Code ladder, Robbers

Question

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.

Answer 1

Retina, 2 bytes, 10 numbers, by Martin Ender.

Original code:

1

This replaces the empty input by the number 1.

Code that produces 1 - 10 (showing newlines as \n for convenience):

1 : \n1
2 : \n1\n
3 : \n11\n
4 : \n111\n
5 : \n1111\n
6 : \n11111\n
7 : \n111111\n
8 : \n1111111\n
9 : \n11111111\n
10: \n111111111\n

For 2 and onwards, we count the number of empty strings in the previous stage. For 2, there is an empty string before and after the 1. For 3 there is an empty string before the ones, in between the ones and after the ones. This works all the way up to 10.

Try it online

Answer 2

Perl, 12 bytes, 10 numbers, by ais523

Original code:

____;say__-9

Code that produces 1 - 10:

wait;say$!-9
wait;say$!-8
wait;say$!-7
wait;say$!-6
wait;say$!-5
wait;say$!-4
wait;say$!-3
wait;say$!-2
wait;say$!-1
wait;say$!-0

Explanation:

wait;         # waits for a child process to end (there is no child so it just returns)
     say      # put the following on the screen:
        $!    # current value of the C errno. In this case, ECHILD (No child processes)
              # which has a value of 10
          -9  # minus 9

Answer 3

Python 2, 9 bytes, 10 numbers, xnor

print 8/8
print 18/8
print 28/8
print 38/8
print 38/7
print 38&7
print 31&7
print 31&72
print 31&73
print 31&74

Answer 4

JavaScript, 30 bytes, 10 numbers, by ETHproductions.

Original code:

1: alert(Array(2).join(3).length)

Other numbers

 2: alert(Array(2).join(31).length)
 3: alert(Array(2).join(314).length)
 4: alert(Array(2).join(3141).length)
 5: alert(Array(2).join(31415).length)
 6: alert(Array(2).join(314159).length)
 7: alert(Array(2).join(3141592).length)
 8: alert(Array(2).join(31415926).length)
 9: alert(Array(2).join(314159265).length)
10: alert(Array(2).join(3141592653).length)

Explanation

Array(2) creates an Array with two empty cells. The .join method joins all the elements in an Array using a delimiter, which converts it into a String. In the Original Code, the delimiter was 3. This meant that the two empty cells in the Array are joined together used 3 as their delimiter. This evaluates "3", without anything on either side of the 3 because the Array is empty. And since the length of "3" is 1, this value is outputted to the console.

For the other numbers, I add 1 more digit to the delimiter that is joining the elements of the Array. Hence, this increases its length, thus alerting with increasing values. Note that the numerical value of the delimiter is not important, only its length is.

Snack Snippet!

Note: I used console.log instead of alert in my Snack Snippet for obvious reasons. Also, output from the console is display in the snippet.

console.log(Array(2).join(3).length)
console.log(Array(2).join(31).length)
console.log(Array(2).join(314).length)
console.log(Array(2).join(3141).length)
console.log(Array(2).join(31415).length)
console.log(Array(2).join(314159).length)
console.log(Array(2).join(3141592).length)
console.log(Array(2).join(31415926).length)
console.log(Array(2).join(314159265).length)
console.log(Array(2).join(3141592653).length)

Answer 5

Octave, 17 bytes, 10 numbers, by Stewie Griffin

original:
_i_(__i__(2_5_))

solution:
fix( 'i'-(2*52))
fix( 'j'-(2*52))
fix( 'k'-(2*52))
fix( 'l'-(2*52))
fix( 'm'-(2*52))
fix( 'n'-(2*52))
fix( 'o'-(2*52))
fix( 'p'-(2*52))
fix( 'q'-(2*52))
fix( 'r'-(2*52))

Answer 6

Python 2, 17 bytes, 10 numbers, by xnor

print len(['  '])
print len(*['  '])
print len(*['   '])
print len(*['    '])
print len(*['     '])
print len(*['      '])
print len(*['       '])
print len(*['        '])
print len(*['         '])
print len(*['          '])

repl.it

Answer 7

Perl, 14 bytes, 10 numbers, by Dada

Original code:

say_!"___"%""_

Code that produces 1 - 10:

1. say"!"+1#"%""r
2. say"!"+2#"%""r
3. say"!"+3#"%""r
4. say"!"+4#"%""r
5. say"!"+5#"%""r
6. say"!"+6#"%""r
7. say"!"+7#"%""r
8. say"!"+8#"%""r
9. say"!"+9#"%""r
10. say"1"+9#"%""r

I have a feeling this isn't what you were going for.

Answer 8

JavaScript, 22 bytes, 10 numbers, by Arnauld

Original program:

alert(0_6_4_>_0_2_0_7)

Cracked:

1. alert(0+6^42>101210^7)
2. alert(0+5^42>101210^7)
3. alert(0+4^42>101210^7)
4. alert(0+3^42>101210^7)
5. alert(0+2^42>101210^7)
6. alert(0+1^42>101210^7)
7. alert(0+0^42>101210^7)
8. alert(0+0^42>101210^8)
9. alert(0+0^42>101210^9)
10. alert(0+3^42>101210^9)

The hard part is to deal with the >, which has a very low precedence. I tried changing it to an >>, but it's easiest to just absorb it by making it uselessly calculate a boolean false (the 42>101210 section) which is numerically 0, and just generate the numbers from 1 to 10 with operators with an even lower precedence than >. I used ^, bitwise xor, because it can generate the numbers from 1 to 10 fairly easily (conveniently, the 6 and 7 in the original input let me simplify the expression down to 6^7 which is 1).

