huftis

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...

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-...

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.

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.

BTW, even Hacker News has an RSS feed, at https://news.ycombinator.com/rss. (If you’re on Android, I recommend using the Materialistic app instead, https://play.google.com/store/apps/details?id=io.github.hidr...)

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).

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.)

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 …

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.)

This reminds me of the following old puzzle.

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. :)