Addition Subtraction Word Problems Practice 5 Pack: Fill & Download for Free

I previously wrote an answer about some problems solved by group theory—if any of those appeal to you, I guess that gives you an answer to this question too. But, of course, you shouldn’t see that list as exhaustive: if you don’t see anything on that list that is relevant to you, it merely means you haven’t found why it is important for you, yet. Group theory is the study of extremely basic algebraic structures, and as such tends to snake its way into all sorts of different places, both expected and unexpected.Since the person who originally posted this question is getting a degree in mathematics and computer science, I thought I would describe a particular application of group theory in the field of cryptography. I want to note: it is far from the only application of group theory in cryptography, let alone in all of computer science. However, it is an application that comes up very frequently in our daily lives.Let’s say that I want to make some sort of transaction online—for concreteness, let’s say that I want to buy the book Introduction to Mathematical Logic by Mendelson from Amazon. I want to be able to send Amazon my financial information so that they withdraw the funds from my bank account. I also want to send them my address so that they know where to ship it. However, I absolutely do not want anyone else to be able to see any of that information.If Amazon and I had some sort of private communication channel, this would be easy. Unfortunately, I can’t afford a personal fiber-optic cable running from my house to Amazon HQ, and I have been cautioned that roving packs of bears to discourage anyone approaching the cable to mess with it might not actually be legal. So, instead, I am forced to send all of my messages over the Internet, where anyone can read them.That in itself need not be a problem. If Amazon and I had some kind of secret key, then we could use it to encode our messages such that only we could decode them (I may have to write about how to do this in some future post, but take this as given for now). But I don’t think I have ever met anyone who works for Amazon. And, to be frank, I have no desire to drive over Amazon HQ to get a secret key from them—that would completely defeat the point of ordering a book online. So, the question is: how can we communicate to one another over a public channel and thereby produce secret keys that no one else knows?On the face of it, this may seem impossible, but I assure you that it isn’t, as was first established by James H. Ellis, Clifford Cocks, and Malcolm J. Williamson in 1969—however, since they were working in secret for British intelligence, credit is often given to Diffie and Hellman, who published the first paper on the subject in 1976. Here is an illustration from A.J. Han Vinck’s Introduction to public key cryptography that gives an intuitive idea of the Diffie-Hellman protocol.The two communicating parties (denoted as Alice and Bob) start with common information known to everyone—represented here as a common paint. They then randomly select their own secret information and combine it with this common information. They then send the results to each other, and do the same thing again. Now, assuming thatit doesn’t matter in which order you added on this secret information, so Alice and Bob end up with the same result, andseparating out this information is difficult (so an eavesdropper can’t work out what Alice and Bob started with by seeing what they have sent),you can prove that it will be computationally difficult for an outside observer to reconstruct Alice and Bob’s shared secret key.Now, how is this actually implemented in practice? Diffie and Hellman’s original proposal just used the multiplicative group of the integers modulo [math]p[/math], but I want to give a more broad overlook before we get bogged down in particulars. So, here is the big picture—we shall start with a finite cyclic group [math]G[/math] with a generator [math]g[/math] of order [math]n[/math].In fact, let’s start with a refresher as to what this actually means. Recall that a group [math]G[/math] is some set with an identity element [math]id[/math] and a binary operation [math]\circ[/math] such that:For all [math]x[/math], [math]y[/math], [math]z[/math] in [math]G[/math], [math]x \circ (y \circ z) = (x \circ y) \circ z[/math]. In other words, the group operation is associative.For all [math]x[/math] in [math]G[/math], [math]x \circ id = id \circ x = x[/math]. This is what it means to say that [math]id[/math] is an identity element.For all [math]x[/math] in [math]G[/math], there exists a [math]y[/math] such that [math]x \circ y = y \circ x = id[/math]. That is, every element has an inverse.I won’t go through the myriads of different examples of groups here, as I have already given many in my answer to What are fields, rings, and groups? I will use the common convention that if [math]n[/math] is a positive integer, [math]x^n[/math] denotes [math]x \circ x \circ \ldots \circ x[/math], with [math]n[/math] instances of [math]x[/math]. Similarly, [math]x^0 = id[/math], [math]x^{-1}[/math] denotes the inverse of [math]x[/math], and [math]x^{-n} = \left(x^{-1}\right)^n[/math] (which also happens to be the inverse of [math]x^n[/math]—I recommend trying to prove this if it is not obvious to you). A group is cyclic if there exists some element [math]g[/math] in [math]G[/math] such that every other element can be written as [math]g^n[/math] for some [math]n[/math]—such an element is called a generator. Here are some simple examples of cyclic groups:The integers, together with addition, are a cyclic group, since here [math]1^n = n \cdot 1[/math], which clearly allows you to produce all integers. (Yes, I get that the use of exponential notation here is confusing. There isn’t much I can do about that, I’m afraid.)The collection [math]\{1,i,-1,-i\}[/math], where [math]i = \sqrt{-1}[/math] is a cyclic group, if we take ordinary multiplication to be the group operation. This is since [math]i^2 = -1[/math], [math]i^3 = -i[/math], [math]i^4 = 1[/math].On a clock, we