complexity theory – Can polynomial reduction from several to one be done for a specific instance of problem?

Let's say I reduce the problem $ A in L $ at $ B in K $ , with a function $ f: Sigma ^ {*} rightarrow Gamma ^ {*} $ such as $ w in L Leftrightarrow f (w) in K $ . I know if I want to solve $ A $, given a polynomial time algorithm for $ B $, I just have to transform $ A $ at $ B $ and solve $ B $. This can be thought of as:

The reduction must be carried out from an arbitrary instance of $ A $ to a legal authority $ B $

My question is, should I reduce to arbitrary example of $ B $ or some example of $ B $? That is to say. TQBNF reduction to generalized geography is performed in a valid graph instance, but there are many other valid generalized geography instances.

Does quantum computing convert an O (2 ^ n) algorithm into a polynomial execution time and how?

For example, if it is one of $ O (2 ^ n) $ algorithm that goes from 0 to $ 2 ^ n – $ 1 and check if the 1-bit number is divisible by 3, 5, 7, 9, 11, does quantum computing reduce it to a non-exponential time and how?

Equation resolution – Polynomial root search on finite field extension

I would like to know if there is a method on Mathematica, a third-party coded resource or a library that can calculate roots on $ F_q $, this is an extension field of $ F_p $then $ q = p ^ n $. For example, you can call Roots (poly, symbol, module) but only works on $ F_p $.

An example of what interests me about:

Entry: two polynomials $ p (x), r (x) in F_ {p} ^ n $ or $ p (q (x)) = r (x) $ for unknown $ q (x) in F_ {p} ^ n $.

Otput: the polynomial $ q (x) in F_p ^ n $ such as $ p (q (x)) = r (x) $.

This problem seems strongly related to the composition of polynomials in the finite fields, but I need a general method to calculate the composition polynomial. $ q (x) in F_ {p} ^ n $.

Theory of complexity – Restriction: an instance decision in polynomial time is necessary to "solve a problem"?

I am reading the book "Third Edition of Combinatorial Optimization (Bernhard Korte Jens Vygen)".

(the last version is the sixth.)

There are descriptions in this book that I do not understand

Not all binary strings are instances of the Hamiltonian circuit, but only those
representing an undirected graph. For the most interesting decision problems, the
positions are an appropriate subset of 0-1 strings. We demand that we can decide in
polynomial time if an arbitrary string is an instance or not:

The decision problem is the pair P = (X, Y), where X is a decidable language in polynomial time.

Why a decision problem required that decides in polynomial time whether an arbitrary string is an instance or not?

I can find no reason for this restriction in the book.

Factoring after long polynomial division

I've got the code to do a long polynomial division, but I want my students to also consider the solution, if possible.enter the description of the image here

Field Theory – Irreducible cubic polynomial in $ mathbb {Q}[x]$ n has no roots in $ mathbb {Q} ( sqrt {2}, 5 ^ {1/4}) $

I am working on the following problem:

Consider the field extension $ F = mathbb {Q} ( sqrt {2}, 5 ^ {1/4}) $ of $ mathbb {Q} $. Let $ f (x) $ to be an irreducible cubic polynomial $ mathbb {Q} (x) $. Prove it $ F $ does not contain roots of $ f (x) $.

I know that since $ f (x) $ an irreducible degree $ 3 $ polynomial in $ mathbb {Q} (x) $, $ f (x) $ does not have roots in $ mathbb {Q} $. In the specified extension $ F $ of $ mathbb {Q} $we are only adjoining $ sqrt {2} $ and $ 5 ^ {1/4} $. Thus, it remains to show that $ f (x) $ can not have a root that is a power of $ sqrt {2} $ or a power of $ 5 ^ {1/4} $.

I think I can see how to argue this partially. I believe $ f (x) $ can not have a root that is a strange power of $ 5 ^ {1/4} $from the start, because the corresponding minimal polynomial for such an element should be $ 4 $, already contradicting the degree of $ f (x) $. So, the problem has been reduced to showing that $ f (x) $ can not have a root that is a power of $ sqrt {2} $ or a power of $ sqrt {5} $. But I'm not sure how to argue these two remaining points. Does this have to do with the polynomial of such an element being of degree $ 2 $, which should be a factor $ f (x) $ ? And that contradicts $ f (x) $ to be irreducible?

Any help would be appreciated. Thank you!

learning theory – dimension VC of the class of polynomial classifiers of degree $ n $

The idea is that a degree polynomial $ n $ at most $ n $ roots, and so can change the signs at most $ n $ time. So no degree polynomial $ n + 1 $ can form an alternating pattern + – + -… or – + – + … length $ n + 2 $. This shows that the VC dimension is at most $ n + 1 $.

On the other hand, for any set of $ n + 1 $ pairs $ (x_1, y_1), ldots, (x_ {n + 1}, y_ {n + 1}) $there is a degree polynomial $ n $ which interpolates them, given by the interpolation formula of Lagrange. Using $ y_i = pm 1 $, you can easily show that all together $ n + 1 $ points is broken. Therefore, the VC dimension is exactly $ n + 1 $.

Factoring a polynomial – Mathematica Stack Exchange

How to factor a polynomial dealing with non x symbols $ (a, b) $ as constants?

Factor (a / b (a / b – 2) x ^ 2 – 2 x (1 – a / b) a ^ 2 / b + a ^ 4 / b ^ 2)

does not work without other conditions. However works for whole $ (a, b). $

It seems to work, obtained right after the submission of the question.

ClearAll (a, b);

A = a / b (a / b – 2);

B = -2 (1 – a / b) a ^ 2 / b;

C1 = a ^ 4 / b ^ 2;

Factor (A x ^ 2 + B x + C1)

numerical methods – Determination of numerically stable polynomial coefficients

I have several polynomials $ P_j (x) = sum_ {k = 0} ^ n alpha_ {jk} x ^ k $ and for each one of them I have $ n + 1 $ couples $ (x_ {ji}, y_ {ji}) $ (such as $ y_ {ji} = P_j (x_i) $), and I want to find the coefficient vectors $ boldsymbol alpha_j $. I know I can use Lagrange interpolation or Vandermonde systems resolution:

$$ V ( mathbf x_j) boldsymbol alpha_j = mathbf y_j $$

S. T.

$$ V ( mathbf x_j) _ {ik} = x_ {ji} ^ k $$

But the interpolation of Lagrange is not numerically stable and the number of conditions of my matrix Vandermonde $ kappa (V ( mathbf x_j)) $ is somewhere between $ 10 ^ {13} $ and $ 10 ^ {17} $ according to the polynomial $ P_j $which poses problems to my application.

What stable method is available to me for this purpose?

factoring – lower bound of a degree 2 polynomial

I want to know the conditions to guarantee:

Let $ alpha> 0, beta> 0, gamma geq 1 $ and consider the polynomial of degree 2 $ P (x, y) = alpha ^ 2 x ^ 2 + beta ^ 2 y ^ 2 – 2 alpha beta gamma xy $. Is it possible to lower the P (x, y) terminal as follows:
$$
P (x, y) geq C (Ax-By) ^ 2,
$$

for some positive constants $ A, B, C $ it depends on $ alpha, beta, $ and $ gamma $

Thanks in advance

S