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01:10
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May look like some type of yellow sugar, but it's actually sulfur.
Although it's yellow as solid, it melts into a cherry-red liquid, which burns with blue flame.
02:13
@user430580 BTW some lines in this image are a little wrong, the examples are all correct but for example it says :0x2008 = 8200 (1 rotation to the left)" while it is actually 1 rotation to the right
Here is a new image with correct labeling. And there are a lot of more examples if you start looking for left rotations instead of right, at least in the 7 digit range
New python script with correct labelling: pastebin.com/rLk4Y7g9
If there are any larger than 0x0795040 = 7950400 that are reasonably possible to find with this code, I'm not sure. But I suspect there is no reason why they shouldn't go on forever. And brute force search is probably not the optimal method :P
02:41
Imagine some poor dude 2000 years ago, trying to find those by hand :D if it were me, I would probably be too lazy to even find 0x391 = 913
 
8 hours later…
10:23
Wow, I was honestly not having too much hope, but I left this script running overnight and woke up to this baby! 0x036915494 = 915494036 (3 rotations to the left)
@Dave Hello, welcome to the chat, always happy to see new faces!
And the script is still running BTW, looking forward to more findings!
 
3 hours later…
13:18
@user430580 well played
@user430580 it's a nice idea, ask if there's any way to predict such occurrences... maybe i'll get to it... BTW i don't mind if you ask a question about it if you're interested, i claim no rights on the concept or something , lolz
@user430580 yes i bet one can at least optimize the brute force search. for instance if you are looking at 0x2000 ... there's no sense comparing to decimal numbers that are manifestly smaller, 2, 20, 200... so one basic check one can make is to ensure you compare a decimal larger than the (apparent) decimal that the hex looks like
although you're just rotating it right, so maybe it's a bit trickier to implement this optimization, nvm :)
 
6 hours later…
19:24
@PM2Ring Hello, welcome to the chat! Always happy to see new faces pop in!
@Amit Please consider asking such a question, I have read some of your other questions and see that you definitely have a better way with words and would certainly phrase the question better that I could
@Amit thanks! :P
Hi. Thanks. I just noticed this chat room in a link on my meta question: physics.meta.stackexchange.com/q/14801/123208
@Amit's hex rotation thing reminded me of an old joke.
@Amit I am open to any new ideas. BTW this thing has been grinding for about 20 hours and 0x036915494 = 915494036 is still the highest so far, no new findings
Please tell us the old joke, we love jokes there
Q: Why do programmers get Halloween and Christmas mixed up? A: Because 31 OCT = 25 DEC.
4
Haha I have seen this particular joke about a hundred billion times, but I still laugh every time :D
:)
In every base, there are numbers (related to reciprocals of primes) that have a cyclic pattern. One famous example in decimal is 142857, which comes from 1/7 = 0.142857142857...
A cyclic number is an integer for which cyclic permutations of the digits are successive integer multiples of the number. The most widely known is the six-digit number 142857, whose first six integer multiples are 142857 × 1 = 142857 142857 × 2 = 285714 142857 × 3 = 428571 142857 × 4 = 571428 142857 × 5 = 714285 142857 × 6 = 857142 == Details == To qualify as a cyclic number, it is required that consecutive multiples be cyclic permutations. Thus, the number 076923 would not be considered a cyclic number, because even though all cyclic permutations are multiples, they are not consecutive integer...
19:40
Thanks for the link, it definitely gets us closer to finding the solution (if it exists)!