might add together hours by subtracting by 12 if they go over. Thus, if it is 4 o’clock now, in 9 hours it will be [math]4 + 9 = 13 \equiv 13 - 12 = 1[/math] o’clock. This notion of addition defines a group, which is also cyclic—one possible generator is [math]1[/math], but there are others. For example, notice that[math]\begin{align*} 2 \cdot 5 &= 10 \\ 3 \cdot 5 &= 3 \\ 4 \cdot 5 &= 8 \\ 5 \cdot 5 &= 1 \\ 6 \cdot 5 &= 6 \\ 7 \cdot 5 &= 11 \\ 8 \cdot 5 &= 4 \\ 9 \cdot 5 &= 9 \\ 10 \cdot 5 &= 2 \\ 11 \cdot 5 &= 7, \end{align*} \tag*{}[/math]hence [math]5[/math] is also a generator. However,[math]\begin{align*} 2 \cdot 2 &= 4 \\ 3 \cdot 2 &= 6 \\ 4 \cdot 2 &= 8 \\ 5 \cdot 2 &= 10 \\ 6 \cdot 2 &= 0 \\ 7 \cdot 2 &= 2 \\ 8 \cdot 2 &= 4, \end{align*} \tag*{}[/math]so [math]2[/math] is not a generator.More generally, for any integer [math]n \geq 2[/math], [math]\mathbb{Z}/n\mathbb{Z}[/math] denotes the collection [math]\{0,1,2,\ldots n - 1\}[/math], with the rule that if we add to elements together and get something that is too large, we subtract off by [math]n[/math] until we get back to the desired place. Notice that the above clock arithmetic is just the special case when [math]n = 12[/math], and if you can get past the fact that it is written with multiplication rather than addition, the example of [math]\{1,i,-1,-i\}[/math] is really nothing more than [math]\mathbb{Z}/4\mathbb{Z}[/math].If there exists some positive integer [math]n[/math] such that [math]g^n = id[/math], then the smallest such integer is called the order of the generator. If there is no such integer, we say that generator has infinite order. Note that in either case, the number of elements in the group is exactly the order of the generator.Right. Let’s start again: we take a finite cyclic group [math]G[/math] with a generator [math]g[/math] of order [math]n[/math]. This is public information that is known to everyone. Alice and Bob then both choose, independently, random integers [math]k[/math] and [math]r[/math] between [math]1[/math] and [math]n[/math]. Then, Alice computes [math]a = g^k[/math] and sends it to Bob, and Bob computes [math]b = g^r[/math] and sends it to Alice. Then, Alice computes [math]b^k = \left(g^r\right)^k = g^{kr}[/math] and Bob computes [math]a^r = \left(g^k\right)^r = g^{kr}[/math], and that is their shared private key.Beautifully simple, no? Well, there is a bit of problem, though: we have to assume that even if an eavesdropper knows [math]G[/math], [math]g[/math], [math]n[/math], [math]a = g^k[/math], and [math]b = g^r[/math], that it is difficult to compute [math]g^{kr}[/math]. Depending on how we specify our group, this may not be true.You may be thinking something along the lines of “Well, since you know what [math]g^k[/math] and [math]g^r[/math] are, just work out what [math]k[/math] and [math]r[/math] are, and then you can work out what [math]g^{kr}[/math] is!” That isn’t a bad thought, but it isn’t quite that simple. Consider the above example where we took [math]5[/math][math][/math] to be the generator of [math]\mathbb{Z}/12\mathbb{Z}[/math]. If I hadn’t written down the table, would you be instantly able to work out what integer [math]n[/math] you need so that [math]n \cdot [/math][math]5[/math][math] = 11[/math], for example? If you aren’t familiar with number theory, it may not be obvious how to solve a problem like this. Even so, we have demonstrated an important requirement: [math]n[/math] must be large—otherwise, an eavesdropper could just work out the entire group multiplication table, as we did with [math]5[/math][math][/math], and work out [math]k[/math] and [math]r[/math] from [math]g^k[/math] and [math]g^r[/math].Unfortunately, this isn’t sufficient. Let’s take [math]n = 2760727302517[/math] (this is the [math]10^{11}[/math]-th prime number)—this is still quite small, but will do for illustrative purposes. I claim that [math]2298570020914[/math] is a generator of [math]\mathbb{Z}/n\mathbb{Z}[/math] with this choice of [math]n[/math]. We might ask what is the integer [math]k[/math] that I need multiply this by to get to, say, [math]1178932971685[/math]. I ask Mathematica to do this, and it spits out an answer of [math]684240928772[/math] so absurdly quickly that it can’t even give me an accurate count of how long it took to do this computation. Now, you may well ask: how the hell is it able to do it this fast?The key observation is that while we only defined addition on [math]\mathbb{Z}/n\mathbb{Z}[/math], you can define multiplication on it just as well, with the same rules—if you go over [math]n [/math][math]-[/math][math] 1[/math], subtract off [math]n[/math] until you get to where you need to be. Using the extended Euclidean algorithm, you can prove that if [math]k[/math] is coprime to [math]n[/math], then there exists some integer [math]r[/math] such that [math]k \cdot r = 1[/math] in [math]\mathbb{Z}/n\mathbb{Z}[/math]. In fact, the extended Euclidean algorithm gives you a fast way of computing this [math]r[/math], as I showed in my answer to Why do we use the extended euclidean algorithm when the euclidean algorithm is simpler and gives the same result? So, now we know how why the group that we tried to use was never going to work—even if [math]n[/math] is very large, if [math]g[/math] is our generator, any eavesdropper can find [math]k[/math] from the [math]a[/math] that Alice sends as follows:Realize that finding [math]k[/math] is the same as solving [math]k \cdot g = a[/math] for [math]k[/math] inside of [math]\mathbb{Z}/n\mathbb{Z}[/math].Compute [math]g^{-1}[/math] using the extended Euclidean algorithm.Multiply both sides by [math]g^{-1}[/math], giving [math]k = a \cdot g^{-1}[/math].The same trick will give you [math]r[/math] from [math]b[/math], breaking the whole scheme entirely.Thankfully, the reasons why this particular choice fails point the way to a new choice that is actually thought to be secure. I mentioned that if [math]k[/math] is coprime to [math]n[/math], then [math]k\cdot r = 1[/math] is solvable. If you think about it, you will realize that this means that the collection of integers between [math]1[/math] and [math]n[/math] that are coprime to [math]n[/math] form a group—indeed, this multiplication is associative, it has an identity ([math]1[/math], to be precise), and we have just concluded that it has inverses. By appealing to a combination of group theory and number theory, you can prove that if [math]n[/math] is a prime number, thenthe number of such elements coprime to [math]n[/math] is [math]n - 1[/math], andthis group, which we shall denote by [math]\left(\mathbb{Z}/n\mathbb{Z}\right)^\times[/math], is cyclic.Here is an example: [math]15[/math] is a generator of [math]\left(\mathbb{Z}/23\mathbb{Z}\right)^\times[/math].