Related
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.
If I only need 1 or 2 digit after the decimal place accuracy, should I use float or still go with decimal(18,2)?
The numeric value in question represents salary.
You should use decimal type or better still, money type, which is specially suited for these needs.
You should not use float as float is an approximate representation of a decimal value.
See MSDN documentation why we should not use float here
The float and real data types are known as approximate data types. The behavior of float and real follows the IEEE 754 specification on approximate numeric data types.
Approximate numeric data types do not store the exact values specified for many numbers; they store an extremely close approximation of the value. For many applications, the tiny difference between the specified value and the stored approximation is not noticeable. At times, though, the difference becomes noticeable.
Because of the approximate nature of the float and real data types, do not use these data types when exact numeric behavior is required, such as in financial applications, in operations involving rounding, or in equality checks. Instead, use the integer, decimal, money, or smallmoney data types.
The main question in deciding whether you want to use a binary decimal or a decimal decimal is not accuracy, really. It's "how would I expect the calculations to proceed".
The main thing you get from decimal is something you can calculate on paper, with fixed-point numbers. E.g.:
13.22
+ 7.3
-----
20.52
It's not really that decimal is more precise than a float (though it can be that as well, for certain applications). The point is it makes the same mistakes you would make on paper - it's a decimal decimal number, not binary.
Or in another line of thought, your inputs are definitely decimal decimal numbers (typically, in a string or such). So if you use a float, you get decimal -> binary -> calculation -> decimal. Binary to decimal means no loss of information (and finite binary number can be exactly represented in a finite decimal number), but the other way around this isn't true - even something like 0.1 has no finite representation in a binary decimal number (just like 1 / 3 has no finite representation in a decimal decimal number, but works fine in say base 6).
This question comes from a homework assignment I was given. You can base your storage system off of one of the three following formats:
DD MM SS.S
DD MM.MMM
DD.DDDDD
You want to maximize the amount of data you can store by using as few bytes as possible.
My solution is based off the first format. I used 3 bytes for latitude: 8 bits for the DD (-90 to 90), 6 bits for the MM (0-59), and 10 bits for the SS.S (0-59.9). I then used 25 bits for the longitude: 9 bits for the DDD (-180 to 180), 6 bits for the MM, and 10 for the SS.S. This solution doesn't fit nicely on a byte border, but I figured the next reading can be stored immediately following the previous one, and 8 readings would use only 49 bytes.
I'm curious what methods others can come up. Is there a more efficient method to storing this data? As a note, I considered an offset based storage, but the problem gave no indication of how much the values may change between readings, so I'm assuming any change is possible.
Your suggested method is not optimal. You are using 10 bits (1024 possible values) to store a value in the range (0..599). This is a waste of space.
If you'll use 3 bytes for latitude, you should map the range [0, 2^24-1] to the range [-90, 90]. Hence each of the 2^24 values represents 180/2^24 degrees, which is 0.086 seconds.
If you want only 0.1 second accuracy, you'll need 23 bits for latitudes and 24 bits for longitudes (you'll get 0.077 seconds accuracy). That's 47 bit total instead of your 49 bits, with better accuracy.
Can we do even better?
The exact number of bits needed for 0.1 second accuracy is log2(180*60*60*10 * 360*60*60*10) < 46.256. Which means that you can use 46256 bits (5782 bytes) to store 1000 (lat,lon) pairs, but the mathematics involved will require dealing with very large integers.
Can we do even better?
It depends. If your data set has concentrations, you can store only some points and relative distances from these points, using less bits. Clustering algorithms should be used.
Sticking to existing technology:
If you used half precision floating point numbers to store only the DD.DDDDD data, you can be a lot more space-efficent, but you'd have to accept an exponent bias of 15, which means: The coordinates stored might not be exact, but at an offset from the original value.
This is due to the way floating point numbers are stored, essentially: A normalized significant is multiplied by an exponent to result in a number, instead of just storing a single value (as in integer numbers, the way you calculated the numbers for your solution).
The next highest commonly used floating point number mechanism uses 32 bits (the type "float" in many programming languages) - still efficient, but larger than your custom format.
If, however, you would design your own custom floating point type as well, and you gradually added more bits, your results would become more exact and it would STILL be more efficient than the solution you first found. Just play around with the number of bits used for significant and exponent, and find out how close your fp approximations come to the desired result in degrees!
Well, if this is for a large number of readings, then you may try a differential approach. Start with an absolute location, and then start saving incremental changes, which should ideally require less bits, depending on the nature of the changes. This is effectively compressing the stream. But somehow I don't think that's what this homework is about.
Here's a little gem directly from my VBE (MS Excel 2007 VBA):
?clng(150*0.85)
127
x = 150*0.85
?clng(x)
128
Can anybody explain this behaviour? IMHO the first expression should yield 128 (.5 rounded to nearest even), or at least should both results be equal.
I think wqw is right, but I'll give the details.
In the statement clng(150 * 0.85), 150 * 0.85 is calculated in extended-precision:
150 = 1.001011 x 2^7
0.85 in double precision =
1.1011001100110011001100110011001100110011001100110011 x 2^-1
Multiply these by hand and you get
1.1111110111111111111111111111111111111111111111111111110001 x 2^6 =
127.4999999999999966693309261245303787291049957275390625
That's 59 bits, which fits comfortably in extended-precision. It's less than 127.5 so rounds down.
In the statement x = 150 * 0.85, that 59 bit value is rounded to 53 bits, giving
1.1111111 x 2^6 = 1111111.1 = 127.5
So it rounds up according to round-half-to-even.
(See my article http://www.exploringbinary.com/when-doubles-dont-behave-like-doubles/ for more information.)
Ahh one of the “fun” things about VBA is the rounding on CInt() etc is what is called bankers rounding. Bankers rounding is where 0.5 values are rounded up or down depending on if the number is an even number so 2.5 rounds to 2, 3.5 to 4 and so on.
More can be found here about rounding
http://www.consultdmw.com/rounding-numbers.htm
This is a bit of a guess, but .85 may not be representable as a floating point number. If it is off by 0.0000000000001 it can still affect rounding in weird ways.
If you use CDec(.85) to force it into decimal mode you don't get that weirdness. This is one of the many reasons why I don't use single/double where accuracy is important.
My theory is VBA/VB6 is using x87 for floating point calculations and this implicitly converts doubles to higher precision if 80 bits for intermediate results. So an assignment to v or explicit cast with CDbl converts intermediate 80-bits value back to 64-bits effectivly rounding it (or truncating it).
Here is some discussion:
Extended (80-bit) double floating point in x87, not SSE2 - we don't miss it?
Both what Kevin and Jonathan have said are true, but Jonathan's answer is more applicable here. If you were dealing with Currency-type numbers instead of floating point, then the Banker's rounding rule would be applied.
How can I do accurate decimal number arithmetic since using floats is not reliable?
I still want to return the answer to a textField.
You can either use int-s (e.g. with "cents" as a unit instead of "dollars"), or use NSDecimalNumber s.
Store your number multiplied by some power of ten of your choice, chosen by the amount of precision you need to the right of the decimal point. We'll call these scaled numbers. Convert entered values that need your precision into scaled numbers. Addition is easy: just add. Multiplication is slightly harder: just multiply two scaled numbers, and then divide by your scale factor. Division is easy: just divide, then multiply by the scale factor. The only inconvenience is you'll have to write your own numeric input/output conversion routines.