hybrid systems modeling, analysis and control · radu grosu vienna university of technology hybrid...

Radu Grosu Vienna University of Technology

Hybrid Systems Modeling, Analysis and Control

Lecture 11 Constraint Satisfaction

Using Interval Approximation

Linear Classifiers with Hard Threshold x 2

: sur

face

-wav

e m

agni

tude

x1: body-wave magnitude

−Linear: A hyperplane. E.g. − 4.9 + 1.7x1 − x2 = 0

−Explosions: − 4.9 + 1.7x1 − x2 > 0

−Earthquake: − 4.9 + 1.7x1 − x2 < 0 Decidable!

! eartquake • nuclear explosion

Decision boundary: Surface that separates two classes

Polynomial Constraints

Theorem (Tarski 1949): The first-order theory of real-closed ordered fields is decidable.

Proof: FOT (R, ≤ ,+,×) admits quantifier elimination.

x2 + y2 − 1 = 0 ∧ y − x2 = 0

x

y

Polynomial Constraints

Theorem (Tarski 1949): The first-order theory of real-closed ordered fields is decidable.

Complexity: ExpExp in # alternations, Exp in # vars

x2 + y2 − 1 = 0 ∧ y − x2 = 0

x

y

What About Certificates?

Certificate: Satisfying valuation (solution) Quantifier-free case: Eg. x2 = 2

Question: How to represent the solution?

−Linear case: Rational numbers. Eg. 3x = 2 ⇒ x = 2 / 3

−Polynomial case: Real algebraic numbers (unintuitive)

−Non-polynomial case f(x) = 0: An approx algorithm

Given: desired precision k ∈NReturns r ∈Q: ∃x. |x − r| ≤ 10−k ∧ f(x) = 0

What About Certificates?

Certificate: Satisfying valuation (solution) Quantifier-free case: Eg. x2 = 2

Question: How to represent the solution?

−Linear case: Rational numbers. Eg. 3x = 2 ⇒ x = 2 / 3

−Polynomial case: Real algebraic numbers (unintuitive)

−Non-polynomial case f(x) = 0: An approx algorithm

Given: desired precision k ∈NReturns r ∈Q: |f(r)| ≤ 10−k

Non-Polynomial Constraints (NPC)

What about: FOT (R, ≤ ,+,×,sin) ?

−Undecidable: A Would allow encoding of polynomial diophantine equations (PDE)

f(x,y) > 0

f(x,y) < 0

x

y

−PDE: Known to be undecidable (Matiyasevich, 1970)

Fixing Precision of NPC

x

y

f(x,y) < 0

f(x,y) > 0


Exponential: In the number of boxes!

Top-Down Algorithm

x

y

f(x,y) < 0

f(x,y) > 0


Top-Down Algorithm

x

y

f(x,y) < 0

f(x,y) > 0


Infinite precision: May not terminate!

Robust Constraints

Real CPS (biology/engineered): Satisfy robust properties

−Satisfiability: Does not change under perturbations

c2

! eartquake • nuclear explosion

c1

x1: body-wave magnitude

x 2: s

urfa

ce-w

ave

mag

nitu

de

Robust Constraints

Constraint ϕ is robust iff

Real CPS (biology/engineered): Satisfy robust properties

−Satisfiability: Does not change under perturbations

x2 ≤ 0 ⇒ x2 ≤ −0.00001 −Not robust:

x2 ≤ 1 ⇒ x2 ≤ +0.99999 −Robust:

−Constraint distance: d(ϕ,ϕ') ! (ϕ ≡upto cst c

ϕ') ? c : ∞

−There is an ε such that

−For all ϕ ' with d(ϕ,ϕ') ≤ ε

ϕ and ϕ ' are equi-satifiable

Quasi Deciadable Problems

A problem is quasi decidable iff:

−There is an algorithm that

−Correctly checks satifiability and −Terminates on all robust instances

Theorem (Ratschan 2002):

−FOT (R, ≤ +, ×, exp, sin, ...) is quasi decidable

Assumptions:

−All variables are bounded (in some interval)

−Shortcut: f = 0 ≡ (f ≤ 0 ∧ f ≥ 0)

Implementation: http://rsolver.sourceforge.net

Not Quasi Deciadable Problems

Theorem (Clarke/Gao 2011):

−δ-Satisfiability is decidable

x

y

f(x,y) < 0

f(x,y) > 0

δ do not know

Algorithm: Quantifier-Free Conjunctive

Box (hyper-rectangle): Cartesian product of intervals

−Given ϕ: Conjunction of ≤ using +, ×, exp, sin

Uses algorithms for both sat and unsat: Why?

