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Mon, 27 Nov 2023
Uncountable sets for seven-year-olds
I was recently talking to a friend whose seven-year old had been reading about the Hilbert Hotel paradoxes. One example: The hotel is completely full when a bus arrives with 53 passengers seeking rooms. Fortunately the hotel has a countably infinite number of rooms, and can easily accomodate 53 more guests even when already full. My friend mentioned that his kid had been unhappy with the associated discussion of uncountable sets, since the explanation he got involved something about people whose names are infinite strings, and it got confusing. I said yes, that is a bad way to construct the example, because names are not infinite strings, and even one infinite string is hard to get your head around. If you're going to get value out of the hotel metaphor, you waste an opportunity if you abandon it for some weird mathematical abstraction. (“Okay, Tyler, now let !!\mathscr B!! be a projection from a vector bundle onto a compact Hausdorff space…”) My first attempt on the spur of the moment involved the guests belonging to clubs, which meet in an attached convention center with a countably infinite sequence of meeting rooms. The club idea is good but my original presentation was overcomplicated and after thinking about the issue a little more I sent this email with my ideas for how to explain it to a bright seven-year-old. Here's how I think it should go. Instead of a separate hotel and convention center, let's just say that during the day the guests vacate their rooms so that clubs can meet in the same rooms. Each club is assigned one guest room that they can use for their meeting between the hours of 10 AM to 4 PM. The guest has to get out of the room while that is happening, unless they happen to be a member of the club that is meeting there, in which case they may stay. If you're a guest in the hotel, you might be a member of the club that meets in your room, or you might not be a member of the club that meets in your room, in which case you have to leave and go to a meeting of one of your clubs in some other room. We can paint the guest room doors blue and green: blue, if the guest there is a member of the club that meets in that room during the day, and green if they aren't. Every door is now painted blue or green, but not both. Now I claim that when we were assigning clubs to rooms, there was a club we missed that has nowhere to meet. It's the Green Doors Club of all the guests who are staying in rooms with green doors. If we did assign the Green Doors Club a guest room in which to meet, that door would be painted green or blue. The Green Doors Club isn't meeting in a room with a blue door. The Green Doors Club only admits members who are staying in rooms with green doors. That guest belongs to the club that meets in their room, and it isn't the Green Doors Club because the guest's door is blue. But the Green Doors Club isn't meeting in a room with a green door. We paint a door green when the guest is not a member of the club that meets in their room, and this guest is a member of the Green Doors Club. So however we assigned the clubs to the rooms, we must have missed out on assigning a room to the Green Doors Club. One nice thing about this is that it works for finite hotels too. Say you have a hotel with 4 guests and 4 rooms. Well, obviously you can't assign a room to each club because there are 16 possible clubs and only 4 rooms. But the blue-green argument still works: you can assign any four clubs you want to the four rooms, then paint the doors, then figure out who is in the Green Doors Club, and then observe that, in fact, the Green Doors Club is not one of the four clubs that got a room. Then you can reassign the clubs to rooms, this time making sure that the Green Doors Club gets a room. But now you have to repaint the doors, and when you do you find out that membership in the Green Doors Club has changed: some new members were admitted, or some former members were expelled, so the club that meets there is no longer the Green Doors Club, it is some other club. (Or if the Green Doors Club is meeting somewhere, you will find that you have painted the doors wrong.) I think this would probably work. The only thing that's weird about it is that some clubs have an infinite number of members so that it's hard to see how they could all squeeze into the same room. That's okay, not every member attends every meeting of every club they're in, that would be impossible anyway because everyone belongs to multiple clubs. But one place you could go from there is: what if we only guarantee rooms to clubs with a finite number of members? There are only a countably infinite number of clubs then, so they do all fit into the hotel! Okay, Tyler, but what happens to the Green Door Club then? I said all the finite clubs got rooms, and we know the Green Door Club never gets a room, so what can we conclude? It's tempting to try to slip in a reference to Groucho Marx, but I think it's unlikely that that will do anything but confuse matters. [ Previously ] [ Update: My friend said he tried it and it didn't go over as well as I thought it might. ] [Other articles in category /math] permanent link Sun, 26 Nov 2023
A Qmail example of dealing with unavoidable race conditions
[ I recently posted about a race condition bug reported by Joe Armstrong and said “this sort of thing is now in the water we swim in, but it wasn't yet [in those days of olde].” This is more about that. ] I learned a lot by reading everything Dan Bernstein wrote about
the design of (I know someone wants to ask what about Postfix? At the time Qmail was released, Postfix was still called ‘VMailer’. The ‘V’ supposedly stood for “Venema” but the joke was that the ‘V’ was actually for “vaporware” because that's what it was.) A few weeks ago I was explaining one of Qmail's data structures to a junior programmer. Suppose a local user queues an outgoing message that needs to be delivered to 10,000 recipients in different places. Some of the deliveries may succeed immediately. Others will need to be retried, perhaps repeatedly. Eventually (by default, ten days) delivery will time out and a bounce message will be delivered back to the sender, listing the recipients who did not receive the delivery. How does Qmail keep track of this information? 2023 junior programmer wanted to store a JSON structure or something. That is not what Qmail does. If the server crashes halfway through writing a JSON file, it will be corrupt and unreadable. JSON data can be written to a temporary file and the original can be replaced atomically, but suppose you succeed in delivering the message to 9,999 of the 10,000 recipients and the system crashes before you can atomically update the file? Now the deliveries will be re-attempted for those 9,999 recipients and they will get duplicate copies. Here's what Qmail does instead. The file in the queue directory is in the following format:
where ■ represents a zero byte. To 2023 eyes this is strange and uncouth, but to a 20th-century system programmer, it is comfortingly simple. When Qmail wants to attempt a delivery to If delivery does succeed, Qmail updates the
The update of a single byte will be done all at once or not at all. Even writing two bytes is riskier: if the two bytes span a disk block boundary, the power might fail after only one of the modified blocks has been written out. With a single byte nothing like that can happen. Absent a catastrophic hardware failure, the data structure on the disk cannot become corrupted. Mail can never be lost. The only thing that can go wrong here is if the local system crashes in between the successful delivery and the updating of the byte; in this case the delivery will be attempted again, to that one user. Addenda
[Other articles in category /prog] permanent link Fri, 24 Nov 2023
Puzzling historical artifact in “Programming Erlang”?
Lately I've been reading Joe Armstrong's book Programming Erlang, and today I was brought up short by this passage from page 208:
Can you guess the obscure bug? I don't think I'm unusually skilled at concurrent systems programming, and I'm certainly no Joe Armstrong, but I thought the problem was glaringly obvious:
I scratched my head over this for quite some time. Not over the technical part, but about how famous expert Joe Amstrong could have missed this. Eventually I decided that it was just that this sort of thing is now in the water we swim in, but it wasn't yet in the primeval times Armstrong was writing about. Sometimes problems are ⸢obvious⸣ because it's thirty years later and everyone has thirty years of experience dealing with those problems. Another exampleI was reminded of a somewhat similar example. Before the WWW came, a sysadmin's view of network server processes was very different than it is now. We thought of them primarily as attack surfaces, and ran as few as possible, as little as possible, and tried hard to prevent anyone from talking to them. Partly this was because encrypted, authenticated communications
protocols were still an open research area. We now have When the Web came along, every sysadmin was thrust into a terrifying new world in which users clamored to write network services that could be talked to at all times by random Internet people all over the world. It was quite a change. [ I wrote more about system race conditions, but decided to postpone it to Monday. Check back then. ] [Other articles in category /prog] permanent link
Math SE report 2023-09: Sense and reference, Wason tasks, what is a sequence?
