175
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In mathematics an exclamation mark ! often means factorial and it comes after the argument.

In programming an exclamation mark ! often means negation and it comes before the argument.

For this challenge we'll only apply these operations to zero and one.

Factorial
0! = 1
1! = 1

Negation
!0 = 1
!1 = 0

Take a string of zero or more !'s, followed by 0 or 1, followed by zero or more !'s (/!*[01]!*/).
For example, the input may be !!!0!!!! or !!!1 or !0!! or 0! or 1.

The !'s before the 0 or 1 are negations and the !'s after are factorials.

Factorial has higher precedence than negation so factorials are always applied first.
For example, !!!0!!!! truly means !!!(0!!!!), or better yet !(!(!((((0!)!)!)!))).

Output the resultant application of all the factorials and negations. The output will always be 0 or 1.

Test Cases

0 -> 0
1 -> 1
0! -> 1
1! -> 1
!0 -> 1
!1 -> 0
!0! -> 0
!1! -> 0
0!! -> 1
1!! -> 1
!!0 -> 0
!!1 -> 1
!0!! -> 0
!!!1 -> 0
!!!0!!!! -> 0
!!!1!!!! -> 0

The shortest code in bytes wins.

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  • 18
    \$\begingroup\$ But 0!=1!, so what's the point of handling multiple factorials? \$\endgroup\$ – boboquack Feb 6 '17 at 9:22
  • 30
    \$\begingroup\$ @boboquack Because that's the challenge. \$\endgroup\$ – Calvin's Hobbies Feb 6 '17 at 9:24
  • 11
    \$\begingroup\$ <?='1'; ... correct 75% of the time in php. \$\endgroup\$ – aslum Feb 6 '17 at 17:17
  • 10
    \$\begingroup\$ I may be wrong here but can't any number with any factorials after it simply be removed and replaced with 1? Like 0!!!! = 1!! = 0!!!!!!!! = 1!!! = 1! = 0! = 1 etc \$\endgroup\$ – Albert Renshaw Feb 6 '17 at 22:33
  • 2
    \$\begingroup\$ @AlbertRenshaw That is correct. \$\endgroup\$ – Calvin's Hobbies Feb 6 '17 at 22:54

47 Answers 47

1
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C#, 88 84 bytes

Saved 4 bytes thanks to TheLethalCoder.

s=>{var c=s.Replace("!","")[0];int b=s.IndexOf(c);return(s.Length>++b?b:b+c-49)%2;};

Previous version:

s=>{var c=s.Replace("!","")[0];int b=s.IndexOf(c),n=s.Length-b>1?1:c-48;return(n+b)%2;};

Full program with commented method and test cases:

using System;

class MathIsFactProgrammingIsNot
{
    static void Main()
    {
        Func<string, int> f =
        s=>
        {
            // removes all the exclamation marks and extracts the 0 or 1 digit
            var c = s.Replace("!","")[0];

            // number of exclamation marks before the digit
            int b = s.IndexOf(c),

            // the number of exclamation marks after the digit increased by 1 (because of the digit)
                n = s.Length - b > 1 ? 1 : c-48;
            // if no exclamation marks are present, converts the digit from the string to an integer

            return (n + b) % 2;   // applies binary negation
        };

        // test cases:
        Console.WriteLine(f("0"));  // 0
        Console.WriteLine(f("1"));  // 1
        Console.WriteLine(f("0!")); // 1
        Console.WriteLine(f("1!")); // 1
        Console.WriteLine(f("!0")); // 1
        Console.WriteLine(f("!1")); // 0
        Console.WriteLine(f("!0!"));    // 0
        Console.WriteLine(f("!1!"));    // 0
        Console.WriteLine(f("0!!"));    // 1
        Console.WriteLine(f("1!!"));    // 1
        Console.WriteLine(f("!!0"));    // 0
        Console.WriteLine(f("!!1"));    // 1
        Console.WriteLine(f("!0!!"));   // 0
        Console.WriteLine(f("!!!1"));   // 0
        Console.WriteLine(f("!!!0!!!!"));   // 0
        Console.WriteLine(f("!!!1!!!!"));   // 0
    }
}
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  • 1
    \$\begingroup\$ n is only used in the return so you can remove declaring it. \$\endgroup\$ – TheLethalCoder Feb 7 '17 at 15:59
1
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Pyth, 11 bytes

s.v:z"0!"\1

Try it online!

