Lockers vs. Crackers: The Five-Element Sequence

Question

The Challenge

A simple "spy versus spy" challenge.

Write a program with the following specifications:

1. The program may be written in any language but must not exceed 512 characters (as represented in a code block on this site).
2. The program must accept 5 signed 32-bit integers as inputs. It can take the form of a function that accepts 5 arguments, a function that accepts a single 5-element array, or a complete program that reads 5 integers from any standard input.
3. The program must output one signed 32-bit integer.
4. The program must return 1 if and only if the five inputs, interpreted as a sequence, match a specific arithmetic sequence of the programmer's choosing, called the "key". The function must return 0 for all other inputs.

An arithmetic sequence has the property that each successive element of the sequence is equal to its predecessor plus some fixed constant a.

For example, 25 30 35 40 45 is an arithmetic sequence since each element of the sequence is equal to its predecessor plus 5. Likewise, 17 10 3 -4 -11 is an arithmetic sequence since each element is equal to its precessor plus -7.

The sequences 1 2 4 8 16 and 3 9 15 6 12 are not arithmetic sequences.

A key may be any arithmetic sequence of your choosing, with the sole restriction that sequences involving integer overflow are not permitted. That is, the sequence must be strictly increasing, strictly decreasing, or have all elements equal.

As an example, suppose you choose the key 98021 93880 89739 85598 81457. Your program must return 1 if the inputs (in sequence) match these five numbers, and 0 otherwise.

Please note that the means of protecting the key should be of your own novel design. Also, probabilistic solutions that may return false positives with any nonzero probability are not permitted. In particular, please do not use any standard cryptographic hashes, including library functions for standard cryptographic hashes.

The Scoring

The shortest non-cracked submission(s) per character count will be declared the winner(s).

If there's any confusion, please feel free to ask or comment.

The Counter-Challenge

All readers, including those who have submitted their own programs, are encouraged to "crack" submissions. A submission is cracked when its key is posted in the associated comments section. If a submission persists for 72 hours without being modified or cracked, it is considered "safe" and any subsequent success in cracking it will be ignored for sake of the contest.

See "Disclaimer" below for details on the updated cracking score policy.

Cracked submissions are eliminated from contention (provided they are not "safe"). They should not be edited. If a reader wishes to submit a new program, (s)he should do so in a separate answer.

The cracker(s) with the highest score(s) will be declared the winners along with the developers of the winning programs.

Please do not crack your own submission.

Best of luck. :)

Leaderboard

Penultimate standings (pending safety of Dennis' CJam 49 submission).

Secure Lockers

1. CJam 49, Dennis
2. CJam 62, Dennis safe
3. CJam 91, Dennis safe
4. Python 156, Maarten Baert safe
5. Perl 256, chilemagic safe
6. Java 468, Geobits safe

Unstoppable Crackers

1. Peter Taylor [Ruby 130, Java 342, Mathematica 146*, Mathematica 72*, CJam 37]
2. Dennis [Pyth 13, Python 86*, Lua 105*, GolfScript 116, C 239*]
3. Martin Büttner [Javascript 125, Python 128*, Ruby 175*, Ruby 249*]
4. Tyilo [C 459, Javascript 958*]
5. freddieknets [Mathematica 67*]
6. Ilmari Karonen [Python27 182*]
7. nitrous [C 212*]

*non-compliant submission

Disclaimer (Updated 11:15 PM EST, Aug 26)

With the scoring problems finally reaching critical mass (given two thirds of the cracked submissions are thus far non-compliant), I've ranked the top crackers in terms of number of submissions cracked (primary) and total number of characters in compliant cracked submissions (secondary).

As before, the exact submissions cracked, the length of the submissions, and their compliant/non-compliant status are all marked so that readers may infer their own rankings if they believe the new official rankings are unfair.

My apologies for amending the rules this late in the game.