Proving there is only one proof?OP asks:
This was actually from back in July, when there was a fairly substantive answer. But it left out what I thought was a simpler, non-substantive answer: For a given theorem !!T!! it's actually quite simple to prove that there is (or isn't) only one proof of !!T!!: just generate all possible proofs in order by length until you find the shortest proofs of !!T!!, and then stop before you generate anything longer than those. There are difficult and subtle issues in provability theory, but this isn't one of them. I say “non-substantive” because it doesn't address any of the possibly interesting questions of why a theorem would have only one proof, or multiple proofs, or what those proofs would look like, or anything like that. It just answers the question given: is it possible to prove that there is only one shortest proof. So depending on what OP was looking for, it might be very unsatisfying. Or it might be hugely enlightening, to discover that this seemingly complicated question actually has a simple answer, just because proofs can be systematically enumerated. This comes in handy in more interesting contexts. Gödel showed that arithmetic contains a theorem whose shortest proof is at least one million steps long! He did it by constructing an arithmetic formula !!G!! which can be interpreted as saying:
If !!G!! is false, it can be proved (in less than one million steps) and our system is inconsistent. So assuming that our axioms are consistent, then !!G!! is true and either:
Which is it? It can't be (1) because there is a proof of !!G!!: simply generate every single proof of one million steps or fewer, and check at the last line of each one to make sure that it is not !!G!!. So it must be (2). What counts as a sequence, and how would we know that it isn't deceiving?This is a philosophical question: What is a sequence, really? And:
And several other related questions that are actually rather subtle: Is a sequence defined by its elements, or by some external rule? If the former how can you know when a sequence is linear, when you can only hope to examine a finite prefix? I this is a great question because I think a sequence, properly construed, is both a rule and its elements. The definition says that a sequence of elements of !!S!! is simply a function !!f:\Bbb N\to S!!. This definition is a sort of spherical cow: it's a nice, simple model that captures many of the mathematical essentials of the thing being modeled. It works well for many purposes, but you get into trouble if you forget that it's just a model. It captures the denotation, but not the sense. I wouldn't yak so much about this if it wasn't so often forgotten. But the sense is the interesting part. If you forget about it, you lose the ability to ask questions like
If all you have is the denotation, there's only one way to answer this question:
and there is nothing further to say about it. The question is pointless and the answer is useless. Sometimes the meaning is hidden a little deeper. Not this time. If we push down into the denotation, hoping for meaning, we find nothing but more emptiness:
We could keep going down this road, but it goes nowhere and having gotten to the end we would have seen nothing worth seeing. But we do ask and answer this kind of question all the time. For example:
Are sequences !!S_1!! and !!S_2!! the same sequence? Yes, yes, of course they are, don't focus on the answer. Focus on the question! What is this question actually asking? The real essence of the question is not about the denotation, about just the elements. Rather: we're given descriptions of two possible computations, and the question is asking if these two computations will arrive at the same results in each case. That's the real question. Well, I started this blog article back in October and it's still not ready because I got stuck writing about this question. I think the answer I gave on SE is pretty good, OP asked what is essentially a philosophical question and the backbone of my answer is on the level of philosophy rather than mathematics. [ Addendum: On review, I am pleasantly surprised that this section of the blog post turned out both coherent and relevant. I really expected it to be neither. A Thanksgiving miracle! ] Can inequalities be added the way that equations can be added?OP says:
I have a theory that if someone is having trouble with the intuitive meaning of some mathematical property, it's a good idea to turn it into some questions involving allocation of resources, or who has more of some commodity, because human brains are good at monkey tasks like seeing who got cheated when the bananas were shared out. About ten years ago someone asked for an intuitive explanation of why you could add !!\frac a2!! to both sides of !!\frac a2 < \frac b2!! to get !!\frac a2+\frac a2 < \frac a2 + \frac b2!!. I said:
I tried something similar this time around:
Then the baker trades your bag !!b!! for a bigger bag !!d!!.
Someday I'll write up a whole blog article about this idea, that puzzles in arithmetic sometimes become intuitively obvious when you turn them into questions about money or commodities, and that puzzles in logic sometimes become intuitively obvious when you turn them into questions about contract and rule compliance. I don't remember why I decided to replace the djinn with a baker this time around. The cookies stayed the same though. I like cookies. Here's another cookie example, this time to explain why !!1\div 0.5 = 2!!. What is the difference between "for all" and "there exists" in set builder notation?This is the same sort of thing again. OP was was asking about $$B = \{n \in \mathbb{N} : \forall x \in \mathbb{N} \text{ and } n=2^x\}$$ but attempting to understand this is trying to swallow two pills at once. One pill is the logic part (what role is the !!\forall!! playing) and the other pill is the arithmetic part having to do with powers of !!2!!. If you're trying to understand the logic part and you don't have an instantaneous understanding of powers of !!2!!, it can be helpful to simplify matters by replacing the arithmetic with something you understand intuitively. In place of the relation !!a = 2^b!! I like to use the relation “!!a!! is the mother of !!b!!”, which everyone already knows. Are infinities included in the closure of the real set !!\overline{\mathbb{R}}!!This is a good question by the Chip Buchholtz criterion: The answer is much longer than the question was. OP wants to know if the closure of !!\Bbb R!! is just !!\Bbb R!! or if it's some larger set like !![-\infty, \infty]!!. They are running up against the idea that topological closure is not an absolute notion; it only makes sense in the context of an enclosing space. I tried to draw an analogy between the closure and the complement of a set: Does the complement of the real numbers include the number !!i!!? Well, it depends on the context. OP preferred someone else's answer, and I did too, saying:
I try to make things very explicit, but the downside of that is that it makes my answers longer, and shorter is generally better than longer. Sometimes it works, and sometimes it doesn't. Vacuous falsehood - does it exist, and are there examples?I really liked this question because I learned something from it. It brought me up short: “Huh,” I said. “I never thought about that.” Three people downvoted the question, I have no idea why. I didn't know what a vacuous falsity would be either but I decided that since the negation of a vacuous truth would be false it was probably the first thing to look at. I pulled out my stock example of vacuous truth, which is:
This is true, because all rubies are red, but vacuously so because I don't own any rubies. Since this is a vacuous truth, negating it ought to give us a vacuous falsity, if there is such a thing:
This is indeed false. And not in the way one would expect! A more typical false claim of this type would be:
This is also false, in rather a different way. It's false, but not vacuously so, because to disprove it you have to get my belts out of the closet and examine them. Now though I'm not sure I gave the right explanation in my answer. I said:
But is this the right analogy? I could have gone the other way:
Ah well, this article has been drying out on the shelf for a month now, I'm making an editorial decision to publish it without thinking about it any more. [Other articles in category /math/se] permanent link Mon, 23 Oct 2023Katara is taking a Data Structures course this year. The most recent assignment gave her a lot of trouble, partly because it was silly and made no sense, but also because she does not yet know an effective process for writing programs, and the course does not attempt to teach her. On the day the last assignment was due I helped her fix the remaining bugs and get it submitted. This is the memo I wrote to her to memorialize the important process issues that I thought of while we were working on it.
