Why do some numbers lose accuracy when stored as floating point numbers?
For example, the decimal number 9.2 can be expressed exactly as a ratio of two decimal integers (92/10), both of which can be expressed exactly in binary (0b1011100/0b1010). However, the same ratio stored as a floating point number is never exactly equal to 9.2:
32-bit "single precision" float: 9.19999980926513671875
64-bit "double precision" float: 9.199999999999999289457264239899814128875732421875
How can such an apparently simple number be "too big" to express in 64 bits of memory?
In most programming languages, floating point numbers are represented a lot like scientific notation: with an exponent and a mantissa (also called the significand). A very simple number, say 9.2, is actually this fraction:
5179139571476070 * 2 -49
Where the exponent is -49 and the mantissa is 5179139571476070. The reason it is impossible to represent some decimal numbers this way is that both the exponent and the mantissa must be integers. In other words, all floats must be an integer multiplied by an integer power of 2.
9.2 may be simply 92/10, but 10 cannot be expressed as 2n if n is limited to integer values.
Seeing the Data
First, a few functions to see the components that make a 32- and 64-bit float. Gloss over these if you only care about the output (example in Python):
def float_to_bin_parts(number, bits=64):
if bits == 32: # single precision
int_pack = 'I'
float_pack = 'f'
exponent_bits = 8
mantissa_bits = 23
exponent_bias = 127
elif bits == 64: # double precision. all python floats are this
int_pack = 'Q'
float_pack = 'd'
exponent_bits = 11
mantissa_bits = 52
exponent_bias = 1023
else:
raise ValueError, 'bits argument must be 32 or 64'
bin_iter = iter(bin(struct.unpack(int_pack, struct.pack(float_pack, number))[0])[2:].rjust(bits, '0'))
return [''.join(islice(bin_iter, x)) for x in (1, exponent_bits, mantissa_bits)]
There's a lot of complexity behind that function, and it'd be quite the tangent to explain, but if you're interested, the important resource for our purposes is the struct module.
Python's float is a 64-bit, double-precision number. In other languages such as C, C++, Java and C#, double-precision has a separate type double, which is often implemented as 64 bits.
When we call that function with our example, 9.2, here's what we get:
>>> float_to_bin_parts(9.2)
['0', '10000000010', '0010011001100110011001100110011001100110011001100110']
Interpreting the Data
You'll see I've split the return value into three components. These components are:
Sign
Exponent
Mantissa (also called Significand, or Fraction)
Sign
The sign is stored in the first component as a single bit. It's easy to explain: 0 means the float is a positive number; 1 means it's negative. Because 9.2 is positive, our sign value is 0.
Exponent
The exponent is stored in the middle component as 11 bits. In our case, 0b10000000010. In decimal, that represents the value 1026. A quirk of this component is that you must subtract a number equal to 2(# of bits) - 1 - 1 to get the true exponent; in our case, that means subtracting 0b1111111111 (decimal number 1023) to get the true exponent, 0b00000000011 (decimal number 3).
Mantissa
The mantissa is stored in the third component as 52 bits. However, there's a quirk to this component as well. To understand this quirk, consider a number in scientific notation, like this:
6.0221413x1023
The mantissa would be the 6.0221413. Recall that the mantissa in scientific notation always begins with a single non-zero digit. The same holds true for binary, except that binary only has two digits: 0 and 1. So the binary mantissa always starts with 1! When a float is stored, the 1 at the front of the binary mantissa is omitted to save space; we have to place it back at the front of our third element to get the true mantissa:
1.0010011001100110011001100110011001100110011001100110
This involves more than just a simple addition, because the bits stored in our third component actually represent the fractional part of the mantissa, to the right of the radix point.
When dealing with decimal numbers, we "move the decimal point" by multiplying or dividing by powers of 10. In binary, we can do the same thing by multiplying or dividing by powers of 2. Since our third element has 52 bits, we divide it by 252 to move it 52 places to the right:
0.0010011001100110011001100110011001100110011001100110
In decimal notation, that's the same as dividing 675539944105574 by 4503599627370496 to get 0.1499999999999999. (This is one example of a ratio that can be expressed exactly in binary, but only approximately in decimal; for more detail, see: 675539944105574 / 4503599627370496.)