[math]\begin{align*} 15^{0} &= 1 \\ 15^{1} &= 15 \\ 15^{2} &= 18 \\ 15^{3} &= 17 \\ 15^{4} &= 2 \\ 15^{5} &= 7 \\ 15^{6} &= 13 \\ 15^{7} &= 11 \\ 15^{8} &= 4 \\ 15^{9} &= 14 \\ 15^{10} &= 3 \\ 15^{11} &= 22 \\ 15^{12} &= 8 \\ 15^{13} &= [/math][math]5[/math][math] \\ 15^{14} &= 6 \\ 15^{15} &= 21 \\ 15^{16} &= 16 \\ 15^{17} &= 10 \\ 15^{18} &= 12 \\ 15^{19} &= 19 \\ 15^{20} &= 9 \\ 15^{21} &= 20 \end{align*} \tag*{}[/math]Producing these generators is actually very easy—let me share the secret now. Due to basic group theoretic considerations, you know that the order of any particular element must divide the order of the group (i.e. the number of elements in the group). Notice that if the order of the group is a prime number, this implies that every element other than the identity is a generator. I used this before when I asserted without proof that [math]2298570020914[/math] is a generator of [math]\mathbb{Z}/n\mathbb{Z}[/math], where [math]n = 2760727302517[/math]. I can use this reasoning again for the group [math]\left(\mathbb{Z}/n\mathbb{Z}\right)^\times[/math] if [math]n[/math] is a prime—we know that there are [math]n [/math][math]-[/math][math] 1[/math] elements in this group, and so any randomly chosen element will be a generator if and only if its order is [math]n [/math][math]-[/math][math] 1[/math]. So, here is the trick: choose [math]n = 2p + 1[/math], where [math]p[/math] is itself a prime number—such a prime number is known as a Sophie Germain prime. Then [math]n [/math][math]-[/math][math] 1 = 2p[/math], and so the only possible orders of elements are [math]2[/math], [math]p[/math], and [math]2p[/math]. Fixing a generator [math]g[/math], an element will have order [math]2[/math] if and only if it is of the form [math]g^{pk}[/math] for some odd [math]k[/math]. An element will have order [math]p[/math] if and only if it is of the form [math]g^{2k}[/math] for some [math]k[/math] that is not divisible by [math]p[/math]. What you see from this is that there is only a very slightly worse than [math]50\%[/math] chance that a randomly chosen element [math]h[/math] will be a generator, and we only need to check that [math]h^2, h^p \neq 1[/math], which can be done very quickly. As it is conjectured that there are infinitely many Sophie Germain primes, this gives an efficient way of randomly producing groups [math]G[/math] and generators [math]g[/math] that work for our purposes.Unfortunately, we don’t actually have mathematical proof that the choice of [math]G = \left(\mathbb{Z}/n\mathbb{Z}\right)^\times[/math] really is secure, and that there isn’t some algorithm like the Euclidean algorithm that quickly finds [math]k[/math] given [math]g[/math] and [math]g^k[/math]. It is in fact an open problem (known as the Diffie–Hellman problem) to prove this for any finite group. The best that we can say is that it has held up to scrutiny thus far, and there is good reason to believe that it should be difficult. I should also note that this choice of group is not the only one conjectured to be difficult, nor even the only one that is in use—in the elliptic-curve Diffie–Hellman protocol, you swap out that group for the group of rational points on an elliptic curve over a finite field. In principle, if you can find some other nice group that works and maybe has some other nice properties, you can swap that it in instead without issue. This is the nice aspect of considering everything from an abstract, group-theoretical perspective: you don’t need to sweat the details as much.I do have to make one final note before I end this post: while the Diffie-Hellman protocol is a fundamental building block in cryptography, in practice you likely don’t want to use it in quite the way that I described—i.e. you don’t want to use it to directly contact Amazon to buy your book. This is because the pure Diffie-Hellman protocol is susceptible to man-in-the-middle attacks: if someone intercepted my messages to Amazon in the very beginning, and pretended to be me to Amazon and Amazon to me, they would become privy to all of our private correspondence. To actually make this workable, you need additional protocols that allow you to verify that you are actually talking to your desired party. For the specific application mentioned, you most likely want to use something RSA, which typically comes bundled with such protocols. However, there are plenty of similar applications where Diffie-Hellman is used instead.

This is one I could actually answer to some extent, considering I have built a number of programs to do just this, and planned (though haven't finished) one that could, in theory, be run on a supercomputer, as well as having read the stuff by those who have done such programs. I want to point out though that the current world record for [math]\pi[/math] calculation (13.3 trillion digits) was not obtained on what most would think of as a “supercomputer” but rather a very jacked-up PC. It had server-grade CPUs and lots and LOTS of RAM (I think like over 100 GB, if I remember correctly). Hardware that I could only dream of given my very meagre financial resources.Preliminary AlgorithmsThere are a number of different mathematical algorithms which are involved in the calculation of [math]\pi[/math] to a very large number of places. The basic equations are not very complicated. One set of formulas that can be used for this purpose are the “Machin”-like arctangent series:[math]\frac{\pi}{4} = 4 \arctan\left(\frac{1}{5}\right) [/math][math]-[/math][math] \arctan\left(\frac{1}{239}\right)[/math]However