Define a function, s, which takes a number and returns the square root.

No use of library functions, such as Java's Math.sqrt() or PHP's built in sqrt(), allowed.

  • 3
    \$\begingroup\$ I think all solutions will be based on Newton approximation such as codemaestro.com/reviews/9 \$\endgroup\$ – Alexandru Jan 28 '11 at 0:56
  • 6
    \$\begingroup\$ How do you feel about exp(0.5*log(x))? \$\endgroup\$ – dmckee --- ex-moderator kitten Jan 28 '11 at 2:07
  • 4
    \$\begingroup\$ The problem with this puzzle is that it lacks specifications for input range and tolerated error, it's kinda boring if "keep on adding a small number until result is reached" method is allowed, it's the shortest in any language. \$\endgroup\$ – aaaaaaaaaaaa Jan 28 '11 at 2:11
  • 2
    \$\begingroup\$ @jtjacques: Er...yes it does. Try a couple of cases. Make sure that you use the same base for exponentiation and logarithm. \$\endgroup\$ – dmckee --- ex-moderator kitten Jan 28 '11 at 2:23
  • 2
    \$\begingroup\$ @jtjacques: Base agreement is key, many languages use log for the base 10 logarithm, so you might try exp(0.5*ln(x)) to get the natural log or pow10(0.5*log(x)) or similar. \$\endgroup\$ – dmckee --- ex-moderator kitten Jan 28 '11 at 2:38

38 Answers 38


Python - 11 chars

Technically not a library function :)


As a function it's 16 chars

f=lambda x:x**.5
  • \$\begingroup\$ Smart, I like it. \$\endgroup\$ – jtjacques Jan 28 '11 at 1:25
  • 2
    \$\begingroup\$ It's questionable whether this falls under a built-in function (it calls __pow__). But then even * is a built-in. \$\endgroup\$ – moinudin Jan 29 '11 at 11:37
  • \$\begingroup\$ Would have preferred it as a function, but I'll give you the accept for the ingenuity. \$\endgroup\$ – jtjacques Jan 31 '11 at 15:45
  • 19
    \$\begingroup\$ @jtjacques What ingenuity? IMHO, this is the one the most obvious ways around the rules. \$\endgroup\$ – Peter Olson May 16 '11 at 3:48
  • \$\begingroup\$ I think the function has to be called 's', not 'f'. \$\endgroup\$ – DoctorHeckle Jul 8 '15 at 17:56

Python (13 chars)


This is equivalent to f=lambda x:x**.5, but 3 bytes shorter.


Python - 41 chars

Takes a while to run for large numbers :)

while i*i<n:i+=1e-9
print i
  • 1
    \$\begingroup\$ For me, it takes a while even for 2. \$\endgroup\$ – nyuszika7h Apr 30 '14 at 13:44

Java (163)

Implementing a double precision square root calculator by making use of the fast invert square root stuff from quake and a new constant for the 64bit floats. In java. Yay for verbosity.

public double i(double a){double b=a/2;long c=0x5fe6ec85e7de30daL-(Double.doubleToRawLongBits(a)>>1);a=Double.longBitsToDouble(c);return a*(1.5-b*a*a);};s=1/i(x);

J, 6, 5

Using power:


Or, perhaps more mnemonically:


Using the slightly less "cheaty" method, exp(log(x)/2).


Except since exp is the inverse of log, we simply "halve (-:) under (&.) log (^.)"

Not that normally a J programmer would not name a method this short; he'd simply embed in in a larger program.


Haskell, 9 characters


Similar to @gnibbler's Python solution.


LISP (66)

Using Babylonian method.

(defun s(A)
   (do((x 1(*(+ x(/ A x)).5))(n 100(1- n)))((zerop n)x)))

Naive solution, accepts only positive integer inputs.

s n=foldr (\x a->x*x==n||a) False [1..]

Haskell, 39 characters.

  • 6
    \$\begingroup\$ remove spaces? s n=foldr(\x a->x*x==n||a)False[1..] \$\endgroup\$ – Ming-Tang Jan 28 '11 at 3:51

Python - 65

A simple solution using Newton's method.

def s(x):
 while 1e-9<abs(x-t*t):t-=(t*t-x)/2/t
 return t

C with some bad form and deprecated features:

s(int n,int g){return g*g-n?s(n,random()%n):g;}

If your compiler is enough of a liberal hippie, it should be possible to call s(n) and (eventually) receive the desired value.

  • \$\begingroup\$ You can save 8 characters, and allow calling s(n) (at least in GCC), by changing s(int n,int g) to s(n,g). \$\endgroup\$ – Joey Adams Mar 18 '11 at 4:20
  • \$\begingroup\$ Save 2 more chars by changing random() to rand(). \$\endgroup\$ – Paul R Jun 14 '11 at 14:36
return 1.f/InvSqrt(x);

InvSqrt of course courtesy of Quake.

