Of the various types of data that one normally encounters, the ones we are concerned with in the context of scientific computing are the numerical types: integers (or whole numbers) $\ldots,-2,-1,0,1,2,\ldots$, real numbers $0,1,-1.5,2/3,\sqrt 2,\log 10,\ldots$, and complex numbers $1+2i,\sqrt 3-\sqrt 5i,\ldots$. Computer memory is organized to give only a certain amount of space to represent each number, in multiples of bytes , each containing 8 bits . Typical values are 4 bytes for an integer, 4 or 8 bytes for a real number, and 8 or 16 bytes for a complex number.
Since only a certain amount of memory is available to store a number, it is clear that not all numbers of a certain type can be stored. For instance, for integers only a range is stored. In the case of real numbers, even storing a range is not possible since any interval $[a,b]$ contains infinitely many numbers. Therefore, any representation of real numbers will cause gaps between the numbers that are stored. Calculations in a computer are sometimes described as finite precision arithmetic . Since many results are not representible, any computation that results in such a number will have to be dealt with by issuing an error or by approximating the result. In this chapter we will look at the ramifications of such approximations of the `true' outcome of numerical calculations.
For detailed discussions, see the book by Overton [Overton:754book] ; it is easy to find online copies of the essay by Goldberg [goldberg:floatingpoint] . For extensive discussions of round-off error analysis in algorithms, see the books by Higham [Higham:2002:ASN] and Wilkinson [Wilkinson:roundoff] .
In scientific computing, most operations are on real numbers. Computations on integers rarely add up to any serious computation load . It is mostly for completeness that we start with a short discussion of integers.
Integers are commonly stored in 16, 32, or 64 bits, with 16 becoming less common and 64 becoming more and more so. The main reason for this increase is not the changing nature of computations, but the fact that integers are used to index arrays. As the size of data sets grows (in particular in parallel computations), larger indices are needed. For instance, in 32 bits one can store the numbers zero through $2^{32}-1\approx 4\cdot 10^9$. In other words, a 32 bit index can address 4 gigabytes of memory. Until recently this was enough for most purposes; these days the need for larger data sets has made 64 bit indexing necessary.
When we are indexing an array, only positive integers are needed. In general integer computations, of course, we need to accommodate the negative integers too. We will now discuss several strategies for implementing negative integers. Our motivation here will be that arithmetic on positive and negative integers should be as simple as on positive integers only: the circuitry that we have for comparing and operating on bitstrings should be usable for (signed) integers.
There are several ways of implementing negative integers. The simplest solution is to reserve one bit as a sign bit , and use the remaining 31 (or 15 or 63; from now on we will consider 32 bits the standard) bits to store the absolute magnitude. By comparison, we will call the straightforward interpretation of bitstring unsigned integers.
\begin{equation} \begin{array} {|c|rrrrrr|} \hline \hbox{bitstring}& 00\cdots0&\ldots&01\cdots1& 10\cdots0&\ldots&11\cdots1\\ \hline \hbox{interpretation as unsigned int}& 0&\ldots&2^{31}-1& 2^{31}&\ldots&2^{32}-1\\ \hline \hbox{interpretation as signed integer}& 0&\cdots&2^{31}-1& -0&\cdots&-(2^{31}-1)\\ \hline \end{array} \end{equation}
This scheme has some disadvantages, one being that there is both a positive and negative number zero. This means that a test for equality becomes more complicated than simply testing for equality as a bitstring. More importantly, in the second half of the bitstrings, the interpretation as signed integer decreases , going to the right. This means that a test for greater-than becomes complex; also adding a positive number to a negative number now has to be treated differently from adding it to a positive number.
Another solution would be to let an unsigned number $n$ be interpreted as $n-B$ where $B$ is some plausible base, for instance $2^{31}$.
\begin{equation} \begin{array} {|c|rrrrrr|} \hline \hbox{bitstring}& 00\cdots0&\ldots&01\cdots1& 10\cdots0&\ldots&11\cdots1\\ \hline \hbox{interpretation as unsigned int}& 0&\ldots&2^{31}-1& 2^{31}&\ldots&2^{32}-1\\ \hline \hbox{interpretation as shifted int}& -2^{31}&\ldots&-1& 0&\ldots&2^{31}-1\\ \hline \end{array} \end{equation}
This shifted scheme does not suffer from the $\pm0$ problem, and numbers are consistently ordered. However, if we compute $n-n$ by operating on the bitstring that represents $n$, we do not get the bitstring for zero. To ensure this desired behaviour, we instead rotate the number line with positive and negative numbers to put the pattern for zero back at zero.
The resulting scheme, which is the one that is used most commonly, is called 2's complement . Using this scheme, the representation of integers is formally defined as follows.
If $0\leq m\leq 2^{31}-1$, the normal bit pattern for $m$ is used.
For $-2^{31}\leq n\leq -1$, $n$ is represented by the bit pattern for $2^{32}-|n|$.
Some observations:
There is no overlap between the bit patterns for positive and negative integers, in particular, there is only one pattern for zero.
The positive numbers have a leading bit zero, the negative numbers have the leading bit set.
If you have a positive number $n$, you get $-n$ by flipping all the bits, and adding 1.
Exercise
For the `naive' scheme and the 2's complement scheme for negative numbers, give pseudocode for the comparison test $m<n$, where $m$ and $n$ are integers. Be careful to distinguish between all cases of $m,n$ positive, zero, or negative.
Adding two numbers with the same sign, or multiplying two numbers of any sign, may lead to a result that is too large or too small to represent. This is called overflow ; see section for the corresponding floating point phenomenon. The following exercise lets you explore the behaviour of an actual program. If you program this in C, it is worth noting that while you probably get an outcome that is understandable, the behaviour of overflow in the case of signed quantities is actually undefined under the C standard.
Exercise
Investigate what happens when you perform such a calculation. What does your compiler say if you try to write down a nonrepresentible number explicitly, for instance in an assignment statement?
In exercise above you explored comparing two integers. Let us now explore how subtracting numbers in two's complement is implemented. Consider $0\leq m\leq 2^{31}-1$ and $1\leq n\leq 2^{31}$ and let us see what happens in the computation of $m-n$.
Suppose we have an algorithm for adding and subtracting unsigned 32-bit numbers. Can we use that to subtract two's complement integers? We start by observing that the integer subtraction $m-n$ becomes the unsigned addition $m+(2^{32}-n)$.
Case: $m<{n}$. In this case, $m-n$ is negative and $1\leq |m-n|\leq 2^{31}$, so the bit pattern for $m-n$ is that of $2^{32}-(n-m)$. Now, $2^{32}-(n-m)=m+(2^{32}-n)$, so we can compute $m-n$ in 2's complement by adding the bit patterns of $m$ and $-n$ as unsigned integers.
