Where does the lg(lg(N)) factor come from in Schönhage–Strassen's run time?

According to page 53 of Modern Computer Arithmetic (pdf), all of the steps in the Schönhage–Strassen Algorithm cost $O(N \cdot \lg(N))$ except for the recursion step which ends up costing $O(N\cdot \lg(N) \cdot \lg(\lg(N)))$.

I don't understand why the same inductive argument used to show the cost of the recursive step doesn't work for $O(N\cdot \lg(N))$.

• Assume that, for all $X

• So the recursive step costs $d \cdot \sqrt{N} F(\sqrt{N})$, and we know this is less than $d \cdot \sqrt{N} c \cdot \sqrt{N} \lg(\sqrt{N}) = \frac{c \cdot d}{2} \cdot N \cdot \lg(N)$ by the inductive hypothesis.

• If we can show that $d

• I'm pretty sure recursion overhead is negligible, so $d \approx 1$ and we have $\frac{c}{2} N \cdot \lg(N)$ left to do the rest. Easy: everything else is $O(N \cdot \lg(N))$ so we can pick a $c$ big enough for it to fit in our remaining time.

Basically, without digging into the details that will contradict this somehow, it looks like things would work out if we assumed the algorithm costs $O(N \cdot \log(N))$. The same thing seems to happen if I expand the recursive invocations then sum it all up... so where is the penalty coming from?

My best guess is that it has to do with the $\lg(\lg(N))$ levels of recursion, since that's how many times you must apply a square root to get to a constant size. But how do we know each recursive pass is not getting cheaper, like in quickselect?

For example, if we group our $N$ initial items into words of size $O(\lg(N))$, meaning we have $O(N/\lg(N))$ items of size $O(\lg(N))$ to multiply when recursing, shouldn't that only take $O(N/\lg(N) \cdot \lg(N) \cdot \lg(\lg(N)) \cdot \lg(\lg(\lg(N)))) = O(N \cdot \lg(\lg(N)) \cdot \lg(\lg(\lg(N))))$ time to do. Not only is that well within the $N \cdot \lg(N)$ limit, it worked even though I used the larger $N\cdot \lg(N)\cdot \lg(\lg(N))$ cost for the recursive steps (for "I probably made a mistake" values of "worked").

My second guess is that there's some blowup at

The sticking point is the size of the integers in the recursive step, which is not quite $\sqrt{n}$, but rather twice as large, since the product of two $t$-bit integers has size $2t$. Assuming that we break the original $n$-bit integer into $\sqrt{n}$ parts of length $\sqrt{n}$, the running time recursion is
$$ T(n) = n\log n + \sqrt{n}T(2\sqrt{n}), $$
and by expanding it, we get
$$
\begin{align*}
T(n) &= n\log n + \sqrt{n}T(2\sqrt{n}) \\ &=
n\log n + 2n\log (2\sqrt{n}) + 2^{1/2} n^{3/4} T(2^{3/2}n^{1/4}) \\ &=
n\log n + 2n\log (2\sqrt{n}) + 4n\log (2^{3/2} n^{1/4}) + 2^{5/4} n^{7/8} T(2^{7/4} n^{1/8}) \\ &=
n\log n + 2n\log (2\sqrt{n}) + 4n\log (2^{3/2} n^{1/4}) + 2^3 n \log (2^{7/4} n^{1/8}) + \cdots,
\end{align*}
$$
so each level of the recursion takes time proportional to $n\log n$. Since there are $\log\log n$ levels, we get the quoted running time. For more information, check out these lecture notes.

StackExchange Computer Science Q#28379, answer score: 5