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I came across this slides which states following two theorems:

Theorem There are $2^{2^{n-1}}$ different self-dual functions of $n$ variables.

Theorem Let $f$ be a self-dual function of $n$ variables, and let $|f|$ be the number of inputs a for which $f(a) = 1$, then $|f| = 2^{n-1}$

However, it does not give any proof. Also, I tried to deduce it on my own but not moving with anything. How can these theorems be proved?

rohitt
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Mahesha999
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2 Answers2

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A boolean function is self-dual, if it is negated by negating all inputs.

$$f(x_1, x_2, ... x_n) = \lnot f(\lnot x_1, \lnot x_2, ... \lnot x_n)$$

This implies that for each of the $2^n$ input combinations, there is another combination with the opposite function value. Therefore, the function table must have $2^{n-1}$ rows with function value $true$ and the same number of rows with function value $false$.

As the self-dual functions are fully defined by $2^{n-1}$ of the $2^n$ truth table entries, the output column can be interpreted as binary number with $2^{n-1}$ bits. This leads to $2^{2^{n-1}}$ different self-dual functions.

Example for clarification
taken from chapter 5 of
Tsutomu Sasao: Switching Theory for Logic Synthesis

The dual function of $(x \land y) \lor (z \land w)$ is $$\overline{(\bar x \land \bar y) \lor (\bar z \land \bar w)} = (x \lor y) \land (z \lor w)$$ The identity expressed in De Morgan's Theorem 1:
The negation of a disjunction is the conjunction of the negations.

Axel Kemper
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  • According to this link, "A function is said to be Self dual if and only if its dual is equivalent to the given function" and to find duals we replace (1) AND with OR and vice versa and (2) 0 with 1 and vice versa. How this is equivalent to your definition of "a self dual function is a function negated by negating all inputs" . – RajS Jul 12 '19 at 12:10
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    @anir: I added an example to make it clearer. – Axel Kemper Jul 12 '19 at 21:30
  • A boolean function is self-dual, if it is negated by negating all inputs. - I guess this is an equivalent definition to "swapping ANDs and ORs leads to the same function" but where does that come from? I tried using DeMorgan's but I couldn't get far. – sprajagopal Jan 27 '21 at 14:16
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As a related note to the question above, the dual of a Boolean function is the complement of the original function, but with all inputs complemented. Thus, a self-dual boolean function would be such that its complement is the same as the function when its inputs are complemented.

A relatively easy way to prove this is based on the fact that each Boolean function has a unique SOP and POS representation. Say we have a function $f$, of $N$ boolean variables, $A_1, A_2, ... A_N$. Then

$$f(A_1, A_2, ...A_N) = \sum_{i\in I} m_i = \prod_{j\in I^C} M_j $$

where $m_i$ is a minterm which is part of that function's 'on-set' ($I$), $M_j$ is a maxterm and part of the function's 'off-set' ($I^C$). Note that the set $I$ is a subset of integer values between 0 and $2^N - 1$.

From de Morgan's law, we can easily prove that $m_i = M_i'$, that is, the i-th minterm is the complement of the i-th maxterm.

Using the classical definition of a Boolean dual (exchange $\cdot$ and $+$, 0s and 1s), the dual of the function $f$ is:

$$f^D(A_1, A_2, ... A_N) = \prod_{i\in I} m_i^D $$

where $m_i^D$ is the dual of the i-th minterm. It can be shown that $m_i^D = M_{2^N - i - 1}$ (for e.g. for a function of 2 Boolean variables, $m_1 = A_1'A_2$ $\Longrightarrow m_1^D = A_1' + A_2 = (A_1 A_2')' = m_2' = M_2$). Thus,

$$f^D(A_1, A_2, ... A_N) = \prod_{i\in I} m_i^D = \prod_{i\in I} m'_{2^N-i-1} $$ which, from de Morgan's law, gives us $$f^D(A_1, A_2, ... A_N) = (\sum_{i\in I} m_{2^N-i-1})'$$

$\because$ the i-th minterm is the same as the $(2^N - i - 1)$-th minterm with complemented inputs,
$$f^D(A_1, A_2, ... A_N) = (f(A_1', A_2', ... A_N'))'$$

Thus, for a self-dual function, $$f(A_1, A_2, ... A_N) = (f(A_1', A_2', ... A_N'))'$$

SPfromECE
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