## Mathematics is fact. Programming is not

163

19

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.

16But 0!=1!, so what's the point of handling multiple factorials? – boboquack – 2017-02-06T09:22:54.217

26@boboquack Because that's the challenge. – Calvin's Hobbies – 2017-02-06T09:24:17.793

78+1 just for the title! – ojdo – 2017-02-06T11:53:48.177

1What type can the output have? (String, char, integer, boolean, float...) – Zgarb – 2017-02-06T12:47:36.920

10<?='1'; ... correct 75% of the time in php. – aslum – 2017-02-06T17:17:03.383

1@Zgarb The type doesn't matter as long as it would be represented as plain 0 or 1 (not 0.0, not True, etc.) – Calvin's Hobbies – 2017-02-06T19:57:50.000

10I 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 – Albert Renshaw – 2017-02-06T22:33:17.057

2@AlbertRenshaw That is correct. – Calvin's Hobbies – 2017-02-06T22:54:58.433

1@AlbertRenshaw - my answer is just one (humble) example of how and why this works for this challenge. – ElPedro – 2017-02-06T23:09:51.033

Do we have to handle invalid inputs in any way? – theonlygusti – 2017-02-07T12:10:24.650

@theonlygusti since he didn't say anything about invalid inputs, you don't have to worry about them (that's the standart rules of codegolf). – Dada – 2017-02-07T15:04:36.787

You need to talk to Alan Turing about the title. – Craig Hicks – 2017-02-09T19:21:32.617

You could take this one step further and require !! to mean the double factorial etc.

– Tobias Kienzler – 2017-02-10T07:06:29.793

I finally get it. – djechlin – 2017-02-12T02:15:26.183

@TobiasKienzler That wouldn't change the results. – Martin Ender – 2017-02-14T12:11:49.277

@MartinEnder It sure would: `5!! = 5•3•1 = 15` while `(5!)! = 120! &gt; 6.6e198`. – Tobias Kienzler – 2017-02-14T13:17:10.663

@TobiasKienzler You don't ever get a value other than `0` or `1` as an argument to the (multi-)factorial though. – Martin Ender – 2017-02-14T13:25:46.687

@MartinEnder Ah yes, of course. The difference only matters when any integer is permitted. And in contrast to the recursive factorial the double, triple etc. factorial would not overflow so quickly :) – Tobias Kienzler – 2017-02-14T13:51:46.647

+1 for the great title! – sergiol – 2017-09-24T14:43:37.493

Can we output `True` for `1` and `False` for `0`? – caird coinheringaahing – 2018-01-14T18:05:47.163

41

## Mathematica, 25 17 bytes

``````Input[]/.!x_:>1-x
``````

Takes input from a user prompt. Assumes Mathematica's notebook environment for implicit printing. To make it a command-line script, wrap it in `Print[...]` or to make it an argumentless function (which then takes input from the prompt), append `&`.

Mathematica has both of the required operators (with the required precedence), so we can just "eval" the input (which is done automatically by `Input[]`), but the logical negation operator doesn't work on integers (so it will remain unevaluated). If there's a `!x` left in the result, we replace it with `1-x`.

A couple of fun facts about the evaluation:

1. Mathematica actually also has the double factorial operator `!!`, which computes `n*(n-2)*(n-4)*...`, but applied to `0` or `1` it still gives `1`, so it doesn't matter that `0!!!!!` will actually be parsed as `((0!!)!!)!`.
2. Even though Mathematica leaves `!0` and `!1` unevaluated, it does know that `!` is self-inverse, so it will automatically cancel all pairs of leading `!`. After the `ToExpression` we're always left with one of `0`, `1`, `!0`, `!1`.

2Since when was a REPL snippet allowed by default? – LegionMammal978 – 2017-02-07T12:26:08.407

2@LegionMammal978 Apparently since December 2015, but I keep forgetting about it. To be fair, it's not a "snippet" in that it doesn't assume that the input is already stored somewhere in memory. And assuming the notebook environment is then not very different from having a language with implicit output. – Martin Ender – 2017-02-07T12:28:00.323

Just curious, could a meta link be provided? (Trying to find information there is stressful, yet another problem of the SE Q&A format...) – LegionMammal978 – 2017-02-08T02:04:10.240

Pure ksh solution `x=\${x/[01]!*/1};echo \$((\$x))` - not allowed to post a proper answer :( – user17752 – 2017-02-09T01:45:50.833

27

# [Bash] + Unix utilities, 21 17 bytes

``````sed s/.!!*\$/1/|bc
``````

This must be saved in a file and run as a program. If you try to enter the command directly from the command line, it won't work because !! is expanded due to history substitution being enabled in bash's interactive mode. (Alternatively, you can turn history substitution off with `set +H`.)

