Calculate the cube root of a number

Question

The goal of this code golf is to create a program or function that calculates and outputs the cube root of a number that's given as input.
The rules:

• No external resources
• No use of built-in cube root functions.
• No use of methods/operators that can raise a number to a power (that includes square root, 4th root, etc.).
• Your function/program must be able to accept floating-point numbers and negative numbers as input.
• If the cube root is a floating-point number, then round it to 4 numbers after the decimal point.
• This is a code golf, the shortest code in bytes wins.

Test cases:

27 --> 3
64 --> 4
1  --> 1
18.609625 --> 2.65
3652264 --> 154
0.001 --> 0.1
7  --> 1.9129

You can use all test cases above to test negative numbers (-27 --> -3, -64 --> -4 ...)

Answer 1

Haskell - 35

c n=(iterate(\x->(x+n/x/x)/2)n)!!99

Example runs:

c 27  =>  3.0
c 64  =>  4.0
c 1  =>  1.0
c 18.609625  =>  2.6500000000000004  # only first 4 digits are important, right?
c 3652264  =>  154.0
c 0.001  =>  0.1
c 7  =>  1.9129311827723892
c (-27)  =>  -3.0
c (-64)  =>  -4.0

Moreover, if you import Data.Complex, it even works on complex numbers, it returns one of the roots of the number (there are 3):

c (18:+26)  =>  3.0 :+ 1.0

The :+ operator should be read as 'plus i times'

Answer 2

SageMath, (69) 62 bytes

However, don't ever believe it will give you the result, it's very difficult to go randomly through all the numbers:

def r(x):
 y=0
 while y*y*y-x:y=RR.random_element()
 return "%.4f"%y

if you didn't insist on truncating:

def r(x):
 y=0
 while y*y*y-x:y=RR.random_element()
 return y

SageMath, 12 bytes, if exp is allowed

Works for all stuff: positive, negative, zero, complex, ...

exp(ln(x)/3)

Answer 3

J: 16 characters

Loose translation of the Haskell answer:

-:@((%*~)+])^:_~

Test cases:

   -:@((%*~)+])^:_~27
3
   -:@((%*~)+])^:_~64
4
   -:@((%*~)+])^:_~1
1
   -:@((%*~)+])^:_~18.609625
2.65
   -:@((%*~)+])^:_~3652264
154
   -:@((%*~)+])^:_~0.001
0.1
   -:@((%*~)+])^:_~7
1.91293

It works like this:

     (-:@((% *~) + ])^:_)~ 27
↔ 27 (-:@((% *~) + ])^:_) 27
↔ 27 (-:@((% *~) + ])^:_) 27 (-:@((% *~) + ])) 27
↔ 27 (-:@((% *~) + ])^:_) -: ((27 % 27 * 27) + 27)
↔ 27 (-:@((% *~) + ])^:_) 13.5185
↔ 27 (-:@((% *~) + ])^:_) 27 (-:@((% *~) + ])) 13.5185
↔ 27 (-:@((% *~) + ])^:_) -: ((27 % 13.5185 * 13.5185) + 13.5185)
↔ 27 (-:@((% *~) + ])^:_) 6.83313
...

In words:

half =. -:
of =. @
divideBy =. %
times =. *
add =. +
right =. ]
iterate =. ^:
infinite =. _
fixpoint =. iterate infinite
by_self =. ~

-:@((%*~)+])^:_~ ↔ half of ((divideBy times by_self) add right) fixpoint by_self

Not one of the best wordy translations, since there's a dyadic fork and a ~ right at the end.

Answer 4

Python - 62 bytes

x=v=input()
exec"x*=(2.*v+x*x*x)/(v+2*x*x*x or 1);"*99;print x

Evaluates to full floating point precision. The method used is Halley's method. As each iteration produces 3 times as many correct digits as the last, 99 iterations is a bit of overkill.

