bazzargh

well, I wrote a reply, twice, and deleted it. I don't think a human would make this mistake, but explaining why is kinda long winded, so I let my words stand as self-explanatory. But here goes.

The AI talks about the movies like they are familiar with Singer's work. But if they were... they would known he was up-and-coming in the 90s, a decade too late to have helmed a blockbuster whose flop would have killed his career? They would have known that Singer made superhero movies _before_ Nolan not just after like the conversation suggests?

Even bringing up Superman IV is weird; it was such a dud it never gets shown, so unless they old enough to see it at the time -like me- (so would notice the 20 year gap!) or were an avid fan (so would know more about the movies!) it wouldn't come up.

On the other hand, for machines all things are equally obscure, they all happened yesterday, and are relevant if cosine similarity says they are.

"There's no way I would be able to detect if this was AI from the examples alone."

The hallucination in the second sample - that Bryan Singer was involved in Superman IV: The Quest for Peace - would have rung alarm bells. It just seemed to be free associating on words in the conversation.

that's more than a little unfair - MGK was famous for making this stuff back then, and was the person people would steal these memes _from_.

yes, you can't stick with a 32 bit account any more, needs 64.

It'd be nice to be able to permalink to the view for a specific location

While that'd be useful for things like telling someone else what buses they can get, planning a trip, etc, what I actually tried to do was to link to the Killermont Street bus station in Glasgow; Aztec Camera had a sentimental song about late night departures from there - leaving your friends and heading south to look for work - that runs:

    As the ships and the steel
    Slip away to the cry of "Compete"
    There's a message for us
    We can get there by bus
    From Killermont Street

And indeed you can...buses all over the place. But I can't link to it :)

Our experience has been the direct opposite. The way to get the most out of cursor was to leave it on auto - we saw it average around 3.9 cents per request under the old contract (per-seat pricing) and more like 39 cents per request under the new (single pooled cost for everyone). Composer came in a little higher, more like 50c/r, while the claude models were up into the dollars. Meanwhile, if you use Claude-the-app, there is _no_ cheap model and the default switched to Opus in April, resulting in increased costs across the board.

We have both Claude and Cursor here, as well as agents running GPT, things in AWS Bedrock, etc and its my team handle the bills...when people use Cursor on auto, costs are under control, but there's always a dozen or so whale users who'll switch models manually and blow through the budget like it's not there.

Another thing: "better cost controls". There was for example no way for us to disable Fable in Claude, but we could in Cursor. Again, the opposite experience.

I quite enjoyed their Omega Mart in Vegas. Was there for work and gambling, shows, and gun ranges are really not my thing, glad there was something different. (I did wander round the strip but enjoyed the hike in Red Rock Canyon more)

When we went we got the priority tickets (can't remember the exact name of the deal) because there were no normal slots left that day; that does get you discounts on stuff but awkwardly the discount is at the gift shop _outside_ and it wasn't clear whether there were things at the fake supermarket checkout _inside_ that you couldn't get outside. It also gets you the card you boop for the scavenger hunt which makes you more involved in the whole thing. So yeah, they got us for some extra cash but it wasn't so bad as a one-off.

You would find things in there that were already close to QM and relativity. The Michelson-Morley experiment was 1887 and Lorentz transformations came along in 1889. The photoelectric effect (which Einstein explained in terms of photons in 1905) was also discovered in 1887. William Clifford (who _died_ in 1889) had notions that foreshadowed general relativity: "Riemann, and more specifically Clifford, conjectured that forces and matter might be local irregularities in the curvature of space, and in this they were strikingly prophetic, though for their pains they were dismissed at the time as visionaries." - Banesh Hoffmann (1973)

Things don't happen all of a sudden, and being able to see all the scientific papers of the era its possible those could have fallen out of the synthesis.

I'm saying it's a tautology because it's just a binary representation of the set. Suppose we have 8 players, with x and y being 2 and 4: set the 2nd and 4th bits (ie 2^2 & 2^4) and you have 00001010.

But to lay it out: every positive integer is a sum of powers of 2. (this is obvious, since every number is a sum of 1s, ie 2^0). But also every number is a sum of _distinct_ powers of 2: if there are 2 identical powers 2^a+2^a in the sum, then they are replaced by 2^(a+1), this happens recursively until there are no more duplicated powers of 2.

It remains to show that each number has a unique binary representation, ie that there are no two numbers x=2^x1+2^x2+... and y=2^y1+2^y2+... that have the same sum, x=y, but from different powers. Suppose we have a smallest such number, and x1 y1 are the largest powers in each set. Then x1 != y1 because then we can subtract it from both numbers and get an _even smaller_ number that has distinct representations, a contradiction. Then either x1 < y1 or y1 < x1. Suppose without loss of generality that it's the first (we can just swap labels). then x<=2^(x1+1)-1 (just summing all powers of 2 from 1..x1) but y>=2^y1>=2^(x1+1)>x, a contradiction.

or, tl;dr just dealing with the case of 2 powers: we want to disprove that there exists a,b,c,d such that

2^a + 2^b = 2^c + 2^d, a>b, c>d, and (a,b) != (c,d).

Suppose a = c, then subtract 2^a from both sides and we have 2^b = 2^d, so b=d, a contradiction.

Suppose a>c; then a >= c+1.

2^c + 2^d < 2^c + 2^c = 2^(c+1).

so

2^c + 2^d <= 2^(c+1) - 1 < 2^(c+1) + 2^b <= 2^a + 2^b

a contradiction.