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workin progress The M in CoPaR From Partition Refjnement to Minimization Hans-Peter Deifel Oberseminar 09.06.2020 1 / 26 The M in CoPaR From Partition Refjnement to Minimization Hans-Peter Deifel Oberseminar 09.06.2020 1 / 26 workin

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The M in CoPaR

From Partition Refjnement to Minimization Hans-Peter Deifel Oberseminar 09.06.2020

workin progress

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The M in CoPaR

From Partition Refjnement to Minimization Hans-Peter Deifel Oberseminar 09.06.2020

workin progress

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CoPaR Coalgebraic Partition Refjner

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Partition Refjnement

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Partition Refjnement

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Partition Refjnement

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Partition Refjnement

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Partition Refjnement

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Partition Refjnement

Difgerent partition refjnement algorithms: Pajge-Tarjan Hopcroft Markov-Chain lumping Color refjnement ⇒ Used in countless minimization tools

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Partition Refjnement

Difgerent partition refjnement algorithms: Pajge-Tarjan Hopcroft Markov-Chain lumping Color refjnement ⇒ Used in countless minimization tools

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CoPaR Coalgebraic Partition Refjner

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CoM Coalgebraic Minimizer

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Coalgebras X FX c

states successor map type functor

Type Functor Set Set

Labelled Transition Systems Deterministic Automata Markov Chains

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Coalgebras X FX c

states successor map type functor

Type Functor F : Set → Set FX = P(A × X)

Labelled Transition Systems

2 × XA

Deterministic Automata

R(X)

Markov Chains

. . .

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Behavioural Equivalence

X FX Y FY c

states successor map type functor

d h

homomorphism

Fh Identifjed by Coalgebra Homomorphism FX = P(A × X)

Bisimilarity

2 × XA

Language Equivalence

R(X)

Weighted Bisimilarity

. . .

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Coalgebraic Partition Refjnement

Generic Algorithm Functor F F-Coalgebra Refjnement Interface for F Partition

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Functor Encoding

Encoding for Functor F

Abstract Type A of labels Encoding map ♭X : FX → B Bags (A × X)

Example: P(−)

A = 1 ♭X({x1, x2, x3}) = {(∗, x1), (∗, x2), (∗, x3)}

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Functor Encoding

Encoding for Functor F

Abstract Type A of labels Encoding map ♭X : FX → B Bags (A × X)

Example: R(−)

A = R ♭X(f) = {(r, x) | x ∈ X, f(x) = r = 0}

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Functor Encoding

Encoding for Functor F

Abstract Type A of labels Encoding map ♭X : FX → B Bags (A × X)

Not Natural

Let f ∈ R2, f(0) = 0.5, f(1) = 0.5 f {(0.5, 0), (0.5, 1)} (∗ → 1.0) {(0.5, 1), (0.5, 1)}

♭2 R(!) B(id,!) ♭1/

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Functor Encoding

Encoding for Functor F

Abstract Type A of labels Encoding map ♭X : FX → B Bags (A × X)

Refjnement Interface

Type W (abstract, could be ints, reals, trees, …) init: F1 × BA → W update: BA × W → W × F3 × W

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Syntax

Input

DX q1: {q2: 0.5, q3: 0.5} q2: {q1: 0.5, q3: 0.5} q3: {q3: 1}

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Syntax

Input

DX q1: {q2: 0.5, q3: 0.5} q2: {q1: 0.5, q3: 0.5} q3: {q3: 1}

Output

Block 0: q1, q2, q3

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Syntax

Input

DX q1: {q2: 0.5, q3: 0.5} q2: {q1: 0.5, q3: 0.5} q3: {q3: 1}

Cooler Output

DX q1: {q1: 1}

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Minimization

Two tasks: Computing the (encoding of) the quotient coalgebra Removing unreachable states

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Minimization

Two tasks: Computing the (encoding of) the quotient coalgebra Removing unreachable states

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Quotient Encoding: Goal

X FX B(A × X) Y FY B(A × Y )

c d ♭X ♭Y q Fq ?

quotient map input coalgebra

  • utput

coalgebra encoding

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Solution: Minimization Interface

merge: B(A) → B(A) FX B(A) F2 B(A)

filS · ♭ FχS merge fil{1} · ♭

and B(A) 1 B(A)

merge {} {}

filS(t)(a) =

x∈S t(a, x)

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Minimization Interface: Examples

Powerset

merge(ℓ)(∗) = min(1, ℓ(∗))

Monoid-Valued

merge(ℓ) =

  • {Σℓ}

Σℓ = 0 {}

  • therwise

Polynomial

merge = id

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Victory?