Case: $m>n$. Here we observe that $m+(2^{32}-n)=2^{32}+m-n$. Since $m-n>0$, this is a number $>2^{32}$ and therefore not a legitimate representation of a negative number. However, if we store this number in 33 bits, we see that it is the correct result $m-n$, plus a single bit in the 33-rd position. Thus, by performing the unsigned addition, and ignoring the overflow bit , we again get the correct result.
Some operations are very simple to do in binary: multiplying by 2 corresponds to shifting all the bits one to the left, and dividing by 2 corresponds to a bitshift to the right.
At least, that's the way it is with unsigned integers.
Exercise Are there extra complications when you use bitshifts to multiply or divide by 2 in 2's-complement?
In this section we will look at how real numbers are represented in a computer, and the limitations of various schemes. The next section will then explore the ramifications of this for arithmetic involving computer numbers.
In the mathematical sciences, we usually work with real numbers, so it's convenient to pretend that computers can do this too. However, since numbers in a computer have only a finite number of bits, most real numbers can not be represented exactly. In fact, even many fractions can not be represented exactly, since they repeat; for instance, $1/3=0.333\ldots$. An illustration of this is given in appendix .
Exercise
Some programming languages allow you to write loops with not just an integer, but also with a real number as `counter'. Explain why this is a bad idea. Hint: when is the upper bound reached?
Whether a fraction repeats depends on the number system. (How would you write $1/3$ in ternary, base 3, arithmetic?) In binary computers this means that fractions such as $1/10$, which in decimal arithmetic are terminating, are repeating. Since decimal arithmetic is important in financial calculations, some people care about this kind of arithmetic right; see section for what was done about it.
Real numbers are stored using a scheme that is analogous to what is known as `scientific notation', where a number is represented as a significant and an exponent , for instance $6.022\cdot 10^{23}$, which has a significant 6022 with a radix point after the first digit, and an exponent 23 . This number stands for \begin{equation} 6.022\cdot 10^{23}= \left[ 6\times 10^0+0\times 10^{-1}+2\times10^{-2}+2\times10^{-3} \right] \cdot 10^{23}. \end{equation} We introduce a base , a small integer number, 10 in the preceding example, and 2 in computer numbers, and write numbers in terms of it as a sum of $t$ terms: \begin{equation} x = \pm 1 \times \left[ d_1\beta^0+d_2\beta^{-1}+d_3\beta^{-2}+\cdots+d_t\beta^{-t+1}b\right] \times \beta^e = \pm \Sigma_{i=1}^t d_i\beta^{1-i} \times\beta^e \end{equation} where the components are
the sign bit : a single bit storing whether the number is positive or negative;
$\beta$ is the base of the number system;
$0\leq d_i\leq \beta-1$ the digits of the mantissa or significant -- the location of the radix point (decimal point in decimal numbers) is implicitly assumed to the immediately following the first digit;
$t$ is the length of the mantissa;
$e\in [L,U]$ exponent; typically $L<{0}<{U} $ and $L\approx-U$.
Note that there is an explicit sign bit for the whole number; the sign of the exponent is handled differently. For reasons of efficiency, $e$ is not a signed number; instead it is considered as an unsigned number in excess of a certain minimum value. For instance, the bit pattern for the number zero is interpreted as $e= L$.
Let us look at some specific examples of floating point representations. Base 10 is the most logical choice for human consumption, but computers are binary, so base 2 predominates there. Old IBM mainframes grouped bits to make for a base 16 representation.
$\beta$ | $t$ | $L$ | $U$\\ |
---|
{r|r|r|r|r}
Of these, the single and double precision formats are by far the most common. We will discuss these in section and further.
Decimal numbers are not relevant in scientific computing, but they are useful in financial calculations, where computations involving money absolutely have to be exact. Binary arithmetic is at a disadvantage here, since numbers such as $1/10$ are repeating fractions in binary. With a finite number of bits in the mantissa , this means that the number $1/10$ can not be represented exactly in binary. For this reason, binary-coded-decimal schemes were used in old IBM mainframes, and are in fact being standardized in revisions of IEEE 754 [ieee754-webpage] ; see also section .
In BCD schemes, one or more decimal digits are encoded in a number of bits. The simplest scheme would encode the digits $0\ldots9$ in four bits. This has the advantage that in a BCD number each digit is readily identified; it has the disadvantage that about $1/3$ of all bits are wasted, since 4 bits can encode the numbers $0\ldots15$. More efficient encodings would encode $0\ldots999$ in ten bits, which could in principle store the numbers $0\ldots1023$. While this is efficient in the sense that few bits are wasted, identifying individual digits in such a number takes some decoding. For this reason, BCD arithmetic needs hardware support from the processor, which is rarely found these days; one example is the IBM Power architecture, starting with the IBM Power6 .
There have been some experiments with ternary arithmetic (see http://en.wikipedia.org/wiki/Ternary_computer and http://www.computer-museum.ru/english/setun.htm ), however, no practical hardware exists.
Since we use only a finite number of bits to store floating point numbers, not all numbers can be represented. The ones that can not be represented fall into two categories: those that are too large or too small (in some sense), and those that fall in the gaps. Numbers can be too large or too small in the following ways.
[Overflow] The largest number we can store is \begin{equation} (\beta-1)\cdot1+(\beta-1)\cdot\beta\inv+\cdots +(\beta-1)\cdot\beta^{-(t-1)}=\beta-1\cdot\beta^{-(t-1)}, \end{equation} and the smallest number (in an absolute sense) is $-(\beta-\beta^{-(t-1)}$; anything larger than the former or smaller than the latter causes a condition called overflow .
[Underflow]The number closest to zero is $\beta^{-(t-1)}\cdot \beta^L$. A computation that has a result less than that (in absolute value) causes a condition called underflow . (See section for gradual underflow .)
The occurrence of overflow or underflow means that your computation will be `wrong' from that point on. Underflow will make the computation proceed with a zero where there should have been a nonzero; overflow is represented as \indextermtt{Inf}, short of `infinite'. Computing with Inf is possible to an extent: adding two of those quantities will again give Inf . However, subtracting them gives \indextermtt{NaN}: `not a number'.
Exercise
For real numbers $x,y$, the quantity $g=\sqrt{(x^2+y^2)/2}$ satisfies \begin{equation} g\leq \max\{|x|,|y|\} \end{equation} so it is representable if $x$ and $y$ are. What can go wrong if you compute $g$ using the above formula? Can you think of a better way?