Input/output:

27 -> 3.0
64 -> 4.0
1 -> 1.0
18.609625 -> 2.65
3652264 -> 154.0
0.001 -> 0.1
7 -> 1.91293118277
0 -> 1.57772181044e-30
-2 -> -1.25992104989

Answer 5

Perl, 92 bytes

sub a{$x=1;while($d=($x-$_[0]/$x/$x)/3,abs$d>1e-9){$x-=$d}$_=sprintf'%.4f',$x;s/\.?0*$//;$_}

• The function a returns a string with the number without an unnecessary fraction part or insignificant zeroes at the right end.

Result:

              27 --> 3
             -27 --> -3
              64 --> 4
             -64 --> -4
               1 --> 1
              -1 --> -1
       18.609625 --> 2.65
      -18.609625 --> -2.65
         3652264 --> 154
        -3652264 --> -154
           0.001 --> 0.1
          -0.001 --> -0.1
               7 --> 1.9129
              -7 --> -1.9129
 0.0000000000002 --> 0.0001
-0.0000000000002 --> -0.0001
               0 --> 0
              -0 --> 0

Generated by

sub test{
    my $a = shift;
    printf "%16s --> %s\n", $a, a($a);
    printf "%16s --> %s\n", "-$a", a(-$a);
}
test 27;
test 64;
test 1;
test 18.609625;
test 3652264;
test 0.001;
test 7;
test "0.0000000000002";
test 0;

The calculation is based on Newton's method:

Answer 6

Javascript (55)

function f(n){for(i=x=99;i--;)x=(2*x+n/x/x)/3;return x}

BONUS, General formulation for all roots
function f(n,p){for(i=x=99;i--;)x=x-(x-n/Math.pow(x,p-1))/p;return x}

For cube root, just use f(n,3), square root f(n,2), etc... Example : f(1024,10) returns 2.

Explanation
Based on Newton method :

Find : f(x) = x^3 - n = 0, the solution is n = x^3
The derivation : f'(x) = 3*x^2

Iterate :
x(i+1) = x(i) - f(x(i))/f'(x(i)) = x(i) + (2/3)*x + (1/3)*n/x^2

Tests

[27,64,1,18.609625,3652264,0.001,7].forEach(function(n){console.log(n + ' (' + -n + ') => ' + f(n) + ' ('+ f(-n) +')')})

27 (-27) => 3 (-3)
64 (-64) => 4 (-4)
1 (-1) => 1 (-1)
18.609625 (-18.609625) => 2.65 (-2.65)
3652264 (-3652264) => 154 (-154)
0.001 (-0.001) => 0.09999999999999999 (-0.09999999999999999)
7 (-7) => 1.912931182772389 (-1.912931182772389) 

Answer 7

PHP - 81 bytes

Iterative solution:

$i=0;while(($y=abs($x=$argv[1]))-$i*$i*$i>1e-4)$i+=1e-5;@print $y/$x*round($i,4);

Answer 8

APL - 31

(×X)×+/1,(×\99⍴(⍟|X←⎕)÷3)÷×\⍳99

Uses the fact that cbrt(x)=e^(ln(x)/3), but instead of doing naive ⋆ exponentiation, it computes e^x using Taylor/Maclaurin series.

Sample runs:

⎕: 27
3
⎕: 64
4
⎕: 1
1
⎕: 18.609625
2.65
⎕: 3652264
154
⎕: 0.001
0.1
⎕: 7
1.912931183
⎕: ¯27
¯3
⎕: ¯7
¯1.912931183

Seeing as there's a J answer in 16 characters, I must be really terrible at APL...

Answer 9

Java, 207 182 181

Sometimes when I play golf I have two many beers and play really really bad

class n{public static void main(String[]a){double d=Double.valueOf(a[0]);double i=d;for(int j=0;j<99;j++)i=(d/(i*i)+(2.0*i))/3.0;System.out.println((double)Math.round(i*1e4)/1e4);}}

Iterative Newton's Method of Approximation, runs 99 iterations.

