I'm aware that double is the default data-type in MATLAB.
When you compare two double numbers that have no floating part, MATLAB is accurate upto the 17th digit place in my testing.
a=12345678901234567 ; b=12345678901234567; isequal(a,b) --> TRUE
a=123456789012345671; b=123456789012345672; isequal(a,b) --> printed as TRUE
I have found a conservative estimate to be use numbers (non-floating) upto only 13th digit as other functions can become unreliable after it (such as ismember, or the MEX functions ismembc etc).
Is there a similar cutoff for floating values? E.g., if I use shares-outstanding for a company which can be very very large with decimal places, when do I start losing decimal accuracy?
a = 1234567.89012345678 ; b = 1234567.89012345679 ; isequal(a,b) --> printed as TRUE
a = 123456789012345.678 ; b = 123456789012345.677 ; isequal(a,b) --> printed as TRUE
isequal may not be right tool to use for comparing such numbers. I'm more concerned about up to how many places should I trust my decimal values once the integer part of a number starts growing?
It's usually not a good idea to test the equality of floating-point numbers. The behavior of binary floating-point numbers can differ drastically from what you may expect from base-10 decimals. Consider the example:
>> isequal(0.1, 0.3/3)
ans =
0
Ultimately, you have 53 bits of precision. This means that integers can be represented exactly (with no loss in accuracy) up to the number 253 (which is a little over 9 x 1015). After that, well:
>> (2^53 + 1) - 2^53
ans =
0
>> 2^53 + (1 - 2^53)
ans =
1
For non-integers, you are almost never going to be representing them exactly, even for simple-looking decimals such as 0.1 (as shown in that first example). However, it still guarantees you at least 15 significant figures of precision.
This means that if you take any number and round it to the nearest number representable as a double-precision floating point, then this new number will match your original number at least up to the first 15 digits (regardless of where these digits are with respect to the decimal point).
You might want to use variable precision arithmetics (VPA) in matlab. It computes expressions exactly up to a given digit count, which may be quite large. See here.
Check out the MATLAB function flintmax which tells you the maximum consecutive integers that can be stored in either double or single precision. From that page:
flintmax returns the largest consecutive integer in IEEE® double
precision, which is 2^53. Above this value, double-precision format
does not have integer precision, and not all integers can be
represented exactly.
Related
I was testing a operation like this:
[input] 3.9/0.1 : 4.1/0.1
[output] 39 40
don't know why 4.1/0.1 is approximated to 40. If I add a round(), it will go as expected:
[input] 3.9/0.1 : round(4.1/0.1)
[output] 39 40 41
What's wrong with the first operation?
In this Q&A I go into detail on how the colon operator works in MATLAB to create a range. But the detail that causes the issue described in this question is not covered there.
That post includes the full code for a function that imitates exactly what the colon operator does. Let's follow that code. We start with start = 3.9/0.1, which is exactly 39, and stop = 4.1/0.1, which, due to rounding errors, is just slightly smaller than 41, and step = 1 (the default if it's not given).
It starts by computing a tolerance:
tol = 2.0*eps*max(abs(start),abs(stop));
This tolerance is intended to be used so that the stop value, if within tol of an exact number of steps, is still used, if the last step would step over it. Without a tolerance, it would be really difficult to build correct sequences using floating-point end points and step sizes.
However, then we get this test:
if start == floor(start) && step == 1
% Consecutive integers.
n = floor(stop) - start;
elseif ...
If the start value is an exact integer, and the step size is 1, then it forces the sequence to be an integer sequence. Unfortunately, it does so by taking the number of steps as the distance between floor(stop) and start. That is, it is not using the tolerance computed earlier in determining the right stop! If stop is slightly above an integer, that integer will be in the range. If stop is slightly below an integer (as in the case of the OP), that integer will not be part of the range.
It could be debated whether MATLAB should round the stop number up in this case or not. MATLAB chose not to. All of the sequences produced by the colon operator use the start and stop values exactly as given by the user. It leaves it up to the user to ensure the bounds of the sequence are as required.
