# Supplemental: Randomness

## Primality Testing

A key component of many cryptography algorithms
is finding large prime numbers, with hundreds or even thousands of
digits. A typical approach is to randomly choose a large number and
then apply a *primality test* to determine whether it is prime – the
prime number theorem states that
approximately \(1/m\) of the \(m\)-bit numbers are prime, so we only
expect to check \(m\) numbers before finding one that is prime [1]. As
long as the primality test itself is efficient with respect to \(m\),
the expected time to find a prime number with \(m\) bits is polynomial
in \(m\).

Formally, we wish to decide the following language:

A positive integer \(m\) is a member of \(\PRIMES\) if it has no positive factors other than \(1\) and \(m\).

Is the language \(\PRIMES\) in \(\P\)? Let us consider the following simple algorithm to decide the language:

This algorithm does \(\O(m)\) mod operations. Is this efficient? The size of the input \(m\) is the number of bits required to represent \(m\), which is \(\O(\log m)\). Thus, the number of operations is \(\O(m) = \O(2^{\log m})\), which is exponential in the size of \(m\).

We can improve the algorithm by observing that if \(m\) has a factor within the interval \([2, m-1]\), it must have a factor within \([2, \sqrt{m}]\) – otherwise, we would obtain a product larger than \(m\) when we multiplied its factors together. Thus, we need only iterate up to \(\sqrt{m}\):

This algorithm does \(\O(\sqrt{m})\) mod operations. However, this still is not polynomial; we have:

To be efficient, a primality-testing algorithm must have runtime \(\O(\log^k m)\) for some constant \(k\). Neither of the algorithms above meet this threshold.

In fact, there is a known efficient algorithm for primality testing,
the AKS primality test. Thus, it is
indeed the case that \(\PRIMES \in \P\). However, this algorithm is
somewhat complicated, and its running time is high enough to preclude
it from being used in practice. Instead, we consider a randomized
primality test that is efficient and works well in practice for most
inputs. The algorithm we construct relies on the *extended Fermat’s
little theorem*.

Let \(n \in \N\) be a natural number such that \(n \ge 2\). Let \(a\)
be a *witness* in the range \(1 \le a \le n - 1\). Then:

If \(n\) is prime, then \(a^{n-1} \equiv 1 \pmod{n}\) for any witness \(a\).

If \(n\) is composite and \(n\) is not a

*Carmichael number*, then \(a^{n-1} \equiv 1 \pmod{n}\) for at most half the witnesses \(1 \le a \le n - 1\).

We postpone discussion of Carmichael numbers for the moment. Instead, we take a look at some small cases of composite numbers to see that the extended Fermat’s little theorem holds. We first consider \(n = 6\). We have:

We see that \(a^{n-1} \equiv 1 \pmod{n}\) for only the single witness \(a = 1\). Similarly, we consider \(n = 9\):

Here, there are two witnesses \(a\) where \(a^{n-1} \equiv 1 \pmod{n}\), out of eight total. Finally, we consider \(n = 7\):

Since \(7\) is prime, we have \(a^{n-1} \equiv 1 \pmod{n}\) for all witnesses \(a\).

The extended Fermat’s little theorem leads directly to a simple,
efficient randomized algorithm for primality testing. This *Fermat
primality test* is as follows:

The modular exponentiation in this algorithm can be done with \(\O(\log m)\) multiplications using a divide and conquer strategy, and each multiplication can be done efficiently. Thus, this algorithm is polynomial with respect to the size of \(m\).

As for correctness, we have:

If \(m\) is prime, then the algorithm always accepts \(m\). In other words:

\[\Pr[\text{the Fermat test accepts $m$}] = 1\]If \(m\) is composite and not a Carmichael number, then the algorithm rejects \(m\) with probability at least \(\frac{1}{2}\). In other words:

\[\Pr[\text{the Fermat test rejects $m$}] \ge \frac{1}{2}\]

Thus, if the algorithm accepts \(m\), we can be fairly confident that \(m\) is prime. And as we will see shortly, we can repeat the algorithm to obtain higher confidence that we get the right answer.

