The infinite power tower

Question

The challenge

Quite simple, given an input x, calculate it's infinite power tower!

x^x^x^x^x^x...

For you math-lovers out there, this is x's infinite tetration.

Keep in mind the following:

x^x^x^x^x^x... = x^(x^(x^(x^(x...)))) != (((((x)^x)^x)^x)^x...)

Surprised we haven't had a "simple" math challenge involving this!*

Assumptions

x will always converge.
Negative and complex numbers should be able to be handled
This is code-golf, so lowest bytes wins!
Your answers should be correct to at least 5 decimal places

Examples

Input >> Output

1.4 >> 1.8866633062463325
1.414 >> 1.9980364085457847
[Square root of 2] >> 2
-1 >> -1
i >> 0.4382829367270323 + 0.3605924718713857i
1 >> 1
0.5 >> 0.641185744504986
0.333... >> 0.5478086216540975
1 + i >> 0.6410264788204891 + 0.5236284612571633i
-i >> 0.4382829367270323 -0.3605924718713857i
[4th root of 2] >> 1.239627729522762

*(Other than a more complicated challenge here)

I don’t think this tower converges at x = −2 or x = −0.5. — Anders Kaseorg, Commented Jul 25, 2017 at 7:37
@AndersKaseorg I agree, though all programs seem to have the same converging answer. Why don't they converge? — Graviton, Commented Jul 25, 2017 at 7:38
x = −2 gets attracted to a 8-cycle and x = −0.5 gets attracted to a 6-cycle. (My program still gives an answer in these cases, but it’s one of the points in the cycle and not a fixed point; this doesn’t indicate convergence.) — Anders Kaseorg, Commented Jul 25, 2017 at 7:44
@AndersKaseorg Aha very interesting. You wouldn't happen to know why '8' for -2 and '6' for -0.5? Just out of curiosity of course. — Graviton, Commented Jul 25, 2017 at 7:47
You can run the iterations just as easily as I can, but here’s a picture: commons.wikimedia.org/wiki/File:Tetration_period.png — Anders Kaseorg, Commented Jul 25, 2017 at 7:50

Adám · Accepted Answer · 2017-07-25 11:39:53Z

16

APL (Dyalog), 4 bytes

*⍣≡⍨

Try it online!

* power

⍣ until

≡ stable

⍨ selfie

answered Jul 25, 2017 at 11:39

Adám

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The Fifth Marshal · Accepted Answer · 2019-09-02 15:12:43Z

12

Pyth, 4 3 bytes

^{crossed out 4 is still regular 4 ;(}

u^Q

Try it online

How it works

u       first repeated value under repeated application of G ↦
 ^QG        input ** G
    Q   starting at input

edited Sep 2, 2019 at 15:12

The Fifth Marshal

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answered Jul 25, 2017 at 7:25

Anders Kaseorg

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2

\$\begingroup\$ You don't need the last G, it will get auto-filled. \$\endgroup\$
– FryAmTheEggman
Commented Jul 25, 2017 at 13:10
\$\begingroup\$ @FryAmTheEggman Right, thanks! \$\endgroup\$
– Anders Kaseorg
Commented Jul 25, 2017 at 20:14

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ბიმო · Accepted Answer · 2017-07-25 08:18:55Z

7

Haskell, 100 63 bytes

For inputs that don't converge (eg. -2) this won't terminate:

import Data.Complex
f x=until(\a->magnitude(a-x**a)<1e-6)(x**)x

Thanks a lot @ØrjanJohansen for teaching me about until and saving me 37 bytes!

Try it online!

edited Jul 25, 2017 at 8:18

answered Jul 25, 2017 at 7:57

ბიმო

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1

\$\begingroup\$ You can shorten this a lot with the until function. Try it online! \$\endgroup\$
– Ørjan Johansen
Commented Jul 25, 2017 at 8:10
\$\begingroup\$ Neat! Did not know until, thanks a lot. \$\endgroup\$
– ბიმო
Commented Jul 25, 2017 at 8:16

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0xffcourse · Accepted Answer · 2017-07-25 13:13:59Z

7

Python 3, 40 39 35 bytes

Thanks @Ørjan Johansen for a byte: d>99 instead of d==99: 1 more iteration for a lesser byte-count
Thanks @Uriel for 4 bytes: wise utilization of the fact that x**True evaluates to x in x**(d>99or g(x,d+1)). The expression in the term evaluates to True for depth greater than 99 and thus returns the passed value.

