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I am reading Depth Reduction of Arithmetic Formula form the survey of Ramprasad Saptharishi. Now in the proof of depth reduction due to Brent, 74 that

Let $f$ be an n-variate degree d polynomial computed by an arithmetic formula $\Phi$ of size s. Then, $f$ can also be computed by a formula $\Phi^{\prime}$ of size $s^{\prime}=\operatorname{poly}(s, n, d)$ and depth $O(\log s)$

In the proof they have written that

Consider the first node in this path $V$ such that the size of the formula rooted at $v$ is smaller than $\frac{2s}{3}$. Let $\Phi_v$ refer to the sub-formula rooted at $V$. By the choice of the path from the root, we have $$\frac{s}3\leq |\Phi_v|\leq \frac{2s}{3}$$

Now how are we getting the $\frac{s}3\leq |\Phi_v|$ inequality?

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You do not cite the part of the survey that is actually relevant for getting the $s/3$ lower bound:

Starting from the root, walk down to the leaves by always taking the child with a larger sub-tree under it. Consider the first node in this path $v$ such that the size of the formula rooted at v is smaller than $2s/3$.

$v$ cannot be the root since the subtree rooted at the root has size $s$. Let $w$ be the predecessor of $v$. By definition, the subtree rooted at $w$ contains more than $2s/3$ nodes. Since it has most two children and $v$ is the child with the largest subtree under it, it has more than half the nodes under $w$, that is, more than $s/3$.

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  • $\begingroup$ See also Emil's linked thread since indeed, you may not only have a value precisely between $s/3$ and $2s/3$ but between $\lceil s/3 \rceil$ and $\lceil 2s/3 \rceil$. $\endgroup$
    – holf
    Nov 8 at 8:53

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