Test case runs:

``````for x in 0 1 '0!' '1!' '!0' '!1' '!0!' '!1!' '0!!' '1!!' '!!0' '!!1' '!0!!' '!!!1' '!!!0!!!!' '!!!1!!!!'; do ./excl <<<"\$x"; done

0
1
1
1
1
0
0
0
1
1
0
1
0
0
0
0
``````

The old version works, this one doesn't – Cows quack – 2017-02-06T09:37:15.347

I used the TIO link – Cows quack – 2017-02-06T09:40:45.463

@KritixiLithos It worked fine when I tried it on my Linux box. The problem was apparently that TIO requires a newline at the end of the simulated input line. It's a confusing situation, so I took out the TIO link. If you want to try it out there, here's the link again (but be sure to include a newline at the end of the input if you change the input to test it out): https://tio.run/nexus/bash#@1@cmqJQrK@nqKilom@oX5OU/P@/oqKiIRArcn3Ny9dNTkzOSAUA

– Mitchell Spector – 2017-02-06T09:53:54.320

2

But what if someone has run `mkdir -p 's/.!!'{bunch,of,different,directories}\\$/1`? Then you'll get Pathname Expansion and Sed will be attempting to read directories as though they were files, instead of reading standard input, and it won't output anything! :)

– Wildcard – 2017-02-07T09:13:14.133

@Wildcard Yes, that's true, of course. You can avoid this issue by writing `\*` instead of `*` in the regex, at the cost of 1 byte, but it seems to be considered acceptable on PPCG to assume that the program is being run in a directory without strangely-named files that expressions like s/.!!*\$/1/ would expand to. – Mitchell Spector – 2017-02-07T09:23:10.633

Of course. :) Just commenting. In general, and in Production scripts, I wish people would play a little less fast and loose with assumptions relating to glob characters...but it's perfectly appropriate for code golf. Consider my comment a token of my wish that some reader will actually learn when quotes should be used in the shell.

– Wildcard – 2017-02-07T09:29:34.037

1@Wildcard I agree completely. In production scripts, I always use quotes in situations like this. (In this case, I would actually put double-quotes around the argument to sed, rather than just escaping the *. It's easier to read than using backslashes, and it avoids the possibility of missing some special character.) – Mitchell Spector – 2017-02-07T09:32:58.747

@MitchellSpector, ha—I always use single quotes unless I specifically need parameter expansion or command substitution. Or unless I'm nesting the command inside `find ... -exec sh -c 'for f; do ... ; done' sh {} +` or similar. Notably, history expansion is not prevented by double quotes. :) (Note: I'm surprised to see you don't have an account on Unix & Linux SE!)

– Wildcard – 2017-02-07T09:37:29.460

@Wildcard You're right -- this one would actually go in single quotes. I glanced at it quickly, saw the \$, and commented as if it used parameter expansion, but it doesn't. – Mitchell Spector – 2017-02-07T09:39:23.483

21

## Retina, 2015 14 bytes

Thanks to Leo for saving 1 byte.

``````0!
1
!!

^1|!0
``````

Try it online!

### Explanation

``````0!
1
``````

Turn `0!` into `1`. We don't care about any other trailing `!`s, the resulting number is the same as if we had applied all factorials.

``````!!

``````

Cancel pairs of negations. This may also cancel some factorials, but that's irrelevant.

``````^1|!0
``````

Count the number of matches of this regex, which is either `1` or `0` and gives the desired result.

Alternate solution for same bytecount: `\d.+`... – Cows quack – 2017-02-06T09:15:35.620

@KritixiLithos Found a way to avoid that altogether. – Martin Ender – 2017-02-06T10:07:46.957

You can remove the `^` before `!0` – Leo – 2017-02-06T12:53:21.960

@Leo Whoops, thank you. – Martin Ender – 2017-02-06T12:53:58.940

16

## Grime, 14 12 9 bytes

``````e`\0!~\!_
``````

Try it online!

