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Say you have N input Boolean function, let's use a parity tree for the example. The function outputs a one or a zero depending on the values of the N inputs. Are the N inputs considered the preimage of the function?

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  • $\begingroup$ @kelalaka; Thank you for replying, does that mean the answer to my question is "yes" , or is the answer "no" ? ;-) $\endgroup$ – William Hird Dec 24 '18 at 20:02
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Functions don't have preimages; outputs do. Your Boolean function has domain the set of $2^N$ binary $N$-tuples, that is, the set $$\big\{(000\ldots 00),~ (000\ldots 01),~ (000\ldots 10),~ (000\ldots 11),~ \cdots , (111\ldots 10),~ (111\ldots 11)\big\}$$ and range (a.k.a. the set of outputs) the set $\{0, 1\}$. The outputs have preimages which are defined as the set of all inputs that are mapped onto that particular output by the specified function; in this instance the Exclusive-OR of the $N$ bits. For example, with $N=3$,

  • the preimage of $0$ is the set $\big\{(000),~ (011), ~ (101),~ (110)\big\}$
  • the preimage of $1$ is the set $\big\{(001),~ (010), ~ (100),~ (111)\big\}$

Notice that the preimages are a partition of the domain; every element of the domain is necessarily a member of one (and only one) of the preimages.

In short, the answer to your question

Are the N inputs considered the preimage of the function?

is No, they are not.

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An $n(=N)$ input Boolean function $f$ has $2^n$ input space, and output 0 or 1. Formally;

$$f:\{0,1\}^n \mapsto \{0,1\}$$

The pre-image of $\{0,1\}$ under $f$ is the all input space $\{0,1\}^n$.

The pre-image of $0$ under $f$ is the set $$f^{-1}[0]=\{ x \in \{0,1\}^n |\; f(x) = 0\}$$

and the pre-image of $1$ under $f$ is the set $$f^{-1}[1]= \{ x \in \{0,1\}^n |\; f(x) = 1\}$$

And;

$$f^{-1}[0] \cup f^{-1}[1] = \{0,1\}^n $$

Don't confuse pre-image with the inverse function.

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  • $\begingroup$ I'm not a computer scientist. All your mathematical formulas make me dizzy! I'm looking for a "Boolean" answer to my question: Yes or NO. Merry Christmas by the way :-) $\endgroup$ – William Hird Dec 24 '18 at 20:41
  • $\begingroup$ No, don't delete anything ! You can show future generations how really smart people cant answer simple questions sometimes ! $\endgroup$ – William Hird Dec 24 '18 at 20:49
  • $\begingroup$ A function must map all of its input space into the domain. But the function doesn't have to be onto. The pre-image is in your sense is the input space. $\endgroup$ – kelalaka Dec 24 '18 at 21:06

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