On a a vaguely related note, here's a little digit pattern puzzle (in decimal) that I noticed many years ago.
There's a 2 digit number that if you divide it by 3, reverse the digits, then add 1, you get back to the original number. What's the number?
20:16
It's 42 (42/3 = 14, then reverse the digits to get 41, then add 1 to get 42 back again) that took me way too long to find :P
Do you know if there would be a formula to find it faster than exhaustive search?
20:34
@user430580 thanks... I hope I'll get to it when I have the time
@PM2Ring That's good one!
@user430580 It's a similar but less crazy idea, they don't compare how numbers "look" in different bases... :P
@user430580 you can write an equation in two variables and just iterate over the integers, it's exhaustive search but it's only 10 combinations
'cause once you get one digit you got the other, it just needs to satisfy the equation and be an integer
actually looking at the form of the equation you got even less options
try to write the condition and you'll see :P
20:57
@user430580 Sure.
The "trick" is to work backwards, to turn the division into multiplication.
Let $a, b$ be the digits.
So $3(10a+b) - 1 = 10b+a$
Thus $30a + 3b - 10b - a = 1$
$29a - 7b = 1$
That's just a linear Diophantine equation, so you can solve it using Euclid's Greatest Common Divisor algorithm..
But the answer should be pretty obvious, by inspection.
heyyy that's what I did :P
Even using Euclid, the first step is to divide 29 by 7, which gives 29 = 7×4 + 1, and we're done.
wait but is it a mistake where
$$ 3(10(a-1)+b) = 10b+a $$
?
first we need to subtract one, then reverse...
I'm working backwards, from 14.
So I triple, subtract 1, then reverse.
that looks like working from the middle, i'm confused i guess
21:07
Sorry. But hey, it's a puzzle. A little bit of confusion is to be expected. ;)
oh i get it, you've started from the reversed number ok
Yesterday I got an upvote on a SO answer from 2017 that I'd totally forgotten about. It involves sequences of pairs of numbers from 1 to n. Here's an example for n=7. Can you see the pattern?
[2, 4, 7, 2, 3, 6, 4, 5, 3, 1, 7, 1, 6, 5]
@Interested111 Hello, welcome to the chat! :)
@Amit In my mind, these two are really similar, even though the second idea does not operate on different bases, but that's just for me. Here's the big disclaimer: my mathematical intuition is somewhere between "zero" and "very bad" :P
@PM2Ring yeah I think there are exactly $n$ numbers between each pair of numbers with value $n$
@PM2Ring Thanks, you explained it really well. Looks simple to understand, but to find it out myself, it would probably take me quite a long time, and result in lots of frustration :P
21:14
Exactly! Here's the only solution for n=3 (a sequence & its reflection are considered to be the same sequence).
[3, 1, 2, 1, 3, 2]
nice... I suppose someone proved such a sequence exists for all $n$?
OakBot Online.
@Michael Hello and thanks for OakBot!
/mood
@user430580 I'm feeling a bit grumpy.
/mood helpful
21:15
@user430580 I am now helpful. :D
/imagine dragons
@Amit For all n where n = 0 or 1, mod 4.
I'm going to use this prompt instead: Imagine a scene featuring four dragons. The first dragon perched atop a majestic mountain peak is a glossy black dragon with webbed wings spread wide against the cloudy sky. The second dragon, a fiery red one, is poised near a molten lava river deep within a volcano. The third dragon is a magnificent green dragon, wandering in a lush forest. The last dragon is a pure white dragon, calmly floating on a serene, icy glacier. They are all mighty creatures of ...
their own elements, showcasing their strength in their natural environments.
@user430580 You're welcome! ;)
They're called Langford sequences, or Langford pairings.
21:16
@PM2Ring ah, very nice!
@Amit Haha just look what I did about an hour ago:
in Sandbox, 1 hour ago, by user430580
/imagine dragons
@user430580 :D
5
A: Langford numbers pairing using using Python

