Fourier Transform in C#

I am fascinated by the Fourier transform; specifically: its ability to decompose a wave into a series of fundamental waves. Mathematicians might prefer to see it described it more formally, but here I'll be trying to make it understandable.

Obviously, for a wave to have motion, time needs to pass. If we freeze time then our wave cannot oscillate. To say a wave has frequency, we need first to observe a period in which the wave has motion. The longer we observe its movement the more we can mathematically extract from it, but for now let's simply say that we need to acquire a range of samples measured as the wave oscillates. From this point, armed with our array of floating point numerical samples, we menacingly approach the rabbit hole.

The remainder of this post will focus on a single practical aspect of the Fourier transform: how to interpret the result, especially its magnitude. I'm using the MathNet.Numerics package from Nuget, which provides the handy method Fourier.Forward(samples, FourierOptions.NoScaling).
Note: you need to specifically pass FourierOptions.NoScaling if you want control of the output.

Since I grew up in an age when Compact Discs were seen as technological marvels, I am going to use the standard 44.1 kHz as my sample rate for the following example. A full second's worth of samples (44100) might seem a reasonable place to start, but as we'll see later, it's by no means a requirement. The chosen sample rate theoretically limits the upper frequency to 22050 Hz, and even though we pass in 44100 samples to Fourier.Forward we get back a symmetrical result with the first 22050 data points carrying the frequencies up to 22050 Hz, and the second 22050 carrying them all in again reverse. We can safely ignore the second half of the output for this post. If our input signal (encoded in the 44100 samples) was a pure sinusoidal signal (of any frequency up to 22050 Hz) oscillating between +1 and -1, it will contribute 22050 units to the frequency bin associated with the signal. Those units can be found in the Complex32.Magnitude property. Multiplying the amplitude of the input signal by any constant A will scale the magnitude to A x 22050. However, if we pass in a fraction of the full second's worth of samples - let's say \(\frac{4}{5}\), which is 35280 instead of 44100 - then we see the number of units fall to \(\sqrt{\frac{4}{5}}A\frac{samplerate}{2}\). And yes, the number would rise if the fraction was greater than one.

It's probably important to note that we haven't changed the sample rate: that's still 44100 and it still means we cannot get any higher frequencies than 22050 Hz. What we've done is alter the resolution. With \(\frac{4}{5}\) of the original resolution each frequency bin would be \(\frac{5}{4}\) as wide so that we could still accommodate all 22050 Hz, but in fewer distinct bins. It's not a net loss, though; it's a trade off: we've lost resolution in the frequency domain but we've gained resolution in the time domain (there will be \(\frac{5}{4}\) as many Fourier transforms in any given time period). Of course, this now means we cannot simply read off the frequency from the bin number; now we need to scale the bin number by our resolution factor. Staying with the \(\frac{4}{5}\) example: 800 Hz would now be found in bin 640, which is \(\frac{4}{5}\) of 800.

So... why did we choose to use the Complex32.Magnitude property, and what does it mean? Each of the data points we get as output from Fourier.Forward is a complex number, having both a Complex32.Real and an Complex32.Imaginary part. Besides these two orthogonal vectors pointing in the real and imaginary axes, another way to interpret them is as a single vector that has two properties: Complex32.Magnitude and Complex32.Phase. Together, these allow us to imagine the frequency bin as a clock with a single spinning hand of variable length. If the hand is long: there's more signal; if it's short: there is less. And the angle of the hand shows us the offset of the signal (relative to convention).

Input Shift	Re	Im	Mag	Phase
+0	0	-22050	22050	\(-\frac{π}{2}\)
+\(\frac{π}{2}\)	22050	0	22050	0
+π	0	22050	22050	\(\frac{π}{2}\)
+\(\frac{3π}{2}\)	-22050	0	22050	π

Finally, how did we get 22050 for the maximum Complex32.Magnitude given a maximum amplitude of 1 (and minimum amplitude of -1) for our input signal? Well, that's how the Fourier transform works: it essentially adds up the 22050 vector results of pairing the input signal with the known pure sinusoidal signal.

