Este é um desafio de policiais e ladrões. Este é o tópico do policial. o linha do ladrão está aqui .

Como policial, você deve escolher qualquer sequência do OEIS e escrever um programa p que imprima o primeiro número inteiro dessa sequência. Você também deve encontrar algumas strings s . Se você inserir s em algum lugar em p , esse programa deverá imprimir o segundo número inteiro da sequência. Se você inserir s + s no mesmo local em p , este programa deverá imprimir o terceiro número inteiro a partir da sequência.s + s + s no mesmo local imprimirá o quarto, e assim por diante. Aqui está um exemplo:

Python 3, sequência A000027

print(1)

A cadeia oculta é dois bytes .

A cadeia é +1, porque o programa print(1+1)imprimirá o segundo número inteiro em A000027, o programa print(1+1+1)imprimirá o terceiro número inteiro etc.

Os policiais devem revelar a sequência, o programa original p e o comprimento da sequência oculta s . Os ladrões quebram uma submissão encontrando qualquer sequência com esse comprimento e o local para inseri-la e criar a sequência. A sequência não precisa corresponder à solução pretendida para ser um crack válido, nem o local em que está inserida.

Regras

• Sua solução deve funcionar para qualquer número na sequência, ou pelo menos até um limite razoável, onde falha, devido a restrições de memória, excesso de número inteiro / pilha etc.

• O ladrão vencedor é o usuário que racha mais entradas, sendo o desempate quem atinge esse número de rachaduras primeiro.

• O policial vencedor é o policial com a menor corda s que não está quebrada. O desempatador é o menor p . Se não houver envios sem rachaduras, o policial que teve uma solução sem rachaduras por mais vitórias.

• Para ser declarada segura, sua solução deve permanecer sem quebra por 1 semana e, em seguida, ter a sequência oculta (e o local para inseri-la) revelada.

• s não pode ser aninhado, deve concatenar de ponta a ponta. Por exemplo, se s fosse 10, cada iteração seria executada em 10, 1010, 101010, 10101010...vez de10, 1100, 111000, 11110000...

• É aceitável iniciar no segundo termo da sequência e não no primeiro.

• Se sua sequência tiver um número finito de termos, passar do último termo poderá resultar em um comportamento indefinido.

• Todas as soluções criptográficas (por exemplo, verificando o hash da substring) são proibidas.

• Se s contiver caracteres não ASCII, você também deverá especificar a codificação que está sendo usada.

MATL , sequência A005206 . Rachado por SamYonnou

voOdoO

Experimente online!

A cadeia oculta tem 8 bytes .

Python 2 , sequência A138147 ( quebrada )

print 10

Experimente online!

A cadeia oculta é de 7 bytes . A sequência é a seguinte:

10, 1100, 111000, 11110000, 1111100000, ...

Barril , sequência A000045

0.

A cadeia oculta é ≤ 6 bytes (para estar em conformidade com as regras de cracking atualizadas)

Cracked

Brain-Flak, sequence A000290 (Square numbers)

((()))({}<>)

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The hidden string is 6 bytes.

Fun fact:

I "discovered" this property while playing this brain-flak based game. The hidden string was a randomly generated item that I discovered was very useful.

Java 8+, sequence A010686 (cracked by xnor)

Lambda function.

()->System.out.println(1);

The hidden string is ≤ 5 bytes

Python 3, sequence A096582

This is really trivial, as I haven't tried Cops and Robbers challenges before.

print(100)

The hidden string is 3 bytes.

Pyret, sequence A083420, Cracked

fold({(b,e):(2 * b) + 1},1,[list: ])

The hidden string has 4 bytes or fewer.

Python 3, sequence A014092 - (cracked)

from sympy import isprime, primerange
from itertools import count
r=1
print(r)

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The hidden sequence is 82 bytes.

My intended code (which doesn't rely on the Goldbach Conjecture) was:

i=(n for n in count(2)if all(not isprime(n-x) for x in primerange(1,n)))
r=next(i)
#

Cracked by NieDzejkob, who uses the Goldbach Conjecture to solve it in a magical 42 characters. Great job!

Forth (gforth), A000042 - (cracked)

1 .

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The hidden sequence is 5 bytes, and it can easily handle hundreds of terms.

A one-byte solution that breaks due to integer overflow is also possible. In fact, I'd say it is embarrassingly trivial. While the challenge text, under some interpretations, might allow you to call that a crack, I urge you not to.

V, sequence A000290. Cracked by Cows quack

é*Ø.

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The hidden string is 5 bytes.

Desktop Calculator, sequence A006125

1n

The hidden string ≤ 12 bytes.

Brain-Flak, sequence A000984 (central binomial coefficients)

(())({{}}{})

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The hidden string has 36 bytes or fewer.

Python 3, A268575

from itertools import product
S,F,D=lambda*x:tuple(map(sum,zip(*x))),lambda f,s:(v for x in s for v in f(x)),lambda s:{(c-48>>4,c&15)for c in map(ord,s)}
W=D("6@AQUVW")
print(len(W))

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The hidden sequence is 102 bytes.

