定向超立方体的领导者选举算法

Question

I'm stuck with some problem where I have to design a leader election algorithm for an oriented hypercube. This should be done by using a tournament with a number of rounds equal to the dimension D of the hypercube. In each stage d, with 1 <= d < D two candidate leaders of neighbouring d-dimensional hypercubes should compete to become the single candidate leader of the (d+1)-dimensional hypercube that is the union of their respective hypercubes.

Answer 1

It has been a long time since I studied distributed systems, but I hope this helps a little :)

The problem: Leader election in a network with a hypercube tolopogy. In this topology, every node has exactly D neighbors (where D is the hypercube's dimension). Since the hypercube is oriented , the nodes know which of their links corresponds to every dimension. Also, I assume that all nodes are labeled with some unique id, as typical with this kind of problems.

If I understand well your solution guidelines, the algorithm seems to be simple: a node has exactly D states. In every state 1<=d<=D it communicates with its neighbor along the d axis. This communication consists of sending it the id of the best candidate it is aware of (when d=1 this is simply its own id), and comparing it with the id received by the peer. Now the node knows the best id of the sub-cube of order d (defined by axes 1,2...,d) it belongs to. This way, at step D all nodes are aware of the global best, and the algorithm completes with a consensus.

For example, assume the following topology(D=2) and id values:

   (1)    (2)
    1-----2
    |     |
    |     |
    4-----3
   (4)    (3)

The ids in parentheses indicate the best ids known by each node so far.

The first iteration (d=1) works along the horizontal axis, and terminates as follows:

   (2)    (2)
    1-----2
    |     |
    |     |
    4-----3
   (4)    (4)

The second (and last) iteration (d=2) works along the vertical axis, and terminates as follows:

   (4)    (4)
    1-----2
    |     |
    |     |
    4-----3
   (4)    (4)

The node number 4 has been chosen, as required (highest id).

Message count complexity

For every edge in the hypercube we have exactly 2 messages (one on each direction). Since the formula for the number of edges in a hypercube of dimension D is E=D*2^(D-1), we have exactly M=D*2^D messages. In order to compute the complexity as a function of N (the number of nodes), recall that N = 2^D, so M=N*log N.