I'm trying to understand how floating point number arithmetic plays a role in computer science when using the binary system. I came across an excerpt from What Every Computer Scientist Should Know About Floating-Point Arithmetic which defines normalized numbers as unique floating-point numbers with the leading significand being non-zero. It goes on to say...
When β
= 2, p = 3, e min = -1 and e max = 2 there are 16 normalized floating-point numbers, as shown in Figure D-1.
Where β is the base, p is the precision, e min is the minimum exponent, and e max is the maximum exponent.
My attempt at understanding how he came to the conclusion of there being 16 normalized floating-point numbers was to multiply together the possible number of significands β^p and the possible number of exponents e max - e min + 1. My result was 32 possible normalized floating-point values. I am unsure of how to get the correct result of 16 normalized floating-point values as was declared in the paper above. I assumed negative floating-point values were excluded, however, I did not include them in my calculations.
This question is more geared toward mathematical formulae. But it will help me to better understand how floating-point arithmetic works in computer science.
I would like to know how to get the correct result of 16 normalized floating-point numbers and why.
Since the first bit is always 1, with 3 bits for the mantissa you have only two bits to vary, yielding 4 different mantissa values. Combined with 4 different exponent values that's 16. I haven't looked at the paper though.
My attempt at understanding how he came to the conclusion of there being 16 normalized floating-point numbers was to multiply together the possible number of significands β^p and the possible number of exponents e max - e min + 1
This is correct except that the number of possible significands is not βp in binary with an implicit leading 1. In these conditions, the number of possible significands is βp-1, encoded over p-1 bits.
In other words, the missing values for the possible significands have already been taken advantage of when the encoding reserved, say, 52 bits to encode a precision of 53 binary digits.
Related
To get a binary representation from a natural number like 20, we divide this number by 2 and so on until we cannot divide by 2 anymore. To get a binary representation from a decimal number like 0.4512, we multiply this number by 2 repeated times.
What is the logic explanation why with these two systems we get a binary representation?
Thanks
It is based on the fact that numbers are coded in binary.
If the number A is an integer, A is rewritten as A=∑i=0n-1ai×2i=an-1×2n-1+an-2×2n-2+...+a1×2+a0
where ai=0 or 1.
It is easy to see that is A is even, a0=0, and if it is odd, a0=1. So we already have the least significant bit a0.
Now, if we divide A by two, a0 disappears and we have
A/2=an-1×2n-2+an-2×2n-3+...+a2×2+a1
We can determine this way a1 depending on the parity of A/2. and we continue, we get all the bits of A.
Fractional numbers are expressed according to negative powers of 2. If A=0.a-1a-2...a-n, A=a-1/2+a-2/4+...+a-n/2^n
If we multiply it by two, 2×A=a-1+a-2/2+...+a-n/2^n-1. If 2×A≥1, we must have a-1=1, otherwise a-1=0. And we can determine other bits is a similar way by successive multiplications by two.
I am aware of the textbook method where we multiply the mantissa by 2 , take its integer part as the next bit, multiply the fractional part by 2 and repeat, until we get zero or reach our desired precision.
Is there an efficient algorithm to convert mantissa from base 10 to base 2 than the above mentioned algorithm?
The algorithm you proposed runs in O(n) time where n is the number of bits desired. We cannot do better than this, because we have to somehow calculate all n desired bits of the output, so any algorithm must be at least O(n), or else the output could not possibly contain all the desired information.
I'm reading this article about exponent bias in floating point numbers and it says the following:
n IEEE 754 floating point numbers, the exponent is biased in the
engineering sense of the word – the value stored is offset from the
actual value by the exponent bias. Biasing is done because exponents
have to be signed values in order to be able to represent both tiny
and huge values, but two's complement, the usual representation for
signed values, would make comparison harder. To solve this problem the
exponent is biased before being stored, by adjusting its value to put
it within an unsigned range suitable for comparison. By arranging the
fields so that the sign bit is in the most significant bit position,
the biased exponent in the middle, then the mantissa in the least
significant bits, the resulting value will be ordered properly,
whether it's interpreted as a floating point or integer value. This
allows high speed comparisons of floating point numbers using fixed
point hardware.
