I just took my midterm but couldn't answer this question.
Can someone please give a couple of examples of the language and construct a grammar for the language or at least show me how i will go about it?
Also how to write grammar for L
:
L = {an bm | n,m = 0,1,2,..., n <= 2m } ?
Thanks in advance.
Before read my this answer you should read first: Tips for creating Context free grammars.
What is you language L = {an bm | n,m = 0,1,2,..., n <= 2m } description?
Language description:
The language L is consist of set of all strings in which symbols a
followed by symbols b
, where number of symbol b
are more than or equals to half of number of a
's.
To understand more clearly:
In pattern an bm, first symbols a
come then symbol b
. total number of a
's is n
and number of b
's is m
. The inequality equation says about relation between n
and m
. To understand the equation:
given: n <= 2m
=> n/2 <= m means `m` should be = or > then n/2
=> numberOf(b) >= numberOf(a)/2 ...eq-1
So inequality of n and m says:
numberOf(b) must be more than or equals to half of numberOf(a)
Some example strings in L:
b numberOf(a)=0 and numberOf(b)=1 this satisfy eq-1
bb numberOf(a)=0 and numberOf(b)=2 this satisfy eq-1
So in language string any number of b
are possible without a
's. (any string of b) because any number is greater then zero (0/2 = 0).
Other examples:
m n
--------------
ab numberOf(a)=1 and numberOf(b)=1 > 1/2
abb numberOf(a)=1 and numberOf(b)=2 > 1/2
abbb numberOf(a)=1 and numberOf(b)=3 > 1/2
aabb numberOf(a)=2 and numberOf(b)=2 > 2/2 = 1
aaabb numberOf(a)=3 and numberOf(b)=2 > 3/2 = 1.5
aaaabb numberOf(a)=4 and numberOf(b)=2 = 4/2 = 2
Points to be note:
all above strings are possible because number of b
's are either equal(=) to half of the number of a
or more (>).
and interesting point to notice is that total a
's can also be more then number of b
's, but not too much. Whereas number of b
's can be more then number of a
's by any number of times.
Two more important case are:
only a
as a string not possible.
note: null ^
string is also allowed because in ^
, numberOf(a) = numberOf(b) = 0
that satisfy equation.
At once, it look that writing grammar is tough but really not...
According to language description, we need following kinds of rules:
rule 1: To generate ^
null string.
N --> ^
rule 2: To generate any number of b
B --> bB | b
Rule 3: to generate a
's:
(1) Remember you can't generate too many a
's without generating b
's.
(2) Because b
's are more then = to half of a
's; you need to generate one b
for every alternate a
(3) Only a
as a string not possible so for first (odd) alternative you need to add b
with an a
(4) Whereas for even alternative you can discard to add b
(but not compulsory)
So you overall grammar:
S --> ^ | A | B
B --> bB | b
A --> aCB | aAB | ^
C --> aA | ^
here S
is start Variable.
In the above grammar rules you may have confusion in A --> aCB | aAB | ^
, so below is my explanation:
A --> aCB | aAB | ^
^_____^
for second alternative a
C --> aA <== to discard `b`
and aAB to keep b
let us we generate some strings in language using this grammar rules, I am writing Left most derivation to avoid explanation.
ab S --> A --> aCB --> aB --> ab
abb S --> A --> aCB --> aB --> abB --> abb
abbb S --> A --> aCB --> aB --> abB --> abB --> abbB --> abbb
aabb S --> A --> aAB --> aaABB --> aaBB --> aabB --> aabb
aaabb S --> A --> aCB --> aaAB --> aaaABB --> aaaBB --> aaabB --> aaabb
aaaabb S --> A --> aCB --> aaAB --> aaaCBB --> aaaaABB --> aaaaBB
--> aaaabB
--> aaaabb
One more for non-member string:
according to language a5 b2 = aaaaabb
is not possible. because 2 >= 5/2 = 2.5 ==> 2 >= 2.5 inequality fails. So we can't generate this string using grammar too. I try to show below:
In our grammar to generate extra a
's we have to use C variable.
S --> A
--> aCB
--> aaAB
--> aa aCB B
--> aaa aA BB
--> aaaa aCB BB
---
^
here with first `a` I have to put a `b` too
While my answer is done but I think you can change A
's rules like:
A --> aCB | A | ^
Give it a Try!!
EDIT:
as @us2012 commented: It would seem to me that then, S -> ^ | ab | aaSb | Sb
would be a simpler description. I feel this question would be good for OP and other also.
OP's language:
L = {an bm | n,m = 0,1,2,..., n <= 2m}.
@us2012's Grammar:
S -> ^ | ab | aaSb | Sb
@us2012's question:
Whether this grammar also generates language L?
Answer is Yes!
The inequality in language between number of a
's = n
and number of b
= m is n =< 2m
We can also understand as:
n =< 2m
that is
numberOf(a) = < twice of numberOf(b)
And In grammar, when even we add one or two a
's we also add one b
. So ultimately number of a
can't be more then twice of number of b
.
Grammar also have rules to generate. any numbers of b
's and null ^
strings.
So the simplified Grammar provided by @us2012 is CORRECT and also generates language L exactly.
Notice: The first solution came from derivation as I written in am linked answer, I started with language description then tried to write some basic rules and progressively I could write complete grammar.
Whereas @us2012's answer came by aptitude, you can gain the aptitude to write grammar by reading others' solutions and writing your own for some - just like how you learn programming.