A lot of sci-fi suffers from space ships hanging in the air in exactly the way that bricks don’t. Trek does it too.

To be fair, they have the advantage of 400mm lenses. You can take a full-frame picture of a basketball from 10m away with those.

To get every password, you’d have to exploit the password manager process itself. The manager asks you to approve every single password it hands out and you would know something is wrong if the extension starts asking for lots of passwords.

The separation keeps the memory where the passwords are stored away from the browser. No malicious code executing inside the browser can access it. Also, the protocol between the extension and the manager can be really simple and (hopefully) easy to get right without making exploitable mistakes.

It’s the Swiss cheese principle. The attacker has to break out of the website sandbox, get into the extension to copy the secret keys that are needed to impersonate the extension in the connection to the password manager, and exploit the password manager through that connection in order to get to the passwords. If any step fails (the holes in the cheese slices don’t align), the attack doesn’t get through.

At least you’re limiting exposure with managers like KeePassXC. The manager runs in a separate process and communicates with the extension via a local connection. You have to approve every password given out by the manager. So a malicious actor can’t just ask for every password under the sun. They could still read the contents of the password field once the extension has filled it if they manage to circumvent the restrictions set by the browser. But that’s no different from when you enter the password manually.

Greetings, sentient!

The second buy can even be the nice one. If you’re unsure how much use the tool will get, buy cheap then upgrade after it breaks.

10800 to 18000°Ra for the Americans

It’s a sequence of numbers where the next number is a fixed multiple of the previous one.

1 2 4 8 16 32 … Is a sequence with a ratio of 2.

96 48 24 12 6 3 … Is the sequence from the question with a ratio of 1/2. You see that elements 2 and 4 (counting from zero) sum to 30 and elements 3 and 5 sum to 15.

In general, the sequence can be expressed as a_k = c r^k for some starting value c and the ratio r.

Your daily weather forecast likely runs on FORTRAN. It’s quite terrible code in many places because the people writing it are not software engineers but meteorologists, mathematicians, or physicists with little to no formal training in software design writing a million-line behemoth.

And FORTRAN adds to the suck because it is superbly verbose, lacks generics, has a few really bad language design decisions carried over from the 60’s, and a thoroughly half-assed object model tacked on. As a cherry on top, the compilers are terrible because nobody uses the language anymore – especially the more recent features (2003 and later).

Experience 112 reacts to the time between sessions. When you start playing after a few days you get a comment where the main character wonders where you’ve been.

Rocket.Chat is a Slack-like environment under MIT license with apps for iOS and Android

But even when they do, I feel that, after rejecting, I get the same banner again the next time I visit the site. I bet that doesn’t happen when you accept tracking.

gitolite for when a shared git folder is not enough but you don’t want a full gitlab. Provides SSH-based access management for a bunch of git repos.

x = 15

Denote the origin of the circle O and the points A, B, C clockwise starting from the left. From the isosceles triangle OAB we get 2 r sin(alpha/2) = 24, where alpha is the angle between OA and OB.

Construct the line orthogonal to OB that goes through C. The length of the line, h, between C and the intersection is h = 7 sin(beta) = x sin(90 - alpha). Denote the lengths of the parts of OB a and b, where a is connected to B. We have a + b = r

Use Thales circle theorem to find that the triangle ABA’ completes the red shape, with A’ on the circle opposite to A. That means that the angle between A’A and A’B is alpha/2, but A’OB is also an isosceles triangle. So the angle on the other side, beta, has to be the same. Thus, beta = alpha/2.

Now, put everything together: a = 7 cos (alpha/2), b = h cot(90 - alpha) = 7 sin(alpha/2) tan(alpha), r = 12 / sin(alpha/2).

a + b = r <=> cos(alpha/2) sin(alpha/2) + sin^2(alpha/2) tan(alpha) = 12 / 7

1/2 sin(alpha) + 1/2(1 - cos(alpha)) tan(alpha) = 12/7 <=> tan(alpha) = 24/7

From the identity for h we know that x = 7 sin(alpha/2) / cos(alpha). Insert alpha = arctan(24/7)

Beltalowda best at rock tossing! Not Olympic discipline, sasa que. Damn inyalowda.

I’ve been a four-star programmer a few times. Imagine a blocked symmetric matrix where the rows and columns are indexed by triples (u,v,w). The entries are zero whenever u != u’ or v != v’, and because of symmetry you only store entries with w <= w’. But the range of v depends on the value of u and the range of w on the value of v. So you do

double ****mat = calloc (UMAX, sizeof(*mat));
for (int u = 0; u < UMAX; ++u) {
  mat[u] = calloc (u + 1, sizeof(**mat));
  for (int v = 0; v <= u; ++v) {
    mat[u][v] = calloc (v + 1, sizeof(***mat));
    for (int w = 0; w <= v; ++w) {
      mat[u][v][w] = calloc (w + 1, sizeof(****mat));
      for (int ww = 0; ww <= w; ++ww)
        mat[u][v][w][ww] = some_function (u, v, w, ww);
    }
  }
}

and weep a little. In reality, this gets a bit optimized by allocating a single chunk of memory and carving that up into the pointer and data arrays, so everything is reasonably close together in memory.