In none of these cases will the computation abort: the processor will continue, unless you tell it otherwise. The `otherwise' consists of you telling the compiler to generate an interrupt , which halts the computation with an error message. See section .
The general definition of floating point numbers, equation \eqref{eq:floatingpoint-def}, leaves us with the problem that numbers have more than one representation. For instance, $.5\times10^{2}=.05\times 10^3$. Since this would make computer arithmetic needlessly complicated, for instance in testing equality of numbers, we use \indextermsub{normalized}{floating point numbers}. A number is normalized if its first digit is nonzero. The implies that the mantissa part is $\beta> x_m\geq 1$.
A practical implication in the case of binary numbers is that the first digit is always 1, so we do not need to store it explicitly. In the IEEE 754 standard, this means that every floating point number is of the form \begin{equation} 1.d_1d_2\ldots d_t\times 2^{\mathrm exp} \end{equation} and only the digits $d_1d_2\ldots d_t$ are stored.
Another implication of this scheme is that we have to amend the definition of underflow (see section above): instead of having a smallest number $-(\beta-\beta^{-(t-1)}$, any number less than $1\cdot\beta^L$ now causes underflow. Trying to compute a number less than that in absolute value is sometimes handled by using unnormalized floating point numbers , a process known as gradual underflow . In this case, a special value of the exponent indicates that the number is no longer normalized. In the case IEEE standard arithmetic (section ) this is done through a zero exponent field.
However, this is typically tens or hundreds of times slower than computing with regular floating point numbers . At the time of this writing, only the IBM Power6 has hardware support for gradual underflow.
Let us consider a real number that is not representable in a computer's number system.
An unrepresentable number is approximated either by ordinary rounding , rounding up or down, or truncation . This means that a machine number $\tilde x$ is the representation for all $x$ in an interval around it. With $t$ digits in the mantissa , this is the interval of numbers that differ from $\bar x$ in the $t+1$st digit. For the mantissa part we get: \begin{equation} \begin{cases} x\in \left[\tilde x,\tilde x+\beta^{-t+1}\right)&\hbox{truncation}\\ x\in \left[\tilde x-\frac12 \beta^{-t+1},\tilde x+\frac12 \beta^{-t+1}\right) &\hbox{rounding} \end{cases} \end{equation}
If $x$ is a number and $\tilde x$ its representation in the computer, we call $x-\tilde x$ the representation error or absolute representation error , and $\frac{x-\tilde x}{x}$ the relative representation error . Often we are not interested in the sign of the error, so we may apply the terms error and relative error to $|x-\tilde x|$ and $|\frac{x-\tilde x}{x}|$ respectively.
Often we are only interested in bounds on the error. If $\epsilon$ is a bound on the error, we will write \begin{equation} \tilde x = x\pm\epsilon \defined |x-\tilde x|\leq\epsilon \Leftrightarrow \tilde x\in[x-\epsilon,x+\epsilon] \end{equation} For the relative error we note that \begin{equation} \tilde x =x(1+\epsilon) \Leftrightarrow \left|\frac{\tilde x-x}{x}\right|\leq \epsilon \end{equation}
Let us consider an example in decimal arithmetic, that is, $\beta=10$,
and with a 3-digit mantissa: $t=3$. The number $x=1.256$ has a
representation that depends on whether we round or truncate: $\tilde
x_{\mathrm{round}}=1.26$, $\tilde x_{\mathrm{truncate}}=1.25$.
The error is in the 4th digit: if $\epsilon=x-\tilde x$
then $|\epsilon|<\beta^{t-1}$.
Exercise The number in this example had no exponent part. What are the error and relative error if there had been one?
Often we are only interested in the order of magnitude of the representation error, and we will write $\tilde x=x(1+\epsilon)$, where $|\epsilon|\leq\beta^{-t}$. This maximum relative error is called the \indexterm{machine precision}, or sometimes machine epsilon \index{machine epsilon|see{machine precision}}. Typical values are: \begin{equation} \begin{cases} \epsilon\approx10^{-7}&\hbox{32-bit single precision}\\ \epsilon\approx10^{-16}&\hbox{64-bit double precision} \end{cases} \end{equation} Machine precision can be defined another way: $\epsilon$ is the smallest number that can be added to $1$ so that $1+\epsilon$ has a different representation than $1$. A small example shows how aligning exponents can shift a too small operand so that it is effectively ignored in the addition operation: \begin{equation} \begin{array} {cll} &1.0000&\times 10^0\\ +&1.0000&\times 10^{-5}\\ \hline \end{array} \quad\Rightarrow\quad \begin{array} {cll} &1.0000&\times 10^0\\ +&0.00001&\times 10^0\\ \hline =&1.0000&\times 10^0 \end{array} \end{equation} Yet another way of looking at this is to observe that, in the addition $x+y$, if the ratio of $x$ and $y$ is too large, the result will be identical to $x$.
The machine precision is the maximum attainable accuracy of computations: it does not make sense to ask for more than 6-or-so digits accuracy in single precision, or 15 in double.
Exercise
% Write a small program that computes the machine epsilon. Does it make any difference if you set the compiler optimization levels low or high?
Exercise
The number $e\approx 2.72$, the base for the natural logarithm, has various definitions. One of them is \begin{equation} e=\lim_{n\rightarrow\infty} (1+1/n)^n. \end{equation} Write a single precision program that tries to compute $e$ in this manner. Evaluate the expression for an upper bound $n=10^k$ with $k=1,\ldots,10$. Explain the output for large $n$. Comment on the behaviour of the error.
Some decades ago, issues like the length of the mantissa and the rounding behaviour of operations could differ between computer manufacturers, and even between models from one manufacturer. This was obviously a bad situation from a point of portability of codes and reproducibility of results. The IEEE 754 standard% % codified all this, for instance stipulating 24 and 53 bits for the mantissa in single and double precision arithmetic, using a storage sequence of sign bit, exponent, mantissa.
The standard also declared the rounding behaviour to be correct rounding : the result of an operation should be the rounded version of the exact result. There will be much more on the influence of rounding (and truncation) on numerical computations, below.
Above (section ), we have seen the phenomena of overflow and underflow, that is, operations leading to unrepresentable numbers. There is a further exceptional situation that needs to be dealt with: what result should be returned if the program asks for illegal operations such as $\sqrt{-4}$? The IEEE 754 standard has two special quantities for this: Inf and NaN for `infinity' and `not a number'. Infinity is the result of overflow or dividing by zero, not-a-number is the result of, for instance, subtracting infinity from infinity. If NaN appears in an expression, the whole expression will evaluate to that value. The rule for computing with Inf is a bit more complicated [goldberg:floatingpoint] .
