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Wouldn't it be neat if programming functions could be inverted, just like the mathematical function they implement?

Write a function (or program) that takes one input x in any form, that outputs ln(x).
When the bytes of the program are reordered/reversed so that the first byte now is the last byte, it should take one input x in any form and output e^x instead.

  • Your answer must have at least 3 correct significant figures.
  • Approximations are fine, as long as they have at least 3 correct significant figures.
  • Your code must be in the same programming language both forwards and backwards.

Let's say this program implements ln(x):

abc你好

Then this program has to implement e^x:

\xBD\xA5\xE5\xA0\xBD\xE4cba

Gold star if you use a language without float support.

This is a weird form of code-golf, so the shortest program wins.

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  • 4
    \$\begingroup\$ "Wouldn't it be neat if programming functions could be inverted, just like the mathematical function they implement?" Some languages (e.g. J and Mathematica) can actually do this for some functions. \$\endgroup\$ Commented Feb 6, 2016 at 11:29
  • \$\begingroup\$ Additionally, K2 could approximate an inverse for an arbitrary monadic pure function via its "function inverse" overload of dyadic and triadic ?, which used the secant method. \$\endgroup\$
    – JohnE
    Commented Feb 6, 2016 at 15:19
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    \$\begingroup\$ "at least 3 correct significant figures" - over what range? \$\endgroup\$
    – TLW
    Commented Feb 6, 2016 at 16:48
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    \$\begingroup\$ I realize it's far too late now, but I think this would have been a really nice challenge had comments been disallowed. \$\endgroup\$
    – Alex A.
    Commented Feb 6, 2016 at 21:45
  • \$\begingroup\$ I actually thought of that when I came up with this challenge @AlexA. but forgot about it while writing the post :P Also that would've made "normal" languages like java, c++ etc basically impossible. \$\endgroup\$
    – Drathier
    Commented Feb 7, 2016 at 12:59

19 Answers 19

75
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Haskell, 11 bytes

f=log
pxe=f

and in reverse order:

f=exp
gol=f

This works without the "comment" trick. Instead each version defines an additional, but unused function (pxe/ gol).

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    \$\begingroup\$ +1 for gol=f. \$\endgroup\$ Commented Feb 6, 2016 at 13:22
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    \$\begingroup\$ This is also a valid solution in Julia. \$\endgroup\$
    – Rainer P.
    Commented Feb 6, 2016 at 23:28
42
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APL, 3 bytes

*⊣⍟

This is a function train. Monadic * returns e^x, monadic returns ln(x). is a dyadic function that returns its left argument. Thus, *⊣⍟ is equivalent to just *, and the reverse ⍟⊣* is equivalent to just .

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22
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Jelly, 5 4 bytes

Yay, my first Jelly answer. :) Input is via command-line argument.

Jelly has its own code page so each character is one byte.

eÆÆl

Try it online!

Reversed:

lÆÆe

Try it online!

Explanation

The Æ on its own is an unrecognised token, so it acts the same as a linefeed. That means in either case the main link is only Æl or Æe which is the 2-character built-in for exp() or ln() and is by default performed on the first command-line argument.

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9
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Javascript, 18 bytes

Math.log//pxe.htaM
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6
  • \$\begingroup\$ Don't you need a return() or console.log() around it? \$\endgroup\$
    – AAM111
    Commented Feb 6, 2016 at 20:32
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    \$\begingroup\$ @OldBunny2800 It evaluates to a function, which should be permissible. \$\endgroup\$
    – Neil
    Commented Feb 6, 2016 at 20:50
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    \$\begingroup\$ Math.ln||pxe.htaM will probably also work. \$\endgroup\$ Commented Feb 7, 2016 at 12:35
  • \$\begingroup\$ @SuperJedi224 Thanks, that helped me spot the error in my answer! \$\endgroup\$
    – Neil
    Commented Feb 7, 2016 at 12:50
  • \$\begingroup\$ @Neil I hadn't even noticed that \$\endgroup\$ Commented Feb 7, 2016 at 18:09
7
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Seriously, 5 bytes

,_.e,

Input, ln, output, then exp on an empty stack (does nothing), and input (does nothing since input is exhausted). Try it online!

Reversed:

,e._,

Try it online!

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1
  • \$\begingroup\$ The language name is hilarious \$\endgroup\$
    – Wasif
    Commented Mar 22, 2021 at 10:02
6
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Mathematica, 19 bytes

1&#@pxE+0&0+Log@#&1

Reversed:

1&#@goL+0&0+Exp@#&1

This was interesting to golf! Mathematica has no line comments / implicit string endings, so I couldn't take the simple route. Instead, I used the fact that 0 + x == x, 0 x == 0, and that 1 x == x, no matter what x is! Testing:

In[1]:= (1&#@pxE+0&0+Log@#&1)[x]

Out[1]= Log[x]

In[2]:= (1&#@goL+0&0+Exp@#&1)[x]

         x
Out[2]= E
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0
5
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Julia, 7 bytes

log#pxe

This is an anonymous function. Assign it to a variable to call it. Evaluates to builtins log or exp plus a comment.

