00:01
So, the first thing is for the equation context -free grammar for palin.
00:19
So, the grammar for generating palin counts over the alphabet a, b can be defined as a string, a beam, first is s, second is s, a, third is s, d, fourth f, a, s, a, and fifth f, d, s, d.
00:53
So, here f stands for start symbol, and rule one allows an empty string, and rule two and three generate single character, which is a and b.
01:15
Rule four and five, we are generating palindrom by surrounding an existing palindrom with the same character on both ends, here and this one, a, a, b, b.
01:32
Just for the first step, and the second is terminal and non -terminal symbols.
01:55
Here terminal symbol is a, b, and non -terminal symbol is a.
02:12
So, we start by passing the palindrom a, b, a, a, a, b, a.
02:29
So, we pass the palindrom a, b, a, a, b, a and we are using the defined grammar, in which we create first is s, then start, then is a, s, a.
02:47
This is the rule for as we said before, a, b, s, b, a.
02:57
This is the big s sign...