# Looping Turing Machines: Accepting & Non-Accepting

do we have turing machine that loops on accepting states and unacepting states

Yes, a Turing machine can be designed to loop on both accepting and non-accepting states. Here’s how each type of looping behavior can be implemented and understood:

Looping Turing Machines: Accepting & Non-Accepting

May 15, 2024

do we have turing machine that loops on accepting states and unacepting states

Yes, a Turing machine can be designed to loop on both accepting and non-accepting states. Here’s how each type of looping behavior can be implemented and understood:

**Looping on Accepting States**:- A Turing machine loops on an accepting state when it is designed to continue processing without halting, even after reaching a condition that would typically signify successful completion of its task. This could be useful in situations where continuous monitoring or repetitive processing is required.
- The looping is generally implemented by having a transition from the accepting state back to itself or to another state that eventually leads back to the accepting state without encountering a halt instruction. This can be done while reading specific symbols that maintain the loop.

**Looping on Non-Accepting States**:- Looping on non-accepting states usually indicates a problem, such as an error in the machine’s design or an input that leads to an infinite loop without satisfying the conditions needed to reach an accepting state. This might happen, for example, if the machine keeps moving back and forth between a set of states without being able to progress to an accepting condition.
- This type of looping is typically unintended, unlike the deliberate design of a looping accepting state. It can be seen as analogous to an infinite loop in computer programming where the program never reaches a termination condition.

Here’s a simple example to demonstrate each type:

### Turing Machine Example:

**Symbols**: {0, 1, □}**Alphabet**: {0, 1}**States**: {q0, q1, q2, q3, q4}**Transitions**:- δ(q0, 0) = {(q1, 0, R)} (Start state, reads 0, moves to q1)
- δ(q1, 0) = {(q1, 0, R)} (Loops on 0s)
- δ(q1, 1) = {(q2, 1, R)} (Transition to check 1s)
- δ(q2, 1) = {(q2, 1, R)} (Loops on 1s, accepting loop)
- δ(q1, □) = {(q3, □, S)} (Reaches end of tape, moves to non-accepting stop)
- δ(q3, □) = {(q4, □, S)} (Non-accepting, halts)
- δ(q2, □) = {(q2, □, R)} (Loops indefinitely when reaching tape end on 1s, accepting loop)

**Final States**: {q2}

In this example:

**q2**is an accepting state that loops if it continues to read 1s or reaches the end of the tape (accepting loop).**q3**transitions to q4 and halts, which would typically not loop, but you could design a non-accepting loop similarly if required by adjusting transitions.

This demonstrates both a functional infinite loop for continuous processing in an accepting state and a typical transition to a halt in a non-accepting state. Looping on non-accepting states is generally avoided unless for a specific purpose such as error handling or waiting for further input.

is this recognizable ?

In the context of Turing machines and formal language theory, the term “recognizable” refers to a type of language. A language is **recognizable** (or **recursively enumerable**) if there exists a Turing machine that will accept any string from the language, possibly looping indefinitely on strings not in the language, but never incorrectly accepting a string that doesn’t belong to the language.

To determine if a language is recognizable, you look for the following characteristics in its Turing machine:

**Acceptance of Valid Strings**: The machine must eventually halt in an accepting state for all strings that are part of the language.**Behavior on Invalid Strings**: The machine can either halt in a non-accepting state or loop indefinitely for strings that are not part of the language.

From the example I provided in the previous message:

- The Turing machine is designed to accept strings of
`0`

s followed by`1`

s (i.e., 0*1*). It halts in an accepting state (`q2`

) if it reads a sequence of one or more`1`

s after any number of`0`

s and then reaches the end of the input. - For strings not following this pattern, it halts in a non-accepting state or could potentially loop indefinitely if not designed properly.

This machine’s design, where it accepts by reaching an accepting state and halts, and rejects by halting in a non-accepting state or potentially looping indefinitely (though not explicitly shown in the example), indicates that the language it recognizes is indeed recognizable. The machine does not incorrectly accept any invalid strings and halts or loops on all others, fulfilling the criteria for a recognizable language.

1( گرافهای دوری گرافهای هسند که یالهای آنها تشکل نها یک دور یدهند. زبان L، مجموعه گرافهای غر جهتداری مانند G است که در بر گرندهی حداقل یک گراف دوری با تعداد رأسهای چهار هسند. با توجه به ان تعریف، کدام گزاره دققتر است؟ 𝑃 ∈ 𝐿 یا 𝑃𝑁 ∈ 𝐿؟ ادعای خود را ابت کنید. )۱۰ نمره( in engliahs

In English, the query discusses a specific property of graphs related to cycle graphs and then asks which statement is more accurate, whether 𝑃 is in 𝐿 or 𝑃𝑁 is in 𝐿, and to justify the claim. Here’s the translation and breakdown of the query:

The term “cycle graphs” refers to graphs whose edges form exactly one cycle. The language 𝐿 consists of undirected graphs G that contain at least one cycle graph with exactly four vertices. Given this definition, the question is to determine which statement is more accurate: 𝑃 ∈ 𝐿 or 𝑃𝑁 ∈ 𝐿, and to provide a justification.