An inventory of the meaning of all bit patterns in IEEE 754 single precision is given in figure . Recall from section above that for normalized numbers the first nonzero digit is a 1, which is not stored, so the bit pattern $d_1d_2\ldots d_t$ is interpreted as $1.d_1d_2\ldots d_t$.
Exercise
Every programmer, at some point in their life, makes the mistake of storing a real number in an integer or the other way around. This can happen for instance if you call a function differently from how it was defined.
void a(double x) {....} int main() { int i; .... a(i) .... }What happens when you print x in the function? Consider the bit pattern for a small integer, and use the table in figure to interpret it as a floating point number. Explain that it will be an unnormalized number .
These days, almost all processors adhere to the IEEE 754 standard. Early generations of the NVidia Tesla \indexac{GPU}s were not standard-conforming in single precision. The justification for this was that single precision is more likely used for graphics, where exact compliance matters less. For many scientific computations, double precision is necessary, since the precision of calculations gets worse with increasing problem size or runtime. This is true for the sort of calculations in chapter , but not for others such as \indexacp{LBM}
The IEEE 754 standard also declares that a processor should be able to switch its rounding behaviour between ordinary rounding, rounding up or down (sometimes phrased as `towards plus infinity' and `towards minus infinity' respectively) or truncation. In C99, the API for this is contained in \indextermtt{fenv.h}:
#include <fenv.h> int roundings[] = {FE_TONEAREST, FE_UPWARD, FE_DOWNWARD, FE_TOWARDZERO}; rchoice = .... int status = fesetround(roundings[rchoice]);In Fortran2003 the function IEEE_SET_ROUNDING_MODE is available in the IEEE_ARITHMETIC module.
Setting the rounding behaviour can serve as a quick test for the stability of an algorithm: if the result changes appreciably between two different rounding strategies, the algorithm is likely not stable.
The behaviour on overflow can also be set to generate an exception . In C, you specify this with a library call:
#include <fenv.h> int main() { ... feenableexcept(FE_DIVBYZERO | FE_INVALID | FE_OVERFLOW);Interrupts can also be enabled by compiler options.
Various operations can give a result that is not representable as a real number. Processors will represent such a result as NaN : `not a number', and will just as happily compute with these quantities as with legitimate floating point numbers.
The following operations produce a NaN:
Real operations with a complex result, such as the root of a negative number, the logarithm of a non-positive number, or the inverse sine of a number greater than 1 in absolute value.
Indeterminate operations, such as $0/0$, $0\times\inf$, or $\inf-\inf$.
Operations where at least one operand is already a NaN.
Since the processor can continue computing with such a number, it is referred to as a quiet NaN . By contrast, some NaN quantities can cause the processor to generate an interrupt or exception . This is called a signalling NaN .
There are uses for a signalling NaN. You could for instance fill allocated memory with such a value, to indicate that it is uninitialized for the purposes of the computation. Any use of such a value is then a program error, and would cause an exception.
The 2008 revision % of IEEE 754 suggests using the most significant bit of a NaN as the is_quiet bit to distinguish between quiet and signalling NaNs.
Numbers that are too large or too small to be represented, leading to overflow and underflow, are uncommon: usually computations can be arranged so that this situation will not occur. By contrast, the case that the result of a computation between computer numbers (even something as simple as a single addition) is not representable is very common. Thus, looking at the implementation of an algorithm, we need to analyze the effect of such small errors propagating through the computation. This is commonly called round-off error analysis .
The IEEE 754 standard, mentioned in section , does not only declare the way a floating point number is stored, it also gives a standard for the accuracy of operations such as addition, subtraction, multiplication, division. The model for arithmetic in the standard is that of correct rounding % % % : the result of an operation should be as if the following procedure is followed:
The exact result of the operation is computed, whether this is representable or not;
This result is then rounded to the nearest computer number.
If this statement sounds trivial or self-evident, consider subtraction as an example. In a decimal number system with two digits in the mantissa, the computation $1.0-\fp{9.4}{-1}=1.0-0.94=0.06=\fp{0.6}{-2}$. Note that in an intermediate step the mantissa $.094$ appears, which has one more digit than the two we declared for our number system. The extra digit is called a guard digit .
Without a guard digit, this operation would have proceeded as
$1.0-\fp{9.4}{-1}$, where $\fp{9.4}{-1}$ would be rounded to $0.9$,
giving a final result of $0.1$, which is almost double the correct result.
Exercise Consider the computation $1.0-\fp{9.5}{-1}$, and assume again that numbers are rounded to fit the 2-digit mantissa. Why is this computation in a way a lot worse than the example?
One guard digit is not enough to guarantee correct rounding. An analysis that we will not reproduce here shows that three extra bits are needed [Goldberg:arithmetic] .In 2008, the IEEE 754 standard was revised to include the behaviour of \indexacf{FMA} operations, that is, operations of the form a*b+c . This operation has twofold motivation:
First, the FMA is potentially more accurate than a separate multiplication and addition, since it can use higher precision for the intermediate results, for instance by using the 80-bit extended precision \index{precision!extended|see{extended precision}} format.
The standard here defines correct rounding to be that the result of this combined computation should be the rounded correct result. A naive implementation of this operations would involve two roundings: one after the multiplication and one after the addition .
Exercise
Can you come up with an example where correct rounding of an FMA is considerably more accurate than rounding the multiplication and addition separately? Hint: let the c term be of opposite sign as a*b , and try to force cancellation in the subtraction.
Secondly, FMA instructions are a way of getting higher performance: through pipelining we asymptotially get two operations per cycle. An FMA unit is then cheaper to construct than separate addition and multiplication units. Fortunately, the FMA occures frequently in practical calculations.
Exercise
Can you think of some linear algebra operations that features FMA operations?
See section 1.2.1.2 for historic use of FMA in processors.
Addition of two floating point numbers is done in a couple of steps. First the exponents are aligned: the smaller of the two numbers is written to have the same exponent as the larger number. Then the mantissas are added. Finally, the result is adjusted so that it again is a normalized number.
As an example, consider $1.00+2.00\times 10^{-2}$. Aligning the exponents, this becomes $1.00+0.02=1.02$, and this result requires no final adjustment. We note that this computation was exact, but the sum $1.00+2.55\times 10^{-2}$ has the same result, and here the computation is clearly not exact: the exact result is $1.0255$, which is not representable with three digits to the mantissa.
In the example $6.15\times 10^1+3.98\times 10^1=10.13\times 10^1=1.013\times 10^2\rightarrow 1.01\times 10^2$ we see that after addition of the mantissas an adjustment of the exponent is needed. The error again comes from truncating or rounding the first digit of the result that does not fit in the mantissa: if $x$ is the true sum and $\tilde x$ the computed sum, then $\tilde x=x(1+\epsilon)$ where, with a 3-digit mantissa $|\epsilon|<10^{-3}$.