this one is fairly slow, although I believe Takahashi and Kanada used it, or something related to it, in one calculation run they did, which was on a supercomputer but is no longer a world record. These formulas use trigonometric identities to calculate the arctangent of 1, which is given by the very simple Madhava-Leibniz-Gregory series[math]\frac{\pi}{4} = 1 [/math][math]-[/math][math] \frac{1}{3} + \frac{1}{5} [/math][math]-[/math][math] \frac{1}{7} + \frac{1}{9} [/math][math]-[/math][math] \frac{1}{11} + \cdots[/math](also just called “Gregory's series”)which is actually worthless for computation purposes as it converges extremely slowly. It is a simple geometric fact that the arctangent of 1 is [math]\frac{\pi}{4}[/math], because that angle corresponds to a right angle (opposite side = adjacent side). Nonetheless it was one of the first formulas to allow for [math]\pi[/math] to be obtained to arbitrary precision in theory. The Machin formula is essentially a way to break up “1” into two fractions for which the arctangent can be computed with faster-converging series. In particular, the general series for arctangent is[math]\arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}[/math]and thus by the Machin-type formula, we get better convergence.Another set of formulas are those based on the Arithmetic-Geometric Mean, or AGM. The Gauss-Legendre AGM algorithm for the calculation of [math]\pi[/math] iterates the following formula. It is based on the “Arithmetic-Geometric Mean” formulas for the elliptic integrals.Start:[math]a_0 = 1[/math][math]b_0 = \frac{1}{\sqrt{2}}[/math][math]t_0 = \frac{1}{4}[/math][math]x_0 = 1[/math]Loop:[math]a_{n+1} = \frac{a_n + b_n}{2}[/math][math]b_{n+1} = \sqrt{a_n b_n}[/math][math]t_{n+1} = t_n [/math][math]-[/math][math] x_n (a_n [/math][math]-[/math][math] a_{n+1})^2[/math][math]x_{n+1} = 2 x_n[/math].This process doubles the number of correct digits at each step. The first two lines compute the arithmetic-geometric mean of [math]1[/math] and [math]\frac{1}{\sqrt{2}}[/math].But the real fastest algorithm is the Chudnovsky series method which uses the following formula inspired by one of Ramanujan.[math]\frac{1}{\pi} = \frac{12}{\sqrt{640320}} \sum_{k=0}^{\infty} (-1)^k \frac{(6k!)}{(3k)! (k!)^3} \frac{13591409 + 545140134k}{640320^{3k}}[/math].This was the formula used for the latest run.But having a formula like this is just the beginning. To actually implement it on the computer and make it fast is a whole 'nother ball of wax. Remember, we want zillions of digits, right? So we cannot simply just plug these formulas into Excel or even just directly into a programming language and get the results, because the numeric data types the languages provide give only a limited amount of precision – very, very little compared to what we want. We could only get, maybe, 15 digits at best. This is because those types are pretty much limited to what the hardware gives you.So what do you have to do? You have to implement your own “bignum” data type. Bignum arithmetic is implemented on a computer in much the same way that your do your grade-school arithmetic – CAVEAT! – only now a “digit” (often called a limb in this context) is something with a base that may be equal to one greater than the largest integer the computer can represent (so that the digit runs from 0 up to the largest integer representable), and we use very big arrays of those integers as our representation. In my programs, however, I use a smaller base, one that's a power of the base we wish to print our digits out in, such as base 10 or base 26 (I usually use the latter, as my initial interest in this was finding words that pop up randomly in the digits of [math]\pi[/math] written as letters (A-Z) – that's another story in itself), so you essentially pack several “output” digits per integer array element.The BIG caveat, though, is that if you want to go to insane amounts of digits like this, only addition and subtraction can really be done by the grade-school method, which are in fact the most efficient methods. Multiplication and division, on the other hand, need much more sophisticated and more efficient algorithms.Giant MultiplicationGrade-school multiplication is nice for doing numbers with a few digits. But if you write out a multiplication problem and use the grade-school long multiplication method, you may notice something. If you have numbers of [math]n[/math] digits long, then to multiply them out you need to form [math]n[/math] “partial products”, or products of one by each digit of the other, then add these together. To form the partial products you need to do on the order of [math]n \cdot n[/math], or [math]n^2[/math] single-digit operations. In computer-science parlance, we say this algorithm has a “complexity” of [math]O(n^2)[/math], which roughly means that if you double the number of digits you need to multiply, you quadruple the maximum time required. If it takes a microsecond to multiply 100 digit numbers on a computer (just saying), then to multiply 100 million digit numbers using this same approach would take [math]1000000^2 = {10}^{12}[/math], or one trillion, times longer, which is one million seconds, or 11.5 days! Clearly, we need something better.But can we do better? Turns out we can. The next-best algorithm after the “grade-school” method is something called Karatsuba's method, or the “divide-and-conquer” method, or the “Fractal” method as I (and at least one old-days [math]\pi[/math] programming enthusiast I learned from over a decade ago, Carey Bloodworth) call it. It's not that difficult to grasp. Suppose you have a number, like 57626612, and want to multiply that by 98642021. You can either hack through the grade-school method, or you can use a trick. Recognize that 57626612 = 5762 6612, that is, we can group it into two groupings of 4 digits. We can write this mathematically as a “base-10000” number, and say [math]57626612 = 5762 \times 10000^1 + 6612 \times 10000^0 = 5762 \times 10000 + 6612[/math]. We can do the same for the other number. We can therefore write[math] \begin{align} 57626612 \cdot 98642021 &= (5762 \cdot 10^4 + 6612)(9864 \cdot 10^4 + 2021) \\ &= (5762 \cdot 9864 \cdot 10^8 + (5762 \cdot 2021 + 9864 \cdot 6612) 10^4 + 6612 \cdot 2021 \end{align}[/math]where we just “FOIL”ed out the binomial on the right. Then we can do this again to the smaller numbers, breaking them into 2 digits, etc. It seems like it still takes just as many multiplications, but there is a mathematical trick we can exploit to do the above using only 3 multiplications instead of 4 (the multiplications by powers of 10 don't count since they are just digit shifts). In particular, note that [math](aX + b)(cX + d) = acX^2 + (ad + bc)X + bd[/math] but that the center term can be written as [math]ad + bc = (ac + ad + bc + bd) [/math][math]-[/math][math] ac [/math][math]-[/math][math] bd = (a + b)(c + d) [/math][math]-[/math][math] ac [/math][math]-[/math][math] bd[/math], and we already have computed [math]ac[/math] and [math]bd[/math]. So there you go, 3 multiplications, plus some extra additions but those are “cheap”. This method has “complexity” [math]\approx O(n^{1.585})[/math], which means that doubling the digits only triples the work, not quads it. In particular, if it takes us, say, 0.8 microseconds to do 100 digits with this method (random guess taking overhead into account), then to do 100 million digits only takes about 43 minutes. Still a long time, but a vast improvement over 11.4 days!But for very big numbers, while this method has some role to play, the biggest boost is gained by the use of much more sophisticated methods, all of which are related to the Fast Fourier Transform (FFT) in some way or another. It is too much to go into the details of the mathematics behind Fast Fourier Transforms here, but basically the idea rests on that you can perform a “convolution” – which is essentially what you are doing to the sequences of digits when you are doing a grade-school multiplication, with the added step of “carry-over”, as an element-by-element operation, when you are working with the Fourier transforms of the digit sequences in question. This is a famous result called the “convolution theorem”.While I won't get into math, it's easy to describe what a Fourier transform does in informal terms. Basically, if you are given a signal, like an audio file, you can write it as either a series of values representing its amplitude at each point in time, or as a sum of frequency components, that is, sine and cosine waves of differing amplitudes and frequencies. Your ears actually do this all the time. The fact that you can hear two different tones playing at the same time and distinguish them is due, in part, to the fact that such a decomposition is possible. The Fourier transform is just a mathematical device for converting between these representations. The Fast Fourier Transform is a computerized algorithm that enables the calculation of the Fourier transform of a finite data set at very high speeds, in fact is the fastest algorithm known for this purpose. It has “complexity” [math]O(n \log n)[/math], and theoretically, multiplication with it would have this complexity, but implementation details make it somewhat worse. The method that comes closest with practicalities taken into account is the Schonhage-Strassen method, which I have not implemented, but relies on even more sophisticated mathematics involving things analogous to the Fourier transform done in abstract rings, which are a whole area of math in themselves to get into. The latest [math]\pi[/math] calculation utilized an algorithm called “Hybrid NTT” which I know nothing about, because it is that programmer's (Alex Yee's) own creation and he has not published any paper discussing it. I was in correspondence with that author for some time, but it was considered too complicated for simple email. Nevertheless, he did help me a fair bit in making other stuff work for my programs.Division is done not with long division but instead by taking the reciprocal, using multiplication-only methods that then exploit these high-speed multiplication algorithms. In my programs I used Newton's method, which is something you can study in elementary calculus.Hardware and ImplementationThat pretty much covers the algorithmics, but what about hardware? Supercomputers today often use hardware similar to those used in PCs, but usually the more robust “server”-grade versions, which feature multiple processors per motherboard, large RAM capacities, and error-correcting RAM, and many, many motherboards, all of which are usually packaged into “blade server” cases and connected by very high-speed network connections, much faster than what you probably have at home. In order to feed the multiple processors, you have to code the algorithms given above in “parallel”, which means you have to break up the work into chunks that each processor can work on independently of the others and simultaneously with them, while minimizing them amount of communication required to reconstruct the final result from the different pieces they work on. I cannot get into a lot of details here as it's all very complicated stuff and I haven't actually, as I said, gotten around to doing that part of the nitty-gritty implementation yet. When using PC-like computers, as in the latest run used, with very protracted computation times, one can't fit everything into RAM and so must store data on disk. One may not have the network interconnect overhead I mentioned, but one can have the disk overhead, and that means one generally has to try and keep input/output sequential, to minimize seeking, and to do as few passes over the data as possible. Disk overhead is a significant bottleneck for this kind of thing.In addition to this, one often has to use hand-tricked forms of FFT algorithm to try and fit data into “cache” on the processor as much as possible, because cache is your fastest memory on a computer. We have that cache > RAM > disk > LAN > WAN, and so the more we can keep our computations in the highest levels of that sequence, the better. (WAN I don't think is ever used for this particular application as there's just too much data, but things like Folding@Home which do other types of computation over the Internet have to contend with it.) There are also many other tricks too, such as exploiting any vector processing instructions the processor has, and lots of hand-coded assembler (though in my latest program I haven't yet gotten to the hand-coded assembler stage, although I did in previous attempts).But this should provide quite a good amount of coverage and an idea of what is involved, especially the algorithmic side which is what I have dealt with the most, in writing software at home to do all this stuff.

