### Turing Recognizable Languages

**Turing recognizable languages**, also known as **recursively enumerable (RE) languages**, are a class of formal languages that can be recognized by a Turing machine. Here’s a detailed explanation of what it means for a language to be Turing recognizable:

#### Definition

A language ( L ) is Turing recognizable if there exists a Turing machine ( M ) such that:

- If a string ( w ) is in ( L ), then ( M ) eventually enters an accepting state on input ( w ).
- If a string ( w ) is not in ( L ), then ( M ) either rejects ( w ) or runs forever (i.e., does not halt).

In other words, a Turing machine can be used to recognize (or enumerate) the strings in the language, but it might not halt for strings that are not in the language.

#### Characteristics

**Existence of a Recognizer**:

- There is a Turing machine that can recognize the language, meaning it can identify when an input string belongs to the language by eventually accepting it.

**Non-Deterministic Halting**:

- For strings that do not belong to the language, the machine may either reject them or run indefinitely without halting.

**Enumerability**:

- A Turing recognizable language can be enumerated by a Turing machine. This means there exists a Turing machine that will list all the strings in the language, possibly with repetition, and no strings that are not in the language.

#### Examples

**Halting Problem**:

- The set of Turing machines and their inputs that halt is Turing recognizable. Given a Turing machine and an input, you can simulate it to see if it halts (accept), but if it doesn’t halt, you cannot be sure when or if it will ever halt.

**Valid Programs**:

- Consider the set of syntactically valid programs in a given programming language. A Turing machine can recognize this set by parsing and checking the syntax of input strings, accepting valid programs and running indefinitely or rejecting invalid ones.

#### Difference from Turing Decidable (Recursive) Languages

**Turing Decidable (Recursive) Languages**:- A language ( L ) is Turing decidable if there exists a Turing machine ( M ) that accepts ( L ) and halts on all inputs (both those in the language and those not in the language).
- For decidable languages, the machine will either accept or reject every input string after a finite amount of time.
**Turing Recognizable (Recursively Enumerable) Languages**:- A language ( L ) is Turing recognizable if there exists a Turing machine ( M ) that accepts ( L ), but ( M ) may not halt on inputs not in ( L ).

#### Formal Definition

Formally, a language ( L \subseteq \Sigma^* ) (where ( \Sigma^* ) is the set of all strings over an alphabet ( \Sigma )) is Turing recognizable if there exists a Turing machine ( M ) such that:

- ( M ) accepts ( w ) if ( w \in L )
- If ( w \notin L ), ( M ) either rejects ( w ) or runs forever

#### Summary

- Turing recognizable languages are those for which a Turing machine can recognize and accept all strings in the language.
- Such a machine might not halt for strings not in the language, leading to non-decisive behavior.
- This class of languages is broader than the class of Turing decidable languages, encompassing more complex and potentially non-terminating processes.

Understanding Turing recognizable languages is fundamental in the theory of computation as it delineates the boundary between what can be recognized by machines and what can be decisively decided.

**Undecidable Languages**:- A language is undecidable if there is no Turing machine that can decide it, meaning there is no Turing machine that will always halt and correctly determine whether an arbitrary string belongs to the language or not.

**Turing Recognizable (Recursively Enumerable) Languages**:- A language is Turing recognizable if there exists a Turing machine that will accept any string in the language but may not halt for strings not in the language.

در این جا تفاوت این دو مطرح شده است:

### Key Points

**Undecidable but Turing Recognizable**:- Some languages are undecidable but still Turing recognizable. This means there is no Turing machine that can always decide membership for any input string, but there is a Turing machine that will accept and halt for all strings that belong to the language.
**Example**: The Halting Problem is a classic example. The language of pairs (M,w)(*M*,*w*), where M*M*is a Turing machine that halts on input w*w*, is undecidable. However, it is Turing recognizable because if M*M*does halt on w*w*, a Turing machine can simulate M*M*on w*w*and accept if M*M*halts. If M*M*does not halt, the recognizing Turing machine will run forever.

**Undecidable and Not Turing Recognizable**:- Some languages are not even Turing recognizable. For such languages, there is no Turing machine that will recognize the strings in the language.
**Example**: The complement of the Halting Problem is an example. The language of pairs (M,w)(*M*,*w*) where M*M*does not halt on input w*w*is not Turing recognizable. There is no Turing machine that can recognize this language because, in the case where M*M*might halt eventually, the machine recognizing the complement would have to run forever to determine this, which it cannot do by definition.

### Summary

**Undecidable but Turing Recognizable**: There exist languages that are undecidable but still Turing recognizable. These are languages where a Turing machine can accept strings that belong to the language but might not halt for strings that do not belong.**Undecidable and Not Turing Recognizable**: There also exist languages that are undecidable and not Turing recognizable. For these languages, there is no Turing machine that can recognize all strings that belong to the language.

در اینجا دارریم برسی میکنیم که ایا ان ماشین تورینگ پزیرنده است یا خیر یعنی زبان ان تهی میباشد یا خیر از پیش میدانیم که این ماشین تورینگ زبان آن تهی بود

### Part (a)

**Statement**: A language A*A* is Turing-recognizable if and only if A≤mATM*A*≤*m**A**TM*.

#### Definitions

**Turing-recognizable**: A language A*A*is Turing-recognizable if there exists a Turing machine M*M*such that for any string w*w*:- If w∈A
*w*∈*A*, M*M*accepts w*w*. - If w∉A
*w*∈/*A*, M*M*either rejects w*w*or runs forever.

- If w∈A
**Many-one reduction (≤m≤**: A language A*m*)*A*is many-one reducible to a language B*B*(A≤mB*A*≤*m**B*) if there exists a computable function f*f*such that for any string w*w*:- w∈A ⟺ f(w)∈B
*w*∈*A*⟺*f*(*w*)∈*B*.

- w∈A ⟺ f(w)∈B

#### Proof

**If**: Assume A≤mATM*A*≤*m**A**TM*. This means there is a computable function f*f* such that for any string w*w*: w∈A ⟺ f(w)∈ATM*w*∈*A*⟺*f*(*w*)∈*A**TM*

- ATM
*A**TM* is Turing-recognizable (since the Halting Problem is Turing-recognizable). - Given f
*f*is computable, we can construct a Turing machine M*M*that:- Takes w
*w*as input. - Computes f(w)
*f*(*w*). - Checks if f(w)∈ATM
*f*(*w*)∈*A**TM*. - Accepts if f(w)∈ATM
*f*(*w*)∈*A**TM*, otherwise continues running.

- Takes w

Since ATM*A**TM* is Turing-recognizable, this means M*M* is also Turing-recognizable, thus A*A* is Turing-recognizable.

**Only if**: Assume A*A* is Turing-recognizable. This means there is a Turing machine M*M* such that:

- If w∈A
*w*∈*A*, M*M*accepts w*w*. - If w∉A
*w*∈/*A*, M*M*either rejects w*w*or runs forever.

To reduce A*A* to ATM*A**TM*, define a function f*f* as follows:

- f(w)
*f*(*w*) is a Turing machine Mw*M**w* that simulates M*M*on input w*w*.

Thus: w∈A ⟺ Mw∈ATM*w*∈*A*⟺*M**w*∈*A**TM*

Therefore, A≤mATM*A*≤*m**A**TM*, completing the proof.

### Part (b)

**Statement**: A language A*A* is decidable if and only if A≤mATM*A*≤*m**A**TM* and A‾≤mATM*A*≤*m**A**TM*.

#### Definitions

**Decidable**: A language A*A*is decidable if there exists a Turing machine M*M*such that for any string w*w*:- If w∈A
*w*∈*A*, M*M*accepts w*w*and halts. - If w∉A
*w*∈/*A*, M*M*rejects w*w*and halts.

- If w∈A
**Many-one reduction (≤m≤**: As defined above.*m*)

#### Proof

**If**: Assume A≤mATM*A*≤*m**A**TM* and A‾≤mATM*A*≤*m**A**TM*. This means there are computable functions f*f* and g*g* such that for any string w*w*: w∈A ⟺ f(w)∈ATM*w*∈*A*⟺*f*(*w*)∈*A**TM* w∈A‾ ⟺ g(w)∈ATM*w*∈*A*⟺*g*(*w*)∈*A**TM*

Construct a Turing machine M*M* that:

- Given w
*w*, computes f(w)*f*(*w*) and g(w)*g*(*w*). - Simulates the Turing machine for f(w)
*f*(*w*) to check if it halts:- If it halts and accepts, w∈A
*w*∈*A*. - If it halts and rejects, w∈A‾
*w*∈*A*.

- If it halts and accepts, w∈A
- Since both f(w)∈ATM
*f*(*w*)∈*A**TM* and g(w)∈ATM*g*(*w*)∈*A**TM* are decidable, M*M*will halt and decide A*A*.

Therefore, A*A* is decidable.

**Only if**: Assume A*A* is decidable. This means there exists a Turing machine M*M* such that:

- If w∈A
*w*∈*A*, M*M*accepts w*w*and halts. - If w∉A
*w*∈/*A*, M*M*rejects w*w*and halts.

To reduce A*A* and A‾*A* to ATM*A**TM*:

- Define a function f
*f*such that f(w)*f*(*w*) is a Turing machine that simulates M*M*on w*w*and accepts if M*M*accepts. - Define a function g
*g*such that g(w)*g*(*w*) is a Turing machine that simulates M*M*on w*w*and rejects if M*M*rejects.

Thus: w∈A ⟺ f(w)∈ATM*w*∈*A*⟺*f*(*w*)∈*A**TM* w∈A‾ ⟺ g(w)∈ATM*w*∈*A*⟺*g*(*w*)∈*A**TM*

Therefore, A≤mATM*A*≤*m**A**TM* and A‾≤mATM*A*≤*m**A**TM*, completing the proof.

### Conclusion

- A language A
*A*is Turing-recognizable if and only if A≤mATM*A*≤*m**A**TM*. - A language A
*A*is decidable if and only if A≤mATM*A*≤*m**A**TM* and A‾≤mATM*A*≤*m**A**TM*.

یعنی باید این صفر استار و یک استار را به ان کاهش داد

یعنی اگر بتوان ان را به این کاهش داد به تورنگ رکوگنایزر تبدیل خواهد شد

در این حالت داریم برسی میکنیم که ایا میتواند که خودش را بگیرد یا خیر.

این مهم است در هر دو حالت در این جا به مثال نقض رسیده ایم

Consider the language L*L* which consists of all Turing machines M*M* such that ⟨M⟩∉L(M)⟨*M*⟩∈/*L*(*M*), where ⟨M⟩⟨*M*⟩ is the encoding of the Turing machine M*M*. We need to determine whether L*L* is decidable or not.

### Definitions and Clarifications

**Turing Machine Encoding**: ⟨M⟩⟨*M*⟩ denotes the encoding (or description) of the Turing machine M*M*.**Language L(M)**: The language L(M)*L*(*M*)*L*(*M*) is the set of all strings that the Turing machine M*M*accepts.**Self-Membership Condition**: The condition ⟨M⟩∉L(M)⟨*M*⟩∈/*L*(*M*) means that the encoding of the machine M*M*is not in the language recognized by M*M*.

### Language L*L*

The language L*L* is defined as follows: L={⟨M⟩∣⟨M⟩∉L(M)}*L*={⟨*M*⟩∣⟨*M*⟩∈/*L*(*M*)}

### Analysis

To determine whether L*L* is decidable, we need to consider whether there exists a Turing machine that can decide for any given Turing machine M*M*, whether ⟨M⟩∉L(M)⟨*M*⟩∈/*L*(*M*).

#### Step-by-Step Proof

**Assume L**:*L*is Decidable- Suppose there exists a Turing machine D
*D*that decides L*L*. This means D*D*can determine, for any ⟨M⟩⟨*M*⟩, whether ⟨M⟩∉L(M)⟨*M*⟩∈/*L*(*M*).

- Suppose there exists a Turing machine D
**Construct a Contradiction**:- Let’s construct a Turing machine M′
*M*′ that uses D*D*to decide whether its own encoding ⟨M′⟩⟨*M*′⟩ is in its own language. - Define M′
*M*′ such that it behaves as follows:- M′
*M*′ takes its own encoding ⟨M′⟩⟨*M*′⟩ as input. - M′
*M*′ uses the decider D*D*to check whether ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′). - If D
*D*accepts (indicating ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′)), then M′*M*′ accepts ⟨M′⟩⟨*M*′⟩. - If D
*D*rejects (indicating ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′)), then M′*M*′ does not accept ⟨M′⟩⟨*M*′⟩.

- M′

- Let’s construct a Turing machine M′
**Self-Referential Paradox**:- Now we analyze the behavior of M′
*M*′ on its own encoding ⟨M′⟩⟨*M*′⟩. - There are two cases to consider:
**Case 1**: If ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′):- According to the definition of M′
*M*′, if D*D*determines ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′), M′*M*′ should accept ⟨M′⟩⟨*M*′⟩. But this contradicts the assumption that ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′).