Answer 1

CJam, 62 characters

"ḡꬼ쏉壥떨ሤ뭦㪐ꍡ㡩折量ⶌ팭뭲䯬ꀫ郯⛅彨ꄇ벍起ឣ莨ຉᆞ涁呢鲒찜⋙韪鰴ꟓ䘦쥆疭ⶊ凃揭"2G#b129b:c~

Stack Exchange is prone to mauling unprintable characters, but copying the code from this paste and pasting it in the CJam interpreter works fine for me.

How it works

After replacing the Unicode string with an ASCII string, the following code gets executed:

" Push 85, read the integers from STDIN and collect everything in an array.               ";

85l~]

" Convert the array of base 4**17 digits into and array of base 2 digits.                 ";

4H#b2b

" Split into chunks of length 93 and 84.                                                  ";

93/~

" Do the following 611 times:

    * Rotate array A (93 elements) and B one element to the left.
    * B[83] ^= B[14]
    * T = B[83]
    * B[83] ^= B[0] & B[1] ^ A[23]
    * A[92] ^= A[26]
    * Rotate T ^ A[92] below the arrays.
    * A[92] ^= A[0] & A[1] ^ B[5].                                                        ";

{(X$E=^:T1$2<:&^2$24=^+\(1$26=^_T^@@1$2<:&^3$5=^+@}611*

" Discard the arrays and collects the last 177 generated bits into an array.              ";

;;]434>

" Convert the into an integer and check if the result is 922 ... 593.                     ";

2b9229084211442676863661078230267436345695618217593=

This approach uses Bivium-B (see Algebraic analysis of Trivium-like ciphers), a weakened version of the stream cipher Trivium.

The program uses the sequence of integers as initial state, updates the state 434 times (354 rounds achieve full diffusion) and generates 177 bit of output, which it compares to those of the correct sequence.

Since the state's size is precisely 177 bits, this should suffice to uniquely identify the initial state.

Example run

$ echo $LANG
en_US.UTF-8
$ base64 -d > block.cjam <<< IgThuKHqrLzsj4nlo6XrlqjhiKTrrabjqpDqjaHjoanmipjvpb7itozuoIDtjK3rrbLul7bkr6zqgKvvjafpg6/im4XlvajqhIfrso3uprrotbfvmL/hnqPojqjguonhhp7mtoHujLPuipzlkaLpspLssJzii5npn6rpsLTqn5PkmKbspYbnlq3itorlh4Pmj60iMkcjYjEyOWI6Y34=
$ wc -m block.cjam
62 block.cjam
$ cjam block.cjam < block.secret; echo
1
$ cjam block.cjam <<< "1 2 3 4 5"; echo
0

Answer 2

CJam, 91 characters

q~]KK#bD#"᫖࿼듋ޔ唱୦廽⻎킋뎢凌Ḏ끮冕옷뿹毳슟夫΢眘藸躦䪕齃噳卤"65533:Bb%"萗縤ᤞ雑燠Ꮖ㈢ꭙ㈶タ敫䙿娲훔쓭벓脿翠❶셭剮쬭玓ୂ쁬䈆﹌⫌稟"Bb=

Stack Exchange is prone to mauling unprintable characters, but copying the code from this paste and pasting it in the CJam interpreter works fine for me.

How it works

After replacing the Unicode string with integers (by considering the characters digits of base 65533 numbers), the following code gets executed:

" Read the integers from STDIN and collect them in an array.                               ";

q~]

" Convert it into an integer by considering its elements digits of a base 20**20 number.   ";

KK#b

" Elevate it to the 13th power modulus 252 ... 701.                                        ";

D#
25211471039348320335042771975511542429923787152099395215402073753353303876955720415705947365696970054141596580623913538507854517012317194585728620266050701%

" Check if the result is 202 ... 866.                                                      ";

20296578126505831855363602947513398780162083699878357763732452715119575942704948999334568239084302792717120612636331880722869443591786121631020625810496866=

Since 13 is coprime to the totient of the modulus (the totient is secret, so you'll just have to trust me), different bases will generate different results, i.e., the solution is unique.