I've also found this explanation from wikipedia's article about offset binary:
This has the consequence that the "zero" value is represented by a 1
in the most significant bit and zero in all other bits, and in general
the effect is conveniently the same as using two's complement except
that the most significant bit is inverted. It also has the consequence
that in a logical comparison operation, one gets the same result as
with a two's complement numerical comparison operation, whereas, in
two's complement notation a logical comparison will agree with two's
complement numerical comparison operation if and only if the numbers
being compared have the same sign. Otherwise the sense of the
comparison will be inverted, with all negative values being taken as
being larger than all positive values.
I don't really understand what kind of comparison they are talking about here. Can someone please explain using a simple example?
'Comparison' here refers to the usual comparison of numbers by size: 5 > 4, etc. Suppose floating-point numbers were stored with as
[sign bit] [unbiased exponent] [mantissa]
For example, if the exponent is a 2's complement 3-bit binary number and the mantissa is a 4-bit unsigned binary number, you'd have
1 010 1001 = 4.5
1 110 0111 = 0.21875
You can see that the first is bigger than the second, but to figure this out, the computer would have to calculate 1.001 x 2^2 and 0.111 x 2^(-2) and then compare the resulting floating-point numbers. This is already complex with floating-point hardware, and if there is no such hardware for this computer, then...
So the number is stored as
[sign bit] [biased exponent] [mantissa]
Using the same 3-bit binary number for the exponent (this time biased; see a related question) and unsigned 4-bit mantissa, we have
1 101 1001 = 4.5
1 001 0111 = 0.21875
But now comparison is very easy! You can treat the two numbers as integers 11011001 and 10010111 and see that the first is obviously bigger: obvious even to a computer, as integer comparisons are easy. This is why biased exponents are used.
I've come across two different precision formulas for floating-point numbers.
⌊(N-1) log10(2)⌋ = 6 decimal digits (Single-precision)
and
N log10(2) ≈ 7.225 decimal digits (Single-precision)
Where N = 24 Significant bits (Single-precision)
The first formula is found at the top of page 4 of "IEEE Standard 754 for Binary Floating-Point Arithmetic" written by, Professor W. Kahan.
The second formula is found on the Wikipedia article "Single-precision floating-point format" under section IEEE 754 single-precision binary floating-point format: binary32.
For the first formula, Professor W. Kahan says
If a decimal string with at most 6 sig. dec. is converted to Single and then converted back to the same number of sig. dec.,
then the final string should match the original.
For the second formula, Wikipedia says
...the total precision is 24 bits (equivalent to log10(224) ≈ 7.225 decimal digits).
The results of both formulas (6 and 7.225 decimal digits) are different, and I expected them to be the same because I assumed they both were meant to represent the most significant decimal digits which can be converted to floating-point binary and then converted back to decimal with the same number of significant decimal digits that it started with.
Why do these two numbers differ, and what is the most significant decimal digits precision that can be converted to binary and back to decimal without loss of significance?
These are talking about two slightly different things.
The 7.2251 digits is the precision with which a number can be stored internally. For one example, if you did a computation with a double precision number (so you were starting with something like 15 digits of precision), then rounded it to a single precision number, the precision you'd have left at that point would be approximately 7 digits.
The 6 digits is talking about the precision that can be maintained through a round-trip conversion from a string of decimal digits, into a floating point number, then back to another string of decimal digits.
So, let's assume I start with a number like 1.23456789 as a string, then convert that to a float32, then convert the result back to a string. When I've done this, I can expect 6 digits to match exactly. The seventh digit might be rounded though, so I can't necessarily expect it to match (though it probably will be +/- 1 of the original string.