Demand and Supply: Part 2

In the previous post, there was a table of coefficients for average cost (AC), total cost (TC) and marginal cost (MC). Let's decompose TC into two components: one that doesn't vary as quantity (Q) varies and another that does. We'll call the components total fixed cost (TFC) and total variable cost (TVC). We can also decompose AC into average fixed cost (AFC) and average variable cost (AVC)

	AC	AFC	AVC	TC	TFC	TVC	MC
Q-1	a	a	0				0
Q0	b	0	b	a	a	0	b
Q1	c	0	c	b	0	b	2c
Q2	d	0	d	c	0	c	3d
Q3				d	0	d

AFC: \(p = \frac{a}{q}\)
AVC: \(p = b + cq + dq^2\)
TFC: \(p = a\)
TVC: \(p = bq + cq^2 + dq^3\)

Note how \(AC = AFC + AVC\) and \(TC = TFC + TVC\)
Note too how MR is unrelated to TFC, but has the same relation to both TC and TVC.

Demand and Supply: Part 1

Let's use a little bit of mathematics to assist our understanding of demand and supply, focusing here on the slightly more complicated concept of supply (demand functions identically, but we don't often decompose revenue into fixed and variable components). In the study of economics we often encounter three "curves" that describe how the price of an item is related to the quantity (Q) a firm is willing to supply: average cost (AC), total cost (TC) and marginal cost (MC). These three functions are mathematically related by two equations: \(AC = \frac{TC}{Q}\) (which also means that \(TC = ACxQ\)) and \(\frac{dTC}{dQ} = MC\) (which also means that \(TC_q = \int_0^q MC dQ\)). It's too soon to explain, but these are not quantities the firm is willing to supply, but rather the quantities the firm is able to supply.

A third degree polynomial function should be sufficient for this demonstration. In the table below, I've displayed the coefficients of each power of Q, highlighting the "fixed" component of TC (at Q⁰) and AC (at Q^-1) noting that MC has no such "fixed" component.

	AC	TC	MC
Q-1	a		0
Q0	b	a	b
Q1	c	b	2c
Q2	d	c	3d
Q3		d

AC: \(p = \frac{a}{q} + b + cq + dq^2\)
TC: \(p = a + bq + cq^2 + dq^3\)
MC: \(p = b + 2cq + 3dq^2\)

If the firm has any fixed costs (e.g. monthly rental payments on commercial property), they would be represented by the coefficient \(a\); if it has variable costs (e.g. monthly wages payable to labour) they would be represented by the coefficients \(b\),\(c\) and \(d\) (these are the costs that vary as Q is varied.

Eigenv{ector;alue}s

There are some nice instructional videos on YouTube that explain eigenvectors and eigenvalues. What follows is a brief summary as well as a caveat or two.

We start with a square transformation matrix M, which can have one or more eigenvectors. If you restrict eigenvalues to real numbers (i.e. if you exclude imaginary numbers) then it's possible to have a transformation matrix without any real eigenvalues (e.g. in a pure rotation). It's also possible to have just one eigenvalue but an infinite number of eigenvectors (e.g. common scale in all axes) ... but I've probably distracted you too much already...

Anyway:
What's an eigenvector? It's a vector that only changes (stretches/squeezes) in scale along its span when transformed by M.
What's a span? That's a radial line to any vector from the origin of the coordinate system.
Each eigenvector has a corresponding scalar eigenvalue λ that, when the two are multiplied, has the same result as M being multiplied by that eigenvector.
It's a rather circular definition so far:
\(\matrix{M}\times\vec{v} = λ\times\vec{v}\)
We already know about a transformation matrix that represents a uniform scale of λ, and that's \(λ\times\matrix{I}\) (it's the identity matrix, but instead of 1s running down the diagonal, it's got λs). Now we have a starting point for our calculation:
\(\matrix{M}\times\vec{v} = (λ\times\matrix{I})\times\vec{v}\)
As tempting as it looks, we should never divide by a vector. Instead, what we want next is the value of λ that results in:
\((\matrix{M} - (λ\times\matrix{I}))\times\vec{v} = \vec{0}\)
Let's let:
\(\matrix{T} = \matrix{M} - (λ\times\matrix{I})\)
Ordinarily, it would now be trivial to solve for the eigenvector using:
\(\matrix{T}\times\vec{v} = \vec{0}\)
\(=> \matrix{T}^{-1}\times\vec{0} = \vec{v}\)
But we cannot do that in this instance, because:
\(det(\matrix{T}) = 0\)
And that means that T is not invertible. But surely that means we've missed a step? Nobody said we need the determinant to be zero. What's happening? I'll explain: if \(\vec{v} = \vec{0}\) then anything multiplied by \(\vec{v}\) will also equal \(\vec{0}\).
If we restrict our eigenvector to non-zero vectors then the only way we can get a zero vector after transformation is if the transformation matrix collapses a dimension: i.e. the determinant (or scale) of the transform must be zero. That's how we get our constraint.