Unless someone can exploit the small exponent (13), the most efficient way of breaking this lock is to factorize the modulus (see RSA problem). I chose a 512-bit integer for the modulus, which should withstand 72 hours of factorization attempts.

Example run

$ echo $LANG
en_US.UTF-8
$ base64 -d > lock.cjam <<< cX5dS0sjYkQjIgHuiJHhq5bgv7zrk4velOWUse6zjuCtpuW7veK7ju2Ci+uOouWHjOG4ju+Rh+uBruWGleyYt+u/ueavs+6boOyKn+Wkq86i55yY6Je46Lqm5KqV6b2D5Zmz75Wp5Y2kIjY1NTMzOkJiJSIB6JCX57ik4aSe74aS6ZuR54eg4Y+G44ii6q2Z44i244K/5pWr5Jm/5aiy7ZuU7JOt67KT7rO26IS/57+g4p2275+K7IWt5Ymu7Kyt546T4K2C7IGs5IiG77mM4quM56ifIkJiPQ==
$ wc -m lock.cjam
91 lock.cjam
$ cjam lock.cjam < lock.secret; echo
1
$ cjam lock.cjam <<< "1 2 3 4 5"; echo
0

Answer 3

Python - 128

Let's try this one:

i=input()
k=1050809377681880902769L
print'01'[all((i>1,i[0]<i[4],k%i[0]<1,k%i[4]<1,i[4]-i[3]==i[3]-i[2]==i[2]-i[1]==i[1]-i[0]))]

(Expects the user to input 5 comma-separated numbers, e.g. 1,2,3,4,5.)

Answer 4

Java : 468

Input is given as k(int[5]). Bails early if not evenly spaced. Otherwise, it takes a bit figuring out if all ten hashes are correct. For large numbers, "a bit" can mean ten seconds or more, so it might dissuade crackers.

//golfed
int k(int[]q){int b=q[1]-q[0],i,x,y,j,h[]=new int[]{280256579,123883276,1771253254,1977914749,449635393,998860524,888446062,1833324980,1391496617,2075731831};for(i=0;i<4;)if(q[i+1]-q[i++]!=b||b<1)return 0;for(i=1;i<6;b=m(b,b/(i++*100),(1<<31)-1));for(i=0;i<5;i++){for(j=1,x=b,y=b/2;j<6;x=m(x,q[i]%100000000,(1<<31)-1),y=m(y,q[i]/(j++*1000),(1<<31)-1));if(x!=h[i*2]||y!=h[i*2+1])return 0;}return 1;}int m(int a,int b,int c){long d=1;for(;b-->0;d=(d*a)%c);return (int)d;}

// line breaks
int k(int[]q){
    int b=q[1]-q[0],i,x,y,j,
    h[]=new int[]{280256579,123883276,1771253254,1977914749,449635393,
                  998860524,888446062,1833324980,1391496617,2075731831};
    for(i=0;i<4;)
        if(q[i+1]-q[i++]!=b||b<1)
            return 0;
    for(i=1;i<6;b=m(b,b/(i++*100),(1<<31)-1));
    for(i=0;i<5;i++){
        for(j=1,x=b,y=b/2;j<6;x=m(x,q[i]%100000000,(1<<31)-1),y=m(y,q[i]/(j++*1000),(1<<31)-1));
        if(x!=h[i*2]||y!=h[i*2+1])
            return 0;
    }
    return 1;
}
int m(int a,int b,int c){
    long d=1;for(;b-->0;d=(d*a)%c);
    return (int)d;
}

Answer 5

Python, 147

Edit: shorter version based on Dennis' comment. I updated the sequence too to avoid leaking any information.

def a(b):
    c=1
    for d in b:
        c=(c<<32)+d
    return pow(7,c,0xf494eca63dcab7b47ac21158799ffcabca8f2c6b3)==0xa3742a4abcb812e0c3664551dd3d6d2207aecb9be

Based on the discrete logarithm problem which is believed to be uncrackable, however the prime I'm using is probably too small to be secure (and it may have other issues, I don't know). And you can brute-force it of course, since the only unknowns are two 32-bit integers.

