# Print the tetration

Tetration, represented as $$\{}^ba\$$, is repeated exponentiation. For example, $$\{}^32\$$ is $$\2^{2^2}\$$, which is $$\16\$$.

Given two numbers $$\a\$$ and $$\b\$$, print $$\{}^ba\$$.

## Test cases

1 2 -> 1
2 2 -> 4
5 2 -> 3125
3 3 -> 7625597484987
etc.


Scientific notation is acceptable.

Remember, this is , so the code with the smallest number of bytes wins.

• What kind of numbers? Positive integers?
– xnor
Oct 22 '16 at 0:42
• Related Oct 22 '16 at 0:43
• Exponentiation is non-associative. You should include at least one test cade with b > 2. Oct 22 '16 at 1:02
• @Dennis 3 3 -> 7625597484987 Oct 22 '16 at 1:12
• @RosLuP No, 3^3^3 automatically means 3^(3^(3)). See en.wikipedia.org/wiki/Order_of_operations, where it says "Stacked exponents are applied from the top down, i.e., from right to left." Nov 17 '16 at 16:10

# Dyalog APL, 3 bytes

*/⍴


TryAPL.

## Explanation

*/⍴  Input: b (LHS), a (RHS)
⍴  Create b copies of a
*/   Reduce from right-to-left using exponentation

• Hey, somebody who's beating @Dennis! Now that's rare! (; :P Dec 3 '16 at 22:30

# J, 5 4 bytes

^/@#


This is literally the definition of tetration.

## Usage

   f =: ^/@#
3 f 2
16
2 f 1
1
2 f 2
4
2 f 5
3125
4 f 2
65536


## Explanation

^/@#  Input: b (LHS), a (RHS)
#  Make b copies of a
^/@   Reduce from right-to-left using exponentation

• Ok a^^b is above reversed b^^a...
– user58988
Oct 22 '16 at 15:16
• @RosLuP Yes, J and APL evaluate from right-to-left, so 2 ^ 2 ^ 2 is evaluated as 2 ^ (2 ^ 2) and so on Oct 24 '16 at 10:53

## Python, 30 bytes

f=lambda a,b:b<1or a**f(a,b-1)


Uses the recursive definition.

## Haskell, 19 bytes

a%b=iterate(a^)1!!b


Iterates exponentiating starting at 1 to produce the list [1,a,a^a,a^a^a,...], then take the b'th element.

Same length directly:

a%0=1;a%b=a^a%(b-1)


Point-free is longer:

(!!).(iterate1).(^)


# Mathematica, 16 bytes

Power@@Table@##&


# Explanation

Table@##


Make b copies of a.

Power@@...


Exponentiation.

# Jelly, 4 bytes

x*@/


### How it works

x*@/  Main link. Arguments: a, b

x     Repeat [a] b times.
*@/  Reduce the resulting array by exponentation with swapped arguments.


# Python, 33 bytes

lambda a,b:eval('**'.join([a]*b))


This evaluates to an unnamed function, that takes the string representation of a number and a number. For example:

>>> f=lambda a,b:eval('**'.join([a]*b))
>>> f('5',2)
3125
>>>


If mixing input formats like this does not count, there is also this 38 byte version:

lambda a,b:eval('**'.join([str(a)]*b))

• What a cool method!
– xnor
Oct 22 '16 at 0:49

# 05AB1E, 94 2 bytes

-2 thanks to ovs.

Gm


Try it online! Beats all other answers

Gm  # full program
G   # for N in [1, 2, 3, ...,
# ..., implicit input...
G   # ...minus 1]...
m  # push...
# implicit input...
m  # to the power of...
# implicit input...
# (implicit) or top of stack if not first iteration
# implicit output

• Welcome to the site and impressive first answer! I've edited your TIO link so that it includes an example input. Be sure to check out other recently active challenges as well as our Tips for golfing in 05AB1E page to see if you can shave any bytes off your answer! Sep 23 '20 at 0:28
• Thanks! I did actually end up checking out that page before submitting this answer :) Sep 23 '20 at 0:33
• Does Gm work?
– ovs
May 29 at 21:47
• @ovs (－‸ლ) of course it does. Thank you! May 29 at 21:49

# Perl, 19 bytes

Includes +1 for -p

Give numbers on separate lines on STDIN

tetration.pl
2
3
^D


tetration.pl

#!/usr/bin/perl -p
$_=eval"$_**"x<>.1


# R, 39 bytes

Recursive function:

f=function(a,b)ifelse(b>0,a^f(a,b-1),1)


# Factor, 29 26 bytes

[ 1 rot [ dupd ^ ] times ]


Try it online!

