XOR reduction bools

Question

Description :

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.

Example :

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.

Answer 1

JavaScript (ES6), 18 bytes

s=>eval(s.join`^`)

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Answer 2

Java (JDK 10), 12 bytes

s->s.sum()%2

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If a string is really required, then 20 bytes:

s->s.chars().sum()%2

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The "string" answer uses the fact that a space is codepoint 32, which mod 2 returns 0.

Answer 3

Python 2, 17 bytes

lambda a:sum(a)%2

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Answer 4

05AB1E, 2 bytes

OÉ

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Explanation:

• O: Take the sum of the input-array
• É: Evaluates sum % 2 == 1, returning 1 if the sum is odd, 0 otherwise

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3 bytes:

A leading Ç can be added if the space-delimited string input was still mandatory, instead of a boolean-array.

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• 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.

Answer 5

JavaScript (Node.js), 23 bytes

a=>a.reduce((c,d)=>c^d)

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Answer 6

MATL, 3 2 bytes

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!).

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Explanation

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

Answer 7

SNOBOL4 (CSNOBOL4), 44 bytes

A	X =X + INPUT	:S(A)
	OUTPUT =REMDR(X,2)
END

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Answer 8

Perl 5 -p040, 10 bytes

Assumes the input list can't be empty

$\^=0+$_}{

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Answer 9

Java 10, 41 bytes

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.

Answer 10

Jelly, 5 3 2 bytes

^/

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-2 bytes thanks to user202729!

Answer 11

Add++, 8 bytes

L~,€Os2%

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Answer 12

(Traditional) APL, 3 or 4 bytes

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:

≠/⎕

Answer 13

Befunge-98 (FBBI), 8 bytes

#.~+2%#@

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Answer 14

Japt, 2 bytes

r^

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Does exactly what it says on the tin, reduce the array by XORing.

Answer 15

Cubix, 11 bytes

i?+<^<@Oa1<

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It's been a while, so it feels good to write a Cubix answer!

Answer 16

Whispers v2, 38 bytes

> Input
> 2
>> ∑1
>> 3%2
>> Output 4

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I feel like 3 answers is too many, but I wanted to get Whispers out again. I'm still working on an XOR approach

Answer 17

C (gcc), 41 39 38 bytes

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;}

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Answer 18

Ruby, 42 bytes

->s{s.split(' ').reduce(0){|a,b|a^b.to_i}}

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Answer 19

APL (Dyalog Unicode), 2 bytes^SBCS

Tacit prefix function.

≠/

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/ is reduction and ≠ is XOR because XOR only gives 1 if its arguments are unequal.

Answer 20

Julia 0.6, 12 bytes

b->⊻(b...)

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⊻ 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 :

Julia 0.6, 22 bytes

b->sum(Int.([b...]))%2

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