Code ladder, Cops

Question

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: _.

abc____i

Leaderboard

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>

Answer 1

Hexagony, 18 Bytes, 10 numbers, SAFE

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

.__{_]5[$@.;=@$!!1

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 ;@.

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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.

Answer 2

Retina, 2 bytes, 10 numbers, Cracked

_1

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.

Answer 3

Octave, 55 bytes, 10 numbers, cracked

(o__(O_o_(@(__o)o__-O}_)_(0<O,{_(_o_O-1)+1_@(_1}_)(__o_

_ is the unknown character.

Solution

(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:

(plus_one=@(n,f)(@(boolean,outcomes)outcomes{2-boolean}())(0<n,{@()f(n-1)+1,@()1}))(n,f)

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

Answer 4

Python 2, 9 bytes, 10 numbers, cracked

print 8/8

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

Answer 5

Perl, 12 bytes, 10 numbers, Cracked!

Underscores represent unknown characters.

____;say__-9

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.)

Answer 6

Perl, 46 bytes, 10 numbers, safe

The problem

__b_b_\__}_b_b_b_0_;
$b[0]=10;$b{0}=1;say$b[0]

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};
$b[0]=10;$b{0}=1;say$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};
$b[0]=10;$b{0}=9;say$b[0]

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

Explanation

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.

Answer 7

JavaScript, 30 bytes, 10 numbers, cracked

alert(Array(_)________.length)

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

Answer 8

Perl, 14 bytes, 10 numbers, Cracked

say_!"___"%""_

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 :

say"!"=~y"%""c

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

However, ais523 found another way :

say"!"+1#"%""r

Answer 9

JavaScript, 22 bytes, 10 numbers, cracked

Probably rather easy to crack.

alert(__14_337__xc_de)

_ being a hidden character

Answer 10

Jelly, 7 bytes, 10 numbers, cracked

“1‘ỌȮḊ‘

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.

Answer 11

Octave, 17 bytes, 10 numbers, Cracked

_od(3_13_13_7_1_)

Original solution

mod(3*1361357,10)
...
mod(3*1361357,17)
mod(3*1361397,17)
mod(9*1361397,17)

_ is the hidden character.

Answer 12

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

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

□□5:**-□-.@

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:

g45:**-2-.@
g45:**-1-.@
g45:**-1\.@
g45:**-1\+.@
g45:**-2\+.@
...
g45:**-7\+.@

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.

Answer 13

R, 21 bytes, 10 numbers Cracked

__i___________i______

Works for 10 numbers. _ is hidden character.

Original solution:

which(letters%in%"a")
which(letters%in%"b")
etc.

Answer 14

Ruby, 16 bytes, 10 numbers, cracked by xsot

x=##/=#%#
)
###x

# is any character.

Answer 15

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

alert(_to__"_Xc0__0_B6____Zp=="))

Version intended to post for generating 1

alert(_to__"_Xc0__0_Bf____Zp=="))

Unrevealed characters are marked with _

alert(btoa|"0Xc0"-0xBf|!("Zp=="))
alert(btoa|"0Xc0"-0xBe|!("Zp=="))
alert(btoa|"0Xc0"-0xBd|!("Zp=="))
alert(btoa|"0Xc0"-0xBc|!("Zp=="))
alert(btoa|"0Xc0"-0xBb|!("Zp=="))
alert(btoa|"0Xc0"-0xBa|!("Zp=="))
alert(btoa|"0Xc0"-0xB9|!("Zp=="))
alert(btoa|"0Xc0"-0xB8|!("Zp=="))
alert(btoa|"0Xc0"-0xB7|!("Zp=="))
alert(btoa|"0Xc0"-0xB6|!("Zp=="))

Answer 16

Python, 10+ numbers, 61 bytes, Cracked!

Here was the code I posted:

try:x
except:print(__import__('sys').??c??n??()[????b????e???

The original code was:

try:x
except:print(__import__('sys').exc_info()[2].tb_lineno)

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!

Answer 17

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')))

...

