Department of Informatis
On the Deidability of
the Reahability
Problem for GSPDIs
Researh Report No.
359
Gerardo Shneider
Isbn 82-7368-317-6
Issn 0806-3036
June 2007
Problem for GSPDIs
Gerardo Shneider
Department of Informatis, University of Oslo
PO Box 1080 Blindern, N-0316 Oslo, Norway
E-mail: {gerardo at i.uio.no}
Abstrat
Polygonal hybrid systems (SPDIs) are a sublass of hybrid sys-
tems whosedynamis isdenedbyonstantdierentialinlusions, for
whihthe reahabilityproblemisdeidable. Thedeidabilityresultis
based, among otherthings, onthefatthat atrajetoryannot enter
and leave a given region through the same edge. An SPDI satisfy-
ing the above restrition is said to have the goodness property. In a
previous work we have given amisleading proofskethof deidability
of reahability for SPDIs when relaxing goodness. In this work we
give a ounter-example to suh proof and we give an algorithm for
semi-deiding reahability ofsuh lassof systems.
1 Introdution
An interesting andstilldeidable (w.r.treahability)lass of hybridsystems
is the so-alledPolygonal Hybrid System (SPDI for short, [ASY01, ASY07,
Sh02 ℄) whih is a sublass of hybrid systems onthe plane whose dynamis
is dened by onstant dierential inlusions. SPDIs are a generalization of
PCDs(deterministisystemswithPiee-wiseConstantDerivatives)forwhih
it has been shown that the reahability problem is deidable for the pla-
nar ase [MP93℄ but undeidablefor three and higher dimensions [AMP95℄.
Slight extensions of suh deidable lasses have been proved to be undeid-
able or equivalent to a problem for whih deidability or undeidability is
not known [AS02, MP05℄.
The onstrutive proof for deiding reahability on SPDI given in [ASY01℄
(see also [ASY07℄ and [Sh02, Chap. 5℄) relies, among other things, on the
the SPDI(loationof theorrespondingautomaton)doesnotallowatraje-
torytotraverseanyedgeofthepolygondeningtheregioninbothdiretions.
Tehnially this means that the diretor vetor of eah edge annot be ob-
tained asapositivelinearombinationofthe vetors dening thedynamis.
An SPDI withoutthe goodness property is alled GeneralSPDI or GSPDI
forshort. Wehavewronglylaimedin[Sh02,Chap. 9℄thatthe reahability
problemforGSPDIisdeidable. Theproofskethwasondutedby proving
that any GSPDI an be redued to a set of SPDIs, preserving reahabil-
ity. The proofsketh, aspresented, is notompletelywrong butinomplete,
letting the deidability onlusion to be still inonlusive. Unfortunately
we have disovered suh mistake in September 2002, just few months after
the nal print of the thesis. We onsidered it was not worth publishing a
refutation of the result at that moment sine there was no researh being
onduted in that diretion then. We revived our interest on the subjet
again only reently due to the publiation of the paper [MP05℄, in whih
the frontier between deidable and undeidable hybrid systems is revisited,
to rene previous result given in [AS02℄. The deidability of reahability of
GSPDIs would have ontributed to narrowthe undeidability frontier; with
the result presented here we letit stillopen, unfortunately.
Inthispaperweprovideaounter-exampletothelaimofthedeidabilityof
thereahabilityproblemforGSPDIsgivenin[Sh02 ,Chap. 9℄,whihremain
thus an open problem. We prove, indeed, that GSPDI reahability annot
be redued to SPDI reahability. We rephrase the results given in [Sh02℄
to give a semi-deidable algorithm for solving the reahability problem for
GSPDIs.
The paper isorganized as follows. In next setionwe explain informallythe
problems arising when relaxing goodness while in Setion 3 we give some
preliminaries,providinguseful notationanddenitionsand reallingthedef-
initionofSPDI.InSetion4wepresentGSPDIs. Setion5isonernedwith
the analysis of trajetories, providing some results needed to establish the
semi-deisionalgorithmforreahability presented inSetion6. Weonlude
in the lastsetion.
2 On Goodness
In this setion we disuss informally why goodness is good for deiding the
reahability problem of SPDI and what are the problems when relaxing it.
More formaldenitions willbe given inSetion 3.
See Fig. 1 for an example of a good and a 'bad' region (here 'bad' stands
gure we an see a good region, where the two vetors
a
and
b
determine
the impossibilityof atrajetorytoenter and leavethe region
P
through thesame edge of the polygon delimiting the region. On the other hand, the
gureontherightshows abadregion: Both
e 2 ande 5 anberossedinboth
diretions by atrajetoryentering and leaving
P
.e 1
e 2
e 3
e 6
e 1
e 2
e 5
e 6
e 4
e 3
e 4
e 5
P P
y
x α
β a b
(a) (b)
y
x a
b
β α
Figure 1: a)A goodregion. b)A bad region.
2.1 Why Goodness is Good?
The algorithmpresentedin[Sh02 ℄fordeidingreahabilityonSPDIheavily
depends on the pre-proessing of trajetory segments to guarantee that it
is possible to list all the possible sets of signatures, i.e., those sequenes
of edges of the SPDI traversed by all the possible trajetories between two
points. This isof ourse not possible in generalas there are innitely many
suhtrajetories. However, aqualitativeanalysisallowstoprovethatindeed
there are a nite number of types of signatures, that are kind of abstrat
signatures that preserve the reahability property.