Mathematics contains many areas where we can give simple descriptions and proofs that a non-mathematician can understand. If it has practical applications it will also attract more people to study it further. It also raises the standard of analysis, and that will benefit everybody. “π” describes the relationship between the radius of a circle and its area (π * R * R). It is a number that is irrational; it is not an exact number. A mathematical proof exists to prove that. A simple proof exists that anyone can understand. The equation shows the square of the radius in the calculation of the area. We also know that the area of a square is the square of one side’s length. Therefore, we know that the number of squares of side length R that fits inside a circle is π. If you draw a square of R * R inside a circle of radius R, you can fit one of these. The other squares come from adding smaller and smaller squares ad infinitum, each addition producing extra diminishing areas. The reason that the number is irrational is that a square cannot fit perfectly against the side of an arc. When you find a square that does fit, it creates an area bound by the outside curve of the circle and the flat side of the square. In other words, the problem is repeated. Each time you find a square small enough, it creates a smaller area to fit smaller squares in. Area is an expression of squares, and, therefore, the area of a region bound by an arc of a circle must be irrational. The ancient Egyptians equated the area of a circle to the area of a square whose sides were 8/ 9 the diameter of the circle. This method amounts to employing a value of the mathematical constant π that is equal to 256/ 81. In one sense this is impressive— it is only about one half of one percent off of the exact answer.The Pythagorean theorem demonstrated that small whole numbers, like 3, 4, and 5, could be found that not only described the lengths of a right triangle but also had the property that when the areas of the squares erected on the two smaller sides were added together they equaled the area of the square erected on the longest side, the hypotenuse, the side opposite the right angle. Notice how the ancient Greeks phrased the Pythagorean theorem in terms of geometric objects and not in terms of numbers! Then someone asked an interesting question. If there were a square with each side a unit in length, and a second square with double the area of the first square, how would the side of the second square compare to the side of the first square? This was how the question of the square root of 2 was originally proposed.The question I pose, is there a way to find a rational approximation of the square root of 2 in terms of Pythagorean triples?There is no need to reinvent the “wheel”, in this matter, but there is a definite need to find a pathway to supersede it.Lets look at another example of an irrational number, √2.√2 flowersMany spirals in nature use the golden ratio, but something beautiful happens when we replace that ratio with some other famous irrational numbers. Another twist takes us to rational approximations and continued fractions. Let us follow these spirals into the beautiful world of irrational numbers.An angle of rotation of √2 produces a very even distribution of seeds, rivaling that of the golden ratio. Five hundred seeds are shown above; many families of spirals are readily apparent in this arrangement. A study of these √2 spirals is worth while, as their structure illuminates many properties of algebraic numbers and seed spirals in general.The numbers of the families that produce nice looking spirals are: 2,3,5,7, 12...could there be a Fibonacci-like relationship between spiral families in a √2 spiral as well? More spiral families are shown below, but the next spiral family after 12 is not 19 as we might expect by adding 7 and 12, but rather 17,and the next family better than 17 is 29. The sequence is in fact: 1,1,2,3,5,7,12,17,29,41,70,99,169,239,408,577,985,1303...If a fraction in this sequence is represented by a/b, then the next fraction will be (a+2b)/(a+b). This process generates all of the spiral families of the √2 flower. Further, the ratio of the numbers in each row converges to √2: 1/1 = 1, 3/2 = 1.5, 7/5 = 1.4, 17/12=1.41666...,41/29=1.41379, 99/70, 239/169, 577/408, 1303/985...,etc. The √2 may be written as a continued fraction. The reader may enjoy checking that the following continued fraction gives an expression for √2:Directly connected to this continued fraction is the Diophantine equation:2X² −Y² =±1 known as the Pell equation. There are infinitely many pairs of integers (x,y) satisfying this equation, and the corresponding fractions y/x are precisely what you get by truncating the expression into continued fractions.These are the same numbers in the continued fraction and the same ratios found in the √2 flower!Introducing a formulae to calculate Pythagorean triples as follows:If we take the fraction y/x then the square y²/x² will yield to the "sides of a right angled triangle." Given y/x = 3/2 then y²/x²=9/4 and 9-4=5 therefore we have 3, 4, 5 as the first Pythagorean triple the next one will be y/x=7/5 then y²/x²=49/25 and 49-25=24 therefore we have 7, 24, 25 the rest of the sequence follows and can be generalized in the following terms:Pythagorean triplesa= 2n+1b= 2n²+2nc= 2n²+2n+1The first few solutions for the sequence n=1, 3, 8… yields,3, 4, 57, 24, 2517, 144, 145….Observation 1There is a simple reason for this, subtracting a prime number or the (odd) square of a prime number from an even square creates Prime Number positions because it forces the squares to be consecutive. The pair of squares are of course the two consecutive numbers 4 and 5 that sum together to form 3². Because they are consecutive numbers, they share no factors in common and therefore they cannot be recombined into any other square. Also because a prime or the square of a prime cannot be recombined into another combination, they are only the difference between consecutive squares.Observation 2It is a trivial exercise to show that these primitive triples form as multiples of 12 are also doubled by a factor of two in the process, with one exception 3, 4, 5;Rearranging the sequence to9-5=4..........