Answer 18

Octave, 17 bytes, 10 numbers, Cracked

_i_(__i__(2_5_))

Unrevealed characters are marked with _.

Intended solution:

    fix(asind(2/59))
    fix(asind(3/59))
    fix(asind(4/59))
    fix(asind(5/59))
    fix(asind(6/59))
    fix(asind(7/59))
    fix(asind(8/59))
    fix(asind(9/59))
    fix(asind(9/55))
    fix(asind(9/50))

Answer 19

Octave, 19 bytes, 10 numbers, cracked

__sca__1_)___'-_6_'

_ is the hidden character.

Intended solution:

pascal(10)('a'-96)'

Answer 20

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!

Answer 21

JavaScript, 22 bytes, 10 numbers, cracked

alert(0_6_4_>_0_2_0_7)

_ is the hidden character.

Hint about the intended solution

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

Answer 22

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!

Answer 23

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.

Answer 24

Python 3, 19 bytes, 10 numbers, cracked

print(??bin()?????)

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

Answer 25

JavaScript 21 Bytes, 10 Numbers Cracked

alert(b_oa_"3____1"))

Unrevealed characters are marked with _

Cracked

My Version:

alert(btoa|"3"&("1"))
alert(btoa|"3"^("1"))
alert(btoa|"3"^("0"))
alert(btoa|"3"^("7"))
alert(btoa|"2"^("7"))
alert(btoa|"1"^("7"))
alert(btoa|"0"^("7"))
alert(btoa|"0"^("8"))
alert(btoa|"0"^("8"))
alert(btoa|"2"^("8"))

Answer 26

Python 3, 16 bytes, 10 numbers, cracked

print(?%??f?r?t)

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.

Answer 27

Hexagony, 15 bytes, 10 numbers, Cracked!

[!?>!@!__\!?!!!

_ can be replaced by any characters.

Oh my.. the code is getting angry due to cheap solutions! Will this keep standing when the wall of ! and ? are guarding the most hearted secret?

Last submission even if this has got any cheap solutions :)

Intended solution

For 1 to 9, there are easy changes:

[!?>!@!1#\!?!!! With 1~9 replacing the 1

However for 10,

1!?>!@!9#\!?!!! Which prints the 1 and then use a numerical jump so we run in a reversed fashion, which in turn goes to ?! prints a 0 and stops. Really wanted to make some that changes IPs. By the way, all 3 submissions are with the intention to use # to print an 1 and then a 0. See if I come up with other nice ideas - if yes I hope more people would find it interesting to play in hexagons!

Answer 28

><> Fish 23 bytes cracked

"H__;v+2__?
  _  >l?!;n

_ is the intended missing characters. Hopefully a bit more complicated than my last one but probably still easy to bypass.

Answer 29

PHP, 38 bytes, 10 numbers (cracked)

<?php__n_p___(_n=___(__=1___echo___?>

_ marks hidden characters.

May be too easy, might have left too much freedom. Have fun!

Answer 30

​O​C​a​m​l​ ​(​i​n​t​e​r​p​r​e​t​e​d​)​,​ ​5​1​ ​b​y​t​e​s​,​ ​1​0​ ​n​u​m​b​e​r​s​ (Cracked)

​ ​B​y​ ​"​i​n​t​e​r​p​r​e​t​e​d​"​ ​I​ ​m​e​a​n​ ​t​h​a​t​ ​i​t​ ​s​h​o​u​l​d​ ​b​e​ ​r​u​n​ ​w​i​t​h​ ​t​h​e​ ​c​o​m​m​a​n​d​ ​​o​c​a​m​l​​ ​r​a​t​h​e​r​ ​t​h​a​n​ ​c​o​m​p​i​l​e​d​ ​t​o​ ​n​a​t​i​v​e​ ​c​o​d​e​ ​w​i​t​h​ ​​o​c​a​m​l​o​p​t​​.​ ​ ​T​h​e​ ​h​i​d​d​e​n​ ​c​h​a​r​a​c​t​e​r​ ​i​s​ ​​^​​.​

^^t^r^^^^-^^^^-^^r^^^^=^^^^^^^r^^^^r^^^^^^^(^^^0/0)

Answer 31

Acc!!, 10 bytes, 9 numbers, cracked

?????????_

Works for 1 to 9. ? is a hidden character.