Something we discussed that I forgot to include in the memo that we discussed is: After you fix something significant, or add significant new functionality, make a checkpoint copy of the entire source code. This can be as simple as simply copying it all into separate folder. That way, when you are fixing the next thing, if you mess up and break everything, it's easy to get back to a known-good state. The computer is really clumsy to use for many tasks, but it's just great at keeping track of information, so exploit that when you can. I think CS curricula should have a class that focuses specifically on these issues, on the matter of how do you actually write software? But they never do. [Other articles in category /prog] permanent link When I first set up this blog I didn't know how long I would do it, so instead of thinking about how I wanted it to look, I just took the default layout that came with Blosxom, and figured that I would change it when I got around to it. Now I am getting around to it. The primary problem is that the current implementation (with nested tables!) performs badly, especially on mobile devices and especially on pages with a lot of MathJax. A secondary issue is that it's troublesome to edit. People sometimes laugh at how it looks like a 2006 design (which it is). I don't much care about this. I like the information-dense layout, which I think is distinctive and on-brand. I would like a redesign that fixes these two problems. The primary goal is to get rid of the nested tables and replace the implementation with something that browsers can handle better, probably something based on CSS Flexbox. It doesn't have to look very different, but it does need to be straightforward enough that I can make the next ten years of changes without a lot of research and experimentation. If you know someone interested in doing this, please email me a referral or have them get in touch with me. If you are interested in doing it yourself, please send me a proposal. Include a cost estimate, as this would be paid work. Please do not advertise this on Hacker News, as that would run the risk of my getting 100 proposals, and that would be 50 times as many as I need. Thanks. [Other articles in category /meta] permanent link Sun, 22 Oct 2023We used to have a cat named Chase. To be respectful we would sometimes refer to him as “Mr. Cat”. And sometimes I amused myself by calling him “Señor Gato”. Yesterday I got to wondering: Where did Spanish get “gato”, which certainly sounds like “cat”, when the Latin is fēles or fēlis (like in “feline’)? And similarly French has chat. Well, the real question is, where did Latin get fēles? Because Latin also has cattus, which I think sounds like a joke. You're in Latin class, and you're asked to translate cat, but you haven't done your homework, so what do you say? “Uhhhh… ‘cattus’?” But cattus is postclassical Latin, replacing the original word fēles no more than about 1500 years ago. The word seems to have wandered all over Europe and Western Asia and maybe North Africa, borrowed from one language into another, and its history is thoroughly mixed up. Nobody is sure where it came from originally, beyond “something Germanic”. The OED description of cat runs to 600 words and shrugs its shoulders. I learned recently that such words (like brinjal, the eggplant) are called Wanderworts, wandering words. [Other articles in category /lang/etym] permanent link Sat, 21 Oct 2023The other day I was looking into vindaloo curry and was surprised to learn that the word “vindaloo” is originally Portuguese vin d'alho, a wine and garlic sauce. Amazing. In Japanese, squashes are called kabocha. (In English this refers to a specific type of squash associated with Japan, but in Japanese it's more generic.) Kabocha is from Portuguese again. The Portuguese introduced squashes to Japan via Cambodia, which in Portuguese is Camboja. [Other articles in category /lang/etym] permanent link Fri, 20 Oct 2023
The discrete logarithm, shorter and simpler
I recently discussed the “discrete logarithm” method for multiplying integers, and I feel like I took too long and made it seem more complicated and mysterious than it should have been. I think I'm going to try again. Suppose for some reason you found yourself needing to multiply a lot of powers of !!2!!. What's !!4096·512!!? You could use the conventional algorithm: $$ \begin{array}{cccccccc} & & & & 4 & 0 & 9 & 6 \\ × & & & & & 5 & 1 & 2 \\ \hline % & & & & 8 & 1 & 9 & 2 \\ & & & 4 & 0 & 9 & 6 & \\ & 2 & 0 & 4 & 8 & 0 & & \\ \hline % & 2 & 0 & 9 & 7 & 1 & 5 & 2 \end{array} $$ but that's a lot of trouble, and a simpler method is available. You know that $$2^i\cdot 2^j = 2^{i+j}$$ so if you had an easy way to convert $$2^i\leftrightarrow i$$ you could just convert the factors to exponents, add the exponents, and convert back. And all that's needed is a simple table: \begin{array}{rr} 0 & 1\\ 1 & 2\\ 2 & 4\\ 3 & 8\\ 4 & 16\\ 5 & 32\\ 6 & 64\\ 7 & 128\\ 8 & 256\\ 9 & 512\\ 10 & 1\,024\\ 11 & 2\,048\\ 12 & 4\,096\\ 13 & 8\,192\\ 14 & 16\,384\\ 15 & 32\,768\\ 16 & 65\,536\\ 17 & 131\,072\\ 18 & 262\,144\\ 19 & 524\,288\\ 20 & 1\,048\,576\\ 21 & 2\,097\,152\\ \vdots & \vdots \\ \end{array} We check the table, and find that $$4096\cdot512 = 2^{12}\cdot 2^9 = 2^{12+9} = 2^{21} = 2097152.$$ Easy-peasy. That is all very well but how often do you find yourself having to multiply a lot of powers of !!2!!? This was a lovely algorithm but with very limited application. What Napier (the inventor of logarithms) realized was that while not every number is an integer power of !!2!!, every number is an integer power of !!1.00001!!, or nearly so. For example, !!23!! is very close to !!1.00001^{313\,551}!!. Napier made up a table, just like the one above, except with powers of !!1.00001!! instead of powers of !!2!!. Then to multiply !!x\cdot y!! you would just find numbers close to !!x!! and !!y!! in Napier's table and use the same algorithm. (Napier's original table used powers of !!0.9999!!, but it works the same way for the same reason.) There's another way to make it work. Consider the integers mod !!101!!, called !!\Bbb Z_{101}!!. In !!\Bbb Z_{101}!!, every number is an integer power of !!2!!! For example, !!27!! is a power of !!2!!. It's simply !!2^7!!, because if you multiply out !!2^7!! you get !!128!!, and !!128\equiv 27\pmod{101}!!. Or: $$\begin{array}{rcll} 14 & \stackrel{\pmod{101}}{\equiv} & 10\cdot 101 & + 14 \\ & = & 1010 & + 14 \\ & = & 1024 \\ & = & 2^{10} \end{array} $$ Or: $$\begin{array}{rcll} 3 & \stackrel{\pmod{101}}{\equiv} & 5844512973848570809\cdot 101 & + 3 \\ & = & 590295810358705651709 & + 3 \\ & = & 590295810358705651712 \\ & = & 2^{69} \end{array} $$ Anyway that's the secret. In !!\Bbb Z_{101}!! the silly algorithm that quickly multiplies powers of !!2!! becomes more practical, because in !!