Haskell, sequence A083318 (cracked)

f=length[2]

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The hidden string is 5 bytes. The sequence goes:

1, 3, 5, 9, 17, 33, 65, 129, 257, 513, ...

Brain-Flak, sequence A000578 (Cube numbers)

((())<>)

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The hidden string is 16 bytes

Cracked

cQuents, sequence A003617

=1#2:pZ

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The hidden string is 1 byte.

Unefunge-98 (PyFunge), sequence A000108

1.@

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The hidden sequence is 19 bytes.

MATL, sequence A000796. Cracked by SamYonnou

'pi'td1_&:_1)Y$J)

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The hidden string has 3 bytes.

Python 3, A008574, cracked by xnor

print(1)

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Insert 4 bytes to complete A008574. A008574: a(0)=1, thereafter a(n) = 4n.

AsciiDots, 36 bytes, A019523

/>1#*$_#\
*^-{+})./
|/\/\/\
\/\/\/\&

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The hidden string has 12 bytes.

Haskell, sequence A014675, cracked by nimi

main=print$uncurry(!!)([2],0)

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The hidden sequence is 35 bytes.

Here's my intended solution:

 main=print$uncurry(!!)$((l:n)->(l>>=flip take[2,1],n+1))([2],0)
                       ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

VDM-SL, sequence A000312

let m={1|->{0}}in hd reverse[x**x|x in set m(1)&x<card m(1)]

The hidden string has 33 bytes or fewer

Haskell, A000045 (Fibonacci) -- Cracked

f = head [0, 1]

I've got a solution with a whopping 23 bytes. I don't expect this to be safe for long, but it was super fun to come up with.

Solution:

I thought Haskell would be a fun language to try this challenge with -- the natural thing is to do to end up adding a function call every time, but if the sequence can't be written recursively in terms of the last term only, you run into some trickiness with Haskell's strictness and function application.
Khuldraeseth na'Barya found a super clever way to do this with an applicative functor. I did something much less brilliant, using where-hacking:

f = head [b,a+b]where[a,b]=[0,1] ^^^^^^^^^^^^^^^^^^
(This is actually 18 bytes. My less-golfed 23 byte version, where I'd totally forgot about pattern matching, used [last a,sum a]where a= instead.)

Java 8+, 1044 bytes, sequence A008008 (Safe)

class c{long[]u={1,4,11,21,35,52,74,102,136,172,212,257,306,354,400,445,488,529,563,587,595,592,584,575,558,530,482,421,354,292,232,164,85,0,-85,-164,-232,-292,-354,-421,-482,-530,-558,-575,-584,-592,-595,-587,-563,-529,-488,-445,-400,-354,-306,-257,-212,-172,-136,-102,-74,-52,-35,-21,-11,-4,-1},v={0,0,0,0,0,0,0,0,0,0,0,0,-1,0,-1,0,-1,0,0,0,0,0,0,0,-1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,-1,0,0,0,0,0,0,0,-1,0,-1,0,-1,0,0,0,0,0,0,0,0,0,0,0,1},w={1,0,0,-1,5};long d=1,e=1;void f(long a,long b){long[]U=u,V=v,W,X;while(a-->0){U=g(U);w=h(v,w);}W=h(v,U);while(b-->0){V=g(V);v=h(v,v);}X=h(V,u);if(w[0]!=v[0]){int i,j,k=0;u=new long[i=(i=W.length)>(j=X.length)?i:j];for(;k<i;k++)u[k]=(k<i?W[k]:0)-(k<j?X[k]:0);d*=e++;}}long[]g(long[]y){int s=y.length,i=1;long[]Y=new long[s-1];for(;i<s;){Y[i-1]=y[i]*i++;}return Y;}long[]h(long[]x,long[]y){int q=x.length,r=y.length,i=0,j;long[]z=new long[q+r-1];for(;i<q;i++)if(x[i]!=0)for(j=0;j<r;)z[i+j]+=x[i]*y[j++];return z;}c(){f(3,0);System.out.println(u[0]/d);}public static void main(String[]args){new c();}}

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Can be solved using a hidden string of size 12. Can definitely be golfed more, but there is no way this is actually winning. I just wanted to contribute out of respect for the number 8008.

Note: before anyone complains that the sequence is hard-coded, I've tested this up to the first term that diverges from the hard-coding (13th term = 307) and it gets it correctly albeit slowly. This is also why it's using long instead of int, otherwise it overflows before that term.

Update (Jul 12 2019): updated to be a bit more performant. Computes the 13th term in 30 seconds on my computer now instead of 5 minutes.

Update (Jul 17 2019): fixed bugs in for loop bounds for the g function, and array length bounds in the bottom of the f function. These bugs should have eventually caused problems, but not early enough to get caught by just checking the output. In either case, since the presence of these bugs 5 days into the game might have confused some people enough into being unable to solve this puzzle, I am totally fine with extending the "safe" deadline until July 24th for this submission.