Now that we've transformed the third component into a fractional number, adding 1 gives the true mantissa.
Recapping the Components
Sign (first component): 0 for positive, 1 for negative
Exponent (middle component): Subtract 2(# of bits) - 1 - 1 to get the true exponent
Mantissa (last component): Divide by 2(# of bits) and add 1 to get the true mantissa
Calculating the Number
Putting all three parts together, we're given this binary number:
1.0010011001100110011001100110011001100110011001100110 x 1011
Which we can then convert from binary to decimal:
1.1499999999999999 x 23 (inexact!)
And multiply to reveal the final representation of the number we started with (9.2) after being stored as a floating point value:
9.1999999999999993
Representing as a Fraction
9.2
Now that we've built the number, it's possible to reconstruct it into a simple fraction:
1.0010011001100110011001100110011001100110011001100110 x 1011
Shift mantissa to a whole number:
10010011001100110011001100110011001100110011001100110 x 1011-110100
Convert to decimal:
5179139571476070 x 23-52
Subtract the exponent:
5179139571476070 x 2-49
Turn negative exponent into division:
5179139571476070 / 249
Multiply exponent:
5179139571476070 / 562949953421312
Which equals:
9.1999999999999993
9.5
>>> float_to_bin_parts(9.5)
['0', '10000000010', '0011000000000000000000000000000000000000000000000000']
Already you can see the mantissa is only 4 digits followed by a whole lot of zeroes. But let's go through the paces.
Assemble the binary scientific notation:
1.0011 x 1011
Shift the decimal point:
10011 x 1011-100
Subtract the exponent:
10011 x 10-1
Binary to decimal:
19 x 2-1
Negative exponent to division:
19 / 21
Multiply exponent:
19 / 2
Equals:
9.5
Further reading
The Floating-Point Guide: What Every Programmer Should Know About Floating-Point Arithmetic, or, Why don’t my numbers add up? (floating-point-gui.de)
What Every Computer Scientist Should Know About Floating-Point Arithmetic (Goldberg 1991)
IEEE Double-precision floating-point format (Wikipedia)
Floating Point Arithmetic: Issues and Limitations (docs.python.org)
Floating Point Binary
This isn't a full answer (mhlester already covered a lot of good ground I won't duplicate), but I would like to stress how much the representation of a number depends on the base you are working in.
Consider the fraction 2/3
In good-ol' base 10, we typically write it out as something like
0.666...
0.666
0.667
When we look at those representations, we tend to associate each of them with the fraction 2/3, even though only the first representation is mathematically equal to the fraction. The second and third representations/approximations have an error on the order of 0.001, which is actually much worse than the error between 9.2 and 9.1999999999999993. In fact, the second representation isn't even rounded correctly! Nevertheless, we don't have a problem with 0.666 as an approximation of the number 2/3, so we shouldn't really have a problem with how 9.2 is approximated in most programs. (Yes, in some programs it matters.)
Number bases
So here's where number bases are crucial. If we were trying to represent 2/3 in base 3, then
(2/3)10 = 0.23
In other words, we have an exact, finite representation for the same number by switching bases! The take-away is that even though you can convert any number to any base, all rational numbers have exact finite representations in some bases but not in others.
To drive this point home, let's look at 1/2. It might surprise you that even though this perfectly simple number has an exact representation in base 10 and 2, it requires a repeating representation in base 3.
(1/2)10 = 0.510 = 0.12 = 0.1111...3
Why are floating point numbers inaccurate?
Because often-times, they are approximating rationals that cannot be represented finitely in base 2 (the digits repeat), and in general they are approximating real (possibly irrational) numbers which may not be representable in finitely many digits in any base.
While all of the other answers are good there is still one thing missing:
It is impossible to represent irrational numbers (e.g. π, sqrt(2), log(3), etc.) precisely!
And that actually is why they are called irrational. No amount of bit storage in the world would be enough to hold even one of them. Only symbolic arithmetic is able to preserve their precision.