### Analyzing the Problem:

**Definitions**:**Cycle Graph**: A cycle graph is a graph that forms a single closed loop. In terms of vertices and edges, a cycle graph of four vertices has exactly four edges, with each vertex being connected in a closed loop.

**Language 𝐿**:- The language 𝐿 is defined as the set of undirected graphs that include at least one subgraph which is a cycle graph with exactly four vertices.

**𝑃 ∈ 𝐿 vs. 𝑃𝑁 ∈ 𝐿**:**𝑃 ∈ 𝐿**implies that graph 𝑃 is a member of the set 𝐿, which would mean that 𝑃 includes at least one subgraph that is a cycle graph with four vertices.**𝑃𝑁 ∈ 𝐿**implies that the negation or the complement of graph 𝑃 (depending on interpretation) is a member of 𝐿. This could mean a graph that does not contain 𝑃 as a subgraph, or a graph obtained by some form of negation operation on 𝑃, fulfills the condition of having a cycle graph with four vertices.

### Justification:

To determine which statement is more accurate, we would need specific details about graph 𝑃. However, based on the general definitions:

- If 𝑃 itself includes a cycle graph with exactly four vertices, then 𝑃 ∈ 𝐿 is true.
- If 𝑃 does not include such a cycle graph but its complement or a graph excluding 𝑃 (interpreted as 𝑃𝑁) does include such a cycle, then 𝑃𝑁 ∈ 𝐿 would be true.

**Example Justification**:

- Suppose 𝑃 is a graph with no cycles or cycles of lengths other than four. In this case, 𝑃 ∉ 𝐿. However, if adding or modifying 𝑃 results in creating a cycle of four vertices (in the complement graph or another related graph), then 𝑃𝑁 ∈ 𝐿.

### Conclusion:

The accuracy of 𝑃 ∈ 𝐿 versus 𝑃𝑁 ∈ 𝐿 hinges entirely on the specific structure of 𝑃. Without additional information about 𝑃, the general analysis requires assuming or presenting scenarios where each statement might be true. Typically, questions like these in academic settings might aim to test understanding of graph properties, language definitions, and logical deductions based on provided or hypothesized graph structures.

2( الف(بستهبودنمجموعهزبانهایدستهیPرانسبتبهعملیاتهای*)Star(ومکملگری نشان دهید. )۱۰ نمره( ب( بسته بودن مجموعه زبانهای دستهی NP را نسبت به عملیاتهای *)Star( و اشتراک نشان دهید. )۱۰ نمره(

The query is about demonstrating closure properties of languages in the class P and NP with respect to certain operations like the Kleene star (*) and complementation (for P) and the Kleene star (*) and intersection (for NP). Here’s how to address these two parts:

### Part A: Closure properties of languages in class P

**Closure under Kleene Star**:**Definition**: If 𝐿*L*is a language, 𝐿∗*L*∗ is the set of all strings that can be formed by concatenating zero or more strings from 𝐿*L*.**Demonstration for P**: If 𝐿*L*is in P, there is a polynomial-time deterministic Turing machine (DTM) that decides 𝐿*L*. To decide 𝐿∗*L*∗, a DTM can be constructed that, for any given input, checks if the input can be segmented into one or more substrings each of which belongs to 𝐿*L*. Using dynamic programming (similar to the CYK algorithm), this can be done in polynomial time, maintaining the input string’s membership decision within P.

**Closure under Complementation**:**Definition**: The complement of a language 𝐿*L*, denoted 𝐿‾*L*, contains all strings not in 𝐿*L*.**Demonstration for P**: If 𝐿*L*is in P, there is a polynomial-time DTM that decides 𝐿*L*. To decide 𝐿‾*L*, simply take the DTM for 𝐿*L*and swap its accepting and rejecting states. This machine will decide 𝐿‾*L*in polynomial time, proving that the complement of a language in P is also in P.

### Part B: Closure properties of languages in class NP

**Closure under Kleene Star**:**Demonstration for NP**: If 𝐿*L*is in NP, there is a nondeterministic Turing machine (NTM) that decides 𝐿*L*in polynomial time. To decide 𝐿∗*L*∗, an NTM can be constructed that nondeterministically guesses breakpoints in the input string to segment it into parts, each of which is checked (via the machine for 𝐿*L*) to belong to 𝐿*L*. This process remains within polynomial time constraints of the NTM, hence 𝐿∗*L*∗ is also in NP.

**Closure under Intersection**:**Definition**: If 𝐿1*L*1 and 𝐿2*L*2 are languages, their intersection contains strings that belong to both 𝐿1*L*1 and 𝐿2*L*2.**Demonstration for NP**: Assume 𝐿1*L*1 and 𝐿2*L*2 are in NP. An NTM to decide 𝐿1∩𝐿2*L*1∩*L*2 can be constructed by running the NTM for 𝐿1*L*1 and, if it accepts, subsequently running the NTM for 𝐿2*L*2 on the same input. Since both machines run in polynomial time relative to their input lengths, the combined machine also runs in polynomial time relative to the input length, showing that 𝐿1∩𝐿2*L*1∩*L*2 is in NP.

These demonstrations highlight fundamental characteristics of P and NP concerning how they handle basic language operations, reflecting their computational complexities and capabilities.