- According to the definition of M′
**Case 2**: If ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′):- According to the definition of M′
*M*′, if D*D*determines ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′), M′*M*′ should reject ⟨M′⟩⟨*M*′⟩. But this contradicts the assumption that ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′).

- According to the definition of M′

- Now we analyze the behavior of M′
**Conclusion**:- In both cases, we encounter a contradiction. This contradiction arises from the assumption that D
*D*exists and can decide L*L*. - Therefore, the language L
*L*cannot be decidable, as no such decider D*D*can exist.

- In both cases, we encounter a contradiction. This contradiction arises from the assumption that D

### Conclusion

The language L*L*, defined as the set of all Turing machines M*M* such that ⟨M⟩∉L(M)⟨*M*⟩∈/*L*(*M*), is **not decidable**. This is proven by constructing a self-referential paradox that shows that assuming the decidability of L*L* leads to a contradiction.

از این استفاده کرده ایم برای این جا

**Self-Referential Paradox**:- Now we analyze the behavior of M′
*M*′ on its own encoding ⟨M′⟩⟨*M*′⟩. - There are two cases to consider:
**Case 1**: If ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′):- According to the definition of M′
*M*′, if D*D*determines ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′), M′*M*′ should accept ⟨M′⟩⟨*M*′⟩. But this contradicts the assumption that ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′).

- According to the definition of M′
**Case 2**: If ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′):- According to the definition of M′
*M*′, if D*D*determines ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′), M′*M*′ should reject ⟨M′⟩⟨*M*′⟩. But this contradicts the assumption that ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′).

- According to the definition of M′

- Now we analyze the behavior of M′

**Self-Referential Paradox**:- Now we analyze the behavior of M′
*M*′ on its own encoding ⟨M′⟩⟨*M*′⟩. - There are two cases to consider:
**Case 1**: If ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′):- According to the definition of M′
*M*′, if D*D*determines ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′), M′*M*′ should accept ⟨M′⟩⟨*M*′⟩. But this contradicts the assumption that ⟨M′⟩∉L(M′)⟨*M*′⟩∈/*L*(*M*′).

- According to the definition of M′
**Case 2**: If ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′):- According to the definition of M′
*M*′, if D*D*determines ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′), M′*M*′ should reject ⟨M′⟩⟨*M*′⟩. But this contradicts the assumption that ⟨M′⟩∈L(M′)⟨*M*′⟩∈*L*(*M*′).

- According to the definition of M′

- Now we analyze the behavior of M′

**Self-Referential Paradox**

### Problem Statement

Let L(M)*L*(*M*) be a language containing all Turing machines T*T* such that L(T)≠∅*L*(*T*)=∅. Prove that L(M)*L*(*M*) is recognizable (Turing-recognizable).

### Definitions

**Turing-recognizable language**: A language L*L*is Turing-recognizable if there exists a Turing machine M*M*such that for every string w*w*:- If w∈L
*w*∈*L*, M*M*accepts w*w*. - If w∉L
*w*∈/*L*, M*M*either rejects w*w*or runs forever.

- If w∈L
**L(T)≠∅**: This means there exists at least one string that the Turing machine T*L*(*T*)=∅*T*accepts.

### Proof

To prove that L(M)*L*(*M*) is recognizable, we need to construct a Turing machine M*M* that recognizes L(M)*L*(*M*). This machine M*M* should accept any Turing machine encoding ⟨T⟩⟨*T*⟩ if L(T)≠∅*L*(*T*)=∅.

#### Constructing the Recognizer M*M*

**Input**: The encoding of a Turing machine ⟨T⟩⟨*T*⟩.**Simulation**: We need to determine whether L(T)≠∅*L*(*T*)=∅, i.e., whether there exists some input w*w*such that T*T*accepts w*w*.**Enumerate Inputs**:- Enumerate all possible input strings w1,w2,w3,…
*w*1,*w*2,*w*3,…. - Simultaneously simulate T
*T*on all these inputs using dovetailing. This means alternating steps among the different simulations to ensure all are advanced fairly.

- Enumerate all possible input strings w1,w2,w3,…

#### Dovetailing Process

**Initialize**: Start with the first input w1*w*1.**Step through Simulations**:- On the first step, simulate T
*T*on w1*w*1 for one step. - On the second step, simulate T
*T*on w1*w*1 for another step and start simulating T*T*on w2*w*2 for one step. - Continue this process, interleaving the simulations of T
*T*on all enumerated inputs w1,w2,w3,…*w*1,*w*2,*w*3,….

- On the first step, simulate T

#### Acceptance Condition

- If any simulation T(wi)
*T*(*w**i*) accepts, halt and accept the input ⟨T⟩⟨*T*⟩. - If no simulation ever accepts, the machine will run forever, which is consistent with the definition of a Turing-recognizable language.

### Detailed Steps of the Turing Machine M*M*

**Input**: ⟨T⟩⟨*T*⟩**Initialize**: Set i=1*i*=1.**Loop**:- For each j=1
*j*=1 to i*i*:- Simulate T
*T*on wj*w**j* for one step. - If T
*T*accepts wj*w**j*, then halt and accept ⟨T⟩⟨*T*⟩.

- Simulate T
- Increment i
*i*and repeat.

- For each j=1

### Pseudocode for M*M*

`vbnetCopy code````
Input: ⟨T⟩
i = 1
while true do
for j = 1 to i do
Simulate T on w_j for one step
if T accepts w_j then
accept ⟨T⟩
i = i + 1
```

### Conclusion

By constructing the Turing machine M*M* that uses dovetailing to simulate T*T* on all possible inputs, we can recognize the language L(M)*L*(*M*) consisting of all Turing machine encodings ⟨T⟩⟨*T*⟩ where L(T)≠∅*L*(*T*)=∅. This ensures that M*M* will accept ⟨T⟩⟨*T*⟩ if T*T* accepts at least one string, proving that L(M)*L*(*M*) is Turing-recognizable.

این گونه باید روی دیتا ی ورودی این را اجرا کنیم

پس باید زبان هایی را پیدا کرد که بتواند ان را دیساید کند .

### Problem Statement

Consider the language L*L* consisting of pairs ⟨G,A⟩⟨*G*,*A*⟩, where G*G* is a context-free grammar and A*A* is a non-terminal symbol in G*G* such that A*A* does not appear in any derivation of any string in the language L(G)*L*(*G*). Prove that L*L* is decidable.

### Definitions

**Context-Free Grammar (CFG)**: A grammar G=(V,Σ,R,S)*G*=(*V*,Σ,*R*,*S*) where V*V*is a finite set of variables (non-terminal symbols), ΣΣ is a finite set of terminal symbols, R*R*is a finite set of production rules, and S*S*is the start symbol.**Derivation**: A sequence of applications of production rules starting from the start symbol S*S*that produces strings in the language L(G)*L*(*G*).**Language L(G)**: The set of all strings that can be derived from the start symbol S*L*(*G*)*S*using the production rules of the grammar G*G*.

### Goal

Prove that the language L={⟨G,A⟩∣G is a CFG and A does not appear in any derivation of any string in L(G)}*L*={⟨*G*,*A*⟩∣*G* is a CFG and *A* does not appear in any derivation of any string in *L*(*G*)} is decidable.

### Approach

To show that L*L* is decidable, we need to construct a Turing machine that decides whether a given pair ⟨G,A⟩⟨*G*,*A*⟩ belongs to L*L*.

### Steps to Construct the Decider

**Input**: The input is a pair ⟨G,A⟩⟨*G*,*A*⟩, where G=(V,Σ,R,S)*G*=(*V*,Σ,*R*,*S*) is a context-free grammar and A∈V*A*∈*V*is a non-terminal symbol.**Check Reachability**:- A non-terminal A
*A*is reachable if there is a derivation sequence starting from the start symbol S*S*that eventually leads to a string containing A*A*. - To determine reachability, we use a reachability analysis:
- Initialize a set Reachable
*R**e**a**c**hab**l**e*containing the start symbol S*S*. - Repeatedly apply the production rules to see if new non-terminals can be added to the Reachable
*R**e**a**c**hab**l**e*set. - If A
*A*is added to Reachable*R**e**a**c**hab**l**e*during this process, then A*A*is reachable from S*S*.

- Initialize a set Reachable

- A non-terminal A
**Check for Appearances in Derivations**:- After determining the set of reachable non-terminals, perform a check to see if A
*A*can appear in any derivation of strings in L(G)*L*(*G*). - This involves checking each production rule to ensure that no sequence of derivations involves A
*A*.

- After determining the set of reachable non-terminals, perform a check to see if A

### Detailed Algorithm

**Input**: ⟨G,A⟩⟨*G*,*A*⟩**Initialize**: Reachable={S}*R**e**a**c**hab**l**e*={*S*} where S*S*is the start symbol.**Reachability Analysis**:- Repeat until no new non-terminals are added to Reachable
*R**e**a**c**hab**l**e*:- For each production X→α
*X*→*α*in R*R*:- If X∈Reachable
*X*∈*R**e**a**c**hab**l**e*and any non-terminal in α*α*is not in Reachable*R**e**a**c**hab**l**e*, add all non-terminals in α*α*to Reachable*R**e**a**c**hab**l**e*.

- If X∈Reachable

- For each production X→α

- Repeat until no new non-terminals are added to Reachable
**Check for Non-Terminal A**:*A*- If A∈Reachable
*A*∈*R**e**a**c**hab**l**e*:- Reject (since A
*A*can appear in some derivation).

- Reject (since A
- Else:
- Accept (since A
*A*does not appear in any derivation of any string in L(G)*L*(*G*)).

- Accept (since A

- If A∈Reachable

### Pseudocode for the Decider

`markdownCopy code````
Input: ⟨G, A⟩
G = (V, Σ, R, S)
1. Initialize Reachable = {S}
2. Repeat
3. For each production X → α in R
4. If X ∈ Reachable
5. For each non-terminal B in α
6. If B ∉ Reachable
7. Add B to Reachable
8. Until no new non-terminals are added to Reachable
9. If A ∈ Reachable
10. Reject ⟨G, A⟩
11. Else
12. Accept ⟨G, A⟩
```

### Explanation

**Reachability Analysis**: We determine which non-terminals can be derived starting from S*S*.**Decision**: If A*A*is reachable, it means A*A*can appear in some derivation of a string in L(G)*L*(*G*), so we reject. If A*A*is not reachable, A*A*does not appear in any derivation, so we accept.

### Conclusion

The language L*L* is decidable because we can construct a Turing machine that effectively performs reachability analysis on the given context-free grammar G*G* and decides whether A*A* can appear in any derivation. This Turing machine halts on all inputs, thus proving that L*L* is decidable.

این گونه باید همه را قابل دسترس کنیم

چون که در پایان یک سی اف جی به ما میدهد میدانیم که هر چیزی در باره. ی سی اف ج یها تصمیم پزیر میباشد

### Problem Statement

Prove that the language A={⟨G⟩∣G is a CFG over {0,1}∗ and L(G)∩L(1∗)=∅}*A*={⟨*G*⟩∣*G* is a CFG over {0,1}∗ and *L*(*G*)∩*L*(1∗)=∅} is decidable.

### Definitions

**Context-Free Grammar (CFG)**: A grammar G=(V,Σ,R,S)*G*=(*V*,Σ,*R*,*S*) where V*V*is a finite set of variables (non-terminal symbols), ΣΣ is a finite set of terminal symbols, R*R*is a finite set of production rules, and S*S*is the start symbol.**Language L(G)**: The set of all strings that can be derived from the start symbol S*L*(*G*)*S*using the production rules of the grammar G*G*.**Language L(1∗)**: The set of all strings consisting solely of the symbol ‘1’, i.e., L(1∗)={1,11,111,…}*L*(1∗)*L*(1∗)={1,11,111,…}.

### Goal

To prove that A*A* is decidable, we need to construct a Turing machine that decides whether a given CFG G*G* over {0,1}∗{0,1}∗ generates any strings consisting solely of the symbol ‘1’ (i.e., any string in L(1∗)*L*(1∗)).

### Approach

To show that A*A* is decidable, we will:

- Construct a CFG G′
*G*′ that generates the intersection of L(G)*L*(*G*) and L(1∗)*L*(1∗). - Check whether L(G′)
*L*(*G*′) is empty.

### Steps to Construct the Decider

**Input**: The input is an encoding of a CFG ⟨G⟩⟨*G*⟩, where G=(V,Σ,R,S)*G*=(*V*,Σ,*R*,*S*) and Σ={0,1}Σ={0,1}.**Construct CFG for L(1∗)**:*L*(1∗)- Define a CFG G1
*G*1 that generates L(1∗)*L*(1∗):- G1=({A},{1},{A→1A∣ϵ},A)
*G*1=({*A*},{1},{*A*→1*A*∣*ϵ*},*A*)

- G1=({A},{1},{A→1A∣ϵ},A)

- Define a CFG G1
**Intersection of CFGs**:- Use the construction to obtain a CFG G′
*G*′ that generates the intersection L(G)∩L(G1)*L*(*G*)∩*L*(*G*1). - Since L(G1)=L(1∗)
*L*(*G*1)=*L*(1∗), G′*G*′ generates L(G)∩L(1∗)*L*(*G*)∩*L*(1∗).

