Online Motion Planning MA-INF 1314
Summersemester 17 Escape Paths
Elmar Langetepe
University of Bonn
Juli 11th, 2017
Escape Path situation
Try to escapefrom an partially unknown environment The adversary manipulates the environment
Leave the area as soon as possible Lost in a forestBellman 1956 Escape pathsfor known region R Single deterministic path
Leave area from any starting point Adversary translates and rotates R Minimize the length of successful path Geometric argumentations
Only known for few shapes
Simple examples
Obviously: The diameter of any regionR is always an escape path!
Theorem:The shortest escape path for a circle of radiusr is a line segment of length 2r.
Proof: Assume there is a better escape path! Contradiction!
C T
S
r
Also for semicircles
Theorem:The shortest escape path for a semicircle of radiusr is a line segment of length 2r.
Proof: Assume there is a better escape path! Contradiction!
S T
O I
T0 S0
X
r
More generally for a rhombus with angle 60
◦Theorem:The shortest escape path for a rhombus of diameterL with angleα= 60◦ is a line segment of lengthL.
Proof: Assume there is a better escape path! Contradiction!
A
B E
D C
X Y
S T
p1
p2
α= 60◦ L
Fatness definition!
Definition: Fatness w.r.t. diameter! Rhombus-Fat!
Corollary: The shortest escape path for rhombus-fat convex set of diameterL is a line segment of lengthL.
Proof: Assume there is a better escape path! Contradiction!
A
B E
D α= 60◦
L
Convex = diameter?
Equilateral triangle: Besicovitch
Zig-Zag escape path with length≈0.9812
More generally from Coulton and Movshovich (2006) Isosceles triangle forα and bα
bα is diameter!
O= (0,0) α
bα
L1:Y = tanα X
L2:Y = tanα(bα−X)
(b,0) =V (x, y) P = (x,3y) 1
Q= (x,0)
L:Y = 3 tanα(bα−X)
lα
rα
i)
Convex = diameter?
Construct symmetric Zig-Zag path of small length Asssume length 1.
O= (0,0) α
bα
L1:Y = tanα X
L2:Y = tanα(bα−X)
(b,0) =V (x, y) P = (x,3y) 1
Q= (x,0)
L:Y = 3 tanα(bα−X)
lα
rα
i)
Convex = diameter?
Extract triangle
1
x = b1α x = b1
α
β
β
x bα−x
d= 3y f 1
ii) O
P
Q V
Convex = diameter?
Finally we determinebα: y= tanα
bα−b1α
and x= b1
α andx2+ (3y)2= 1 which gives bα=
r
1 + 1
9 tan2α.
O= (0,0) α
bα
L1:Y = tanα X
L2:Y = tanα(bα−X)
(b,0) =V (x, y) P = (x,3y) 1
Q= (x,0)
L:Y = 3 tanα(bα−X)
lα
rα
i)
Further constraint for α
There should be no better Zig-Zag path forTα!
LineL3 :Y = tan(2α) runs in parallel withL2. This means
−3 tanα= tan 2α or tanα= q5
3.
O= (0,0)
bα
L3:Y = tan 2α X
(b,0) =V (x, y) P = (x,3y) 1
Q= (x,0)
L:Y = 3 tanα(bα−X)
lα
rα
α α L03
Besicovitch triangles
Theorem:For anyα∈[arctan(
q5
3),60◦] there is a symmetric Zig-Zag path of lenght 1 that is an escape path ofTα smaller than the diameterbα.
bα= q
1 +9 tan12α
α= 60◦:bα= q28
27
bα:= 1 =⇒q
27
28 <1 is Zig-Zag path length Optimality? Yes!
Different performance measures
Set Lm of m line segmentssi of unknown length |si| Dark corridors, escape, digging for oil
Test corridors successively
sj1 up to a certain distancex1, thensj2 for another distance x2 and so on
s1 s2 s3 s4 s5 s6 s7 x1 x2
x3 x4
x5 (i)
More information
Assume distribution is known!
f1 ≥f2 ≥ · · · ≥fm order of the length given Extreme cases! Good strategies!
s1s2s3s4s5 s6 s7
(ii) (iii)s1 s2s3s4 s5 s6 s7 x x x x x x x
More information
f1 ≥f2 ≥ · · · ≥fm order of the length given Check i arbitrary segments with length fi: minii·fi is the best strategy
f1f2f3 f4f5f6 f7 fi
(iv) i·fi
Known length in general
f1 ≥f2 ≥ · · · ≥fm order of the length given Check i arbitrary segments with length fi: minii·fi is a reasonable strategy
C(Fm,A) travel cost for algorithmA maxTrav(Fm) := minAC(Fm,A)
f1f2f3 f4f5f6 f7 fi
(iv) i·fi
Optimal strategy for this case
Theorem:For a set of sorted distancesFm (i.e.
f1≥f2 ≥ · · · ≥fm) we have
maxTrav(Fm) := min
i i·fi. Proof:
f1f2f3 f4f5f6 f7 fi
(iv) i·fi
Optimal strategy for this case
Theorem:For a set of sorted distancesFm (i.e.
f1≥f2 ≥ · · · ≥fm) we have
maxTrav(Fm) := min
i i·fi.
Proof:
Arbitrary strategy A
Less than minii·fi means less than j·fj for anyj Visiting depth d1≥d2 ≥ · · · ≥dm
Not reached f1 byd1,
not reached f2, since d1+d2<2f2 and d2 ≤d1 and so on Not successful!
minii·fi always sufficient!
Online Strategy
Fm with f1 ≥f2 ≥ · · · ≥fm not known Compete against maxTrav(Fm) := minii·fi
Dovetailing strategy: Rounds c = 1,2,3,4, . . . For any round c from left to right:
Path length of segment i is extendedup to distancec
i
Online Strategy
Dovetailing strategy: Rounds c = 1,2,3,4, . . . For any round c from left to right:
Path length of segment i is extendedup to distancec
i
1 1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10
10 11
2 3 4
4
5 6 6 6
7 8
8
8
9 f3= 3 9
Online Strategy!
Theorem:Hyperbolic traversal algorithm solves the multi-segment escape problem for any listFm with maximum traversal cost bounded by
D·(maxTrav(Fm) ln(min(m,maxTrav(Fm))) for some constantD.
Proof:(W.l.o.g.Fm integers)
Let minii·fi =j·fj for somej c with c =j·fj exists (Round c) Overoll cost:
m
X
t=1
jc t
k≤
min(m,c)
X
t=1
c t ≤c+
Z min(m,c)
1
c
t dt =c(1+ln min(m,c)).
Matches Lower bound!
Theorem:For any deterministic online strategyAthat solves the multi-segment escape problem we can construct input sequences Fm(A,C) so thatA has cost at leastd·C ln min(C,m) and maxTrav(Fm(C,A))≤C holds for some constantd and arbitrarily large valuesC.
d1+/m=:f1(A, C)
d4< f40(C) =C4 f10(C) =C1
f2(A, C) :=d2+/m
f7(A, C) :=f70(C) f8(A, C) :=f80(C) f3(A, C) :=d3+/m
Matches Lower bound! Proof!
C is given! fi0(C) = Ci (not yet fixed) Wait until cost Pm
i=1di ≥d ·Cln min(C,m) for some d Fix the scenario as shown below!
1 2 3 4 5 6 7 8
d1+/m=:f1(A, C)
d4< f40(C) =C4 f10(C) =C1
f2(A, C) :=d2+/m
f7(A, C) :=f70(C) f8(A, C) :=f80(C) f3(A, C) :=d3+/m