If A ∉ NP then A ∈ co-NP.

True, false or we don’t know?

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# Tag: coNP

## If A ∉ NP then A ∈ co-NP. True, false or we don’t know?

## Why isn’t SAT in coNP?

## complexity theory – Is it possible that Co-NP = P but NP != P

## number theory – Must a decision problem in $NP$ have a complement in $Co-NP$, if I can verify the solutions to in polynomial-time?

## Algorithm

## Question

## complexity theory – Definition of understanding of NP and co-NP

## $ NP cup co-NP $ is closed under $ cap $ implies $ NP = co-NP $

## complexity theory: is NP $ cap $ coNP less difficult than NP-complete?

## complexity theory: is the problem P vs NP, a problem NP or co-NP?

## complexity theory – Complement of languages and coNP

## Are there "complete" languages in $ coNP -NP $?

If A ∉ NP then A ∈ co-NP.

True, false or we don’t know?

I understand why NP=coNP if SAT is in coNP (How do I prove that SAT in coNP implies NP=coNP?).

But I’m missing why the following machine doesn’t turing recognize the complementary of SAT:

Given a turing machine M that recognizes SAT, the following turing machine recognizes coSAT:

- Run M on the input word w.
- If M accepts – reject.
- If M rejects – accept.

Because coSAT is the language of all unsatisfiable formulas, a formula is unsatisfiable if it doesn’t have a satisfiable interpretation, which is exactly the opposite of what M outputs.

What am I missing in here?

Suppose there exists an algorithm that takes as input an unsatisfiable SAT formula, and never fails to verify it in polynomial time. However, when the input is a satisfiable formula, it doesn’t work (let’s say time and memory problems, the algorithm gets lost in the reasoning).

Then Co-NP = P because we can verify unsatisfiable instances in polynomial time. The certificate of unsatisfiability is the formula itself.

But then is it still possible that P != NP?

Goldbach’s Conjecture says every even integer $>$ $2$ can be expressed as the sum of two primes.

Let’s say $N$ is our input and its $10$. Which is an integer > 2 and is not odd.

1.Create list of numbers from $1,to~N$

2.Use prime-testing algorithm for creating a second list of prime numbers

3.Use my 2_sum solver that allows you to use primes twice that sum up to $N$

```
for j in range(list-of-primes)):
if N-(list-of-primes(j)) in list-of-primes:
print('yes')
break
```

4.Verify solution efficently

```
if AKS-primality(N-(list-of-primes(j))):
if AKS-primality(list-of-primes(j)):
print('Solution is correct')
```

5.Output

```
yes 7 + 3
Solution is correct
```

If the conjecture is true, then the answer will always be Yes. Does that mean it can’t be in $Co-NP$ because the answer is always Yes?

From some of the texts I read, a definition of NP is: "An equivalent definition of NP is the set of decision problems that a non-deterministic Turing machine can solve in polynomial time." and that we have the following, where $ n $ is the length of the entry:

$$

text {NP} = bigcup text {NTIME} (n ^ k)

$$

This means that one way to show that a problem is $ in text {NP} $ is if we can build a non-deterministic TM $ N $ using a polynomial time checker $ V $ in a certificate $ C $or:

TM $ N $: in entrance $ x $ Instance of the problem:

1. Guess in a non-deterministic way a certificate $ C $ dice $ x $

2. if V accepts $ C $to accept

But if I use the definition $ text {NP} = bigcup text {NTIME} (n ^ k) $, this would not imply that $ text {co-NP} subseteq text {NP} $since I can build a TM $ N & # 39; $ which can recognize co-NP:

TM $ N & # 39; $: in entrance $ x $ Instance of the problem:

1. Guess in a non-deterministic way a certificate $ C $ dice $ x $

2. if V accepts $ C $ for any branch, reject

3. if all branches reject the guessed certificates, accept

In this case, since all branches of $ N & # 39; $operate in polynomial time, $ N & # 39; $ You should also be able to solve problems for co-NP in nondeterministic polynomial time.

But since you are still not sure if NP = co-NP, how should I understand the definition? $ text {NP} = bigcup text {NTIME} (n ^ k) $?

The question is as indicated in the title. I would appreciate any help.

I can't see how closeness can imply $ NP co-NP cup subseteq NP cap co-NP $

I am taking a complexity class now, and I struggle to understand the concept of "hardness":

Assume that $ L in textit {NP} cap textit {coNP} $. In means that under the assumption $ NP neq coNP $, $ L $ It cannot be NP-complete. The formal meaning is that not all languages in NP can be reduced to $ L $but it means that $ L $ Is it easier to solve than NP-complete language (in the sense that it is more likely to have a non-exponential algorithm that decides)?

Is it plausible that the optimal algorithm for $ L $ is it exponential? (For 3-SAT there is a known assumption, ETH, which as far as I understand states that the optimal algorithm for this must be exponential).

A few years ago, a YouTube channel called hackerdashery, made an extraordinary YouTube video explaining **P vs NP**, semi-vulgarized:

https://www.youtube.com/watch?v=YX40hbAHx3s

However, at 7 minutes and 56 seconds, he talks about why it is so difficult to prove that **P $ = $ public notary** or that **P $ neq $ public notary**, and claims to try things **is** in fact a **public notary** problem (in the video, however, he writes that it is actually a **with P** trouble).

Therefore, it is demonstrating things, in particular **P $ = $ public notary** or **P $ neq $ public notary**, a **public notary** or **with P** trouble? If so, why? And where could I find more about this?

By definition, any language (decision problem) $ L $ is defined as a subset of $ {0,1 } ^ * $, where $ {0,1 } $ is the alphabet

$ L ^ c $ It is said to be the complement of language, and seems to be defined as follows: $ forall w, w en L implies w notin L ^ c $ Y $ w notin L implies w in L ^ c $.

Equivalently, according to the book by Arora and Barak "Computational complexity: a modern approach", $ L ^ c = {0,1 } ^ * – L $.

I will demonstrate my question with an example: the definition of complexity class $ coNP $ It is, according to the book by Arora and Barak, $ coNP = {L: L ^ c in NP } $.

But if we take a specific case, let's say

$ SAT = {f (x_1, x_2, cdots, x_n): exists x

in {0,1 } ^ n f (x) = 1 } $,

so he $ coNP $ the counterpart is supposed to be

$ overline {SAT} = {f (x_1, x_2, cdots, x_n): forall x in {0,1 } ^ n f (x) = 0 } $

Meaning, $ overline {SAT} $ It contains all Boolean formulas that are false for all entries.

However, according to the previous definitions $ (L ^ c = {0,1 } ^ * – L) $, many random words must be included in the language $ overline {SAT} $ which are not even "well-formed" Boolean formulas to begin with.

Should the complement be on all Boolean formulas instead of $ {0,1 } ^ * $?

It is the complement of each language always defined on a subset of words that are "well formed" instead of having been defined on $ {0,1 } ^ *? $

Suppose $ coNP neq NP $

Language B would be called "complete" in $ coNP-NP $ Yes:

- $ B in coNP – NP $
- $ A in coNP-NP implies A leq_pB $

Is there a "complete" language in $ coNP – NP $?

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