12:25 AM
Lol, not as much as Scratch ;)
11

Scratch 3.0, 7 blocks/62 bytes Try it online Scratch! As SB Syntax: when gf clicked ask()and wait say(round((answer)/([sqrt v]of(2 It's always fun to usual visual languages!

So uh I just put my username through Google translate using auto detect language as the source and English as the target language
And it thinks "Lyxal" is in Azerbaijani
And it makes the output "Lyhal"
So I put that back into the detect language box and it says it's in Arabic
And it says the English translation is "to solve"
So I guess solving code golf challenges is an inherent part of my username now

12:48 AM

1:22 AM
@a'_' apparently so

2 hours later…
3:14 AM
0

Here is the data, the questions are below In[0] import numpy as np import pandas as pd import matplotlib.pyplot as plt In[0] np.random.seed(42) def offset_data(data): if np.min(data) < 0: return np.add(data, (np.abs(np.min(data))) + 1.0) else: return data height, wi...

2 hours later…
4:58 AM
0

A "simple circular" word is a word whose chords do not intersect. The chords of a word may be seen by laying out the alphabet in a circle, and then connecting the word's consecutive letters. Examples of Simple Circular Words ROLE LAKE BALMY Failing Example A word fails to be simple ...

5 hours later…
9:35 AM
0

strSub Swingingapp (Substring Swapping) Intro/Background I was looking at a Flybuys card at coles recently and noticed that if I took something like this: Source, randomly found on the Internet And swapped some words around, I'd get this: Sorry about the bad quality edit, I did it on mobi...

2 hours later…
11:38 AM
How is the name of a USB key determined? On a mac for example my USB key comes up as R2D2 and on windows Star Wars :)

Basically the computer decides.

but what is it reading from the key?

The name of the key is read from the key.

@a'_' any idea how it reads different names in that case?

@Anush It's because the public keys in that case are different, therefore different names are read.

11:46 AM
as in the cryptographic public keys are different between mac and windows?

CMC: Find two decryption systems that are invented before this CMC. Find a string so that this decrypted in one yields R2D2, and decryption in the other yields Star Wars.

:)

2 hours later…
1:25 PM
@Anush Windows might be reading the label from autorun.inf?

@Neil oh that's an interesting idea
I wonder which stackexchange site accepts questions on this topic :)

sounds like Super User would be most relevant

Thanks... I can't work out if it's just a waste of time :)

1 hour later…
2:59 PM
They were all in the beta stage. but Catija says the same can happen to non-beta sites

@LuisMendo ah well then..

2 hours later…
4:52 PM
-2

The question and solutions are available here: https://ioinformatics.org/page/ioi-2019/51 There's a folder containing "MakeFile" for tests, I do not know what to do with it. Can someone please help me run the checker against my Java solution?

3 hours later…
8:10 PM
is it possible in python to make 64 bit random bit arrays and then do fast Hamming distance computations using XOR and popcount?
or should I just give up on Python :)
the first part is easy if one can make random 64 bit unsigned integers

random.randint(0, 2**64-1)?

@El'endiaStarman Is that then a 64 bit int that I can perform XOR on ?
it's not at all clear to me what python ints are

Yep, you can definitely xor those.

ok then bin(x).count('1') I guess
as popcount

In [97]: import random
In [98]: x = random.randint(0, 2**64-1)
In [99]: y = random.randint(0, 2**64-1)
In [100]: x, y, x^y
Out[100]: (13737186597515989731, 1134597505236996627, 12761701636154360048)

8:25 PM
I wonder how fast it will be
can you try bin(x).count('1') too?

Yeah, I don't know of a much better way than bin(x).count('1'). If you're looking for pure speed, Python is not necessarily the right choice anyway.
In [101]: list(map(lambda z: bin(z).count('1'), _))
Out[101]: [34, 34, 20]

stackoverflow.com/questions/109023/… has some tricks but they seem to be 32 bit mostly
@El'endiaStarman right.. I am looking for speed so have to weigh up the cost of learning a new language :)
I should really learn julia

In [109]: for z in (x, y, x^y): print(bin(z)[2:].zfill(64))
1011111010100100010010100110110101111110001100101001001011100011
0000111110111110111001100110101101101110101100100001001000010011
1011000100011010101011000000011000010000100000001000000011110000

1 hour later…
9:38 PM
1

This challenge is about the number of orderings which contain at most \$n\$ classes and at most \$k\$ of the \$k^{\text{th}}\$ class. One way to represent such an ordering is as a sequence of positive integers with these three restrictions: A number may only appear after all lower numbers have...

2 hours later…
11:55 PM
@Anush x=rand(UInt64,1000);count_ones.(x.⊻x') generates 1000 random numbers, xors all (1,000,000) pairs together, and gets the popcounts quite quickly