Answer 6

Java : 342

int l(int[]a){String s=""+(a[1]-a[0]);for(int b:a)s+=b;char[]c=new char[11];for(char x:s.toCharArray())c[x<48?10:x-48]++;for(int i=0;i<11;c[i]+=48,c[i]=c[i]>57?57:c[i],i++,s="");for(int b:a)s+=new Long(new String(c))/(double)b;return s.equals("-3083.7767567702776-8563.34366442527211022.4345579010483353.1736981951231977.3560837512646")?1:0;}

Here's a string-based locker that depends on both the input character count and the specific input. The sequence might be based on obscure pop culture references. Have fun!

A bit ungolfed:

int lock(int[]a){
    String s=""+(a[1]-a[0]);
    for(int b:a)
        s+=b;
    char[]c=new char[11];
    for(char x:s.toCharArray())
        c[x<48?10:x-48]++;
    for(int i=0;i<11;c[i]+=48,
                     c[i]=c[i]>57?57:c[i],
                     i++,
                     s="");
    for(int b:a)
        s+=new Long(new String(c))/(double)b;
    return s.equals("-3083.7767567702776-8563.34366442527211022.4345579010483353.1736981951231977.3560837512646")?1:0;
}

Answer 7

Javascript 125

This one should be cracked pretty quickly. I'll follow up with something stronger.

function unlock(a, b, c, d, e)
{
    return (e << a == 15652) && (c >> a == 7826) && (e - b == d) && (d - c - a == b) ? 1 : 0;
}

Answer 8

Ruby, 175

a=gets.scan(/\d+/).map(&:to_i)
a.each_cons(2).map{|x,y|x-y}.uniq[1]&&p(0)&&exit
p a[2]*(a[1]^a[2]+3)**7==0x213a81f4518a907c85e9f1b39258723bc70f07388eec6f3274293fa03e4091e1?1:0

Unlike using a cryptographic hash or srand, this is provably unique (which is a slight clue). Takes five numbers via STDIN, delimited by any non-digit, non-newline character or characters. Outputs to STDOUT.

Answer 9

Mathematica 80 67

f=Boole[(p=NextPrime/@#)-#=={18,31,6,9,2}&&BitXor@@#~Join~p==1000]&

Running:

f[{1,2,3,4,5}] (* => 0 *)

Probably pretty easy to crack, might also have multiple solutions.

Update: Improved golfing by doing what Martin Büttner suggested. Functionality of the function and the key hasn't changed.

Answer 10

GolfScript (116 chars)

Takes input as space-separated integers.

~]{2.5??:^(&}%^base 2733?5121107535380437850547394675965451197140470531483%5207278525522834743713290685466222557399=

Answer 11

C 459 bytes

SOLVED BY Tyilo -- READ EDIT BELOW

int c (int* a){
int d[4] = {a[1] - a[0], a[2] - a[1], a[3] - a[2], a[4] - a[3]};
if (d[0] != d[1] || d[0] != d[2] || d[0] != d[3]) return 0;
int b[5] = {a[0], a[1], a[2], a[3], a[4]};
int i, j, k;
for (i = 0; i < 5; i++) { 
for (j = 0, k = 2 * i; j < 5; j++, k++) {
k %= i + 1;
b[j] += a[k];
}
}
if (b[0] == 0xC0942 - b[1] && 
b[1] == 0x9785A - b[2] && 
b[2] == 0x6E772 - b[3] && 
b[3] == 0xC0942 - b[4] && 
b[4] == 0xB6508 - b[0]) return 1;
else return 0;
}

We need someone to write a C solution, don't we? I'm not impressing anybody with length, I'm no golfer. I hope it's an interesting challenge, though!

