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

Write a code that has the same number of bytes as the cop post and where the revealed characters are identical. The code must produce 1 in its initial state, and produce the numbers 2, 3 .. when single alterations are made.

The winner will be the robber that cracked the most posts.

Example post:

MyLang, 9 bytes, 8 numbers (including link to answer), Cop's username.

Original code:

abc1efghi

Code that produces 1 - 8:

1: abc1efghi
2: abc2efghi
3: abc2efghij
4: kbc2efghij
5: kbc23fghij
6: kbc23fghi
7: kbc63fghi
8: kbc63fhi

Explanations are encouraged. The exact format of the robber post is optional.

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85 Answers 85

Reset to default
1 2
3
2
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Python, 10+ numbers, 61 bytes, Flp.Tkc

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

For subsequent numbers, add a newline to the start of the program.

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1
  • \$\begingroup\$ Good job! I knew it wasn't hard to crack, I just wanted a funny way to print numbers: "hey, let's throw an error and print the line it's on " \$\endgroup\$
    – FlipTack
    Commented Nov 20, 2016 at 10:29
1
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Pyth, 4 bytes, 9 numbers, isaacg

Original code:

."1Q

Which is a packed string containing the character 1. The other programs:

."2Q
."3Q
."4Q
."5Q
."6Q
."7Q
."8Q
."9Q

I thought this would also contain some unprintables, but if you insert a l in the beginning of each code, you can see that the length of each string is always 1 (only containing the digit).

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1
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Python, 28 bytes, 10 numbers, by CrazyPython

print (0+1+0*1000)+(0*0*0*4)
print (1+1+0*1000)+(0*0*0*4)
print (2+1+0*1000)+(0*0*0*4)
...
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1
  • \$\begingroup\$ My original solution had odr and stuff.. welp you cracked it; congrats. \$\endgroup\$
    – noɥʇʎԀʎzɐɹƆ
    Commented Nov 14, 2016 at 3:23
1
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><>, 5 bytes, 10 numbers, Teal pelican

>21n;
>22n;
>23n;
>24n;
>25n;
>26n;
>27n;
>28n;
>29n;
>2an;
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1
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Javascript 19 bytes 10 numbers, by Shaun H

Original:

alert((_1*1_10__1_)

Code that produces 1 - 10:

alert(( 1*1%10)/1 )
alert(( 2*1%10)/1 )
alert(( 3*1%10)/1 )
alert(( 4*1%10)/1 )
alert(( 5*1%10)/1 )
alert(( 6*1%10)/1 )
alert(( 7*1%10)/1 )
alert(( 8*1%10)/1 )
alert(( 9*1%10)/1 )
alert(( 9*1%10)+1 )
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1
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Python 2, 26 bytes, 10 numbers by DLosc, non-competing

Non-competing because Sp3000 beat me to it.

print r___e(_______+_[___]

print range(19)[1]#+_[___]
print range(19)[2]#+_[___]
print range(19)[3]#+_[___]
print range(19)[4]#+_[___]
print range(19)[5]#+_[___]
print range(19)[6]#+_[___]
print range(19)[7]#+_[___]
print range(19)[8]#+_[___]
print range(19)[9]#+_[___]
print range(19)[-9]#+_[___]
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1
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R, 21 bytes, 10 numbers, Gregor

  • Cop's: __i___________i______
  • Rob's: write(a<-1+0,file="")

Of course there's a bunch of possibility as a can be any letter (and is not needed at all)...