Recursive lambda with a max-depth 100 i.e. for a depth 100 returns the same value. Actually is convergency-agnostic, so expect the unexpected for numbers with non-converging values for the function.

g=lambda x,d=0:x**(d>99or g(x,d+1))

Try it online!

edited Jul 25, 2017 at 13:13

answered Jul 25, 2017 at 8:23

0xffcourse

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1

\$\begingroup\$ In the tio link, you can replace complex('j') with 1j \$\endgroup\$
– Mr. Xcoder
Commented Jul 25, 2017 at 8:26
1

\$\begingroup\$ d>99 does one more iteration and is shorter. \$\endgroup\$
– Ørjan Johansen
Commented Jul 25, 2017 at 8:43
1

\$\begingroup\$ save 4 bytes with g=lambda x,d=0:x**(d>99or g(x,d+1)), x**True evaluates to x \$\endgroup\$
– Uriel
Commented Jul 25, 2017 at 12:57
\$\begingroup\$ @Uriel, That is really smart..... Thanks!!! \$\endgroup\$
– 0xffcourse
Commented Jul 25, 2017 at 13:09

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daniero · Accepted Answer · 2017-07-25 19:07:57Z

6

Python 3, 37 30 27 bytes

-7 bytes from @FelipeNardiBatista.
-3 bytes from from @xnor

I don't remember much of Python anymore, but I managed to port my Ruby answer and beat the other Python 3 answer :D

lambda x:eval('x**'*99+'1')

Try it online!

edited Jul 25, 2017 at 19:07

answered Jul 25, 2017 at 12:58

daniero

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1

\$\begingroup\$ FYI, it appears that f-strings were first introduced in Python 3.6: see python.org/dev/peps/pep-0498 . (This would explain why your code didn't work for me in 3.5.2.) Just thought I'd mention this in case anyone else was confused. \$\endgroup\$
– mathmandan
Commented Jul 25, 2017 at 16:01
1

\$\begingroup\$ You don't need to substitute in the value of x, eval('x**'*99+'1') works \$\endgroup\$
– xnor
Commented Jul 25, 2017 at 18:51
\$\begingroup\$ @xnor doh, of course it does :) thanks \$\endgroup\$
– daniero
Commented Jul 25, 2017 at 19:08
\$\begingroup\$ @xnor Neat -- I applied the same thing in my Ruby answer and it somehow fixed it :) \$\endgroup\$
– daniero
Commented Jul 25, 2017 at 19:17
1

\$\begingroup\$ +1, I am slapping myself for forgetting the existence of eval.... :D \$\endgroup\$
– 0xffcourse
Commented Jul 26, 2017 at 5:29

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alephalpha · Accepted Answer · 2017-07-25 07:23:44Z

4

Mathematica, 12 bytes

#//.x_:>#^x&

Takes a floating‐point number as input.

answered Jul 25, 2017 at 7:23

alephalpha

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miles · Accepted Answer · 2017-07-25 08:52:07Z

4

J, 5 bytes

^^:_~

Try it online!

Explanation

First, I'll show what command is being executed after parsing the ~ at the end, and the walk-through will be for the new verb.

(^^:_~) x = ((x&^)^:_) x

((x&^)^:_) x  |  Input: x
      ^:_     |  Execute starting with y = x until the result converges
  x&^         |    Compute y = x^y

answered Jul 25, 2017 at 8:52

miles

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\$\begingroup\$ The J solution is really nice here. To break down your first line in finer grain, is it correct to say that the following happens: (^^:_) creates a new dyadic verb via the power conj, then self adverb ~ makes that verb monadic, so that when given an argument x it's expanded to x (^^:_) x. the left x subsequently "sticks", giving ((x&^)^:_) x per your note, and only the right argument changes during iteration? \$\endgroup\$
– Jonah
Commented Jul 25, 2017 at 14:30
1

\$\begingroup\$ @Jonah Sure, when giving two arguments to a dyad with power, x u^:n y, the left argument is bonded with the dyad to form a monad that is nested n times on y. x u^:n y -> (x&u)^:n y -> (x&u) ... n times ... (x&u) y \$\endgroup\$
– miles
Commented Jul 25, 2017 at 14:36

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Charlie · Accepted Answer · 2017-07-25 11:27:47Z

C# (.NET Core), 79 78 bytes

x=>{var a=x;for(int i=0;i++<999;)a=System.Numerics.Complex.Pow(x,a);return a;}

Try it online!

