## Mathematics is fact. Programming is not

184

22

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.

21But 0!=1!, so what's the point of handling multiple factorials? – boboquack – 2017-02-06T09:22:54.217

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

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

11<?='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! > 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

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

45

## 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.

3Since 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=${x/[01]!*/1};echo$(($x)) - not allowed to post a proper answer :( – None – 2017-02-09T01:45:50.833 29 # Bash + Unix utilities, 21 17 bytes sed s/.!!*$/1/|bc

Verify the test cases online!

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.)

The old version works, this one doesn't – user41805 – 2017-02-06T09:37:15.347

I used the TIO link – user41805 – 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

### 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)

let f =

s=>+eval(s.replace(/.!+$/,1)) ;[ "0", "1", "0!", "1!", "!0", "!1", "!0!", "!1!", "0!!", "1!!", "!!0", "!!1", "!0!!", "!!!1", "!!!0!!!!", "!!!1!!!!" ].map( s => console.log(s, '=>', f(s)) ) I can only save 10 bytes from your non-regex method, so it's still too long: f=([c,...s])=>1/c?c|s>'':1-f(s). – Neil – 2017-02-06T13:41:06.213 @Neil Since it's much better than my 1st attempt anyway, I took the liberty to include your suggestion. – Arnauld – 2017-02-06T14:19:33.227 Ha, I had the same idea but you golfed it better. :) – Devsman – 2017-02-08T15:23:44.093 12 # Retina, 13 bytes A somewhat weird approach, but it's short and it works. 0$
!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

2Really lovely solution, nice work! :) – Martin Ender – 2017-02-06T23:32:35.487

11

# 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?
$- last two links as a monad: M - Maximal indexes - the "0" and "1" characters are greater than "!", - so this results in a list of one item [i] where - i is the 1-based index of the 0 or 1 character. Ḃ - %2 (vectorises) - [i%2], so a 0 if we need to logically negate and a 1 if not - hence we check equality with e rather than inequality. 10 # 05AB1E, 9 bytes Code: .V¹'!ÜgG_ Uses the CP-1252 encoding. Try it online! or Verify all test cases! Explanation: .V # Evaluate the input as 05AB1E code. This computes the factorial part. '!Ü # Remove trailing exclamation marks.. ¹ # ..from the first input g # Get the length of the resulting string G # Do the following length - 1 times: _ # Negate the number 10 # Ruby, 12+1 = 392415 13 bytes Uses the -n flag. Thanks to @GB for -9 bytes! p~/!*$|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 8 # 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<34. – user41805 – 2017-02-06T18:18:08.960 Alas *a%2 is shorter than *a-48 – user41805 – 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<34;a++) can be shortened to for(;*a++<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++<34;)c^=1;c=*a?c:*--a&1^!c;} – None – 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 ## Haskell, 39 bytes 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 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 '!').
$1+ If so, drop the duplicate, increment a counter, and repeat. v Otherwise move to the second line, leaving the digit value on the stack. >0~ Read one more character and check if less than 0 (i.e. EOF). * Multiple by the digit value, making it zero if not followed by EOF. + Add to the exclamation count. 2% Modulo 2 the result. ! Then not that value. .@ And finally write to stdout and exit. 6 # Haskell, 27 bytes f('!':b)=1-f b f"0"=0 f _=1 Try it online! Each leading ! complements the output for the rest of the expression, done as 1-. We keep flipping until we hit a digit. If the remaining is just "0", the result is 0. Otherwise, it's a 1 or is followed by one or more !, so the result is 1. 5 ## Ruby, 22 21 20 bytes ->s{(s=~/!*$|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 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;} 2c^=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 # Haskell, 67 65 bytes f s|foldr(\_->not)(last selem"1!")$fst.span(<'0')$s="1"|1<3="0" Try it online! Usage: f "!!!0!!!!" Saved two bytes thanks to @nimi. 3 ## CJam, 12 11 bytes r_W='0=!\~; Try it online! Test suite (prints a 1 for each correct test case). r e# Read input. _W='0= e# Duplicate and check whether the string ends in '0'. This is the e# only case in which the factorial part results in 0. ! e# Negate this to get the actual result of the factorial part. \ e# Swap with the input. ~ e# Evalute the input as CJam code. The leading ! will apply the logical e# negations to the factorial result. The 0 or 1 will then push a junk value e# which is potentially negated a few times as well, by the factorials. ; e# Discard the junk value. 3 # Brainfuck, 115 bytes >,[->++++[<-------->-]<[--------------->,[<[-]+>-]<<[->-[>+<+]>[-<+>]<<]>>++++++[-<++++++++>]<.>>+<]>-[<<+>,>[-]]<] Try it online! ## Ungolfed: % 0: inverter count % 1: result % 2: if/else flag; tmpspace in inner loop 0 >1,[ ->2++++[<-------->-]<1 subtract 33 (!) [ % we've reached the number --------------- % now it's either 0 or 1 % check next char; If it's not 0 then it's '!' % 0! = 1! = 1!...! so we only need to determine if at least one ! exists >2, [<[-]+>-]<1 % apply inversions <0 [->1 % invert cell 1 once each iteration % cell 1 is 0 or 1 - % cell 1 is 255 or 1 [>+<+] % cell 1 is 0; cell 2 is 1 iff cell 1 should be 1 >2[-<+>]<1 % cell 1 is 1 or 0 <0] % print result >1>++++++[-<++++++++>]<1. >>2+< % tape={0 r 0 1} ] >2-[ % we haven't seen the number yet <<0+>1,>2 % add to inverter count [-] ]<1 ] 3 # APL (Dyalog Extended), 23 bytes ⍎⌽('!(\d.*)'⎕R'\1~')⍣≡⍞ Repeatedly shift any !s to the right with the fixpoint ⍣≡ !!!!0! -> !!!0!~ -> !!0!~~ -> !0!~~~ -> 0!~~~~ Then reverse with ⌽, and execute with ⍎ Try it online! 2 # PHP 7.1, 58555437 35 bytes Note: uses IBM-850 encoding echo!!$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 $argn available) 2 ## Batch, 62 bytes @set/ps= @set s=%s:0!=1% @set s=%s:!!=% @cmd/cset/a%s:1!=1% Takes input on STDIN. Batch actually understands leading !s correctly for this challenge, but the trailing !s need to be dealt with, which takes three steps: • Change 0! to 1 • Delete pairs of !! (this is safe for the !!s before the digit too) • Delete any remaining trailing ! (which by now can only be after a 1) 2 # sed, 3633 31 bytes Pure sed, no bc / shell utils. Works on GNU sed < 4.3; 33 bytes on BSD and GNU 4.3+. s/.!!*$/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

