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

If A ∉ NP then A ∈ co-NP.

True, false or we don’t know?

## Why isn’t SAT in coNP?

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:

1. Run M on the input word w.
2. If M accepts – reject.
3. 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?

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

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?

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

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.

## Algorithm

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
``````

## Question

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?

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

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

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

The question is as indicated in the title. I would appreciate any help.
I can't see how closeness can imply $$NP co-NP cup subseteq NP cap co-NP$$

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

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).

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

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

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?

## complexity theory – Complement of languages ​​and coNP

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 $${0,1 } ^ *?$$

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

Suppose $$coNP neq NP$$

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

1. $$B in coNP – NP$$
2. $$A in coNP-NP implies A leq_pB$$

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