37
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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.

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  • 8
    \$\begingroup\$ For anyone attempting to find a good sequence, the OEIS has a webcam that selects random sequences. \$\endgroup\$ – Giuseppe Jul 10 at 14:23
  • 1
    \$\begingroup\$ If I claim "the hidden string has length 10 or less", my answer is not cracked, and my hidden string actually has length 8, what is my score? Or is it simply not allowed to claim a larger length than your actual length? \$\endgroup\$ – Luis Mendo Jul 11 at 8:37
  • \$\begingroup\$ @LuisMendo I would probably say that claiming a larger length than your actual length is not allowed. Is there any reason you'd want to? That would probably just make it easier for robbers. \$\endgroup\$ – DJMcMayhem Jul 11 at 19:09
  • \$\begingroup\$ @DJMcMayhem Probably no reason, other than cause confusion. But I agree it's better not to allow that. (The claimed length in my answer exactly corresponds to my hidden string) \$\endgroup\$ – Luis Mendo Jul 12 at 14:09

27 Answers 27

20
\$\begingroup\$

MATL, sequence A005206. Cracked by SamYonnou

voOdoO

Try it online!

The hidden string has 8 bytes.

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  • 28
    \$\begingroup\$ voodoo??? I'm convinced MATL is black magic now \$\endgroup\$ – Giuseppe Jul 10 at 17:22
  • 2
    \$\begingroup\$ Cracked \$\endgroup\$ – SamYonnou Jul 15 at 17:03
8
\$\begingroup\$

Python 2, sequence A138147 (cracked)

print 10

Try it online!

The hidden string is 7 bytes. The sequence goes:

10, 1100, 111000, 11110000, 1111100000, ...
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  • 1
    \$\begingroup\$ Cracked. \$\endgroup\$ – hobbs Jul 11 at 6:46
  • 2
    \$\begingroup\$ Since the challenge specifies that any string up to the given length can be used to crack, this can also be done relatively trivially with just a 2 byte string \$\endgroup\$ – Theo Jul 11 at 14:59
  • 2
    \$\begingroup\$ @Theo how? As I understand it, the string needs to be inserted repeatedly end to end, not nested \$\endgroup\$ – DreamConspiracy Jul 11 at 15:09
  • 1
    \$\begingroup\$ @DreamConspiracy oh you're probably right, I took "If you insert s into the same location in p" to mean that you could nest them. \$\endgroup\$ – Theo Jul 11 at 15:10
7
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Keg, sequence A000045

0.

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

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  • \$\begingroup\$ Correct, the rules say any string up to that length. \$\endgroup\$ – DJMcMayhem Jul 10 at 14:42
  • 1
    \$\begingroup\$ No TIO for Keg :-( \$\endgroup\$ – Luis Mendo Jul 10 at 23:47
  • 1
    \$\begingroup\$ My program was invalid. I rewrote my program. \$\endgroup\$ – A _ Jul 11 at 6:44
  • 1
    \$\begingroup\$ Cracked. \$\endgroup\$ – tomsmeding Jul 11 at 8:10
  • 2
    \$\begingroup\$ @Jono2906 You can ask Dennis to include it. But better wait a few days, based on the comments in the chat room \$\endgroup\$ – Luis Mendo Jul 11 at 11:20
5
\$\begingroup\$

Cracked

Brain-Flak, sequence A000290 (Square numbers)

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

Try it online!

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.

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5
\$\begingroup\$

Java 8+, sequence A010686 (cracked by xnor)

Lambda function.

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

The hidden string is ≤ 5 bytes

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  • 1
    \$\begingroup\$ Cracked? \$\endgroup\$ – xnor Jul 11 at 23:12
  • \$\begingroup\$ That is indeed one of the 2 possible strings I came up with @xnor. The other was ^2<<1 but that's the same thing. \$\endgroup\$ – Benjamin Urquhart Jul 11 at 23:50
4
\$\begingroup\$

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.

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4
\$\begingroup\$

Pyret, sequence A083420, Cracked

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

The hidden string has 4 bytes or fewer.

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  • 1
    \$\begingroup\$ Cracked. You're new here; welcome to Code Golf SE! How are you liking it so far? \$\endgroup\$ – Khuldraeseth na'Barya Jul 11 at 15:14
  • \$\begingroup\$ Nice! Thanks, I’m learning as I go! \$\endgroup\$ – MLavrentyev Jul 11 at 15:26
4
\$\begingroup\$

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!

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  • 1
    \$\begingroup\$ Did you assume the Goldbach conjecture? \$\endgroup\$ – FryAmTheEggman Jul 12 at 20:38
  • \$\begingroup\$ Cracked \$\endgroup\$ – NieDzejkob Jul 12 at 20:43
4
\$\begingroup\$

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.

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4
\$\begingroup\$

V, sequence A000290. Cracked by Cows quack

é*Ø.

Try it online!

The hidden string is 5 bytes.

