#### Computational Physics Basics: Floating Point Numbers

Posted 25th May 2021 by

In a previous contribution, I have shown you that computers are naturally suited to store finite length integer numbers. Most quantities in physics, on the other hand, are real numbers. Computers can store real numbers only with finite precision. Like storing integers, each representation of a real number is stored in a finite number of bits. Two aspects need to be considered. The precision of the stored number is the number of significant decimal places that can be represented. Higher precision means that the error of the representation can be made smaller. Bur precision is not the only aspect that needs consideration. Often, physical quantities can be very large or very small. The electron charge in SI units, for example, is roughly $1.602\times10^{-19}$C. Using a fixed point decimal format to represent this number would require a large number of unnecessary zeros to be stored. Therefore, the *range* of numbers that can be represented is also important.

In the decimal system, we already have a notation that can capture very large and very small numbers and I have used it to write down the electron charge in the example above. The scientific notation writes a number as a product of a mantissa and a power of 10. The value of the electron charge (without units) is written as

$$1.602\times10^{-19}.$$

Here 1.602 is the mantissa (or the significand) and -19 is the exponent. The general form is

$$m\times 10^n.$$

The mantissa, $m$, will always be between 1 and 10 and the exponent, $n$, has to be chosen accordingly. This format can straight away be translated into the binary system. Here, any number can be written as

$$m\times2^n,$$

with $1\le m<2$. Both $m$ and $n$ can be stored in binary form.

#### Memory layout of floating-point numbers, the IEEE 754 standard

In most modern computers, numbers are stored using 64 bits but some architectures, like smartphones, might only use 32 bits. For a given number of bits, a decision has to be made on how many bits should be used to store the mantissa and how many bits should be used for the exponent. The IEEE 754 standard sets out the memory layout for various size floating-point representations and almost all hardware supports these specifications. The following table shows the number of bits for mantissa and exponent for some IEEE 754 number formats.

Bits | Name | Sign bit | Mantissa bits, m | Exponent bits, p | Exponent bias | Decimal digits |
---|---|---|---|---|---|---|

16 | half-precision | 1 | 10 | 5 | 15 | 3.31 |

32 | single precision | 1 | 23 | 8 | 127 | 7.22 |

64 | double precision | 1 | 52 | 11 | 1023 | 15.95 |

128 | quadruple precision | 1 | 112 | 15 | 16383 | 34.02 |

The layout of the bits is as follows. The first, most significant bit represents the sign of the number. A 0 indicates a positive number and a 1 indicates a negative number. The next $p$ bits store the exponent. The exponent is not stored as a signed integer, but as an unsigned integer with offset. This offset, or bias, is chosen to be $2^p – 1$ so that a leading zero followed by all ones corresponds to an exponent of 0.

The remaining bits store the mantissa. The mantissa is always between 1 and less than 2. This means that, in binary, the leading bit is always equal to one and doesn’t need to be stored. The $m$ bits, therefore, only store the fractional part of the mantissa. This allows for one extra bit to improve the precision of the number.

**Example**

#### Infinity and NaN

The IEEE 754 standard defines special numbers that represent infinity and the not-a-number state. Infinity is used to show that a result of a computation has exceeded the allowed range. It can also result from a division by zero. Infinity is represented by the maximum exponent, i.e. all $p$ bits of the exponent are set to 1. In addition, the $m$ bits of the mantissa are set to 0. The sign bit is still used for infinity. This means it is possible to store a `+Inf`

and a `-Inf`

value.

**Example**

The special state `NaN`

is used to store results that are not defined or can’t otherwise be represented. For example, the operation $\sqrt{-1}$ will result in a not-a-number state. Similar to infinity, it is represented by setting the $p$ exponent bits to 1. To distinguish it from infinity, the mantissa can have any non-zero value.

#### Subnormal Numbers

As stated above, all numbers in the regular range will be represented by a mantissa between 1 and 2 so that the leading bit is always 1. Numbers very close to zero will have a small exponent value. Once the exponent is exactly zero, it is better to explicitly store all bits of the mantissa and allow the first bit to be zero. This allows even smaller numbers to be represented than would otherwise be possible. Extending the range in this way comes at the cost of reduced precision of the stored number.