I don't think there's an obvious way to crack this one, and I eagerly await all attempts! I know this solution to be unique. Very minimal obfuscation, mostly to meet length requirements. This can be tested simply:

int main(){
    a[5] = {0, 0, 0, 0, 0} /* your guess */
    printf("%d\n", c(a));
    return 0;
}

P.S. There's a significance to a[0] as a number in its own right, and I'd like to see somebody point it out in the comments!

EDIT:

Solution: 6174, 48216, 90258, 132300, 174342

A note about cracking:

While this is not the method used (see the comments), I did happen to crack my own cipher with a very easy bruteforce. I understand now it is vitally important to make the numbers large. The following code can crack any cipher where upper_bound is a known upper bound for a[0] + a[1] + a[2] + a[3] + a[4]. The upper bound in the above cipher is 457464, which can be derived from the system of equations of b[] and some working-through of the algorithm. It can be shown that b[4] = a[0] + a[1] + a[2] + a[3] + a[4].

int a[5];
for (a[0] = 0; a[0] <= upper_bound / 5; a[0]++) {
    for (a[1] = a[0] + 1; 10 * (a[1] - a[0]) + a[0] <= upper_bound; a[1]++) {
        a[2] = a[1] + (a[1] - a[0]);
        a[3] = a[2] + (a[1] - a[0]);
        a[4] = a[3] + (a[1] - a[0]);
        if (c(a)) {
            printf("PASSED FOR {%d, %d, %d, %d, %d}\n", a[0], a[1], a[2], a[3], a[4]);
        }
    }
    printf("a[0] = %d Checked\n", a[0]);
}

With a[0] = 6174, this loop broke my work in a little under a minute.

Answer 12

Python27, 283 182

Alright, I am very confident in my locker, however it is quite long as I've added 'difficult to reverse' calculations to the input, to make it well - difficult to reverse.

import sys
p=1
for m in map(int,sys.argv[1:6]):m*=3**len(str(m));p*=m<<sum([int(str(m).zfill(9)[-i])for i in[1,3,5,7]])
print'01'[p==0x4cc695e00484947a2cb7133049bfb18c21*3**45<<101]

edit: Thanks to colevk for the further golfing. I realized during editing that there was a bug as well as a flaw in my algorithm, maybe I'll have better luck next time.

Answer 13

Mathematica 142 146

EDIT: key wasn't unique, added 4 chars, now it is.

n=NextPrime;
f=Boole[
    FromDigits /@ (
        PartitionsQ[n@(237/Plus@##) {1, ##} + 1] & @@@ 
            IntegerDigits@n@{Plus@##-37*Log[#3],(#1-#5)#4}
    ) == {1913001154,729783244}
]&

(Spaces and newlines added for readability, not counted & not needed).

Usage:

f[1,2,3,4,5]   (* => 0 *)

Answer 14

Cracked by @Dennis in 2 hours

---

Just a simple one to get things started - I fully expect this to be quickly cracked.

Pyth, 13

h_^ZqU5m-CGdQ

Takes comma separated input on STDIN.

Run it like this (-c means take program as command line argument):

$ echo '1,2,3,4,5' | python3 pyth.py -c h_^ZqU5m-CGdQ
0

Fixed the program - I had not understood the spec.

This language might be too esoteric for this competition - If OP thinks so, I will remove it.

