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 MM is a Turing machine that halts on input ww, is undecidable. However, it is Turing recognizable because if MM does halt on ww, a Turing machine can simulate MM on ww and accept if MM halts. If MM 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 MM does not halt on input ww is not Turing recognizable. There is no Turing machine that can recognize this language because, in the case where MM 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 AA is Turing-recognizable if and only if A≤mATMA≤mATM.
Definitions
- Turing-recognizable: A language AA is Turing-recognizable if there exists a Turing machine MM such that for any string ww:
- If w∈Aw∈A, MM accepts ww.
- If w∉Aw∈/A, MM either rejects ww or runs forever.
- Many-one reduction (≤m≤m): A language AA is many-one reducible to a language BB (A≤mBA≤mB) if there exists a computable function ff such that for any string ww:
- w∈A ⟺ f(w)∈Bw∈A⟺f(w)∈B.
Proof
If: Assume A≤mATMA≤mATM. This means there is a computable function ff such that for any string ww: w∈A ⟺ f(w)∈ATMw∈A⟺f(w)∈ATM
- ATMATM is Turing-recognizable (since the Halting Problem is Turing-recognizable).
- Given ff is computable, we can construct a Turing machine MM that:
- Takes ww as input.
- Computes f(w)f(w).
- Checks if f(w)∈ATMf(w)∈ATM.
- Accepts if f(w)∈ATMf(w)∈ATM, otherwise continues running.
Since ATMATM is Turing-recognizable, this means MM is also Turing-recognizable, thus AA is Turing-recognizable.
Only if: Assume AA is Turing-recognizable. This means there is a Turing machine MM such that:
- If w∈Aw∈A, MM accepts ww.
- If w∉Aw∈/A, MM either rejects ww or runs forever.
To reduce AA to ATMATM, define a function ff as follows:
- f(w)f(w) is a Turing machine MwMw that simulates MM on input ww.
Thus: w∈A ⟺ Mw∈ATMw∈A⟺Mw∈ATM
Therefore, A≤mATMA≤mATM, completing the proof.
Part (b)
Statement: A language AA is decidable if and only if A≤mATMA≤mATM and A‾≤mATMA≤mATM.
Definitions
- Decidable: A language AA is decidable if there exists a Turing machine MM such that for any string ww:
- If w∈Aw∈A, MM accepts ww and halts.
- If w∉Aw∈/A, MM rejects ww and halts.
- Many-one reduction (≤m≤m): As defined above.
Proof
If: Assume A≤mATMA≤mATM and A‾≤mATMA≤mATM. This means there are computable functions ff and gg such that for any string ww: w∈A ⟺ f(w)∈ATMw∈A⟺f(w)∈ATM w∈A‾ ⟺ g(w)∈ATMw∈A⟺g(w)∈ATM
Construct a Turing machine MM that:
- Given ww, 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∈Aw∈A.
- If it halts and rejects, w∈A‾w∈A.
- Since both f(w)∈ATMf(w)∈ATM and g(w)∈ATMg(w)∈ATM are decidable, MM will halt and decide AA.
Therefore, AA is decidable.
Only if: Assume AA is decidable. This means there exists a Turing machine MM such that:
- If w∈Aw∈A, MM accepts ww and halts.
- If w∉Aw∈/A, MM rejects ww and halts.
To reduce AA and A‾A to ATMATM:
- Define a function ff such that f(w)f(w) is a Turing machine that simulates MM on ww and accepts if MM accepts.
- Define a function gg such that g(w)g(w) is a Turing machine that simulates MM on ww and rejects if MM rejects.
Thus: w∈A ⟺ f(w)∈ATMw∈A⟺f(w)∈ATM w∈A‾ ⟺ g(w)∈ATMw∈A⟺g(w)∈ATM
Therefore, A≤mATMA≤mATM and A‾≤mATMA≤mATM, completing the proof.
Conclusion
- A language AA is Turing-recognizable if and only if A≤mATMA≤mATM.
- A language AA is decidable if and only if A≤mATMA≤mATM and A‾≤mATMA≤mATM.
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 LL which consists of all Turing machines MM such that ⟨M⟩∉L(M)⟨M⟩∈/L(M), where ⟨M⟩⟨M⟩ is the encoding of the Turing machine MM. We need to determine whether LL is decidable or not.
Definitions and Clarifications
- Turing Machine Encoding: ⟨M⟩⟨M⟩ denotes the encoding (or description) of the Turing machine MM.
- Language L(M)L(M): The language L(M)L(M) is the set of all strings that the Turing machine MM accepts.
- Self-Membership Condition: The condition ⟨M⟩∉L(M)⟨M⟩∈/L(M) means that the encoding of the machine MM is not in the language recognized by MM.
Language LL
The language LL is defined as follows: L={⟨M⟩∣⟨M⟩∉L(M)}L={⟨M⟩∣⟨M⟩∈/L(M)}
Analysis
To determine whether LL is decidable, we need to consider whether there exists a Turing machine that can decide for any given Turing machine MM, whether ⟨M⟩∉L(M)⟨M⟩∈/L(M).
Step-by-Step Proof
- Assume LL is Decidable:
- Suppose there exists a Turing machine DD that decides LL. This means DD can determine, for any ⟨M⟩⟨M⟩, whether ⟨M⟩∉L(M)⟨M⟩∈/L(M).
- Construct a Contradiction:
- Let’s construct a Turing machine M′M′ that uses DD 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 DD to check whether ⟨M′⟩∉L(M′)⟨M′⟩∈/L(M′).
- If DD accepts (indicating ⟨M′⟩∉L(M′)⟨M′⟩∈/L(M′)), then M′M′ accepts ⟨M′⟩⟨M′⟩.
- If DD rejects (indicating ⟨M′⟩∈L(M′)⟨M′⟩∈L(M′)), then M′M′ does not accept ⟨M′⟩⟨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 DD 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′).
- Case 2: If ⟨M′⟩∈L(M′)⟨M′⟩∈L(M′):
- According to the definition of M′M′, if DD 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′).
- Case 1: If ⟨M′⟩∉L(M′)⟨M′⟩∈/L(M′):
- Conclusion:
- In both cases, we encounter a contradiction. This contradiction arises from the assumption that DD exists and can decide LL.