FX B(A × X) B(A × Y ) B(A)(Y ) FY B(A × Y ) B(A)(Y )

♭ B(A×q) group Fq merge(Y ) ♭ ungroup

curry ·B(swap) monoid-valued for B-monoid 15 / 26

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Encoding Assumption

Nope! We additionally need the following assumption on ♭: FX B(A × X) B(A) F2 B(A × 2)

♭X Fχ{x} fil{x} ♭2 fil{1}

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Encoding Assumption: Counterexample

An encoding for Powerset

A = 1 + 1 ♭2({x1, . . . , xn}) = ({(inj1 ∗, x1) . . . , (inj1 ∗, xn)}) ♭X=2({x1, . . . , xn}) = ({(inj2 ∗, x1) . . . , (inj2 ∗, xn)}) ⇒ Lawful merge possible! But: Ruled out by assumption

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Encoding Assumption: Counterexample

An encoding for Powerset

A = 1 + 1 ♭2({x1, . . . , xn}) = ({(inj1 ∗, x1) . . . , (inj1 ∗, xn)}) ♭X=2({x1, . . . , xn}) = ({(inj2 ∗, x1) . . . , (inj2 ∗, xn)}) ⇒ Lawful merge possible! But: Ruled out by assumption

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Victory!

FX B(A × X) B(A × Y ) B(A)(Y ) FY B(A × Y ) B(A)(Y )

♭ Fq B(A×q) group merge(Y ) ♭ ungroup

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Victory!

FX B(A × X) B(A × Y ) B(A)(Y ) FY B(A × Y ) B(A)(Y )

♭ Fq B(A×q) group merge(Y ) ♭ ungroup

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Minimization

Two tasks: Computing the (encoding of) the quotient coalgebra Removing unreachable states

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Minimization

Two tasks: Computing the (encoding of) the quotient coalgebra Removing unreachable states

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Reachability: Problems

“Two tasks” all the way down: Pointed Coalgebras Computing the Reachable Subcoalgebra

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Reachability: Problems

“Two tasks” all the way down: Pointed Coalgebras Computing the Reachable Subcoalgebra

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Reachability: Problems

“Two tasks” all the way down: Pointed Coalgebras ⇒ Simple matter of programming™ Computing the Reachable Subcoalgebra

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Reachability: Problems

“Two tasks” all the way down: Pointed Coalgebras ⇒ Simple matter of programming™ Computing the Reachable Subcoalgebra

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Computing the Reachable Subcoalgebra

In Set: Standard graph search on canonical graph

Canonical Graph

does not factorize through For coalgebra:

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Computing the Reachable Subcoalgebra

In Set: Standard graph search on canonical graph

Canonical Graph

τX : FX → PX τX(t) = {x ∈ X | 1 t − → FX does not factorize through F(X \ {x}) Fi − → FX} For coalgebra: C

c

− → FC

τC

− → PC

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Goal

We want: Canonical graph of original coalgebra and canonical graph of encoded coalgebra to be the same! C FC PC B(A × C)

c τ ♭ τ

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Direction: τ(t) ⊆ τ(♭(t))

Questment: Encoding contains canonical graph? Should be true™

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Direction: τ(t) ⊆ τ(♭(t))

Questment: Encoding contains canonical graph? ⇒ Should be true™

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Counterexample for τ(t) ⊇ τ(♭(t))

Functor: R(−)

Encoded Coalgebra

a b c d 0.5 0.5

Canonical Graph

a b d

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Requirement: Sub-Naturality

Encoding must be sub-natural: X FX B(A × X) Y FY B(A × Y )

f ♭X Ff B(A×f) ♭Y

now works!

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Requirement: Sub-Naturality

Encoding must be sub-natural: X FX B(A × X) Y FY B(A × Y )

f ♭X Ff B(A×f) ♭Y

⇒ now works!

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Question: Connection?

Assumption 1

FX B(A × X) B(A) F2 B(A × 2)

♭X Fχ{x} fil{x} ♭2 fil{1}

Assumption 2

X FX B(A × X) Y FY B(A × Y )

f ♭X Ff B(A×f) ♭Y

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