Formally, let us consider the computation of $s=x_1+x_2$, and we assume that the numbers $x_i$ are represented as $\tilde x_i=x_i(1+\epsilon_i)$. Then the sum $s$ is represented as \begin{equation} \begin{array} {rl} \tilde s&=(\tilde x_1+\tilde x_2)(1+\epsilon_3)\\ &=x_1(1+\epsilon_1)(1+\epsilon_3)+x_2(1+\epsilon_2)(1+\epsilon_3)\\ &\approx x_1(1+\epsilon_1+\epsilon_3)+x_2(1+\epsilon_2+\epsilon_3)\\ &\approx s(1+2\epsilon) \end{array} \end{equation} under the assumptions that all $\epsilon_i$ are small and of roughly equal size, and that both $x_i>0$. We see that the relative errors are added under addition.
Floating point multiplication, like addition, involves several steps. In order to multiply two numbers $m_1\times\beta^{e_1}$ and $m_2\times\beta^{e_2}$, the following steps are needed.
The exponents are added: $e\leftarrow e_1+e_2$.
The mantissas are multiplied: $m\leftarrow m_1\times m_2$.
The mantissa is normalized, and the exponent adjusted accordingly.
For example: $\fp{1.23}{0}\times\fp{5.67}1=\fp{0.69741}1\rightarrow \fp{6.9741}0\rightarrow\fp{6.97}0$.
Exercise
Analyze the relative error of multiplication.
Subtraction behaves very differently from addition. Whereas in addition errors are added, giving only a gradual increase of overall roundoff error, subtraction has the potential for greatly increased error in a single operation.
For example, consider subtraction with 3 digits to the mantissa: $1.24-1.23=0.01\rightarrow \fp{1.00}{-2}$. While the result is exact, it has only one significant digit . To see this, consider the case where the first operand $1.24$ is actually the rounded result of a computation that should have resulted in $1.235$. In that case, the result of the subtraction should have been $\fp{5.00}{-3}$, that is, there is a 100\% error, even though the relative error of the inputs was as small as could be expected. Clearly, subsequent operations involving the result of this subtraction will also be inaccurate. We conclude that subtracting almost equal numbers is a likely cause of numerical roundoff.
There are some subtleties about this example. Subtraction of almost equal numbers is exact, and we have the correct rounding behaviour of IEEE arithmetic. Still, the correctness of a single operation does not imply that a sequence of operations containing it will be accurate. While the addition example showed only modest decrease of numerical accuracy, the cancellation in this example can have disastrous effects. You'll see an example in section .
Exercise
Consider the iteration \begin{equation} x_{n+1}=f(x_n) = \ \begin{cases} 2x_n&\hbox{if $2x_n<1$}\\ 2x_n-1&\hbox{if $2x_n\>1$}\\ \end{cases} \end{equation} Does this function have a fixed point, $x_0\equiv f(x_0)$, or is there a cycle $x_1=f(x_0),\,x_0\equiv x_2=f(x_1)$ et cetera?
Now code this function and see what happens with various starting points $x_0$. Can you explain this?
Another effect of the way floating point numbers are treated is on the associativity of summation . While this operation is mathematically associative, this is no longer the case in computer arithmetic.
Let's consider a simple example, showing how this can be caused by the rounding behaviour of floating point numbers. Let floating point numbers be stored as a single digit for the mantissa, one digit for the exponent, and one guard digit ; now consider the computation of $4+6+7$. Evaluation left-to-right gives: \begin{equation} \begin{array} {l@{{}\Rightarrow{}}lp{3in}} (4\cdot10^0 + 6\cdot10^0)+7\cdot10^0&10\cdot10^0+7\cdot10^0&addition\\ &1\cdot 10^1 + 7\cdot10^0&rounding\\ &1.0\cdot 10^1 + 0.7\cdot10^1&using guard digit\\ &1.7\cdot 10^1\\ &2\cdot10^1&rounding \end{array} \end{equation} On the other hand, evaluation right-to-left gives: \begin{equation} \begin{array} {l@{{}\Rightarrow{}}lp{3in}} 4\cdot10^0 + (6\cdot10^0 + 7\cdot10^0)&4\cdot 10^0 + 13\cdot10^0&addition\\ &4\cdot 10^0 + 1\cdot10^1&rounding\\ &0.4\cdot 10^1 + 1.0\cdot10^1&using guard digit\\ &1.4\cdot 10^1\\ &1\cdot10^1&rounding \end{array} \end{equation} The conclusion is that the sequence in which rounding and truncation is applied to intermediate results makes a difference. You can also observe that starting with the smaller numbers gives a more accurate result. In section you'll see a more elaborate example of this principle.
Usually, the evaluation order of expressions is determined by the definition of the programming language, or at least by the compiler. In section we will see how in parallel computations the associativity is not so uniquely determined.
From the above, the reader may got the impression that roundoff errors only lead to serious problems in exceptional circumstances. In this section we will discuss some very practical examples where the inexactness of computer arithmetic becomes visible in the result of a computation. These will be fairly simple examples; more complicated examples exist that are outside the scope of this book, such as the instability of matrix inversion. The interested reader is referred to [Wilkinson:roundoff,Higham:2002:ASN] .
As a practical example, consider the quadratic equation $ax^2+bx+c=0$ which has solutions $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Suppose $b>0$ and $b^2\gg 4ac$ then $\sqrt{b^2-4ac}\approx b$ and the `$+$' solution will be inaccurate. In this case it is better to compute $x_-=\frac{-b-\sqrt{b^2-4ac}}{2a}$ and use $x_+\cdot x_-=c/a$.
Exercise
Write a program that computes the roots of the quadratic equation, both the `textbook' way, and as described above.
Let $b=1$ and $a=-c$, and $4ac\downarrow 0$ by taking progressively smaller values for $a$ and $c$.
Print out the computed root, the root using the stable computation, and the value of $f(x)=ax^2+bx+c$ in the computed root.
The previous example was about preventing a large roundoff error in a single operation. This example shows that even gradual buildup of roundoff error can be handled in different ways.
Consider the sum $\sum_{n=1}^{10000}\frac{1}{n^2}=1.644834$ and assume we are working with single precision, which on most computers means a machine precision of $10^{-7}$. The problem with this example is that both the ratio between terms, and the ratio of terms to partial sums, is ever increasing. In section we observed that a too large ratio can lead to one operand of an addition in effect being ignored.