We now return to the problem of Carmichael numbers. A *Carmichael
number* is a composite number \(n\) such that \(a^{n-1} \equiv 1
\pmod{n}\) for all witnesses \(a\) that are *relatively prime* to \(n\),
i.e. \(\gcd(a, n) = 1\). This implies that for a Carmichael number, the
Fermat test reports with high probability that the number is prime,
despite it being composite. We call a number that passes the Fermat
test with high probability a *pseudoprime*, and the Fermat test is
technically a randomized algorithm for deciding the following
language:

Carmichael numbers are much rarer than prime numbers, so for many applications, the Fermat test is sufficient to determine with high confidence that a randomly chosen number is prime. On the other hand, if the number is chosen by an adversary, then the Fermat test is unsuitable, and a more complex randomized algorithm such as the Miller-Rabin primality test must be used instead.

### The Miller-Rabin Test

The Miller-Rabin test is designed around the fact that for prime \(m\), the only square roots of 1 modulo \(m\) are -1 and 1. More specifically, if \(x\) is a square root of 1 modulo \(m\), we have

By definition of modular arithmetic, this means that

for some integer \(k\) (i.e. \(m\) evenly divides the difference of \(x^2\) and 1). We can factor \(x^2-1\) to obtain:

When \(m\) is composite, so that \(m = pq\) for integers \(p,q > 1\), then it is possible for \(p\) to divide \(x+1\) and \(q\) to divide \(x-1\), in which case \(pq = m\) divides their product \((x+1)(x-1)\). However, when \(m\) is prime, the only way for the equation to hold is if \(m\) divides either \(x+1\) or \(x-1\); otherwise, the prime factorization of \((x+1)(x-1)\) does not contain \(m\), and by the fundamental theorem of arithmetic, it cannot be equal to a number whose prime factorization contains \(m\) [2]. Thus, we have either \(x+1 = am\) for some integer \(a\), or \(x-1 = bm\) for some \(b \in \Z\). The former gives us \(x \equiv -1 \pmod{m}\), while the latter results in \(x \equiv 1 \pmod{m}\)., This, if \(m\) is prime and \(x^2 \equiv 1 \pmod{m}\), then either \(x \equiv 1 \pmod{m}\) or \(x \equiv -1 \pmod{m}\).

Euclid’s lemma more directly states that if \(m\) is prime and \(m\) divides a product \(ab\), then \(m\) must divide either \(a\) or \(b\).

The Miller-Rabin test starts with the Fermat test: choose a witness \(1 \le a \le m-1\) and check whether \(a^{m-1} \equiv 1 \pmod{m}\). If this does not hold, then \(m\) fails the Fermat test and therefore is not prime. If it does indeed hold, then the test checks the square root \(a^{\frac{1}{2}(m-1)}\) of \(a^{m-1}\) to see if it is 1 or -1. If it is 1, the test checks the square root \(a^{\frac{1}{4}(m-1)}\) of \(a^{\frac{1}{2}(m-1)}\) to see if it is 1 or -1, and so on. The termination conditions are as follows:

The test finds a square root of 1 that is not -1 or 1. By the reasoning above, \(m\) must be composite.

The test finds a square root of 1 that is -1. The reasoning above only holds for square roots of 1, so the test cannot continue by computing the square root of -1. In this case, \(m\) may be prime or composite.

The test reaches some \(r\) for which \(1/2^r\cdot(m-1)\) is no longer an integer. Then it cannot compute \(a\) to this power modulo \(m\). In this case, \(m\) may be prime or composite.

To compute these square roots, we first extract powers of 2 from the number \(m-1\) (which is even for odd \(m\), the cases of interest):

Then \(ExtractPowersOf2(m-1)\) computes \(d\) and \(s\) such that

where \(\gcd(d, 2) = 1\). Then we have \(a^{2^{s-1}d}\) is the square root of \(a^{2^sd}\), since

The full Miller-Rabin algorithm is as follows:

This algorithm is efficient: \(a^d\) can be computed using \(\O(\log d) = \O(\log m)\) multiplications, and it takes another \(\O(\log m)\) multiplications to compute \(a^{2d}, a^{4d}, \dots, a^{2^sd}\) since \(s = \O(\log m)\). Each multiplication is efficient as it is done modulo \(m\), so the entire algorithm takes polynomial time in the size of \(m\).

As for correctness, we have:

If \(m\) is prime, then the algorithm accepts \(m\) with probability 1.

If \(m\) is composite, then the algorithm rejects \(m\) with probability at least \(\frac{3}{4}\).