Although if you would limit your math needs to rational numbers only the problem of precision becomes manageable. You would need to store a pair of (possibly very big) integers a and b to hold the number represented by the fraction a/b. All your arithmetic would have to be done on fractions just like in highschool math (e.g. a/b * c/d = ac/bd).
But of course you would still run into the same kind of trouble when pi, sqrt, log, sin, etc. are involved.
TL;DR
For hardware accelerated arithmetic only a limited amount of rational numbers can be represented. Every not-representable number is approximated. Some numbers (i.e. irrational) can never be represented no matter the system.
There are infinitely many real numbers (so many that you can't enumerate them), and there are infinitely many rational numbers (it is possible to enumerate them).
The floating-point representation is a finite one (like anything in a computer) so unavoidably many many many numbers are impossible to represent. In particular, 64 bits only allow you to distinguish among only 18,446,744,073,709,551,616 different values (which is nothing compared to infinity). With the standard convention, 9.2 is not one of them. Those that can are of the form m.2^e for some integers m and e.
You might come up with a different numeration system, 10 based for instance, where 9.2 would have an exact representation. But other numbers, say 1/3, would still be impossible to represent.
Also note that double-precision floating-points numbers are extremely accurate. They can represent any number in a very wide range with as much as 15 exact digits. For daily life computations, 4 or 5 digits are more than enough. You will never really need those 15, unless you want to count every millisecond of your lifetime.
Why can we not represent 9.2 in binary floating point?
Floating point numbers are (simplifying slightly) a positional numbering system with a restricted number of digits and a movable radix point.
A fraction can only be expressed exactly using a finite number of digits in a positional numbering system if the prime factors of the denominator (when the fraction is expressed in it's lowest terms) are factors of the base.
The prime factors of 10 are 5 and 2, so in base 10 we can represent any fraction of the form a/(2b5c).
On the other hand the only prime factor of 2 is 2, so in base 2 we can only represent fractions of the form a/(2b)
Why do computers use this representation?
Because it's a simple format to work with and it is sufficiently accurate for most purposes. Basically the same reason scientists use "scientific notation" and round their results to a reasonable number of digits at each step.
It would certainly be possible to define a fraction format, with (for example) a 32-bit numerator and a 32-bit denominator. It would be able to represent numbers that IEEE double precision floating point could not, but equally there would be many numbers that can be represented in double precision floating point that could not be represented in such a fixed-size fraction format.
However the big problem is that such a format is a pain to do calculations on. For two reasons.
If you want to have exactly one representation of each number then after each calculation you need to reduce the fraction to it's lowest terms. That means that for every operation you basically need to do a greatest common divisor calculation.
If after your calculation you end up with an unrepresentable result because the numerator or denominator you need to find the closest representable result. This is non-trivil.
Some Languages do offer fraction types, but usually they do it in combination with arbitary precision, this avoids needing to worry about approximating fractions but it creates it's own problem, when a number passes through a large number of calculation steps the size of the denominator and hence the storage needed for the fraction can explode.
Some languages also offer decimal floating point types, these are mainly used in scenarios where it is imporant that the results the computer gets match pre-existing rounding rules that were written with humans in mind (chiefly financial calculations). These are slightly more difficult to work with than binary floating point, but the biggest problem is that most computers don't offer hardware support for them.
It is possible to print to several hundred decimal places a square root in bc, as it is in C. However in C it is only accurate to 15. I have checked the square root of 2 to 50 decimal places and it is accurate but what is the limit in bc? I can't find any reference to this.
To how many decimal places is bc accurate?
bc is an arbitrary precision calculator. Arbitrary precision just tells us how many digits it can represent (as many as will fit in memory), but doesn't tell us anything about accuracy.
However in C it is only accurate to 15
C uses your processor's built-in floating point hardware. This is fast, but has a fixed number of bits to represent each number, so is obviously fixed rather than arbitrary precision.
Any arbitrary precision system will have more ... precision than this, but could of course still be inaccurate. Knowing how many digits can be stored doesn't tell us whether they're correct.
However, the GNU implementation of bc is open source, so we can just see what it does.
