# Calculate the Super-Logarithm

This should be a simple challenge.

Given a number n >= 0, output the super-logarithm (or the log*, log-star, or iterated logarithm, which are equivalent since n is never negative for this challenge.) of n.

This is one of the two inverse functions to tetration. The other is the super-root, which is in a related question.

### Examples

Input       Output
0           0
1           0
2           1
3           2
4           2
...
15          2
16          3
...
3814279     3
3814280     4


### Rules

• You do not need to support decimals, though you may.
• You need to support input of at least 3814280 = ceiling(e^e^e).
• You may not hard-code the values like 3814280. (Your program must theoretically support higher numbers.) I want an algorithm to be implemented.
• Shortest code wins.

Related OEIS

# Jelly, 8 bytes

ÆlÐĿĊḊi1


### Background

We start by successively taking natural logarithms of the input and the subsequent results until the result no longer changes. This works because the extension of the natural logarithm to the complex plane has a fixed point; if z = e-W(-1) ≈ 0.318 + 1.337i – where W denotes the Lambert W function – we have log(z) = z.

For input n, after computing [n, log(n), log(log(n)), …, z], we first apply the ceiling function to each of the results. Jelly's implementation (Ċ) actually computes the imaginary part of complex number instead, but we're not interested in these anyway.

Once the kth application of log yields a value less than or equal to 1, Ċ will return 1 for the first time. The 0-based index of that first 1 is the desired result.

The straightforward implementation (compute 1-based index, decrement) fails because of edge case 0, which does not have a 1 in its list of logarithms. In fact, for input 0, the sequence of logarithms is

[0, None]


This is because Jelly's logarithm (Æl) is overloaded; it first tries math.log (real logarithm), then cmath.log (complex logarithm), and finally "gives up" and returns None. Fortunately, Ċ is similarly overloaded and simply returns it argument if it cannot round up or take an imaginary part.

Likewise, input 1 returns

[1, 0, None]


which may create problems in other approaches that do or do not involve Ċ.

One way to fix this problem is apply Ḋ (dequeue; removes first element) to the array of logarithms. This maps

0ÆlÐĿ -> [0, None]    -> [None]
1ÆlÐĿ -> [1, 0, None] -> [0, None]


so neither list has a 1 now. This way, finding the index of the first 1 will return 0 (not found), which is the desired output for inputs 0 and 1.

### How it works

ÆlÐĿĊḊi1  Main link. Argument: n (non-negative integer)

ÐĿ      Apply the following link until the results are no longer unique.
Æl          Natural logarithm.
Return the array of all unique results.
Ċ     Round all resulting real numbers up to the nearest integer. This takes
the imaginary part of complex numbers and does nothing for non-numbers.
Ḋ    Dequeue; remove the first item (n) of the array of results.


This is one of the only three atoms in Jelly that are overloaded in a non-obvious manner.

