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Recently,I have been trying to understand how the Binary Extended Euclidean Algorithm works at the processor level. This question is all about finding an Inverse element in GF(2^m) with polynomial basis.
Generally I came across the Extended Euclidean Algorithm for evaluating an inverse element but the fact is that it involves too many addition and multiplication operations. The Binary EEA algorithm requires just bit shifting operations (equivalent to division by 2--logical shift right). The algorithm is in this link, page number 8.
In step 3 and 5 of this algorithm, every iteration shifts the parameters u and b by 1 bit to the right adding zero to the MSB at the same time. The loop ends when u == 1 and returns b. My question is how many primitive operations does a processor (say a 32 bit processor for example) perform in step 3 or step 5 of every iteration?
I came across barrel shifter and I am quite confused about how fast the shifting takes place. Should I really consider these primitive operations or should I ignore them if because the shifting may be faster?
It would really help me a lot if someone would show the primitive operations for the case where the size of u is 194 bits.
In case you might be wondering about the denominator x in step 3 and 5 of the algorithm, its the polynomial representation and x means nothing but 10 in binary and parameter u is an N-bit binary number.
There is no generic answer to this question: you can use portable code that will be tedious to optimize or highly machine specific code that will be even more complicated to optimize without breaking.
If you want real performance, you have to use MMX/AVX registers on the maximum width you can get your hands on. Intel provides lightweight wrappers on low-level instructions as macros and inline functions.
Always use unsigned types for your shifting operations to avoid unnecessary steps.
Usually ther is a "right shift" assembly OP code which is able to right shift a register a given number of bits. Such an operation takes one cycle.
This assumes thet your value is already loaded to the register however.
The best answer anyway: Implement this algorithm in a low level language (C, C++) and look at the assembly code produced by the compiler.
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Example:
int max = a > b ? a : b;
int min = a + b - max;
What determines whether this will work? The processor? The hardware? The language? Help me understand this at as deep a level as possible.
The processor IS the hardware (at least for the purposes of this question).
The language is purely a way for you to express things in such a way as to allow it to convert it to what the processor itself expects. The role of the language here would be to define what "int" means, what arithmetic operators are/do, and what their exceptional behavior is. In the low-level languages (like C/C++), it leaves several things to be "implementation defined", like the overflow behavior of integers. Other languages (like Python) may define "int" to be an abstract (not a hardware) concept and thereby change some of the rules (like detecting overflows and doing custom behavior).
If the language leaves something implementation defined and the implementation offloads that decision to the hardware, then the hardware is what defines the behavior of your code.
The high level programming language provides a way for humans to describe what they want to happen. A compiler reduces that down into a language the processor understands, (ultimately) machine code. The instruction set for a particular processor is designed to be useful for doing tasks, general purpose processors for general purpose tasks including the ones you have described. Unlike pencil and paper math where if we need another column another power of ten, 99+1 = 100 for example two digits wide going in, 3 digits coming put. Processors have a fixed with for their registers, that doesnt mean you cant get creative, but the language and the resources (memory, disk space, etc) have limits. And the processor either directly in the logic or the compiler implementing the right sequence of instructions, can and will detect an overflow if you ask it to, in general. Some processors harder than others and some processors are not general purpose enough, but I dont think we need to worry about those, the one you are reading this web page in definitely can handle this.
Computers(hardware) represent numbers in two's complement. Check this for details of two's complement, and why computers use it.
In two's complement signed numbers(not floating ones for now, for sake of simplicity) have a sign bit as most significant bit. For example:
01111111
Represents 127 in two's complement. And
10000000
represents -128. In both example, the first bit is sign bit, if it's 0, then the number is positive, else negative.
8-bit signed numbers can represent numbers between -128 and 127, so if you add 127 and 3, you won't get 130, you will get -126 because of overflow. Let's see why:
01111111
00000011
+________
10000010 which is a negative number, -126 in two's complement.
How hardware understand if an overflow occurred? In addition for example, if you add two positive numbers and the result gets negative, it means overflow. And if you add two negative numbers and result gets positive it means overflow again.
