How can I construct a context free grammar for the following language:
L = {a^l b^m c^n d^p | l+n==m+p; l,m,n,p >=1}
I started by attempting:
S -> abcd | aAbBcd | abcCdD | aAbcdD | AabBcCd
and then A = something else... but I couldn't get this working.
.
I was wondering how can we remember how many c's shud be increased for the no. of b's increased?
For example:
string : abbccd
The grammar is :
S1 -> a S1 d | S2
S2 -> S3 S4
S3 -> a S3 b | epsilon
S4 -> S5 S6
S5 -> b S5 c | epsilon
S6 -> c S6 d | epsilon
Rule 1 adds equal number of a's and d's.
Rule 3 adds equal number of a's and b's.
Rule 5 adds equal number of b's and c's.
Rule 6 adds equal number of c's and d's
The rules also ensure that the ordering of the alphabets are maintained according to the language given.
How's about this:
S1 -> a S2 d # at least one a and d
S2 -> a S2 d
S2 -> S3 S4 # no more d, split into ab and bc parts
S2 -> S4 S5 # no more a, split into bc and cd parts
S3 -> a S3 b
S3 -> # already ensured at least one a and b
S4 -> b S4 c
S4 -> b c # at least one b and c
S5 -> c S5 d
S5 -> # already ensured at least one c and d
The key to this is how you group... (i.e. "parts" rather than non-terminals.)
来源：https://stackoverflow.com/questions/12930010/construct-context-free-grammar