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1
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R, 8 bytes, 10 numbers by Tensibai

Original code:

_a___+__

Modified:

max(0+1)
max(1+1)
...
max(4+5)
max(5+5)
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3
  • 1
    \$\begingroup\$ That's the idea, I assume that it's ok as the problem didn't specify if the program should print or return, so cat or max will give the same thing for 1+0 to 1+9. Good job ;) \$\endgroup\$
    – Tensibai
    Commented Nov 15, 2016 at 12:41
  • 1
    \$\begingroup\$ @Tesnbai Awesome! I've not used R before so I wasn't sure on IO rules... Thanks for confirming! \$\endgroup\$
    – Dom Hastings
    Commented Nov 15, 2016 at 12:44
  • 1
    \$\begingroup\$ Well, on a script call, this one won't output anything, the result will only be printed in an interactive session, but as the original challenge doesn't really specify it, I assume it's ok :) \$\endgroup\$
    – Tensibai
    Commented Nov 15, 2016 at 12:47
1
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ASMD, 8 bytes, 10 numbers, by Oliver

  • For 1, 1t-t+++C
  • For 2, 1t-t+++C{
  • For 3, 1t-t+++C{{
  • …

Works with the latest commit at the time of posting. C claims to be implemented, but it isn’t.

Each program works as follows:

1              Push 1.             1
 t             Triplicate.         1 1 1
  -            Subtract.           1 0
   t           Triplicate.         1 0 0 0
    +          Add.                1 0 0
     +         Add.                1 0
      +        Add.                1
       C       Do nothing.         1
        {{{    Increment n times.  4
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1
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Perl, 31 bytes, 10 numbers, by Gabriel Benamy

Original code:

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

Code that produces 1 to 10 (note: these answers each have a leading space that doesn't show up in StackExchange's markdown):

  1. $e=$a+q<1_$>;1 while!$e ;say$e
  2. $e=$a+q<1_$>+1 while!$e ;say$e
  3. $e=$a+q<1_$>+2 while!$e ;say$e
  4. $e=$a+q<1_$>+3 while!$e ;say$e
  5. $e=$a+q<1_$>+4 while!$e ;say$e
  6. $e=$a+q<1_$>+5 while!$e ;say$e
  7. $e=$a+q<1_$>+6 while!$e ;say$e
  8. $e=$a+q<1_$>+7 while!$e ;say$e
  9. $e=$a+q<1_$>+8 while!$e ;say$e
  10. $e=$a+q<1_$>+9 while!$e ;say$e

I'm genuinely unsure whether you intended something like this and were aiming for misdirection, or whether this solution is completely different from the intended one.

The main trick I used is Perl's ability to get rid of unwanted characters by using a q, thus changing them into custom quotation marks (while allowing me to drop something that looks like 1 as an integer into the string that's actually quoted). It's not a trick that's normally that useful when golfing, because Perl has quote marks that use fewer bytes, but it comes in handy for challenges like this.

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1
  • \$\begingroup\$ Holy crap?? This is not even remotely what I was going for, but really nicely done! Let me see if I can add one more character to give you another challenge! \$\endgroup\$
    – Gabriel Benamy
    Commented Nov 15, 2016 at 16:22
1
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Perl, 7 bytes, 10 numbers, ais523

^F means a "Control-F" character. $^F is a magic variable that equals 2.

say$^F-1
say$^F*1
say$^F+1
say$^F+2
say$^F+3
say$^F+4
say$^F+5
say$^F+6
say$^F+7
say$^F+8
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  • \$\begingroup\$ Unless "control-f" is 2 bytes, this is only 7 bytes, not 8. \$\endgroup\$
    – Riley
    Commented Nov 15, 2016 at 21:30
  • \$\begingroup\$ I miscounted. The original was meant to be 7 bytes. \$\endgroup\$
    – user62131
    Commented Nov 15, 2016 at 22:19
1
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Ruby, 81 bytes, 10 numbers, histocrat

Yay comments...

require'digest/md5'
p Digest::MD5.digest('')&&0+1#')[n=0].ord^'straYNpraq'[n].ord

0+1 ... 9+1 as usual.