# Jelly, 9 bytes

Æl>1$Ð¿L’  Try it online! Test suite. (Slightly modified.) ### Explanation Æl>1$Ð¿L’
Ð¿    while loop, collect all intermediate results.
>1$condition: z>1 Æl body: natural logarithm. L length of the array containing all intermediate results, meaning number of iterations ’ minus one.  # Javascript, 4527 26 bytes l=a=>a>1&&1+l(Math.log(a))  Here is test suite (3rd rev) Thanks @LeakyNun for saving 1 byte with conditional and then converting function to lambda, and @Neil for pointing out false is ok return value for <=1 (changed test to be == instead of ===) • I was doing it without es6, but yeah that would be 1 byte shorter, thanks. – CShark Jul 19 '16 at 20:07 • Why would you not use lambda? – Leaky Nun Jul 19 '16 at 20:08 • no good reason, i just haven't used it that much, so it's not my first instinct – CShark Jul 19 '16 at 20:09 • Apparently we're allowed to return false instead of 0 (as it auto-converts to 0 in an integer expression) in which case you can drop the |0. – Neil Jul 19 '16 at 20:17 • That would save 1 byte, but what do you mean by "it auto-converts to 0"? What is "it"? – CShark Jul 19 '16 at 20:21 # C, 38 bytes f(double n){return n>1?1+f(log(n)):0;}  Pretty self-explanatory. Try it on ideone. # Mathematica, 21 bytes If[#>1,1+#0@Log@#,0]&  Recursive anonymous function. Takes an integer as input and returns its super-logarithm as output. Just uses the given definition. • I actually looked ahead of time to see if there was a built-in. I was surprised when there wasn't. :D – mbomb007 Jul 19 '16 at 20:32 # Dyalog APL, 13 bytes Direct translation of OP: {⍵≤1:0⋄1+∇⍟⍵}  TryAPL online! # Pyth, 10 bytes L&>b1hy.lb  Test suite. This defines a function. • I don't see any output in your test suite. Just a bunch of empty lines in the output. – mbomb007 Jul 19 '16 at 19:13 • @mbomb007 Fixed. – Leaky Nun Jul 19 '16 at 19:19 • Way cooler: tl.u?>N1.l ;-) – Jakube Jul 19 '16 at 21:52 • @Jakube You could post that! – Leaky Nun Jul 20 '16 at 2:53 ## Haskell, 23 bytes l x|x>1=1+l(log x)|1<2=0  Usage example: l 3814280 -> 4. # Python 3, 45 bytes import math s=lambda x:x>1and-~s(math.log(x))  For x <= 1, this returns False (which is == 0 in Python). • Yes, False can be used for 0. – mbomb007 Jul 19 '16 at 20:22 • Also, you beat my naive implementation (by using and rather than if else). Grats. – mbomb007 Jul 19 '16 at 20:24 ## 05AB1E, 16 13 bytes [Dî2‹#¼žr.n]¾  Explanation  # implicit input n [ ] # infinite loop Dî2‹# # break if n rounded up is less than 2 ¼ # else, increase counter žr.n # set next n = log(n) ¾ # push counter and implicitly print  Try it online # J, 211918 16 bytes Saved 2 bytes to Leaky Nun, 1 byte to Galen Ivanov, and 2 bytes to FrownyFrog! 2#@}.(0>.^.)^:a:  Try it online! ## Test cases ls =: >:@$:@^.0:@.(<:&1)
ls 0
0
ls 1
0
ls 2
1
ls 3
2
ls 4
2
ls 15
2
ls 16
3
ls 3814280
4

• Here's my 18 bytes solution: 2#@}.^.^:(0<])^:a: (I started sovling what turned out to be a dup of this problem.) – Galen Ivanov Feb 4 '18 at 10:18
• 2#@}.(0>.^.)^:a: seems to work. – FrownyFrog Feb 4 '18 at 20:19
• Not sure if it’s equivalent though. – FrownyFrog Feb 4 '18 at 20:47

# MATL, 15 12 bytes

0ZetG>~}x@q


Try it online! Or verify all test cases (slightly modified version to handle several inputs).

### How it works

Starting with 0, apply iterated exponentiation until exceeding the input. The output is the number of iterations minus 1.

0       % Push 0
       % Do...while loop
Ze    %   Exponential
t     %   Duplicate
G     %   Push input
>~    %   Is current value less than or equal to the input? If so: next iteration
}       % Finally (code executed at the end of the last iteration)
x     %   Delete
@q    %   Iteration index minus 1
% Implicitly end loop
% Implicitly display stack


# Julia, 17 bytes

!x=x>1&&1+!log(x)


Try it online!

# MATLAB / Octave, 44 bytes

function a=g(n);a=0;if n>1;a=1+g(log(n));end

Tried to do it all as one anonymous function, but I forgot that MATLAB/Octave continues to evaluate expressions even if they are multiplied by a boolean false (zero) value:

f=@(n)(n>1)*(1+f(log(n)))

• Yes, it would be nice to have a short-circuiting product :-) – Luis Mendo Jul 21 '16 at 22:41

# R, 38 37 bytes

f=function(x)if(x>1)1+f(log(x))else 0


Thanks @user5957401 for the extra byte!

Test cases:

> f(0)
[1] 0
> f(1)
[1] 0
> f(2)
[1] 1
> f(3)
[1] 2
> f(4)
[1] 2
> f(3814279)
[1] 3
> f(3814280)
[1] 4

• I think you can save a byte by using a literal if, else statement. i.e. if(x>1)1+f(log(x))else 0 is one byte shorter. – user5957401 Aug 18 '16 at 14:54

# R, 34 bytes

f=pryr::f(if(n>1,1+f(log(n)),0))


Try it online!

A non-recursive approach is possible: 36 bytes and takes input from stdin.

n=scan()
while((n=log(n))>0)F=F+1
+F


# Java 7, 47 bytes

int c(double n){return n>1?1+c(Math.log(n)):0;}


Try it online.

The recursive Java 7 style method above is 2 bytes shorter than an iterative Java 8 style lambda:

n->{int c=0;for(;n>1;c++)n=Math.log(n);return c;}


Try it online.