Briey, the aboveis ahieved by performingthe followingsteps.
1. Simpliationoftrajetorysegments: straighteningthemandremoving
self-rossings. Givenanarbitrarytrajetorysegmentfromonepointto
another, weshowhowtoget apieewise onstantderivativetrajetory
segment withoutself-rossing.
2. Abstration of trajetory segmentsintosignatures, onsidering the se-
queneoftraversededges. ThisresultisbasedonthePoinarémap[HS74,
NS60℄, that relates
n
-dim ontinuous-time systems with(n − 1)
-dimdisrete-time systems.
3. Fatorization of signatures ina onvenientway, having onlysequenes
of edges and simple yles. This fatorization allows to have a nie
representation of signatures.
4. Abstration of fatorized signatures into types of signatures, that are
signatureswithouttakingintoaountthenumberoftimeseahsimple
yle isiterated.
Many of the lemmas for proving that the above provides anite number of
typessignaturesritiallydependonthegoodnessassumption,whihpropa-
gatethisdependenytotheonstrutiveproofgivenfordeidingreahability
of SPDIs.
2.2 Why Relaxing Goodness is not so Good?
The mainquestion nowis,howmuh doweneed todepend onthe goodness
assumption to prove deidability of reahability of SPDIs? In other words,
let us onsider the new lass of polygonal hybrid systems, GSPDI, obtained
by relaxinggoodnessinSPDI.Isreahabilitystilldeidable? Fromtheabove
disussion weare letwith the followingtwoalternatives:
1. Adapt the proofs of deidability for SPDIs to GSPDIs. This would
implytorestate the proofstomakethem independent of the goodness
assumption.
2. ProvideaompletelynewdeidabilityproofforGSPDI.Thiswillprob-
ablyneedtouse dierenttehniquesand resultsthantheones usedfor
SPDIs.
The rst alternative above seems the most straightforward and easy to do.
However, aswewillshowlateritisnotpossibletoredueGSPDIreahability
toSPDIreahability. Thisisdonebyprovingthatitisnotingeneralpossible
tosimplifyertaintrajetoriesenteringandleavingagivenregionthroughthe
same edge, totrajetoriesbehaving as in SPDIs. One of the main problems
whenrelaxinggoodnessisthataregionannotbebi-partitionedanymoreinto
twosemi-planeswerealltheedgesinonesemi-planean betraversedonlyin
one diretion,w.r.t. theregion,andalltheedgesinthe othersemi-planean
be traversed only in the other diretion. That is, the goodness assumption
permit a ertain 'ontiguity' of entry edges and exit edges belonging to two
the reahability algorithm depend on this ontiguity. If we relax goodness,
weshouldbeabletore-proveallsuhresultswithoutassumingtheontiguity
of entry and exit edges.
This letus with the seond alternative. Unfortunately, to date we have not
sueeded inproviding a proof of deidability (nor of undeidability) to the
reahabilityproblem onGSPDIs.
On the other hand and as stated in the introdution, we willshow that we
an relax the goodness assumption as to give a terminating semi-deision
algorithmfor reahability analysis on GSPDIs.
3 Preliminaries
This setion is more tehnial, realling the main denitions and onepts
needed tounderstandthe restofthe paper. Foramoredetailedpresentation
see [ASY07, Sh02℄.
3.1 SPDI
Let
a = (a 1 , a 2 ), x = (x 1 , x 2 ) ∈ R 2 and α, β ∈ R
. The inner produt of two
vetors
a = (a 1 , a 2 )
andx = (x 1 , x 2 )
is dened asa · x = a 1 x 1 + a 2 x 2. We
denote by
x ˆ
the vetor(x 2 , −x 1 )
obtained fromx
by rotating lokwise by the angleπ/2
. Notie thatx · x ˆ = 0
.An angle
∠ b a onthe plane, dened by two non-zero vetors a , b
isthe set of
allpositivelinearombinationsx = α a + β b
,with α, β ≥ 0
,and α + β > 0
.
We an always assume that
b
is situated in the ounter-lokwise diretion froma
.Denition 1. A polygonal dierential inlusion system (SPDI) is dened
by giving a nite partition
P
of the plane into onvex polygonal sets (alledregions), and assoiating with eah
P ∈ P
a ouple of vetorsa P and b P.
Let φ(P ) = ∠ b a P P, we have that for eah x ∈ P
, x ˙ ∈ φ(P )
.
φ(P ) = ∠ b a P P, we have that for eah x ∈ P
, x ˙ ∈ φ(P )
.
Let
E(P )
bethe set of edgesofP
. Wesay thate ∈ E(P )
isanentry ofP
iffor all
x ∈ e
and for allc ∈ φ(P )
,x + c ǫ ∈ P
for someǫ > 0
. We say thate
is an exit ofP
if the same ondition holds for someǫ < 0
. We denote byIn
(P ) ⊆ E(P )
the set of all entries ofP
and by Out(P ) ⊆ E(P )
the set ofall exitsof
P
.Assumption 1. All the edges in
E(P )
are either entries or exits, that is,E(P ) =
In(P ) ∪
Out(P )
. We say then that all the regions in an SPDI aregood or that they have the goodness property.
