I'm trying to write a basic digit counter (an integer is inputted and the number of digits of that integer is outputted) for positive integers. This is my general formula:
dig(x) := Math.floor(Math.log(x,10))
I tried implementing the equivalent of dig(x) in Ruby, and found that when I was computing dig(1000) I was getting 2 instead of 3 because Math.log was returning 2.9999999999999996 which would then be truncated down to 2. What is the proper way to handle this problem? (I'm assuming this problem can occur regardless of the language used to implement this approach, but if that's not the case then please explain that in your answer).
To get an exact count of the number of digits in an integer, you can do the usual thing: (in C/C++, assuming n is non-negative)
int digits = 0;
while (n > 0) {
n = n / 10; // integer division, just drops the ones digit and shifts right
digits = digits + 1;
}
I'm not certain but I suspect running a built-in logarithm function won't be faster than this, and this will give you an exact answer.
I thought about it for a minute and couldn't come up with a way to make the logarithm-based approach work with any guarantees, and almost convinced myself that it is probably a doomed pursuit in the first place because of floating point rounding errors, etc.
From The Art of Computer Programming volume 2, we will eliminate one bit of error before the floor function is applied by adding that one bit back in.
Let x be the result of log and then do x += x / 0x10000000 for a single precision floating point number (C's float). Then pass the value into floor.
This is guaranteed to be the fastest (assuming you have the answer in numerical form) because it uses only a few floating point instructions.
Floating point is always subject to roundoff error; that's one of the hazards you need to be aware of, and actively manage, when working with it. The proper way to handle it, if you must use floats is to figure out what the expected amount of accumulated error is and allow for that in comparisons and printouts -- round off appropriately, compare for whether the difference is within that range rather than comparing for equality, etcetera.
There is no exact binary-floating-point representation of simple things like 1/10th, for example.
(As others have noted, you could rewrite the problem to avoid using the floating-point-based solution entirely, but since you asked specifically about working log() I wanted to address that question; apologies if I'm off target. Some of the other answers provide specific suggestions for how you might round off the result. That would "solve" this particular case, but as your floating operations get more complicated you'll have to continue to allow for roundoff accumulating at each step and either deal with the error at each step or deal with the cumulative error -- the latter being the more complicated but more accurate solution.)
If this is a serious problem for an application, folks sometimes use scaled fixed point instead (running financial computations in terms of pennies rather than dollars, for example). Or they use one of the "big number" packages which computes in decimal rather than in binary; those have their own round-off problems, but they round off more the way humans expect them to.
Related
I have a program which calculates probability values
(p-values),
but it is entering a very large negative number into the
exp function
exp(-626294.830) which evaluates to zero instead of the very small
positive number that it should be.
How can I get this to evaluate as a very small floating point number?
I have tried
Math::BigFloat,
bignum, and
bigrat
but all have failed.
Wolfram Alpha says that exp(-626294.830) is 4.08589×10^-271997... zero is a pretty close approximation to that ;-) Although you've edited and removed the context from your question, do you really need to work with such tiny numbers, or perhaps there is some way you could optimize your algorithm or scale your numbers?
Anyway, you are correct that code like Math::BigFloat->new("-626294.830")->bexp seems to take quite some time, even with the support of use Math::BigFloat lib => 'GMP';.
The only alternative I can offer at the moment is Math::Prime::Util::GMP's expreal, although you need to specify a precision to it.
use Math::Prime::Util::GMP qw/expreal/;
use Math::BigFloat;
my $e = Math::BigFloat->new(expreal(-626294.830,272000));
print $e->bnstr,"\n";
__END__
4.086e-271997
But on my machine, even that still takes ~20s to run, which brings us back to the question of potential optimization in other places.
Floating point numbers do not have infinite precision. Assuming the number is represented as an IEEE 754 double, we have 52 bits for a fraction, 11 bits for the exponent, and one bit for the sign. Due to the way exponents are encoded, the smallest positive number that can be represented is 2^-1022.
