## Wednesday, March 29, 2017

### 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.

## Saturday, March 04, 2017

### 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:
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.

## Wednesday, December 28, 2016

### 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

## Monday, November 02, 2015

### 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

## Saturday, October 03, 2015

### ForEach

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

## Saturday, May 17, 2014

### 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.

## Saturday, May 10, 2014

### 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!