6 - Parikh’s Theorem Flashcards
draw the inclusions and how we prove it for pharik, SLS, presburger
everything is the same:
Presburger-definable = SLS (Ginsbur & Spanier)
SLS = Ψ(REG) (easy)
Ψ(REG) ⊆ Ψ(CF) (by definition, REG ⊆ CF)
Ψ(CF) ⊆ SLS (Parikh)
phariks theorem
Consider a CFG G. Ψ(L(G)) is Semi-linear
Parikh image of G is a finite union of the parik images of subtrees (something like that)
reachability in petri nets is decidable?
yes (Mayr ‘81) EXPACE-hard
NP-complete for communication free Petri nets (Esparza ‘98)
Verma, Seidl
Ψ(CF) ⊆ Presburguer-Definable
2 steps:
1) Translate given CFG G into a communication-free petri net Ng
2) Turn the communication-free Petri net Ng into an ∃PA formula φg
Construction for part (1)
We are interested in the parikh image, therefore the petri net will have 1 marking on initial state, and 1 transition for each expansion rule, with weights. Ex:
S-> aSa | bSb | c
yelds 1 p for each rule, 1 transition for each symbol, weight 2 for a, for example
Construction for part (2) 2 constraints 3 kinds of subformulas 1) For each non-terminal A ∈ V, tehere is an equation 2) Ensure Xa is consistent with yp 3) Express requirement b
2-counter machines (important)
Automaton with finite set of counters C
Σ = {inc,dec,zero} x C
Counter machines are petri nets that can actively test for zero
Theorem (Minsky ‘67):
2-counter machines are turing-complete
Theorem (Ibarra ‘79)
For any n-counter machine with r-reversals, there is a n*(1+r/2)-counter machine which counters does only 1 reversal each
closed under intersection ??
not sure if this question is for parikh…
direciton 1:
- slsl can have presb forumla
direction 2:
- quant el, slsl for base case and the use intersection
intersection
2 parts
rought part
a and b represent the sets that they do (escape)