-3 thanks to @Bubbler

• May 31 at 1:55

# Element, 11 bytes

__2:':1[^]


Try it online!

This is just "straightforward" exponentiation in a loop.

__2:':1[^]
__              take two values as input (x and y)
2:'           duplicate y and send one copy to the control stack
:          make y copies of x
1         push 1 as the initial value
[ ]      loop y times
^       exponentiate
     print result


# JavaScript (ES7), 24 bytes

f=(a,b)=>b?a**f(a,b-1):1


The ES6 version is 33 bytes:

f=(a,b)=>b?Math.pow(a,f(a,b-1)):1

• Save 1 byte: f=a=>b=>b?a**f(a,b-1):1
– user58826
Apr 6 '17 at 18:42
• @programmer5000 no, the function call would have to be f(a)(b-1) inside, adding 1 byte and cancelling out the effect.
– user100690
Jun 1 at 7:52
• @ophact You're right, but that's a 4-year-old comment... However, you can really get -3 bytes by calling only the inner function.
– FZs
Jun 5 at 19:11

# dc, 35 29 bytes:

?dsdsa?[ldla^sa1-d1<b]dsbxlap


Here is my first complete program in dc.

for($b=$p=$argv[1];++$i<$argv[2];)$p=$b**$p;echo$p;  # GameMaker Language, 52 50 bytes d=a=argument0;for(c=1;c<b;c++)d=power(a,d)return d  • This is my 300th answer :o Oct 23 '16 at 20:35 • GameMaker wtf? lol Oct 27 '17 at 23:21 • @SimplyBeautifulArt Yes, and while I'm at it I'll take off 2 bytes for you. Oct 28 '17 at 2:44 • Lol, nice. =) Have my +1, seems simple enough and I understand it. Oct 28 '17 at 19:11 • @SimplyBeautifulArt Appreciated Oct 29 '17 at 3:38 # Perl, 40 bytes map{$a=$ARGV[0]**$a}0..$ARGV[1];print$a;


Accepts two integers as input to the function and outputs the result

• Use pop to get $ARGV[1], then use "@ARGV" to get $ARGV[0]. Use say instead of print (option -M5.010 or -E is free). But still, ARGV is terribly long. A -p program almost always wins Oct 22 '16 at 6:56

# Actually, 6 bytes

nⁿ)Y


Try it online!

Input is taken as b\na (\n is a newline)

Explanation:

nⁿ)Y
n       a copies of b
ⁿ)Y  while stack changes between each call (fixed-point combinator):
ⁿ       pow
)      move top of stack to bottom (for right-associativity)


# CJam, 9 bytes

q~)*{\#}*


Try it online!

### Explanation

q~          e# Take input (array) and evaluate
)         e# Pull off last element
*        e# Array with the first element repeated as many times as the second
{  }*   e# Reduce array by this function
\#     e# Swap, power


# Wonder, 21 bytes

f\@@[#0?^#1f#1-#0 1?1


Uses the recursive approach. Usage:

f\@@[#0?^#1f#1-#0 1?1];f 2 3


# Bonus solution, 22 bytes

@@:^ -#0 1(genc ^#1)#1


A slightly unconventional approach. Usage:

t\@@+>#[^;#1]tk -#0 1rpt#1;t 2 3


More readable:

@@
iget
- #0 1
(genc ^#1) #1


Assuming a^^b:

Generates an infinite list of tetrated a; for a=2, this list would look something like [2 4 16 65536...]. Then indexes at b-1 because Wonder is zero-indexed.

# Flurry-nii, 14 bytes

{}{{}({})}{{}}


Try it online!

Takes two numbers a b from stdin, and outputs as the return value.

