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Is Multiplying the Inverse Better or Worse?
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Closed 7 years ago.
Sometimes I encounter in arithmetic operations expression like this: n*(1/k).
Such expression can be presented in simpler manner: n/k.
I could imagine that in certain situations the former could be more descriptive if (1/k) represents well known ingredient but it is not always the case.
What about performance gains/losses? What about precision?
Is there any hidden reason that some developers use n*(1/k) form?
To me i find that using n*(1/k) will be having a lesser accurate answer because of the reason that when the control solves an produces the result of (1/k) the will be situations that may cause rounding off or trimming of the result that may lead to the loss of accuracy.And during the multiplication process the magnitude of the loss increases.Hence as far as i am concerned i will say n/k is better
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I am making some software that need to work with integers.
Also I need to apply some formula to those integers, repeatedly over time (example, do x/=z several times in a row for a indefinite amount).
All tools, algorithms and formulas I could think or find, or don't work with integers at all, or work as approximations at best.
For example the x/=z several times in a row for example, you can theoretically calculate what x will be in the 10th time by doing x = x/(z^10), but that will be wrong if the result is fractional, you can use floor(x/(z^10)), but the result will STILL be wrong.
Plotting software that I found also don't have integers at all, or has "floor()/ceil()" functions support, at best, and still the result would fall in the problem of the previous paragraph.
So how I do it?
Here's something to get you going for the iteration of x/=z:
(that should have ended in "all three terms are 0 with regard to integer division")
Now if x or z are negative, you can try and see whether this still holds; I did not invest the time to make the necessary case distinctions, but they should be fairly analogous.
As Karoly Horvath mentions in a comment, without a clear specification of the kinds of functions for which you would like to find a shortcut to replace iterative evaluation, helping you out won't be possible since there are uncountably many functions over the integers, and the same approach won't work for all of them.
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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.
Recently I realized I have been doing too much branching without caring the negative impact on performance it had, therefore I have made up my mind to attempt to learn all about not branching. And here is a more extreme case, in attempt to make the code to have as little branch as possible.
Hence for the code
if(expression)
A = C; //A and C have to be the same type here obviously
expression can be A == B, or Q<=B, it could be anything that resolve to true or false, or i would like to think of it in term of the result being 1 or 0 here
I have come up with this non branching version
A += (expression)*(C-A); //Edited with thanks
So my question would be, is this a good solution that maximize efficiency?
If yes why and if not why?
Depends on the compiler, instruction set, optimizer, etc. When you use a boolean expression as an int value, e.g., (A == B) * C, the compiler has to do the compare, and the set some register to 0 or 1 based on the result. Some instruction sets might not have any way to do that other than branching. Generally speaking, it's better to write simple, straightforward code and let the optimizer figure it out, or find a different algorithm that branches less.
Jeez, no, don't do that!
Anyone who "penalize[s] [you] a lot for branching" would hopefully send you packing for using something that awful.
How is it awful, let me count the ways:
There's no guarantee you can multiply a quantity (e.g., C) by a boolean value (e.g., (A==B) yields true or false). Some languages will, some won't.
Anyone casually reading it is going observe a calculation, not an assignment statement.
You're replacing a comparison, and a conditional branch with two comparisons, two multiplications, a subtraction, and an addition. Seriously non-optimal.
It only works for integral numeric quantities. Try this with a wide variety of floating point numbers, or with an object, and if you're really lucky it will be rejected by the compiler/interpreter/whatever.
You should only ever consider doing this if you had analyzed the runtime properties of the program and determined that there is a frequent branch misprediction here, and that this is causing an actual performance problem. It makes the code much less clear, and its not obvious that it would be any faster in general (this is something you would also have to measure, under the circumstances you are interested in).
After doing research, I came to the conclusion that when there are bottleneck, it would be good to include timed profiler, as these kind of codes are usually not portable and are mainly used for optimization.
An exact example I had after reading the following question below
Why is it faster to process a sorted array than an unsorted array?
I tested my code on C++ using that, that my implementation was actually slower due to the extra arithmetics.
HOWEVER!
For this case below
if(expression) //branched version
A += C;
//OR
A += (expression)*(C); //non-branching version
The timing was as of such.
Branched Sorted list was approximately 2seconds.
Branched unsorted list was aproximately 10 seconds.
My implementation (whether sorted or unsorted) are both 3seconds.
This goes to show that in an unsorted area of bottleneck, when we have a trivial branching that can be simply replaced by a single multiplication.
It is probably more worthwhile to consider the implementation that I have suggested.
** Once again it is mainly for the areas that is deemed as the bottleneck **
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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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Regex: Determine if two regular expressions could match for the same input?
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Closed 10 years ago.
I have two regexps. I need to determine if it is possible to build string of given length that matches these two regexps simultaneously. I need algorithm to do that.
String's length wouldn't exceed 20 characters.
It depends. For perl compatible regular expressions (pcre), this is not generally possible, as they are turing complete: you cannot even be sure that matching always terminates.
For the original, "clean" form of reguler languages as defined in the Chomsky-hierarchy, it is known that they are closed under intersection, this is already discussed in this thread.
As soon as you have the NFA for the intersection, it is easy to check whether any string matches it - if thera is a path from the start to the end of your NFA, then the string for this path is the string you are searching for, for DFAs, an algorithm is given here, it should be simple to adapt it to NFAs.