What, you mean you wanted an accurate result?

  • \$\begingroup\$ Well to the limit of the return type, and also one which does not use a library. \$\endgroup\$ – jtjacques Jan 28 '11 at 1:08
def sqrt_newton(x):
 f,g,w,d=lambda a:a*a-x,lambda a:2.0*a,x/2.0,1e-4
 while abs(f(w))>d:w-=(f(w)/2.0/w)
 return w

reduced version of https://gist.github.com/713104

  • 3
    \$\begingroup\$ And the size is ... \$\endgroup\$ – user unknown Nov 8 '11 at 5:35

C++ (61)

Uses Heron's Method:

f64 s(f64*x,u64 n=9){*x=(x[1]/*x+*x)/2;return n?s(x,--n):*x;}

Recursion, pointers, optional arguments, and ternary operators FTW!


f64 x[2] = {12.0, 12.0};
std::cout << s(x);

C++ (35)

f64 s(f64 x){return exp(log(x)/2);}

I promise I didn't use sqrt()!


ActionScript3 (53)

Using Newton's method. (Hooray 4 IEEE 754)

function s(b,d=2){return b==d*d?d:s(b,d-(d*d-b)/2/d)}

I know I'm TOO late but I just wanted to write something I did :(


GolfScript - 21

It uses the Babylonian method and works only with integers. According to my tests, it's good for up to 56 digits.


Usage: 1000 s -> 31

I think this is the shortest solution so far that doesn't call some kind of power function or external library.


k (17 chars)

Iterative (implementation of Babylonian method):


Iterates until two successive values are equal


sqrt[1234] = {{0.5*y+x%y}[x]/x}1234

Also the mandatory xexp[;0.5] for 10.


JavaScript, 121

No, it doesn't even come close, but it is more optimal than other solutions and doesn't use Math.pow with fractions.

s=n=>{for(var c=Math.pow(10,(''+Math.floor(n)).length),v=0,l;(l=v*v<n),c>1e-10;l!=v*v<n?c/=10:0)v*v<n?v+=c:v-=c;return v}
  • \$\begingroup\$ You're missing the function name there. Should be something like function s(n). \$\endgroup\$ – nyuszika7h Apr 30 '14 at 10:41
  • \$\begingroup\$ @nyuszika7h: It’s a function literal. \$\endgroup\$ – Ry- Apr 30 '14 at 14:28
  • \$\begingroup\$ But you would need to assign that to a variable (which is the same "penalty" as using my previous suggestion). I don't see Python solutions like lambda x: ... either, I see solutions like f=lambda x: .... \$\endgroup\$ – nyuszika7h Apr 30 '14 at 14:30
  • \$\begingroup\$ @nyuszika7h: Their loss. This code is a function. \$\endgroup\$ – Ry- Apr 30 '14 at 14:30
  • 1
    \$\begingroup\$ Also, sorry if it came off as being rude, I didn't mean to be. \$\endgroup\$ – nyuszika7h Apr 30 '14 at 14:46

Mathematica 27


Golf-Basic 84, 9 characters


As a function, 15 characters:

i`A:Return A^.5

JavaScript, 36 (or 14 without the function)

function s(n){return Math.pow(n,.5)}

AWK, 8


Following solution can handle all values except 0

AWK, 6


Befunge 98 - 18


This program takes an input number from the user, and ends by pushing the integer square root on the stack (note that you said square root, but didn't specify whether floating point was necessary). Here is a function (well, closest thing to one) (requiring free access to cell 00) (15 chars):


In these I expect the byte to be squared is in the current cell. It needs 8 cells to the right empty and it will have answers in the 4 cells starting from where the number was. These are:

  1. integer result
  2. a alternative result (would be the same or one higher if those multiplied is closer to the argument)
  3. remainder
  4. flag for negative remainder

Extended BrainFuck: 310


It turns into the following:

BrainFuck: 390 (the same as above run through the compiler)


It's Newtons method starting at guess 16 and goes downwards. It stops when the last iteration didn't make a different integer result. This is actualy from the sqrt macro of EBF since I use it to implement print-string operator |. Here's the part from EBF source ungolfed:

; sqrt ^0  uses  9 cells that need to be empty
; will be fuzzy after calling because of divmod
; returns ^0 result
;         ^1 same as ^0 or it increased by 1. typically will ^2*^3 be closerto the requested argument than ^2*^2
;         ^2 remainder
;         ^3 indicates if remainder is negative
  check for wrong usage !diff:diff
  $guess+     ;initial guess is 16, but
  $mod 15+    ; its in mod and incremented (rounded up)
  $guess(     ; worst case is 1 with 5 iterations
      $cur &roundivmod ; remainder wil now be in mod