Example 1. In Fig.1-(a), region
P
(withφ(P ) = ∠ b a) is good,sine all are
entry orexitedges. Fig. 1-(b) shows aregionthat isnot good: edges
e 2 and
e 5 are not inIn(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.The following is a very simple example of an SPDI: a swimmer trying to
esape froma whirlpoolina river.
Example. The dynamis
x ˙
of the swimmer around the whirlpool is ap- proximated by the piee-wise dierential inlusion dened as follows. Thezoneoftherivernearbythewhirlpoolisdividedinto8regions
R 1 , . . . , R 8. To
eahregion
R i weassoiate apair of vetors ( a i , b i )
meaningthat x ˙
belongs
to theirpositivehull: a 1 = b 1 = (1, 5)
,a 2 = b 2 = (−1, 1 2 )
,a 3 = (−1, 11 60 )
and
b 3 = (−1, − 1 4 )
, a 4 = b 4 = (−1, −1)
, a 5 = b 5 = (0, −1)
, a 6 = b 6 = (1, −1)
,
a 7 = b 7 = (1, 0)
, a 8 = b 8 = (1, 1)
. The orresponding SPDI is illustrated in
Fig. 2.
R 3
R 7
R 1 R 5
e 3
R 4
R 6
e 4
e 5
e 2 R 2
e 6 e 7
e 8
e 1
R 8
Figure2: The SPDI of the swimmer.
Given anSPDI, we xa one-dimensionaloordinatesystem oneah edge to
representpointslayingonedges. Fornotationalonveniene, wewilluse
e
todenote both the edge and its one-dimensional representation. Aordingly,
we write
x ∈ e
orx ∈ e
,to meanpointx
inedgee
with oordinatex
intheone-dimensional oordinate system of
e
. The same onvention is applied tosets of pointsof
e
represented asintervals(e.g.,x ∈ I
orx ∈ I
,whereI ⊆ e
)and to trajetories (e.g.,
ξ
startinginx
orξ
startinginx
).Now, let
P ∈ P
,e ∈
In(P )
ande ′ ∈
Out(P )
. ForI ⊆ e
,Succ ee ′ (I)
isthe set of all points in
e ′ reahable from some point in I
by a trajetory
segment
ξ : [0, t] → R 2 in P
(i.e., ξ(0) ∈ I ∧ ξ(t) ∈ e ′ ∧ Sig(ξ) = ee ′). Given
I = [l, u]
, Succ ee ′ (I) = F (I ∩ S) ∩ J
, whereS
and J
are intervals,F ([l, u]) = hf l (l), f u (u)i
andf landf u are anefuntions(afuntionf : R → R
isane
I = [l, u]
,Succ ee ′ (I) = F (I ∩ S) ∩ J
, whereS
andJ
are intervals,F ([l, u]) = hf l (l), f u (u)i
andf landf u are anefuntions(afuntionf : R → R
isane
f : R → R
isanei
f (x) = ax + b
witha > 0
).For
I ⊆ e ′, Pre ee ′ (I)
isthe set of points ine
that an reah apointinI
by a
trajetorysegment in
P
. Wehave that:Pre ee ′ = Succ −1 ee ′ and Pre σ = Succ −1 σ .
3.1.2 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= j ≤ k
.
1 ≤ i 6= j ≤ k
.Let
Succ σ (I ) = F (I ∩ S) ∩ J
withF = hf l , f u i
.Assumption 2. None of the two funtions
f l , f u is the identity.
Let
l ∗ and u ∗ bethe x-points1 off l andf u,respetively,andS ∩ J = hL, Ui
.
f l andf u,respetively,andS ∩ J = hL, Ui
.
S ∩ J = hL, Ui
.It an beshown that asimple yle is of one of the following types:
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
.1
Thex-point
x ∗
is omputedbysolvingalinearequationf (x ∗ ) = x ∗
, whih anbeniteorinnite.
but the leftmost one stays inside, that is,
L ≤ l ∗ ≤ U < u ∗.
The lassiation above provides useful information about the qualitative
behavior of trajetories. Any trajetory that enters a yle of type DIE will
eventuallyquititafteranitenumberof turns. Iftheyle isoftypeSTAY,
all trajetories that happen to enter it will keep turning inside it forever.
In allother ases, some trajetories will turn for a while and then exit, and
others will ontinue turning forever. This information is ruial for solving
the reahability problemfor SPDIs.
To nish this setion we reall the representation theorem for SPDIs that
allows tofatorize thesignatures (step3 inSetion2.1)ina onvenient way.
Given a sequene
w
,ε
denotes the empty sequene whereasf irst(w)
andlast(w)
are the rst and lastelements of the sequene respetively. An edge signatureσ
an be expressed as a sequene of edges and yles of the formr 1 s k 1 1 r 2 s k 2 2 . . . r n s k n n r n +1, where
1. For all
1 ≤ i ≤ n + 1
,r i is asequene of pairwise dierent edges;
2. Forall
1 ≤ i ≤ n
,s i isasimple yle (i.e., withoutrepetition ofedges)
repeated
k i times;
This is summarized by the followingrepresentation theorem for SPDIs that
not only guarantees the existene of the aboverepresentation for SPDIs but
also provides aonstrutive way of doing so[Sh02 , Theorem 17℄.
