Department of Informatis
Stati Analysis of
SPDIs for State-Spae
Redution
Researh Report No.
336
Gordon Pae
Gerardo Shneider
Isbn 82-7368-291-9
Issn 0806-3036
April 2006
Redution
Gordon Pae
∗
Gerardo Shneider
†
April 2006
Abstrat
Polygonal hybrid systems (SPDI) are a sublass of planar hybrid
automata whih an be represented by pieewiseonstant dierential
inlusions. Thereahabilityproblemaswellastheomputationofer-
tainobjets ofthephaseportrait, namelytheviability,ontrollability
and invariane kernels, for suh systems is deidable. In this paper
we show how to ompute another objet of an SPDI phase portrait,
namely semi-separatrix urves and show how the phase portrait an
be used for reduing the state-spae for optimizing the reahability
analysis.
1 Introdution
Hybrid systems ombining disreteand ontinuous dynamisarise asmath-
ematialmodels ofvariousartiialand naturalsystems,and asapproxima-
tions to omplex ontinuous systems. They have been used in various do-
mains, inludingavionis, robotis andbioinformatis. Reahabilityanalysis
has beentheprinipalresearhquestionintheveriationofhybridsystems,
even ifitisawell-known resultthatformost non-trivialsublassesofhybrid
systems reahability and most veriation questions are undeidable. Vari-
ous deidable sublasses have, subsequently, been identied, inludingtimed
[AD94 ℄ and retangular automata [HKPV95℄, hybrid automata with linear
∗
Dept. of Computer Siene and AI, University of Malta, Msida, Malta. E-mail:
gordon.paeum.edu.mt
†
Dept. ofInformatisUniv. ofOslo, P.O. Box1080Blindern, N-0316Oslo,Norway.
E-mail: gerardoi.uio.no
and polygonaldierentialinlusionsystems (SPDIs)[ASY01℄.
Compared to reahability veriation, qualitative analysis of hybrid sys-
tems is a relatively negleted area [ALQ
+
01b, DV95, KdB01, MS00, SP02,
SJSL00℄. Typial qualitative questions inlude: Are there `sink' regions
where a trajetory an never leave one it enters the region?; Whih are
the basins of attration of suh regions?; Are there regions in whih every
pointinthe regionis reahable fromevery other pointinthe region without
leaving it?. To answer suh questions one usually gives a olletion of ob-
jets haraterizingthese sets, hene providinguseful informationabout the
qualitative behavior of the hybrid system. The set of all suh objets for a
given system is alled the phase portrait of the system.
Deningandonstrutingphaseportraitsofhybridsystemshasbeendiretly
addressed for PCDs in [MS00℄, and for SPDIs in[ASY02℄. In this paper we
present a a new element of the phase portrait for SPDIs, and disuss how
the phase portrait an be used to redue the size of an SPDI, as an aid to
veriation.
Roughly speaking, anSPDI (Fig. 1)isa nitepartition
P
of the plane (intoonvexpolygonalareas),and,foreah
P ∈ P
anassoiatedpair ofvetorsa P
and
b P. TheSPDI behaviourisdened by thedierentialinlusionx ˙ ∈ ∠ b a P P
for
x ∈ P
, where∠ b a denotes the angle on the plane between the vetors a
and
b
.In [ASY01℄ it has been proved that edge-to-edge and polygon-to-polygon
reahabilityinSPDIs isdeidable by exploitingthe topologialproperties of
the plane. The proedure is not based on the omputation of the reah-set
but rather on the exploration of a nite number of types of qualitative be-
haviors obtained from the edge-signatures of trajetories (the sequenes of
their intersetions with the edges of the polygons). Suh types of signatures
may ontain loops whih an be very expensive (or impossible) to explore
naively. However, it has been shown that loops have strutural properties
that are exploited by the algorithmto eiently ompute the eet of suh
loops. In summary,the novelty oftheapproahisthe ombinationofseveral
tehniques, namely,(i)therepresentation ofthe two-dimensionalontinuous
dynamisasaone-dimensionaldisretedynamialsystem,(ii)the harater-
ization ofthe set of qualitativebehaviorsofthe latter asanite set of types
ofsignatures, and(iii)theaeleration oftheiterationsinthease ofyli
signatures.
Givenayle onaSPDI,wean speakaboutanumberof kernels pertaining
tothatyle. Theviability kernelisthelargestsetofpointsintheylewhih
mayloopforeverwithintheyle. Theontrollability kernelisthelargestset
R1 R2 R3
R6 R5 R4
e1 e4
e0
I
e2 e6
I’
e5 e3
Figure 1: An SPDI and itstrajetory segment.
bereahedfromanyother). Aninvariantset isaset ofpointssuhthateah
pointmustkeep rotatingwithintheset forever. Theinvariane kernel isthe
largest of suh sets. The information gathered for omputing reahability
turns out to be useful for omputing viability,ontrollability and invariane
kernels of suh systems. Algorithms for omputing these kernels have been
presented in[ASY02,Sh04 ℄andimplementedinthetoolsetSPeeDI
+
[PS06℄.
The ontributionof this paperisthreefold. We start by givinganalgorithm
to ompute semi-separatrix urves (or simply, semi-separatries) of SPDIs.
Separatriesareonvexpolygonsdissetingtheplaneintotwomutuallynon-
reahable subsets. The notion of separatrix an be relaxed, obtainingsemi-
separatrixurves,suhthatsomepointsinonesetmaybereahablefromthe
other set, but not vie-versa. We then show how the kernels an be used to
answer reahability questions diretly. We also show how semi-separatries
anbeused tooptimizethereahabilityalgorithmforSPDIsby reduingthe
numberofstatesoftheSPDIgraph. Theoptimizationisbasedontopologial
properties of the plane (and inpartiular, that of SPDIs).
The paper is strutured as follows. In the next setion we introdue the
neessary theoretial bakground, inluding the denition of SPDI, kernels
and semi-separatriesaswellashowtoomputesuhphaseportraitobjets.
In Setion 3 we show how the semi-separatries an be used for reduing
the state-spaeof the reahabilitygraphwhereas inSetion4wepresent the
optimization done by using the kernels.