I hope that would be a nice example for how these things are happening in hardware level.
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im new to assembly language and i know many codes.
However im working with 8086 emulator with only works with 16 bit numbers.
this is a home work that im really stuck in :
How can i write an assembly code with do the following :
1-get 20 , maximum 6-digits decimal numbers and store them in an array.
2- sort the array in ascending order.
its really hard for me to understand how to manage registers and stack for this long numbers.
every help will be appreciated in advance .
In order to sort 32bit numbers (or broader) with 16bit registers you have to compare the upper part of each number separately.
Assume we have these random two 32 bit numbers (shown in hex) 4567afdf and 321abc09.
Now when you look at them as 16 bit values they look like this:
4567 afdf
321a bc09
As you can easily see, the upper 16 bits you can compare individually.
If the upper 16 bits are higher or lower, then you know that the lower part doesn't matter anymore and you sort them accordingly.
If the upper 16 bits are equal, then you compare the lower 16 bits and if they are also equal, both numbers are equal => no sort needed, otherwise you shuffle them accordingly. Since the upper 16 bits are also equal, you don't even need to shuffle them.
If the upper 16bits are different, you still have to shuffle the lower 16 bits accordingly, as they might be different.
The basics of this approach can be used for an arbitrary number of bits not just 32bits. Generally when you have a seemingly hard problem, you should try to think of the easy examples and how you can solve it. Then you can extend it to more complicated cases.
EDIT:
An alternative approach would be, if you have strings of decimal numbers and you want to sort them based on the string representation instead of the numbers.
In this case, you can do it as follows
If the length of the two number strings are differnt, the shorter one is the lower number.
if the length is equal, then you can look at each digit individually (starting with the first digit) until you hit a non-equal digit or the string end. If you reaced the end of the string, the numbers are the same, otherwise you kn ow which one is higher/lower.
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I have a math problem that I can't solve: I don't know how to find the value of n so that
365! / ((365-n)! * 365^n) = 50%.
I am using the Casio 500ms scientific calculator but I don't know how.
Sorry because my question is too easy, I am changing my career so I have to review and upgrade my math, the subject that I have neglected for years.
One COULD in theory use a root-finding scheme like Newton's method, IF you could take derivatives. But this function is defined only on the integers, since it uses factorials.
One way out is to recognize the identity
n! = gamma(n+1)
which will effectively allow you to extend the function onto the real line. The gamma function is defined on the positive real line, though it does have singularities at the negative integers. And of course, you still need the derivative of this expression, which can be done since gamma is differentiable.
By the way, a danger with methods like Newton's method on problems like this is it may still diverge into the negative real line. Choose poor starting values, and you may get garbage out. (I've not looked carefully at the shape of this function, so I won't claim for what set of starting values it will diverge on you.)
Is it worth jumping through the above set of hoops? Of course not. A better choice than Newton's method might be something like Brent's algorithm, or a secant method, which here will not require you to compute the derivative. But even that is a waste of effort.
Recognizing that this is indeed a problem on the integers, one could use a tool like bisection to resolve the solution extremely efficiently. It never requires derivatives, and it will work nicely enough on the integers. Once you have resolved the interval to be as short as possible, the algorithm will terminate, and take vary few function evaluations in the process.
Finally, be careful with this function, as it does involve some rather large factorials, which could easily overflow many tools to evaluate the factorial. For example, in MATLAB, if I did try to evaluate factorial(365):
factorial(365)
ans =
Inf
I get an overflow. I would need to move into a tool like the symbolic toolbox, or my own suite of variable precision integer tools. Alternatively, one could recognize that many of the terms in these factorials will cancel out, so that
365! / (365 - n)! = 365*(365-1)*(365-2)*...*(365-n+1)
The point is, we get an overflow for such a large value if we are not careful. If you have a tool that will not overflow, then use it, and use bisection as I suggested. Here, using the symbolic toolbox in MATLAB, I get a solution using only 7 function evaluations.