## Explanation

This matches the input against a pattern, printing `1` for match and `0` for no match.

``````e`\0!~\!_
e`         Match entire input against this pattern:
!      not
\0       a sole 0
~     xor
\!   exclamation mark
_  followed by this pattern matched recursively.
``````

The idea is this. If the input begins with a digit, then the recursive part `\!_` always fails, and `\0!` succeeds unless we have a single `0`. Their xor succeeds unless the input is a single `0`. If the input begins with a `!`, then `\0!` always succeeds, and `\!_` succeeds if the recursive match succeeds. Their xor succeeds exactly when the recursive match fails, thus negating it.

14

# Python, -44- 42 bytes

Saved 2 bytes thanks to Zgarb!

``````lambda x:(x[-1]=='0')^len(x.rstrip('!'))%2
``````

Step by step:

1. `x[-1]!='0'`
if `x` ends with `1` or `!``x` doesn't end with `0`, the factorial portion must have value `1`, else `0`
2. `^len(x.rstrip('!'))%2`
exploit xor's property as a "conditional not". The condition in this case is if the length of initial `!`s is odd. However, `.rstrip` doesn't remove the number from the string so the length calculated is offset by 1, therefore the condition is inverted
3. The offset by 1 in step 2 is corrected by changing `!=` to `==` in step 1. Zgarb suggested using a difference comparison operator rather than applying another inversion, saving 2 bytes.

Try it online!

Fails on an input of `!!0`; it is currently returning `1`. – Value Ink – 2017-02-06T09:45:00.100

@ValueInk should be working now – busukxuan – 2017-02-06T09:55:28.633

1`lambda x:(x[-1]=='0')^len(x.rstrip('!'))%2` avoids the extra inversion. – Zgarb – 2017-02-06T11:18:24.543

@Zgarb Cool, thanks! – busukxuan – 2017-02-06T11:32:23.653

13crossed out 44 is still regular 44 – Riker – 2017-02-06T15:23:48.960

@EasterlyIrk What do you mean? – busukxuan – 2017-02-08T01:42:27.677

3I think he's stating that a crossed out 44 in the font used doesn't look crossed out... :) The cross out portion overlaps with the horizontal portion of the 4s. – JeffC – 2017-02-09T16:12:47.270

14

# Brainfuck - way to many bytes (232 bytes)

Clearly the wrong language for winning in code golf. Mainly I noticed a lack of anyone using this esolang. There is a good online interpreter bf interpeter or you can actually watch what the program does using this bf visualizer.

``````>>>>>,[>+++[<---------------->-]<<<<<<[-]+>[-]>>>>[-[<<[>+<<<<->>>[<<+>>-] ]<<[>>+<<-]<[>>+<<[-]]>>>>>[-]]<<<<<[>>>++<<<-]>+>>>>[-]]<<<<-[>>+<<[-]]>>>>,]<<->[<[-]+>[-]]<<[<[-]>>[<<+>>[-]]+<<[->>-<<]>-]>>[-]+++[<++++++++++++++++>-]<.
``````

3You crazy man!! – Almo – 2017-02-08T23:20:52.037

Love it, can you do that in malbolge? XD – Stefan Nolde – 2017-02-10T04:36:45.677

13

## JavaScript (ES6), 4341 29 bytes

``````s=>+eval(s.replace(/.!+\$/,1))
``````

### Non-regex method (41 31 bytes)

Below is my initial approach. It's slightly more interesting, but significantly longer still a bit longer even after a significant optimization by Neil (10 bytes saved).

``````f=([c,...s])=>1/c?c|s>'':1-f(s)
``````

### Test cases

``````let f =

s=>+eval(s.replace(/.!+\$/,1))

;[
"0", "1", "0!", "1!", "!0", "!1", "!0!", "!1!", "0!!",
"1!!", "!!0", "!!1", "!0!!", "!!!1", "!!!0!!!!", "!!!1!!!!"
].map(
s => console.log(s, '=>', f(s))
)``````

I can only save 10 bytes from your non-regex method, so it's still too long: `f=([c,...s])=&gt;1/c?c|s&gt;'':1-f(s)`. – Neil – 2017-02-06T13:41:06.213

@Neil Since it's much better than my 1st attempt anyway, I took the liberty to include your suggestion. – Arnauld – 2017-02-06T14:19:33.227

Ha, I had the same idea but you golfed it better. :) – Devsman – 2017-02-08T15:23:44.093

10

# Retina, 13 bytes

A somewhat weird approach, but it's short and it works.