A(positive)ane funtion
f : R → R
issuhthatf(x) = ax + b
witha > 0
.An ane multivalued funtion
F : R → 2 R, denoted F = hf l , f u i
, is dened
by
F (x) = hf l (x), f u (x)i
wheref l andf u areaneandh·, ·i
denotesaninter-
h·, ·i
denotesaninter-val. For notational onveniene, we do not make expliit whether intervals
are open, losed,left-open orright-open, unless requiredfor omprehension.
Foran interval
I = hl, ui
we have thatF (hl, ui) = hf l (l), f u (u)i
. The inverseof
F
is dened byF −1 (x) = {y | x ∈ F (y)}
. The universal inverse ofF
isdened by
F ˜ −1 (I) = I ′ if and only if I ′ is the greatest non-empty interval
suh that for all
x ∈ I ′, F (x) ⊆ I
.
It is not diult to show that
F −1 = hf u −1 , f l −1 i
and similarly thatF ˜ −1 = hf l −1 , f u −1 i
, provided thathf l −1 , f u −1 i 6= ∅
. Notie thatifI
is asingletonthenF ˜ −1 is dened only if f l = f u. These lasses of funtions are losed under
omposition.
A trunated ane multivalued funtion (TAMF)
F : R → 2 R is dened
by an ane multivalued funtion
F
and intervalsS ⊆ R + and J ⊆ R + as
follows:
F (x) = F (x) ∩ J
ifx ∈ S
, otherwiseF (x) = ∅
. For onvenienewe write
F (x) = F ({x} ∩ S) ∩ J
. For an intervalI
,F (I) = F (I ∩ S) ∩ J
and
F −1 (I) = F −1 (I ∩ J ) ∩ S
. The universal inverse ofF
is dened byF ˜ −1 (I) = I ′ if and onlyif I ′ is the greatest non-empty intervalsuhthat for
all
x ∈ I ′, F (x) ⊆ I
and F (x) = F (x)
.
We say that
F
is normalized ifS = DomF = {x | F (x) ∩ J 6= ∅}
(thus,S ⊆ F −1 (J)
)andJ = ImF = F(S)
.The following theorem states that TAMFs are losed under omposition
[ASY01℄.
Theorem 1. The omposition of two TAMFs
F 1 (I) = F 1 (I ∩ S 1 ) ∩ J 1 and
F 2 (I) = F 2 (I ∩ S 2 ) ∩ J 2, is the TAMF (F 2 ◦ F 1 )(I) = F (I) = F (I ∩ S) ∩ J
,
where
F = F 2 ◦ F 1, S = S 1 ∩ F 1 −1 (J 1 ∩ S 2 )
and J = J 2 ∩ F 2 (J 1 ∩ S 2 )
.
2.1 SPDI
An angle
∠ b a onthe plane, dened by two non-zero vetors a, b
isthe set of
allpositivelinearombinations
x = α a + β b
,withα, β ≥ 0
,andα + β > 0
.We an always assume that
b
is situated in the ounter-lokwise diretion froma
.A polygonal hybrid system 1
(SPDI) is dened by giving a nite partition
P
of the plane into onvex polygonal sets, and assoiating with eah
P ∈ P
a1
In theliterature thenamespolygonal dierential inlusion andsimple planar dier-
ouple of vetors
a P and b P. Let φ(P ) = ∠ b a P P. The SPDI is determined by
φ(P ) = ∠ b a P P. The SPDI is determined by
˙
x ∈ φ(P )
forx ∈ P
.Let
E(P )
be the set of edges ofP
. We say thate
is an entry ofP
if forall
x ∈ e
and for allc ∈ φ(P )
,x + cǫ ∈ P
for someǫ > 0
. We say thate
is an exit of
P
if the same ondition holds for someǫ < 0
. We denote byin
(P ) ⊆ E(P )
the set ofallentries ofP
andby out(P ) ⊆ E(P )
the set of allexits of
P
.Assumption 1. All the edges in
E(P )
are either entries or exits, that is,E(P ) =
in(P ) ∪
out(P )
.A trajetory segment of an SPDI is a ontinuous funtion
ξ : [0, T ] → R 2
whihis smootheverywhere exept inadisrete set ofpoints,and suh that
for all
t ∈ [0, T ]
, ifξ(t) ∈ P
andξ(t) ˙
is dened thenξ(t) ˙ ∈ φ(P )
. Thesignature, denoted
Sig(ξ)
, is the ordered sequene of edges traversed by thetrajetory segment, that is,
e 1 , e 2 , . . .
, whereξ(t i ) ∈ e i and t i < t i +1. If
T = ∞
,a trajetory segment isalled a trajetory.Example 1. Consider the SPDI illustrated inFig. 1. For sake of simpliity
wewillonlyshowthe dynamisassoiatedtoregions
R 1 toR 6 inthe piture.
For eah region
R i, 1 ≤ i ≤ 6
, there is a pair of vetors (a i , b i )
, where:
a 1 = (45, 100), b 1 = (1, 4)
,a 2 = b 2 = (1, 10)
,a 3 = b 3 = (−2, 3)
,a 4 = b 4 = (−2, −3)
,a 5 = b 5 = (1, −15)
,a 6 = (1, −2), b 6 = (1, −1)
.A trajetory segment starting on interval
I ⊂ e 0 and nishing in interval
I ′ ⊆ e 4 isdepited.
Denition1. We saythatasignature
σ
isfeasible ifandonlyifthere existsa trajetory segment
ξ
with signatureσ
, i.e.,Sig(ξ) = σ
.From this denition, it immediately follows that extending an unfeasible
signature, an nevermake it feasible:
Proposition1. Ifa signature
σ
isnotfeasible,thenneitherisany extensionof the signature for any signatures
σ ′ and σ ′′, the signature σ ′ σσ ′′ is not
σ ′ σσ ′′ is not
feasible.