3( زبانSAT-SATباتعریفزررادرنظربگرید: SAT-SAT = { ⟨φ⟩ | حداقل دو مقداردهی مخلف وجود دارد که به ازای آنها عبارت منطقی φ برقرار یشود { ابت کنید SAT-SAT یک زبان NP-Complete است. )۲۰ نمره( in englis h

The question asks about the language called SAT-SAT, defined as follows:

**SAT-SAT** = { ⟨φ⟩ | there exist at least two different assignments that satisfy the logical formula φ }.

The task is to demonstrate that SAT-SAT is an NP-complete language.

### Explanation and Proof:

**Definition of NP-Completeness**:- A language is
**NP-complete**if:- It is in NP.
- Every problem in NP can be polynomially reduced to it.

- A language is
**SAT-SAT is in NP**:- To show that SAT-SAT is in NP, we need to demonstrate that there exists a nondeterministic polynomial time (NTM) verifier for it. For SAT-SAT, given a formula φ and a certificate (a solution), the verifier needs to check if there are at least two distinct satisfying assignments.
- A verifier for SAT-SAT could take a logical formula φ and two distinct assignments as certificates. It would then verify that both assignments satisfy φ. Since checking whether an assignment satisfies a formula is in P (polynomial time), and since we are only adding a step to check a second assignment, the process remains polynomial in time. Thus, SAT-SAT is in NP.

**NP-Completeness of SAT-SAT**:- To prove NP-completeness, we must show a polynomial-time reduction from a known NP-complete problem to SAT-SAT. One straightforward choice is the SAT problem itself, which is the problem of determining if there exists any assignment that satisfies a given logical formula.
**Reduction from SAT to SAT-SAT**:- Given a SAT instance φ, construct a new formula φ’ such that φ’ = φ ∨ (x ∧ ¬x).
- Here, x is a new variable not used in φ.
- The construction is polynomial in time (adding only a few symbols to φ).
- If φ is satisfiable (has at least one satisfying assignment in the SAT problem), then φ’ has at least two satisfying assignments: one where x is true and another where x is false, in addition to whatever satisfies φ.
- If φ is not satisfiable, then neither is φ’ because the added part (x ∧ ¬x) is always false, so φ’ cannot have two distinct satisfying assignments.

- This shows that any instance of the SAT problem can be transformed into an instance of SAT-SAT in polynomial time such that the transformation preserves the yes/no answer (if φ is satisfiable, φ’ is in SAT-SAT; if φ is not satisfiable, φ’ is not in SAT-SAT).

Therefore, since SAT-SAT is in NP and every problem in NP can be polynomially reduced to SAT-SAT (as demonstrated by the reduction from SAT), SAT-SAT is NP-complete. This proves its status as an NP-complete language, showing that it is computationally intensive and significant in the study of computational complexity.

4( فرض کنید مجموعهی }𝑛𝑥 , … ,𝑥2 ,𝑥1{ = 𝑀 وجود دارد. S، یک مجموعهی mای از زرمجموعههای M است که اجتماع آنها برابر مجموعهی M یشود. برای مال، اگر }1,2,3,4,5,6{=𝑀 باشد، S یتواند }}5,6{,}5 ,4{,}4{,}3 ,1,2{,}1,2{{=𝑆 باشد. یک K-SET از مجموعهی S، تعداد k عضو از S است که اجتماع آنها برابر M خواهد شد. برای مال، یک 𝑇𝐸𝑆 − 3 از مجموعهی S یتواند برابر }}6 ,5{ ,}5 ,4{ ,}3 ,2 ,1{{ = 𝑆 باشد. ابت کنید مسئلهی K-SET با ورودیهای ⟩𝑘 ,𝑆 ,𝑀⟨ یک مسئلهی NP-Complete است )یتوانید از مسئلهی VERTEX-COVER استفاده کنید(. )۲۰ نمره(

The problem described here is essentially the K-SET problem: given a set 𝑀*M* and a collection 𝑆*S* of its subsets whose union equals 𝑀*M*, the task is to determine if there exists a subset of 𝑆*S* consisting of 𝑘*k* sets whose union also equals 𝑀*M*. The challenge is to prove that the K-SET problem is NP-complete, and you’re encouraged to use the VERTEX-COVER problem in your reduction.

### Definition and NP-Completeness Proof

**Step 1: K-SET is in NP**

- To show that K-SET is in NP, we need to demonstrate that any certificate (a solution) for the problem can be verified in polynomial time.
- Given a certificate, which is a selection of 𝑘
*k*subsets from 𝑆*S*, we can verify in polynomial time whether the union of these 𝑘*k*subsets equals 𝑀*M*. This verification involves checking each element of 𝑀*M*against the union of the selected subsets, a process that is polynomial in the size of 𝑀*M*and 𝑆*S*.

