Suppose we have the set of floating point number with "m" bit mantissa and "e" bits for exponent. Suppose more over we want to approximate a function "f".
From the theory we know that usually a "range reduced function" is used and then from such function we derive the global function value.
For example let x = (sx,ex,mx) (sign exp and mantissa) then...
log2(x) = ex + log2(1.mx) so basically the range reduced function is "log2(1.mx)".
I have implemented at present reciprocal, square root, log2 and exp2, recently i've started to work with the trigonometric functions. But i was wandering if given a global error bound (ulp error especially) it is possible to derive an error bound for the range reduced function, is there some study about this kind of problem? Speaking of the log2(x) (as example) i would lke to be able to say...
"ok i want log2(x) with k ulp error, to achieve this given our floating point system we need to approximate log2(1.mx) with p ulp error"
Remember that as i said we know we are working with floating point number, but the format is generic, so it could be the classic F32, but even for example e=10, m = 8 end so on.
I can't actually find any reference that shows such kind of study. Reference i have (i.e. muller book) doesn't treat the topic in this way so i was looking for some kind of paper or similar. Do you know any reference?
I'm also trying to derive such bound by myself but it is not easy...
There is a description of current practice, along with a proposed improvement and an error analysis, at https://hal.inria.fr/ensl-00086904/document. The description of current practice appears consistent with the overview at https://docs.oracle.com/cd/E37069_01/html/E39019/z4000ac119729.html, which is consistent with my memory of the most talked about problem being the mod pi range reduction of trigonometric functions.
I think IEEE floating point was a big step forwards just because it standardized things at a time when there were a variety of computer architectures, so lowering the risks of porting code between them, but the accuracy requirements implied by this may have been overkill: for many problems the constraint on the accuracy of the output is the accuracy of the input data, not the accuracy of the calculation of intermediate values.
Related
I am trying to calculate the Mean Squared Error in Vitis HLS. I am using hls::pow(...,2) and divide by n, but all I receive is a negative value for example -0.004. This does not make sense to me. Could anyone point the problem out or have a proper explanation for this??
Besides calculating the mean squared error using hls::pow does not give the same results as (a - b) * (a - b) and for information I am using ap_fixed<> types and not normal float or double precision
Thanks in advance!
It sounds like an overflow and/or underflow issue, meaning that the values reach the sign bit and are interpreted as negative while just be very large.
Have you tried tuning the representation precision or the different saturation/rounding options for the fixed point class? This tuning will depend on the data you're processing.
For example, if you handle data that you know will range between -128.5 and 1023.4, you might need very few fractional bits, say 3 or 4, leaving the rest for the integer part (which might roughly be log2((1023+128)^2)).
Alternatively, if n is very large, you can try a moving average and calculate the mean in small "chunks" of length m < n.
p.s. Getting the absolute value of a - b and store it into an ap_ufixed before the multiplication can already give you one extra bit, but adds an instruction/operation/logic to the algorithm (which might not be a problem if the design is pipelined, but require space if the size of ap_ufixed is very large).
I seem to see people asking all the time around here questions about comparing floating-point numbers. The canonical answer is always: just see if the numbers are within some small number of each other…
So the question is this: Why would you ever need to know if two floating point numbers are equal to each other?
In all my years of coding I have never needed to do this (although I will admit that even I am not omniscient). From my point of view, if you are trying to use floating point numbers for and for some reason want to know if the numbers are equal, you should probably be using an integer type (or a decimal type in languages that support it). Am I just missing something?
A few reasons to compare floating-point numbers for equality are:
Testing software. Given software that should conform to a precise specification, exact results might be known or feasibly computable, so a test program would compare the subject software’s results to the expected results.
Performing exact arithmetic. Carefully designed software can perform exact arithmetic with floating-point. At its simplest, this may simply be integer arithmetic. (On platforms which provide IEEE-754 64-bit double-precision floating-point but only 32-bit integer arithmetic, floating-point arithmetic can be used to perform 53-bit integer arithmetic.) Comparing for equality when performing exact arithmetic is the same as comparing for equality with integer operations.
