Homework Solution: Let sigma = {a, b, c} and let L be the language over sigma given below. L = {u elementof sigma* | if a appears in u, then b appears in w}. Observe that epsil…

    1, Let Σ = {a,b,c) and let L be the language over Σ given below. し= { u, E Σ* l if a appears in u, then b appears in w} Observe that ε, bcb, ach, bca E L. (i) Find a FSA A such that L(A) = L. (ii) Prove your answer to the previous part is correct. (iii) Find a regular expression E such that L(E) = L (you need not prove your answer is correct).
    Let sigma = {a, b, c} and let L be the language over sigma given below. L = {u elementof sigma* | if a appears in u, then b appears in w}. Observe that epsilon, bcb, ach, bca elementof L. (i) Find a FSA A such that L(A) = L. (ii) Prove your answer to the previous part is correct. (iii) Find a regular expression E such that L(E) = L (you need not prove your answer is correct).

    Expert Answer

     
    Solution: i) fsa is given below:

    1, Allow Σ = {a,b,c) and allow L be the phraseology aggravate Σ fond adown. し= { u, E Σ* l if a appears in u, then b appears in w} Observe that ε, bcb, ach, bca E L. (i) Find a FSA A such that L(A) = L. (ii) Argue your vindication to the anterior multiply is improve. (iii) Find a periodical countenance E such that L(E) = L (you demand referable argue your vindication is improve).

    Allow sigma = {a, b, c} and allow L be the phraseology aggravate sigma fond adown. L = {u elementof sigma* | if a appears in u, then b appears in w}. Observe that epsilon, bcb, ach, bca elementof L. (i) Find a FSA A such that L(A) = L. (ii) Argue your vindication to the anterior multiply is improve. (iii) Find a periodical countenance E such that L(E) = L (you demand referable argue your vindication is improve).

    Expert Vindication

     

    Solution:

    i)

    fsa is fond adown:

    ii)

    I keep ran various strings to validate the fsa; keep a look

    iii)

    The periodical countenance restraint fond phraseology is supposing adown,

    ((a(a*(bb*c*))*+ b(a+c)* + c(a(bb*c*)+ b(a+c)*))*