Cops and Robbers: Redacted Primality (Cops' Thread)

Question

^{This challenge was originally sandboxed by Magic Octopus Urn; I adopted and posted it with his permission.}

This is the cops' thread. The robbers' thread is here.

The challenge

• Step One: Write a piece of code (function or full program) which checks for primality.
• Step Two: Remove pieces of your code by replacing characters with the symbol █.
• Step Three: Post the redacted code on the cops thread.
• Step Four: Wait for your code to be cracked and try to crack other's code.

For example, the Groovy code {it.isPrime()} could become {██.is█████()}. (This one would be stupidly easy to crack; also, I know, .isPrime() is not a Groovy method.)

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Scoring

You must include your program's score in its submission. The score is defined as the ratio of redacted characters to characters. So if your program had 20 characters and 5 were redacted, your score would be 0.25. The Groovy code above would have a score of 0.5.

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Rules

• Your program only needs to handle positive integers. It should output a truthy value if the number is prime and a falsy value otherwise. Please specify in your answer what it outputs.
• Your code may not contain any comments or unnecessary whitespace.
• No hashing or cryptographic obfuscation.
• Your code may be no more than 50% redacted (at least 1/2 the characters must be shown). This means that the highest possible score is 0.5.
• If your answer is not cracked within a week you may mark it safe and edit in the intended crack.

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Winning

The winner will be the lowest-scoring uncracked answer within two weeks of posting. In the case of a tie, whichever has the most votes will win. This thread is always open to more submissions, but the winner chosen after two weeks will be permanent.

Answer 1

Functoid, score = 14 / 223 ≈ 0.062780 [safe]

Y(yG(BK██)(B(S(BS(C(BC(C(BB(B(v
S(CB█)(█C█B>vK  BSBB())█K(BS(S?
>(KZ)(C(C(Bv>██        >   v██<
█)Sg3I)$; @>B(BS(b(C(BBI)Iv>(█g
())I)))I)IBB(C(b(CB(C())))<v)█C
I))I))0)))(C(BC(B(BB)(C(BBv>)))
]I))))I))>    >)█   $;@   >I)(B

Takes input as command-line argument and outputs True (prime) or False, try it online!

Hint (added 4 days after posting):

The first and 4th █ are a red herring: The intended (and most likely every) solution's IP will follow the first line and reach the ? character.

Solution

Y(yG(BKL2)(B(S(BS(C(BC(C(BB(B(v
S(CBO)( C B>vK  BSBB())OK(BS(S?
>(KZ)(C(C(Bv>O)        >   vY <
^)Sg3I)$; @>B(BS(b(C(BBI)Iv>(yg
())I)))I)IBB(C(b(CB(C())))<v)-C
I))I))0)))(C(BC(B(BB)(C(BBv>)))
]I))))I))>    >)2   $;@   >I)(B

Try it online!

Explanation

Due to the randomness that comes from ? it's not possible to flatten the program. Here's the flat program with a question mark where a random expression will be:

Y(yG(BKL2)(B(S(BS(C(BC(C(BB(B(?(yg(KZ)(C(C(BB(BS(b(C(BBI)I))))(C(BC(b(C(BBI)I)))I))(C-))))I))I))0)))(C(BC(B(BB)(C(BBI)(B]I))))I)))2$;@

Full program:

Y{trial_division}      --  fix-point of {trial_division}
                 2     --  apply 2 (begin division with 2)
                  $    --  apply argument (work with the supplied input)
                   ;   --  print result as boolean
                    @  --  terminate program

The {trial_division}:

y                         -- recursive function with two arguments x,y
 G                        -- | base predicate: x >= y
  (BKL2)                  -- | base function:  BKL2 x y
                             |  ->             K(L2) x y
                             |  ->             L2 y
                             |  ->             2 <= y
        {recursive_call}  -- | recursive call

{recursive_call}, taking arguments f (self-reference), x and y (note 0 is the same as False)