**Step 2: Reducing VERTEX-COVER to K-SET**

**VERTEX-COVER Problem**: Given a graph 𝐺=(𝑉,𝐸)*G*=(*V*,*E*) and an integer 𝑘*k*, does there exist a subset of vertices 𝑉′⊆𝑉*V*′⊆*V*such that ∣𝑉′∣≤𝑘∣*V*′∣≤*k*and every edge in 𝐸*E*has at least one endpoint in 𝑉′*V*′?**Reduction Approach**:**Construct 𝑀**: Let 𝑀*M**M*be the set of all edges in 𝐺*G*, i.e., 𝑀=𝐸*M*=*E*.**Construct 𝑆**: For each vertex 𝑣*S**v*in 𝑉*V*, create a subset 𝑠𝑣*s**v* of 𝑀*M*that contains all edges incident to 𝑣*v*. Thus, 𝑆={𝑠𝑣∣𝑣∈𝑉}*S*={*s**v*∣*v*∈*V*}.**Mapping the Problem**: A vertex cover in 𝐺*G*corresponds to a selection of vertices whose associated edge sets (in 𝑆*S*) cover all edges in 𝑀*M*. The task of finding a vertex cover of size 𝑘*k*in 𝐺*G*thus translates to finding 𝑘*k*subsets in 𝑆*S*whose union is 𝑀*M*.

**Proof of Correctness**:- If there exists a vertex cover of size 𝑘
*k*in 𝐺*G*, then the subsets corresponding to these vertices will cover all edges, meaning their union equals 𝑀*M*. Thus, there is a corresponding K-SET solution. - Conversely, if there is a solution to the K-SET problem (i.e., 𝑘
*k*subsets of 𝑆*S*whose union is 𝑀*M*), then the vertices corresponding to these subsets form a vertex cover in 𝐺*G*because they cover all edges.

- If there exists a vertex cover of size 𝑘

This reduction shows that solving VERTEX-COVER directly informs the solution to K-SET, and since VERTEX-COVER is a known NP-complete problem, the ability to polynomially reduce it to K-SET implies that K-SET is also NP-complete.

### Conclusion

By demonstrating that K-SET is in NP and that any instance of the VERTEX-COVER problem can be reduced to K-SET in polynomial time, we prove that K-SET is NP-complete. This establishes the complexity and challenges associated with solving or approximating the K-SET problem in general computational settings.

5( زبانH-CLIQUEباتعریفزررادرنظربگرید: H-CLIQUE = { ⟨G⟩ | m یک گراف بدون جهت است که دارای یک زرگراف کامل با حداقل گرههای | 𝑚2 | است که G تعداد گره های G است { ابت کنید H-CLIQUE یک زبان NP-Complete است. )۲۰ نمره

The problem described is the H-CLIQUE problem, defined as follows:

**H-CLIQUE** = { ⟨G⟩ | G is an undirected graph that contains a complete subgraph (clique) with at least ⌈∣𝑉(𝐺)∣2⌉⌈2∣*V*(*G*)∣⌉ vertices, where 𝑉(𝐺)*V*(*G*) denotes the set of vertices in G }.

To prove that H-CLIQUE is NP-complete, we’ll go through the following steps:

### Step 1: H-CLIQUE is in NP

- A language is in NP if there exists a nondeterministic polynomial time algorithm (or equivalently, a polynomial-time verifier) that can verify a given solution.
- For H-CLIQUE, the certificate would be a subset of vertices 𝑆⊆𝑉(𝐺)
*S*⊆*V*(*G*) where ∣𝑆∣≥⌈∣𝑉(𝐺)∣2⌉∣*S*∣≥⌈2∣*V*(*G*)∣⌉. - The verifier checks if:
- All pairs of vertices in 𝑆
*S*are connected by an edge, which confirms that 𝑆*S*forms a clique. - The size of 𝑆
*S*is at least ⌈∣𝑉(𝐺)∣2⌉⌈2∣*V*(*G*)∣⌉.

- All pairs of vertices in 𝑆
- Both of these checks can be done in polynomial time relative to the size of 𝐺
*G*, thereby confirming that H-CLIQUE is in NP.

### Step 2: NP-Completeness of H-CLIQUE

To prove NP-completeness, we need to perform a polynomial-time reduction from a known NP-complete problem to H-CLIQUE. We’ll use the CLIQUE problem for this reduction, which is defined as:

**CLIQUE** = { ⟨G, k⟩ | G is an undirected graph containing a complete subgraph of at least k vertices }.

CLIQUE is a well-known NP-complete problem.

#### Reduction from CLIQUE to H-CLIQUE:

- Given an instance ⟨𝐺,𝑘⟩⟨
*G*,*k*⟩ of the CLIQUE problem, construct an instance ⟨𝐺′⟩⟨*G*′⟩ for H-CLIQUE where:- 𝐺′
*G*′ is the same as 𝐺*G*. - We choose 𝑘=⌈∣𝑉(𝐺′)∣2⌉
*k*=⌈2∣*V*(*G*′)∣⌉ directly in the problem specification.

- 𝐺′
**Analysis**:- If 𝐺
*G*has a clique of size 𝑘*k*, then 𝐺′*G*′ will satisfy the condition for H-CLIQUE if 𝑘≥⌈∣𝑉(𝐺′)∣2⌉*k*≥⌈2∣*V*(*G*′)∣⌉. - Conversely, if 𝐺′
*G*′ contains a clique of size ⌈∣𝑉(𝐺′)∣2⌉⌈2∣*V*(*G*′)∣⌉, then 𝐺*G*contains a clique of size 𝑘*k*as 𝑘*k*is set to ⌈∣𝑉(𝐺′)∣2⌉⌈2∣*V*(*G*′)∣⌉.