Searching sorted or structured data. Floating-point values can be used as keys for searching, in which case testing for equality is necessary to determine that the sought item has been found. (There are issues if NaNs may be present, since they report false for any order test.)
Avoiding poles and discontinuities. Functions may have special behaviors at certain points, the most obvious of which is division. Software may need to test for these points and divert execution to alternate methods.
Note that only the last of these tests for equality when using floating-point arithmetic to approximate real arithmetic. (This list of examples is not complete, so I do not expect this is the only such use.) The first three are special situations. Usually when using floating-point arithmetic, one is approximating real arithmetic and working with mostly continuous functions. Continuous functions are “okay” for working with floating-point arithmetic because they transmit errors in “normal” ways. For example, if your calculations so far have produced some a' that approximates an ideal mathematical result a, and you have a b' that approximates an ideal mathematical result b, then the computed sum a'+b' will approximate a+b.
Discontinuous functions, on the other hand, can disrupt this behavior. For example, if we attempt to round a number to the nearest integer, what happens when a is 3.49? Our approximation a' might be 3.48 or 3.51. When the rounding is computed, the approximation may produce 3 or 4, turning a very small error into a very large error. When working with discontinuous functions in floating-point arithmetic, one has to be careful. For example, consider evaluating the quadratic formula, (−b±sqrt(b2−4ac))/(2a). If there is a slight error during the calculations for b2−4ac, the result might be negative, and then sqrt will return NaN. So software cannot simply use floating-point arithmetic as if it easily approximated real arithmetic. The programmer must understand floating-point arithmetic and be wary of the pitfalls, and these issues and their solutions can be specific to the particular software and application.
Testing for equality is a discontinuous function. It is a function f(a, b) that is 0 everywhere except along the line a=b. Since it is a discontinuous function, it can turn small errors into large errors—it can report as equal numbers that are unequal if computed with ideal mathematics, and it can report as unequal numbers that are equal if computed with ideal mathematics.
With this view, we can see testing for equality is a member of a general class of functions. It is not any more special than square root or division—it is continuous in most places but discontinuous in some, and so its use must be treated with care. That care is customized to each application.
I will relate one place where testing for equality was very useful. We implement some math library routines that are specified to be faithfully rounded. The best quality for a routine is that it is correctly rounded. Consider a function whose exact mathematical result (for a particular input x) is y. In some cases, y is exactly representable in the floating-point format, in which case a good routine will return y. Often, y is not exactly representable. In this case, it is between two numbers representable in the floating-point format, some numbers y0 and y1. If a routine is correctly rounded, it returns whichever of y0 and y1 is closer to y. (In case of a tie, it returns the one with an even low digit. Also, I am discussing only the round-to-nearest ties-to-even mode.)
If a routine is faithfully rounded, it is allowed to return either y0 or y1.
Now, here is the problem we wanted to solve: We have some version of a single-precision routine, say sin0, that we know is faithfully rounded. We have a new version, sin1, and we want to test whether it is faithfully rounded. We have multiple-precision software that can evaluate the mathematical sin function to great precision, so we can use that to check whether the results of sin1 are faithfully rounded. However, the multiple-precision software is slow, and we want to test all four billion inputs. sin0 and sin1 are both fast, but sin1 is allowed to have outputs different from sin0, because sin1 is only required to be faithfully rounded, not to be the same as sin0.
However, it happens that most of the sin1 results are the same as sin0. (This is partly a result of how math library routines are designed, using some extra precision to get a very close result before using a few final arithmetic operations to deliver the final result. That tends to get the correctly rounded result most of the time but sometimes slips to the next nearest value.) So what we can do is this:
For each input, calculate both sin0 and sin1.
Compare the results for equality.
If the results are equal, we are done. If they are not, use the extended precision software to test whether the sin1 result is faithfully rounded.
