2
$\begingroup$

Now, I have the following conjecture.

$\text{MOD}_{2017}(x_1, \ldots, x_k)$ is computable by a bounded depth polynomial size circuit in the basis $\{\neg, \text{MAJ}\}$.

However, I am at a loss at how to realize/see this. Could anybody help me out?

$\endgroup$
1
  • 2
    $\begingroup$ If you have no idea why it is true, then why do you conjecture this? $\endgroup$ Commented Mar 23, 2017 at 8:41

1 Answer 1

2
$\begingroup$

Consider what happens when you feed into a majority gate the the inputs $x_1,\ldots,x_k$ and $k-1$ zeroes. You get an $\mathsf{AND}$ gate. If you feed into it $k-3$ zeroes, you get a gate corresponding to $x_1+\cdots+x_k \geq k-1$. Using similar ideas, you can construct a constant-depth circuit over the basis $\{\lnot,\mathsf{MAJ}\}$ that computes $x_1+\cdots+x_k = \ell$ and so $\mathsf{MOD}_{2017}$.

If you're not allowed to use constants but there is a tie-breaking rule for $\mathsf{MAJ}$, then you can implement constants using $\mathsf{MAJ}(x_1,\lnot x_1)$, whose constant value will depend on the tie-breaking rule.

If you're not allowed to use constants and every majority gate needs to have an odd number of inputs, then it is generally not possible to construct such a circuit. Indeed, suppose that you plug $x_i = x_1$ for all $i>1$, and that $k$ is chosen so that the output is constant whatever the value of $x_1$ is (this depends on your definition of $\mathsf{MOD}$, which you haven't supplied). On the other hand, you can prove by induction that every gate computes either $x_1$ or $\lnot x_1$, and in particular not a constant function

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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