- Therefore, the language LL cannot be decidable, as no such decider DD can exist.
Conclusion
The language LL, defined as the set of all Turing machines MM 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 LL 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 DD 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′).
- Case 2: If ⟨M′⟩∈L(M′)⟨M′⟩∈L(M′):
- According to the definition of M′M′, if DD 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′).
- Case 1: If ⟨M′⟩∉L(M′)⟨M′⟩∈/L(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 DD 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′).
- Case 2: If ⟨M′⟩∈L(M′)⟨M′⟩∈L(M′):
- According to the definition of M′M′, if DD 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′).
- Case 1: If ⟨M′⟩∉L(M′)⟨M′⟩∈/L(M′):
Self-Referential Paradox
Problem Statement
Let L(M)L(M) be a language containing all Turing machines TT such that L(T)≠∅L(T)=∅. Prove that L(M)L(M) is recognizable (Turing-recognizable).
Definitions
- Turing-recognizable language: A language LL is Turing-recognizable if there exists a Turing machine MM such that for every string ww:
- If w∈Lw∈L, MM accepts ww.
- If w∉Lw∈/L, MM either rejects ww or runs forever.
- L(T)≠∅L(T)=∅: This means there exists at least one string that the Turing machine TT accepts.
Proof
To prove that L(M)L(M) is recognizable, we need to construct a Turing machine MM that recognizes L(M)L(M). This machine MM should accept any Turing machine encoding ⟨T⟩⟨T⟩ if L(T)≠∅L(T)=∅.
Constructing the Recognizer MM
- 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 ww such that TT accepts ww.
- Enumerate Inputs:
- Enumerate all possible input strings w1,w2,w3,…w1,w2,w3,….
- Simultaneously simulate TT on all these inputs using dovetailing. This means alternating steps among the different simulations to ensure all are advanced fairly.
Dovetailing Process
- Initialize: Start with the first input w1w1.
- Step through Simulations:
- On the first step, simulate TT on w1w1 for one step.
- On the second step, simulate TT on w1w1 for another step and start simulating TT on w2w2 for one step.
- Continue this process, interleaving the simulations of TT on all enumerated inputs w1,w2,w3,…w1,w2,w3,….
Acceptance Condition
- If any simulation T(wi)T(wi) 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 MM
- Input: ⟨T⟩⟨T⟩
- Initialize: Set i=1i=1.
- Loop:
- For each j=1j=1 to ii:
- Simulate TT on wjwj for one step.
- If TT accepts wjwj, then halt and accept ⟨T⟩⟨T⟩.
- Increment ii and repeat.
- For each j=1j=1 to ii:
Pseudocode for MM
vbnetCopy codeInput: ⟨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 MM that uses dovetailing to simulate TT 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 MM will accept ⟨T⟩⟨T⟩ if TT 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 LL consisting of pairs ⟨G,A⟩⟨G,A⟩, where GG is a context-free grammar and AA is a non-terminal symbol in GG such that AA does not appear in any derivation of any string in the language L(G)L(G). Prove that LL is decidable.
Definitions
- Context-Free Grammar (CFG): A grammar G=(V,Σ,R,S)G=(V,Σ,R,S) where VV is a finite set of variables (non-terminal symbols), ΣΣ is a finite set of terminal symbols, RR is a finite set of production rules, and SS is the start symbol.
- Derivation: A sequence of applications of production rules starting from the start symbol SS that produces strings in the language L(G)L(G).
- Language L(G)L(G): The set of all strings that can be derived from the start symbol SS using the production rules of the grammar GG.
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 LL is decidable, we need to construct a Turing machine that decides whether a given pair ⟨G,A⟩⟨G,A⟩ belongs to LL.
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∈VA∈V is a non-terminal symbol.
- Check Reachability:
- A non-terminal AA is reachable if there is a derivation sequence starting from the start symbol SS that eventually leads to a string containing AA.
- To determine reachability, we use a reachability analysis:
- Initialize a set ReachableReachable containing the start symbol SS.
- Repeatedly apply the production rules to see if new non-terminals can be added to the ReachableReachable set.
- If AA is added to ReachableReachable during this process, then AA is reachable from SS.
- Check for Appearances in Derivations:
- After determining the set of reachable non-terminals, perform a check to see if AA 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 AA.
Detailed Algorithm
- Input: ⟨G,A⟩⟨G,A⟩
- Initialize: Reachable={S}Reachable={S} where SS is the start symbol.
- Reachability Analysis:
- Repeat until no new non-terminals are added to ReachableReachable:
- For each production X→αX→α in RR:
- If X∈ReachableX∈Reachable and any non-terminal in αα is not in ReachableReachable, add all non-terminals in αα to ReachableReachable.
- For each production X→αX→α in RR:
- Repeat until no new non-terminals are added to ReachableReachable:
- Check for Non-Terminal AA:
- If A∈ReachableA∈Reachable:
- Reject (since AA can appear in some derivation).
- Else:
- Accept (since AA does not appear in any derivation of any string in L(G)L(G)).
- If A∈ReachableA∈Reachable:
Pseudocode for the Decider
markdownCopy codeInput: ⟨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 SS.
- Decision: If AA is reachable, it means AA can appear in some derivation of a string in L(G)L(G), so we reject. If AA is not reachable, AA does not appear in any derivation, so we accept.
Conclusion
The language LL is decidable because we can construct a Turing machine that effectively performs reachability analysis on the given context-free grammar GG and decides whether AA can appear in any derivation. This Turing machine halts on all inputs, thus proving that LL 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 VV is a finite set of variables (non-terminal symbols), ΣΣ is a finite set of terminal symbols, RR is a finite set of production rules, and SS is the start symbol.
- Language L(G)L(G): The set of all strings that can be derived from the start symbol SS using the production rules of the grammar GG.
- Language L(1∗)L(1∗): The set of all strings consisting solely of the symbol ‘1’, i.e., L(1∗)={1,11,111,…}L(1∗)={1,11,111,…}.
Goal
To prove that AA is decidable, we need to construct a Turing machine that decides whether a given CFG GG 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 AA 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 G1G1 that generates L(1∗)L(1∗):
- G1=({A},{1},{A→1A∣ϵ},A)G1=({A},{1},{A→1A∣ϵ},A)
- Define a CFG G1G1 that generates L(1∗)L(1∗):
- Intersection of CFGs:
- Use the construction to obtain a CFG G′G′ that generates the intersection L(G)∩L(G1)L(G)∩L(G1).