  B (S(BS(C(BC(C(BB(B{divides}I))I))0))) (C(BC(B(BB)(C(BBI)(B]I))))I) f x y
->       (C(BC(C(BB(B{divides}I))I))0) x y (BC(B(BB)(C(BBI)(B]I)))) f I x y)
->       (C(BC(C(BB(B{divides}I))I))0) x y (BC(B(BB)(C(BBI)(B]I)))) f I x y)
->            (C(BB(B{divides}I))I) x y 0  (BC(B(BB)(C(BBI)(B]I)))) f I x y)
->            (C(BB(B{divides}I))I) x y 0  (   B(BB)(C(BBI)(B]I))   f x I y)
->                   {divides}      x y 0  (         C(BBI)(B]I)    f x y  )
->              if x `divides` y then 0 else         C(BBI)(B]I)    f x y
->                                                    f (B]I x)  y
->                                                    f (] x) y
->                                                    f (x+1) y

{divides} is ?(yg(KZ)(C(C(BB(BS(b(C(BBI)I))))(C(BC(b(C(BBI)I)))I))(C-))) where ? is randomly chosen (depending on the random direction) from:

• Y
• S(CBO)(CBO)
• S(SB(KO))(BBSBKO)

These are all equivalent to each other, so {divides} becomes the fix-point of:

y                       -- recursive function with two arguments x,y
 g                      -- | base predicate: x > y
  (KZ)                  -- | base function:  KZ x y
                        -- |  ->              0 == y
      {recursive_call}  -- | recursive call

{recursive_call} is a pretty obfuscated expression that basically just does f x (y-x)

Answer 2

8086 DOS COM, 87 bytes, score 19/87 ~= 0.2183

Cracked by NieDzejkob

1└╣██1█╛ü ¼<█t<< u≈¼<█t█,0|/<██+ô≈ßô☺├δδâ√█|█╞█S█Y╣██9┘t█ë╪1╥≈±○╥t█Aδ∩╞█S█N┤█║S█═!├A
$

This is a COM program; expects number as a command line argument, outputs Y or N. Limit: 65535 because 16 bit processor (sizeof(int) would be 2). Newline is 0x0D 0x0A on this platform. Yes you count 20 █ instead of 19 █. One of them's a real █ and has not been substituted. Muhahaha.

The space in position 10 is actually a NUL byte. The symbol for NUL is the same as space in the old VGA font.

Answer 3

Swift 4, score 26 / 170 ≈ 0.153, safe

func p(n:Int)->Bool{func █(_ █:Int,_ █:Int)->Int{var h=(1...l).map{$0██m██
while(m=h.count,m██).1{h=[Int](h[█...])};return m}
return █>██&(█.███).index█j█n██0)>=█0}█=██l}

Try it online!

Intended Crack

func p(n:Int)->Bool{func j(_ l:Int,_ k:Int)->Int{var h=(1...l).map{$0},m=l
while(m=h.count,m>k).1{h=[Int](h[k...])};return m}
return n>1&&(2..<n).index{j(n,$0)>=$0}==nil}

Ungolfed

func p(n:Int)->Bool{
  func j(_ l:Int,_ k:Int)->Int{    // Modulus function (l mod k)
    var h=(1...l).map{$0},m=l      //  Create an array h of size l
    while(m=h.count,m>k).1{        //  While h has more than k elements:
      h=[Int](h[k...])             //   Remove k elements from h
    }
    return m                       //  Return the length of h (equal to k if l divides k)
  }
  return n>1&&                     // Test if n > 1
  (2..<n).index{j(n, $0)>=$0}==nil //  and no number from 2 to n-1 divides n
}

Answer 4

brainfuck, 37/540 bytes (score: 0.06851) (Cracked by Nitrodon)