- If 𝐺

This reduction is polynomial in time since it merely involves passing 𝐺*G* unchanged and setting 𝑘*k* based on the size of 𝑉(𝐺)*V*(*G*). This transformation preserves the yes/no nature of the problem, thus confirming that if CLIQUE is hard, so is H-CLIQUE.

### Conclusion

By establishing that H-CLIQUE is in NP and demonstrating a polynomial-time reduction from the CLIQUE problem, we have proven that H-CLIQUE is NP-complete. This categorizes H-CLIQUE as computationally intensive and significant within computational complexity theory, particularly for understanding the limits of algorithmic approaches to graph-based problems.

5( زبانH-CLIQUEباتعریفزررادرنظربگرید: H-CLIQUE = { ⟨G⟩ | m یک گراف بدون جهت است که دارای یک زرگراف کامل با حداقل گرههای | 𝑚2 | است که G تعداد گره های G است { ابت کنید H-CLIQUE یک زبان NP-Complete است. )۲۰ نمره( پاسخ

To demonstrate that the H-CLIQUE problem is NP-complete, we need to both establish that it is in NP and that it is as hard as any other problem in NP. The definition of H-CLIQUE given in your query is:

**H-CLIQUE** = { ⟨G⟩ | G is an undirected graph that contains a complete subgraph (clique) with at least ⌈∣𝑉(𝐺)∣2⌉⌈2∣*V*(*G*)∣⌉ vertices, where 𝑉(𝐺)*V*(*G*) is the set of vertices in G }.

### Step 1: Showing H-CLIQUE is in NP

A language or problem is in NP if a solution (in this case, a subgraph that is a clique) can be verified in polynomial time given a certificate (the claimed clique).

**Certificate**: A subset of vertices 𝑆⊆𝑉(𝐺)*S*⊆*V*(*G*) that supposedly forms a clique.**Verification**:- Verify that every pair of vertices in 𝑆
*S*is connected by an edge. This check involves examining (∣𝑆∣2)(2∣*S*∣) pairs, which is polynomial with respect to the size of 𝐺*G*. - Verify that ∣𝑆∣≥⌈∣𝑉(𝐺)∣2⌉∣
*S*∣≥⌈2∣*V*(*G*)∣⌉.

- Verify that every pair of vertices in 𝑆
- Both steps are clearly polynomial in the number of vertices of 𝐺
*G*, confirming that H-CLIQUE is in NP.

### Step 2: Proving NP-completeness of H-CLIQUE

NP-completeness requires showing a polynomial-time reduction from a known NP-complete problem to the problem in question (H-CLIQUE). We’ll use the CLIQUE problem, which is known to be NP-complete.

**CLIQUE** = { ⟨G, k⟩ | G is an undirected graph that contains a clique of at least 𝑘*k* vertices }.

#### Reduction from CLIQUE to H-CLIQUE

Given an instance ⟨𝐺,𝑘⟩⟨*G*,*k*⟩ of the CLIQUE problem, we need to construct an instance ⟨𝐺′⟩⟨*G*′⟩ for H-CLIQUE such that 𝐺′*G*′ has a solution if and only if 𝐺*G* has a clique of size 𝑘*k*. Here’s how:

**Construction**:- Let 𝐺′=𝐺
*G*′=*G*. - Select 𝑘
*k*such that 𝑘=⌈∣𝑉(𝐺′)∣2⌉*k*=⌈2∣*V*(*G*′)∣⌉.

- Let 𝐺′=𝐺

This simple construction doesn’t alter 𝐺*G* but reinterprets the size of the clique being sought. The key challenge is showing that a CLIQUE of size 𝑘*k* in 𝐺*G* translates to an H-CLIQUE in 𝐺′*G*′:

- If 𝐺
*G*contains a clique of size 𝑘*k*as per CLIQUE, and 𝑘≥⌈∣𝑉(𝐺′)∣2⌉*k*≥⌈2∣*V*(*G*′)∣⌉ (as chosen), then 𝐺′*G*′ also contains a clique of size at least ⌈∣𝑉(𝐺′)∣2⌉⌈2∣*V*(*G*′)∣⌉, fulfilling H-CLIQUE’s requirement. - Conversely, if 𝐺′
*G*′ (which is just 𝐺*G*) has a clique of size ⌈∣𝑉(𝐺′)∣2⌉⌈2∣*V*(*G*′)∣⌉, this clique is also a valid solution for the CLIQUE instance ⟨𝐺,𝑘⟩⟨*G*,*k*⟩ where 𝑘=⌈∣𝑉(𝐺′)∣2⌉*k*=⌈2∣*V*(*G*′)∣⌉.

### Conclusion

This reduction is polynomial in time since it requires no change to the graph and only recalculates 𝑘*k*. Thus, if CLIQUE is hard, so is H-CLIQUE. By showing this reduction and verifying that H-CLIQUE is in NP, we conclude that H-CLIQUE is NP-complete. This categorizes H-CLIQUE as a computationally demanding problem, central to understanding computational complexity, especially in graph theory and algorithm design.