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    \$\begingroup\$ Same answer works for R \$\endgroup\$
    – Dason
    Commented Feb 6, 2016 at 20:30
4
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Python2, 73 bytes

io: stdin/stdout

from math import*;print log(input())#))(tupni(pxe tnirp;*tropmi htam morf

inverse:

from math import*;print exp(input())#))(tupni(gol tnirp;*tropmi htam morf
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1
  • \$\begingroup\$ You can shave 10 characters off by using __import__("math"). instead of \$\endgroup\$
    – TLW
    Commented Feb 6, 2016 at 16:43
3
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CJam, 11 bytes

rdmle#eemdr

Test it here.

Reversed:

rdmee#elmdr

Test it here.

Basically the same comment-trick as the OP's Python answer. e# starts a comment. rd reads the input and ml or me computes the logarithm or exponential.

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Brachylog, 3 bytes

*₁≡

Try it online!

Initially, I had hoped to use ~*, but although *~ computes e^x and successfully ignores the trailing tilde, ~* fails for all integer inputs and hits a float overflow on most non-integer inputs.

Forwards:

       The output
  ≡    is
*₁     the natural logarithm of
       the input.

Backwards:

       The output is
  *    Euler's number to the power of
       the input
≡      passed through the identity predicate
 ₁     with an extraneous subscript.

This uses the identity predicate because, although trailing tildes are tolerated, leading subscripts are not. (If they were, the Brachylog answer would be *₁ alone, which is just the normal builtin for natural log.)

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2
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Vitsy, 5 bytes

This is a program that exits on an error.

EL^rE
E   E  Push java.lang.Math.E
 L     Push log_(top) (input) (ln(input))
  ^    Push (top)^(input)  (e^(input))
   r   Reverse the stack

This program exits on an error with ln(input) on the stack.

Try it online! (note that I have put N to have visible output)

Then it's inverse:

Er^LE

This program exits on an error with e^(input) on the stack.

Try it online!

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2
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Fuzzy Octo Guacamole, 7 bytes

non-competing, FOG is newer than the challenge

EZO@pZE

This is the equivalent of a function in FOG. It assumes the input is on the stack. This can be assigned to a function by the code "EZO@pZE""f"o, where f is any single-char name you want to assign. Then use it like any other command. Example: "EZO@pZE"'f'o^f.

Explanation:

EZO@pZE
E       # Push E (2.718....)
 Z      # Reverse stack (it is now [e, input])
  O     # log(x, y) which is ln(input)
   @    # Exit. (implicit output) Nothing after this gets run.
    p   # x^y (power function)
     Z  # Reverse stack
      E # Push E.

Reversed:

EZp@OZE
E       # Push E (2.718....)
 Z      # Reverse stack (it is now [e, input])
  O     # x^y (power function)
   @    # Exit. (implicit output) Nothing after this gets run.
    p   # log(x, y) which is ln(input)
     Z  # Reverse stack
      E # Push E.
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1
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Matl, 5 bytes

Yl%eZ

Yl: log Ze: exp %: comment

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1
1
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Pyth, 12 bytes

Finds ln(input())

.lQ) " Q1n.^

Finds e^input()

^.n1Q " )Ql.

Spaces stop implicit printing of strings, each version calculates it then creates a string with the remaining characters.

ln(x) mode here

e^x mode here

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1
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𝔼𝕊𝕄𝕚𝕟, 8 chars / 10 bytes

МŬï//ïŦМ

Try it here (Firefox only).Try reverse here (Firefox only).

Just 2 builtins separated by a comment.

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1
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Jolf, 9 bytes

Program 1: exp of input

amoj"jOma
a         print
 moj      e^j
    "jOma  the rest of the line is captured as a string; implicit printing is restricted.

Program 2: ln of input

amOj"joma
a         print
 mOj      ln(j)
    "joma  the rest of the line is captured as a string; implicit printing is restricted.

Bonus points for being a case-insensitive palindrome? Try it here!

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1
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J, 8 bytes

The natural logarithm is ^., and exponential ^. The problem is, . can only modify a valid verb, otherwise, a spelling error will occur. Thus, we can't use the left argument trick in the APL answer, becuase ^.[^ would cause an error when reversed, as ^[.^ creates an invalid verb. So, we must use comments; but NB. is so long :( Fortunately, they both end with ., so&ldots; there's that.

Logarithm:

^.NB.BN^

Exponential:

^NB.BN.^

You can enter them for yourself online!

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1
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Runic Enchantments, 9 bytes

i'lA@Ae'i

Try it online!

An ungodly uninteresting program. @ insures termination of the implied entry point at the left, everything after is unexecuted. I tried really hard to re-use the ' or A instructions, but to no avail, even at larger program sizes. The required explicit entry point for multi-line programs essentially precludes it.

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0
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Java 8, 198 182 30 bytes

d->Math.log(d)//)d(pxe.htaM<-d

Try it online.

and reversed:

d->Math.exp(d)//)d(gol.htaM<-d

Try it online.

Uses the comment trick (//) with built-ins for Math.log and Math.exp.

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