computing square root by hand -- no sqrt (golf, beginner)

[boo_radley]

I ran across this Dr. Math faq and thought it would make a nifty intro to perl golf( in that the algorithm is pretty simple).

your script should take 2 inputs on the command line, the first being the number we're looking for the root of, and a first 'best guess' Again, don't use sqrt in your code. Stop guessing when you have a reasonable idea of what the answer is, and print it out.

Your code isn't responsible for error checking or verifying that the output's correct (verify it with a calculator if you like).
here's an outline of the algorithm described in the first section of the above url:

First, start by guessing a square root value. It helps if
your guess is a good one but it will work even if it is a
terrible guess. We will guess that 2 is the square root of 12.
In step two, we divide 12 by our guess of 2 and we get 6.
In step three, we average 6 and 2: (6+2)/2 = 4
Now we repeat step two with the new guess of 4. So 12/4 = 3
Now average 4 and 3: (4+3)/2 = 3.5
Repeat step two: 12/3.5 = 3.43
Average: (3.5 + 3.43)/2 = 3.465

some sample input/output :

C:\>sqrt 701 25
26.4764404223228

C:\>sqrt 12 2
3.46428571428571

C:\>sqrt 120 10
10.9544511505092

All of these have have a precision of 3 digits.

my answer, at 66 characters, below :

($n,$_)=@ARGV;while(1){$_=($_+$n/$_)/2;($l^$_)||print&&exit;$l=$_}

[japhy]

Golfing to 59:

($n,$_)=@ARGV;{$l^($_=($_+$n/$_)/2)||exit print;$l=$_;redo}

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area)
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

[jynx]

a reinterpretation of japhy's solution gives 54:

$_=pop;$l^$_?$l=$_:die"$_\n"while$_=($_+$ARGV[0]/$_)/2

jynx

update: this works at 53:

$_=pop;$l^$_?$l=$_:die"$_\n"while($_+=$ARGV[0]/$_)/=2

update2: using the comma instead of quotes for die it's 52:

$_=pop;$l^$_?$l=$_:die$_,$/while($_+=$ARGV[0]/$_)/=2

[Matts]

How's 18?

print $ARGV[0]**.5

;-)

[Juerd]

print $ARGV[0]**.5

This is the best answer I've seen in this thread :) It's clear, short and very precise...
How about some golfing while you're at it?

die+shift()**.5,$/

I know that this is 18 strokes like yours, but this one prints a newline ;)

- Yes, I reinvent wheels.
- Spam: Visit eurotraQ.

[jynx]

Since you're only using one argument you can call it with one argument and shorten it to 16:

die pop()**.5,$/

but that's bending the rules...
jynx

[MeowChow]

How's 17? The space is unnecessary.

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

[jynx]

While playing around, i thought i would (jokingly) try to golf Matt's solution and found something interesting. Using subtraction instead of bit-xor is more accurate. So, cleaning up my current mess of solutions and using subtraction for accuracy, it's now 49 chars (52 to print a carriage return):

#23456789_123456789_123456789_123456789_123456789_123456789_
$_=pop;$l=$_,($_+=$ARGV[0]/$_)/=2while$l-$_;print

jynx

update:an approximation method:

# 44
#23456789_123456789_123456789_123456789_123456789_
$a=pop;($a+=$ARGV[0]/$a)/=2for-9..99;print$a

[particle]

mine at 57:

#0        1         2         3         4         5         6
#123456789012345678901234567890123456789012345678901234567890
 $_=pop,$n=pop;{$l^($_=($_+$n/$_)/2)||die$_,$/;$l=$_;redo}

~Particle *accelerates*

[MeowChow]

38 chars, note that the second argument is entirely superfluous:

($_+=$ARGV[0]/$_)/=2for($_=1)x99;print

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

[cLive ;-)]

What short memories you all have!

cLive ;-)

--
seek(JOB,$$LA,0);

[pepik_knize]

44 using both arguments, but no \n. Add 3 for -l to get it.


0        1         2         3         4
12345678901234567890123456789012345678901234
$b=pop;$b=($b+$ARGV[0]/$b)/2for 0..9;print$b


Pepik