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2
  • \$\begingroup\$ You're quite lucky, that barely fits... \$\endgroup\$
    – ETHproductions
    Commented Nov 16, 2016 at 0:17
  • \$\begingroup\$ Ha, I should've left in a linebreak. \$\endgroup\$
    – histocrat
    Commented Nov 16, 2016 at 14:23
1
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Python 2, 33 bytes, 9 numbers, CrazyPython

Original code:

#__________________
riny
________

Code to print 1:

# my comment
riny=1
riny
print +1

And then just increment the 1 from there on out.

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2
  • \$\begingroup\$ Try solving it without exploiting an unintended easy solution. \$\endgroup\$
    – noɥʇʎԀʎzɐɹƆ
    Commented Nov 16, 2016 at 0:52
  • \$\begingroup\$ @CrazyPython Do you mean "without literal newlines", or there something further? \$\endgroup\$
    – xnor
    Commented Nov 16, 2016 at 0:53
1
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Alternate crack to DLosc, 26 bytes, 10 numbers

original:

print r___e(3_4____+_[___]

crack

>>> print r'10e(3642345+'[0+0]
1
>>> print r'10e(3642345+'[0+7]
2
>>> print r'10(3642345+'[0+7]
3
>>> print r'103642345+'[0+7]
4
>>> print r'10642345+'[0+7]
5
>>> print r'10642345+'[0+2]
6
>>> print r'10742345+'[0+2]
7
>>> print r'10842345+'[0+2]
8
>>> print r'10942345+'[0+2]
9
>>> print r'10942345+'[0:2]
10

programs 2 to 5's purpose is to place a number next to 10, so that for 10, we can change [0+2] to [0:2], and get the number 10. the rest just change the number to get the other numbers

nice challenge :)

Intended solution?

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3
  • \$\begingroup\$ This is essentially the intended solution, just with a few details changed. Nice work! \$\endgroup\$
    – DLosc
    Commented Nov 16, 2016 at 1:22
  • \$\begingroup\$ @DLosc I take it that the original program was print r'10e(3541234+'[0+7]? \$\endgroup\$
    – Destructible Lemon
    Commented Nov 16, 2016 at 1:51
  • \$\begingroup\$ No--see edit to cop submission. \$\endgroup\$
    – DLosc
    Commented Nov 16, 2016 at 1:56
1
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Hexagony, 12 bytes, 10 numbers, Sunny Pun

[!_>!@!1)@!_
[!_>!@!2)@!_
[!_>!@!3)@!_
[!_>!@!4)@!_
[!_>!@!5)@!_
[!_>!@!6)@!_
[!_>!@!7)@!_
[!_>!@!8)@!_
[!_>!@!9)@!_
[!__!@!9)@!_

Try it online!

Unfortunately, this is again a cheap solution.

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0
1
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Hexagony, 18 bytes, 10 numbers, Riley

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

Try it online!

Certainly not the intended solution.

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1
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Acc!!, 46 bytes, 10 numbers, DLosc

1
Count i while 0^_-i  {
Write 49
}
Write _+48

The first character goes 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. This uses a Python golfing trick that 0**_ is like +(not _). So if _ (which is set by the first line) is zero, the loop will execute once, while if it is nonzero the loop will execute zero times.

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1
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><>, 23 bytes, Teal pelican

"H"|;v+2i~?
  _  >l?!;n

For the remaining 9 numbers, replace the i with 0, 1, ..., 8 successively.

Try it online!

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1
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QBasic, 12 bytes, 10 numbers, by DLosc

Cop:

_I_T__O_(3))

Crack:

?I+T-(O<(3))
?I+1-(O<(3))
?I+2-(O<(3))
?I+3-(O<(3))
?I+4-(O<(3))
?I+5-(O<(3))
?I+6-(O<(3))
?I+7-(O<(3))
?I+8-(O<(3))
?I+9-(O<(3))

Still downloading QB64. But I think this should work. Not sure whether it is the original, though.

QBasic uses bitwise operators as logical operators, instead of having both of them. So true is -1.

