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

Acc!!, 10 bytes, 9 numbers, cracked

?????????_

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

Answer 2

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 3

Python 2, 17 bytes, 10 numbers, cracked

print len([????])

The hidden character is ?.

Answer 4

Pyth, 4 bytes, 9 numbers - Cracked

.__Q

Not super difficult, but should hopefully be fun.

Answer 5

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 6

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 7

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 8

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 9

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 10

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 11

><> 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 12

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 13

​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 14

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 15

R, 22 bytes, 10 numbers, Cracked

_et_______)____is_____

_ is a hidden character.

See the redux version.

Answer 16

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 17

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 18

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 19

JavaScript, 15 bytes, 10 numbers, cracked

alert(_+"+____)

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

Answer 20

Octave, 21 bytes 9 numbers, cracked

_a__repmat(_one___,__

_ is a hidden character.

My idea:

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

Answer 21

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 22

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 23

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 24

Java, 70 bytes, 10 Numbers, Cracked

__________________________ ____(_____________________________)________

_ is the hidden character

Edit: major flaw fixed

Answer 25

><> 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 26

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 27

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 28

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 29

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 30

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.