Answer 32

05AB1E, 11 bytes, Cracked!

3628801__0_

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

Answer 33

Octave, 25 bytes, 9 numbers. Cracked

__a__repmat(__one___,__)_

_ is a hidden character.

Answer 34

Octave, 24 bytes, 9 numbers, cracked

_a__repmat(__one___,__)_

_ is a hidden character.

(Inspired by @LuisMendo's challenge.)

Answer 35

JavaScript, 9 bytes, 10 numbers, Cracked

alert(__)

_ is the hidden character.

Answer 36

Octave, 26 bytes, 10 numbers. Cracked

__a__repmat(_+one___1+__))

_ is a hidden char.

New variation...

---

The intended solution was:

>> eval(repmat('+ones',1+''))
ans =  1

>> eval(repmat('+ones',1+2'))
ans =  3

>> eval(repmat('+ones',1+9'))
ans =  10

Answer 37

Octave, 21 bytes, 10 numbers, Cracked

disp(fpr___f__e_(_'_)

This works for 1-10. _ are hidden characters. This will be my last and hopefully best post.

Intended solution:

disp(fpr= -fopen('')) and disp(fpr=1-fopen('')) for 2...10

Explanation:

disp(fprintf('_')) will actually print _1 if executed, so fpr___f was chosen to throw people off. The f was needed in fopen so there was only one additional space. fopen('') returns -1 since it fails to open a file with an empty name. fpr= -fopen('') makes fpr = 1.

Answer 38

Pyth, 4 bytes, 9 numbers - Cracked

.__Q

Not super difficult, but should hopefully be fun.

Answer 39

Python 2, 17 bytes, 10 numbers, cracked

print len([????])

The hidden character is ?.

Answer 40

Hexagony, 7 bytes, 10 numbers, Cracked

1<@|!__

Works for 1 to 10, _ are hidden characters. I hope I didn't overlook any cheap way to work around the intended solution (I'm looking forward to clever ways to work around the intended solution ;)).

You can try Hexagony online over here.

Answer 41

Perl, 31 bytes - 10 numbers - Cracked by ais523

__e_$a__<__$_;1 while_$__;say$_

_ is the hidden character.

I am not posting my solution here, because the crack is in a completely different direction from what I did, so I'll be creating a new challenge for that one. The cracked solution is below:

 $e=$a+q<1_$>;1 while!$e ;say$e  #Prints 1 (note the leading space)
 $e=$a+q<1_$>+1 while!$e ;say$e  #Prints 2
 $e=$a+q<1_$>+2 while!$e ;say$e  #Prints 3
 $e=$a+q<1_$>+3 while!$e ;say$e  #Prints 4
 $e=$a+q<1_$>+4 while!$e ;say$e  #Prints 5
 $e=$a+q<1_$>+5 while!$e ;say$e  #Prints 6
 $e=$a+q<1_$>+6 while!$e ;say$e  #Prints 7
 $e=$a+q<1_$>+7 while!$e ;say$e  #Prints 8
 $e=$a+q<1_$>+8 while!$e ;say$e  #Prints 9
 $e=$a+q<1_$>+9 while!$e ;say$e  #Prints 10

Answer 42

R, 22 bytes, 10 numbers, Cracked

_et_______)____is_____

_ is a hidden character.

See the redux version.

Answer 43

C++, Linux platform, 10 numbers, 468 bytes, Cracked

Ok so previously my puzzle played with technically undefined behavior (and the gimmick was stabilizing it) and had some transcription error. I have improved my process so I expect there aren't transcription errors. This time I am not playing with undefined behavior at all and would rather just reject invoking it. I think I can justify this as being completely defined by either the language or the platform. $ is the replacement character as I needed real _ characters in this puzzle.