Note: Matrices are either invertible or singular. A matrix is invertible iff the determinant is non-zero, and it is singular iff the determinant is zero. Our transform T must be singular. That means: it collapses down by a dimension.

So, if the determinant of our 2X2 transformation matrix is zero:
\(det\begin{bmatrix}a-λ & b \\ c & d-λ \\\end{bmatrix} = 0\)
Then we know (from the Leibniz formula of a determinant) \((a-λ) \times (d-λ) - c \times b = 0\)
And we can expand to:
\(λ^2 - λ(a+d) + (a \times d) - (c \times b) = 0\)
That's just a quadratic, easily solved (but take care with complex numbers, i does appear in rotation matrices). So we have our eigenvalue(s) λ and (by substitution) also T. But how do we find the eigenvectors if we cannot take the inverse of T? Well, you can try a technique like row reduction. But don't expect a unique solution... in fact, because the transform has a determinant of zero, the eigenvector is strictly a ratio of X:Y in two dimensions (or X:Y:Z in three dimensions, etc.) and you're going to find the lines overlap; in jargon: they are linearly dependent.

Vinyl LP Record

In this post I will detail one 12" stereophonic 33⅓ vinyl LP record. Initially I was hoping to be able to image the grooves cut into the surface and recover the audio track visually, but the resolution of a borrowed microscope demonstrated that the task was beyond its capabilities. Unfortunately that's left me to issue a caveat from the outset: reader beware; no experimental fact checking has taken place...

First, some mathematical properties:
Rotations per minute: 33⅓ = 100/3
Rotations per second: => 5/9ths
Outermost groove radius: 146.05mm
Innermost groove radius: 60.3mm
This gives us about 85.75mm of playable radius.
We're told that each side can contain up to 23 minutes (1380 seconds) of audio, the playback of which would require the disc to rotate 767 times (5/9 x 1380).
Fitting 767 rotations into 85.75mm gives us roughly 0.112mm separation between adjacent edges of the groove.
Finally, we define the spiral travelled by the stylus at time t as some initial offset less 0.112mm for each 2πθ of rotation as it moves from the outermost to the innermost radius.
let r(t) = 146.05 - 0.112 * (5 / 9) * t;
Let's switch to using µm, as things are starting to get tiny!

Note: the groove on a stereophonic record is a 90 degree cut rotated 45 degrees so that it looks like a V cut into the vinyl. The inner surface (closest to the center of the record) modulates one of the stereo audio tracks, while the outer surface (closest to the edge of the record) modulates the other. To increase likelihood of the stylus tracking the groove the V should never be shallower (or, equivalently, narrower) than 25µm.

We'll divide the 112µm groove into a handful of regions:

1. First, our mandatory 25µm notch, which we later split into an inner 12.5µm and an outer 12.5µm.
2. An optional (but recommended) gap
3. Inner and outer channels for modulating the left and right stereo data

Mathematically we say the groove width is made up of these components: 112µm - 25µm - gap = 2 * audio;
This tells us: once we've picked a size for the gap, we will know the maximum width for each of our audio channels; we don't need to refer to the gap size again once we've worked out the audio channel widths.

Now it's time to scale our input audio signals... For simplicity, let's ignore the bitness of a digital audio signal and assume that the signal amplitude fits into the range from +1 to -1. Note that the 0 point maps exactly onto the half-channel.
*Another assumption: positive samples have shallower grooves*
Remember: at time t, the stylus can be found at radius = r(t) [defined above].
let inner and outer be the left and right audio samples (numbers between -1 and +1) at time t. Combined with the rotation of the disc (giving us θ at time t) these four equations below define the groove geometry...
Surface: inner radial offset = (audio / 2) * (inner - 1) - 12.5µm
Surface: outer radial offset = (audio / 2) * (1 - outer) + 12.5µm
Groove: radial offset = (audio / 2) * (inner - outer) / 2
Groove: vertical offset = (audio / 2) * ((inner + outer) / 2 - 1) - 12.5µm

The image below is a stylized aid for identifying the described parts of a groove, as seen looking down the groove from head on. The image shows the inner channel modulated to a positive audio sample and the outer channel modulated to a negative audio sample. The stylus is offset to the right as a result of this stereo modulation.