f = #(n) vpa(factorial(sym(365))/(factorial(sym(365 - n))*365^sym(n)));
f(0)
ans =
1.0
f(365)
ans =
1.4549552156187034033714015903853e-157
f(182)
ans =
0.00000000000000000000000095339164972764493041114884521295
f(91)
ans =
0.000004634800180846641815683109605743
f(45)
ans =
0.059024100534225072005461014516788
f(22)
ans =
0.52430469233744993108665513602619
f(23)
ans =
0.49270276567601459277458277166297
Or, if you can't take an option like that, but do have a tool that can evaluate the log of the gamma function, AND you have a rootfinder available as MATLAB does...
f = #(n) exp(gammaln(365+1) - gammaln(365-n + 1) - n*log(365));
fzero(#(n) f(n) - .5,10)
ans =
22.7677
As you can see here, I used the identity relating gamma and the factorial function, then used the log of the gamma function, in MATLAB, gammaln. Once all the dirty work was done, then I exponentiated the entire mess, which will be a reasonable number. Fzero tells us that the cross-over occurs between 22 and 23.
If a numerical approximation is ok, ask Wolfram Alpha:
n ~= -22.2298272...
n ~= 22.7676903...
I'm going to assume you have some special reason for wanting an actual algorithm, even though you only have one specific problem to solve.
You're looking for a value n where...
365! / ((365-n)! * 365^n) = 0.5
And therefore...
(365! / ((365-n)! * 365^n)) - 0.5 = 0.0
The general form of the problem is to find a value x such that f(x)=0. One classic algorithm for this kind of thing is the Newton-Raphson method.
[EDIT - as woodchips points out in the comment, the factorial is an integer-only function. My defence - for some problems (the birthday problem among them) it's common to generalise using approximation functions. I remember the Stirling approximation of factorials being used for the birthday problem - according to this, Knuth uses it. The Wikipedia page for the Birthday problem mentions several approximations that generalise to non-integer values.
It's certainly bad that I didn't think to mention this when I first wrote this answer.]
One problem with that is that you need the derivative of that function. That's more a mathematics issue, though you can estimate the derivative at any point by taking values a short distance either side.
You can also look at this as an optimisation problem. The general form of optimisation problems is to find a value x such that f(x) is maximised/minimised. In your case, you could define your function as...
f(x)=((365! / ((365-n)! * 365^n)) - 0.5)^2
Because of the squaring, the result can never be negative, so try to minimise. Whatever value of x gets you the smallest f(x) will also give you the result you want.
There isn't so much an algorithm for optimisation problems as a whole field - the method you use depends on the complexity of your function. However, this case should be simple so long as your language can cope with big numbers. Probably the simplest optimisation algorithm is called hill-climbing, though in this case it should probably be called rolling-down-the-hill. And as luck would have it, Newton-Raphson is a hill-climbing method (or very close to being one - there may be some small technicality that I don't remember).
[EDIT as mentioned above, this won't work if you need an integer solution for the problem as actually stated (rather than a real-valued approximation). Optimisation in the integer domain is one of those awkward issues that helps make optimisation a field in itself. The branch and bound is common for complex functions. However, in this case hill-climbing still works. In principle, you can even still use a tweaked version of Newton-Raphson - you just have to do some rounding and check that you don't keep rounding back to the same place you started if your moves are small.]
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I was asked the above question in an interview and interviewer is very certain of the answer. But i am not sure. Can anyone help me here?
Sure. The obvious brute force method is just a big lookup table with one entry for every possible value of the input number. That's not very practical if the number is very big, but is still enough to prove it's possible.
Edit: the notion has been raised that this is complete nonsense, and the same could be said of essentially any algorithm.
To a limited degree, that's a fair statement -- but the limitations are so severe that for most algorithms it remains utterly meaningless.
My original point (at least as well as I remember it) was that population counting is about equivalent to many other operations like addition and subtraction that we normally assume are O(1).
At the hardware level, circuitry for a single-cycle POPCNT instruction is probably easier than for a single-cycle ADD instruction. Just for one example, for any practical size of data word, we can use table lookups on 4-bit chunks in parallel, then add the results from those pieces together. Even using fairly unlikely worst-case assumptions (e.g., separate storage for each of those tables) this would still be easy to implement in a modern CPU -- in fact, it's probably at least somewhat simpler than the single-cycle addition or subtraction mentioned above1.
