# §1.6: Highlight: Almost linear functions and the BLR Test

In linear algebra there are two equivalent definitions of what it means for a function to be linear:

Definition 29 A function $f : {\mathbb F}_2^n \to {\mathbb F}_2$ is linear if either of the following equivalent conditions hold:

1. $f(x+y) = f(x) + f(y)$ for all $x, y \in {\mathbb F}_2^n$;
2. $f(x) = a \cdot x$ for some $a \in {\mathbb F}_2^n$; i.e., $f(x) = \sum_{i \in S} x_i$ for some $S \subseteq [n]$.

In an exercise you are asked to verify that the conditions are indeed equivalent. If we encode the output of $f$ by $\pm 1 \in {\mathbb R}$ in the usual way then the “linear” functions $f : {\mathbb F}_2^n \to \{-1,1\}$ are precisely the $2^n$ parity functions $(\chi_S)_{S \subseteq [n]}$.

Let’s think of what it might mean for a function $f : {\mathbb F}_2^n \to {\mathbb F}_2$ to be approximately linear. Definition 29 suggests two possibilities:

1. $f(x+y) = f(x) + f(y)$ for most pairs $x, y \in {\mathbb F}_2^n$;
2. there is some $S \subseteq [n]$ such that $f(x) = \sum_{i \in S} x_i$ for most $x \in {\mathbb F}_2^n$.

Are these equivalent? The proof of (2) $\Rightarrow$ (1) in Definition 29 is “robust”: it easily extends to show (II) $\Rightarrow$ (I) (see the exercises). But the natural proof of (1) $\Rightarrow$ (2) in Definition 29 does not have this robustness property. The goal of this section is to show that (I) $\Rightarrow$ (II) nevertheless holds.

Motivation for this problem comes from an area of theoretical computer science called property testing which we will discuss in more detail in Chapter 7. Imagine that you have “black-box” access to a function $f : {\mathbb F}_2^n \to {\mathbb F}_2$, meaning that the function $f$ is unknown to you but you can “query” its value on inputs $x \in {\mathbb F}_2^n$ of your choosing. The function $f$ is “supposed” to be a linear function and you would like to try to verify this.

The only way you can be certain $f$ is indeed a linear function is to query its value on all $2^n$ inputs; unfortunately, this is very expensive. The idea behind “property testing” is to try to verify that $f$ has a certain property — in this case, linearity — by querying its value on just a few random inputs. In exchange for efficiency, we need to be willing to only approximately verify the property.

Definition 30 If $f$ and $g$ are boolean-valued functions we say they are $\epsilon$-close if $\mathrm{dist}(f,g) \leq \epsilon$. If $\mathcal{P}$ is a property of $n$-bit boolean functions, we say that $f$ is $\epsilon$-close to $\mathcal{P}$ if $f$ is $\epsilon$-close to some $g$ satisfying $\mathcal{P}$.

In particular, in property testing we take property (II) above to be the notion of “approximately linear”: we say $f$ is $\epsilon$-close to being linear if $\mathrm{dist}(f,g) \leq \epsilon$ for some truly linear $g(x) = \sum_{i \in S} x_i$.

Blum, Luby, and Rubinfeld [BLR90] showed that indeed (I) $\Rightarrow$ (II) holds, giving the following “test” for the property of linearity that makes just $3$ queries:

BLR Test Given query access to $f : {\mathbb F}_2^n \to {\mathbb F}_2$:

1. Choose ${\boldsymbol{x}} \sim {\mathbb F}_2^n$ and $\boldsymbol{y} \sim {\mathbb F}_2^n$ independently.
2. Query $f$ at ${\boldsymbol{x}}$, $\boldsymbol{y}$, and ${\boldsymbol{x}} + \boldsymbol{y}$.
3. “Accept” if $f({\boldsymbol{x}}) + f(\boldsymbol{y}) = f({\boldsymbol{x}} + \boldsymbol{y})$.