Theorem 1. Given an SPDI, let
σ = e 1 . . . e p be an edge signature, then it
analwaysbewrittenas
σ A = r 1 s k 1 1 . . . r n s k n n r n +1,whereforany1 ≤ i ≤ n+1
,
r i is a sequene of pairwise dierentedges andfor all 1 ≤ i ≤ n
, s i isa sim-
1 ≤ i ≤ n
,s i isa sim-
ple yle (i.e., without repetition of edges).
This representation of signatures is the base to obtain types of signatures
(step 4 in Setion 2.1) with the following good properties [Sh02, Lemma
20℄.
Lemma 2. Given an SPDI, let
σ = e 0 . . . e p be a feasible signature, then its
type,
type(σ) = r 1 , s 1 , . . . , r n , s n , r n +1 satises the following properties.
P 1 For every 1 ≤ i 6= j ≤ n + 1
, r i and r j are disjoint;
r j are disjoint;
P 2 For every 1 ≤ i 6= j ≤ n
, s i and s j are dierent.
s j are dierent.
The above is the base for the argument on the niteness of dierent types
of signatures to take into aount in the reahability algorithm and thus to
termination of SPDI reahability.
Thegoodnessrestrition(Assumption1)wasoriginallyintroduedtosimplify
treatment oftrajetoriesto guarantee, amongother things,that eahregion
an be partitioned into entry and exit edges inan ordered way, fat used in
the proof of deidability of the reahability problem. We will study in this
setion what happens when goodness is relaxed. First notie that without
goodness there are edges that are neither of entry nor of exit as shown in
Fig. 1. This naturallyleads to the followingdenition.
Denition 2. An edge
e ∈ P
is an inout edge ofP
ife
is neither an entrynor an exit edge of
P
.Asalreadyexplainedinprevioussetions,the abovedenitionisthebasefor
obtaininga new lass of polygonal hybrid systems whih generalizes SPDI.
Denition 3. An SPDI without the goodness restrition is alled a general
SPDI (GSPDI).
Thus, inGSPDIs there are three kinds of edges: inouts, entries and exits.
Self-rossing of trajetorysegments of SPDIs an be eliminatedwhih allow
us to onsider only non-rossing trajetory (segments). The proof given in
[Sh02 , Chap. 4, Se. 4.2.2℄ an be extended to deal with the ase when
the self-rossing trajetories involve inout edges, so the result still holds for
GSPDIs. Thus in what follows we will onsider only trajetory segments
without self-rossings.
Notie that on GSPDIs a trajetory an interset an edge at an innite
number of pointsbeause it an slide at it. Thus, a trae is not anymore a
sequene of points but rather asequene of intervals.
Denition4. The traeof a trajetory
ξ
isthesequenetrace(ξ) = I 0 I 1 . . .
of the intersetion intervals of
ξ
withthe set of edges, thatis,I i ⊆ (ξ ∩ E )
.A point interval
I = [ x , x ]
will be sometimes written asx
whenever no onfusion mightarise.Denition 5. An edge signature (or simply a signature) of a GSPDI is
a sequene of edges. The edge signature of a trajetory
ξ
,Sig(ξ)
, is theorderedsequeneoftraversededgesbythetrajetorysegment,thatis,
Sig (ξ) = e 0 e 1 . . .
, withtrace(ξ) = I 0 I 1 . . .
andI i ⊆ e i. The region signature of ξ
is
the sequene
RSig(ξ) = P 0 P 1 . . .
of traversed regions, that is,e i ∈
In(P i )
.Notie that in many ases the intervals of a trae are in fat points. We
say that a trajetory with edge signature
Sig (ξ) = e 0 e 1 . . . e i . . .
and traetrace(ξ) = I 0 I 1 . . . I i . . .
interval-rossesedgee i if I i is not apoint.
(a’) (a)
x 0
x f x f
x 0
e e
Figure3: (a): A properinout edge; (b): A slidingedge.
Givenatrajetorysegment,wewillmakethe dierenebetween proper inout
edges and sliding edges.
Denition 6. Let
ξ
be a trajetory segment from pointx 0 ∈ e 0 to x f ∈ e f,
with edge signature
Sig (ξ) = e 0 . . . e i . . . e n, and e i ∈ E(P )
be an edge of P
.
We say that
e i is a sliding edge of P
for ξ
if ξ
interval-rosses e i, otherwise
e
is saidto be a properinout edge ofP
forξ
.Wesaythatatrajetorysegment
ξ
slidesonanedgee
ife
isaslidingedgeofP
forξ
andξ
is said to be a sliding trajetory if there is at least one slidingedge
e ∈ Sig (ξ)
.Example 2. In Fig. 3-(a),
e
is aproperinout edge. Edgee
onFig. 3-(b)isa sliding edge.
5 Simpliation of GSPDI's Trajetory Segments
Inthissetionweshowthatinmanyasesitispossibletosimplifytrajetory
segments eliminatinginoutedges, but not always. Werst start by showing
that the good properties of the representation theorem for SPDIs are not
validany longer for GSPDIs, explaining why inouts edges are not desirable
in areahability analysis.
Proposition1. Property
P 2 oftherepresentationtheoremforSPDIs(Lemma 2) does not hold in general for GSPDIs.