Answer 9

JavaScript, 9 bytes, Hedi

Original code:

alert(+1)

Modifications:

alert(1+1)   // 2
alert(2+1)   // 3
alert(3+1)   // 4
alert(4+1)   // 5
alert(5+1)   // 6
alert(6+1)   // 7
alert(7+1)   // 8
alert(8+1)   // 9
alert(9+1)   // 10

Answer 10

JavaScript, 22 bytes, 10 numbers by Arnauld

Original code:

alert(__14_337__xc_de)

Modified:

alert(00140337-0xc0de) // 1
alert(00140337-0xc0dd) // 2
alert(00140337-0xc0dc) // 3
alert(00140337-0xc0db) // 4
alert(00140337-0xc0da) // 5
alert(00140337-0xc0d9) // 6
alert(00140337-0xc0d8) // 7
alert(00140337-0xc0d7) // 8
alert(00140337-0xc0d6) // 9
alert(00140337-0xc0d5) // 10

I'm guessing this isn't the intended solution, but I'm hoping it's not too far off...

Answer 11

Octave, 55 bytes, 10 numbers, flawr

(o=@(O,o)(@(O,o)o{2-O} ) (0<O,{ (yo=O-1)+1,@()1}))(1,o)

The final 1 can be cycled up to 9, and then change the +1 to +2.

Answer 12

Python 3, 16 bytes, 10 numbers, by Sp3000

This was fun.

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

print(1%0xfor t)
print(2%0xfor t)
...
print(8%0xfor t)
print(84%0xfor t)
print(85%0xfor t)

Exploits the fact (which I didn't know till now) that a hex literal ends as soon as a character that isn't 0-9a-fA-F is encountered--thus 0xfor is a very sneaky way to write 15 or. After that, it was a matter of finding the right value congruent to 10 modulo 15.

Answer 13

Ruby, 81 bytes, 10 numbers, histocrat

Original code:

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

Cracked:

x=/$/=~%q
)

p x

This was a tricky one. Many characters that were revealed turned out to be red herrings! / isn't division but part of a regex literal. % isn't mod but part of an uncommon string literal syntax. This program simply prints the length of a string delimited by newlines. The subsequent programs can be obtained by inserting additional characters in the second line.

Answer 14

Octave, 32 bytes, 10 numbers, by Luis Mendo

Original code:

nnz(isprime(floor(s=3):i*-2i))+1

Modified codes:

2:  nnz(isprime(floor(s=3):i*-3i))+1
3:  nnz(isprime(floor(s=3):i*-5i))+1
4:  nnz(isprime(floor(s=3):i*-9i))+1
5:  nnz(isprime(floor(s=3):i*-9i))+2
6:  nnz(isprime(floor(s=3):i*-9i))+3
7:  nnz(isprime(floor(s=3):i*-9i))+4
8:  nnz(isprime(floor(s=3):i*-9i))+5
9:  nnz(isprime(floor(s=3):i*-9i))+6
10: nnz(isprime(floor(s=3):i*-9i))+7

There are many ways to make modifications here (for instance s=2 and +0 in the beginning).

Edit: A more probable alternative:

nnz(isprime(floor(s=2):i*-2i))+0
nnz(isprime(floor(s=2):i*-2i))+1
nnz(isprime(floor(s=2):i*-2i))+2
nnz(isprime(floor(s=2):i*-2i))+3
nnz(isprime(floor(s=2):i*-2i))+4
nnz(isprime(floor(s=2):i*-2i))+5
nnz(isprime(floor(s=2):i*-2i))+6
nnz(isprime(floor(s=2):i*-2i))+7
nnz(isprime(floor(s=2):i*-2i))+8
nnz(isprime(floor(s=2):i*-2i))+9

Answer 15

05AB1E, 11 bytes, 10 numbers, by Oliver

Original code:

 1: 3628801zï0>

Other numbers

 2: 3628801zï0>>
 3: 3628801zï0>>>
 4: 3628801zï0>>>>
 5: 3628801zï0>>>>>
 6: 3628801zï0>>>>>>
 7: 3628801zï0>>>>>>>
 8: 3628801zï0>>>>>>>>
 9: 3628801zï0>>>>>>>>>
10: 3628801zï0>>>>>>>>>>

Explanation

3628801                        # push this number
       z                       # find its reciprocal
        ï                      # round it to the nearest integer (becomes 0)
         0                     # push 0
          >>>>                 # increment by 1 for each `>` present (this is for 4)
                               # implicit output

To summarise it, push 3628801, take its reciprocal and round that to the nearest integer (evaluates to 0). Then increment the stack. TADA!

Try it online! For output = 1 Then add as many >s as you want depending on what number you want to output.

Answer 16

JavaScript, 12 bytes, 10 numbers, Hedi

Original code:

alert(+[+1])

2 through 10:

alert(+[1+1])
alert(+[2+1])
alert(+[3+1])
alert(+[4+1])
alert(+[5+1])
alert(+[6+1])
alert(+[7+1])
alert(+[8+1])
alert(+[9+1])

There's only so much you can do with this much code...

