# Example of concrete statement which requires probabilistic algorithm

In my paper I would like to include an example of easy concrete finitary statement which can be easily verified by probabilistic algorithm to any reasonable confidence level, but which looks completely hopeless to prove rigorously.

A candidate example is the statement "$10^{1500} + 2329$ is prime". This can be easily checked with (probabilistic) Miller–Rabin primarity test, while deterministic AKS primality test is too slow to work in this region. However, I am not sure if some other methods (based on elliptic curves, etc.) could be used to rigorously and efficiently prove that this particular number is prime.

Does there exists $n$ such that Miller–Rabin primarity test works efficiently for numbers of the order $10^n$, but all rigorous methods are expected to work "forever"?

Alternatively, can you suggest any other concrete statement which works? For example, based on polynomial identity testing? That is, statement "$P=0$" (or "$P \neq 0$") for a concrete polynomial P, simple enough to be included in the paper explicitly.

• The last version of pari/gp, which contains a very nice implementation of ECPP by Jared Asuncion, rigorously proves the primality of this number in 12 minutes on my laptop. – Aurel Jun 19 '18 at 11:05
• What is the meaning of an algorithm running "too slowly" or "forever" for your purposes? For example, do you think having to wait 2 days for the algorithm to run is too slow? (If so I would say you just need to be more patient.) What about a week? A year? Two years? None of these wait times really qualify as anything like "forever". – KConrad Jun 19 '18 at 12:13
• Aurel - thank you. I had a guess that this number is not big enough for rigorous proof of primality to be really difficult. That is why I am asking how big numbers I should choose, or maybe use some statement of completely different nature instead... – Bogdan Jun 19 '18 at 15:28
• KConrad - a short answer is "the harder the better". Of course, there is no way to actually PROVE that any given problem would take this many years, but, for example, everyone agrees that factor a 1000-gidit number (which, for example, is the product of 2 random primes) is hard. For what n proving primarily of n-digit number is expected to be similarly hard? Would probabilistic algorithms still work for this n? – Bogdan Jun 19 '18 at 15:56
• Primality testing is much easier than factorisation. The deterministic AKS algorithm has complexity $O((\log n)^6)$ or probably even less by now, for an integer $n.$ – kodlu Jun 19 '18 at 22:16

First of all, there is a conjecture in computer science that BPP = P, i.e., anything that can be done in randomized polynomial time can also be done in deterministic polynomial time. The hope is that you can replace true-random trials with repeated trials with a cryptographically secure pseudo-random number generator. No one can prove that such PRNGs exist, but there are many candidates that are secure as far as anyone knows. A more refined version of the conjecture suggests a quadratic speedup, so that it's still polynomial but not the same degree.

The situation with primality testing is in keeping with this same general picture. The time complexity of Miller-Rabin is $O(d^{2+\epsilon})$, while the time complexity of the rigorous ECPP test is (I am told) $O(d^{4+\epsilon})$. Now ECPP is not Miller-Rabin with repeated trials, but the extended Riemann hypothesis implies that Miller-Rabin with repeated trials would also give you a $O(d^{4+\epsilon})$ time algorithm. My guess is that ECPP is a little slower than ERH-enhanced Miller-Rabin, but it has the advantage that it primality is completely unconditional. In fact, ECPP is also a randomized algorithm, but its answer is always rigorous; the only uncertainty is that it is probably fast. (This is more formally called ZPP and is also called "Las Vegas randomized".)

This page shows current (or current as of a few years ago?) records for finding probable primes using Miller-Rabin. Meanwhile this page shows current records for proving primality using ECPP. A convenient comparison case is given by looking at Wagstaff primes, which are primes of the form $(2^n + 1)/3$. These primes are similar to Mersenne primes in some ways, except that there is no known deterministic primality test like Lucas-Lehmer which is used for the Mersenne case. (In fact Lucas-Lehmer is itself similar to Miller-Rabin and is about as fast; they are both simply refinements of Fermat's Little Theorem.) So, comparing records, $(2^{13,372,531}+1)/3$ was witnessed as probably prime in September 2013, while $(2^{83,339}+1)/3$ was proven to be prime by ECPP one year later, in September 2014. I don't have a specific comparison of computational resources used, but qualitatively it's moot. Taking the $O(d^{4+\epsilon})$ time estimate, it would presumably take about a billion times as much computational power to prove that $(2^{13,372,531}+1)/3$ is prime as to prove that $(2^{83,339}+1)/3$ is prime. In the nearly four years since the smaller number was shown to be prime, the number of digits in ECPP records has grown by a factor of 1.4, which is roughly a factor of 4 in increased computational resources. (Curiously there are 2 or 3 more Wagstaff probable primes in this range, and I'm not sure why they haven't yet been ECPP certified.) I suppose that rigorously proving that $(2^{13,372,531}+1)/3$ is prime is within reach of the US federal budget. But that too is surely moot because a small fraction of that same money could be spent on finding yet bigger probable Wagstaff primes.

• Thank you! I was not aware about these webpages. Yes, statement "$(2 ^{13,372,531} +1)/3$ is prime" is the example I was asking for. – Bogdan Jun 20 '18 at 10:07

Take two binary vectors $x$ and $y$ of bitlength $n.$ Party X has $x$ and party Y has $y$. They are trusted parties (trust each other) separated by a one way reliable communication channel. Let $n$ be arbitraryily large.

The goal is for party X to transmit $x$ to party Y and for party Y to decide whether $x$ is indeed equal to $y$.

A deterministic algorithm will require $O(n)$ bits to be transmitted. The randomized algorithm below requires $O(\log n)$ bits to be transmitted.

Party X uniformly at random chooses a prime $p$ from $[2,n^2]$ (since there are $\sim \frac{n^2}{2\log n}$ such primes by the PNT, this can be done using $O(\log n)$ random bits by first setting the last bit to be 1, to ensure an odd number is selected. Party X can test primality, after choosing uniformly an odd number from $[2,n^2]$ this has cost $O((\log n)^c)$ for $c\leq 6.$

Party X now encodes $p$ and $N(x) \pmod p~$, where $N(x)$ is the integer whose binary expansion is the bitvector $x$, in binary and transmits these $O(\log n)$ bits to Party Y. So party Y can reconstruct the prime and the remainder. It can then compute $x \pmod y$ and check equality.

If $x \neq y$ then the residue is different with high probability $p,$ where $$p \geq 1-\frac{\log n^2}{n}=1-\frac{2\log n}{n}.$$ For large $n$ this is essentially $1.$

This can be argued by observing that the residue mod $p$ is the same for the distinct numbers if $p$ divides $|N(x)-N(y)|.$

So the probability that party Y accepts $x$ as equal to $y$ when $x\neq y$ can be made very small by a randomized algorithm, choosing a number of primes at random, instead of just one.

Reference: This is from Hromkovic's book, Algorithmics for Hard Problems, section 5.2.

Adapted from Leonid Levin, https://www.cs.bu.edu/fac/lnd/expo/gdl.htm : Take a coin and flip it 2 million times. The resulting string of T and H has Kolmogorov complexity no smaller than 1 million bits. Almost certainly true, but can't be proved by any finite decision procedure.