Explanation

s.v:z"0!"\1
   :         Replace...
    z        in the input...
     "0!"    the string "0!"...
         \1  with the string "1".
 .v          Evaluate the result. Since only the first expression is evaluated,
             anything after the number will be ignored.
s            Convert the result to an integer and implicitly print it.
             This is necessary because ! returns True/False, not 0/1.
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1
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PowerShell, 28 bytes

+($args-replace'\d!+',1|iex)

Try it online!

Explanation

Replace any digit followed by 1 or more ! with 1 (to solve the factorials), then just execute the remaining string which is valid code. The result will be boolean so I use unary + to convert it to a number.

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1
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F#, 69 48 Bytes

let rec f=function|'!'::t->1-f t|'0'::[]->0|_->1

Try it online!

Tests

let test (input,expected) = 
  let result = input |> Seq.toList |> f
  if result = expected then
    printfn "%s = %d" input result
  else
    printfn "Error at %s" input

[
  ("0", 0)
  ("1", 1)
  ("0!", 1)
  ("1!", 1)
  ("!0", 1)
  ("!1", 0)
  ("!0!", 0)
  ("!1!", 0)
  ("0!!", 1)
  ("1!!", 1)
  ("!!0", 0)
  ("!!1", 1)
  ("!0!!", 0)
  ("!!!1", 0)
  ("!!!0!!!!", 0)
  ("!!!1!!!!", 0)
] |> Seq.iter test
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1
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Vim + bc, 18 bytes

:s/\d!\+/1␊V!bc␊

is a literal newline

Explanation

:s/\d!\+/1␊  Replace any digit followed by factorials with 1 ("0!!!" -> "1"; "1!!" -> "1")
V!bc␊        Evaluate the not-operators using the `bc` command, similarly to this answer

Vim, 36 bytes

:s/\d!\+/1␊:s/!!//g␊:s/!1/0␊:s/!0/1␊

is a literal newline

Explanation

:s/\d!\+/1␊  Replace any digit followed by factorials with 1 ("0!!!" -> "1"; "1!!" -> "1")
:s/!!//g␊    Remove all double nots ("!!!!!" -> "!"; "!!!!!!" -> "")
:s/!1/0␊     Replace !1 with 0
:s/!0/1␊     Replace !0 with 1
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1
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C (gcc), 42 bytes

c;e(char*a){c=*a<34?!e(a+1):*a-48|1[a]%2;}

Try it online!

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1
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Japt, 11 bytes

+OxUe/\d!/1

Try it online!

How it works

+OxUe/\d!/1

   Ue        Apply recursive replace to input string...
     /\d!/1    which replaces digit+! to 1
 Ox          Eval the result as vanilla JS
+            Cast the result (String or Boolean) to Number

The idea is similar to Arnauld's JS answer though I came up with it independently.

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0
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RProgN, 31 bytes

~'(!*1?)0?(!*)'{L`x=L2%x+0>1*}R

Explained

~'(!*1?)0?(!*)'{L`x=L2%x+0>1*}R
~                               # Zero Space Segment
 '(!*1?)0?(!*)'                 # A pattern string, matching any number of !'s with optionally a 1, optionally an uncaptured 0, and any number of !'s
               {             }  # An anonymous function, which takes two arguments. The last !'s and the optional 1 with the first 1's.
                L               # Get the length of the last !'s
                 `x=            # Set 'x' to equal it.
                    L2%         # Get the length of the first !'s with the optional 1, mod 2, giving us the boolean portion.
                       x+       # Add x
                         0>1*   # If the total is larger than 0, converted to a number. If there are any leading !'s, this will always be 1, otherwise, it will be the boolean of the left handside.
                              R # Replace the input string via the function matching the first pattern.

Try it online!

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0
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Pyth, 13 bytes

Code

s.v:z"\d!+""1

There may be a way to shave off a couple of bytes, but alas.

Explanation

s                # Cast to an integer (Python's int()).
 .v              # Evaluate (Python's eval()). This handles the negations.
   :             # Regex substitution. The following three expressions are its arguments.
    z            # Argument 1: what to replace in. This is equal to the (unevaluated) input string.
     "\d!+"      # Argument 2: what to replace. This is a regex that matches a number followed by one or more !'s.
           "1    # Argument 3: what to replace to. The string "1" (ending quote not needed in Pyth).

You can check it out here or run the test suite here. I have no earthly idea how to (or if one actually can) use the test suite feature to run tests as opposed to just evaluating a bunch of inputs at once, but if someone else knows, I'm all ears.