−Due to undecidability failure to prove sat

−Box B: Providing an interval for each variable

Quasi-Satisfiability Algorithm

−Decide: Sat or Unsat

−Does not imply unsat and vice versa −Later on can use information from each other

Search for Sat

Satisfiability: Statement over one valuation −Good search method: Suffices

−Local search examples: Newton-type methods

−Iteratively refine the grid

−Local search: Start from this points

x

y

f(x,y) < 0

f(x,y) > 0

−Approximation errors: Floating-point rounding OK

Search for Unsat: Branch and Bound

Non-Satisfiability: Statement over an uncountable set

−Symbolic representation: Necessary

S = test(ϕ,B) if (S = unsat) return S else

−Use: test(ϕ,B) ∈{unsat, unknown}

let B = B1 ∪B2 non-overlapping

Algorithm BB(ϕ,B): Either returns unsat or runs forever

if (BB(ϕ,B1) = BB(ϕ,B2) = unsat) return unsat

Can be interleaved: With satisfiability test

Test for Unsat

Works recursively: On the structure of ϕ −Special case: One single equality

−Input: f(x1,...,xn) = 0, B = intervals (I1,...,In)

{f(x1,...,xn) | x1 ∈I1,...,xn ∈In} ⊆ [f] (I1,...,In)

if (0 ∉[f](I1,...,In)) return unsat else unknown

Compute: [f] (I1,...,In), an inclusion of f, s.t:

Example:

x × y + 1 = 0

x ∈[2,3], y ∈[4,7] B = ([2,3],[4,7])

[2,3] [4,7]

x y

[8,21]

[×] [1,1]

[9,22]

[+]

1

Interval Arithmetic: Addition

Function [+] on intervals [a,a] and [b,b] is an inclusion:

[a,a] [+] [b,b] =

∅ if [a,a] = ∅ ∨ [b,b] = ∅[a + b,a + b] otherwise

⎧⎨⎪

⎩⎪

Example:

[8,21] [+] [1,1] = [9,22] [8,21] [1,1]

[9,22]

[+]

Proof: + does not have any local minimum Corners suffice

Interval Arithmetic: Multiplication

Function [×] on intervals [a,a] and [b,b] is an inclusion:

[a,a] [×] [b,b] =

∅ if [a,a] = ∅ ∨ [b,b] = ∅ [min{ab,ab,ab,ab}, max{ab,ab,ab,ab}] else

⎧⎨⎪

⎩⎪

Example:

[2,3] [×] [4,7] = [8,21] [2,3] [4,7]

[8,21]

[×]

Proof: × does not have any local minimum Corners suffice

Multiplication Improved

Do we really need all corners?

b ≥ 0 0 ∈[b,b] b ≤ 0a ≥ 0 [ab, ab] [ab,ab] [ab,ab]

0 ∈[a,a] [ab,ab] [min{ab,ab}, max{ab,ab}] [ab,ab]a ≤ 0 [ab,ab] [ab,ab] [ab,ab]

Example:

[8,21] [+] [1,1] = [9,22] [2,3] [4,7]

[8,21]

[×]

−Observation: Fixed monotonicity in quadrants

−For: [a,a] = [0,0]∨ [b,b] = [0,0] return [0,0] else

Other Operations

What about exp, sin, cos?

−Observation: Exploit monotonicity!

Interval arithmetic: Given term t in vars x1 ∈I1,...,xn ∈In

[t](I1,...,In) =

Ii if t = xi

[c,c] if t = c

[s]([t1](I1,...,In),...,[tk ](I1,...,In)) if t = s(t1,..., tk )s ∈{+,×,exp,...}

⎧

⎨⎪⎪

⎩⎪⎪

Fundamental theorem of interval arithmetic:

−For a term e, its interval extension [e] is an inclusion.

Properties of Inclusions

Width of a box: w([a1,b1] × ... × [an,bn]) = max

i∈{1,...,n}| bi − ai |

−Thin: iff w(B) = 0 ⇒ w([f](B)) = 0 An inclusion [f] of a function f is:

where: f(B) ! {f(x) | x ∈B}, [A] ! [inf A, sup A]

−Convergent: iff ∀B1 ⊇ ... ⊇Bn.limn→∞ w(Bn) = 0 ⇒ w([f](Bn)) = 0

−Monotonic: iff B1 ⊆ B2 ⇒ [f](B1) ⊆ [f](B2)

−Optimal: iff [f](B) = [f(B)]

Theorem:

If f: Rn → R does not contain any variable more than once then [f] is optimal.