**Example**

#### Floating Point Numbers in Python, C++, and JavaScript

Both Python and JavaScript exclusively store floating-point numbers using 64-bit precision. In fact, in JavaScript, **all** numbers are stored as 64-bit floating-point, even integers. This is the reason for the fact that integers in JavaScript only have 53 bits. They are stored in the mantissa of the 64-bit floating-point number.

C++ offers a choice of different precisions

Type | Alternative Name | Number of Bits |
---|---|---|

`float` |
single precision | usually 32 bits |

`double` |
double precision | usually 64 bits |

`long double` |
extended precision | architecture-dependent, not IEEE 754, usually 80 bits |

#### Spherical Blast Wave Simulation

Posted 2nd December 2020 by Holger Schmitz

Here is another animation that I made using the Vellamo fluid code. It shows two very simple simulations of spherical blast waves. The first simulation has been carried out in two dimensions. The second one shows a very similar setup but in three dimensions.

You might have seen some youtube videos on blast waves and dimensional analysis on the Sixty Symbols channel or on Numberphile. The criterion for the dimensional analysis given in those videos is true for strong blast waves. The simulation that I carried out, looks at the later stage of these waves when the energy peters out and the strong shock is replaced by a compression wave that travels at a constant velocity. You can still see some of the self-similar behaviour of the Sedov-Taylor solution during the very early stages of the explosion. But after the speed of the shock has slowed down to the sound speed, the compression wave continues to travel at the speed of sound, gradually losing its energy.

The video shows the energy density over time. The energy density includes the thermal energy as well as the kinetic energy of the gas.

For those of you who are interested in the maths and the physics, the code simulates the Euler equations of a compressible fluid. These equations are a model for an ideal adiabatic gas. For more information about the Euler equations check out my previous post.

#### The Euler Equations: Sod Shock Tube

Posted 30th September 2020 by Holger Schmitz

In the last post, I presented a simple derivation of the Euler fluid equations. These equations describe hydrodynamic flow in the form of three conservation equations. The three partial differential equations express the conservation of mass, the conservation of momentum, and the conservation of energy. These fundamental conservation equations are written in terms of fluxes of the densities of the conserved quantities.

In general, it is impossible to solve the Euler equations for an arbitrary problem. This means that in practice when we want to find the hydrodynamic flow in a particular situation, we have to use computers and numerical methods to calculate an approximation to the solution.

Numerical simulation codes approximate the continuous mathematical solution by storing the values of the functions at discrete points. The values are stored using a finite precision format. On most CPUs nowadays numbers will be stored as 64-bit floating-point, but for codes running on GPUs, this might be reduced to 32 bits. In addition to the discretisation in space, the equations are normally integrated using discrete steps in time. All of these factors mean that the computer only stores part of the information of the exact function.

These discretisations mean that the continuous differential equations have to be turned into a prescription how to update the discrete values from one time step to the next. For each differential equation, there are numerous ways to translate them into a numerical algorithm. Each algorithm will make different approximations and introduce different errors into the system. In general, it is impossible for a numerical algorithm to reproduce the solutions of the mathematical equations exactly. This will have implications for the physics that will be simulated with the code. Talking in terms of the Euler equations, some numerical algorithms will have problems capturing shocks, while others might smear out the solutions. Even others might cause the density to turn negative in places. Making sure that physical invariants, such as mass, momentum, and energy are conserved by the algorithm is often an important aspect when designing a numerical scheme.

### The Sod Shock Tube

One of the standard tests for numerical schemes simulating the Euler equations is the Sod shock tube problem. This is a simple one-dimensional setup that is initialised with a single discontinuity in the density and energy density. It then develops a left-going rarefaction wave, a right-going shock and a somewhat slower right-going contact discontinuity. The shock tube has been proposed as a test for numerical schemes by Gary A. Sod 1978 [1]. It was then picked up by others like Philip Roe [2] or Bram van Leer [3] who made it popular. Now, it is one of the standard tests for any new solver for the Euler equations. The advantage of the Sod Shock Tube is that it has an analytical solution that can be compared against. I won’t go into the details of deriving the solution in this post. You can find a sketch of the procedure on Wikipedia.

Here, I want to present an example of the shock tube test using an algorithm by Kurganov, Noelle, and Petrova [4]. The advantage of the algorithm is that it is relatively easy to implement and that is straightforward to extend to multiple dimensions. I have implemented the algorithm in my open-source fluid code Vellamo. At the current stage, the code can only simulate the Euler equations, but it can run in 1d, 2d, or 3d. It is also parallelised so that it can run on computer clusters It uses the Schnek library to manage the computational grids, their distribution onto multiple CPUs, and the communication between the individual processes.