I chose to iterate until i=999 because if I iterated until 99 some examples did not reach the required precision. Example:

Input:                      (0, 1)
Expected output:            (0.4382829367270323, 0.3605924718713857)
Output after 99 iterations: (0.438288569331222,  0.360588154553794)
Output after 999 iter.:     (0.438282936727032,  0.360592471871385)

As you can see, after 99 iterations the imaginary part failed in the 5th decimal place.

Input:                      (1, 1)
Expected output:            (0.6410264788204891, 0.5236284612571633)
Output after 99 iterations: (0.64102647882049,   0.523628461257164)
Output after 999 iter.:     (0.641026478820489,  0.523628461257163)

In this case after 99 iterations we get the expected precision. In fact, I could iterate until i=1e9 with the same byte count, but that would make the code considerably slower

1 byte saved thanks to an anonymous user.

\$\begingroup\$ +1 For the complex class I didn't even know that existed. \$\endgroup\$
– TheLethalCoder
Commented Jul 25, 2017 at 11:29 — TheLethalCoder, Commented Jul 25, 2017 at 11:29
\$\begingroup\$ @TheLethalCoder neither did I until I googled it. :-) \$\endgroup\$
– Charlie
Commented Jul 25, 2017 at 11:30 — Charlie, Commented Jul 25, 2017 at 11:30

Dennis · Accepted Answer · 2017-07-25 07:15:22Z

3

Jelly, 5 bytes

³*$ÐL

Try it online!

answered Jul 25, 2017 at 7:15

Dennis

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daniero · Accepted Answer · 2017-07-25 22:25:51Z

2

Ruby, 21 20 bytes

->n{eval'n**'*99+?1}

Disclaimer: It seems that Ruby returns some weird values when raising a complex number to a power. I assume it's out of scope for this challenge to fix Ruby's entire math module, but otherwise the results of this function should be correct. Edit: Applied the latest changes from my Python 3 answer and suddenly it somehow gives the same, expected results :)

Try it online!

edited Jul 25, 2017 at 22:25

answered Jul 25, 2017 at 12:33

daniero

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\$\begingroup\$ Take out the space after the eval. \$\endgroup\$
– Value Ink
Commented Jul 25, 2017 at 21:41
\$\begingroup\$ Your original version failed on the complex test case because it evaled the string "0+1i**0+1i**0+1i**...", which parses in the wrong way since ** has higher precedence than +. \$\endgroup\$
– Ørjan Johansen
Commented Jul 26, 2017 at 0:39
\$\begingroup\$ @ØrjanJohansen huh, you're right. I guess I was fooled by the fact that #inspect and #to_s return different values. Before submitting the initial answer I did some testing in irb and saw that e.g entering Complex(1,2) in the REPL would give (1+2i), including the parentheses. When stringifying the value however the parentheses are not included, so the precedence, as you point out, messed it up. \$\endgroup\$
– daniero
Commented Jul 26, 2017 at 2:37
\$\begingroup\$ I thought eval use was forbidden. \$\endgroup\$
– V. Courtois
Commented Jul 26, 2017 at 9:07
\$\begingroup\$ @V.Courtois Ok. But it's not. \$\endgroup\$
– daniero
Commented Jul 26, 2017 at 10:10

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

TI-BASIC, 16 bytes

The input and output are stored in Ans.

Ans→X
While Ans≠X^Ans
X^Ans
End

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jul 25, 2017 at 22:26

kamoroso94

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Giuseppe · Accepted Answer · 2017-07-25 19:24:30Z

1

R, 36 33 bytes

-3 bytes thanks to Jarko Dubbeldam

Reduce(`^`,rep(scan(,1i),999),,T)

Reads from stdin. Reduces from the right to get the exponents applied in the correct order.

Try it (function)

Try it (stdin)

edited Jul 25, 2017 at 19:24

answered Jul 25, 2017 at 18:36

Giuseppe

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1

\$\begingroup\$ scan(,1i) works. Similar to how scan(,'') works. \$\endgroup\$
– JAD
Commented Jul 25, 2017 at 19:20
\$\begingroup\$ @JarkoDubbeldam of course! sometimes my brain doesn't work. \$\endgroup\$
– Giuseppe
Commented Jul 25, 2017 at 19:24

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quant2016 · Accepted Answer · 2017-07-25 21:31:42Z

1

Javascript, 33 bytes

f=(x,y=x)=>(x**y===y)?y:f(x,x**y)

answered Jul 25, 2017 at 21:31

quant2016

1111 bronze badge

\$\begingroup\$ JavaScript doesn't handle imaginary numbers. \$\endgroup\$
– kamoroso94
Commented Jul 25, 2017 at 22:24

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Cinaski · Accepted Answer · 2017-07-28 09:20:54Z

1

MATL, 20 10 bytes

cut down to half thanks to @LuisMendo

t^`Gw^t5M-

Try it online!