If we look at your number e^-626294.830, we can do a change of base and see that it equals 2^(log_2 e · -626294.830) = 2^-903552.445, which is significantly smaller than 2^-1022. Approximating your number as zero is therefore correct.
Instead of calculating this value using arbitrary-precision numerics, you are likely better off solving the necessary equations by hand, then coding this in a way that does not require extreme precision. For example, it is unlikely that you need the exact value of e^-626294.830, but perhaps just the magnitude. Then, you can calculate the logarithm instead of using exp().
I've got:
SELECT x(point), y(point) WHERE x(point) = 3.69334468807005
x and y are of type double precision.
I see that this value is in the table indeed, however running the query in PostgreSQL does not return anything. Why could that be the case? Maybe due to a precision problem?
Thanks!
When dealing with floating point numbers (both single or double precision) then doing an exact compare is futile in 99% of all cases. This is true not only for PostgreSQL but for all computer languages using FP arithmetic.
The three reasons are, that the internal representation of a double can contain much more digits than displayed and that at the same time many numbers cannot be expressed exactly using FP (0.1 is an often cited example) and that therefore all "displayed" values are truncated to something a human can comprehend (i.e. nothing like "0.099999999999999999999999999" instead of "0.1").
Therefore it is necessary to to avoid direct comparison as soon as one of the numbers to be compared has been calculated (rounding errors) or has been converted from a string. Instead some "range" must be admitted like
where x between 3.69334468807004 and 3.69334468807006 -- note the different numbers
The only valid cases for direct comparison are cases where the value has been just copied previously. A fictive example would be:
SELECT x, y, f1(x,y), f2(x,y), ... INTO TEMP temp_xy FROM points;
SELECT * FROM points p JOIN temp_xy t on p.x = t.x and p.y = t.y;
x and y have been just copied, therefore they can be used as a join criteria.
Edit A good starter for this and some more non-intuitive problems with floats is this article.
Old school answer: "Don't compare floating point numbers solely for equality." (Elements of Programming Style, Kernighan and Plauger, 1978)
Why? Because comparing two floats for equality will always work under certain circumstances, but it will almost never work under slightly different circumstances. That's due to the nature of floating-point numbers, not to programmer skill.
The canonical article for floating-point math is What Every Computer Scientist Should Know About Floating-Point Arithmetic.
In your case, you might be able to adapt the relative difference function from this C language FAQ. (Scroll down, look for RelDif().)
You could certainly test if it is a precision problem, just expand the WHERE clause of your statement to be a range, and tighten that range (by adding more precision) until you have your record or can confirm it is related to precision:
SELECT x(point), y(point)
WHERE x(point) > 3.69
AND x(point) < 3.70
The other thing I would look at is perhaps using some other form of key when filtering your data. Does your table have some sort of natural key you could use or maybe just add an auto-incremented field for use a primary key?
I have also seen indexes behave badly when functions are involved. Are there any indexes on this table?
I compute this simple sum on Matlab:
2*0.04-0.5*0.4^2 = -1.387778780781446e-017
but the result is not zero. What can I do?
Aabaz and Jim Clay have good explanations of what's going on.
It's often the case that, rather than exactly calculating the value of 2*0.04 - 0.5*0.4^2, what you really want is to check whether 2*0.04 and 0.5*0.4^2 differ by an amount that is small enough to be within the relevant numerical precision. If that's the case, than rather than checking whether 2*0.04 - 0.5*0.4^2 == 0, you can check whether abs(2*0.04 - 0.5*0.4^2) < thresh. Here thresh can either be some arbitrary smallish number, or an expression involving eps, which gives the precision of the numerical type you're working with.
EDIT:
Thanks to Jim and Tal for suggested improvement. Altered to compare the absolute value of the difference to a threshold, rather than the difference.
Matlab uses double-precision floating-point numbers to store real numbers. These are numbers of the form m*2^e where m is an integer between 2^52 and 2^53 (the mantissa) and e is the exponent. Let's call a number a floating-point number if it is of this form.
