## Look And Say

### March 15, 2011

The “Look and Say” sequence, Sloane number A005150, begins 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, …. Each term is constructed from its predecessor by stating the frequency and number of each group of like digits. For instance, the term after 1211 is “one 1, one 2, and two 1s”, or 111221. This sequence was studied by the British mathematician John Horton Conway, and has some fascinating properties; see MathWorld or follow the references at Sloane’s if you are interested.

Your task is to write a program to compute the look and say sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

### 22 Responses to “Look And Say”

1. Martin Oldfield said

Here’s a quick solution in ghci:

Prelude> import Data.List
Prelude Data.List> let n = concatMap (\x -> (show \$ length x) ++ [head x]) . group
Prelude Data.List> take 10 \$ iterate n “1”
[“1″,”11″,”21″,”1211″,”111221″,”312211″,”13112221″,”1113213211″,”31131211131221″,”13211311123113112211”]

```import Data.List

lookAndSay :: Integer -> [Integer]
lookAndSay = iterate (read . f . group . show) where
f = (>>= \x -> show (length x) ++ take 1 x)
```
3. […] today’s Programming Praxis exercise, our task is to generate the Look and Say sequence (1, 11, 21, 1211, […]

4. Dave Webb said

A Python solution:

```import itertools

def lookandsay(number):
return "".join(str(sum(1 for x in g)) + k for k,g in itertools.groupby(number))

seq = ['1']
for _ in xrange(10):
seq.append(lookandsay(seq[-1]))
print seq
```
5. Graham said

Using the `groupby()` function from Python’s itertools module:

```from itertools import groupby

def next_iter(s):
return ''.join([str(len(list(g))) + k for k, g in groupby(s)])

if __name__ == "__main__":
s = '1'
for _ in xrange(10):
print s
s = next_iter(s)
```
6. Graham said

Looks like Dave beat me to it :-) I should have refreshed the page before posting.

7. Leonel said

Clojure solution:

(defn lookandsay [s]
(mapcat (juxt count first)
(partition-by identity s)))
(take 10 (iterate lookandsay ))

8. Jussi Piitulainen said

A listless solution, integer to integer, in Scheme.

```(define (say-next n)
(let loop ((m (quotient n 10))
(digit (remainder n 10))
(count 1)
(result 0)
(length 1))
(if (= m digit 0)
result
(let ((m1 (quotient m 10))
(digit1 (remainder m 10)))
(if (= digit1 digit)
(loop m1 digit (+ count 1) result length)
(loop m1 digit1 1
(+ (* count 10 length)
(* digit length)
result)
(* length 100)))))))
```

Output using iterate from that Standard Prelude.

```guile> (iterate 10 say-next 1)
(1 11 21 1211 111221 312211 13112221 1113213211
31131211131221 13211311123113112211)
```
9. Mike said

Because Sloane’s is a catalogue of integer sequences, I wrote a generator that return successive integers of the sequence.

My first version is basically the same as Webb’s and Graham’s above, so I also did a second version using the regular expression library.

I was somewhat suprised that the second version appears to be slightly faster, based on using the timeit module.

```# version 1: Uses itertools.groupby().
from itertools import groupby

def look_and_say1(n):
'''generate 'look and say' sequence.

>>> list(islice(look_and_say1(13),5))
[13, 1113, 3113, 132113, 1113122113]
'''

s = list(str(n))

while True:
yield int(''.join(s))
s = [d for k,g in groupby(s) for d in (str(len(list(g))), k)]

# version 2: Uses re.sub().
import re

def look_and_say2(n):
'''generate 'look and say' sequence.

>>> list(islice(look_and_say2(13),5))
[13, 1113, 3113, 132113, 1113122113]
'''

def say(mo):
g = mo.group()
return '{}{}'.format(len(g),g)

s = str(n)
pat = re.compile(r'([1-9])\1*')

while True:
yield int(s)
s = pat.sub(say ,s)

#testing
from itertools import islice, izip

def test(limit=10):
'''compares output of both generators.

>>> test()
Ok
'''
for n in range(1,9):
for s,t in islice(izip(look_and_say1(n),look_and_say2(n)), limit):
if s != t:
print s, '!=', t
return
print 'Ok'

if __name__ == "__main__":
import doctest
doctest.testmod()
```
10. Rainer said

My try in REXX

```txt = '1'
MAXLEN = 20

say lookandsay(txt)
exit

lookandsay: procedure expose MAXLEN
parse arg seq
outseq = seq
do while length(seq) < MAXLEN
new = ''
prv = ''
cnt = 0
do x = 1 to length(seq)
n = substr(seq, x, 1)
if n \= prv then do
if cnt > 0 then new = new||cnt||prv
prv = n
cnt = 0
end
if length(new) > MAXLEN then leave
cnt = cnt + 1
end
if cnt > 0 then new = new||cnt||prv
outseq = outseq new
seq = new
end
return outseq
```
11. […] autre petit exercice de Programming Praxis, programmer la suite Look and Say. Ce n’est pas très difficile : import Data.List import […]

