+1 vote
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Consider the following CFG.
S → aSa | bSb | a | b | ε
For the above CFG, the total number of strings generated whose length is less than or equal to 8 [exclude the empty string] is _____________.

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0

You could do it this way:

The number of length one strings (L1) = 2.
The number of length two strings (L2) = 2.

These can be trivially found out.

Now for L3, we fix the first and last positions as a_a or b_b. Now, in the blank there has to be a string of length 1, for which we have two options. Therefore, the total number of strings are (a_a)*2 + (b_b)*2 = 4.

We can extend this logic for all the further substrings.

L4 = a__a + b__b where the blank will contain strings of length 2.
L5 =  a___a + b___b where the blank will contain strings of length 3.
L6 = a____a + b____b where the blank will contain strings of length 4 and so on.

When you add all these, you should get 60.
by Loyal (7k points)
0

Good derivation bro....

To make it complete add that

(n) = T(2)*T(n-2) for n>2

## Given grammar is produce palindrome of string. by Boss (36.9k points)