All numbers used in calculations must be floating-point numbers. Often, this can be done exactly, as with 2 and 0.5 in your expression. But for other numbers, most notably most numbers with digits after the decimal point, this is not possible, and an approximation has to be used. What happens in this case is that the number is rounded to the nearest floating-point number.
So, whenever you write something like 0.04 in Matlab, you're really saying "Get me the floating-point number that is closest to 0.04. In your expression, there are 2 numbers that need to be approximated: 0.04 and 0.4.
In addition, the exact result of operations like addition and multiplication on floating-point numbers may not be a floating-point number. Although it is always of the form m*2^e the mantissa may be too large. So you get an additional error from rounding the results of operations.
At the end of the day, a simple expression like yours will be off by about 2^-52 times the size of the operands, or about 10^-17.
In summary: the reason your expression does not evaluate to zero is two-fold:
Some of the numbers you start out with are different (approximations) to the exact numbers you provided.
The intermediate results may also be approximations of the exact results.
What you are seeing is quantization error. Matlab uses doubles to represent numbers, and while they are capable of a lot of precision, they still cannot represent all real numbers because there are an infinite number of real numbers. I'm not sure about Aabaz's trick, but in general I would say there isn't anything you can do, other than perhaps massaging your inputs to be double-friendly numbers.
I do not know if it is applicable to your problem but often the simplest solution is to scale your data.
For example:
a=0.04;
b=0.2;
a-0.2*b
ans=-6.9389e-018
c=a/min(abs([a b]));
d=b/min(abs([a b]));
c-0.2*d
ans=0
EDIT: of course I did not mean to give a universal solution to these kind of problems but it is still a good practice that can make you avoid a few problems in numerical computation (curve fitting, etc ...). See Jim Clay's answer for the reason why you are experiencing these problems.
I'm pretty sure this is a case of ye olde floating point accuracy issues.
Do you need 1e-17 accuracy? Is this merely a case of wanting 'pretty' output?
In that case, you can just use a formatted sprintf to display the number of significant digits you want.
Realize that this is not a matlab problem, but a fundamental limitation of how numbers are represented in binary.
For fun, work out what .1 is in binary...
Some references:
http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems
http://www.mathworks.com/support/tech-notes/1100/1108.html
Just asked by my 5 year old kid: what is the biggest number in the computer?
We are not talking about max number for a specific data types, but the biggest number that a computer can represent.
Infinity is not allowed.
UPDATE my kid always wants to print as
well, so lets say the computer needs
to print this number and the kid to
know that its a big number. Of course,
in practice we won't print because
theres not enough trees.
This question is actually a very interesting one which mathematicians have devoted a fair bit of thought to. You can read about it in this article, which is a fascinating and accessible read.
Briefly, a guy named Tibor Rado set out to find some really big, but still well-defined, numbers by defining a sequence called the Busy Beaver numbers. He defined BB(n) to be the largest number of steps any Turing Machine could take before halting, given an input of n symbols. Note that this sequence is by its very nature not computable, so the numbers themselves, while well-defined, are very difficult to pin down. Here are the first few:
BB(1) = 1
BB(2) = 6
BB(3) = 21
BB(4) = 107
... wait for it ...
BB(5) >= 8,690,333,381,690,951
No one is sure how big exactly BB(5) is, but it is finite. And no one has any idea how big BB(6) and above are. But at least these numbers are completely well-defined mathematically, unlike "the largest number any human has ever thought of, plus one." ;)
So how about this:
The biggest number a computer can represent is the most instructions a program small enough to fit in its available memory can perform before halting.
Squared.
No, wait, cubed. No, raised to the power of itself!
Dammit!
Bits are not numbers. You, as a programmer, give them the meaning you want, possibly numbers.
Now, I decide that 1 represents "the biggest number ever thought by a human plus one".
Errr this is a five year old?