This is a decided contrast to many other algorithms. For one obvious example, let's consider sorting. For even the most trivial sort most people can imagine -- 2 items, 8 bits apiece, we're already at a 64 kilobyte lookup table to get constant complexity. Long before we can do even a fairly trivial sort (e.g., 100 items) we need a lookup table that contains far more data items than there are atoms in the universe.
Looking at it from the opposite direction, it's certainly true that at some point, essentially nothing is O(1) any more. Let's consider the most trivial operations possible. For an N-bit CPU, bitwise OR is normally implemented as a set of N OR gates in parallel. Unlike addition, there's no interaction between one bit and another, so for any practical size of CPU, this easy to execute in a single instruction.
Nonetheless, if I specify a bit-wise OR in which each operand is 100 petabits, there's nothing even approaching a practical way to do the job with constant complexity. Using the usual method of parallel OR gates, we end up with (among other things) 300 petabits worth of input and output lines -- a number that completely dwarfs even the number of pins on the largest CPUs.
On reasonable hardware, doing a bitwise OR on 100 petabit operands is going to take a while (not to mention quite a bit of hard drive space). If we increase that to 200 petabit operands, the time is likely to (about) double -- so from that viewpoint, it's an O(N) operation. Obviously enough, the same is going to be true with the other "trivial" operations like addition, subtraction, bit-wise AND, bit-wise XOR, and so on.
Nonetheless, unless you have very specific instructions to say you're going to be dealing with utterly immense operands, you're typically going to treat every one of these as a constant complexity operation. Looked at in these terms, a POPCNT instruction falls about halfway between bit-wise AND/OR/XOR on one hand, and addition/subtraction on the other, in terms of the difficulty to execute in fixed time.
1. You might wonder how it could possibly be simpler than an add when it actually includes an add after doing some other operations. If so, kudos -- it's an excellent question.
The answer is that it's because it only needs a much smaller adder. For example, a 64-bit CPU needs one half-adder and 63 full-adders. In the simple implementation, you carry out the addition bit-wise -- i.e., you add bit-0 of one operand to bit-0 of the other. That generates an output bit, and a carry bit. That carry bit becomes an input to the addition for the next pair of bits. There are some tricks to parallelize that to some degree, but the nature of the beast (so to speak) is bit-serial.
With a POPCNT instruction, we have an addition after doing the individual table lookups, but our result is limited to the size of the input words. Given the same size of inputs (64 bits) our final result can't be any larger than 64. That means we only need a 6-bit adder instead of a 64-bit adder.
Since, as outlined above, addition is basically bit-serial, this means that the addition at the end of the POPCNT instruction is fundamentally a lot faster than a normal add. To be specific, it's logarithmic on the operand size, whereas simple addition is roughly linear on the operand size.
If the bit size is fixed (e.g. natural word size of a 32- or 64-bit machine), you can just iterate over the bits and count them directly in O(1) time (though there are certainly faster ways to do it). For arbitrary precision numbers (BigInt, etc.), the answer must be no.
Some processors can do it in one instruction, obviously for integers of limited size. Look up the POPCNT mnemonic for further details.
For integers of unlimited size obviously you need to read the whole input, so the lower bound is O(n).
The interviewer probably meant the bit counting trick (the first Google result follows): http://www.gamedev.net/topic/547102-bit-counting-trick---new-to-me/
I was asked this question in an interview.
If you had two numbers represented in the binary form and stored as a string. How would you perform simple addition. This was the easy part. (my solution: run through the shortest one and keep track of carry, repeat for the remaining)
The difficult part was when he asked me:
how would you use hardware to make the process faster.
Any suggestion SO community?
I'd say, convert them to proper integers, and use the hardware (ALU) to perform the addition, then convert the result back to a string if needed.
Converting the numbers to an integer variable and letting the CPU do the addition immediately springs to mind. You can then divide the number back into bits if you so choose to.