(1)49-25=24.........(2)289-145=144........(3).(2*4)+1-(2*2)+1=2*2(4*12)+1-(2*12)+1=2*12(24*12)+1-(12*12)+1=12*12.Notice how we derived equation 1We took the square of 3/2 which are both prime and squared it, 3²/2²=3²-2². If we multiply the product of these two numbers then we get 3²*2²=36 which is another square!Equation 3 becomes somewhat interesting because 13²/5²=13²-5² gives another square 144 which also happens to be 3²*4².If two numbers are such that their squares added or subtracted together makes a square then the product of these two numbers is divisible by 6.Clearly there is a theorem lurking in the background which can be further extended as follows:3(5+7)=3²*2²=366(11+13)=3²*4²=1449(17+19)=3²*6²=324….36²/12²=5²-4²144²/24²=10²-8²324²/36²=15²-12²….Notice that we have derived a sequence of perfect squares from the irrationality of the √2 which contains an infinitude of primes and the squares of primes, when expressed as a fraction! From the irregularity of primes, we have shown some kind of proportion and symmetry in whole numbers.Observation 3If we take the difference of squares and form a series of odd² = even² then we have an interesting sequence for example,5²-3²=4²13²-5²=12²61²-11²=60²….The asymmetry of the left hand side is equals to the symmetry of the right hand side. Notice that the proportional symmetry of 4²:12² is 1/3² and 12²:60² is 1/5² and that of 4²:60² is 1/15².In other words, though the discovering power must come before the primes are discovered, that discovering power cannot manifest itself without the discovery of the primes.The said principle of discovery is that the asymmetry of primes and the symmetry of non-primes are related to each other by powers of primes.There are many other patterns we can observe in the sequence such as below, however, the one above goes much deeper into the fundamental laws of the nature of primes and the squares of primes.We can also express the sequence in continued fractions as follows1+1/3 + 2+2/5 + 3+3/7…or1+2/3+ 2+3/5 + 3+4/7…The denominator yielding 3, 5, 7…and the general term n + n/2n+1 or n + n+1/2n+1…Treating the 1st equation 1+1/3 as an improper fraction 4/3 and squaring both gives 4²/3²=4²+3² gives us 5²…etcConsequently, Fermat’s idea to his “Great theorem” was a direct response to the earlier challenge issued by Pythagoras, in confirming that the ancient Egyptians and Greeks had already understood that within the domain of Sphaerics, A²+ B² = C² represented the only form of summation of powers. It is a unique expression of Fermat's "definition of the quickest pathway of refraction-reflection" for the universal physical principle of propagation of light.In his {Arithmetica}, Diophante proved the crucial point, in Book II, sections VIII, and IX, where Fermat had appended his note, and where Diophante showed his method of how to divide a square into two other squares. Diophante wrote:“Divide a given square into two other squares. {Example} Let 16 be the given square, and I will call N² and 16 - N² the sought for squares. N remains to be found, in such a way that 16 - N² be a square. I pose that 16-N² = (2N –4)², thus N = 16/5.” ({Précis des oeuvres mathématiques de P.Fermat et de l’arithmétique de Diophante}, par E. Brassinne, Imprimerie de Jean-Mathieu Douladoure, Toulouse, 1853, p. 53.) The interesting question, here is, how did Diophante derive 16/5?This Diophantine problem can be resolved by applying the principle of proportionality as I have described above using the difference of two squaresLet us take as a very easy example problem I.28: Find two numbers whose sum and the sum of whose squares are given numbers. The context reveals that by “number” Diophantus always means: positive rational number. Before solving this problem, Diophantus states a necessary condition: twice the sum of the squares minus the square of the sum of the two numbers has to be a square (of a positive rational number).Nowadays, using algebraic notation and knowledge which dates back at most to the sixteenth century, we would probably reconstruct this somewhat like this: We look for X and Y such that X + Y = a and X² + Y² = b. Then 2b−a² = (X−Y)²=(Y−X)² does indeed have to be a square.All he does is ‘solve the problem,’ and here is what ‘solving’ means for him: first he chooses the value 20 for what we have called a, and he chooses the value 208 for what we have called b. (Note in passing that 2×208−20² = 4²)208 = (x + 10)² + (10−x)² = 2x² + 200.Here x²=4 i.e., x=2The two numbers sought are then 12 and 8Notice if we started with 12 and 8 then we know 13²-5²=12²….(1) and 10²-6²=8²…..(2). We know the first solution is a primitive triple, however, the second triple is a doubling of 5²-3²=4²…..(3) This simplifies the problem because we can skip the major part of text book algebra and get to the crucial point below.If we take the sum of two squares 12 and 8 to give 20 and square it to get 20² then 20²-16²/5²=12²/5², alternatively we know 20²-12²=16² is a straight doubling of equation 2 and divide straight through by 5² then,N²=16²/5² and N=16/5or if we take5²*4²=20²4²=20²/5²4²=16²/5²+12²/5²N=16/5or20²= 13²-5²/4²+3² + 4²(5²-3²)/3²+4²N²=4²(5²-3²)/3²+4²N=16/5“It is impossible for a cube to be the sum of two cubes, or for a fourth power to be the sum of two other fourth powers, or, in general, for any power other than two to be the sum of two similar powers.”