Building on your answer to Exercise 6.11, discuss two possible implementations of EN(e) and
SN(s). Why must the intersection of EN(e) and SN(s) be non-null for a given e that matches s (the
intersection rule)? Does this apply in your possible implementations?
What will be an ideal response?
As mentioned in Chapter 6, a common mechanism for implementing the two functions is to map on to an
underlying Distributed Hash Table (DHT). For example, consider Scribe, a publish-subscribe system
implemented over Pastry. Scribe is a topic-based publish-subscribe system. In Scribe, EN(e) and SN(s) are
identical operations and are implemented by the underlying hash function applied to the topic name. This
ensures that rendezvous-nodes are chosen at random from the underlying nodes in the associated overlay.
Meghdood [Cao and Singh 2005] offers a second example of how to implement rendezvous-based
routing. Meghdood is a content-based publish-subscribe system and, as such, the designers were interested in
determining rendezvous nodes not just from the topic or one field but from the entire content. This is a much
harder problem. The solution is to map on to a Content Addressable Network (or CAN). A CAN is an overlay
network representing an n-dimensional space where each node is responsible for a partition of this space.
Meghdood provides a particular subscription language which effectively defines this multi-dimensional space
(the space of all possible subscriptions) with the underlying CAN then providing the associated partitioning of
this space. The node responsible for each partition is exactly equivalent to the rendezvous node. Please refer
to the Meghdood paper for details.
It should be fairly obvious why the intersection rule is required. In all rendezvous-based routing
algorithms, subscriptions are sent to one or more rendezvous nodes with the subscriptions stored at these
nodes. For a publish event to be routed to all subscriptions, for any event, the calculated set of nodes offered
by EN(e) must overlap with the corresponding calculation of SN(s) to find a node that will then match to the
subscription and forward it accordingly.
In the first implementation, the intersection rule is trivially met as EN=SN. For the second, the argument
the rule is also met. In Megdhood, an event maps to a CAN region which cover all possible subscriptions that
can map to that event (known as the event region).
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