## Is "one representative" sufficient or do many similar things have to be tested?

This is (again) a question of methodology …

Let's say we're testing a service that returns `Article`s given `id`s, i.e. `List`

` getArticles(List ids);`. The appropriate location will also be `null` if the ID is invalid.

First fill the database with SQL:

``````INSERT INTO articles VALUES (2001, "aaa"), (2002, "bbbb"), (2003, "cc"), (2004, "ddddd");
``````

Then test as follows. How should I do that (see details in the comments below)?

``````void testGetArticle() {
// NOTE: WHICH WAY?

// way1: one existing id, one nonexisting id.
var ids = Arrays.newArrayList(2002, 9999);

// way2: many existing ids, many nonexisting ids.
var ids = Arrays.newArrayList(1000, 1001, 2002, 2003, 2004, 9997, 9998, 9999);

List`````` a = xx.getArticle(ids);
assertNull(a.get(0));    assertEquals(a.get(1), ...);  ...etc... // assert each of the results
}
``````

Thank you for all suggestions!

## Statistics – Are 4 to 4 points sufficient to clearly determine an oblique normal distribution?

The normal PDF is defined as:

$$phi (x | mu, sigma) = frac {1} { sigma sqrt {2 pi}} e ^ {- frac {(x- mu) ^ {2}} {2 sigma ^ 2}}$$

The cdf is given by:

$$Phi (x) = int _ {- infty} ^ {x} phi (t) dt$$

If we get the value of $$phi (x)$$ at three different points this is enough to clearly determine $$mu$$ and $$sigma$$,

The general slant normal distribution is defined as:

$$f (x) = frac {2} { omega} phi left ( frac {x- xi} { omega} right) Phi left ( alpha left ( frac {x- xi} { omega} right) right)$$

Four points are enough to determine $$xi$$. $$omega$$ and $$alpha$$?

## What level of physical destruction is sufficient to ensure that an SSD is not readable?

My organization upgraded some printers and put the internal SSD hard drives out of operation by passing the memory chips through a band saw, cutting each chip in half and, in some cases, tearing entire sections off the greenboard.

These printers have been used so that they are likely to contain PHI / HIPAA information.

I seek advice on whether this method of destruction was sufficient or not.

I don't think so, but I want additional resources.

I have published what I have found so far as an answer, as this may be the answer to my question, but I hope for other contributions.

## US Citizens – Is a 1-Hour Connection from MAD, Norwegian to Delta in LAX Sufficient?

Norwegian is a super discount airline that has no partner alliances or codeshares. I have the feeling that these are separate tickets,

That is the bad thing. You must consult with the person who sells these tickets if she offer a kind of connection insurance. But without something like that Norwegian is just not responsible for that knocks Effects of a slightly late arrival.

Delta owes you nothing if you miss the flight. You didn't show up and there's no way to blame this Norwegian. Norwegian does not owe you money for the effects of a (slightly) late arrival. This is the difference between thru ticketing and separate ticketing.

You can pounce on Delta's mercy; Sometimes airlines have rules for "flat tires", e.g. if you miss a flight through no fault of your own. However, this would be your fault if you booked such a close connection.

## Transit – is MAD a 1-hour connection from Norwegian to Delta sufficient?

I might give you a 1:10 chance of making this connection. Maybe a little higher if you have Global Entry / Mobile Passport and / or TSA PreCheck or if the inbound flight is early. Significantly lower if the incoming flight is delayed, even by only 10 minutes.

If the inbound flight doors open on time and you have checked in for your delta flight in advance, you have 45 minutes to:

• Get out of the inbound flight. Depending on the seat, this can take between a few seconds and ~ 10 minutes.
• Come to immigration. Again somewhere between a few seconds and a few minutes, depending on when you came.
• Come through immigration. Depending on the length of the queue, between a few minutes and more than 30 minutes
• Go to Terminal 2 or 3. Maybe 5 minutes
• Get through security in this terminal. Between a few minutes and more than 30 minutes, depending on several factors.
• Go to your gate – and be there 15 minutes before departure, otherwise your seat will probably be taken.

Possible? Yes. Probably? No…

If you have checked in your baggage or for some reason have not been able to check in online, your chances of winning decrease even further, as you have to be at the Delta Terminal 45 minutes before departure or 30 minutes before departure to drop off your luggage.