\Bbb Z_{101}!!, every number is a power of !!2!!. What works for !!101!! works in other cases larger and more interesting. It doesn't work to replace !!101!! with !!7!! (try it and see what goes wrong), but we can replace it with !!107, 797!!, or !!297779!!. The key is that if we want to replace !!101!! with !!n!! and !!2!! with !!a!!, we need to be sure that there is a solution to !!a^i=b\pmod n!! for every possible !!b!!. (The jargon term here is that !!a!! must be a “primitive root mod !!n!!”. !!2!! is a primitive root mod !!101!!, but not mod !!7!!.) Is this actually useful for multiplication? Perhaps not, but it does have cryptographic applications. Similar to how multiplying is easy but factoring seems difficult, computing !!a^i\pmod n!! for given !!a, i, n!! is easy, but nobody knows a quick way in general to reverse the calculation and compute the !!i!! for which !!a^i\pmod n = m!! for a given !!m!!. When !!n!! is small we can simply construct a lookup table with !!n-1!! entries. But if !!n!! is a !!600!!-digit number, the table method is impractical. Because of this, Alice and Bob can find a way to compute a number !!2^i!! that they both know, but someone else, seeing !!2^i!! can't easily figure out what the original !!i!! was. See Diffie-Hellman key exchange for more details. [ Also previously: Percy Ludgate's weird variation on this ] [Other articles in category /math] permanent link Sun, 15 Oct 2023[ Addendum 20231020: This came out way longer than it needed to be, so I took another shot at it, and wrote a much simpler explanation of the same thing that is only one-third as long. ] A couple days ago I discussed the weird little algorithm of Percy Ludgate's, for doing single-digit multiplication using a single addition and three scalar table lookups. In Ludgate's algorithm, there were two tables, !!T_1!! and !!T_2!!, satisfying the following properties: $$ \begin{align} T_2(T_1(n)) & = n \tag{$\color{darkgreen}{\spadesuit}$} \\ T_2(T_1(a) + T_1(b)) & = ab. \tag{$\color{purple}{\clubsuit}$} \end{align} $$ This has been called the “Irish logarithm” method because of its resemblance to ordinary logarithms. Normally in doing logarithms we have a magic logarithm function !!\ell!! with these properties: $$ \begin{align} \ell^{-1}(\ell(n)) & = n \tag{$\color{darkgreen}{\spadesuit}$} \\ \ell^{-1}(\ell(a) + \ell(b)) & = ab. \tag{$\color{purple}{\clubsuit}$} \end{align} $$ (The usual notation for !!\ell(x)!! is of course “!!\log x!!” or “!!\ln x!!” or something of that sort, and !!\ell^{-1}(x)!! is usually written !!e^x!! or !!10^x!!.) The properties of Ludgate's !!T_1!! and !!T_2!! are formally identical, with !!T_1!! playing the role of the logarithm function !!\ell!! and !!T_2!! playing the role of its inverse !!\ell^{-1}!!. Ludgate's versions are highly restricted, to reduce the computation to something simple enough that it can be implemented with brass gears. Both !!T_1!! and !!T_2!! map positive integers to positive integers, and can be implemented with finite lookup tables. The ordinary logarithm does more, but is technically much more difficult. With the ordinary logarithm you are not limited to multiplying single digit integers, as with Ludgate's weird little algorithm. You can multiply any two real numbers, and the multiplication still requires only one addition and three table lookups. But the cost is huge! The tables are much larger and more complex, and to use them effectively you have to deal with fractional numbers, perform table interpolation, and worry about error accumulation. It's tempting at this point to start explaining the history and use of logarithm tables, slide rules, and so on, but this article has already been delayed once, so I will try to resist. I will do just one example, with no explanation, to demonstrate the flavor. Let's multiply !!7!! by !!13!!.
If I were multiplying !!7.236!! by !!13.877!!, I would be willing to accept all these costs, and generations of scientists and engineers did accept them. But for !!7.0000×13.000 = 91.000!! the process is ridiculous. One might wonder if there wasn't some analogous technique that would retain the small, finite tables, and permits multiplication of integers, using only integer calculations throughout. And there is! Now I am going to demonstrate an algorithm, based on logarithms, that exactly multiplies any two integers !!a!! and !!b!!, as long as !!ab ≤ 100!!. Like Ludgate's and the standard algorithm, it will use one addition and three lookups in tables. Unlike the standard algorithm, the tables will be small, and will contain only integers. Here is the table of the !!\ell!! function, which corresponds to Ludgate's !!T_1!!: $$ \begin{array}{rrrrrrrrrrr} {\tiny\color{gray}{1}} & 0, & 1, & \color{darkblue}{69}, & 2, & 24, & 70, & \color{darkgreen}{9}, & 3, & 38, & 25, \\ {\tiny\color{gray}{11}} & 13, & \color{darkblue}{71}, & \color{darkgreen}{66}, & 10, & 93, & 4, & 30, & 39, & 96, & 26, \\ {\tiny\color{gray}{21}} & 78, & 14, & 86, & 72, & 48, & 67, & 7, & 11, & 91, & 94, \\ {\tiny\color{gray}{31}} & 84, & 5, & 82, & 31, & 33, & 40, & 56, & 97, & 35, & 27, \\ {\tiny\color{gray}{41}} & 45, & 79, & 42, & 15, & 62, & 87, & 58, & 73, & 18, & 49, \\ {\tiny\color{gray}{51}} & 99, & 68, & 23, & 8, & 37, & 12, & 65, & 92, & 29, & 95, \\ {\tiny\color{gray}{61}} & 77, & 85, & 47, & 6, & 90, & 83, & 81, & 32, & 55, & 34, \\ {\tiny\color{gray}{71}} & 44, & 41, & 61, & 57, & 17, & 98, & 22, & 36, & 64, & 28, \\ {\tiny\color{gray}{81}} & \color{darkred}{76}, & 46, & 89, & 80, & 54, & 43, & 60, & 16, & 21, & 63, \\ {\tiny\color{gray}{91}} & 75, & 88, & 53, & 59, & 20, & 74, & 52, & 19, & 51, & 50\hphantom{,} \\ \end{array} $$ (If we only want to multiply numbers with !!1\le a, b \le 9!! we only need the first row, but with the full table we can also compute things like !!7·13=91!!.) Like !!T_2!!, this is not really a two-dimensional array. It just a list of !!100!! numbers, arranged in rows to make it easy to find the !!81!!st number when you need it. The small gray numerals in the margin are a finding aid. If you want to look up !!\ell(81)!! you can see that it is !!\color{darkred}{76}!! without having to count up !!81!! elements. This element is highlighted in red in the table above. Note that the elements are numbered from !!1!! to !!100!!, whereas all the other tables in these articles have been zero-indexed. I wondered if there was a good way to fix this, but there really isn't. !!\ell!! is analogous to a logarithm function, and the one thing a logarithm function really must do is to have !!\log 1 = 0!!. So too here; we have !!\ell(1) = 0!!. We also need an !!\ell^{-1}!! table analogous to Ludgate's !!T_2!!: $$ \begin{array}{rrrrrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, & 27, & 54, & 7, \\ {\tiny\color{gray}{10}} & 14, & 28, & 56, & 11, & 22, & 44, & 88, & 75, & 49, & 98, \\ {\tiny\color{gray}{20}} & 95, & 89, & 77, & 53, & 5, & 10, & 20, & 40, & 80, & 59, \\ {\tiny\color{gray}{30}} & 17, & 34, & 68, & 35, & 70, & 39, & 78, & 55, & 9, & 18, \\ {\tiny\color{gray}{40}} & \color{darkblue}{36}, & 72, & 43, & 86, & 71, & 41, & 82, & 63, & 25, & 50, \\ {\tiny\color{gray}{50}} & 100, & 99, & 97, & 93, & 85, & 69, & 37, & 74, & 47, & 94, \\ {\tiny\color{gray}{60}} & 87, & 73, & 45, & 90, & 79, & 57, & 13, & 26, & 52, & 3, \\ {\tiny\color{gray}{70}} & 6, & 12, & 24, & 48, & 96, & \color{darkgreen}{91}, & \color{darkred}{81}, & 61, & 21, & 42, \\ {\tiny\color{gray}{80}} & 84, & 67, & 33, & 66, & 31, & 62, & 23, & 46, & 92, & 83, \\ {\tiny\color{gray}{90}} & 65, & 29, & 58, & 15, & 30, & 60, & 19, & 38, & 76, & 51\hphantom{,} \\ \end{array} $$ Like !!