If we sum the series in the sequence it is given, we observe that the first term is 1, so all partial sums ($\sum_{n=1}^N$ where $N<10000$) are at least 1. This means that any term where $1/n^2<10^{-7}$ gets ignored since it is less than the machine precision. Specifically, the last 7000 terms are ignored, and the computed sum is $1.644725$. The first 4 digits are correct.
However, if we evaluate the sum in reverse order
we obtain the exact result in single precision. We are still adding
small quantities to larger ones, but now the ratio will never be as
bad as one-to-$\epsilon$, so the smaller number is never ignored.
To see this,
consider the ratio of two terms subsequent terms:
\begin{equation}
\frac{n^2}{(n-1)^2}=\frac{n^2}{n^2-2n+1}=\frac1{1-2/n+1/n^2}
\approx 1+\frac2n
\end{equation}
Since we only sum $10^5$ terms and the machine precision is $10^{-7}$,
in the addition $1/n^2+1/(n-1)^2$ the second term will not be wholly
ignored as it is when we sum from large to small.
Exercise There is still a step missing in our reasoning. We have shown that in adding two subsequent terms, the smaller one is not ignored. However, during the calculation we add partial sums to the next term in the sequence. Show that this does not worsen the situation.
The lesson here is that series that are monotone (or close to monotone) should be summed from small to large, since the error is minimized if the quantities to be added are closer in magnitude. Note that this is the opposite strategy from the case of subtraction, where operations involving similar quantities lead to larger errors. This implies that if an application asks for adding and subtracting series of numbers, and we know a priori which terms are positive and negative, it may pay off to rearrange the algorithm accordingly.
Exercise
The sine function is defined as \begin{equation} \begin{array} {rcl} \sin(x) &=& x-\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\\ &=& \sum_{i\geq 0}^\infty (-1)^{2i}\frac{x^{2i+1}}{(2i+1)!}. \end{array} \end{equation} Here are two code fragments that compute this sum (assuming that x and nterms are given): \begin{multicols} {2} \unitindent=0pt \scriptsize
double term = x, sum = term; for (int i=1; i<=nterms; i+=2) { term *= - x*x / (double)((i+1)*(i+2)); sum += term; } printf("Sum: %e\n\n",sum);\columnbreak
double term = x, sum = term; double power = x, factorial = 1., factor = 1.; for (int i=1; i<=nterms; i+=2) { power *= -x*x; factorial *= (factor+1)*(factor+2); term = power / factorial; sum += term; factor += 2; } printf("Sum: %e\n\n",sum);\end{multicols}
Explain what happens if you compute a large number of terms for $x>1$.
Does either code make sense for a large number of terms?
Is it possible to sum the terms starting at the smallest? Would that be a good idea?
Can you come with other schemes to improve the computation of $\sin(x)$?
We will now consider an example where we can give a direct argument that the algorithm can not cope with problems due to inexactly represented real numbers.
Consider the recurrence $y_n=\int_0^1 \frac{x^n}{x-5}dx = \frac1n-5y_{n-1}$. This is easily seen to be monotonically decreasing; the first term can be computed as $y_0=\ln 6 - \ln 5$.
Performing the computation in 3 decimal digits we get:
computation | correct result\\ |
---|
{lll}
We see that the computed results are quickly not just inaccurate, but actually nonsensical. We can analyze why this is the case.
If we define the error $\epsilon_n$ in the $n$-th step as \begin{equation} \tilde y_n-y_n=\epsilon_n, \end{equation} then \begin{equation} \tilde y_n=1/n-5\tilde y_{n-1}=1/n+5n_{n-1}+5\epsilon_{n-1} = y_n+5\epsilon_{n-1} \end{equation} so $\epsilon_n\geq 5\epsilon_{n-1}$. The error made by this computation shows exponential growth.
Sometimes we can make statements about the numerical precision of a problem even without specifying what algorithm we use. Suppose we want to solve a linear system, that is, we have an $n\times n$ matrix $A$ and a vector $b$ of size $n$, and we want to compute the vector $x$ such that $Ax=b$. (We will actually considering algorithms for this in chapter .) Since the vector $b$ will the result of some computation or measurement, we are actually dealing with a vector $\tilde b$, which is some perturbation of the ideal $b$: \begin{equation} \tilde b = b+\Delta b. \end{equation} The perturbation vector $\Delta b$ can be of the order of the machine precision if it only arises from representation error, or it can be larger, depending on the calculations that produced $\tilde b$.
We now ask what the relation is between the exact value of $x$, which we would have obtained from doing an exact calculation with $A$ and $b$, which is clearly impossible, and the computed value $\tilde x$, which we get from computing with $A$ and $\tilde b$. (In this discussion we will assume that $A$ itself is exact, but this is a simplification.)
Writing $\tilde x= x+\Delta x$, the result of our computation is now \[ A\tilde x = \tilde b \] or \begin{equation} A(x+\Delta x)=b+\Delta b. \end{equation} Since $Ax=b$, we get $A\Delta x=\Delta b$. From this, we get (see appendix for details) \begin{equation} \left \{ \begin{array} {rl} \Delta x&=A\inv \Delta b\\ Ax&=b \end{array} \right\} \Rightarrow \left\{ \begin{array} {rl} \|A\| \|x\|&\geq\|b\| \\ \|\Delta x\|&\leq \|A\inv\| \|\Delta b\| \end{array} \right. \Rightarrow \frac{\|\Delta x\|}{\|x\|} \leq \|A\| \|A\inv\| \frac{\|\Delta b\|}{\|b\|} \end{equation} The quantity $\|A\| \|A\inv\|$ is called the \indexterm{condition number} of a matrix. The bound \eqref{eq:xbound} then says that any perturbation in the right hand side can lead to a perturbation in the solution that is at most larger by the condition number of the matrix $A$. Note that it does not say that the perturbation in $x$ needs to be anywhere close to that size, but we can not rule it out, and in some cases it indeed happens that this bound is attained.
Suppose that $b$ is exact up to machine precision, and the condition number of $A$ is $10^4$. The bound \eqref{eq:xbound} is often interpreted as saying that the last 4 digits of $x$ are unreliable, or that the computation `loses 4 digits of accuracy'.
Equation \eqref{eq:xbound} can also be interpreted as follows: when we solve a linear system $Ax=b$ we get an approximate solution $x+\Delta x$ which is the exact solution of a perturbed system $A(x+\Delta x)=b+\Delta b$. The fact that the perturbation in the solution can be related to the perturbation in the system, is expressed by saying that the algorithm exhibits \indexterm{backwards stability}.
The analysis of the accuracy of linear algebra algorithms is a field of study in itself; see for instance the book by Higham [Higham:2002:ASN] .