## Proof by contradiction – only checking a right neighbor in a sequence of pairs of different integers is sufficient to identify the first local maximum

I'm trying to find out if my evidence is valid. I think it makes sense intuitively, but I'm afraid I'm missing something. Any help would be appreciated!

``````A peak element is an element that is greater than its neighbors.

Given an input array nums, where nums(i) ≠ nums(i+1), find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that nums(-1) = nums(n) = -∞.

Example 1:

Input: nums = (1,2,3,1)
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = (1,2,1,3,5,6,4)
Output: 1 or 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.
``````
``````public int findPeakElement(int() nums) {
for (int i = 0; i < nums.length-1; i++) {
if (nums(i) > nums(i + 1)) {
return i;
}
}
return nums.length - 1;
}
``````

## Why don't you have to check the left neighbor for every element?

Assume a contradiction we are going through `nums`to discover another climax and we come across and element at the index `i` its right neighbor at the index `i+1` is strictly smaller. If the element at the index `i` were Not a peak then the element at the index `i-1` should be bigger. Then we have that

``````nums(i-1) > nums(i) > nums(i+1)
``````

This then implies that `nums(i+1)` is the last element in a strictly decreasing sequence of elements (which we have seen) whose start must be a peak (either the sequence starts at the index) `0` or it starts at the index `k, 0 < k < i`). This contradicts our assumption, so the first element whose right neighbor is strictly smaller is a local peak.

## Database Design – Does the ComplaintMessages entity need to be related to the User and Admin entities, or is it sufficient to relate to the Complaint entity? (ER-model)

I have doubts about the ER model of a scenario in which a user can file complaints.

The user enters the title and description of the complaint and submits the information. A site administrator can view and respond to all user complaints.

The user then also has access to the administrator's response, and the user should be able to respond to that administrator's response, similar to a chat scenario in which the administrator also has access to the user's new complaint message, allowing more than one are notifications.

I have doubts in which ER model this scenario is correct for 1 or 2. Do you know whether it is also necessary to relate the ComplaintMessages entity to the User and Admin entities?

1] It is not necessary to link the "ComplaintMessages" entity with the "User" and "Admin" entities.

2] Or is it also necessary to relate the ComplaintMessages entity to the User and Admin entities?

## Airports – Is 1hour37min sufficient for transit to Air NZ in LAX?

We have booked tickets for next April to New Zealand from JFK. Initially, the flight left JFK (Alaska Airlines) at 4:55 p.m. and landed in LAX at 8:05 p.m. The stopover lasted 2 hours and 25 minutes. The flight took off at 10:30 p.m. (Air NZ). Since then, the departure to New Zealand has advanced to 10:05 p.m. And the flight from JFK now arrives at 8:28 pm – so our stay is only 1 hour 37 minutes. Are we doing this to make it ??? Most likely no checked bags, but even if we do, they should be checked in directly at JFK so we don't have to pick them up at LAX. So we check in with the app for the NZ flight, so no check-in counter is required. The distance from each terminal is 2200 feet. We also have to remove security in LAX. Alaska Airlines will not move us to an earlier flight without charging us \$ 200 + per ticket (x4) because they think it is enough time.

## Periodic Markov chain contains all sufficient large multiplies of the period.

The part I do not get when the proof introduces $$M$$, How do we know there exists $$Md_x$$ and $$(M + 1) d_x$$, What is the main idea behind to show that $$D_x$$ contains all sufficient large multiplies of $$D_x$$?

## User Behavior – Is the Gmail undo pattern sufficient for mass mailing? What are the better alternatives?

We're in the process of creating an email client that sends email campaigns to thousands of users (Mailchimp is a comparable product). There are ideas and opinions, and one of them adds an undo option to the mailing.

The owner of this approach was convinced that this would increase user confidence. I defend an opposing idea that adds an extra step that reviews the content and recipients of the emails and asks for final approval.

As my solution adds friction to the flow, it gives the user's activities a higher level of confidence.

Now everything is suppositions above and we will test them with users, but I want to hear more about the subject, especially the personal opinions.

I use Gmail every day and personally see the undo function cause more stress in my life. It has happened once or twice that I undo the e-mail and edited a few details, but if I could not, it would not be a disaster either.

On the other hand, I think when a user sends a mass mail, the user needs to understand better what he / she is up to.

What are your thoughts? Are there any other alternatives?

PS: I have already read this question about the same functionality, while I understand the assumptions here that I would like to hear more.