Answer 17

Jelly, 7 bytes, 10 numbers, Jonathan Allan

Original code:

“1‘ỌȮḊ‘

Then:

“1‘‘ỌȮḊ‘
“1‘‘‘ỌȮḊ‘
“1‘‘‘‘ỌȮḊ‘
“1‘‘‘‘‘ỌȮḊ‘
“1‘‘‘‘‘‘ỌȮḊ‘
“1‘‘‘‘‘‘‘ỌȮḊ‘
“1‘‘‘‘‘‘‘‘ỌȮḊ‘
“1‘‘‘‘‘‘‘‘‘ỌȮḊ‘
“1‘‘‘‘‘‘‘‘‘ỌvḊ‘

You can check the last one on TryItOnline!

---

Another way would be

“2‘ỌȮḊ‘
“3‘ỌȮḊ‘
...
“9‘ỌȮḊ‘
“9‘ỌvḊ‘

Answer 18

Octave, 21 bytes, 10 numbers, by Stewie Griffin

The fpr was a nice misguidance=) But I love the (1') -> 2(1') -> 2+(1') sequence!

original:
disp(fpr___f__e_(_'_)     % original
disp(fprintf= e=(1'))     % my solution
disp(fprintf= e=2(1'))
disp(fprintf= e=2+(1'))
disp(fprintf= e=3+(1'))
disp(fprintf= e=4+(1'))
disp(fprintf= e=5+(1'))
disp(fprintf= e=6+(1'))
disp(fprintf= e=7+(1'))
disp(fprintf= e=8+(1'))
disp(fprintf= e=9+(1'))

Answer 19

Hexagony, 7 bytes, 10 numbers, by Martin Ender

Original:

1<@|!__

Code that produces 1 - 10:

1<@|!).
2<@|!).
3<@|!).
4<@|!).
5<@|!).
6<@|!).
7<@|!).
8<@|!).
9<@|!).
9<@|!).!

1: Try it Online!
9: Try it Online!
10: Try it Online!

Here is the hex for the first 9 programs (where # is the number that you want to print):

 # <
@ | !
 ) .

The memory edge is just set to the value you want to print, deflected to the SE and printed.

And the hex for 10:

  9 < @
 | ! ) .
! . . . .
 . . . .
  . . . 

Here, I added a character to the end which increased the hex size. First, the memory edge is set to 9, the pointer is deflected to the SE, memory is incremented, wraps around, the pointer is deflected to the SW, the value is printed and the program ends.

---

Here is a second solution:

1<@|!$)
2<@|!$)
3<@|!$)
...
8<@|!$)
9<@|!$)
9#@|!$)

1: Try it Online!
9: Try it Online!
10: Try it Online!

The hex for 1 - 9 (this time * is the number to print):

 * <
@ | !
 $ )

This works the same as 1 - 9 above.

The hex for 10:

 9 #
@ | !
 $ )

This hits 9 then # switched the instruction pointer to the lower right one, memory is incremented to 10, $ jumps over the !, the pointer is reversed back east where it jumps over ) and ends.

Answer 20

C#, 90 bytes, 10 numbers, Scepheo

Original:

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

First:

using System;class D{public D(){Console.Write(+1);}};class C{static void Main(){new D();}}

Then:

using System;class D{public D(){Console.Write(1+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(2+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(3+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(4+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(5+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(6+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(7+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(8+1);}};class C{static void Main(){new D();}}
using System;class D{public D(){Console.Write(9+1);}};class C{static void Main(){new D();}}

Answer 21

Python 2, 111 bytes, 10 numbers, feersum

Original, then the program for 1:

print +int('''3EE5EE6EE0EE2EE5EE6EE4EE9EE1EE5EE9EE4EE6EE7EE5EE7E''')%int('''2EE0EE9EE8EE9EE2EE3EE5EE0EE3EE6''')
print +int('''39951562603020050069949998519154798549467371659275''')%int('''2000009008009002\x3005000\x3006''')

The other programs are just appending 1-9 before the +. With numbers of this size the problem is fairly simple, since we have a lot of freedom in the lower digits of the larger number. To make things a bit easier, a few of the 3s in the modulus have been turned into \x30, which is the code point for 0.

Answer 22

Perl, 31 bytes, 10 numbers, by Gabriel Benamy

I'm pretty sure this is the intended answer this time (up to trivial variations, such as using '<' rather than "<").

Original code (must be saved to a file and run via giving the file at the command line, it doesn't matter what the name is):

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

Crack for 1:

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

To crack for the other values, append a newline to the file each time (e.g. the above prints 2 with one newline appended, 3 with two newlines appended, and so on). The program works by reading its own source code from disk (this isn't a quine competition, so you're allowed to do that!) and printing the line number of the line just beyond the end.

I found this via deducing that it probably started with a keyword or built-in function with a four letter name, and searching the list of keywords and builtin functions for functions matching ..e.. As soon as you realise that open is a possibility, everything is straightforward from there (the unknown character before the final semicolon is in retrospect very suspicious).