• Green: maximum and minimum audio modulation points (where audio signal is +1 or -1)
• Orange: the imaginary "zero" line
• Light blue: surface of the inner modulation area
• Red: surface of the outer modulation area
• Black: stylus position
• Dark gray: uncut portion of vinyl LP record
• Light gray: mandatory unusable 25µm notch

Questions:

• do we map left to inner and right to outer?
• do we map positive to high (shallower V) and negative to low (deeper V)
• what's an appropriate audio channel width? if we know the maximum modulation in the 45 degree vector, we can scale it by SQRT(2) to find the channel width.
• is it appropriate to map the zero sample point to half of the audio channel width? it might

Numerical Integration Basics

Consider the equation \(y = 4\). If we plot this on two axes, x and y, it has no slope; it's just a horizontal line. We could also write the equation as \(y = 0x + 4\) because the value of y is unaffected by the value of x. We could also write it as \(y = f(x)\).

The integral of a function can be used to find the area between the function and the x-axis and is written \(I = \int_a^b f(x) dx\). It's essentially the sum of tiny changes in x (\(dx\)) multiplied by their corresponding y values (\(f(x)\)). The equation reminds us that we need to restrict ourselves to a lower and an upper value of x (otherwise the area would be infinite).

Let's choose a = x = 0 (the y-axis) and b = x = 3 as those two bounds.

\(I = \int_0^3 4 dx\)
\(I = \left[4 x\right]_0^3\)
\(I = \left(4\times3\right) - \left(4\times0\right)\)
\(I = 12\)

We're not restricted to 0 as the lower bound, but it exposes an interesting property: the area under a-to-b is equal to the area under 0-to-b less the area under 0-to-a. Let's try the same integral as above, but from 0-to-1.

\(I = \int_0^1 4 dx\)
\(I = \left[4 x\right]_0^1\)
\(I = \left(4\times1\right) - \left(4\times0\right)\)
\(I = 4\)

We might then subtract 4 from 12 to arrive at 8, but why not do this in one go?

\(I = \int_1^3 4 dx\)
\(I = \left[4 x\right]_1^3\)
\(I = \left(4\times3\right) - \left(4\times1\right)\)
\(I = 8\)

Singleton Pattern vs. C# Static Class

I ran into an old foe the other day when we were discussing interview questions; which to choose: standard singleton pattern or static class? At the most basic level, both constructs are useful in limiting the number of instances of an object that can be created. There are numerous subtle differences at the implementation level, but the single biggest difference is at an object-oriented level: static classes do not support polymorphism. This stems from a combination of factors at the language level; in short, a static class cannot inherit from a base class (abstract or concrete) nor can it implement any interface methods. Conversely, you can neither define static virtual methods on a class, nor can you define them on an interface. So, if you're ever using dependency injection - a technique that relies heavily on polymorphism - you will likely find that static classes will be inadequate. There are a number of other runtime level differences, but these are all secondary to the polymorphism issue. The simplest is that a static class can never be instantiated, and all methods are executed against the type instance. Contrast this with the singleton, of which exactly one instance is created, and where methods are executed against that instance. Next time I choose, I'm going anti-static for all but the simplest "utility" classes.

Cisco 7960 SIP Phone

I bought a Cisco 7960 IP phone (preloaded with SIP firmware) to connect with my experimental asterisk server running on a Raspberry Pi. Setting it up wasn't that hard, once I stopped following the bad advice that abounded on Google search results. I'm partially documenting my own experience here - not to help, but to hinder. You don't need a TFTP server (and therefore you don't need to set any hard-to-reach DHCP options) to distribute configuration files, you can do everything you need through the phone's keypad. If you're going to make changes, you need to perform the "Unlock Configuration" steps each time.

Unlock the Configuration

1. Press Settings
2. Scroll down to 9: Unlock
3. Enter the password "cisco"

Set the TFTP server IP address manually

1. Press Settings
2. Scroll down to 3: Network Configuration
3. Scroll down to TFTP Server
4. Press Edit
5. Enter the IP address of the TFTP server

Configure SIP on the Phone

1. Press Settings
2. Scroll down to SIP Configuration
3. Select the line you want to edit
4. Confirm the following entries exist (or add them) [They should match what you've got setup in sip.config on the Asterisk server).

Setup SIP on the Asterisk Server

1. Add a new block to sip.conf. Make sure the extension and secret match the values you provided when configuring SIP on the phone. [1234] context=home secret=secret etc.
2. Add a dialplan rule or two to extensions.conf. Make sure the context in sip.conf matches the section name in extensions.conf. exten => 1234,1,Dial(SIP/1234)
asterisk -r then sip reload and dialplan reload

ADSL2+ Sync Rate

If you have ever called up your ISP support team to complain, you may have heard them quote a number called SNR to describe the quality of your connection.

