So - We have a code Kata email every friday - usually posed by the boolean frog (pascal.nationality == french)

I think the interaction on our list around these is awsome so I think it is worth sharing - and here is why 

most Friday's Pascal posts a problem and we all attempt to solve it - Mostly in C Sharp but sometimes in F sharp ruby etc (Am on a mac and do you think I can find the effing sharp key :-( ) 

Anyway 

This weeks puzzle was 

Excel can solve this Kata very easily, wonder how easy it is in C#/Ruby/F#/JS.

Shortest (correct) answer wins!

/*
Given a list of nullable decimals, calculate the average of each consecutive pair of decimal values.
Example:
 
{6, 2, 0, 4, 3, null, 2, 6, 8, 4}        =>            {4, 2, 1.5, 4, 6}   
{6, 2, 0, 4, 3, null, 2, 6, 8, 4, 99}    =>            {4, 2, 1.5, 4, 6, 99}
 
*/
void Main()
{
       var rand = new Random();
       int max = rand.Next(0, 48);
       IEnumerable serie = Enumerable.Range(1, max).Select (i => (decimal?)rand.Next(0, 10));
 
       var result = CalculateAverages(serie);
 
       result.Dump();
}
 
decimal[] CalculateAverages(IEnumerable serie)
{
       // YOUR CODE HERE
       return new decimal[]{};
}

So the question is how do you solve it??

Being the resident ruby geek my solution was - ahem - in ruby there were some nice other solutions but how do you solve it???

Before you try some things to note

1. Null values are treated as 0 in the list passed in as parameters.
2. The list passed in can be made of either integers, decimals (floats in Ruby) or null values.
3. If all values are null, then 0 will be the average for all 2 consecutive value pairs.
4. The list can have 0, 1 or more numbers.
5. The list can have either an even or odd count.
6. The result should not contain any null values.

I'll post some of the answers here tomorrow - But if you want to try to win (Remember the shortest answer to the Calculate average method is what we are after) then feel free to leave a comment and if you win you will get - ummm - some kudos - -)

good luck and answers here (Not quite the quoted tomorrow - Ah well )

Mal 

5-08-2011