### How it works

It works basically like xnor's Haskell answer: repeat (a^) b times to 1. a is left on the stack, so that it can be referenced in each iteration.

// In Church numeral, x a → a^x
main = b power_of_a 1

power_of_a = \x. x a
= \x. x (push pop)  // assumes a is on the top of the stack,
// pops and takes its value and pushes it back


# GolfScript, 9 bytes

~])*{\?}*


Try it online!

~]          # Put a and b in an array
)         # Remove b from the array
*        # Make b copies of a
{  }*   # Execute this block for each number in the array
\?     # Exponentiation


Just {?} would be (3^3)^3 instead of 3^(3^3).

# Vyxalr, 4 bytes

This is lyxal's solution

‹(⁰e


Try it Online!

‹(⁰e
‹     Decrement b
(  ) For loop (second parentheses is implicit) using b-1
⁰   Push a
e  And raise it to the power of the accumulator (initially a)


## Vyxal, 6 5 bytes

Saved 1 byte thanks to lyxal

ʁ•⁽eḭ


Try it Online!

This one is a worse solution. a is inputted as a singleton list.

ʁ•λe;ḭ
ʁ      Make a range [0,b) (only the length of the range matters (b))
•     Reshape [a] to that (make a list of b a's)
ḭ Reduce from the right
λe;  using exponentiation

• 5 bytes May 30 at 22:27
• 4 bytes May 30 at 22:53

# K (ngn/k), 7 bytes

(*/#)/#


Try it online!

Takes two inputs in the order of [hyperexponent; base] and returns the result.

K does not have exponentiation built-in, so I simulate n^m with "product */ of m copies # of n" instead. Tetration is basically the same, in the sense that it is a fold by exponentiation. K's fold is left to right, and */# accepts [exponent; base], so the direction of exponentiation is correctly y^(y^(y^...)).

Although it does not use exponentiation built-in, it fails to solve this challenge because 4^(4^4) is way too big to handle with built-in integer sizes.

# Red, 3826 37 bytes

Thanks to @9214 for telling me about the loop command for -12 bytes.

+11 bytes to fix invalidities brought out by @Bubbler.

func[a b][x: a loop b - 1[x: a ** x]]


Try it online!

• Tip: you can simplify it down to 27 with loop.
– 9214
Jun 4 at 14:54
• @9214 Thanks, that is exactly the command I needed, and apparently, it returns a value, so I can even simplify it down to 26 bytes! Jun 4 at 15:07
• Gah, I said 27 because of the missed space after 1...
– 9214
Jun 4 at 15:09
• Is this answer valid without func[a b] part? Because it looks like, say, omitting lambda a,b: part in Python, which is invalid (makes the answer a snippet with a and b hardcoded). Jun 9 at 0:12
• @Bubbler Fixed. I think I understand now how function submissions work. Thanks for pointing that out! Jun 9 at 17:54

# Pyth, 6 bytes

u^QGE1


Try it online.

### Explanation

          (implicit: input a to Q)
1    Start from 1.
u   E     b times,
^GQ      raise the previous number to power a.



## Minkolang 0.15, 12 11 bytes

nnDI1-[;]N.


Try it here!

### Explanation

nn             Read two integers from input
D            Pop top of stack and duplicate next element that many times
I1-         Push length of stack, minus 1
[        Pop top of stack and repeat for loop that many times
;       Pop b, a and push a^b
]      Close for loop
N.    Output as number and stop.


## Racket 51 bytes

(define ans 1)(for((i b))(set! ans(expt a ans)))ans


Ungolfed:

(define (f a b)
(define ans 1)
(for((i b))
(set! ans
(expt a ans)))
ans)


Testing:

(f 1 2)
(f 2 2)
(f 5 2)
(f 3 3)


Output:

1
4
3125
7625597484987


# Scala, 45 bytes

Seq.fill(_:Int)(_:Double)reduceRight math.pow


Ungolfed:

(a:Int,b:Double)=>Seq.fill(a)(b).reduceRight(math.pow)


Build a sequence of as with b elements, and apply math.pow from right to left.

# TI-Basic, 19 bytes

Prompt A,B
A
For(C,2,B
A^Ans
End
`