#include <unistd.h>
#include <iostream>
#include <fstream>

$$$$$$$$$$$$$ {
    std::ofstream *garbalgase()
    {
        std::ofstream *golgi = new std::ofstream;
        golgi->open("/dev/null");
        return golgi;
    }

    std::ostream * const jawbone = garbalgase();
    $$$$$ int value = $_$$$$$$;
}

int main()
{
    $$$$$ $$$$$$$$/$$int vl = value;
    if (fork() == 0)
    {
        $$$$$$$$$$$<std::ostream $*>($jawbone) = &std::cout;
        vl -= 4;
        _exit(0);
    }
    (*jawbone) << (vl - 9) << std::endl;
}

Edit: Stock solution is based on an incorrect understanding of the rules where you may insert one character to get the 1. As it's already cracked I won't be correcting it. Good luck finding the stock solution though.

Answer 44

Python 2, 26 bytes, 10 numbers, cracked

Let's try this again.

print r___e(3______+_[___]

Works for 1-10; _ is the hidden character.

Answer 45

Java, 10+ digits, 1446 bytes SAFE

Revenge of the Unicode

\u___9\u___D\u___0\u___F\u___2\u___4\u___0\u___A\u___1\u___6\u___1\u___E\u___9\u___F\u___E\u___A\u___B\u___3\u___C\u___1\u___3\u___3\u___0\u___D\u___B\u___0\u___5\u___2\u___C\u___9\u___3\u___0\u___3\u___4\u___1\u___4\u___9\u___3\u___0\u___6\u___F\u___9\u___4\u___0\u___D\u___1\u___9\u___E\u___8\u___3\u___4\u___2\u___9\u___E\u___7\u___B\u___D\u___1\u___9\u___4\u___8\u___2\u___F\u___7\u___3\u___0\u___5\u___8\u___3\u___5\u___0\u___4\u___9\u___F\u___E\u___B\u___6\u___9\u___C\u___5\u___4\u___5\u___3\u___3\u___2\u___9\u___0\u___4\u___F\u___2\u___0\u___6\u___4\u___0\u___D\u___0\u___E\u___5\u___7\u___0\u___6\u___9\u___C\u___5\u___4\u___5\u___3\u___3\u___2\u___9\u___0\u___4\u___F\u___2\u___8\u___9\u___B\u___3\u___5\u___E\u___E\u___D\u___9\u___3\u___3\u___E\u___3\u___8\u___1\u___2\u___5\u___4\u___3\u___5\u___3\u___2\u___5\u___4\u___3\u___E\u___7\u___5\u___4\u___A\u___1\u___6\u___1\u___9\u___F\u___6\u___9\u___C\u___5\u___4\u___5\u___3\u___3\u___2\u___9\u___0\u___4\u___F\u___2\u___1\u___3\u___3\u___5\u___3\u___3\u___8\u___9\u___E\u___3\u___5\u___4\u___8\u___6\u___4\u___C\u___1\u___9\u___B\u___E\u___5\u___7\u___0\u___0\u___2\u___9\u___E\u___4\u___3\u___4\u___2\u___5\u___1\u___D\u___8\u___E\u___5\u___7\u___0\u___6\u___9\u___C\u___5\u___F\u___5\u___4\u___0\u___5\u___4\u___3\u___4\u___2\u___5\u___1\u___D\u___8\u___6\u___4\u___9\u___9\u___E\u___0\u___2\u___9\u___E\u___4\u___8\u___7\u1__1\u___7\u___5\u___7\u___1\u___7\u___9\u___B\u___D\u___D

No comments this time :P and it only works on the Oracle Java :P

_ as redaction like usual

Solution:

\u0069\u006D\u0070\u006F\u0072\u0074\u0020\u006A\u0061\u0076\u0061\u002E\u0069\u006F\u002E\u002A\u003B\u0063\u006C\u0061\u0073\u0073\u0020\u004D\u007B\u0070\u0075\u0062\u006C\u0069\u0063\u0020\u0073\u0074\u0061\u0074\u0069\u0063\u0020\u0076\u006F\u0069\u0064\u0020\u006D\u0061\u0069\u006E\u0028\u0053\u0074\u0072\u0069\u006E\u0067\u005B\u005D\u0061\u0029\u0074\u0068\u0072\u006F\u0077\u0073\u0020\u0045\u0078\u0063\u0065\u0070\u0074\u0069\u006F\u006E\u007B\u0046\u0069\u006C\u0065\u0044\u0065\u0073\u0063\u0072\u0069\u0070\u0074\u006F\u0072\u0020\u0066\u0064\u0020\u003D\u0020\u006E\u0065\u0077\u0020\u0046\u0069\u006C\u0065\u0044\u0065\u0073\u0063\u0072\u0069\u0070\u0074\u006F\u0072\u0028\u0029\u003B\u0073\u0075\u006E\u002E\u006D\u0069\u0073\u0063\u002E\u0053\u0068\u0061\u0072\u0065\u0064\u0053\u0065\u0063\u0072\u0065\u0074\u0073\u002E\u0067\u0065\u0074\u004A\u0061\u0076\u0061\u0049\u004F\u0046\u0069\u006C\u0065\u0044\u0065\u0073\u0063\u0072\u0069\u0070\u0074\u006F\u0072\u0041\u0063\u0063\u0065\u0073\u0073\u0028\u0029\u002E\u0073\u0065\u0074\u0028\u0066\u0064\u002C\u0031\u0029\u003B\u006E\u0065\u0077\u0020\u0050\u0072\u0069\u006E\u0074\u0053\u0074\u0072\u0065\u0061\u006D\u0028\u006E\u0065\u0077\u0020\u0046\u0069\u006C\u0065\u004F\u0075\u0074\u0070\u0075\u0074\u0053\u0074\u0072\u0065\u0061\u006D\u0028\u0066\u0064\u0029\u0029\u002E\u0070\u0072\u0069\u006E\u0074\u0028\u0027\u1111\u0027\u0025\u0027\u0111\u0027\u0029\u003B\u007D\u007D

Deobfuscated:

import java.io.*;class M{public static void main(String[]a)throws Exception{FileDescriptor fd = new FileDescriptor();sun.misc.SharedSecrets.getJavaIOFileDescriptorAccess().set(fd,1);new PrintStream(new FileOutputStream(fd)).print('\u1111'%'\u0111');}}

Entire Thing relies on some dirty Oracle Java classes trickery to link a Filedescriptor object to the 1 Filedescriptor which is stdout

Got the hack from stackoverflow

Answer 46

Perl, 31 bytes, 10 numbers Cracked by ais523

One more go at this, before I reveal my intended solution.

__e_$a,_<__$_;1 while_$a_;say$_

Original code:

open$a,"<",$0;1 while<$a>;say$.  #Prints 1

The code must be saved as a file and not run from the command line, but the file name is not important, as it gets stored in $0. This code reads the file as long as there is text left, and prints $. which contains the number of lines in the original code. Adding a newline pretty much anywhere not inside a word or variable will increase $. by 1.

open$a,"<",$0;
1 while<$a>;say$.   #Prints 2
------
open$a,"<",$0;

1 while<$a>;say$.   #Prints 3
------

et cetera

Answer 47

OCaml (interpreted), 51 bytes, 10 numbers (Cracked)

By "interpreted" I mean that it should be run with the command ocaml rather than compiled to native code with ocamlopt.

The hidden character is ^.

^^t^r^^^^-^^^^-^^r^^^^=^^^^^^^r^^^^r^^^^^^^(^^^^/0)

Answer 48

Octave, 16 bytes, 10 numbers (cracked)

As a cop, it's my duty to keep this site clean and remove any explicit content. I found¹ a piece of code referring to a diabolical body part that I've decided to censor here using underscores, _.

ev_l('P_NIS'-2_)

^{1: Well, I found it on the screen after I wrote it.}

Answer 49

QBasic, 12 bytes, 10 numbers, cracked

All right, let's try this:

_INT(_O_(3))

Works for 1-10. _ is the hidden character.