Given anSPDI
S
, letE
bethe set of edgesofS
, then wean dene agraphG S where nodes orrespond to edges of S
and suh that there exists an ar
from one node toanother if there exists a trajetory segment from the rst
edge to the seond one withouttraversing any other edge. More formally:
Denition 2. Given an SPDI
S
, the underlying graphofS
(or simply thegraph of
S
), is a graphG S = (N G , A G )
, withN G = E
andA G = {(e, e ′ ) |
∃ξ, t . ξ(0) ∈ e ∧ ξ(t) ∈ e ′ ∧ Sig (ξ) = ee ′ }
. We say that a sequenee 0 e 1 . . . e k
of nodes in
G S is a path whenever (e i , e i +1 ) ∈ A G for 0 ≤ i ≤ k − 1
.
0 ≤ i ≤ k − 1
.and paths in itsorresponding graph.
Lemma 2. If
ξ
is a trajetory segment ofS
with edge signatureSig (ξ) = σ = e 0 . . . e p, it follows that σ
is a path in G S.
Remark. Notiethattheonverseoftheabovelemmaisnottrueingeneral.
It is possible tond a ounter-example wherethere exists apath from node
e
toe ′, but it does not exist a trajetory segment form edge e
to edge e ′ on
the SPDI.
Lemma 3. If
σ = e 0 . . . e p isafeasible signature,thenσ
isa pathin G S.
2.2 Suessors and predeessors
Given an SPDI, we x a one-dimensional oordinate system on eah edge
torepresent pointslayingonedges [ASY01℄. Fornotationalonveniene, we
indistintlyuseletter
e
todenotethe edgeoritsone-dimensionalrepresenta- tion. Aordingly, we writex ∈ e
orx ∈ e
, tomean pointx
in edgee
withoordinate
x
inthe one-dimensional oordinatesystem ofe
. The same on-ventionisappliedtosetsof pointsof
e
represented asintervals(e.g.,x ∈ I
orx ∈ I
,whereI ⊆ e
) and to trajetories (e.g.,ξ
starting inx
orξ
startingin
x
).Now, let
P ∈ P
,e ∈
in(P )
ande ′ ∈
out(P )
. ForI ⊆ e
,Succ e,e ′ (I)
is theset of allpointsin
e ′ reahablefromsome pointinI
by atrajetorysegment
ξ : [0, t] → R 2 in P
(i.e., ξ(0) ∈ I ∧ ξ(t) ∈ e ′ ∧ Sig (ξ) = ee ′). It has been
P
(i.e.,ξ(0) ∈ I ∧ ξ(t) ∈ e ′ ∧ Sig (ξ) = ee ′). It has been
shown [ASY01℄ that
Succ e,e ′ is a TAMF.
Example 2. Let
e 1 , . . . , e 6 be as in Fig. 1 and I = [l, u]
. We assume a
one-dimensional oordinate system. We have:
F e 1 e 2 (I) = l
4 , 9 20 u
, S = [0, 10] , J =
0, 9 2
F e 2 e 3 (I) = [l + 1, u + 1] , S = [0, 9] , J = [1, 10]
F e 3 e 4 (I) = 3
2 l, 3 2 u
, S =
0, 20 3
, J = [0, 10]
F e 4 e 5 (I) = 2
3 l, 2 3 u
, S = [0, 10] , J =
0, 20 3
F e 5 e 6 (I) =
l − 2
3 , u − 2 3
, S = 2
3 , 10
, J =
0, 28 3
F e 6 e 1 (I) = [l, 2u] , S = [0, 10] , J = [0, 10]
with
Succ e i e i +1 (I) = F e i e i +1 (I ∩ S i ) ∩ J i+1, for 1 ≤ i ≤ 5
, and Succ e 6 e 1 (I) = F e 6 e 1 (I ∩ S 6 ) ∩ J 1.
Given asequene
w = e 1 , e 2 , . . . , e n, Theorem 1impliesthat the suessor of
I
alongw
dened asSucc w (I ) = Succ e n − 1 ,e n ◦ . . . ◦ Succ e 1 ,e 2 (I)
is aTAMF.Example 3. Let
σ = e 1 · · · e 6 e 1. It results that Succ σ (I) = F (I ∩ S) ∩ J
,
where:
F (I) = l
4 + 1 3 , 9
10 u + 2 3
(1)
S = [ 37 25 e −16 , 10]
andJ = [ 1 3 , 29 3 ]
are omputed usingTheorem 1.For
I ⊆ e ′, Pre e,e ′ (I)
is the set of points in e
that an reah a point in
I
by a trajetory segment in P
. The ∀
-predeessor Pre f (I )
is dened in a
similar way to
Pre(I)
using the universal inverse instead of justthe inverse:For
I ⊆ e ′, Pre f ee ′ (I)
is the set of points in e
suh that any suessor of
suh points are in
I
by a trajetory segment inP
. Both denitions an beextended straightforwardly to signatures
σ = e 1 · · · e n: Pre σ (I)
and Pre f σ (I)
.
Therefore, the suessor operator has two inverse operators.
Example 4. Let
σ = e 1 . . . e 6 e 1 be as in Fig. 1 and I = [l, u]
. Now,
Pre e i e i+1 (I) = F e −1 i e i+1 (I ∩ J i +1 ) ∩ S i,for 1 ≤ i ≤ 5
,and Pre e 6 e 1 (I) = F e −1 6 e 1 (I ∩ J 1 ) ∩ S 6,where:
1 ≤ i ≤ 5
,andPre e 6 e 1 (I) = F e −1 6 e 1 (I ∩ J 1 ) ∩ S 6,where:
F e −1 1 e 2 (I) = 20
9 l, 4u
F e −1 2 e 3 (I) = [l − 1, u − 1]
F e −1 3 e 4 (I) = 2
3 l, 2 3 u
F e −1 4 e 5 (I) = 3
2 l, 3 2 u
F e −1 5 e 6 (I) =
l + 2
3 , u + 2 3
F e −1 6 e 1 (I) = l
2 , u
Besides,
Pre σ (I) = F −1 (I ∩ J) ∩ S
, whereF −1 (I) = [ 10 9 l − 20 27 , 4u − 4 3 ]
.Similarly,weompute
Pre f σ (I) = ˜ F −1 (I ∩J )∩S
,whereF ˜ −1 (I) =
4l − 4 3 , 10 9 u − 20 27
.