Answer 15

Lua 105

I suspect it won't be long before it's cracked, but here we go:

function f(a,b,c,d,e)
   t1=a%b-(e-2*(d-b))
   t2=(a+b+c+d+e)%e
   t3=(d+e)/2
   print(t1==0 and t2==t3 and"1"or"0")
end

(spaces added for clarity, but are not part of count)

Answer 16

Perl - 256

sub t{($z,$j,$x,$g,$h)=@_;$t="3"x$z;@n=(7,0,split(//,$g),split(//,$h),4);@r=((2)x6,1,1,(2)x9,4,2,2,2);$u=($j+1)/2;for$n(0..$#r+1){eval{substr($t,$j,1)=$n[$n]};if($@){print 0; return}$j+=$r[$n]*$u}for(1..$x){$t=pack'H*',$t;}eval$t;if($@||$t!~/\D/){print 0}}

I had to put in a lot of error handling logic and this can definitely be golfed down a lot more. It will print a 1 when you get the right five numbers. It will hopefully print a 0 for everything else (might be errors or nothing, I don't know). If anyone wants to help improve the code or golf it more, feel free to help out!

---

Call with:

t(1,2,3,4,5);

Answer 17

Ruby - 130

Based on Linear Feedback Shift Register. Inputs by command line arguments.
Should be unique based on the nature of LFSRs. Clue: ascending and all positive.

Will give more clues if no one solves it soon.

x=($*.map{|i|i.to_i+2**35}*'').to_i
(9**8).times{x=((x/4&1^x&1)<<182)+x/2}
p x.to_s(36)=="qnsjzo1qn9o83oaw0a4av9xgnutn28x17dx"?1:0

Answer 18

Ruby, 249

a=gets.scan(/\d+/).map(&:to_i)
a.each_cons(2).map{|x,y|x-y}.uniq[1]&&p(0)&&exit
r=(a[0]*a[1]).to_s(5).tr'234','(+)'
v=a[0]<a[1]&&!r[20]&&(0..3).select{|i|/^#{r}$/=~'%b'%[0xaa74f54ea7aa753a9d534ea7,'101'*32,'010'*32,'100'*32][i]}==[0]?1:0rescue 0
p v

Should be fun. Who needs math?

Answer 19

CJam, 49 characters

"腕옡裃䃬꯳널֚樂律ࡆᓅ㥄뇮┎䔤嬣ꑙ䘿휺ᥰ籃僾쎧諯떆Ἣ餾腎틯"2G#b[1q~]8H#b%!

Try it online.

How it works

" Push a string representing a base 65536 number and convert it to an integer.            ";

"腕옡裃䃬꯳널֚樂律ࡆᓅ㥄뇮┎䔤嬣ꑙ䘿휺ᥰ籃僾쎧諯떆Ἣ餾腎틯"2G#b

" Prepend 1 to the integers read from STDIN and collect them into an array.               ";

[1q~]

" Convert that array into an integer by considering it a base 2**51 number.               ";

8H#b

" Push the logical NOT of the modulus of both computed integers.                          ";

%!

The result will be 1 if and only if the second integer is a factor of the first, which is a product of two primes: the one corresponding to the secret sequence and another that doesn't correspond to any valid sequence. Therefore, the solution is unique.

Factorizing a 512 bit integer isn't that difficult, but I hope nobody will be able to in 72 hours. My previous version using a 320 bit integer has been broken.

Example run

$ echo $LANG
en_US.UTF-8
$ base64 -d > flock512.cjam <<< IuiFleyYoeijg+SDrOqvs+uEkNaa76a/5b6L4KGG4ZOF76Gi46WE64eu4pSO5JSk5ayj6pGZ5Ji/7Zy64aWw57GD5YO+7I6n6Kuv65aG7qK04byr6aS+6IWO7rSn7YuvIjJHI2JbMXF+XThII2IlIQ==
$ wc -m flock512.cjam
49 flock512.cjam
$ cjam flock512.cjam < flock512.secret; echo
1
$ cjam flock512.cjam <<< "1 2 3 4 5"; echo
0

Answer 20

Javascript 958

Converts the inputs to a number of data types and performs some manipulations relevant to each data type along the way. Should be fairly easily reversed for anyone that takes the time.