5( زبانH-CLIQUEباتعریفزررادرنظربگرید: H-CLIQUE = { ⟨G⟩ | m یک گراف بدون جهت است که دارای یک زرگراف کامل با حداقل گرههای | 𝑚2 | است که G تعداد گره های G است { ابت کنید H-CLIQUE یک زبان NP-Complete است. )۲۰ نمره( پاسخ

The problem described, H-CLIQUE, involves proving that it is NP-complete. H-CLIQUE is defined as follows:

**H-CLIQUE** = {⟨G⟩ | G is an undirected graph that contains a complete subgraph (clique) with at least ⌈∣𝑉(𝐺)∣2⌉⌈2∣*V*(*G*)∣⌉ vertices, where 𝑉(𝐺)*V*(*G*) is the set of vertices in 𝐺*G*}.

To demonstrate that H-CLIQUE is NP-complete, we’ll first prove that it belongs to the complexity class NP

) PCP حالتی از را در نظر بگیرید که در آن، طول هر رشته بالای هر دومینو با طول هر رشته پایین هر دومینو برابر باشند. برای مثال در نظر بگیرید: 𝑳 = {𝒂𝒃 𝒂𝒂 , 𝒂 𝒃 , 𝒂𝒃𝒂𝒂 𝒃𝒃𝒃𝒂} در این دسته PCP گیرد. نشان دهید این حالت از مسأله قرار می پذیر است.

The problem you are referring to involves a special case of the **Post Correspondence Problem (PCP)**, where each domino’s top and bottom strings are of equal length. Let’s delve into how to demonstrate that even this restricted version of PCP remains undecidable.

### Background on PCP:

The classical Post Correspondence Problem posed by Emil Post in 1946 is a decision problem that can be stated as follows:

- Given a finite set of dominoes, each with a top string and a bottom string, determine whether there is a sequence of these dominoes that can be arranged such that the concatenation of the top strings is equal to the concatenation of the bottom strings.

### Special Case: Equal-Length PCP

In the equal-length version of PCP, each domino has its top and bottom strings of the same length. You provided an example set 𝐿*L*:

- [𝑎𝑏𝑎𝑎],[𝑎𝑏],[𝑎𝑏𝑎𝑎𝑏𝑏𝑏𝑏𝑎][
*ab**aa*],[*a**b*],[*abaa**bbbba*]

### Proving Undecidability:

To prove that this restricted version of PCP is undecidable, we need to show that there is no general algorithm that can solve this problem for all possible sets of dominoes. Here’s a step-by-step approach:

**Reduction from General PCP**:- In the general PCP, there is no constraint on the lengths of the top and bottom strings of each domino.
- To reduce the general PCP to the equal-length PCP, consider padding the shorter string of each domino (top or bottom) with a special character (say, a delimiter like ‘#’) until both strings are of equal length. This does not alter the essence of the problem but transforms any instance of the general PCP into an instance of the equal-length PCP.

**Inherent Complexity**:- The general PCP is known to be undecidable. This means that there is no algorithm that can decide whether any arbitrary set of dominoes has a matching sequence of concatenated top and bottom strings.
- By showing that every instance of the general PCP can be transformed into an instance of the equal-length PCP while preserving the problem’s constraints and challenges, we demonstrate that the equal-length PCP is also undecidable. The key here is that the transformation process (padding) is systematic and can be applied universally.

**Implication**:- If there were an algorithm to solve the equal-length PCP, then the same algorithm could be applied to solve the general PCP (by first converting the general PCP problem into the equal-length format). Since the general PCP is undecidable, the existence of such an algorithm for the equal-length PCP would contradict the undecidability of the general PCP.

**Conclusion**:- Since we can transform any instance of the general PCP (undecidable) into an instance of the equal-length PCP without changing the nature of the problem, the equal-length PCP must also be undecidable.

This proof relies on the fact that the complexity of the general PCP, an established undecidable problem, transfers directly to the equal-length PCP via a straightforward transformation (padding). Thus, solving the equal-length PCP is just as hard as solving the general PCP, confirming its status as an undecidable problem.

HALT TM is undecidable. can you give me python code that show me this

The statement that the Halting Problem for Turing machines (HALT_TM) is undecidable is a fundamental theorem in theoretical computer science, proven by Alan Turing in 1936. This theorem states that there is no general algorithm that can determine whether any arbitrary Turing machine will halt or run forever given a specific input.

However, it’s important to note that demonstrating this undecidability through a Python program is not straightforward or even possible in the traditional sense because the proof itself is theoretical, relying on a logical paradox (akin to the diagonalization argument used by Cantor for real numbers or Russell’s paradox in set theory).

What we can do instead is simulate a simplified scenario that illustrates the core concept behind why the halting problem is undecidable. We’ll use Python to implement a simplified version of Turing’s proof by contradiction, creating a scenario that leads to a logical paradox if we assume an algorithm could decide the halting problem.

