## Math TricksBack sometime around freshman or sophomore year of high school I was having some troubles. I would occasionally see some hexadecimal, octal, or binary and not really understand it at all. I mean I'm sure I understood bit operations and using logical operators to set bit flags and the like. But I didn't understand the binary numbering system or how to convert between bases. Then one day I saw a program on the discovery channel about computers. In it they discussed binary and powers of two. My little world suddenly got bigger. Just the right neurons went off in my head. The next thing I know, I'm calculating the number of nodes in a full tree and the number of nodes at the base. Then I wrote a program to convert between bases (from 2 to 36). You can find it on my projects page. Unfortunately, it was built using an old Borland compiler for windows 3.1. Borland is now Inprise. And windows 3.1 is extinct as far as I'm concerned. Anyway, it was great because I now understood all this stuff. And this along with my curiosity about register variables, incidentally, is what led me to learn about x86 assembly, logic design, boolean algebra, and different numbering systems. It's too bad I didn't have an assembler. It also allowed me to start thinking about powers of two and the neat math tricks you can do with it. One day I came up with this. (I'm sure there's a name for it, but I don't know what it is.) If you take a binary number that's all ones and add one more, all the ones become zero and you get a single one at a higher decimal place. So, 01111 +00001 ------ 10000 What this says in powers of two is that 2^0+2^1+2^2+2^3+2^4 + 1 == 2^5. In other words, 2^N - 1 == sum(2^n,n=0..N-1), where N is the number of decimals. Hm, well that's kind of neat. Then I pushed a little further. Let's say I try this in base 4. I want to do basically the same thing, but I'll have to multiply by 3 before I add one. 03333 +00001 ------ 10000 What this says in powers of four is that 3*(4^0+4^1+4^2+4^3+4^4) + 1 == 4^5. In other words, m^N - 1 ------- == sum(m^n,n=0..N-1) m - 1 Here m is the base. Is this useful? I have no idea. But it's certainly not obvious. If someone had put that right under your nose, you probably would have no idea where it came from. But I didn't just whip it up out of nowhere. I started with a simple, concrete binary example. Then I generalized to an arbitrary number of bits. Then I generalized to an arbitrary base. With each generalization, I've got a more powerful little formula. What's really cool is this works with real numbers too (m doesn't have to be an integer, but N does). In fact, it also works with complex numbers. If you're wondering why, just multiply both sides by (m-1), and the answer will be obvious. OK, let's try another. This one actually is useful. When I was a freshman in high school, I was writing a basic program to solve some optimization problems. At some point I needed a way to sum integers from 1 to N. Now this seems really easy, but I didn't like the idea that it was going to be O(n). It seemed there must be a way to do it closed form in O(1). There is actually an interesting story I was told about this. There's this guy named Gauss. Perhaps you've heard of him? Maybe you've heard of Gaussian elimination? He's responsible for a lot of other math too. Smart guy. In fact, he had such a reputation that Napoleon supposedly spared his little town as he took over Europe. When Gauss was six, he was getting in his teacher's hair; so, she told him to go sum the integers from one to one-hundred. He came back only a few minutes later with the answer. Now Gauss was a little smarter than me. I think it took me a few hours to figure out how to do this. But it's pretty simple. If you want to find a formula to do it, just count the number of 100s. That is, you've got one to start, but you can make another by adding 1 and 99 and another by adding 2 and 98 and so on. After you've counted them, multiply the number you counted by 100 to get the total sum. Now generalize. I'll just give it to you. If n is the number you want to sum to, then n*(n+1) ------- == sum(k,k=1..n) 2 As I said, this actually does come in handy. A few years later, I noticed my calculus book also had formulas for the summation of i^2 and i^3. That is, sum(i^3,i=1..n). You can't just sit down and think of a quick way to come up with a closed form solution for anything greater than i^1, or at least I couldn't. But it is relatively easy to find these formulas. I realized all you have to do is apply some simple algebra techniques and solve a matrix equation. Basically, for i^2, you have to evaluate the summation for several points. You can then get a system of equations and solve using standard matrix techniques (like Gaussian elimination). For i^3, it's basically the same thing, but you need one more equation, and so on. So I wrote a little Java program to do this. After I solved for several of them, I realized you can find these formulas recursively. I don't know why it works, but it's very easy to find a closed form solution for sum(i^m,i=0..N) for any positive integer, m, extremely quickly. There is a program for it on my projects page. OK, just one more. This is a little trick I discovered while working on my master's degree. The DSP I was using has a circular addressing mode. You can set properties of special registers so that they use modular arithmetic; they wrap around to the beginning or end (of a buffer) when you reach the modulus or before zero. Let's say your modulus is 5. Then the following are true. Say the == as congruent. 