Competitive Exploration of Rectilinear Polygons - Semantic Scholar
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2006
Competitive Exploration of Rectilinear Polygons Hammar, Mikael Elsevier 367-378 http://hdl.handle.net/2043/11304 Downloaded from DSpace Repository, DSpace Institution's institutional repository
Competitive Exploration of Re tilinear Polygons
Mikael Hammar
Bengt J.
Nilsson
y
y
Mia Persson
Abstra t
Exploring a polygon with robots, when the robots do not have knowledge of the surroundings
an be viewed as an online problem. Typi al for online problems is that de isions must be made based on past events without omplete information about the future. In our ase the robots do not have omplete information about the environment. Competitive analysis an be used to measure the performan e of methods solving online problems. The ompetitive ratio of su h a method is the ratio between the method's performan e and the performan e of the best method having full knowledge of the future. We are interested in obtaining good bounds on the ompetitive ratio of exploring polygons and prove onstant ompetitive strategies and lower bounds for exploring a simple re tilinear polygon in the L1 metri .
1 Introdu tion Exploring an environment is an important and well studied problem in roboti s. In many realisti situations the robots do not possess omplete knowledge about the environment, e.g., they may not have a map of the surroundings [1, 2, 4, 6, 7, 8, 9℄. The sear h of the robots an be viewed as an online problem sin e the robots' de isions about the sear h are based only on the part of the environment that they have seen so far. We use the framework of ompetitive analysis to measure the performan e of an online sear h strategy S . The ompetitive ratio of S is de ned as the maximum of the ratio of the distan e traveled by the robot that moves the farthest using S to the optimal distan e of the sear h. We are interested in obtaining good bounds for the ompetitive ratio of exploring a re tilinear polygon. The sear h is modeled by a path or losed tour followed by one or more point sized robots inside the polygon, given a starting point for the sear h. The only information that the robots have about the surrounding polygon is the part of the polygon that they together have seen so far. For the ase of exploration with one robot, Deng et al. [4℄ show a deterministi strategy having
ompetitive ratio two for this problem if distan e is measured a
ording to the L1 -metri . Hammar et al. [5℄ prove a strategy with ompetitive ratio 5=3 and Kleinberg [7℄ proves a lower bound of 5=4 for the ompetitive ratio of any deterministi strategy. We will show a deterministi strategy obtaining a
ompetitive ratio of 3=2 for sear hing a re tilinear polygon in the L1 -metri with one robot. Department of Computer S ien e, Salerno University, Baronissi (SA) - 84081, Italy. email:
hammardia.unisa.it
email:
{Bengt.Nilsson,Mia.Persson}ts.mah.se
y Te hnology and So iety, Malmö University College, S-205 06 Malmö, Sweden.
1
We also present a lower bound of 2 for the orresponding path exploration problem for one robot in re tilinear polygons and ompetitive results on exploration with two and three robots. The paper is organized as follows. In the next se tion we present some de nitions and preliminary results. In Se tion 3 we give an overview of the strategy by Deng et al. [4℄. Se tion 4 ontains an improved strategy for single robot exploration giving a ompetitive ratio of 3=2. In se tions 5 and 6 we onsider path exploration and exploration with multiple robots.
2 Preliminaries We will hen eforth always measure distan e a
ording to the L1 metri , i.e., the distan e between two points p and q is de ned by
jjp; qjj = jp
x
j j
qx + py
j
qy ;
where px and qx are the x- oordinates of p and q and py and qy are the y - oordinates. We de ne the x-distan e between p and q to be jjp; q jjx = jpx qx j and the y -distan e to be jjp; q jjy = jpy qy j. If C is a polygonal urve, then the length of C , denoted length (C ), is de ned the sum of the distan es between onse utive pairs of segment end points in C . Let P be a simple re tilinear polygon. Two points in P are said to see ea h other, or be visible to ea h other, if the line segment onne ting the points lies in P. Let p be a point somewhere inside P. A wat hman route through p is de ned to be a losed urve C that passes through p su h that every point in P is seen by some point on C . The shortest wat hman route through p is denoted by SWR p . It an be shown that the shortest wat hman route in a simple polygon is a losed polygonal urve [3℄. Sin e we are only interested in the L1 length of a polygonal urve we an assume that the urve is re tilinear, that is, the segments of the urve are all axis parallel. Note that the shortest re tilinear wat hman route through a point p is not ne essarily unique. For a point p in P we de ne four quadrants with respe t to p. Those are the regions obtained by
utting P along the two maximal axis parallel line segments that pass through p. The four quadrants are denoted Q1 (p), Q2 (p), Q3 (p), and Q4 (p) in anti- lo kwise order from the top right quadrant to the bottom right quadrant. We let Qi;j (p) denote the union of Qi (p) and Qj (p). Consider a re ex vertex of P. The two edges of P onne ting at the re ex vertex an ea h be extended inside P until the extensions rea h a boundary point. The segments thus onstru ted are
alled extensions and to ea h extension a dire tion is asso iated. The dire tion is the same as that of the ollinear polygon edge as we follow the boundary of P in lo kwise order. We use the four ompass dire tions north , west , south , and east to denote the dire tion of an extension. Lemma 2.1 (Chin, Ntafos [3℄)
A losed urve is a wat hman route for
has at least one point to the right of every extension of
P
P
if and only if the urve
.
Our rst obje tive is to present a ompetitive online strategy that enables a robot to follow a losed
urve from the start point s in P and ba k to s with the urve being a wat hman route for P. An extension e splits P into two sets Pl and Pr with Pl to the left of e and Pr to the right. We say a point p is to the left of e if p belongs to Pl . To the right is de ned analogously. As a further de nition we say that an extension e is a left extension with respe t to a point p, if p lies to the left of e, and an extension e dominates another extension e0 , if all points of P to the right 2
f
0
prin ipal proje tion f
frontier
f
f
ext (f )
0 f
=
v f
s
C
(a)
ext (f )
0
C
(b)
fr
v
fl
p
0
0
fl
q
( )
fr s
(d)
Figure 1: Illustrating de nitions.
of e are also to the right of e0 . By Lemma 2.1 we are only interested in the extensions that are left extensions with respe t to the starting point s sin e the other ones already have a point (the point s) to the right of them. So without loss of larity when we mention extensions we will always mean extensions that are left extensions with respe t to s.
3 An Overview of GO Consider a re tilinear polygon P that is not a priori known to the robot. Let s be the robot's initial position inside P. For the starting position s of the robot we asso iate a point f 0 on the boundary of P that is visible from s and all f 0 the prin ipal proje tion point of s. For instan e, we an hoose f 0
to be the rst point on the boundary that is hit by an upward ray starting at s. Let f be the end point of the boundary that the robot sees as we s an the boundary of P in lo kwise order; see Figure 1(a). The point f is alled the urrent frontier. Let C be a polygonal urve starting at s. Formally a frontier f of C is a vertex of the visibility polygon, VP(C ) of C adja ent to an edge e of VP(C ) that is not an edge of P. Extend e until it hits a point q on C and let v be the vertex of P that is rst en ountered as we move along the line segment [q; f ℄ from q to f . We denote the left extension with respe t to s asso iated to the vertex v by ext (f ); see Figures 1(b) and ( ). Deng et al. [4℄ introdu e an online strategy alled greedy-online, GO for short, to explore a simple re tilinear polygon P in the L1 metri . If the starting point s lies on the boundary of P, their strategy, we all it BGO, goes as follows: from the starting point s an the boundary lo kwise and establish the rst frontier f . Move to the losest point on ext (f ) and establish the next frontier. Continue in this fashion until all of P has been seen and move ba k to the starting point. Deng et al. show that a robot using strategy BGO to explore a re tilinear polygon follows a tour with shortest length, i.e., BGO has ompetitive ratio one. They also present a similar strategy, alled IGO, for the ase when the starting point s lies in the interior of P. For IGO they show a ompetitive ratio of two, i.e., IGO spe i es a tour that is at most twi e as long as the shortest wat hman route through s. IGO shoots a ray upwards to establish a prin ipal proje tion point f 0 and then s ans the boundary
lo kwise to obtain the frontier. Next, it pro eeds exa tly as BGO, moving to the losest point on the extension of the frontier, updating the frontier, and repeating the pro ess until all of the polygon has been seen. 3
It is lear that BGO ould just as well s an the boundary anti- lo kwise instead of lo kwise when establishing the frontiers and still have the same ompetitive ratio. Hen e, BGO an be seen as two strategies, one s anning lo kwise and the other anti- lo kwise. We an therefore parameterize the two strategies so that BGO(p; orient ) is the strategy beginning at some point p on the boundary and s anning with orientation orient where orient is either lo kwise w or anti- lo kwise aw . Similarly for IGO, we an not only hoose to s an lo kwise or anti- lo kwise for the frontier but also hoose to shoot the ray giving the rst prin ipal proje tion point in any of the four ompass dire tions north, west, south, or east. Thus IGO in fa t be omes eight dierent strategies that we an parameterize as IGO(p; dir ; orient ) and the parameter dir an be one of north , south , west , or east . We further de ne partial versions of GO starting at boundary and interior points. Strategies PBGO(p; orient ; region ) and PIGO(p; dir ; orient ; region ) apply GO until either the robot has explored all of region or the robot leaves the region region . The strategies return as result the position of the robot when it leaves region or when region has been explored. Note that PBGO(p; orient ; P) and PIGO(p; dir ; orient ; P) are the same strategies as BGO(p; orient ) and IGO(p; dir ; orient ) respe tively ex ept that they do not move ba k to p when all of P has been seen.