Proof: Let
ξ
be a trajetory with signatureSig(ξ) = σ = e 0 . . . e i . . . e n . . .
of a given GSPDI. The proposition states that it is not possible in general
to write
σ
in the formσ A = r 1 s k 1 1 . . . r n s k n n r n +1 with the properties stated
ounter-example shouldallowtoobtain asignature onsistingof alokwise
spiral followed by aounter-lokwise spiral(orvie-versa) and thenbak to
the rst spiral. In suh a ase it is possible to nd two simple yles whih
are repeated inthe type of signature. Let us onsider the GSPDI of Fig. 4.
To let it simple we do not write down the dynamis of the regions and we
assume that they are as to allow the segments of trajetories shown in the
pituretobewell-dened. InsuhaGSPDIitispossibletoobtainthefollow-
ing type of signature:
r 1 s 1 r 2 s 2 r 3 s 3 . . .
, wheres 1 = (abcd)
,s 2 = (dcba)
, ands 3 = (abcd)
. Sines 1 = s 3,thenproperty P 2 ofLemma2isnotsatised.
a b
c
d
Figure4: Counter-example for Proposition1.
The followinglemmapresentssometypialaseswhereitispossibletoelim-
inate properinout edges.
Lemma 3. Let
ξ
be a trajetory segmentx 0 ∈ e 0 to x f ∈ e f with edge
signature
Sig(ξ) = e 0 . . . e i . . . e n. If e i is a proper inout edge then in some
ases there exists a trajetory segment
ξ ′ from x 0 to x f that traverses e i in
x f that traverses e i in
at most one sense (that is,
e i is either an entry or an exit, but no both).
Proof Sketh: In Fig. 5-(a) we illustrate a typial ase where edge
e i is a
properinoutedge. Afterastraightforwardalgebraivetormanipulation,on
in Fig. 5-(a')is obtained.
x f
(a)
x 0
x f
e e
(a')
x 0
Figure5: Inout ase.
Note that the above does not establish ompleteness of a redution from
GSPDIs intoSPDIs reahability sinethere areases wherethe above isnot
possible. Wehave then the followingresult.
Proposition 2. Given a GSPDI, assume there exists a trajetory segment
from points
x 0 ∈ e 0 to x f ∈ e f, traversing inout edges in both diretions.
Then it is, in general, not possible to nd a trajetory segment whose edge
signature ontains no proper inout edges (traversed in both diretions), be-
tween them.
Proof: TheGSPDIofFig. 6presentsatypialexampleofaninoutedge(
e 2)
whihannot bediretly eliminated astopreservethat
x f isreahablefrom
x 0. Tokeep the explanation simple we donot present here a formalGSPDI as ounter-example. The example, however, sheds some light onthe kind of
GSPDIregionsservingasounter-examples. Itsuestotakeanytrajetory
with a dynamis suh that the angle is slightly less than 180 degrees. The
trajetorymusttraverse aninout edgefollowingthe
b
vetorand entersinto theregionbyfollowingthea
vetor. Thetrajetorymustnotrossitself.Weshow now howto eliminatesliding edges.
Lemma 4. Let
ξ
be a trajetory segmentx 0 ∈ e 0 to x f ∈ e f with edge
signature
Sig(ξ) = e 0 . . . e i . . . e n. Ife i is aslidingedge for ξ
thenthere exists
ξ
thenthere existsa trajetory segment
ξ ′ from x 0 to x f that does not slide on edge e i.
x f that does not slide on edge e i.
x 0
e 5
P b
a
e 1
x f
e 2
e 3
e 6
e 4
Figure6: A GSPDI with a non-eliminatinginout edge.
Proof Sketh: Slidingedgesanariseinfourdierentases(withouttaking
intoaount thesymmetriases); theyare shown inFig.14-(a)to(d). The
orresponding primed gures (Fig. 14-(a') to (d')) show the transformation
done in order to avoid sliding on edge
e
. The reason why the above trans-formation ispossible isbeause inallthe ases the new obtainedsegment of
trajetory an be expressed as a positivelinear ombination of two suitable
existing segments of trajetory. Suh two segments are the sliding segment,
and another segment of trajetory with starting point at the beginning or
the end of the sliding segment.
As a onsequene we have the following result.
Proposition 3 (Existeneofanon-slidingtrajetory). If there existsa slid-
ing trajetorysegment from points
x 0 ∈ e 0 tox f ∈ e f thentherealways exists
a non-slidingtrajetory segment between them.
Proof: By indution on the number
n
of sliding edges of the signature ofthe trajetorysegment using Lemma 4 inthe indution step.
Weusually eliminaterst properinout edges (whenpossible)and next slid-
ing. In fat, the number of sliding edges is not guaranteed to derease if
sliding edges are eliminated before proper inout edges as shown in the fol-
lowingexample.
Example3. InFig. 7-(a)atrajetorysegmentthatslidesatedge
e ′isshown.
After eliminatingthe sliding atedge
e ′,a newsliding edge isintrodued(e
).
This is shown in Fig. 7-(b). However, if proper inout edges are eliminated
same gure.
x f x f x f
(a)
x 0
e e ′
(b)
x 0
e e ′
()
x 0
e e ′
Figure7: Eliminationorder of inout edges.
