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96 MOTION PLANNING FOR A MOBILE ROBOT
T
S
Figure 3.9 Illustration for Theorem 3.3.4.
defined hit point. Now, move from Q along the already generated path segment
in the direction opposite to the accepted local direction, until the closest hit point
j
on the path is encountered; say, that point is H . We are interested only in those
cases where Q is involved in at least one local cycle—that is, when MA passes
j
point Q more than once. For this event to occur, MA has to pass point H at
j
least as many times. In other words, if MA does not pass H more than once, it
cannot pass Q more than once.
According to the Bug2 procedure, the first time MA reaches point H j it
approaches it along the M-line (straight line (Start, Target))—or, more precisely,
along the straight line segment (L j−1 ,T ). MA then turns left and starts walking
around the obstacle. To form a local cycle on this path segment, MA has to
j
return to point H again. Since a point can become a hit point only once (see the
j
proof for Lemma 3.3.4), the next time MA returns to point H it must approach
it from the right (see Figure 3.9), along the obstacle boundary. Therefore, after
j
having defined H , in order to reach it again, this time from the right, MA must
somehow cross the M-line and enter its right semiplane. This can take place in
one of only two ways: outside or inside the interval (S, T ). Consider both cases.
1. The crossing occurs outside the interval (S, T ). This case can correspond
only to an in-position configuration (see Definition 3.3.2). Theorem 3.3.4,
therefore, does not apply.
2. The crossing occurs inside the interval (S, T ). We want to prove now
that such a crossing of the path with the interval (S, T ) cannot produce
local cycles. Notice that the crossing cannot occur anywhere within the
j
interval (S, H ) because otherwise at least a part of the straight-line seg-
j
ment (L j−1 ,H ) would be included inside the obstacle. This is impossible