Answer 23

Octave, 25 bytes, 10 numbers, by Luis Mendo

mean(repmat(1,ones(1,1)))
mean(repmat(2,ones(1,1)))
mean(repmat(3,ones(1,1)))
mean(repmat(4,ones(1,1)))
mean(repmat(5,ones(1,1)))
mean(repmat(6,ones(1,1)))
mean(repmat(7,ones(1,1)))
mean(repmat(8,ones(1,1)))
mean(repmat(9,ones(1,1)))

Answer 24

아희(Aheui), 19 bytes, 10 numbers, by JHM

Original code:

봃法희
반자뭉

Modified:

2
봃法희
반반뭉
3
봃法희
반밬뭉
4
봃法희
반밭뭉
5
봃法희
반발뭉
6
봃法희
반타뭉
7
봃法희
반밝뭉
8
봃法희
반밣뭉
9
봃法희
반밢뭉
10
봃法희
반다뭉

Answer 25

Acc!!, 10 bytes, 9 numbers, by DLosc

Original code :

Write 49+_

To produce the numbers 1 to 9 :

Write 49+_
Write 49+1
Write 49+2
Write 49+3
Write 49+4
Write 49+5
Write 49+6
Write 49+7
Write 49+8

49 is the ascii code of 1. _ contains 0 by default (so 49+_ = 1). And Write prints the character corresponding to the ascii code of its argument. Pretty straight forward.

Answer 26

Octave, 24 bytes, 9 numbers, by flawr

max(repmat(1,ones(),1))%
max(repmat(2,ones(),1))%
max(repmat(3,ones(),1))%
max(repmat(4,ones(),1))%
max(repmat(5,ones(),1))%
max(repmat(6,ones(),1))%
max(repmat(7,ones(),1))%
max(repmat(8,ones(),1))%
max(repmat(9,ones(),1))%

Octave, 24 bytes 9 10 numbers

Note: This crack is for the first version of the cop post (posted the wrong snippet), with the revealed code looking like this: ___a__repmat(__one__),__). The correct version is above.

Original code:

+mean(repmat(1,ones(),1))

Modified versions that print 2-10

1+mean(repmat(1,ones(),1))
2+mean(repmat(1,ones(),1))
3+mean(repmat(1,ones(),1))
4+mean(repmat(1,ones(),1))
5+mean(repmat(1,ones(),1))
6+mean(repmat(1,ones(),1))
7+mean(repmat(1,ones(),1))
8+mean(repmat(1,ones(),1))
9+mean(repmat(1,ones(),1))

Answer 27

Octave, 21 bytes, 9 numbers, by flawr

I really enjoyed this one... Clever :)

a='repmat(1one'(1,8)
a='repmat(2one'(1,8)
a='repmat(3one'(1,8)
a='repmat(4one'(1,8)
a='repmat(5one'(1,8)
a='repmat(6one'(1,8)
a='repmat(7one'(1,8)
a='repmat(8one'(1,8)
a='repmat(9one'(1,8)

Answer 28

JavaScript, 15 bytes, 10 numbers, ETHProductions

Original code:

alert( +"+1"+0)  // Prints 1

Programs:

alert( +"+1"+1)  // Prints 2
alert( +"+1"+2)  // Prints 3
alert( +"+1"+3)  // Prints 4
alert( +"+1"+4)  // Prints 5
alert( +"+1"+5)  // Prints 6
alert( +"+1"+6)  // Prints 7
alert( +"+1"+7)  // Prints 8
alert( +"+1"+8)  // Prints 9
alert( +"+1"+9)  // Prints 10

Answer 29

Python 2, 49 bytes, 10 numbers, AlexRacer

Maybe (just maybe) the intended solution, but if so, plenty of bytes could have been golfed off:

y=10-6;s='0901100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0902100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0903100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0904100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0905100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0906100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0907100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0908100000';print int(s[8-8+y-2+0+0:y])
y=10-6;s='0909100000';print int(s[8-8+y-2+0+0:y])
y=10-4;s='0909100000';print int(s[8-8+y-2+0+0:y])

Answer 30

Befunge-93, 11 bytes, 10 numbers, James Holderness

905:**-8-.@
905:**-7-.@
905:**-6-.@
905:**-5-.@
905:**-4-.@
905:**-3-.@
905:**-2-.@
905:**-1-.@
905:**-0-.@
905:**-~-.@

Try it online!

Probably not the intended solution.

What I did was get rid of the pesky 25 (5:*) by multiplying it by 0. If we then put a 9 at the beginning then the outputs will be 9 - N where need to generate N from a single command. The only issue is -1 to get 10, but trying to read an integer if there's no input conveniently pushes a -1.

Answer 31

05AB1E, 5 bytes, 10 numbers, by Adnan.

Original code:

…[==1

1-10:

…[==1
…[==2
…[==3
…[==4
…[==5
…[==6
…[==7
…[==8
…[==9
…[==T

Explanation:

…[==   3 char string
    1  Push number

Answer 32

05AB1E, 6 bytes, 10 numbers, Adnan

Original code:

¼ [==q

Keep adding ¼'s to output up to 10:

¼¼ [==q
¼¼¼ [==q
¼¼¼¼ [==q
¼¼¼¼¼ [==q
¼¼¼¼¼¼ [==q
¼¼¼¼¼¼¼ [==q
¼¼¼¼¼¼¼¼ [==q
¼¼¼¼¼¼¼¼¼ [==q
¼¼¼¼¼¼¼¼¼¼ [==q

Try it online

How it works:

¼        # Increment counter_variable to count_of(¼), it's printed at some point...
  [      # Infinite loop start
   ==    # Print nothing because stack is empty
     q   # Quit

Answer 33

Python 3, 19 bytes, 10 numbers, Sp3000

Newlines are a real bummer.

print(##bin()

0+1)

0+1 ... 9+1.

