Can the majority of $n$ bits be computed by a depth 2 formula all of whose gates compute the majority of $m$ bits where $m=O(n^c)$ for a constant $c<1$? Such a formula contains $m+1$ gates and $m^2$ leaves so $c$ must be at least $1/2$. I assume that the leaves can only be labeled by variables (without negations), but it would be also interesting to know the answer if we allow also negated variables and constants.

Two examples of such formulas are given below.

$n=7$, $m=5$:

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$n=9$, $m=7$: enter image description here

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    $\begingroup$ Just to clarify: the leaves are variables, or do you also allow constants or negative literals? $\endgroup$ Jan 10 '15 at 14:11
  • $\begingroup$ What does the function $m(n)$ measure? The smallest possible such $m$? $\endgroup$ Jan 10 '15 at 19:08
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    $\begingroup$ Thank you for the feedback! I've simplified the question as well as added two examples. $\endgroup$ Jan 11 '15 at 10:58
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    $\begingroup$ @Alexander: this could be somehow related. $\endgroup$
    – Stasys
    Jan 11 '15 at 11:33
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    $\begingroup$ While this is likely not relevant for the asymptotic growth of $m$, I note there is a parity constraint: $\mathrm{MAJ}_m$-circuits with $m$ odd compute self-dual monotone functions, hence they can only express $\mathrm{MAJ}_n$ for $n$ odd. $\endgroup$ Jan 11 '15 at 14:34

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