2.3 Qualitative analysis of simple edge-yles
Let
σ = e 1 · · · e k e 1 be a simple edge-yle, i.e., e i 6= e j for all 1 ≤ i 6=
1 ≤ i 6=
j ≤ k
. LetSucc σ (I) = F (I ∩ S) ∩ J
withF = hf l , f u i
(we suppose thatthis representation is normalized). We denote by
D σ the one-dimensional
disrete-time dynamialsystem dened by Succ σ, that isx n +1 ∈ Succ σ (x n )
.
x n +1 ∈ Succ σ (x n )
.Assumption 2. None of the two funtions
f l , f u is the identity.
Let
l ∗ and u ∗ bethe xpoints2 off l and f u, respetively,andS ∩ J = hL, Ui
.
f l and f u, respetively,andS ∩ J = hL, Ui
.
S ∩ J = hL, Ui
.A simple yle is of one of the following types [ASY01℄:
STAY. Theyleisnotabandonedneitherbytheleftmostnortherightmost
trajetory,that is,
L ≤ l ∗ ≤ u ∗ ≤ U
.DIE. Therightmosttrajetoryexitstheylethroughtheleft(onsequently
the leftmost one also exits) or the leftmost trajetory exits the yle
through the right (onsequently the rightmost one alsoexits), that is,
u ∗ < L ∨ l ∗ > U
.EXIT-BOTH. Theleftmost trajetoryexitsthe yle through the leftand
the rightmost one through the right,that is,
l ∗ < L ∧ u ∗ > U
.EXIT-LEFT. Theleftmosttrajetoryexitstheyle (throughtheleft)but
the rightmost one stays inside, that is,
l ∗ < L ≤ u ∗ ≤ U
.EXIT-RIGHT. Therightmosttrajetoryexitstheyle(throughtheright)
but the leftmostone stays inside, that is,
L ≤ l ∗ ≤ U < u ∗.
Example 5. Let
σ = e 1 · · · e 6 e 1. We have S ∩ J = hL, Ui = [ 1 3 , 29 3 ]
. The
xpoints of Eq. (1) are suh that
1
3 < l ∗ = 11 25 < u ∗ = 20 3 < 29 3
. Thus,σ
is aSTAY.
The lassiation above gives us some useful information about the quali-
tative behavior of trajetories. Any trajetory that enters a yle of type
DIE will eventually quit it after a nite number of turns. If the yle is of
type STAY, alltrajetories that happen to enter it willkeep turning inside
it forever. In allotherases, some trajetories willturn fora whileand then
exit, andothers willontinue turningforever. Thisinformationisruialfor
proving deidabilityof the reahability problem.
Example 6. Considerthe SPDIof Fig.1. Fig. 2showspart ofthe reahset
of the interval
[8, 10] ⊂ e 0,answering positively tothe reahabilityquestion:
Is
[1, 2] ⊂ e 4 reahable from [8, 10] ⊂ e 0? Fig. 2 has been automatially
generated by theSPeeDitoolboxwe havedeveloped forreahabilityanalysis
of SPDIs basedon the results of [ASY01 ℄.
2
Thexpoint
x ∗
isomputedbysolvingtheequationf (x ∗ ) = x ∗
,wheref (·)
ispositiveane.
I’
I
Figure 2: Reahability analysis.
The above result doesnot allowusto diretlyanswerother questions about
thebehavioroftheSPDIsuhasdetermineforagivenpoint(orsetofpoints)
whether: (a) there exists (at least) one trajetory that remainsinthe yle,
and (b)itispossibletoontrolthe systemtoreahany otherpoint. Inorder
to do this, we need to further study the properties of the system around
simple edge-yles.
2.4 Kernels
We an nowpresent how toompute the invariane, ontrollabilityand via-
bility kernelsof anSPDI. Proofsare omittedbut for further details,refer to
[ASY02℄ and [Sh04 ℄. In the following, for
K
a subset ofR 2 and σ
a yli
signature, we dene
K σ as follows:
K σ = [ k i=1
(
int(P i ) ∪ e i )
(2)where
P i is suh that e i−1 ∈
in(P i )
,e i ∈
out(P i )
and int(P i )
isP i'sinterior.
2.4.1 ViabilityKernel
Wenow reall the denition of viability kernel[Aub01℄.
Denition 3. A trajetory
ξ
is viable inK
ifξ(t) ∈ K
for allt ≥ 0
.K
is a viability domain if for every
x ∈ K
, there exists at least one trajetoryFigure3: (a) ViabilityKernels; (b) ControllabilityKernels
ξ
, withξ(0) = x
, whih is viable inK
. The viability kernel ofK
, denotedViab(K )
, is the largest viability domainontained inK
.For
I ⊆ e 1 wedenePre σ (I)
tobethe setof allx ∈ R 2 forwhihthereexists
a trajetorysegment
ξ
startinginx
,that reahes somepointinI
, suh thatSig(ξ)
is a sux ofe 2 . . . e k e 1. It is easy to see that Pre σ (I)
is a polygonal
subset of the plane whih an be alulated using the following proedure.
We start by dening:
Pre e (I) = {x | ∃ξ : [0, t] → R 2 , t > 0 . ξ(0) = x ∧ ξ(t) ∈ I ∧ Sig (ξ) = e}
and apply this operation
k
times:Pre σ (I) = S k
i=1 Pre e i (I i ) with I 1 = I
,
I k = Pre e k ,e 1 (I 1 )
and I i = Pre e i ,e i+1 (I i +1 )
, for 2 ≤ i ≤ k − 1
.
The followingresult providesa non-iterativealgorithmiproedurefor om-
puting the viability kernel of
K σ onan SPDI:
Theorem 4. If
σ
is not DIE,Viab(K σ ) = Pre σ (S)
, otherwiseViab(K σ ) =
∅
.Example 7. Fig. 3-(a) shows all the viability kernels of the SPDI given in
Example 1. There are 4 yles with viability kernels in the piture two of
the kernels are overlapping.
