5
$\begingroup$

This is homework and I'm looking for a push in the right direction. Proofs were never something I was properly taught, so now they're kind of a weak point.

Here's the problem:

The following grammar generates numbers in binary notation ($C$ is the start symbol):

$\qquad \begin{align}C &\to C 0 \mid A 1 \mid 0 \\ A &\to B 0 \mid C 1 \mid 1 \\ B &\to A 0 \mid B 1 \end{align}$

  1. Prove that the alternating sums of the digits of the generated numbers are multiples of $3$. The alternating sum of $w=w_0\dots w_n$ is defined as $\sum_{i=0}^n (-1)^i \cdot w_i$. As an example, $C$ generates $1001$ via $C \Rightarrow A1 \Rightarrow B01 \Rightarrow A001 \Rightarrow 1001$ with alternating sum of $0$; clearly, $0$ is a multiple of $3$.

  2. Prove that all such numbers (i.e., numbers whose alternating sum is a multiple of 3) are generated by the grammar.

I'm thinking I need to show that the grammar can only generate strings which are made up of repeated subsequences of digits which always add up to 0, 3, or -3. But I'm not sure how to show that it can only generate those three subsequences.

I also have worked out these thoughts:

  • Consider that any even number of consecutive 1s is irrelevant, as they cancel each other out.

  • Consider that all zeros are in of themselves irrelevant, as they add nothing.

  • Consider then that the only relevant pattern is that of alternating 1s and zeros, and where this pattern starts and ends.

$\endgroup$
  • $\begingroup$ Welcome! See here for our homework policy. $\endgroup$ – Raphael Sep 2 '12 at 21:59
4
$\begingroup$

Rough hint: play around with the many equivalent models you have for regular languages.

For 1), I find problematic that the contribution of an already generated digit changes as more symbols are added to the front. Things would be easier if the grammar was right-linear.

Convert the given grammar to an NFA and convert that one to a grammar with the simple construction, which always yields right-linear grammars. (You should have proven algorithms for both.) Determinising/minimising the intermediate automaton might also help the proof.

Then, you have to figure out an induction hypothesis on sentences that works and implies the statement (for words), and perform induction over the whole set of sentences (by the length of the derivation).

For example, if a non-terminal $C$ can only end the derivation by deriving a $1$, then for any sentence of the form $wC$, either $w$ has even length and its alternating digit sum plus one divides three, or it has odd length.

For 2), build an automaton that you can easily show generates all such words.

For instance, use a counter construction, that is count the alternating digit sum modulo three in the states.

Now you can show equivalence of this automaton and the one derived from the grammar in 1).

$\endgroup$
2
$\begingroup$

Here is a hint: figure out what each of the other two symbols $A,B$ generate. You can then prove part (1) by induction. This will also allow you to come up with the explicit construction needed to prove part (2).

$\endgroup$
  • $\begingroup$ I'm sorry; I read that a few times and reread the problem but I can't get anything helpful out of that. I'm just not sufficiently experienced with this to infer the content of the gaps in this hint. $\endgroup$ – temporary_user_name Sep 3 '12 at 4:05
  • $\begingroup$ Aerovistae, I'm not going to solve the problem for you. You'll just have to keep on thinking. Try some examples of strings generated by $A,B$ and see if they have anything in common. You might find useful the function $\pi(w_0\ldots w_n) = \sum_{i=0}^n (-1)^{n-i} w_i \mod{3}$. $\endgroup$ – Yuval Filmus Sep 3 '12 at 5:22
1
$\begingroup$

For 1, I'd work by induction. That is I'd find a property of the strings generated by A, B and C and show that it holds for all construction. For example for C the property is that the alternating sum is a multiple of 3.

  • it holds for the third production: the alternating sum of 0 is a multiple of 3
  • it holds for the first production: the alternating sum of xxx0 is a multiple of 0 if the alternating sum of xxx is 0
  • trying to show that it holds for the second production will give you an hint about which property you want for A

For 2, I'd first try to show that the strings generated for A and B don't have an alternating sum multiple of 3. Then I'd find a characterization of the strings not generated by A, B, C and show that they don't have an alternating sum multiple of 3 either.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.