4 The Strategy CGO We present a new strategy ompetitive-greedy-online (CGO) that explores two quadrants simultaneosly without using up too mu h distan e. We assume that s lies in the interior of P sin e otherwise we
an use BGO and a hieve an optimal route. The strategy uses two frontier points simultaneously to improve the ompetitive ratio. However, to initiate the exploration, the strategy begins by performing a s an of the polygon boundary to de ide in whi h dire tion to start the exploration. This is to minimize the loss in i ted upon us by our hoi e of initial dire tion. The initial s an works as follows: onstru t the visibility polygon VP(s) of the initial point s. Consider the set of edges in VP(s) not oin iding with the boundary of P. The end points of these edges de ne a set of frontier points ea h having an asso iated left extension. Let e denote the left extension that is furthest from s (distan e being measured orthogonally to the extension). Let l be the in nite line through e. We rotate the view point of s so that Q3 (s) and Q4 (s) interse t l whereas Q1 (s) and Q2 (s) do not. Hen e, e is a horizontal extension lying below s. The initial dire tion of exploration is upwards through Q1 (s) and Q2 (s). The two frontier points used by the strategy are obtained as follows: the left frontier fl is established by shooting a ray towards the left for the left prin ipal proje tion point fl0 and then s an the boundary in lo kwise dire tion for fl ; see Figure 1(d). The right frontier fr is established by shooting a ray towards the right for the right prin ipal proje tion point fr0 and then s an the boundary in anti- lo kwise dire tion for fr ; see Figure 1(d). To ea h frontier point we asso iate a left extension ext (fl ) and a right extension ext (fr ) with respe t to s. The strategy CGO, presented in pseudo ode below makes use of three dierent substrategies: CGO-0, CGO-1, and CGO-2, that ea h takes are of spe i ases that an o
ur. Subsequently we will prove the orre tness and ompetitive ratio for ea h of the substrategies. Our strategy ensures that whenever it performs one of the substrategies this is the last time that the outermost while-loop is exe uted. Hen e, the loop is repeated only when the strategy does not enter any of the spe i ed substrategies. The loop will lead the strategy to follow a straight line and we will maintain the invariant during the while-loop that all of the region Q3;4 (p) \ Q1;2 (s) has been 4
fl
=
fl
u
=
fl
u
s
u
s
(a)
s
(b)
( )
Figure 2: Illustrating the key point u.
seen. We distinguish four lasses of extensions. A is the lass of extensions e whose de ning edge is above e, B is the lass of extensions e whose de ning edge is below e. Similarly, L is the lass of extensions e whose de ning edge is to the left of e, and R is the lass of extensions e whose de ning edge is to the right of e. For on iseness, we use C1 C2 as a shorthand for the Cartesian produ t C1 C2 of the two
lasses C1 and C2 . We de ne two key verti es u and v together with their extensions ext (u) and ext (v ) that are useful to establish the orre t substrategy to enter. The vertex u lies in Q2 (s) and v in Q1 (s). If ext (fl ) 2 A[B , then u is the vertex issuing ext (fl ) and ext (u) = ext (fl ). If ext (fl ) 2 L and ext (fl ) rosses the verti al line through s, then u is the vertex issuing ext (fl ) and again ext (u) = ext (fl ). If ext (fl ) 2 L does not
ross the verti al line through s, then u is the leftmost vertex of the bottommost edge visible from the robot, on the boundary going from fl lo kwise until we leave Q2 (s). The extension ext (u) is the left extension issued by u, and hen e, ext (u) 2 A; see Figures 2(a), (b), and ( ). The vertex v is de ned symmetri ally in Q1 (s) with respe t to fr . Ea h of the substrategies is presented in sequen e and for ea h of them we prove that if CGO exe utes the substrategy, then the ompetitive ratio of CGO is bounded by 3=2. Let FR s be the
losed route followed by strategy CGO starting at an interior point s. Let FR s (p; q; orient ) denote the subpath of FR s followed in dire tion orient from point p to point q , where orient an either be w ( lo kwise) or aw (anti- lo kwise). Similarly, we de ne the subpath SWR s (p; q; orient ) of SWR s . We denote by SP (p; q ) a shortest re tilinear path from p to q inside P. We begin by establishing two simple but useful lemmas. Lemma 4.1
If
t
is a point on some tour
SWR
s , then
length (SWR ) length (SWR t
:
s)
SWRs passes through t, the route is a wat hman route through t. But sin e the shortest wat hman route through t, the lemma follows.
Proof: Sin e
SWR
t
is
2
SWR S1 = S \ Q1 2 (s) \ Q3 4(s) S4 = S \ Q1 4 (s) length (SWR ) 2 maxfjjs; pjj g + 2 maxfjjs; pjj g + 2 maxfjjs; pjj g + 2 maxfjjs; pjj g: 1 2 3 4
Lemma 4.2 Let S be a set S2 = S 2;3 (s), S3 = S
\Q
s
SWR
of points that are en losed by some tour
p2S
;
, and
y
x
p 2S
s , and let
;
,
. Then
;
p 2S
y
p 2S
x
en loses all the points in S and sin e we al ulate length a
ording to the L1 metri , the smallest tour en losing the points is the smallest re tangle ontaining them. The length of the re tangle's perimeter is as stated, proving the result. 2 Proof:
s
5
The stru ture of the following proofs are very similar to ea h other. In ea h ase we will establish a point t that we an ensure is passed by SWR s and that either lies on the boundary of P or an be viewed as to lie on the boundary of P. We then onsider the tour SWRt and ompare its length the length of FR s . By Lemma 4.1 we know that length (SWR t ) length (SWR s ), hen e the dieren e in length between FR s and SWRt is an upper bound on the loss produ ed by CGO.
