Копи: Прихована підрядок OEIS


37

Це виклик копів та грабіжників. Це нитка копа. У потоці грабіжника тут .

Як поліцейський, ви повинні вибрати будь-яку послідовність з OEIS і написати програму p, яка друкує перше ціле число з цієї послідовності. Ви також повинні знайти деякі рядки s . Якщо ви десь вставите s в p , ця програма повинна надрукувати друге ціле число з послідовності. Якщо ви вставите s + s в те саме місце в p , ця програма повинна надрукувати третє ціле число з послідовності. s + s + s в тому самому місці буде друкувати четверте, і так далі, і так далі. Ось приклад:

Python 3, послідовність A000027

print(1)

Прихований рядок - два байти .

Рядок є +1, оскільки програма print(1+1)надрукує друге ціле число в A000027, програма print(1+1+1)надрукує третє ціле число тощо.

Копи повинні розкрити послідовність, оригінальну програму p та довжину прихованої рядки s . Грабіжники розтріскують подання, знаходячи будь-який рядок до цієї довжини та місце, де його вставити, щоб створити послідовність. Рядок не повинен відповідати призначеному рішенню, щоб бути дійсним тріском, а також розташуванню, в яке він вставлений.

Правила

  • Ваше рішення повинно працювати на будь-яке число в послідовності або принаймні до розумного обмеження, коли воно не справляється із обмеженнями пам’яті, переповненням цілого чи стека тощо.

  • Розбійник-переможець - це той користувач, який тріщить більшість матеріалів, і тим самим той, хто першим дійшов до цієї кількості тріщин.

  • Виграючий коп - це поліцейський з найкоротшим рядком , який не тріщиться. Tiereaker - найкоротший p . Якщо немає неподілених подань, поліцейський, який мав рішення, що не було розірвано, виграє найдовше.

  • Для того, щоб оголосити його безпечним, ваше рішення повинно залишатися нерозкритим протягом 1 тижня, а потім відкрити приховану рядок (і місце для її вставки).

  • s може не бути вкладеним, він повинен об'єднатись у кінець. Наприклад, якщо s був 10, кожна ітерація буде йти , 10, 1010, 101010, 10101010...а не10, 1100, 111000, 11110000...

  • Припустимо починати з другого члена послідовності, а не з першого.

  • Якщо у вашій послідовності є обмежена кількість термінів, проходження минулого терміну може призвести до невизначеної поведінки.

  • Усі криптографічні рішення (наприклад, перевірка хешу підрядки) заборонені.

  • Якщо s містить будь-які символи, що не належать до ASCII, ви також повинні вказати кодування, яке використовується.


8
Для тих, хто намагається знайти гарну послідовність, OEIS має веб-камеру, яка вибирає випадкові послідовності.
Джузеппе

1
Якщо я стверджую, що "прихована рядок має довжину 10 або менше", моя відповідь не тріскається, а моя прихована рядок насправді має довжину 8, яка моя оцінка? Або просто заборонено претендувати на більшу довжину, ніж ваша фактична довжина?
Луїс Мендо

@LuisMendo Я, напевно, можу сказати, що претензії на більшу довжину, ніж ваша фактична довжина, не допускаються. Чи є якась причина, яку ви хочете зробити? Це, мабуть, просто полегшило б грабіжників.
DJMcMayhem

@DJMcMayhem Напевно, немає ніяких причин, крім того, щоб викликати плутанину. Але я згоден, краще цього не допускати. (Заявлена ​​довжина у моїй відповіді точно відповідає моїй прихованій рядку)
Луїс Мендо

Відповіді:



8

Python 2 , послідовність A138147 ( тріщини )

print 10

Спробуйте в Інтернеті!

Прихований рядок - 7 байт . Послідовність йде:

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


2
Оскільки виклик вказує, що будь-який рядок до заданої довжини може бути використаний для розтріскування, це також можна зробити порівняно тривіально лише за допомогою 2-байтної рядки
Тео,

2
@То як? Як я розумію, рядок потрібно вставляти кілька разів у кінці, а не
вкладати

1
@DreamConspiracy ох, ти, мабуть, правий, я взяв "Якщо ви вставите s в те саме місце в p", що означає, що ви можете їх вкласти.
Тео

7

Кег , послідовність A000045

0.

Прихований рядок становить ≤ 6 байт (для відповідності оновленим правилам крекінгу)


Правильно, правила говорять про будь-який рядок до такої довжини .
DJMcMayhem

1
Немає TIO для Keg :-(
Луїс Мендо

1
Моя програма була недійсною. Я переписав свою програму.


2
@ Jono2906 Ви можете попросити Денніса включити його. Але краще почекати кілька днів, грунтуючись на коментарях у чаті
Луїс Мендо





4

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!


1
Did you assume the Goldbach conjecture?
FryAmTheEggman















2

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


1
That VDM-SL link doesn't seem to work.
Nic Hartley

@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
Expired Data

1
This works.
A̲̲

1
@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!
Expired Data

1
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
Expired Data

2

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.)


1
Cracked. Welcome back!
Khuldraeseth na'Barya

2

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.


1
I guess your submission is the first uncracked one!
Embodiment of Ignorance

1
Added a solution
SamYonnou

1
You should probably also edit the header to say that it's safe
Unrelated String

1

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.


1
That apparently useless b is fairly suspicious…
Fatalize


1

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.


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?)
Unrelated String

1
@UnrelatedString You need an assembly reference to System.CodeDom, also it creates files so it won't work on tio
Embodiment of Ignorance

1
@UnrelatedString Or you can run it with .net framework, so you don't need the assembly reference to System.CodeDom
Embodiment of Ignorance

0

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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