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Given two CFG's G1 = (V1, T1, P1, S1) for language L1(G1)
G2 = (V2, T2, P2, S2) for language L2(G2)
Rename variables until V1 intersect V2 is Phi.
We can easily get CFG's for the following languages:
L1 union L2 = L3(G3)
G3 = (V1 union V2 union {S3}, T1 union T2, P1+P2+P3, S3)
S3 -> S1 | S2
L1 concatenated L2 = L4(G4)
G4 = (V1 union V2 union {S4}, T1 union T2, P1+P2+P4, S4)
S4 -> S1S2
L1 star = L5(G5)
G5 = (V1 union V2 union {S5}, T1 union T2, P1+P2+P5, S5)
S5 -> S5S1 | epsilon
L2 substituted for terminal "a" in L1 = L6(G6)
G6 = (V1 union V2, T1-{a} union T2, P6+P2, S1)
P6 is P1 with every occurrence of "a" replaced with S2.
Notice that L1 intersect L2 may not be a CFG.
i i j i j i
Example: L1={a b c | i,j>0} L2={a b c | i,j>0} are CFG's
i i i
but L1 intersect L2 = {a b c | i>0} which is not a CFG.
The complement of L1 may not be a CFG.
The difference of two CFG's may not be a CFG.
The intersection of a Context Free Language with a Regular Language
is a Context Free Language.
As a supplement, the following shows how to take a CFG and possibly
use 'yacc' or 'bison' to build a parser for the grammar.
The steps are to create a file xxx.y that includes the grammar,
and a file that is a main program that runs the parser.
An example grammar is coded in acfg.y
and a sample main program is coded in yymain.c
One possible set of commands to run the sample is:
bison acfg.y
gcc yymain.c acfg.tab.c
a.out
abaa
The output from this run shows the input being read and the rules
being applied:
read a
read b
read a
A -> ba
read a
read end of line \n converted to 0
S -> a
S -> aAS
accepted
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