- Use the construction to obtain a CFG G′
**Emptiness Check**:- Determine whether L(G′)
*L*(*G*′) is empty. - Use the algorithm for checking the emptiness of the language generated by a CFG.

- Determine whether L(G′)

### Detailed Algorithm

**Input**: ⟨G⟩⟨*G*⟩**Construct G1**:*G*1- G1=({A},{1},{A→1A∣ϵ},A)
*G*1=({*A*},{1},{*A*→1*A*∣*ϵ*},*A*)

- G1=({A},{1},{A→1A∣ϵ},A)
**Construct G′**:*G*′- Use the product construction to create G′
*G*′ such that L(G′)=L(G)∩L(G1)*L*(*G*′)=*L*(*G*)∩*L*(*G*1).

- Use the product construction to create G′
**Check Emptiness of G′**:*G*′- Determine if L(G′)
*L*(*G*′) is empty using the standard CFG emptiness check:- Remove non-productive symbols (symbols that do not derive any terminal string).
- Remove unreachable symbols (symbols that cannot be derived from the start symbol).
- If the start symbol of G′
*G*′ is removed, then L(G′)*L*(*G*′) is empty.

- Determine if L(G′)
**Decision**:- If L(G′)
*L*(*G*′) is empty, accept ⟨G⟩⟨*G*⟩. - If L(G′)
*L*(*G*′) is not empty, reject ⟨G⟩⟨*G*⟩.

- If L(G′)

### Pseudocode for the Decider

`mathematicaCopy code````
Input: ⟨G⟩
1. Construct CFG G1 for L(1*):
G1 = ({A}, {1}, {A → 1A | ε}, A)
2. Construct CFG G' for L(G) ∩ L(G1) using product construction
3. Check if L(G') is empty:
- Initialize WorkList with the start symbol of G'
- Mark productive symbols
- Remove non-productive symbols
- Mark reachable symbols
- Remove unreachable symbols
- If start symbol is not in remaining symbols, L(G') is empty
4. If L(G') is empty, accept ⟨G⟩
5. Otherwise, reject ⟨G⟩
```

### Conclusion

The language A*A* is decidable because we can construct a Turing machine that:

- Constructs a CFG G′
*G*′ representing the intersection L(G)∩L(1∗)*L*(*G*)∩*L*(1∗). - Checks if L(G′)
*L*(*G*′) is empty using standard CFG emptiness techniques. This Turing machine halts on all inputs, thus proving that A*A*is decidable.

### Problem Statement

Consider a variant of a Pushdown Automaton (PDA) that, instead of one stack, uses two stacks. Prove that the computational power of this automaton is equivalent to that of a Turing machine.

### Definitions

Pushdown Automaton (PDA): A type of automaton that employs a stack for its computation. It can be described as a 7-tuple $(Q,\mathrm{\Sigma},\mathrm{\Gamma},\delta ,{q}_{0},{Z}_{0},F)$ where:

- $Q$ is a finite set of states.
- $\mathrm{\Sigma}$ is the input alphabet.
- $\mathrm{\Gamma}$ is the stack alphabet.
- $\delta $ is the transition function.
- ${q}_{0}$ is the start state.
- ${Z}_{0}$ is the initial stack symbol.
- $F$ is the set of accepting states.

Two-Stack PDA: A PDA with two stacks instead of one. It can be described as an 8-tuple $(Q,\mathrm{\Sigma},\mathrm{\Gamma},\delta ,{q}_{0},{Z}_{0},F,{Z}_{1})$ where:

- $Q,\mathrm{\Sigma},\mathrm{\Gamma},\delta ,{q}_{0},{Z}_{0},F$ are as defined above.
- ${Z}_{1}$ is the initial symbol of the second stack.

Turing Machine: An abstract computational model that can simulate any algorithm. It consists of an infinite tape, a tape head, a state register, and a transition function.

### Goal

To prove that a two-stack PDA (2-PDA) is equivalent in computational power to a Turing machine.

### Proof Outline

We will prove this in two parts:

- Simulating a Turing machine using a two-stack PDA.
- Simulating a two-stack PDA using a Turing machine.

### Part 1: Simulating a Turing Machine with a Two-Stack PDA

#### Intuition

A Turing machine has a tape that can move left and right, and it can read and write symbols on the tape. We can simulate this behavior using two stacks.

#### Simulation Strategy

- Use one stack to represent the tape content to the left of the current tape head position.
- Use the other stack to represent the tape content to the right of the current tape head position.
- The top of the left stack represents the symbol immediately to the left of the tape head, and the top of the right stack represents the symbol under the tape head.

#### Steps

- Initialization:
- Push the initial tape content to the right stack.
- Leave the left stack empty initially.

- Simulation of Moves:
- Move Right:
- Pop the top symbol of the right stack (this symbol is currently under the tape head).
- Push this symbol to the left stack.

- Move Left:
- Pop the top symbol of the left stack (this symbol is now under the tape head).
- Push this symbol to the right stack.

- Move Right:
- Read and Write:
- To read the symbol under the tape head, simply look at the top of the right stack.
- To write a new symbol, pop the top of the right stack (discard the old symbol) and push the new symbol onto the right stack.

### Part 2: Simulating a Two-Stack PDA with a Turing Machine

#### Intuition

A Turing machine can simulate the behavior of two stacks by using multiple tapes or a single tape with a special encoding.

#### Simulation Strategy

- Use the Turing machine’s tape to simulate the two stacks by dividing the tape into two sections.
- Use a special separator symbol to delineate the end of one stack and the start of the other.

#### Steps

Encoding:

- Use a special symbol (e.g., ‘#’) as a delimiter between the two stacks.
- Encode the stacks as a string of symbols on the tape with the delimiter separating them.

Operations:

- Push to Stack 1: Move the tape head to the delimiter, shift the delimiter and subsequent symbols to the right, and write the new symbol in place of the delimiter.
- Pop from Stack 1: Move the tape head to the delimiter, shift the delimiter and subsequent symbols to the left.
- Push to Stack 2: Move the tape head to the right end of the tape, write the new symbol, and update the delimiter.
- Pop from Stack 2: Move the tape head to the right end of the tape, read and remove the symbol, and update the delimiter.

### Conclusion

We have shown that a two-stack PDA can simulate a Turing machine by treating the stacks as the left and right parts of the tape. Conversely, we have demonstrated that a Turing machine can simulate a two-stack PDA by using its tape to represent the two stacks with appropriate encoding and operations. Therefore, the computational power of a two-stack PDA is equivalent to that of a Turing machine.

### Problem Statement

Consider a variant of a Turing machine that, instead of deciding its next move based solely on the current cell’s value, makes decisions based on the values of the current cell and the cells to the left and right. For example, if the values of these three cells are 011, the machine transitions to state q*q*. Prove that the computational power of this Turing machine is equivalent to that of a standard Turing machine.

### Definitions

**Standard Turing Machine (STM)**: A machine that reads the value of the current tape cell, writes a new value, transitions to a new state, and moves the tape head left or right based on its transition function.**Three-cell Turing Machine (3CTM)**: A machine that reads the values of the current tape cell, the cell to its left, and the cell to its right, then writes a new value to the current cell, transitions to a new state, and moves the tape head left or right based on its transition function.

### Goal

Prove that the computational power of the 3CTM is equivalent to that of an STM. Specifically, show that any computation performed by a 3CTM can be simulated by an STM and vice versa.

### Approach

**Simulate an STM using a 3CTM**.**Simulate a 3CTM using an STM**.

### Part 1: Simulating a Standard Turing Machine with a Three-cell Turing Machine

#### Intuition

A 3CTM has more information at each step (it sees three cells at a time), so it can directly simulate an STM, which only sees one cell at a time.

#### Simulation Strategy

- The 3CTM can easily replicate the behavior of an STM by ignoring the additional information from the left and right cells.

#### Steps

**State and Transition Function**:- For each transition in the STM, the 3CTM will look at the middle cell and use the same transition function as the STM, ignoring the values of the left and right cells.

**Implementation**:- If the STM reads a symbol a
*a*and transitions to state q*q*, writes symbol b*b*, and moves left or right, the 3CTM can perform the same operation by focusing on the middle cell’s value and ignoring the others.

- If the STM reads a symbol a

### Part 2: Simulating a Three-cell Turing Machine with a Standard Turing Machine

#### Intuition

An STM has less information at each step, so to simulate a 3CTM, the STM needs to encode the additional information (the values of the left and right cells) into its states.

#### Simulation Strategy

- The STM will simulate the 3CTM by encoding the values of the left and right cells into an extended state.

#### Steps

**Extended State Definition**:- Define the state of the STM to encode the current state of the 3CTM and the values of the left and right cells.
- If the 3CTM has states Q
*Q*and the tape alphabet ΓΓ, the STM’s state space will be Q×Γ×Γ*Q*×Γ×Γ to encode the values of the current cell and its neighbors.

**Transition Function**:- For each transition in the 3CTM, the STM will:
- Read the current cell.
- Use the extended state (which encodes the left and right cell values) to determine the next state, the symbol to write, and the direction to move.
- Update its extended state to reflect the new left, current, and right cell values.

- For each transition in the 3CTM, the STM will:
**Implementation**:- The STM simulates the 3CTM’s step-by-step behavior by maintaining and updating the extended state that includes information about neighboring cells.
- The STM will:
- Read the current cell and the previous extended state.
- Determine the new extended state based on the 3CTM’s transition function.
- Write the appropriate symbol and move the tape head accordingly.

### Example Simulation

Suppose the 3CTM has a transition (q,a,b,c)→(q′,d,R)(*q*,*a*,*b*,*c*)→(*q*′,*d*,*R*), meaning in state q*q* with current cell a*a*, left cell b*b*, and right cell c*c*, it writes d*d*, moves right, and transitions to state q′*q*′.

**Initial State**:- The STM starts in state (q,b,c)(
*q*,*b*,*c*) if the current state is q*q*, the left cell is b*b*, and the right cell is c*c*.

- The STM starts in state (q,b,c)(
**Read and Transition**:- The STM reads a
*a*and transitions to state (q′,d,c′)(*q*′,*d*,*c*′), where c′*c*′ is the new right cell value after moving right.

- The STM reads a
**State Update**:- The STM updates its state and the tape to reflect the new configuration.

### Conclusion

By showing that an STM can simulate a 3CTM through extended states and transitions, and a 3CTM can directly simulate an STM, we conclude that the computational power of a three-cell Turing machine is equivalent to that of a standard Turing machine. This demonstrates that both models can perform the same class of computations, thus having the same computational power.

یعنی هر بار میتواند به سه تا از داده ها ی ان ها دسترسی داشته باشد یعنی از روی تیپ بخواند

### Problem Statement

Consider a Turing machine that has a restriction: the symbol it writes on the tape cannot be the same as the symbol it reads from the tape. Does this restriction reduce the computational power of the Turing machine?

### Definitions

**Standard Turing Machine (STM)**: A machine that reads a symbol from the tape, writes a new symbol (which can be the same as the symbol read), moves the tape head left or right, and transitions to a new state based on its transition function.**Restricted Turing Machine (RTM)**: A machine that reads a symbol from the tape and must write a different symbol. The transition function for the RTM can be defined as δ:Q×Γ→Q×(Γ∖{s})×{L,R}*δ*:*Q*×Γ→*Q*×(Γ∖{*s*})×{*L*,*R*}, where s*s*is the symbol read from the tape.

### Goal

Determine whether the computational power of the RTM is the same as that of the STM, i.e., whether the restriction of not writing the same symbol read reduces the computational power of the Turing machine.

### Analysis

To prove that the restriction does not reduce the computational power, we need to show that for any computation performed by a standard Turing machine, there exists an equivalent restricted Turing machine that can perform the same computation.

### Approach

**Simulate an STM with an RTM**: Construct an RTM that simulates the behavior of an STM, ensuring that the RTM adheres to the restriction of not writing the same symbol it reads.

#### Simulation Strategy

- Introduce additional symbols to the tape alphabet of the RTM to manage the restriction.
- Use state transitions and auxiliary symbols to emulate writing the same symbol indirectly.

### Steps for Simulation

**Introduce Auxiliary Symbols**:- For each symbol a∈Γ
*a*∈Γ, introduce an auxiliary symbol a′*a*′ to the tape alphabet. The auxiliary symbols are distinct and not part of the original tape alphabet. - The RTM uses these auxiliary symbols to indirectly represent writing the same symbol.

- For each symbol a∈Γ
**State Encoding**:- Augment the state space to remember the intention of writing a particular symbol while managing the read/write restriction.

**Simulation Details**:- When the STM wants to write the same symbol it reads, the RTM will write an auxiliary symbol and enter an intermediate state.
- The intermediate state will immediately convert the auxiliary symbol back to the original symbol in the next step, effectively simulating the behavior of the STM.

### Example

Suppose the STM has a transition (q,a)→(q′,a,R)(*q*,*a*)→(*q*′,*a*,*R*), meaning in state q*q* and reading symbol a*a*, it writes a*a*, transitions to state q′*q*′, and moves right.

#### Simulated with RTM:

**Read a**:*a*- RTM reads a
*a*in state q*q*. - Instead of writing a
*a*(which is restricted), the RTM writes a′*a*′ (an auxiliary symbol) and transitions to a new state q1*q*1 (an intermediate state).

- RTM reads a
**Convert Auxiliary Symbol**:- In state q1
*q*1 and reading a′*a*′, the RTM writes a*a*and transitions to state q′*q*′, then moves right.

- In state q1

### Formal Construction

**Define the States**:- For each state q
*q*in the STM, introduce corresponding states q*q*and q1*q*1 in the RTM.

- For each state q
**Transition Function**:- Define the transition function for the RTM to handle the auxiliary symbols and ensure compliance with the restriction.

### Pseudocode for RTM Transition Function

`rustCopy code````
For each (q, a) -> (q', a, R) in STM:
RTM (q, a) -> (q_1, a', R) // Write auxiliary symbol and go to intermediate state
RTM (q_1, a') -> (q', a, R) // Convert auxiliary symbol back to original
For each (q, a) -> (q', b, R) in STM, where a != b:
RTM (q, a) -> (q', b, R) // Direct transition as no restriction violation
```

