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).

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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

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 2

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 3

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

____________0110______00____00______

Answer 4

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 5

Python 2, 111 bytes, 10 numbers (cracked)

How about some random digits?

The hidden character is E.

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

Answer 6

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 7

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 8

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 9

QBasic, 13 bytes, 10 numbers, cracked

__I_T__O_(3))

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

Answer 10

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 11

Aceto, 5 bytes, 9 numbers, safe

_ are unrevealed chars

__xx_

Try it online!

Answer 12

Python, 28 bytes, 10 numbers, cracked!

_____ (__________)_(_____*4)_

Intended solution:

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

Answer 13

Java 10+ numbers 126 tokens

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

_ is redacted

Answer 14

Python 2, 26 bytes, 10 numbers, cracked

print r___e(_______+_[___]

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

Answer 15

Python 2, 33 bytes, 9 numbers

#__________________
riny
________

_{(how mysterious?)}

Answer 16

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 17

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 18

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 19

Python 2, 109 bytes, 10 numbers

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

Answer 20

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 21

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 22

Acc!!, 46 bytes, 10 numbers, cracked

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

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

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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 23

TI-Basic, 13 tokens, 10 numbers, cracked

DelVar A1→θ
__>A
θ

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

Answer 24

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 25

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 26

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)

Answer 27

ASMD, 8 bytes, Cracked!

____+++C

Works from 1-10.

Answer 28

ASMD, 10 bytes Cracked!

1_**_**_*÷

Hidden characters are denoted by _.

This should be really easy to crack.

Works from 1 to 10.