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When identical agents engage in a pursuit with an intree assignment graph, since there is a stationary agent, it is relatively simple to guarantee that
by choosing an appropriate
:
Lemma 11
Unit speed pursuit of
Dubins car agents with an intree assignment graph has the property
and will rendezvous in finite time if the agents maintain their targets in the windshields of span
with
.
PROOF. When
, one agent is stationary and one agent is moving;
guarantees
. When
, for an edge
in an intree, we may express
similary as (8) (note the speeds are all
):
|
(23) |
When the target agent
is the root of the intree, the second term in (23) is zero since the root has no target and does not move. That is,
for at least one edge
of the assignment graph. Let this edge be
. Summing over all assignment edges yields
|
(24) |
is equal to
|
(25) |
Assuming
,
and
, (25) is guaranteed by
|
(26) |
which is equivalent to
. The finite time guarantee follows the argument from Corollary 9.
Next: Cycle plus branches
Up: Guaranteed Rendezvous of Identical
Previous: The cyclic case
Jingjin Yu
2011-01-18