I am reading a survey on polynomial identity testing by Saxena. On page 6 he writes that PIT for depth-3 circuits of the form $\prod \sum \prod$ is trivial. He gives no citation and as such I believe this should be folklore. However, I am unfamiliar with the lore and haven't come up with a "trivial" randomized algorithm via Schwartz-Zippel.

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    $\begingroup$ Given a circuit C of that form, it is identically zero iff one of its factors is zero. If given as a whitebox, you can read the factors off of the circuit description, and then it is reduced to the problem of PIT for sparse polynomials (i.e. $\Sigma \Pi$). As a black box, you can first use polynomial factorization to get the factors. I'm not sure I'd call this trivial, but it reduces - trivially in the whitebox case, nontrivally but by a well-known result in the blackbox case - to a previously solved problem. $\endgroup$ Commented Apr 13 at 21:26
  • $\begingroup$ Wait what I said about blackbox factorization isn't known. Multivariable factorization is equivalent to PIT (Kopparty-Saraf-Shpilka). $\endgroup$ Commented Apr 14 at 0:03
  • $\begingroup$ I missed the part where this was talking about the whitebox case. I've never heard of that equivalence between blackbox multivariable factorization and PIT, I'll have to look into that. Thanks! $\endgroup$
    – Anakin Dey
    Commented Apr 17 at 4:31

1 Answer 1


If you can solve PIT for a circuit class $\mathscr{C}$, this generally lifts to a solution of PIT for $\Pi \mathscr{C}$, where $\Pi \mathscr{C}$ is the set of polynomials $f$ that can be expressed as $f = \prod_i f_i$ with $f_i \in \mathscr{C}$.

In the white-box case, the PIT algorithm for $\Pi \mathscr{C}$ is exactly what Josh pointed out in the comments. The product $f$ is nonzero iff all the factors $f_i$ are nonzero, and you can see the individual factors $f_i$, so you simply run your $\mathscr{C}$-PIT algorithm on each factor.

The black-box case is easy, once the right definitions are in place. Up to some mild loss in parameters, you can assume that your black-box PIT algorithm is given by a hitting set generator. This is a polynomial map $\mathcal{G} : \mathbb{F}^\ell \to \mathbb{F}^n$ with the property that for every polynomial $h \in \mathscr{C}$, the composition $h\circ \mathcal{G}$ is zero iff $h$ itself is zero. If a generator $\mathcal{G}$ hits $\mathscr{C}$, then $\mathcal{G}$ also hits $\Pi \mathscr{C}$, as a product of nonzero polynomials remains nonzero. (To obtain a PIT algorithm from a generator, test the composed polynomial $f \circ \mathcal{G}$ by brute force.)

For the specific case in the original question, Klivans and Spielman (STOC 2001) constructed a hitting set generator for sparse polynomials. The same generator hits $\Pi \Sigma \Pi$ circuits.


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