# 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.

• What sort of challenge is this? Mar 5, 2014 at 12:01
• How about exp(log(x)*y)? Mar 5, 2014 at 12:04
• Is an answer for integers only acceptable? Since these are the first replies. Mar 5, 2014 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, 2014 at 16:47
• Unfortunately the prohibition on loops rules out fun mathematical solutions like Taylor expansion. Jul 31, 2014 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
• use replicate y x instead of take y $repeat x Mar 5, 2014 at 18:38 • I was convinced that you could save characters by writing your second function pointfree. As it turns out f=(product.).flip replicate is exactly the same number of chars. – Kaya Mar 6, 2014 at 2:17 • @mniip It doesn't matter, this isn't code golf. Dec 10, 2014 at 14:44 # Python If y is a positive integer def P(x,y): return reduce(lambda a,b:a*b,[x]*y) # JavaScript (ES6), 31 // Testable in Firefox 28 f=(x,y)=>eval('x*'.repeat(y)+1) Usage: > f(2, 0) 1 > f(2, 16) 65536 Explanation: The above function builds an expression which multiply x y times then evaluates it. I'm surprised to see that nobody wrote a solution with the Y Combinator, yet... thus: # Python2 Y = lambda f: (lambda x: x(x))(lambda y: f(lambda v: y(y)(v))) pow = Y(lambda r: lambda (n,c): 1 if not c else n*r((n, c-1))) No loops, No vector/list operations and No (explicit) recursion! >>> pow((2,0)) 1 >>> pow((2,3)) 8 >>> pow((3,3)) 27 • Uh, I've just seen right now KChaloux's Haskell solution that uses fix, upvoting him... Mar 6, 2014 at 16:35 # C# : 45 Works for integers only: int P(int x,int y){return y==1?x:x*P(x,y-1);} • Beat me to it :-) I think you could save a few bytes by writing return --y?x:x*P(x,y); instead Mar 5, 2014 at 12:13 • But this isn't code-golf... Mar 5, 2014 at 12:26 • @oberon winning criteria was not clear when this was posted. Things have moved on. Mar 5, 2014 at 13:47 • @steveverrill Sorry. Mar 5, 2014 at 13:54 • Also in C# --y would be an int which is not the same as a bool like in other languages. Mar 7, 2014 at 9:39 # bash & sed No numbers, no loops, just an embarrasingly dangerous glob abuse. Preferably run in an empty directory to be safe. Shell script: #!/bin/bash rm -f xxxxx* eval touch$(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, 2014 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. Mar 12, 2014 at 21:49
• This is brilliant. Wish I'd thought of it for my R post. Jul 31, 2014 at 4:47

# 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. Mar 7, 2014 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.

# JavaScript

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

# 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. Mar 5, 2014 at 17:49
• @ugoren oh sure, you're right Mar 5, 2014 at 18:31
• This is a popular interview question :) "How can you write pow(n, x) better than O(n)?" Mar 6, 2014 at 0:28

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$x} • Matching Golfscript in length is a feat to achieve. Mar 7, 2014 at 0:29 Similar logic as many others, in PHP: <?=array_product(array_fill(0,$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. Mar 5, 2014 at 12:28 • @SilviuBurcea True, editing. Mar 5, 2014 at 12:30 • @SilviuBurcea operator =/= function Mar 5, 2014 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 ... Mar 5, 2014 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. Mar 6, 2014 at 1:34
• Yep, the goto version is a loop, but isn't tail recursion also essentially a loop? Mar 6, 2014 at 2:21

# Python

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