We say
K
is ontrollable if for any two pointsx
andy
inK
there exists atrajetorysegment
ξ
startinginx
thatreahes anarbitrarilysmallneighbor- hoodofy
withoutleavingK
. More formally:Denition 4. A set
K
is ontrollable if∀x, y ∈ K, ∀δ > 0, ∃ξ : [0, t] → R 2 , t > 0 . (ξ(0) = x ∧ |ξ(t) − y| < δ ∧ ∀t ′ ∈ [0, t] . ξ(t ′ ) ∈ K )
. Theontrollabilitykernel of
K
, denotedCntr(K)
, isthelargest ontrollablesubset ofK
.For agiven yli signature
σ
, we deneC D (σ)
as follows:C D (σ) =
hL, U i
ifσ
is EXIT-BOTHhL, u ∗ i
ifσ
is EXIT-LEFThl ∗ , Ui
ifσ
is EXIT-RIGHThl ∗ , u ∗ i
ifσ
is STAY∅
ifσ
is DIE(3)
For
I ⊆ e 1 letusdeneSucc σ (I)
asthesetofallpointsy ∈ R 2forwhihthere
exists a trajetory segment
ξ
starting in some pointx ∈ I
, that reahesy
,suhthat
Sig (ξ)
isaprex ofe 1 . . . e k. Thesuessor Succ σ (I)
isapolygonal
subset of the plane whihan beomputed similarlyto
Pre σ (I)
. DeneC (σ) = (Succ σ ∩ Pre σ )(C D (σ))
Weompute the ontrollability kernel of
K σ asfollows:
Theorem 5.
Cntr (K σ ) = C (σ)
.Example 8. Fig. 3-(b) shows all the ontrollability kernels of the SPDI
given in Example 1. There are 4 yles with ontrollability kernels in the
piture two of the kernels are overlapping.
The followingresult whihrelatesontrollabilityand viabilitykernels,states
that theviabilitykernelofagivenyle istheloalbasinofattration ofthe
orresponding ontrollability kernel.
Proposition 2. Any viable trajetory in
K σ onverges to Cntr (K σ )
.
Let
Cntr l (K σ )
be the losed urve obtained by taking the leftmost traje-tory and
Cntr u (K σ )
be the losed urve obtained by taking the rightmosttrajetorywhihanremaininsidethe ontrollabilitykernel. Inotherwords,
Cntr l (K σ )
andCntr u (K σ )
are the two polygons dening the ontrollability kernel.A non-empty ontrollability kernel
Cntr (K σ )
of a given yli signatureσ
partitionsthe plane intothree disjointsubsets: (1)the ontrollabilitykernel
itself,(2)thesetofpointslimitedby
Cntr l (K σ )
(andnotinludingCntr l (K σ )
)and(3)thesetofpointslimitedby
Cntr u (K σ )
(andnotinludingCntr u (K σ )
).Denition 5. We dene the inner of
Cntr (K σ )
(denoted byCntr in (K σ )
) tobe the subset dened by (2) above if the yle is ounter-lokwise or to be
the subset dened by (3) if it is lokwise. The outer of
Cntr(K σ )
(denotedby
Cntr out (K σ )
) is dened to be the subset whih is not the inner nor theontrollability itself.
Remark: Notie that an edge in the SPDI may be split into parts by the
ontrollabilitykernel part inside, part onthe kernel and part outside. In
suh ases, we an generate a dierent SPDI, with the same dynamis but
with theedge splitintoparts, suh thateahpart isompletelyinside, onor
outsidethekernel. Althoughthesignatureswillobviouslyhange,itistrivial
toprovethatthe behaviourofthe SPDIremainsidentialtotheoriginal. To
simplify presentation, inthe rest of the paper, we willassume that alledges
are either ompletely inside, on or ompletelyoutside the kernels. We note
that in pratie splittingis not neessary sine we an just onsider parts of
edges.
Proposition 3. Given two edges
e
ande ′, one lying ompletely inside a
ontrollability kernel, and the other outside or on the same ontrollability
kernel, suhthat
ee ′ isfeasible,thenthere existsapoint ontheontrollability
kernel, whih is reahablefrom e
and from whih e ′ is reahable.
Proof. Let
e ⊆ Cntr in (K σ )
. Let us assume thate ′ ⊆ Cntr (K σ )
; sineee ′
is feasible, by the Jordan urve theorem [Hen79℄, the trajetory must ross
Cntr l (K σ )
orCntr u (K σ )
at least one. Assume the rst holds, then thereexists
x ∈ Cntr l (K σ )
suh thatexe ′ is feasible. If e ′ ⊆ Cntr out (K σ )
the proof
is onduted in a similar way as the previous ase by using the denition
of ontrollability kernel: every point inside the kernel is reahable from any
other point inthe kernel.
2.4.3 Invariane Kernel
In general,an invariant set is a set of points suh that for any point in the
set, every trajetorystartinginsuhpointremainsintheset foreverand the
invariane kernel is the largest of suh sets. In partiular, for SPDI, given
a yli signature, aninvariant set is a set of points whih keep rotating in
the yle forever and the invariane kernel is the largest of suh sets. More
Denition6. A set
K
issaidtobe invariantifforanyx ∈ K
thereexistsatleast one trajetory starting in it and every trajetory starting in
x
is viablein
K
. Given a setK
, its largest invariant subset is alled the invarianekernel of
K
and isdenoted byInv (K σ )
.We need some preliminary denitions before showing how to ompute the
kernel. The extended
∀
-predeessor of an output edgee
of a regionR
is theset of pointsin
R
suh that every trajetory segment startingin suh pointreahes
e
withouttraversinganyotheredge. More formally,letR
bearegionand
e
be anedge in out(R)
, then thee
-extended∀
-predeessor ofI
,Pre f e (I)
is dened as:
Pre f e (I) = {x | ∀ξ . (ξ(0) = x ⇒ ∃t ≥ 0 . (ξ(t) ∈ I ∧ Sig (ξ[0, t]) = e))}.
It is easy to see that
Pre f σ (I)
is a polygonal subset of the plane whih anbe alulated using the following proedure. First ompute
Pre f e i (I)
for all1 ≤ i ≤ k
and then apply this operationk
times:Pre f σ (I) = S k
i =1 Pre f e i (I i )
with
I 1 = I
,I k = Pre f e k e 1 (I 1 )
andI i = Pre f e i e i +1 (I i+1 ),
for2 ≤ i ≤ k − 1
. Weompute the invariane kernel of
K σ asfollows:
Theorem6.If
σ
isSTAYthenInv(K σ ) = Pre f σ ( Pre f σ (J))
,otherwiseInv(K σ ) =
∅
.Example 9. Fig. 4-(a) shows the unique invariane kernels of the SPDI
given in Example 1.
