Note: This challenge is finished. Submissions are still welcome but can not win.

This is the cops' thread. The robbers' thread goes here.

Write a code that outputs the integer 1. If you add, remove or substitute a single character (of your choosing), the code should output the integer 2. Change one more character (the same or another), and the code should output 3. Continue like this as far as you can, but maximum up to 10. Default output formats such as ans = 1 are accepted. You can ignore output to STDERR (or equivalent).

You must reveal the language, byte count of your initial code, the number of integers it works for, as well as an optional number of characters of the initial code. Note: You don't have to reveal any characters, but remember that revealing characters might make it harder for the robbers as they must use the same character in the same position. You can choose which character you use to denote unrevealed characters (for instance underscore), but make sure to specify this.

Cops can provide the uncracked code after one week and call the submission "SAFE". The winning submission will be the shortest uncracked submission that produces the number 10. If no uncracked submissions are able to print 10, the shortest code that produces 9 will win, and so on. Note that the robbers don't have to make the same changes as you do, and they don't have to reproduce the exact code (unless you reveal all characters). They must only reproduce the output.

Submissions posted later than November 24th are welcome but not eligible for the win (because there will likely be fewer robbers around).

Example post:

The following post is a submission in the language MyLang, it is 9 bytes long, and it works for numbers 1 - 8.

MyLang, 9 bytes, 8 numbers

This submission works for 1 - 8. Unrevealed characters are indicated with an underscore: _.



Disclaimer: The leaderboard is not tested and uncracked submissions might not appear in the list.

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><style>table th,table td{padding: 5px;}th{text-align: left;}.score{text-align: right;}table a{display: block;}.main{float: left;margin-right: 30px;}.main h3,.main div{margin: 5px;}.message{font-style: italic;}#api_error{color: red;font-weight: bold;margin: 5px;}</style> <script>QUESTION_ID=99546;var safe_list=[];var uncracked_list=[];var n=0;var bycreation=function(x,y){return (x[0][0]<y[0][0])-(x[0][0]>y[0][0]);};var byscore=function(x,y){return (x[0][1]>y[0][1])-(x[0][1]<y[0][1]);};function u(l,o){jQuery(l[1]).empty();l[0].sort(o);for(var i=0;i<l[0].length;i++) l[0][i][1].appendTo(l[1]);if(l[0].length==0) jQuery('<tr><td colspan="3" class="message">none yet.</td></tr>').appendTo(l[1]);}function m(s){if('error_message' in s) jQuery('#api_error').text('API Error: '+s.error_message);}function g(p){jQuery.getJSON('//api.stackexchange.com/2.2/questions/' + QUESTION_ID + '/answers?page=' + p + '&pagesize=100&order=desc&sort=creation&site=codegolf&filter=!.Fjs-H6J36w0DtV5A_ZMzR7bRqt1e', function(s){m(s);s.items.map(function(a){var he = jQuery('<div/>').html(a.body).children().first();he.find('strike').text('');var h = he.text();if (!/cracked/i.test(h) && (typeof a.comments == 'undefined' || a.comments.filter(function(b){var c = jQuery('<div/>').html(b.body);return /^cracked/i.test(c.text()) || c.find('a').filter(function(){return /cracked/i.test(jQuery(this).text())}).length > 0}).length == 0)){var m = /^\s*((?:[^,;(\s]|\s+[^-,;(\s])+).*(0.\d+)/.exec(h);var e = [[n++, m ? m[2]-0 : null], jQuery('<tr/>').append( jQuery('<td/>').append( jQuery('<a/>').text(m ? m[1] : h).attr('href', a.link)), jQuery('<td class="score"/>').text(m ? m[2] : '?'), jQuery('<td/>').append( jQuery('<a/>').text(a.owner.display_name).attr('href', a.owner.link)) )];if(/safe/i.test(h)) safe_list.push(e);else uncracked_list.push(e);}});if (s.items.length == 100) g(p + 1);else{var s=[[uncracked_list, '#uncracked'], [safe_list, '#safe']];for(var i=0;i<2;i++) u(s[i],byscore);jQuery('#uncracked_by_score').bind('click',function(){u(s[0],byscore);return false});jQuery('#uncracked_by_creation').bind('click',function(){u(s[0],bycreation);return false});}}).error(function(e){m(e.responseJSON);});}g(1);</script><link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/Sites/codegolf/all.css?v=7509797c03ea"><div id="api_error"></div><div class="main"><h3>Uncracked submissions</h3><table> <tr> <th>Language</th> <th class="score">Score</th> <th>User</th> </tr> <tbody id="uncracked"></tbody></table><div>Sort by: <a href="#" id="uncracked_by_score">score</a> <a href="#" id="uncracked_by_creation">creation</a></div></div><div class="main"><h3>Safe submissions</h3><table> <tr> <th>Language</th> <th class="score">Score</th> <th>User</th> </tr> <tbody id="safe"></tbody></table></div>