We now show that if the BLR Test accepts $f$ with high probability then $f$ is close to being linear. The proof works by directly relating the acceptance probability to the quantity $\sum_{S} \widehat{f}(S)^3$; see equation \eqref{eqn:BLR} below.

Theorem 31 Suppose the BLR Test accepts $f : {\mathbb F}_2^n \to {\mathbb F}_2$ with probability $1 – \epsilon$. Then $f$ is $\epsilon$-close to being linear.

Proof: In order to use the Fourier transform we encode $f$’s output by $\pm 1 \in {\mathbb R}$; thus the acceptance condition of the BLR Test becomes $f({\boldsymbol{x}})f(\boldsymbol{y}) = f({\boldsymbol{x}} + \boldsymbol{y})$. Since $\tfrac{1}{2} + \tfrac{1}{2} f({\boldsymbol{x}})f(\boldsymbol{y})f({\boldsymbol{x}}+\boldsymbol{y}) = \begin{cases} 1 & \text{if f({\boldsymbol{x}})f(\boldsymbol{y}) = f({\boldsymbol{x}} + \boldsymbol{y})}, \\ 0 & \text{if f({\boldsymbol{x}})f(\boldsymbol{y}) \neq f({\boldsymbol{x}} + \boldsymbol{y}),} \end{cases}$ we conclude \begin{align*} 1 – \epsilon = \mathop{\bf Pr}[\text{BLR accepts $f$}] &= \mathop{\bf E}_{{\boldsymbol{x}}, \boldsymbol{y}}[\tfrac{1}{2} + \tfrac{1}{2} f({\boldsymbol{x}})f(\boldsymbol{y})f({\boldsymbol{x}}+\boldsymbol{y})] \\ &= \tfrac{1}{2} + \tfrac{1}{2} \mathop{\bf E}_{{\boldsymbol{x}}}[f({\boldsymbol{x}}) \cdot \mathop{\bf E}_{\boldsymbol{y}}[f(\boldsymbol{y})f({\boldsymbol{x}}+\boldsymbol{y})]]\\ &= \tfrac{1}{2} + \tfrac{1}{2} \mathop{\bf E}_{{\boldsymbol{x}}}[f({\boldsymbol{x}}) \cdot (f * f)({\boldsymbol{x}})] \tag{by definition}\\ &= \tfrac{1}{2} + \tfrac{1}{2} \sum_{S \subseteq [n]} \widehat{f}(S) \widehat{f * f}(S) \tag{Plancherel} \\ &= \tfrac{1}{2} + \tfrac{1}{2} \sum_{S \subseteq [n]} \widehat{f}(S)^3 \tag*{(Theorem 28).} \end{align*} We rearrange this equality and then continue: \begin{align} 1 – 2\epsilon &= \sum_{S \subseteq [n]} \widehat{f}(S)^3 \label{eqn:BLR}\\ &\leq \max_{S \subseteq [n]} \{\widehat{f}(S)\} \cdot \sum_{S \subseteq [n]} \widehat{f}(S)^2 \nonumber\\ &= \max_{S \subseteq [n]} \{\widehat{f}(S)\} \tag*{(Parseval).} \end{align} But $\widehat{f}(S) = \langle f, \chi_S \rangle = 1 – 2\mathrm{dist}(f, \chi_S)$ (Proposition 9). Hence there exists some $S^* \subseteq [n]$ such that $1 – 2\epsilon \leq 1 – 2\mathrm{dist}(f,\chi_{S*})$; i.e., $f$ is $\epsilon$-close to the linear function $\chi_{S*}$. $\Box$

In fact, for small $\epsilon$ one can show that $f$ is more like $(\epsilon/3)$-close to linear, and this is sharp. See the exercises.