How about something along the lines of: "I'd love to tell you but the number is so big and would take so long to say, I'd die before I finished telling you".
// wait to see
for(;;)
{
printf("9");
}
roughly 2^AVAILABLE_MEMORY_IN_BITS
EDIT: The above is for actually storing a number and treats all media (RAM, HD, cloud etc.) as memory. Subtracting the OS footprint (measured in KB) doesn't make "roughly" less accurate...
If you want to "represent" a number in a meaningful way, then you probably want to go with what the CPU provides: unsigned 32 bit integers (roughly 4 Gigs) or unsigned 64 bit integers for most computers your kid will come into contact with.
NOTE for talking to 5-year-olds: Often, they just want a factoid. Give him a really big and very accurate number (lots of digits), like 4'294'967'295. Then, once the glazing leaves his eyes, try to see how far you can get with explaining how computers represent numbers.
EDIT #2: I once read this article: Who Can Name the Bigger Number that should provide a whole lot of interesting information for your kid. Obviously he's not your normal five-year-old. So this might get you started in a cool direction about numbers and computation.
The answer to life (and this kids question): 42
That depends on the datatype you use to represent it. The computer only stores bits (0/1). We, as developers, give the bits meaning. (65 can be a number or the letter A).
For example, I can define my datatype as 1^N where N is unsigned and represented by an array of bits of arbitrary size. The next person can come up with 10^N which would be ten times larger than my biggest number.
Sure, there would be gaps but if you don't need them, that doesn't matter.
Therefore, the question is meaningless since it doesn't have context.
Well I had the same question earlier this day, so thought why not to make a little c++ codes to see where the computer gonna stop ...
But my laptop wasn't with me in class so I used another, well the number was to big but it never ends, i'll run it again for a night then i'll share the number
you can try the code is stupid
#include <stdlib.h>
#include <stdio.h>
int main() {
int i = 0;
for (i = 0; i <= i; i++) {
printf("%i\n", i);
i++;
}
}
And let it run till it stops ^^
The size will obviously be limited by the total size of hard drives you manage to put into your PC. After all, you can store a number in a text file occupying all disk space.
You can have 4x2Tb drives even in a simple box so around 8Tb available. if you store as binary, then the biggest number is 2 pow 64000000000000.
If your hard drive is 1 TB (8'000'000'000'000 bits), and you would print the number that fits on it on paper as hex digits (nobody would do that, but let's assume), that's 2,000,000,000,000 hex digits.
Each page would contain 4000 hex digits (40 x 100 digits). That's 500,000,000 pages.
Now stack the pages on top of each other (let's say each page is 0.004 inches / 0.1 mm thick), then the stack would be as 5 km (about 3 miles) tall.
I'll try to give a practical answer.
Common Lisp number crunching is particularly powerful. It has something called "bignums" which are integers that can be arbitrarily large, limited by the amount of available.
See: http://en.wikibooks.org/wiki/Common_Lisp/Advanced_topics/Numbers#Fixnums_and_Bignums
Don't know much about theory, but I far as I understood from your question, is: what is the largest number that the computer can represent (and I add: in a reasonable time, and not printing "9" until the Earth will "be eaten by the Sun"). And I put my PC to make one simple calculation (in PHP or whatever language): echo pow(2,1023) - resulting: 8.9884656743116E+307. So I guess this is the largest number that my PC can calculate. On the other side, I think the respresentation of the largest negative number can be: -0,(0)1
LE: That computed value was obataind through PHP, but I tried to figure out what's the largest number that my windows calculator can compute, and it is pow(2, 33219) = 8.2304951207588748764521361245002E+9999. Now I guess this is the largest number my PC can handle.
I think you should be very proud that your 5 year old is already asking questions like this.
And you should continue to promote that! This is truly amazing! With that said, I would say that saying Infinity does not
count is thinking incorrectly about what numbers mean in computer memory.
I feel like this way of thinking is a handicap.
