24
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While bored in high-school (when I was half my current age...), I found that f(x) = x(x-1) had some interesting properties, including e.g. that the maximum f for 0 ≤ x is f(e), and that the binding energy per nucleon of an isotope can be approximated as 6 × f(x ÷ 21)...

Anyway, write the shortest function or program that calculates the xth root of x for any number in your language's domain.

Examples cases

For all languages

     -1   >       -1
   ¯0.2   >    -3125
   ¯0.5   >        4
    0.5   >     0.25
      1   >        1
      2   >    1.414
      e   >    1.444 
      3   >    1.442
    100   >    1.047
  10000   >    1.001

For languages that handle complex numbers

   -2   >        -0.7071i
    i   >            4.81         
   2i   >    2.063-0.745i
 1+2i   >   1.820-0.1834i
 2+2i   >   1.575-0.1003i

For languages that handle infinities

-1/∞   >   0    (or ∞ or ̃∞)
   0   >   0    (or 1 or ∞)
 1/∞   >   0
   ∞   >   1
  -∞   >   1

For languages that handle both infinities and complex numbers

 -∞-2i   >   1      (or ̃∞)

̃∞ denotes directed infinity.

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  • 1
    \$\begingroup\$ Here is a Wolfram Alpha plot for positive real x. If you omit the x limits in the query, Wolfram Alpha will include negative values of x where the function value depends on a choice of "branch" for the complex logarithm (or for a similar complex function). \$\endgroup\$ – Jeppe Stig Nielsen Feb 24 '16 at 14:36
  • \$\begingroup\$ What about for languages that do not handle power of decimals? \$\endgroup\$ – Leaky Nun Mar 31 '16 at 3:24
  • 1
    \$\begingroup\$ @KennyLau Feel free to post with a note that says so, especially if the algorithm would work, had the language supported it. \$\endgroup\$ – Adám Mar 31 '16 at 5:37

61 Answers 61

38
\$\begingroup\$

TI-BASIC, 3 bytes

Ans×√Ans

TI-BASIC uses tokens, so Ans and ×√ are both one byte.

Explanation

Ans is the easiest way to give input; it is the result of the last expression. ×√ is a function for the x'th root of x, so for example 5×√32 is 2.

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  • 8
    \$\begingroup\$ As far as I am aware ans would count as hardcoding inputs into variables and does not seem to be an accepted input method for code-golf. In that case, please make a full program or a function. \$\endgroup\$ – flawr Feb 24 '16 at 9:14
  • 4
    \$\begingroup\$ @flawr I can see what you're saying but it seems it's always been done like this. Maybe it warrants a meta post? \$\endgroup\$ – NinjaBearMonkey Feb 24 '16 at 16:08
  • 3
    \$\begingroup\$ Ans is STDIN/STDOUT for TI-Basic. \$\endgroup\$ – Timtech Feb 24 '16 at 22:43
  • 5
    \$\begingroup\$ stdin and stdout are text streams, usually for interactive text input and output. Ans is not interactive, unlike some other functions in TI-BASIC, which are interactive. \$\endgroup\$ – Olathe Feb 25 '16 at 5:21
  • 7
    \$\begingroup\$ @flawr The reason Ans is usually accepted is because its value is set by any expression (expressions are separated by :). Therefore something like 1337:prgmXTHROOT would input 1337, which looks a lot like input via CLAs in a normal language. \$\endgroup\$ – lirtosiast Mar 5 '16 at 5:06
23
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Jelly, 2 bytes

Try it online!