12. Benoît Huron said

```import Data.List
import Data.Digits

lookAndSay :: [Integer]
lookAndSay = map (unDigits 10) \$ iterate next 
where next = concatMap count . group
count lst@(x:xs) = [fromIntegral (length lst), x]
```
13. Aaron said

in scala:

def lookAndSay(previous:List[Int]) : Stream[List[Int]] = {
def next(num : List[Int]) : List[Int] = num match {
case Nil => Nil
case head :: tail => {
val size = (num takeWhile{_ == head}).size
}
}
val x = next(previous)
x #:: lookAndSay(x)
}

lookAndSay(1::Nil) take 10 foreach (println _)

14. Vitaliy Isikov said

Here’s a JS solution. I’ve only been programming for about a year, mostly JS and PHP. Going to attempt a PHP version next, then subsequently, ajax-ify it.

Look and Say Sequence

var start = false;
var result;
function lookSay(int, repeat) {
for (t=1; t<=repeat; t++) {
if (!start) {
start = int;
result = start+"”;
}
newNumber = “”;
digit = start.charAt(0);
count = 1;
for (var i = 1; i < start.length; i++) {
if (digit == start.charAt(i)) {
count++;
}
else {
newNumber = newNumber + count + digit;
digit = start.charAt(i);
count = 1;
}
}
start = newNumber + count + digit;
if (t == repeat) {
result+=start;
document.getElementById('result').innerHTML=result;
start = false;
result = '';
}
else {
result +=start+"”;
}
}
}

function returnResult() {
s = document.getElementById(‘start’).value;
r = document.getElementById(‘repeat’).value;
lookSay(s, r);
}

Start Integer:
Repeat Amount:
Submit

15. Toby said

A Scheme solution using lists is:

```; Copyright (C) 2011 Toby Thain, toby@telegraphics.com.au

(define (look-and-say terms)
(define (next-value lst)
(let count-digit ((this-digit (car lst))
(count 1)
(rest (cdr lst)))
(cond
((null? rest)
(list count this-digit))
((eq? this-digit (car rest))
(count-digit this-digit (+ count 1) (cdr rest)))
(else
(append (list count this-digit)
(count-digit (car rest) 1 (cdr rest)))))))
(let step ((value '(1))
(n terms))
(if (zero? n)
'()
(cons value (step (next-value value) (- n 1))))))
```
16. Jean-Pascal Billaud said

A ruby version:

sloane = 
puts sloane.inspect

8.times.each do |iteration|
new_sloane = []
rep_digit = 1

last_sum = sloane.inject(0) do |sum, digit|
next sum + 1 if digit == rep_digit

new_sloane << rep_digit
new_sloane << sum
rep_digit = digit
1
end

new_sloane << rep_digit
new_sloane << last_sum

puts sloane.replace(new_sloane).reverse.inspect
end
[/sourcecode]

17. yaq said

```(define (build-next items)
(define (build-next-iter items number freq next)
(cond ((and (pair? items) (equal? (car items) number))
(build-next-iter (cdr items) number (+ freq 1) next))
((and (pair? items) (not (equal? (car items) number)))
(build-next-iter (cdr items) (car items) 1 (cons next (cons freq number))))
(else
(cons next (cons freq number)))))
(flatten (build-next-iter items (car items) 0 '())))

(define (look-and-say-from s)
(cons-stream s (look-and-say-from (build-next s))))

(define look-and-say-stream (look-and-say-from (cons 1'())))

18. Khanh Nguyen said

Mine in F#

```let look_and_say (s:string) =
let next  = s.ToCharArray()
|> List.ofArray
|> List.fold (fun acc x ->
match acc with
| [] -> [(x, 1)]
| (a,b)::t ->
match compare x a with
| 0 -> (a,b+1)::t
| _ -> (x,1)::acc
) []
|> List.rev
|> List.map (fun (a,b) -> (string b) + (string a))
|> Seq.ofList
|> String.concat ""
Some(s, next)

let Sloan = Seq.unfold look_and_say "1"

```
19. John said

For the curious, Factor looks like this:

: look-and-say ( n — n’ )
number>string [ = ] monotonic-split [
[ length ] [ first ] bi "%d%c" sprintf
] map concat string>number ;

20. rohit said

is anybody have solution in c plz upload here

21. Toby said

Rohit, check my website link here for a version in C. Please note that it is NOT written for clarity. But it is in C, and it works. Increase the ‘M’ constant if you want more numbers in the series; for example, 1000000 gives you the first 49 terms.