The latter is due to the fact that when \(m\) is composite, \(m\) passes the Miller-Rabin test for at most \(\frac{1}{4}\) of the witnesses \(1 \le a \le m-1\). Unlike the Fermat test, there are no exceptions to this, making the Miller-Rabin test a better choice in many applications.

*Polynomial identity testing* is the problem of determining whether
two polynomials \(p(x)\) and \(q(x)\) are the same, or equivalently,
whether \(p(x) - q(x)\) is the zero polynomial.

Let \(d\) be an upper bound on the degree of \(p(x)\) and \(q(x)\). Show that for a randomly chosen integer \(a \in [1, 4d]\), if \(p(x) \ne q(x)\), then \(\Pr[p(a) \ne q(a)] \ge \frac{3}{4}\).

**Hint:**A nonzero polynomial \(r(x)\) of degree at most \(d\) can have at most \(d\)*roots*\(s_i\) such that \(r(s_i) = 0\).Devise an efficient, randomized algorithm \(A\) to determine whether \(p(x)\) and \(q(x)\) are the same, with the behavior that:

if \(p(x) = q(x)\), then

\[\Pr[A \text{ accepts } p(x),q(x)] = 1\]if \(p(x) \ne q(x)\), then

\[\Pr[A \text{ rejects } p(x),q(x)] \ge \frac{3}{4}\]

Assume that a polynomial can be efficiently evaluated on a single input.

## Multiplicative Chernoff Bounds

Like Markov’s inequality, *Chernoff bounds* are a form of
concentration bounds that allow us to reason about the deviation of a
random variable from its expectations. There are multiple variants of
Chernoff bounds; we restrict ourselves to the following
“multiplicative” versions.

Let \(X = X_1 + \dots + X_n\) be the sum of independent indicator random variables, where the indicator \(X_i\) has the expectation

Let \(\mu\) be the expected value of \(X\):

Here, we’ve applied linearity of expectation to relate the expectation of \(X\) to that of the indicators \(X_i\). The Chernoff bounds then are as follows.

Let \(X = X_1 + \dots + X_n\), where the \(X_i\) are independent indicator random variables with \(\Ex[X_i] = p_i\), and \(\mu = \sum_i p_i\). Suppose we wish to bound the probability of \(X\) exceeding its expectation \(\mu\) by at least a factor of \(1 + \delta\), where \(\delta > 0\). Then

Let \(X = X_1 + \dots + X_n\), where the \(X_i\) are independent indicator random variables with \(\Ex[X_i] = p_i\), and \(\mu = \sum_i p_i\). Suppose we wish to bound the probability of \(X\) being below its expectation \(\mu\) by at least a factor of \(1 - \delta\), where \(0 < \delta < 1\). Then

These inequalities can be unwieldy to work with, so we often use the following simpler, looser bounds.

Let \(X = X_1 + \dots + X_n\), where the \(X_i\) are independent indicator random variables with \(\Ex[X_i] = p_i\), and \(\mu = \sum_i p_i\). Suppose we wish to bound the probability of \(X\) exceeding its expectation \(\mu\) by at least a factor of \(1 + \delta\), where \(\delta > 0\). Then

Let \(X = X_1 + \dots + X_n\), where the \(X_i\) are independent indicator random variables with \(\Ex[X_i] = p_i\), and \(\mu = \sum_i p_i\). Suppose we wish to bound the probability of \(X\) being below its expectation \(\mu\) by at least a factor of \(1 - \delta\), where \(0 < \delta < 1\). Then

Before we proceed to make use of these bounds, we first prove that the unsimplified upper-tail and lower-tail bounds hold [3].

Refer to the appendix for proofs of the simplified bounds.

To demonstrate the upper-tail Chernoff bound, we make use of the
*Chernoff bounding technique* – rather than reasoning about the
random variable \(X\) directly, we instead reason about \(e^{tX}\),
since small deviations in \(X\) turn into large deviations in
\(e^{tX}\). We have that \(X \ge (1+\delta)\mu\) exactly when \(e^{tX}
\ge e^{t(1+\delta)\mu}\) for any \(t \ge 0\); we obtain this by
raising \(e^t \ge 1\) to the power of both sides of the former
inequality. Then

where the latter step follows from Markov’s inequality. Continuing, we have

We now make use of the fact that the \(X_i\) (and therefore the
\(e^{tX_i}\)) are independent. For **independent** random variables
\(Y\) and \(Z\), we have \(\Ex[YZ] = \Ex[Y]\cdot\Ex[Z]\) (see the
appendix for a proof). Thus,