How it works

*İ    Main link. Input: n

 İ    Inverse; yield 1÷n.
*     Power (fork); compute n ** (1÷n).
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  • \$\begingroup\$ Jelly doesn't have a stack. A dyad follow by a monad in a monadic chain behaves like APL's forks. \$\endgroup\$ – Dennis Feb 23 '16 at 4:46
  • 3
    \$\begingroup\$ No, J's ^% is a hook (which do not exist in Dyalog APL), not a fork. Jelly and APL code is difficult to compare since Jelly is left-to-right. The nearest equivalent would be ÷*⊢ (also a fork), which computes (1/x)**x because of the different direction. Since Jelly's atoms aren't overloaded (they are either monadic or dyadic, but never both), there can be monadic 1,2,1- and 2,1-forks. \$\endgroup\$ – Dennis Feb 23 '16 at 4:58
  • \$\begingroup\$ Thanks for the clarification. Naturally, I'm quite intrigued by Jelly (which I still think should be named ȷ or something similar.) \$\endgroup\$ – Adám Feb 23 '16 at 5:07
17
\$\begingroup\$

Javascript (ES2016), 11 bytes

x=>x**(1/x)

I rarely ever use ES7 over ES6.

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  • 2
    \$\begingroup\$ x=>x**x**-1 also works, again for 11 bytes. \$\endgroup\$ – Neil Feb 23 '16 at 8:51
  • 7
    \$\begingroup\$ All hail the new exponentiation operator! \$\endgroup\$ – mbomb007 Feb 23 '16 at 21:33
15
\$\begingroup\$

Python 3, 17 bytes

lambda x:x**(1/x)

Self-explanatory

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  • 7
    \$\begingroup\$ I quite like lambda x:x**x**-1, but it's not shorter. \$\endgroup\$ – seequ Feb 23 '16 at 6:32
  • 1
    \$\begingroup\$ @Seeq Your expression is the same length, but it has the advantage of working in both Python 2 and 3. \$\endgroup\$ – mathmandan Feb 23 '16 at 17:54
  • 1
    \$\begingroup\$ Python 2's shortest is lambda x:x**x**-1, so it is the same in 2 and 3. \$\endgroup\$ – mbomb007 Feb 23 '16 at 21:31
  • \$\begingroup\$ I couldn't find this answer for ages and was really annoyed when I did. \$\endgroup\$ – user63571 Feb 5 '17 at 21:09
12
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Haskell, 12 11 bytes

Thanks @LambdaFairy for doing some magic:

(**)<*>(1/) 

My old version:

\x->x**(1/x)
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  • 4
    \$\begingroup\$ (**)<*>(1/) is 11 bytes. \$\endgroup\$ – Lambda Fairy Feb 24 '16 at 1:32
  • \$\begingroup\$ @LambdaFairy Thanks! Do you mind explaining? It looks like you are doing some magic with partially applied functions but as I am quite new to Haskell I do not really understand how this works=) \$\endgroup\$ – flawr Feb 24 '16 at 9:02
  • \$\begingroup\$ This uses the fact that a 1-argument function can be considered an applicative functor (the "reader monad"). The <*> operator takes an applicative that produces a function, and an applicative that produces a value, and applies the function to the value. So in this case, a mind-bending way to apply a 2-argument function to a 1-argument function. \$\endgroup\$ – MathematicalOrchid Feb 24 '16 at 16:43
  • 2
    \$\begingroup\$ The function <*> takes 3 arguments, two functions f and g and an argument x. It is defined as (<*>) f g x = f x (g x), i.e. it applies f to x and g x. Here it's partially applied to f and g leaving out x, where f = (**) and g = (1/) (another partially applied function (a section) that calculates the reciprocal value of it's argument). So ( (**)<*>(1/) ) x is (**) x ((1/) x) or written in infix: x ** ((1/) x) and with the section resolved: x ** (1/x). -- Note: <*> is used in function context here and behaves differently in other contexts. \$\endgroup\$ – nimi Feb 24 '16 at 16:52
  • \$\begingroup\$ @nimi So it's the equivalent of the S combinator i.e. S(**)(1/)? \$\endgroup\$ – Neil Mar 2 '16 at 13:03
10
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J, 2 bytes

^%

Try it online!.

