Exercise 297. ▶ [HM26] (Emo Welzl.) Let $G_k(z) = \sum_n C_{n,p+q-1}(z/3)^{n-q}(2z/3)^p$ be the generating function for stopping time $t = 2p + q$ when $Y_0 = q$ in the proof of Theorem U. a) Find a closed form for $G_k(z)$, using formulas from Section 7.2.1.6. b) Explain why $G_k(1)$ is less than 1. c) Evaluate and interpret the quantity $G'_k(1)/G_k(1)$. d) Use Markov's inequality to bound the probability that $Y_t = 0$ for some $t \le N$. e) Show that Corollary W follows from this analysis.

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Solve time: 3m45s

Solution

Equation (131) defines the quantity used for the flushing decision by replacing the old target value $M_t$ with a value farther in the future. The update is not a reassignment of the measured agility values; it is a change in the next time at which the agility test will be performed.

Suppose the current value of the counter is $M$ and the increment chosen by the reluctant-doubling schedule is $\Delta t$. The new target value is obtained by the single assignment

$$ M_t \leftarrow M+\Delta t . $$

The value of $M$ is the current time of the measurement, so the right-hand side must be formed from the old value of $M$ before the replacement is made. After the replacement, the next test is postponed until the counter reaches this new value of $M_t$.

The reluctant-doubling sequence determines the successive values of $\Delta t$. Its beginning is

$$ 1,1,2,1,1,2,4,1,\ldots , $$

so the successive targets are obtained by adding these increments to the current values of $M$. For example, if the present value is $M=0$, the first four updates give

$$ M_t=0+1=1, $$

$$ M_t=1+1=2, $$

$$ M_t=2+2=4, $$

and

$$ M_t=4+1=5. $$

Thus every execution of the update uses the current $M$ as its base and advances the next checking point by the selected $\Delta t$.

The agility threshold in Table 4 is the quantity against which AGILITY is compared when $M$ reaches $M_t$. The threshold depends on the number $\Delta f$ of flushes already performed. From the schedule in Table 4, the threshold is

$$ \operatorname{AGILITY}_{\rm threshold}

\frac{\psi}{2^{\Delta f}} . $$

Therefore the test performed after reaching $M_t$ is

$$ \operatorname{AGILITY}<\frac{\psi}{2^{\Delta f}} . $$

For $\Delta f=0$, the threshold is

$$ \psi . $$

For $\Delta f=1$, it is

$$ \frac{\psi}{2}. $$

For $\Delta f=2$, it is

$$ \frac{\psi}{4}. $$

In general, after $\Delta f$ previous flushes, the threshold has been divided by $2^{\Delta f}$. Hence the complete threshold specified by Table 4 is

$$ \boxed{\frac{\psi}{2^{\Delta f}}}. $$

This completes the proof. ∎

TAOCP 7.2.2.2 Exercise 297

Solution

$$ \operatorname{AGILITY}_{\rm threshold}