Again, this is a special case for using floating-point arithmetic. But it is one where testing for equality serves very well; the final test program runs in a few minutes instead of many hours.
The only time I needed, it was to check if the GPU was IEEE 754 compliant.
It was not.
Anyway I haven't used a comparison with a programming language. I just run the program on the CPU and on the GPU producing some binary output (no literals) and compared the outputs with a simple diff.
There are plenty possible reasons.
Since I know Squeak/Pharo Smalltalk better, here are a few trivial examples taken out of it (it relies on strict IEEE 754 model):
Float>>isFinite
"simple, byte-order independent test for rejecting Not-a-Number and (Negative)Infinity"
^(self - self) = 0.0
Float>>isInfinite
"Return true if the receiver is positive or negative infinity."
^ self = Infinity or: [self = NegativeInfinity]
Float>>predecessor
| ulp |
self isFinite ifFalse: [
(self isNaN or: [self negative]) ifTrue: [^self].
^Float fmax].
ulp := self ulp.
^self - (0.5 * ulp) = self
ifTrue: [self - ulp]
ifFalse: [self - (0.5 * ulp)]
I'm sure that you would find some more involved == if you open some libm implementation and check... Unfortunately, I don't know how to search == thru github web interface, but manually I found this example in julia libm (a variant of fdlibm)
https://github.com/JuliaLang/openlibm/blob/master/src/s_remquo.c
remquo(double x, double y, int *quo)
{
...
fixup:
INSERT_WORDS(x,hx,lx);
y = fabs(y);
if (y < 0x1p-1021) {
if (x+x>y || (x+x==y && (q & 1))) {
q++;
x-=y;
}
} else if (x>0.5*y || (x==0.5*y && (q & 1))) {
q++;
x-=y;
}
GET_HIGH_WORD(hx,x);
SET_HIGH_WORD(x,hx^sx);
q &= 0x7fffffff;
*quo = (sxy ? -q : q);
return x;
Here, the remainder function answer a result x between -y/2 and y/2. If it is exactly y/2, then there are 2 choices (a tie)... The == test in fixup is here to test the case of exact tie (resolved so as to always have an even quotient).
There are also a few ==zero tests, for example in __ieee754_logf (test for trivial case log(1)) or __ieee754_rem_pio2 (modulo pi/2 used for trigonometric functions).
After a call to sincos(x,&s,&c) from the unix m math library it would be natural to get the tangent as s/c. Is this safe or there may be (ill) cases in which the (supposedly) more expensive tan(x) should be preferred due to precision issues?
If the error in sin(x) is es, and the error in cos(x) is ec, then the error in t=(sin(x)/cos(x) is
et = abs(t)*sqrt((es/sin(x))^2 + (ec/cos(x))^2)
The error in sin, cos and tan should be right around the precision of the number representation times the value, perhaps a bit or two off. Since the error will track with the precision of the number, the above equation reduces to
et = sqrt(es^2 + ec^2)
and es and ec should be close to each other. So, the total error from using sin/cos as opposed to a new calculation of the tangent should be about a factor of sqrt(2) greater than the error from calculating the tangent directly.
That assumes the error in sin, cos and tan are all about the same, but that's a pretty good assumption these days. There is a small wobble in error over the range of the function, but that should be handled using some extended precision bits during the calculation, and should not show up significantly in the values you see.
Whether this matters, of course, depends on the situation.
See http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm for a decent quick-and-dirty intro to error propagation, which is a good way to tackle this sort of question.
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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.]
I am interested in use or created an script to get error rounding reports in algorithms.
I hope the script or something similar is already done...
I think this would be usefull for digital electronic system design because sometimes it´s neccesary to study how would be the accuracy error depending of the number of decimal places that are considered in the design.
This script would work with 3 elements, the algorithm code, the input, and the output.
This script would show the error line by line of the algorithm code.
It would modify the algorith code with some command like roundn and compare the error of the output.