The bc_sqrt function uses an iterative approximation (Newton's method, although the same technique was apparently known by the Babylonians in at least 1,000BC).
This approximation is just run, improving each time, until two consecutive guesses differ by less than the precision requested. That is, if you ask for 1,000 digits, it'll keep going until the difference is at most in the 1,001st digit.
The only exception is when you ask for an N-digit result and the original number has more than N digits. It'll use the larger of the two as its target precision.
Since the convergence rate of this algorithm is faster than one digit per iteration, there seems little risk of two consecutive iterations agreeing to some N digits without also being correct to N digits.
I created Cassandra table with column type: DataType.FLOAT.
Execute my SQL using CqlSession:
CqlSessionBuilder builder = CqlSession.builder();
builder.addContactPoint(new InetSocketAddress(properties.getHost(), properties.getPort()));
builder.withLocalDatacenter(properties.getDatacenter());
builder.withAuthCredentials(properties.getUsername(), properties.getPassword());
builder.build();
But when I insert float numbers, it's rounded up:
12334.9999 -> 12335.0.
0.999999 -> 0.999999
12345.9999 -> 12346.0
It seems like Cassandra rounds the float and consider the number of all digits, not only after the point.
What are the options to solve this problem? I know that I can use Decimal datatype, but may be you have other solution?
I actually covered this issue with Apache Cassandra and DataStax Astra DB in an article I wrote last month:
The Guerilla Guide to Building E-commerce Product Services with DataStax Astra DB
So the problem here, is that FLOAT is a fixed floating point precision type. This means that when the numeric values are converted from base-10 (decimal) to base-2 (binary), each one of the 32 binary precision points must have a value (zero or one, obviously). It's during this conversion process between base-2 and base-10 that rounding errors occur. The likelihood of a rounding error increases as the value does (on either side of the decimal point).
What are the options to solve this problem? I know that I can use Decimal datatype, but may be you have other solution?
Well, you mentioned the best solution (IMO), which to use a DECIMAL to store the value. This works, because DECIMAL is an arbitrary floating point type. The values in a DECIMAL type are stored in base-10, so there's no conversion necessary and only the required precision is used.
Before arbitrary precision types came along, we used to use INTEGERs for things that had to be accurate. The first E-commerce team I worked on stored product prices in the DB as pennies, to prevent the rounding issue.
Yes, both INT and FLOAT are fixed precision types, but an INT stores whole numbers, and all of its precision points can be used for that. Therefore the usage patterns of the bits are quite different. While both INT and FLOAT allocate a bit for the "sign" (+/-), with floating point numbers the remaining 31 precision points are pre-allocated for the full numeric value and its exponent.
So your example of 12334.9999 is essentially stored in Cassandra like this:
123349999 x 10^-4
And of course, that's stored in binary, which I won't include here for brevity.
tl;dr;
Basically FLOATs use fixed precision to store values as a formula (significand and exponent) in base-2, and the conversion back to base-10 makes rounding errors likely.
You're right, use a DECIMAL type. When you need to be exact, that's the only real solution.
If you're interested, here are two additional SO answers which provide more detail on this topic:
Double vs. BigDecimal?
What is the difference between the float and integer data type when the size is the same?
Consider the following terminating decimal numbers.
3.1^2 = 9.61
3.1^4 = 92.3521
3.1^8 = 8528.91037441
The following shows how Mathematica treats these expressions
In[1]:= 3.1^2
Out[1]= 9.61
In[2]:= 3.1^4
Out[2]= 92.352
So far so good, but
In[3]:= 3.1^8
Out[3]= 8528.91
doesn't provide enough precision.
So let's try N[], NumberForm[], and DecimalForm[] with a precision of 12
In[4]:= N[3.1^8,12]
Out[4]= 8528.91
In[5]:= NumberForm[3.1^8,12]
Out[5]= 8528.91037441
In[6]:= DecimalForm[3.1^8,12]
Out[6]= 8528.91037441
In this case DecimialForm[] and NumberForm[] work as expected, but N[] only provided the default precision of 6, even though I asked for 12. So DecimalForm[] or NumberForm[] seem to be the way to go if you want exact results when the inputs are terminating decimals.