An interesting property of invariane kernels is that the limitsare inluded
in the invarianekernel, i.e.
[l ∗ , u ∗ ] ⊆ Inv(K σ )
. In other words:Proposition 4. The set delimited by the polygons dened by the interval
[l ∗ , u ∗ ]
is an invariane set of STAY yles.In [ASY02℄ it has been proved that for
σ
a STAY yle, then (1)C(σ)
isinvariant and (2) there exists a neighborhood
K
ofC(σ)
suh that any vi-able trajetory starting in
K
onverges toC (σ)
. From this, the denitionof invariane kernel and theorem 6 it follows the following result relating
ontrollabilityand invarianekernels.
Proposition 5. If
σ = e 1 . . . e n e 1 is STAY then Cntr(K σ ) ⊆ Inv(K σ )
.
Figure 4: (a) Invariane Kernel;(b) All the Kernels
Example 10. Fig. 4-(b) shows the viability, ontrollability and invariane
kernels of the SPDIgivenin Example1. Forany pointin theviability kernel
of a yle there exists a trajetory whih will onverge to its ontrollability
kernel (proposition2). Itispossibletoseein thepiture that
Cntr (·) ⊂ Inv (.)
(proposition 5). All the above pitures has been obtained with the toolbox
SPeeDI
+
[PS06℄.
Inasimilarwayasfortheontrollabilitykernel,wedene
Inv l (K σ )
,Inv u (K σ )
,the inner
Inv in (K σ )
and outerInv out (K σ )
of aninvarianekernel.2.5 Semi-Separatrix Curves
In this setionwedene the notionofseparatrix urves,whihare urves on
R 2 disseting the plane into two mutually non-reahable subsets. We relax the notion of separatrix obtaining semi-separatrix urves suh that some
pointsin one set may bereahable fromthe other set, but not vie-versa.
Wedene rst the abovenotions for the plane independently of SPDIs.
Denition 7. Let
K ⊆ R 2. A separatrix in K
is a losed urve γ
parti-
tioning
K
into three setsK A, K B and γ
itself, suh that K A ∩ K B ∩ γ = ∅
,
K = K A ∪ K B ∪ γ
and the following onditions hold:
γ
itself, suh thatK A ∩ K B ∩ γ = ∅
,K = K A ∪ K B ∪ γ
and the following onditions hold:1. For any point
x 0 ∈ K A and trajetory ξ
, with ξ(0) = x 0, there is no t
t
suh that
ξ(t) ∈ K B; and
2. For any point
x 0 ∈ K B and trajetory ξ
, with ξ(0) = x 0, there is no t
t
suh that
ξ(t) ∈ K A.
If only one of the above onditions holds then we say that the urve is a
semi-separatrix. If only ondition 1 holds, then we say that
K A is the inner
of
γ
(writtenγ in) and K B is the outer of γ
(written γ out). If only ondition
γ
(writtenγ out). If only ondition
2 holds,
K B is the inner and K B is the outer of γ
.
γ
.Remark: Notie that,asinthe ase of theontrollabilitykernel, anedge of
the SPDI maybesplit intotwoby asemi-separatrixpartinside,and part
outside. As before, we an split the edge into parts, suh that eah part is
ompletely inside,or ompletelyoutside the semi-separatrix.
The set of all the separatries of
R 2 is denoted by Sep( R 2 )
, or simply Sep
.
The above notions are extended toSPDIs straightforwardly.
Now,let
σ = e 1 . . . e n e 1 beasimpleyle,∠ b a i i (1 ≤ i ≤ n
)bethe dynamisof
1 ≤ i ≤ n
)bethe dynamisoftheregionsforwhih
e i isanentryedgeandI = [l, u]
andintervalonedgee 1.
Rememberthat
Succ e 1 e 2 (I) = F (I ∩ S)∩ J
,whereF = [a 1 l +b 1 , a 2 u +b 2 ]
. Letl
bethe vetor orrespondingtothe point one 1 with loaloordinates l
and
l ′ bethe vetor orrespondingtothe pointone 2 with loaloordinates F (l)
e 2 with loaloordinates F (l)
(similarly,wedene
u
andu ′ forF (u)
). WedenerstSucc b e 1 1 (I ) = {x | l ′ = αx + l, 0 < α < 1}
andSucc a e 1 1 (I) = {x | u ′ = αx + u, 0 < α < 1}
. Weextend
these denitions in a straight way to any (yli) signature
σ = e 1 . . . e n e 1,
denoting themby
Succ b σ (I)
andSucc a σ (I)
,respetively;weanomputethem similarlyasforPre
. Wheneverappliedtothex-pointI ∗ = [l ∗ , u ∗ ]
,wedenoteSucc b σ (I ∗ )
andSucc a σ (I ∗ )
byξ σ l andξ σ urespetively. Intuitively,ξ l σ(ξ σ u)denotes
ξ l σ(ξ σ u)denotes
thepiee-wiseanelosedurvedenedbytheleftmost(rightmost)x-point
l ∗ (u ∗).
Weshow now howto identify semi-separatries for simple yles.
Theorem7. GivenanSPDI, let
σ
beasimpleyle,thenthefollowinghold:1. If
σ
is EXIT-RIGHT thenξ σ l is a semi-separatrix urve (ltering tra- jetories from left to right);
2. If
σ
isEXIT-LEFTthenξ σ u is a semi-separatrixurve (ltering traje-
tories from right to left);
3. If
σ
is STAY, then the two polygons dening the invariane kernel(
Inv l (K σ )
andInv u (K σ )
),are semi-separatries.leftby
ξ σ l,whihisapiee-wiseanelosedurve,partitioningR 2 into
three disjoint sets:
K B, the right part of ξ σ l; K A, the left part of
K A, the left part of
ξ σ l; and ξ σ l itself. By Jordan's theorem, any trajetory may pass from
K B to K A if and only if it ross ξ l σ. However, by denition of EXIT-
K B to K A if and only if it ross ξ l σ. However, by denition of EXIT-
ξ l σ. However, by denition of EXIT-
RIGHT,thisisonlypossiblefrom
K A toK B butnot vie-versa. Hene
ξ σ l is a semi-separatrixurve.