Remark. Slidingis noteasy totreatingeneralsine anedgealways belong
to two dierent regions with dierent dynamis. Thus a trajetory may be
'allowed' to slide by one of the dynamis but not by the other. We do not
analyze this in more detail, for our purposes we assume that at an inout
edge a trajetory an slide if at least one of the dynamis allows so. This
assumption does not aet the reahability analysis.
About the ordering between edges. We nish this setion with an
informal disussion about the importane of the 'ontiguous' order between
entry and exit edges onSPDIs.
In SPDIsedges of aregionan bebi-partitionedintoentryand exitedges in
a ontiguous way (see Fig. 8) having as a onsequene an orderingbetween
edges. This is not longer the ase in GSPDIs.
P
b a
In
Out
Figure 8: Ordering of edges onan SPDI (allthe edges
e
satisfya e ˆ > 0
).order to preserve the `positiveanity' (and hene the monotoniity) of the
suessor funtions. Given a region
R
with dierentialinlusion∠ b a,lete
be
anentryedgeand
e 1 ande 2 twoexitedgesofR
. Fore
wehosethe diretion
R
. Fore
wehosethe diretion(given by a diretor vetor
e
) that satises the inequality
ˆ a e > 0
(see Fig.11). The same for
e 1 and e 2. As a onsequene we obtain an ordering like
the one shown in Fig. 8.
Note that on a GSPDI (see Fig. 9(a)), the property that for any edge
e
,ˆ
a e > 0
isnot longervalidsine anedgean beofentryand ofexitand thenthe ordering an hange. In spite of that, one an inout edge is 'onverted'
into an entry (or exit) then we an have the notation of onsidering the
ordering of entry edges going ounter-lokwise and lokwise for exit edges
(see Fig. 9(b)).
(a) (b)
P
a
In
In In
In
Out Out
Out Out
b
P
a
I n
I n I nout
I nout
I nout
I nout Out
b I nout
Figure9: (a)A GSPDI; (b) Ordering after xing input and output edges.
Eventhoughthedenitionofedgeandregionsignaturesaswellasedgeyle
ontinue to hold, it is not the ase for region yle. We an have a region
signature
P 1 · · · P i · · · P k P 1 that is not a region yle. The reason is that in
GSPDIs a trajetory an enter a regionthrough two dierent edges without
forming a yle.
Thuswehavethataregionsignature
P 1 · · · P i · · · P k P 1 isaregion yle if the
edge signature
e 1 · · · e k e 1,with e i ∈
Out(P i )
for all1 ≤ i ≤ k
, forms anedge
yle.
In Fig. 10 the following is a region yle:
P 1 P 2 P 3 P 4 P 2 P 5 P 1. Notie that
P 2 P 3 P 4 P 2 isregion yle forSPDIs but not for the given GSPDI.
P 2
P 3
P 4
P 1
P 5
e 1
e 2
e 3
e 4
e 5
e 6
Figure10: A regionyle.
6 Reahability Analysis for GSPDIs
In this setion we `topologially' rephrase and prove the results of [Sh02,
Chap. 4,5℄ that use the ontiguity between entry and exit edges in their
proofs. Wealsore-provesoundnessofExit-LEFTandExit-STAYalgorithms
andattheendwegiveasemi-deisionalgorithmforGPSDIreahability. We
have informallyexplainedin Setion2.2why we need to doso.
6.1 Proof of Lemmas without using the Contiguity As-
sumption
The onlyresults that use the ontiguity order between entry and exit edges
are Lemmas20,Lemma 26and Corollary27of [Sh02 ℄. Lemma20hasbeen
repeated hereinSetion3asLemma2,whihaswe have seendoesnot hold
ingeneralfor GSPDIs(Proposition1). However, afterxing allthe edgesas
either of entry or exit, we an prove the result holds sine it behaves as an
SPDI, modulothe ontiguity of entry and exit edges.
Weprovethenthese threeresultswithoutusingthe orderbetween entryand
exit edges. We restate Lemma 2 ([Sh02, Lemma 20℄) for property
P 2, for the ase whenGPSDI istransformedastoxinout edgesasentries orexits.
Lemma 5. Given a GSPDI where edges has been xed as entry or exit, let
σ = e 0 . . . e p be a feasible signature, then its type, type(σ) = r 1 , s 1 , . . . , r n ,
(a) (b)
P
e e
P
e
ˆ a
e
a a
ˆ a
Figure11: (a)
ˆ a e > 0
; (b)a e ˆ < 0
.s n , r n+1 satisesthe followingproperty, P 2: For every 1 ≤ i 6= j ≤ n
, s i and
1 ≤ i 6= j ≤ n
,s i and
s j are dierent.