### Conclusion

By introducing auxiliary symbols and intermediate states, the restricted Turing machine (RTM) can simulate the behavior of a standard Turing machine (STM). This demonstrates that the computational power of the RTM is equivalent to that of the STM, meaning the restriction does not reduce the computational power of the Turing machine.

### Problem Statement

Consider a Turing machine with a two-dimensional (2D) tape that is infinite in all directions. Prove that the computational power of this machine is equivalent to that of a standard Turing machine (STM), which has a one-dimensional (1D) tape.

### Definitions

**Standard Turing Machine (STM)**: A machine with a single infinite tape that is linear (one-dimensional), a tape head that reads and writes symbols on the tape, and can move left or right.**Two-Dimensional Turing Machine (2DTM)**: A machine with a two-dimensional infinite tape, a tape head that can read and write symbols, and can move up, down, left, or right.

### Goal

To prove that the computational power of a 2DTM is equivalent to that of an STM, we need to show that any computation performed by a 2DTM can be simulated by an STM and vice versa.

### Approach

**Simulate a Standard Turing Machine with a Two-Dimensional Turing Machine**.**Simulate a Two-Dimensional Turing Machine with a Standard Turing Machine**.

### Part 1: Simulating an STM with a 2DTM

#### Intuition

A 2DTM has more space and more directions to move (up, down, left, right) compared to an STM. Thus, it can easily simulate the linear tape and left/right movements of an STM.

#### Simulation Strategy

- Use a single row of the 2D tape to represent the 1D tape of the STM.
- The 2DTM will perform all operations within this single row, moving left and right as needed.

#### Steps

**Initialization**:- The 2DTM starts with its tape head on a designated row (say row 0), which simulates the 1D tape of the STM.

**Read and Write**:- The 2DTM reads and writes symbols only on row 0.
- The transition function is the same as the STM, where the 2DTM interprets left and right moves along row 0.

**State Transitions**:- The 2DTM transitions between states exactly as the STM would, based on the symbols read from row 0.

### Part 2: Simulating a 2DTM with an STM

#### Intuition

To simulate a 2DTM with an STM, we need to encode the 2D tape onto the 1D tape of the STM. This can be done using a special encoding scheme.

#### Simulation Strategy

- Map the 2D tape coordinates (i,j)(
*i*,*j*) to positions on the 1D tape. - Use a delimiter or a specific encoding to separate rows and columns.

#### Steps

**Encoding Scheme**:- Use an interleaving method or a delimiter-based encoding to map the 2D coordinates (i,j)(
*i*,*j*) to a single linear index. - For example, use pairs of indices (i,j)(
*i*,*j*) and a delimiter to separate the rows and columns.

- Use an interleaving method or a delimiter-based encoding to map the 2D coordinates (i,j)(
**Read and Write**:- The STM maintains a mapping between the 2D tape positions and the corresponding positions on the 1D tape.
- When the 2DTM would move up, down, left, or right, the STM adjusts the mapping and moves accordingly.

**State Transitions**:- The STM interprets the encoded tape positions and transitions between states as the 2DTM would, based on the decoded symbols.

### Detailed Example

#### Encoding the 2D Tape on a 1D Tape

- Represent the 2D tape as a sequence of rows, each row being a sequence of cells.
- Use a unique delimiter (e.g., a special symbol) to separate cells within a row and another delimiter to separate rows.

For example, the 2D tape:

`cssCopy code````
a b c
d e f
g h i
```

Can be encoded as a 1D tape:

`cssCopy code````
a,b,c|d,e,f|g,h,i
```

Where `|`

separates rows and `,`

separates cells within a row.

#### Simulating Movements

**Move Right**:- The STM moves right to the next cell within the same row.

**Move Left**:- The STM moves left to the previous cell within the same row.

**Move Down**:- The STM moves right to the next row (skipping over the delimiter).

**Move Up**:- The STM moves left to the previous row (skipping over the delimiter).

### Pseudocode for STM Simulation of 2DTM

**Initialization**:- Encode the initial 2D tape onto the 1D tape.

**State Transitions**:- For each state transition of the 2DTM, update the corresponding positions and states on the 1D tape.
- Adjust the tape head position based on the encoded mapping.

### Conclusion

By demonstrating that a 2DTM can simulate an STM using a single row of its tape and that an STM can simulate a 2DTM using an encoding scheme to map the 2D tape onto a 1D tape, we conclude that the computational power of a Turing machine with a two-dimensional tape is equivalent to that of a standard Turing machine. Thus, the additional dimensions do not increase the computational power of the Turing machine.

### a. One-letter Automaton with One Stack

We will describe a one-letter automaton with a single one-letter stack that accepts the language L={bman∣m≥n≥1}*L*={*b**m**a**n*∣*m*≥*n*≥1}.

#### Components of the Automaton

**States**: Q={q0,q1,q2,qf}*Q*={*q*0,*q*1,*q*2,*q**f*}**Input Alphabet**: Σ={a,b}Σ={*a*,*b*}**Stack Alphabet**: Γ={Z,X}Γ={*Z*,*X*}- Z
*Z*: Stack bottom marker. - X
*X*: Symbol that can be pushed or popped from the stack.

- Z
**Initial State**: q0*q*0**Initial Stack Symbol**: Z*Z***Accepting States**: {qf}{*q**f*}

#### Transition Function

**Start State q0**:*q*0- On reading b
*b*, push X*X*onto the stack and move to q1*q*1.- (q0,ϵ,Z)→(q0,Z)(
*q*0,*ϵ*,*Z*)→(*q*0,*Z*) - (q0,b,Z)→(q1,XZ)(
*q*0,*b*,*Z*)→(*q*1,*XZ*) - (q1,b,X)→(q1,XX)(
*q*1,*b*,*X*)→(*q*1,*XX*)

- (q0,ϵ,Z)→(q0,Z)(

- On reading b
**State q1**(reading b*q*1*b*‘s and pushing X*X*onto the stack):- On reading b
*b*, push X*X*onto the stack and stay in q1*q*1. - On reading a
*a*, pop X*X*from the stack and move to q2*q*2.- (q1,a,X)→(q2,ϵ)(
*q*1,*a*,*X*)→(*q*2,*ϵ*)

- (q1,a,X)→(q2,ϵ)(

- On reading b
**State q2**(reading a*q*2*a*‘s and popping X*X*from the stack):- On reading a
*a*, pop X*X*from the stack and stay in q2*q*2. - If stack reaches the bottom marker Z
*Z*, move to qf*q**f*.- (q2,a,X)→(q2,ϵ)(
*q*2,*a*,*X*)→(*q*2,*ϵ*) - (q2,ϵ,Z)→(qf,Z)(
*q*2,*ϵ*,*Z*)→(*q**f*,*Z*)

- (q2,a,X)→(q2,ϵ)(

- On reading a

#### Description

**Initial Phase**: The automaton starts in q0*q*0. It reads b*b*‘s and pushes X*X*onto the stack until it encounters the first a*a*, transitioning to q2*q*2.**Middle Phase**: In q2*q*2, it reads a*a*‘s and pops X*X*from the stack.**Final Phase**: If the number of a*a*‘s read is less than or equal to the number of b*b*‘s read, the stack will not be empty until all a*a*‘s are processed. If the stack reaches the bottom marker Z*Z*, it means m≥n*m*≥*n*. If the stack is empty after processing all a*a*‘s, the automaton transitions to qf*q**f*, accepting the string.

### b. Power of One-Letter Automaton with Four Stacks

We need to show that a one-letter automaton with four one-letter stacks is equivalent in computational power to a Turing machine.

#### Simulation of Turing Machine with Four Stacks

To simulate a Turing machine using a one-letter automaton with four stacks, we will use the four stacks to represent different parts of the Turing machine’s tape and its head movements.

**Stacks**:- Stack 1: Represents the tape content to the left of the head.
- Stack 2: Represents the tape content to the right of the head.
- Stack 3: Tracks the head’s current position.
- Stack 4: Keeps auxiliary information for the current state.

#### Simulation Strategy

- Use stacks to store and retrieve symbols as needed to emulate the tape’s left and right movements.
- Manage state transitions and symbol replacements by appropriately pushing and popping from the stacks.

#### Steps

**Initialization**:- Stack 1: Empty (represents the left part of the tape initially).
- Stack 2: Initialized with the input string (represents the right part of the tape).
- Stack 3: Tracks the position of the head.
- Stack 4: Stores the current state.

**Tape Movements**:**Move Right**: Pop from Stack 2, push to Stack 1, update Stack 3.**Move Left**: Pop from Stack 1, push to Stack 2, update Stack 3.

**Read and Write**:- Read the top symbol of the stack corresponding to the head’s position (Stack 3).
- Write a new symbol by updating the appropriate stack.

### c. Power of One-Letter Automaton with Three Stacks

Now, we need to show that a one-letter automaton with three one-letter stacks is equivalent in computational power to a Turing machine.

#### Simulation Strategy for Three Stacks

- We can reduce the number of stacks from four to three by combining the functions of some stacks.

**Stacks**:- Stack 1: Represents the tape content to the left of the head.
- Stack 2: Represents the tape content to the right of the head.
- Stack 3: Combines tracking the head’s current position and auxiliary information for the current state.

#### Steps

**Initialization**:- Stack 1: Empty (left part of the tape).
- Stack 2: Initialized with the input string (right part of the tape).
- Stack 3: Tracks both the head’s position and the current state.

**Tape Movements**:**Move Right**: Pop from Stack 2, push to Stack 1, update Stack 3.**Move Left**: Pop from Stack 1, push to Stack 2, update Stack 3.

**Read and Write**:- Read the top symbol of the stack corresponding to the head’s position (part of Stack 3).
- Write a new symbol by updating the appropriate stack.

#### Conclusion

By showing that a one-letter automaton with four one-letter stacks can simulate a Turing machine and then reducing the number of stacks to three while maintaining the ability to simulate the Turing machine’s behavior, we demonstrate that the computational power of a one-letter automaton with three stacks is equivalent to that of a Turing machine. Thus, the one-letter automaton with three stacks can perform any computation that a Turing machine can, establishing their equivalence in computational power.

### Statement Analysis

Determine the truth or falsity of each statement, providing reasoning for each.

### a) If it is proven that an 𝑁𝑃*NP*-complete problem is in 𝑃*P*, then all 𝑁𝑃*NP* problems are in 𝑃*P*.

**True**

**Reason**: If an 𝑁𝑃*NP*-complete problem is proven to be in 𝑃*P*, it means there is a polynomial-time algorithm for solving that problem. Since 𝑁𝑃*NP*-complete problems are the hardest problems in 𝑁𝑃*NP*, and any 𝑁𝑃*NP* problem can be reduced to an 𝑁𝑃*NP*-complete problem in polynomial time, this would imply that all 𝑁𝑃*NP* problems can also be solved in polynomial time. Thus, 𝑃=𝑁𝑃*P*=*NP*.

### b) If it is proven in the future that the 33-SAT problem can be solved in 𝑜(𝑛6)*o*(*n*6) time, then all 𝑁𝑃*NP* problems can be solved in 𝑜(𝑛6)*o*(*n*6) time.