1 + 1 == 2 2 + 2 == 4 4 + 4 == 3 0 + 5 == 0 (i.e., 5 == 0) 1 + 4 == 0 -1 == 4 == 9 Under modulus 5, the last three integers, -1, 4, and 9 are congruent. They're equivalent under modular arithmetic. Let's say you have an array or buffer of length 5, elements 0 to 4. When you do processing in real-time, it's too expensive to be moving memory around. When you fill the buffer, you want to leave what's already there and start overwriting data at the front (or back depending on what you're doing). Suppose you're averaging three elements at a time. Then, you'll want to do the following. (buf[0] + buf[1] + buf[2]) / 3 (buf[1] + buf[2] + buf[3]) / 3 (buf[2] + buf[3] + buf[4]) / 3 (buf[3] + buf[4] + buf[0]) / 3 (buf[4] + buf[0] + buf[1]) / 3 (buf[0] + buf[1] + buf[2]) / 3 (buf[1] + buf[2] + buf[3]) / 3 Notice how I wrapped right around back to the start? I'm skipping over a lot of details. You aren't averaging the same numbers every cycle of 5. By the time you wrap back around, there is new data placed in the start of the buffer. For example, DMA (Direct Memory Access) might be employed to keep the buffer fresh with new data. In this way you can process a stream of never ending input. And it's incredibly fast. There are no conditional branches, just a jump to the beginning of a never ending loop. Now the problem. You need to calculate those indices. Many languages have a modulus operator. But it's not just some simple arithmetic to find the index. Underneath, there's some pretty heavy stuff going on. You could use an if statement to detect the boundaries. if(ind > (5-1)) ind = ind - 5; if(ind < 0) ind = ind + 5; This is bad for several reasons. (1) With conditional branches, you can turn a nice tight loop running at 2 instructions per cycle (.5 CPI) into something more like .2 instructions per cycle (20 CPI) in a highly pipelined architecture. (2) When you're writing lots of numerical code, constantly checking the index after every calculation is going to make the code hard to read. (3) The code above assumes the index never exceeds 9 or -5. (4) Doing comparisons is ugly and requires temporary variables. For example, which is greater, 3 or -1? The trick I discovered, and I suspect this is how it's implemented in hardware, is to use a bitwise AND. First, restrict your modulus to a power of 2 -- let's say 8. By ANDing the index with the modulus less one, the wrapping will be done correctly with a single instruction. (3 + 3) & (8 - 1) == 6 (4 + 4) & (8 - 1) == 0 (7 + 2) & (8 - 1) == 1 (0 + 8) & (8 - 1) == 0 (i.e., 8 == 0) (1 + 8) & (8 - 1) == 1 0111 +0010 ----- 1001 1001 &0111 ----- 0001 The binary arithmetic is the third equality done as an example. This also works with negative numbers in the twos complement system. Try it! And it doesn't matter how large the index gets, it will always work. In a highly pipelined, superscalar architecture, that AND may come completely free. The only down side is you need the array to be of length, power of 2, but this is a fairly common theme in DSP. Don't let that stop you from using circular buffers in non-DSP code. If I have intrigued you at all with this math, you might also want to study up on twos complement. There's probably more involved than you understand. For example, you might have known that you can negate a number with a bitwise NOT (flip the bits) and then adding one. But did you know you can find the base ten number in almost the same way as you do with a positive number? 0101 == 0*2^3 + 1*2^2 + 0*2^1 + 1*2^0 == 5 1011 == -1*2^3 + 0*2^2 + 1*2^1 + 1*2^0 == -5 You might want to study why that works. Read up on 10s, 9s, 2s, and 1s complement. I hope you have gotten out of this the process of coming up with these little tricks. Start with something concrete and easy to understand. Try a few examples. Once you get it, generalize just a little. Then you have a little formula you can play with. Think about what it means, and see if you can generalize a little more. This is one of the traits that made Newton so brilliant. From a few very simple examples he could generalize to amazingly complicated math. Now you try. See if you can mathematically describe the number of powers of ten in a power of two from just the exponent. So 2^0, 2^1, 2^2, 2^3 would all be zero tens. And 2^4, 2^5, 2^6 would be one ten. So f(0) == 0, f(1) == 0, f(2) == 0, f(3) == 0, f(4) == 1, f(5) == 1, f(6) == 1. Make sure you see a pattern before you try to come up with any mathematical description. And by the way, don't ask me for the answer. I don't know it. I just remember coming up with something once a long time ago. |