- Since L(G1)=L(1∗)L(G1)=L(1∗), G′G′ generates L(G)∩L(1∗)L(G)∩L(1∗).
- Emptiness Check:
- Determine whether L(G′)L(G′) is empty.
- Use the algorithm for checking the emptiness of the language generated by a CFG.
Detailed Algorithm
- Input: ⟨G⟩⟨G⟩
- Construct G1G1:
- G1=({A},{1},{A→1A∣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(G1).
- 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′)L(G′) is empty using the standard CFG emptiness check:
- Decision:
- If L(G′)L(G′) is empty, accept ⟨G⟩⟨G⟩.
- If L(G′)L(G′) is not empty, reject ⟨G⟩⟨G⟩.
Pseudocode for the Decider
mathematicaCopy codeInput: ⟨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 AA 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 AA 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 where:
- is a finite set of states.
- is the input alphabet.
- is the stack alphabet.
- is the transition function.
- is the start state.
- is the initial stack symbol.
- 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 where:
- are as defined above.
- 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 qq. 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 aa and transitions to state qq, writes symbol bb, and moves left or right, the 3CTM can perform the same operation by focusing on the middle cell’s value and ignoring the others.
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 QQ and the tape alphabet ΓΓ, the STM’s state space will be Q×Γ×ΓQ×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 qq with current cell aa, left cell bb, and right cell cc, it writes dd, 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 qq, the left cell is bb, and the right cell is cc.
- Read and Transition:
- The STM reads aa and transitions to state (q′,d,c′)(q′,d,c′), where c′c′ is the new right cell value after moving right.
- 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.
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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 ss 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∈Γa∈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.
- 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 qq and reading symbol aa, it writes aa, transitions to state q′q′, and moves right.
Simulated with RTM:
- Read aa:
- RTM reads aa in state qq.
- Instead of writing aa (which is restricted), the RTM writes a′a′ (an auxiliary symbol) and transitions to a new state q1q1 (an intermediate state).
- Convert Auxiliary Symbol:
- In state q1q1 and reading a′a′, the RTM writes aa and transitions to state q′q′, then moves right.
Formal Construction
- Define the States:
- For each state qq in the STM, introduce corresponding states qq and q1q1 in the RTM.
- 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 codeFor 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.
- 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 codea b c
d e f
g h i
Can be encoded as a 1D tape:
cssCopy codea,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={bman∣m≥n≥1}.
Components of the Automaton
- States: Q={q0,q1,q2,qf}Q={q0,q1,q2,qf}
- Input Alphabet: Σ={a,b}Σ={a,b}
- Stack Alphabet: Γ={Z,X}Γ={Z,X}
- ZZ: Stack bottom marker.
- XX: Symbol that can be pushed or popped from the stack.
- Initial State: q0q0
- Initial Stack Symbol: ZZ
- Accepting States: {qf}{qf}
Transition Function
- Start State q0q0:
- On reading bb, push XX onto the stack and move to q1q1.
- (q0,ϵ,Z)→(q0,Z)(q0,ϵ,Z)→(q0,Z)
- (q0,b,Z)→(q1,XZ)(q0,b,Z)→(q1,XZ)
- (q1,b,X)→(q1,XX)(q1,b,X)→(q1,XX)
- On reading bb, push XX onto the stack and move to q1q1.
- State q1q1 (reading bb‘s and pushing XX onto the stack):
- On reading bb, push XX onto the stack and stay in q1q1.
- On reading aa, pop XX from the stack and move to q2q2.
- (q1,a,X)→(q2,ϵ)(q1,a,X)→(q2,ϵ)
- State q2q2 (reading aa‘s and popping XX from the stack):
- On reading aa, pop XX from the stack and stay in q2q2.
- If stack reaches the bottom marker ZZ, move to qfqf.
- (q2,a,X)→(q2,ϵ)(q2,a,X)→(q2,ϵ)
- (q2,ϵ,Z)→(qf,Z)(q2,ϵ,Z)→(qf,Z)
Description
- Initial Phase: The automaton starts in q0q0. It reads bb‘s and pushes XX onto the stack until it encounters the first aa, transitioning to q2q2.
- Middle Phase: In q2q2, it reads aa‘s and pops XX from the stack.
- Final Phase: If the number of aa‘s read is less than or equal to the number of bb‘s read, the stack will not be empty until all aa‘s are processed. If the stack reaches the bottom marker ZZ, it means m≥nm≥n. If the stack is empty after processing all aa‘s, the automaton transitions to qfqf, 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 𝑁𝑃NP-complete problem is in 𝑃P, then all 𝑁𝑃NP problems are in 𝑃P.
True
Reason: If an 𝑁𝑃NP-complete problem is proven to be in 𝑃P, it means there is a polynomial-time algorithm for solving that problem. Since 𝑁𝑃NP-complete problems are the hardest problems in 𝑁𝑃NP, and any 𝑁𝑃NP problem can be reduced to an 𝑁𝑃NP-complete problem in polynomial time, this would imply that all 𝑁𝑃NP problems can also be solved in polynomial time. Thus, 𝑃=𝑁𝑃P=NP.
b) If it is proven in the future that the 33-SAT problem can be solved in 𝑜(𝑛6)o(n6) time, then all 𝑁𝑃NP problems can be solved in 𝑜(𝑛6)o(n6) time.
True
Reason: If 33-SAT, which is an 𝑁𝑃NP-complete problem, can be solved in 𝑜(𝑛6)o(n6) time, then any problem in 𝑁𝑃NP 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(n6) time, the combined process would still be within 𝑜(𝑛6)o(n6) time. Therefore, all 𝑁𝑃NP problems would be solvable in 𝑜(𝑛6)o(n6) time.
c) If one 𝑁𝑃NP problem is solved in polynomial time, all 𝑁𝑃NP problems can be solved in polynomial time.