**True**

**Reason**: If 33-SAT, which is an 𝑁𝑃*NP*-complete problem, can be solved in 𝑜(𝑛6)*o*(*n*6) time, then any problem in 𝑁𝑃*NP* can be reduced to 33-SAT in polynomial time. Since the reduction takes polynomial time and 33-SAT can be solved in 𝑜(𝑛6)*o*(*n*6) time, the combined process would still be within 𝑜(𝑛6)*o*(*n*6) time. Therefore, all 𝑁𝑃*NP* problems would be solvable in 𝑜(𝑛6)*o*(*n*6) time.

### c) If one 𝑁𝑃*NP* problem is solved in polynomial time, all 𝑁𝑃*NP* problems can be solved in polynomial time.

**False**

**Reason**: Solving one 𝑁𝑃*NP* problem in polynomial time does not necessarily imply that all 𝑁𝑃*NP* problems can be solved in polynomial time. Only if the problem solved is 𝑁𝑃*NP*-complete does this implication hold true. The statement lacks this crucial detail. Therefore, the general statement is false.

### d) Consider a problem 𝐴*A*; if it can be reduced to the 33-SAT problem in polynomial time and the 33-SAT problem can be reduced to 𝐴*A* in polynomial time, then problem 𝐴*A* is 𝑁𝑃*NP*-complete.

**True**

**Reason**: To prove that problem 𝐴*A* is 𝑁𝑃*NP*-complete, two conditions must be met:

- 𝐴
*A*is in 𝑁𝑃*NP*. - Any 𝑁𝑃
*NP*problem can be reduced to 𝐴*A*in polynomial time.

Given that 𝐴*A* can be reduced to 33-SAT in polynomial time and vice versa, 𝐴*A* and 33-SAT are polynomially equivalent. Since 33-SAT is 𝑁𝑃*NP*-complete and all 𝑁𝑃*NP* problems can be reduced to 33-SAT, the same reduction can be used to reduce all 𝑁𝑃*NP* problems to 𝐴*A*. Hence, 𝐴*A* is 𝑁𝑃*NP*-complete.

### Summary

**a) True**: If an 𝑁𝑃*NP*-complete problem is in 𝑃*P*, then 𝑃=𝑁𝑃*P*=*NP*.**b) True**: If 33-SAT can be solved in 𝑜(𝑛6)*o*(*n*6) time, then all 𝑁𝑃*NP*problems can be solved in 𝑜(𝑛6)*o*(*n*6) time.**c) False**: Solving one 𝑁𝑃*NP*problem in polynomial time does not imply that all 𝑁𝑃*NP*problems can be solved in polynomial time unless it is 𝑁𝑃*NP*-complete.**d) True**: If problem 𝐴*A*can be reduced to 33-SAT and vice versa in polynomial time, then 𝐴*A*is 𝑁𝑃*NP*-complete.

### Problem Statement

Consider the problem of determining whether a graph 𝐺′*G*′ is a subgraph of a graph 𝐺*G*. Prove that this problem is 𝑁𝑃*NP*-complete.

### Definitions

**Subgraph Isomorphism Problem**: Given two graphs 𝐺*G*and 𝐺′*G*′, determine whether 𝐺′*G*′ is isomorphic to a subgraph of 𝐺*G*.**Class 𝑁𝑃**: A problem is in 𝑁𝑃*NP**NP*if a proposed solution can be verified in polynomial time by a deterministic Turing machine.**Class 𝑁𝑃**: A problem is 𝑁𝑃*NP*-complete*NP*-complete if it is both in 𝑁𝑃*NP*and 𝑁𝑃*NP*-hard. A problem is 𝑁𝑃*NP*-hard if every problem in 𝑁𝑃*NP*can be reduced to it in polynomial time.

### Goal

To prove that the subgraph isomorphism problem is 𝑁𝑃*NP*-complete, we need to show two things:

- The problem is in 𝑁𝑃
*NP*. - The problem is 𝑁𝑃
*NP*-hard.

### Step 1: Prove the Problem is in 𝑁𝑃*NP*

A problem is in 𝑁𝑃*NP* if a proposed solution can be verified in polynomial time.

#### Verification

- Given a mapping of the vertices of 𝐺′
*G*′ to a subset of vertices in 𝐺*G*, we can check in polynomial time whether this mapping preserves the adjacency (i.e., whether the edges in 𝐺′*G*′ correspond to edges in 𝐺*G*). - This verification can be done by checking each edge of 𝐺′
*G*′ and ensuring it maps to an edge in 𝐺*G*.

**Verification Algorithm**:

- For each vertex 𝑢′
*u*′ in 𝐺′*G*′:- Map 𝑢′
*u*′ to a vertex 𝑢*u*in 𝐺*G*.

- Map 𝑢′
- For each edge (𝑢′,𝑣′)(
*u*′,*v*′) in 𝐺′*G*′:- Check if there is an edge (𝑢,𝑣)(
*u*,*v*) in 𝐺*G*corresponding to the mapped vertices.

- Check if there is an edge (𝑢,𝑣)(
- If all edges of 𝐺′
*G*′ are verified to have corresponding edges in 𝐺*G*, the mapping is correct.

The verification process involves checking each vertex and edge of 𝐺′*G*′, which takes polynomial time in the size of 𝐺′*G*′ and 𝐺*G*.

Therefore, the problem is in 𝑁𝑃*NP*.

### Step 2: Prove the Problem is 𝑁𝑃*NP*-hard

To prove the problem is 𝑁𝑃*NP*-hard, we need to show that any problem in 𝑁𝑃*NP* can be reduced to this problem in polynomial time. We will use a known 𝑁𝑃*NP*-complete problem for the reduction. The **Clique Problem** is a suitable choice because it is known to be 𝑁𝑃*NP*-complete.

#### Clique Problem

**Input**: A graph 𝐺*G*and an integer 𝑘*k*.**Output**: Determine if 𝐺*G*contains a clique of size 𝑘*k*(a subset of 𝑘*k*vertices, each pair of which is connected by an edge).

#### Reduction from Clique Problem to Subgraph Isomorphism

- Given an instance of the Clique Problem (𝐺,𝑘)(
*G*,*k*):- Construct a graph 𝐺′
*G*′ which is a complete graph 𝐾𝑘*K**k* on 𝑘*k*vertices. - The question is now whether 𝐺′
*G*′ (which is 𝐾𝑘*K**k*) is a subgraph of 𝐺*G*.

- Construct a graph 𝐺′

**Reduction Algorithm**:

- Construct 𝐺′
*G*′ as a complete graph 𝐾𝑘*K**k* with 𝑘*k*vertices. - Check if 𝐺′
*G*′ is isomorphic to a subgraph of 𝐺*G*using the subgraph isomorphism problem.

If 𝐺*G* contains a clique of size 𝑘*k*, then 𝐾𝑘*K**k* (which is 𝐺′*G*′) will be a subgraph of 𝐺*G*. Conversely, if 𝐾𝑘*K**k* is a subgraph of 𝐺*G*, then 𝐺*G* contains a clique of size 𝑘*k*.

This reduction can be done in polynomial time because constructing 𝐾𝑘*K**k* and checking for subgraph isomorphism are both polynomial-time operations.

Therefore, since the Clique Problem is 𝑁𝑃*NP*-complete and it can be reduced to the subgraph isomorphism problem in polynomial time, the subgraph isomorphism problem is 𝑁𝑃*NP*-hard.

### Conclusion

Since we have shown that the subgraph isomorphism problem is in 𝑁𝑃*NP* and 𝑁𝑃*NP*-hard, it follows that the subgraph isomorphism problem is 𝑁𝑃*NP*-complete.

### Problem Statement

Prove that the complement of the language REG-EQREG-EQ is in the class 𝑁𝑃*NP*.

### Definitions

**REG-EQ**: The language REG-EQREG-EQ consists of pairs ⟨𝐿1,𝐿2⟩⟨*L*1,*L*2⟩, where 𝐿1*L*1 and 𝐿2*L*2 are regular languages represented by regular expressions without the use of the Kleene star (*) operator, and 𝐿1=𝐿2*L*1=*L*2.**Complement of REG-EQ (REG-EQ‾REG-EQ)**: The language consisting of pairs ⟨𝐿1,𝐿2⟩⟨*L*1,*L*2⟩ where 𝐿1*L*1 and 𝐿2*L*2 are not equivalent.**Class 𝑁𝑃**: A problem is in 𝑁𝑃*NP**NP*if a proposed solution can be verified in polynomial time by a deterministic Turing machine.

### Goal

Prove that REG-EQ‾REG-EQ is in 𝑁𝑃*NP*. This involves showing that there exists a polynomial-time verification algorithm for a non-equivalence witness between 𝐿1*L*1 and 𝐿2*L*2.

### Approach

To prove REG-EQ‾REG-EQ is in 𝑁𝑃*NP*, we need to show that there is a polynomial-time verifiable certificate for non-equivalence of 𝐿1*L*1 and 𝐿2*L*2.

### Key Idea

If 𝐿1*L*1 and 𝐿2*L*2 are not equivalent, there exists at least one string 𝑤*w* such that 𝑤*w* is in 𝐿1*L*1 and not in 𝐿2*L*2 or vice versa. This string 𝑤*w* serves as a certificate for non-equivalence.

### Steps to Show REG-EQ‾∈𝑁𝑃REG-EQ∈*NP*

**Certificate for Non-Equivalence**:- The certificate is a string 𝑤
*w*that witnesses the non-equivalence of 𝐿1*L*1 and 𝐿2*L*2.

- The certificate is a string 𝑤
**Verification Algorithm**:- Given the certificate 𝑤
*w*and the regular expressions for 𝐿1*L*1 and 𝐿2*L*2:- Verify that 𝑤
*w*belongs to exactly one of 𝐿1*L*1 or 𝐿2*L*2, but not both. - This can be done by checking membership of 𝑤
*w*in both 𝐿1*L*1 and 𝐿2*L*2 using the regular expressions.

- Verify that 𝑤

- Given the certificate 𝑤

### Detailed Steps

**Input**: Pair ⟨𝐿1,𝐿2⟩⟨*L*1,*L*2⟩ and certificate 𝑤*w*.**Parse Regular Expressions**: Convert the regular expressions for 𝐿1*L*1 and 𝐿2*L*2 into their equivalent deterministic finite automata (DFA) forms. This step is polynomial in the size of the regular expressions because conversion of a regular expression to a DFA is polynomial when no Kleene star is involved.**Check Membership**:- Simulate the DFA for 𝐿1
*L*1 on 𝑤*w*to check if 𝑤∈𝐿1*w*∈*L*1. - Simulate the DFA for 𝐿2
*L*2 on 𝑤*w*to check if 𝑤∈𝐿2*w*∈*L*2.

- Simulate the DFA for 𝐿1
**Verification**:- If 𝑤∈𝐿1
*w*∈*L*1 and 𝑤∉𝐿2*w*∈/*L*2, or 𝑤∈𝐿2*w*∈*L*2 and 𝑤∉𝐿1*w*∈/*L*1, then 𝑤*w*is a valid certificate proving 𝐿1≠𝐿2*L*1=*L*2.