Proof: In order to prove property
P 2 we prove that, given a simple yle
s i = e ′ , . . . , e
,thesequene ofedgesee ′ annotourafterleavings i (heneit
annotourinanyothersimpleyle
s j,with1 ≤ i < j ≤ n
). Afteryling
k i times yle s i is abandoned by edge e
(guaranteed by onstrution). Let
s i is abandoned by edge e
(guaranteed by onstrution). Let
P
be a region s.t.e ∈
In(P )
and onsider the unfolding of the last iterationand its ontinuation(see Fig.12-(a)):
. . . , e, e ′ , . . . , e, e ′′ , . . .
where
e ′′ = f irst(r i+1 )
,e ∈
In(P )
ande ′ , e ′′ ∈
Out(P )
(e ′ 6= e ′′). Let x 2 be
thelastpointvisitedonedge
e
beforeleavingyles i andx ′′ 2 betherstpoint
on edge
e ′′ after leavings i (see Fig. 12-(b)). Segment x 2 x ′′
x 2 x ′′
2
of the trajetorysegment divides region
P
intotwosubregionsP 1 and P 2 andedge e
intotwo
e
intotwosegments
e l x 2 and x 2 e u. By the non-rossing hypothesis (and monotoniity
on edges) after leaving s i the only aessible part of edge e
is the segment
x 2 e u ∈ e
. ByJordan'surve theorem theonly way toreahedge e ′ fromany
s i the only aessible part of edge e
is the segment
x 2 e u ∈ e
. ByJordan'surve theorem theonly way toreahedge e ′ fromany
pointin
x 2 e u ∈ e
isbyrossingx 2 x ′′ 2 orbyrossingoneofthe edgesofregion
P 2. The rst ase is not possible sine itwould ontradit the hypothesis of
non-rossing trajetory and in the seond ase the sequene
ee ′ would not
belongto the trajetory segment.
Remark. Note that for our purposes it is irrelevant whether property
P 1
holds or not, sine it does not aet the niteness argument. This isdue to
0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000 0000000000000
1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111 1111111111111
000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000 000000000000
111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111 111111111111
x 1
x 2
e
P x ′ 1
e ′ e ′′
x ′′ 2
(a)
e l e
x 2
x 3
e u
(b)
P 2
x ′′ 2 e ′′
e ′
P 1
Figure12: (a): Simpleyle
s i anditsontinuationthroughedgee
;(b) Edge
e ′ annot be reahed frompointx 3 withoutinterseting x 2 x ′′
x 3 withoutinterseting x 2 x ′′
2
the fat that a type of signature is nite if the number of simple yles are
not repeated, whihis stated in
P 2.
In what follows we use the following notation. Whenever we partition the
spae into two regions
P L and P R by the line dened by a segment of line
xy
,P L is the semi-spae of all the points that are a left rotation of xy ~
and
P R isthe semi-spae orresponding tothe pointsthat are a right rotationof
the samevetor. With
f(x) ↓
wemeanthatf
isdened atx
andf (x) ↑
willmean that
f
isundened atx
.Next we will(topologially)rephrase [Sh02 , Lemma 26℄and [Sh02 , Corol-
lary 27℄and we provethem both.
Lemma 6. Let
P
be a region,e ∈
In(P )
,e 1 , e 2 ∈
Out(P )
,hl i , u i i
be anysubinterval of
he l i , e u i i
andf i (x) = F e,e c i (x)
.1. Let
P
be partitioned into two regionsP L and P R by the line dened by
xl 1, then the following holds: if e 2 ∈ P L, f 2 (x) ↓
and l 1 < f 1 (x)
then
u 2 < f 2 (x)
;
f 2 (x) ↓
andl 1 < f 1 (x)
thenu 2 < f 2 (x)
;2. Let the plane be partitioned into two subspaes
P L and P R by the line
denedby
xl 2, thenthefollowingholds: ife 1 ∈ P R, f 1 (x) ↓
andf 2 (x) <
f 1 (x) ↓
andf 2 (x) <
u 2 then f 1 (x) < l 1.
Proof:
(a) (b)
b a b a
P
e x
f 2 (x)
e 1
e 2
l 2
u 2
u 1
A
l 1
f 1 (x)
P L
P R
P
e x
f 2 (x) ↑
e 1
u 1
l 1
f 1 (x)
P L
P R
e 2
u 2
l 2
Figure13: Lemma6'-1. (a) When
f 2 l (x) ↓
; (b) The asef 2 l (x) ↑
.1. Remember that the line dened by
e 2 is ordered and that u 2, A
and
f 2 (x)
belongs to it. We have then that e 2 ∈ P L (and hene u 2 ∈ P L)
A
andf 2 (x)
belongs to it. We have then thate 2 ∈ P L (and hene u 2 ∈ P L)
and that
f 2 (x) ∈ P R (by onstrutionof the partition). We have then
that u 2 < A
and A < f 2 (x)
, that implies u 2 < f 2 (x)
. See Fig. 13(a).
2. This ase is symmetri tothe previous one.
Corollary 7. Let
P
bea region,e ∈
In(P )
,e 1 , e 2 ∈
Out(P )
,f i (x) = F e,e c i (x)
be anane funtionand
F i (hx, yi) = F i (hx, yi ∩ S i ) ∩J i bea trunated ane
multi-valued funtion (with
F i = [f i l , f i u ]
andJ i = hL i , U i i
).1. Let
P
be partitioned into two regionsP L and P R by the line dened
by
xL 1, then the following holds: If e 2 ∈ P L and L 1 < f 1 l (x)
then
F 2 (hx, yi) = ∅
;
L 1 < f 1 l (x)
thenF 2 (hx, yi) = ∅
;2. Let
P
be partitioned into two regionsP L and P R by the line dened
by
xL 2, then the following holds: if e 1 ∈ P R and f 2 u (y) < U 2 then
f 2 u (y) < U 2 then
F 1 (hx, yi) = ∅
.Proof:
1. If
f 2 l (x)
is undened, then it is obviousthatF 2 (hx, yi) = ∅
. Iff 1 l (x)
isdened, thenthe resultfollowsdiretlyfromLemma6-1and denition
of
F i (hx, yi)
.2. Symmetri tothe above ase using Lemma 6-2.
