kotlin2

Digital ocean doesn't charge traffic costs. I'm not sure if that was used in the article, but DO can provide significant savings for high-bandwidth services.

People, even those at other big tech companies, unironically refer to working at Google as "retiring." Not sure 'big tech' is a great example to use when talking about productivity.

Is there hard data on how deadly they are vs. other auto manufacturers? There is definitely a narrative that the cars are dangerous, but I'd like to see that quantified.

You can also use green hydrogen in place of blue hydrogen, e.g. when making ammonia.

I think LoC has been determined to be a bad productivity metric because it’s gameable and incentivizes bad behavior. However, that doesn’t apply here as the engineers didn’t know they were going to be evaluated by LoC a priori. I struggle to think of cases were lines-changed wouldn’t be correlated with the productivity of engineers working in a consistent environment if they weren’t trying to game the system.

A world where your cash is held in British pounds? Why do you accept forex fluctuations but not security fluctuations?

I’ve worked in big companies and I’ve worked in small companies. The one constant has been that writing code is required to make product changes. No code likely means no changes. Maybe LOC != productivity makes sense when algorithms are of great importance, like in situations where a genius algorithm can unlock tons of value. Machine learning would be one example. However, most work is feature work, which typically has a straightforward path from start to finish. In product feature work, it’s unclear to me what activity would use time productively that didn’t result in lines of code.

Judging the sizes of the stacked pages from the last 60 days of code would probably be a pretty good indication of productivity. Perfect? No. Good enough? Maybe.

Equity in Meta is liquid net worth, unless you're a materially important share holder. The stocks you hold right now can be converted into cash easy at any point in time. In what world is that nearly equivalent to cash?

Large public companies aren't innovators. Innovations requires risk. Meta is a huge part of retirement portfolios and pensions, which are extremely risk averse. There is an expectation that the company is going to do an efficient job at extracting profits. If Meta isn't meeting that expectation, the market reaction is to be expected .

Companies are valued based on future cash flows. A stable company typically has a P/E ratio of 20. Meta’s is around 10, which is an indication the market believes profits will cut in half and then stabilize.

That seems really inconvenient to be honest.

I don’t think it’s valid to use “only” with “e.g”. “e.g” means it’s an example, which implies the existence of other cases that satisfy the criteria. “Only” implies some uniqueness of the subject.

Why not try to service it yourself? It could be fun to take it apart and reassemble it.

I think the backups end up stored in a GCS bucket in your account and can be removed manually or with an object lifecycle.

What languages allow such a construct? It seems like it would be super confusing if these two code samples produced different values:

    # One
    a = d["a"]["b"]["c"]
    
    # Two
    a = d["a"]["b"]
    b = a["c"]

Ok, but what about the other 95% of emissions. We could just ignore what Russia does and still make a big impact.

If d["a"]["b"] is 42, then how could d["a"]["b"]["c"] also be 42? What you want doesn't make sense semantically. Normally, we'd expect these two statements to be equivalent

d["a"]["b"]["c"] == (d["a"]["b"])["c"]

Now do an associative array containing another associative array.

I mean what was the difference between your outstanding mortgage principal and the value of the homes at the time of default?

How much did you owe after the properties were sold?

In most states, lenders have recourse. So if you stop paying your mortgage, the bank will foreclose on your home and then come after your other assets to make up the difference in what you owe vs. what the home is currently worth.

You could delist yourself from the phone book.

Exactly. So your claim that:

> So you can remove the ones that do not halt by inspecting them one by one and developing a specific algorithm for each one that determines if it halts or not.

Is impossible. You can’t, in general, inspect Turing machines one-by-one to determine if they halt.

That's true. The part that "feels" weird is that there is no algorithm that could perform the separation into halting / non-halting subsets. Choosing elements from a set based on uncomputable properties almost feels like an extension of the axiom of choice.

If you could inspect any Turing machine and devise a specific algorithm that determines if it halts, then you yourself would be an algorithm that solves the generic halting problem.

In the proof you just provided, you have a step "remove ones which do not halt". By which process do you do that? By the halting-problem, you can't actually make that selection. This ties back into the fact that BB(n) is not computable for all n. I guess what I'm looking for is a proof that the number exists without violating the halting problem. Intuitively, we feel BB(n) should exist because there are finite turing machines with n states. But I'm not sure how to express that rigorously. Or maybe relying on a solution to the halting problem doesn't matter in terms of a proof.

I didn’t realize the initial state of the tape was part of the definition of a Turing machine. I was picturing having the same operations applied to different tapes, but that’s not the definition so I’m wrong.

I'm unfamiliar with BB(n), so this raises an interesting question for me. How do we know that BB(n) exists for all n? This is relevant to this post because the author's definition of omega relies on BB(n) being defined for all n. Obviously there are numbers that are not computable, but it seems one needs to be careful that the numbers discussed are sure to exist.