>>>>>+>,[>++++++[-<-------->]<+>,]<[-[█<█<]++++++++++<]>[-]>>██[>█>>█>]+[<]<<[<]>█<<+>>[>]█>[>]█+[<]<<[<]>-█>]>>[->]<[-[[<]<]++++++++++<]>[-]>[<█]>]>[>]<[[█]<]<<<<<[<]<<██>>[>]<█[->+<]<█>>[>]<[-[[<]<]++++++++++<]>███>[<<]>[[[>]>████[<]<[-[[<]<]++++++++++<]>[-]>[█<]>]>[>]<[[-]>+[>]<-<[<]<]+<<<<<[<]>[[>]+[[>]>]>[>]>[-<+>]<[<]<[>+[<]>>-<<<<<[[<]<]>>███████>[[█]>]<]<[[<]<]<[█]>]>>>[[>]<->>]]>[[>]>]<<[[[█]<]<]<<<[█]<<█>>>[>]█[-[[<]<]++++++++++<]>>[[>]+[------->++<]>.+.+++++.[---->+<]>+++.>>]>[>]+[------->++<]>++.++.---------.++++.--------.

Try it online!

Prints "prime" if prime, "not prime" if composite. Technically works for arbitrary integers but times out on TIO for numbers above 6000

Answer 5

Python 3, 388 bytes, .155, Cracked

Last-minute crack. Yes, this is the Miller-Rabin test.

I suppose probabilistic tests are allowed, uncertainty 2^-100

Well, a great hint in the previous sentence though

Made return value 0 as COMPOSITE and 1 as PROBABLY PRIME

*368 > 388 : Fixed the problem when z<4

import ██████
def f(z):
 if z<4:return z>>1
 d,s,n,e,c=██z,0,z,0,50
 while not ██1:d//=2;s+=1
 while n>0:n//=2;e+=1
 ███████████()
 while c>0:
  a=0
  while a<2or a>z-█:
   a,b=0,e
   while b>0:a=a*2+██████████████(0,1);b-=█
  x,r=███(█,█,z),██s
  if ██x and x!=██z:
   while r>0:
    x,r=███(█,█,z),██r
    if not ██x:return 0
    elif x==██z:break
   else:return 0
  c-=█
 else:return 1

Solution:

import random
def f(z):
 if z<4:return z>>1
 d,s,n,e,c=~-z,0,z,0,50
 while not d&1:d//=2;s+=1
 while n>0:n//=2;e+=1
 random.seed()
 while c>0:
  a=0
  while a<2or a>z-1:
   a,b=0,e
   while b>0:a=a*2+random.randint(0,1);b-=1
  x,r=pow(a,d,z),~-s
  if ~-x and x!=~-z:
   while r>0:
    x,r=pow(x,2,z),~-r
    if not ~-x:return 0
    elif x==~-z:break
   else:return 0
  c-=1
 else:return 1

Answer 6

Mathematica, 97 bytes, score 0.2989690722 (Cracked)

f[x_]:=(██ToString███████████████;StringMatchQ[████Infinity,RegularExpression@"█\█\█{█\█+, ███"])

Strings! Regex! Primes?

There is such a thing as a primality checking regex, but that's not whats happening here.

This has been cracked, but the way I intended was quite different, so I won't reveal the intended solution yet.

Answer 7

Jelly, score 0.(142857) (cracked)

25██26█966836897364918299█0█1█65849159233270█02█837903312854349029387313█ị██v

Try it online!

Repost of my other answer, this time with a few more bytes revealed to avoid unintended cheats.

Answer 8

095, score 0.20512820512 [Safe]

1id#█#=(DD#█#█{d_█%(█D0█]D}██s]D1.=[1s]

Prints 1 if prime, 0 if composite

Solution:

1id#2#=(DD#2#-{d_.%(rD0R]D}drs]D1.=[1s]

Answer 9

Octave, Score: 0.15 (86 bytes)

I revealed several more characters. I thought the winning criterion was highest score, not lowest.

@(x)eval([(str2num(cell2mat([cellstr(reshape('0█1███1█0█0█00',████))])')█'█')','(x)'])

Try it online!

Good luck =)

Answer 10

Node JavaScript, score: 0.4

Here's where it works. Full program that takes input from first command line argument, and outputs to stdout.

Hopefully, a not-so-difficult solution to start this off.