The video shows four simulations run with different grid resolutions. At the highest resolution of 10000 grid points, the results are more or less identical to the analytic solution. The left three panels show the density $\rho$, the momentum density $\rho v$, and the energy density $E$. These are the quantities that are actually simulated by the code. The right three panels show quantities derived from the simulation, the temperature $T$, the velocity $v$, and the pressure $p$.

Starting from the initial discontinuity at $x=0.5$ you can see a rarefaction wave moving to the left. The density $\rho$ has a negative slope whereas the momentum density $\rho v$ increases in this region. In the plot of the velocity, you can see a linear slope, indicating that the fluid is being accelerated by the pressure gradient. The expansion of the fluid causes cooling and the temperature $T$ decreases inside the rarefaction wave.

While the rarefaction wave moves left, a shock wave develops on the very right moving into the undisturbed fluid. All six graphs show a discontinuity at the shock. The density jumps $\rho$ as the fluid is compressed and the velocity $v$ rises from 0 to almost 1 as the expanding fluid pushes into the undisturbed background. The compression also causes a sudden increase in pressure $p$ and temperature $T$.

To the left of the shock wave, you can observe another discontinuity in some but not all quantities. This is what is known as a contact discontinuity. To the left of the discontinuity, the density $\rho$ is high and the temperature $T$ is low whereas to the right density $\rho$ is low and the temperature $T$ is high. Both effects cancel each other when calculating the pressure which is the same on both sides of the discontinuity. And because there is no pressure difference, the fluid is not accelerated either, and the velocity $v$ is the same on both sides as well.

If you look at the plots of the velocity $v$ and the pressure $p$ you can’t make out where the contact discontinuity is. In a way, this is quite extraordinary because these two quantities are calculated from the density $\rho$, the momentum density $\rho v$ and the energy density $E$. All of these quantities have jumps at the location of the contact discontinuity. The level to which the derived quantities stay constant across the discontinuity is an indicator of the quality of the numerical scheme.

The simulations with lower grid resolutions show how the scheme degrades. At 1000 grid points, the results are still very close to the exact solution. The discontinuities are slightly smeared out. If you look closely at the temperature profile, you can see a slight overshoot on the high-temperature side of the contact discontinuity. At 100 grid points, the discontinuities are more smeared out. The overshoot in the temperature is more pronounced and now you can also see an overshoot at the right end of the rarefaction wave. This overshoot is present in most quantities.

As a rule of thumb, a numerical scheme for hyperbolic differential equations has to balance accuracy against numerical diffusion. In regions where the solution is well behaved, high order schemes will provide very good accuracy. Near discontinuities, however, a high order scheme will tend to produce artificial oscillations. These artefacts can create non-physical behaviour when, for instance, the numerical solution predicts a negative density or temperature. A good scheme detects the conditions where the high order algorithm fails and falls back to a lower order in those regions. This will naturally introduce some numerical diffusion into the system.

At the lowest resolution with 50 grid points, the discontinuities are even more spread out. Nonetheless, at later times in the simulation, all discontinuities can be seen and the speed at which these discontinuities move compares well with the exact result.

I hope you have enjoyed this article. If you have any questions, suggestions, or corrections please leave a comment below. If you are interested in contributing to Vellamo or Schnek feel free to contact me via Facebook or LinkedIn.

[1] Sod, G.A., 1978. *A survey of several finite difference methods for systems of nonlinear hyperbolic conservation laws.* Journal of computational physics, **27**(1), pp.1-31.

[2] Roe, P.L., 1981. Approximate Riemann solvers, parameter vectors, and difference schemes. Journal of computational physics, 43(2), pp.357-372.

[3] Van Leer, B., 1979. Towards the ultimate conservative difference scheme. V. A second-order sequel to Godunov’s method. Journal of Computational Physics, 32(1), pp.101-136.

[4] Kurganov, A., Noelle, S. and Petrova, G., 2001. Semidiscrete central-upwind schemes for hyperbolic conservation laws and Hamilton–Jacobi equations. SIAM Journal on Scientific Computing, 23(3), pp.707-740.

