# Write a function that takes (x, y) and return x to the power of y WITHOUT Loops [closed]

This is a really neat short challenge.

Write a function or a procedure that takes two parameters, x and y and returns the result of xy WITHOUT using loops, or built in power functions.

The winner is the most creative solution, and will be chosen based on the highest number of votes after 3 days.

## closed as unclear what you're asking by Sriotchilism O'Zaic, Laikoni, NoOneIsHere, Stephen, Embodiment of IgnoranceMar 11 at 16:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What sort of challenge is this? – VisioN Mar 5 '14 at 12:01
• How about exp(log(x)*y)? – squeamish ossifrage Mar 5 '14 at 12:04
• Is an answer for integers only acceptable? Since these are the first replies. – mmumboss Mar 5 '14 at 12:59
• Looks like the answers so far either use recursion or lists of repeated 'x's. I'm wracking my brains trying to think of another way (particularly something that allows a non-integer y). – BenM Mar 5 '14 at 16:47
• Unfortunately the prohibition on loops rules out fun mathematical solutions like Taylor expansion. – shadowtalker Jul 31 '14 at 3:15

## APL (7)

{×/⍵/⍺}


Left argument is base, right argument is exponent, e.g.:

     5 {×/⍵/⍺} 6
15625


Explanation:

• ⍵/⍺ replicates ⍺ ⍵ times, e.g. 5 {⍵/⍺} 6 -> 5 5 5 5 5 5
• ×/ takes the product, e.g. ×/5 5 5 5 5 5 -> 5×5×5×5×5×5 -> 15625
• Hm.. This can be written in 5 characters in J, exactly the same method. */@$~ – seequ Jul 30 '14 at 18:32 • @Sieg 4 even, if you allow exponent on the left, base on the right. – ɐɔıʇǝɥʇuʎs Feb 20 '15 at 17:16 • I had the flip adverb because I thought it's not allowed. – seequ Feb 20 '15 at 18:44 • @Seeq 4 in Dyalog APL: ×/⍴⍨ – Adám Jun 16 '16 at 23:30 # C#: Floating point exponents OK, this solution is quite fragile. You can easily break it by throwing ridiculously huge numbers like 6 at it. But it works beautifully for things like DoublePower(1.5, 3.4), and doesn't use recursion!  static double IntPower(double x, int y) { return Enumerable.Repeat(x, y).Aggregate((product, next) => product * next); } static double Factorial(int x) { return Enumerable.Range(1, x).Aggregate<int, double>(1.0, (factorial, next) => factorial * next); } static double Exp(double x) { return Enumerable.Range(1, 100). Aggregate<int, double>(1.0, (sum, next) => sum + IntPower(x, next) / Factorial(next)); } static double Log(double x) { if (x > -1.0 && x < 1.0) { return Enumerable.Range(1, 100). Aggregate<int, double>(0.0, (sum, next) => sum + ((next % 2 == 0 ? -1.0 : 1.0) / next * IntPower(x - 1.0, next))); } else { return Enumerable.Range(1, 100). Aggregate<int, double>(0.0, (sum, next) => sum + 1.0 / next * IntPower((x - 1) / x, next)); } } static double DoublePower(double x, double y) { return Exp(y * Log(x)); }  • "ridiculously huge numbers like 6" I enjoyed that. – DavidC Mar 5 '14 at 21:59 • Surely use of Enumerable functions is relying on looping that was forbidden in the question or is it ok because the loop is inside framework methods? – Chris Mar 7 '14 at 9:42 # C++ How about some template meta programming? It bends what little rules there were, but it's worth a shot: #include <iostream> template <int pow> class tmp_pow { public: constexpr tmp_pow(float base) : value(base * tmp_pow<pow-1>(base).value) { } const float value; }; template <> class tmp_pow<0> { public: constexpr tmp_pow(float base) : value(1) { } const float value; }; int main(void) { tmp_pow<5> power_thirst(2.0f); std::cout << power_thirst.value << std::endl; return 0; }  • but this isn't a function, is a compile-time value, isn't it? :O – Paula_plus_plus Mar 6 '14 at 8:49 • Well, a constructor is a function, and template parameters are almost like function arguments... right? =) – erlc Mar 7 '14 at 13:47 • @PaperBirdMaster Yeah... that's why I admitted to some rule bending. I thought I was going to submit something besides tail-recursion, but I just submitted compile time tail recursion, haha. Close enough though, right? – astephens4 Mar 7 '14 at 16:19 • @astephens4 close enough, I love it :3 – Paula_plus_plus Mar 7 '14 at 23:46 ## Python def power(x,y): return eval(((str(x)+"*")*y)[:-1])  Doesn't work for noninteger powers. • I like this one. – CodyBugstein Mar 5 '14 at 20:06 • Why are you adding a separator without using join? eval('*'.join([str(x)] * y)). – Bakuriu Mar 6 '14 at 7:45 • Was this code-trolling? – gerrit Mar 6 '14 at 14:51 • Would like to also note that python has the ** operator, so you could've eval()d that. – Riking Mar 6 '14 at 21:14 • @Riking: that'd be an inbuilt, though. – Hovercouch Mar 6 '14 at 21:50 # Haskell - 25 chars f _ 0=1 f x y=x*f x (y-1)  Following Marinus' APL version: f x y = product$ take y $repeat x  With mniip's comment and whitespace removed, 27 chars: f x y=product$replicate y x