``````0\$
!1
!!

^\d
``````

With the first two lines we replace an ending `0` with `!1`: with this replacement we now know that the part of our string from the digit onwards is equal to 1.

Next two lines, remove pairs of `!`: double negation erases itself, and we already accounted for factorial with the previous step.

Last line, match a digit at the start of the string and return the number of matches: if the negations have all been eliminated we'll find a match (and as we said before we know this is equal to 1), if there's still a negation this won't match.

Try it online!

1Wouldn't the final digit necessarily always be a 1? In that case, you could use `1` rather than `\d`. – None – 2017-02-06T22:39:15.473

1@ais523 no, because the first part will only replace an ending 0, so for example the input `0!` will stay unchanged until the last line – Leo – 2017-02-06T23:07:57.500

1Really lovely solution, nice work! :) – Martin Ender – 2017-02-06T23:32:35.487

10

# Jelly, 5 bytes

``````VeMḂ\$
``````

Try it online!

Monadic function expecting a string. Inputs with leading `!`s cause a `1` to be printed to STDOUT along the way, so the TIO link I give is a test harness that prints the input-output pairs beneath the first line of output.

### How?

``````VeMḂ\$ - Monadic link: string
V     - eval the string
- the implicit input of 0 causes !...! to evaluate to 1 (which gets printed),
- the result is the evaluation of the rest: "0"=0; "0!"=1; "1"=1; "1!"=1; ...
e    - exists in?
M   -     Maximal indexes - the "0" and "1" characters are greater than "!",
- so this results in a list of one item [i] where
- i is the 1-based index of the 0 or 1 character.
Ḃ  -     %2 (vectorises) - [i%2], so a 0 if we need to logically negate and a 1 if not
- hence we check equality with e rather than inequality.
``````

10

# Brainfuck, 85 72 (84) bytes

``````,[>-[-----<->]<++[>++++[-<++++>]+<[[+],[[-]>-<]]]>[<<+[-->]>[<],>-]<]<+.
``````

to return numerically, or

``````,[>-[-----<->]<++[>++++[-<++++>]+<[[+],[[-]>-<]]]>[<<+[-->]>[<],>-]<]-[-----<+>]<--.
``````

for ASCII text. > may also be prefixed to avoid memory wrapping.

Try it online!

``````Loops over the input.
On 1, ends.
On "!", toggles bool a stored as 0 or 255.
On "0", toggles if there is no trailing bit, then ends.

Memory labels  | BOOL | INPUT | FLAG |

,                   first input
[                     # loop on INPUT
>-[-----<->]<++     subtract 49 == "1"

[                     # case not "1"
>++++[-<++++>]      add 16 since 49 take 16 == "!"

+                   set FLAG
<                   move to INPUT
[                     # case "0"
[+],                clear and new INPUT
[                     # case "0!"
[-]>-<              clear INPUT and FLAG
]
]
]

>                   move to FLAG
[                     # case "!" or "0" without tail
<<+[-->]>[<]        not the BOOL
,                   take new input
>-                  clear FLAG
]
<                   move to INPUT
]

+.                    return 0 or 1
``````

Or for text response, replace the last line with

``````-[-----<+>]<--.       add 49 for "0" or "1" conversion and return
``````

9

# 05AB1E, 9 bytes

Code:

``````.V¹'!ÜgG_
``````

Uses the CP-1252 encoding. Try it online! or Verify all test cases!

Explanation:

``````.V         # Evaluate the input as 05AB1E code. This computes the factorial part.
'!Ü     # Remove trailing exclamation marks..
¹        # ..from the first input
g    # Get the length of the resulting string
G   # Do the following length - 1 times:
_  #   Negate the number
``````

9

# Ruby, 12+1 = 392415 13 bytes

Uses the `-n` flag. Thanks to @GB for -9 bytes!

``````p~/!*\$|0\$/%2
``````

Since you only check the length, you can delete the trailing zero, instead of checking "!0" first and a single zero after that. – G B – 2017-02-07T08:10:57.180

@GB that's a wonderful idea! However, I found a solution that's even shorter by modifying my regex to look for the position of `0` or end-of-line – Value Ink – 2017-02-07T09:25:51.927

Then you could just check for trailing '!' or zero or end of line: p~/!+\$|0\$|\$/%2 is just 14 bytes. – G B – 2017-02-07T09:39:26.520

And then "0\$|\$" could become "0?\$" to save another byte. – G B – 2017-02-07T14:51:46.080

1Better yet `!*\$` is shorter by two! – Value Ink – 2017-02-07T18:03:55.287

Cool. Are you sure it works for all test cases? I can't check right now. – G B – 2017-02-07T22:12:28.843

@GB yes, it works. It's just using the same principle as what your suggestion was, but instead of `0` being the optional one, `!` is. – Value Ink – 2017-02-07T23:24:39.037

9

# Perl, 20 bytes

19 bytes of code + `-p` flag.