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0
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Haskell, 42

There must be a better way to do this...

f(h:t)|h=='!'=1-f t|h=='1'=1|t==[]=0|1<2=1
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0
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C, 56 bytes

c=1;f(char*a){c=-c;*a&16?c+=1-*a-1[a],c&=2,c/=2:f(a+1);}

Hint: only the last two bits count.

The basic idea was to use least significant bit for storing result of factorial, and next bit for storing the negation, then xoring the two.

c=0;
for(;*a<34;a++)c^=2; // invert the 2nd bit at each negation
while(*a)c|=*a++; // '0' ends with bits 00, '1' and '!' ends with bits 01, so this OR will let the first bit to resut of factorial (LSB) and leave the 2nd bit unchanged
c=((c>>1)^c)&1; // apply the negation (2nd bit) on the factorial (1st bit)

But it makes our intentions too clear. First, we don't need a loop for the factorial, and we can allways take 2 char, the 2nd being eventually a NULL terminator will have neutral 00 end bits. This is much like the answer Mathematics is fact. Programming is not from Ahemone, but longer and less elegant so far.

c=0;
while(*a++<34)c^=2; // invert the 2nd bit at each negation
c|=*a,c|=*--a; // '0' and NULL ends with bits 00, '1' and '!' ends with bits 01, so this OR will let the first bit to resut of factorial (LSB) and leave the 2nd bit unchanged
c=((c>>1)^c)&1; // apply the negation (2nd bit) on the factorial (1st bit)

C isn't going to win anyway, so let's trade some golf for some obfuscation: replace the last expression with something else, assuming 2-complement: -x == (~x+1) and observe how the last two bits evolve

- ...00 -> ...11+1 -> ...00
- ...01 -> ...10+1 -> ...11
- ...10 -> ...01+1 -> ...10
- ...11 -> ...00+1 -> ...01

We see that the LSB is unchanged via c=-c, and the 2nd bit becomes the xor of last two bits. So we can just pick this second bit with c>>=1,c&=1 or c&=2,c/=2;

Of course, the bit inversion ~x is useless, just adding+1 has the same effect.
But there is a reason behind it:
what if we would replace the XOR flip/flop with negated op?
at each neg -...01 becomes ...11 et vice et versa
If we then subtract 1, we have either ...00 or ...10 at the end of the loop.
We are back to our original solution.

c=1;
while(*a++<34)c=-c;
c-=1;
c|=*a,c|=*--a;
c=-c;
c&=2,c/=2;

And let's see what happens if we add the factorial bit instead of ORing:

...00 becomes 00 or 01 or 10 in case of '0' , '0!'||'1' , '1!'.
...10 becomes 10 or 11 or 00.
So using + gives the same parity than | on last two bits, even if we accidentally add a bit twice du to '1!' case.

Now we just have to roll the final c-=c inside the loop, and replace the + by - for getting our obfuscated solution.

Ah and also use recursion to take a functional style disguise, but of course with non reentrant, ugly static variable assignment side effect, else there would be no "advantage" to code in C ;)

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0
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JavaX, 77 74 bytes

!7p{print(repeatMultiReplace3(args[0],splitAtSpace("0! 1 1! 1 !0 1 !1 0";}

Run with, e.g.: " java -jar x30.jar 1006862 '!0' "

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0
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SmileBASIC, 63 bytes

INPUT S$WHILE"#">S$[0]N=!N
S$[0]="
WEND?(VAL(S$)||LEN(S$)>1)!=N

I don't think this is the best way...

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0
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Java (OpenJDK 8), 109 bytes

s->{int l=s.length(),n=(l>1?s.split("[01]")[0]:s).length();return l<2?s:(l-n<2&&s.charAt(n)<'1')==n%2>0?1:0;}

Try it online!

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0
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Cubically, 42 bytes

UD3(L2~:7=3)6+13=7!6{<0%6&}~:7=3?6L2-6=0%6

Try it online!

For some unknown reason it exits with an error when the digit is 1, but it outputs the correct result regardless. The following is my basic algorithm:

  1. Start with 0
  2. Invert for each !
  3. If digit is 1 invert again and output (any number of trailing ! are irrelevant), ending the program
  4. If there is a ! after the digit, invert
  5. Output (any further ! are irrelevant)

And a more thorough explanation:

UD3                            Sets RIGHT to 33 and LEFT to 15

(L2~:7=3)6                     Flips TOP between 0 and 15 each time the input is ASCII 33 '!'