However, if the colon operator hadn't special-cased the sequence of integers, the result would have been less surprising in this case. Let's add a very small number to the start value, so it's not an integer:
>> a = 3.9/0.1 : 4.1/0.1
a =
39 40
>> b = 3.9/0.1 + eps(39) : 4.1/0.1
b =
39.0000 40.0000 41.0000
Floating-point numbers suffer from loss of precision when represented with a fixed number of bits (64-bit in MATLAB by default). This is because there are infinite number of real numbers (even within a small range of say 0.0 to 0.1). On the other hand, a n-bit binary pattern can represent a finite 2^n distinct numbers. Hence, not all the real numbers can be represented. The nearest approximation will be used instead, resulted in loss of accuracy.
The closest representable value for 4.1/0.1 in the computer as a 64-bit double precision floating point number is actually,
4.1/0.1 ≈ 40.9999999999999941713291207...
So, in essence, 4.1/0.1 < 41.0 and that's what you get from the range. If you subtract, for example, 41 - 4.1/0.1 = 7.105427357601002e-15. But when you round, you get the closest value of 41.0 as expected.
The representation scheme for 64-bit double-precision according to the IEEE-754 standard:
The most significant bit is the sign bit (S), with 0 for positive numbers and 1 for negative numbers.
The following 11 bits represent exponent (E).
The remaining 52 bits represents fraction (F).
The eps routine in MATLAB essentially returns the positive distance between floating point numbers. It can take an optional argument, too.
My question: How does MATLAB calculate this value? (Does it use a lookup table, or does it use some algorithm to calculate it at runtime, or something else...?)
Related: how could it be calculated in any language providing bit access, given a floating point number?
WIkipedia has quite the page on it
Specifically for MATLAB it's 2^(-53), as MATLAB uses double precision by default. Here's the graph:
It's one bit for the sign, 11 for the exponent and the rest for the fraction.
The MATLAB documentation on floating point numbers also show this.
d = eps(x), where x has data type single or double, returns the positive distance from abs(x) to the next larger floating-point number of the same precision as x.
As not all fractions are equally closely spaced on the number line, different fractions will show different distances to the next floating-point within the same precision. Their bit representations are:
1.0 = 0 01111111111 0000000000000000000000000000000000000000000000000000
0.9 = 0 01111111110 1100110011001100110011001100110011001100110011001101
the sign for both is positive (0), the exponent is not equal and of course their fraction is vastly different. This means that the next floating point numbers would be:
dec2bin(typecast(eps(1.0), 'uint64'), 64) = 0 01111001011 0000000000000000000000000000000000000000000000000000
dec2bin(typecast(eps(0.9), 'uint64'), 64) = 0 01111001010 0000000000000000000000000000000000000000000000000000
which are not the same, hence eps(0.9)~=eps(1.0).
Here is some insight into eps which will help you to write an algorithm.
See that eps(1) = 2^(-52). Now, say you want to compute the eps of 17179869183.9. Note that, I have chosen a number which is 0.1 less than 2^34 (in other words, something like 2^(33.9999...)). To compute eps of this, you can compute log2 of the number, which would be ~ 33.99999... as mentioned before. Take a floor() of this number and add it to -52, since eps(1) = 2^(-52) and the given number 2^(33.999...). Therefore, eps(17179869183.9) = -52+33 = -19.
If you take a number which is fractionally more than 2^34, e.g., 17179869184.1, then the log2(eps(17179869184.1)) = -18. This also shows that the eps value will change for the numbers that are integer powers of your base (or radix), in this case 2. Since eps value only changes at those numbers which are integer powers of 2, we take floor of the power. You will be able to get the perfect value of eps for any number using this. I hope it is clear.
MATLAB uses (along with other languages) the IEEE754 standard for representing real floating point numbers.
In this format the bits allocated for approximating the actual1 real number, usually 32 - for single or 64 - for double precision, are grouped into: 3 groups
1 bit for determining the sign, s.
