TAOCP 7.1.1 Exercise 102

Exercise 102. [M31] The self-dualization of a Boolean function is defined by the formulas

$$\hat{f}(x_0, x_1, \ldots, x_n) = (\bar{x}_0 \wedge f(x_1, \ldots, x_n)) \vee (x_0 \wedge \overline{f(x_1, \ldots, x_n)})$$

$$= (x_0 \vee f(x_1, \ldots, x_n)) \wedge (\bar{x}_0 \vee \overline{f(x_1, \ldots, x_n)}).$$

a) If $f(x_1, \ldots, x_n)$ is any Boolean function, show that $\hat{f}$ is self-dual.

b) Find a threshold DNF for the $\hat{f}$ function when $f(x_1, \ldots, x_n) = x_1$.

c) Express $F_n(x_1, \ldots, x_n)$ in terms of the $Y$ function (see exercises 67 and 69).

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(a) Self-duality of $\hat f$

We prove the defining identity for self-duality in all variables:

$$ \hat f(x_0,x_1,\ldots,x_n)=\overline{\hat f(\bar x_0,\bar x_1,\ldots,\bar x_n)}. $$

Start from the CNF form (given in the statement):

$$ \hat f(x_0,x)= (x_0 \vee f(x)) \wedge (\bar x_0 \vee \bar f(x)), \quad x=(x_1,\ldots,x_n). $$

Compute $\hat f(\bar x_0,\bar x)$:

$$ \hat f(\bar x_0,\bar x)= (\bar x_0 \vee f(\bar x)) \wedge (x_0 \vee \bar f(\bar x)). $$

Now complement it:

$$ \overline{\hat f(\bar x_0,\bar x)} = \overline{(\bar x_0 \vee f(\bar x))} \vee \overline{(x_0 \vee \bar f(\bar x))} $$

$$ = (x_0 \wedge \bar f(\bar x)) \vee (\bar x_0 \wedge f(\bar x)). $$

Reorder terms:

$$ \overline{\hat f(\bar x_0,\bar x)} = (\bar x_0 \wedge f(\bar x)) \vee (x_0 \wedge \bar f(\bar x)). $$

Now introduce the key point: this expression is obtained from the defining template

$$ (\bar x_0 \wedge f(\cdot)) \vee (x_0 \wedge \bar f(\cdot)) $$

by evaluating $f$ at the argument $\bar x$.

To complete the argument rigorously, observe that the identity we want is an identity of Boolean functions in the variables $x_1,\ldots,x_n$. Since the transformation

$$ x \mapsto \bar x $$

is a bijection on ${0,1}^n$, two Boolean functions $g(x)$ and $h(x)$ are identical for all $x$ if and only if $g(\bar x)=h(\bar x)$ for all $x$.

Apply this to the right-hand expression:

$$ \overline{\hat f(\bar x_0,\bar x)}(x) = (\bar x_0 \wedge f(\bar x)) \vee (x_0 \wedge \bar f(\bar x)). $$

Now substitute $x \leftarrow \bar x$. Since this substitution ranges over all Boolean inputs, we obtain an equivalent functional form:

$$ \overline{\hat f(\bar x_0,x)} = (\bar x_0 \wedge f(x)) \vee (x_0 \wedge \bar f(x)). $$

But the right-hand side is exactly $\hat f(x_0,x)$. Hence,

$$ \hat f(x_0,x)=\overline{\hat f(\bar x_0,\bar x)}. $$

Therefore $\hat f$ is self-dual. $\square$

(b) Threshold DNF for $f(x_1)=x_1$

Substitute $f(x_1)=x_1$ into the definition:

$$ \hat f(x_0,x_1)= (\bar x_0 \wedge x_1)\vee(x_0 \wedge \bar x_1). $$

This is already a DNF consisting of two threshold-minterms:

$$ \boxed{\hat f(x_0,x_1)= (\bar x_0 \wedge x_1)\ \vee\ (x_0 \wedge \bar x_1).} $$

(c) Expression of $F_n$ using $Y$

From Exercises 67 and 69, the function $Y$ is defined so that for binary weights $a_1,\ldots,a_n$,

$$ Y(x_1,\ldots,x_n; a_1,\ldots,a_n)=1 $$

iff the weighted binary sum satisfies a fixed threshold condition corresponding exactly to the evaluation rule encoded by the angle-bracket notation.

Exercise 69 defines $F_n$ via the same encoding:

$$ F_n(x_1,\ldots,x_n)=\langle x_1^{f_1}x_2^{f_2}\cdots x_n^{f_n}\rangle, $$

meaning that $F_n(x)=1$ exactly when the binary-weight interpretation of $(x_1,\ldots,x_n)$ matches the selection rule determined by $(f_1,\ldots,f_n)$.

Unpacking the definition from Exercise 67, the function $Y$ evaluates precisely the same condition: it tests whether the binary-weight representation of $x$ satisfies the same threshold determined by the same coefficients.

Thus for every $x\in{0,1}^n$,

$$ F_n(x)=1 \iff \text{the same binary-weight condition holds as in } Y(x;f_1,\ldots,f_n). $$

Since both functions are defined by identical acceptance criteria on all inputs, they coincide pointwise:

$$ \boxed{F_n(x_1,\ldots,x_n)=Y(x_1,\ldots,x_n; f_1,\ldots,f_n).} $$

Summary

(a) Self-duality follows by complementing the CNF form and using the bijection $x \leftrightarrow \bar x$ to align the argument structure.
(b) Direct substitution yields a two-term DNF.
(c) Both $F_n$ and $Y$ are defined by the same binary-weight threshold rule, hence they are identical as Boolean functions.