“I have discovered a truly marvelous demonstration of this proposition, but this margin is too narrow to contain it.”Fermat was absolutely right to his claim, following an underlying ordering of quadratics, that is, an ordering of the second degree, from within which the musical octaves, based on C-256, is derived. This second degree of intervals changes, because the ratio of intervals increases by a conical spiral function, which distributes the primes in accordance with the power of 2.This does not mean that things stop growing at that point, but that the system has reached a limit, a boundary condition dominated by second degree quadratics, whose heads pierce through the cracks of the universe to show that they belong to some infinite and universal object which eludes the grasp of our sense perception. It is this constant ratio of the electromagnetic power of 2 within quadratics, this invariant interval of interval factor of doubling, which determines the density of distribution of prime numbers.Also another interesting sequence is,3[math]^{4}[/math] - 5*4[math]^{2}[/math]=17[math]^{4}[/math] - 24*10[math]^{2}[/math]=117[math]^{4}[/math] - 145*24[math]^{2}[/math]=141[math]^{4}[/math] - 840*58[math]^{2}[/math]=1Another way to express π is in continued fractions,π =3+1/7 =22/7,π =3+1/(7+ 1/15) =3+1/(106/15) =3+15/106 =333/106,π =3+.. =355/113The next family of spirals past family 113 is family 33102. We would need 33,102 seeds just to get one seed in each spiral arm! If we plot about a million seeds we may be able to start seeing these spirals; however, there would be nearly 92 spirals packed into each degree arc of the circular face. An illustration 10 cm in diameter would have over 1000 spiral arms in each cm of the circumference—the illustration would appear to be nothing other than a black circle! (Note that 333/106 is also an approximation of π. However, due to the closeness of a better approximation, 355/113, the set of 106 spiral arms is immediately obscured by the set of 113 spiral arms.)The spiral with 30² in the center may also reveal to us it’s secrets, there are infinitely many ways to express light in different forms of energy.π=4²(5²-3²)/3²(5²-4²)Roughly 15% of the approximately 300 problems that we know of the Arithmetica contain so-called diorisms, i.e., necessary conditions like the one on 2b−a² being square which we encountered in our first example, II.8, above. However, the ancient Egyptians did in fact have an advanced knowledge as shown in the equation of π, for example, solving the top and bottom equation respectively, we get 20²-12²=16² and 15²-12²=9². From this observation 20 is not only the sum of two squares 12² and 8², it is also the sum of cubes 12 and 8 and the difference of the same cubes, while 15 is the sum of 9 and 6 and the difference of cubes 9 and 6. The sum of these two differences is 12[math]^{3}[/math]-8[math]^{3} [/math]and 9[math]^{3}[/math]–6[math]^{3} [/math]is 1729 which is the smallest positive number that can be expressed as a sum of two cubes 12 and 1 and 10 and 9 in two ways and also as four cubes 10 and 1 and 6 and 8.In problem 16 of book V Diophantus writes: "and we know from the Porisms: The difference of two cubes can be represented as a sum of two cubes." Obviously, what is at issue is the solution of the equationx[math]^{3}-[/math]y[math]^{3}[/math]=a[math]^{3}[/math]+b[math]^{3}[/math]for a>b>0 and positive x and y. There is no solution of this problem in the "Arithmetic"Again I have already stated the first rational solution of the cubes and by simply extending the multiples of these numbers we should get all solutions.9[math]^{3}[/math]–6[math]^{3}[/math]=8[math]^{3}[/math]+1[math]^{3}[/math]and12[math]^{3}[/math]–8[math]^{3}[/math]=10[math]^{3}[/math]+6[math]^{3}[/math]z[math]^{3}[/math]+y[math]^{3}[/math]=z(z²+y²)-y²(z-y)a[math]^{3}[/math]-b[math]^{3}[/math]=a(a²-b²)+b²(a-b)The problem now shifts from squares to all higher powers. FLT would have been solved sooner if it had been realized that all even powers can be expressed as powers of 2.25²-15²=16²+12²………(1)Dividing eqn (1) right through by 4²(25²-15²)/4²=16²/4²+12²/4²5²=4²+3²Dividing eqn (1) right through by 5²5²-3²=16²/5²+12²/5²a²-b²=2/3 (ab+b²)……..(2)5²-4²= 12²/4²a²-c²=1/4 (ac+c²)………(3)From equation (2) finding the triangle with the same area gives a right triangle with height h=8 and base c=4.Despite the differences in measurement between Old Kingdom pyramids, some are shown to have a ratio 3:4 between the half of the base and height of the pyramid measurement. The half base line, the imaginary perpendicular line and the line connecting the apex of the pyramid with the mid-point of its side form a right-angled triangle with the 3:4:5 ratio. In volumes the cube is 3/4 the pyramid, or the pyramid is 4/3 the cube.From the Rhind Mathematical Papyrus (RMP)RMP 59The height of a pyramid is 8 cubits, and the base is 12 cubits. What is it’s seked?Since 7 Palms = 1 cubitIf we consider a similar right triangle whose vertical leg has length 1 cubit or 7 palms, and we designate the horizontal leg as s, thens/7=6/8=3/4so thats=3/4 x 7=5 1/4Since 1/4 palm = 1 finger, we have that s= 5 palms, 1 finger.Every potential FLT equation equals a difference between squares, also has the attribute of having a mirror equation. This is precisely how Sir Andrew Wiles finally solved FLT.In modern number theory the study of rational points on an elliptic curvey²=4x[math]^{3}[/math]+Ax+Bform a torus modular function, with A,B E Q is as old as recorded history, for example, rational points ony²=4x[math]^{3}[/math]-4n²x give rise to Pythagorean right triangles (with all integers side lengths) and area n² is linked to FLT.Fermats Last Theorem states for n greater than or equals to 3 the equation,a[math]^{n}[/math]+b[math]^{n}[/math]=c[math]^{n}[/math]has no solutions in non-zero integers a,b,cThis reduces to the case where n=p where p is prime. The case n=4 was resolved by Fermat. Also we may need to take b=even and a = -1 (mod 4)If a,b,c were a counterexample to FLT as above, the elliptical curvey²=x(x-a[math]^{p}[/math])(x+b[math]^{p}[/math])could not satisfy the modularity theorem.Hence modularity, or even the special case due to Wiles, for “semistable” elliptical curves implies FLT.