8 (or 11) bits for exponent, e.
23 (or 52) bits for the fraction, f.
Then a real number, n, is approximated by the following three - term - relation:
n = (-1)s * 2(e - bias) * (1 + fraction)
where the bias offsets negatively2 the values of the exponent so that they describe numbers between 0 and 1 / (1 and 2) .
Now, the gap reflects the fact that real numbers does not map perfectly to their finite, 32 - or 64 - bit, representations, moreover, a range of real numbers that differ by abs value < eps maps to a single value in computer memory, i.e: if you assign a values val to a variable var_i
var_1 = val - offset
...
var_i = val;
...
val_n = val + offset
where
offset < eps(val) / 2
Then:
var_1 = var_2 = ... = var_i = ... = var_n.
The gap is determined from the second term containing the exponent (or characteristic):
2(e - bias)
in the above relation3, which determines the "scale" of the "line" on which the approximated numbers are located, the larger the numbers, the larger the distance between them, the less precise they are and vice versa: the smaller the numbers, the more densely located their representations are, consequently, more accurate.
In practice, to determine the gap of a specific number, eps(number), you can start by adding / subtracting a gradually increasing small number until the initial value of the number of interest changes - this will give you the gap in that (positive or negative) direction, i.e. eps(number) / 2.
To check possible implementations of MATLAB's eps (or ULP - unit of last place , as it is called in other languages), you could search for ULP implementations either in C, C++ or Java, which are the languages MATLAB is written in.
1. Real numbers are infinitely preciser i.e. they could be written with arbitrary precision, i.e. with any number of digits after the decimal point.
2. Usually around the half: in single precision 8 bits mean decimal values from 1 to 2^8 = 256, around the half in our case is: 127, i.e. 2(e - 127)
2. It can be thought that: 2(e - bias), is representing the most significant digits of the number, i.e. the digits that contribute to describe how big the number is, as opposed to the least significant digits that contribute to describe its precise location. Then the larger the term containing the exponent, the smaller the significance of the 23 bits of the fraction.
Are numeric variables following a documented standard on TI calculators ?
I've been really surprised noticing on my TI 83 Premium CE that this test actually returns true (i.e. 1) :
0.1 -> X
0.1 -> Y
0.01 -> Z
X*Y=Z
I was expecting this to fail, assuming my calculator would use something like IEEE 754 standard to represent floating points numbers.
On the other hand, calculating 2^50+3-2^50 returns 0, showing that large integers seems use such a standard : we see here the big number has a limited mantissa.
TI-BASIC's = is a tolerant comparison
Try 1+10^-12=1 on your calculator. Those numbers aren't represented equally (1+10^-12-1 gives 1E-12), but you'll notice the comparison returns true: that's because = has a certain amount of tolerance. AFAICT from testing on my calculator, if the numbers are equal when rounded to ten significant digits, = will return true.
Secondarily,
TI-BASIC uses a proprietary BCD float format
TI floats are a BCD format that is nine bytes long, with one byte for sign and auxilliary information and 14 digits (7 bytes) of precision. The ninth byte is used for extra precision so numbers can be rounded properly.
See a source linked to by #doynax here for more information.
These two long numbers are the same except for the last digit.
test = [];
test(1) = 33777100285870080;
test(2) = 33777100285870082;
but the last digit is lost when the numbers are put in the array:
unique(test)
ans = 3.3777e+16
How can I prevent this? The numbers are ID codes and losing the last digit is screwing everything up.
Matlab uses 64-bit floating point representation by default for numbers. Those have a base-10 16-digit precision (more or less) and your numbers seem to exceed that.
Use something like uint64 to store your numbers:
> test = [uint64(33777100285870080); uint64(33777100285870082)];
> disp(test(1));
33777100285870080
> disp(test(2));
33777100285870082
This is really a rounding error, not a display error. To get the correct strings for output purposes, use int2str, because, again, num2str uses a 64-bit floating point representation, and that has rounding errors in this case.
