Is L = { αγβ | α, β ∈ {a, b}* & γ ∈ {a, b}*1{a, b}* & |α| = |β| ≥ |γ| } context-free?
Multi tool use
$begingroup$
L = { αγβ | α, β ∈ {a, b}* & γ ∈ {a, b}1{a, b} & |α| = |β| ≥ |γ| }.
So i need to find a contex-free grammar for L ?
The thing is, i am not sure how to start.
I belive I need to simplify the language.
Any tips how can I do that, so the grammar would come easyly ?
context-free-grammar
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
$endgroup$
add a comment |
$begingroup$
L = { αγβ | α, β ∈ {a, b}* & γ ∈ {a, b}1{a, b} & |α| = |β| ≥ |γ| }.
So i need to find a contex-free grammar for L ?
The thing is, i am not sure how to start.
I belive I need to simplify the language.
Any tips how can I do that, so the grammar would come easyly ?
context-free-grammar
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
$endgroup$
add a comment |
$begingroup$
L = { αγβ | α, β ∈ {a, b}* & γ ∈ {a, b}1{a, b} & |α| = |β| ≥ |γ| }.
So i need to find a contex-free grammar for L ?
The thing is, i am not sure how to start.
I belive I need to simplify the language.
Any tips how can I do that, so the grammar would come easyly ?
context-free-grammar
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
$endgroup$
L = { αγβ | α, β ∈ {a, b}* & γ ∈ {a, b}1{a, b} & |α| = |β| ≥ |γ| }.
So i need to find a contex-free grammar for L ?
The thing is, i am not sure how to start.
I belive I need to simplify the language.
Any tips how can I do that, so the grammar would come easyly ?
context-free-grammar
context-free-grammar
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
edited Jan 17 at 18:21
Paul Keseru
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
asked Jan 17 at 17:55
Paul KeseruPaul Keseru
475
475
add a comment |
add a comment |
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