co.combinatorics – Reference for a combinatoric problem about ‘the center of gravity of some lattice points is still a lattice point’

Fix natural numbers $d$ and $k$, here $kge 2$. Say $n$ is $(k,d)-good$, if for any $n$ given $d$-dimensional lattice points $P_1,P_2, … ,P_n$, we can always choose $k$ points $P_{i_1}, … ,P_{i_k}$ of them such that the center of gravity $frac{P_{i_1}+ … +P_{i_k}}{k}$of them is still a lattice point. The least such $n$ is denoted by $n=n(k,d)$.

For example, $n(2,d)=2^d+1$, $n(3,1)=5$, $n(3,2)=9$.

I am going to research on this question, so I want to know is there any results/references for it. I will be grateful for your help.

Are any ultra-wide angle digital point and shoot cameras made?

The GoPro Hero 2 has a wide angle lens with 170 degrees and 10 megapixels. The latest edition black model has other bits and pieces to make it more versatile. Priced around $400 up.

Another possibility would be to get a Pentax K-01 with the 10-17mm fisheye to 28mm equivalent zoom (love this lens.) This would be smaller than a full on DSLR but still much heavier than a point and shoot and not pocketable.

I too would love to have a compact fisheye to wide angle zoom compact. Hey makers – there is a niche market out there. That way, just the 2 cameras do almost everything a normal photographer would wish to do.

I have not seen reviews for the GoPro in terms of IQ, but I could not imagine it is anywhere near a DSLR.

nikon – Times when I don’t want auto color correction or auto white point setting

enter image description hereThere are times when I have taken pictures in a forest on a sunny day, and the light filtering through the leaves provides a nice green cast that I want to preserve. However, my Nikon Coolpix S9900 and Nikon D5500 always want to neutralize the cast. The solution doesn’t seem as clear cut as selecting a different white balance setting from the D5500’s Shooting Menu. And that seems like a somewhat cumbersome approach, even if there were to option to “preserve forest color cast”.

Does anyone have suggestions that work and are simple to preserve the color cast in such situations?

encryption – Multi-party computation: how to avoid a single point of weakness?

I’m trying to understand how to eliminate a single point of weakness given the problems MPC claims to solve.

Take a simple example where some secret is split between servers B and C. If either B or C are breached the whole key cannot be known nor derived.

However, server A makes a request to B and C. They liaise together in order for server A to end up with the complete secret.

Surely all an attacker needs to do is breach server A to initiate the process from there to retrieve the secret, regardless of whether the computation happened in other places. Or is there some step I’m missing?

real analysis – Properties of good kernels (Convolution of good kernel with function continous at a point)

Let $(K_t)_{t>0}$ be a family of nonnegative functions such that

$$(i) int_{mathbb{R}^n}K_t(x)text{d}{y}=1 ;;;text{for all} ;;;t>0$$
$$(ii) ;text{For every} ;epsilon>0 ;;;text{we have} ;;;lim_{t to 0}int_{|x|>epsilon}K_t(x)text{d}{y}=0$$
Show that if $fin L^1(mathbb{R}^n)$ is continous at $x_0$ then we have
$$f(x_0)=lim_{t to 0}(f*K_t)(x_0)$$
My attempt goes as follows. Choose $delta>0$ such that $left|f(x_0-y)-f(x_0)right|leqepsilon$ for $|y|<delta$
begin{align*}
bigg|(f*K_t)(x_0)-f(x_0)bigg|&=left|int_{mathbb{R}^n}f(x_0-y)K_t(y)text{d}{y}-f(x_0)right|\
&=left|int_{mathbb{R}^n}Big(f(x_0-y)-f(x_0)Big)K_t(y)text{d}{y}right|\
&leqint_{|y|<delta}left|f(x_0-y)-f(x_0)right|K_t(y)text{d}{y}+int_{|y|geqdelta}left|f(x_0-y)-f(x_0)right|K_t(y)text{d}{y}\
&leqepsilon+int_{|y|geqdelta}left|f(x_0-y)-f(x_0)right|K_t(y)text{d}{y}
end{align*}

But I don’t know how to use property (ii) for the second integral because $f$ isn’t necessarily bounded.

A very similar question has been asked here, but I am not sure if the solution there is complete.

pr.probability – Poisson point process in polar coordinates

Let $D = mathbb{R^+} times (mathbb{R}backslash {0})$
Let $mu(dt times dx)$ be a $sigma$-finite measure on the Borel $sigma$-algebra $sigma(D)$.
Let $M(dt times dx)$ be the Poisson random measure with intensity measure $mu$, i.e. for $B in sigma(D)$, $M(B)$ is a Poisson random variable with intensity $int_B mu (dt times dx)$.
(or more generally $E int f(t,x)dN = int f(t,x)d mu$ etc.)

Question:
Is it possible to express $M$ in polar coordinates?
What is the transformed intensity measure?
What I mean is this:
Taking $E = mathbb{R^+} times ((-pi/2, pi/2) backslash {0})$, how do I define a Poisson random measure $Q$ on $sigma(E)$, so that for a set $B in sigma(D) cap sigma(E)$, $M(B)$ and $Q(B)$ have the same distribution. Let $q(dt times dx)$ be the intensity measure of $Q$. What is the expression for $q$ in terms of $mu$?

Finally, if $mu(dt, dx) = dt times m(dx)$ (i.e., $mu$ is homogeneous in time, does $q$ have the same property? ie can $q$ be written as $q(dt times d theta) = d t times r(d theta)$?

References?
Many thanks in advance!!

combinatorics – Number of Rectangles in a Region bounded by points that contain a specific point

I want to find the number of rectangles that there are in a region, that do not include any of a given set of points, but do include another specific point. Below is an example, where the blue point is the point that must be included, and the red points cannot be included in any of the rectangles. Is there any formula, or summation, that I could use to find this number out? I know of this problem, which is similar to mine, except that I have more conditions, and I cannot find out how to apply this formula with my restrictions.

image

In the example, the green dot must be in all rectangles, red dots cannot be in any rectangles, the blue rectangle is an example of a valid rectangle, and purple rectangle is example of invalid rectangle. The rectangles can extend all the way up to the black boundaries, which are also given.

multivariable calculus – Finding a point on the surface of $f(x, y, z) = 0$ where the tangent plane is perpendicular to a given line; what am I doing wrong?

I have to solve the following question; I believe my reasoning is correct, but I am getting the wrong answer so I’m not sure what’s happening.

Find a point on the surface $x^2 + y^2 + 4z^2 = 36$ where the tangent plane is perpendicular to the line with parametric equations
$$x = -4t -1, text{ } y = 2t+1 , text{ } z = 8t-3$$

My reasoning is as follows: define $f(x, y, z) = x^2 + y^2 + 4z^2$ and let $(a, b, c)$ be the desired point. We want the normal vector of the tangent plane at $(a, b, c)$ to be parallel to $langle -4, 2, 8 rangle$. But the normal vector of the tangent plane is given by the gradient, so

$$ langle 2a, 2b, 8c rangle = k langle -4, 2, 8 rangle $$

for some constant $k$. This gives the system
$$
begin{align*}
a & = -2k\
b & = k\
c & = k
end{align*}$$

Plugging in $(a, b, c)$ into $x^2 + y^2 + 4z^2 = 36$, we get that $k = pm sqrt 6$, so the the two desired points are $(-2 sqrt 6, sqrt 6, sqrt 6)$ and $(2 sqrt 6, -sqrt 6, -sqrt 6)$. However, these answers are incorrect; the correct ones are $(-4, 2, 2)$ and $(4, -2, -2).$ Can you please tell me what I am doing wrong?