Complexity Theory – Polynomial Close Closure

Having trouble generalizing when a class of complexity $$D$$ is closed under polynomial reduction.

For example, consider the following examples:

1. $$bigcup_ {c ge 1} mathsf {DTime} (2 ^ {cn ^ 5})$$
2. $$bigcup_ {c ge 1} mathsf {DTime} (5 ^ {cn})$$
3. $$bigcup_ {c ge 1} mathsf {DTime} (n ^ {c log ^ 5n})$$

$$P$$ is contained in all these classes, so if we are given a problem $$A$$ who belongs to one of these classes, and a reduction of another problem $$B$$ at $$A$$the polynomial reduction will not affect the overall execution time of the decision. Can it be generalized to a higher class than $$P$$?

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Transitive closure vs. accessibility in graphics

I am faced with the most curious situation with [my current information of] transitive closure algorithms. Specifically, what follows is not an algorithm to determine the transitive closure of a graph `G (V, E)` and the time complexity is not `O (| V | + | E |)`:

Use Kosaraju to calculate SCCs (strongly connected components) – O (| V | + | E |)

Create a DAG from SCCs as (super) nodes – O (| V | + | E |)

In a single super-node, the reachability of all the nodes is identical and equal to the union of the nodes in the containing supernode, plus the supernodes accessible in the DAG-O (| V | + | E |) to compute the in the DAG and O (| V | * | V |) total to compute the accessibility sets for all nodes

These accessibility sets make to constitute the transitive closure of the graph altogether, no?
But there are not any `O (| V | + | E |)` transitive closure algorithms around.
What do I miss?

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Commutative algebra – What is the difference between total integral closure and integral closure?

I have been advised here to make this question a new question:

What is the difference between total integral closure and integral closure (geometrically, in the context of rigid analytical geometry)? I have read in several places that the difference is essentially between rank 1 and higher rank evaluations, but how can we clarify that?

It is surprisingly difficult even to find a definition of total integral closure …

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Closed convex cone – equivalence of the definition via closure and via infinite sums

I have a set $$P$$ of points in a Banach space. Consider the following two cones:

• Closing the set of all non negative linear (negative) combinations of $$P$$. (I.e., topological closure of $${ sum_ {i = 1} ^ n a_ip_i: a_i geq0, p_i in P }$$.)
• The set of all non-negative infinite linear combinations of $$P$$. (That is to say., $${ sum_i a_ip_i: a_i geq0, p_i in P }$$ or $$i$$ can go on infinite sets, and we only consider sums that converge absolutely.)

Are these sets equal?

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Prove that \$ K = k (t) \$ is the fractional body of R and \$ k[t]\$ is the full closure of \$ R \$ in \$ K \$

assume $$k$$ is a field and leaves $$t = overline {x} / overline {y}$$ in the field of integral domain fractions $$R = k[x,y]/ (x ^ 2-y ^ 3)$$. Prove it $$K = k (t)$$ is the fraction field of R and $$k$$

angular – Is there a way to hide a closure container in a model?

I may cache DOM child / grandchild items with * ngIf, but the parent and grandparent elements have formatting that remains even when the child / grandchildren have been removed from the DOM (the limits remain). Is it possible to hide parent / grandparent elements to completely remove formatting when children / grandchildren are deleted?

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Find the closure and derived set of \$ A \$ in some topologies.

Let $$X =[0,1] subset mathbb {R}$$, let $$mathscr {B}$$ to be a base for the topology $$mathscr {T} _ {1}$$ sure $$X$$ given by $$mathscr {B} = { {0 }, {1 } } cup {(0,2 ^ {- k}) ,: , k in mathbb {Z} _ { ge0} },$$and let $$mathscr {T} _ {2}$$ to be a topology on $$X$$ given by $$mathscr {T} _ {2} = left {G subset X ,: , frac {1} {2} not in G right } cup {G subset X ,: , (0,1) subset G }.$$Yes $$A = left { left ( frac {1} {4}, frac {1} {2} right) right$$ is a subset of the product space $$X ^ { ast} = (X, mathscr {T} _ {1}) times (X, mathscr {T} _ {2})$$, find the closure $$overline {A}$$and the derived set $$A$$ in $$X ^ { ast}$$.

Find $$A$$, I've watched the $$A = left { frac {1} {4} right } times { frac {1} {2}$$ and using the formula for $$A$$ in the product space like $$A = left ( left { frac {1} {4} right } times { frac {1} {2} } right) & # 39; = left (, overline { left { frac {1} {4} right }} times { frac {1} {2} } & # 39; right) cup left ( left { frac {1} {4} right } times overline { { frac {1} {2} }} , right) = left ( left[Frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[Frac{1}{4}1right)times{frac{1}{2}}[Frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[Frac{1}{4}1right)times{frac{1}{2}}[frac{1}{4}1droite)times{frac{1}{2}}droite)cupleft(left(frac{1}{4}1droite)times{frac{1}{2}}right)=left[frac{1}{4}1droite)times{frac{1}{2}}[frac{1}{4}1right)times{frac{1}{2}}right)cupleft(left(frac{1}{4}1right)times{frac{1}{2}}right)=left[frac{1}{4}1right)times{frac{1}{2}}$$

But the solution is $$A = left ( frac {1} {4}, 1 right) times { frac {1} {2}$$.

Where am I wrong? or can not use this formula for a subset of a point in the product space?

ct.category Theory – Closure of presentable objects within finite limits

In a locally presentable category $$cal E$$, there are arbitrarily large regular cardinals $$lambda$$ such as $$lambda$$-presentable (a.k.a. $$lambda$$-compact) are closed under withdrawals. Namely, the withdrawal functor $${ cal E} ^ {( to leftarrow)} to cal E$$ is a right assistant, so accessible. So, he keeps $$lambda$$-presentable objects for arbitrarily large $$lambda$$so just check that the $$lambda$$objects -presentable in $${ cal E} ^ {( to leftarrow)}$$ are the ones that are punctual in $$cal E$$. (A version of this argument is given in this answer in the case of finished products.)

Of course, "arbitrarily big" means that for any cardinal $$mu$$ there is a regular cardinal $$lambda> mu$$ with this property. A stronger statement would be that this is true for all big enough regular cardinals $$lambda$$, that is, invert the quantifiers and say that there is a $$mu$$ such as all regular cardinals $$lambda> mu$$ have this property $$lambda$$-The presentable objects are closed under withdrawals). Is this stronger affirmation true?

Note that this is certainly not true all regular cardinals $$lambda$$ have this property; counterexamples can be found here.

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abstract algebra – For \$ Omega \$ an algebraic closure of k \$, why is \$ k[x]/ (f) otimes_k Omega cong Omega ^ m \$?

Fix a field $$k$$ and an algebraic closure $$Omega$$ of $$k$$. For a monic, irreducible, separable polynomial $$f in k[x]$$, let $$f$$ to divide $$Omega$$ as

$$f (x) = prod_ {i = 1} ^ m (x – a_i).$$

It was said that the following holds:

$$frac {k[x]} {(f)} otimes_k Omega con prod_ {i = 1} ^ m frac { Omega[x]} {(x-a_i)} cong Omega ^ m.$$

I can see the Chinese rest theorem here, but what is not clear to me is the resolution of the tensor because I am not yet comfortable working with them. Why is the first isomorphism true?

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I like Google Plus, but its closure is imminent. Sigh

Google Plus is one of my favorites because it sells your privacy, but they are out of business soon. Sigh…