False
Reason: Solving one 𝑁𝑃NP problem in polynomial time does not necessarily imply that all 𝑁𝑃NP problems can be solved in polynomial time. Only if the problem solved is 𝑁𝑃NP-complete does this implication hold true. The statement lacks this crucial detail. Therefore, the general statement is false.
d) Consider a problem 𝐴A; if it can be reduced to the 33-SAT problem in polynomial time and the 33-SAT problem can be reduced to 𝐴A in polynomial time, then problem 𝐴A is 𝑁𝑃NP-complete.
True
Reason: To prove that problem 𝐴A is 𝑁𝑃NP-complete, two conditions must be met:
- 𝐴A is in 𝑁𝑃NP.
- Any 𝑁𝑃NP problem can be reduced to 𝐴A in polynomial time.
Given that 𝐴A can be reduced to 33-SAT in polynomial time and vice versa, 𝐴A and 33-SAT are polynomially equivalent. Since 33-SAT is 𝑁𝑃NP-complete and all 𝑁𝑃NP problems can be reduced to 33-SAT, the same reduction can be used to reduce all 𝑁𝑃NP problems to 𝐴A. Hence, 𝐴A is 𝑁𝑃NP-complete.
Summary
- a) True: If an 𝑁𝑃NP-complete problem is in 𝑃P, then 𝑃=𝑁𝑃P=NP.
- b) True: If 33-SAT can be solved in 𝑜(𝑛6)o(n6) time, then all 𝑁𝑃NP problems can be solved in 𝑜(𝑛6)o(n6) time.
- c) False: Solving one 𝑁𝑃NP problem in polynomial time does not imply that all 𝑁𝑃NP problems can be solved in polynomial time unless it is 𝑁𝑃NP-complete.
- d) True: If problem 𝐴A can be reduced to 33-SAT and vice versa in polynomial time, then 𝐴A is 𝑁𝑃NP-complete.
Problem Statement
Consider the problem of determining whether a graph 𝐺′G′ is a subgraph of a graph 𝐺G. Prove that this problem is 𝑁𝑃NP-complete.
Definitions
- Subgraph Isomorphism Problem: Given two graphs 𝐺G and 𝐺′G′, determine whether 𝐺′G′ is isomorphic to a subgraph of 𝐺G.
- Class 𝑁𝑃NP: A problem is in 𝑁𝑃NP if a proposed solution can be verified in polynomial time by a deterministic Turing machine.
- Class 𝑁𝑃NP-complete: A problem is 𝑁𝑃NP-complete if it is both in 𝑁𝑃NP and 𝑁𝑃NP-hard. A problem is 𝑁𝑃NP-hard if every problem in 𝑁𝑃NP can be reduced to it in polynomial time.
Goal
To prove that the subgraph isomorphism problem is 𝑁𝑃NP-complete, we need to show two things:
- The problem is in 𝑁𝑃NP.
- The problem is 𝑁𝑃NP-hard.
Step 1: Prove the Problem is in 𝑁𝑃NP
A problem is in 𝑁𝑃NP 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 𝐺G, we can check in polynomial time whether this mapping preserves the adjacency (i.e., whether the edges in 𝐺′G′ correspond to edges in 𝐺G).
- This verification can be done by checking each edge of 𝐺′G′ and ensuring it maps to an edge in 𝐺G.
Verification Algorithm:
- For each vertex 𝑢′u′ in 𝐺′G′:
- Map 𝑢′u′ to a vertex 𝑢u in 𝐺G.
- For each edge (𝑢′,𝑣′)(u′,v′) in 𝐺′G′:
- Check if there is an edge (𝑢,𝑣)(u,v) in 𝐺G corresponding to the mapped vertices.
- If all edges of 𝐺′G′ are verified to have corresponding edges in 𝐺G, 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 𝐺G.
Therefore, the problem is in 𝑁𝑃NP.
Step 2: Prove the Problem is 𝑁𝑃NP-hard
To prove the problem is 𝑁𝑃NP-hard, we need to show that any problem in 𝑁𝑃NP can be reduced to this problem in polynomial time. We will use a known 𝑁𝑃NP-complete problem for the reduction. The Clique Problem is a suitable choice because it is known to be 𝑁𝑃NP-complete.
Clique Problem
- Input: A graph 𝐺G and an integer 𝑘k.
- Output: Determine if 𝐺G contains a clique of size 𝑘k (a subset of 𝑘k 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 𝐾𝑘Kk on 𝑘k vertices.
- The question is now whether 𝐺′G′ (which is 𝐾𝑘Kk) is a subgraph of 𝐺G.
Reduction Algorithm:
- Construct 𝐺′G′ as a complete graph 𝐾𝑘Kk with 𝑘k vertices.
- Check if 𝐺′G′ is isomorphic to a subgraph of 𝐺G using the subgraph isomorphism problem.
If 𝐺G contains a clique of size 𝑘k, then 𝐾𝑘Kk (which is 𝐺′G′) will be a subgraph of 𝐺G. Conversely, if 𝐾𝑘Kk is a subgraph of 𝐺G, then 𝐺G contains a clique of size 𝑘k.
This reduction can be done in polynomial time because constructing 𝐾𝑘Kk and checking for subgraph isomorphism are both polynomial-time operations.
Therefore, since the Clique Problem is 𝑁𝑃NP-complete and it can be reduced to the subgraph isomorphism problem in polynomial time, the subgraph isomorphism problem is 𝑁𝑃NP-hard.
Conclusion
Since we have shown that the subgraph isomorphism problem is in 𝑁𝑃NP and 𝑁𝑃NP-hard, it follows that the subgraph isomorphism problem is 𝑁𝑃NP-complete.
Problem Statement
Prove that the complement of the language REG-EQREG-EQ is in the class 𝑁𝑃NP.
Definitions
- REG-EQ: The language REG-EQREG-EQ consists of pairs ⟨𝐿1,𝐿2⟩⟨L1,L2⟩, where 𝐿1L1 and 𝐿2L2 are regular languages represented by regular expressions without the use of the Kleene star (*) operator, and 𝐿1=𝐿2L1=L2.
- Complement of REG-EQ (REG-EQ‾REG-EQ): The language consisting of pairs ⟨𝐿1,𝐿2⟩⟨L1,L2⟩ where 𝐿1L1 and 𝐿2L2 are not equivalent.
- Class 𝑁𝑃NP: A problem is in 𝑁𝑃NP if a proposed solution can be verified in polynomial time by a deterministic Turing machine.