Answer 34

JavaScript, 21 Bytes, 10 Numbers, Shaun H

Original:

alert(b_oa_"3____1"))

First:

alert(btoa|"3"&("1")) 

Then:

alert(btoa|"3"-("1")) 
alert(btoa|"3"-("0")) 
alert(btoa|"4"-("0")) 
alert(btoa|"5"-("0")) 
alert(btoa|"6"-("0")) 
alert(btoa|"7"-("0")) 
alert(btoa|"8"-("0")) 
alert(btoa|"9"-("0")) 
alert(btoa|"9"-(~"0"))

//Using console.log instead of alert

console.log(btoa|"3"&("1"))
console.log(btoa|"3"-("1"))
console.log(btoa|"3"-("0"))
console.log(btoa|"4"-("0"))
console.log(btoa|"5"-("0"))
console.log(btoa|"6"-("0"))
console.log(btoa|"7"-("0"))
console.log(btoa|"8"-("0"))
console.log(btoa|"9"-("0"))
console.log(btoa|"9"-(~"0"))

Answer 35

Python 2, 26 bytes, 10 numbers, DLosc

Original:

print r___e(3______+_[___]

Solve:

print range(3,)[1]#+_[___]
print range(3,)[-1]#+_[___]
print range(3,4)[-1]#+_[___]
print range(3,5)[-1]#+_[___]
print range(3,6)[-1]#+_[___]
print range(3,7)[-1]#+_[___]
print range(3,8)[-1]#+_[___]
print range(3,9)[-1]#+_[___]
print range(3^9)[-1]#+_[___]
print range(2^9)[-1]#+_[___]

Yeah, I'm fairly sure this still isn't the intended solve.

Answer 36

Befunge-93, 11 bytes, 10 numbers, James Holderness

"15:**-6-.@
"15:**-6-2.@
"15:**-6-2+.@
"15:**-6-3+.@
...
"15:**-6-9+.@

A more complete crack of James' Befunge cop, after Martin's here. Because Befunge-93 has a limited board, the IP actually waits until it gets to the end of the board before wrapping around, unlike other 2D languages such as ><>. This means that a wraparound char push with " ends up pushing the values of the empty cells too, which by default is 32 for space.

Hence here at the start we push a bunch of chars, with 32 on top, then 1:5**- on that gives 32 - (1*(5*5)) = 32 - 25 = 7. Subtract another 6, and we have 1 to start off the ladder.

Answer 37

JavaScript, 33 Bytes, 10 Numbers, Shaun H

Original:

alert(_to__"_Xc0__0_B6____Zp=="))

First:

alert(atob|"0Xc0"/0xB6||("Zp=="))

Then:

alert(atob|"0X1c0"/0xB6||("Zp=="))
alert(atob|"0X2c0"/0xB6||("Zp=="))
alert(atob|"0X2F0"/0xB6||("Zp=="))
alert(atob|"0X3F0"/0xB6||("Zp=="))
alert(atob|"0X4F0"/0xB6||("Zp=="))
alert(atob|"0X4F0"/0xB0||("Zp=="))
alert(atob|"0X5F0"/0xB0||("Zp=="))
alert(atob|"0X5F0"/0xA0||("Zp=="))
alert(atob|"0X5F0"/0x90||("Zp=="))

// Using console.log instead of alert

console.log(atob|"0Xc0"/0xB6||("Zp=="))
console.log(atob|"0X1c0"/0xB6||("Zp=="))
console.log(atob|"0X2c0"/0xB6||("Zp=="))
console.log(atob|"0X2F0"/0xB6||("Zp=="))
console.log(atob|"0X3F0"/0xB6||("Zp=="))
console.log(atob|"0X4F0"/0xB6||("Zp=="))
console.log(atob|"0X4F0"/0xB0||("Zp=="))
console.log(atob|"0X5F0"/0xB0||("Zp=="))
console.log(atob|"0X5F0"/0xA0||("Zp=="))
console.log(atob|"0X5F0"/0x90||("Zp=="))

---

The intended version:

alert(_to__"_Xc0__0_Bf____Zp=="))

First:

alert(atob|"0Xc0"/0xBf||("Zp=="))

Then:

alert(atob|"0X1c0"/0xBf||("Zp=="))
alert(atob|"0X2c0"/0xBf||("Zp=="))
alert(atob|"0X2c0"/0xB0||("Zp=="))
alert(atob|"0X3c0"/0xB0||("Zp=="))
alert(atob|"0X4c0"/0xB0||("Zp=="))
alert(atob|"0X4F0"/0xB0||("Zp=="))
alert(atob|"0X5F0"/0xB0||("Zp=="))
alert(atob|"0X5F0"/0xA0||("Zp=="))
alert(atob|"0X5F0"/0x90||("Zp=="))

// Using console.log instead of alert

console.log(atob|"0Xc0"/0xBf||("Zp=="))
console.log(atob|"0X1c0"/0xBf||("Zp=="))
console.log(atob|"0X2c0"/0xBf||("Zp=="))
console.log(atob|"0X2c0"/0xB0||("Zp=="))
console.log(atob|"0X3c0"/0xB0||("Zp=="))
console.log(atob|"0X4c0"/0xB0||("Zp=="))
console.log(atob|"0X4F0"/0xB0||("Zp=="))
console.log(atob|"0X5F0"/0xB0||("Zp=="))
console.log(atob|"0X5F0"/0xA0||("Zp=="))
console.log(atob|"0X5F0"/0x90||("Zp=="))

Intended crack?