As we discussed in section , and as you saw in the above example of summing a series, addition in computer arithmetic is not associative \index{floating point arithmetic!associativity of}. A similar fact holds for multiplication. This has an interesting consequence for parallel computations: the way a computation is spread over parallel processors influences the result.
As a very simple example, consider computing the sum of four numbers $a+b+c+d$. On a single processor, ordinary execution corresponds to the following associativity: \begin{equation} ((a+b)+c)+d. \end{equation} On the other hand, spreading this computation over two processors, where processor 0 has $a,b$ and processor 1 has $c,d$, corresponds to \begin{equation} ((a+b)+(c+d)). \end{equation} Generalizing this, we see that reduction operations will most likely give a different result on different numbers of processors. (The MPI standard declares that two program runs on the same set of processors should give the same result.) It is possible to circumvent this problem by replace a reduction operation by a gather operation to all processors, which subsequently do a local reduction. However, this increases the memory requirements for the processors.
There is an intriguing other solution to the parallel summing problem. If we use a mantissa of 4000 bits to store a floating point number, we do not need an exponent, and all calculations with numbers thus stored are exact since they are a form of fixed-point calculation [Kulish:dotproduct,Kulisch:2011:VFE] . While doing a whole application with such numbers would be very wasteful, reserving this solution only for an occasional inner product calculation may be the solution to the reproducibility problem.
From the above discussion it should be clear that some simple statements that hold for mathematical real numbers do not hold for floating-point numbers. For instance, in floating-point arithmetic \begin{equation} (a+b)+c\not=a+(b+c). \end{equation} This implies that a compiler can not perform certain optimizations without it having an effect on round-off behaviour . In some codes such slight differences can be tolerated, for instance because the method has built-in safeguards. For instance, the stationary iterative methods of section damp out any error that is introduced.
On the other hand, if the programmer has written code to account for round-off behaviour, the compiler has no such liberties. This was hinted at in exercise above. We use the concept of value safety to describe how a compiler is allowed to change the interpretation of a computation. At its strictest, the compiler is not allowed to make any changes that affect the result of a computation.
Compilers typically have an option controlling whether optimizations are allowed that may change the numerical behaviour. For the Intel compiler that is -fp-model=... . On the other hand, options such as -Ofast are aimed at performance improvement only, and may affect numerical behaviour severely. For the Gnu compiler full 754 compliance takes the option -frounding-math whereas -ffast-math allows for performance-oriented compiler transformations that violate 754 and/or the language standard.
These matters are also of importance if you care about reproducibility of results. If a code is compiled with two different compilers, should runs with the same input give the same output? If a code is run in parallel on two different processor configurations? These questions are very subtle. In the first case, people sometimes insist on bitwise reproducibility , whereas in the second case some differences are allowed, as long as the result stays `scientifically' equivalent. Of course, that concept is hard to make rigorous.
Here are some issues that are relevant when considering the influence of the compiler on code behaviour and reproducibility.
Re-association
Foremost among changes that a compiler can
make to a computation is
re-association
, the technical
term for grouping $a+b+c$ as $a+(b+c)$. The
\indextermbus{C}{language
standard} and the
C++ language standard
prescribe
strict left-to-right evaluation of expressions without parentheses, so
re-association is in fact not allowed by the standard. The
Fortran language standard
has no such prescription,
but there the compiler has to respect the evaluation order that is
implied by parentheses.
A common source of re-association is loop unrolling ; see section . Under strict value safety, a compiler is limited in how it can unroll a loop, which has implications for performance. The amount of loop unrolling, and whether it's performed at all, depends on the compiler optimization level, the choice of compiler, and the target platform.
A more subtle source of re-association is parallel execution; see section . This implies that the output of a code need not be strictly reproducible between two runs on different parallel configurations.
Constant expressions
It is a common compiler optimization to compute constant expressions
during compile time. For instance, in
float one = 1.; ... x = 2. + y + one;the compiler change the assignment to x = y+3. . However, this violates the re-association rule above, and it ignores any dynamically set rounding behaviour.
Expression evaluation
In evaluating the expression $a+(b+c)$, a processor will generate an
intermediate result for $b+c$ which is not assigned to any variable.
Many processors are able to assign a higher
precision of the intermediate result
.
A compiler can have a flag to dictate whether to use this facility.
Behaviour of the floating point unit
Rounding behaviour (truncate versus round-to-nearest) and treatment of
gradual underflow
may be controlled by library
functions or compiler options.
Library functions
The IEEE 754 standard only prescribes simple operations;
there is as yet no standard that treats sine or log functions.
Therefore, their implementation may be a source of variability.
For more discussion, see [Lionel:reproducibility] .
The example in section made visible some of the problems of computer arithmetic: rounding can cause results that are quite wrong, and very much dependent on evaluation order. A number of algorithms exist that try to compensate for these problems, in particular in the case of addition. We briefly discuss Kahan summation [Kahan:1965:summation] , named after William Kahan , which is one example of a compensated summation algorithm.
Algorithm: $\mathrm{sum}\leftarrow0$ $\mathrm{correction}\leftarrow0$While: there is another \mathrm{input}{ $\mathrm{oldsum}\leftarrow\mathrm{sum}$ $\mathrm{input}\leftarrow\mathrm{input}-\mathrm{correction}$ $\mathrm{sum}\leftarrow\mathrm{oldsum}+\mathrm{input}$ $\mathrm{correction}\leftarrow(\mathrm{sum}-\mathrm{oldsum})-\mathrm{input}$}Exercise
Go through the example in section , adding a final term 3; that is compute $4+6+7+3$ and $6+7+4+3$ under the conditions of that example. Show that the correction is precisely the $3$ undershoot when 17 is rounded to 20, or the $4$ overshoot when 14 is rounded to 10; in both cases the correct result of 20 is computed.
Different languages have different approaches to declaring integers and floating point numbers.
In Fortran, variable declarations can take various forms. For instance, it is possible for a type identifier to declare the number of bytes that it takes to store a variable: INTEGER*2 , REAL*8 . One advantage of this approach is the easy interoperability with other languages, or the MPI library.
Often it is possible to write a code using only INTEGER , REAL , and use compiler flags to indicate the size of an integer and real number in bytes.
More sophisticated, modern versions of Fortran can indicate the number of digits of precision a floating point number needs to have:
integer, parameter :: k9 = selected_real_kind(9) real(kind=k9) :: r r = 2._k9; print *, sqrt(r) ! prints 1.4142135623730The `kind' values will usually be 4,8,16 but this is compiler dependent.
C99 and Fortran2003
Recent standards of the C and Fortran
languages incorporate the C/Fortran interoperability standard, which
can be used to declare a type in one language so that it is compatible
with a certain type in the other language.