To add more explanation to #rubenvb's solution, your values are greater than flintmax for IEEE 754 double precision floating-point, i.e, greater than 2^53. After this point not all integers can be exactly represented as doubles. See also this related question.
I use mod() to compare if a number's 0.01 digit is 2 or not.
if mod(5.02*100, 10) == 2
...
end
The result is mod(5.02*100, 10) = 2 returns 0;
However, if I use mod(1.02*100, 10) = 2 or mod(20.02*100, 10) = 2, it returns 1.
The result of mod(5.02*100, 10) - 2 is
ans =
-5.6843e-14
Could it be possible that this is a bug for matlab?
The version I used is R2013a. version 8.1.0
This is not a bug in MATLAB. It is a limitation of floating point arithmetic and conversion between binary and decimal numbers. Even a simple decimal number such as 0.1 has cannot be exactly represented as a binary floating point number with finite precision.
Computer floating point arithmetic is typically not exact. Although we are used to dealing with numbers in decimal format (base10), computers store and process numbers in binary format (base2). The IEEE standard for double precision floating point representation (see http://en.wikipedia.org/wiki/Double-precision_floating-point_format, what MATLAB uses) specifies the use of 64 bits to represent a binary number. 1 bit is used for the sign, 52 bits are used for the mantissa (the actual digits of the number), and 11 bits are used for the exponent and its sign (which specifies where the decimal place goes).
When you enter a number into MATLAB, it is immediately converted to binary representation for all manipulations and arithmetic and then converted back to decimal for display and output.
Here's what happens in your example:
Convert to binary (keeping only up to 52 digits):
5.02 => 1.01000001010001111010111000010100011110101110000101e2
100 => 1.1001e6
10 => 1.01e3
2 => 1.0e1
Perform multiplication:
1.01000001010001111010111000010100011110101110000101 e2
x 1.1001 e6
--------------------------------------------------------------
0.000101000001010001111010111000010100011110101110000101
0.101000001010001111010111000010100011110101110000101
+ 1.01000001010001111010111000010100011110101110000101
-------------------------------------------------------------
1.111101011111111111111111111111111111111111111111111101e8
Cutting off at 52 digits gives 1.111101011111111111111111111111111111111111111111111e8
Note that this is not the same as 1.11110110e8 which would be 502.
Perform modulo operation: (there may actually be additional error here depending on what algorithm is used within the mod() function)
mod( 1.111101011111111111111111111111111111111111111111111e8, 1.01e3) = 1.111111111111111111111111111111111111111111100000000e0
The error is exactly -2-44 which is -5.6843x10-14. The conversion between decimal and binary and the rounding due to finite precision have caused a small error. In some cases, you get lucky and rounding errors cancel out and you might still get the 'right' answer which is why you got what you expect for mod(1.02*100, 10), but In general, you cannot rely on this.
To use mod() correctly to test the particular digit of a number, use round() to round it to the nearest whole number and compensate for floating point error.
mod(round(5.02*100), 10) == 2
What you're encountering is a floating point error or artifact, like the commenters say. This is not a Matlab bug; it's just how floating point values work. You'd get the same results in C or Java. Floating point values are "approximate" types, so exact equality comparisons using == without some rounding or tolerance are prone to error.
>> isequal(1.02*100, 102)
ans =
1
>> isequal(5.02*100, 502)
ans =
0
It's not the case that 5.02 is the only number this happens for; several around 0 are affected. Here's an example that picks out several of them.
x = 1.02:1000.02;
ix = mod(x .* 100, 10) ~= 2;
disp(x(ix))
To understand the details of what's going on here (and in many other situations you'll encounter working with floats), have a read through the Wikipedia entry for "floating point", or my favorite article on it, "What Every Computer Scientist Should Know About Floating-Point Arithmetic". (That title is hyperbole; this article goes deep and I don't understand half of it. But it's a great resource.) This stuff is particularly relevant to Matlab because Matlab does everything in floating point by default.