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

// Using console.log instead of alert

console.log(atob|"0Xc0"-0xBf||("Zp=="))
console.log(atob|"0Xc0"-0xBe||("Zp=="))
console.log(atob|"0Xc0"-0xBd||("Zp=="))
console.log(atob|"0Xc0"-0xBc||("Zp=="))
console.log(atob|"0Xc0"-0xBb||("Zp=="))
console.log(atob|"0Xc0"-0xBa||("Zp=="))
console.log(atob|"0Xc0"-0xB9||("Zp=="))
console.log(atob|"0Xc0"-0xB7||("Zp=="))
console.log(atob|"0Xc0"-0xB6||("Zp=="))

Answer 38

Python 3, 56 bytes, 6 numbers, by Flp.Tkc

Cop's post:

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

Crack:

print(0+1)#__(str(ord(x))for x in ________).___________)
print(1+1)#__(str(ord(x))for x in ________).___________)
print(2+1)#__(str(ord(x))for x in ________).___________)
...

Answer 39

Ocaml, 51 bytes, 10 numbers, by feersum

Original code:

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

Cracks for 1-10:

1. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/0)
2. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/1)
3. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/2)
4. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/3)
5. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/4)
6. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/5)
7. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/6)
8. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/7)
9. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/8)
10. let(r, (-/)(*- r*)) =0,(+);;r;; print_int(1 -/9)

The hard part is to get rid of the /0, which would normally cause an error. There's just enough room to define a new operator (-/), which I defined to be an alias for addition (+). Then I could construct the answer with 1-/0, 1-/1, 1-/2, and so on. The rest of the solution fits the other required characters into place with comment marks (* *) and statement marks ;;, and mostly just cancels itself out (although I had to define r in order to handle the first r in the program and in order for ;;r;; to correctly be a no-op).

Answer 40

OCaml, 51 bytes, 10 numbers, by feersum

Original code:

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

Code that produces 1-10:

1. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/0/0)
2. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/1/0)
3. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/2/0)
4. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/3/0)
5. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/4/0)
6. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/5/0)
7. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/6/0)
8. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/7/0)
9. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/8/0)
10. let r,((*-^^^^-^^r*)/)=0,(+);;r;; print_int( 1/9/0)

The same trick still works; there's rather less room, because now I have to redefine division to mean addition rather than using a new operator, but there's still enough.

Answer 41

R, 22 bytes, 10 numbers by Gregor

Original code:

_et_______)____is_____

Cracked:

For 1

#et
cat(+1)#___is_____

For 2

#et
cat(1+1)#___is_____

For the rest just increment the 1's

found a short enough print to sneak it in a gap with newline and comment.

Answer 42

Octave, 16 bytes, 10 numbers by Stewe Griffin

Original code:

ev_l('P_NIS'-2_)

Cracked:

eval('PFNIS'-28)
eval('PFNIR'-28)
eval('PFNIQ'-28)
eval('PFNIP'-28)
eval('PFNIO'-28)
eval('PFNIN'-28)
eval('PFNIM'-28)
eval('PFNzM'-28)
eval('PFNGM'-28)
eval('PFNGN'-28)

After finding a way around the 8 without a zero it was easy. Trick was to use the 1 as a exponent which is a no-op and then contineru with +
relevant part unobfuscated here:

4*2-1
eval('PFNIM'-28)
4*2^1
eval('PFNzM'-28)
4*2+1
eval('PFNGM'-28)

Answer 43

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.

Answer 44

Pyth, 3 bytes, 10 numbers, by Yotam Salmon

Original code was __Q, with_ unknown.

1.Q: Implicit print 1, throw error when trying to evaluate empty input. Same answer for 2-9 (with the number 1 replaced by 2-9), and the code T.Q for 10 (since T is initialized to 10).

Answer 45

Octave, 19 bytes, 10 numbers, by flawr

Probably not the intended solution, but hey.

'psca'(1 )+'1'-160'
'qsca'(1 )+'1'-160'
...
'ysca'(1 )+'1'-160'

Answer 46

Octave, 26 bytes, 10 numbers, by LuisMendo

original
__a__repmat(_+one___1+__))

solution:
mean(repmat(0+ones, 1+0 ))
mean(repmat(1+ones, 1+0 ))
mean(repmat(2+ones, 1+0 ))
...

Answer 47

Octave, 17 bytes, 10 numbers, by flawr

From

_od(3_13_13_7_1_)

To the original code:

mod(3113+1387,11) #1

The rest of the numbers:

mod(3113+1388,11) #2
mod(3113+1389,11) #3
mod(3114+1389,11) #4
mod(3115+1389,11) #5
mod(3116+1389,11) #6
mod(3117+1389,11) #7
mod(3118+1389,11) #8
mod(3119+1389,11) #9
mod(3119+1379,11) #10

Try it online! for the first case

Explanation

mod(x, y) computes the modulo of x and y. Here is the docs for mod.

Process

I don't know Octave, I've never used it, but I was interested in this particular sumbission. So I decided to try to crack it. First, I used this website to help me find an Octave command that is three characters long and has an od in the end. I found it to be mod. Then I began experimenting with the numbers in its online interpreter. After some time, I ended up with

mod(3_13+13_7,1_)

after which time I decided to create a python script (on TIOv2!) to brute-force the solution. The script printed the values for the _ whenever first_param%second_param==1. I first ran the python script for 10 (the second paramter to the mod function). I got tons of hits. Then I ran it again for 11. I got only a few hits this time, after which I decided to use 11 as the second parameter and the following values for the first parameter 3113+1387 just out of randomness. Now I got the original code. From there, I began incrementing the last digit of the numbers, so that I did 1387->1388->1389 and then 3113->3114->3115->3116->3117->3118->3119. Now I got the first 9 numbers.

mod(3119+1389,11) #evaluates to 9

Now 10 was the only one remaining. I tried changing the 1389 to 1399 to see if it becomes 10, but instead it resulted to 8. So I changed it to 1379, and it resulted in 10!