The \(X_i\) are indicators, with \(\Pr[X_i = 1] = p_i\) and \(\Pr[X_i = 0] = 1 - p_i\). When \(X_i = 1\), \(e^{tX_i} = e^t\), and when \(X_i = 0\), \(e^{tX_i} = e^0 = 1\). Thus, the distribution of \(e^{tX_i}\) is

We can then compute the expectation of \(e^{tX_i}\):

Plugging this into the upper bound that resulted from Markov’s inequality, we get

We can further simplify this expression by observing that \(1+x \le e^x\) for all \(x\):

Using \(x = p_i(e^t-1)\), we have

Applying this to our previously computed upper bound, we get

We can now choose \(t\) to minimize the exponent, which in turn minimizes the value of the exponential itself. Let \(f(t) = e^t - 1 - t(1+\delta)\). Then we compute the derivative with respect to \(t\) and set that to zero to find an extremum:

Computing the second derivative \(f''(\ln(1+\delta)) = e^{\ln(1+\delta)} = 1+\delta\), we see that it is positive since \(\delta > 0\), therefore \(t = \ln(1+\delta)\) is a minimum. Substituting it into our earlier expression, we obtain

The proof of the lower-tail Chernoff bound follows similar reasoning as that of the upper-tail bound. We have:

Here, we can apply Markov’s inequality to the random variable \(e^{-tX}\) since it is nonnegative, unlike \(-X\). Continuing, we have

In the last step, we make use of the fact that the \(X_i\) and therefore the \(e^{-tX_i}\) are independent. The distribution of \(e^{-tX_i}\) is

Then the expectation of \(e^{-tX_i}\) is:

Plugging this into the bound we’ve computed so far, we get

As with the upper-tail, we use the fact that \(1+x \le e^x\) for all \(x\) to simplify this, with \(x = p_i(e^{-t}-1)\):

We choose \(t\) to minimize the exponent. Let \(f(t) = e^{-t} - 1 + t(1-\delta)\). Then we compute the derivative with respect to \(t\) and set that to zero to find an extremum:

Computing the second derivative \(f''(-\ln(1-\delta)) = e^{-(-\ln(1-\delta))} = 1-\delta\), we see that it is positive since \(0 < \delta < 1\), therefore \(t = \ln(1-\delta)\) is a minimum. Substituting it into our earlier expression, we obtain

In lieu of applying Chernoff bounds to the algorithm for estimating \(\pi\), we consider a different example of flipping a coin. If we flip a biased coin with probability \(p\) of heads, we expect to see \(np\) heads. Let \(H\) be the total number of heads, and let \(H_i\) be an indicator variable corresponding to whether the \(i\)th flip is heads. We have \(\Pr[H_i = 1] = p\), and \(\Ex[H] = np\) by linearity of expectation.

Suppose the coin is fair. What is the probability of getting at least six heads out of ten flips? This is a fractional deviation of \(\delta = 6/5 - 1 = 0.2\), and applying the upper-tail Chernoff bound gives us:

What is the probability of getting at least 60 heads out of 100 flips, which is the same fractional deviation \(\delta = 0.2\) from the expectation? Applying the upper tail again, we get

Thus, the probability of deviating by a factor of \(1+\delta\) decreases significantly as the number of samples increases. For this particular example, we can compute the actual probabilities quite tediously from exact formulas for a binomial distribution, which yields 0.37 for getting six heads out of ten flips and 0.03 for 60 heads out of 100 flips. However, this approach becomes more and more expensive as \(n\) increases, and the Chernoff bound produces a reasonable result with much less work.

### Polling Analysis with Chernoff Bounds

Recall that in polling, a confidence level \(1-\gamma\) and margin of error \(\varepsilon\) requires

or equivalently

Formally, we define indicator variables \(X_i\) as

for each person \(i\) in the set that we poll. Then \(X = X_1 + \dots + X_n\) is the sum of independent indicator variables, with

For an arbitrary margin of error \(\varepsilon\), we want to bound the probability

In the second-to-last step, we use the fact that \(X = x\) and \(X > x\) are disjoint events, since a random variable maps each sample point to a single value, and that \((X \ge x) = (X = x) \cup (X > x)\).