Here is the unGolfed:

class n{
    public static void main(String a[]){
        //assuming the input value is the first parameter of the input
        //arguments as a String, get the Double value of it
        double d=Double.valueOf(a[0]);
        //Newton's method needs a guess at a starting point for the 
        //iterative approximation, there are much better ways at 
        //going about this, but this is by far the simplest. Given
        //the nature of the problem, it should suffice fine with 99 iterations
        double i=d;

        //make successive better approximations, do it 99 times
        for(int j=0;j<99;j++){
            i=( (d/(i*i)) + (2.0*i) ) / 3.0;
        }
        //print out the answer to standard out
        //also need to round off the double to meet the requirements
        //of the problem.  Short and sweet method of rounding:
        System.out.println( (double)Math.round(i*10000.0) / 10000.0 );
    }
}

Answer 10

TI-Basic, 26 24 bytes

Input :1:For(I,1,9:2Ans/3+X/(3AnsAns:End

Answer 11

Js 57 bytes

f=(x)=>eval('for(w=0;w**3<1e12*x;w++);x<0?-f(-x):w/1e4')

f=(x)=>eval('for(w=0;w**3<1e12*x;w++);x<0?-f(-x):w/1e4')
document.getElementById('div').innerHTML += f(-27) + '<br>'
document.getElementById('div').innerHTML += f(-64) + '<br>'
document.getElementById('div').innerHTML += f(-1) + '<br>'
document.getElementById('div').innerHTML += f(-18.609625) + '<br>'
document.getElementById('div').innerHTML += f(-3652264) + '<br>'
document.getElementById('div').innerHTML += f(-0.001) + '<br>'
document.getElementById('div').innerHTML += f(-7) + '<br><hr>'
document.getElementById('div').innerHTML += f(27) + '<br>'
document.getElementById('div').innerHTML += f(64) + '<br>'
document.getElementById('div').innerHTML += f(1) + '<br>'
document.getElementById('div').innerHTML += f(18.609625) + '<br>'
document.getElementById('div').innerHTML += f(3652264) + '<br>'
document.getElementById('div').innerHTML += f(0.001) + '<br>'
document.getElementById('div').innerHTML += f(7) + '<br>'

<div id="div"></div>

Answer 12

Javascript: 73/72 characters

This algorithm is lame, and exploits the fact that this question is limited to 4 digits after the decimal point. It is a modified version of the algorithm that I suggested in the sandbox for the purpose of reworking the question. It counts from zero to the infinite while h*h*h<a, just with a multiplication and division trick to handle the 4 decimal digits pecision.

function g(a){if(a<0)return-g(-a);for(h=0;h*h*h<1e12*a;h++);return h/1e4}

Edit, 4 years later: As suggested by Luis felipe De jesus Munoz, by using ** the code is shorter, but that feature was not available back in 2014 when I wrote this answer. Anyway, by using it, we shave an extra character:

function g(a){if(a<0)return-g(-a);for(h=0;h**3<1e12*a;h++);return h/1e4}

Answer 13

Javascript - 157 characters

This function:

• Handle negative numbers.
• Handle floating-pointing numbers.
• Execute quickly for any input number.
• Has the maximum precision allowed for javascript floating-point numbers.

function f(a){if(p=q=a<=1)return a<0?-f(-a):a==0|a==1?a:1/f(1/a);for(v=u=1;v*v*v<a;v*=2);while(u!=p|v!=q){p=u;q=v;k=(u+v)/2;if(k*k*k>a)v=k;else u=k}return u}

Ungolfed explained version:

function f(a) {
  if (p = q = a <= 1) return a < 0 ? -f(-a)      // if a < 0, it is the negative of the positive cube root.
                           : a == 0 | a == 1 ? a // if a is 0 or 1, its cube root is too.
                           : 1 / f (1 / a);      // if a < 1 (and a > 0) invert the number and return the inverse of the result.

  // Now, we only need to handle positive numbers > 1.