\$\endgroup\$
  • 2
    \$\begingroup\$ This doesn't print the first term of the sequence A000290 which is 0... \$\endgroup\$ – Expired Data Jul 12 at 9:42
  • 1
    \$\begingroup\$ @ExpiredData Hmm. I could easily edit my answer to start at 0 without affecting the byte count, but there are several other answers that start on the second term too. I don't really want to invalidate those \$\endgroup\$ – DJMcMayhem Jul 12 at 15:04
  • \$\begingroup\$ Cracked in 4 \$\endgroup\$ – Kritixi Lithos Jul 15 at 17:09
3
\$\begingroup\$

Desktop Calculator, sequence A006125

1n

The hidden string ≤ 12 bytes.

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3
\$\begingroup\$

Brain-Flak, sequence A000984 (central binomial coefficients)

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

Try it online!

The hidden string has 36 bytes or fewer.

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3
\$\begingroup\$

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.

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3
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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, ...
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  • 1
    \$\begingroup\$ Nice one! Cracked \$\endgroup\$ – nimi Jul 19 at 11:48
2
\$\begingroup\$

Brain-Flak, sequence A000578 (Cube numbers)

((())<>)

Try it online!

The hidden string is 16 bytes

\$\endgroup\$
  • \$\begingroup\$ cracked \$\endgroup\$ – H.PWiz Jul 11 at 15:05
2
\$\begingroup\$

Cracked

cQuents, sequence A003617

=1#2:pZ

Try it online!

The hidden string is 1 byte.

\$\endgroup\$
2
\$\begingroup\$

Unefunge-98 (PyFunge), sequence A000108

1.@

Try it online!

The hidden sequence is 19 bytes.

\$\endgroup\$
2
\$\begingroup\$

MATL, sequence A000796. Cracked by SamYonnou

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

Try it online!

The hidden string has 3 bytes.

\$\endgroup\$
2
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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.

\$\endgroup\$
  • \$\begingroup\$ Cracked. \$\endgroup\$ – xnor Jul 15 at 8:44
  • \$\begingroup\$ @xnor so fast... T_T \$\endgroup\$ – tsh Jul 15 at 8:47
2
\$\begingroup\$

AsciiDots, 36 bytes, A019523

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

Try it online!

The hidden string has 12 bytes.

\$\endgroup\$
2
\$\begingroup\$

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

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  • \$\begingroup\$ My guess. \$\endgroup\$ – nimi Jul 14 at 16:24
2
\$\begingroup\$

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

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  • 1
    \$\begingroup\$ That VDM-SL link doesn't seem to work. \$\endgroup\$ – Nic Hartley Jul 11 at 20:37
  • \$\begingroup\$ @nichartley it's a link to the language manual as a PDF.. so maybe you're viewing it on something that can't view PDFs \$\endgroup\$ – Expired Data Jul 11 at 20:48
  • 1
    \$\begingroup\$ This works. \$\endgroup\$ – A _ Jul 13 at 4:13
  • 1
    \$\begingroup\$ @UnrelatedString I've added a binding to the set comprehension which should force you to use the 35 byte plan I initially had, also the hd reverse should make it pop the last element. I noticed 2 bytes I could golf from the string so it's now 33. GL if you're still trying! \$\endgroup\$ – Expired Data Jul 17 at 7:00
  • 1
    \$\begingroup\$ 13 bytes :o I'm not very good at this challenge then haha. In VDM you can define consecutive let statements for the same variable and reference the previous one my solution used that if that clue helps. M is a map just to make things more confusing @unrelatedstring \$\endgroup\$ – Expired Data Jul 18 at 6:33
2
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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.)

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2
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ I guess your submission is the first uncracked one! \$\endgroup\$ – Embodiment of Ignorance Jul 26 at 13:42
  • 1
    \$\begingroup\$ Added a solution \$\endgroup\$ – SamYonnou Jul 26 at 20:30
  • 1
    \$\begingroup\$ You should probably also edit the header to say that it's safe \$\endgroup\$ – Unrelated String Jul 27 at 7:05
1
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ That apparently useless b is fairly suspicious… \$\endgroup\$ – Fatalize Jul 17 at 7:13
  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ – Fatalize Jul 17 at 8:56
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Could you provide instructions for running this? I tried running it on TIO but it just errored out instead of printing 2: .code.tio(1,284): error CS0103: The name 'CodeDomProvider' does not exist in the current context .code.tio(1,390): error CS0246: The type or namespace name 'CompilerParameters' could not be found (are you missing a using directive or an assembly reference?) \$\endgroup\$ – Unrelated String Aug 3 at 19:10
  • 1
    \$\begingroup\$ @UnrelatedString You need an assembly reference to System.CodeDom, also it creates files so it won't work on tio \$\endgroup\$ – Embodiment of Ignorance Aug 4 at 2:53
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    \$\begingroup\$ @UnrelatedString Or you can run it with .net framework, so you don't need the assembly reference to System.CodeDom \$\endgroup\$ – Embodiment of Ignorance Aug 4 at 2:54
0
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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.

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