#### The Euler Equations

Posted 3rd September 2020 by Holger Schmitz

Leonhard Euler was not only brilliant but also a very productive mathematician. In this post, I want to talk about one of the many things named after the Swiss genius, the Euler equations. These should not be confused with the Euler identity, that famous relation involving $e$, $i$, and $\pi$. No, the Euler equations are a set of partial differential equations that can be used to describe fluid flow. They are related to the Navier-Stokes equations which are at the heart of one of the unsolved Millenium Problems of the Clay Mathematics Institute.

In this short series of posts, I want to present a phenomenological derivation of the equations in one dimension. This derivation is not mathematically rigorous, but will hopefully give you an understanding of the physics that the equations are describing.

### Conservation Laws

One way to understand the Euler equations is to write them in the form of conservation laws. Each equation represents a basic conservation law of physics. A central concept in these conservation equation is **flux**. In a 1d description, the flux of a quantity is a measure for the amount of that quantity that passes through a point. In 2d it’s the amount passing through a line and in 3d through a surface, but let’s stick to 1d during this article.

Take some quantity $N$. To make it concrete, imagine something like the total mass or number of atoms in a volume. I am using the word *volume* loosely. In 1d a volume is the same as an interval. Let’s call the flux of this quantity $F$. The flux can be different at different locations, so $F$ is a function of position $x$, in other words, the flux is $F(x)$.

Next, consider a small volume between $x$ and $x+h$. Here $h$ is the width of the volume and you should think of it as a small quantity. The rate of change of the quantity $N$ within this volume is determined by the difference between the flux into the volume and the flux out of the volume. If we take positive flux to mean that the quantity (mass, number of atoms, stuff) is moving to the right, then we can write down the following equation.

$$\frac{dN}{dt} = F(x) – F(x+h)$$

Here the left-hand side of the equation represents the rate of change of $N$. I have not written it here but, of course, $N$ depends on the position $x$ as well as the with of the volume $h$. The next step to turn this into a conservation equation is to divide both sides by $h$.

$$\frac{d}{dt}\left(\frac{N}{h}\right) = -\frac{F(x+h) – F(x)}{h}$$

If you remember your calculus lessons, you might already see where this is going. We now make $h$ smaller and smaller and look at the limit when $h \to 0$. The right hand side of the equation will give us the derivative of $F$,

$$\lim_{h \to 0} \frac{F(x+h) – F(x)}{h} = \frac{dF}{dx}$$

The amount of stuff $N$ in the volume will shrink and shrink as $h$ decreases, but because we are dividing by $h$ the ratio $N/h$ will converge to a sensible value. Let’s call this value $n$,

$$n = \lim_{h \to 0}\frac{N}{h}.$$

The quantity $n$ is called the density of $N$. Now the conservation law is simply written as

$$\frac{\partial n}{\partial t} = -\frac{\partial F}{\partial x}.$$

This equation expresses the change of the density of the conserved quantity through the derivative of the flux of that quantity.

### The Equations

Now that the basic form of the conservation equations is established, let’s look at some quantities that are conserved and that can be used to express fluid flow. The first conservation law to look at is the conservation of mass. Note again that the conservation equation is written in terms of the density of the conserved quantity. The mass density is the mass contained in a small volume divided by that volume. It is often abbreviated by the Greek letter $\rho$ (rho).

In order to find out what the mass flux is, consider the fluid moving with velocity $v$ through some point. During a small time interval, $\Delta t$ the fluid will move by a small distance $\Delta x$. All the mass $M$ contained in the interval of width $\Delta x$ will cross the point. This mass can be calculated to be

$$M = \rho \Delta x.$$

The flux is the amount of mass crossing the point divided by the time it took the mass to cross that point, $F_{\rho} = M / \Delta t$. In other words,

$$F_{\rho} = \rho \frac{\Delta x}{\Delta t} = \rho v.$$

This now lets us write the mass conservation equation,

$$\frac{\partial \rho}{\partial t} = -\frac{\partial }{\partial x}(\rho v).$$

What does this equation mean? Let me explain this with two examples. First, imagine a constant flow velocity $v$ but an increasing density profile $\rho(x)$. At any given point the density will decrease because the density profile will constantly move towards the right without changing shape. The lower density that was previously located at somewhat left of $x$ will move to the point $x$ thus decreasing the density there.