Mathematicians will never be able to write out ALL the digits of pi or eulers number, BUT we FULLY understand it.
Pi, as an example, is perfectly represented by infinite this series: (Pi / 4) = 1 - 1/3 + 1/5 - 1/7 + 1/9 - …
Just because you literally can’t go to inf. or print every single digit in a console means nothing.
You could have printed the symbol representing pi and therefore capturing the inf. series.
Computer Algebra Systems (CAS) represent numbers symbolically all the time. Pi, for instance,
may be a Symbolic object in memory (the binary in memory did not DIRECTLY represent the number. It represents an "mathematical algorithm" for producing the answer to arbitrary precision).
Then you do some math with it, transforming from one expression to the next.
At no point in time did we not represent the number COMPLETELY.
At the end, you can do 2 things with this:
A) Evaluate the expression, turning it into a number of some kind (or Matrix or whatever).
BUT this number could very well be an approximation (say like 20 digits of pi).
B) Keep it in its symbolic form for reference. Obviously we don’t like staring at symbols because we
need to eventually turn the nobs on the apparatii.
NOTE: sometimes you can get a finite (non-irrational) number perfectly represented in memory (like number 1)
by taking limits or going to inf. Not literally having an inf. number in memory, but symbolically representing it.
Just throw this in Wolfram alpha: Lim[Exp[-x], x --> Inf]; It gives you the number 0. Which is EXACT.
In short:
It was the HUMANS need to have some binary in memory that DIRECTLY represented the number that caused
the number to degrade. Symbolically it was perfectly represented. You could design some algorithm that
just continues to calculate the next digits of pi or eulers number giving you an arbitrary amount of precision (Now, this is obviously not practical of course).
I hope this was at least somewhat useful or interesting to you, even if you disagree =)
Depends on how much the computer can handle. Although there are some times when the computer can handle numbers greater than (2^(bits-1)-1)... For example:
My computer is 64 bit (9223372036854775807), however the calculator that comes with the computer itself can handle numbers of up to 10^9999.
Many other supercomputers can exceed these limits, and the one with the most memory (bits) might as well be the one with the record (current largest number that can be held by computers).
Or, if it comes to visually seeing it on computers, you can just make a program that, on monitor, repeats writing 9 and not skips that line to form an ever-growing bunch of 9. :P
go on chrome then go on three dots above and click them then go on tools and then go on developer tool click on console and type Number.MAX_VALUE
I have some inputs on my site representing floating point numbers with up to ten precision digits (in decimal). At some point, in the client side validation code, I need to compare a couple of those values to see if they are equal or not, and here, as you would expect, the intrinsics of IEEE754 make that simple check fails with things like (2.0000000000==2.0000000001) = true.
I may break the floating point number in two longs for each side of the dot, make each side a 64 bit long and do my comparisons manually, but it looks so ugly!
Any decent Javascript library to handle arbitrary (or at least guaranteed) precision float numbers on Javascript?
Thanks in advance!
PS: A GWT based solution has a ++
There is the GWT-MATH library at http://code.google.com/p/gwt-math/.
However, I warn you, it's a GWT jsni overlay of a java->javascript automated conversion of java.BigDecimal (actually the old com.ibm.math.BigDecimal).
It works, but speedy it is not. (Nor lean. It will pad on a good 70k into your project).
At my workplace, we are working on a fixed point simple decimal, but nothing worth releasing yet. :(
Use an arbitrary precision integer library such as silentmatt’s javascript-biginteger, which can store and calculate with integers of any arbitrary size.
Since you want ten decimal places, you’ll need to store the value n as n×10^10. For example, store 1 as 10000000000 (ten zeroes), 1.5 as 15000000000 (nine zeroes), etc. To display the value to the user, simply place a decimal point in front of the tenth-last character (and then cut off any trailing zeroes if you want).
Alternatively you could store a numerator and a denominator as bigintegers, which would then allow you arbitrarily precise fractional values (but beware – fractional values tend to get very big very quickly).