Load more and filter (not necessarily in this order)

The problem exposing might seems trivial to you. It still isn’t clear for me which solution to choose.

Saying that you have a input text where you can type anything. Beneath it a list of ten element and a load more button. Once the user clicks on “load more”, ten more elements are added to the list.

My question is, when typing a text to filter, if there is 100 potentials results in the database, should we show the 10 first and a load more button or 20 ? since the user already clicked once on load more, we could expect him to want it to stay as is.

Thanks,

Do colors red and green necessarily signify opposites

In some cultures red signifies stop and green go – stop and go can be considered opposites. Is it really different in other cultures?

But suppose I wanted to show a screen with only two buttons, possibly of different shapes, each for a different kind of action. So action 1 and action 2 are not opposites. Is it a good idea to make one of the buttons green and the other red?

is the following construction using a PRF necessarily yields a PRF?

given a PRF – F, is $ G_s(x)=F_{F_s(x)}(x)$ necessarily a PRF?

First I thought how to tackle this problem. First I tried using a hybrid argument: $ |\mathbb P[F_{F_s(x)}(x)] -\mathbb P[f(x)]|\leq |\mathbb P[F_{F_s(x)}(x)] -\mathbb P[F_{f(x)}(x)]| + |\mathbb P[F_{f(x)}(x)] -\mathbb P[f(x)]|$ so I can contradict the possibility that the first part of the is not negligible but the second part of the sum is problematic. Then I tried to think of counter examples. But it seems to me that because I dont know what the key of F will be if I could break G then I could break F with one single query.. So i’m pretty stuck and would appreciate help.
note: by mistake I asked the same question yesterday from a user i don’t have access to so please refer to this one because the former is not relevant.

Why isn’t DIV necessarily in P?

In my formal languages class, we discussed DIV, defined as following. DIV = { : a, b ∈ N and a has a divisor d for some 1 < d ≤ b} (<> means encoded, let’s say as binary) We were told that it isn’t known whether DIV is in P and were tasked to prove it was in NP. I naively and mistakenly assumed that DIV was in P because of the following algorithm: On input For all 1

I thought that this algorithm would run in polynomial time because we do b many divisions at worst. Division is polynomial time, therefore, b many divisions is also polynomial time. (also note, b < a, or DIV is trivially true, where d = a).

However, i was told that this algorithm is not polynomial time with respect to the input. I don’t really understand this part. Something along the lines of since a and b are encoded in binary, the input is of order O(log n). And that means our b many divisions is actually O(b) * O(divisions), and that O(b) is 2^(log b). However, isn’t 2^(log base 2 of b) the same as b? How is that not polynomial time?

How to show user timetable’s items width do not necessarily match with time

I am currently designing a timetable for a web app which will show appointments as they are scheduled.

Problem is, as I think that in order to make the user understand the scheduled appointment beforehand there is some important information that has to be displayed (Title, time, customer), appointments that are too short in duration will have to have a different width than the time they represent.

My question is, how do I make the user understand that the appointment’s width does not necessarily match the time?

I came up with this idea, but I think this still is not straightforward enough…

enter image description here

I’m open to any kind of suggestions!

Thanks!

Is the Fourier transform of a measurable function as a tempered distribution necessarily a complex Borel measure?

Let $ u\in\mathcal{S}'(\mathbb{R}^n)$ . Suppose that $ u$ is also a measurable function on $ \mathbb{R}^n$ . Is it true that the Fourier transform $ \hat{u}$ as a tempered distribution is always a complex Borel measure?

I believe that this is the case with more assumptions on $ u$ . For instance, say $ u$ is integrable. fixing a bounded open set $ \Omega$ in $ \mathbb{R}^n$ , for any a test function $ \phi\in \mathcal{D}(\Omega)$ , $ $ |\langle \hat{u},\phi\rangle|=\big|\int_{\mathbb{R}^n}u(x)\int_{\Omega}\phi(p)e^{-ix\cdot p}dp\big|\le \|u\|_{1}\|\phi\|_{\infty}|\Omega|$ $ and from there on the Riesz–Markov–Kakutani representation theorem could be used.

But is $ \hat{u}$ a complex Borel measure under weaker assumptions on $ u$ , or none at all other than that $ u$ is a measurable function?

Why is this incidence matrix necessarily positive definite?

A recently asked question here was solved with the claim that any symmetric square matrix $ M$ of the following form is positive definite:

All of the off-diagonal elements are the same positive integer $ k$ .

Each diagonal element is a positive integer $ n_i \gt k$ . The diagonal elements may or may not be equal to one another.

(The matrix arises as the product of a particular incidence matrix with its transpose.)

Why is a matrix of this form positive definite?