Cops: The Hidden OEIS Substring

Question

This is a Cops and Robbers challenge. This is the cop's thread. The robber's thread is here.

As a cop, you must pick any sequence from the OEIS, and write a program p that prints the first integer from that sequence. You must also find some string s. If you insert s somewhere into p, this program must print the second integer from the sequence. If you insert s + s into the same location in p, this program must print the third integer from the sequence. s + s + s in the same location will print the fourth, and so on and so forth. Here's an example:

Python 3, sequence A000027

print(1)

The hidden string is two bytes.

The string is +1, because the program print(1+1) will print the second integer in A000027, the program print(1+1+1) will print the third integer, etc.

Cops must reveal the sequence, the original program p, and the length of the hidden string s. Robbers crack a submission by finding any string up to that length and the location to insert it to create the sequence. The string does not need to match the intended solution to be a valid crack, nor does the location it's inserted at.

Rules

• Your solution must work for any number in the sequence, or at least until a reasonable limit where it fails do to memory restrictions, integer/stack overflow, etc.

• The winning robber is the user who cracks the most submissions, with the tiebreaker being who reached that number of cracks first.

• The winning cop is the cop with the shortest string s that isn't cracked. Tiebreaker is the shortest p. If there are no uncracked submissions, the cop who had a solution uncracked for longest wins.

• To be declared safe, your solution must stay uncracked for 1 week and then have the hidden string (and location to insert it) revealed.

• s may not be nested, it must concatenated end to end. For example, if s was 10, each iteration would go 10, 1010, 101010, 10101010... rather than 10, 1100, 111000, 11110000...

• It is acceptable to start on the second term of the sequence rather than the first.

• If your sequence has a finite number of terms, going past the last term is allowed to result in undefined behavior.

• All cryptographic solutions (for example, checking the hash of the substring) are banned.

• If s contains any non-ASCII characters, you must also specify the encoding being used.

Answer 1

MATL, sequence A005206. Cracked by SamYonnou

voOdoO

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The hidden string has 8 bytes.

Answer 2

Python 2, sequence A138147 (cracked)

print 10

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The hidden string is 7 bytes. The sequence goes:

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

Answer 3

Keg, sequence A000045

0.

The hidden string is ≤6 bytes(in order to conform the updated cracking rules)

Answer 4

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.

Answer 5

Java 8+, sequence A010686 (cracked by xnor)

Lambda function.

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

The hidden string is ≤ 5 bytes

Answer 6

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.

Answer 7

Pyret, sequence A083420, Cracked

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

The hidden string has 4 bytes or fewer.

Answer 8

V, sequence A000290. Cracked by Cows quack

é*Ø.

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The hidden string is 5 bytes.

Answer 9

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!

Answer 10

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.

Answer 11

Desktop Calculator, sequence A006125

1n

The hidden string ≤ 12 bytes.

Answer 12

Brain-Flak, sequence A000984 (central binomial coefficients)

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

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The hidden string has 36 bytes or fewer.

Answer 13

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.

Answer 14

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, ...

Answer 15

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

Answer 16

Cracked

cQuents, sequence A003617

=1#2:pZ

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The hidden string is 1 byte.

Answer 17

Brain-Flak, sequence A000578 (Cube numbers)

((())<>)

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The hidden string is 16 bytes

Answer 18

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.

Answer 19

MATL, sequence A000796. Cracked by SamYonnou

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

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The hidden string has 3 bytes.

Answer 20

Unefunge-98 (PyFunge), sequence A000108

1.@

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The hidden sequence is 19 bytes.

Answer 21

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)
                       ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Answer 22

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.

Answer 23

AsciiDots, 36 bytes, A019523

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

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The hidden string has 12 bytes.

Answer 24

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.)

Answer 25

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

≜ṗ↔ṗb&w

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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.

Answer 26

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.

Answer 27

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.