## Greatest Common Divisors; Euclidean Algorithm.

### January 5, 2013

Recently, I’ve been learning to program in a new language and have been doing Project Euler problems for practice — of course, as the name suggests, most of these problems deal explicitly with problems which must be solved (efficiently) with mathematical techniques.  Two of the most common algorithms I’ve used are: Prime Testing, GCD Finder.  I’ll post about the former later, but the latter is an interesting problem in its own right:

Initial Problem.  Given two natural (positive whole) numbers called $m, n$, can we find some other natural number that divides both of them?

This problem is a first step.  It’s nice to be able to write numbers as a multiple of some other number; for example, if we have 16 and 18, we may write them as $8\times 2$ and $9\times 2$, thus giving us some insight as to the relationship between these two numbers. In that case it may be easy to see, but perhaps if you’re given the numbers 46629 and 47100, you may not realize right away that these numbers are $99\times 471$ and $100\times 471$ respectively.  This kind of factorization will reveal "hidden" relationships between numbers.

So, given two numbers, how do we find if something divides both of them — in other words, how do we find the common divisors of two numbers?  If we think back to when we first began working with numbers (in elementary school, perhaps) the first thing to do would be to note that 1 divides every number.  But that doesn’t help us all that much, as it turns out, so we go to the next number: if both numbers are even, then they have 2 as a common factor.  Then we "factor" both numbers by writing them as $2\times\mbox{ something}$ and then attempt to keep dividing things out of the something.  We then move onto 3, skip 4 (since this would just be divisible by 2 twice), go onto 5, then 7, then…and continue for the primes.  This gives a prime factorization, but we have to note that if, say, 2 and 5 divide some number, then so does 10.  These latter divisors are the composite factors.

This seems excessive, but it is sometimes the only way one can do it.

Anecdote!: On my algebra qualifying exam, there was a question regarding a group of order 289 which required us to see if 289 was prime or not; if not, we were to factor it.  We were not allowed calculators, so what could we do?  Try everything.  Note that we only need to try up to the square root of the number (which we could estimate in other ways), but it’s still a number of cases.  If you check, none of the following numbers divide into 289: 2, 3, 5, 7, 11, 13.  At this point, I was about to give up and call it a prime, but, for whatever reason, I decided to try 17.  Of course, as the clever reader will have pointed out, $289 = 17\times 17$.  It is not prime.  There was, luckily, only one student who thought it was prime, but it points out how the algorithm above is not entirely trivial if one does not have access to a computer or calculator.

Once we have a common divisor, or a set of common divisors, a natural thing to want to do is to find the biggest (we already have the smallest, 1) since in this way we can write our numbers with the largest common factor multiplied by some other number.  It will, in effect, make things prettier.

Real Problem.  Find the greatest divisor which is common to two natural numbers, $m, n$.

If you were just learning about this kind of thing, you may spout out the following solution: find all of the common divisors, then pick the greatest.  While this is not especially efficient, it is a solution.  Unfortunately, even for small numbers, this gets out of hand quickly.  For example, 60 and  420 have the following common divisors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.  This takes a while to compute by hand.

Even if we were to find prime factors, this would be $60 = 2^2 \times 3\times 5$ and $420 = 2^2 \times 3 \times 5\times 7$, which gives us that they share a number of prime factors.  A bit of thinking gives us that we take all of the prime factors they "share" and multiply them together to get the greatest common divisor.  This is another potential solution which is much faster than simply listing out all of the common divisors.  Unfortunately, this falls prey to the same kind of trap that other prime-related problems do: it is, at times, especially difficult to factor large composite numbers.  For example, the "reasonably small" number 49740376105597 has a prime factorization of $2741 \times 37813 \times 479909$; this is not at all efficient to factor if one does not have a computer or a specialized calculator with a factoring algorithm on it.  As a mean joke, you may ask your friend to factor something like 1689259081189, which is actually the product of the 100,000th and 100,001st prime — that is, they would need to test 99,999 primes before getting to the one which divides the number.  If they divided by one prime per second (which is quite fast!) this would take them 1 day, 3 hours, and 46 minutes.  Not especially effective, but it will eventually get the job done.