---

Finally, the crack used my intended approach. :P

Answer 50

Aceto, 5 bytes, 9 numbers, safe

_ are unrevealed chars

__xx_

Try it online!

Answer 51

아희(Aheui), 19 bytes (6 chars + 1 newline), 10 numbers

봃法희
반_뭉

Works for 1 to 10. _ is the hidden character. Shouldn't be that hard.

Bonus: It works for 0 too.

Try it online!

Answer 52

Pyth, 3 bytes, 10 numbers (Cracked)

I think it's a very simple one, and will easily get cracked.
So, it goes like this: (hidden characters: _)

__Q

Answer 53

Octave, 21 bytes 9 numbers, cracked

_a__repmat(_one___,__

_ is a hidden character.

My idea:

a=[repmat(bone,0),9]

Answer 54

JavaScript, 15 bytes, 10 numbers, cracked

alert(_+"+____)

Another short one, this time with an unfinished string >:-)

Answer 55

JavaScript, 12 bytes, 10 numbers, Cracked

Let's play with arrays:

alert(_[__])

_ is a hidden character.

---

I wanted robbers to think that the first hidden character was a +, but i didn't realize that would work.

Intended solution :

alert(-[~0])
alert(-[~1])
alert(-[~2])
...
alert(-[~9])

Answer 56

Python 2, 49 bytes, 10 numbers, cracked

y=10__;s='09________';print int(s[8__________:y])

_ is the hidden character.

---

Cracked by boboquack

The initial code was

y=10-0;s='0987654321';print int(s[8+y:]+s[y-1:y])

Answer 57

C, 10+ numbers, 216 bytes, Cracked

#include <stdio.h>

FILE *a, *b;

void x() { ____________ int x; x = __; ________a___d____x_;}
void y() { ____________ int x; fprintf(b,"%d\n",x);}
int main() {a=fopen("/dev/null", "w");b=stdout;x();y();return 0;}

If your solution doesn't work with -O1 I will reject it.

Cracked by Dada. Original file:

#include <stdio.h>

FILE *a, *b;

void x() { /**/volatile int x; x = 07; fprintf(a,"%d\n", x);}
void y() { /**/volatile int x; fprintf(b,"%d\n", x);}
int main() {a=fopen("/dev/null", "w");b=stdout;x();y();return 0;}

Answer 58

PowerShell, 7 bytes, 10 numbers, Cracked

Works for 1 through 10. Hidden characters are _.

___(__0

Intended solution: +!@()+0 and swap out the 0 on the end for successively larger numbers.

Answer 59

Java, 70 bytes, 10 Numbers, Cracked

__________________________ ____(_____________________________)________

_ is the hidden character

Edit: major flaw fixed

Answer 60

><> Fish 5 bytes 10 numbers Cracked by Emigna

_2__;

It should be easy to guess but I threw in a little curve ball to make it slightly harder.

Answer 61

Javascript 19 bytes 10 numbers Cracked

alert((_1*1_10__1_)

Unrevealed characters are marked with _

Cracked! Guess i should have blotted better

alert((11*1+10)%10)
alert((11*2+10)%10)
alert((11*3+10)%10)
alert((11*4+10)%10)
alert((11*5+10)%10)
alert((11*6+10)%10)
alert((11*7+10)%10)
alert((11*8+10)%10)
alert((11*9+10)%10)
alert((11*9+10)%11)

Answer 62

QBasic, 13 bytes, 10 numbers, cracked

__I_T__O_(3))

Works for 1-10. _ is the hidden character.

Answer 63

JavaScript (ES6), 63 bytes, 10 numbers, cracked

This challenge is quite fun, so here's one last submission. :-)

eval([...(n=0,"l?0?a?(?x?")].sort(_=>[...????+[]][n++]).join``)

? is a hidden character.

NB: similarly to my two other entries, this code is supposed to print (not return) 1.

Answer 64

R, 8 bytes, 10 numbers, Cracked

_a___+__

Original solution (as the cracked is a little different, but works):

cat(1+0)
cat(1+1)
...
cat(1+8)
cat(1+9)

Answer 65

Perl, 26 bytes, 10 numbers Cracked by ais523 and Sunny Pun

____;@a=($_..10);say$a[__]

I'm almost positive there's an alternate solution to this problem, but let's see if anyone else can find it because I haven't been able to see it.