It's not quite as simple as a single number, though.

ADSL2+ uses multiple "channels" at once to transfer data between the ISP and your home modem. In the case of my ASUS modem, there are 512 channels: there is 2.208 MHz of available spectrum on the copper wire; these are split into equally spaced 4.3125 kHz wide channels.

However, not every channel (or bin) is created equally: there is almost certainly some electrical "noise" or congestion on the cables. Some bins are reserved for normal telephone connections, others for upstream and downstream communication with the server, with a few reserved as guards to keep sections separated.

I pay for a 4 Mb/s connection, so let's assume the ISP allows me 1672 bits' worth of downstream sync rate*. It is up to the modem and ISP to negotiate how to allocate the sync rate across the available bins. My modem surveys the ratio of signal to noise (SNR) across each bin in the downstream set of channels. A rule of thumb is that each bit requires about 3dB of SNR to reliably decode the signal without errors. Bins with higher SNR will therefore have higher bit capacity for bits.

My modem allows me to connect with SSH and look at three files, which are ordinarily used by the administrative web server interface to present me a way to easily change modem settings and view diagnostics:

• /var/tmp/spectrum-bpc-ds
• /var/tmp/spectrum-bpc-us
• /var/tmp/spectrum-snr

Let's do some very simple mathematics with the data in these files that won't be too difficult to follow:

Data taken from the SNR file:
1672 bits to allocate
268 channels into which they must be allocated
10,523.27 dB (the sum of SNR across the available reception channels)

Assumptions:
3 dB/bit required to convert received signal without errors

1) 1672 bits * 3 dB/bit = 5016 dB SNR required to decode the signal

2) 10,523.27 dB - 5016 dB = 5507.27 dB unused SNR

3) 5507.27 dB / 268 channels = 20.54951 dB unused SNR per channel

4) Now, we iterate each of the 268 channels in turn. I'll demonstrate with the first channel, where the SNR was 24.21 dB.

a) 24.21 dB - 20.54951 dB = ~3.66 dB to be used for data reception

b) 3.66 dB / 3 dB/bit = ~1.22 bits

c) round 1.22 bits to a whole number: 1

It turns out that this simple algorithm gets us remarkably close to the numbers of bits per channel (BPC) in the other files, which are arrived at by the modem's own algorithms. The difference could even be accounted for by (lack of, or outdated) "bitswap", where bins are periodically re-checked for SNR and reallocated to make the best use of the spectrum as it evolves over time.


* Ignore - for now - that those numbers are not equivalent, they are effectively the same thing quoted for different units of time
 

OpenVPN

This article on Ars Technica inspired me to try and setup a VPN, but it seemed to lack a couple of extra steps that were required to run the server on a Raspberry Pi behind a router. The official OpenVPN documentation was helpful but long-winded, and I resorted to a few askubuntu answers in the end.

1. Certificate names

   client-no-pass doesn't have to be replaced with the client host name, but can be (pretty much) any identifier for the client. The important thing is to sign all certificates with the same CA key.

2. VPN topology

   The default topology of net30 is perhaps not as easy to comprehend as subnet, and subnet works with Ubuntu, Windows and iPhone clients.

3. Default gateway

   On the server, I needed to push redirect-gateway, push "route xxx.xxx.xxx.0 255.255.255.0" and consequently push "dhcp-option DNS xxx.xxx.xxx.1" so that the client would be able to route to the server's local network, and the tun0 device would be able to do DNS lookups.

4. Firewall rules

   Besides enabling the forwarding of IPv4 so that the Raspberry Pi acted as a router, I need to add three firewall rules in iptables: one for NAT masquerading, and two for accepting new and related established connections.

5. DNS setup for clients

   The dhcp-options that were pushed to the client were ignored, which resulted in the client being unable to resolve any DNS names to IP addresses. This was only a problem on Ubuntu clients, but the script /etc/openvpn/update-resolv-conf was provided when I installed openvpn; all I had to do was reference it from client-name.config as a pair of lines:

   up /etc/openvpn/update-resolv-conf
   down /etc/openvpn/update-resolv-conf
   

Grub

It's nice when the Grub bootloader remembers your last choice and reuses it as the default value next time your machine boots. Using Linux, put the following in /etc/default/grub:

GRUB_DEFAULT=saved
GRUB_SAVEDEFAULT=true

Then run:

sudo update-grub

SubscribeOn / ObserveOn

Once you have an instance of an IObservable<T> object, Reactive Extensions (Rx) provides at least two points where you can choose an IScheduler which will determine the threading context under which the later operations will run. Those operations can be broken down into two phases: subscription and observation.

