Hidden Inversions (Robbers' Thread)

Question

This is a cops-and-robbers puzzle, the cops' thread can be found here.

Your task will be to find an anagram of the provided programs in the cops' thread that performs its left inverse.

Once you crack an answer post the solution as an answer below and notify the original answerer.

You will be scored on the number of programs you are the first to crack.

Answer 1

Python 3, 46 bytes, Lynn

lambda x0223334566789:(x0223334566789*89)//178

Answer 2

Python 2, 225 bytes, orlp

p=90930641353124136621573325641513715557077985927835294018496194596645372722158;q=101979812089012306249375934082966806799688507587087308196267706260111970225882#--223444799
lambda n:pow(n,pow(65537,(p*q-2*(p+q))/4,p*q),~p*~q)

Guess I got lucky after guessing random prime divisors all day...

^{(Default c4.8xlarge spot limit is 4, but I managed to bump it to 10 last year. Had to tweak the FAAS config from 16 slaves to 6 though (+3 mpi, 1 master). 20m polyselect, 12h 50m sieving, 2h 25m linalg, 30m sqrt. Total cost ~$70. At least @orlp was nice enough to pick a solvable size, but I'm not doing this again! Thanks to @IlmariKaronen for the last step, and yes I'm joking about the guessing :P)}

Answer 3

Python 2, 83 bytes, orlp

Original:

#((()))****+,,---/2289;==oppppppqqqqqw~~
lambda n:pow(n,65537,10998167423251438693)

Crack:

p=3207399658;q=3428998126#--11
lambda n:pow(n,pow(65537,(p*q-2*(p+q))/4,p*q),~p*~q)

Try it online!

RSA cracking done by Wolfram Alpha. ;)

Answer 4

Python 3, 80 bytes, Wolfram

from bisect import*
q=int(input())
print(bisect([(h+1)**2 for h in range(q)],q))

This was really hard to crack! I use the bisect library, which is included in the Python 3 distribution. The bisect function takes a sorted list and an element, and returns the rightmost index where the element could be inserted to maintain order. We just give it the length-q list of squares starting from 1 and the element q.

Answer 5

Javascript, 21 bytes, Arnauld

Original

b=>Math.pow(b,torc=3)

Crack

o=>Math.cbrt(o,pbw=3)

Returns the cube root.

Answer 6

7, 9 bytes, ais523

00000000: 0173 dc25 7e13 dcb6 1f                   .s.%~....

Because brute force always wins, and 9! is only 362880

Answer 7

Processing.js, 59 bytes, Kritixi Lithos

Original:

float igetuwebaoli(int p){return p*(((17*-4*-3)))+0+0;}//,,

Crack:

int loabewuteg(float p,i){return (i+0**+0,(p/17/(-4*-3)));}

Well, that was easy enough. The hardest part was figuring out where to stick the extra commas and asterisks. Fortunately, it seems that Processing allows extra unused function parameters as well as C-style comma expressions.

Answer 8

JavaScript (ES6), 63 bytes, SLuck49

Original:

x=>eval(atob`eCp4KzEvLyAgfXBModLS4TvEn4wp1iys9YRRKC85KLIhNMC=`)

Crack:

x=>eval(atob`CgpNYXRoLnBvdyh4LTEsMC41KSAvLw4589CEIKKMRefipyz=`)

The base64 code above decodes to:

Math.pow(x-1,0.5) //...

where the ... stands for a bunch of random garbage that is ignored by the JS interpreter, since it's in a comment.

I found this solution by trial and error. In the end, the only really tricky part were the two newlines at the beginning of the code, needed to make the rest line up properly and to get the M in Math to base64-encode into something that was available in the original character set. I first tried spaces, but " M" base64-encodes into "ICBN" and I needed the only available B to encode ".po" later in the code. "0+M", "1*M", "1?M" or any other similar no-op prefixes I could think of didn't work either, but newlines did.

I suspect this may not be exactly the intended solution, but whatever — it works. :)

Demo:

var f = x=>eval(atob`eCp4KzEvLyAgfXBModLS4TvEn4wp1iys9YRRKC85KLIhNMC=`)
var g = x=>eval(atob`CgpNYXRoLnBvdyh4LTEsMC41KSAvLw4589CEIKKMRefipyz=`)
for (var i = -0; i <= 10; i++) console.log(i, '->', f(i), '->', g(f(i)))

Answer 9

Brain-Flak, 26 bytes, Wheat Wizard

Original (adds 13)

((((()()())){}[()]){}{}{})

Crack (subtracts 13)

([(((()())()){}){}{}](){})

Answer 10

J, 8 bytes, miles

[:]-:[+:

Simple swapping of +: for -: (double for halve).

Answer 11

Javascript, 15 bytes, insertusernamehere

Original

t=>(!!0+~~t+~0)

Crack

t=>(~~!!~0+t+0)

Answer 12

Python 2, 47 bytes, Wheat Wizard

lambda x:sorted(a**2for a in range(x)).index(x)

Answer 13

JavaScript (ES6), 46 bytes, SLuck49

Original (calculates ln(x+1))

x=>Math.log(x+(+String(t=985921996597669)[5]))

Crack

x=>Math[(lg=19979699+55686).toString(9+25)](x)

I never would have cracked this if I hadn't realized that the inverse is a Math built-in. (lg=19979699+55686).toString(9+25) is just a convoluted way of returning "expm1".

Answer 14

J, 10 bytes, miles

1%:@*~>:[<

I have to write something here because the answer is too short.

Answer 15

J, 29 bytes, Zgarb

Original

5#.[:,(3 5&#:(-$]-)7)#.inv"0]

Crack

[:(](07-5)"3 #.-:&#$,)5#.inv]

Try it online!

Another crack equivalent is

[:((3 ]7-5)#.-:&#$,)5#.inv"0]

Explanation

[:(](07-5)"3 #.-:&#$,)5#.inv]  Input: integer n
                            ]  Get n
                      5        The constant 5
                       #.inv   Get the digits of n in base 5
[:(                  )         Operate on those digits D
                    ,            Flatten D (does nothing since it is already a list)
                  #              Get the length of D
               -:&               Halve it
                   $             Reshape D to half its length (only the base 2 digits)
    (07-5)"3                     The constant 2 with rank 3
             #.                  Convert the front-half of D to a decimal from base 2
   ]                             Return the right result