This is my first code-golf and my first time using MATL so i'm sure it could be easily outgolfed.

edited Jul 28, 2017 at 9:20

answered Jul 26, 2017 at 7:47

Cinaski

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\$\begingroup\$ Welcome to the site, and nice first answer! A few suggestions: XII is equivalent to t. You can also get rid of XH and H using the automatic clipboard M, that is, ttt^`yw^t5M-]bb-x. And in the last part, instead of deleting the unwanted values you can use &, which tells the implicit display function to only show the top. So, you can use ttt^`yw^t5M-]& and save a few bytes. \$\endgroup\$
– Luis Mendo
Commented Jul 27, 2017 at 15:01
\$\begingroup\$ Also, the first t is not needed, and using G instead of another t you can avoid & and thus leave ] implicit: t^`Gw^t5M-. Hey, we've reduced byte count by a half! \$\endgroup\$
– Luis Mendo
Commented Jul 27, 2017 at 15:09
\$\begingroup\$ @LuisMendo Thanks for the great tips! I have a lot to learn about MATL, but I really like it. \$\endgroup\$
– Cinaski
Commented Jul 28, 2017 at 9:11
\$\begingroup\$ Glad to hear that! \$\endgroup\$
– Luis Mendo
Commented Jul 28, 2017 at 9:11

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emanresu A · Accepted Answer · 2021-08-29 03:51:33Z

1

Vyxal `r`, 4 bytes

‡?eẊ

Try it Online!

answered Aug 29, 2021 at 3:51

emanresu A

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Julian · Accepted Answer · 2021-08-29 21:47:21Z

1

PowerShell Core, 165 140 bytes

param($a)$u=1;for($r=$a;$u-join'+'|iex){$r=[Numerics.Complex]::Pow($a,($p=$r))
$u=($p.Real,($p|% I*),-$r.Real,-($r|% I*))|% *g '.0000000'}$r

Try it online!

-7 bytes thanks to mazzy!

55 bytes without complex numbers

edited Aug 29, 2021 at 21:47

answered Aug 29, 2021 at 8:49

Julian

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1

\$\begingroup\$ [Numerics.Complex] instead [System.Numerics.Complex]? \$\endgroup\$
– mazzy
Commented Aug 29, 2021 at 14:52
1

\$\begingroup\$ Thanks @mazzy ! I followed your advice :) \$\endgroup\$
– Julian
Commented Aug 29, 2021 at 21:19

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Sean · Accepted Answer · 2017-07-25 17:50:06Z

0

Perl 6, 17 bytes

{[R**] $_ xx 999}

Try it online!

R** is the reverse-exponentiation operator; x R** y is equal to y ** x. [R**] reduces a list of 999 copies of the input argument with reverse exponentiation.

answered Jul 25, 2017 at 17:50

Sean

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Stack Exchange Network

The infinite power tower

The challenge

Assumptions

Examples

17 Answers 17

APL (Dyalog), 4 bytes

Pyth, 4 3 bytes

How it works

Haskell, 100 63 bytes

Python 3, 40 39 35 bytes

Python 3, 37 30 27 bytes

Mathematica, 12 bytes

J, 5 bytes

Explanation

C# (.NET Core), 79 78 bytes

Jelly, 5 bytes

Ruby, 21 20 bytes

TI-BASIC, 16 bytes

R, 36 33 bytes

Javascript, 33 bytes

MATL, 20 10 bytes

Vyxal `r`, 4 bytes

PowerShell Core, 165 140 bytes

Perl 6, 17 bytes

Your Answer

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code-golf
math
or ask your own question.

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The infinite power tower

The challenge

Assumptions

Examples

17 Answers 17

APL (Dyalog), 4 bytes

Pyth, 4 3 bytes

How it works

Haskell, 100 63 bytes

Python 3, 40 39 35 bytes

Python 3, 37 30 27 bytes

Mathematica, 12 bytes

J, 5 bytes

Explanation

C# (.NET Core), 79 78 bytes

Jelly, 5 bytes

Ruby, 21 20 bytes

TI-BASIC, 16 bytes

R, 36 33 bytes

Javascript, 33 bytes

MATL, 20 10 bytes

Vyxal r, 4 bytes

PowerShell Core, 165 140 bytes

Perl 6, 17 bytes

Your Answer

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Hot Network Questions

Vyxal `r`, 4 bytes

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