CGO Establish the exploration dire tion by performing the initial s an of the polygon boundary 0 0 2 Establish the left and right prin ipal proje tion points fl and fr for Q2 (s) and Q1 (s) respe tively 3 while Q1;2 (s) is not ompletely seen do 3.1 Obtain the left and right frontiers, fl and fr 3.2 if fl lies in Q2 (s) and fr lies in Q1 (s) then 3.2.1 Update verti es u and v as des ribed in the text 3.2.2 if (ext (u); ext (v )) 2 LR or (ext (u); ext (v )) 2 AR [ LA and ext (u) rosses ext (v ) then 3.2.2.1 Go to the losest horizontal extension elseif (ext (u); ext (v )) 2 BR [ LB or (ext (u); ext (v )) 2 AR [ LA and ext (u) does not
ross ext (v ) then 3.2.2.2 Apply substrategy CGO-1 Strategy 1
elseif
ext (u); ext (v )) 2 AA [ AB [ BA [ BB
(
then
Apply substrategy CGO-2
3.2.2.3
endif else
Apply substrategy CGO-0
3.2.3
endif endwhile 4
if
4.1
P is not ompletely seen then Apply substrategy CGO-0
endif End
CGO
We start by presenting CGO-0, that does the following: Let p be the urrent robot position. If Q1 (p) is ompletely seen from p then we run PIGO(p; north ; aw ; P) and move ba k to the starting point s, otherwise Q2 (p) is ompletely seen from p and we run PIGO(p; north ; w ; P) and move ba k to the starting point s. Lemma 4.3
If the strategy applies substrategy
CGO-0,
then
length (FR
s)
=
length (SWR
:
s)
Proof: Assume that CGO-0 realizes that when FR s rea hes the point p, then Q1 (p) is ompletely seen from p. The other ase, that Q2 (p) is ompletely seen from p is symmetri . Sin e the path FR s (s; p; orient ) that the strategy has followed when it rea hes point p is a straight line, the point p is the urrently topmost point of the path. Hen e, we an add a verti al spike issued by the boundary point immediately above p, giving a new polygon P0 having p on the boundary and furthermore with the same shortest wat hman route through p as P. This means that performing strategy IGO(p; north ; orient ) in P yields the same result as performing BGO(p; orient ) in P0 , p being
6
a boundary point in P0 , and orient being either w or aw . The tour followed is therefore a shortest wat hman route through the point p in both P0 and P. Also the point p lies on an extension with respe t to s, by the way p is de ned, and it is the
losest point to s su h that all of Q1 (s) has been seen by the path FR s (s; p; orient ) = SP (s; p). Hen e, there is a route SWR s that ontains p and by Lemma 4.1 length (SWRp ) length (SWR s ). The tour followed equals FR s = SP (s; p) [ SWRp (p; s; aw ); and we have that length (FR s ) = length (SWR p ) length (SWRs ); and sin e FRs annot be stri tly shorter than SWRs the equality holds whi h on ludes the proof. 2 Next we present CGO-1. Let u and v be the key verti es as de ned earlier. The strategy does the following: if (ext (u); ext (v )) 2 LA [ LB , we mirror the polygon P at the verti al line through s and swap the names of u and v . Hen e, (ext (u); ext (v )) 2 AR [ BR. We ontinue moving upwards updating fr and v until either all of Q1 (s) has been seen or ext (v ) no longer rosses the verti al line through s. If all of Q1 (s) has been seen then we explore the remaining part of P using PIGO(p; east ; aw ; P), where p is the urrent robot position. If ext (v ) no longer rosses the verti al line through s then we either need to ontinue the exploration by moving to the right or return to u and explore the remaining part of the polygon from there. If jjs; pjjy + jjs; ujjx jjs; v jjx we hoose to return to u. If ext (u) 2 A we run PBGO(u; aw ; P) and if ext (u) 2 B we use PBGO(u; w ; P); see Figure 3. Otherwise, jjs; pjjy + jjs; ujjx > jjs; v jjx and in this
ase we move to the losest point v 0 on ext (v ). By de nition, the extension of v is either in A or B in this ase. If ext (v ) 2 B then v = v 0 and we hoose to run PBGO(v; aw ; P). Otherwise, ext (v ) 2 A. If Q1 (v 0 ) is seen from v 0 then the entire quadrant has been explored and we run PIGO(v 0 ; east ; aw ; P) to explore the remainder of the polygon. If Q1 (v 0 ) is not seen from v 0 then there are still things hidden from the robot in Q1 (v ). We explore the rest of the quadrant using PBGO(v 0 ; north ; aw ; Q1 (v )) rea hing a point q where a se ond de ision needs to be made. If v is seen from the starting point and jjs; q jjx jjs; v jj, we go ba k to v and run PBGO(v; aw ; P), otherwise we run PIGO(q; east ; w ; P) from the interior point q ; see Figure 5. If v is not seen from the starting point s then we go ba k to v and run PBGO(v; aw ; P). To nish the substrategy CGO-1 our last step is to return to the starting point s. Lemma 4.4
If the strategy applies substrategy
CGO-1,
then
length (FR
s)
23 length (SWR ): s
FRs rea hes the point Hen e, we have the same situation as in the proof of Lemma 4.3 and using the same proof te hnique it follows that length (FR s ) = length (SWR s ). Assume for the se ond ase that CGO-1 de ides to go ba k to u, i.e., that jjs; pjjy +jjs; ujjx jjs; v jjx ; see Figures 3(a) and (b). The tour followed equals one of
Proof: We handle ea h ase separately. Assume for the rst ase that when
p, then
Q1 (p) is ompletely visible.
FR
s
=
SP (s; p) [ SP (p; u) [ SWR SP (s; p) [ SP (p; u) [ SWR
u
[ SP (u; s) u; r; w ) [ SP (r; s)
u(
where r is the last interse tion point of FR s with the horizontal line through s. Using that jjs; pjjy + jjs; ujjx jjs; vjjx it follows that the length of FRs in both ases is bounded by
length (FR
s)
=
jjs; pjj + jjp; ujj + length (SWR
u)
+
7
jju; sjj
=
length (SWR
u)
jj jj
+ 2 s; p
y
jj jj
+ 2 s; u
x
p
p
FR
FR
v
s
v
s
u u r
s
SWR
s
SWR
u
(a)
u
(b)
Figure 3: Illustrating the ases in Lemma 4.4 when jjs; pjjy + jjs; ujjx
jj jj s; v
x
.
x
.
v p
FR
u
p
v
FR
u
s
r
r
SWR
0
v
s
s
SWR
v
(a)
s v
(b)
Figure 4: Illustrating the proof of Lemma 4.4 when jjs; pjjy +jjs; ujjx
length (SWR
s)
+
jjs; pjj
y
+
jjs; ujj
x
+
jjs; vjj
3
x
2
>
jj jj s; v
length (SWR
:
s)
The inequalities follow from the assumption together with Lemmas 4.1 and 4.2. Assume for the third ase that CGO-1 goes to the right, i.e., that jjs; pjjy + jjs; ujjx > jjs; v jjx . We begin by handling the dierent sub ases that are independent of whether s sees v ; see Figures 4(a) and (b). The tour followed equals one of
FR Sin e jjs; v jjx
s
SP (s; v) [ SWR (v; r; aw ) [ SP (r; s) SP (s; v ) [ SWR (v ; r; aw ) [ SP (r; s) v
=
0
v0
0
jjs; v jj the length of FR is in both sub ases bounded by length (FR ) length (SWR ) + 2jjs; vjj < length (SWR ) + jjs; pjj 23 length (SWR ); = s
0
x
s
s
x
s
s
The inequalities follow from Lemmas 4.1 and 4.2.
8
y
+
jjs; ujj
x
+
jjs; vjj
x
Assume now that CGO-1 goes to the right, i.e., that jjs; pjjy + jjs; ujjx > jjs; v jjx and that v is indeed seen from s; see Figures 5(a) and (b). The tour followed in this ase is one of
v; q; w ) [ SP (q; v ) [ SWR (v; r; aw ) [ SP (r; s) () [ SP (v; s) where q is the resulting lo ation after exploring Q1 (v ). Here we use that v is seen from s, and hen e, that the initial s an guarantees that there is a point t of SWR in Q3 4 (s) su h that jjs; tjj jjs; v jj ,
FR
thus
s
FR
=
SP (s; v) [ SWR SP (s; v) [ SWR
v(
v
v
s
s
;
y
x
is bounded by
length (FR
s)
fjjs; vjj; jjs; qjj g ) + jjs; v jj + jjs; v jj + jjs; q jj < length (SWR ) + jjs; v jj + jjs; tjj + jjs; q jj + jjs; ujj
=
length (SWR length (SWR
v)
+ 2 min
x
s
y
s
y
x
x
y
x
x
3 2
length (SWR
:
s)
On the other hand, when v is not seen from s, the tour follows the path marked with () above; see Figure 5( ). Thus, the polygon boundary obs ures the view from s to v , and hen e, there is a point q 0 on the boundary su h that the shortest path from s to v 0 ontains q 0 . The path our strategy follows between s and v 0 is a shortest path and we an therefore assume that it also passed through q 0 . We use that jjs; q 0 jjx jjs; v jjx jjs; q jjx to get the bound.
length (FR
s)
jj length (SWR ) + jjs; vjj + jjs; qjj 3 length (SWR ) + jjs; vjj + jjs; ujj + jjs; qjj length (SWR ): 2
length (SWR
=
s; v
x
s)
23 length (SWR ): s
. The other ase is proved symmet-
passes through v ; see Figures 6(a), (b), and ( ). The tour followed equals
[ SP (r; s)
w ) [ SP (q; v ) [ SWR (v; r; aw ) [ SP (r; s) : [ SP (v; s) where r is the last interse tion point of FR with the horizontal line through s. The length of FR FR
( )
jj jj
s
r
s v
s; v
(d)
Figure 6: Illustrating the ases in the proof of Lemma 4.5.
