Reimplementing square root [duplicate]

Question

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.

Answer 1

Python - 11 chars

Technically not a library function :)

input()**.5

As a function it's 16 chars

f=lambda x:x**.5

Answer 2

Python (13 chars)

f=.5.__rpow__

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

Answer 3

Python - 41 chars

Takes a while to run for large numbers :)

n=input()
i=0
while i*i<n:i+=1e-9
print i

Answer 4

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);

Answer 5

J, 6, 5

Using power:

^&0.5

Or, perhaps more mnemonically:

^&1r2

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.

Answer 6

Haskell, 9 characters

s=(**0.5)

Similar to @gnibbler's Python solution.

Answer 7

LISP (66)

Using Babylonian method.

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

Answer 8

Naive solution, accepts only positive integer inputs.

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

Haskell, 39 characters.

Answer 9

Python - 65

A simple solution using Newton's method.

def s(x):
 t=1.0
 while 1e-9<abs(x-t*t):t-=(t*t-x)/2/t
 return t

Answer 10

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.

Answer 11

return 1.f/InvSqrt(x);

InvSqrt of course courtesy of Quake.

What, you mean you wanted an accurate result?

Answer 12

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

Answer 13

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!

---

Usage:

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

Answer 14

C++ (35)

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

I promise I didn't use sqrt()!

Answer 15

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 :(

Answer 16

GolfScript - 21

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

{1{.2$\/+2/}99*\;}:s;

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.

Answer 17

k (17 chars)

Iterative (implementation of Babylonian method):

{{.5*y+x%y}[x]/x}

Iterates until two successive values are equal

Example:

sqrt[1234] = {{0.5*y+x%y}[x]/x}1234
1b

Also the mandatory xexp[;0.5] for 10.

Answer 18

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}

Answer 19

Mathematica 27

f@x_:=Nest[(#+x/#)/2&,1.,9]

Answer 20

Golf-Basic 84, 9 characters

i`Ad`A^.5

As a function, 15 characters:

i`A:Return A^.5

Answer 21

JavaScript, 36 (or 14 without the function)

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

Answer 22

AWK, 8

1,$0^=.5

Following solution can handle all values except 0

AWK, 6

$0^=.5

Answer 23

Befunge 98 - 18

&00pv  
0*::<+1vj!`g0

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):

00p::*00g`!jv1+

Answer 24

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

>+4>15+4<[>>+<+<[->[->->>+<<]>[-<+>>]<+<<]>[-]>[-]3<[->+>>+3<]>[-<+>]4>[-<+3<+4>]<<[->-[>+>>]>[+[-<+>]>+>>]5<]3>[-3<+3>]<[->>+>+3<]<[->+4>+5<]>+[>>[-<]<[>]<-]3>+<[[-]>[-]>[-4<+4>]5<+<+4>]>[->[-]<[-3<+3>]]3<[->+<]<<[->>+4<->>]<+<[-[+3>[-]>>[-]4<-]>[-5>+4<]<]>[->]<<]<[-]5>[-4<+<+5>]>>[-6<+6>]<[-3<+3>]<<[-<<+>>]

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
{sqrt
  check for wrong usage !diff:diff
  :in:guess:temp:cur:div:mod:res:indicator
  @in
  $guess+     ;initial guess is 16, but
  $mod 15+    ; its in mod and incremented (rounded up)
  $guess(     ; worst case is 1 with 5 iterations
      $cur+$temp+
      $guess(-$temp[->-$mod+$cur]>[@cur-$temp+$div]$cur+)
      $temp(-)$cur(-)
      $in(-$guess+$cur+)
      $guess(-$in+)
      $mod(-$div+$guess+)
      $cur &roundivmod ; remainder wil now be in mod
      $mod(-$res+)
      $cur(-$mod+$guess-)
      $temp+
      $guess[-[+$div(-)$res(-)$temp-]>[-@temp$indicator+$cur]$temp]>[-@temp>]
  )
  $in(-)
  $mod(-$guess+$in+)
  $indicator(-$guess+)
  $res(-$cur+)
  $div(-$temp+)
  $in
  !indicator!res!mod!div!cur!temp!guess!in
}

;; 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
{roundivmod
    &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
{divmod[->-[>+>>]>[+[-<+>]>+>>]<<<<<]}

Answer 25

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)

invocation:

scala> q(12345)
res36: D = 111.10805572305925

Answer 26

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)

Output:

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.

Answer 27

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 

Commented:

/s{
    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
    }loop   
    pop exch pop
}def

Answer 28

bash 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 $x1 ] ;do
    x0=$x1
    printf -v x1 "%u" $(( (${x0}000 + ${1}00000000000/${x0} )/2 ))
    printf -v x1 "%.0f" ${x1:0:${#x1}-3}.${x1:${#x1}-3}
    done
    x1=0000$x1
    printf "%.3f\n" ${x1:0:${#x1}-4}.${x1:${#x1}-4}
}

sqrt 2000000
1414.214

sqrt 1414214
1189.207

echo $((1189207000/1414214))
840

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

The same, but working under zsh, dash, ksh, bash and ash (busybox):

sqrt() {
    _val=$1
    set -- 6 2
    x1=$(eval echo \$$((1+(_val&1))))00000 x0=1
    while [ $x0 -ne $x1 ] ;do
    x0=$x1
    x1=000$(( (${x0}000 + ${_val}000000000/${x0} )/2 ))
    x1=$(printf "%.0f" $(echo $x1|sed 's/\(...\)$/.\1/'))
      done
    printf "%.3f\n" $(echo $x1|sed 's/\(...\)$/.\1/')
}

Answer 29

dc, 46 chars

Babylonian method - iterative with given precision:

Fk?sp?sn1[ddstlnr/+2/pdlt-[0r-]sbd0>blp<a]dsax

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.

Explanation:

• 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.

Answer 30

CoffeeScript (12)

s=(n)->n**.5

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