Answer 48

C, 216 bytes, 10 numbers, by Joshua

Original code :

#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;}

Cracked code :

#include <stdio.h>

FILE *a, *b;

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

Change the x = 0 to the number you want to print minus 1 : x=0 for 1, x=1 for 2, x=2 for 3 etc, until x=9 for 10.

Answer 49

PowerShell, 7 bytes, 10 numbers, by TimmyD

Original code :

___(__0

Cracked code :

+1#(__0
+2#(__0
+3#(__0
...
+9#(__0
1+9#(__0

Answer 50

Java, 70 bytes, 10(+) numbers, by masterX244

//Submission v
__________________________ ____(_____________________________)________
//Original code v
class A{public static void main(String[]a){System.out.print(')'-40);}}//1

This makes use of the character codes for numbers. The char ) is 41. So when you subtract 40 from it, it evaluates to 1. By incrementing the char by 1, we effectively increase the output by 1 too.

Try it online in Ideone!. Note: in Ideone, you need to have public before class and the name of the class should be HelloWorld

Other numbers:

class A{public static void main(String[]a){System.out.print('*'-40);}}//2
class A{public static void main(String[]a){System.out.print('+'-40);}}//3
class A{public static void main(String[]a){System.out.print(','-40);}}//4
class A{public static void main(String[]a){System.out.print('-'-40);}}//5
class A{public static void main(String[]a){System.out.print('.'-40);}}//6
class A{public static void main(String[]a){System.out.print('/'-40);}}//7
class A{public static void main(String[]a){System.out.print('0'-40);}}//8
class A{public static void main(String[]a){System.out.print('1'-40);}}//9
class A{public static void main(String[]a){System.out.print('2'-40);}}//10
...
class A{public static void main(String[]a){System.out.print('~'-40);}}//86

Edit: fixed mistake

Edit 2: Yes, this can go upto 86 and maybe even more!

Answer 51

Java, 126 bytes, 10 numbers, masterX244

Yes we have seen Unicode escapes many times here at PPCG.

cl\u0061ss Z\u007bpublic st\u0061tic void main(String[]a){System. out.print(/*_"___42*/0+1/*_00____00____".*/)/*______0)_*/;}}

s/0+1/1+1/ ... 9+1.

Answer 52

QBasic, 13 bytes by DLosc

?+INT(LOG(3))  'Fixed thanks to feersome
?1+INT(LOG(3))
?2+INT(LOG(3))
?3+INT(LOG(3))
?4+INT(LOG(3))
?5+INT(LOG(3))
?6+INT(LOG(3))
?7+INT(LOG(3))
?8+INT(LOG(3))
?9+INT(LOG(3))

Try QBasic at archive.org.

Seems like the intended solution is:

? INT(LOG(3))    '1
? INT(LOG(13))   '2
? CINT(LOG(13))   '3

but that can't possibly work because the natural log function grows too quickly. There are no sequences of this form that get past 4.

Answer 53

Python 2, 26 bytes, 10 numbers, DLosc

print range(11)[1or+ [  ]]
print range(11)[2or+ [  ]]
print range(11)[3or+ [  ]]
print range(11)[4or+ [  ]]
print range(11)[5or+ [  ]]
print range(11)[-5or+ [  ]]
print range(11)[-4or+ [  ]]
print range(11)[-3or+ [  ]]
print range(11)[-2or+ [  ]]
print range(11)[-1or+ [  ]]

I thought I'd try this with comments or newlines or anything. Definitely not the intended solution...

Answer 54

JavaScript (ES6), 63 bytes, 10 numbers by Arnauld

Original code:

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

Modified:

eval([...(n=0,"le0ta1(rx)")].sort(_=>[...8/73+[]][n++]).join``)
eval([...(n=0,"le0ta2(rx)")].sort(_=>[...8/73+[]][n++]).join``)
...
eval([...(n=0,"le0ta9(rx)")].sort(_=>[...8/73+[]][n++]).join``)
eval([...(n=0,"le0taa(rx)")].sort(_=>[...8/73+[]][n++]).join``)

This was a lot of fun to get to the bottom of, even though my first approach was off!

Answer 55

Perl, 31 bytes, 10 numbers, by Gabriel Benamy

Original code:

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

Cracked:

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

The principle using q<> I used last time doesn't seem to quite work here, so I used one of the intended uses of <> brackets (if a fairly obscure one) to create the string 1$ via globbing. Using += rather than = for the assignment to $a converts it to an integer, thus just 1, and then the solution continues the same way as my previous crack.

Fun fact: random sequences of letters in void context (such as the) are no-ops in Perl if they don't happen to spell a keyword.