  • \$\begingroup\$ Not sure I understand... given an arbitrary code in CJam that produces 1, how can I prevent the robbers from adding ) repeatedly to generate the rest of the numbers? The same would be valid for quite a few languages \$\endgroup\$ – Luis Mendo Nov 12 '16 at 12:48
  • 2
    \$\begingroup\$ If that's possible for any program that outputs 1 then it appears CJam is a bad choice of language for this challenge. There's no way to prevent robbers from doing that. \$\endgroup\$ – Stewie Griffin Nov 12 '16 at 12:50
  • 3
    \$\begingroup\$ @LuisMendo Well, it will certainly make this more interesting... \$\endgroup\$ – LegionMammal978 Nov 12 '16 at 12:58
  • 1
    \$\begingroup\$ @DanielJour It can be modifiable up to any number, but the maximum number the robbers need to find is 10. That rule is in place because many submissions can probably be extended to infinity (in theory), so scoring based on the highest achieved number wouldn't make sense. \$\endgroup\$ – Stewie Griffin Nov 13 '16 at 16:25
  • 1
    \$\begingroup\$ You may want to try only disqualifying an entry if the header contains cracked in some form. This is what the redesign userscript currently does. \$\endgroup\$ – ETHproductions Nov 15 '16 at 20:51

88 Answers 88


Hexagony, 18 Bytes, 10 numbers, SAFE

This submission works for 1 - 10. Unrevealed characters are indicated with an underscore: _.


You can try Hexagony online over here.

My solution:

1:   .<[{8]5[$@.;=@$!!10
2:   .<[{8]5[$@);=@$!!10
3:   2<[{8]5[$@);=@$!!10
4:   3<[{8]5[$@);=@$!!10
5:   4<[{8]5[$@);=@$!!10
6:   5<[{8]5[$@);=@$!!10
6:   7<[{8]5[$@);=@$!!10
8:   7<[{8]5[$@);=@$!!10
9:   8<[{8]5[$@);=@$!!10
10:  9<[{8]5[$@);=@$!!10

Hex for output 1:

Try it Online!

Full Hex:
  . < [ 
 { 8 ] 5
[ $ @ . ;
 = @ $ ! 
  ! 1 0

Important parts:
  . < .
 . 8 . 5
. $ @ . ;
 . . $ .
  . 1 .
  1. At the < the memory edge is 0, so it turns up.
  2. Hits 1
  3. Jumps to 5
  4. Jumps over 8, but is reversed at < and gets the 8 on the way back.
  5. Hits 5 again
  6. Jumps over 1
  7. Hits the < at this point, the memory value is 1585 which, mod 256, happens to be ASCII 1
  8. Finally prints and exits with ;@.

Hex for output 2:

Try it Online!

Important parts:
  . < .
 . 8 . 5
. $ @ ) ;
 . . $ .
  . 1 .

This follows the same path, but on the way back it hits a ) which increments the memory edge to 1586, or 2.