Dependency Problem

Interval arithmetic forgets equality between variables:

[x − x]([-1,1]) = [x]([-1,1]) [−] [x]([-1,1])

Diff ways of writing the same exp, diff results:

[a × (b + c)](B) ⊆ [a × b + a × c)](B)

There are systematic methods for dealing with the dependency problem: [Neumaier 1990, Moore 2009]

[x × y + x]([1,2],[-1,-1]) = ([x]([1,2]) [×] [y]([-1,-1])) [+] [x]([1,2])

= [-1,1] [−] [-1,1] = [-2,2] ⊃ [0,0]

= ([1,2] [×] [-1,-1]) [+] [1,2]

= [-2,-1] [+] [1,2] = [-1,1] ⊃ [0,0]

Wrapping Effect

Rotation matrix:

At =cos(t) sin(t)−sin(t) cos(t)

⎛

⎝⎜

⎞

⎠⎟

Appears in several contexts especially ODE solving

Exponential blow-up

−Smaller t makes things worse

−In the limit box eplodes by a

−A factor of e2π ≈ 535 per revolution

Partial solution: Coordinate transfomations

Efficiency of Branch-And-Bound

Curse of dimensionality:

−Halving boxes in n-dim: 2n boxes

−Branching does not scale in problem dim

−Can one deduce more without branching?

Example

x − 99 = 0 ∧ y − 99 = 0, x ∈[0,100], y ∈[0,100]

x

y

100

100

Contractor Method (Davis 1987,Clearly 1987, Jaulin 2001)

Istead of an inclusion use a contractor:

−For given constraint with solution p ⊆ Rn and box B ⊆ Rn

−Find box C ⊆ Rn such that (B∩p) ⊆ C ⊆ B

x − 99 = 0 ∧ y − 99 = 0, x ∈[0,100], y ∈[0,100]

x

y

100

100


x

y

100

100




x − 99 = 0 ∧ y − 99 = 0, x ∈[0,100], y ∈[0,100]





x − 99 = 0 ∧ y − 99 = 0, x ∈[0,100], y ∈[0,100]

A contractor ⌢p of a constraint p is called optimal iff:

−For every box B: ⌢p(B) = (B∩p)

−In practice: Optimal up to rounding

Computing Contractors: Sum

sum ! {(x,y,z) | x + y = z}

Contractor:

Need hull of projection to 3 axes within B = Ix × Iy × Iz

i π1(sum∩ Ix × Iy × Iz ) = Ix ∩ (Iz − Iy )

i π2(sum∩ Ix × Iy × Iz ) = Iy ∩ (Iz − Ix )

i π3(sum∩ Ix × Iy × Iz ) = Iz ∩ (Ix + Iy )

i s⌢um(Ix × Iy × Iz ) = Ix ∩ (Iz − Iy ) × Iy ∩ (Iz − Ix ) × Iz ∩ (Ix + Iy )

Computing Contractors: Product

prod ! {(x,y,z) | x × y = z}

where:

Need hull of projection to 3 axes within B = Ix × Iy × Iz

i π1(prod∩ Ix × Iy × Iz ) = Ix ∩ (Iz / Iy )

i π2(prod∩ Ix × Iy × Iz ) = Iy ∩ (Iz / Ix )

i π3(prod∩ Ix × Iy × Iz ) = Iz ∩ (Ix × Iy )

i [a,b] / [c,d] = {z | ∃x∃y. y × z = x ∧ x ∈[a,b]∧ y ∈[c,d]}

Contractor:

i p⌢rod(Ix × Iy × Iz ) = Ix ∩ (Iz / Iy ) × Iy ∩ (Iz / Ix ) × Iz ∩ (Ix × Iy )

i This is not an interval! [1,1] / [-1,1]

i sin, exp: similar

Contractors: Conjunction c1∧...∧cn

C = {c1,...,cn}

while (∃c ∈C. ⌢c(B) ≠ B) B =

⌢c(B)


C = {c1,...,cn}

while (∃c ∈C. ⌢c(B) ≠ B) B =

⌢c(B)

Termination? Optimal?


C = {c1,...,cn}

while (∃c ∈C. ⌢c(B) ≠ B) B =

⌢c(B)

Termination? Optimal? x − y = 0 ∧ x + y = 0

Contractors: Constraints with General Terms

Optimal?

Example: x × y + sin(z) = 0

i x × y = v1 ∧ sin(z) = v2 ∧ v1 + v2 = v3 ∧ v3 = 0

i Primitive constraints: x + y = z, x × y = z, sin(x) = x,...

i No,but at least as good as interval arithmetic

i RSolver implementation, dReal implementation

hybrid systems modeling, analysis and control · radu grosu vienna university of technology hybrid...

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