Here is a Python simulation that attempts to create a function that hypothetically would decide the halting problem, and then shows why such a function leads to a contradiction:

`pythonCopy code````
def hypothetical_halting_solver(turing_machine, input):
"""
This function is a placeholder for the hypothetical solver of the halting problem.
It's supposed to return True if the Turing machine halts on the given input,
and False otherwise. This function cannot exist in reality for arbitrary Turing machines.
"""
# As we cannot actually implement this, we return a placeholder value
# In reality, this function's behavior is undecidable
pass
def paradox_machine(input):
"""
This Turing machine takes its own description as input and simulates what would happen if
the hypothetical halting solver claims it does not halt.
"""
if hypothetical_halting_solver(paradox_machine, input):
# Simulate an infinite loop
while True:
pass
else:
# Halt immediately, which creates a paradox if the solver says it won't halt
return
# Trying to use the hypothetical solver on the paradox machine
try:
if hypothetical_halting_solver(paradox_machine, paradox_machine):
print("Paradox machine halts.")
else:
print("Paradox machine does not halt.")
except RecursionError:
print("This leads to a paradox, as expected from undecidability.")
```

### Explanation:

**hypothetical_halting_solver()**: This function is supposed to decide if a Turing machine halts on a given input. In reality, no such function can exist for all Turing machines, as proved by Turing.**paradox_machine()**: This machine is designed to create a paradox. It uses`hypothetical_halting_solver()`

on itself. If the solver says it halts, the machine goes into an infinite loop (a contradiction). If the solver says it does not halt, it halts immediately (another contradiction).

### Limitations:

- The provided Python code cannot run as intended since
`hypothetical_halting_solver()`

cannot be implemented. The`pass`

statement is just a placeholder. - The paradox and recursion shown reflect the logical reasoning behind Turing’s proof rather than an executable demonstration.

This script is fundamentally theoretical and is intended to illustrate the logic behind the proof rather than to provide a practical demonstration of the undecidability of the halting problem.

PROOF We let R be a TM that decides REGULARTM and construct TM S to decide ATM. Then S works in the following manner. S = “On input ⟨M, w⟩, where M is a TM and w is a string: 1. Construct the following TM M2. M2 = “On input x: 1. Ifxhastheform0n1n,accept. 2. If x does not have this form, run M on input w and accept if M accepts w.” 2. Run R on input ⟨M2⟩. 3. If R accepts, accept; if R rejects, reject.” can you give me this python code ?

The proof you’ve described leverages the concept of reduction to demonstrate that a problem (in this case, determining if a Turing machine recognizes a regular language, REGULARTM) is undecidable by reducing another known undecidable problem (the acceptance problem for Turing machines, ATM) to it. This method is often used in theoretical computer science to show the computational limits of decision problems.

To simulate this in Python, we need to understand that we cannot actually build a Turing machine nor fully simulate its operation for arbitrary inputs and machines. However, we can create a Python representation that captures the essence of the logic used in the proof.

Here’s a Python program that conceptually aligns with the proof structure:

`pythonCopy code````
def decides_REGULARTM(TM_description):
"""
Hypothetical function that decides if a Turing machine description defines a TM
that recognizes a regular language.
"""
# This function cannot exist in reality for arbitrary Turing machines.
pass
def ATM_to_REGULARTM(TM, w):
"""
This function attempts to reduce the acceptance problem (ATM) to determining
if a TM recognizes a regular language (REGULARTM).
"""
# Define a new TM, M2, as per the proof outline
def M2(x):
# Step 1: Check if the input x has the form 0^n 1^n
if all(c == '0' for c in x[:len(x)//2]) and all(c == '1' for c in x[len(x)//2:]) and len(x) % 2 == 0:
return True # Accept if the input is in the form 0^n 1^n
else:
# Step 2: If x does not have this form, simulate M on input w
return TM(w) # Accept if TM accepts w
# Step 2: We hypothetically run a decider for REGULARTM on M2
if decides_REGULARTM(M2):
return True # Accept if M2 is a TM for a regular language
else:
return False # Reject otherwise
# Example usage
def TM(w):
"""
Example Turing machine that might accept or reject based on input w.
This function should represent some decision logic.
"""
return w == "specific_string" # Arbitrary acceptance condition for illustration
# Example input
w = "specific_string"
result = ATM_to_REGULARTM(TM, w)
print("The reduction decides:", result)
```

### Explanation:

**decides_REGULARTM(TM_description)**: This is a hypothetical function because no algorithm can decide for all Turing machines whether they recognize a regular language. It’s here to illustrate the assumption in your proof.**ATM_to_REGULARTM(TM, w)**: This function represents the reduction. It constructs a Turing machine`M2`

described in your proof. This TM accepts inputs of the form 0^n1^n unconditionally (a regular language pattern), and for other inputs, it simulates the TM`M`

on input`w`

.**TM(w)**: A placeholder Turing machine function that represents some decision logic, arbitrarily accepting a specific string.

### Limitations:

- The Python script does not execute the operations of a real Turing machine nor does it handle infinite loops or real machine states.
- The functions like
`decides_REGULARTM`

are placeholders to represent undecidable problems, meant only to illustrate the logical structure of the reduction in the proof.

This program is strictly illustrative and represents a high-level view of how one might set up such a proof conceptually, rather than an executable simulation of actual Turing machine operations.

that we change that init tm ?

If you’re asking how to modify the initial setup of the Turing machine (TM) or create a new one within the framework of the given proof scenario, the key lies in adjusting the definition of the TM `M2`

to handle a new functionality or a new input. This change might involve altering how `M2`

processes its input or interacts with the Turing machine `M`

that was given in the original input. Let’s explore how you might adjust the TM in the Python code to reflect different behaviors or requirements.