Hex for output 3-9:

Try it Online!

Important parts:
  2 < [
 . . ] .
. $ . ) .
 . @ . !
  . 1 .
  1. Hits the 2
  2. Now the memory edge is positive when it gets to <, so it turn down.
  3. The ] changes the instruction pointer, but is immediately comes back with [
  4. ) increments to 3
  5. ! Prints 3
  6. $ is left over from the first two numbers so we jump over the end (@)
  7. 1 changes the memory edge, but that doesn't matter now.
  8. < reflects the pointer back.
  9. Again 1 doesn't matter because we hit @ to end the program.

Retina, 2 bytes, 10 numbers, Cracked


Works for 1 to 10, _ is a hidden character. This shouldn't be too hard, but I hope it provides a somewhat interesting puzzle. :)

You can try Retina online over here.


Octave, 55 bytes, 10 numbers, cracked


_ is the unknown character.


(o=@(O,o)(@(O,o)o{2-O}())(0<O,{@()o(O-1)+1,@()1}))(0,o) %then changing the very last 0 to 1,2,3 e.t.c.

Given x, this does recursively calculate x+1. It is mainly composed of two anonymous functions. One provides an if statement to anchor the recursion:

if_ = @( boolean, outcomes) outcomes{ 2 - boolean}();

This is just abusing the fact that a boolean values evaluates to 0 or 1. This function accepts a boolean value, and a cell array of two functions, and evaluates one or the other of these two functiosn depending on the boolean value. The second part is the actual recursion:

plus_one = @(n,f) if_(0<n ,{@()f(n-1)+1, @()1})

As an anyonmous function is anonymous, you cannot directly access it from itsefl. That why we need a second argument f first. Later we will provide a handle to the function instelf as a second argument, so a final function would looks like so:

plus_one_final = @(n)plus_one(n,plus_one);

So in this notation my submission becomes:


I asked about recursion anchors for anonymous functions in MATLAB a while ago on stackoverflow.

  • 19
    \$\begingroup\$ o_O O____o O_O o_O \$\endgroup\$ – TuxCrafting Nov 13 '16 at 18:51
  • \$\begingroup\$ I'm not so sure whether this is really that difficult to crack=) \$\endgroup\$ – flawr Nov 13 '16 at 21:49
  • \$\begingroup\$ Well, it sure as hell isn't easy to crack! Could maybe do it with pen and paper(!) \$\endgroup\$ – Stewie Griffin Nov 14 '16 at 22:11
  • \$\begingroup\$ I take that as a compliment :) I think you will like my solution, but I'm not gonna reveal anything until it is cracked / it is safe. \$\endgroup\$ – flawr Nov 14 '16 at 22:51
  • 4
    \$\begingroup\$ Cracked \$\endgroup\$ – feersum Nov 15 '16 at 0:23

Python 2, 9 bytes, 10 numbers, cracked

print 8/8

No hidden chars. Can you crack it without brute forcing?

  • \$\begingroup\$ Cracked - that was fun :) \$\endgroup\$ – Sp3000 Nov 16 '16 at 11:18

Perl, 12 bytes, 10 numbers, Cracked!

Underscores represent unknown characters.


Probably fairly easy, and it wouldn't surprise me if there were multiple solutions. Still, it might be fun to crack.

(The intended solution was the same as the crack. This is fundamentally just a problem about assigning 10 to a variable in four characters, which is surprisingly difficult in Perl; unlike many golfing languages, it doesn't have a variable that helpfully starts at 10.)

  • 1
    \$\begingroup\$ Cracked. At first I thought it was something like $_=1;say;#-9, but I couldn't figure out how to get 10. \$\endgroup\$ – Riley Nov 14 '16 at 15:45

Perl, 46 bytes, 10 numbers, safe

The problem


The shorter problems tend to get cracked quickly, so I thought I'd try a longer one. The longer ones also tend to get cracked if people leave enough of a gap to sneak something naughty like say or exit in, so all the gaps here are short. Hidden characters are represented using _.