For example, consider the following code:
#include <iostream>
#include <iomanip>
int main() {
double init = 987.23456789;
for (int i = 0; i < 100; i++) {
float f = init + i / 100.0;
std::cout << std::setprecision(10) << std::setw(20) << f;
}
}
This produces a table like the following:
987.2345581 987.2445679 987.2545776 987.2645874
987.2745972 987.2845459 987.2945557 987.3045654
987.3145752 987.324585 987.3345947 987.3445435
987.3545532 987.364563 987.3745728 987.3845825
987.3945923 987.404541 987.4145508 987.4245605
987.4345703 987.4445801 987.4545898 987.4645386
987.4745483 987.4845581 987.4945679 987.5045776
987.5145874 987.5245972 987.5345459 987.5445557
987.5545654 987.5645752 987.574585 987.5845947
987.5945435 987.6045532 987.614563 987.6245728
987.6345825 987.6445923 987.654541 987.6645508
987.6745605 987.6845703 987.6945801 987.7045898
987.7145386 987.7245483 987.7345581 987.7445679
987.7545776 987.7645874 987.7745972 987.7845459
987.7945557 987.8045654 987.8145752 987.824585
987.8345947 987.8445435 987.8545532 987.864563
987.8745728 987.8845825 987.8945923 987.904541
987.9145508 987.9245605 987.9345703 987.9445801
987.9545898 987.9645386 987.9745483 987.9845581
987.9945679 988.0045776 988.0145874 988.0245972
988.0345459 988.0445557 988.0545654 988.0645752
988.074585 988.0845947 988.0945435 988.1045532
988.114563 988.1245728 988.1345825 988.1445923
988.154541 988.1645508 988.1745605 988.1845703
988.1945801 988.2045898 988.2145386 988.2245483
If we look through this, we can see that the first six significant digits always follow the pattern precisely (i.e., each result is exactly 0.01 greater than its predecessor). As we can see in the original double, the value is actually 98x.xx456--but when we convert the single-precision float to decimal, we can see that the 7th digit frequently would not be read back in correctly--since the subsequent digit is greater than 5, it should round up to 98x.xx46, but some of the values won't (e.g,. the second to last item in the first column is 988.154541, which would be round down instead of up, so we'd end up with 98x.xx45 instead of 46. So, even though the value (as stored) is precise to 7 digits (plus a little), by the time we round-trip the value through a conversion to decimal and back, we can't depend on that seventh digit matching precisely any more (even though there's enough precision that it will a lot more often than not).
1. That basically means 7 digits, and the 8th digit will be a little more accurate than nothing, but not a whole lot--for example, if we were converting from a double of 1.2345678, the .225 digits of precision mean that the last digit would be with about +/- .775 of the what started out there (whereas without the .225 digits of precision, it would be basically +/- 1 of what started out there).
what is the most significant decimal digits precision that can be
converted to binary and back to decimal without loss of significance?
The most significant decimal digits precision that can be converted to binary and back to decimal without loss of significance (for single-precision floating-point numbers or 24-bits) is 6 decimal digits.
Why do these two numbers differ...
The numbers 6 and 7.225 differ, because they define two different things. 6 is the most decimal digits that can be round-tripped. 7.225 is the approximate number of decimal digits precision for a 24-bit binary integer because a 24-bit binary integer can have 7 or 8 decimal digits depending on its specific value.
7.225 was found using the specific binary integer formula.
dspec = b·log10(2) (dspec
= specific decimal digits, b = bits)
However, what you normally need to know, are the minimum and maximum decimal digits for a b-bit integer. The following formulas are used to find the min and max decimal digits (7 and 8 respectively for 24-bits) of a specific binary integer.
dmin = ⌈(b-1)·log10(2)⌉ (dmin
= min decimal digits, b = bits, ⌈x⌉ = smallest integer ≥ x)
dmax = ⌈b·log10(2)⌉ (dmax
= max decimal digits, b = bits, ⌈x⌉ = smallest integer ≥ x)
To learn more about how these formulas are derived, read Number of Decimal Digits In a Binary Integer, written by Rick Regan.
This is all well and good, but you may ask, why is 6 the most decimal digits for a round-trip conversion if you say that the span of decimal digits for a 24-bit number is 7 to 8?
The answer is — because the above formulas only work for integers and not floating-point numbers!
Every decimal integer has an exact value in binary. However, the same cannot be said for every decimal floating-point number. Take .1 for example. .1 in binary is the number 0.000110011001100..., which is a repeating or recurring binary. This can produce rounding error.
Moreover, it takes one more bit to represent a decimal floating-point number than it does to represent a decimal integer of equal significance. This is because floating-point numbers are more precise the closer they are to 0, and less precise the further they are from 0. Because of this, many floating-point numbers near the minimum and maximum value ranges (emin = -126 and emax = +127 for single-precision) lose 1 bit of precision due to rounding error. To see this visually, look at What every computer programmer should know about floating point, part 1, written by Josh Haberman.
Furthermore, there are at least 784,757 positive seven-digit decimal numbers that cannot retain their original value after a round-trip conversion. An example of such a number that cannot survive the round-trip is 8.589973e9. This is the smallest positive number that does not retain its original value.