### Updating the Turing Machine `M2`

within the Python Simulation

To demonstrate, I’ll show you how to modify the TM `M2`

so that it has a new behavior based on some hypothetical new criteria or changes in its input processing strategy:

`pythonCopy code````
def ATM_to_REGULARTM(TM, w, new_behavior=False):
"""
Reduces ATM to REGULARTM with an option to modify behavior of M2.
:param TM: A Turing machine simulation function
:param w: Input string to the Turing machine TM
:param new_behavior: Boolean flag to trigger new behavior in M2
"""
def M2(x):
if new_behavior:
# New behavior: Suppose we want M2 to accept based on a new pattern or condition
if x == "new_pattern":
return True # Accept if x matches a specific new pattern
elif not any(c.isdigit() for c in x): # Example of a new check
return False # Reject if x contains no digits (arbitrary new rule)
# Existing behavior remains as originally defined
if all(c == '0' for c in x[:len(x)//2]) and all(c == '1' for c in x[len(x)//2:]) and len(x) % 2 == 0:
return True # Accept if the input is in the form 0^n 1^n
else:
# If x does not have this form, simulate TM on input w
return TM(w) # Accept if TM accepts w
# Hypothetical run of a decider for REGULARTM on M2
if decides_REGULARTM(M2):
return True # Accept if M2 is a TM for a regular language
else:
return False # Reject otherwise
# Example Turing machine function
def TM(w):
"""
Example Turing machine logic that accepts a specific string.
"""
return w == "specific_string"
# Example usage with new behavior triggered
w = "specific_string"
result = ATM_to_REGULARTM(TM, w, new_behavior=True)
print("The reduction decides with new behavior:", result)
```

### Explanation of Changes

**New Behavior in**:`M2`

- The function
`M2`

now includes an additional parameter`new_behavior`

which, when set to`True`

, triggers a new set of rules or processing logic within the TM. - This can be used to simulate a change in how
`M2`

processes its inputs, such as accepting or rejecting based on new patterns or conditions that weren’t part of the original design.

- The function
**Flexible Simulation**:- By adding flags or parameters that adjust the behavior of
`M2`

, you can explore how changes in the underlying Turing machine affect the outcome of the reduction, and hence, the decidability of the problem.

- By adding flags or parameters that adjust the behavior of

This approach allows you to use the simulation to explore theoretical concepts in computability and undecidability, demonstrating how minor changes in a TM’s definition can impact the overall system’s behavior. Remember, though, that these simulations are purely illustrative and cannot fully model the complexity of real Turing machine operations or their theoretical implications.

can you give me PCP too

The Post Correspondence Problem (PCP) is a classical undecidable problem in computational theory. The problem involves finding a match between the top and bottom sequences of dominos from a given set, such that when arranged in a sequence, the concatenated sequence of symbols on the top of the dominos matches the concatenated sequence on the bottom. Here is how the problem is typically defined:

**Problem Definition:** Given a set of dominos, each with a top string 𝑡𝑖*t**i* and a bottom string 𝑏𝑖*b**i*, determine if there is a sequence in which these dominos can be arranged so that the concatenated top strings equal the concatenated bottom strings.

**Python Simulation:** To illustrate PCP, we can create a simple Python function that tries to find a solution to a given instance of PCP using a brute-force approach. This won’t solve PCP for all cases due to its undecidability, but it can demonstrate the problem with small examples.

Here’s a Python script to simulate PCP with a very naive approach:

`pythonCopy code````
from itertools import product
def check_pcp(pairs):
"""
Brute-force search for a solution to the Post Correspondence Problem.
This function takes a list of tuples representing the dominos.
:param pairs: List of tuples (top_string, bottom_string)
:return: True if a solution exists, False otherwise
"""
# Check sequences of increasing length
for length in range(1, 10): # Arbitrary limit to prevent infinite loop
# Generate all possible sequences of this length
for sequence in product(pairs, repeat=length):
top_concat = ''.join(domino[0] for domino in sequence)
bottom_concat = ''.join(domino[1] for domino in sequence)
if top_concat == bottom_concat:
print(f"Solution found: {sequence}")
return True
print("No solution found within the limit.")
return False
# Example dominos set
dominos = [("ab", "aa"), ("a", "b"), ("abaa", "bbba")]
check_pcp(dominos)
```

### Explanation:

**Functionality**: The`check_pcp`

function tries to find a sequence of dominos such that the concatenated top strings are equal to the concatenated bottom strings.**Brute-force Approach**: It generates all possible sequences of dominos up to a specified length and checks each sequence to see if it meets the PCP criteria.**Limitations**: This code sets a limit on the maximum length of sequences it checks (`range(1, 10)`

). In practice, PCP solutions might require longer sequences or might not exist at all, and this approach won’t find them if they do. Additionally, the brute-force method is computationally expensive and not feasible for large sets of dominos or longer sequences due to the exponential growth of possibilities.

This script is purely illustrative and highlights the inherent challenge in solving PCP. It effectively demonstrates the concept for small examples, helping to understand the nature of this undecidable problem, but it is not a general solution to PCP.