We now have events that are in the right form to apply a Chernoff bound, with \(\delta = \varepsilon n/\mu\). We first apply the simplified upper-tail bound to the first term, obtaining

We want this to be less than some value \(\beta\):

Unfortunately, this expression includes \(p\), the quantity we’re trying to estimate However, we know that \(p \le 1\), so we can set

to ensure that \(\Pr[X \ge (1+\varepsilon n/\mu)\mu] < \beta\).

We can also apply the simplified lower-tail bound to the second term in our full expression above:

As before, we want this to be less some value \(\beta\):

We again observe that \(p \le 1\), so we can set

to ensure that \(\Pr[X \le (1-\varepsilon n/\mu)\mu] < \beta\).

To bound both terms simultaneously, we can choose \(\beta=\gamma/2\), so that

To achieve this, we require

For example, for a 95% confidence level and a margin of error of ±2%, we have \(\varepsilon=0.02\) and \(\gamma=0.05\). Plugging those values into the result above, we need no more than

samples to achieve the desired confidence level and margin of error. Observe that this does not depend on the total population size!

## Probabilistic Complexity Classes

The Fermat primality test is an example of a
*one-sided-error randomized algorithm* – an input that is pseudoprime
is always accepted, while a non-pseudoprime is sometimes accepted and
sometimes rejected. We can define complexity classes corresponding to
decision problems that have efficient one-sided-error randomized
algorithms.

\(\RP\) is the class of languages that have efficient one-sided-error
randomized algorithms with no *false positives*. A language \(L\) is
in \(\RP\) if there is an efficient randomized algorithm \(A\) such
that:

if \(x \in L\), \(\Pr[A \text{ accepts } x] \ge c\)

if \(x \notin L\), \(\Pr[A \text{ rejects } x] = 1\)

Here, \(c\) must be a constant greater than 0. Often, \(c\) is chosen to be \(\frac{1}{2}\).

\(\coRP\) is the class of languages that have efficient
one-sided-error randomized algorithms with no *false negatives*. A
language \(L\) is in \(\coRP\) if there is an efficient randomized
algorithm \(A\) such that:

if \(x \in L\), \(\Pr[A \text{ accepts } x] = 1\)

if \(x \notin L\), \(\Pr[A \text{ rejects } x] \ge c\)

Here, \(c\) must be a constant greater than 0. Often, \(c\) is chosen to be \(\frac{1}{2}\).

\(\RP\) stands for *randomized polynomial* time. If a language \(L\) is in
\(\RP\), then its complement language \(\overline{L}\) is in \(\coRP\) – an
algorithm for \(L\) with no false positives can be converted into an
algorithm for \(\overline{L}\) with no false negatives. The Fermat test
produces no false negatives, so \(\mathrm{PSEUDOPRIMES} \in \coRP\).
Thus, the language

is in \(\RP\).

The constant \(c\) in the definition of \(\RP\) and \(\coRP\) is arbitrary.
With any \(c > 0\), we can *amplify* the probability of success by
repeatedly running the algorithm. For instance, if we have a
randomized algorithm \(A\) with no false positives for a language \(L\),
we have:

We can construct a new algorithm \(B\) as follows:

\(B\) just runs \(A\) twice on the input, accepting if at least one run of \(A\) accepts; \(B\) only rejects if both runs of \(A\) reject. Thus, the probability that \(B\) rejects \(x\) is:

Here, we have used the fact that the two runs of \(A\) are independent, so the probability of \(A\) rejecting twice is the product of the probabilities it rejects each time. This gives us:

or equivalently:

Repeating this reasoning, if we modify \(B\) to run \(A\) \(k\) times, we get:

By applying a form of Bernoulli’s inequality, \((1 + x)^r \le e^{rx}\), we get:

If we want a particular lower bound \(1 - \varepsilon\) for acceptance of true positives, we need

This is a constant for any constants \(c\) and \(\varepsilon\). Thus, we can amplify the probability of accepting a true positive from \(c\) to \(1 - \varepsilon\) for any \(\varepsilon\) with a constant \(\ln(1/\varepsilon)/c\) number of repetitions. The same reasoning can be applied to amplify one-sided-error algorithms with no false negatives.

One-sided-error randomized algorithms are a special case of
*two-sided-error randomized algorithms*. Such an algorithm for a
language \(L\) can produce the wrong result for an input \(x\) whether or
not \(x \in L\). However, the probability of getting the wrong result is
bounded by a constant.