``````s/\d!+/1/;\$_=0+eval
``````

Try it online!

Perl's negation returns `undef` or `1`, so I use `0+` to numerify the result `0+undef` returns `0`. Besides that, not much to say about the code.

2Just wrote exactly this. Have a +1. – primo – 2017-02-07T09:26:11.567

@primo Glad to see that for once I'm not 20 bytes behind you! Thanks :) – Dada – 2017-02-07T09:59:05.740

9

# C, 686261 53 bytes

``````c;e(char*a){for(c=1;*a<34;a++)c^=1;c=a[1]?c:*a&1^!c;}
``````

Squeezed out a few more bytes with some abuse

Try it online!

1I think you can remove the `int` from the function and you can change the `*a==33` to `*a&lt;34`. – Cows quack – 2017-02-06T18:18:08.960

Alas `*a%2` is shorter than `*a-48` – Cows quack – 2017-02-06T18:22:20.093

Thanks for the tip. I was also able to eliminate another character by removing the brackets around the return and assigning it. – Ahemone – 2017-02-06T22:30:51.010

I am pretty sure `for(;*a&lt;34;a++)` can be shortened to `for(;*a++&lt;34;)` saving 1 byte – Albert Renshaw – 2017-02-06T22:34:22.583

Unfortunately not, as the conditional statement it will always execute and so will push the pointer too far ahead for the return dereference. – Ahemone – 2017-02-06T22:40:10.410

-1 byte: `c;e(char*a){for(c=1;*a++&lt;34;)c^=1;c=*a?c:*--a&amp;1^!c;}` – Rogem – 2018-01-14T22:07:28.940

6

# Perl 6, 3228 23 bytes

``````{m/(\!)*(1|0.)*/.sum%2}
``````

### How it works

``````{                     }  # A lambda.
{m/            /      }  # Match the lambda argument against the regex:
(\!)*                 #   Zero or more `!`.
#     (First capture will be an array with one element per negation).
(1|0.)*          #   A `1`, or a `0` and another character, zero or more times.
#     (Second capture will be a one-element array if the factorial
#     part evaluates to 1, and an empty array otherwise.)
.sum     # Add the lengths of the two captures,
%2   # and return that sum modulo 2.
``````

6

# Befunge, 24 bytes

``````~"!"-:#v_\$1+
*+2%!.@>0~`
``````

Try it online!

This starts by counting the number of `!` characters read from stdin. The first character that isn't a `!` will either be a `0` or `1`, but in the process of testing for `!` we will have subtracted 33, making it either 15 or 16. We then read one more character, that will either be an `!` or EOF, and compare if that is less than 0 (i.e. EOF).

Taking those three data points - the exclamation count (c), the digit value, (d), and the end-of-file condition (e) - we can calculate the result as follows:

``````!((c + d*e) % 2)
``````

Multiplying the digit value by the end-of-file condition means it will be converted to zero if the digit was followed by a `!`, thus giving it the same modulo 2 value as a `1` (which remember has been converted to 16). But before applying the modulo 2, we add the initial exclamation count, which effectively toggles the modulo 2 result as many times as their were `!` prefixes. And finally we not the result since our baseline values for `0` and `1` are the opposite of what we need.

Looking at the code in more detail:

``````~                Read a character from stdin.
"!"-            Subtract 33 (ASCII for '!').
:  _        Make a duplicate and check if zero (i.e. is it a '!').
\$1+     If so, drop the duplicate, increment a counter, and repeat.
v         Otherwise move to the second line, leaving the digit value on the stack.
>0~`      Read one more character and check if less than 0 (i.e. EOF).
*                Multiple by the digit value, making it zero if not followed by EOF.
+               Add to the exclamation count.
2%             Modulo 2 the result.
!            Then not that value.
.@          And finally write to stdout and exit.
``````

5

``````f('!':b)="10"!!read[f b]
f[a]=a
f _='1'
``````

Defines a function `f`, which takes a string and returns a character. Try it online!

## Explanation

There are three cases: input begins with `!`, input has length 1, and everything else.