Real Problem, With Efficiency: Find the greatest divisor which is common to two natural numbers, $m,n$, but do so in an efficient manner (we’ve all got deadlines!).

We need to sit down and think about this now.  We need an entirely new idea.  We note, at least, that for the two numbers $m,n$ that one of them must be larger than the other (or else the problem is trivial).  One thing to try would be to see if the smaller one goes into the larger one (for example, above we had 60 going into 420, which gave us the easy solution that 60 must be the greatest common divisor).  If not, maybe we can see how much is left over.  That is, if $m$ is the larger number,

$m = a_{1}n + r_{1}$

where here $a_{1}$ is the number of times $n$ goes into $m$ without exceeding it, and $r_{1}$ is the "remainder"; if it’s equal to 0, then $n$ evenly divides into $m$, and otherwise it is less than $n$ (or else we could divide an additional $n$ into $m$).

Using this, if $r_{1}\neq 0$, we may write $m - a_{1}n = r_{1}$; this means that, in particular, $r_{1}$ divides $m$ and $a_{1}n$, so it is a factor of $m$ and of $a_{1}n$.  But it may not actually be a factor of $n$; so let’s see how many times it goes into $n$.  Using the same process…

$n = a_{2}r_{1} + r_{2}$

and by rearranging, we have that $n - a_{2}r_{1}$ is divisible by $r_{2}$.  So, $n$ is divisible by $r_{2}$, but we aren’t sure if $r_{1}$ is divisible by $r_{2}$…if it were, we would be able to say that $r_{2}$ was a common divisor of $m$ and $n$ (why?).  That’s something at least.

The cool thing about our algorithm here is that, because $a_{1}n + r_{1} = m$ we have that either $r_{1} = 0$ and we’re done with the algorithm, or $r_{1} > 0$ and we may form a new equation $n = a_{2}r_{1} + r_{2}$; this equation has, on the left-hand side, the number $n$ which is less than the previous equation’s left-hand side, which was $m$.  Continuing this process, we will have $r_{1}, r_{2}, \dots$ on the left-hand side, each of which is less than the one which came before it.  Because $r_{i} \geq 0$ for any of the remainders, eventually it will become 0 (why?) and this algorithm will terminate.  That is, we will have found some $r_{i}$ which is a common divisor for both $n, m$; specifically, it will be the $r_{i}\neq 0$ such that $r_{i+1} = 0$ (or, it may simply be $n$ if $n$ divides $m$).

This algorithm, called the Euclidean Algorithm, actually does more "automatically": it not only finds a common divisor, but actually finds the greatest common divisor of $m,n$, which, from now on, we will denote $\gcd(m,n)$.  The "proof" of this is simply noting that $\gcd(m,n) = \gcd(n,r_{1}) = \gcd(r_{1},r_{2}) = \cdots = \gcd(r_{i-1},r_{i}) = r_{i}$ (we noted this above without making reference to the gcd, but the reader should attempt to go through all the same steps using the idea of the gcd).

So.  If you have two natural numbers, $n,m$, you divide them, find the remainder, write the equation, then continue as above until you get a 0 remainder.  Then you pick the remainder directly before you got 0 as your gcd (or, you pick the smaller number if one number divides the other).  Pretty simple algorithm, but is it efficient?

Without going into formal "efficiency" definitions, "yes", it is quite efficient.  To prove it, let’s take an "average" example using the "large" numbers 1337944608 and 4216212.  We note that (by pen and paper, or by using a standard calculator) that

1337944608 = 317(4216212) + 1405404.

Next, we note that

4216212 = 3(1405404) + 0

which instantly gives us the solution $\gcd(4216212, 1337944608) = 1405404$.  That’s pretty awesome.  Note that this was an especially quick trial, but even the "worst" ones are relatively quick.