My solution

sub b{\@_}*b=b$b{0};

To print 2, 3, etc., up to 9, keep changing the number assigned to $b{0} in the second line (i.e. $b{0}=2, $b{0}=3, etc.). The program for 9 looks like this:

sub b{\@_}*b=b$b{0};

Then to produce 10, comment out the first line by prepending a # character to it.


The first thing to note is that the solution isn't really golfed apart from removing whitespace: if we lay it out with more readable whitespace, we get this:

sub b { \@_ }
*b = b $b{0};
$b[0] = 10;
$b{0} = 1;
say $b[0];

Normally, when you access the arguments of a subroutine in Perl, you do so via copying them out of @_. There's a good reason for this: @_ aliases the arguments the subroutine is given (so, for example, (sub { $_[0] = 3 })->($x) will assign to $x), something that isn't normally desirable.

Although @_ might seem magical, it's actually just using a standard feature of the Perl internals (which is readily available from XS but only comes up in a few weird cases in pure Perl, such as @_ itself): an array doesn't store its elements directly, but rather by reference. Thus, when we call b in the second line below, Perl generates an array (calling it @_) whose first element is a reference to the same storage that $b{0} uses. (Hash values are also stored by reference; $_[0] and $b{0} are both referencing the same storage at this point.) Because @_ isn't doing anything special from an internals point of view, we can take a reference to it just like we could with any other array, causing it to outlive the subroutine it's defined in.

Perl variables also refer to data storage by reference. A long time ago, people used to use code like *x = *y; to set $x as an alias to $y (via making them reference the same thing), likewise @x as an alias to @y, %x as an alias to %y, and so on. That rather breaks the invariant that variables with similar names don't have to act similarly, so modern Perl provides an alternative; assigning a reference to a typeglob overrides only the variable that matches the type of the reference (so *x = \%y would alias %x to point to the same storage as %y but leave, say, $x alone). This syntax notably doesn't care about whether the storage you're aliasing to has a name, so when we assign the return value of b (which is an array reference that's keeping the array formerly called @_ alive) to *b, what happens is that @b is changed to alias the argument list to the call to b (while leaving %b unchanged). This means, notably, that $b[0] and $b{0} now point to the same storage, and assigning to one will therefore change the other. Everything from then on is completely straightforward.

The Perl documentation doesn't really talk about this sort of detail, so I'm not surprised anyone got the crack; the nature of @_ as not quite being like other arrays isn't something that's really emphasised, and most coding styles aim to minimize the effects that this has rather than amplifying them.

  • 1
    \$\begingroup\$ I'm so intrigued by this. I've learnt about pseudo-hashes and refreshed my memory on referencing and de-referencing but I can't seem to figure it out! \$\endgroup\$ – Dom Hastings Nov 17 '16 at 21:42
  • 1
    \$\begingroup\$ @DomHastings, ... and I've been fooling around with various demarcation characters for q and s and y and m (mostly trying to get them to end after the $b[0] assignment), but nothing's working for me (yet). \$\endgroup\$ – msh210 Nov 22 '16 at 22:15
  • \$\begingroup\$ You were both on the wrong track, but admittedly, there's not much help given by the language or by my submission as to what the right track is (I had to leave one clue to avoid leaving too large a gap, but there are plenty of reasons that a program might contain a backslash, it's not always related to references). \$\endgroup\$ – user62131 Nov 23 '16 at 5:38
  • \$\begingroup\$ I feel like I was frustratingly close. I had sub b{\@_} cemented in my mind, and, although I experimented with *b I couldn't get it! Thanks for the explanation. I might have glossed over it in your explanation, but why does sub b{\@_}*b=b$b[0] not do the same? \$\endgroup\$ – Dom Hastings Nov 23 '16 at 7:21
  • \$\begingroup\$ You're trying to make the array element and the hash element share memory, so you need to mention both. The array element's implicitly mentioned when you assign an array reference (\@_) to *b, but you have to mention the hash element explicitly yourself. With *b=b$b[0], you're basically just aliasing the new $b[0] (after changing where @b points) to the $b[0] that existed at the start of the program, which is useless. \$\endgroup\$ – user62131 Nov 23 '16 at 7:29

JavaScript, 30 bytes, 10 numbers, cracked


Shouldn't be too hard, but hopefully it's just hard enough to provide a challenge. :) Unrevealed characters are marked with _.