\(\BPP\) is the class of languages that have efficient two-sided-error randomized algorithms. A language \(L\) is in \(\BPP\) if there is an efficient randomized algorithm \(A\) such that:

if \(x \in L\), \(\Pr[A \text{ accepts } x] \ge c\)

if \(x \notin L\), \(\Pr[A \text{ rejects } x] \ge c\)

Here, \(c\) must be a constant greater than \(\frac{1}{2}\), so that the algorithm produces the correct answer the majority of the time. Often, \(c\) is chosen to be \(\frac{2}{3}\) or \(\frac{3}{4}\).

\(\BPP\) stands for *bounded-error probabilistic polynomial* time. Given
the symmetric definition of a two-sided error algorithm, it is clear
that the class \(\BPP\) is closed under complement – if a language \(L
\in \BPP\), then \(\overline{L} \in \BPP\) as well.

Languages in \(\RP\) and in \(\coRP\) both trivially satisfy the conditions for \(\BPP\); for instance, we have the following for \(\RP\):

if \(x \in L\), \(\Pr[A \text{ accepts } x] \ge c\)

if \(x \notin L\), \(\Pr[A \text{ rejects } x] = 1 \ge c\)

Thus, \(\RP \cup \coRP \subseteq \BPP\). By similar reasoning, we can relate \(\P\) to these classes as follows:

As with one-sided-error algorithms, the probability of success for two-sided-error randomized algorithms can be amplified arbitrarily, as we will see shortly.

One-sided-error and two-sided-error randomized algorithms are known as
*Monte Carlo* algorithms. Such algorithms have a bounded runtime, but
they may produce the wrong answer. Contrast this with *Las Vegas*
algorithms, which always produce the correct answer, but where the
runtime bound only holds *in expectation*. We can define a complexity
class for languages that have Las Vegas algorithms with expected
runtime that is polynomial in the size of the input.

\(\ZPP\) is the class of languages that have efficient Las Vegas algorithms. A language \(L\) is in \(\ZPP\) if there is a randomized algorithm \(A\) such that:

If \(x \in L\), \(A\) always accepts \(x\).

If \(x \notin L\), \(A\) always rejects \(x\).

The expected runtime of \(A\) on input \(x\) is \(\O(\abs{x}^k)\) for some constant \(k\).

There is a relationship between Monte Carlo and Las Vegas algorithms: if a language \(L\) has both a one-sided-error algorithm with no false positives and a one-sided-error algorithm with no false negatives, then it has a Las Vegas algorithm, and vice versa. In terms of complexity classes, we have \(L\) is in both \(\RP\) and \(\coRP\) if and only if \(L\) is in \(\ZPP\). This implies that

We demonstrate this as follows. Suppose \(L\) is in \(\RP \cap \coRP\). Then it has an efficient, one-sided-error algorithm \(A\) with no false positives such that:

if \(x \in L\), \(\Pr[A \text{ accepts } x] \ge \frac{1}{2}\)

if \(x \notin L\), \(\Pr[A \text{ accepts } x] = 0\)

Here, we have selected \(c = \frac{1}{2}\) for concreteness. \(L\) also has an efficient, one-sided algorithm \(B\) with no false negatives such that:

if \(x \in L\), \(\Pr[B \text{ rejects } x] = 0\)

if \(x \notin L\), \(\Pr[B \text{ rejects } x] \ge \frac{1}{2}\)

We can construct a new algorithm \(C\) as follows:

Since \(A\) only accepts \(x \in L\) and \(C\) only accepts when \(A\) does, \(C\) only accepts \(x \in L\). Similarly, \(B\) only rejects \(x \notin L\), so \(C\) only rejects \(x \notin L\). Thus, \(C\) always produces the correct answer for a given input.

To show that \(L \in \ZPP\), we must also demonstrate that the expected runtime of \(C\) is polynomial. For each iteration \(i\) of \(C\), we have:

if \(x \in L\), \(\Pr[A \text{ accepts } x \text{ in iteration } i] \ge \frac{1}{2}\)

if \(x \notin L\), \(\Pr[B \text{ rejects } x \text{ in iteration } i] \ge \frac{1}{2}\)

Thus, if \(C\) gets to iteration \(i\), the probability that \(C\) terminates in that iteration is at least \(\frac{1}{2}\). We can model the number of iterations as a random variable \(X\), and \(X\) has the probability distribution:

This is similar to a *geometric distribution*, where

for some \(p \in (0, 1]\). The expected value of such a distribution is \(\Ex[Y] = 1/p\). This gives us:

Thus, the expected number of iterations of \(C\) is no more than two, and since \(A\) and \(B\) are both efficient, calling them twice is also efficient.

