,”13211311123113112211”]
My Haskell solution (see for a version with comments):
[…] today’s Programming Praxis exercise, our task is to generate the Look and Say sequence (1, 11, 21, 1211, […]
A Python solution:
Using the
groupby()function from Python’s itertools module:
Looks like Dave beat me to it :-) I should have refreshed the page before posting.
Clojure solution:
(defn lookandsay [s]
(mapcat (juxt count first)
(partition-by identity s)))
(take 10 (iterate lookandsay [1]))
A listless solution, integer to integer, in Scheme.
Output using iterate from that Standard Prelude.
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.
My try in REXX
[…] autre petit exercice de Programming Praxis, programmer la suite Look and Say. Ce n’est pas très difficile : import Data.List import […]
My Haskell solution
in scala:
def lookAndSay(previous:List[Int]) : Stream[List[Int]] = {
def next(num : List[Int]) : List[Int] = num match {
case Nil => Nil
case head :: Nil => 1 :: head :: Nil
case head :: tail => {
val size = (num takeWhile{_ == head}).size
List(size, head) ::: next(num.drop(size))
}
}
val x = next(previous)
x #:: lookAndSay(x)
}
lookAndSay(1::Nil) take 10 foreach (println _)
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
A Scheme solution using lists is:
A ruby version:
sloane = [1]
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]
Mine in F#
Here is a solution in Factor:
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 ;
is anybody have solution in c plz upload here
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.