Using this snippet to calculate score.

require███████████2<<2██>n█████rin█(b████████x)█████(9211█6830)██(a,c)=>a[c[0]██████████(c)]███████);d=███+n;q=████27775██9564,[502█59,█6])[█]);a=██q>0████r(n=q█a█&█-q-██)a██n%q?██0██(137152█8270,22288)(a)

Answer 11

Japt, 19 bytes, 0.315789... score, Safe

I don't know if I obscured more of this than I needed to, costing myself a better score.

█h8575¥█
█UâÊ█Ê█ █2

View solution (Explanation coming soon)

Answer 12

Jelly, score 0.(142857)

25██26█9668368973649182992051██5849159233270202█837903312854349029387313█████

Try it online!

Takes a command-line argument.

False = 0
True = 1

Answer 13

JavaScript, 103 bytes, score 0.1923

x=>{if(x<4)return(!0);for(y=x>>>Math.log10(p=████;--y-1;(p=x/y%1)████if(██&&(███))break████return(███)}

Returns a boolean.

Unintended crack

Answer 14

Brain-Flak, Score: 35/134 = 0.2612 (cracked!)

(({████){██[████)█>(({}))<>}<>{}███{}((██({}))█████{}]██)({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})██[██()██(()█[()]██{}██}{}<>{})

Returns 1 for prime, 0 for composite.

This is a very hard language to try this challenge in, as the formatting is so restricted that it takes effort not to make it obvious what the missing character is.

This is a very hard language to solve this challenge in, as it is ridiculously hard to read.

Answer 15

Javascript, score 0.1894093686354379

let t=[2,3,3,3,3,3,3,5,7,5,7,5,7,7,11,12,13,11,13,13,1,2,17,13,2,3,17,19,23,29,19,19,41,23,23,29,23,"","",29,7,31,31,524,31,37,33,34,41]; function r(a, b) {█████████████████████████████████████████████████████████████};function l(t){let a=0;let b=[];while(true){b.push(t[a]);█████████████;if(!t[a]){return█████};function p(v) {let i=0;let a=r(2,v██);for (i in a){if(v%(█████████a█i██)==0){return false;}};return true;};function f(v){if(l(t).indexOf(v)!=-1){return true;}else{return p(v)};};

Good luck. :p

call f with the prime you want to check.

Answer 16

><>, score 0.096, cracked by Jo King

:1@v>~~:?1n;█$-1<█?=2:}*{█@:$@:

Intended crack:

:1@v>~~:?1n;
$-1<^?=2:}*{%@:$@:

Answer 17

Java 1.4+, 24/145 (0.16551724137)

class X{public static void main(String[]args){System.out.println(new String(████████[Integer.parseInt(args[0])]).matches("█████████████")?███);}}

Try it online!

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Weirdest way I've seen to prime check in Java by far lol.

Answer 18

C, 34/76 = 0.447368, Safe

int p(int n){int r███████2;██r███r++)███+███n;████&███r));return███████n██;}

Having this many blanks means that I will be much more likely to get an unintended crack than the intended one.

Solution:

int p(int n){int r=1,e=n%2;for(;(r++)*(r++)<n;e=e&&(n%r));return e?n>1:n<3;}

explanation:

e holds a boolean value of whether or not the number is not prime (with a few special case exceptions). r iterates through the odd numbers less than or equal to the square root of n. return e?n>1:n<3; handles the special cases when n is 1 or 2.

Answer 19

M, score: 4/22 = .1818..., cracked by Dennis

███“;;█»VOḣ2S⁵++3Ọ;”Pv

This may end up with an unintended crack, we'll have to see. It did.

Dennis' solutions is

ÆPø“;;“»VOḣ2S⁵++3Ọ;”Pv

Try it online!

I will leave my solution hidden for someone to crack. My hint to Dennis on his robber submission was the word "zoo".

Answer 20

Pyth, score: ~0.(461538) (13 bytes) (Cracked)

█]█Q ██Q{█M█P

Try to crack it here!