6.2 Soundness of Exit-STAY and Exit-LEFT
Weprove nowsoundnessof theExit-STAYand Exit-LEFTalgorithmwhose
proofs rely onthe results proved in the previous setion.
Let
A = Succ b s (L)
and onsider the linedened byAL
. This line partitionthe spaeinto
P L and P R as before.
Exit-STAY
funtion
Exit ST AY (I, s, ex)
←− ∅
Soundness Byhypothesis,
L < l ∗ < u ∗ < U
. Hene, foralli
,I ˜ i = h ˜ l i , u ˜ i i ⊆ hL, U i
, heneI i = ˜ I i and by Corollary7 we havethat Succ i s,ex (I) = ∅
.
Termination Trivial.
Exit-LEFT:
funtion
Exit LEF T (I, s, ex)
←− Succ s,ex ( Succ s,f (hL, max{u, u ∗ }i))
Soundness By hypothesis,
l ∗ < L < u ∗ ≤ U
. Thus, there exists a naturalnumber
n
s.t.l ˜ n ≤ L
and for alli
,u i = ˜ u i ≤ U
. Let's onsider thefollowingtwoases:
1. If
ex ∈ P RthenEx = ∅
(bydenitionofExit-LEFT)andSucc s,ex (I i ) =
∅
foranyi
(byCorollary7-2),soSucc s,ex ( Succ s,f (hL, max{u, u ∗ }i)) =
∅
;2. If
ex ∈ P L, we onsider two ases:
(a) If
u < u ∗thenforalli
,u i = ˜ u i ≤ u ∗andthen∪ m>0 Succ m s,f (I) =
Succ s,f (L, u ∗ )
, thus Ex = Succ s,ex (Succ s,f (L, u ∗ ))
;
∪ m>0 Succ m s,f (I) =
Succ s,f (L, u ∗ )
, thusEx = Succ s,ex (Succ s,f (L, u ∗ ))
;(b) If
u ∗ < u
then for alli
,u i = ˜ u i ≤ u
and∪ m>0 Succ m s,f (I) = Succ s,f (L, u)
. Consequently,Ex = Succ s,ex (Succ s,f (L, u))
;Frombothaseswehavethat
Ex = Succ s,ex (Succ s,f (hL, max{u, u ∗ }i))
.Termination Trivial.
6.3 A semi-deision algorithm for reahability analysis
of GSPDIs
Fromthe aboveresultswehavethatthemainalgorithmforreahabilitymay
beapplied toGSPDIs after performingertain pre-proessing steps.
Before presentingasound (butinomplete)algorithmforreahabilityanaly-
sisof GSPDIsweneed thefollowingnotation. Given aGSPDI
H
,wedenoteby
H red = {H 1 , . . . , H n }
the set of alltheSPDIs obtained afterxing alltheinout edges of
H
as inputs oroutputs, onsideringall the possible permuta-tions.
ThereahabilityalgorithmforaGSPDI
H
,Reah(H, x 0 , x f )
,onsistsofthefollowingsteps:
1. Detet allthe inout edges;
2. Generate the set of SPDIs
H red = {H 1 , . . . , H n }
;3. Apply the reahability algorithmfor SPDIs to eah
H i (1 ≤ i ≤ n
).
4. Ifthereexistsatleastone
H i ∈ H redsuhthatReah(H i , x 0 , x f ) = Yes
then Reah
(H, x 0 , x f ) = Yes
, otherwise we do not know.Wehave then the followingresult about termination of GSPDI reahability.
Lemma 8. Reah
(H, x 0 , x f )
always terminate.Proof: Theresult followsfromtheterminationofsteps 1and 2oftheabove
algorithm, as well as from that of Reah
(H i , x 0 , x f )
(for allH i ∈ H red,
1 ≤ i ≤ n
).Wenishthissetionwiththemainresultofourpaper,whihfollowsfromall
thepreviousresults,statingthatweansemi-deidereahabilityforGSPDIs.
Theorem 9. Given aGSPDI
H
, ifReah(H i , x 0 , x f ) = Yes
forsomeH i ∈ H red, then Reah(H, x 0 , x f ) = Yes
. On the other hand, if for allH i ∈ H red,
Reah
(H i , x 0 , x f ) = No
, then Reah(H, x 0 , x f )
is inonlusive.soundness of the algorithm for SPDIs [Sh02 , Se. 5.2℄, inluding the new
proofgiveninSetion6.2onsideringtheuseofnon-ontiguousentryandexit
edges. ThefatthatreahabilityisinonlusivewheneverReah
(H i , x 0 , x f ) = No
for allH i ∈ H red follows from Proposition2.
7 Final Disussion
In this work we have provided a ounter-example to aprevious proof of the
deidability of the reahability problem for GSPDIs given in [Sh02 , Chap.
9℄,whihremainthusanopen problem. Wehaverephrased theresultsgiven
in above mentioned work in order to give a semi-deidable algorithm for
solving the reahability problemfor suhlass of systems.
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(a’)
e
(d') (d)
x 0 x 0
e
x f x f
(') ()
x 0
e e
x 0
(b') (b)
e
x f x f
e x 0
(a)