Answer 56

Perl, 26 bytes, 10 numbers by Gabriel Benamy

Original code:

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

Cracked:

$_=0;@a=($_..10);say$a[+1]
$_=1;@a=($_..10);say$a[+1]
$_=2;@a=($_..10);say$a[+1]
$_=3;@a=($_..10);say$a[+1]
$_=4;@a=($_..10);say$a[+1]
$_=5;@a=($_..10);say$a[+1]
$_=6;@a=($_..10);say$a[+1]
$_=7;@a=($_..10);say$a[+1]
$_=8;@a=($_..10);say$a[+1]
$_=9;@a=($_..10);say$a[+1]

In fact I tested this on my Mac using print instead of say with perl -e... However I do think they should work basically the same.

This is incrementing the "base" of the array and say the second element of the array. Replacing _s by other characters should work too.

Answer 57

Perl, 26 bytes, 10 numbers, by Gabriel Benamy

Original code:

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

Cracked:

1. ____;@a=($|..10);say$a[+1]
2. ____;@a=($|..10);say$a[1+1]
3. ____;@a=($|..10);say$a[2+1]
4. ____;@a=($|..10);say$a[3+1]
5. ____;@a=($|..10);say$a[4+1]
6. ____;@a=($|..10);say$a[5+1]
7. ____;@a=($|..10);say$a[6+1]
8. ____;@a=($|..10);say$a[7+1]
9. ____;@a=($|..10);say$a[8+1]
10. ____;@a=($|..10);say$a[9+1]

I didn't even need the blanks at the start (and as ____ is a legal but pointless Perl statement, I left them as is). The only subtlety here is finding a variable that starts at 0, but there are plenty to choose from; I chose $|.

Answer 58

C++, 10 numbers, 468 bytes, by Joshua

Original program:

#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;
}

My crack (prints 1):

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

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

    std::ostream * const jawbone = garbalgase();
    /*$*/ int value = '_',vl=1;
}

int main()
{
    std:: cout<<+1/1;int vl = value;
    if (fork() == 0)
    {
        //$$$$$$$$$<std::ostream $*>($jawbone) = &std::cout;
        vl -= 4;
        _exit(0);
    }
    (*jawbone) << (vl - 9) << std::endl;
}

To get a crack for 2 up to 10, change the +1/1 to 1+1/1, 2+1/1, 3+1/1, and so on.

I originally tried to comment out much more of the program than this, but it turns out that most of it is naturally harmless and cancels itself out. (Because it happens to use _exit rather than exit, we can put the number to print out into the stdout buffer in both threads; the parent thread won't have time to flush it before it exits.) The hardest part is dealing with the } character just before int main(); you can't make it the closing brace of a function without -fpermissive, but you can make it the closing brace of a calling convention block (in this case I used extern "C++", the default, to basically cancel the braces out).

I also accidentally declared vl twice, but oh well. It doesn't break the crack, just shows how many spare characters there are here.

Answer 59

COW, 51 bytes, 10 numbers, by Gabriel Benamy

MoO ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ OOM // incomplete
MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM //original code outputs 1

Other numbers:

MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM //2
MoO MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM //3
MoO MoO MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM //4
MoO MoO MoO MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM //5
MoO MoO MoO MoO MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ Mo_ OOM //6
MoO MoO MoO MoO MoO MoO MoO Mo_ Mo_ Mo_ Mo_ Mo_ OOM //7
MoO MoO MoO MoO MoO MoO MoO MoO Mo_ Mo_ Mo_ Mo_ OOM //8
MoO MoO MoO MoO MoO MoO MoO MoO MoO Mo_ Mo_ Mo_ OOM //9
MoO MoO MoO MoO MoO MoO MoO MoO MoO MoO Mo_ Mo_ OOM //10

Explanation

MoO //increment current memory block value by 1.
OOM //print current memory block as output to STDOUT
Mo_ //gets ignored since the compiler doesn't recognise it

Try it online!

Cows don't moo, Cows quack!

Answer 60

Hexagony, 15 bytes, 10 numbers, Sunny Pun

\!?\1<<3@_!?_<3
\!?\2<<3@_!?_<3
\!?\3<<3@_!?_<3
\!?\4<<3@_!?_<3
\!?\5<<3@_!?_<3
\!?\6<<3@_!?_<3
\!?\7<<3@_!?_<3
\!?\8<<3@_!?_<3
\!?\9<<3@_!?_<3
\!?\9)<<3@_!?_<3

Try it online!

I'm afraid this wasn't sufficiently well constrained (as this is almost definitely nowhere near the intended solution).

Answer 61

Python 2, 26 bytes, 10 numbers DLosc

Original: print r___e(3_4____+_[___]

print range(304)[ 1+#[
0 ]

Where the 0 character is incremented up through 9.

Answer 62

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.

Answer 63

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.

Answer 64

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.

Answer 65

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.

Answer 66

><>, 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!

Answer 67

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.

Answer 68

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))

Answer 69

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).

Answer 70

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)
...

Answer 71

><>, 5 bytes, 10 numbers, Teal pelican

>21n;
>22n;
>23n;
>24n;
>25n;
>26n;
>27n;
>28n;
>29n;
>2an;

Answer 72

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 )

Answer 73

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]#+_[___]

Answer 74

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)...

Answer 75

R, 8 bytes, 10 numbers by Tensibai

Original code:

_a___+__

Modified:

max(0+1)
max(1+1)
...
max(4+5)
max(5+5)

Answer 76

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

Answer 77

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.

Answer 78

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

Answer 79

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?

Answer 80

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.

Answer 81

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. :)

Answer 82

Python 2, 110 bytes, 10 numbers CrazyPython

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

Answer 83

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.

Answer 84

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.

Answer 85

ASMD, 10 bytes (non-competing), 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.