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1
  • \$\begingroup\$ Well that was unexpected. No, definitely not the original. ;) \$\endgroup\$
    – DLosc
    Commented Nov 18, 2016 at 18:09
1
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QBasic, 12 bytes, 10 numbers, by DLosc

Cop:

_INT(_O_(3))

Crack:

?INT(LOG(3))
?INT(LOG(13))
?INT(LOG(33))
?INT(LOG(73))
?INT(LOG(173))
?INT(LOG(573))
?INT(LOG(1573))
?INT(LOG(5573))
?INT(LOG(9573))
?INT(LOG(39573))
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1
  • \$\begingroup\$ Finally, the intended solution! \$\endgroup\$
    – DLosc
    Commented Nov 18, 2016 at 18:48
0
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Hexagony, 15 bytes, 10 numbers, Sunny Pun

[!?>!@!1)\!?!!!
[!?>!@!2)\!?!!!
[!?>!@!3)\!?!!!
[!?>!@!4)\!?!!!
[!?>!@!5)\!?!!!
[!?>!@!6)\!?!!!
[!?>!@!7)\!?!!!
[!?>!@!8)\!?!!!
[!?>!@!9)\!?!!!
$!?>!@!9)\!?!!!

Try it online!

This might be the intended solution this time, but I'm not entirely sure. This one was definitely a lot more challenging in any case. :)

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1
  • \$\begingroup\$ I laughed into tears because I changed the @ in my mind to a `\` as that looks less repetitive to the last post... Anyway I am impressed to have forced a change to the 1st byte in the code :) \$\endgroup\$
    – Sunny Pun
    Commented Nov 16, 2016 at 13:01
0
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Python 2, 110 bytes, 10 numbers CrazyPython

from hashlib import *
h=md5()
h.update('*F[`')
if h.hexdigest()=='88f4002e27b0902c7018359da9bc5b44':print(0+1)
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0
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TI-Basic, 13(?) bytes, 10 numbers, Timtech

DelVar A1→θ
BL>A
θ

The 1 goes up to 9, and then we change it to Xmax. I'm too good at remembering useless information I don't even want to know.

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3
  • \$\begingroup\$ Nice try! Almost correct, but you cannot assume variables like Xmax will be set to certain values. tibasicdev.wikidot.com/graphscreen \$\endgroup\$
    – Timtech
    Commented Nov 17, 2016 at 12:06
  • \$\begingroup\$ @Timtech Yes I can. See for example codegolf.stackexchange.com/a/52967/30688 \$\endgroup\$
    – feersum
    Commented Nov 17, 2016 at 21:10
  • \$\begingroup\$ Alright then, whatever. I made the mistake of thinking you had to add characters... didn't realize substitutions xD \$\endgroup\$
    – Timtech
    Commented Nov 17, 2016 at 21:26
0
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PHP, 38 bytes, 10 numbers, by Leo

Original code:

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

Cracks for 1-10:

  1. <?php "n p ( n= ( =1"; echo +1?>
  2. <?php "n p ( n= ( =1"; echo 1+1?>
  3. <?php "n p ( n= ( =1"; echo 2+1?>
  4. <?php "n p ( n= ( =1"; echo 3+1?>
  5. <?php "n p ( n= ( =1"; echo 4+1?>
  6. <?php "n p ( n= ( =1"; echo 5+1?>
  7. <?php "n p ( n= ( =1"; echo 6+1?>
  8. <?php "n p ( n= ( =1"; echo 7+1?>
  9. <?php "n p ( n= ( =1"; echo 8+1?>
  10. <?php "n p ( n= ( =1"; echo 9+1?>

Unfortunately, you left way too much space after the echo and it became possible to sneak a +1 in there.

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0
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ASMD, 10 bytes, by Oliver

Original:

1_**_**_*÷

Code that produces 1 - 10:

1t**t**t*÷
1t**t**t*÷{
1t**t**t*÷{{
...
1t**t**t*÷{{{{{{{{{

Did I miss something? I had to mess around with the interpreter to get it to work on my system, and this seems too easy.

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