The Cursed Computer Iceberg Meme
suricrasia.online
suricrasia.online
I have the feeling that the whole iceberg only exists for us to click on this very article (as both share the same domain)
I think it is silly, but there are people anxious about it, which is not true about Pascals Wager.
There's reason to be anxious about Pascal's Wager: assuming one holds a charitable view of the historicity of religious texts then Pascal's Wager is very rational - but one might feel social pressure to not be religious - or more likely: social pressure to hold a particular religion) - and just procrastination to get-around-to-researching-this-whole-religion-thing, so the uncertainty and lack of confidence in ones' own actions can lead to anxiety like that.
Actually, it's only dangerous if one believes in it. Which is why it can't really work: a superintelligent AI would assume nobody is gullible enough to actually believe it, like, for real.
Decoders that try to restart on unexpected byte sequences: https://john-millikin.com/%F0%9F%A4%94/case-report-surugaya-...
GHC 7.2 using Unicode to represent file paths on Linux (where paths are arbitrary bytes), an approach which was further explored in GHC 7.4 by using reserved codepoints.
\u vs \U escapes in languages that added Unicode support before the codepoint space was expanded beyond 16 bits.
[0]: https://www.eveonline.com/news/view/about-the-boot.ini-issue
1. mitigate ASAP (pull or fix the update)
2. Make customers whole (anyone affected should be fixed no matter the hassle or cost - paying for Geek Squad is brilliant)
3. Determine how testing could have caught it and didn’t (does your CI include rebooting? Nonstandard configurations?)
4. Remove overlapping filenames
5. Be open about all the above
Immutable distributed histories are all susceptible to the “encoded illegal content” and I don’t know how they’ll end up resolving it.
for i in range(1, 100):
divisible_by_3 = i % 3 == 0
divisible_by_5 = i % 5 == 0
fizz = 'Fizz' if divisible_by_3 else ''
buzz = 'Buzz' if divisible_by_5 else ''
num = '' if divisible_by_3 or divisible_by_5 else i
print "{0}{1}{2}".format(fizz, buzz, num)
for (let i = 1; i <= 100; i++) {
console.log(
(i % 3 == 0 ? "Fizz" : "") + (i % 5 == 0 ? "Buzz" : "")
|| i
);
}
import functools
def fizzbuzz(n):
if not n % 3: yield (r := "fizz")
if not n % 5: yield (r := "buzz")
if not "r" in locals(): yield n
compose = lambda f, g: lambda *args: f(g(*args))
functions = "".join, functools.partial(map, str), fizzbuzz
transform = functools.reduce(compose, functions)
print("\n".join(map(transform, range(1, 100))))I was not disappointed.
> Meanwhile, we want there to be at most one place that outputs fizz, buzz, or the string representation, and each test to be performed only once.
These do intuitively seem like the properties of an elegant solution.
fizzbuzz n = case (n^4 `mod` 15) of
1 -> show n
6 -> "fizz"
10 -> "buzz"
0 -> "fizzbuzz"
main :: IO ()
main = print $ map fizzbuzz [1..30]I think it is an elegant solution.
import itertools
# this block is just for simple functional pipes
class B:
def __init__(self, f): self.f = f
class Pipe (B): __ror__ = lambda self, x: self.f(x)
class Map (B): __ror__ = lambda self, x: map (self.f, x)
class MapValues(B): __ror__ = lambda self, it: it | Map(lambda kv: (kv[0], self.f(kv[1])))
# here the actual FizzBuzz
(
range(1, 100)
| Map(lambda i: (i, itertools.compress(('Fizz', 'Buzz'), (i % 3 == 0, i % 5 == 0))))
| MapValues(''.join)
| Map(lambda kv: kv[1] or kv[0])
| Pipe(list)
)
[1, 2, 'Fizz', 4, 'Buzz', 'Fizz', 7, 8, 'Fizz', 'Buzz', 11, 'Fizz', 13, 14, 'FizzBuzz', 16, 17, ... ]
Seeing it done with a GBA is not surprising.
I couldn't find that on the link?
This the text editor?
>LISP, or manual garbage collect
This Lisp Machine garbage collection = reboot?
(The lowest few pages of memory are protected in the same way to increase the likelihood that a NUL pointer correctly blows up in your face when you try to access its memory.)