Goal
Prove that REG-EQ‾REG-EQ is in 𝑁𝑃NP. This involves showing that there exists a polynomial-time verification algorithm for a non-equivalence witness between 𝐿1L1 and 𝐿2L2.
Approach
To prove REG-EQ‾REG-EQ is in 𝑁𝑃NP, we need to show that there is a polynomial-time verifiable certificate for non-equivalence of 𝐿1L1 and 𝐿2L2.
Key Idea
If 𝐿1L1 and 𝐿2L2 are not equivalent, there exists at least one string 𝑤w such that 𝑤w is in 𝐿1L1 and not in 𝐿2L2 or vice versa. This string 𝑤w serves as a certificate for non-equivalence.
Steps to Show REG-EQ‾∈𝑁𝑃REG-EQ∈NP
- Certificate for Non-Equivalence:
- The certificate is a string 𝑤w that witnesses the non-equivalence of 𝐿1L1 and 𝐿2L2.
- Verification Algorithm:
- Given the certificate 𝑤w and the regular expressions for 𝐿1L1 and 𝐿2L2:
- Verify that 𝑤w belongs to exactly one of 𝐿1L1 or 𝐿2L2, but not both.
- This can be done by checking membership of 𝑤w in both 𝐿1L1 and 𝐿2L2 using the regular expressions.
- Given the certificate 𝑤w and the regular expressions for 𝐿1L1 and 𝐿2L2:
Detailed Steps
- Input: Pair ⟨𝐿1,𝐿2⟩⟨L1,L2⟩ and certificate 𝑤w.
- Parse Regular Expressions: Convert the regular expressions for 𝐿1L1 and 𝐿2L2 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 𝐿1L1 on 𝑤w to check if 𝑤∈𝐿1w∈L1.
- Simulate the DFA for 𝐿2L2 on 𝑤w to check if 𝑤∈𝐿2w∈L2.
- Verification:
- If 𝑤∈𝐿1w∈L1 and 𝑤∉𝐿2w∈/L2, or 𝑤∈𝐿2w∈L2 and 𝑤∉𝐿1w∈/L1, then 𝑤w is a valid certificate proving 𝐿1≠𝐿2L1=L2.
Pseudocode for Verification Algorithm
vbnetCopy codeInput: ⟨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⟩⟨L1,L2⟩ and 𝑤w.
Conclusion
Since we can verify a certificate (string 𝑤w witnessing the non-equivalence of 𝐿1L1 and 𝐿2L2) in polynomial time, the complement of REG-EQREG-EQ is in 𝑁𝑃NP. 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 𝐴A and 𝐵B are two 𝑁𝑃NP-complete languages, then 𝐴∩𝐵A∩B is 𝑁𝑃NP-complete.
False
Counterexample:
- Languages: Consider 𝐴=SATA=SAT and 𝐵=TAUTB=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 𝑁𝑃NP-complete.
- TAUT is in 𝑐𝑜𝑁𝑃coNP, not 𝑁𝑃NP.
- Intersection:
- 𝐴∩𝐵=SAT∩TAUTA∩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∩𝐵A∩B is empty. Therefore, SAT∩TAUT=∅SAT∩TAUT=∅.
- Complexity:
- The empty set is trivially in 𝑃P, but it is not 𝑁𝑃NP-complete unless 𝑃=𝑁𝑃P=NP.
Therefore, 𝐴∩𝐵A∩B is not necessarily 𝑁𝑃NP-complete if 𝐴A and 𝐵B are 𝑁𝑃NP-complete.
b) If 𝐴A and 𝐵B are two 𝑁𝑃NP-complete languages, then 𝐴∪𝐵A∪B is 𝑁𝑃NP-complete.
True
Proof:
- Membership in 𝑁𝑃NP:
- To show 𝐴∪𝐵∈𝑁𝑃A∪B∈NP, consider an instance 𝑥x and a nondeterministic Turing machine 𝑀M.
- 𝑀M guesses whether 𝑥x should be verified for membership in 𝐴A or 𝐵B.
- If 𝑥x is guessed to be in 𝐴A, 𝑀M runs the verifier for 𝐴A. If 𝑥x is guessed to be in 𝐵B, 𝑀M runs the verifier for 𝐵B.
- Since both verifiers run in polynomial time, 𝑀M also runs in polynomial time.
- NP-hardness:
- To show 𝐴∪𝐵A∪B is 𝑁𝑃NP-hard, consider an arbitrary 𝑁𝑃NP problem 𝐶C.
- Since 𝐴A and 𝐵B are 𝑁𝑃NP-complete, there are polynomial-time reductions from 𝐶C to 𝐴A and from 𝐶C to 𝐵B.
- Let 𝑓f be the polynomial-time reduction from 𝐶C to 𝐴A.
- Let 𝑔g be the polynomial-time reduction from 𝐶C to 𝐵B.
- Construct a new reduction ℎh from 𝐶C to 𝐴∪𝐵A∪B as follows:
- For an instance 𝑥x of 𝐶C, compute 𝑓(𝑥)f(x) and 𝑔(𝑥)g(x).
- Define ℎ(𝑥)=𝑓(𝑥)h(x)=f(x) if 𝑓(𝑥)∈𝐴f(x)∈A, otherwise ℎ(𝑥)=𝑔(𝑥)h(x)=g(x).
- This reduction works because:
- If 𝑥∈𝐶x∈C, then either 𝑓(𝑥)∈𝐴f(x)∈A or 𝑔(𝑥)∈𝐵g(x)∈B (or both), so ℎ(𝑥)∈𝐴∪𝐵h(x)∈A∪B.
- If 𝑥∉𝐶x∈/C, then neither 𝑓(𝑥)∈𝐴f(x)∈A nor 𝑔(𝑥)∈𝐵g(x)∈B, so ℎ(𝑥)∉𝐴∪𝐵h(x)∈/A∪B.
- Since ℎh is a polynomial-time reduction, 𝐴∪𝐵A∪B is 𝑁𝑃NP-hard.
- Conclusion:
- Since 𝐴∪𝐵A∪B is in 𝑁𝑃NP and 𝑁𝑃NP-hard, it is 𝑁𝑃NP-complete.
Therefore, if 𝐴A and 𝐵B are 𝑁𝑃NP-complete languages, 𝐴∪𝐵A∪B is 𝑁𝑃NP-complete.