- If 𝑤∈𝐿1

### Pseudocode for Verification Algorithm

`vbnetCopy code````
Input: ⟨L_1, L_2⟩ (as regular expressions) and string w
1. Convert L_1 to DFA D1
2. Convert L_2 to DFA D2
3. Check if w ∈ L_1 by simulating D1 on w
4. Check if w ∈ L_2 by simulating D2 on w
5. If (w ∈ L_1 and w ∉ L_2) or (w ∈ L_2 and w ∉ L_1) then
Accept (witness is valid)
Else
Reject (witness is invalid)
```

### Polynomial Time Complexity

**Conversion**: Converting a regular expression (without Kleene star) to a DFA is polynomial in the size of the regular expression.**Simulation**: Simulating a DFA on a string is linear in the length of the string and the size of the DFA.- The overall verification algorithm runs in polynomial time relative to the size of ⟨𝐿1,𝐿2⟩⟨
*L*1,*L*2⟩ and 𝑤*w*.

### Conclusion

Since we can verify a certificate (string 𝑤*w* witnessing the non-equivalence of 𝐿1*L*1 and 𝐿2*L*2) in polynomial time, the complement of REG-EQREG-EQ is in 𝑁𝑃*NP*. This shows that REG-EQ‾∈𝑁𝑃REG-EQ∈*NP*.

### Statement Analysis

Determine the truth or falsity of each statement, providing reasoning or a counterexample for each.

### a) If 𝐴*A* and 𝐵*B* are two 𝑁𝑃*NP*-complete languages, then 𝐴∩𝐵*A*∩*B* is 𝑁𝑃*NP*-complete.

**False**

**Counterexample**:

**Languages**: Consider 𝐴=SAT*A*=SAT and 𝐵=TAUT*B*=TAUT.**SAT**: The language of satisfiable Boolean formulas.**TAUT**: The language of tautologies (Boolean formulas that are true under all assignments).

**NP-completeness**:**SAT**is 𝑁𝑃*NP*-complete.**TAUT**is in 𝑐𝑜𝑁𝑃*co**NP*, not 𝑁𝑃*NP*.

**Intersection**:- 𝐴∩𝐵=SAT∩TAUT
*A*∩*B*=SAT∩TAUT. - Since TAUTTAUT contains only formulas that are always true and SATSAT contains formulas that are true under at least one assignment, their intersection 𝐴∩𝐵
*A*∩*B*is empty. Therefore, SAT∩TAUT=∅SAT∩TAUT=∅.

- 𝐴∩𝐵=SAT∩TAUT
**Complexity**:- The empty set is trivially in 𝑃
*P*, but it is not 𝑁𝑃*NP*-complete unless 𝑃=𝑁𝑃*P*=*NP*.

- The empty set is trivially in 𝑃

Therefore, 𝐴∩𝐵*A*∩*B* is not necessarily 𝑁𝑃*NP*-complete if 𝐴*A* and 𝐵*B* are 𝑁𝑃*NP*-complete.

### b) If 𝐴*A* and 𝐵*B* are two 𝑁𝑃*NP*-complete languages, then 𝐴∪𝐵*A*∪*B* is 𝑁𝑃*NP*-complete.

**True**

**Proof**:

**Membership in 𝑁𝑃**:*NP*- To show 𝐴∪𝐵∈𝑁𝑃
*A*∪*B*∈*NP*, consider an instance 𝑥*x*and a nondeterministic Turing machine 𝑀*M*. - 𝑀
*M*guesses whether 𝑥*x*should be verified for membership in 𝐴*A*or 𝐵*B*. - If 𝑥
*x*is guessed to be in 𝐴*A*, 𝑀*M*runs the verifier for 𝐴*A*. If 𝑥*x*is guessed to be in 𝐵*B*, 𝑀*M*runs the verifier for 𝐵*B*. - Since both verifiers run in polynomial time, 𝑀
*M*also runs in polynomial time.

- To show 𝐴∪𝐵∈𝑁𝑃
**NP-hardness**:- To show 𝐴∪𝐵
*A*∪*B*is 𝑁𝑃*NP*-hard, consider an arbitrary 𝑁𝑃*NP*problem 𝐶*C*. - Since 𝐴
*A*and 𝐵*B*are 𝑁𝑃*NP*-complete, there are polynomial-time reductions from 𝐶*C*to 𝐴*A*and from 𝐶*C*to 𝐵*B*. - Let 𝑓
*f*be the polynomial-time reduction from 𝐶*C*to 𝐴*A*. - Let 𝑔
*g*be the polynomial-time reduction from 𝐶*C*to 𝐵*B*. - Construct a new reduction ℎ
*h*from 𝐶*C*to 𝐴∪𝐵*A*∪*B*as follows:- For an instance 𝑥
*x*of 𝐶*C*, compute 𝑓(𝑥)*f*(*x*) and 𝑔(𝑥)*g*(*x*). - Define ℎ(𝑥)=𝑓(𝑥)
*h*(*x*)=*f*(*x*) if 𝑓(𝑥)∈𝐴*f*(*x*)∈*A*, otherwise ℎ(𝑥)=𝑔(𝑥)*h*(*x*)=*g*(*x*).

- For an instance 𝑥
- This reduction works because:
- If 𝑥∈𝐶
*x*∈*C*, then either 𝑓(𝑥)∈𝐴*f*(*x*)∈*A*or 𝑔(𝑥)∈𝐵*g*(*x*)∈*B*(or both), so ℎ(𝑥)∈𝐴∪𝐵*h*(*x*)∈*A*∪*B*. - If 𝑥∉𝐶
*x*∈/*C*, then neither 𝑓(𝑥)∈𝐴*f*(*x*)∈*A*nor 𝑔(𝑥)∈𝐵*g*(*x*)∈*B*, so ℎ(𝑥)∉𝐴∪𝐵*h*(*x*)∈/*A*∪*B*.

- If 𝑥∈𝐶
- Since ℎ
*h*is a polynomial-time reduction, 𝐴∪𝐵*A*∪*B*is 𝑁𝑃*NP*-hard.

- To show 𝐴∪𝐵
**Conclusion**:- Since 𝐴∪𝐵
*A*∪*B*is in 𝑁𝑃*NP*and 𝑁𝑃*NP*-hard, it is 𝑁𝑃*NP*-complete.

- Since 𝐴∪𝐵

Therefore, if 𝐴*A* and 𝐵*B* are 𝑁𝑃*NP*-complete languages, 𝐴∪𝐵*A*∪*B* is 𝑁𝑃*NP*-complete.

### Showing SATSAT is 𝑁𝑃*NP*-Complete

To demonstrate that SATSAT (the Boolean satisfiability problem) is 𝑁𝑃*NP*-complete, we need to show two things:

**SATSAT is in 𝑁𝑃**.*NP***SATSAT is 𝑁𝑃**.*NP*-hard

### 1. SATSAT is in 𝑁𝑃*NP*

A problem is in 𝑁𝑃*NP* if a proposed solution can be verified in polynomial time by a deterministic Turing machine.

**Verification Process for SATSAT**:

- Given a Boolean formula 𝜙
*ϕ*in conjunctive normal form (CNF) and a satisfying assignment 𝜎*σ*, the verifier can check if 𝜎*σ*indeed satisfies 𝜙*ϕ*. - The verifier evaluates 𝜙
*ϕ*by substituting the variables in 𝜙*ϕ*with the values in 𝜎*σ*. - If 𝜙
*ϕ*evaluates to true, then 𝜎*σ*is a valid satisfying assignment.

**Time Complexity**:

- Each clause in 𝜙
*ϕ*can be checked in constant time. - Since there are a polynomial number of clauses, the overall verification process runs in polynomial time.

Thus, SATSAT is in 𝑁𝑃*NP*.

### 2. SATSAT is 𝑁𝑃*NP*-hard

To show that SATSAT is 𝑁𝑃*NP*-hard, we need to reduce any problem in 𝑁𝑃*NP* to SATSAT in polynomial time. Stephen Cook’s proof of the Cook-Levin theorem provides this reduction by showing that any problem in 𝑁𝑃*NP* can be translated to an instance of SATSAT.

**Sketch of Cook-Levin Theorem**:

- For any problem 𝐿∈𝑁𝑃
*L*∈*NP*, there exists a nondeterministic Turing machine 𝑀*M*that decides 𝐿*L*in polynomial time. - We can construct a Boolean formula 𝜙
*ϕ*that represents the computation of 𝑀*M*on an input 𝑥*x*. - The formula 𝜙
*ϕ*is constructed in such a way that it is satisfiable if and only if 𝑀*M*accepts 𝑥*x*.

**Construction of the Formula 𝜙 ϕ**:

**Encoding Turing Machine Computation**:- Represent the computation of 𝑀
*M*using Boolean variables to encode the states, tape symbols, and head positions at each step of the computation. - Use a polynomial number of variables, given that 𝑀
*M*runs in polynomial time.

- Represent the computation of 𝑀
**Construct Clauses**:- Encode the initial configuration of 𝑀
*M*. - Encode the transition function of 𝑀
*M*to ensure valid transitions between configurations. - Encode the accepting condition of 𝑀
*M*to ensure that 𝑀*M*reaches an accepting state.

- Encode the initial configuration of 𝑀
**Formula 𝜙**:*ϕ*- Combine the clauses to form a CNF formula 𝜙
*ϕ*. - 𝜙
*ϕ*is satisfiable if and only if there exists an accepting computation for 𝑀*M*on 𝑥*x*.

- Combine the clauses to form a CNF formula 𝜙

**Reduction**:

- Given any problem 𝐿∈𝑁𝑃
*L*∈*NP*, construct the formula 𝜙*ϕ*representing the computation of the polynomial-time verifier for 𝐿*L*. - This construction can be done in polynomial time, resulting in a polynomial-time reduction from 𝐿
*L*to SATSAT.

### Conclusion

Since SATSAT is in 𝑁𝑃*NP* and 𝑁𝑃*NP*-hard, SATSAT is 𝑁𝑃*NP*-complete. This conclusion is foundational in computational complexity theory and is the basis for proving the 𝑁𝑃*NP*-completeness of many other problems through polynomial-time reductions.

Consider the language Common-SubsCommon-Subs which consists of all inputs of the form ⟨𝐷,𝑀,𝑘⟩⟨*D*,*M*,*k*⟩ where 𝐷*D* is a collection of subsets of 𝑀*M*, and 𝑀*M* has at least one 𝑘*k*-element subset that intersects with every member of 𝐷*D*. Prove that Common-SubsCommon-Subs is 𝑁𝑃*NP*-complete.

### Definitions

**Collection 𝐷**: A collection of subsets of 𝑀*D**M*.**Subset 𝑆⊆𝑀**: A 𝑘*S*⊆*M**k*-element subset of 𝑀*M*is considered a hitting set for 𝐷*D*if 𝑆*S*intersects with every subset in 𝐷*D*.**Language Common-SubsCommon-Subs**: The set of all inputs ⟨𝐷,𝑀,𝑘⟩⟨*D*,*M*,*k*⟩ such that there exists a 𝑘*k*-element subset 𝑆⊆𝑀*S*⊆*M*that intersects with every subset in 𝐷*D*.

### Goal

Prove that Common-SubsCommon-Subs is 𝑁𝑃*NP*-complete.

### Approach

To show that Common-SubsCommon-Subs is 𝑁𝑃*NP*-complete, we need to prove two things:

- Common-SubsCommon-Subs is in 𝑁𝑃
*NP*. - Common-SubsCommon-Subs is 𝑁𝑃
*NP*-hard.

### 1. Common-SubsCommon-Subs is in 𝑁𝑃*NP*

A problem is in 𝑁𝑃*NP* if a proposed solution can be verified in polynomial time by a deterministic Turing machine.

**Verification Process for Common-SubsCommon-Subs**:

- Given a candidate 𝑘
*k*-element subset 𝑆*S*of 𝑀*M*, verify that 𝑆*S*intersects with every subset in 𝐷*D*. - This involves checking, for each subset 𝑑∈𝐷
*d*∈*D*, if 𝑆∩𝑑≠∅*S*∩*d*=∅.

**Verification Algorithm**:

- For each subset 𝑑∈𝐷
*d*∈*D*:- Check if 𝑆∩𝑑≠∅
*S*∩*d*=∅.

- Check if 𝑆∩𝑑≠∅
- If all checks pass, accept; otherwise, reject.

**Time Complexity**:

- Checking intersection for each subset 𝑑
*d*with 𝑆*S*can be done in 𝑂(𝑘⋅∣𝑑∣)*O*(*k*⋅∣*d*∣) time. - Since there are ∣𝐷∣∣
*D*∣ subsets to check, the overall verification process runs in polynomial time.

Thus, Common-SubsCommon-Subs is in 𝑁𝑃*NP*.

### 2. Common-SubsCommon-Subs is 𝑁𝑃*NP*-hard

To show that Common-SubsCommon-Subs is 𝑁𝑃*NP*-hard, we need to reduce a known 𝑁𝑃*NP*-complete problem to Common-SubsCommon-Subs in polynomial time. We will use the **Hitting Set** problem, which is known to be 𝑁𝑃*NP*-complete.

#### Hitting Set Problem

**Input**: A collection 𝐷*D*of subsets of a finite set 𝑀*M*, and an integer 𝑘*k*.**Output**: Determine if there exists a subset 𝑆⊆𝑀*S*⊆*M*of size at most 𝑘*k*that intersects with every subset in 𝐷*D*.

#### Reduction from Hitting Set to Common-SubsCommon-Subs

- Given an instance ⟨𝐷,𝑀,𝑘⟩⟨
*D*,*M*,*k*⟩ of the Hitting Set problem, construct the corresponding instance ⟨𝐷,𝑀,𝑘⟩⟨*D*,*M*,*k*⟩ of Common-SubsCommon-Subs. - The instance remains the same because the problem definitions are equivalent: finding a 𝑘
*k*-element hitting set is the same as finding a 𝑘*k*-element subset that intersects with every subset in 𝐷*D*.

**Reduction Algorithm**:

- Take the instance ⟨𝐷,𝑀,𝑘⟩⟨
*D*,*M*,*k*⟩ of the Hitting Set problem. - Directly use this instance as the instance for the Common-SubsCommon-Subs problem.

**Correctness of the Reduction**:

- If the Hitting Set instance ⟨𝐷,𝑀,𝑘⟩⟨
*D*,*M*,*k*⟩ has a solution, then the corresponding Common-SubsCommon-Subs instance ⟨𝐷,𝑀,𝑘⟩⟨*D*,*M*,*k*⟩ also has a solution. - If the Common-SubsCommon-Subs instance ⟨𝐷,𝑀,𝑘⟩⟨
*D*,*M*,*k*⟩ has a solution, then the Hitting Set instance ⟨𝐷,𝑀,𝑘⟩⟨*D*,*M*,*k*⟩ also has a solution.

Since the reduction is trivial and can be performed in polynomial time, Common-SubsCommon-Subs is 𝑁𝑃*NP*-hard.

### Conclusion

Since we have shown that Common-SubsCommon-Subs is in 𝑁𝑃*NP* and 𝑁𝑃*NP*-hard, we conclude that Common-SubsCommon-Subs is 𝑁𝑃*NP*-complete.

### Problem Statement

Prove that the language 𝐿*L* consisting of all DFAs (Deterministic Finite Automata) whose languages are infinite is decidable.

### Definitions

**Language 𝐿**: The set of all DFAs 𝐴*L**A*such that the language 𝐿(𝐴)*L*(*A*) recognized by 𝐴*A*is infinite.**Decidable Language**: A language is decidable if there exists a Turing machine that will accept all strings in the language and reject all strings not in the language, and it always halts.

### Approach

To prove that 𝐿*L* is decidable, we need to construct a decision procedure (algorithm) that, given a DFA, determines whether the language recognized by the DFA is infinite.

### Steps to Prove Decidability

**Understanding the Problem**:- A DFA has an infinite language if and only if there exists some string that can be repeated infinitely while still being accepted by the DFA.
- This can be determined by checking for cycles that can be traversed from some state reachable from the start state.

**Key Insight**:- If a DFA accepts a language that is infinite, it must have a cycle in its transition graph that can be visited infinitely often starting from some reachable state.

**Algorithm to Decide 𝐿**:*L*- Given a DFA 𝐴=(𝑄,Σ,𝛿,𝑞0,𝐹)
*A*=(*Q*,Σ,*δ*,*q*0,*F*):- Use a breadth-first search (BFS) or depth-first search (DFS) to detect reachable states from the start state 𝑞0
*q*0. - For each reachable state, check for cycles in the transition graph.
- If a reachable state with a cycle is found, check if there exists a path from this state to an accepting state. If so, the DFA recognizes an infinite language.
- If no such cycles are found, the DFA recognizes a finite language.

- Use a breadth-first search (BFS) or depth-first search (DFS) to detect reachable states from the start state 𝑞0

- Given a DFA 𝐴=(𝑄,Σ,𝛿,𝑞0,𝐹)

### Detailed Algorithm

**Reachability Analysis**:- Use BFS or DFS starting from the initial state 𝑞0
*q*0 to mark all reachable states.

- Use BFS or DFS starting from the initial state 𝑞0
**Cycle Detection**:- For each reachable state 𝑞
*q*:- Check if there is a cycle starting and ending at 𝑞
*q*. - This can be done using a DFS. If during the DFS from 𝑞
*q*we encounter 𝑞*q*again, a cycle is detected.

- Check if there is a cycle starting and ending at 𝑞

- For each reachable state 𝑞
**Check Acceptance**:- For each state involved in the cycle, check if there exists a path to an accepting state 𝑓∈𝐹
*f*∈*F*. - If such a path exists, the DFA recognizes an infinite language.

- For each state involved in the cycle, check if there exists a path to an accepting state 𝑓∈𝐹

### Pseudocode for the Algorithm

`pythonCopy code````
def is_infinite_language(DFA):
Q, Σ, δ, q0, F = DFA # States, Alphabet, Transition function, Start state, Accepting states
# Step 1: Use BFS or DFS to find all reachable states from q0
reachable = set()
stack = [q0]
while stack:
state = stack.pop()
if state not in reachable:
reachable.add(state)
for symbol in Σ:
stack.append(δ[state][symbol])
# Step 2: For each reachable state, check for cycles using DFS
def has_cycle(state):
visited = set()
stack = [(state, set())] # (current state, path)
while stack:
current_state, path = stack.pop()
if current_state in path:
return True
if current_state not in visited:
visited.add(current_state)
new_path = path | {current_state}
for symbol in Σ:
stack.append((δ[current_state][symbol], new_path))
return False
# Step 3: Check if there exists a cycle with a path to an accepting state
for state in reachable:
if has_cycle(state):
# Check for a path from cycle to an accepting state
stack = [state]
visited = set()
while stack:
current_state = stack.pop()
if current_state in F:
return True # Infinite language
if current_state not in visited:
visited.add(current_state)
for symbol in Σ:
stack.append(δ[current_state][symbol])
return False # Finite language
def is_DFA_in_L(DFA):
return is_infinite_language(DFA)
```

