TAOCP 6.3 Exercise 31

Exercise 31. [M21] [M21] Solve the recurrence (16).

Verified: no
Solve time: 4m43s

Let $A_n$ be the expected cost of an $M$-ary digital search tree built from $n$ random keys, and let $P(z)$ be its Poisson transform.

1. Poissonized recurrence

As in the proposed solution, Poissonization with parameter $z$ gives independent subtree sizes

$$ N_0,\dots,N_{M-1} \sim \mathrm{Poisson}(z/M), $$

and hence the exact functional equation

$$ P(z)=z+M P!\left(\frac{z}{M}\right). $$

This step is correct.

2. Correct normalization

The key step is to remove the linear scaling by defining

$$ Q(z) := \frac{P(z)}{z}, \qquad z>0. $$

Substitute $P(z)=zQ(z)$ into the functional equation:

$$ zQ(z)=z+M\cdot \frac{z}{M}Q!\left(\frac{z}{M}\right). $$

Cancel $z$:

$$ Q(z)=1+Q!\left(\frac{z}{M}\right). $$

This reduction is exact.

3. Exact solution of the functional equation

We now solve

$$ Q(z)=1+Q(z/M). $$

Iterate $k$ times:

$$ Q(z)=k+Q!\left(\frac{z}{M^k}\right). $$

Choose $k=\lfloor \log_M z \rfloor$, so that

$$ \frac{z}{M^k} \in [1,M). $$

Define

$$ {\log_M z} := \log_M z - \lfloor \log_M z \rfloor. $$

Then

$$ Q(z)=\log_M z + Q!\left(M^{{\log_M z}}\right). $$

Define the bounded function

$$ \Phi(t) := Q(M^t), \qquad t \in [0,1), $$

and extend it periodically by $\Phi(t+1)=\Phi(t)$.

Then the solution is

$$ Q(z)=\log_M z + \Phi({\log_M z}), $$

where $\Phi$ is a bounded 1-periodic function determined by the initial condition of $Q$ on $[1,M)$.

Crucial correction

There is no constant term $\gamma_M$. Any constant can be absorbed into the periodic function $\Phi$, since constants correspond to shifts of a periodic function.

4. Poissonized asymptotics

Multiply back:

$$ P(z)=zQ(z)=z\log_M z + z,\Phi({\log_M z}). $$

Thus the correct structure is

$$ P(z)=z\log_M z + z,\Phi(\log_M z) , $$

where $\Phi$ is bounded and 1-periodic (in $\log_M z$).

This shows explicitly the oscillatory linear term, which was missing in the flawed solution.

5. Depoissonization (justified form)

For digital search trees, $P(z)$ is a standard JS-admissible function (analytic Poisson transform with at most polynomial growth and smooth Mellin behavior). Under these conditions, depoissonization yields

$$ A_n = P(n) + O(\log n), $$

and in fact the periodic fluctuations transfer directly from Poisson to fixed size.

Substituting $z=n$,

$$ A_n = n\log_M n + n,\Phi(\log_M n) + O(\log n). $$

6. Final form and interpretation

The expected cost satisfies

$$ A_n = n\log_M n + n,\Phi(\log_M n) + O(\log n), $$

where:

$\Phi$ is a bounded 1-periodic function,
the term $n,\Phi(\log_M n)$ represents genuine digital periodic fluctuations, not a constant correction,
no constant $\gamma_M$ exists at this level of precision.

7. Summary of corrections

The functional equation solution is not “constant plus error”, but includes a periodic function of $\log_M z$.
The linear term in $P(z)$ is not $\gamma_M z$, but $z,\Phi(\log_M z)$.
Depoissonization preserves this periodic structure; it does not remove it.
The correct asymptotic form necessarily includes oscillations of order $n$.

Final answer

$$ \boxed{ A_n = n\log_M n + n,\Phi(\log_M n) + O(\log n), } $$

where $\Phi$ is a bounded 1-periodic function.