Next consider rational numbers with infinite repeating decimals like 1/3.
In[7]:= N[1/3,20]
Out[7]= 0.33333333333333333333
In[9]:= NumberForm[1/3, 20]
Out[9]=
1/3
In[9]:= DecimalForm[1/3, 20]
Out[9]=
1/3
Unlike the previous case, N[] seems to be the proper way to go here, whereas NumberForm[] and DecimalForm[] do not respect precisions.
Finally consider irrational numbers like Sqrt[2] and Pi.
In[10]:= N[Sqrt[2],20]
Out[10]= 1.4142135623730950488
In[11]:= NumberForm[Sqrt[2], 20]
Out[11]=
sqrt(2)
In[12]:= DecimalForm[Sqrt[2], 20]
Out[12]=
sqrt(2)
In[13]:= N[π^12,30]
Out[13]= 924269.181523374186222579170358
In[14]:= NumberForm[Pi^12,30]
Out[14]=
π^12
In[15]:= DecimalForm[Pi^12,30]
Out[15]=
π^12
In these cases N[] works, but NumberForm[] and DecimalForm[] do not. However, note that N[] switches to scientific notation at π^13, even with a larger precision. Is there a way to avoid this switch?
In[16]:= N[π^13,40]
Out[16]= 2.903677270613283404988596199487803130470*10^6
So there doesn't seem to be a consistent way of formulating how to get decimal numbers with requested precisions and at the same time avoiding scientific notation. Sometimes N[] works, othertimes DecimalForm[] or NumberForm[] works, and at othertimes nothing seems to work.
Have I missed something or are there bugs in the system?
It isn't a bug because it is designed purposefully to behave this way. Precision is limited by the precision of your machine, your configuration of Mathematica, and the algorithm and performance constraints of the calculation.
The documentation for N[expr, n] states it attempts to give a result with n‐digit precision. When it cannot give the requested precision it gets as close as it can. DecimalForm and NumberForm work the same way.
https://reference.wolfram.com/language/ref/N.html explains the various cases behind this:
Unless numbers in expr are exact, or of sufficiently high precision, N[expr,n] may not be able to give results with n‐digit precision.
N[expr,n] may internally do computations to more than n digits of precision.
$MaxExtraPrecision specifies the maximum number of extra digits of precision that will ever be used internally.
The precision n is given in decimal digits; it need not be an integer.
n must lie between $MinPrecision and $MaxPrecision. $MaxPrecision can be set to Infinity.
n can be smaller than $MachinePrecision.
N[expr] gives a machine‐precision number, so long as its magnitude is between $MinMachineNumber and $MaxMachineNumber.
N[expr] is equivalent to N[expr,MachinePrecision].
N[0] gives the number 0. with machine precision.
N converts all nonzero numbers to Real or Complex form.
N converts each successive argument of any function it encounters to numerical form, unless the head of the function has an attribute such as NHoldAll.
You can define numerical values of functions using N[f[args]]:=value and N[f[args],n]:=value.
N[expr,{p,a}] attempts to generate a result with precision at most p and accuracy at most a.
N[expr,{Infinity,a}] attempts to generate a result with accuracy a.
N[expr,{Infinity,1}] attempts to find a numerical approximation to the integer part of expr.
I'm using Python3's decimal module. Is the underlying arithmetic done using the processors floating point types, or does it use integers? The notion that the results are 'exact' and of arbitrary precision suggests to me that integer maths is used below the surface.
Indeed it is integer math, not float math for sure. Roughly speaking every float is two parts - before and after the decimal dot (integer and the remainder). Thanks to that the calculations are done using integer arithmetic and hence are not rounded up so they are staying precise even if you sum up a very large value with a very small fraction.
This comes at a price - the number of operations is significantly larger and it is not always necessary to be so precise at all times. That is why most of the calculations are done using float arithmetic that may cause a loss of precision when there are many arithmetic operations on floats or there are significant differences between the values (e.g. 10^10 ratio and more). There is a separate field of computer science: numerical analysis or numerical methods that study the clever ways to get the most of the speed of float calculations while maintaining highest precision possible.