### Explanation

**Reachable States**: First, identify all states that can be reached from the initial state 𝑞0*q*0. This helps in narrowing down the states to consider for cycles.**Cycle Detection**: For each reachable state, use DFS to detect cycles. If a state can reach itself through a path of transitions, a cycle is detected.**Acceptance Check**: For states involved in cycles, ensure that there exists a path to an accepting state. This confirms that the cycle contributes to the infinite language.

### Conclusion

The algorithm above provides a decision procedure to determine if the language recognized by a given DFA is infinite. Since the algorithm always halts and correctly decides whether a DFA’s language is infinite, the language 𝐿*L* is decidable. Thus, we have shown that the problem of determining if a DFA’s language is infinite is in the class of decidable languages.

**Algorithm to Decide 𝐿 L**:

- Given a DFA 𝐴=(𝑄,Σ,𝛿,𝑞0,𝐹)
*A*=(*Q*,Σ,*δ*,*q*0,*F*):- Use a breadth-first search (BFS) or depth-first search (DFS) to detect reachable states from the start state 𝑞0
*q*0. - For each reachable state, check for cycles in the transition graph.
- If a reachable state with a cycle is found, check if there exists a path from this state to an accepting state. If so, the DFA recognizes an infinite language.
- If no such cycles are found, the DFA recognizes a finite language

- Use a breadth-first search (BFS) or depth-first search (DFS) to detect reachable states from the start state 𝑞0

we can finde that cycles too

### Problem Statement

Prove that the language 𝐿*L* consisting of pairs ⟨𝐵,𝐺⟩⟨*B*,*G*⟩ where 𝐺*G* is a context-free grammar and 𝐵*B* is a variable in 𝐺*G* such that 𝐵*B* is used in the derivation of at least one string produced by 𝐺*G* is decidable.

### Definitions

**Language 𝐿**: The set of pairs ⟨𝐵,𝐺⟩⟨*L**B*,*G*⟩ where 𝐺=(𝑉,Σ,𝑅,𝑆)*G*=(*V*,Σ,*R*,*S*) is a context-free grammar, 𝐵∈𝑉*B*∈*V*, and 𝐵*B*is used in the derivation of at least one string produced by 𝐺*G*.**Decidable Language**: A language is decidable if there exists a Turing machine that will accept all strings in the language and reject all strings not in the language, and it always halts.

### Goal

To prove that 𝐿*L* is decidable, we need to construct a decision procedure (algorithm) that, given a pair ⟨𝐵,𝐺⟩⟨*B*,*G*⟩, determines whether 𝐵*B* is used in the derivation of at least one string produced by 𝐺*G*.

### Approach

To show that 𝐿*L* is decidable, we need to determine if there exists a string in the language generated by 𝐺*G* that uses the variable 𝐵*B* in its derivation. This involves:

- Constructing a decision procedure to check if 𝐵
*B*can appear in any derivation starting from the start symbol 𝑆*S*. - Utilizing graph reachability techniques to analyze the production rules of the grammar 𝐺
*G*.

### Steps to Prove Decidability

**Construct Dependency Graph**:- Construct a directed graph where each variable 𝐴
*A*in 𝐺*G*is a node. - Add a directed edge 𝐴→𝐶
*A*→*C*if there exists a production 𝐴→𝛼*A*→*α*in 𝑅*R*such that 𝐶*C*appears in 𝛼*α*.

- Construct a directed graph where each variable 𝐴
**Check Reachability**:- Perform a reachability analysis starting from the start symbol 𝑆
*S*to see if 𝐵*B*can be reached. - This can be done using a depth-first search (DFS) or breadth-first search (BFS).

- Perform a reachability analysis starting from the start symbol 𝑆
**Decision Procedure**:- If 𝐵
*B*is reachable from 𝑆*S*in the dependency graph, accept the pair ⟨𝐵,𝐺⟩⟨*B*,*G*⟩. - Otherwise, reject the pair ⟨𝐵,𝐺⟩⟨
*B*,*G*⟩.

- If 𝐵

### Detailed Algorithm

**Input**: A pair ⟨𝐵,𝐺⟩⟨*B*,*G*⟩, where 𝐺=(𝑉,Σ,𝑅,𝑆)*G*=(*V*,Σ,*R*,*S*) and 𝐵∈𝑉*B*∈*V*.**Construct Dependency Graph**:- Initialize an empty directed graph 𝐺′
*G*′ with nodes corresponding to the variables in 𝑉*V*. - For each production 𝐴→𝛼
*A*→*α*in 𝑅*R*:- For each variable 𝐶
*C*in 𝛼*α*, add an edge from 𝐴*A*to 𝐶*C*in 𝐺′*G*′.

- For each variable 𝐶

- Initialize an empty directed graph 𝐺′
**Perform Reachability Analysis**:- Initialize a set
`reachable`

to keep track of variables that can be reached from 𝑆*S*. - Initialize a stack with 𝑆
*S*. - While the stack is not empty:
- Pop a variable 𝑋
*X*from the stack. - For each variable 𝑌
*Y*such that there is an edge 𝑋→𝑌*X*→*Y*in 𝐺′*G*′:- If 𝑌
*Y*is not in`reachable`

:- Add 𝑌
*Y*to`reachable`

. - Push 𝑌
*Y*onto the stack.

- Add 𝑌

- If 𝑌

- Pop a variable 𝑋

- Initialize a set
**Check if 𝐵**:*B*is Reachable- If 𝐵
*B*is in`reachable`

, accept the pair ⟨𝐵,𝐺⟩⟨*B*,*G*⟩. - Otherwise, reject the pair ⟨𝐵,𝐺⟩⟨
*B*,*G*⟩.