2. Symmetri tothe previous ase.
3. Follows diretly from the denition of invarianekernel, sine any tra-
jetorywithinitialpointin
Inv(K σ ) ∪ Inv in (K σ )
annotleaveInv(K σ )
.If the trajetory yles lokwise it annot traverse
Inv l (K σ )
and if ityles ounter-lokwise it annot traverse
Inv u (K σ )
. In both ases nopoint on
Inv out (K σ )
an be reahed. Symmetrially,trajetories start- ing inInv(K σ ) ∪ Inv out (K σ )
annot reah any point onInv in (K σ )
.Remark: In the ase of STAY yles,
ξ σ l and ξ σ u are also semi-separatries.
Notie that in the above result, omputing a semi-separatrix depends only
onone simpleyle,and the orrespondingalgorithmisthenredued tond
simple yles in the SPDI and heking whether it is STAY, EXIT-RIGHT
or EXIT-LEFT.
Example 11. Fig. 5 shows all the semi-separatries of the SPDI given in
Example 1. The small arrows traversing the semi-separatries show the in-
ner and outer of eah semi-separatrix: a trajetory may traverse the semi-
separatrix followingthe diretion of the arrow, but not vie-versa.
The following two results relatefeasible signatures and semi-separatries.
Proposition 6. If, for some semi-separatrix
γ
,e ∈ γ in and e ′ ∈ γ out, then
the signature
ee ′ is not feasible.
Proof. Diretly fromthe denition of semi-separatrix.
Proposition 7. If, for some semi-separatrix
γ
, and signatureσ
(of at leastlength 2), then, if head
(σ) ∈ γ in and last(σ) ∈ γ out, σ
is notfeasible.
σ
is notfeasible.Proof. The proofproeedsby indutiononsequene
σ
. Thebase ase,whenσ
isoflength2,reduestoproposition6. Now,assumingthattheproposition istrue forsignatures oflengthn
,weare requiredtoprove thatitisalsotruefor signatures of length
n + 1
. Consider the signatureσ ′ = ee ′ σe ′′, with
Case 1:
e ′ ∈ γ in. Thesignaturee ′ σe ′′satisestheonditionsandisoflength
n
. Therefore, theindutiveproperty applies,andwean onludethat
e ′ σe ′′ is not feasible. However, sine any extension of an unfeasible
n
. Therefore, theindutiveproperty applies,andwean onludethate ′ σe ′′ is not feasible. However, sine any extension of an unfeasible
signature is itself unfeasible,itfollows that
σ ′ is not feasible.
Case 2:
e ′ ∈ γ out. The signature ee ′ is unfeasible by proposition 6. There-
fore,being anextensionof ee ′, σ ′ isalsounfeasible(proposition1).
ee ′, σ ′ isalsounfeasible(proposition1).
3 State-Spae Redution Using Semi-Separatries
Semi-separatriespartitionthe statespaeintotwo parts 3
oneonerosses
suh a border, all states outside the region an be ignored. We present a
tehnique, whih, given an SPDI and a reahability question, enables us to
disard portionsof the state spae basedon this information. The approah
is based onidentifyinginert states(edges in the SPDI) whih annotplaya
role in the reahability analysis.
Denition 8. Given an SPDI
S
, a set of semi-separatriesΓ ⊆ Sep
, asoure edge
e0
and a destination edgee1
, an edgee
is said to be inert if itlies outside a semi-separatrix inside whih lies
e0
, or it lies inside a semi-separatrix outside whih lies
e1
:3
Wedon'tonsiderthesemi-separatrixitself.
inert
Γ
e 0→e 1 = {e : E | ∃γ ∈ Γ · e0 ∈ γ in ∧ e ∈ γ out }
∪ {e : E | ∃γ ∈ Γ · e1 ∈ γ out ∧ e ∈ γ in }
Weanprovethat theseinertedgesan neverappear inafeasible signature:
Lemma 8. Given an SPDI
S
, a set of semi-separatriesΓ
, a soure edgee0
and a destination edgee1
, and a feasible signaturee0σe1
inS
. No inertedge from inert
Γ
e0→e1
may appear ine0σe1
.Proof. From the denition of inert states, it follows that eitherboth
e0
ande1
are inert, or neither is. If both are inert, then for someγ
,e0 ∈ γ in and
e1 ∈ γ out. But if this were so, then e0σe1
isunfeasible by proposition7. We
an thus onsider onlyinert edges in
σ
.Let
e
be an inert edge appearing inσ
. Therefore,e0σe1 = e0σ 1 eσ 2 e1
. Bydenition of inert edges,
e
an either be inert beause (i) it lies outside asemi-separatrix inside whih lies
e0
, or (ii) it lies inside a semi-separatrix outside whih liese1
.Case 1: Let
γ ∈ Γ
beasemi-separatrixsuhthate0 ∈ γ in ande ∈ γ out. But
by proposition 7,
e0σ 1 e
is not feasible. Hene, neither ise0σ 1 eσ 2 e1
.Case 2: Let
γ ∈ Γ
be asemi-separatrixsuh thate ∈ γ in and e1 ∈ γ out. By
proposition 7,
eσ 2 e1
is not feasible, and hene, neitherise0σ 1 eσ 2 e1
.It thus follows that
e0σe1
is not feasible.Given anSPDI, we an redue the state spae by disarding inert edges.
Denition 9. Givenan SPDI
S
, a set of semi-separatriesΓ
, a soure edgee0
and a destination edgee1
, we dene the redued SPDIS e0→e1 Γ to be the
same as
S
but without the inert edges.Clearly, the resultingSPDI is smaller than the originalone.
Proposition 8. For any SPDI
S
, aset of semi-separatriesΓ
, andedgese0
and
e1
,S
does not have less edges thanS e Γ 0→e 1.
