Protocols for Secure Communication in Wireless Sensor Networks

5.2. Routing on Spanning Trees 137
Algorithm 3 Variant: Forward a message to all covering children
1: on receive message DATA〈t,s,d, p〉 from j:
2: if d ∈ cover (t)
u then
3: /* handle message */
4: end if
5: ∀c : c = v ∧ d ∈ T [t].reach(c): forward DATA〈t,s,d, p〉 to c
6: if T [t].parent ∈ {u,v} then
7: forward DATA〈t,s,d, p〉 to T [t].parent
8: end if
9:
condition
D1 ∪ D2 ⊆ D1 ⊎ D2 ,
i.e. all addresses contained in either D1 or D2 are also contained in the resulting
set. For some address spaces, it may be impractical to represent subsets of that
space both efficiently and accurately. We therefore admit some “fuzziness” in
the union operation, i.e. we allow
D1 ⊎ D2 \ (D1 ∪ D2) = /0 .
This means that additional addresses may be in the combined covers that are
not covered by either of the components.
In the following, we show how to instantiate this general scheme with concrete
addressing schemes.
Tree Addressing
Within a tree, there is a canonical addressing scheme, assuming a static order
on the children of each node. Starting from the root, each node can be given a
unique index address
i0.i1.i2. ... .ih
where i j ∈ IN denotes the index of a child of the current node. If two such
addresses have the same prefix, they are located in the same subtree. Note that
the root’s address is the empty index sequence ε. Addressing nodes in this way
is efficient since this scheme requires no further state information except for
the static order on children.
Index addressing has the drawback that the same node has a different address
in every tree, therefore an address is only valid for the lifetime of the tree. Also,
the sender has to know the index address of the destination node in order to send
a message. Getting the current index address of the destination to the sender
requires additional overhead.