Turn a 2D grid into a 'diamond' with LINQ - is it possible?

Question

The other day I needed an algorithm to turn a 2D grid into a diamond (by effectively rotating 45 degrees), so I could deal with diagonal sequences as flat enumerables, like so:

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

My algorithm is as follows:

public static IEnumerable<IEnumerable<T>> RotateGrid<T>(IEnumerable<IEnumerable<T>> grid)
    {
        int bound = grid.Count() - 1;
        int upperLimit = 0;
        int lowerLimit = 0;

        Collection<Collection<T>> rotated = new Collection<Collection<T>>();

        for (int i = 0; i <= (bound * 2); i++)
        {
            Collection<T> row = new Collection<T>();

            for (int j = upperLimit, k = lowerLimit; j >= lowerLimit || k <= upperLimit; j--, k++)
            {
                row.Add(grid.ElementAt(j).ElementAt(k));
            }

            rotated.Add(row);

            if (upperLimit == bound)
                lowerLimit++;

            if (upperLimit < bound)
                upperLimit++;
        }

        return rotated;
    }

Can this be achieved in a more concise format using LINQ?

.. or even just a more concise format? :)

Answer 1

Here's what I came up with:

void Main()
{
    var lists = new string[] { "123", "456", "789" };

    foreach (var seq in RotateGrid(lists))
        Console.WriteLine(string.Join(", ", seq));
}

public IEnumerable<IEnumerable<T>> RotateGrid<T>(IEnumerable<IEnumerable<T>> grid) 
{
    int rows = grid.Count();
    int cols = grid.First().Count();
    return 
        from i in Enumerable.Range(0, rows + cols - 1)
        select (
            from j in Enumerable.Range(0, i + 1)
            where i - j < rows && j < cols
            select grid.ElementAt(i - j).ElementAt(j)
        );
}

Output:

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

This gets a lot cleaner and more performant if you can assume IList<T>s instead of just IEnumerable<T>. I'm thinking of a more efficient approach (not using .ElementAt) that would work with IEnumerable` also, which I will post if I manage to write it.

Update:

Here's my more practical version, that still manages to shoe-horn in a fair amount of linq. It's a reasonably efficient algorithm, in that it only creates an enumerator once for each IEnumerable.

public IEnumerable<IEnumerable<T>> RotateGrid<T>(IEnumerable<IEnumerable<T>> grid) 
{
    var enumerators = new LinkedList<IEnumerator<T>>();
    var diagonal = 
        from e in enumerators
        where e.MoveNext()
        select e.Current;
    foreach (var row in grid)
    {
        enumerators.AddFirst(row.GetEnumerator());
        yield return diagonal.ToArray();
    }

    T[] output;
    while (true)
    {
        output = diagonal.ToArray();
        if(output.Length == 0) yield break;
        yield return output;
    }
}