Given a string of space separated binary digits or space separated booleans or an array of binary digits or array of booleans . Your job is to find the xor of each until you end up with one answer either 0 or 1. The inputs will always be valid and will only be either 0 or 1.
1 0 0 1 0 --> 0
1 0 1 1 1 0 0 1 0 0 0 0 --> 1
This is code golf so shortest code will win. Good luck.
s->s.sum()%2
If a string is really required, then 20 bytes:
s->s.chars().sum()%2
The "string" answer uses the fact that a space is codepoint 32, which mod 2 returns 0.
OÉ
Explanation:
O: Take the sum of the input-arrayÉ: Evaluates sum % 2 == 1, returning 1 if the sum is odd, 0 otherwise3 bytes:
A leading Ç can be added if the space-delimited string input was still mandatory, instead of a boolean-array.
Ç: Push the ASCII values of all characters, and implicitly convert it to a list. 0 1 would become [48, 32, 49] in that case. The O (sum) and É (is odd?) will still act the same.
so
Input can be a numeric vector of the form [1 0 0 1 0], or a string such as '10010' (thanks to @Giuseppe for noticing!).
The code is so simple that it hardly needs an explanation, but here it goes.
s % Implicit input: numeric vector (or string). Sum of the numbers. (For string
% input, the ASCII codes are summed. Character '1' is odd, '0' is even, and
% space is even too, so the parity is the same as with numeric vector input)
o % Parity. Implicit display
space=32 and doesn't change the parity.
\$\endgroup\$
U to convert from string to numeric vector, but indeed it can be removed
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Apr 13, 2018 at 15:14
b->{var r=1<0;for(var c:b)r^=c;return r;}
The variable r is initially set to false, since we are XORing b[0] with it. Try it online here.
#q~+2% instead of #.~+2%#@ and output via exit code to save 2 bytes. (In TIO you need to expand the debug tab if you want to see the exit code)
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Apr 21, 2018 at 18:44
> Input
> 2
>> ∑1
>> 3%2
>> Output 4
I feel like 3 answers is too many, but I wanted to get Whispers out again. I'm still working on an XOR approach
TIO, courtesy Adám
≠/⍎⎕
Analysis:
⎕ - Accept input.
⍎ - "unquote" it (if it's a quoted string, this will convert it to a numeric vector. If it's already a numeric vector, this is effectively a null op)
/ - reduction - apply the operator to the left to each successive item in the vector to the right
≠ - "not equal" - when applied strictly to boolean arguments, this is functionally identical to XOR (which is not implemented as a separate operator in APL)
If it may be assumed that the input will be a numeric vector instead of a quoted string, then the 'unquote' can be removed, saving one byte:
≠/⎕
⎕ and default output happens in Output and ⍞ output happens in Debug). For convenience, you can define your workspace in Header, Code and Footer, but only Code is counted into the char/byte count: Try it online!
\$\endgroup\$
Tacit prefix function.
≠/
/ is reduction and ≠ is XOR because XOR only gives 1 if its arguments are unequal.
b->⊻(b...)
⊻ is the xor symbol, and b... distributes an array as individual elements before sending it to the xor function (since xor needs multiple arguments passed separately, not a single array argument).
A bit more interestingly, a version that accepts space separated (/comma-separated/unseparated) boolean values as a string, and returns the xor result :
b->sum(Int.([b...]))%2
Saved a few bytes with inspiration from ceilingcat.
Saved another byte thanks to Jonathan Frech
r;f(char*s){for(r=0;*s;)r^=*s++;r&=1;}