I would define the error as
Errorrounding = Output(without rounding) - Output round
For instance I have the next algorithm
calculation1 = input*constan1 + constan2 %line 1 of the algorithm
output = exp(calculation1) %line 2 of the algorithm
Where 'input' is the input of n elements vector and 'output' is the output and 'constan1' and 'constan2' are constants.
n is the number of elements of the input vector
So, I would put my algorithm in the script and it generated in a automatic way the next algorithm:
input_round = roundn(input,-1*mdec)
calculation1 = input*constant1+constant2*ones(1,n)
calculation1_round = roundn(calculation1,-1*mdec)
output=exp(calculation1_round)
output_round= roundn(output,-1*mdec)
where mdec is the number of decimal places to consider.
Finally the script give the next message
The rounding error at line 1 is #Errorrounding_calculation1
Where '#Errorrounding' would be the result of the next operation Errorrounding_calculation1 = calculation1 - calculation1_round
The rounding error at line 2 is #Errorrounding_output
Where 'Errorrounding_output' would be the result of the next operation Errorrounding_output = output - output_round
Does anyone know if there is something similar already done, or Matlab provides a solution to deal with some issues related?
Thank you.
First point: I suggest reading What Every Computer Scientist Should Know About Floating-Point Arithmetic by David Goldberg. It should illuminate a lot of issues regarding floating-point computations that will help you understand more of the intricacies of the problem you are considering.
Second point: I think the problem you are considering is a lot more complicated than you realize. You are interested in the error introduced into a calculation due to the reduced precision from rounding. What you don't realize is that these errors will propagate through your computations. Consider your example:
output = input*C1 + C2
If each of the three operands is a double-precision floating-point number, they will each have some round-off error in their precision. A bound on this round-off error can be found using the function EPS, which tells you the distance from one double-precision number to the next largest one. For example, a bound on the relative error of the representation of input will be 0.5*eps(input), or halfway between it and the next largest double-precision number. We can therefore estimate some errors bounds on the three operands as follows:
err_input = 0.5.*eps(input); %# Maximum round-off error for input
err_C1 = 0.5.*eps(C1); %# Maximum round-off error for C1
err_C2 = 0.5.*eps(C2); %# Maximum round-off error for C2
Note that these errors could be positive or negative, since the true number may have been rounded up or down to represent it as a double-precision value. Now, notice what happens when we estimate the true value of the operands before they were rounded-off by adding these errors to them, then perform the calculation for output:
output = (input+err_input)*(C1+err_C1) + C2+err_C2
%# ...and after reordering terms
output = input*C1 + C2 + err_input*C1 + err_C1*input + err_input*err_C1 + err_C2
%# ^-----------^ ^-----------------------------------------------------^
%# | |
%# rounded computation difference
You can see from this that the precision round-off of the three operands before performing the calculation could change the output we get by as much as difference. In addition, there will be another source of round-off error when the value output is rounded off to represent it as a double-precision value.
So, you can see how it's quite a bit more complicated than you thought to adequately estimate the errors introduced by precision round-off.
This is more of an extended comment than an answer:
I'm voting to close this on the grounds that it isn't a well-formed question. It sort of expresses a hope or wish that there exists some type of program which would be interesting or useful to you. I suggest that you revise the question to, well, to be a question.
You propose to write a Matlab program to analyse the numerical errors in other Matlab programs. I would not use Matlab for this. I'd probably use Mathematica, which offers more sophisticated structural operations on strings (such as program source text), symbolic computation, and arbitrary precision arithmetic. One of the limitations of Matlab for what you propose is that Matlab, like all other computer implementations of real arithmetic, suffers rounding errors. There are other languages which you might choose too.
What you propose is quite difficult, and would probably require a longer answer than most SOers, including this one, would be happy to contemplate writing. Happily for you, other people have written books on the subject, I suggest you start with this one by NJ Higham. You might also want to investigate matters such as interval arithmetic.
Good luck.