Subscription refers to the operations that modify the behaviours of an IObservable<T> (e.g. Throttle, Merge, Concat) The IScheduler for the subscription phase is assigned by the SubscribeOn() method.

Observation refers to the callback operations that are invoked as the result of observing an IObservable<T> (e.g. OnNext, OnCompleted, OnError) The IScheduler for the observation phase is assigned by the ObserveOn() method.

While not the subject of this post, you can also choose an IScheduler implementation to convert any IEnumerable<T> to an IObservable<T> with the ToObservable method. This is relevant when the IEnumerable<T> has the potential to take a long time to be enumerated.

6 Simple Rules for Async/Await

1. If a method A calls another method B that returns a Task (or Task<T>) then the calling method does not block on the completion of that task, unless it either:
1. awaits that task, or
2. waits on the result of that task
2. The calling method cannot await a task unless it is declared with the async modifier, which causes the compiler to build a state machine (similar to the IEnumerable iterator) for that awaitable method.
3. In point 1 above, the behaviour is independent of whether or not B is marked with the async keyword
4. It is this non-blocking behaviour of A that allows the calling thread to proceed past the point where B is invoked. It's even possible for the thread's call stack to unwind, and for the thread to go back to reading the Windows message queue if it was the UI thread.
5. The compiler will only warn you "because this call is not awaited ..." if B was marked with async.
6. It is expected that A will do something with the Task returned by B; at the very least there should be some code to check that the Task did not throw any exceptions. If - instead of B - we have an async method C that returns void then we do not present A with any opportunity to monitor the completion of C. Unobserved exceptions thrown during the execution of C could indicate corrupted program state and can be configured to terminate the application in much the same way that an unhandled exception in synchronous code can unwind a stack fully and terminate the process.

iSCSI Enterprise Target

In the presence of conflicting information (lots of it) this is how I got iSCSI Enterprise Target running on Ubuntu 16.04 (LTS). First, install the required software with

sudo apt-get install iscsitarget

man ietd indicates that the configuration file is located at /etc/ietd.conf but it isn't, so instead:

sudo nano /etc/iet/ietd.conf

If you're following the "Creating an Open Source SAN" chapter of "Pro Ubuntu Server Administration" by Sander van Vugt, be careful when adding your target not to include a space between the comma and the type

Target iqn.2008-08.com.sandervanvugt:mytarget
    Lun 0 Path=/dev/sdb,Type=fileio
    Lun 1 Path=/dev/sdc, Type=fileio

One step that's missing from this chapter is to enable it (why isn't it enabled?).

sudo nano /etc/default/iscsitarget

Ensure this line is present in the configuration

ISCSITARGET_ENABLE=true

Then exit nano, saving changes, and restart the iscsitarget service

sudo /etc/init.d/iscsitarget restart

If everything has gone to plan, you should be able to see these files

cat /proc/net/iet/session
cat /proc/net/iet/volume

Otherwise, have a look at the log

tail /var /log/syslog

Sudoku

It's hard to find a newspaper without one, and I personally find that Sudoku solutions are an interesting problem. Randomly filling a grid of 81 cells with numbers from 1-9 wouldn't take long, but the end result probably wouldn't be a Sudoku solution, where each row, each column, and each box, all need to have the numbers 1-9 exactly once.

Instead of doing it randomly, we could try a constrained approach: only pick numbers from the set that would form an allowable Sudoku solution. We could start off by defining the set of all values in row M, all values in column N, and all values in box MN. The union of all three sets would indicate the values that have already been picked; therefore they're the numbers we couldn't re-use if we were to make a valid Sudoku solution. \( R(m) \cup C(n) \cup B(m,n) \) is the numbers that have been used, \( (R(m) \cup C(n) \cup B(m,n))' \) are the numbers available to us (this last set is the equivalent of \(R(m)' \cap C(n)' \cap B(m,n)'\)).