11
0
t
p
p
p
s s
(a)
x s
(b)
( )
x
Figure 7: Illustrating the proof of Theorem 2.
From this we dedu e that the strategy CGO presented previously is 3- ompetitive for path exploration. We ontinue to show that any strategy for path exploration must be at least 2- ompetitive. Theorem 2
There is no deterministi strategy for path exploration of a re tilinear polygon that has
ompetitive ratio
2
for any
> 0.
Proof: We onstru t a ounterexample as in Figure 7. The starting point for the robot is at the lower
left orner of the polygon and it essentially sees only the two walls adja ent to it; see Figure 7(a). The robot now has to move to one of the extensions that it sees. These are both at distan e 1 from the starting point. Assume without loss of generality that it moves to the horizontal extension, then the robot realizes that there is a further horizontal extension at distan e Æ above it. The robot has the option of either ontinuing upwards until it has visited all the horizontal extensions (of whi h the
orresponding frontier point is only visible from the previous extension and the extensions are only separated by a distan e of Æ ); see Figure 7(b). This option will make the robot move a distan e of 1=Æ upwards until it rea hes the last extension and moves to the verti al extension where it realizes that there is a not h at point x (whi h lies at distan e Æ below the starting point s) for ing it to move ba k down to this point. The total distan e moved is 2=Æ + 1 + Æ . The optimal path is to move to the verti al extension rst, visit the not h at x and then move upwards until all horizontal extensions have been visited, requiring only a distan e of 1 + 2Æ + 1=Æ to be moved. The ratio be omes 2=Æ + 1 + Æ 1=Æ + 1 + 2Æ
=2
Æ + 3Æ 2 1 + Æ + 2Æ 2
2
=4 < 1=4. On the other hand, if the robot at some point after moving to the rst horizontal extension de ides to move to the verti al extension it then realizes that it has to move ba k to the not h at x. If it de ides to ontinue upwards until all horizontal extensions have been visited we have the previous situation. If the robot de ides at some point to move downwards to visit the not h at x, then it has to move ba k if Æ
12
up again to visit the next horizontal extension whi h also turns out to be the last one and the robot terminates at point t; see Figure 7( ). Assume that the robot has moved a verti al distan e of D 1 when it de ides to move down and visit the not h at x. The total distan e moved is then 3D + 1 + 3Æ whereas the optimal path has length at most D + 1 + 3Æ and the ratio be omes 3D + 1 + 3Æ
D + 1 + 3Æ
if Æ
=3
2 + 6Æ
D + 1 + 3Æ
=2 < 1=2, thus on luding the proof.
2
2
6 Exploration with Multiple Robots We now look at the situation when several robots together are required to explore a re tilinear polygon. Again we look at the tour variant, i.e., ea h robot must terminate the axploration at the starting point. We give upper and lower bounds for the situation with two and three robots all starting at the same point. the measure that we optimize on is the length of the longest tour that any of the robots follow. Let OPTsk be the tour of the robot that moves the longest length of all the k robots. Sin e a single robot an follow ea h of the tours that the k robots follow and thus get a wat hman route, we have that length (OPTsk ) length (SWRs )=k . We prove lower bounds on the ompetitive ratio of any exploration strategy using two or three robots. Theorem 3
There are no deterministi strategies for exploring a re tilinear polygon with two or three
robots having smaller ompetitive ratio than 3/2.
Proof: We rst show the lower bound for two robots and later extend it for three robots. The lower
bound is based on essentially the same ounterexample that Kleinberg uses for the lower bound for a single robot [7℄. The initial polygon is given in Figure 8(a) and onsists of a square with not hes in the orners. The length of the sides of the square is 2. Ea h of the two robots has to move a distan e of 2 before it gets to a orner of the polygon and furthermore at most two omplete orners an be seen by the robots. Hen e pla ing a not h in one of the orners that is not yet ompletely seen requires one of the robots to move 6 units whereas the optimal motion an be done with only 4 units, thus proving the result. The same proof a tually goes through for three robots on e you realize that independently of how the three robots start their exploration we an for e one of them to move 6 units pla ing at most three not hes as shown in Figure 8(b). 2 We also show a strategy for two robots that has ompetitive ratio 2. We all the strategy TGO (two-robot GO) sin e it is based on the GO-strategy of Deng et al ; [4℄. The two robots ea h run IGO(s; north ; orient ), one with orient = w and the other with orient = aw until the two robots have seen the omplete polygon after whi h they both move ba k to the starting point. Theorem 4
The strategy TGO is 2- ompetitive for exploration of a re tilinear polygon with two
robots.
13
not h
s
not h
s
not h not h (a)
(b)
Figure 8: Illustrating the proof of Theorem 3. Proof: Let r be the interse tion point of SWRs and the verti al axis issuing from s upwards. The interse tion point r lies at distan e D 0 from s; see Figure 9(a). We an view the strategy TGO as rst moving the two robots from s to r and then separating, one moving lo kwise and the other
ounter lo kwise, following SWR r in two dire tions until the robots have seen all of the polygon and move ba k to s. Consider now the robot that moves the farthest. Let t be the point of interse tion between the robot's walk and the last extension that it visits before it realizes that the whole polygon is explored. The robot then moves the distan e L = D + length (SWR r (r; t; dir )) + length (SP (t; s)); where dir is the dire tion that the robot moves. Suppose that we follow the tour SWRs from s in the dire tion that visits the point r before t. We
an assume that t is a point on SWRs sin e it is an interse tion point with an extension. Assume without loss of generality that this dire tion is lo kwise. We thus have that
L
=
D + length (SWRr (r; t; dir )) + length (SP (t; s))
length (SWR (s; r; w )) + length (SWR s
=
length (SWR ) 2 2length (OPT )
r; t; w )) + length (SWRs (t; s; w ))
s(
s
s
whi h proves the result. That the analysis is tight follows from the example in Figure 9(b).
2
7 Con lusions We have presented onstant ompetitive strategies and lower bounds to explore a re tilinear simple polygon in the L1 metri with one or more robots. Unfortunately none of our results are tight so obvious open problems are to redu e the gaps between the lower bounds and the upper bounds. Espe ially exploration using k robots, for an arbitrary number k , needs to be investigated. 14
r
SWR
s
s
SP (t; s) t
(a)
(b)
Figure 9: Illustrating the proof of Theorem 4.
Referen es [1℄ Margrit Betke, Ronald L. Rivest, Mona Singh. Pie emeal Learning of an Unknown Environment. Ma hine Learning, 18(23):231254, 1995. [2℄ K-F. Chan, T. W. Lam. An on-line algorithm for navigating in an unknown environment. International Journal of Computational Geometry & Appli ations, 3:227244, 1993. [3℄ W. Chin, S. Ntafos. Optimum Wat hman Routes. 1988.
Information Pro essing Letters, 28:3944,
[4℄ X. Deng, T. Kameda, C.H. Papadimitriou. How to Learn an Unknown Environment I: The Re tilinear Case. Journal of the ACM, 45(2):215245, 1998. [5℄ M. Hammar, B.J. Nilsson, S. S huierer. Improved Exploration of Re tilinear Polygons. Nordi Journal of Computing, 9(1):3253, 2002. [6℄ F. Hoffmann, C. I king, R. Klein, K. Kriegel. The Polygon Exploration Problem. Journal on Computing, 31(2):577600, 2001. [7℄ J. M. Kleinberg. On-line sear h in a simple polygon. In Dis rete Algorithms, pages 815, 1994.
SIAM
Pro . of 5th ACM-SIAM Symp. on
[8℄ Aohan Mei, Yoshihide Igarashi. An E ient Strategy for Robot Navigation in Unknown Environment. Inform. Pro ess. Lett., 52:5156, 1994. [9℄ C. H. Papadimitriou, M. Yannakakis. Shortest Paths Without a Map. 84(1):127150, 1991.