Showing SATSAT is 𝑁𝑃NP-Complete
To demonstrate that SATSAT (the Boolean satisfiability problem) is 𝑁𝑃NP-complete, we need to show two things:
- SATSAT is in 𝑁𝑃NP.
- SATSAT is 𝑁𝑃NP-hard.
1. SATSAT is in 𝑁𝑃NP
A problem is in 𝑁𝑃NP if a proposed solution can be verified in polynomial time by a deterministic Turing machine.
Verification Process for SATSAT:
- Given a Boolean formula 𝜙ϕ in conjunctive normal form (CNF) and a satisfying assignment 𝜎σ, the verifier can check if 𝜎σ indeed satisfies 𝜙ϕ.
- The verifier evaluates 𝜙ϕ by substituting the variables in 𝜙ϕ with the values in 𝜎σ.
- If 𝜙ϕ evaluates to true, then 𝜎σ is a valid satisfying assignment.
Time Complexity:
- Each clause in 𝜙ϕ 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 𝑁𝑃NP.
2. SATSAT is 𝑁𝑃NP-hard
To show that SATSAT is 𝑁𝑃NP-hard, we need to reduce any problem in 𝑁𝑃NP to SATSAT in polynomial time. Stephen Cook’s proof of the Cook-Levin theorem provides this reduction by showing that any problem in 𝑁𝑃NP can be translated to an instance of SATSAT.
Sketch of Cook-Levin Theorem:
- For any problem 𝐿∈𝑁𝑃L∈NP, there exists a nondeterministic Turing machine 𝑀M that decides 𝐿L in polynomial time.
- We can construct a Boolean formula 𝜙ϕ that represents the computation of 𝑀M on an input 𝑥x.
- The formula 𝜙ϕ is constructed in such a way that it is satisfiable if and only if 𝑀M accepts 𝑥x.
Construction of the Formula 𝜙ϕ:
- Encoding Turing Machine Computation:
- Represent the computation of 𝑀M 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 𝑀M runs in polynomial time.
- Construct Clauses:
- Encode the initial configuration of 𝑀M.
- Encode the transition function of 𝑀M to ensure valid transitions between configurations.
- Encode the accepting condition of 𝑀M to ensure that 𝑀M reaches an accepting state.
- Formula 𝜙ϕ:
- Combine the clauses to form a CNF formula 𝜙ϕ.
- 𝜙ϕ is satisfiable if and only if there exists an accepting computation for 𝑀M on 𝑥x.
Reduction:
- Given any problem 𝐿∈𝑁𝑃L∈NP, construct the formula 𝜙ϕ representing the computation of the polynomial-time verifier for 𝐿L.
- This construction can be done in polynomial time, resulting in a polynomial-time reduction from 𝐿L to SATSAT.
Conclusion
Since SATSAT is in 𝑁𝑃NP and 𝑁𝑃NP-hard, SATSAT is 𝑁𝑃NP-complete. This conclusion is foundational in computational complexity theory and is the basis for proving the 𝑁𝑃NP-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 𝐷D is a collection of subsets of 𝑀M, and 𝑀M has at least one 𝑘k-element subset that intersects with every member of 𝐷D. Prove that Common-SubsCommon-Subs is 𝑁𝑃NP-complete.
Definitions
- Collection 𝐷D: A collection of subsets of 𝑀M.
- Subset 𝑆⊆𝑀S⊆M: A 𝑘k-element subset of 𝑀M is considered a hitting set for 𝐷D if 𝑆S intersects with every subset in 𝐷D.
- Language Common-SubsCommon-Subs: The set of all inputs ⟨𝐷,𝑀,𝑘⟩⟨D,M,k⟩ such that there exists a 𝑘k-element subset 𝑆⊆𝑀S⊆M that intersects with every subset in 𝐷D.
Goal
Prove that Common-SubsCommon-Subs is 𝑁𝑃NP-complete.
Approach
To show that Common-SubsCommon-Subs is 𝑁𝑃NP-complete, we need to prove two things:
- Common-SubsCommon-Subs is in 𝑁𝑃NP.
- Common-SubsCommon-Subs is 𝑁𝑃NP-hard.
1. Common-SubsCommon-Subs is in 𝑁𝑃NP
A problem is in 𝑁𝑃NP if a proposed solution can be verified in polynomial time by a deterministic Turing machine.
Verification Process for Common-SubsCommon-Subs:
- Given a candidate 𝑘k-element subset 𝑆S of 𝑀M, verify that 𝑆S intersects with every subset in 𝐷D.
- This involves checking, for each subset 𝑑∈𝐷d∈D, if 𝑆∩𝑑≠∅S∩d=∅.
Verification Algorithm:
- For each subset 𝑑∈𝐷d∈D:
- Check if 𝑆∩𝑑≠∅S∩d=∅.
- If all checks pass, accept; otherwise, reject.
Time Complexity:
- Checking intersection for each subset 𝑑d with 𝑆S 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 𝑁𝑃NP.
2. Common-SubsCommon-Subs is 𝑁𝑃NP-hard
To show that Common-SubsCommon-Subs is 𝑁𝑃NP-hard, we need to reduce a known 𝑁𝑃NP-complete problem to Common-SubsCommon-Subs in polynomial time. We will use the Hitting Set problem, which is known to be 𝑁𝑃NP-complete.
Hitting Set Problem
- Input: A collection 𝐷D of subsets of a finite set 𝑀M, and an integer 𝑘k.
- Output: Determine if there exists a subset 𝑆⊆𝑀S⊆M of size at most 𝑘k that intersects with every subset in 𝐷D.
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 𝑘k-element hitting set is the same as finding a 𝑘k-element subset that intersects with every subset in 𝐷D.
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 𝑁𝑃NP-hard.
Conclusion
Since we have shown that Common-SubsCommon-Subs is in 𝑁𝑃NP and 𝑁𝑃NP-hard, we conclude that Common-SubsCommon-Subs is 𝑁𝑃NP-complete.
Problem Statement
Prove that the language 𝐿L consisting of all DFAs (Deterministic Finite Automata) whose languages are infinite is decidable.