How it works

^%  Monadic verb. Argument: y

 %  Inverse; yield 1÷y.
^   Power (hook); compute y ** (1÷y).
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  • \$\begingroup\$ I was gonna write this answer. I'm too slow at this. \$\endgroup\$ – Bijan Mar 19 '17 at 22:02
  • 1
    \$\begingroup\$ @Bijan Over a year too slow, it seems. :P \$\endgroup\$ – Dennis Mar 19 '17 at 22:21
  • \$\begingroup\$ I see, I've only been golfing for a week now. \$\endgroup\$ – Bijan Mar 19 '17 at 22:37
9
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Pyth, 3 bytes

@QQ

Trivial challenge, trivial solution...

(noncompeting, 1 byte)

@

This uses the implicit input feature present in a version of Pyth that postdates this challenge.

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  • \$\begingroup\$ Does this solution predate the feature of implicit input? \$\endgroup\$ – Leaky Nun Apr 7 '16 at 23:39
  • \$\begingroup\$ @KennyLau Yes, by a long time. But I've edited the one-byte solution in anyway. \$\endgroup\$ – Doorknob Apr 8 '16 at 4:40
8
\$\begingroup\$

JavaScript ES6, 18 bytes

n=>Math.pow(n,1/n)
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8
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Java 8, 18 bytes

n->Math.pow(n,1/n)

Java isn't in last place?!?!

Test with the following:

import java.lang.Math;

public class Main {
  public static void main (String[] args) {
    Test test = n->Math.pow(n,1/n);
    System.out.println(test.xthRoot(6.0));
  }
}

interface Test {
  double xthRoot(double x);
}
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  • \$\begingroup\$ It's the fact that it's a function \$\endgroup\$ – CalculatorFeline Feb 29 '16 at 21:30
6
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Java, 41 bytes

float f(float n){return Math.pow(n,1/n);}

Not exactly competitive because Java, but why not?

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  • 1
    \$\begingroup\$ Welcome to PPCG! I think you might be missing a return type on this function. \$\endgroup\$ – a spaghetto Feb 23 '16 at 18:28
  • \$\begingroup\$ Oops, got sloppy. A Java 8 answer already beat this one of course... \$\endgroup\$ – Darrel Hoffman Feb 23 '16 at 23:18
6
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MATL, 5 bytes

t-1^^

Try it online!

t       % implicit input x, duplicate
 -1     % push -1
   ^    % power (raise x to -1): gives 1/x
    ^   % power (raise x to 1/x). Implicit display
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6
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Mathematica, 8 7 4 7 bytes

#^#^-1&

More builtin-only answers, and now even shorter! Nope. By definition, the next answer should be 13 bytes. (Fibonacci!) The pattern is still broken. :(

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  • 1
    \$\begingroup\$ #^#^-1& saves 1 byte. \$\endgroup\$ – njpipeorgan Feb 23 '16 at 15:16
  • \$\begingroup\$ NOW it is golfed. \$\endgroup\$ – Adám Feb 23 '16 at 23:16
  • 1
    \$\begingroup\$ NOW it is golfed. \$\endgroup\$ – CalculatorFeline Mar 10 '16 at 2:43
  • 1
    \$\begingroup\$ When Mthmtca is released, we are going to rule this board. \$\endgroup\$ – Michael Stern Mar 16 '16 at 3:48
  • 1
    \$\begingroup\$ Surely just Surd isn't valid as it requires two arguments? \$\endgroup\$ – LLlAMnYP Jun 7 '16 at 14:16
5
\$\begingroup\$