+13=7!6                        If the next character is NOT ASCII 15+33 '0':
       {<0%6&}                   output 1 if TOP is 15, 0 otherwise, then exit

~:7=3?6                        If the next character is ASCII 33 '!':
       L2                        flip TOP 

-6=0%6                         Output 1 if TOP is 0, 0 otherwise
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  • \$\begingroup\$ Each %6 can be shortened to % thanks to language updates. \$\endgroup\$ – MD XF Oct 8 '17 at 21:44
0
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Befunge-98 (PyFunge), 22 20 bytes

Golfed off 2 bytes because I realized that if I used k instead of j to do the logical notting to account for a 0 without a factorial, I didn't have to mod the ASCII value of 0 or 1 by 2. I did have to move the code because removing the 2% left a hole between the $~ and #< previously.

-kv$!~:'!
#<k!.@.$$~

Try it online!

Explanation

The first line:

-kv$         Does nothing because the top of the stack is 0
    !        Nots the top, so we start with a value of 1
     ~       Gets a character from input
-     :'!    Pushes (that character - the value of '!')
 kv          If that value is not 0 (the character was not '!'), go to the next line
   $!        If it is, Throw away the extra '!' and not the number below it (originally 1)
             Repeat from the ~

Now that we're on the second line, we have notted 1 for each ! before the number. If there is a ! after the number, we can just print this value, but if there isn't we need to adjust for whether or not the number is 0.

The second line:

 <            Directs the IP left.
#             Doesn't skip anything because it's at the beginning of a line
         ~    If we reached EOF (no factorials), the ~ will reverse the IP's direction
       $$     No EOF: Drop the '!' we just read along with the number
     @.               Print (1 notted the appropriate amount of times) and end
#<            EOF: Wrap around and skip the arrow
  k!               Not the top n + 1 times, where n is the ASCII value of 0 or 1 (48 or 49)
                       If the number is 1, it will be notted 50 times (even),
                       yielding no change.
                       If the number is 0, we not it 49 times (odd), which makes up for
                       starting with a 1 at the beginning.
    .@             Print this value and exit
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  • \$\begingroup\$ Can you save a couple of bytes by outputting via exit code? \$\endgroup\$ – Jo King Jan 15 '18 at 1:30
0
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Deorst, 42 bytes

o1:ER'[01]!+'gs''p@
'(!!)+'gst!0gso0@t!1gs

Try it online!

Regex based solution. Input must be quoted ('0')

How it works

Example input: '!!!0!!!!'

o1                     - Push '1';      STACK = ['!!!0!!!!', '1']
  :                    - Duplicate;     STACK = ['!!!0!!!!', '1', '1']
   ER                  - Reverse;       STACK = ['1', '1', '!!!0!!!!']
     '[01]!+'          - Push '[01]!+'; STACK = ['1', '1', '!!!0!!!!', '[01]!+']
             gs        - Regez replace; STACK = ['1', '!!!1']
               ''p     - Push '';       STACK = ['1', '!!!1', '']
                  @    - Swap;          STACK = ['1', '', '!!!1']
'(!!)+'                - Push '(!!)+';  STACK = ['1', '', '!!!1', '(!!)+']
       gs              - Regex replace; STACK = ['1', '!1']
         t!0           - Push '!0';     STACK = ['1', '!1', '!0']
            gs         - Regex replace; STACK = ['!1']
              o0       - Push '0';      STACK = ['!1', '0']
                @      - Swap;          STACK = ['0', '!1']
                 t!1   - Push '!1';     STACK = ['0', '!1', '!1']
                    gs - Regex replace; STACK = ['0']

Alternative, 27 bytes

t!1@t0$gs''p@
t!!gs'^\d'gcB

Try it online!

Requires input to be quoted ('0'). Based on Leo's Retina answer

How it works

t!1           - Push '!1';         STACK = ['!!!0!!!!', '!1']
   @          - Swap;              STACK = ['!1', '!!!0!!!!']
    t0$       - Push '0$';         STACK = ['!1', '!!!0!!!!', '0$']
       gs     - Regex replace;     STACK = ['!!!0!!!!']
         ''p  - Push empty string; STACK = ['!!!0!!!!', '']
            @ - Swap;              STACK = ['', '!!!0!!!!']
t!!           - Push '!!';         STACK = ['', '!!!0!!!!', '!!']
   gs         - Regex replace;     STACK = ['!0']
     '^\d'    - Push '^\d';        STACK = ['!0', '^\d']
          gcB - Count occurrences; STACK = [0]
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protected by Community Apr 21 '18 at 6:59

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