``````f('!':b)=    -- If input has head '!' and tail b,
"10"!!      -- we index into the string "10"
read[f b]  -- using f b converted to int. This essentially inverts f b.
f[a]=        -- If input has only one character, we know it's a digit,
a           -- so we can just return it.
f _=         -- In all other cases, we know the input is a digit followed by !s,
'1'         -- so we can return '1'.
``````

Switch from String to Integer as the return type: `f('!':b)=[1,0]!!f b;f"0"=0;f _=1`. – nimi – 2017-02-06T20:24:46.230

5

``````f('!':b)=1-f b
f"0"=0
f _=1
``````

Try it online!

Each leading `!` complements the output for the rest of the expression, done as `1-`. We keep flipping until we hit a digit. If the remaining is just `"0"`, the result is 0. Otherwise, it's a `1` or is followed by one or more `!`, so the result is 1.

5

## Ruby, 22 21 20 bytes

``````->s{(s=~/!*\$|0\$/)%2}
``````

Explanation:

• First case, I got some '!' at the end, remove them, get length modulo 2.
• Second case, no '!', if last character is zero then remove it, get length modulo 2
• If the last character is 1, back to the first case

(-1 byte stealing @Value Ink's idea)

Awesome, I looked at this puzzle for 10 minutes but had not much time and then forgot about it. Now spotted it again in active questions and was delighted to see such a nice approach. – akostadinov – 2017-02-13T07:54:28.480

4

# Jelly, 8 bytes

``````œr”!LḂ=V
``````

Try it online!

This is a function (monadic link) that takes one argument and returns via its return value. (It also often writes junk to standard output as a side effect, but we don't care about that.)

## Explanation

``````œr”!LḂ=V
œr”!      Take {the input}, with all trailing ! deleted
L     Take the length of this
Ḃ    Take the parity of that length
=   Return 0 if unequal, 1 if equal to:
V    the value of {the input} when eval'ed as a niladic Jelly program
``````

First, note that as the input always consists of some number of `!`, followed by a digit, followed by more `!`, that if we delete the trailing `!` and take the length, we'll end up with one plus the number of leading `!` in the program. Taking the parity of this will return 0 if there were an odd number of `!`, or 1 if there were an even number of `!`. Comparing to 0 is a "not" function, whereas comparing to 1 is the identity function; thus `œr”!LḂ=` effectively implements the "treat leading `!` as NOT operators" part of the question.

As for the second half, handling factorials, `!` is a factorial operation in Jelly, so if the program has no leading `!`, we can solve the problem directly with a simple `eval` (`V`). If the program does have leading `!`, those will be interpreted as taking the factorial of 0 (possibly multiple times), producing a return value of 1, which will be printed to standard output and discarded once a digit is seen; thus, they have no impact on the return value of the function that's my submission to the question.

Very nice and great explanation. – ElPedro – 2017-02-06T21:48:17.753

+1 given for parity – Barett – 2017-02-13T19:42:52.750

4

# Java 7, 10582 81 bytes

``````int a(char[]a){int b=0,c=0;for(;a[b++]<34;c^=1);return(b<a.length?1:a[b-1]&1)^c;}
``````

Try it online!

### Old regex-ish solution

``````int a(String a){a=a.replace("0!","1").replaceAll("1.*","1");int b=a.length()-1;return b%2^a.charAt(b)&1;}
``````

2`c^=1` is super clever. That's an unused operator if ever I've seen one. – Addison Crump – 2017-02-09T08:19:49.640

4

# Python, 38 bytes

``````lambda s:(s[1::2]>s[::2])^ord(s[-1])%2
``````

TryItOnline!

An unnamed function taking an input string `s` and returning an integer `0` or `1`.

`s[1::2]` is a slice of the input string that starts at index 1 and has a step size of two:
`'Like this' -> 'ieti'`

`s[::2]` is similar but starts at the default index of 0:
`'Like this' -> 'Lk hs'`

The test `(s[1::2]>s[::2])` checks if the 0-based index of the `'0'` or `'1'` is odd, i.e. if we need to complement.
This works because the ordering of strings is checked lexicographically with any non-empty string greater than the empty string, and with ASCII ordering, so `'1'>'0'>'!'`. This is a byte shorter than the simpler `s.index(max(s))%2`.

The `ord(s[-1])%2` checks to see if the last character is not a `'0'` (for valid input), and results in an integer (whereas the same length `(s[-1]!='0')` would return a boolean).
This works because the last character of the input, `s[-1]`, will be a `'0'`, `'1'`, or `'!'` which have ASCII code points 48, 49, and 33 respectively, which are 0, 1, and 1 modulo 2.

The `^` then performs a bitwise exclusive or operation on the two above values, returning an integer since one input, the right one, is an integer. If the left is True the complement of the right is returned, if the left is False the right is returned, as required.

3

