# Homework Solution: Does DIVISION preserve regularity? Decidability? What if L1 is regular and L2 is arbitrary?…

Does DIVISION preserve regularity? Decidability? What if L1 is regular and L2 is arbitrary?
The definition of SHUFFLE operator is the same as above Define a DIVISION operator on languages as follows: L_1/L_2 = {w|w elementof sigma* and exist v elementof L_2 ReverseElement wv elementof L_1}

Suppose that M1=(Q,Σ,δ,q0,F1)M1=(Q,Σ,δ,q0,

Does DIVISION spare connectedness? Decidability? What if L1 is orderly and L2 is harsh?

The restriction of SHUFFLE operator is the selfselfcongruous as aloft Define a DIVISION operator on languages as follows: L_1/L_2 = {w|w elementof sigma* and insist v elementof L_2 ReverseElement wv elementof L_1}

Suppose that M1=(Q,Σ,δ,q0,F1)M1=(Q,Σ,δ,q0,F1) is the DFA which accepts L1L1. Then a DFA which accepts L2L2 is M2=(Q,Σ,δ,q0,F2)M2=(Q,Σ,δ,q0,F2), where:

F2={q∈Q:∃x∈L2.δ(q,x)∈F1}F2={q∈Q:∃x∈L2.δ(q,x)∈F1}

Now here’s the caveat: this scrutiny depict is referable constructive! To deflect this into an algorithm requires some cem to specify or recognise strings in L2L2, and mind, L2L2 can be anylanguage.

If L1L1 is matter munificent and L2L2 is orderly, L1/L2L1/L2 is constantly matter-munificent (scrutiny left as an use), excluding it is referable necessarily orderly. (To view why, meditate L2={ϵ}L2={ϵ}.)

EDIT I’ve normal referableiced that the restriction in the doubt is contrariant from the regular restriction of “right quotient”. The regular restriction is:

A/B={x∈Σ∗:∃y∈B.xy∈A}A/B={x∈Σ∗:∃y∈B.xy∈A}

The restriction in the doubt is:

A÷B={x∈Σ∗:∀y∈B.xy∈A}A÷B={x∈Σ∗:∀y∈B.xy∈A}

My reply us ce the regular restriction, consequently these are rule results. The replys are congruous ce the restriction in the doubt, bar that the view is partially contrariant. (View David Smith’s criticise.)