15
Theoret. Comput. S i.,
http://dspace.org
Faculty of Technology and Society
Articles /TS
2006
Competitive Exploration of Rectilinear Polygons Hammar, Mikael Elsevier 367-378 http://hdl.handle.net/2043/11304 Downloaded from DSpace Repository, DSpace Institution's institutional repository
Competitive Exploration of Re tilinear Polygons
Mikael Hammar
Bengt J.
Nilsson
y
y
Mia Persson
Abstra t
Exploring a polygon with robots, when the robots do not have knowledge of the surroundings
an be viewed as an online problem. Typi al for online problems is that de isions must be made based on past events without omplete information about the future. In our ase the robots do not have omplete information about the environment. Competitive analysis an be used to measure the performan e of methods solving online problems. The ompetitive ratio of su h a method is the ratio between the method's performan e and the performan e of the best method having full knowledge of the future. We are interested in obtaining good bounds on the ompetitive ratio of exploring polygons and prove onstant ompetitive strategies and lower bounds for exploring a simple re tilinear polygon in the L1 metri .
1 Introdu tion Exploring an environment is an important and well studied problem in roboti s. In many realisti situations the robots do not possess omplete knowledge about the environment, e.g., they may not have a map of the surroundings [1, 2, 4, 6, 7, 8, 9℄. The sear h of the robots an be viewed as an online problem sin e the robots' de isions about the sear h are based only on the part of the environment that they have seen so far. We use the framework of ompetitive analysis to measure the performan e of an online sear h strategy S . The ompetitive ratio of S is de ned as the maximum of the ratio of the distan e traveled by the robot that moves the farthest using S to the optimal distan e of the sear h. We are interested in obtaining good bounds for the ompetitive ratio of exploring a re tilinear polygon. The sear h is modeled by a path or losed tour followed by one or more point sized robots inside the polygon, given a starting point for the sear h. The only information that the robots have about the surrounding polygon is the part of the polygon that they together have seen so far. For the ase of exploration with one robot, Deng et al. [4℄ show a deterministi strategy having
ompetitive ratio two for this problem if distan e is measured a
ording to the L1 -metri . Hammar et al. [5℄ prove a strategy with ompetitive ratio 5=3 and Kleinberg [7℄ proves a lower bound of 5=4 for the ompetitive ratio of any deterministi strategy. We will show a deterministi strategy obtaining a
ompetitive ratio of 3=2 for sear hing a re tilinear polygon in the L1 -metri with one robot. Department of Computer S ien e, Salerno University, Baronissi (SA) - 84081, Italy. email:
hammardia.unisa.it
email:
{Bengt.Nilsson,Mia.Persson}ts.mah.se
y Te hnology and So iety, Malmö University College, S-205 06 Malmö, Sweden.
1
We also present a lower bound of 2 for the orresponding path exploration problem for one robot in re tilinear polygons and ompetitive results on exploration with two and three robots. The paper is organized as follows. In the next se tion we present some de nitions and preliminary results. In Se tion 3 we give an overview of the strategy by Deng et al. [4℄. Se tion 4 ontains an improved strategy for single robot exploration giving a ompetitive ratio of 3=2. In se tions 5 and 6 we onsider path exploration and exploration with multiple robots.
2 Preliminaries We will hen eforth always measure distan e a
ording to the L1 metri , i.e., the distan e between two points p and q is de ned by
jjp; qjj = jp
x
j j
qx + py
j
qy ;
where px and qx are the x- oordinates of p and q and py and qy are the y - oordinates. We de ne the x-distan e between p and q to be jjp; q jjx = jpx qx j and the y -distan e to be jjp; q jjy = jpy qy j. If C is a polygonal urve, then the length of C , denoted length (C ), is de ned the sum of the distan es between onse utive pairs of segment end points in C . Let P be a simple re tilinear polygon. Two points in P are said to see ea h other, or be visible to ea h other, if the line segment onne ting the points lies in P. Let p be a point somewhere inside P. A wat hman route through p is de ned to be a losed urve C that passes through p su h that every point in P is seen by some point on C . The shortest wat hman route through p is denoted by SWR p . It an be shown that the shortest wat hman route in a simple polygon is a losed polygonal urve [3℄. Sin e we are only interested in the L1 length of a polygonal urve we an assume that the urve is re tilinear, that is, the segments of the urve are all axis parallel. Note that the shortest re tilinear wat hman route through a point p is not ne essarily unique. For a point p in P we de ne four quadrants with respe t to p. Those are the regions obtained by
utting P along the two maximal axis parallel line segments that pass through p. The four quadrants are denoted Q1 (p), Q2 (p), Q3 (p), and Q4 (p) in anti- lo kwise order from the top right quadrant to the bottom right quadrant. We let Qi;j (p) denote the union of Qi (p) and Qj (p). Consider a re ex vertex of P. The two edges of P onne ting at the re ex vertex an ea h be extended inside P until the extensions rea h a boundary point. The segments thus onstru ted are
alled extensions and to ea h extension a dire tion is asso iated. The dire tion is the same as that of the ollinear polygon edge as we follow the boundary of P in lo kwise order. We use the four ompass dire tions north , west , south , and east to denote the dire tion of an extension. Lemma 2.1 (Chin, Ntafos [3℄)
A losed urve is a wat hman route for
has at least one point to the right of every extension of
P
P
if and only if the urve
.
Our rst obje tive is to present a ompetitive online strategy that enables a robot to follow a losed
urve from the start point s in P and ba k to s with the urve being a wat hman route for P. An extension e splits P into two sets Pl and Pr with Pl to the left of e and Pr to the right. We say a point p is to the left of e if p belongs to Pl . To the right is de ned analogously. As a further de nition we say that an extension e is a left extension with respe t to a point p, if p lies to the left of e, and an extension e dominates another extension e0 , if all points of P to the right 2
f
0
prin ipal proje tion f
frontier
f
f
ext (f )
0 f
=
v f
s
C
(a)
ext (f )
0
C
(b)
fr
v
fl
p
0
0
fl
q
( )
fr s
(d)
Figure 1: Illustrating de nitions.
of e are also to the right of e0 . By Lemma 2.1 we are only interested in the extensions that are left extensions with respe t to the starting point s sin e the other ones already have a point (the point s) to the right of them. So without loss of larity when we mention extensions we will always mean extensions that are left extensions with respe t to s.
3 An Overview of GO Consider a re tilinear polygon P that is not a priori known to the robot. Let s be the robot's initial position inside P. For the starting position s of the robot we asso iate a point f 0 on the boundary of P that is visible from s and all f 0 the prin ipal proje tion point of s. For instan e, we an hoose f 0
to be the rst point on the boundary that is hit by an upward ray starting at s. Let f be the end point of the boundary that the robot sees as we s an the boundary of P in lo kwise order; see Figure 1(a). The point f is alled the urrent frontier. Let C be a polygonal urve starting at s. Formally a frontier f of C is a vertex of the visibility polygon, VP(C ) of C adja ent to an edge e of VP(C ) that is not an edge of P. Extend e until it hits a point q on C and let v be the vertex of P that is rst en ountered as we move along the line segment [q; f ℄ from q to f . We denote the left extension with respe t to s asso iated to the vertex v by ext (f ); see Figures 1(b) and ( ). Deng et al. [4℄ introdu e an online strategy alled greedy-online, GO for short, to explore a simple re tilinear polygon P in the L1 metri . If the starting point s lies on the boundary of P, their strategy, we all it BGO, goes as follows: from the starting point s an the boundary lo kwise and establish the rst frontier f . Move to the losest point on ext (f ) and establish the next frontier. Continue in this fashion until all of P has been seen and move ba k to the starting point. Deng et al. show that a robot using strategy BGO to explore a re tilinear polygon follows a tour with shortest length, i.e., BGO has ompetitive ratio one. They also present a similar strategy, alled IGO, for the ase when the starting point s lies in the interior of P. For IGO they show a ompetitive ratio of two, i.e., IGO spe i es a tour that is at most twi e as long as the shortest wat hman route through s. IGO shoots a ray upwards to establish a prin ipal proje tion point f 0 and then s ans the boundary
lo kwise to obtain the frontier. Next, it pro eeds exa tly as BGO, moving to the losest point on the extension of the frontier, updating the frontier, and repeating the pro ess until all of the polygon has been seen. 3
It is lear that BGO ould just as well s an the boundary anti- lo kwise instead of lo kwise when establishing the frontiers and still have the same ompetitive ratio. Hen e, BGO an be seen as two strategies, one s anning lo kwise and the other anti- lo kwise. We an therefore parameterize the two strategies so that BGO(p; orient ) is the strategy beginning at some point p on the boundary and s anning with orientation orient where orient is either lo kwise w or anti- lo kwise aw . Similarly for IGO, we an not only hoose to s an lo kwise or anti- lo kwise for the frontier but also hoose to shoot the ray giving the rst prin ipal proje tion point in any of the four ompass dire tions north, west, south, or east. Thus IGO in fa t be omes eight dierent strategies that we an parameterize as IGO(p; dir ; orient ) and the parameter dir an be one of north , south , west , or east . We further de ne partial versions of GO starting at boundary and interior points. Strategies PBGO(p; orient ; region ) and PIGO(p; dir ; orient ; region ) apply GO until either the robot has explored all of region or the robot leaves the region region . The strategies return as result the position of the robot when it leaves region or when region has been explored. Note that PBGO(p; orient ; P) and PIGO(p; dir ; orient ; P) are the same strategies as BGO(p; orient ) and IGO(p; dir ; orient ) respe tively ex ept that they do not move ba k to p when all of P has been seen.
