show that this decidable set $C$ exists

I came across this problem which says that given disjoint sets $ A$ and $ B$ s.t $ \bar{A}$ and $ \bar{B}$ are both computably enumerable (c.e.), there exists a decidable set $ C$ s.t. $ A \subseteq C$ and $ A \cap B = \emptyset$ .

I tried out an attempt, but there’s a part of the attempt of which am not so sure… So here’s my attempt:

Since both $ \bar{A}$ and $ \bar{B}$ are c.e. and the fact that $ A$ and $ B$ are disjoint, we have $ \bar{A}-\bar{B}=B$ . We claim that $ B$ is also c.e. (and this is the part which am not sure of). To show that $ \bar{A}-\bar{B}=B$ is c.e., we can have this enumerator $ E$ for $ \bar{A}-\bar{B}$ :

$ E$ : ignore any input

enumerate words $ \{s_1,s_2,s_3,…,s_k\} \in \Sigma^*$ lexicographically from $ k=1$ to $ \infty$

run recognizer $ M_{\bar{A}}$ on input words $ \{s_1,s_2,s_3,…,s_k\}$ for $ k$ steps

run recognizer $ M_{\bar{B}}$ on input words $ \{s_1,s_2,s_3,…,s_k\}$ for $ k$ steps

if both $ M_{\bar{A}}$ and $ M_{\bar{B}}$ accept an input word $ s \in \{s_1,s_2,s_3,…,s_k\}$ , print $ s$

Assuming that $ E$ correctly enumerates $ \bar{A}-\bar{B}$ , since $ B=\bar{A}-\bar{B}$ , we have that $ B$ is c.e.. Since both $ B$ and $ \bar{B}$ are c.e., it follows that $ B$ is decidable.

So to construct $ C$ , we can just set $ \bar{B}=C$ . It follows that $ A \subseteq C$ and $ C$ is decidable.

Is this attempt correct or am I missing something?

Determine whether there exists a path in a directed acyclical graph that reaches all nodes without revisiting a node

For this I came up with a DFS recursion.

Do DFS from any node and keep doing it until all nodes are Exhausted. I.E. pick the next unvisited node once you cant keep recursing.

The element with the highest post number or the last element you visit should be the first element in your topological ordering.

Now do another DFS recursion that executes on every node called DFS_find:

DFS_find(Node): if (node has no neighbors): return 1; otherwise: return 1 + the maximum of DFS_find(Node) for all neighboring nodes

Execute DFS_find(Node) on the first node in your topological ordering. If it returns a number equal to the number of vertices, then a directed path that crosses every node once, exists. Otherwise it does not.

How can I prove whether or not this algorithm is correct?

I think this may be a little less time efficient than the classical way to just do a topological sort and then check if each consecutive pair has an edge between them.

Problem in NP: $EQ1 = \{(p_1,…,p_n): \exists x_1,…,x_m\in Z \ p_1(x_1,…,x_m)=…=p_n(x_1,…,x_m)=0. \}$

I have to following problem to show is in NP class.

$ EQ1 = \{(p_1,…,p_n): \exists x_1,…,x_m\in Z \ p_1(x_1,…,x_m)=…=p_n(x_1,…,x_m)=0. \}$

Here $ p_1,…,p_n$ are polynomials in m variables with integer coefficients.

I know how to proof EQ1 is in NP, but I confess I have not understood what is the instance accepted by the problem EQ1 (all polynomials are unsoddisfacible?)

What is the time complexity of determining whether a solution $x$ exists to $x^k \equiv c \pmod{N}$ if we know the factorization of $N$?

Suppose we are given an integer $ c$ and positive integers $ k, N$ , with no further assumptions on relationships between these numbers. We are also given the prime factorization of $ N$ . These inputs are written in binary. What is the best known time complexity for determining whether there exists an integer $ x$ such that $ x^k \equiv c \pmod{N}$ ?

We are given the prime factorization of $ N$ because this problem is thought to be hard on classical computers even for k = 2 if we do not know the factorization of $ N$ .

This question was inspired by this answer, where D.W. stated that the nonexistence of a solution to $ x^3 \equiv 5 \pmod{7}$ can be checked by computing the modular exponentiation for $ x = 0,1,2,3,4,5,6$ , but that if the exponent had been 2 instead of 3, we could have used quadratic reciprocity instead. This lead to my discovery that there are a large number of other reciprocity laws, such as cubic reciprocity, quartic reciprocity, octic reciprocity, etc. with their own Wikipedia pages.

Is any key signing party directory – or a mean to facilitate such meetings, exists?

I need to develop my web of trust. I don’t live in or near a metropolitan area and as such it is a bit difficult to find possible local people to sign. I assume I must not be alone in that context.

My question: is there any directory/listings of upcoming gpg-signing party per area, or any existing infrastructure to facilitate such meetings? Or alternative ways to find / meet people who can sign?