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Let $f : \{1,2\}^n \rightarrow \mathbb{Z}_3$ be a function from the multiplicative subgroup of order $2$ of $\mathbb{Z}_3$ over $n$ variables ($\{1,2\}^n$) to $\mathbb{Z}_3$, such that each coordinate is chosen uniformly at random.

Can we bound the probability that $f$ has $m$ monomials in the polynomial $p(x)$ represents it (the one with the lowest degree,I assume $p(x)$ was obtained by interpretation process )?

My intuition tells me this number has to be roughly $2^n$ (i.e. all the monomials appear), but I struggle saying something quantitative.

Larry a.
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  • It's easy to prove that there always exists a polynomial $p(x) : \mathbb{Z}_2^n\rightarrow \mathbb{Z}_3$ that coincides with $f(x)$. Is this what you want, or are you trying to find the expected degree of $p(x)$ here? – Dark Malthorp Apr 28 '20 at 18:52
  • I am trying to find the expected number of monomials in $p(x)$ (and the expected degree, but most importantly the number of monomials) – Larry a. Apr 28 '20 at 18:55
  • Hm. That's a tough question. There is the issue that two distinct polynomials over a finite field can be equal at all points (for example, in $\mathbb{Z}_2$, $p(x) = x^2 - x$ is $0$ for all $x$, but this is not the $0$ polynomial!), so I'm guessing that you want the one with lowest degree? – Dark Malthorp Apr 28 '20 at 18:57
  • Yes, basically $p(x)$ will be obtained by an interpolation process, so I assume I will hit the lowest degree representation – Larry a. Apr 28 '20 at 19:00
  • @Dark Malthorp I edited the question – Larry a. Apr 28 '20 at 19:04
  • The problem is that if you treat these actually as elements of those two rings, there's no way to map elements of $\mathbb{Z}_2$ to elements of $\mathbb{Z}_3$ that are nonzero. There's no nontrivial homomorphism between the two rings. – Matt Samuel Apr 28 '20 at 19:19
  • What if I map the multiplicative subgroup in $\mathbb{Z}_2$? I.e. I assume the input is -1,1 and the output is in $\mathbb{Z}_3$ – Larry a. Apr 28 '20 at 19:26
  • @MattSamuel - you bring up a good point. It's not clear what is meant by a polynomial here (I for some reason thought it was simply a function from one finite field into another, silly me for not reading carefully). – Dark Malthorp Apr 28 '20 at 19:41
  • @Larrya. I apologize if I'm being bothersome; I genuinely think there's a good question in here somewhere but I still can't figure out what it is. As Matt pointed out, the only polynomials from $Z_2^n$ to $Z_3^n$ are constant. are you instead asking about polynomials from ${0,1}^n\rightarrow {0,1,2}$, as subsets of $\mathbb{Z}$ (or $\mathbb{Q}$ or $\mathbb{R}$)? Making a polynomial only makes sense if you ignore the algebraic structure of $Z_2$ and $Z_3$. – Dark Malthorp Apr 28 '20 at 19:50
  • @DarkMalthorp I don't want to ignore the structure of $\mathbb{Z}_3$, but rather to consider the polynomials from ${-1,1}^n$ to $\mathbb{Z}_3$, or: a polynomial from the multiple subgroup of order 2 in $\mathbb{Z}_3$. Does it make sense? – Larry a. Apr 28 '20 at 19:55
  • @DarkMalthorp I edited it this way, I think it makes sense now. Thanks for the help so far! – Larry a. Apr 28 '20 at 20:12
  • The title and body of the question don't match. In the title you ask about the expected number of monomials; in the body you want to bound the probability that $f$ has $m$ monomials. Please clarify. – joriki Apr 28 '20 at 20:39
  • I think the question is ok (at least in its current form). My intuition suggests to me that I would expect one third of the possible terms to be missing (in average). This is simply because each and every one of the $2^n$ terms in a minimal degree polynomial corresponding to a given function is equally likely to have any member of $\Bbb{Z}_3$ as its coefficient. Meaning that for each term there is $1/3$ probability that it is missing. Making the expected number of terms $2^{n+1}/3$. – Jyrki Lahtonen Apr 28 '20 at 20:39
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    For example when $n=1$ we only need the polynomials $0$ (no terms), $1,2,x,2x$ (a single term) and $1+x,2+x,1+2x,2+2x$ (2 terms). A total of $12$ terms split over the nine different functions - an average of $12/9=4/3$ as promised. – Jyrki Lahtonen Apr 28 '20 at 20:43
  • And it makes no difference whether you are interested in the functions with domain ${1,2}^n$ or ${0,1}^n$. As long as each coordinate has exactly two possible values we can reduce the polynomial by a single variable quadratic, making the lowest degree polynomial matching the given function only have monomials with exponents $\in{0,1}$. – Jyrki Lahtonen Apr 28 '20 at 20:48
  • Thanks, I will accept that as an answer – Larry a. Apr 29 '20 at 15:01
  • I will flesh it out later today. That is, unless somebody else beats me to it. – Jyrki Lahtonen Apr 30 '20 at 04:19

1 Answers1

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Unless I misunderstood the question is about representing functions from the set $$S=\{1,2\}^n\subset\Bbb{Z}_3^n$$ to $\Bbb{Z}_3$ as polynomials in $R_n:=\Bbb{Z}_3[x_1,x_2,\ldots,x_n]$. It is well known and easy to see that to any function $f:\Bbb{Z}_3^n\to\Bbb{Z}_3$ there exists a polynomial $P\in R_n$ such that $P(a_1,a_2,\ldots,a_n)=f(a_1,a_2,\ldots,a_n)$ for all $(a_1,a_2,\ldots,a_n)\in\Bbb{Z}_3^n$. It follows that the same applies to functions $f:S\to \Bbb{Z}_3$ (extend $f$ outside of $S$ any which way you want, and you still find a matching polynomial $P$).

On the set $S$ the polynomials $x_i^2-1$ give rise to the zero function, so if $$I(S)=\langle x_i^2-1\mid i=1,2,\ldots,n\rangle_{R_n}\subset R_n$$ is the ideal they generate, then we can replace $P$ with the lowest degree element in its coset $P+I$ without changing the function $f:S\to\Bbb{Z}_3$. Using the congruences $x_i^2\equiv1\pmod I$ we can thus find a polynomial $P'\in P+I$ such that $P'$ is in the $\Bbb{Z}_3$-span $D$ of $$ D=\langle \prod_{j\in J}x_j\mid J\subseteq \{1,2,\ldots,n\}\rangle_{\Bbb{Z}_3} $$ all the $2^n$ products $x_J:=\prod_{j\in J}x_j$ of distinct variables $x_j$.

The space of functions $F(S):=\{f:S\to\Bbb{Z}_3\}$ also has dimension $|S|=2^n$, so to each function $f\in F(S)$ we can associate a unique polynomial $P\in D$.

After this longish setting up of the scene the question can be formulated unambíguously and answered easily! Namely, the question is what is the expected number of terms $x_J, J\subseteq \{1,2,\ldots,n\}$ in a uniformly chosen random element of $D$?

The answer follows from the observation that the term $x_J$ appears in the sum $\sum_{J}a_Jx_J$ if and only if the coefficient $a_J$ is non-zero. Because the choice of $a_J$ is uniform over $\Bbb{Z}_3$, this happens with probability $2/3$.

Therefore the expected number of terms is $$2^n\cdot\frac23=\frac{2^{n+1}}3.$$

Jyrki Lahtonen
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