Wednesday, August 23, 2017

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


Sunday, August 13, 2017

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

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);