Let's assume we're looking at an empty cell, in an empty grid. \(R(m)=\emptyset, C(n)=\emptyset, B(m,n)=\emptyset\), therefore we can choose anything from \(\emptyset'\). The world is our oyster. Let's pick 1. We could have picked anything, but 1's a good place to start. In the adjacent cell, \(R(m)\) and \(B(m,n)\) would both yield \(\{1\}\) so we'd be forced to pick anything from \(\{1\}'\). Let's pick 2. Continuing in ascending order (and increasing level of constraint) we'd finally reach the 9th cell where \(R(m)=\{1,2,3,4,5,6,7,8\}\) and \(B(m,n)=\{7,8\}\). The only number we can pick is 9. Our top row is complete.

Continuing at the first column of the second row, and proceeding in the same fashion, we keep filling cells. It's too easy, something's got to break.

Ah. On the second row, at the seventh colum, we reach Box(2,7). We've already written \(\{4,5,6,1,2,3\}\) into the row, and \(\{7,8,9\}\) into the box. What can we do now?

The answer is to backtrack. We didn't have to choose the lowest value from \(R(m)' \cap C(n)' \cap B(m,n)'\) every time; in fact - more often than not - we had a wide array of options. So, we backtrack. That involves reverting the operation on the current cell (leaving it blank), moving to the previous cell, making sure it's blank (but we remembered which value(s) we'd already tried), and then trying another value. We don't need to limit ourselves to going back just one cell, either. In fact, in this case, we need to revert the last three cells. A second row that looks like \(\{4,5,6,7,8,9,1,2,3\}\) is a valid partial solution to a Sudoku problem. We use backtracking even more on the third row, producing \(\{7,8,9,1,2,3,4,5,6\}\). Still: a valid partial solution.

It turns out that our set-based constraints, plus the backtracking ability, give us the power to solve any Sudoku problem, even when the starting point is ambiguous (in the case of the empty grid).

1 2 3 4 5 6 7 8 9
4 5 6 7 8 9 1 2 3
7 8 9 1 2 3 4 5 6
2 1 4 3 6 5 8 9 7
3 6 5 8 9 7 2 1 4
8 9 7 2 1 4 3 6 5
5 3 1 6 4 2 9 7 8
6 4 2 9 7 8 5 3 1
9 7 8 5 3 1 6 4 2

ForEach

var forEach = new TransformManyBlock<IEnumerable<int>, int>(x => x);

Geostationary

Satellites in a geostationary orbit appear to maintain a fixed position relative to a position on the surface of the Earth. To accomplish this, they need to rotate at the same angular speed that the Earth spins, which makes one full rotation every sidereal day. There are 23.93446122 hours (or 86,164.0604 minutes) in one sidereal day. A full revolution of 2π radians in 86,164.0604 seconds gives us \(\omega = 7.29212 \times 10^{-5} \ radians \cdot s^{-1}\). This is our target angular speed if we want to keep a satellite overhead the same spot.

Earth's standard gravitational parameter is its mass M x G (the gravitational constant), which is roughly μ = 398,600.4418. The standard gravitational parameters for planets are better known than either M or G individually (the best way we have at our disposal to weigh extremely massive objects is by observing their gravitational effect on other bodies like orbiting satellites). Acceleration due to gravity is \(\mu \over r^2\). The equatorial radius of Earth is 6,378.14 km. If we plug in that figure, we'd see acceleration on the surface is the familiar 0.009798285 km/s. 422 km above the surface (in the orbital sphere of the ISS), it's just 0.008619904 km/s (87.97%). The further out we go, the lower the acceleration and the longer the orbital period.

To find out how far away we have to place a geostationary satellite, consider that it will need to be in a circular orbit with an orbital period of one sidereal day. In one second, it will move \(7.29212 \times 10^{-5} radians\); therefore \(r \times (1 - cos(7.29212 \times 10^{-5}))\) will be the distance that it falls towards the Earth in that time. We know from the SUVAT equation \(s = ut + {1 \over 2}at^2\) that in a second, with no initial velocity, that the distance travelled is half the acceleration. We now have an identity \(r (1 - \cos(7.29212 \times 10^{-5})) = {\mu \over r^2}\). Rearrange it and solve: \[r = \sqrt[3]{\mu \over {2 - 2 cos(7.29212 \times 10^{-5})}}\] It turns out that the altitude of 42164.15974 km above the center of the Earth (35786.02274 km above the equatorial surface) is where you'll find geostationary satellites - and they can only exist in that one orbital plane.

Gravity / ISS

Let's say we're interested in finding how fast the International Space Station has to move in order not to fall back to earth (i.e. to stay in orbit). The earth has an equatorial radius of about 6384km and the satellite is about 422km above the equator when it's overhead on its inclined orbital path. For the sake of simplicity we will ignore the oblate spheroid shape of the earth, and the replace the elliptical orbit with a circular one, 6806km from the center of the earth.