- If 𝐵

### Pseudocode for the Algorithm

`pythonCopy code````
def is_variable_used_in_derivation(B, G):
V, Σ, R, S = G # Variables, Alphabet, Productions, Start symbol
# Step 1: Construct Dependency Graph
dependency_graph = {variable: set() for variable in V}
for A, production in R.items():
for symbol in production:
if symbol in V:
dependency_graph[A].add(symbol)
# Step 2: Perform Reachability Analysis
reachable = set()
stack = [S]
while stack:
X = stack.pop()
if X not in reachable:
reachable.add(X)
for Y in dependency_graph[X]:
if Y not in reachable:
stack.append(Y)
# Step 3: Check if B is reachable from S
return B in reachable
def is_in_L(B, G):
return is_variable_used_in_derivation(B, G)
# Example usage:
V = {'S', 'A', 'B', 'C'}
Σ = {'a', 'b'}
R = {
'S': ['AB', 'b'],
'A': ['a', 'BC'],
'B': ['b'],
'C': ['S', 'a']
}
S = 'S'
G = (V, Σ, R, S)
print(is_in_L('B', G)) # Should return True if 'B' is used in any derivation from 'S'
```

### Explanation

**Dependency Graph**: Constructing the dependency graph helps to visualize which variables depend on which other variables in the derivations.**Reachability Analysis**: Using a stack to keep track of reachable variables ensures that we explore all possible derivations starting from the start symbol.**Decision**: The reachability check effectively determines if the variable 𝐵*B*can be used in any derivation from the start symbol 𝑆*S*.

### Conclusion

By constructing a dependency graph and performing reachability analysis, we can determine if a variable 𝐵*B* is used in the derivation of at least one string produced by the grammar 𝐺*G*. This decision procedure always halts and correctly identifies whether 𝐵*B* is used, proving that the language 𝐿*L* is decidable.

### teps to Prove Decidability

**Construct Dependency Graph**:- Construct a directed graph where each variable 𝐴
*A*in 𝐺*G*is a node. - Add a directed edge 𝐴→𝐶
*A*→*C*if there exists a production 𝐴→𝛼*A*→*α*in 𝑅*R*such that 𝐶*C*appears in 𝛼*α*.

- Construct a directed graph where each variable 𝐴
**Check Reachability**:- Perform a reachability analysis starting from the start symbol 𝑆
*S*to see if 𝐵*B*can be reached. - This can be done using a depth-first search (DFS) or breadth-first search (BFS).

- Perform a reachability analysis starting from the start symbol 𝑆
**Decision Procedure**:- If 𝐵
*B*is reachable from 𝑆*S*in the dependency graph, accept the pair ⟨𝐵,𝐺⟩⟨*B*,*G*⟩. - Otherwise, reject the pair ⟨𝐵,𝐺⟩⟨
*B*,*G*⟩.

- If 𝐵

we can simply have just a graph by htat to decide it

we just need a decider for this

### Problem Statement

Prove that the language 𝑅*R*, which describes languages that include at least one string 𝑤*w* such that “111” is a substring of 𝑤*w*, is decidable.

### Definitions

**Language 𝑅**: The set of all regular expressions 𝑅*R**R*over an alphabet ΣΣ such that the language described by 𝑅*R*contains at least one string 𝑤*w*with “111” as a substring.

### Goal

To prove that 𝑅*R* is decidable, we need to construct a decision procedure (algorithm) that, given a regular expression 𝑅*R*, determines whether there exists at least one string in the language described by 𝑅*R* that contains “111” as a substring.

### Approach

To show that 𝑅*R* is decidable, we can use the following steps:

- Convert the regular expression 𝑅
*R*to an equivalent deterministic finite automaton (DFA) 𝐴*A*. - Construct a new DFA 𝐴′
*A*′ that accepts exactly those strings containing “111” as a substring. - Check if the intersection of the languages of 𝐴
*A*and 𝐴′*A*′ is non-empty. - If the intersection is non-empty, then 𝑅
*R*describes a language containing at least one string with “111” as a substring.

### Detailed Steps

**Convert Regular Expression 𝑅**:*R*to DFA 𝐴*A*- Given a regular expression 𝑅
*R*, there are standard algorithms to convert 𝑅*R*to an equivalent nondeterministic finite automaton (NFA) 𝑁*N*, and from 𝑁*N*to an equivalent DFA 𝐴*A*. - The resulting DFA 𝐴
*A*accepts the same language as 𝑅*R*.

- Given a regular expression 𝑅
**Construct DFA 𝐴′**:*A*′ for Strings Containing “111”- Construct a DFA 𝐴′
*A*′ that accepts any string containing the substring “111”. - The DFA 𝐴′
*A*′ will have states representing whether the last few characters seen are building up to “111”:- 𝑞0
*q*0: Initial state, no “1”s seen yet. - 𝑞1
*q*1: Seen “1”. - 𝑞2
*q*2: Seen “11”. - 𝑞3
*q*3: Seen “111” (accepting state).

- 𝑞0
- Transitions:
- From 𝑞0
*q*0: On “1” go to 𝑞1*q*1, on other characters stay in 𝑞0*q*0. - From 𝑞1
*q*1: On “1” go to 𝑞2*q*2, on other characters go to 𝑞0*q*0. - From 𝑞2
*q*2: On “1” go to 𝑞3*q*3, on other characters go to 𝑞0*q*0. - From 𝑞3
*q*3: On any character stay in 𝑞3*q*3.

- From 𝑞0

- Construct a DFA 𝐴′
**Intersection of 𝐴**:*A*and 𝐴′*A*′- Construct a DFA 𝐴′′
*A*′′ that recognizes the intersection of the languages of 𝐴*A*and 𝐴′*A*′. - This can be done using the product construction:
- The states of 𝐴′′
*A*′′ are pairs of states from 𝐴*A*and 𝐴′*A*′. - The transition function of 𝐴′′
*A*′′ is defined such that it simulates both 𝐴*A*and 𝐴′*A*′ simultaneously.

- The states of 𝐴′′

- Construct a DFA 𝐴′′
**Check Emptiness of the Intersection**:- Check if the language recognized by 𝐴′′
*A*′′ is non-empty. - This can be done by searching for any accepting state in 𝐴′′
*A*′′ that is reachable from the initial state.

- Check if the language recognized by 𝐴′′

### Pseudocode for the Algorithm

`pythonCopy code````
def convert_regex_to_dfa(regex):
# Implement or use a library function to convert regex to NFA and then to DFA
nfa = regex_to_nfa(regex)
dfa = nfa_to_dfa(nfa)
return dfa
def construct_dfa_for_111():
# Construct DFA for strings containing "111"
dfa_111 = DFA(
states={'q0', 'q1', 'q2', 'q3'},
alphabet={'0', '1'},
transition={
('q0', '0'): 'q0', ('q0', '1'): 'q1',
('q1', '0'): 'q0', ('q1', '1'): 'q2',
('q2', '0'): 'q0', ('q2', '1'): 'q3',
('q3', '0'): 'q3', ('q3', '1'): 'q3'
},
start_state='q0',
accept_states={'q3'}
)
return dfa_111
def intersect_dfa(dfa1, dfa2):
# Implement the product construction for intersection of two DFAs
states = set((q1, q2) for q1 in dfa1.states for q2 in dfa2.states)
alphabet = dfa1.alphabet.intersection(dfa2.alphabet)
transition = {}
for (q1, a1), t1 in dfa1.transition.items():
for (q2, a2), t2 in dfa2.transition.items():
if a1 == a2:
transition[((q1, q2), a1)] = (t1, t2)
start_state = (dfa1.start_state, dfa2.start_state)
accept_states = set((q1, q2) for q1 in dfa1.accept_states for q2 in dfa2.accept_states)
return DFA(states, alphabet, transition, start_state, accept_states)
def is_language_nonempty(dfa):
# Check if there is any path to an accept state in the DFA
visited = set()
stack = [dfa.start_state]
while stack:
state = stack.pop()
if state in dfa.accept_states:
return True
if state not in visited:
visited.add(state)
for symbol in dfa.alphabet:
next_state = dfa.transition.get((state, symbol))
if next_state and next_state not in visited:
stack.append(next_state)
return False
def is_in_R(regex):
dfa = convert_regex_to_dfa(regex)
dfa_111 = construct_dfa_for_111()
intersection_dfa = intersect_dfa(dfa, dfa_111)
return is_language_nonempty(intersection_dfa)
# Example usage:
regex = '1*01*1*' # Example regular expression
print(is_in_R(regex)) # Should return True if the language of the regex contains a string with "111" as a substring
```

### Explanation

**Convert Regular Expression to DFA**: The function`convert_regex_to_dfa`

converts a given regular expression to its equivalent DFA.**Construct DFA for “111”**: The function`construct_dfa_for_111`

constructs a DFA that recognizes strings containing the substring “111”.**Intersection DFA**: The function`intersect_dfa`

constructs the DFA for the intersection of the languages of two given DFAs using the product construction.**Check Non-Emptiness**: The function`is_language_nonempty`

checks if the intersection DFA accepts any strings, indicating that the original language contains at least one string with “111” as a substring.

### Conclusion

By constructing a DFA that recognizes strings containing “111”, intersecting it with the DFA derived from the given regular expression, and checking if the resulting DFA accepts any strings, we can decide whether the language described by the regular expression contains a string with “111” as a substring. Thus, the language 𝑅*R* is decidable.

### Problem Statement

Given two languages 𝐴*A* and 𝐵*B* that are Turing-recognizable (recursively enumerable), determine whether the language 𝐴−𝐵*A*−*B* is decidable (Turing-decidable) or undecidable (Turing-undecidable).

### Definitions

**Turing-Recognizable Language**: A language is Turing-recognizable if there exists a Turing machine that will accept all strings in the language and either reject or run forever on strings not in the language.**Decidable Language**: A language is decidable if there exists a Turing machine that will accept all strings in the language and reject all strings not in the language, and it always halts.**Language Difference**: 𝐴−𝐵*A*−*B*is the set of strings that are in 𝐴*A*but not in 𝐵*B*, i.e., 𝐴−𝐵={𝑥∣𝑥∈𝐴 and 𝑥∉𝐵}*A*−*B*={*x*∣*x*∈*A*and*x*∈/*B*}.

### Goal

Determine the decidability of the language 𝐴−𝐵*A*−*B*.

### Proof Outline

To determine whether 𝐴−𝐵*A*−*B* is decidable, we need to explore the properties of 𝐴*A* and 𝐵*B* being Turing-recognizable and understand the implications for the language 𝐴−𝐵*A*−*B*.

**Understanding 𝐴**:*A*and 𝐵*B*- 𝐴
*A*is Turing-recognizable: There exists a Turing machine 𝑀𝐴*M**A* that accepts all strings in 𝐴*A*. - 𝐵
*B*is Turing-recognizable: There exists a Turing machine 𝑀𝐵*M**B* that accepts all strings in 𝐵*B*.

- 𝐴
**Language Difference 𝐴−𝐵**:*A*−*B*- The language 𝐴−𝐵
*A*−*B*consists of all strings that are in 𝐴*A*but not in 𝐵*B*.

- The language 𝐴−𝐵

Prove that the problem of determining whether a Turing machine 𝑀*M* accepts only strings that are palindromes is undecidable.

### Definitions

**Palindrome**: A string that reads the same forwards and backwards.**Turing Machine 𝑀**: A computational model that accepts or rejects strings based on a set of states and transitions.*M***Language of 𝑀**: The set of all strings that 𝑀*M**M*accepts.

### Goal

Prove that the problem of deciding whether a Turing machine 𝑀*M* accepts only palindromes is undecidable.

### Approach

To show that the problem is undecidable, we can use a reduction from a known undecidable problem. We will reduce the **Halting Problem** to our problem.

### Halting Problem

**Input**: A Turing machine 𝑀*M*and an input string 𝑤*w*.**Question**: Does 𝑀*M*halt on input 𝑤*w*?- The Halting Problem is known to be undecidable.

### Reduction Strategy

We will show that if we could decide whether a Turing machine accepts only palindromes, we could solve the Halting Problem, which is a contradiction since the Halting Problem is undecidable.

### Construction

Given an instance of the Halting Problem ⟨𝑀,𝑤⟩⟨*M*,*w*⟩, construct a new Turing machine 𝑀′*M*′ such that:

- If 𝑀
*M*halts on input 𝑤*w*, 𝑀′*M*′ accepts all strings. - If 𝑀
*M*does not halt on input 𝑤*w*, 𝑀′*M*′ accepts only palindromes.

### Construction of 𝑀′*M*′

**Modify 𝑀**to create 𝑀′*M**M*′ as follows:- 𝑀′
*M*′ simulates 𝑀*M*on input 𝑤*w*. - If 𝑀
*M*halts on 𝑤*w*, 𝑀′*M*′ transitions to a state where it accepts all strings. - If 𝑀
*M*does not halt on 𝑤*w*, 𝑀′*M*′ accepts only palindromes.

- 𝑀′

### Detailed Construction

- Construct 𝑀′
*M*′ with the following behavior:- On input 𝑥
*x*, 𝑀′*M*′ simulates 𝑀*M*on 𝑤*w*. - If 𝑀
*M*halts on 𝑤*w*, 𝑀′*M*′ enters a special state where it accepts any input 𝑥*x*. - If 𝑀
*M*does not halt on 𝑤*w*, 𝑀′*M*′ checks if 𝑥*x*is a palindrome. If 𝑥*x*is a palindrome, 𝑀′*M*′ accepts 𝑥*x*; otherwise, it rejects 𝑥*x*.

- On input 𝑥

### Formal Definition of 𝑀′*M*′

**States**: 𝑄′=𝑄∪{𝑞pal,𝑞accept-all}*Q*′=*Q*∪{*q*pal,*q*accept-all}**Transitions**:- For each transition in 𝑀
*M*, add the corresponding transition in 𝑀′*M*′ for simulating 𝑀*M*on 𝑤*w*. - Add transitions from the halting states of 𝑀
*M*on 𝑤*w*to 𝑞accept-all*q*accept-all. - In state 𝑞accept-all
*q*accept-all, 𝑀′*M*′ accepts any input. - From the start state, simulate 𝑀
*M*on 𝑤*w*. If 𝑀*M*halts, transition to 𝑞accept-all*q*accept-all. - In state 𝑞pal
*q*pal, check if the input is a palindrome. If it is, accept; otherwise, reject.

- For each transition in 𝑀

### Proof of Undecidability

**Reduction**:- Assume there exists a Turing machine 𝐻
*H*that decides whether a Turing machine accepts only palindromes. - Construct 𝑀′
*M*′ as described. - Use 𝐻
*H*to check if 𝑀′*M*′ accepts only palindromes. - If 𝐻
*H*says 𝑀′*M*′ accepts only palindromes, conclude that 𝑀*M*does not halt on 𝑤*w*. - If 𝐻
*H*says 𝑀′*M*′ does not accept only palindromes, conclude that 𝑀*M*halts on 𝑤*w*.

- Assume there exists a Turing machine 𝐻
**Contradiction**:- The ability to decide whether 𝑀′
*M*′ accepts only palindromes would solve the Halting Problem. - Since the Halting Problem is undecidable, no such Turing machine 𝐻
*H*can exist.

- The ability to decide whether 𝑀′

### Conclusion

By constructing a Turing machine 𝑀′*M*′ based on an instance ⟨𝑀,𝑤⟩⟨*M*,*w*⟩ of the Halting Problem and showing that deciding whether 𝑀′*M*′ accepts only palindromes would solve the Halting Problem, we have proven that the problem of determining whether a Turing machine accepts only palindromes is undecidable.