Unexpected Corollary!:  For $n,m$ natural numbers, if $\gcd(n,m) = k$ then there exists integers $a,b$ such that $an + bm = k$.

This is more useful than you might think at first glance, and we’ll get into why in a later post, but what’s nice about this corollary is that it comes "for free" from the Euclidean algorithm.  Note that, since $k$ divides $n, m$, it suffices to prove this corollary for $an + bm = 1$ where $n, m$ have $\gcd(n,m) = 1$.  The proof uses induction on the number of steps of the Euclidean algorithm for those numbers, but for those of you who are more experienced and know modular arithmetic, you may enjoy the following simple proof:

"Clever" Proof of the Corollary: Let $m > n$ (for equality, the proof is easy).  We will only care about remainders in this proof, so we will look at some numbers modulo $m$.  Consider

$r_{1} = n\mod m$

$r_{2} = 2n\mod m$

$\vdots$

$r_{m-1} = (m-1)n\mod m$

Note there are exactly $m-1$ remainders here and that the remainder $0$ never occurs (since $m,n$ are relatively prime).  Suppose that $r_{i} \neq 1$ for each of the $i$; that is, the remainder 1 does not ever show up in this list.  By the pigeon-hole principle (as there are $m - 1$ remainders but only $m -2$ possible values for the remainders) we must have that $r_{i} = r_{j}$ for some $i\neq j$.  That is, we have

$in \mod m = jn \mod m$

which implies

$(i-j)n\mod m = 0$

but this is impossible, since it implies that either $m = 0$ or $n$ is some integer multiple of $m$, but $m > 0$ and we have assumes $m,n$ are relatively prime.  Hence, the remainder $1$ must occur.  That is, $r_{c} = 1$ for some $c$ and

$cn \mod m = 1.$

But what does this mean?  It means that there is some integer $a$ such that $am - cn = 1$.  To make this prettier, let $b = -c$ and we find that there exists $a,b$ integers such that $am + bn = 1$, as required.  $\Box$

Pretty slick, no?

## Factorials and Exponentials.

### November 6, 2011

I’ve been working on a problem (here is a partial paper with some ideas) that’s really easy for any calculus student to understand but quite difficult for even wolfram alpha to work out some cases.  Here’s the idea:

We know that $\lim_{n\to\infty}\dfrac{e^n}{n!} = 0$.  It’s not hard to reason this out (there are some relatively obvious inequalities, etc.), but I wanted to know what happened if we considered something like:

$\lim_{n\to\infty}\dfrac{e^{e^n}}{n!}$.

It turns out, this goes to infinity.  Maybe this is not so surprising.  But, to balance this out, I thought maybe I could add another factorial on the bottom.  What about

$\lim_{n\to\infty}\dfrac{e^{e^n}}{(n!)!}$

where this double factorial is just $n!$ with another factorial at the end.  It turns out, this one goes to 0.

The problem here is that after $((n!)!)!$, Mathematica doesn’t seem to be able to handle the sheer size of these numbers.  Consequently, I only have a few values for this.  I’ve included everything I have in a google-doc PDF (the only way I can think to share this PDF), and I’m looking for suggestions.  Here’s some things I thought of:

• Stirling’s formula.  Unfortunately, this starts to get very complicated very quickly, and if you consider subbing it in for even $((n!)!)!$ it can take up a good page of notes.  It also doesn’t reduce as nicely as I’d like.
• Considering the Gamma function.  It may be easier to work with compositions of the gamma function since it is not discrete and we may be able to use some sort of calculus-type things on it.
• Number Crunching.  For each of these cases, it seems like there is a point where either the numerator or the denominator "clearly" trumps over the other; this is not the "best" method to use, but it will give me some idea of which values potentially go to infinity and which go to zero.
• Asymptotics.  I’m not so good at discrete math or asymptotics, so there may be some nice theorems (using convexity, maybe?) in that field that I’ve just never seen before.  Especially things like: if $f\sim g$ then $f\circ f \sim g\circ g$ under such-and-such a condition.

Feel free to comment below if you think of anything.