Update (Jul 18 2019): After some testing I have confirmed that overflows start after the 4th term in the sequence and start affecting the validity of the output after the 19th term. Also in the program as it is written here, each consecutive term takes roughly 5 times longer than the previous to compute. The 15th term takes about 14 minutes on my computer. So actually computing the 19th term using the program as written would take over 6 days.

Also, here is the code with sane spacing/indentation so it is a bit easier to read if people don't have an IDE with auto-formatting on hand.

class c {

  long[] u = {1, 4, 11, 21, 35, 52, 74, 102, 136, 172, 212, 257, 306, 354, 400, 445, 488, 529, 563, 587, 595, 592, 584,
      575, 558, 530, 482, 421, 354, 292, 232, 164, 85, 0, -85, -164, -232, -292, -354, -421, -482, -530, -558, -575,
      -584, -592, -595, -587, -563, -529, -488, -445, -400, -354, -306, -257, -212, -172, -136, -102, -74, -52, -35,
      -21, -11, -4, -1},
      v = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0,
          0, 1, 0, 1, 0, 1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},
      w = {1, 0, 0, -1, 5};

  long d = 1, e = 1;

  void f(long a, long b) {
    long[] U = u, V = v, W, X;
    while (a-- > 0) {
      U = g(U);
      w = h(v, w);
    }
    W = h(v, U);
    while (b-- > 0) {
      V = g(V);
      v = h(v, v);
    }
    X = h(V, u);
    if (w[0] != v[0]) {
      int i, j, k = 0;
      u = new long[i = (i = W.length) > (j = X.length) ? i : j];
      for (; k < i; k++)
        u[k] = (k < i ? W[k] : 0) - (k < j ? X[k] : 0);
      d *= e++;
    }
  }

  long[] g(long[] y) {
    int s = y.length, i = 1;
    long[] Y = new long[s - 1];
    for (; i < s;) {
      Y[i - 1] = y[i] * i++;
    }
    return Y;
  }

  long[] h(long[] x, long[] y) {
    int q = x.length, r = y.length, i = 0, j;
    long[] z = new long[q + r - 1];
    for (; i < q; i++)
      if (x[i] != 0)
        for (j = 0; j < r;)
          z[i + j] += x[i] * y[j++];
    return z;
  }

  c() {
    f(3, 0);
    System.out.println(u[0] / d);
  }

  public static void main(String[] args) {
    new c();
  }
}

Solution

f(1,v[0]=1); right before the System.out.println
The program works by computing the n'th Taylor expansion coefficient at 0. Where the original function is a quotient of polynomials, represented by u and v which I got from here, except that in the linked document the denominator is not multiplied out, and nowhere do they say that you have to compute the Taylor series, I stumbled on that by accident and then confirmed via another source.
The calculation is done via repeated application of the quotient rule for derivatives.
The incorrect first term of v, the entire array w and a few other things like the function f having any arguments are thrown in to mess with people.

Brachylog, 7 bytes (Brachylog SBCS), A114018 (Cracked)

≜ṗ↔ṗb&w

Crack it online!

The string has 2 or fewer bytes.

Fatalize's solution, ẹb, is the original string which I had intended. Note that ẹk also works, for the same reasons. In addition to the issue with 9001 beheading to 001=1, it actually turns out that b on a number just won't fail, because all single digit numbers behead to 0, including 0 itself.

C# (.NET Core), A003678, 29727 bytes (Safe)

using System;using System.Linq;using System.CodeDom.Compiler;class P{static void Main(){Int32 z=0;\u0049nt32 T(\u0049nt32 i){i--;var \u0064="";for(;i>0;\u0069/=5)d=i%5+d;return d.Aggre\u0067a\u0074e(0,(a,b)=>a*5+b%48*2%5)+1;}System.D\u0069agn\u006fs\u0074ics.Pro\u0063ess.\u0053tart(CodeDo\u006dProvi\u0064er.\u0043reateP\u0072ovider("CSharp").Co\u006dpi\u006ceAssembly\u0046romSource(new 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The hidden sequence is 4 bytes or less.

The way the program works is by decoding the long string into a program. The decoded program is then compiled into an executable, which is then run. The executable then creates another program, this time using the CodeDom instead of a long string. Finally, the last program outputs the result. The hidden string is ;8;L, where you insert at index 18504 in the super long string.

Prolog (SWI), 28 bytes, A011557, safe

+ 1.
?- +X,X<2,write(X),X>2.

Try it online!

(I'm not really sure what counts as a full program for Prolog, but this works as a program on TIO.)

The hidden string is 5 bytes or less.

I'm a bit surprised this survived a week... The hidden string is

 + 0.

, inserted after + 1. (note the leading newline). Try it online. Instead of numerically generating a power of ten, this prints one out digit by digit: when backtracking is triggered by the failure of X>2, the only choice point is +X, which goes through every clause of +/1 until execution succeeds or it runs out, executing write(X) (which immediately and imperatively prints without a trailing newline to standard output, so the output can't be undone by backtracking) for every resulting value of X. X<2 is just there to prevent the 1-byte solution 0.