Right Quotient

Question

$L_{1}=\left\{a^{n}b^{n}c^{n}|n>=0\right\}\\
L_{2}=\left\{b^{i}c^{j}|i,j>=0\right\}\\
Find \ out\ L_{1}/L_{2}$

Comments

L1 = $\left\{\epsilon,abc,aabbcc,aaabbbccc........\right\}$

L2 = $\left\{\epsilon,b,c,bc,bbcc,bbbccc........\right\}$

Case i)

Now take a string in L1 say aabbcc and L2 say c

L1/L2= aabbc

Similarly aaabbbcc,aaaabbbbccc.......$\Rightarrow a^{n} b^{n} c^{n-1}$ if L2 is $\epsilon$ everything is accepted

$a^{l} b^{m} c{n} | l=m>=n $              ---1

Case ii)

Now take a string in L1 say aabbcc and L2 say bbcc or bcc

L1/L2= aa or aab

Similarly aa,aaa,aaa .....,aab,aaabb.......$\Rightarrow a^{n}  \ or \ a^{n+1} b^{n} $

$ \  a^{l} \ or \ a^{l} b^{m} c^{n} | l>=m  \ and\ n=0\  $               ------ 2

Can someone verify this ??

Answer 1

• L₁= {aⁿbⁿcⁿ | >=0}
• L₂ = {b*c*}

L₁ / L₂ =  {$\epsilon$ , abc/$\epsilon$ , abc /b , abc/c , aabbcc/ $\epsilon$ , aabbcc/b, aabbcc/bb, aabbcc/c, aabbcc/cc, aabbcc/bbc, aabbcc/bbcc........}​​

 = {$\epsilon$ abc, ab, aabbcc, aabbc, aabb, aab, aa......}

= {a* , aⁿbⁿcⁿ, aⁿbⁿ , aⁿbⁿc^n-1, ....}

= {aⁿbⁿcⁿ| >=0 } - {bⁿcⁿ| >=0} -{cⁿ| >=0}$\cup$ $\epsilon$ $\cup${ aⁿbⁿ| >=0 } $\cup$ {aⁿ| >=0} $\cup$ many more

Comments

@nitish L1/L2 = {a^xb^yc^z | y<=x, z<=y, x,y,z>=0} it's not correct because L1/L2 can accept aaaabbbcc
though this string can't be generated by L1 after apending any string from L2.

See my updated solution for correct answer. 

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@Anu007 sir,

i was incorrect, but now i think correct RE should be

L1/L2 = {aⁿbⁿc^m | m<=n , m,n>=0} ⋃ {aⁿb^m |m<=n, m,n>=0}

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hmmm leave that gate not ask to write it.

Answer 2

L = L1/L2

L = {x | such that xy belongs to L1 and y belongs to L2}

It means L will consider those strings from the language L1 which have suffix from the language L2, and L removes the suffix part and keep it in the language.

$L1 = {a^nb^nc^n | n>=0}$
$L2 = {b^*c^*}$

L1/L2 = {$a^n b^n c^n$ | n>=0} U {$a^n$ | n>=1} U {$a^n b^n$ | n>=1} U {$a^nb^nc^*$}

Strings in L1/L2 = eps, a, ab, abc, aabbc, aabbcc etc..

Comments

What about L containing a, aa, ab, aabbc ???

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first my answer was a*b*c* later I changed it to a^nb^nc^n, i think correct answer is a*b*c*.

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check now!

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Hello manu your answer is almost correct , but in last part i think

it should be {a^qb^qc^r | q>=1 , 0<=r<q}

like abc/c gives ab , aabbcc/c gives aabbc , aaabbbccc/c gives aaabbbcc and aaabbbccc/cc gives aaabbbc , hope you got..

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@Rupendra

i think, it should be correct, check it please..

L1/L2 = {a^xb^yc^z | y<=x, z<=y, x,y,z>=0}

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Exactly nitish.

Even i forget to include in my last comment that strings like aaab , aaaabb are also supported.

Which you take cared. As i'm able to see , it's perfect.

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thanks @nitish and Ruperndra, I updated my solution.

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@Rupendra I updated my solution last one was incorrect. as 
L1/L2 = {a^xb^yc^z | y<=x, z<=y, x,y,z>=0} it's not correct because L1/L2 can accept aaaabbbcc 
though this string can't be generated by L1 after apending any string from L2.

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You're right manu but your answer is yet incorrect.

It's accepting strings like abcccc ....i mean aⁿb^bc^* is wrong..

your language is not accepting strings like aaabbbc , aaabb , which are also part of language.

Let me try to write it.

L={aⁿbⁿcⁿ |n>=0} U {aⁿ |n>=1} U {aⁿbⁿc^m | n>=1,0<=m<=n}U{aⁿb^m| n>=1 ,0<=m<=n}

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what about $\left \{a^{k}b^ic^j | (j\leq k) \cap (i\leq k)\right \}$?

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wrong.

what's about relation b/w i and j?

like aaabcc is accepted according to your language which is wrong..

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Yup, missed that. Thanks. I think ur last comment covers all the cases.

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@Manu Thakur

mine was incorrect but yous is also incorrect,...

now final correct will be

L1/L2 = {aⁿbⁿc^m | m<=n , m,n>=0} $\bigcup$ {aⁿb^m |m<=n, m,n>=0}