# 5.1 The hiring problem

## 5.1-1

Show that the assumption that we are always able to determine which candidate is best in line 4 of procedure $\text{HIRE-ASSISTANT}$ implies that we know a total order on the ranks of the candidates.

A total order is a partial order that is a total relation $(\forall a, b \in A:aRb \text{ or } bRa)$. A relation is a partial order if it is reflexive, antisymmetric and transitive.

Assume that the relation is good or better.

• Reflexive: This is a bit trivial, but everybody is as good or better as themselves.
• Transitive: If $A$ is better than $B$ and $B$ is better than $C$, then $A$ is better than $C$.
• Antisymmetric: If $A$ is better than $B$, then $B$ is not better than $A$.

So far we have a partial order.

Since we assume we can compare any two candidates, then comparison must be a total relation and thus we have a total order.

## 5.1-2 $\star$

Describe an implementation of the procedure $\text{RANDOM}(a, b)$ that only makes calls to $\text{RANDOM}(0, 1)$. What is the expected running time of your procedure, as a function of $a$ and $b$?

As $(b - a)$ could be any number, we need at least $\lceil \lg(b - a) \rceil$ bits to represent the number. We set $\lceil \lg(b - a) \rceil$ as $k$. Basically, we need to call $\text{RANDOM}(0, 1)$ $k$ times. If the number represented by binary is bigger than $b - a$, it's not valid number and we give it another try, otherwise we return that number.

RANDOM(a, b)
range = b - a
bits = ceil(log(2, range))
result = 0
for i = 0 to bits - 1
r = RANDOM(0, 1)
result = result + r << i
if result > range
return RANDOM(a, b)
else return a + result


The expectation of times of calling procedure $\text{RANDOM}(a, b)$ is $\frac{2^k}{b - a}$. $\text{RANDOM}(0, 1)$ will be called $k$ times in that procedure.

The expected running time is $\Theta(\frac{2^k}{b - a} \cdot k)$, $k$ is $\lceil \lg(b - a) \rceil$. Considering $2^k$ is less than $2 \cdot (b - a)$, so the running time is $O(k)$.

## 5.1-3 $\star$

Suppose that you want to output $0$ with probability $1 / 2$ and $1$ with probability $1 / 2$. At your disposal is a procedure $\text{BIASED-RANDOM}$, that outputs either $0$ or $1$. It outputs $1$ with some probability $p$ and $0$ with probability $1 - p$, where $0 < p < 1$, but you do not know what $p$ is. Give an algorithm that uses $\text{BIASED-RANDOM}$ as a subroutine, and returns an unbiased answer, returning $0$ with probability $1 / 2$ and $1$ with probability $1 / 2$. What is the expected running time of your algorithm as a function of $p$?

There are 4 outcomes when we call $\text{BIASED-RANDOM}$ twice, i.e., $00$, $01$, $10$, $11$.

The strategy is as following:

• $00$ or $11$: call $\text{BIASED-RANDOM}$ twice again
• $01$: output $0$
• $10$: output $1$

We can calculate the probability of each outcome:

• $\Pr\{00 | 11\} = p^2 + (1 - p)^2$
• $\Pr\{01\} = (1 - p)p$
• $\Pr\{10\} = p(1 - p)$

Since there's no other way to return a value, it returns $0$ and $1$ both with probability $1 / 2$.

The pseudo code is as follow:

UNBIASED-RANDOM
while true
x = BIASED-RANDOM
y = BIASED-RANDOM
if x != y
return x


This algorithm actually uses the equivalence of the probability of occurrence of $01$ and $10$, and subtly converts the unequal $00$ and $11$ to $01$ and $10$, thus eliminating the probability that its probability is not equivalent.

Each iteration is a Bernoulli trial, where "success" means that the iteration does return a value.

We can view each iteration as a Bernoulli trial, where "success" means that the iteration returns a value.

\begin{aligned} \Pr\{\text{success}\} & = \Pr\{0\text{ is returned}\} + \Pr\{1\text{ is returned}\} \\ & = 2p(1 - p). \end{aligned}

The expected number of trials for this scenario is $1 / (2p(1 - p))$. Thus, the expected running time of $\text{UNBIASED-RANDOM}$ is $\Theta(1 / (2p(1 - p))$.