  // Start u and v with 1, and double v until it becomes a power of 2 greater than the given number.
  for (v = u = 1; v * v * v < a; v *= 2);

  // Bisects the u-v interval iteratively while u or v are changing, which means that we still did not reached the precision limit.
  // Use p and q to keep track of the last values of u and v so we are able to detect the change.
  while (u != p | v != q) {
    p = u;
    q = v;
    k = (u + v) / 2;
    if (k * k * k > a)
      v=k;
    else
      u=k
  }

  // At this point u <= cbrt(a) and v >= cbrt(a) and they are the closest that is possible to the true result that is possible using javascript-floating point precision.
  // If u == v then we have an exact cube root.
  // Return u because if u != v, u < cbrt(a), i.e. it is rounded towards zero.
  return u
}

Answer 14

PHP, 61

Based on Newton's method. Slightly modified version of Michael's answer:

for($i=$x=1;$i++<99;)$x=(2*$x+$n/$x/$x)/3;echo round($x,14);

It works with negative numbers, can handle floating point numbers, and rounds the result to 4 numbers after the decimal point if the result is a floating point number.

Working demo

Answer 15

Befunge 98 - Work in progress

This language does not support floating point numbers; this attempts to emulate them. It currently works for positive numbers that do not start with 0 after the decimal point (mostly). However, it only outputs to 2 decimal places.

&5ka5k*&+00pv
:::**00g`!jv>1+
/.'.,aa*%.@>1-:aa*

It works by inputting the part before the decimal point, multiplying that by 100000, then inputting the part after the point and adding the two numbers together. The second line does a counter until the cube is greater than the inputted number. Then the third line extracts the decimal number from the integer.

If anyone can tell me why the third line only divides by 100 to get the correct values, please tell me.

IOs:

27.0       3 .0
64.0       4 .0
1.0        1 .0
18.609625  2 .65
0.001      0 .1
7.0        1 .91

0.1        0 .1

Answer 16

Smalltalk, 37

credit goes to mniip for the algorithm; Smalltalk version of his code:

input in n; output in x:

1to:(x:=99)do:[:i|x:=2*x+(n/x/x)/3.0]

or, as a block

[:n|1to:(x:=99)do:[:i|x:=2*x+(n/x/x)/3.0].x]

Answer 17

GameMaker Language, 51 bytes

for(i=x=1;i++<99;1)x=(2*x+argument0/x/x)/3;return x

Answer 18

Haskell: 99C

Can't beat @mniip in cleverness. I just went with a binary search.

c x=d 0 x x
d l h x
 |abs(x-c)<=t=m
 |c < x=d m h x
 |True=d l m x
 where m=(l+h)/2;c=m*m*m;t=1e-4

Ungolfed:

-- just calls the helper function below
cubeRoot x = cubeRoot' 0 x x

cubeRoot' lo hi x
    | abs(x-c) <= tol = mid           -- if our guess is within the tolerance, accept it
    | c < x = cubeRoot' mid hi x      -- shot too low, narrow our search space to upper end
    | otherwise = cubeRoot' lo mid x  -- shot too high, narrow search space to lower end
    where
        mid = (lo+hi)/2
        cubed = mid*mid*mid
        tol = 0.0001

Answer 19

J 28

*@[*(3%~+:@]+(%*~@]))^:_&|&1

Using Newtons method, finding the root of x^3 - X the update step is x - (x^3 - C)/(3*x^2), where x is the current guess, and C the input. Doing the maths on this one yields the ridiculously simple expression of (2*x+C/x^2) /3 . Care has to be taken for negative numbers.

Implemented in J, from right to left:

1. | Take abs of both arguments, pass them on
2. ^:_ Do until convergence
3. (%*~@]) is C / x^2 (*~ y is equivalent to y * y)
4. +:@] is 2 x
5. 3%~ divide by three. This yields the positive root
6. *@[ * positive_root multiplies positive root with the signum of C.