Example 12. The shaded (light blue) areas of Fig. 6 (a) and (b) are the
subsets of the SPDI (edges of the reahability graph) eliminated by the re-
dutionpresented in thissetion,whenansweringthe question: Isinterval
I ′
reahable from
I
?I I’
(a)
I’
I
(b)
Figure6: Redution using Semi-separatries
Finally,weprovethathekingreahabilityonthereduedSPDIisequivalent
to hekingreahability onthe original SPDI:
Theorem 9. Given an SPDI
S
, a set of semi-separatriesΓ
, and edgese0
and
e1
, then,e1
is reahablefrome0
inS
if and only ife1
is reahable frome0
inS e0→e1 Γ .
Proof. The proof is split into two parts: that reahability in the redued
SPDI implies reahability in the original automaton (soundness) and vie-
versa (ompleteness).
Soundness: Assume that
e1
is reahable frome0
inS e Γ 0→e 1. Then, there
must exista feasiblesignature
σ
inS e Γ 0→e 1 whihstarts one0
and ends
at
e1
. Sine every SPDI edge inS e Γ 0→e 1 is alsoin S
, and the dynamis
of thetwo systemsare idential, itfollows that
σ
isalsoafeasiblepathin
S
. Therefore,e1
isalso reahable frome0
inS
.Completeness: Now assume that
e1
is reahable frome0
inS
. By deni-tion of reahability, there exists a feasible signature
e0σe1
inS
. Byproposition 8, no inert edge may appear in
e0σe1
. Therefore,e0σe1
is also a feasible signature in
S e Γ 0→e 1, whih in turn implies that e1
is
reahable from
e0
inS e 0→e 1.
ity question,we an redue the size of the SPDI to beveried. Thisenables
ustoverify SPDIsmuhmoreeiently. It isimportanttonote thatmodel-
heking an SPDI requires identiation of simple loops, whih means that
the alulation of the semi-separatries is not more expensive than the ini-
tialpassof themodel-hekingalgorithm. Furthermore,wean perform this
analysis onlyone for an SPDI and store the information tobe used in any
reahabilityanalysis onthatSPDI. Redution,however, an onlybeapplied
one we know the soure and destination states.
4 State-Spae Redution Using Kernels
4.1 State-spae redution using kernels
We have already shown that any invariant set, is essentially a pair of semi-
separaties. In partiular, the invariane kernel is a largest invariantset for
apartiularloop,weanuse theresultspresented insetion3toabstrat an
SPDI by using invarianekernels. We nowturn our attention tostate spae
redution using ontrollabilitykernels:
Denition 10. Givenan SPDI
S
, a loopσ
, a soure edgee0
and a destina-tion edge
e1
, an edgee
issaid to be redundant if it lies on the opposite sideof a ontrollability kernel as both
e0
ande1
:redundant
σ
e 0→e 1 = {e : E | ∃e0, e1 ∈ Cntr in (σ) ∪ Cntr (σ) ∧ e ∈ Cntr out (σ)}
∪ {e : E | ∃e0, e1 ∈ Cntr out (σ) ∪ Cntr (σ) ∧ e ∈ Cntr in (σ)}
Wean prove that we an dowithout these edges tohek feasibility:
Lemma 10. Given an SPDI
S
, a loopσ
, a soure edgee0
, a destination edgee1
, and a feasible signaturee0σe1
then there exists a feasible signaturee0σ ′ e1
suh thatσ ′ ontains no redundant edge from redundantσ e0→e1
.
Proof. Let
e0σe1
beafeasiblesignaturewhihontainssomeredundantedgefrom the set redundant
σ
e 0→e 1
. Without loss of generality, we assume thate0, e1 ∈ Cntr out (σ) ∪ Cntr(σ)
. Letf0
andf 1
be, respetively, the rst and last redundant edges inσ
. By denition of redundant edges, it follows thatf 0, f 1 ∈ Cntr in (σ)
. The path we are following isthus:Sine
f0
(f 1
)istherst(last)redundantedge,itfollowsthatthelastelementof
σ 1 (the rst element of σ 3) is inside the ontrollability kernel. Using
proposition 3, it follows that there exists a point p
on the ontrollability
kernel reahablefrom the lastelementof σ 1 (apointq
onthe ontrollability
kernel from whih the rst element of σ 3 is reahable). Sine all points on
p
on the ontrollability kernel reahablefrom the lastelementofσ 1 (apointq
onthe ontrollability
kernel from whih the rst element of σ 3 is reahable). Sine all points on
theontrollabilitykernelaremutuallyreahable,itfollowsthat
q
isreahablefrom
p
along some disrete pathσ 2 ′ ompletely within the kernel. We have
thusobtained ashorterdisretepath
e0σ 1 σ 2 ′ σ 3 e1
whihisfeasibleand whihontains noredundant edges.
GivenanSPDI,weanreduethestatespaebydisardingredundantedges.
Denition 11. Given an SPDI
S
, a loopσ
, a soure edgee0
and a desti-nation edge
e1
, we dene the redued SPDIS e σ 0→e 1 to be the same as S
but
without redundant edges.
Clearly, the resultingSPDI is smaller than the originalone.
Proposition9. ForanySPDI
S
,aloopσ
, asoureedgee0
andadestination edgee1
,S
does not have less edges thanS e0→e1 σ .
Finally,weprovethathekingreahabilityonthereduedSPDIisequivalent
to hekingreahability onthe original SPDI:
Theorem 11. Givenan SPDI
S
, witha setof loopsσ
, a soure edgee0
anda destination edge
e1
, then,e1
is reahable frome0
inS
if and only ife1
isreahable from
e0
inS e σ 0→e 1.
Proof. The theorem follows immediatelyfrom proposition 10.
Given aloopwhih hasa ontrollabilitykernel, wean thusreduethe state
spae to explore. In pratie, we apply this state spae redution for eah
ontrollability kernel in the SPDI. One a loop in the SPDI is identied, it
is straightforward to apply the redution algorithm.
4.2 Immediate answers to reahability questions
By denition of the ontrollability kernel, any two points inside it are mu-
tually reahable. This an be used to answer ertain reahability questions
simply by inspeting the ontrollability kernel: if both the soure and des-
tination edge lie (possibly partially) within the same ontrollability kernel,
then, there exists atrajetory from the soureto the destination edge.