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Figure 3.2 The path of the robot (dashed lines) under algorithm Bug1. ob1 and ob2 are obstacles, H1 and H2 are hit points, L1 and L2 are leave points.
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Bug1 Procedure 1. From point Li 1 , move toward point T (Target) along the straight line until one of these occurs: (a) Point T is reached. The procedure stops. (b) An obstacle is encountered and a hit point, Hi , is de ned. Go to Step 2. 2. Using the local direction, follow the obstacle boundary. If point T is reached, stop. Otherwise, after having traversed the whole boundary and having returned to Hi , de ne a new leave point Li = Qm . Go to Step 3. 3. Based on the contents of registers R2 and R3 , determine the shorter way along the boundary to point Li , and use it to reach Li . Apply the test for target reachability. If point T is not reachable, the procedure stops. Otherwise, set i = i + 1 and go to Step 1.
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Analysis of Algorithm Bug1
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Lemma 3.3.1. Under Bug1 algorithm, when MA leaves a leave point of an obstacle in order to continue toward point T , it will never return to this obstacle again. Proof: Assume that on its way from point S to point T , MA does meet some obstacles. We number those obstacles in the order in which MA encounters them. Then the following sequence of distances appears: D, d(H1 ), d(L1 ), d(H2 ), d(L2 ), d(H3 ), d(L3 ), . . .
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If point S happens to be on an obstacle boundary and the line (S, T ) crosses that obstacle, then D = d(H1 ). According to our model, if MA s path touches an obstacle tangentially, then MA needs not walk around it; it will simply continue its straight-line walk toward point T . In all other cases of meeting an ith obstacle, unless point T lies on an obstacle boundary, a relation d(Hi ) > d(Li ) holds. This is because, on the one hand, according to the model, any straight line (except a line that touches the obstacle tangentially) crosses the obstacle at least in two distinct points. This is simply a re ection of the nite thickness of obstacles. On the other hand, according to algorithm Bug1, point Li is the closest point from obstacle i to point T . Starting from Li , MA walks straight to point T until (if ever) it meets the (i + 1)th obstacle. Since, by the model, obstacles do not touch one another, then d(Li ) > d(Hi+1 ). Our sequence of distances, therefore, satis es the relation d(H1 ) > d(L1 ) > d(H2 ) > d(L2 ) > d(H3 ) > d(L3 ) . . . (3.6)
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where d(H1 ) is or is not equal to D. Since d(Li ) is the shortest distance from the ith obstacle to point T , and since (3.6) guarantees that algorithm Bug1 monotonically decreases the distances d(Hi ) and d(Li ) to point T , Lemma 3.3.1 follows. Q.E.D. The important conclusion from Lemma 3.3.1 is that algorithm Bug1 guarantees to never create cycles. Corollary 3.3.1. Under Bug1, independent of the geometry of an obstacle, MA de nes on it no more than one hit and no more than one leave point. To assess the algorithm s performance in particular, we will be interested in the upper bound on the length of paths that it generates an assurance is needed that on its way to point T , MA can encounter only a nite number of obstacles. This is not obvious: While following the algorithm, MA may be looking at the target not only from different distances but also from different directions. That is, besides moving toward point T , it may also rotate around it (see Figure 3.3). Depending on the scene, this rotation may go rst, say, clockwise, then counterclockwise, then again clockwise, and so on. Hence we have the following lemma. Lemma 3.3.2. Under Bug1, on its way to the Target, MA can meet only a nite number of obstacles. Proof: Although, while walking around an obstacle, MA may sometimes be at distances much larger than D from point T (see Figure 3.3), the straightline segments of its path toward the point T are always within the same circle of radius D centered at point T . This is guaranteed by inequality (3.6). Since,
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