-2

I have come up with:

S→0SX | 1SY | 2SZ | SS | ϵ
X→1
Y→0 | 2
Z→1

I think I am wrong. Any directions?

Desperado
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    We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – Raphael Jul 12 '17 at 21:34
  • Why do you think you're wrong? Have you tried proving your grammar correct? – Raphael Jul 12 '17 at 21:35
  • I made this and I checked for some strings and they are all fine , but still I think if I am missing something. – Desperado Jul 12 '17 at 21:51
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    First try to prove that any string $x \in L$ is derivable by your grammar. Then try to prove that if $x$ is derivable by your grammar then it is in $L$ (that is number of 1s is equal to the total number of 0s and 2s). If you prove both statements correctly then you are on the right track. – fade2black Jul 12 '17 at 21:59
  • Also to simplify your work you may get rid off $X$, $Y$, and $Z$, leave only $S$, like $S \rightarrow 0S1 \ | \ 1S0 \ | \ 1S2 \ | \ 2S1 \ | \ SS \ | \epsilon$ – fade2black Jul 12 '17 at 22:08

1 Answers1

2

These observations may help when using induction:
1) Since the number of $1$s is equal to the sum of numbers of $0$s and $2$s, any string in $L$ is of even length.
2) Starting from $S$ any number of steps of a derivation without using $S \rightarrow SS \ | \ \epsilon$ results in a string $a_1a_2 \dots S \dots a_{n-1}a_n$ where number of $1$s is equal to the sum of $0$s and $2$s.
3) If $x = a_1a_2 \dots a_{n-1}a_n \in L$ ($|x|$ is even and number of $1$s in $x$ is equal to the total number of $0$s and $2$s) then $x$ has at least one of substrings $12, 21, 01$ or $10$ which is derived from $S \rightarrow 0S1 \ | \ 1S0 \ | \ 1S2 \ | \ 2S1$.

fade2black
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  • Suggested observation is very good. I have checked and seems to be working but still if you see anything wrong or some case where it fails please tell. Or should I consider my solution correct ? – Desperado Jul 13 '17 at 02:44
  • I think your grammar is correct though you may eliminate the redundant rules. Did you prove it? You should demonstrate that your grammar realy generates the language. Simply writing the grammar is not enough. – fade2black Jul 13 '17 at 02:52
  • Please point me to somewhere to know how to prove . – Desperado Jul 13 '17 at 03:12