4 The Strategy CGO We present a new strategy ompetitive-greedy-online (CGO) that explores two quadrants simultaneosly without using up too mu h distan e. We assume that s lies in the interior of P sin e otherwise we
an use BGO and a hieve an optimal route. The strategy uses two frontier points simultaneously to improve the ompetitive ratio. However, to initiate the exploration, the strategy begins by performing a s an of the polygon boundary to de ide in whi h dire tion to start the exploration. This is to minimize the loss in i ted upon us by our hoi e of initial dire tion. The initial s an works as follows: onstru t the visibility polygon VP(s) of the initial point s. Consider the set of edges in VP(s) not oin iding with the boundary of P. The end points of these edges de ne a set of frontier points ea h having an asso iated left extension. Let e denote the left extension that is furthest from s (distan e being measured orthogonally to the extension). Let l be the in nite line through e. We rotate the view point of s so that Q3 (s) and Q4 (s) interse t l whereas Q1 (s) and Q2 (s) do not. Hen e, e is a horizontal extension lying below s. The initial dire tion of exploration is upwards through Q1 (s) and Q2 (s). The two frontier points used by the strategy are obtained as follows: the left frontier fl is established by shooting a ray towards the left for the left prin ipal proje tion point fl0 and then s an the boundary in lo kwise dire tion for fl ; see Figure 1(d). The right frontier fr is established by shooting a ray towards the right for the right prin ipal proje tion point fr0 and then s an the boundary in anti- lo kwise dire tion for fr ; see Figure 1(d). To ea h frontier point we asso iate a left extension ext (fl ) and a right extension ext (fr ) with respe t to s. The strategy CGO, presented in pseudo ode below makes use of three dierent substrategies: CGO-0, CGO-1, and CGO-2, that ea h takes are of spe i ases that an o
ur. Subsequently we will prove the orre tness and ompetitive ratio for ea h of the substrategies. Our strategy ensures that whenever it performs one of the substrategies this is the last time that the outermost while-loop is exe uted. Hen e, the loop is repeated only when the strategy does not enter any of the spe i ed substrategies. The loop will lead the strategy to follow a straight line and we will maintain the invariant during the while-loop that all of the region Q3;4 (p) \ Q1;2 (s) has been 4
fl
=
fl
u
=
fl
u
s
u
s
(a)
s
(b)
( )
Figure 2: Illustrating the key point u.
seen. We distinguish four lasses of extensions. A is the lass of extensions e whose de ning edge is above e, B is the lass of extensions e whose de ning edge is below e. Similarly, L is the lass of extensions e whose de ning edge is to the left of e, and R is the lass of extensions e whose de ning edge is to the right of e. For on iseness, we use C1 C2 as a shorthand for the Cartesian produ t C1 C2 of the two
lasses C1 and C2 . We de ne two key verti es u and v together with their extensions ext (u) and ext (v ) that are useful to establish the orre t substrategy to enter. The vertex u lies in Q2 (s) and v in Q1 (s). If ext (fl ) 2 A[B , then u is the vertex issuing ext (fl ) and ext (u) = ext (fl ). If ext (fl ) 2 L and ext (fl ) rosses the verti al line through s, then u is the vertex issuing ext (fl ) and again ext (u) = ext (fl ). If ext (fl ) 2 L does not
ross the verti al line through s, then u is the leftmost vertex of the bottommost edge visible from the robot, on the boundary going from fl lo kwise until we leave Q2 (s). The extension ext (u) is the left extension issued by u, and hen e, ext (u) 2 A; see Figures 2(a), (b), and ( ). The vertex v is de ned symmetri ally in Q1 (s) with respe t to fr . Ea h of the substrategies is presented in sequen e and for ea h of them we prove that if CGO exe utes the substrategy, then the ompetitive ratio of CGO is bounded by 3=2. Let FR s be the
losed route followed by strategy CGO starting at an interior point s. Let FR s (p; q; orient ) denote the subpath of FR s followed in dire tion orient from point p to point q , where orient an either be w ( lo kwise) or aw (anti- lo kwise). Similarly, we de ne the subpath SWR s (p; q; orient ) of SWR s . We denote by SP (p; q ) a shortest re tilinear path from p to q inside P. We begin by establishing two simple but useful lemmas. Lemma 4.1
If
t
is a point on some tour
SWR
s , then
length (SWR ) length (SWR t
:
s)
SWRs passes through t, the route is a wat hman route through t. But sin e the shortest wat hman route through t, the lemma follows.
Proof: Sin e
SWR
t
is
2
SWR S1 = S \ Q1 2 (s) \ Q3 4(s) S4 = S \ Q1 4 (s) length (SWR ) 2 maxfjjs; pjj g + 2 maxfjjs; pjj g + 2 maxfjjs; pjj g + 2 maxfjjs; pjj g: 1 2 3 4
Lemma 4.2 Let S be a set S2 = S 2;3 (s), S3 = S
\Q
s
SWR
of points that are en losed by some tour
p2S
;
, and
y
x
p 2S
s , and let
;
,
. Then
;
p 2S
y
p 2S
x
en loses all the points in S and sin e we al ulate length a
ording to the L1 metri , the smallest tour en losing the points is the smallest re tangle ontaining them. The length of the re tangle's perimeter is as stated, proving the result. 2 Proof:
s
5
The stru ture of the following proofs are very similar to ea h other. In ea h ase we will establish a point t that we an ensure is passed by SWR s and that either lies on the boundary of P or an be viewed as to lie on the boundary of P. We then onsider the tour SWRt and ompare its length the length of FR s . By Lemma 4.1 we know that length (SWR t ) length (SWR s ), hen e the dieren e in length between FR s and SWRt is an upper bound on the loss produ ed by CGO.
