The minimal path in an information network has 3 points ( and therefore the path has the order k=3 )

• one starting point
• one mid point
• one target point

and we can specify each path by a  sequence of figures  like 123  or 231  etc.

An important question when  analyzing networks is, how many different paths exist in a complete network ( without feedback loops ) and all its sub-networks.

• in a network of order 3 exist 6 different paths.

We consider only open paths; the number of closed paths is just twice the number of the open paths.

Two equations from mathematical combinatorics are used:

The faculty of a natural number n is  noted  as n! = 1*2*3* …n

Examples:

3! = 1*2*3 = 6

4! = 1*2*3*4 = 24

Important to note is the definition:   0!  = 1

If we want to calculate, in how many different ways we can form different groups  of k numbers,  if n different numbers are given, then we can calculate that with following equation:

This equation is valid, if it the order of the k chosen numbers does not matter.

If the order matters, we have to multiply this equation with k! , because there exist so many different arrangenents  of  k numbers.

And then we get the number of different paths in a complete network of order n,  each path with k  different  points

The complete network has the order 7.

• In the first row are the orders of the sub-networks
• In the first column are the orders of the paths, 3 being the minimal possible order.

From this table we can deduce probabilities, that a certain path will be used.

For instance, if we know, that the complete network has the order n = 5 and that there will be taken a path of order k= 3, then the probability for it is 1/60

For a network of order n=3 and a path of order k=3, the probability is 1/6. That is 10 times higher, than for a network of order n=5

From that observation we can deduce a criteria, to find out, if we really use a complete network of order n.

We have only to note, how many times a path of order k exists. If for instance, we find out, that  paths of order 3 exist 50 times, then the order of the complete network must be at least n= 5

If we have found only 4 points, then at least 1 point is still missing, to get the complete network. And in this way we can go on with the other orders of paths.