how to rotate diagonal of matrix symmetrically using coding

		a				A
			b		B
				c C
			B		d
		E				e

Let's say you have the diagonal patch as a-b-c-d-e in above 7*7 matrix

and after rotation it gets as A-B-C-D-E

length = 5

start = [1,2] = a

Now, to rotate it symmetrically, use the symmetric property as 

- iterate length/2 times and in every iteration shift the 2 elements, eventually shifting length/2 * 2 = total elements

But How?

Please have a look at following diagram:

length = 5, so total iterations = length/2 = 2

in 1st iteration, shift as shown by bigger arrow

in 2nd iteration, shift as shown by smaller arrow

center element remain as it is.

1)  Why length/2 iterations?

=> Idea is we are basically starting with start of diagonal, and by making use of length we deal with the other element of the diagonal e.g. [1,2] = a and [1+len-1, 2+len-1] = [5, 6] = e

So, idea of accessing is 2 elements in ONE iteration is clear.

2) Now, how to re-position these 2 elements for the desired rotation?

=> gap = length - 1

     current_element = [i, j]

     new_element = [i, j + gap]  #just shift right to correct column

     current_symmetric_element = [l, m]      # l = i + gap, m = j + gap

     new_symmetric_element = [l, m - gap] #shift left to correct column

3) The step 2 is for just one element, but we need to slide down for rows, finally it gets as follows:

[i,j] = start_element, length = length of diagonal

[l,m] = end_of_diagonal

gap = length - 1

for x in 0 to length/2:

   print i + x, j + (gap - x)

   print l - x, j - (gap -x)

if length is odd:

   print i + length/2, j + length/2

Python code:

i, j, l = 2, 3, 5

for i1 in range(0, l):

  print i + i1, j + i1

print

print

#we need to use symmetricity to rotate i.e. we just need to iterate half the times of length which is l here

#in 1st iteration we will shift the extreme elements and in 2nd remaining 2

#this also works in generic way i.e. iterate half the times of lenght and deal with 2 elemts at a time which in total orders the total elements in diagonal

for x in range(0,l/2):

  print i+x, j + (l - 1 - x)

  print i+l-1-x, j+l-1-(l-1-x)

if not l%2 == 0:

  print i+l/2, j+l/2