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I have to implement a circuit following the boolean equation A XOR B XOR C, however the XOR gates I am using only have two inputs (I am using the 7486 XOR gate to be exact, in case that makes a difference)... is there a way around this?

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    $\begingroup$ Of course there is no such thing as A XOR B XOR C. That symbol is ambiguous in the absence of a well-defined rule of order of operations. XOR is a binary operator. So you must specify either (A XOR B) XOR C or A XOR (B XOR C). Logically, the results are equivalent, but your choice of association tells you how to build it from ordinary XOR gates, right? $\endgroup$
    – MPW
    Commented Mar 23, 2015 at 1:26

3 Answers 3

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Take the output of A XOR B and pipe it into an XOR having C as the other input. (This implements (A XOR B) XOR C, and XOR is associative.

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Exclusive or $\oplus$ is associative (and commutative), so you can use any order and any pairing.

$A \oplus B \oplus C = (A \oplus B) \oplus C = A \oplus (B \oplus C) $.

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use two gates. input $A$ and $B$ to the first gate, then input the output of the first gate and $C$ to the second gate.

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