We have shown that \(\RP \cap \coRP \subseteq \ZPP\). We will proceed to show that \(\ZPP \subseteq \RP\), meaning that if a language has an efficient Las Vegas algorithm, it also has an efficient one-sided-error algorithm with no false positives. Assume that \(C\) is a Las Vegas algorithm for \(L \in \ZPP\), with expected runtime of no more than \(\abs{x}^k\) steps for an input \(x\) and some constant \(k\). We construct a new algorithm \(A\) as follows:

\(A\) is clearly efficient in the size of the input \(x\). It also has no false positives, since it only accepts if \(C\) accepts within the first \(2\abs{x}^k\) steps, and \(C\) always produces the correct answer. All that is left is to show that if \(x \in L\), \(\Pr[A \text{ accepts } x] \ge \frac{1}{2}\) (again, we arbitrarily choose \(c = \frac{1}{2}\)).

Let \(S\) be the number of steps before \(C\) accepts \(x\). By assumption, we have

\(A\) runs \(C\) for \(2\abs{x}^k\) steps, so we need to bound the probability that \(C\) takes more than \(2\abs{x}^k\) steps on \(x\). We have:

Here, we applied Markov’s inequality in the second step. Then:

Thus, \(A\) is indeed a one-sided-error algorithm with no false positives, so \(L \in \RP\). A similar construction can be used to demonstrate that \(L \in \coRP\), allowing us to conclude that \(\ZPP \subseteq \RP \cap \coRP\). Combined with our previous proof that \(\RP \cap \coRP \subseteq \ZPP\), we have \(\RP \cap \coRP = \ZPP\).

One final observation we will make is that if a language \(L\) is in \(\RP\), then it is also in \(\NP\). Since \(L \in \RP\), there is an efficient, one-sided-error randomized algorithm \(A\) to decide \(L\). \(A\) is allowed to make random choices, and we can model each choice as a coin flip, i.e. being either 0 or 1 according to some probability distribution. We can represent the combination of these choices in a particular run of \(A\) as a binary string \(c\). This enables us to write an efficient verifier \(V\) for \(L\) as follows:

If \(x \in L\), then \(\Pr[A \text{ accepts } x] \ge \frac{1}{2}\), so at least half the possible sequences of random choices lead to \(A\) accepting \(x\). On the other hand, if \(x \notin L\), then \(\Pr[A \text{ rejects } x] = 1\), so all sequences of random choices lead to \(A\) rejecting \(x\). Thus, \(V\) accepts at least half of all possible certificates when \(x \in L\), and \(V\) rejects all certificates when \(x \notin L\), so \(V\) is a verifier for \(L\). Since \(A\) is efficient, \(V\) is also efficient.

We summarize the known relationships between complexity classes as follows:

These relationships are represented pictorially, with edges signifying containment, as follows:

Here, \(\coNP\) is the class of languages whose complements are in \(\NP\):

We do not know if any of the containments above are strict, and we do
not know the relationships between \(\NP\), \(\coNP\), and \(\BPP\). It is
commonly **believed** that \(\P\) and \(\BPP\) are equal and thus \(\BPP\)
is contained in \(\NP \cap \coNP\), that neither \(\P\) nor \(\BPP\) contain
all of \(\NP\) or all of \(\coNP\), and that \(\NP\) and \(\coNP\) are not
equal. However, none of these conjectures has been proven [4].

It is known, however, that if \(\P = \NP\), then \(\P = \BPP\). This is because \(\BPP\) is contained within the polynomial hierarchy, and one of the consequences of \(\P = \NP\) is the “collapse” of the hierarchy. The details are beyond the scope of this text.

## Amplification for Two-Sided-Error Algorithms

Previously, we saw how to amplify the probability of success for one-sided-error randomized algorithms. We now consider amplification for two-sided-error algorithms. Recall that such an algorithm \(A\) for a language \(L\) has the following behavior:

if \(x \in L\), \(\Pr[A \text{ accepts } x] \ge c\)

if \(x \notin L\), \(\Pr[A \text{ rejects } x] \ge c\)

Here, \(c\) must be a constant that is strictly greater than \(\frac{1}{2}\).