  • \$\begingroup\$ Nice one! I had a swell time solving this one, it was easy, yet challenging at the same time. \$\endgroup\$ – user41805 Nov 12 '16 at 16:41

Perl, 14 bytes, 10 numbers, Cracked


Works for 1 to 10. _ are hidden characters.

I think this shouldn't be too hard to crack. I have an harder one, for 22 bytes, I'll post it if this one is cracked.

Original code :


And replace the "!" by a string of the length of the number you wish to print, for instance !, !!, !!!, etc.

However, ais523 found another way :

  • 2
    \$\begingroup\$ Cracked. I suspect this isn't what you were going for at all. (I was trying to do something using regex, but this was much simpler.) \$\endgroup\$ – user62131 Nov 12 '16 at 23:06

JavaScript, 22 bytes, 10 numbers, cracked

Probably rather easy to crack.


_ being a hidden character

  • 1
    \$\begingroup\$ I'd be shocked if xc_de was anything but a red herring \$\endgroup\$ – ETHproductions Nov 13 '16 at 21:02
  • \$\begingroup\$ Cracked! \$\endgroup\$ – Dom Hastings Nov 14 '16 at 15:39

Octave, 17 bytes, 10 numbers, Cracked


Original solution


_ is the hidden character.


Jelly, 7 bytes, 10 numbers, cracked


No wildcards.

The crack achieved (to use an eval with an argument) was, as many seem to be in this thread, not the intended one.

The intended crack was:

“1‘ỌȮḊ‘ - (prints 1)
“1‘     - code page index list of characters "1": [49]
   Ọ    - cast to ordinals: ['1']
    Ȯ   - print (with no line feed) and return input: effectively prints "1"
        -     (but if left at this would then implicitly print another "1")
     Ḋ  - dequeue: []
      ‘ - increment (vectorises): []
        - implicit print: prints ""

“1‘ỌŒḊ‘ - (prints 2)
“1‘Ọ    - as above: ['1']
    ŒḊ  - depth: 1
      ‘ - increment: 2
        - implicit print: prints "2"

“1‘ỌŒḊ‘‘ - (prints 3)
“1‘ỌŒḊ‘  - as above: 2
       ‘ - increment: 3
         - implicit print: prints "3"

... keep adding an increment operator to print 4 - 10.
  • \$\begingroup\$ I would be able to crack it if the '‘' was on the next line. So close... :) \$\endgroup\$ – user41805 Nov 13 '16 at 8:57
  • \$\begingroup\$ Cracked but could you explain what happens for 10, because I got lucky on this one while trying things that might work. \$\endgroup\$ – Hedi Nov 13 '16 at 18:17
  • \$\begingroup\$ @Hedi Ninja'd me too quickly, I was working on it. \$\endgroup\$ – Erik the Outgolfer Nov 13 '16 at 18:24
  • \$\begingroup\$ @Hedi - yours was an unintended crack. The way 10 is working for you is, I believe, by evaluating the jelly code in the string 9 with an argument of 0 (the default value of the input), which you then dequeue (no effect) and then increment. \$\endgroup\$ – Jonathan Allan Nov 13 '16 at 18:33
  • \$\begingroup\$ ...actually I think the way 10 is working for you is by evaluating the jelly code in the string 9 with an argument of [] - the default value of the input, 0, dequeued - which you then increment., Like “1‘‘‘‘‘‘‘‘‘Ọv0Ḋ¤‘ \$\endgroup\$ – Jonathan Allan Nov 13 '16 at 18:41

Befunge-93, 11 bytes, 10+ numbers, Cracked

This submission works for at least 1 - 10. Unrevealed characters are indicated with .


