Approximation algorithms for indefinite quadratic form maximization with linear constraints

Consider the following program: \begin{align} \max_x ~& x^TQx \ \mbox{s.t.} ~& Ax \geq b \end{align} where $ Q$ is a symmetric (possibly indefinite) matrix and the inequality is element-wise and constrains feasible solutions to a convex polytope.

This is NP-hard to solve, but what are known approximation results?

A relevant result is given by (Kough 1979). It is shown that this program can be optimized using Benders’ decomposition to within $ \epsilon$ of the optimum. However, the paper does not seem to clearly specify what this means, or the complexity of the procedure.

I believe the $ \epsilon$ -approximation is in the usual sense employed in the field of mathematical programming, that is, is $ OPT$ is the optimal value of the program, $ ALG$ is the result of the above procedure and $ MIN$ is the minimal value attainable by a feasible solution, $ $ \frac{ALG-MIN}{OPT-MIN} \geq (1-\epsilon). $ $ Or something of the sort.

Questions:

  • Is the mentioned procedure a polynomial-time algorithm?
  • Are there known polynomial-time algorithms yielding approximations to the above program in the traditional sense, i.e. $ ALG \geq \alpha OPT $ for some $ \alpha < 1$ , constant or not.

Kough, Paul F. “The indefinite quadratic programming problem.” Operations Research 27.3 (1979): 516-533.

Can still enter the UK using my indefinite leave to remain status although the immigration officer stamped my passport with a Leave to enter

On May 13, 1983 I was granted a Leave to Remain in the UK for an indefinite period visa and since then I have maintained this status by not staying outside the UK for more than two years. During my last visit on July 29, 2018 the border force immigration officer stamped my passport with a Leave to enter for six months and not as a returning resident. I am planning to travel back to the UK next month. Do I need a visa or I can still enter the UK using my indefinite leave to remain status

An indefinite Integral Problem with algebric numerator and trigonometric denominator

$ $ \int \frac{x^2+(n(n-1))}{(xsinx +ncosx )^2 } dx$ $ I know this is an homework problem, but I really couldn’t think of any way to solve it. Like DI Method (No go) , What kind of substitution as denominator is trigonometric whereas Numerator is algebric. Thought of n(n-1) can come by double differentiating but.. like how would we have it here … etc confusing and weird thoughts. Please help me out

is a while loop always an indefinite loop?

Building on this example and others found elsewhere we have two types of loops, definite and indefinite:

definite being a loop that has its number of iterations known before it is executed. E.g. a for loop:

for x in range(1,10): 

and indefinite loops which don’t have the number of iterations known before it is executed. E.g. this while loop:

while answer <> "Yes":     answer = input("are you an idiot?") 

I’m checking some teaching materials that have an example of a while loop that looks definite to me as by looking at the code you know exactly when it is going to end:

number = 1 while  number <= 10:     print(number)     number = number +1 

Is this example a definite loop even though we have to do a little work to work out the for loop equivalent?

What SAN value is ‘one fifth of current sanity’ measured against to determine indefinite insanity?

The rules for Indefinite Insanity state that it kicks in after a character loses…

a fifth or more of current Sanity points in one game “day”

Given that SAN loss might be occurring at various intervals throughout the day, what is the ‘current sanity’ against which ‘one fifth’ is measured? Is it the character’s SAN at the start of the day?

A problem with indefinite matrix

Let $ A$ be a real symmmetric matrix of the type $ n\times n$ , positive semidefinite of the rank $ k$ , where $ k<n$ . Let $ K:=\left [ \begin {array}{ll} A & b\ b^T & d \end{array} \right ], $ where $ d\neq 0$ and $ b$ is a some matrix of the type $ n \times 1$ , such that is not semidefinite (is indefinite).

Then at least one principal minor of $ K$ of some size is negative. I suppose that there is a principal minor of the type $ (k+1) \times (k+1)$ which is negative. I don’t know how to prove it.

A necessary and sufficient condition for the existence of a single-valued indefinite integral

Suppose $ \Omega$ is a domain with $ n$ holes $ A_1, A_2, \cdots, A_n$ . $ \gamma_i$ is a circle around each $ A_i$ . Then $ \gamma_1, \gamma_2, …, \gamma_n$ is a homology basis for the region $ \Omega$ . Let $ \gamma$ be some cycle (a sum of closed curves) in $ \Omega$ . Then $ $ \int_\gamma fdz = \sum_{i=1}^n c_i\int_{\gamma_i}fdz, $ $ where $ c_i = n(\gamma_i, a)$ for some $ a \in A_i$ . The number $ \int_{\gamma_i} fdz$ are called the peridods of the indefinite integral.

Why is it true that the vanishing of the periods is a necessary and sufficient condition for the existence of a single-valued indefinite integral? What is the meaning of that?