The BLR Test shows that given black-box access to $f : {\mathbb F}_2^n \to \{-1,1\}$, we can “test” whether $f$ is close to some linear function $\chi_S$ using just $3$ queries. The test does not reveal which linear function $\chi_S$ is close to (indeed, determining this takes at least $n$ queries; see the exercises). Nevertheless, we can still determine the value of $\chi_S(x)$ with high probability for every $x \in {\mathbb F}_2^n$ of our choosing using just $2$ queries. This property is called local correctability of linear functions.

Proposition 32 Suppose $f : {\mathbb F}_2^n \to \{-1,1\}$ is $\epsilon$-close to the linear function $\chi_S$. Then for every $x \in {\mathbb F}_2^n$, the following algorithm outputs $\chi_S(x)$ with probability at least $1-2\epsilon$:

1. Choose $\boldsymbol{y} \sim {\mathbb F}_2^n$.
2. Query $f$ at $\boldsymbol{y}$ and $x+\boldsymbol{y}$.
3. Output $f(\boldsymbol{y})f(x+\boldsymbol{y})$.

We emphasize the order of quantifiers here: if we just output $f(x)$ then this will equal $\chi_S(x)$ for most $x$; however, the above “local correcting” algorithm determines $\chi_S(x)$ (with high probability) for every $x$.

Proof: Since $\boldsymbol{y}$ and $x+\boldsymbol{y}$ are both uniformly distributed on ${\mathbb F}_2^n$ (though not independently) we have $\mathop{\bf Pr}[f(\boldsymbol{y}) \neq \chi_S(\boldsymbol{y})] \leq \epsilon$ and $\mathop{\bf Pr}[f(x+\boldsymbol{y}) \neq \chi_S(x+\boldsymbol{y})] \leq \epsilon$ by assumption. By the union bound, the probability of either event occurring is at most $2\epsilon$; when neither occurs, $f(\boldsymbol{y}) f(x+\boldsymbol{y}) = \chi_S(\boldsymbol{y}) \chi_S(x + \boldsymbol{y}) = \chi_S(x)$ as desired. $\Box$

### 7 comments to §1.6: Highlight: Almost linear functions and the BLR Test

• Zachary Hamaker

In the lead-in to Prop 32, you have “…we can still still determine…”. Also, in Chrome $f($ renders in an unpleasant fashion, for what it’s worth. Thanks!

-Zach

• Ryan O'Donnell

Thanks! Yeah, I’m not too satisfied with how $\widehat{f}(S)$ renders in Chrome, either, but I guess we’ll have to live with it.

• miforbes

Does this fourier-analytic proof of soundness of the BLR test extend to testing linear maps F_2^n->F_2^m, for arbitrary n,m? I know that the majority-correction proof of BLR works for any groups.

• Ryan O'Donnell

Hi Michael, not as far as I am aware. Hopefully, I will be writing about the “1%” version of the $\mathbb{F}_2^n \to \mathbb{F}_2^m$ problem — i.e., Polynomial Freiman-Ruzsa — in a later chapter.

• Noam Lifshitz

In I and II ,I think it is supposed to be
“for almost all” instead of “most” because
if we take $f\left(x\right)=\begin{cases} 1 & x\neq0,x_{1}=0\\ 0 & \mbox{ otherwise} \end{cases} then for most x ‘s f\left(x\right)=0x$ but f\left(x+y\right)=f\left(x\right)+f\left(y\right) only if one of x,y,x+y is zero.

• Noam Lifshitz

In I and II ,I think it is supposed to be
“for almost all” instead of “most” because if we take $f\left(x\right)=\begin{cases} 1 & x\neq0,x_{1}=0\\ 0 & \mbox{ otherwise} \end{cases}$ then for most $x$ ‘s $f\left(x\right)=0x$ but $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ only if one of $x,y,x+y$ is zero.

• Ryan O'Donnell

I see what you mean. I guess by “most” I, informally, meant “almost all”. But you’re right, it’s probably clearly if I actually write that. Thanks.