``````f s|foldr(\_->not)(last s`elem`"1!")\$fst.span(<'0')\$s="1"|1<3="0"
``````

Try it online! Usage: `f "!!!0!!!!"`

Saved two bytes thanks to @nimi.

2

# PHP 7.1, 58555437 35 bytes

Note: uses IBM-850 encoding

``````echo!!\$argn[-1]^strspn(\$argn,~Ì)%2;
``````

Run like this:

``````echo '!!!0!!!!' | php -nR 'echo!!\$argn[-1]^strspn(\$argn,~Ì)%2;';echo
> 0
``````

# Explanation

``````echo
strspn(\$a=\$argv[1],~Ì) # Count the number of leading exclamation marks.
% 2                    # Make 0 (even) or 1 (odd).
^ !!\$a[-1];            # Negate with factorial part (truthy value of the
# last char):
# - "0" is considered falsy.
# - "1" or "!" is considered truthy.
``````

# Tweaks

• Saved 3 bytes by using IBM-850 encoding
• Saved a byte by changing the regex slightly
• Saved 17 bytes, new version without long function names and return
• Saved 2 bytes by using `-R` (which makes `\$argn` available)

2

## Batch, 62 bytes

``````@set/ps=
@set s=%s:0!=1%
@set s=%s:!!=%
@cmd/cset/a%s:1!=1%
``````

Takes input on STDIN. Batch actually understands leading `!`s correctly for this challenge, but the trailing `!`s need to be dealt with, which takes three steps:

• Change `0!` to `1`
• Delete pairs of `!!` (this is safe for the `!!`s before the digit too)
• Delete any remaining trailing `!` (which by now can only be after a `1`)

2

# IBM/Lotus Notes Formula - 77 bytes

``````@Eval(@Left(a;@If(@Like(a;"%1%");"1";"0"))+@If(@Ends(a;"!");"1";@Right(a;1)))
``````

There is no TIO for Notes Formula so a screenshot of all test cases is shown below:

How it works

`@Eval()` evaluates a string as an expression

First we check if the input string in field (input) `a` contains `1` or `0` and take all characters to the left of whichever it is which will be a string of `!` characters. We don't care how many. `@Eval()` will take care of that.

Next we look to see if there is a `!` at the end of the string. If there is we append `1` to the `!` string (`0!` and `1!` are both 1 - it doesn't matter how many `!` characters there are at the end) otherwise we append the last character unchanged because it is not a `!` and could be either a `1` or a `0`.

We now have a string containing the leading inversions plus a number defined by whether there are any factorial characters so we can feed this to `@Eval()` and get the results above.

2

## CJam, 12 11 bytes

``````r_W='0=!\~;
``````

Try it online! Test suite (prints a `1` for each correct test case).

``````r      e# Read input.
_W='0= e# Duplicate and check whether the string ends in '0'. This is the
e# only case in which the factorial part results in 0.
!      e# Negate this to get the actual result of the factorial part.
\      e# Swap with the input.
~      e# Evalute the input as CJam code. The leading `!` will apply the logical
e# negations to the factorial result. The 0 or 1 will then push a junk value
e# which is potentially negated a few times as well, by the factorials.
;      e# Discard the junk value.
``````

2

# sed, 3633 31 bytes

Pure sed, no bc / shell utils. Works on GNU sed < 4.3; 33 bytes on BSD and GNU 4.3+.