\ell^{-1}!! and !!T_2!!, this is just a list of !!100!! numbers in order. As the notation suggests, !!\ell^{-1}!! and !!\ell!! are inverses. We already saw that the first table had !!\ell(81)=\color{darkred}{76}!! and !!\ell(1) = 0!!. Going in the opposite direction, we see from the second table that !!\ell^{-1}(76)= \color{darkred}{81}!! (again in red) and !!\ell^{-1}(0)=1!!. The elements of !!\ell!! tell you where to find numbers in the !!\ell^{-1}!! table. Where is !!17!! in the second table? Look at the !!17!!th element in the first table. !!\ell(17) = 30!!, so !!17!! is at position !!30!! in the second table. Before we go too deeply into how these were constructed, let's try the !!7×13!! example we did before. The algorithm is just !!\color{purple}{\clubsuit}!!: $$ \begin{align} % \ell^{-1}(\ell(a) + \ell(b)) & = ab\tag{$\color{purple}{\clubsuit}$} \\ 7·13 &= \ell^{-1}(\ell(7) + \ell(13)) \\ &= \ell^{-1}(\color{darkgreen}{9} + \color{darkgreen}{66}) \\ &= \ell^{-1}(75) \\ &= \color{darkgreen}{91} \end{align} $$ (The relevant numbers are picked out in green in the two tables.) As promised, with three table lookups and a single integer addition. What if the sum in the middle exceeds !!99!!? No problem, the !!\ell^{-1}!! table wraps around, so that element !!100!! is the same as element !!0!!: $$ \begin{align} % \ell^{-1}(\ell(a) + \ell(b)) & = ab\tag{$\color{purple}{\clubsuit}$} \\ 3·12 &= \ell^{-1}(\ell(3) + \ell(12)) \\ &= \ell^{-1}(\color{darkblue}{69} + \color{darkblue}{71}) \\ &= \ell^{-1}(140) \\ &= \ell^{-1}(40) &\text{(wrap around)}\\ &= \color{darkblue}{36} \end{align} $$ How about that. (This time the relevant numbers are picked out in blue.) I said this only computes !!ab!! when the product is at most !!100!!. That is not quite true. If you are willing to ignore a small detail, this algorithm will multiply any two numbers. The small detail is that the multiplication will be done mod !!101!!. That is, instead of the exact answer, you get one that differs from it by a multiple of !!101!!. Let's do an example to see what I mean when I say it works even for products bigger than !!100!!: $$ \begin{align} % \ell^{-1}(\ell(a) + \ell(b)) & = ab\tag{$\color{purple}{\clubsuit}$} \\ 16·26 &= \ell^{-1}(\ell(16) + \ell(26)) \\ &= \ell^{-1}(4 + 67) \\ &= \ell^{-1}(71) \\ &= 12 \end{align} $$ This tell us that !!16·26 = 12!!. The correct answer is actually !!16·26 = 416!!, and indeed !!416-12 = 404!! which is a multiple of !!101!!. The reason this happens is that the elements of the second table, !!\ell^{-1}!!, are not true integers, they are mod !!101!! integers. Okay, so what is the secret here? Why does this work? It should jump out at you that it is often the case that an entry in the !!\ell^{-1}!! table is twice the previous entry: $$\ell^{-1}(1+n) = 2\cdot \ell^{-1}(n)$$ In fact, this is true everywhere, if you remember that the numbers are not ordinary integers but mod !!101!! integers. For example, the number that follows !!64!!, in place of !!64·2=128!!, is !!27!!. But !!27\equiv 128\pmod{101}!! because they differ by a multiple of !!101!!. From a mod !!101!! point of view, it doesn't matter wther we put !!27!! or !!128!! after !!64!!, as they are the same thing. Those two facts are the whole secret of the !!\ell^{-1}!! table:
Certainly !!\ell^{-1}(0) = 2^0 = 1!!. And every entry in the !!\ell^{-1}!! is twice the previous one, if you are thinking in mod !!101!!. The two secrets are actually one secret: $$\ell^{-1}(n) = 2^n\pmod{101}.$$ This is why the multiplication algorithm works. Say we want to multiply !!7!! and !!13!! again. We look up !!7!! and !!13!! in !!\ell!!, and find !!\ell(7)=9!! and !!\ell(13)=66!!. What this is really telling us is that $$ \begin{align} 7 & = 2^{9\hphantom6} \pmod{101} \\ 13 & = 2^{66} \pmod{101} \\ \end{align} $$ so that multiplying !!7\cdot13!! mod !!101!! is the same as multiplying $$2^9\cdot 2^{66}.$$ But multiplying exact powers of !!2!! is easy, since you just add the exponents: !!2^9\cdot2^{66} = 2^{9+66} = 2^{75}!!, whether you are doing it in regular numbers or mod !!101!! numbers. And the !!\ell^{-1}!! table tells us directly that !!2^{75} = 91\pmod{101}!!. The !!\ell!! function, which is analogous to the regular logarithm, is called a discrete logarithm. What's going on with Percy Ludgate's algorithm? It's a sort of compressed, limited version of the discrete logarithm. I had a hope that maybe we could reimplement Ludgate's thing by basing it more directly on discrete logarithms. Say we had the !!\ell^{-1}!! table encoded in a wheel of some sort, with the !!100!! entries in order around the rim. There's a “current position” !!p!!, initially !!0!!, and a “current number” !!2^p!!, initially !!1!!. On the same axle as the wheel, mount a gear with exactly 100 teeth. We can easily turn the wheel exactly !!q!! positions by taking a straight bar with !!q!! teeth and using it to turn the gear, which turns the wheel. We easily multiply the current number by !!2!! just by turning the wheel one position clockwise. Multiplying by !!7!! isn't too hard, just turn the wheel !!9!! positions clockwise. We can do this by constructing a short bar with exactly !!9!! teeth and using it to turn the gear. Or maybe we have a meshing gear with !!9!! teeth, on another axle, which we give one full turn. Either way, if the current number was !!5!! before, it's !!35!! after. Multiplying by !!3!! is rather more of a pain, because we have to turn the wheel !!69!! positions, so we need a bar or a meshing gear with 69 teeth. (We could get away with one with only !!31!! teeth, if we could turn the wheel the other way, but that seems like it might be more complicated. Hmm, I suppose it would work to use a meshing gear with 31 teeth that engages a second gear (with any number of teeth) that engages the main gear.) Anyway I took a look to see if there were any better tables do use, and the answer is: maybe! If, instead of a table of !!2^n!!, we use a table of !!26^n!!, then the brass wheel approach performs a little better: $$ \begin{array}{rrrrrrrrrr} {\tiny\color{gray}{0}} & 1, & 26, & 70, & 2, & 52, & 39, & 4, & 3, & 78, & 8, \\ {\tiny\color{gray}{10}}& 6, & 55, & 16, & 12, & 9, & 32, & 24, & 18, & 64, & 48, \\ {\tiny\color{gray}{20}}& 36, & 27, & 96, & 72, & 54, & 91, & 43, & 7, & 81, & 86, \\ {\tiny\color{gray}{30}}& 14, & 61, & 71, & 28, & 21, & 41, & 56, & 42, & 82, & 11, \\ {\tiny\color{gray}{40}}& 84, & 63, & 22, & 67, & 25, & 44, & 33, & 50, & 88, & 66, \\ {\tiny\color{gray}{50}}& 100, & 75, & 31, & 99, & 49, & 62, & 97, & 98, & 23, & 93, \\ {\tiny\color{gray}{60}}& 95, & 46, & 85, & 89, & 92, & 69, & 77, & 83, & 37, & 53, \\ {\tiny\color{gray}{70}}& 65, & 74, & 5, & 29, & 47, & 10, & 58, & 94, & 20, & 15, \\ {\tiny\color{gray}{80}}& 87, & 40, & 30, & 73, & 80, & 