Here's the formula that you should be using for floating-point number precision that will give you 6 decimal digits for round-trip conversion.
dmax = ⌊(b-1)·log10(2)⌋ (dmax
= max decimal digits, b = bits, ⌊x⌋ = largest integer ≤ x)
To learn more about how this formula is derived, read Number of Digits Required For Round-Trip Conversions, also written by Rick Regan. Rick does an excellent job showing the formulas derivation with references to rigorous proofs.
As a result, you can utilize the above formulas in a constructive way; if you understand how they work, you can apply them to any programming language that uses floating-point data types. All you have to know is the number of significant bits that your floating-point data type has, and you can find their respective number of decimal digits that you can count on to have no loss of significance after a round-trip conversion.
June 18, 2017 Update: I want to include a link to Rick Regan's new article which goes into more detail and in my opinion better answers this question than any answer provided here. His article is "Decimal Precision of Binary Floating-Point Numbers" and can be found on his website www.exploringbinary.com.
Do keep in mind that they are the exact same formulas. Remember your high-school math book identity:
Log(x^y) == y * Log(x)
It helps to actually calculate the values for N = 24 with your calculator:
Kahan's: 23 * Log(2) = 6.924
Wikipedia's: Log(2^24) = 7.225
Kahan was forced to truncate 6.924 down to 6 digits because of floor(), bummer. The only actual difference is that Kahan used 1 less bit of precision.
Pretty hard to guess why, the professor might have relied on old notes. Written before IEEE-754 and not taking into account that the 24th bit of precision is for free. The format uses a trick, the most significant bit of a floating point value that isn't 0 is always 1. So it doesn't need to be stored. The processor adds it back before it performs a calculation. Turning 23 bits of stored precision into 24 of effective precision.
Or he took into account that the conversion from a decimal string to a binary floating point value itself generates an error. Many nice round decimal values, like 0.1, cannot be perfectly converted to binary. It has an endless number of digits, just like 1/3 in decimal. That however generates a result that is off by +/- 0.5 bits, achieved by simple rounding. So the result is accurate to 23.5 * Log(2) = 7.074 decimal digits. If he assumed that the conversion routine is clumsy and doesn't properly round then the result can be off by +/-1 bit and N-1 is appropriate. They are not clumsy.
Or he thought like a typical scientist or (heaven forbid) accountant and wants the result of a calculation converted back to decimal as well. Such as you'd get when you trivially look for a 7 digit decimal number whose conversion back-and-forth does not produce the same number. Yes, that adds another +/- 0.5 bit error, summing up to 1 bit error total.
But never, never make that mistake, you always have to include any errors you get from manipulating the number in a calculation. Some of them lose significant digits very quickly, subtraction in particular is very dangerous.
I have faced an interview question related to embedded systems and C/C++. The question is:
If we multiply 2 signed (2's complement) 16-bit data, what should be the size of resultant data?
I've started attempting it with an example of multiplying two signed 4-bit, so, if we multiply +7 and -7, we end up with -49, which requires 7 bits. But, I could not formulate a general relation.
I think I need to understand binary multiply deeply to solve this question.
First, n bits signed integer contains a value in the range -(2^(n-1))..+(2^(n-1))-1.
For example, for n=4, the range is -(2^3)..(2^3)-1 = -8..+7
The range of the multiplication result is -8*+7 .. -8*-8 = -56..+64.
+64 is more than 2^6-1 - it is 2^6 = 2^(2n-2) ! You'll need 2n-1 bits to store such POSITIVE integer.
Unless you're doing proprietary encoding (see next paragraph) - you'll need 2n bits:
One bit for the sign, and 2n-1 bits for the absolute value of the multiplication result.
If M is the result of the multiplication, you can store -M or M-1 instead. this can save you 1 bit.
This will depend on context. In C/C++, all intermediates smaller than int are promoted to int. So if int is larger than 16-bits, then the result will be a signed 32-bit integer.
However, if you assign it back to a 16-bit integer, it will truncate leaving only bottom 16 bits of the two's complement of the new number.
So if your definition of "result" is the intermediate immediately following the multiply, then the answer is the size of int. If you define the size as after you've stored it back to a 16-bit variable, then answer is the size of the 16-bit integer type.