+1 vote
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Let $L = \{\text{madeeasy2016}\}$ over $\Sigma = \{m,a,d,e,s,y,2,0,1,6\},$ $L_1=\text{prefix}(L)$ and $L_2= L_1 \Big /{} _{ \Sigma ^*}$. The number of strings in $L_2$ (assume $L_1$ and $L_2$ do not include empty string) are ______.

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Given L1=prefix(L) therefore,

Now L2=L1/∑* ={ x | xy ∈ L1 for some y∈ ∑* } therefore

L2= {a,ad,d,ade,e,de,adee,ee,dee.......} many string but they have given some explanation that L1/∑*=L1 only ....which one  is correct?

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78?
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Ans given is just 12 ...what is ur approach?
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sorry its 12 only....

You have 13 prefixes right...

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Any reason to choose L2=L1/∑*=L1 ,Plz provide some source if possible...

$L =\{\text{madeeasy2016}\}$

$L_1 = \text{prefix}(L)$, excluding empty string

$L_1= \{m,ma,mad,made,madee,madeea,madeeas,madeeasy,\\ \qquad \quad madeeasy2,madeeasy20,madeeasy201,madeeasy2016\}$

In general Right Quotient of $L_1$ with $L_2$ means $L_1\Big /L_2=\{x\;:\;xy\in L_1 \;\text{for some}\;y \in L_2\}$

so In our problem

$L_2=L_1\Big /\Sigma^*=\{x\;:\;xy\in L_1 \;\text{for some}\;y \in \Sigma^*\}$

for $y=\epsilon$, $L_2=L_1 \Big / \Sigma^* = L_1$

Number of strings in $L_2 =$ Number of strings in $L_1 =12$

by Veteran (56.8k points)
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Lets explore the quotient operator:

L1/L2 = ?

? is every 'x' that satisfies x.y L1 where y ∈ L2

Now, consider any arbitrary language L. What is L/∑* ?

So, take...      x.y ∈ L     as 'y' can range from epsilon to every string possible, you will find that x is actually a prefix for L. How? Say y='b', and L contains some string like aabbb. So x becomes 'aabb'. So, if you consider all possible x (which is nothing but the quotient) , you will get prefix(L).

Now, coming to your question, L1/∑* is nothing but prefix(L1). Thus, L2=Prefix(L1) which inturn is prefix for given string. So, just calculate prefixes, that's it :)

by Active (2.2k points)