Answer 21

Pyt, score: 0.288288... [Safe]

Đ2⇹█ŘĐĐŁ███⇹ʀĐ↔Đ5Ș↔⇹██=█ŕĐ↔Đ5Ș↔Đř█⇹█████↔Đ4Ș5Ș⇹██⇹3Ș°04Ș↔█3ȘĐŁ█3Ș05Ș↔█⇹04Ș0↔⇹██=█ŕ↔ŕĐĐŁ██↔██↔ŕŕŕŕ█↔████↔ŕŕŕ██¬¬

Outputs "True" if prime, "False" if not

Forgot to mention that it is a probabilistic test.

Solution:

Đ2⇹⁻ŘĐĐŁ₂`⁻⇹ʀĐ↔Đ5Ș↔⇹Ǥ1=?ŕĐ↔Đ5Ș↔Đř²⇹%∈2*⁻↔Đ4Ș5Ș⇹⁻₂⇹3Ș°04Ș↔+3ȘĐŁ⁺3Ș05Ș↔+⇹04Ș0↔⇹%+=?ŕ↔ŕĐĐŁ⁺⁺↔ł:↔ŕŕŕŕ;↔⁺⁻±?↔ŕŕŕ:;¬¬

This implements the Solovay-Strassen primality test.

Try it online here!

Answer 22

C, 66 bytes, 29 redacted, score 0.439

i;p(n){█████2███████ 0███████2;███;███)if(████)return 0;return 1;}

Just a simple C submission; I'll see how long this one takes before I post a really evil one.

Answer 23

sh + coreutils, score 19 / 143 ~= 0.1328

cracked

e█ec█s█ █c "██████WyAkKHNoIC1jICJg█WNobyBabUZqZEc5eWZIUnlJQ2█2SnlBblhHNG5m██JoYVd3Z0t6SjhkMk1nTFhjSyB8YmFzZTY0IC1kYCIpIC1lcSAxIF0K█b█se6███d`"

TIO

Answer 24

Brain-Flak, score 29 / 140 = 0.207

({}██()██<>){██({}[()])██{}{}███({<({}[()])><>({})<>}{}██████{}██){(({})){({}[()])<>}{}}<>([{}()]{}<>{})<>}(<>██{}({}████)((){[()]██{}██}{})

Try it online!

Outputs 1 for prime and 0 for non-prime.

Answer 25

Tampio (imperative), score: 24/51 = 0.5

Luku on alkuluku,jos ████████████e███████ on █████.

This is an obvious solution, I hope no one here understands Finnish.

Answer 26

Tampio (imperative), score: 26/223 = 0.11659...

Luvun kokonaislukuarvot ovat riippuen siitä,onko se yksi,joko listan,jonka alkioita ovat yksi █████████████████████,alkiot tai █████ liitettynä sen alkutekijöihin.Luku on alkuluku,jos sen kokonaislukuarvojen summa on nolla.

Answer 27

Python 3, score: 0.386363, cracked

p=lambda x,i=2:█████or(x%i and ████████)████

Going for the really low hanging fruit at first. I'll come up with a cheeky answer soon.

user71546 made it "work" with

p=lambda x,i=2:i>=x or(x%i and p(x,i+1))or 0

...but that was unintended. Original code was

p=lambda x,i=2:i>x/2or(x%i and p(x,i+1))or 0

Neither work for x<2, turns out. Oops.

Answer 28

Ruby, 27/73 = 0.369863

def p n;███████(██n.times████a[2..-1].map{|s|█.██n████s}██.█*█|██})█);end

This should be fun.

Answer 29

JavaScript (ES7), 297 bytes, 103 redacted, .347

M=(N,X=N,F=(n,a=█████)=>a>1e-20?█████+F(n,█████████):1,G=(n,a=█████)=>a>1e-20?█████+G(n,███████):n==2?0:G(n-1),H=(n,a=█████)=>a>1e-20?█████-H(n,███████):0,I=n=>████████I(████),J=n=>I(n)*████+H(█████████-1),K=(n,l=n|0)=>(n-l>=.5)+l,L=(a,b)=>██████████(a)+█(b)████,Z=L(--N,N)██)=>L(Z,████M(N,X)██)██

My previous Python answer was too straightforward, so here's an evil one ;)

The logic behind is straightforward though.