Definitions
- Language 𝐿L: The set of all DFAs 𝐴A such that the language 𝐿(𝐴)L(A) recognized by 𝐴A 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 𝐿L 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 𝐿L:
- Given a DFA 𝐴=(𝑄,Σ,𝛿,𝑞0,𝐹)A=(Q,Σ,δ,q0,F):
- Use a breadth-first search (BFS) or depth-first search (DFS) to detect reachable states from the start state 𝑞0q0.
- 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.
- Given a DFA 𝐴=(𝑄,Σ,𝛿,𝑞0,𝐹)A=(Q,Σ,δ,q0,F):
Detailed Algorithm
- Reachability Analysis:
- Use BFS or DFS starting from the initial state 𝑞0q0 to mark all reachable states.
- Cycle Detection:
- For each reachable state 𝑞q:
- Check if there is a cycle starting and ending at 𝑞q.
- This can be done using a DFS. If during the DFS from 𝑞q we encounter 𝑞q again, a cycle is detected.
- For each reachable state 𝑞q:
- Check Acceptance:
- For each state involved in the cycle, check if there exists a path to an accepting state 𝑓∈𝐹f∈F.
- If such a path exists, the DFA recognizes an infinite language.
Pseudocode for the Algorithm
pythonCopy codedef 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 𝑞0q0. 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 𝐿L 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 𝐿L:
- Given a DFA 𝐴=(𝑄,Σ,𝛿,𝑞0,𝐹)A=(Q,Σ,δ,q0,F):
- Use a breadth-first search (BFS) or depth-first search (DFS) to detect reachable states from the start state 𝑞0q0.
- 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
we can finde that cycles too
Problem Statement
Prove that the language 𝐿L consisting of pairs ⟨𝐵,𝐺⟩⟨B,G⟩ where 𝐺G is a context-free grammar and 𝐵B is a variable in 𝐺G such that 𝐵B is used in the derivation of at least one string produced by 𝐺G is decidable.
Definitions
- Language 𝐿L: The set of pairs ⟨𝐵,𝐺⟩⟨B,G⟩ where 𝐺=(𝑉,Σ,𝑅,𝑆)G=(V,Σ,R,S) is a context-free grammar, 𝐵∈𝑉B∈V, and 𝐵B is used in the derivation of at least one string produced by 𝐺G.
- 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 𝐿L is decidable, we need to construct a decision procedure (algorithm) that, given a pair ⟨𝐵,𝐺⟩⟨B,G⟩, determines whether 𝐵B is used in the derivation of at least one string produced by 𝐺G.
Approach
To show that 𝐿L is decidable, we need to determine if there exists a string in the language generated by 𝐺G that uses the variable 𝐵B in its derivation. This involves:
- Constructing a decision procedure to check if 𝐵B can appear in any derivation starting from the start symbol 𝑆S.
- Utilizing graph reachability techniques to analyze the production rules of the grammar 𝐺G.
Steps to Prove Decidability
- Construct Dependency Graph:
- Construct a directed graph where each variable 𝐴A in 𝐺G is a node.
- Add a directed edge 𝐴→𝐶A→C if there exists a production 𝐴→𝛼A→α in 𝑅R such that 𝐶C appears in 𝛼α.
- Check Reachability:
- Perform a reachability analysis starting from the start symbol 𝑆S to see if 𝐵B can be reached.
- This can be done using a depth-first search (DFS) or breadth-first search (BFS).
- Decision Procedure:
- If 𝐵B is reachable from 𝑆S in the dependency graph, accept the pair ⟨𝐵,𝐺⟩⟨B,G⟩.
- Otherwise, reject the pair ⟨𝐵,𝐺⟩⟨B,G⟩.
Detailed Algorithm
- Input: A pair ⟨𝐵,𝐺⟩⟨B,G⟩, where 𝐺=(𝑉,Σ,𝑅,𝑆)G=(V,Σ,R,S) and 𝐵∈𝑉B∈V.
- Construct Dependency Graph:
- Initialize an empty directed graph 𝐺′G′ with nodes corresponding to the variables in 𝑉V.
- For each production 𝐴→𝛼A→α in 𝑅R:
- For each variable 𝐶C in 𝛼α, add an edge from 𝐴A to 𝐶C in 𝐺′G′.
- Perform Reachability Analysis:
- Initialize a set
reachable
to keep track of variables that can be reached from 𝑆S. - Initialize a stack with 𝑆S.
- While the stack is not empty:
- Pop a variable 𝑋X from the stack.
- For each variable 𝑌Y such that there is an edge 𝑋→𝑌X→Y in 𝐺′G′:
- If 𝑌Y is not in
reachable
:- Add 𝑌Y to
reachable
. - Push 𝑌Y onto the stack.
- Add 𝑌Y to
- If 𝑌Y is not in
- Initialize a set
- Check if 𝐵B is Reachable:
- If 𝐵B is in
reachable
, accept the pair ⟨𝐵,𝐺⟩⟨B,G⟩. - Otherwise, reject the pair ⟨𝐵,𝐺⟩⟨B,G⟩.
- If 𝐵B is in
Pseudocode for the Algorithm
pythonCopy codedef 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 𝐵B can be used in any derivation from the start symbol 𝑆S.
Conclusion
By constructing a dependency graph and performing reachability analysis, we can determine if a variable 𝐵B is used in the derivation of at least one string produced by the grammar 𝐺G. This decision procedure always halts and correctly identifies whether 𝐵B is used, proving that the language 𝐿L is decidable.
teps to Prove Decidability
- Construct Dependency Graph:
- Construct a directed graph where each variable 𝐴A in 𝐺G is a node.
- Add a directed edge 𝐴→𝐶A→C if there exists a production 𝐴→𝛼A→α in 𝑅R such that 𝐶C appears in 𝛼α.
- Check Reachability:
- Perform a reachability analysis starting from the start symbol 𝑆S to see if 𝐵B can be reached.
- This can be done using a depth-first search (DFS) or breadth-first search (BFS).
- Decision Procedure:
- If 𝐵B is reachable from 𝑆S in the dependency graph, accept the pair ⟨𝐵,𝐺⟩⟨B,G⟩.
- Otherwise, reject the pair ⟨𝐵,𝐺⟩⟨B,G⟩.
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 𝑅R, which describes languages that include at least one string 𝑤w such that “111” is a substring of 𝑤w, is decidable.
