14
$\begingroup$

Razborov proved that the monotone function matching is not in mP. But can we compute matching using a polynomial size circuit with a few negations? Is there a P/poly circuit with $O(n^\epsilon)$ negations that computes matching? What is the trade-off between the number of negations and the size for matching?

$\endgroup$
21
$\begingroup$

Markov proved that any function of $n$ inputs can be computed with only $\lceil \log (n+1)\rceil$ negations. An efficient constructive version was described by Fisher. See also an exposition of the result from the GLL blog.

More precisely:

Theorem: Suppose $f : \{0,1\}^n \to \{0,1\}^m$ is computed by a circuit $C$ with $g$ gates, then it is also computed by a circuit $C^*$ with $2g + O(n^2 \log^2 n)$ gates and $\lceil \log (n+1) \rceil$ negations.

The main idea is to add for each wire $w$ in $C$ a parellel wire $w'$ in $C^*$ that always carries the complement of $w$. The base case is for the input wires: Fisher describes how to construct an inversion circuit $I(x) = \overline x$ with $O(n^2 \log^2 n)$ gates and only $\lceil \log (n+1) \rceil$ negations. For the AND gates of circuit $C$, we can augment $a = b \land c$ with $a' = b' \lor c'$, and likewise for OR gates. NOT gates in $C$ cost nothing, we just swap the roles of $w$ and $w'$ downstream of the NOT gate. In this way, the entire circuit besides the inverter subcircuit is monotone.

A. A. Markov. On the inversion complexity of a system of functions. J. ACM, 5(4):331–334, 1958.

M. J. Fischer. The complexity of negation-limited networks - A brief survey. In Automata Theory and Formal Languages, 71–82, 1975

$\endgroup$
1
$\begingroup$

How to compute the inversion of $2^n-1$ bits using $n$ negations

Let the bits $x_0, \ldots, x_{2^n-1}$ be sorted in the decreasing order, i.e. $i<j$ implies $x_i \ge x_j$. This can be achieved by a monotone sorting network like the Ajtai–Komlós–Szemerédi sorting network.

We define the inversion circuit for $2^n-1$ bits $I^n(\vec{x})$ inductively: For the base case we have $n=1$ and $I^1_0(\vec{x}) := \lnot x_0$. Let $m=2^{n-1}$. We reduce $I^n$ (for $2m+1$) bits to one $I^{n-1}$ gate (for $m$ bits) and one negation gate using $\land$ and $\lor$ gates. We use negation to compute $\lnot x_m$. For $i<m$ let $y_i := (x_i \land \lnot x_m) \lor x_{m+i}$. We use $I^{n-1}$ to invert $\vec{y}$. Now we can define $I^n$ as follows:

$$I^n_i := \begin{cases} I^{n-1}_i(\vec{y}) \land \lnot x_m & i<m \\ \lnot x_m & i=m \\ I^{n-1}_i(\vec{y}) \lor \lnot x_m & i<m \\ \end{cases}$$

It is easy to verify this inverts $\vec{x}$ by considering the possible values of $x_n$ and using the fact that $\vec{x}$ is decreasing.

From Michael J. Fischer, The complexity of negation-limited networks - a brief survey, 1975.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.