Cheapo Enigma machine (Robbers)

Question

For cops' post, Cheapo Enigma machine (Cops)

A robber's submission consists of a program/function that accepts the output of a cop's code and returns the input for all the outputs provided by that cop's code. (In other words, you must write the inverse function)

You cannot use built-ins that have the sole purpose of hashing or encryption.

Input/Output format

8 bits (0 or 1), or a base-10 integer in range 1-256, 0-255 or -128 to 127. Can use standard I/O or file I/O. Function can also return a value as output. Input and output must belong to the same range (as in, binary, 1-256, 0-255 or -128 to 127), which must also be the same range as that used by the cop.

Scoring

Ratio of the cop's byte count to your byte count. Highest score wins.

You can submit a robber attempt against your own cop code for reference. (Of course this code isn't eligible for winning)

Notification

Please edit the corresponding cop's answer to include your new byte count and the corresponding ratio.

Answer 1

JavaScript by Christoph, 8/25

f=(y,x=0)=>y?f(y/2,x^y):x

(range 0-255)

Sadly, f=(y,x)=>y?f(y/2,x^y):x works for all values except 0.

Technical note

We use y/2 rather than y>>1 to save a byte. This is abusing the fact that any value of y will eventually be rounded to 0 because of arithmetic underflow.

Answer 2

JavaScript by fəˈnɛtɪk, 13/19

x=>((514>>x%2)-x)/2

(range 1-256)

Answer 3

C, by Dave, 64/95 92 85

b,i,e,t[256];r(x){for(;!b;++i,b=e==x)for(srand(i&&e);t[e=rand()%256]++;);return i-1;}

Try it here!

C, shorter version, 64/89 71

i,e,t[256];r(x){for(srand(1);t[e=rand()%256]++||++i,e!=x;);return i-1;}

This one is more implementation specific, but works on TIO. Same length as the PHP solution, I wasn't able to get it any shorter than this.

Try it here!

Answer 4

JavaScript by fəˈnɛtɪk, 13 / 12

x=>x*128%257

Another multiplicative inverse.

Answer 5

JavaScript, 11/13

x=>a.sort().indexOf(x)
for(a=[],i=0;i<256;)a[i]=i++;

Try it online!

Answer 6

Jelly, 22/3 = 7 1/3

⁹ḶDÞḊi

Try it online!

The cop submission by fəˈnɛtɪk was to return the n^th (0-indexed) lexicographically sorted decimal number using the domain [0,255].

I first literally reversed the operation described, ⁹ḶDÞi⁸‘ - takes the lowered range of 256, ⁹Ḷ, and Þ instructs to sort it by a key function of conversion to a decimal list, D; then finds the index of, i, the input, ⁸, and subtracts 1, ‘ (Jelly lists are 1-indexed).

Then I golfed it by dequeuing the sorted list with Ḋ. When an item is not found i returns 0 as required for the removed first element (0), while everything else is found one index earlier, allowing the removal of the decrement, ‘, which in turn gives i implicit input on its right from the left (only) input to the monadic link.

Answer 7

Javascript by Magenta, 32 / 23

x=>x%16*16+(x/16+13)%16

Code basically switches the lower and the upper 4 bits and does a modulo addition on one part.

Answer 8

Javascript by histocrat, 27 / 29

x=>x-65?x-126?x*127%258:131:6

Sadly two hard codings are needed in order to break it. Note that the orginal function doesn't map any value to 130 but maps a value to 256.

Answer 9

PHP, Score 64/71

for(srand(0);$a<256;)$b[$c=rand()%256]++||$d[$c]=$a++;echo$d[$argn]|0;

Luckily PHP's rand just forwards to the stdlib like C. So this works as long as we're using the same stdlib. This means it works on TIO but not on e.g. sandbox.onlinephpfunctions.com. The current version of Dave's code just iterates over a pseudorandom sequence and returns the nth unique value so I guess there might be much shorter answers if a golfing languages also uses the stdlib.

Here's an implementation of Dave's code that does not dependent on the stdlib. That might also help.

Answer 10

Octave, 16/18

@(x)mod(x*171,256)

Try it online!

Answer 11

Jelly, 2/5 (non-competing)

^H$7¡

Try it online to see the whole table.

Answer 12

Jelly, 8/5

^H$7¡

Try it online to see the whole table.

Answer 13

Jelly, 16/6

×171%⁹

Try it online to see the whole table.

Answer 14

Jelly, 11/8

*205%257

Try it online to see the whole table.

In case robber submissions in a different languages are disallowed in the future, I call dibs on the following solution.

g=(x,y=0)=>x-++y**5%257?g(x,y):y

Answer 15

Jelly, 35/7

⁹Ṗ+\%⁹i

Try it online to see the whole table.

Answer 16

C (gcc) by Bijan, 32 / 30

g(x){x=x?g(--x)*205+51&255:0;}

Had a fun time golfing it, thanks ! x= allows to skip return with gcc and tcc (you might want to change your answer to include it). g(--x)*205+51&255 is the inverse.

Answer 17

Jelly, 37/11

O_77×191%⁹Ọ

Uses the same I/O format as the cop. Not sure if that is required.

Try it online to see the whole table.

How it works

For input n, the cop answer computes f(n) := ((((n + 5) % 256 × 2) % 256 + 2) % 256 × 9) % 256. Since % is a linear operator, this is equivalent to f(n) = (((n + 5) × 7 + 2) × 9) % 256. Expanding the right term, we get f(n) = (63n + 333) % 256 = (63n + 77) % 256.

Reversing this is rather straightforward. To undo the addition, we simply have to subtract 77. Also, since 191 × 63 % 256 = 12033 % 256 = 1, it follows that 191 is 63's inverse modulo 256, so multiplying by 191 undoes multiplying by 63. This way, g(n) = (n - 77) × 191 % 256 defines the inverse of f.