1; output := ""; for i from 2 to StringTools:-Length(s) do if (s[i] <> s[i-1]) then output := cat(output, times, s[i-1]); times := 1; # re-assign else times ++; end if; end do; cat(output, times, s[i - 1]); end proc:
Look_and_Say :=proc(n) local value, i; value := "1"; print(value); for i from 2 to n do value := generate_seq(value); print(value); end do; end proc:
- Test:
Look_and_Say(10);</lang>
- Output:
"1"
"11"
"21"
"1211"
"111221"
"312211"
"13112221"
"1113213211"
"31131211131221"
"13211311123113112211"
Mathematica/Wolfram Language
The function: <lang Mathematica> LookAndSay[n_Integer?Positive]:= Map[Reverse, Tally/@ Split @ IntegerDigits @ n, 2] // Flatten // FromDigits</lang>
takes an *arbitrary* positive integer, and generates the next member of the ‘Look and Say’ sequence.
The first example returns the next 12 numbers of the sequence starting with 1:
<lang Mathematica>NestList[LookAndSay, 1, 12] // Column</lang>
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221
The second example returns the next 12 numbers of the sequence starting with 7:
<lang Mathematica>NestList[LookAndSay, 7, 12] // Column</lang>
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
Modula-2
<lang modula2>MODULE LookAndSay; FROM InOut IMPORT WriteString, WriteLn; FROM Strings IMPORT Assign, Length;
CONST
MaxSize = 128; Steps = 14;
VAR
buf1, buf2: ARRAY [0..MaxSize-1] OF CHAR; step: CARDINAL;
PROCEDURE LookSay(in: ARRAY OF CHAR; VAR out: ARRAY OF CHAR);
VAR count, inIdx, outIdx: CARDINAL;
curChar: CHAR;
BEGIN
inIdx := 0;
outIdx := 0;
WHILE in[inIdx] # CHR(0) DO
curChar := in[inIdx];
count := 0;
REPEAT
INC(inIdx);
INC(count);
UNTIL in[inIdx] # curChar;
out[outIdx] := CHR(ORD('0') + count);
out[outIdx+1] := curChar;
outIdx := outIdx + 2;
END;
out[outIdx] := CHR(0);
END LookSay;
BEGIN
Assign("1", buf1);
FOR step := 1 TO Steps DO
WriteString(buf1);
WriteLn();
LookSay(buf1, buf2);
Assign(buf2, buf1);
END;
END LookAndSay.</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 1113122113121113123112111311222112132113213221133122211311221
NewLisp
<lang NewLisp>
- Compute the following number in the sequence
(define (next-number s) (let (n 0 c (first s) res "") (dostring (x s) (if (= (char x) c) ; the iteration variable is the ASCII code (++ n) (begin (setq res (string res n c)) (setq n 1 c (char x))))) (setq res (string res n c)) res))
- Print out the first n+1 numbers, starting from 1
(define (go n) (let (s "1") (println s) (dotimes (x n) (setq s (next-number s)) (println s))))
(go 10) </lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221
Nim
<lang nim>iterator lookAndSay(n: int): string =
var current = "1"
yield current
for round in 2..n:
var ch = current[0]
var count = 1
var next = ""
for i in 1..current.high:
if current[i] == ch:
inc count
else:
next.add $count & ch
ch = current[i]
count = 1
current = next & $count & ch
yield current
for s in lookAndSay(12):
echo s</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
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);
{$IFDEF FPC}
{$MODE DELPHI}
{$ENDIF}
uses
SysUtils;
function LookAndSay(s: string): string; var
item: char; index: integer; count: integer;
begin
Result := ;
item := s[1];
count := 1;
for index := 2 to length(s) do
if item = s[index] then
inc(count)
else
begin
Result := Result + intTostr(count) + item;
item := s[index];
count := 1;
end;
Result := Result + intTostr(count) + item;
end;
var
number: string;
begin
writeln('Press RETURN to continue and ^C to stop.');
number := '1';
while not eof(input) do
begin
write(number);
readln;
number := LookAndSay(number);
end;
end.</lang>
- Output:
% ./LookAndSay Press RETURN to continue and ^C to stop. 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211^C
Even faster imperative Version Improvement: setlength of result,no inttoStr and using pChar But the Code Alignment is very important.
