Start SharePoint virtual machines via command line

Like it or not, there is no better choice except developing SharePoint 2010 using a virtual machine. While VMWare is pretty good, it is also a bloated software. My machine runs two vm-s, trying to simulate a real-world SharePoint environment – one being the DNS and hosting the AD and SQL Server 2008 R2 and another the Windows Server 2008 R2, Visual Studio 2010 and SharePoint 2010 itself. Since I have to stop and start VMWares frequently, I found the following command line script very useful.

net start "VMWare Agent Service"
net start "VMWare Authorization Service"
net start "VMWare DHCP Service"
net start "VMWare NAT Service"
"C:\Program Files (x86)\VMware\VMware Workstation\vmrun" start "C:\VirtualMachines\sqlserver.vmx"
"C:\Program Files (x86)\VMware\VMware Workstation\vmrun" start "C:\VirtualMachines\sharepoint.vmx"

Save the above file as a batch file and store in a directory accessible by PATH.

For stopping, reverse the commands, use “suspend” instead of “start”.

Groovy OLC #2 – Strand Puzzle 1914

I think it was the year 1986 when I participated in an interschool programming competition in Chennai. I wasn’t very good at math, but I did tag along with a couple of my friends who were darn good at it. There were three problems given that we have to solve using GW-BASIC in about 30 minutes. There were probably 100 participants. I did get a peep into a few poor guys inside the computer room who were typing our solutions into the computer blindly and see if they worked. I just remember only one of the problems. I didn’t get the correct solution, but one of my friends – Sowmy Srinivasan, did and win the competition.

Years later, I learned that this was a very interesting puzzle tackled by the genius Srinivasa Ramanujan. In the year 1914, PC Mahalanobis, a Kings college student in England, got hold of a puzzle from the Strand magazine.

“In a given street of houses with consecutive numbers between 50 and 500, find the house number, for which, the sum of numbers on the left is equal to the sum of numbers on the right”

In other words, given ‘n’ number of houses, find the house number ‘m’ that would sum(1..m-1) = sum(m+1..n).

Mahalanobis solved the problem by trial and error and went to Ramanujan to show it. Ramanujan was cooking some vegetables in the pan. The instant he heard the problem, Ramanujan, while still frying his veggies, not only gave the answer to the puzzle, but also provided the generic formula that would give infinite number of solutions. The problem belongs to the area called Continuous Fractions and is very fascinating. The problem can be reduced to x*x – Dy*y = ±1 and has been tackled by ~10th century Indian mathematician Brahmaguputa using a method called chakravala. I will probably write up on the “chakravala” method later. The Ramanujan anecdote is documented in several places on the net and a simple search will provide more interesting details.

So I was working on something in Groovy, I suddenly got reminded of this problem and was thinking if this can be solved as a Groovy OLC. I am going to use the brute-force method rather than solving via the quadratic equation, because it nicely demostrates the List capabilities of Groovy. It does run slow for large numbers, but speed or optimization is not the point here.

So here is Groovy OLC #2 – Ramanujan’s Continuous Fractions (via brute force method)

public test_ramanujan_continous_fraction() {
(1..100).each { n, houses = 1..n -> houses.eachWithIndex {m, i -> if (houses.subList(0,i).sum() == houses.subList(i+1,houses.size()).sum()) println "house# = $m, total_houses = $n" } }

house# = 1, total_houses = 1
house# = 6, total_houses = 8
house# = 35, total_houses = 49

Ignore the first one, as I am not bothered about boundary conditions, though its technically correct.

[6,8] = sum(1..5) = sum(7..8) = 15
[35,49] = sum(1..34) = sum(36..49) = 595

Lets retest it with the actual problem from the Strand magazine:

public test_ramanujan_continous_fraction() {
(50..500).each { n, houses = 1..n -> houses.eachWithIndex {m, i -> if (houses.subList(0,i).sum() == houses.subList(i+1,houses.size()).sum()) println "house# = $m, total_houses = $n" } }

house# = 204, total_houses = 288

This is the actual solution that Ramunjan gave instantly to the Strand’s puzzle.

From Groovy’s point of view, the most interesting thing is the second argument in the first closure – “houses” – is actually a variable defintion which is a function of the first argument, ie houses = f(n) !!!

Groovy OLC #1 – Prime Numbers

It is interesting how formatting conventions keep changing with the programming languages. I’m not talking about naming conventions, but just text formatting conventions aka code-layout conventions.

In Java, you dont commonly see beginning separate braces in a new line, although its okay for the ending braces. While in .NET, its a norm to have curly braces in separate lines. Inline code is also very rare in Java and .NET, except for get and set, but in Groovy/JavaScript, it seems to be a very common occurrence. Some of these conventions evolve as a tradition and some right from the invention. In Pascal, structured code was considered art. In Python block indentations are obligatory. Code format conventions, just like naming conventions are very subjective and sometimes personal and if a team cannot agree on one, its a rough road. Although some modern IDE-s have smartened up and present the personalized formatted code to the developer, while the source control would store a standard version. Some of the preferences are driven by language features too.

Closures, especially compact ones, are very intriguing when presented in a single line and some times very challenging. Solutions to puzzles and problems when rewritten in a single line look very catchy. Those who have coded in Prolog would perhaps realize what it means to say that a piece of code is work of art. You can do a regular findBy logic in Prolog, but doing it “Prolog”-way via predicates is certainly challenging.

So here is a One-Liner-Closure (OLC) that calculates the prime numbers > 2.

Given: a list of natural numbers
Find: all the prime numbers

void find_prime_numbers() {
def list = 1..100
println list.findAll { x -> (2..Math.sqrt(x)).every { x % it != 0 } }

[3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]