Kogasa

teto

cats just do that sometimes

I can sorta see that. An exit code is an exit code. But exit codes, like words, have meanings. Sometimes that meaning is a boolean value, as is the case in e.g. GNU Coreutils conditions: https://www.gnu.org/software/coreutils/manual/coreutils.html#Conditions

Yes, it is used consistently in GNU Coreutils: https://www.gnu.org/software/coreutils/manual/coreutils.html#Conditions

• false: do nothing, unsuccessfully (returns 1)

• true: do nothing, successfully (returns 0)

• test: check file types and compare values is documented as “returns a status of 0 (true) or 1 (false)”

What you’re saying is irrelevant. In the real world, when an exit code is a boolean, 0 is true.

Every time an exit code is a boolean, 0 is true. Shell scripting would be very annoying if this were inconsistent

Even if that were literally what it did, having a StackOverflow button would be pretty cool

I’ve been using a reverse proxy on a Hetzner VPS pointing at my home plex server for years without issue. Maybe this only applies to people running the actual Plex software on a Hetzner VPS?

Content of the image is unrelated to this post FYI

Wayland support for multi-monitor configurations is better than X11. Whatever performance degradation you claim to experience is also a “you” thing, not a “Wayland” thing. There’s an underlying issue with your setup.

Time… travel?

Uhh, yeah.

Maybe it’s the 5G.

Neat, I do have them. It’s probably because most people have a lot of subtle freckles everywhere and any 3 dots forms a triangle or something.

The hot dog says “served with fried potatoes” at least. If the burger doesn’t come with fries that’s criminal. How can you serve a burger without fries

Yes, we are.

Ridiculous prices? This is pretty much normal in my area, which isn’t one of the most expensive places in the country. Boring ass menu though

The direct connection is cool, I just wonder if a P2P connection is actually any better than going through a data center. There’s gonna be intermediate servers right?

Do you need to have Tailscale set up on any network you want to use this on? Because I’m a fan of being able to just throw my domain or IP into any TV and log in

I just use nginx on a tiny Hetzner vps acting as a reverse proxy for my home server. I dunno what the point of Tailscale is here, maybe better latency and fewer network hops in some cases if a p2p connection is possible? But I’ve never had any bandwidth or latency issues doing this

It gets around port forwarding/firewall issues that most people don’t know how to deal with. But putting it behind a paywall kinda kills any chance of it being a benevolent feature.

1000 or so moves

Ok, this nerd-snipes me a bit. The real point of the Towers of Hanoi problem, at least if you’re doing it in a discrete math class, is to count exactly the number of moves for n discs. Call this number f(n).

To make this post easier to read, we will label the discs from 1 to n, with 1 being the bottom disc and n being the top. We will also label the pegs A, B, and C, with A being the starting peg and C being the target peg. This lets us describe moves in a compact notation: 3: A -> C means disc 3 moves from A to C.

For n=0, we vacuously need 0 moves to win.

For n=1, we need just 1 move: 1:A->C.

For n=2, we can do it in 3 moves, so f(2) = 3. The sequence is:

• 2: A->B # unblock the next move
• 1: A->C # move 1 to its final location
• 2: B->C # move 2 to its final location

Now suppose we know f(n) for n between 1 and N, with N >= 2. For n=N+1, we can move N+1: A -> C and attempt to use our strategy for moving the remaining N discs from A to C, shuffling N+1 around as needed. Since it’s the smallest disc, we can always move it to any pillar, and any time we want to move something to the pillar it’s on, it must be moved first. Let’s see how that plays out for N=2:

• 3: A->C # move the N+1 disc
• 2: A->B # start the N-disc strategy
• 3: C->B # unblock the next step
• 1: A->C # next step of N-disc strategy
• 3: B->A # unblock the next step
• 2: B->C # last step of N-disc strategy
• 3: A->C # finish the N+1-disc problem

We hazard a guess that every step of the N-disc strategy is preceded by a move of the N+1 disc, plus one final move. In other words, f(N+1) = 2f(N) + 1. Some careful analysis will justify this guess.

Now we have a recurrence relation for f(n), but how to solve it? A classical technique is “guess the formula magically, then prove it by induction.” It’s certainly doable here if you compute a few values by hand:

f(2) = 3

f(3) = 2(3) + 1 = 7

f(4) = 2(7) + 1 = 15

f(5) = 2(15) + 1 = 31

f(6) = 2(31) + 1 = 63

You’ll probably see it right away: f(n) = 2^n - 1. Indeed, we can prove this by induction: the base case n=2 holds as f(2) = 3 = 2^(2) - 1, and if f(n) = 2^n - 1 then f(n+1) = 2f(n) + 1 = 2(2^n - 1) + 1 = (2^(n+1) - 2) + 1 = 2^(n+1) - 1 as desired.

We may conclude f(12) = 2^(12) - 1 = 4095. In other words, with 12 discs, exactly 4095 moves are required.

–

Bonus: an alternative to the “guess and prove” technique that is generalizable to a broad class of recurrence relations. The technique is called “generating functions.” Given the sequence f(n), we consider the formal power series

F(x) = f(0) + f(1)x + f(2)x^(2) + … + f(n)x^(n) + …

This F(x) is the “ordinary generating function” for the sequence f(n). Strictly speaking, it may not be a well-defined function of x since we haven’t made any assumptions about convergence, but we will assume (for now, and prove later) that the set of such formal power series behaves algebraically much like we expect. Namely, given the recurrence relation above, we can write:

F(x) - f(0) - f(1)x = f(2)x^(2) + f(3)x^(3) + f(4)x^(4) + … + f(n)x^n + …

= (2f(1) + 1)x^(2) + (2f(2) + 1)x^(3) + … + (2f(n-1) + 1)x^(n) + …

= 2x(f(1)x + f(2)x^(2) + … + f(n)x^(n)) + (x^(2) + x^(3) + … + x^(n) + …)

= 2x(F(x) - f(0)) + x^(2)/(1-x)

In our case, we have f(0) = 0, f(1) = 1, f(2) = 3 so we can write more succinctly:

F(x) - x = 2xF(x) + x^(2)/(1-x)

Solving for F,

F(x)(1 - 2x) = x + x^(2)/(1-x)

= x(1 + x/(1-x))

F(x) = x(1 + x/(1-x))/(1 - 2x)

= x/(2x^(2) - 3x + 1)

Ok, great. We’ve found that our generating function, convergence notwithstanding, is that rational function. We can use partial fraction decomposition to write it as

F(x) = 1/(1 - 2x) - 1/(1-x)

which has the advantage of telling us exactly how to compute the coefficients of the Taylor series for F(x). Namely,

1/(1-x) = 1 + x + x^(2) + … + x^(n) + …

1/(1 - 2x) = 1 + 2x + 4x^(2) + … + 2^(n) x^(n) + …

So F(x) = (1-1) + (2-1)x + (4-1)x^(2) + … + (2^(n)-1)x(n) + …

The nth coefficient of the Taylor series for F about 0 is 2^(n)-1, and by the definition of F as the ordinary generating function for f(n), we have f(n) = 2^(n) - 1. (The rigorous justification for ignoring convergence here still needs to be done; for now, this can be seen as a useful magic trick.)