Este es un desafío de policías y ladrones. Este es el hilo del policía. El hilo del ladrón está aquí .

Como policía, debe elegir cualquier secuencia del OEIS y escribir un programa p que imprima el primer entero de esa secuencia. También debes encontrar algunas cadenas s . Si inserta s en algún lugar de p , este programa debe imprimir el segundo entero de la secuencia. Si inserta s + s en la misma ubicación en p , este programa debe imprimir el tercer entero de la secuencia. s + s + s en la misma ubicación imprimirá el cuarto, y así sucesivamente. Aquí hay un ejemplo:

Python 3, secuencia A000027

print(1)

La cadena oculta es de dos bytes .

La cadena es +1, porque el programa print(1+1)imprimirá el segundo entero en A000027, el programa print(1+1+1)imprimirá el tercer entero, etc.

Los policías deben revelar la secuencia, el programa original p y la longitud de la cadena oculta s . Los ladrones descifran una presentación al encontrar cualquier cadena de esa longitud y la ubicación para insertarla para crear la secuencia. La cadena no necesita coincidir con la solución prevista para ser una grieta válida, ni tampoco la ubicación en la que se inserta.

Reglas

• Su solución debe funcionar para cualquier número en la secuencia, o al menos hasta un límite razonable en el que falla en restricciones de memoria, desbordamiento de enteros / pila, etc.

• El ladrón ganador es el usuario que descifra la mayor cantidad de envíos, siendo el desempate el que alcanzó ese número de grietas primero.

• El policía que gana es el policía con la cadena más corta s que no está agrietada. Tiebreaker es el más corto p . Si no hay envíos no descifrados, el policía que tenía una solución descifrada para ganar más tiempo.

• Para ser declarado seguro, su solución debe permanecer sin descifrar durante 1 semana y luego debe revelar la cadena oculta (y la ubicación para insertarla).

• s no puede estar anidado, debe concatenarse de extremo a extremo. Por ejemplo, si s fuera 10, cada iteración iría en 10, 1010, 101010, 10101010...lugar de10, 1100, 111000, 11110000...

• Es aceptable comenzar con el segundo término de la secuencia en lugar del primero.

• Si su secuencia tiene un número finito de términos, pasar el último término puede generar un comportamiento indefinido.

• Todas las soluciones criptográficas (por ejemplo, verificar el hash de la subcadena) están prohibidas.

• Si s contiene caracteres no ASCII, también debe especificar la codificación que se está utilizando.

MAT , secuencia A005206 . Agrietado por SamYonnou

voOdoO

Pruébalo en línea!

La cadena oculta tiene 8 bytes .

Python 2 , secuencia A138147 ( agrietada )

print 10

Pruébalo en línea!

La cadena oculta es de 7 bytes . La secuencia va:

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

Barrilete , secuencia A000045

0.

La cadena oculta es ≤ 6 bytes (para cumplir con las reglas de craqueo actualizadas)

Agrietado

Brain-Flak , secuencia A000290 (números cuadrados)

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

Pruébalo en línea!

La cadena oculta es de 6 bytes .

Hecho de la diversión:

"Descubrí" esta propiedad mientras jugaba este juego basado en el cerebro . La cadena oculta era un elemento generado aleatoriamente que descubrí que era muy útil.

Java 8+, secuencia A010686 ( descifrado por xnor )

Función lambda.

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

La cadena oculta es ≤ 5 bytes

Python 3 , secuencia A096582

Esto es realmente trivial, ya que no he probado los desafíos de policías y ladrones antes.

print(100)

La cadena oculta es de 3 bytes.

Pyret , secuencia A083420 , Agrietada

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

La cadena oculta tiene 4 bytes o menos.

Python 3, sequence A014092 - (cracked)

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

Try it online!

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 .

Try it online!

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.

Try it online!

The hidden string is 5 bytes.

Desktop Calculator, sequence A006125

1n

The hidden string ≤ 12 bytes.

Brain-Flak, sequence A000984 (central binomial coefficients)

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

Try it online!

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))

Try it online!

The hidden sequence is 102 bytes.

Haskell, sequence A083318 (cracked)

f=length[2]

Try it online!

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)

((())<>)

Try it online!

The hidden string is 16 bytes

Cracked

cQuents, sequence A003617

=1#2:pZ

Try it online!

The hidden string is 1 byte.

Unefunge-98 (PyFunge), sequence A000108

1.@

Try it online!

The hidden sequence is 19 bytes.

MATL, sequence A000796. Cracked by SamYonnou

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

Try it online!

The hidden string has 3 bytes.

Python 3, A008574, cracked by xnor

print(1)

Try it online!

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

AsciiDots, 36 bytes, A019523

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

Try it online!

The hidden string has 12 bytes.

Haskell, sequence A014675, cracked by nimi

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

Try it online!

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();}}

Try it online!

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.