For the other example, imagine that the density is constant but the velocity has an increasing profile. Matter to the right will flow faster than matter to the left. This will also decrease the density over time because the existing mass is constantly thinned out.

The next conservation equation expresses the conservation of momentum. Momentum is mass times velocity and therefore momentum density is mass density times velocity, $\rho v$. In the mass conservation equation, you could see that the flux consists only of the conserved density multiplied with the velocity. This is called the convective term and this term is present in all Euler equations. However, the momentum can also change through a force. Without external forces, the only force onto each fluid element is through the pressure $p$. The momentum conservation equation can be written as

$$\frac{\partial }{\partial t}(\rho v) = -\frac{\partial }{\partial x}(\rho v^2 + p).$$

The last conservation equation is the conservation of energy, expressed by the energy density $e$,

$$\frac{\partial e}{\partial t} = -\frac{\partial }{\partial x}(e v + p v).$$

The first term is again the convective term expressing the transport of energy contained in the fluid as the fluid moves. The second term is the work done by the pressure force. Note that the energy density $e$ contains the internal (heat) energy as well as the kinetic energy of the fluid moving as a whole.

Finally, we need to close the set of equations by defining what the pressure $p$ is. This closure depends on the type of fluid you want to model. For an ideal gas, the pressure is related to the other variables through

$$p = (\gamma – 1)\left(E – \frac{\rho v^2}{2}\right).$$

Here $\gamma$ is the adiabatic gas index. The second bracket on the right-hand side is the internal energy, expressed as the total energy minus the kinetic energy of the flow.

### Putting it All Together

In summary, the Euler equations are conservation equations for the mass, momentum, and energy in the system. In 1d, they can be written as

$$\begin{eqnarray}

\frac{\partial }{\partial t}\rho &=& -\frac{\partial }{\partial x}(\rho v) \\

\frac{\partial }{\partial t}(\rho v) &=& -\frac{\partial }{\partial x}(\rho v^2 + p) \\

\frac{\partial }{\partial t}e &=& -\frac{\partial }{\partial x}(e v + p v)

\end{eqnarray}

$$

In higher dimensions, the derivative with respect to $x$ is replaced by the divergence operator $\nabla$.

$$\begin{eqnarray}

\frac{\partial }{\partial t}\rho &=& -\nabla(\rho \mathbf{v}) \\

\frac{\partial }{\partial t}(\rho v) &=& -\nabla(\rho \mathbf{v}\otimes\mathbf{v} + p\mathbf{I}) \\

\frac{\partial }{\partial t}e &=& -\nabla(e \mathbf{v} + p \mathbf{v})

\end{eqnarray}

$$

The $\otimes$ symbol denotes the outer product of the velocity vectors. It multiplies two vectors to produce a matrix. I will not go into the definition of the outer product here. You can read more on Wikipedia. The symbol $\mathbf{I}$ is the identity matrix.

### Summary

The Euler equations are a set of partial differential equations describing fluid flow. When Euler published a simpler version of these equations, they were one of the earliest examples of partial differential equations. Today, the Euler equations and their generalisations, including the Navier-Stokes equations provide the basis for understanding fluid behaviour, turbulence, and the weather.

The Euler equations are difficult to solve and only some special solutions can be written down on paper. For the general case, we need computers to give us answers. In the next post, I will show the results of some simulations I have created with my freely available fluid code Vellamo.

#### Computational Physics Basics: Integers in C++, Python, and JavaScript

Posted 5th August 2020 by Holger Schmitz

In a previous post, I wrote about the way that the computer stores and processes integers. This description referred to the basic architecture of the processor. In this post, I want to talk about how different programming languages present integers to the developer. Programming languages add a layer of abstraction and in different languages that abstraction may be less or more pronounced. The languages I will be considering here are C++, Python, and JavaScript.

### Integers in C++

C++ is a language that is very close to the machine architecture compared to other, more modern languages. The data that C++ operates on is stored in the machine’s memory and C++ has direct access to this memory. This means that the C++ integer types are exact representations of the integer types determined by the processor architecture.