Try it online

I must say I was impressed that two people could come up with independent solutions for this, neither of which were what I was expecting. While Martin got there first, I'm giving the "win" to Sp3000 as their solution is more portable.

This was my intended solution though:


Because a stack underflow in Befunge is interpreted as 0, the g just reads from 0,0 returning the ASCII value of 'g', namely 103. 45:**- subtracts 100, giving you 3. Then 2- gives you 1.

For the third iteration, the - (subtract) is changed to a \ (swap) instruction, so the 3 becomes the topmost stack entry. And in iteration four, a + (add) instruction is inserted, thus adding the 3 to the 1 giving 4.

  • \$\begingroup\$ Cracked. I'm curious to see what solution you had in mind. :) \$\endgroup\$ – Martin Ender Nov 13 '16 at 15:29
  • \$\begingroup\$ @MartinEnder As I commented on your answer, I'd like to leave my intended solution hidden for a while in case someone else wants to attempt a more portable solution. Is that OK? \$\endgroup\$ – James Holderness Nov 13 '16 at 16:19
  • \$\begingroup\$ Of course, that's fine. :) \$\endgroup\$ – Martin Ender Nov 13 '16 at 16:22
  • \$\begingroup\$ "15:**-6-.@ gives 1, but I'm not sure whether the fact that " pushes 32 on top (due to implicit spaces) is an artefact of the TIO interpreter or part of Befunge's spec, because trying a few interpreters it seems not all interpreters do that. Does your intended solution depend on this behaviour? \$\endgroup\$ – Sp3000 Nov 15 '16 at 8:15
  • \$\begingroup\$ @Sp3000 That wasn't my intended solution, but that's perfectly valid Befunge - the interpreters that don't support that are incorrect. If you post an answer with the rest of the sequence I'd definitely consider that a complete crack. \$\endgroup\$ – James Holderness Nov 15 '16 at 11:48

R, 21 bytes, 10 numbers Cracked


Works for 10 numbers. _ is hidden character.

Original solution:


  • \$\begingroup\$ @StewieGriffin This is my first post on this site and I don't know the norms. I have one more R challenge - a bit trickier, I think. Can I add another answer? Or append it to this one? \$\endgroup\$ – Gregor Thomas Nov 14 '16 at 22:51
  • \$\begingroup\$ Adding a new one it's perfectly fine (as a separate one) 😊 welcome to the site 😊 \$\endgroup\$ – Stewie Griffin Nov 14 '16 at 23:04
  • \$\begingroup\$ Did I cracked it here ? \$\endgroup\$ – Tensibai Nov 15 '16 at 9:57
  • \$\begingroup\$ @Tensibai Cracked :) \$\endgroup\$ – Gregor Thomas Nov 15 '16 at 18:07

Ruby, 16 bytes, 10 numbers, cracked by xsot


# is any character.

  • \$\begingroup\$ Cracked! \$\endgroup\$ – xsot Nov 17 '16 at 9:45

Octave, 32 bytes, 10 numbers. Cracked

_n_(isprime(floor(s____i__ i____

_ is a hidden character.

You can try Octave online here.

Original solution:

1: nnz(isprime(floor(sqrt(i):pi')))

2: nnz(isprime(floor('sqrt(i):pi')))

3: nnz(isprime(floor('sqrt(i):pia')))

4: nnz(isprime(floor('sqrt(i):piaa')))


  • \$\begingroup\$ Very nice! Cracked. Not sure if I reproduced your code though? \$\endgroup\$ – Stewie Griffin Nov 12 '16 at 16:50
  • \$\begingroup\$ @StewieGriffin Well done! I should have revealed more characters... :-D \$\endgroup\$ – Luis Mendo Nov 12 '16 at 16:55
  • 1
    \$\begingroup\$ Damn that was clever :) Too bad you couldn't say it was MATLAB... One more character would probably have made it much harder... I enjoyed it though... I used a full 25 minutes :) \$\endgroup\$ – Stewie Griffin Nov 12 '16 at 16:59
  • \$\begingroup\$ @Stewie Yes, you made good use of non-Matlab features! :-) \$\endgroup\$ – Luis Mendo Nov 12 '16 at 17:08

Octave, 17 bytes, 10 numbers, Cracked


Unrevealed characters are marked with _.