On Earth's surface the acceleration due to gravity is \(9.8 m \cdot s^{-2}\). The further you go away, the lower the force of gravity. 422km above the surface we're told that the acceleration is 89% of surface gravity - it's \(8.722 m \cdot s^{-2}\).

Using the SUVAT equation \(s = ut + {1 \over 2}at^2\) we can see that an object dropped from 422km (i.e. \(g = 8.772 m \cdot s^{-2}\)) would fall 4.361m in the first second.

In the same second, we know that the ISS traces out the circular orbit (i.e. it doesn't crash into the Earth). The angular distance it travels (we could use the unit circle for visual confirmation) is \(\arccos({{r - s} \over r})\) or \(\arccos({{6806000 - 4.361} \over 6806000})\) or \(\arccos({6805995.639 \over 6806000})\) or simply 0.001132041 radians.

If it travels 0.001132041 radians in a second, it will take 5550.316851 seconds to perform a complete revolution of 2π. 5550 seconds is 92½ minutes, a figure that's very close to the published figure (on Wikipedia) and that's quite amazing given the rough estimates we've made to get this far. A circle of radius 6806km has a circumference of 13612km. Divide that circumference by the orbital period in seconds and you get \(7704 m \cdot s^{-1}\). You could also try \(r \sin(\theta)\) or \(6806000 \sin(0.001132041)\) which gives the same result. It's moving pretty quickly, but would have to be even quicker if the orbit was any closer to the surface!

Newtonian Reflector

Newtonian reflector telescopes use two mirrors – a figured primary mirror (the objective) which focuses incoming light, and a flat secondary mirror which redirects that light at an angle so that it can be viewed without getting your head stuck in the tube. The secondary mirror is only practically important – a CCD sensor could be placed directly into the path of the light focused by the objective mirror and it would capture the light without requiring an additional reflection. However, the subject of this post is the primary mirror, particularly: it’s shape.

A 2D parabola is defined by the quadratic equation \(y = ax^2 + bx + c\). If we make some simplifying assumptions (such as: its open side faces up; it “rests” on the x-axis; the focus is on the y-axis at height p) then our equation reduces to \(y = 4px^2\). The first order derivative (e.g. slope) of this equation is \({dy \over dx} = 2px\). That means, for a given focal length (and radius from the centre of the mirror), we can estimate the height of the reflective surface from the x-axis, as well as the slope at that point. These two computed values show where light is reflected when it encounters the surface of the objective mirror. Remember that incident light is reflected about the normal vector to the surface:
\(R = D – (2D \cdot N)N\)

To enable a 3D version (a paraboloid) we can take advantage of the knowledge that we only need rotate the parabola (i.e. the shape is rotationally symmetric about the vertical axis). In 3D, I chose to rotate around the z-axis.

It then becomes rather straight forward to take a solid angle of light, to compute the angles at which each incident ray encounters the parabolic reflector surface (parallax will affect closer light sources more than further away sources) and then note the points at which two or more rays converge to a single point. It turns out that the parabolic shape is ONLY able to focus light that is travelling parallel to the reflector’s primary axis. Closer sources converge further back than the stated focal length; sources at infinity converge exactly at the parabola’s focal point. Off-axis sources result in a complex out-of-focus shape when we attempt to capture them on a flat focal plane like a CCD. Looking at the distribution of focal points for a given field of view it was difficult to imagine a single transformation that might enable coma to be completely (and correctly) removed, though clearly a retail market for coma-correcting lenses abounds.

Day-glo

Ever wonder what makes those striking day-glo colors shine brighter than everything around them? Our eyes can only see visible light (the colors of the rainbow) but the sun emits electromagnetic waves across a vastly wider spectrum (which we cannot see with our eyes, of course, and not just because it's not a good idea to stare at the sun). But just because we cannot see them, doesn't mean they aren't there. When photons from the ultraviolet region of the spectrum bump into an object, some of the energy can be reflected. Because we can't see them, they go unnoticed. However, certain chemical combinations transform higher-energy ultraviolet photons into lower-energy visible light photons (a process called fluorescence). When those fluorescent objects start emitting more visible light than the non-fluorescent objects around them, they give the perception of being brighter: day-glo. Interesting fact: whiter-than-white washing detergent is known to employ fluorescence to make clothes appear brighter.