The following integer datatypes exist in C++

Type | Alternative Names | Number of Bits | G++ on Intel 64 bit (default) |
---|---|---|---|

`char` |
at least 8 | 8 | |

`short int` |
`short` |
at least 16 | 16 |

`int` |
at least 16 | 32 | |

`long int` |
`long` |
at least 32 | 64 |

`long long int` |
`long long` |
at least 64 | 64 |

This table does not give the exact size of the datatypes because the C++ standard does not specify the sizes but only lower limits. It is also required that the larger types must not use fewer bits than the smaller types. The exact number of bits used is up to the compiler and may also be changed by compiler options. To find out more about the regular integer types you can look at this reference page.

The reason for not specifying exact sizes for datatypes is the fact that C++ code will be compiled down to machine code. If you compile your code on a 16 bit processor the plain `int`

type will naturally be limited to 16 bits. On a 64 bit processor on the other hand, it would not make sense to have this limitation.

Each of these datatypes is signed by default. It is possible to add the `signed`

qualifier before the type name to make it clear that a signed type is being used. The `unsigned`

qualifier creates an unsigned variant of any of the types. Here are some examples of variable declarations.

char c; // typically 8 bit unsigned int i = 42; // an unsigned integer initialised to 42 signed long l; // the same as "long l" or "long int l"

As stated above, the C++ standard does not specify the exact size of the integer types. This can cause bugs when developing code that should be run on different architectures or compiled with different compilers. To overcome these problems, the C++ standard library defines a number of integer types that have a guaranteed size. The table below gives an overview of these types.

Signed Type | Unsigned Type | Number of Bits |
---|---|---|

`int8_t` |
`uint8_t` |
8 |

`int16_t` |
`uint16_t` |
16 |

`int32_t` |
`uint32_t` |
32 |

`int64_t` |
`uint64_t` |
64 |

More details on these and similar types can be found here.

The code below prints a 64 bit `int64_t`

using the binary notation. As the name suggests, the `bitset`

class interprets the memory of the data passed to it as a bitset. The bitset can be written into an output stream and will show up as binary data.

#include <bitset> void printBinaryLong(int64_t num) { std::cout << std::bitset<64>(num) << std::endl; }

### Integers in Python

Unlike C++, Python hides the underlying architecture of the machine. In order to discuss integers in Python, we first have to make clear which version of Python we are talking about. Python 2 and Python 3 handle integers in a different way. The Python interpreter itself is written in C which can be regarded in many ways as a subset of C++. In Python 2, the integer type was a direct reflection of the `long int`

type in C. This meant that integers could be either 32 or 64 bit, depending on which machine a program was running on.

This machine dependence was considered bad design and was replaced be a more machine independent datatype in Python 3. Python 3 integers are quite complex data structures that allow storage of arbitrary size numbers but also contain optimizations for smaller numbers.

It is not strictly necessary to understand how Python 3 integers are stored internally to work with Python but in some cases it can be useful to have knowledge about the underlying complexities that are involved. For a small range of integers, ranging from -5 to 256, integer objects are pre-allocated. This means that, an assignment such as

n = 25

will not create the number 25 in memory. Instead, the variable `n`

is made to reference a pre-allocated piece of memory that already contained the number 25. Consider now a statement that might appear at some other place in the program.

a = 12 b = a + 13

The value of `b`

is clearly 25 but this number is not stored separately. After these lines `b`

will reference the exact same memory address that `n`

was referencing earlier. For numbers outside this range, Python 3 will allocate memory for each integer variable separately.

Larger integers are stored in arbitrary length arrays of the C `int`

type. This type can be either 16 or 32 bits long but Python only uses either 15 or 30 bits of each of these "digits". In the following, 32 bit `int`

s are assumed but everything can be easily translated to 16 bit.

Numbers between −(2^{30} − 1) and 2^{30} − 1 are stored in a single `int`

. Negative numbers are not stored as two’s complement. Instead the sign of the number is stored separately. All mathematical operations on numbers in this range can be carried out in the same way as on regular machine integers. For larger numbers, multiple 30 bit digits are needed. Mathamatical operations on these large integers operate digit by digit. In this case, the unused bits in each digit come in handy as carry values.

### Integers in JavaScript

Compared to most other high level languages JavaScript stands out in how it deals with integers. At a low level, JavaScript does not store integers at all. Instead, it stores all numbers in floating point format. I will discuss the details of the floating point format in a future post. When using a number in an integer context, JavaScript allows exact integer representation of a number up to 53 bit integer. Any integer larger than 53 bits will suffer from rounding errors because of its internal representation.

const a = 25; const b = a / 2;

In this example, `a`

will have a value of 25. Unlike C++, JavaScript does not perform integer divisions. This means the value stored in `b`

will be 12.5.

