maximum distance in nRm graph, if moving one element to another slot is an edge.
Solution: n - top(m^2/n)
There are 2 parts to the proof.
1) Proving any element in nRm has at least top(m^2/n) common elements with the configuration A having m contiguous elements.
Assume, each slot can have partial elements. each slot will have m/n elements. m contiguous slots will have m^2/n elements. So there will be at least a m-contigous slots with top(m^2/n) elements.
2) Proving that, if the distance between the 2 configurations X & Y is d, then there is a configuration B with same distance d with A.
Do a mapping between X and A such that, a slot in X has an element iff its corresponding slot in A has an element. Do the same mapping from Y to get B.
This implies distance in 12R5 is just 2. Remember 12R5 is the set of ragams without Melakartha restrictions. If you consider Melakartha restrictions it very well be mere 1. No wonder all classical music krithis always sounds the same for a layman.