Perl 5, 10 bytes

9 bytes plus 1 for -p

$_**=1/$_
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5
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R, 19 17 bytes

function(x)x^x^-1

-2 bytes thanks to @Flounderer

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  • \$\begingroup\$ Why not x^(1/x) ? Edit: x^x^-1 seems to work too. \$\endgroup\$ – Flounderer Feb 23 '16 at 20:10
  • \$\begingroup\$ That's a snippet, and apparently people don't like snippets. \$\endgroup\$ – CalculatorFeline Feb 29 '16 at 21:28
  • \$\begingroup\$ @CatsAreFluffy it is the definition of a function. \$\endgroup\$ – mnel Feb 29 '16 at 21:30
5
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Ruby, 15 Bytes

a=->n{n**n**-1}

Ungolfed:

-> is the stabby lambda operator where a=->n is equivalent to a = lambda {|n|}

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5
\$\begingroup\$

NARS APL, 2 bytes

√⍨

NARS supports the function, which gives the ⍺-th root of ⍵. Applying commute (⍨) gives a function that, when used monadically, applies its argument to both sides of the given function. Therefore √⍨ xx √ x.

Other APLs, 3 bytes

⊢*÷

This is a function train, i.e. (F G H) x(F x) G H x. Monadic is identity, dyadic * is power, and monadic ÷ is inverse. Therefore, ⊢*÷ is x raised to 1/x.

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5
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Python 2 - 56 bytes

The first actual answer, if I'm correct. Uses Newton's method.

n=x=input();exec"x-=(x**n-n)/(1.*n*x**-~n);"*999;print x
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  • \$\begingroup\$ Functions are okay. \$\endgroup\$ – CalculatorFeline Feb 24 '16 at 1:18
5
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CJam, 6 bytes

rd_W##

Try it online!

How it works

rd     e# Read a double D from STDIN and push it on the stack.
  _    e# Push a copy of D.
   W   e# Push -1.
    #  e# Compute D ** -1.
     # e# Compute D ** (D ** -1).
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4
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Seriously, 5 bytes

,;ì@^

Try it online!

Explanation:

,;ì@^
,;     input, dupe
  ì@   1/x, swap
    ^  pow
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4
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Pylons, 5 bytes.

ideAe

How it works.

i # Get command line input.
d # Duplicate the top of the stack.
e # Raise the top of the stack to the power of the  second to the top element of the stack.
A # Push -1 to the stack (pre initialized variable).
e # Raise the top of the stack to the power of the second to the top element of the stack.
  # Implicitly print the stack.
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4
\$\begingroup\$

Japt, 3 bytes

UqU

Test it online!

Very simple: U is the input integer, and q is the root function on numbers.

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4
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C++, 48 bytes

#include<math.h>
[](auto x){return pow(x,1./x);}

The second line defines an anonymous lambda function. It can be used by assigning it to a function pointer and calling it, or just calling it directly.

Try it online

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  • \$\begingroup\$ Does ^ not work in C++ as it does in C? \$\endgroup\$ – takra Feb 23 '16 at 23:18
  • 2
    \$\begingroup\$ @minerguy31: ^ is bitwise xor in C (and C++). \$\endgroup\$ – marinus Feb 23 '16 at 23:19
4
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Milky Way 1.6.5, 5 bytes

'1'/h

Explanation

'      ` Push input
 1     ` Push the integer literal
  '    ` Push input
   /   ` Divide the STOS by the TOS
    h  ` Push the STOS to the power of the TOS

x**(1/x)


Usage

$ ./mw <path-to-code> -i <input-integer>
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4
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O, 6 bytes

j.1\/^

No online link because the online IDE doesn't work (specifically, exponentiation is broken)

Explanation:

j.1\/^
j.      push two copies of input
  1\/   push 1/input (always float division)
     ^  push pow(input, 1/input)
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  • \$\begingroup\$ oh hey you did it yay \$\endgroup\$ – phase Mar 19 '16 at 4:20
4
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𝔼𝕊𝕄𝕚𝕟, 5 chars / 7 bytes

Мű⁽ïï

Try it here (Firefox only).

Trivial.