60, & 45, & 59, & 19, & 90, \\ {\tiny\color{gray}{90}}& 17, & 38, & 79, & 34, & 76, & 57, & 68, & 51, & 13, & 35\hphantom{,} \\ \end{array} $$ Multiplying by !!2!! is no longer as simple as turning the wheel one notch clockwise; you have to turn it !!3!! positions counterclockwise. But that seems pretty easy. Multiplying by !!3!! is also rather easy: just turn the wheel !!7!! positions. If the table above is !!T_2!!, then the analogue of Ludgate's !!T_1!! table is: $$ \begin{array}{cccccccccc} \tiny\color{gray}{1} & \tiny\color{gray}{2} & \tiny\color{gray}{3} & \tiny\color{gray}{4} & \tiny\color{gray}{5} & \tiny\color{gray}{6} & \tiny\color{gray}{7} & \tiny\color{gray}{8} & \tiny\color{gray}{9} \\ 0 & 3 & 7 & 6 & 72 & 10 & 27 & 9 & 14 \\ \end{array} $$ That is, if you want to compute !!3·6!!, you start with the wheel in position !!0!!, then turn it by !!T_1(3) = 7!! positions, then by !!T_1(6) = 10!!, and now it's at position !!17!!, where the current number is !!18!!. The numbers in the !!T_1!! table are all pretty small, except that to multiply by !!5!! you have to turn by !!72!! positions, which is kinda awful. Still it's only a little worse than in the powers-of-2 version where to multiply by !!3!! you would have to turn the wheel by !!69!! positions. And overall the powers-of-26 table is better: the sum of the !!9!! entries is only !!148!!, which is optimal; the corresponding sum of the entries for the powers-of-2 table is !!216!!. Who knows, it might work, and even if it didn't work well it might be pretty cool. [Other articles in category /math] permanent link Mon, 02 Oct 2023
Irish logarithm forward instead of backward
Yesterday I posted about the so-called “Irish logarithm”, Percy Ludgate's little algorithm for single-digit multiplication. Hacker News user sksksfpuzhpx said:
and referred to Brian Coghlan's aticle “Percy Ludgate's Logarithmic indices”. Whereas I was reverse-engineering Ludgate's tables with a sort of ad-hoc backtracking search, if you do it right you can do it it more easily with a simple greedy search. Uh oh, I thought, I will want to write this up before I move on to the thing I planned to do next, which made it all the more likely that I never would get to the thing I had planned to do next. But Shreevatsa R. came to my rescue and wrote up the Coghlan thing at least as well as I could have myself. Definitely check it out. Thank you, Shreevatsa! [ Update 20231015: A different kind of all-integer logarithm: the discrete logarithm. ] [ Update 20231020: Better explanation of the discrete logarithm. ] [Other articles in category /math] permanent link Sun, 01 Oct 2023The Wikipedia article on “Irish logarithm” presents this rather weird little algorithm, invented by Percy Ludgate. Suppose you want to multiply !!a!! and !!b!!, where both are single-digit numbers !!0≤a,b≤9!!. Normally you would just look it up on a multiplication table, but please bear with me for a bit. To use Ludgate's algorithm you need a different little table: $$ \begin{array}{rl} T_1 = & \begin{array}{cccccccccc} \tiny\color{gray}{0} & \tiny\color{gray}{1} & \tiny\color{gray}{2} & \tiny\color{gray}{3} & \tiny\color{gray}{4} & \tiny\color{gray}{5} & \tiny\color{gray}{6} & \tiny\color{gray}{7} & \tiny\color{gray}{8} & \tiny\color{gray}{9} \\ 50 & 0 & 1 & 7 & 2 & 23 & 8 & 33 & 3 & 14 \\ \end{array} \end{array} $$ and a different bigger one: $$ \begin{array}{rl} T_2 = & % \left( \begin{array}{rrrrrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, & 3, & 6, & 12, \\ {\tiny\color{gray}{10}} & 24, & 48, & 0, & 0, & 9, & 18, & 36, & 72, & 0, & 0, \\ {\tiny\color{gray}{20}} & 0, & 27, & 54, & 5, & 10, & 20, & 40, & 0, & 81, & 0, \\ {\tiny\color{gray}{30}} & 15, & 30, & 0, & 7, & 14, & 28, & 56, & 45, & 0, & 0, \\ {\tiny\color{gray}{40}} & 21, & 42, & 0, & 0, & 0, & 0, & 25, & 63, & 0, & 0, \\ {\tiny\color{gray}{50}} & 0, & 0, & 0, & 0, & 0, & 0, & 35, & 0, & 0, & 0, \\ {\tiny\color{gray}{60}} & 0, & 0, & 0, & 0, & 0, & 0, & 49\hphantom{,} \end{array} % \right) \end{array} $$ I've formatted !!T_2!! in rows for easier reading, but it's really just a zero-indexed list of !!101!! numbers. So for example !!T_2(23)!! is !!5!!. The tiny gray numbers in the margin are not part of the table, they are counting the elements so that it is easy to find element !!23!!. Ludgate's algorithm is simply: $$ ab = T_2(T_1(a) + T_1(b)) $$ Let's see an example. Say we want to multiply !!4×7!!. We first look up !!4!! and !!7!! in !!T_1!!, and get !!2!! and !!33!!, which we add, getting !!35!!. Then !!T_2(35)!! is !!28!!, which is the correct answer. This isn't useful for paper-and-pencil calculation, because it only works for products up to !!9×9!!, and an ordinary multiplication table is easier to use and remember. But Ludgate invented this for use in a mechanical computing engine, for which it is much better-suited. The table lookups are mechanically very easy. They are simple one-dimensional lookups: to find !!T_1(6)!! you just look at entry !!6!! in the !!T_1!! table, which can be implemented as a series of ten metal rods of different lengths, or something like that. Looking things up in a multiplication table is harder because it is two-dimensional. The single addition in Ludgate's algorithm can also be performed mechanically: to add !!T_1(a)!! and !!T_1(b)!!, you have some thingy that slides up by !!T_1(a)!! units, and then by !!T_1(b)!! more, and then wherever it ends up is used to index into !!T_2!! to get the answer. The !!T_2!! table doesn't have to be calculated on the fly, it can be made up ahead of time, and machined from brass or something, and incorporated directly into the machine. (It's tempting to say “hardcoded”.) The tables look a little uncouth at first but it is not hard to figure out what is going on. First off, !!T_1!! is the inverse of !!T_2!! in the sense that $$T_2(T_1(n)) = n\tag{$\color{darkgreen}{\spadesuit}$}$$ whenever !!n!! is in range — that is when !!0≤ n ≤ 9!!. !!T_2!! is more complex. We must construct it so that $$T_2(T_1(a) + T_1(b)) = ab.