Is close quarters combat with firearms in a confined space considered a hazardous task?

I know firing indoors has the potential to hit walls, hulls and equipment, but from the rules this is potentially an additional armor value if a player or npc has cover. I also see where some combat tasks are hazardous if untrained. What I can’t find is if firing indoors should be considered hazardous and failures (or fumbles) roll on the failure and potentially mishap tables. (Like a hull breach if it penetrates or a ricochet otherwise.)

Is this covered anywhere? I have both the original Traveller 2300 rule books (Referee’s and Players’ guides) and the 2300AD CDROM canon (which includes the Adventurers’ and Director’s guides).

Traveller 2300 was renamed to 2300AD, some changes were made and quite a few rules were expanded or added. I’m only curious about Traveller 2300 independently if rules that cover this are only in one edition or are different between the editions.

Find combinations of values in different arrays close to a sum

could someone help me with this case

Imagine i have 3 arrays (A,B,C) with three values in each of them, where A=[1, 2, 3] B=[4, 5, 2] C=[3, 4, 7]

Now I would like to find the combination of elements from each array which result in a sum of 10, the combination must contain a value (only one) from each array. How can I receive all the combinations in Python?

Thanks!

How to find efficiently the minimum modification to avoid close consecutive numbers?

I have an array of sorted numbers:

arr = [-0.1, 0.0, 0.5, 0.8, 1.2] 

I want the difference (dist below) between consecutive numbers for that array to be above or equal a given threshold. For example, if threshold is 0.25:

dist = [0.1, 0.5, 0.3, 0.4] # must be >=0.25 for all elements 

arr[0] and arr[1] are too close to each other, so one of them must be modified. In this case the desired array would be:

valid_array = [-0.25, 0.0, 0.5, 0.8, 1.2] # all elements distance >= threshold 

In order to obtain valid_array, I want to modify the minimum amount of elements in arr. So I substract 0.15 from arr[0] rather than, say, substract 0.1 from arr[0] and add 0.05 to arr[1]:

[-0.2, 0.05, 0.5, 0.8, 1.2] 

Previous array is also valid, but we have modified 2 elements rather than one. In order to obtain valid_array, I already have a brute force solution which works fine, but it is quite slow for large arrays. My questions are:

What is the time complexity of that brute force solution?

Does a more efficient algorithm even exist?

Edit

First, I need to clarify what I mean by difference, which I define the same way as in here, so out[n] = a[n+1] - a[n]. The fact that all elements in that difference must be above (or equal) threshold implies that valid_array is also sorted.

Second, the number of modifications (which must be minimized) is obtained by comparing elementwise the original arr and valid_array

Try to cut down the usb camera open, read and close time [on hold]

I am now working on a image processing project using opencv with 6 cameras covering 360 degrees, connected to a raspberry pi 3 B+. I need to get a single still image from each camera on each rotation. To do this I open, read and close each camera on every 360 degree rotation. But it takes around 1.5 seconds for each camera to go through its starting sequence and a total time of 10+ seconds is way too long for my project. So I am looking for a solution that will cut down the start-up time for each camera to within 0.2 sec to get a still image from each camera on every rotation. Any ideas?

Below is a test code :

import cv2

import sys

print sys.argv

if len(sys.argv) < 2:

print "Usage : > python video.py [camno1] [camno2]"  sys.exit(0) 

cam = []

cap = []

for i in range(len(sys.argv) – 1):

cam.append(int(sys.argv[i+1]))  cap.append(cv2.VideoCapture(cam[i])) 

while True:

for i in range(len(cam)):      print "cam"+str(i)+" read"      ret, frame = cap[i].read()      if ret :          cv2.imshow("frame"+str(i), frame)  if cv2.waitKey(1) & 0xFF == ord('q'):      break 

for i in range(len(cam)):

cap[i].release() 

cv2.destroyAllWindows()

Does “Prevent computer from sleeping…” remain in effect even if I close the lid?

These are the default settings for when a Mac laptop is plugged in:

enter image description here

Notice how

  1. the display will be turned off automatically after 10 minutes of inactivity, and
  2. the ”Prevent computer from sleeping…” option is not checked, which suggests that the laptop will go to sleep when the display is turned off (after 10 minutes of inactivity).

If I check the “Prevent computer…” option, I can expect that the laptop will not go to sleep after the display is turned off automatically.

But what if I then close the lid? Does this setting also prevent the laptop from going to sleep even then?

I want to make the console close WITHOUT someone having to type but i dont want it to close instanly

Im new to C++ and im following a guide on cin he used a getline() function for the program to wait until the client presses enter to close the program, it works for him, but my program still instantly closes after i put the first input in. (I DO NOT WANT THE USER TO HAVE TO TYPE SOMETHING SO PLEASE DO NOT MARK AS A DUPLICATE!)

#include <iostream> #include  <filesystem> #include <string> using namespace std; int main() {     string santype;     cout << "What type of sandwhich you want kid?" << endl;     cin >> santype;     cout << "Creating " << santype << " Sandwhich" << endl;     cout << "Press Enter To Exit..";     getline(cin, santype); } 

Whole server disconnect for awhile If I close the server software with 50000 clients connecting to it

I have a server software written in .NET listen for clients connect to it and keep them alive. When I suddenly close the server software with ~50000 clients online. The whole server immediately become disconnect for awhile (about 30-90 seconds). This issue does not occur if the software have only few thousand clients online. It become very frustrating because sometimes I need to restart the server software for upgrade. Anyone know why it happened?