And here is the corresponding report, ‘You can log out, but you can never leave: How Amazon manipulates consumers to keep them subscribed to Amazon Prime’ (PDF), which has more information:
https://fil.forbrukerradet.no/wp-content/uploads/2021/01/202...
FYI, KDE/Plasma supports all of those features you mentioned:
1. You can set the launcher/taskbar/panel to auto-hide.
2. You can set up the screen corners (or a keyboard shortcut) to show an Exposé-style view of all running applications/windows.
3. You can have a global menu bar.
You can get pretty close to emulating Unity's interface with Plasma, especially if you use Latte-dock. But there are still certain things that can't be done; the biggest to me is having the menu bar (File, Edit, etc.) in the title bar of all the windows (a global menu at the top works though). You can get a button that then gives you those options, which is close but an extra click.
Yes, it’s annoying. But many sites have RSS feeds that they don’t advertise. For example, almost all sites based on Wordpress have an RSS feed that you can get to by appending ‘/feed/’ to the site’s URL. For example, if the site is hosted at ‘example.org’, ‘example.org/feed/’ will give you the RSS feed. Appending ‘comments/feed/’ instead will give you an RSS feed for the site’s comments.
And for podcasts that are listed on iTunes or SoundCloud, you can use http://getrssfeed.com/ to extract the URL to the RSS feed.
For reading RSS feeds in general, I prefer rss2email (https://github.com/rss2email/rss2email) to normal RSS readers. You get each new post as an e-mail (HTML or plaintext, with support for digest mode).
Ten people are standing in a line. Each person is wearing a black or a white hat and can only see the colours of the hats of the people in front of them. The goal is to guess the colour of one’s own hat. First, the last person in the line shouts out their guess (‘black’ or ‘white’), then the person in front of them shouts out their guess, etc. No other form of communication is allowed (no clapping, no touching, no texting – and no tricks involving, e.g., encoding information in the delay between it being ‘your turn’ and actually shouting out your guess). The group ‘wins’ if at most one person guesses wrong.
The group is allowed to discuss a strategy before taking part in this puzzle (i.e., before being given their hats). Which strategy should they choose to maximise the chance of winning?
They’re only allowed one wrong guess, so this is basically trying to encode 9 bits of information in 1 bit. And here the ordering of the bits actually matters. Still, it’s possible to choose a strategy which gives the group 100% chance of winning! Good luck trying to solve this one. :)
The way huftis stated it, it is indeed impossible.
In the popular, solvable version of the puzzle, the group also gets the information if each guess was wrong or right. It's not at all about encoding "9 bits of information in 1 bit".
No you don’t need information on whether the each guess is right. (And, of course, if the group is playing it correctly, every guess except possible the first one is correct.)
You’re of course right about this not being encoding ‘9 bits of information in 1 bit’. That would be impossible; to encode 9 (arbitrary) bits of information, you need 9 bits.
On the other hand, you must encode information on the colour of 9 hats, and only the last person in the line is free to provide some information. He can’t possibly know the colour of his own hat, so he can only guess – or use his turn to provide 1 bit of information to the rest of the group (while the other people have to shout out their correct hat colour). The difficulty lies in figuring out why the puzzle as stated does not need you to encode 9 bits using 1 bit …
Well it’s easy to do with two people because the first person can just guess the color of the second, since we are allowed one wrong guess. I don’t see how this can be extended.
I assure you that it's possible. To give you a hint without giving everything away: the guess of the first person must be derived from the colour of all other person's hats.
You could think of it as a puzzle in coding theory, and it's not implausible. If you sum up the number of hats that each person sees and the number of guesses they hear, you'll note that everybody has 9 bits of information available. And they're supposed to make 9 correct guesses. It adds up.
Another hint would be to think of the case of 3 persons first. That can be done with patient case analysis, and it's likely to get you an idea for generalisation.
Think about it as if each guess preceding a person is a bit of information, the goal is to come up with a strategy for using that information (you control encoding as well as decoding) to correctly infer another bit.
You can extend to three people if the first "guess" indicates if nr 2 and 3 have the same color or not. But doesn't go further because after the first all guesses must be 100% correct and can't be used for communicating additional information.
It does extend. It's just an XOR/parity trick. The last person shouts out the XOR of all the previous colors (if you prefer: whether the number of visible white hats is even or odd). From there, every other subsequent person knows their color by XORing the hats they see together with the guesses made so far.
For example, the second to last person knows their hat is black (0) if the last person's guess matches the XOR of the hats they see - if it doesn't, their hat flipped it, and thus is white (1). And so on and so forth. Your hat color is everything you see and everything that was said XORed together.
That’s correct. Or, to explain in (slightly) more non-mathematical terms: The last person (‘person 10’) shouts out (e.g.) ‘white’ if the number of white hats in front of him is an odd number (1, 3, 5, 7, 9) and ‘black’ otherwise. Let’s say he shouted ‘white’. The person in front of him looks at the people in front him. If it’s still an odd number of white hats, his must obviously be black; otherwise it must be white.
If his hat is black, his shouts out ‘black’, and the person in front of him knows that the rest of the line (8 people) must still have an odd number of white hats, and he applies the exact same logic. But if the hat of person 9 was white, he would shout out ‘white’, and person 8 would know that the rest of the line (including himself) should now have an even number of white hats.
So, basically the rule is: Define ‘the rest of the line’ to mean the people who have not yet shouted out a colour. When the first person (person 10) shouts out ‘white’, this means that ‘the rest of the line’ has an odd number of white hats (parity: odd). Whenever someone shouts out ‘white’, the parity (even/odd) of ‘the rest of the line’ is flipped.
Using this, each person only has to count the number of white hats of the people in front of them and observe if it is even or odd. If it matches the parity of ‘the rest of the line’, the person’s hat is black, otherwise it’s white. (This also works for the person in front of the line (person 1). He sees no (or 0) white hats, i.e. he sees an even number of white hats.)
Based on this, the Norwegian Consumer Council has filed a legal complaint against Amazon for breaches of the Unfair Commercial Practices Directive: https://www.forbrukerradet.no/siste-nytt/amazon-manipulates-...