JavaScript allows bitwise operations only on 32 bit integers. When a bitwise operation is performed on a number JavaScript first converts the floating point number to a 32 bit signed integer using two’s complement. The result of the operation is subsequently converted back to a floating point format before being stored.

#### Computational Physics Basics: How Integers are Stored

Posted 4th April 2020 by Holger Schmitz

### Unsigned Integers

Computers use binary representations to store various types of data. In the context of computational physics, it is important to understand how numerical values are stored. To start, let’s take a look at non-negative integer numbers. These unsigned integers can simply be translated into their binary representation. The binary number-format is similar to the all-familiar decimal format with the main difference that there are only two values for the digits, not ten. The two values are **0** and **1**. Numbers are written in the same way as decimal numbers only that the place values of each digit are now powers of 2. For example, the following 4-digit numbers show the values of the first four

0 0 0 1 decimal value 2^{0} = 1

0 0 1 0 decimal value 2^{1} = 2

0 1 0 0 decimal value 2^{2} = 4

1 0 0 0 decimal value 2^{3} = 8

The binary digits are called **bits** and in modern computers, the bits are grouped in units of 8. Each unit of 8 bits is called a **byte** and can contain values between 0 and 2^{8} − 1 = 255. Of course, 255 is not a very large number and for most applications, larger numbers are needed. Most modern computer architectures support integers with 32 bits and 64 bits. Unsigned 32-bit integers range from 0 to 2^{32} − 1 = 4, 294, 967, 295 ≈ 4.3 × 10^{9} and unsigned 64-bit integers range from 0 to 2^{64} − 1 = 18, 446, 744, 073, 709, 551, 615 ≈ 1.8 × 10^{19}. It is worthwhile noting that many GPU architectures currently don’t natively support 64-bit numbers.

The computer’s processor contains registers that can store binary numbers. Thus a 64-bit processor contains 64-bit registers and has machine instructions that perform numerical operations on those registers. As an example, consider the addition operation. In binary, two numbers are added in much the same way as using long addition in decimal. Consider the addition of two 64 bit integers 7013356221863432502 + 884350303838366524. In binary, this is written as follows.

01100001,01010100,01110010,01010011,01001111,01110010,00010001,00110110 + 00001100,01000101,11010111,11101010,01110101,01001011,01101011,00111100 --------------------------------------------------------------------------- 01101101,10011010,01001010,00111101,11000100,10111101,01111100,01110010

The process of adding two numbers is simple. From right to left, the digits of the two numbers are added. If the result is two or more, there will be a carry-over which is added to the next digit on the left.

You could add integers of any size using this prescription but, of course, in the computer numbers are limited by the number of bits they contain. Consider the following binary addition of (2^{64} − 1) and 1 .

11111111,11111111,11111111,11111111,11111111,11111111,11111111,11111111 + 00000000,00000000,00000000,00000000,00000000,00000000,00000000,00000001 --------------------------------------------------------------------------- 00000000,00000000,00000000,00000000,00000000,00000000,00000000,00000000

If you were dealing with mathematical integers, you would expect to see an extra digit `1`

on the left. The computer cannot store that bit in the register containing the result but stores the extra bit in a special **carry flag**. In many computer languages, this unintentional overflow will go undetected and the programmer has to take care that numerical operations do not lead to unintended results.

### Signed Integers

The example above shows that adding two non-zero numbers can result in 0. This can be exploited to define negative numbers. In general, given a number *a*, the negative − *a* is defined as the number that solves the equation

*a* + ( − *a*) = 0.

Mathematically, the *N*-bit integers can be seen as the group of integers modulo 2^{N}. This means that for any number *a* ∈ {0, …, 2^{N} − 1} the number − *a* can be defined as

− *a* = 2^{N} − *a* ∈ {0, …, 2^{N} − 1}.

By convention, all numbers whose highest value binary bit is zero are considered positive. Those numbers whose highest value bit is one are considered negative. This makes the addition and subtraction of signed integers straightforward as the processor does not need to implement different algorithms for positive or negative numbers. Signed 32-bit integers range from − 2, 147, 483, 648 to 2, 147, 483, 647, and 64-bit integers range from − 9, 223, 372, 036, 854, 775, 808 to 9, 223, 372, 036, 854, 775, 807.