Definitions
- Language 𝑅R: The set of all regular expressions 𝑅R over an alphabet ΣΣ such that the language described by 𝑅R contains at least one string 𝑤w with “111” as a substring.
Goal
To prove that 𝑅R is decidable, we need to construct a decision procedure (algorithm) that, given a regular expression 𝑅R, determines whether there exists at least one string in the language described by 𝑅R that contains “111” as a substring.
Approach
To show that 𝑅R is decidable, we can use the following steps:
- Convert the regular expression 𝑅R to an equivalent deterministic finite automaton (DFA) 𝐴A.
- Construct a new DFA 𝐴′A′ that accepts exactly those strings containing “111” as a substring.
- Check if the intersection of the languages of 𝐴A and 𝐴′A′ is non-empty.
- If the intersection is non-empty, then 𝑅R describes a language containing at least one string with “111” as a substring.
Detailed Steps
- Convert Regular Expression 𝑅R to DFA 𝐴A:
- Given a regular expression 𝑅R, there are standard algorithms to convert 𝑅R to an equivalent nondeterministic finite automaton (NFA) 𝑁N, and from 𝑁N to an equivalent DFA 𝐴A.
- The resulting DFA 𝐴A accepts the same language as 𝑅R.
- 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”:
- 𝑞0q0: Initial state, no “1”s seen yet.
- 𝑞1q1: Seen “1”.
- 𝑞2q2: Seen “11”.
- 𝑞3q3: Seen “111” (accepting state).
- Transitions:
- From 𝑞0q0: On “1” go to 𝑞1q1, on other characters stay in 𝑞0q0.
- From 𝑞1q1: On “1” go to 𝑞2q2, on other characters go to 𝑞0q0.
- From 𝑞2q2: On “1” go to 𝑞3q3, on other characters go to 𝑞0q0.
- From 𝑞3q3: On any character stay in 𝑞3q3.
- Intersection of 𝐴A and 𝐴′A′:
- Construct a DFA 𝐴′′A′′ that recognizes the intersection of the languages of 𝐴A and 𝐴′A′.
- This can be done using the product construction:
- The states of 𝐴′′A′′ are pairs of states from 𝐴A and 𝐴′A′.
- The transition function of 𝐴′′A′′ is defined such that it simulates both 𝐴A and 𝐴′A′ simultaneously.
- 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.
Pseudocode for the Algorithm
pythonCopy codedef 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 𝑅R is decidable.
Problem Statement
Given two languages 𝐴A and 𝐵B that are Turing-recognizable (recursively enumerable), determine whether the language 𝐴−𝐵A−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−𝐵A−B is the set of strings that are in 𝐴A but not in 𝐵B, 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−𝐵A−B.
Proof Outline
To determine whether 𝐴−𝐵A−B is decidable, we need to explore the properties of 𝐴A and 𝐵B being Turing-recognizable and understand the implications for the language 𝐴−𝐵A−B.
- Understanding 𝐴A and 𝐵B:
- 𝐴A is Turing-recognizable: There exists a Turing machine 𝑀𝐴MA that accepts all strings in 𝐴A.
- 𝐵B is Turing-recognizable: There exists a Turing machine 𝑀𝐵MB that accepts all strings in 𝐵B.
- Language Difference 𝐴−𝐵A−B:
- The language 𝐴−𝐵A−B consists of all strings that are in 𝐴A but not in 𝐵B.
Prove that the problem of determining whether a Turing machine 𝑀M accepts only strings that are palindromes is undecidable.
Definitions
- Palindrome: A string that reads the same forwards and backwards.
- Turing Machine 𝑀M: A computational model that accepts or rejects strings based on a set of states and transitions.
- Language of 𝑀M: The set of all strings that 𝑀M accepts.
Goal
Prove that the problem of deciding whether a Turing machine 𝑀M 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 𝑀M and an input string 𝑤w.
- Question: Does 𝑀M halt on input 𝑤w?
- 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 𝑀M halts on input 𝑤w, 𝑀′M′ accepts all strings.
- If 𝑀M does not halt on input 𝑤w, 𝑀′M′ accepts only palindromes.
Construction of 𝑀′M′
- Modify 𝑀M to create 𝑀′M′ as follows:
- 𝑀′M′ simulates 𝑀M on input 𝑤w.
- If 𝑀M halts on 𝑤w, 𝑀′M′ transitions to a state where it accepts all strings.
- If 𝑀M does not halt on 𝑤w, 𝑀′M′ accepts only palindromes.
Detailed Construction
- Construct 𝑀′M′ with the following behavior:
- On input 𝑥x, 𝑀′M′ simulates 𝑀M on 𝑤w.
- If 𝑀M halts on 𝑤w, 𝑀′M′ enters a special state where it accepts any input 𝑥x.
- If 𝑀M does not halt on 𝑤w, 𝑀′M′ checks if 𝑥x is a palindrome. If 𝑥x is a palindrome, 𝑀′M′ accepts 𝑥x; otherwise, it rejects 𝑥x.
Formal Definition of 𝑀′M′
- States: 𝑄′=𝑄∪{𝑞pal,𝑞accept-all}Q′=Q∪{qpal,qaccept-all}
- Transitions:
- For each transition in 𝑀M, add the corresponding transition in 𝑀′M′ for simulating 𝑀M on 𝑤w.
- Add transitions from the halting states of 𝑀M on 𝑤w to 𝑞accept-allqaccept-all.
- In state 𝑞accept-allqaccept-all, 𝑀′M′ accepts any input.
- From the start state, simulate 𝑀M on 𝑤w. If 𝑀M halts, transition to 𝑞accept-allqaccept-all.
- In state 𝑞palqpal, check if the input is a palindrome. If it is, accept; otherwise, reject.
Proof of Undecidability
- Reduction:
- Assume there exists a Turing machine 𝐻H that decides whether a Turing machine accepts only palindromes.
- Construct 𝑀′M′ as described.
- Use 𝐻H to check if 𝑀′M′ accepts only palindromes.
- If 𝐻H says 𝑀′M′ accepts only palindromes, conclude that 𝑀M does not halt on 𝑤w.
- If 𝐻H says 𝑀′M′ does not accept only palindromes, conclude that 𝑀M halts on 𝑤w.
- 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 𝐻H can exist.
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.