;; helper macros
; roundivmod uses divmod and puts the rounded result in ^0 and indication of rounded in ^1
; and a remainder (which ^1 is an indication is either reduction or inrement) in ^2
    &divmod @cur
    ; *0|n-rem|rem|res|
    <(->>+>+) make double copy of remainder
    ; res|n-rem|0|0|rem|rem
    ; res|0|n-rem|0|rem|rem|n-rem
  end divide

; this does the divmod. compiler is fuzzy after so caller must fix position to calling

Scala 78 chars:

type D=Double
def q(x:D,g:D=9):D=if(math.abs(g*g-x)<.001)g else q(x,(g+x/g)/2)


scala> q(12345)
res36: D = 111.10805572305925

Python, 44

def s(x):t=1.;exec"t=(t+x/t)/2;"*99;return t

Tested with:

for x in [0, (3+5**0.5)/2, 1000, 10**5, 10**11]:
    print x, '->', s(x)


0 -> 1.57772181044e-30
2.61803398875 -> 1.61803398875
1000 -> 31.6227766017
100000 -> 316.227766017
100000000000 -> 316227.766017

Python, 50 49

In case you like recursion:

s=lambda x,t=1.,r=99:r and s(x,t/2+x/t/2,r-1)or t

Same testing code, same output.


Postscript 94 89

Babylonian method. Iterates until successive values are equal.

s{dup 2 div{2 copy div 1 index add .5 mul
2 copy eq{exit}if exch pop}loop pop exch pop}def 


    dup 2 div  % S x
    {  % S x
        2 copy div  % S x S/x 
        1 index add .5 mul  % S x (S/x+x)/2
        2 copy eq { exit } if  % S x (S/x+x)/2
        exch pop  % S (S/x+x)/2:->x
    pop exch pop

228 ( a lot but pure bash only! ;)

For the count:

x=(600000 200000);z=${x[$(($1&1))]} y=1;while [ $y -ne $z ] ;do y=$z;printf -v z "%u" $(((${y}000+${1}00000000000/${y})/2));printf -v z "%.0f" ${z:0:${#z}-3}.${z:${#z}-3};done;z=0000$z;printf "%.3f\n" ${z:0:${#z}-4}.${z:${#z}-4}

As a function:

sqrt() {
    local -a xx=(600000 200000)
    local x1=${xx[$(($1&1))]} x0=1
    while [ $x0 -ne $x1 ] ;do
    printf -v x1 "%u" $(( (${x0}000 + ${1}00000000000/${x0} )/2 ))
    printf -v x1 "%.0f" ${x1:0:${#x1}-3}.${x1:${#x1}-3}
    printf "%.3f\n" ${x1:0:${#x1}-4}.${x1:${#x1}-4}

sqrt 2000000

sqrt 1414214

echo $((1189207000/1414214))

(Nota: 840 x 1189 is A0 paper size in milimeters ;-)

The same, but working under , , , and (busybox):

sqrt() {
    set -- 6 2
    x1=$(eval echo \$$((1+(_val&1))))00000 x0=1
    while [ $x0 -ne $x1 ] ;do
    x1=000$(( (${x0}000 + ${_val}000000000/${x0} )/2 ))
    x1=$(printf "%.0f" $(echo $x1|sed 's/\(...\)$/.\1/'))
    printf "%.3f\n" $(echo $x1|sed 's/\(...\)$/.\1/')
  • \$\begingroup\$ The portable version seems to work fine in zsh too. \$\endgroup\$ – manatwork Dec 16 '13 at 8:31
  • \$\begingroup\$ @manatwork Thanks I'v forgot them! \$\endgroup\$ – F. Hauri Dec 16 '13 at 8:36

dc, 46 chars

Babylonian method - iterative with given precision:


Takes precision and N from stdin. Example invocation (first number is the precision, second number is N:

{ echo 0.0000001 ; echo 489 ; } | dc -e'Fk?sp?sn1[ddstlnr/+2/pdlt-[0r-]sbd0>blp<a]dsax'

It prints all the iterations, just for fun. Printing only last one is trivial change - just move the p behind /+2/ at the end of the program.


  • Fk sets precision to 15 (see the hexa trick here)
  • ?sp and ?sn asks for input and stores it to the registers
  • 1 is the original seed
  • the iteration is computed by ddstlnr/+2/. Then, dlt- computes difference from previous iteration. [0r-]sbd0>b performs absolute value of the difference, and if it is greater than the specified precision, loop continues - lp<a.

CoffeeScript (12)


Thanks to gnibbler's post for the .5 idea, I wouldn't have remembered that by myself.


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