function encrypt(num)
{
    var dateval = new Date(num ^ (1024-1) << 10);

    dateval.setDate(dateval.getDate() + 365);

    var dateString = (dateval.toUTCString() + dateval.getUTCMilliseconds()).split('').reverse().join('');

    var result = "";

    for(var i = 0; i < dateString.length; i++)
        result += dateString.charCodeAt(i);

    return result;
}

function unlock(int1, int2, int3, int4, int5)
{
    return encrypt(int1) == "5549508477713255485850495848483249555749321109774324948324410511470" && encrypt(int2) == "5756568477713252485848495848483249555749321109774324948324410511470" && encrypt(int3) == "5149538477713248485856485848483249555749321109774324948324410511470" && encrypt(int4) == "5356498477713256535853485848483249555749321109774324948324410511470" && encrypt(int5) == "5748568477713251535851485848483249555749321109774324948324410511470" ? 1 : 0;
}

Answer 21

C, 239 (Cracked by Dennis)

Go here for my updated submission.

Could probably be golfed a little more thoroughly. Admittedly, I haven't taken the time to prove the key is unique (it probably isn't) but its definitely on the order of a hash collision. If you crack it, please share your method :)

p(long long int x){long long int i;x=abs(x);
for (i=2;i<x;i++) {if ((x/i)*i==x) return 0;}return 1;}
f(a,b,c,d,e){char k[99];long long int m;sprintf(k,"%d%d%d%d%d",e,d,c,b,a);
sscanf(k,"%lld",&m);return p(a)&&p(b)&&p(c)&&p(d)&&p(e)&&p(m);}

Answer 22

C, 212 by Orby -- Cracked

https://codegolf.stackexchange.com/a/36810/31064 by Orby has at least two keys:

13 103 193 283 373
113 173 233 293 353

Orby asked for the method I used to crack it. Function p checks whether x is prime by checking x%i==0 for all i between 2 and x (though using (x/i)*i==x instead of x%i==0), and returns true if x is a prime number. Function f checks that all of a, b, c, d and e are prime. It also checks whether the number m, a concatenation of the decimal representations of e, d, c, b and a (in that order), is prime. The key is such that a,b,c,d,e and m are all prime.

Green and Tao (2004) show that there exist infinitely many arithmetic sequences of primes for any length k, so we just need to look for these sequences that also satisfy m being prime. By taking long long as being bounded by -9.223372037e+18 and 9.223372037e+18, we know that for the concatenated string to fit into long long, the numbers have an upper bound of 9999. So by using a python script to generate all arithmetic sequences within all primes < 10000 and then checking whether their reverse concatenation is a prime, we can find many possible solutions.

For some reason I came up with false positives, but the two above are valid according to the program. In addition there may be solutions where e is negative and the rest are positive (p uses the modulus of x), but I didn't look for those.

The keys I gave are all arithmetic sequences but Orby's script doesn't appear to actually require the inputs to be an arithmetic sequence, so there may be invalid keys too.

Answer 23

MATLAB: Apparently invalid

Very simple, you just have to generate the right random number.

function ans=t(a,b,c,d,e)
rng(a)
r=@(x)rng(rand*x)
r(b)
r(c)
r(d)
r(e)
rand==0.435996843156676

It can still error out, but that shouldn't be a problem.

Answer 24

MATLAB (with Symbolic Toolbox), 173 characters

This isn't an official entry and won't count towards anyone's cracking score, but it will net you mad bragging rights. ;)

function b=L(S),c=sprintf('%d8%d',S(1),S(2)-S(1));b=numel(unique(diff(S)))==1&&numel(c)==18&&all(c([8,9])==c([18,17]))&&isequal(c,char(sym(sort(c,'descend'))-sym(sort(c))));

The symbolic toolbox is only required to handle subtraction of big integers.

Brute forcing it should be a dog, but if you're familiar with the series it involves, the solution is trivial.

Answer 25

Python 2 (91)

Edit: This isn't allowed because the argument for uniqueness is probabilistic. I give up.