In C, the commonly used type identifiers do not correspond to a standard length. For integers there is short int, int, long int , and for floating point float, double . The sizeof() operator gives the number of bytes used to store a datatype.
The numerical ranges of C integers are defined in \indextermtt{limits.h}, typically giving an upper or lower bound. For instance, INT_MAX is defined to be 32767 or greater.
Floating point types are specified in \indextermtt{float.h}.
C integral types with specified storage exist: constants such as int64_t are defined by typedef in \indextermtt{stdint.h}.
The constant NAN is declared in math.h . For checking whether a value is NaN , use isnan() .
Limits
You can still use the C header limits.h or \indextermtt{climits}, but it's better to use \indextermtt{std::numeric_limits}, which is templated over the types. For instance
std::numerical_limits<int>.max();
In C++,
std::numeric_limits
The same module also has infinity() and signaling_NaN() .
For checking whether a value is NaN , use std::isnan() from cmath in C++.
See further http://en.cppreference.com/w/cpp/numeric/math/nan .
Exceptions
Both the IEEE 754 standard and the C++ language define a concept exception which differ from each other. The 754 exception means the occurence of an operation that has `no outcome suitable for every reasonable application'. This does not necessarily translate to the language-defined exception.
Other systems have been proposed to dealing with the problems of inexact arithmetic on computers. One solution is extended precision arithmetic, where numbers are stored in more bits than usual. A common use of this is in the calculation of inner products of vectors: the accumulation is internally performed in extended precision, but returned as a regular floating point number. Alternatively, there are libraries such as GMPlib [gmplib] that allow for any calculation to be performed in higher precision.
Another solution to the imprecisions of computer arithmetic is `interval arithmetic' [wikipedia:interval-arithmetic] , where for each calculation interval bounds are maintained. While this has been researched for considerable time, it is not practically used other than through specialized libraries [boost:interval-arithmetic] .
When the IEEE 754 standard was drawn up, it was envisioned that processors could have a whole range of precisions. In practice, only single and double precision as defined have been used. However, one instance of extended precision still survives: processors sometimes have 80-bit registers for storing intermediate results. (This goes back to the Intel 80287 co-processor .) This strategy makes sense in \indexac{FMA} instructions, and in the accumulation of inner products.
You can ask `does double precision always give benefits over single precision' and the answer is not always `yes' but rather: `it depends'.
In iterative linear system solving (section , the accuracy is determined by how precise the residual is calculated, not how precise the solution step is done. Therefore, one could do operations such as applying the preconditioner (section ) in reduced precision [Dongarra:mixed-refinement] . This is a form of iterative refinement ; see section .
Certain applications, such as deep learning , do not need full precision. The upcoming Intel Knights Mill processor has half precision , 16-bit floating point, modes for this purpose.
A fixed-point number (for a more thorough discussion than found here, see [YatesFixedPoint] ) can be represented as $\fxp NF$ where $N\geq\beta^0$ is the integer part and $F<1$ is the fractional part. Another way of looking at this, is that a fixed-point number is an integer stored in $N+F$ digits, with an implied decimal point after the first $N$ digits.
Fixed-point calculations can overflow, with no possibility to adjust an exponent. Consider the multiplication $\fxp{N_1}{F_1}\times \fxp{N_2}{F_2}$, where $N_1\geq \beta^{n_1}$ and $N_2\geq \beta^{n_2}$. This overflows if $n_1+n_2$ is more than the number of positions available for the integer part. (Informally, the number of digits of the product is the sum of the digits of the operands.) This means that, in a program that uses fixed-point, numbers will need to have a number of zero digits, if you are ever going to multiply them, which lowers the numerical accuracy. It also means that the programmer has to think harder about calculations, arranging them in such a way that overflow will not occur, and that numerical accuracy is still preserved to a reasonable extent.
So why would people use fixed-point numbers? One important application is in embedded low-power devices, think a battery-powered digital thermometer. Since fixed-point calculations are essentially identical to integer calculations, they do not require a floating-point unit, thereby lowering chip size and lessening power demands. Also, many early video game systems had a processor that either had no floating-point unit, or where the integer unit was considerably faster than the floating-point unit. In both cases, implementing non-integer calculations as fixed-point, using the integer unit, was the key to high throughput.
Another area where fixed point arithmetic is still used is in signal processing. In modern CPUs, integer and floating point operations are of essentially the same speed, but converting between them is relatively slow. Now, if the sine function is implemented through table lookup, this means that in $\sin(\sin x)$ the output of a function is used to index the next function application. Obviously, outputting the sine function in fixed point obviates the need for conversion between real and integer quantities, which simplifies the chip logic needed, and speeds up calculations.
Some programming languages have complex numbers as a native data type, others not, and others are in between. For instance, in Fortran you can declare
COMPLEX z1,z2, z(32) COMPLEX*16 zz1, zz2, zz(36)A complex number is a pair of real numbers, the real and imaginary part, allocated adjacent in memory. The first declaration then uses 8 bytes to store to REAL*4 numbers, the second one has REAL*8 s for the real and imaginary part. (Alternatively, use \n{DOUBLE COMPLEX} or in Fortran90 COMPLEX(KIND=2) for the second line.)
By contrast, the C language does not natively have complex numbers, but both C99 and C++ have a complex.h header file . This defines as complex number as in Fortran, as two real numbers.
Storing a complex number like this is easy, but sometimes it is computationally not the best solution. This becomes apparent when we look at arrays of complex numbers. If a computation often relies on access to the real (or imaginary) parts of complex numbers exclusively, striding through an array of complex numbers, has a stride two, which is disadvantageous (see section 1.3.4.7 ). In this case, it is better to allocate one array for the real parts, and another for the imaginary parts.
Exercise
Suppose arrays of complex numbers are stored the Fortran way. Analyze the memory access pattern of pairwise multiplying the arrays, that is, $\forall_i\colon c_i\leftarrow a_i\cdot b_i$, where a(), b(), c() are arrays of complex numbers.
Exercise
Show that an $n\times n$ linear system $Ax=b$ over the complex numbers can be written as a $2n\times 2n$ system over the real numbers. Hint: split the matrix and the vectors in their real and imaginary parts. Argue for the efficiency of storing arrays of complex numbers as separate arrays for the real and imaginary parts.
Computations done on a computer are invariably beset with numerical error. In a way, the reason for the error is the imperfection of computer arithmetic: if we could calculate with actual real numbers there would be no problem. (There would still be the matter of measurement error in data, and approximations made in numerical methods; see the next chapter.) However, if we accept roundoff as a fact of life, then various observations hold:
Mathematically equivalent operations need not behave identically from a point of stability; see the `abc-formula' example.
Even rearrangements of the same computations do not behave identically; see the summing example.