Intended solution:


  • \$\begingroup\$ cracked (finally=) but probably not with your original approach, right? \$\endgroup\$ – flawr Nov 12 '16 at 23:03
  • \$\begingroup\$ Nice, added my approach in a spoiler tag :) \$\endgroup\$ – Stewie Griffin Nov 12 '16 at 23:05
  • 2
    \$\begingroup\$ Oh your solution is really clever!!! \$\endgroup\$ – flawr Nov 12 '16 at 23:08

Octave, 19 bytes, 10 numbers, cracked


_ is the hidden character.

Intended solution:


  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ – Luis Mendo Nov 13 '16 at 3:49
  • \$\begingroup\$ I couldn't remember any function with sca. Coming from you, I should have thought of matrix functions :-) \$\endgroup\$ – Luis Mendo Nov 13 '16 at 14:51
  • \$\begingroup\$ There is a reason why I made a complete list of Octave function names =) \$\endgroup\$ – flawr Nov 13 '16 at 19:38

05AB1E, 5 bytes, 10 numbers, cracked!

Not very hard, but a fun one :)


_ is a random character. Uses the CP-1252 encoding. Try it online!

  • \$\begingroup\$ I could almost get it to work, but then it enters an infinite loop... \$\endgroup\$ – user41805 Nov 13 '16 at 16:32
  • \$\begingroup\$ Cracked \$\endgroup\$ – acrolith Nov 13 '16 at 16:37
  • \$\begingroup\$ @daHugLenny Hahaha, that's neat! I didn't think of that :). \$\endgroup\$ – Adnan Nov 13 '16 at 17:11

05AB1E, 6 bytes, 10 numbers, cracked

Attempt 2, this time without the three-char string :p.

_ [==_

_ is a random character. Uses the CP-1252 encoding. Try it online!

  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ – milk Nov 13 '16 at 18:54
  • \$\begingroup\$ @milk Nice, that was the intended solution :) \$\endgroup\$ – Adnan Nov 13 '16 at 18:56

JavaScript, 22 bytes, 10 numbers, cracked


_ is the hidden character.

Hint about the intended solution

The character that needs to be changed to generate all numbers is always the same.

  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ – user62131 Nov 14 '16 at 4:18
  • \$\begingroup\$ @ais523 Well done! \$\endgroup\$ – Arnauld Nov 14 '16 at 12:47

JavaScript 21 Bytes, 10 Numbers Cracked


Unrevealed characters are marked with _


My Version:

  • \$\begingroup\$ Cracked \$\endgroup\$ – Hedi Nov 14 '16 at 21:58

Python 3, 19 bytes, 10 numbers, cracked


Unrevealed characters are marked with ?. Tested in Python 3.5.2.

  • \$\begingroup\$ Cracked \$\endgroup\$ – feersum Nov 14 '16 at 21:03

Python 3, 16 bytes, 10 numbers, cracked


Unrevealed characters are marked with ?. This is probably a bit easy since there's only five question marks, but I'm hoping it'll be a fun one.

  • \$\begingroup\$ Cracked \$\endgroup\$ – DLosc Nov 15 '16 at 6:30

C#, 90 bytes, 10 numbers, cracked

using ______________________________________________;class C{static void Main(){_______;}}

I honestly have no idea how hard this is to crack.

Edit: Oops, transcription error. One _ too few after using.

Now cracked by Hedi, who found the intended (barring the class name) solution.