# Scala, 6563 47 bytes

s=>s.takeWhile(34>).length%2^(s.last-48).abs%14

Try it online!

Many thanks to user for -16 chars!!!

Great answer. You can also use ascii values to get 59 bytes

– user – 2020-10-16T13:26:01.440

47 bytes, actually – user – 2020-10-16T13:58:42.020

1

# 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.

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
}
}

1n 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

-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 1 # 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 1 # C (gcc), 42 bytes c;e(char*a){c=*a<34?!e(a+1):*a-48|1[a]%2;} Try it online! 1 # R, 41 bytes +eval(parse(t=gsub("\\d!+",1,scan(,"")))) Try it online! Two and a half years later, finally fixed the bug pointed out by JayCe, by switching * in the regex to a +... The + is just to coerce the logical to numeric. For some reason it's not working when I test it. What about this? – JayCe – 2018-05-18T16:46:36.627 @JayCe did you somehow reassign + to something else? ;-) I'm not having any trouble testing it (on TIO)...do you have a particular example? That approach is, I think different enough that it can be its own answer. – Giuseppe – 2018-05-18T16:53:03.477 You're right, my bad! Will post separately as you're suggesting. That said, we both return 1 for the input 0 so I need to fix this first... – JayCe – 2018-05-18T16:59:24.703 @JayCe goood point. In that case, I need to fix my answer, so I'll be deleting it for the moment. – Giuseppe – 2018-05-18T17:01:53.883 1 # 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. 0 # RProgN, 31 bytes ~'(!*1?)0?(!*)'{Lx=L2%x+0>1*}R ## Explained ~'(!*1?)0?(!*)'{Lx=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! 0 # 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 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

# 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']

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

# Wren, 50 bytes

Fn.new{|a|(a[-1]=="0"?1:0)^a.trimEnd("!").count%2}

Try it online!

0

# GolfScript, 11 bytes

Blatant port (not copying the whole algorithm) of Martin Ender's CJam answer. At least I got the same length.

.)\;48=!\~;

Try it online!

## Explanation

.           # Add a fresh copy of the input
)          # Generate the input without the last item
# and the last item of the input
\         # Swap so that the first output of ) is on top
;        # Remove so that there is only the last item
48=     # Is the item equal to 48 (ASCII '0') ?
!    # If not (0), there are !'s before either 0 or 1
# and the result is obviously 1 (0 negated)
# If yes (1), the last item is going to be 0,
# and the result is 0 (1 negated).
\   # Take the fresh unused copy of the input to the top
~  # Apply the input to the top
; # (Due to postfix evaluation the trailing
# !'s are applied to the number.)
# Discard the applied number, which is garbage

# Implicitly output the result
$$`$$

0

# Rust macros, 69 bytes

macro_rules!f{(!$($x:tt)*)=>{1-f!($($x)*)};(0)=>{0};($($x:tt)*)=>{1}}

Defined a macro f that takes a sequence of tokens (including !, 0 and 1) as input and produces a token sequence that evaluates to the expected result. Port of a Haskell answer.

try it online

## Explanation

macro_rules! f {
(! $($x:tt)*) => {  // match an exclamation mark followed by any number of token trees
1 - f!($($x)*)  // expand to 1- and call the macro recursively
};
(0) => {            // match a 0 that's not followed by anything
0               // replace with a 0
};
($($x:tt)*) => {    // match any number of token trees
1               // replace with 1
}
}

0

# Python 2, 37 bytes

f=lambda s:s<"0"and 1-f(s[1:])or"0"<s

Try it online!

Sometimes returns True/False instead of 0/1, which is allowed I think.

How

• The factorial part can be evaluated simply as "0"<s. This is 0 only when s is 0, and 1 if s is anything else (e.g 1 or 0!!).
• The negation part is evaluated using recursion. If s starts with ! (s<"0"), recurs as 1-f(s[1:]).