Lecture 4:

Teachers: Dietmar Berwanger

Construction of a winning strategy in a finite parity game

  • pick position $q$ of minium priority (say even)
  • Induction Hypothesis: $V \backslash Attr^0(\lbrace q \rbrace)$ splits into winning regions $(W_0’, W_1’)$
  • $W_1’ ⊆ W_1$

  • How about $W_0’ ∪ Attr^0(\lbrace q\rbrace)$?

    • $q ∈ V_0$ and $qE ⊆ W_1’$ OR $q ∈ V_1$ and $qE ∩ W_1’ ≠ ∅$:

      By IH: split $V \backslash (Attr^1(\lbrace q \rbrace) ∪ W_1’)$ into $(W_0’’, W_1’’)$.

      In this case: $W_0’ = W_0, \; W_1 = Attr^1(\lbrace q \rbrace) ∪ W_1’ ∪ W_1’’$: indeed, it suffices to show:

      • $W_0’ ⊆ W_0$
      • $Attr^1(\lbrace q \rbrace) ∪ W_1’ ∪ W_1’’ ⊆ W_1$

    • $q ∈ V_0$ and $qE ∩ (W_0’ ∪ Attr^0(\lbrace q\rbrace)) ≠ ∅$ OR $q ∈ V_1$ and $qE ⊆ W_0’ ∪ Attr^0(\lbrace q\rbrace)$: we’re done, P0 wins

Thm (Parity-Decidability): “Given $G, v$ decide whether P0 has winning strategy from $v$” is in NP.

Given a memoryless strategy $σ:V_0 ⟶ V$ we can decide in PTIME whether it is winning.

$G_{ σ} ≝ (V, E_σ, V_0, V_1)$ with $E_σ$ st
  • for $v ∈ V_1$, $vE_σ = vE$
  • for $v ∈ V_0$, $vE_σ = σ(v)$
$σ$ winning ⟺ in $G_{ σ}$, all reachable from $v_0$ cycles have an even minimum.

And this property can be decided in time $O\Big(\frac {\vert Ω(V) \vert (\vert V \vert + \vert E \vert)} 2\Big)$ for a graph $G ≝ (V, E, Ω)$:

  • Idea: for odd priorities: $1, 3, 5, …, m$:

    consider

    • \[G_k ≝ G_{| \lbrace v ∈ V \; \mid \; Ω(v) ≥ k\rbrace}\]
    • decompose $G_k$ into strong connected components (SCC) (Tarjan algorithm)
    • check existence of a SCC “reachable from $v_0$ in $G$” that contains $w$ with $Ω(w) = k$: if yes, reject

    else: repeat with another value of $k$

\[ParityDcc ∈ NP ∩ coNP\]

Application

Synthesis problem:

  • input: $\overline{x} ∈ Σ^ω$
  • output: $\overline{y} ∈ \Pi^ω$

Specification: Relation

\[R ⊆ (Σ × \Pi)^ω\]

Question: contruct a function $f: Σ^\ast → \Pi$ st

\[(x_1 x_2 ⋯, f(ε) f(x_1) f(x_1 x_2) ⋯) ∈ R\]

Indeed: Here $R = ℒ(A)$ (regular relation, when representable by a 2-tape automaton) for

$𝒜$ a 2-tape automaton:
\[(Q, Σ × \Pi, q_0, δ: Q × (Σ × \Pi) → Q, Acc)\]

Acceptability condition: Reach, Büchi (better for non-determinism), Parity

Variants: decide whether a (possibly computable/finite-state/automaton) solution $f$ exists

Regular function:

\((Q, \underbrace{Σ}_{\text{input}}, \underbrace{\Pi}_{\text{output}}, q_0, δ: Q × Σ → Q × \Pi)\) or Mealy machine: \((Q, Σ, \Pi, q_0, δ: Q × Σ → Q, \underbrace{β: Q × Σ → \Pi}_{\text{generation of output}})\) or Moore machine: \((Q, Σ, \Pi, q_0, δ: Q × Σ → Q, ζ: Q → \Pi)\)

Extending transition output for

\(δ: Q × Σ^\ast → Q\\ δ(q,ε) = q\\ δ(q, w a) = δ(δ(q,w),a)\) where $q ∈ Q, a ∈ Σ, w ∈ Σ^\ast$

Similarly for the output function:

\[β: Q × Σ^\ast → \Pi\\ β(q,ε) = q\\ β(q, w a) = β(δ(q,w),a)\]

Back to the problem: given $Σ, \Pi, 𝒜$ with $ℒ(A) = Spec ⊆ (Σ × \Pi)^ω$.

We want $ℳ ≝ (M, Σ × \Pi, m_0, ν: M × E → M, β: M× E → \Pi)$ st

\[x_1, x_2 …, β(q_0, x_1), β(q_0, x_1 x_2) … β(q_0, x_1 ⋯ x_t) ∈ ℒ(𝒜)\] 
  digraph {
    rankdir=LR;
    invis1[shape=none, label=""];
    invis2[shape=rectangle, label=""];
    invis3[shape=none, label=""];
    invis1 -> invis2[label="Σ"];
    invis2 -> invis3[label="Π"];
  }

  1. Represent the system as a parity game: $G ≝ (V, V_0, E, Ω), v_0$

    with

    \[V = Q × (Σ ∪ \Pi) ∪ \lbrace q_0\rbrace\\ V_0 = Q × Σ \qquad V_1 = Q × \Pi ∪ \lbrace q_0\rbrace\]

    Moves $E$:

    \[E ≝ \lbrace (⟨q, a⟩, \big\langle δ(q, ⟨a,b⟩), b\big\rangle) \; \mid \; q∈ Q, a ∈ Σ, b ∈ \Pi \rbrace \\ ∪ \lbrace (⟨q, b⟩, \langle q, a\rangle) \; \mid \; q∈ Q, a ∈ Σ, b ∈ \Pi \rbrace\\ ∪ \lbrace (q_0, ⟨q_0, a⟩) \; \mid \; a ∈ Σ\rbrace\]

    Priorities:

    \[Ω^G(q, c) = Ω^𝒜(q) \qquad ∀ q ∈ Q, c ∈ Σ ∪ \Pi\]

  2. Finite parity game:

    Solve $G, q_0$: memoryless winning strategy

    • If P1 has a winning strategy: $τ: q_0 ⟼ τ(q_0) = x ∈ Σ$. For every $τ: (q, b) ⟼ τ(q,b) ∈ Σ$

    • If P0 has a winning strategy $σ: Q×Σ ⟶ \Pi$, then \(ν(q, a) = δ(q, (a, σ(q,a)))\\ β(q,a) = σ(q,a) \qquad ∀ q ∈ Q, a∈ Σ\)

      Then every execution $(\overline{x}, \overline{y}) ∈ (Σ × \Pi)^ω$ that follows $ℳ$ is accepted by $𝒜$

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