Unlike in the one-sided case with no false positives, we cannot just run the algorithm multiple times and observe if it accepts at least once. A two-sided-error algorithm can accept both inputs in and not in the language, and it can reject both such inputs as well. However, we observe that because \(c > \frac{1}{2}\), we expect to get the right answer the majority of the time when we run a two-sided-error algorithm on an input. More formally, suppose we run the algorithm \(n\) times. Let \(Y_i\) be an indicator random variable that is 1 if the algorithm accepts in the \(i\)th run. Then:

if \(x \in L\), \(\Ex[Y_i] = \Pr[Y_i = 1] \ge c\)

if \(x \notin L\), \(\Ex[Y_i] = \Pr[Y_i = 1] \le 1 - c\)

Let \(Y = Y_1 + \dots + Y_n\) be the total number of accepts out of \(n\) trials. By linearity of expectation, we have:

if \(x \in L\), \(\Ex[Y] \ge cn > \frac{n}{2}~~~\) (since \(c > \frac{1}{2}\))

if \(x \notin L\), \(\Ex[Y] \le (1 - c)n < \frac{n}{2}~~~\) (since \(1 - c < \frac{1}{2}\))

This motivates an amplification algorithm \(B\) that runs the original algorithm \(A\) multiple times and takes the majority vote:

Suppose we wish to obtain a bound that is within \(\gamma\) of 1:

if \(x \in L\), \(\Pr[B \text{ accepts } x] \ge 1 - \gamma\)

if \(x \notin L\), \(\Pr[B \text{ rejects } x] \ge 1 - \gamma\)

What value for \(n\) should we use in \(B\)?

We first consider the case where \(x \in L\). The indicators \(Y_i\) are independent, allowing us to use Hoeffding’s inequality on their sum \(Y\). \(B\) accepts \(x\) when \(Y \ge \frac{n}{2}\), or equivalently, \(\frac{Y}{n} \ge \frac{1}{2}\). We want

or equivalently,

where \(\varepsilon = c - \frac{1}{2}\). To apply Hoeffding’s inequality, we need something of the form

where \(p = \lrEx{\frac{Y}{n}}\). Unfortunately, we do not know the exact value of \(p\); all we know is that \(p = \lrEx{\frac{Y}{n}} \ge c\). However, we know that because \(p \ge c\),

In general, the event \(X \le a\) includes at most as many sample points as \(X \le b\) when \(a \le b\); the latter includes all the outcomes in \(X \le a\), as well as those in \(a < X \le b\). We thus need only compute an upper bound on \(\lrPr{\frac{Y}{n} \le p - \varepsilon}\), and that same upper bound will apply to \(\lrPr{\frac{Y}{n} \le c - \varepsilon}\). Taking \(\varepsilon = c - \frac{1}{2}\) and applying the lower-tail Hoeffding’s inequality, we get:

We want this to be bound by \(\gamma\):

As a concrete example, suppose \(c = \frac{3}{4}\), meaning that \(A\) accepts \(x \in L\) with probability at least \(\frac{3}{4}\). Suppose we want \(B\) to accept \(x \in L\) at least 99% of the time, giving us \(\gamma = 0.01\). Then:

Thus, it is sufficient for \(B\) to run \(n \ge 37\) trials of \(A\) on \(x\).

We now consider \(x \notin L\). We want \(B\) to reject \(x\) with probability at least \(1 - \gamma\), or equivalently, \(B\) to accept \(x\) with probability at most \(\gamma\):

Similar to before, we have

with \(\varepsilon = c - \frac{1}{2}\). Since \(p = \lrEx{\frac{Y}{n}} \le (1 - c)\), we have

This follows from similar reasoning as earlier: an event \(X \ge a\) contains no more sample points than \(X \ge b\) when \(a \ge b\). With \(\varepsilon = c - \frac{1}{2}\), the upper-tail Hoeffding’s inequality gives us:

We want this to be no more than \(\gamma\), which leads to the same solution as before:

Using the same concrete example as before, if we want \(B\) to reject \(x \notin L\) at least 99% of the time when \(A\) rejects \(x \notin L\) with probability at least \(\frac{3}{4}\), it suffices for \(B\) to run \(n \ge 37\) trials of \(A\) on \(x\).