• use replicate y x instead of take y $repeat x – mniip Mar 5 '14 at 18:38 • I was convinced that you could save characters by writing your second function pointfree. As it turns out f=(product.).flip replicate is exactly the same number of chars. – Kaya Mar 6 '14 at 2:17 • @mniip It doesn't matter, this isn't code golf. – nyuszika7h Dec 10 '14 at 14:44 # Python If y is a positive integer def P(x,y): return reduce(lambda a,b:a*b,[x]*y)  # JavaScript (ES6), 31 // Testable in Firefox 28 f=(x,y)=>eval('x*'.repeat(y)+1)  Usage: > f(2, 0) 1 > f(2, 16) 65536  Explanation: The above function builds an expression which multiply x y times then evaluates it. I'm surprised to see that nobody wrote a solution with the Y Combinator, yet... thus: # Python2 Y = lambda f: (lambda x: x(x))(lambda y: f(lambda v: y(y)(v))) pow = Y(lambda r: lambda (n,c): 1 if not c else n*r((n, c-1)))  No loops, No vector/list operations and No (explicit) recursion! >>> pow((2,0)) 1 >>> pow((2,3)) 8 >>> pow((3,3)) 27  • Uh, I've just seen right now KChaloux's Haskell solution that uses fix, upvoting him... – berdario Mar 6 '14 at 16:35 # C# : 45 Works for integers only: int P(int x,int y){return y==1?x:x*P(x,y-1);}  • Beat me to it :-) I think you could save a few bytes by writing return --y?x:x*P(x,y); instead – squeamish ossifrage Mar 5 '14 at 12:13 • But this isn't code-golf... – Oberon Mar 5 '14 at 12:26 • @oberon winning criteria was not clear when this was posted. Things have moved on. – Level River St Mar 5 '14 at 13:47 • @steveverrill Sorry. – Oberon Mar 5 '14 at 13:54 • Also in C# --y would be an int which is not the same as a bool like in other languages. – Chris Mar 7 '14 at 9:39 # bash & sed No numbers, no loops, just an embarrasingly dangerous glob abuse. Preferably run in an empty directory to be safe. Shell script: #!/bin/bash rm -f xxxxx* eval touch$(printf xxxxx%$2s | sed "s/ /{1..$1}/g")
ls xxxxx* | wc -l
rm -f xxxxx*

• "Preferably run in an empty directory to be safe." :D – Almo Mar 6 '14 at 13:53

# Javascript

function f(x,y){return ("1"+Array(y+1)).match(/[\,1]/g).reduce(function(l,c){return l*x;});}


Uses regular expressions to create an array of size y+1 whose first element is 1. Then, reduce the array with multiplication to compute power. When y=0, the result is the first element of the array, which is 1.