CGO Establish the exploration dire tion by performing the initial s an of the polygon boundary 0 0 2 Establish the left and right prin ipal proje tion points fl and fr for Q2 (s) and Q1 (s) respe tively 3 while Q1;2 (s) is not ompletely seen do 3.1 Obtain the left and right frontiers, fl and fr 3.2 if fl lies in Q2 (s) and fr lies in Q1 (s) then 3.2.1 Update verti es u and v as des ribed in the text 3.2.2 if (ext (u); ext (v )) 2 LR or (ext (u); ext (v )) 2 AR [ LA and ext (u) rosses ext (v ) then 3.2.2.1 Go to the losest horizontal extension elseif (ext (u); ext (v )) 2 BR [ LB or (ext (u); ext (v )) 2 AR [ LA and ext (u) does not
ross ext (v ) then 3.2.2.2 Apply substrategy CGO-1 Strategy 1
elseif
ext (u); ext (v )) 2 AA [ AB [ BA [ BB
(
then
Apply substrategy CGO-2
3.2.2.3
endif else
Apply substrategy CGO-0
3.2.3
endif endwhile 4
if
4.1
P is not ompletely seen then Apply substrategy CGO-0
endif End
CGO
We start by presenting CGO-0, that does the following: Let p be the urrent robot position. If Q1 (p) is ompletely seen from p then we run PIGO(p; north ; aw ; P) and move ba k to the starting point s, otherwise Q2 (p) is ompletely seen from p and we run PIGO(p; north ; w ; P) and move ba k to the starting point s. Lemma 4.3
If the strategy applies substrategy
CGO-0,
then
length (FR
s)
=
length (SWR
:
s)
Proof: Assume that CGO-0 realizes that when FR s rea hes the point p, then Q1 (p) is ompletely seen from p. The other ase, that Q2 (p) is ompletely seen from p is symmetri . Sin e the path FR s (s; p; orient ) that the strategy has followed when it rea hes point p is a straight line, the point p is the urrently topmost point of the path. Hen e, we an add a verti al spike issued by the boundary point immediately above p, giving a new polygon P0 having p on the boundary and furthermore with the same shortest wat hman route through p as P. This means that performing strategy IGO(p; north ; orient ) in P yields the same result as performing BGO(p; orient ) in P0 , p being
6
a boundary point in P0 , and orient being either w or aw . The tour followed is therefore a shortest wat hman route through the point p in both P0 and P. Also the point p lies on an extension with respe t to s, by the way p is de ned, and it is the
losest point to s su h that all of Q1 (s) has been seen by the path FR s (s; p; orient ) = SP (s; p). Hen e, there is a route SWR s that ontains p and by Lemma 4.1 length (SWRp ) length (SWR s ). The tour followed equals FR s = SP (s; p) [ SWRp (p; s; aw ); and we have that length (FR s ) = length (SWR p ) length (SWRs ); and sin e FRs annot be stri tly shorter than SWRs the equality holds whi h on ludes the proof. 2 Next we present CGO-1. Let u and v be the key verti es as de ned earlier. The strategy does the following: if (ext (u); ext (v )) 2 LA [ LB , we mirror the polygon P at the verti al line through s and swap the names of u and v . Hen e, (ext (u); ext (v )) 2 AR [ BR. We ontinue moving upwards updating fr and v until either all of Q1 (s) has been seen or ext (v ) no longer rosses the verti al line through s. If all of Q1 (s) has been seen then we explore the remaining part of P using PIGO(p; east ; aw ; P), where p is the urrent robot position. If ext (v ) no longer rosses the verti al line through s then we either need to ontinue the exploration by moving to the right or return to u and explore the remaining part of the polygon from there. If jjs; pjjy + jjs; ujjx jjs; v jjx we hoose to return to u. If ext (u) 2 A we run PBGO(u; aw ; P) and if ext (u) 2 B we use PBGO(u; w ; P); see Figure 3. Otherwise, jjs; pjjy + jjs; ujjx > jjs; v jjx and in this
ase we move to the losest point v 0 on ext (v ). By de nition, the extension of v is either in A or B in this ase. If ext (v ) 2 B then v = v 0 and we hoose to run PBGO(v; aw ; P). Otherwise, ext (v ) 2 A. If Q1 (v 0 ) is seen from v 0 then the entire quadrant has been explored and we run PIGO(v 0 ; east ; aw ; P) to explore the remainder of the polygon. If Q1 (v 0 ) is not seen from v 0 then there are still things hidden from the robot in Q1 (v ). We explore the rest of the quadrant using PBGO(v 0 ; north ; aw ; Q1 (v )) rea hing a point q where a se ond de ision needs to be made. If v is seen from the starting point and jjs; q jjx jjs; v jj, we go ba k to v and run PBGO(v; aw ; P), otherwise we run PIGO(q; east ; w ; P) from the interior point q ; see Figure 5. If v is not seen from the starting point s then we go ba k to v and run PBGO(v; aw ; P). To nish the substrategy CGO-1 our last step is to return to the starting point s. Lemma 4.4
If the strategy applies substrategy
CGO-1,
then
length (FR
s)
23 length (SWR ): s
FRs rea hes the point Hen e, we have the same situation as in the proof of Lemma 4.3 and using the same proof te hnique it follows that length (FR s ) = length (SWR s ). Assume for the se ond ase that CGO-1 de ides to go ba k to u, i.e., that jjs; pjjy +jjs; ujjx jjs; v jjx ; see Figures 3(a) and (b). The tour followed equals one of
Proof: We handle ea h ase separately. Assume for the rst ase that when
p, then
Q1 (p) is ompletely visible.
FR
s
=
SP (s; p) [ SP (p; u) [ SWR SP (s; p) [ SP (p; u) [ SWR
u
[ SP (u; s) u; r; w ) [ SP (r; s)
u(
where r is the last interse tion point of FR s with the horizontal line through s. Using that jjs; pjjy + jjs; ujjx jjs; vjjx it follows that the length of FRs in both ases is bounded by
length (FR
s)
=
jjs; pjj + jjp; ujj + length (SWR
u)
+
7
jju; sjj
=
length (SWR
u)
jj jj
+ 2 s; p
y
jj jj
+ 2 s; u
x
p
p
FR
FR
v
s
v
s
u u r
s
SWR
s
SWR
u
(a)
u
(b)
Figure 3: Illustrating the ases in Lemma 4.4 when jjs; pjjy + jjs; ujjx
jj jj s; v
x
.
x
.
v p
FR
u
p
v
FR
u
s
r
r
SWR
0
v
s
s
SWR
v
(a)
s v
(b)
Figure 4: Illustrating the proof of Lemma 4.4 when jjs; pjjy +jjs; ujjx
length (SWR
s)
+
jjs; pjj
y
+
jjs; ujj
x
+
jjs; vjj
3
x
2
>
jj jj s; v
length (SWR
:
s)
The inequalities follow from the assumption together with Lemmas 4.1 and 4.2. Assume for the third ase that CGO-1 goes to the right, i.e., that jjs; pjjy + jjs; ujjx > jjs; v jjx . We begin by handling the dierent sub ases that are independent of whether s sees v ; see Figures 4(a) and (b). The tour followed equals one of
FR Sin e jjs; v jjx
s
SP (s; v) [ SWR (v; r; aw ) [ SP (r; s) SP (s; v ) [ SWR (v ; r; aw ) [ SP (r; s) v
=
0
v0
0
jjs; v jj the length of FR is in both sub ases bounded by length (FR ) length (SWR ) + 2jjs; vjj < length (SWR ) + jjs; pjj 23 length (SWR ); = s
0
x
s
s
x
s
s
The inequalities follow from Lemmas 4.1 and 4.2.
8
y
+
jjs; ujj
x
+
jjs; vjj
x
Assume now that CGO-1 goes to the right, i.e., that jjs; pjjy + jjs; ujjx > jjs; v jjx and that v is indeed seen from s; see Figures 5(a) and (b). The tour followed in this ase is one of
v; q; w ) [ SP (q; v ) [ SWR (v; r; aw ) [ SP (r; s) () [ SP (v; s) where q is the resulting lo ation after exploring Q1 (v ). Here we use that v is seen from s, and hen e, that the initial s an guarantees that there is a point t of SWR in Q3 4 (s) su h that jjs; tjj jjs; v jj ,
FR
thus
s
FR
=
SP (s; v) [ SWR SP (s; v) [ SWR
v(
v
v
s
s
;
y
x
is bounded by
length (FR
s)
fjjs; vjj; jjs; qjj g ) + jjs; v jj + jjs; v jj + jjs; q jj < length (SWR ) + jjs; v jj + jjs; tjj + jjs; q jj + jjs; ujj
=
length (SWR length (SWR
v)
+ 2 min
x
s
y
s
y
x
x
y
x
x
3 2
length (SWR
:
s)
On the other hand, when v is not seen from s, the tour follows the path marked with () above; see Figure 5( ). Thus, the polygon boundary obs ures the view from s to v , and hen e, there is a point q 0 on the boundary su h that the shortest path from s to v 0 ontains q 0 . The path our strategy follows between s and v 0 is a shortest path and we an therefore assume that it also passed through q 0 . We use that jjs; q 0 jjx jjs; v jjx jjs; q jjx to get the bound.
length (FR
s)
jj length (SWR ) + jjs; vjj + jjs; qjj 3 length (SWR ) + jjs; vjj + jjs; ujj + jjs; qjj length (SWR ): 2
length (SWR
=
s; v
x
s)
23 length (SWR ): s
. The other ase is proved symmet-
passes through v ; see Figures 6(a), (b), and ( ). The tour followed equals
[ SP (r; s)
w ) [ SP (q; v ) [ SWR (v; r; aw ) [ SP (r; s) : [ SP (v; s) where r is the last interse tion point of FR with the horizontal line through s. The length of FR FR
( )
jj jj
s
r
s v
s; v
(d)
Figure 6: Illustrating the ases in the proof of Lemma 4.5.
