C’est un défi Cops and Robbers. C'est le fil du flic. Le fil du voleur est ici .

En tant que policier, vous devez choisir n’importe quelle séquence de l’ OEIS et écrire un programme p qui affiche le premier entier de cette séquence. Vous devez également trouver des chaînes s . Si vous insérez s quelque part dans p , ce programme doit imprimer le second entier de la séquence. Si vous insérez s + s au même endroit dans p , ce programme doit imprimer le troisième entier de la séquence. s + s + s au même endroit imprimera le quatrième, et ainsi de suite. Voici un exemple:

Python 3, séquence A000027

print(1)

La chaîne cachée est deux octets .

La chaîne est +1, parce que le programme print(1+1)imprimera le deuxième entier dans A000027, le programme print(1+1+1)imprimera le troisième entier, etc.

Les flics doivent révéler la séquence, le programme original p et la longueur de la chaîne cachée s . Les voleurs craquent une soumission en recherchant une chaîne pouvant atteindre cette longueur et l'emplacement où l'insérer pour créer la séquence. Il n'est pas nécessaire que la chaîne corresponde à la solution voulue pour être une fissure valide, ni à l'emplacement où elle est insérée.

Règles

• Votre solution doit fonctionner pour n’importe quel nombre de la séquence, ou au moins jusqu’à une limite raisonnable où elle ne respecte pas les restrictions en matière de mémoire, de dépassement de nombre entier / pile, etc.

• Le voleur gagnant est l'utilisateur qui obtient le plus grand nombre de soumissions, le décisif ayant atteint ce nombre en premier.

• Le flic gagnant est le flic avec la plus courte chaîne s qui ne sont pas fissurés. Tiebreaker est le plus court p . S'il n'y a pas de soumissions non fissurées, le flic qui avait une solution non cassée pour les plus longues victoires.

• Pour être déclaré sûr, votre solution doit rester dans l’écran pendant une semaine, puis la chaîne cachée (et l’emplacement pour l’insérer) seront révélés.

• s ne peut pas être imbriqué, il doit être concaténé bout à bout. Par exemple, si s était 10, chaque itération irait 10, 1010, 101010, 10101010...plutôt que10, 1100, 111000, 11110000...

• Il est acceptable de commencer par le deuxième terme de la séquence plutôt que par le premier.

• Si votre séquence comporte un nombre fini de termes, le fait de dépasser le dernier terme peut entraîner un comportement indéfini.

• Toutes les solutions cryptographiques (par exemple, vérifier le hachage de la sous-chaîne) sont interdites.

• Si s contient des caractères non-ASCII, vous devez également spécifier le codage utilisé.

MATL , séquence A005206 . Fissurée par SamYonnou

voOdoO

Essayez-le en ligne!

La chaîne cachée a 8 octets .

Python 2 , séquence A138147 ( fissuré )

print 10

Essayez-le en ligne!

La chaîne cachée est de 7 octets . La séquence va:

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

Keg , séquence A000045

0.

La chaîne cachée est ≤ 6 octets (afin de se conformer aux règles de craquage mises à jour)

Fissuré

Brain-Flak , séquence A000290 ( Nombres carrés)

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

Essayez-le en ligne!

La chaîne cachée est de 6 octets .

Fait amusant:

J'ai "découvert" cette propriété en jouant à ce jeu basé sur la théorie du cerveau . La chaîne cachée était un élément généré de manière aléatoire que j'ai découvert très utile.

Java 8+, séquence A010686 ( fissurée par xnor )

Fonction lambda.

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

La chaîne cachée est ≤ 5 octets

Python 3 , séquence A096582

C'est vraiment trivial, vu que je n'avais jamais essayé les défis Cops and Robbers.

print(100)

La chaîne cachée est de 3 octets.

Pyret , séquence A083420 , fissurée

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

La chaîne masquée a 4 octets ou moins.

Python 3 , séquence A014092 - ( fissuré )

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

Essayez-le en ligne!

La séquence masquée est de 82 octets .

Mon code prévu (qui ne repose pas sur la conjecture de Goldbach) était:

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 , qui utilise la conjecture de Goldbach pour la résoudre en 42 caractères magiques . Bon travail!

Forth (gforth) , A000042 - ( fissuré )

1 .

Essayez-le en ligne!

La séquence masquée est de 5 octets et peut facilement gérer des centaines de termes.

Une solution sur un octet qui se rompt en raison d'un débordement d'entier est également possible. En fait, je dirais que c'est trivialement embarrassant. Bien que le texte du défi, sous certaines interprétations, puisse vous permettre d'appeler cela une fissure, je vous exhorte à ne pas le faire.

V , séquence A000290 . Craque de vaches

é*Ø.

Essayez-le en ligne!

La chaîne cachée est de 5 octets .

Calculatrice de bureau , séquence A006125

1n

La chaîne cachée ≤ 12 octets .

Flak cérébrale , séquence A000984 (coefficients binomiaux centraux)

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

Essayez-le en ligne!

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.