PM 2RingYes, there are a few faster ways to make Langford sequences. Firstly, here's a fairly simple way. Rather than testing all of the permutations containing the pairs of numbers from 1 to n, we generate the permutations of numbers from 1 to n and then try to build Langford sequences from those permu...

I swear I didn't peek :P
Wow you got 55k rep on SO....
Oops! I made a sign error. That should be n = 0 or n = -1 mod 4.
21:19
so 3 mod 4
multiples of 3 and 4
Wow you have the golden badge on tag [python], you are my hero! (python is my favourite language)
@Amit I'd have a lot more by now, but I got fed up with all the low quality questions on SO. It was always a busy site, but the company made various changes to make the site more friendly, which ended up encouraging quantity over quality. :(
but is it provable that they don't exist for other values of (n mod 4)?
@Amit Yes. And that proof is pretty easy.
@PM2Ring Did it happen to some extent on Phys.SE too you think? (I'm not here as long as you)
21:23
@OakBot Please tell me the Langford sequence for n equal to 7
@user430580 Bad human! You are over quota. Try again in 23 hours.
ok ;_;
@PM2Ring ok, weird that Wikipedia doesn't have a lot about it
/help
> My commands are also listed here: github.com/JavaChat/OakBot/wiki/Commands
@Amit The proof that it's always possible for n=0 or 3 mod 4 is a bit harder. I've read over it, but I don't fully understand it. And at one point it just says: this construction will always produce a valid sequence, without exactly showing how they found that construction. But they do show that the construction does work.
21:26
@PM2Ring Well what's that quote about writing a math paper is the art of covering your tracks... something like that? :P
Thanks, that looks really good
Here's a page with the proofs I mentioned: susam.net/langford-pairing.html
@user430580 There's a Python room on SO. It used to be really busy, but it's gotten a lot quieter over the last couple of years. chat.stackoverflow.com/rooms/6/python
Here's a sequence for 15. My recursive code can easily find solutions for large n, but of course there will be a lot of solutions. [15, 13, 14, 8, 5, 12, 7, 11, 4, 10, 5, 9, 8, 4, 7, 13, 15, 14, 12, 11, 10, 9, 6, 3, 1, 2, 1, 3, 2, 6]
21:45
@PM2Ring Hey, if there are many solutions, I wonder if you can find a Langford magic square if you see what I mean :P
oh I guess you can't 'cause you have to start with the largest number right
so it can't work
I mean every sequence always necessarily starts with the largest number, for a square you have to be able to start with the others
maybe if you allow for cyclic spacings.... :D
lol, cyclic AGAIN
but if you allow for cyclic spacing it's trivial, nvm :P
at least i'm living up to the room's name "thinking out loud" indeed, not always clever thoughts 😂
@Amit Kind of like a Latin Square, but with pairs of numbers in each row & column .
In combinatorics and in experimental design, a Latin square is an n × n array filled with n different symbols, each occurring exactly once in each row and exactly once in each column. An example of a 3×3 Latin square is The name "Latin square" was inspired by mathematical papers by Leonhard Euler (1707–1783), who used Latin characters as symbols, but any set of symbols can be used: in the above example, the alphabetic sequence A, B, C can be replaced by the integer sequence 1, 2, 3. Euler began the general theory of Latin squares. == History == The Korean mathematician Choi Seok-jeong was the...
There are Langford quilts, like this dialectrix.com/langford/quilt/index.html
22:01
pretty indeed
Reminds me of Tom baker's doctor who scarf (if that means anything lol)
@Amit No, the sequence doesn't have to start with the largest number. But I think you're right that a Latin Square type arrangement is impossible.
oh right, I got confused, there are 2n numbers in each row each in range $1\dots n$
so why do you think it's impossible? It looks hopeful suddenly
On a vaguely related note, there's a thing called the Hadamard matrix, which is a binary matrix. In a 8×8 Hadamard matrix, every pair of rows matches in 4 columns, and every pair of columns matches in 4 rows. See puzzling.stackexchange.com/a/126953/36040
@Amit It's just a gut feeling. Also, I don't think it was mentioned on any of the Langford sequence pages I read yesterday. ;)
Note that we can extend a Langford sequence to include zero by appending (0, 0) to the start or end of the normal sequence.
@PM2Ring that gives even more hope! :P
@MoreAnonymous Hello, welcome to the chat, always happy to see new faces! :)
22:17
Tiling a 5×5 grid with eight L-shaped triomino tiles (plus one square tile):
Accidental Swastikas... lol
Don't mention the war
Haha I remember how some time ago I read that the algorithm used to generate gravatars (generic default avatars on SE network like the avatar I have right now) had to be tweaked really hard because it was especially prone to accidentally generate swastikas :D which has to do with the 2-axis symmetry behavior
it doesn't sound so difficult actually, I think as long as there is a symmetry along the vertical axis you won't get a Swastika like figure
or along the horizontal. it actually looks like SE does both
or at least to some extent both, idk :P I need to see more examples
yours is not horizontally symmetric that's clear
22:37
The Nazis stole the swastika from India, where it was a symbol of social harmony... but I guess it could also be interpreted as a symbol of oppression to conform to the expected social order. But in any case, the swastika in India is now tainted by the Nazi association.
True, swastika's current association is historically fairly recent
8
Q: Why have gravatars become reflectionally symmetric instead of rotationally symmetric?

Rand al'ThorAutomatically generated gravatars, for SE users without their own profile picture, used to have 4-fold rotational symmetry. Some time recently, there's been a very subtle change and they now have reflectional symmetry in the y-axis instead. They haven't changed much, but I notice it every time I ...

19
Q: Preventing generation of swastika-like images when generating identicons

F21I am using this PHP script to generate identicons. It uses Don Park's original identicon algorithm. The script works great and I have adapted it to my own application to generate identicons. The problem is that sometimes swastikas are generated. While swastikas have peaceful origins, people do t...

22:58
There were also cases related to similarity to the "SS flag"
Here's one example from the US marines
23:18
@Amit To an extent. A lot of people just got sick of the constant stream of no-effort homework-dump questions. We still get a few of those, but it got pretty bad a few years ago. And they tend to get closed fairly quickly these days.
But also there was drama on the site, involving a clash of some big egos. I don't know the details because it happened before I joined. One prominent member ended up getting quite nasty to various people, including mods and staff. Bad mistake. Mods can ban someone from their site for a year or two, IIRC. But staff have greater powers... That guy is banned over the whole network for a couple of centuries.
It's a shame. We used to have a lot more professors and textbook authors. We even had a Nobel prize winner.
I think I found the user you are talking about
Are there any meta posts about that drama?
@user430580 Sort of. But the mods discourage people discussing past dramas, and asking for details of why someone was banned isn't allowed.
OK sorry, nevermind then
On that particular drama I think there are details on off-SE sites...
The logic is that if someone is banned they should be allowed a fresh start, without them being haunted by their past. But I guess that logic doesn't really apply when someone is banned for life. But still, it'd be bad if people were discussing you on a site where you can't defend yourself.
23:36
good night all
@PM2Ring That makes sense, I'm not yet too familiar with the rules to be honet
*honest
@Amit Good night see you soon!
Let's try the dad joke with bot
I'm hungry!
@user430580 Hi hungry, I'm Oak!

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