Admittedly, my goal was i) not use recursion, ii) make it obscure.

# Mathematica

f[x_, y_] := Root[x, 1/y]


Probably cheating to use the fact that x^(1/y) = y√x

• Not cheating. Smart. – Michael Stern Mar 12 '14 at 21:49
• This is brilliant. Wish I'd thought of it for my R post. – shadowtalker Jul 31 '14 at 4:47

# JavaScript

function f(x,y){return y--?x*f(x,y):1;}


# Golfscript, 8 characters (including I/O)

~])*{*}*


Explanation:

TLDR: another "product of repeated array" solution.

The expected input is two numbers, e.g. 2 5. The stack starts with one item, the string "2 5".

Code     - Explanation                                             - stack
- "2 5"
~        - pop "2 5" and eval into the integers 2 5                - 2 5
]        - put all elements on stack into an array                 - [2 5]
)        - uncons from the right                                   - [2] 5
*        - repeat array                                            - [2 2 2 2 2]
{*}      - create a block that multiplies two elements             - [2 2 2 2 2] {*}
*        - fold the array using the block                          - 32

• Golfscript is always the way to go. – Nit Mar 7 '14 at 0:27

## Ruby

class Symbol
define_method(:**) {|x| eval x }
end

p(:****[$*[0]].*(:****$*[1]).*('*'))


Sample use:

$ruby exp.rb 5 3 125$ ruby exp.rb 0.5 3
0.125


This ultimately is the same as several previous answers: creates a y-length array every element of which is x, then takes the product. It's just gratuitously obfuscated to make it look like it's using the forbidden ** operator.

# C, exponentiation by squaring

int power(int a, int b){
if (b==0) return 1;
if (b==1) return a;
if (b%2==0) return power (a*a,b/2);
return a*power(a*a,(b-1)/2);
}


golfed version in 46 bytes (thanks ugoren!)

p(a,b){return b<2?b?a:1:p(a*a,b/2)*(b&1?a:1);}


should be faster than all the other recursive answers so far o.O

slightly slower version in 45 bytes

p(a,b){return b<2?b?a:1:p(a*a,b/2)*p(a,b&1);}

• For odd b, ~-b/2 == b/2. – ugoren Mar 5 '14 at 17:49
• @ugoren oh sure, you're right – izabera Mar 5 '14 at 18:31
• This is a popular interview question :) "How can you write pow(n, x) better than O(n)?" – Jordan Scales Mar 6 '14 at 0:28

## Haskell - 55

pow x y=fix(\r a i->if i>=y then a else r(a*x)(i+1))1 0


There's already a shorter Haskell entry, but I thought it would be interesting to write one that takes advantage of the fix function, as defined in Data.Function. Used as follows (in the Repl for the sake of ease):

ghci> let pow x y=fix(\r a i->if i>=y then a else r(a*x)(i+1))1 0
ghci> pow 5 3
125


# Q

9 chars. Generates array with y instances of x and takes the product.