  • \$\begingroup\$ d'oh... one too short for write() \$\endgroup\$ – masterX244 Nov 14 '16 at 13:19
  • \$\begingroup\$ Didnt help... wrong gap. And that semicolon after first gap thwarted amother idea \$\endgroup\$ – masterX244 Nov 14 '16 at 19:19
  • \$\begingroup\$ Cracked \$\endgroup\$ – Hedi Nov 15 '16 at 22:09

JavaScript 33 Bytes, 10 Numbers Cracked x2

Oops I post posted my line for generating 10 Which Hedi cracked as though it was for 1


Version intended to post for generating 1


Unrevealed characters are marked with _

  • 1
    \$\begingroup\$ Cracked \$\endgroup\$ – Hedi Nov 16 '16 at 20:18
  • \$\begingroup\$ I added a crack for the intended version. I'll try to find a way to get to your solution for 10: console.log(atob|"0Xc0"-0xB6|("Zp==")) I think \$\endgroup\$ – Hedi Nov 16 '16 at 22:18
  • \$\begingroup\$ I updated my answer with what should be the intended crack. \$\endgroup\$ – Hedi Nov 16 '16 at 22:33

Python, 10+ numbers, 61 bytes, Cracked!

Here was the code I posted:


The original code was:


Basically, it throws an error ('x' is not defined) and then prints the line the error was found on. So, just keep adding newlines at the beginning to increment the number.

I knew it wouldn't be hard to crack - I just wanted a funny way to print numbers - but I wasn't expecting Sp3000 to get it so fast, that's some pro skills!

  • \$\begingroup\$ Cracked - amusing, but yeah there aren't really many options :P \$\endgroup\$ – Sp3000 Nov 20 '16 at 10:27
  • \$\begingroup\$ @Sp3000 yeah, I wanted to make sure nobody could squeeze in deleting what's on STDOUT and printing a number, but I guess I narrowed it down too much. Ah well. \$\endgroup\$ – FlipTack Nov 20 '16 at 10:35
  • \$\begingroup\$ Revealing sys definitely made it a lot simpler, since it gave a nice starting point for searching :P \$\endgroup\$ – Sp3000 Nov 20 '16 at 10:37

05AB1E, 11 bytes, Cracked!


Works from 1-10. _ is a hidden character.

Intended Solution:

3628801R¬0+ # 1
3628801R¬1+ # 2
3628801R¬2+ # 3
3628801R¬3+ # 4
3628801R¬4+ # 5
3628801R¬5+ # 6
3628801R¬6+ # 7
3628801R¬7+ # 8
3628801R¬8+ # 9
3628801R¬9+ # 10

Octave, 24 bytes, 9 numbers, cracked


_ is a hidden character.

(Inspired by @LuisMendo's challenge.)

  • \$\begingroup\$ Oh you're way too quick!!! \$\endgroup\$ – flawr Nov 12 '16 at 22:13

JavaScript, 9 bytes, 10 numbers, Cracked


_ is the hidden character.


Octave, 25 bytes, 9 numbers. Cracked


_ is a hidden character.

  • \$\begingroup\$ @StewieGriffin Sorry!! Only 9. My mistake. I'm really sorry. Edited \$\endgroup\$ – Luis Mendo Nov 12 '16 at 21:29
  • \$\begingroup\$ cracked \$\endgroup\$ – flawr Nov 12 '16 at 21:36
  • 1
    \$\begingroup\$ Sorry =P I'm gonna post a replacement=) \$\endgroup\$ – flawr Nov 12 '16 at 21:45
  • \$\begingroup\$ @Stewie It's been my fault, sorry! I'll post a modified version later, since my original solution is different \$\endgroup\$ – Luis Mendo Nov 12 '16 at 21:46
  • \$\begingroup\$ For the record, my crack was identical to flawr's. I've posted another Octave one here. \$\endgroup\$ – Stewie Griffin Nov 12 '16 at 21:58

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