Explanation:

int c(double n){      // Method with double parameter and integer return-type
return n>1?         //  If the input is larger than 1:
1+                //   Return 1 +
c(Math.log(n))  //   A recursive call with log(input)
:                  //  Else:

n->{                  // Method with double parameter and integer return-type
int c=0;            //  Create a counter, starting at 0
for(;n>1;           //  Loop as long as the input is still larger than 1:
c++)              //   Increase the counter by 1
n=Math.log(n);    //   And update the input to log(input)
return c;}          //  After the loop: return the counter as result

• You might get it shorter with a Java 8 lambda. – mbomb007 Jul 21 '16 at 21:29
• @mbomb007 responding three years later, haha.. (at the time I was only code-golfing in Java 7), but to still answer your question: no, unfortunately a Java 8 lambda is 2 bytes longer than the recursive method. I've added it to my answer, and I've also added an explanation. – Kevin Cruijssen Aug 9 '19 at 11:18
• So you can't do recursive lambdas? – mbomb007 Aug 9 '19 at 14:42
• @mbomb007 No, in Java unfortunately not. In Python, JavaScript, and I think C# .NET as well, recursive lambdas are possible, but in Java not for some reason.. – Kevin Cruijssen Aug 9 '19 at 15:43

## Emacs Lisp, 38 bytes

(defun l(n)(if(> n 1)(1+(l(log n)))0))


Testcases:

(mapcar 'l '(0 1 2 3 4 15 16 3814279 3814280))
;; (0 0 1 2 2 2 3 3 4)


# Jelly, 8 bytes

-Ælß$Ị?‘  Straightforward implementation of the definition. Try it online! or verify all test cases. ### How it works -Ælß$Ị?‘  Main link. Argument: x

Ị    Insignificant; test if |x| ≤ 1.
?   If the result is 1:
-           Return -1.
Else:
$Execute the monadic chain formed by the two links to the left. Æl Apply natural logarithm to x. ß Recursively call the main link. ‘ Increment the result.  ## Perl 5, 35 bytes Very simple, requires -M5.016 (which is free) to enable the __SUB__ keyword for anonymous recursion. sub{$_[0]>1?1+__SUB__->(log pop):0}


Another alternative is

sub{$_[0]>1?1+__SUB__->(log pop):0}  which is 34 bytes, and gives the same output for all inputs > 1, but returns the special false value for inputs <= 1. False is numerically equal to zero, but prints as "" (empty string), so it probably doesn't qualify. • Great answer. You can win 1 byte by doing sub{($_=pop)>1?1+__SUB__->(log):0} though – Dada Jul 22 '16 at 11:09

## CJam (16 bytes)

rd{_1>}{_ml}w],(


Online demo

Simple while loop with pre-condition. (What I really want here is a Golfscript-style unfold operation, but CJam doesn't have one, and floating point in GolfScript is messy and not at all golfy).

• As an aside, this is my 80th answer in math and earned me my second tag badge today. – Peter Taylor Jul 20 '16 at 8:01

# PARI/GP, 24 bytes

Just the straightforward recursion.

f(n)=if(n>1,1+f(log(n)))


Mathematica, 29 bytes

Simple as all hell, and works for comically large as well as negative inputs:

f[x_]:=If[x>1,1+f[Log[x]],0]


# Racket, 61 bytes

(λ(x)(letrec([a(λ(b)(if(> b 1)(+ 1 (a(log b)))0))])(a x)))


# Maple, 32,30 29 bytes

f:=x->if(x>1,1+f(log(x)),0)


Test cases:

> f(0.);
0
> f(1.);
0
> f(2.);
1
> f(3.);
2
> f(4.);
2
> f(3814279.);
3
> f(3814280.);
4


# R, 36 bytes

Slightly different approach from Plannapus

->n;a=0;while(n>1){a=a+1;n=log(n)};a


Uses a right assign to run the code -- so the desired number must precede it. i.e.

10->n;a=0;while(n>1){a=a+1;n=log(n)};a


### Ruby, 29 bytes

l=->n{n<=1?0:1+l[Math.log n]}


# Perl 6, 21 bytes

{($_,*.log...1>=*)-1}  Try it online! The parenthesized expression is a sequence. $_, the argument to the function, is the first element. *.log generates each successive element by taking the log of the previous element. The sequence continues until the ending condition, 1 >= *`, is true: 1 is greater than or equal to the current element. Subtracting 1 from the sequence coerces it to a number: its length.