This format of storing negative numbers is called the **two’s complement** format. The reason for this name becomes obvious when observing how to transform a positive number to its negative.

01100001,01010100,01110010,01010011,01001111,01110010,00010001,00110110 (7013356221863432502) 10011110,10101011,10001101,10101100,10110000,10001101,11101110,11001010 (-7013356221863432502)

To invert a number, first, invert all its bits and then add 1. This simple rule of taking the two’s complement can be easily implemented in the processor’s hardware. Because of the simplicity of this prescription, and the fact that adding a negative number follows the same algorithm as adding a positive number, two’s complement is de-facto the only format used to store negative integers on modern processors.

### Exercises

- Show that taking the two’s complement of an
*N*-bit number*a*does indeed result in the negative −*a*if the addition of two numbers is defined as the addition modulo 2^{N}. - Find out how integers are represented in the programming language of your choice. Does this directly reflect the representation of the underlying architecture? I will be writing another post about this topic soon.
- Most processors have native commands for multiplying two integers. The result of multiplying the numbers in two
*N*-bit registers are stored in two*N*-bit result registers representing the high and low bits of the result. Show that the resulting 2*N*bits will always be enough to store the result. - Show how the multiplication of two numbers can be implemented using only the bit-shift operator and conditional addition based on the bit that has been shifted out of the register. The bit-shift operator simply shifts all bits of a register to the left or right.

#### Interactive Data Visualisation in Schnek

Posted 18th April 2018 by Holger Schmitz

Back in the old days computer were simple devices. When I got my first computer it was a Commodore 64. There were some games you could play but most of the fun to be had was in programming the device. With the **Basic** language you could directly access any place in the memory and manipulate its contents. In fact, this was the way to use most of the features of the C64. Special addresses in the memory would control the sound or the graphics. By simply changing the value you could start the sound generator or switch between the different screen modes. If you wanted to create graphics you would simply write the correct bits directly into the graphics memory and the pixels would instantly appear on the screen.

Of course, when I learnt about fractals and the Mandelbrot set I didn’t waste much time to try out creating these fascinating images on my home computer. On the right you can see what this might have looked like. The image was created using the Vice emulator for the Commodore 64. The Basic program is available on GitHub. It is not the original code that I used back in the past. I don’t have that any more. Instead it is a slightly modified version of the Mandelbrot code written by Joseph Dewey available from here. On the C64 it would have taken about 6 hours to produce the finished picture.

Ever since those days I was fascinated with using maths and physics to create computer images. This is what ultimately drove me into computational physics. Nowadays I run simulation codes on large computers. In almost all cases the software will not produce graphical output directly, but instead it will write lots of data to the hard drive. The data is turned into images during a second processing step. This is sensible because the codes will be run by a batch system on a computer cluster, and it will often run overnight. Creating nice graphics directly during run-time is not very useful in this situation.

On the other hand, I often think of the old times when I could play around with parameters and immediately see what is happening. Even nowadays this feature could still be useful when performing small test runs of the full scale simulations. Test runs are frequently needed to make sure that the parameters of a simulation are correct before starting an expensive run using hundreds or even thousands of cores. For this reason, I have written a small extension to the Schnek simulation library that will create a window and plot a colour plot of any two-dimensional data field.

The extension uses GTK to create a window inside a Schnek Block. This means that you can control the appearance of windows through the setup file of the simulation you are running. You should be able to open multiple windows and display different data in each window. The current implementation is relatively simple and will only work when running a simulation on a single processor. The code can be found on my GitHub repository.

Here are some implementation details. Just like any other UI toolkit, GTK likes to take control. This means that at some point you are supposed to call a function that will not return until the program is ended. Inside that function all the low level user interaction is taken care of. Your code then just sits and waits until the user has chosen to do something and GTK will call into your code. Naturally, this is not ideal for a simulation code. For this reason, I implemented a class that creates a new thread and runs GTK within this thread. At regular intervals the GTK thread checks if the `ColorPlot`

component has updated its data and then proceeds to refresh the window that shows the data. I have manually implemented double buffering to avoid a flashing screen effect. Right now, there is only one single, very ugly, colour map to display the data. An example of a simulation output can be seen in the image on the left.