\tag{$\color{purple}{\clubsuit}$}$$ for all !!a!! and !!b!! of interest, which means that !!0\le a, b\le 9!!. If you look over the table you should see that the entry !!n!! is often followed by !!2n!!. That is, !!T_2(i+1) = 2T_2(i)!!, at least some of the time. In fact, this is true in all the cases we care about, where !!2n = ab!! for some single digits !!a, b!!. The second row could just as well have started with !!24, 48, 96, 192!!, but Ludgate doesn't need the !!96, 192!! entries, so he made them zero, which really means “I don't care”. This will be important later. The algorithm says that if we want to compute !!2n!!, we should compute $$ \begin{align} 2n & = T_2(T_1(2) + T_1(n)) && \text{Because $\color{purple}{\clubsuit}$} \\ & = T_2(1 + T_1(n)) \\ & = 2T_2(T_1(n)) && \text{Because moving one space right doubles the value}\\ & = 2n && \text{Because $\color{darkgreen}{\spadesuit}$} \end{align} $$ when !!0≤n≤9!!. I formatted !!T_2!! in rows of !!10!! because that makes it easy to look up examples like !!T_2(35) = 28!!, and because that's how Wikipedia did it. But this is very misleading, and not just because it makes !!T_2!! appear to be a !!10×10!! table when it's really a vector. !!T_2!! is actually more like a compressed version of a !!7×4×3×3!! table. Let's reformat the table so that the rows have length !!7!! instead of !!10!!: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 5, & 10, & 20, & 40, & 0, \\ {\tiny\color{gray}{28}} & 81, & 0, & 15, & 30, & 0, & 7, & 14, \\ {\tiny\color{gray}{35}} & 28, & 56, & 45, & 0, & 0, & 21, & 42, \\ {\tiny\color{gray}{42}} & 0, & 0, & 0, & 0, & 25, & 63, & 0, \\ {\tiny\color{gray}{49}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{56}} & 35, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{63}} & 0, & 0, & 0, & 49 \\ \end{array} $$ We have already seen that moving one column right usually multiplies the entry by !!2!!. Similarly, moving down by one row is seen to triple the !!T_2!! value — not always, but in all the cases of interest. Since the rows have length !!7!!, moving down one row from !!T_2(i)!! gets you to !!T_2(i+7)!!, and this is why !!T_1(3) = 7!!: to compute !!3n!!, one does: $$ \begin{align} 3n & = T_2(T_1(3) + T_1(n)) && \text{Because $\color{purple}{\clubsuit}$} \\ & = T_2(7 + T_1(n)) \\ & = 3T_2(T_1(n)) && \text{Because moving down triples the value}\\ & = 3n && \text{Because $\color{darkgreen}{\spadesuit}$} \end{align} $$ Now here is where it gets clever. It would be straightforward easy to build !!T_2!! as a stack of !!5×7!! tables, with each layer in the stack having entries quintuple the layer above, like this: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{28}} & 81, & 0, & 0, & 0, & 0, & 0, & 0, \\ \\ {\tiny\color{gray}{35}} & 5, & 10, & 20, & 40, & 0, & 0, & 0, \\ {\tiny\color{gray}{42}} & 15, & 30, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{49}} & 45, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{56}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{63}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ \\ {\tiny\color{gray}{70}} & 25, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{77}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{84}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{91}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{98}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ \end{array} $$ This works, if we make !!T_1(5)!! the correct offset, which is !!7·5 = 35!!. But it wastes space, and the larger !!T_2!! is, the more complicated and expensive is the brass thingy that encodes it. The last six entries of the each layer in the stack are don't-cares, so we can just omit them: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{28}} & 81, \\ \\ {\tiny\color{gray}{29}} & 5, & 10, & 20, & 40, & 0, & 0, & 0, \\ {\tiny\color{gray}{36}} & 15, & 30, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{43}} & 45, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{50}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{57}} & 0, \\ \\ {\tiny\color{gray}{58}} & 25, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{65}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{72}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{79}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{86}} & 0\hphantom{,} \\ \end{array} $$ And to compensate we make !!T_1(5) = 29!! instead of !!35!!: you now move down one layer in the stack by skipping !!29!! entries forward, instead of !!35!!. The table is still missing all the multiples of !!7!!, but we can repeat the process. The previous version of !!T_2!! can now be thought of as a !!29×3!! table, and we can stack another !!29×3!! table below it, with all the entries in the new layer being !!7!! times the original one: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{28}} & 81, \\ \\ {\tiny\color{gray}{29}} & 5, & 10, & 20, & 40, & 0, & 0, & 0, \\ {\tiny\color{gray}{36}} & 15, & 30, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{43}} & 45, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{50}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{57}} & 0, \\ \\ {\tiny\color{gray}{58}} & 25, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{65}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{72}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{79}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{86}} & 0, \\ \\ \hline \\ {\tiny\color{gray}{87}} & 7, & 14, & 28, & 56, & 0, & 0, & 0, \\ {\tiny\color{gray}{94}} & 21, & 42, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{101}} & 63, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{108}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{115}} & 0, \\ \\ {\tiny\color{gray}{116}} & 35, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{123}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{130}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{137}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{144}} & 0, \\ \\ {\tiny\color{gray}{145}} & 0, & 0, & 0, & 0, & 0, & 0, & \ldots \\ \\ \hline \\ {\tiny\color{gray}{174}} & 49\hphantom{,} \\ \end{array} $$ Each layer in the stack has !!29·3 = 87!! entries, so we could take !!T_1(7) = 87!! and it would work, but the last !!28!! entries in every layer are zero, so we can discard those and reduce the layers to !!59!! entries each. $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{28}} & 81, \\ \\ {\tiny\color{gray}{29}} & 5, & 10, & 20, & 40, & 0, & 0, & 0, \\ {\tiny\color{gray}{36}} & 15, & 30, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{43}} & 45, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{50}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{57}} & 0, \\ \\ {\tiny\color{gray}{58}} & 25, \\ \\ \hline \\ {\tiny\color{gray}{59}} & 7, & 14, & 28, & 56, & 0, & 0, & 0, \\ {\tiny\color{gray}{66}} & 21, & 42, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{73}} & 63, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{80}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{87}} & 0, \\ \\ {\tiny\color{gray}{88}} & 35, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{95}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{102}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{109}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{116}} & 0, \\ \\ {\tiny\color{gray}{117}} & 0, \\ \\ \hline \\ {\tiny\color{gray}{118}} & 49\hphantom{,} \\ \end{array} $$ Doing this has reduced the layers from !!87!! to !!59!! elements each, but Ludgate has another trick up his sleeve. The last few numbers in the top layer are !!45, 25,!! and a lot of zeroes. If he could somehow finesse !!45!! and !!25!!, he could trim the top two layers all the way back to only 38 entries each: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{28}} & 81, \\ \\ {\tiny\color{gray}{29}} & 5, & 10, & 20, & 40, & 80, & 0, & 0, \\ {\tiny\color{gray}{36}} & 15, & 30, \\ \\ \hline \\ {\tiny\color{gray}{38}} & 7, & 14, & 28, & 56, & 0, & 0, & 0, \\ {\tiny\color{gray}{45}} & 21, & 42, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{52}} & 63, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{59}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{66}} & 