Test run:

   NB. give it a name:
   c=: *@[*(3%~+:@]+(%*~@]))^:_&|&1
   c 27 64 1 18.609625 3652264 0.001 7
3 4 1 2.65 154 0.1 1.91293

Answer 20

AWK, 53 bytes

{for(n=x=$1;y-x;){y=x;x=(2*x+n/x/x)/3}printf"%.4g",y}

Example usage:

$ awk '{for(n=x=$1;y-x;){y=x;x=(2*x+n/x/x)/3}printf"%.4g",y}' <<< 18.609625 
2.65$

Thanks go to @Mig for the JavaScript solution which this is derived from. It runs surprisingly quickly given that the for loop requires the iteration to stop changing.

Answer 21

C, 69 bytes

i;float g(float x){for(float y=x;++i%999;x=x*2/3+y/3/x/x);return x;}

Just another implementation of Newton's method. Try it online!

Answer 22

Stax, 10 bytes^CP437

╘♀┘A╕äO¶∩'

Run and debug online!

Explanation

Uses the unpacked version to explain.

gpJux*_+h4je
gp              Iterate until a fixed point is found, output the fix point
  Ju            Inverse of square
    x*          Multiplied by input
      _+h       Average of the value computed by last command and the value at current iteration
         4je    Round to 4 decimal digits
  

Answer 23

JAVA Solution

public BigDecimal cubeRoot(BigDecimal number) {

    if(number == null || number.intValue() == 0) return BigDecimal.ZERO;
    BigDecimal absNum = number.abs();
    BigDecimal t;
    BigDecimal root =  absNum.divide(BigDecimal.valueOf(3), MathContext.DECIMAL128);


    do {

        t = root;
        root = root.multiply(BigDecimal.valueOf(2))
                .add(absNum.divide(root.multiply(root), MathContext.DECIMAL128))
                .divide(BigDecimal.valueOf(3), MathContext.DECIMAL128);

    } while (t.toBigInteger().subtract(root.toBigInteger()).intValue() != 0);

    return root.multiply(number.divide(absNum), MathContext.DECIMAL128);
}

Answer 24

Python Solution

def cube_root(num):
    if num == 0:
        return 0

    t = 0
    absNum = abs(num)
    root = absNum/3

    while (t - root) != 0:
        t = root
        root = (1/3) * ((2 * root) + absNum/(root * root))

    return root * (num / absNum)

Answer 25

MMIX, 44 bytes (11 instrs)

E0013FF0 48000002 E8018000 C1FF0100
1001FFFF 14010001 040101FF E401FFF0
11FF01FF 53FFFFFA F8020000

Disassembly

cbrt    SETH  $1,#3FF0      // y = 1.
        BNN   $0,0F         // if(x >= 0) goto loop (ensuring sign is correct)
        ORH   $1,#8000      // y = fnabs(y) (bit tricks!)
0H      SET   $255,$1       // loop: prev = y
        FMUL  $1,$255,$255  // y = prev *. prev
        FDIV  $1,$0,$1      // y = x /. y
        FADD  $1,$1,$255    // y = y +. prev
        INCH  $1,#FFF0      // y = y /. 2. (bit tricks again)
        FCMPE $255,$1,$255  // prev = comp_eps(y, prev)
        PBNZ  $255,0B       // if(prev) goto loop
        POP   2,0           // return {y,x}

It is up to the caller to set a suitable value of rE. Much faster convergence may be arranged by essentially setting y up by dividing the exponent by 3, but that costs another four instructions (though in time it pays really quickly, since three FDIVs cost about the same amount as two DIVs):

cbrt    SLU  $1,$0,1    // 3B010001 shift left to erase sign
        INCH $1,#8020   // E4018020 unbias exponent
        DIV  $1,$1,3    // 1D010103 divide by 3 as integer; approximate cube root
        INCH $1,#7FE0   // E4017FE8 rebias exponent
        SRU  $1,$1,1    // 3F010101 shift right again

and then continue from the second line in the original.

If mispredicted branches are particularly expensive, I would also replace the second and third lines of the original with:

        ZSN $255,$0,1       // 71020001 $255 = ($0 neg? 1 : 0)
        SLU $255,$255,63    // 3B02023F $255 <<= 63
        OR  $1,$1,$255      // C0010102 $1 |= $255

This costs one more instruction, but is branchless.