Poliziotti: The Hidden OEIS Substring

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Questa è una sfida di poliziotti e ladri. Questo è il filo del poliziotto. Il filo del ladro è qui .

Come poliziotto, devi scegliere qualsiasi sequenza dall'OEIS e scrivere un programma p che stampa il primo intero da quella sequenza. È inoltre necessario trovare qualche stringa s . Se si inserisce s da qualche parte in p , questo programma deve stampare il secondo numero intero dalla sequenza. Se si inserisce s + s nella stessa posizione in p , questo programma deve stampare il terzo intero dalla sequenza. s + s + s nella stessa posizione stamperà il quarto, e così via e così via. Ecco un esempio:

Python 3, sequenza A000027

print(1)

La stringa nascosta è di due byte .

La stringa è +1, poiché il programma print(1+1)stamperà il secondo numero intero in A000027, il programma print(1+1+1)stamperà il terzo numero intero, ecc.

I poliziotti devono rivelare la sequenza, il programma originale p e la lunghezza della stringa nascosta s . I ladri decifrano un invio trovando qualsiasi stringa fino a quella lunghezza e la posizione in cui inserirla per creare la sequenza. La stringa non deve necessariamente corrispondere alla soluzione prevista per essere un crack valido, né la posizione in cui è inserita.

Regole

  • La soluzione deve funzionare per qualsiasi numero nella sequenza, o almeno fino a un limite ragionevole in cui non riesce a limitare la memoria, overflow di interi / stack, ecc.

  • Il rapinatore vincente è l'utente che ha la maggior parte delle richieste, con il tiebreaker che ha raggiunto per primo quel numero di crepe.

  • Il poliziotto vincente è il poliziotto con la stringa più corta s che non è spezzata. Tiebreaker è il p più breve . Se non ci sono invii non crackati, il poliziotto che aveva una soluzione non crackata per le vittorie più lunghe.

  • Per essere dichiarata sicura, la soluzione deve rimanere senza crack per 1 settimana e quindi mostrare la stringa nascosta (e la posizione per inserirla).

  • s non può essere nidificato, deve essere concatenato end-to-end. Ad esempio, se s è stato 10, ogni iterazione andrebbe 10, 1010, 101010, 10101010...piuttosto che10, 1100, 111000, 11110000...

  • È accettabile iniziare dal secondo termine della sequenza anziché dal primo.

  • Se la sequenza ha un numero finito di termini, andare oltre l'ultimo termine può comportare un comportamento indefinito.

  • Tutte le soluzioni crittografiche (ad esempio il controllo dell'hash della sottostringa) sono vietate.

  • Se s contiene caratteri non ASCII, è necessario specificare anche la codifica utilizzata.

code-challenge  sequence  integer  cops-and-robbers 
— DJMcMayhem
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8
Per chiunque tenti di trovare una buona sequenza, l'OEIS ha una webcam che seleziona sequenze casuali.
— Giuseppe,
1
Se pretendo che "la stringa nascosta abbia una lunghezza di 10 o meno", la mia risposta non è incrinata e la mia stringa nascosta in realtà ha una lunghezza di 8, qual è il mio punteggio? O semplicemente non è consentito rivendicare una lunghezza maggiore di quella effettiva?
— Luis Mendo,
@LuisMendo Probabilmente direi che non è consentito rivendicare una lunghezza maggiore di quella effettiva. C'è qualche motivo che vorresti? Ciò probabilmente renderebbe più semplice per i ladri.
— DJMcMayhem
@DJMcMayhem Probabilmente nessun motivo, oltre a causare confusione. Ma sono d'accordo che è meglio non permetterlo. (La lunghezza richiesta nella mia risposta corrisponde esattamente alla mia stringa nascosta)
— Luis Mendo,

Risposte:

8

Python 2 , sequenza A138147 ( crackizzato )

print 10

Provalo online!

La stringa nascosta è di 7 byte . La sequenza va:

10, 1100, 111000, 11110000, 1111100000, ...
— XNOR
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1
— Hobbs
2
Poiché la sfida specifica che qualsiasi stringa fino alla lunghezza data può essere utilizzata per craccare, questo può anche essere fatto in modo relativamente banale con solo una stringa da 2 byte
— Theo
2
@Theo come? A quanto ho capito, la stringa deve essere inserita ripetutamente end-to-end, non nidificata
— DreamConspiracy
1
@DreamConspiracy oh, probabilmente hai ragione, ho preso "Se inserisci s nella stessa posizione in p" per significare che potresti annidarli.
— Theo
7

Barilotto , sequenza A000045

0.

La stringa nascosta è ≤ 6 byte (per conformare le regole di cracking aggiornate)

— UN
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Corretto, le regole dicono qualsiasi stringa fino a quella lunghezza .
— DJMcMayhem
1
Nessun TIO per Keg :-(
— Luis Mendo il
1
Il mio programma non era valido. Ho riscritto il mio programma.
— A̲̲
1
— tomsmeding
2
@ Jono2906 Puoi chiedere a Dennis di includerlo. Ma meglio aspettare qualche giorno, in base ai commenti nella chat room
— Luis Mendo l'
4

Python 3 , sequenza A014092 - ( incrinato )

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

Provalo online!

La sequenza nascosta è di 82 byte .

Il mio codice previsto (che non si basa sulla congettura di Goldbach) era:

i=(n for n in count(2)if all(not isprime(n-x) for x in primerange(1,n)))
r=next(i)
#

Incrinato da NieDzejkob , che usa la congettura di Goldbach per risolverlo in 42 magici personaggi. Ottimo lavoro!

— agtoever
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1
Hai assunto la congettura di Goldbach?
— FryAmTheEggman,
— NieDzejkob,
4

Forth (gforth) , A000042 - ( crackizzato )

1 .

Provalo online!

La sequenza nascosta è 5 byte e può gestire facilmente centinaia di termini.

È anche possibile una soluzione a un byte che si interrompe a causa di un overflow di numeri interi. In effetti, direi che è imbarazzantemente banale. Mentre il testo della sfida, sotto alcune interpretazioni, potrebbe permetterti di definirlo una fessura, ti esorto a non farlo.

— NieDzejkob
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Cracked. Mi è piaciuto questo.
— Khuldraeseth na'Barya
4

V , sequenza A000290 . Incrinato dal ciarlatano delle mucche

é*Ø.

Provalo online!

La stringa nascosta è di 5 byte .

— DJMcMayhem
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2
Questo non stampa il primo termine della sequenza A000290 che è 0 ...
— Dati scaduti il
1
@ExpiredData Hmm. Potrei facilmente modificare la mia risposta per iniziare da 0 senza influire sul conteggio dei byte, ma ci sono molte altre risposte che iniziano anche nel secondo termine. Non voglio davvero invalidare quelli
— DJMcMayhem
— Kritixi Lithos
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

— Expired Data
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That VDM-SL link doesn't seem to work.
— Fund Monica's Lawsuit
@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.)

— Rin's Fourier transform
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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.

— SamYonnou
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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.

— Unrelated String
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That apparently useless b is fairly suspicious…
— Fatalize
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— 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.

— Embodiment of Ignorance
fonte
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.

— Unrelated String
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