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4
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Pyke (commit 29), 6 bytes

D1_R^^

Explanation:

D      - duplicate top
 1_    - load -1
   R   - rotate
    ^  - ^**^
     ^ - ^**^
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  • \$\begingroup\$ Can haz link pls? \$\endgroup\$ – cat Feb 24 '16 at 17:33
  • \$\begingroup\$ Oh, I thought you meant there's no implementation available. Yes, the interpreter doesn't have to be hosted, just a link to the repo / source (or docs) will suffice \$\endgroup\$ – cat Feb 24 '16 at 17:36
4
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C# - 18 43 41 bytes

float a(float x){return Math.Pow(x,1/x);}

-2 byes thanks to @VoteToClose

Try it out

Note:

First actual attempt at golfing - I know I could do this better.

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  • \$\begingroup\$ Welcome to the crowd! It is exactly because of newcomers that I make trivial challenges like this. \$\endgroup\$ – Adám Feb 24 '16 at 19:29
  • \$\begingroup\$ Fixed. Thanks for informing me about this \$\endgroup\$ – EnragedTanker Feb 25 '16 at 15:40
  • \$\begingroup\$ @crayzeedude No problem at all. Nice job and again, welcome to PPCG! \$\endgroup\$ – Alex A. Feb 25 '16 at 18:00
  • \$\begingroup\$ Does C# have float? \$\endgroup\$ – Addison Crump Feb 27 '16 at 0:31
  • \$\begingroup\$ Indeed it does. \$\endgroup\$ – EnragedTanker Feb 27 '16 at 18:29
4
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C, 23 bytes

#define p(a)pow(a,1./a)

This defines a macro function p which evaluates to the ath root of a.

Thanks to Dennis for reminding me that gcc doesn't require math.h to be included.

Thanks to @EʀɪᴋᴛʜᴇGᴏʟғᴇʀ for reminding me that the space after the first ) is not needed.

Try it online

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  • \$\begingroup\$ With GCC, you don't need to include math.h. \$\endgroup\$ – Dennis Mar 10 '16 at 3:04
  • \$\begingroup\$ -1 byte: #define p(a)pow(a,1./a) \$\endgroup\$ – Erik the Outgolfer Jun 14 '16 at 8:37
4
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dc, 125 bytes

15k?ddsk1-A 5^*sw1sn0[A 5^ln+_1^+ln1+dsnlw!<y]syr1<y1lk/*sz[si1[li*li1-dsi0<p]spli0<p]so0dsw[lzlw^lwlox/+lw1+dswA 2^!<b]dsbxp

Unlike the other dc answer, this works for all real x greater than or equal to 1 (1 ≤ x). Accurate to 4-5 places after the decimal.

I would have included a TIO link here, but for some reason this throws a segmentation fault with the version there (dc 1.3) whereas it does not with my local version (dc 1.3.95).

Explanation

As dc does not support raising numbers to non-integer exponents to calculate x^(1/x), this takes advantage of the fact that:

Advantage

So, to calculate ln(x), this also takes advantage of the fact that:

Advantage2

whose definite integral from 1 to (b = x) is numerically-approximated in increments of 10^-5 using the following summation formula:

Summation Formula.

The resulting sum is then multiplied by 1/x to get ln(x)/x. e^(ln(x)/x) is then finally calculated using the e^x Maclaurin Series to 100 terms as follows:

e^x Maclaurin Series.

This results in our relatively accurate output of x^(1/x).

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  • 1
    \$\begingroup\$ +1 This has got to be one of the best dc answers out there. I'm bookmarking this! \$\endgroup\$ – Cows quack Mar 19 '17 at 20:43
  • \$\begingroup\$ @KritixiLithos Thank you! I appreciate the kind words. :) \$\endgroup\$ – R. Kap Mar 19 '17 at 20:47
3
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PHP 5.6, 32 30 29 bytes

function($x){echo$x**(1/$x);}

or

function($x){echo$x**$x**-1;}

30->29, thank you Dennis!

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