Edit: It looks like two people managed to circumvent what I was trying to do!

ais523's solution:

____;@a=($|..10);say$a[+1]  #Prints 1
____;@a=($|..10);say$a[1+1] #Prints 2
...
____;@a=($|..10);say$a[9+1] #Prints 10

Sunny Pun's solution:

$_=0;@a=($_..10);say$a[+1]  #Prints 1
...
$_=9;@a=($_..10);say$a[+1]  #Prints 10

Intended solution:

$[=1;@a=($[..10);say$a[$[]  #Prints 1
$[=2;@a=($[..10);say$a[$[]  #Prints 2
$[=3;@a=($[..10);say$a[$[]  #Prints 3
$[=4;@a=($[..10);say$a[$[]  #Prints 4
$[=5;@a=($[..10);say$a[$[]  #Prints 5
$[=5;@a=($[..10);say$a[-$[]  #Prints 6
$[=4;@a=($[..10);say$a[-$[]  #Prints 7
$[=3;@a=($[..10);say$a[-$[]  #Prints 8
$[=2;@a=($[..10);say$a[-$[]  #Prints 9
$[=1;@a=($[..10);say$a[-$[]  #Prints 10

While deprecated, $[ still allows you to modify the indexing of arrays (typically, 0-based). It prints a warning to STDERR, but OP says that any output to STDERR can be ignored.

Answer 66

Perl, 31 bytes, 10 numbers Cracked by ais523

__e_$a__<__$_;1 while_$a_;say$_

My original post was cracked with a solution that wasn't even in the realm of what I was going for, so I'm adding a single extra byte that renders the previous solution (hopefully) unsalvageable.

Once again cracked in a way I haven't accounted for:

the;$a+=<1\$>;1 while!$a ;say$a #Prints 1
the;$a+=<1\$>+1 while!$a ;say$a #Prints 2
...
the;$a+=<1\$>+9 while!$a ;say$a #Prints 10

Answer 67

Perl, 7 bytes, 9 numbers, Cracked!

say$_-_

Underscores represent unknown characters.

The solution (as found by feersum) was to use a literal control-F to replace the first underscore, thus finding the predefined variable that has the value 2.

Answer 68

Python 2, 26 bytes, 10 numbers, cracked by feersum and Destructible Watermelon

Third try's the charm?

print r___e(3_4____+_[___]

Works for 1-10, _ is hidden.

---

Destructible Watermelon got the general idea I was going for. Here's my solution:

print r"10e(3246578+9["[0]
print r"10e(3246578+9["[5]
print r"10e(3246578+9["[4]
print r"10e(3246578+9["[6]
print r"10e(3246578+9["[-6]
print r"10e(3246578+9["[-7]
print r"10e(3246578+9["[-5]
print r"10e(3246578+9["[-4]
print r"10e(3246578+9["[-2]
print r"10e(3246578+9["[:2]

Answer 69

COW, 51 bytes, 10 numbers Cracked by Kritixi Lithos

MoO ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ OOM

Original code:

MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM #Prints 1
MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM #Prints 2
...
MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO Mo_ Mo_ OOM #Prints 10 (the last two operations were red-herrings)

For a Javascript-based interpreter to use for testing, you can go here.

_ is hidden character

Answer 70

Wentel x87, 36 bits/8 = 4.5 bytes, 2 numbers

____________0110______00____00______

Answer 71

Python 2, 111 bytes, 10 numbers (cracked)

How about some random digits?

The hidden character is E.

print +int('''3EE5EE6EE0EE2EE5EE6EE4EE9EE1EE5EE9EE4EE6EE7EE5EE7E''')%int('''2EE0EE9EE8EE9EE2EE3EE5EE0EE3EE6''')

Answer 72

Hexagony, 12 bytes, 10 numbers, Cracked!