---

s=3
for n in input():s+=pow(n,s,7**58)
print s==0x8b5ca8d0cea606d2b32726a79f01adf56f12aeb6e

Takes lists of integers as input, like [1,2,3,4,5].

The loop is meant to operate on the inputs in an annoying way, leaving a tower of sums and exponents. The idea is like discrete log, but with messy complication instead of mathematical simplicity. Maybe the compositeness of the of the modulus is a vulnerability, in which case I could make it something like 7**58+8.

I don't really know how I'd prove that my key is the only one, but the range of outputs is at least 10 times bigger than the range of inputs, so probably? Though maybe only a small fraction of potential outputs are achievable. I could always increase the number of digits at the cost of characters. I'll leave it up to you to decide what's fair.

Happy cracking!

Answer 26

Mathematica - 72

Version 2 of my script, with the same key as the one intended for my version 1.

This basically removes negative prime numbers for NextPrime.

f=Boole[(p=Abs[NextPrime/@#])-#=={18,31,6,9,2}&&BitXor@@#~Join~p==1000]&

Running:

f[{1,2,3,4,5}] (* => 0 *)

Answer 27

Python, 86 characters

a,b,c,d,e=input()
print 1if(a*c^b*e)*d==0xd5867e26a96897a2f80 and b^d==48891746 else 0

Enter the numbers like 1,2,3,4,5.

> python 36768.py <<< "1,2,3,4,5"
0
> python 36768.py <<< "[REDACTED]"
1

Answer 28

CJam, 37 characters (broken)

"煷➻捬渓类ⶥ땙ዶ꾫㞟姲̷ᐂ㵈禙鰳쥛忩蔃"2G#b[1q~]4G#b%!

Try it online.

How it works

See my new answer.

Example run

$ echo $LANG
en_US.UTF-8
$ base64 -d > flock.cjam <<< IueFt+Keu+aNrOa4k+exu+K2peuVmeGLtuq+q+Oen+Wnsu6AhMy34ZCC47WI56aZ6bCz7KWb5b+p6JSDIjJHI2JbMXF+XTRHI2IlIQ==
$ wc -m flock.cjam
37 flock.cjam
$ cjam flock.cjam < flock.secret; echo
1
$ cjam flock.cjam <<< "1 2 3 4 5"; echo
0

Answer 29

Python, 78

(Cracked by Tyilo in 14 mins)

Fun!

def L(a): return 1 if a==[(i-i**6) for i in bytearray(' ','utf-8')] else 0

Okay, it doesn't display properly here :(

Expects a list of five numbers, eg. [1,2,3,4,5]

Answer 30

C, 212 (Cracked)

This is the same idea as my previous submission, golfed more thoroughly, with a bug corrected that passed 0,0,0,0,0 (Thanks to Dennis for pointing out the bug). Compile with -std=c99.

#define L long long
p(L x){x=abs(x);for(L i=2;i<x;i++){if((x/i)*i==x)return 0;}return(x>1);}
f(a,b,c,d,e){char k[99];L m;sprintf(k,"%d%d%d%d%d",e,d,c,b,a);sscanf(k,"%lld",&m);
return p(a)&p(b)&p(c)&p(d)&p(e)&p(m);}

Answer 31

Python 3 (152)

Edit 2: I'm withdrawing this entry because it accepting the correct key seems to depend on hardware or version specifics, which would be unfair to crackers. On online Python 3 interpreters I tried, low-significance bits of the result differed slightly, suggesting the algorithm is sensitive to the exact hardware implementation of arithmetic.

Edit: Made a much longer target to make uniqueness likely given 160 bits of input. Unfortunately, now this target takes over half my characters. This means that this style of submission is unfortunately largely a compression contest.

def f(l,a=[0]):
 for n in l:a+=[(.8+.6j)**(n+a[-1])]
 return a[2::2]==[-0.20857691594748845+0.9907381787645837j,-0.7480510203178323-0.0856921111325014j]

Nothing systematic, just some annoying mangling with complex exponents.