{prd y#x}


Can explicitly cast to float for larger range given int/long x:

{prd y#9h$x}  • Matching Golfscript in length is a feat to achieve. – Nit Mar 7 '14 at 0:29 Similar logic as many others, in PHP: <?=array_product(array_fill(0,$argv[2],$argv[1]));  Run it with php file.php 5 3 to get 5^3 I'm not sure how many upvotes I can expect for this, but I found it somewhat peculiar that I actually had to write that very function today. And I'm pretty sure this is the first time any .SE site sees this language (website doesn't seem very helpful atm). # ABS def Rat pow(Rat x, Int y) = if y < 0 then 1 / pow(x, -y) else case y { 0 => 1; _ => x * pow(x, y-1); };  Works for negative exponents and rational bases. I highlighted it in Java syntax, because that's what I'm currently doing when I'm working with this language. Looks alright. # Pascal The challenge did not specify the type or range of x and y, therefore I figure the following Pascal function follows all the given rules: { data type for a single bit: can only be 0 or 1 } type bit = 0..1; { calculate the power of two bits, using the convention that 0^0 = 1 } function bitpower(bit x, bit y): bit; begin if y = 0 then bitpower := 1 else bitpower := x end;  No loop, no built-in power or exponentiation function, not even recursion or arithmetics! ## J - 5 or 4 bytes Exactly the same as marinus' APL answer. For x^y: */@$~


For y^x:

*/@$ For example:  5 */@$~ 6
15625
6 */@$5 15625  x$~ y creates a list of x repeated y times (same as y $x */ x is the product function, */ 1 2 3 -> 1 * 2 * 3 # Python from math import sqrt def pow(x, y): if y == 0: return 1 elif y >= 1: return x * pow(x, y - 1) elif y > 0: y *= 2 if y >= 1: return sqrt(x) * sqrt(pow(x, y % 1)) else: return sqrt(pow(x, y % 1)) else: return 1.0 / pow(x, -y)  • ** is built-in operator imo. – Silviu Burcea Mar 5 '14 at 12:28 • @SilviuBurcea True, editing. – Oberon Mar 5 '14 at 12:30 • @SilviuBurcea operator =/= function – VisioN Mar 5 '14 at 12:32 • @VisioN true, but the idea was about built-ins. I don't think the OP knows about all these built-in operators ... – Silviu Burcea Mar 5 '14 at 12:34 # Javascript With tail recursion, works if y is a positive integer function P(x,y,z){z=z||1;return y?P(x,y-1,x*z):z}  # Bash Everyone knows bash can do whizzy map-reduce type stuff ;-) #!/bin/bash x=$1
reduce () {
((a*=$x)) } a=1 mapfile -n$2 -c1 -Creduce < <(yes)
echo $a  If thats too trolly for you then there's this: #!/bin/bash echo$(( $( yes$1 | head -n$2 | paste -s -d'*' ) ))  # C Yet another recursive exponentiation by squaring answer in C, but they do differ (this uses a shift instead of division, is slightly shorter and recurses one more time than the other): e(x,y){return y?(y&1?x:1)*e(x*x,y>>1):1;}  # Mathematica This works for integers. f[x_, y_] := Times@@Table[x, {y}]  Example f[5,3]  125 How it works Table makes a list of y x's. Times takes the product of all of them. Another way to achieve the same end: #~Product~{i,1,#2}&  Example #~Product~{i, 1, #2} & @@ {5, 3}  125 ## Windows Batch Like most of the other answers here, it uses recursion. @echo off set y=%2 :p if %y%==1 ( set z=%1 goto :eof ) else ( set/a"y-=1" call :p %1 set/a"z*=%1" goto :eof )  x^y is stored in the environment variable z. ## perl Here's a tail recursive perl entry. Usage is echo$X,$Y | foo.pl: ($x,$y) = split/,/, <>; sub a{$_*=$x;--$y?a():$_}$_=1;
print a


Or for a more functional-type approach, how about:

($x,$y) = split/,/, <>;
$t=1; map {$t *= $x } (1..$y);
print \$t

• "a: stuff goto a if something" looks like a loop. – Glenn Randers-Pehrson Mar 6 '14 at 1:34
• Yep, the goto version is a loop, but isn't tail recursion also essentially a loop? – skibrianski Mar 6 '14 at 2:21

# Python

def getRootOfY(x,y):
return x**y

`