<lang pascal> program LookAndSayDemo(input, output); {$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)).$2 }</lang>
Phix
function lookandsay(string s) string res = "" integer p = s[1], c = 1 for i=2 to length(s) do if p=s[i] then c += 1 else res &= sprintf("%d%s",{c,p}) p = s[i] c = 1 end if end for res &= sprintf("%d%s",{c,p}) return res end function string s = "1" ?s for i=1 to 10 do s = lookandsay(s) ?s end for
- Output:
"1" "11" "21" "1211" "111221" "312211" "13112221" "1113213211" "31131211131221" "13211311123113112211" "11131221133112132113212221"
PHP
<lang php><?php
function lookAndSay($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>
PL/M
<lang plm>100H: BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
COPY$STRING: PROCEDURE (SRC, DEST);
DECLARE (SRC, DEST, I) ADDRESS;
DECLARE (S BASED SRC, D BASED DEST) BYTE;
I = 0;
DO WHILE S(I) <> '$';
D(I) = S(I);
I = I + 1;
END;
D(I) = '$';
END COPY$STRING;
COUNT: PROCEDURE (POS) BYTE;
DECLARE POS ADDRESS, (I, P BASED POS) BYTE;
I = 1;
DO WHILE P(I) = P;
I = I + 1;
END;
RETURN I;
END COUNT;
LOOK$SAY: PROCEDURE (OLD, NEW);
DECLARE (OLD, NEW) ADDRESS;
DECLARE (O BASED OLD, N BASED NEW, C) BYTE;
DO WHILE O <> '$';
C = COUNT(OLD);
N = O;
N(1) = C + '0';
NEW = NEW + 2;
OLD = OLD + C;
END;
N = '$';
END LOOK$SAY;
DECLARE STEPS LITERALLY '15'; DECLARE BUF$SIZE LITERALLY '128';
DECLARE BUFR1 (BUF$SIZE) BYTE INITIAL ('1$'); DECLARE BUFR2 (BUF$SIZE) BYTE; DECLARE I BYTE;
DO I=1 TO STEPS;
CALL PRINT(.BUFR1); CALL PRINT(.(13,10,'$')); CALL LOOK$SAY(.BUFR1, .BUFR2); CALL COPY$STRING(.BUFR2, .BUFR1);
END; CALL EXIT; EOF</lang>
- Output:
1 11 12 1121 122111 112213 12221131 1123123111 12213111213113 11221131132111311231 12221231123121133112213111 1123112131122131112112321222113113 1221311221113112221131132112213121112312311231 11221131122213311223123112312112221131112113213111213112213111 122212311223113212223111213112213111211223123113211231211131132111311222113113
PowerBASIC
This uses the RLEncode function from the PowerBASIC Run-length encoding entry.
<lang powerbasic>FUNCTION RLEncode (i AS STRING) AS STRING
DIM tmp1 AS STRING, tmp2 AS STRING, outP AS STRING DIM Loop0 AS LONG, count AS LONG
FOR Loop0 = 1 TO LEN(i)
tmp1 = MID$(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"
Quackery
<lang Quackery> [ stack ] is instances
[ 1 instances put
$ "" swap
behead swap space join
witheach
[ 2dup != iff
[ rot instances share
number$ join
rot join swap
1 instances replace ]
else
[ drop
1 instances tally ] ]
drop instances release ] is lookandsay ( $ --> $ )
$ "1"
15 times
[ dup echo$ cr
lookandsay ]
echo$ cr</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221 132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211
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(/(.)$0*/, { .chars ~ .[0] }, :g) ... *)[^15];</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
REXX
Programming note: this version works with any string (a null is assumed, which causes 1 to be used).
If a negative number is specified (the number of iterations
to be used for the calculations), only the length of
the number (or character string) is shown.
simple version
<lang rexx>/*REXX program displays the sequence (and/or lengths) for the look and say series.*/ parse arg N ! . /*obtain optional arguments from the CL*/ if N== | N=="," then N= 20 /*Not specified? Then use the default.*/ if !== | !=="," then != 1 /* " " " " " " */
do j=1 for abs(N) /*repeat a number of times to show NUMS*/
if j\==1 then != lookNsay(!) /*invoke function to calculate next #. */
if N<0 then say 'length['j"]:" length(!) /*Also, display the sequence's length.*/
else say '['j"]:" ! /*display the number to the terminal. */
end /*j*/
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ lookNsay: procedure; parse arg x,,$ /*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
(Shown at three-quarter size.)
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 != lookNsay(!) /*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 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ lookNsay: 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]
Sather
<lang sather>class MAIN is
look_and_say!: STR is
current ::= "1";
loop
yield current;
buf ::= #FSTR;
last ::= current[0];
count ::= 0;
loop
ch ::= current.elt!;
if ch /= last then
buf := buf + count + last;
last := ch; count := 1;
else
count := count + 1;
end;
end;
current := (buf + count + last).str;
end;
end;
main is
loop 12.times!;
#OUT+ look_and_say! + "\n";
end;
end;
end;</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211
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
Wren
<lang ecmascript>var lookAndSay = Fn.new { |s|
var res = ""
var digit = s[0]
var count = 1
for (i in 1...s.count) {
if (s[i] == digit) {
count = count + 1
} else {
res = res + "%(count)%(digit)"
digit = s[i]
count = 1
}
}
return res + "%(count)%(digit)"
}
var las = "1" for (i in 1..15) {
System.print(las) las = lookAndSay.call(las)
}</lang>
- Output:
1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221 3113112221232112111312211312113211 1321132132111213122112311311222113111221131221 11131221131211131231121113112221121321132132211331222113112211 311311222113111231131112132112311321322112111312211312111322212311322113212221
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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