11
0
t
p
p
p
s s
(a)
x s
(b)
( )
x
Figure 7: Illustrating the proof of Theorem 2.
From this we dedu e that the strategy CGO presented previously is 3- ompetitive for path exploration. We ontinue to show that any strategy for path exploration must be at least 2- ompetitive. Theorem 2
There is no deterministi strategy for path exploration of a re tilinear polygon that has
ompetitive ratio
2
for any
> 0.
Proof: We onstru t a ounterexample as in Figure 7. The starting point for the robot is at the lower
left orner of the polygon and it essentially sees only the two walls adja ent to it; see Figure 7(a). The robot now has to move to one of the extensions that it sees. These are both at distan e 1 from the starting point. Assume without loss of generality that it moves to the horizontal extension, then the robot realizes that there is a further horizontal extension at distan e Æ above it. The robot has the option of either ontinuing upwards until it has visited all the horizontal extensions (of whi h the
orresponding frontier point is only visible from the previous extension and the extensions are only separated by a distan e of Æ ); see Figure 7(b). This option will make the robot move a distan e of 1=Æ upwards until it rea hes the last extension and moves to the verti al extension where it realizes that there is a not h at point x (whi h lies at distan e Æ below the starting point s) for ing it to move ba k down to this point. The total distan e moved is 2=Æ + 1 + Æ . The optimal path is to move to the verti al extension rst, visit the not h at x and then move upwards until all horizontal extensions have been visited, requiring only a distan e of 1 + 2Æ + 1=Æ to be moved. The ratio be omes 2=Æ + 1 + Æ 1=Æ + 1 + 2Æ
=2
Æ + 3Æ 2 1 + Æ + 2Æ 2
2
=4 < 1=4. On the other hand, if the robot at some point after moving to the rst horizontal extension de ides to move to the verti al extension it then realizes that it has to move ba k to the not h at x. If it de ides to ontinue upwards until all horizontal extensions have been visited we have the previous situation. If the robot de ides at some point to move downwards to visit the not h at x, then it has to move ba k if Æ
12
up again to visit the next horizontal extension whi h also turns out to be the last one and the robot terminates at point t; see Figure 7( ). Assume that the robot has moved a verti al distan e of D 1 when it de ides to move down and visit the not h at x. The total distan e moved is then 3D + 1 + 3Æ whereas the optimal path has length at most D + 1 + 3Æ and the ratio be omes 3D + 1 + 3Æ
D + 1 + 3Æ
if Æ
=3
2 + 6Æ
D + 1 + 3Æ
=2 < 1=2, thus on luding the proof.
2
2
6 Exploration with Multiple Robots We now look at the situation when several robots together are required to explore a re tilinear polygon. Again we look at the tour variant, i.e., ea h robot must terminate the axploration at the starting point. We give upper and lower bounds for the situation with two and three robots all starting at the same point. the measure that we optimize on is the length of the longest tour that any of the robots follow. Let OPTsk be the tour of the robot that moves the longest length of all the k robots. Sin e a single robot an follow ea h of the tours that the k robots follow and thus get a wat hman route, we have that length (OPTsk ) length (SWRs )=k . We prove lower bounds on the ompetitive ratio of any exploration strategy using two or three robots. Theorem 3
There are no deterministi strategies for exploring a re tilinear polygon with two or three
robots having smaller ompetitive ratio than 3/2.
Proof: We rst show the lower bound for two robots and later extend it for three robots. The lower
bound is based on essentially the same ounterexample that Kleinberg uses for the lower bound for a single robot [7℄. The initial polygon is given in Figure 8(a) and onsists of a square with not hes in the orners. The length of the sides of the square is 2. Ea h of the two robots has to move a distan e of 2 before it gets to a orner of the polygon and furthermore at most two omplete orners an be seen by the robots. Hen e pla ing a not h in one of the orners that is not yet ompletely seen requires one of the robots to move 6 units whereas the optimal motion an be done with only 4 units, thus proving the result. The same proof a tually goes through for three robots on e you realize that independently of how the three robots start their exploration we an for e one of them to move 6 units pla ing at most three not hes as shown in Figure 8(b). 2 We also show a strategy for two robots that has ompetitive ratio 2. We all the strategy TGO (two-robot GO) sin e it is based on the GO-strategy of Deng et al ; [4℄. The two robots ea h run IGO(s; north ; orient ), one with orient = w and the other with orient = aw until the two robots have seen the omplete polygon after whi h they both move ba k to the starting point. Theorem 4
The strategy TGO is 2- ompetitive for exploration of a re tilinear polygon with two
robots.
13
not h
s
not h
s
not h not h (a)
(b)
Figure 8: Illustrating the proof of Theorem 3. Proof: Let r be the interse tion point of SWRs and the verti al axis issuing from s upwards. The interse tion point r lies at distan e D 0 from s; see Figure 9(a). We an view the strategy TGO as rst moving the two robots from s to r and then separating, one moving lo kwise and the other
ounter lo kwise, following SWR r in two dire tions until the robots have seen all of the polygon and move ba k to s. Consider now the robot that moves the farthest. Let t be the point of interse tion between the robot's walk and the last extension that it visits before it realizes that the whole polygon is explored. The robot then moves the distan e L = D + length (SWR r (r; t; dir )) + length (SP (t; s)); where dir is the dire tion that the robot moves. Suppose that we follow the tour SWRs from s in the dire tion that visits the point r before t. We
an assume that t is a point on SWRs sin e it is an interse tion point with an extension. Assume without loss of generality that this dire tion is lo kwise. We thus have that
L
=
D + length (SWRr (r; t; dir )) + length (SP (t; s))
length (SWR (s; r; w )) + length (SWR s
=
length (SWR ) 2 2length (OPT )
r; t; w )) + length (SWRs (t; s; w ))
s(
s
s
whi h proves the result. That the analysis is tight follows from the example in Figure 9(b).
2
7 Con lusions We have presented onstant ompetitive strategies and lower bounds to explore a re tilinear simple polygon in the L1 metri with one or more robots. Unfortunately none of our results are tight so obvious open problems are to redu e the gaps between the lower bounds and the upper bounds. Espe ially exploration using k robots, for an arbitrary number k , needs to be investigated. 14
r
SWR
s
s
SP (t; s) t
(a)
(b)
Figure 9: Illustrating the proof of Theorem 4.
Referen es [1℄ Margrit Betke, Ronald L. Rivest, Mona Singh. Pie emeal Learning of an Unknown Environment. Ma hine Learning, 18(23):231254, 1995. [2℄ K-F. Chan, T. W. Lam. An on-line algorithm for navigating in an unknown environment. International Journal of Computational Geometry & Appli ations, 3:227244, 1993. [3℄ W. Chin, S. Ntafos. Optimum Wat hman Routes. 1988.
Information Pro essing Letters, 28:3944,
[4℄ X. Deng, T. Kameda, C.H. Papadimitriou. How to Learn an Unknown Environment I: The Re tilinear Case. Journal of the ACM, 45(2):215245, 1998. [5℄ M. Hammar, B.J. Nilsson, S. S huierer. Improved Exploration of Re tilinear Polygons. Nordi Journal of Computing, 9(1):3253, 2002. [6℄ F. Hoffmann, C. I king, R. Klein, K. Kriegel. The Polygon Exploration Problem. Journal on Computing, 31(2):577600, 2001. [7℄ J. M. Kleinberg. On-line sear h in a simple polygon. In Dis rete Algorithms, pages 815, 1994.
SIAM
Pro . of 5th ACM-SIAM Symp. on
[8℄ Aohan Mei, Yoshihide Igarashi. An E ient Strategy for Robot Navigation in Unknown Environment. Inform. Pro ess. Lett., 52:5156, 1994. [9℄ C. H. Papadimitriou, M. Yannakakis. Shortest Paths Without a Map. 84(1):127150, 1991.
15
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