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Problem 15

# Show that the problem of determining whether a CFG generates all strings in $$1^{*}$$ is decidable. In other words, show that $\left\\{\langle G\rangle \mid G\right.$$is a CFG over$$\\{0,1\\}$$and$$\left.1^{*} \subseteq L(G)\right\\}$ is a decidable language.

Expert verified
In brief, to show that the problem of determining whether a CFG generates all strings in $$1^*$$ is decidable, follow these steps: (1) Check if the given CFG's language only contains 1's, (2) Convert it to Greibach Normal Form if needed, (3) Analyze its productions to ensure all combinations of 1's can be generated, (4) Confirm all non-terminals derive strings of 1's, and (5) Verify there are no useless non-terminal symbols. If these conditions are satisfied, the problem is decidable, and $$1^*\subseteq L(G)$$.
See the step by step solution

## Step 1: Verify if G's language contains only 1's

Begin by checking if the given Context Free Grammar (CFG) G only uses '1' as a terminal symbol. If there are other symbols in the alphabet of the CFG, then it cannot generate all strings in $$1^*$$.

## Step 2: Convert G to Greibach Normal Form (GNF) if needed

If G does not have an existing Greibach Normal Form, convert it to one. GNF is a contexts free grammar where the productions have the following form: A -> aB_1B_2...B_n where A, B_1, B_2, ... , B_n are non-terminal symbols and a is a terminal symbol. In this case, our grammar will only contain the terminal '1'.

## Step 3: Check the productions

Analyze the productions of G to determine if it generates all possible combinations of 1's. Check if at least one of the productions has the form A -> 1B, ensuring that the grammar can generate strings of increasing length by substituting different non-terminal symbols.

## Step 4: Ensure all non-terminals derive strings of 1's

For each non-terminal in G, make sure that it derives some string of 1's. This will guarantee that every production in the grammar contributes to generating strings of 1's.

## Step 5: Check if there are useless non-terminal symbols

Find out if there are any non-terminal symbols with no productions leading to them. These are considered useless non-terminals and should be removed. If any exist, the grammar cannot generate all strings in $$1^*$$ completely.

## Step 6: Determine if the language is decidable

If G meets the conditions mentioned above – the grammar only contains 1's as terminal symbols and generates all possible combinations of 1's – then the given problem is decidable. By following the steps above, we can confirm that determining whether a CFG generates all strings in $$1^*$$ is a decidable problem. If the resulting grammar only contains the terminal symbol '1' in its language and generates all combinations of 1's in its productions, we can conclude that $$1^* \subseteq L(G)$$ as required.

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