A Boolean function F of n variables is self-dual if and only if F(x1, x2, …,…
2014
A Boolean function F of n variables is self-dual if and only if
F(x1, x2, …, xn) = ¬F(¬x1, ¬x2, …, ¬xn).
Which expression gives, for every integer n ≥ 1, the number of self-dual Boolean functions of n variables?
Answer: D. 22⁽ⁿ⁻¹⁾ — ConceptA self-dual Boolean function assigns opposite output values to complementary input vectors: the value at x determines the value at ¬x. For n ≥ 1, the…
- A.
2n
- B.
22ⁿ
- C.
2n²
- D.
22⁽ⁿ⁻¹⁾
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Correct answer: D
Concept
A self-dual Boolean function assigns opposite output values to complementary input vectors: the value at x determines the value at ¬x.
For n ≥ 1, the 2n input vectors form disjoint complementary pairs, and each pair contributes one independent binary choice.
Application
There are 2n possible n-bit input vectors.
No n-bit vector equals its own complement, so these vectors form 2n−1 complementary pairs.
Within each pair {x, ¬x}, choosing F(x) fixes F(¬x) = ¬F(x); therefore each pair allows exactly two assignments.
Independent choices across all pairs give 2 raised to the number of pairs: 22⁽ⁿ⁻¹⁾.
Cross-check with n = 2
The four inputs split into the following two complementary pairs:
Input | Complement |
|---|---|
00 | 11 |
01 | 10 |
Each pair permits two opposite-output assignments, so the total is 2 × 2 = 4. The formula also gives 22⁽²⁻¹⁾ = 22 = 4.
Result: the number of self-dual Boolean functions is 22⁽ⁿ⁻¹⁾.