0, \\ \\ {\tiny\color{gray}{67}} & 35, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{74}} & 0, & 0, \\ \hline \\ {\tiny\color{gray}{76}} & 49\hphantom{,} \\ \end{array} $$ We're now missing !!25!! and we need to put it back. Fortunately the place we want to put it is !!T_1(5) + T_1(5) = 29+29 = 58!!, and that slot contains a zero anyway. And similarly we want to put !!45!! at position !!14+29 = 43!!, also empty: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{28}} & 81, \\ \\ {\tiny\color{gray}{29}} & 5, & 10, & 20, & 40, & 0, & 0, & 0, \\ {\tiny\color{gray}{36}} & 15, & 30, \\ \\ \hline \\ {\tiny\color{gray}{38}} & 7, & 14, & 28, & 56, & 0, & \color{purple}{45}, & 0, \\ {\tiny\color{gray}{45}} & 21, & 42, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{52}} & 63, & 0, & 0, & 0, & 0, & 0, & \color{purple}{25}, \\ {\tiny\color{gray}{59}} & 0, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{66}} & 0, \\ \\ {\tiny\color{gray}{67}} & 35, & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{74}} & 0, & 0, \\ \\ \hline \\ {\tiny\color{gray}{76}} & 49\hphantom{,} \\ \end{array} $$ The arithmetic pattern is no longer as obvious, but property !!\color{purple}{\clubsuit}!! still holds We're not done yet! The table still has a lot of zeroes we can squeeze out. If we change !!T_1(5)!! from !!29!! to !!23!!, the !!5,10,20,40!! group will slide backward to just after the !!54!!, and the !!15, 30!! will move to the row below that. We will also have to move the other multiples of !!5!!. The !!5!! itself moved back by six entries, and so did everything after that in the table, including the !!35!! (from position !!32+29!! to !!32+23!!) and the !!45!! (from position !!14+29!! to !!14+23!!) so those are still in the right places. Note that this means that !!7!! has moved from position !!38!! to position !!32!!, so we now have !!T_1(7) = 32!!. But the !!25!! is giving us trouble. It needed to move back twice as far as the others, from !!29+29 = 58!! to !!23+23 = 46!!, and unfortunately it now collides with !!63!! which is currently at position !!7+7+32 = 46!!. $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & \color{purple}{5}, & \color{purple}{10}, & \color{purple}{20}, & \color{purple}{40}, & 0, \\ {\tiny\color{gray}{28}} & 81, & 0, & \color{purple}{15} & \color{purple}{30}, \\ \\ \hline \\ {\tiny\color{gray}{32}} & 7, & 14, & 28, & 56, & 0, & \color{darkgreen}{45}, & 0, \\ {\tiny\color{gray}{39}} & 21, & 42, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{46}} & {63\atop\color{darkred}{¿25?}} & 0, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{53}} & 0, & 0, & \color{darkgreen}{35}, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{60}} & 0, & 0, & 0, & 0, \\ \\ \hline \\ {\tiny\color{gray}{64}} & 49\hphantom{,} \\ \end{array} $$ We need another tweak to fix !!25!!. !!7!! is currently at position !!32!!. We can't move !!7!! any farther back to the left without causing more collisions. But we can move it forward, and if we move it forward by one space, the !!63!! will move up one space also and the collision with !!25!! will be solved. So we insert a zero between !!30!! and !!7!!, which moves up !!7!! from position !!32!! to !!33!!: $$ \begin{array}{rrrrrrrr} {\tiny\color{gray}{0}} & 1, & 2, & 4, & 8, & 16, & 32, & 64, \\ {\tiny\color{gray}{7}} & 3, & 6, & 12, & 24, & 48, & 0, & 0, \\ {\tiny\color{gray}{14}} & 9, & 18, & 36, & 72, & 0, & 0, & 0, \\ {\tiny\color{gray}{21}} & 27, & 54, & 5, & 10, & 20, & 40, & 0, \\ {\tiny\color{gray}{28}} & 81, & 0, & 15 & 30, \\ \\ \hline \\ {\tiny\color{gray}{32}} & \color{purple}{0}, & \color{darkgreen}{7}, & \color{darkgreen}{14}, & \color{darkgreen}{28}, & \color{darkgreen}{56}, & 45, & 0, \\ {\tiny\color{gray}{39}} & 0, & \color{darkgreen}{21}, & \color{darkgreen}{42}, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{46}} & 25,& \color{darkgreen}{63}, & 0, & 0, & 0, & 0, & 0, \\ {\tiny\color{gray}{53}} & 0, & 0, & 0, & \color{darkgreen}{35}, & 0, & 0, & 0, \\ {\tiny\color{gray}{60}} & 0, & 0, & 0, & 0, \\ \\ \hline \\ {\tiny\color{gray}{64}} & \color{purple}{0}, & \color{purple}{0}, & \color{darkgreen}{49}\hphantom{,} \\ \end{array} $$ All the other multiples of !!7!! moved up by one space, but not the non-multiples !!25!! and !!45!!. Also !!49!! had to move up by two, but that's no problem at all, since it was at the end of the table and has all the space it needs. And now we are done! This is exactly Ludgate's table, which has the property that $$T_2(p + 7q + 23r + 33s) = 2^p3^q5^r7^s$$ whenever !!2^p3^q5^r7^s = ab!! for some !!0≤a,b≤9!!. Moving right by one space multiplies the !!T_2!! entry by !!2!!, at least for the entries we care about. Moving right by seven spaces multiplies the entry by !!3!!. To multiply by !!5!! or !!7!! we move right by or !!23!! or by !!33!!, respectively. These are exactly the values in the !!T_1!! table: $$\begin{align} T_1(2) & = 1\\ T_1(3) & = 7\\ T_1(5) & = 23\\ T_1(7) & = 33 \end{align}$$ The rest of the !!T_1!! table can be obtained by remembering !!\color{darkgreen}{\spadesuit}!!, that !!T_2(T_1(n)) = n!!, so for example !!T_1(6) = 8!! because !!T_2(8) = 6!!. Or we can get !!T_1(6)!! by multiplication, using !!\color{purple}{\clubsuit}!!: multiplying by !!6!! is the same as multiplying by !!2!! and then by !!3!!, which means you move right by !!1!! and then by !!7!!, for a total of !!8!!. Here's !!T_1!! again for reference: $$ \begin{array}{rl} T_1 = & \begin{array}{cccccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 50 & 0 & 1 & 7 & 2 & 23 & 8 & 33 & 3 & 14 \\ \end{array} \end{array} $$ (Actually I left out a detail: !!T_1(0) = 50!!. Ludgate wants !!T_2(T_1(0) + T_1(b)) = 0!! for all !!b!!. So we need !!T_2(T_1(0) + k) = 0!! for each !!k!! in !!T_1!!. !!T_1(0) = 50!! is the smallest value that works. This is rather painful, because it means that the !!66!!-item table aboveis not sufficient. Ludgate has to extend !!T_2!! all the way out to !!101!! items in order to handle the seemingly trivial case of !!0\cdot 0 = T_2(50 + 50)!!. But the last 35 entries are all zeroes, so the the brass widget probably doesn't have to be too much more complicated.) Wasn't that fun? A sort of mathematical engineering or a kind that has not been really useful for at least fifty years. But actually that was not what I planned to write about! (Did you guess that was coming?) I thought I was going to write this bit as a brief introduction to something else, but the brief introduction turned out to be 2500 words and a dozen complicated tables. We can only hope that part 2 is forthcoming. I promise nothing. [ Update 20231002: Rather than the ad-hoc backtracking approach I described here, one can construct !!T_1!! and !!T_2!! in a simpler and more direct way. Shreevatsa R. explains. ] [ Update 20231015: Part 2 has arrived! It discusses a different kind of all-integer logarithm called the “discrete” logarithm. ] [ Update 20231020: I think this is a clearer explanation of the discrete logarithm. Shorter, anyway. ] [Other articles in category /math] permanent link |