``````s/.!!*\$/1/
:
s/!0/1/
s/!1/0/
t
``````

Straightforward enough if you're familiar with `sed`; commented for those who aren't:

``````# Since 0! == 1! == 1 and factorial has precedence, just collapse any trailing "!"
s/.!!*\$/1/
# Define an anonymous label
:
# Invert 0 if needed
s/!0/1/
# Invert 1 if needed
s/!1/0/
# If a change was made, go back to the anonymous label.
t
``````

Test:

``````% cat 109248.sed
s/.!!*\$/1/
:l
s/!0/1/
s/!1/0/
tl
% wc -c 109248.sed
33 109248.sed
% cat cases
0
1
0!
1!
!0
!1
!0!
!1!
0!!
1!!
!!0
!!1
!0!!
!!!1
!!!0!!!!
!!!1!!!!
% sed -f 109248.sed cases
0
1
1
1
1
0
0
0
1
1
0
1
0
0
0
0
% gsed -f 109248.sed cases
0
1
1
1
1
0
0
0
1
1
0
1
0
0
0
0
%
``````

IIRC some (all?) versions of `sed` allow you to use the null string as a label name. If you can get that to work here, it'd save you two bytes. Actually, I'm not sure the label is even needed; unless I've missed something, the first line is idempotent, so you might be able to jump back to the start of the program rather than needing a label. – None – 2017-02-13T05:04:43.943

@ais523 I thought so too, but evidently it doesn't work in BSD versions. The man page says "If no label is specified, branch to the end of the script," and even that wasn't working when I tried. – Kevin – 2017-02-13T09:05:34.990

GNU sed does allow a label to be just `:` (more of a bug taken as feature), in which case both the `t` and `b`! commands jump to the position of the label. Plus, a sed code must work for at least one version of sed, similar to other languages, so you don't need to create code that works for BSD as well. – seshoumara – 2017-02-17T17:31:22.073

2

# Brainfuck, 115 bytes

``````>,[->++++[<-------->-]<[--------------->,[<[-]+>-]<<[->-[>+<+]>[-<+>]<<]>>++++++[-<++++++++>]<.>>+<]>-[<<+>,>[-]]<]
``````

Try it online!

## Ungolfed:

``````% 0: inverter count
% 1: result
% 2: if/else flag; tmpspace in inner loop 0

>1,[
->2++++[<-------->-]<1 subtract 33 (!)
[
% we've reached the number
---------------
% now it's either 0 or 1

% check next char; If it's not 0 then it's '!'
% 0! = 1! = 1!...! so we only need to determine if at least one ! exists
>2,
[<[-]+>-]<1

% apply inversions
<0
[->1
% invert cell 1 once each iteration
% cell 1 is 0 or 1
-          % cell 1 is 255 or 1
[>+<+]     % cell 1 is 0; cell 2 is 1 iff cell 1 should be 1
>2[-<+>]<1 % cell 1 is 1 or 0
<0]

% print result
>1>++++++[-<++++++++>]<1.

>>2+< % tape={0 r 0 1}
]
>2-[ % we haven't seen the number yet
<<0+>1,>2 % add to inverter count
[-]
]<1
]
``````

1

# Bean, 24 bytes

Hexdump:

``````00000000 26 4a c1 53 a0 17 53 d0 80 a0 5d 20 80 0a a1 80  &JÁS .SÐ. ] ..¡.
00000010 81 00 25 3a ae a1 ab 24                          ..%:®¡«\$
00000018
``````

Equivalent JavaScript:

``````+eval(a.replace(/.!+\$/,1))
``````

Sorry for stepping on your toes, Arnauld.

### Explanation:

Takes first line of input as unformatted string in `a`, and replaces any digit followed by one or more `!` with `1`, so that the rest can be `eval`'d by JavaScript.

1

# 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
}
}
``````

1`n` is only used in the return so you can remove declaring it. – TheLethalCoder – 2017-02-07T15:59:17.670

1

# 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.
``````

1

# 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.

1

# 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
``````

You're allowed to take a string as a list of characters, meaning that `r` is a valid solution by itself. That should save you quite some bytes.

– None – 2017-02-13T05:01:48.777

Oh, thanks! That indeed saved a lot of bytes – Lukas Boersma – 2017-02-14T08:06:26.293

0

# 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.
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!

0

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

``````f(h:t)|h=='!'=1-f t|h=='1'=1|t==[]=0|1<2=1
``````

0

# 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.

0

# 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' "

0

# 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 ;)

0

# 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...

0

# 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!

0

# 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
``````

Can you save a couple of bytes by outputting via exit code? – Jo King – 2018-01-15T01:30:49.600

0

# 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:

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
``````

Each `%6` can be shortened to `%` thanks to language updates. – MD XF – 2017-10-08T21:44:06.960

0

# 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]
``````

0

# 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
``````