Deduce the average distance of the nearest neighbor connected network

When the number of nodes tends to infinity, the average distance of the nearest neighbor connected network with an average of 4 is deduced.
Solution: Set the number of nodes to N, define% as the take-up operator, set r:= (N-1)%4 to represent (N-1)/4 remainder.
According to the nature of the nearest neighbor connection network, each node has homogeneity, so we just work out one of the nodes (set to I) and the other nodes of the sum of the distance, and then multiplied by N/2 (divided by 2 because each point pair of distance is repeated added), you can get the sum of the distance of each point pair, The average distance can be calculated by dividing the sum by the total number of point pairs.

Because the topic stipulates that the average degree of the network is 4, referring to the above figure, so the distance node I distance 1 node has 4, the distance is 2 node is 4; and so on, the furthest distance in the former N-1-r node is (n-1-r)/4; The remaining R nodes are less than 4, so the distance is calculated separately, The distance between them and node I is (N+3-R)/4
Then the sum of the distance between node I and other nodes si can be calculated by the following formula:
S_i=4x ((1+ (N-1-R)/4) x (N-1-R)/4)/2+ (n+3-r)/4XR
And the sum of the distances of all the pairs of points is equal to:
s= (ns_i)/2
And then the total number of points to T:
T=n (N-1)/2
Then the average distance d̅ is:
d̅=s/t
D̅= ((N+3-R)/8) (1-r/(N-1)) + (n (4-r)/(N-1)) R/4
Since the value range of R is [0, 3], when n tends to infinity, the upper formula can be simplified to:
Lim┬ (n→∞) d̅= (n+3+r)/8≈N/8