[!_>!_!1_@!_

_ can be replaced by any characters.

Seems that love doesn't belong to here should not stay here, to keep the world more interesting :P

Spent time finding cheap alternatives. So giving out much tips so that it is harder to crack (I hope impossible) with cheap alternatives !_!

Answer 73

Python 2, 109 bytes, 10 numbers

from hashlib import *
h=md5()
h.update('__[_')
if h.hexdigest()=='88f4002e27b0902c7018359da9bc5b44':print(0+1)

Answer 74

Hexagony, 18 bytes, 10 numbers, Cracked!

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

.__{_]5[$@__=@_!!1

You can try Hexagony online over here.

I posted a follow-up using the same indended solution, but with fewer hidden characters here.

Answer 75

Python 3, 56 bytes, 6 numbers, Cracked

Quite a simple concept, should be easily crackable. The 'mystery characters' are indicated with underscores.

print(_______(str(ord(x))for x in ________).___________)

Answer 76

tinylisp, 43 bytes, 10 numbers

((v(d f(q((n)(i n(s 1(s 0(f(___))))1)))))0)

Works for 1-10; _ is hidden.

Answer 77

QBasic, 12 bytes, 10 numbers, cracked

Previous version's red herring created a loophole...

_I_T__O_(3))

Works for 1-10. _ is the hidden character.

Answer 78

Acc!!, 46 bytes, 10 numbers, cracked

?
Count i while ?????????????????
}
Write ????

Works for 1 to 10, with ? as the hidden character.

---

Original solution:

1
Count i while _/10 {
Write 49
0
}
Write 48+_

The initial accumulator value goes from 1 to 9 and then to 99. Whenever it is less than 10, _/10 is 0 and the loop doesn't execute. When it is greater than 10, the inner loop writes a 1 and sets the accumulator to 0, which breaks out of the loop and writes a 0.

Answer 79

TI-Basic, 13 tokens, 10 numbers, cracked

DelVar A1→θ
__>A
θ

_ represents a hidden token. Hopefully this isn't too easy xD

Answer 80

Python, 28 bytes, 10 numbers, cracked!

_____ (__________)_(_____*4)_

Intended solution:

print (ord("a"   )-(24   *4))

Answer 81

ASMD, 10 bytes (non-competing) Cracked!

1_**_**_*÷

Hidden characters are denoted by _.

This should be really easy to crack.

Works from 1 to 10.

Answer 82

Java 10+ numbers 126 tokens

____00________007____________00______________________________t____o_____in_(___"___42________00____00____".___________0)___;__

_ is redacted

Answer 83

Python 2, 26 bytes, 10 numbers, cracked

print r___e(_______+_[___]

Works for 1-10; _ is the hidden character.

Answer 84

ASMD, 8 bytes (non-competing), Cracked!

____+++C

Works from 1-10.

Answer 85

Hexagony, 15 bytes, 10 numbers, Cracked!

\!?___<3__!?_<3

What kind of love (<3) is this? This is the love of Hexagons :P

I believe this is not a hard one but a hearty one (Pun intended).

Answer 86

Ruby, 81 bytes, 10 numbers, cracked by feersum

require'digest/md5'
p Digest::MD5.digest('########')[n=0].ord^'straYNpraq'[n].ord

# is the placeholder character. Is this a standard loophole yet? Not that it's uncrackable I don't think, I've left in some weaknesses intentionally and I'm sure unintentionally.

Answer 87

Python 2, 33 bytes, 9 numbers

#__________________
riny
________

_{(how mysterious?)}

Answer 88

R, 22 bytes, 10 numbers SAFE

_et___e_(_)____is_1___

_ is hidden character

Solution:

set.seed(4);rpois(1,1)
set.seed(4);rpois(1,2)
set.seed(4);rpois(1,3)
set.seed(4);rpois(1,4)
set.seed(4);rpois(1,5)
set.seed(4);rpois(1,6)
set.seed(4);rpois(1,7)
set.seed(6);rpois(1,7)
set.seed(6);rpois(1,8)
set.seed(6);rpois(1,9)