In the survey see section 5.3.2 : Depth reduction for arithmetic circuits for notations.

I follow the proof of the following two identities :

$[u]=\Sigma_{w\in \cal{F}_m}[u:w].[w]$ where $deg(u)\geq m$ and

$[u:v]=\Sigma_{w \in \cal{F}_m} [u:w][w:v]$ where $deg(u)\geq m \gt deg(v)$

By using these identities we are trying to reduce the depth of the circuit.

The idea it to write $[u]$ as product of polynomials that have degree at most half the degree of $[u]$.

Let $\cal{F}(u)=\cal{F}_m$ where $m=deg(u)/2$ and

$\cal{F}(u,v)=\cal{F}_m$ where $m=(deg(u)+deg(v))/2$ .

Then we have

  1. $[u]=\Sigma_{w\in \cal{F}(u)}[u:w].[w]$ and

  2. $[u:v]=\Sigma_{w \in \cal{F}(u,v)} [u:w][w:v]$.

The proof in the survey further expands both the equations writing $[w]$ as $[w_L].[w_R]$ in the first equation and $[w:v]$ as $[w_L].[w_R:v]$ in the second.

However I see no need for this as for the 1st equation we have

degree on RHS = $deg(u)$

degree on LHS : degree of $[u:w] \leq deg(u)-deg(w)$ and degree of $[w] = m = deg(u)/2$ as $w\in \cal{F}(u)$ .

For the second equation :

degree on RHS = $deg(u) - deg(v)$

degree on LHS $deg[u:w] \leq deg(u) -deg(w) = (deg(u)-deg(v))/2$ and $deg[w:v] \leq (deg(u)-deg(v))/2$ as $deg(w) = (deg(u)+deg(v))/2$ for all $w's$.

So my question is why in the proof presented in the survey, they are expanding further. They claim that the degrees are not yet halved if do not expand further but my working above shows that the degrees are halved.

Have I misunderstood anything here ?


1 Answer 1


See that $w\in \mathcal{F}_m$ so $\deg(w)\geq m$ but $\deg(w_L),\deg(w_R)<m$. So if $\deg(w_L)=\deg(w_R)=m-1$ then $\deg(w)$ becomes $2m-2$. If $m$ is large enough then $2m-2 >m$. Hence we can certainly have $\deg(w)>m$. Then $\deg([w])>m$ and all the terms in RHS will not have degree at most $\deg(u)/2$

Similarly think about the second part.


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.