Binge Artist
Bluelighter
^just thinking out loud, BTW.
-
Life Advice & Visual Arts
Posting Rules Bluelight Rules 🌌 Share Your Dreams! 🌌 📸 15 Years of Photo Contests 📸 -
LAVA Moderator: Shinji Ikari
🌟🌟 Social 🌟🌟 😃 LAVA Chit-Chat Thread! 🤪
- Thread starter Rogue Robot
- Start date
Binge Artist
Bluelighter
Each painter's isomorphic image in Z mod 2^17 - 1 is that painter's effect on zero. Profound idea, really. Just lacking a proof...
Perhaps a good start would be to show that the painter permutations are, themselves, isomorphisms; ie, show A(x + y) = A(x) + A(y).
Nope, scratch that; can't be, because A(0) is never 0. Back to the drawing board.
So, the basic question is this: I have 17 functions, commutative under composition. What do I need to show that those 17 functions are elements of a LARGER group isomorphic to Z mod 2^17 - 1?
And why is x_0 = 0 = 000...000 so important?
I think I'm starting to see it. Let f be the "tenative" isomorphism from the "expanded"*** painter permutations to Z mod 2^17 - 1. I'm saying that for all A in {PP}, f(A) = A(0). I think it suffices to show the following two statements:
1. Whenever A(0) = x and B(0) = y, for any A, B in {PP} and x, y in Z mod 2^17 - 1, then AB(0) = BA(0) = x + y. And its easy to show the commutativity part.
2. If D & E are in the "expanded" {PP}, and D(0) = E(0), then D(x) = E(x) for all x in Z mod 2^17 - 1.
***And by expanded painter permutations, I mean all 17 painter permutations, PLUS all their compositions. In other words, the group generated by the painter permutations.
Perhaps a good start would be to show that the painter permutations are, themselves, isomorphisms; ie, show A(x + y) = A(x) + A(y).
Nope, scratch that; can't be, because A(0) is never 0. Back to the drawing board.
So, the basic question is this: I have 17 functions, commutative under composition. What do I need to show that those 17 functions are elements of a LARGER group isomorphic to Z mod 2^17 - 1?
And why is x_0 = 0 = 000...000 so important?
I think I'm starting to see it. Let f be the "tenative" isomorphism from the "expanded"*** painter permutations to Z mod 2^17 - 1. I'm saying that for all A in {PP}, f(A) = A(0). I think it suffices to show the following two statements:
1. Whenever A(0) = x and B(0) = y, for any A, B in {PP} and x, y in Z mod 2^17 - 1, then AB(0) = BA(0) = x + y. And its easy to show the commutativity part.
2. If D & E are in the "expanded" {PP}, and D(0) = E(0), then D(x) = E(x) for all x in Z mod 2^17 - 1.
***And by expanded painter permutations, I mean all 17 painter permutations, PLUS all their compositions. In other words, the group generated by the painter permutations.
Last edited:
RedLeader
Bluelight Crew
You think I am going to give you problems with ordinary isomorphic functions!?!?!
NSFW:
Your solution isn't the same as mine, but this problem does have multiple routes. I'd say put up a finalized version in S&T soon, and when you do, I'll show you my solution and we can compare/contrast them.
Binge Artist
Bluelighter
Lol, but one just doesn't expect a subgroup of permutations to be isomorphic to Z mod anything under addition...
...even an abelian subgroup of permutations.
Just counter-intuitive, I guess.
RedLeader
Bluelight Crew
Dude, you asked for level 5. I gave you level 5.
Binge Artist
Bluelighter
^anyway, I think I found a pretty short & sweet solution (posted in S&T).
It's based on the idea that if one guy painted twice in a row, it would have the same effect as the guy next to him painting once.
It's based on the idea that if one guy painted twice in a row, it would have the same effect as the guy next to him painting once.
amanda_eats_pandas
Bluelighter
First day of school and no class until noon.
Its beginning to look like a good day :]
Its beginning to look like a good day :]
Rogue Robot
Bluelight Crew
I'm really over my trig class. I wear my prof drinks more coffee than I do, and that bothers me. Hyperactive trig != fun.
RedLeader
Bluelight Crew
math == fun
If (You.LikesMath == N) {gtfo!!!;}
Just playing. I should have a coffee. I really, really need to invest in a coffee pot. Tea does not cut it these days.
If (You.LikesMath == N) {gtfo!!!;}
Just playing. I should have a coffee. I really, really need to invest in a coffee pot. Tea does not cut it these days.
Rogue Robot
Bluelight Crew
I'm not math impaired. Seriously. The math that I need to do is very basic, but my professor trying to teach trig just makes my brain explode. ![:\](https://bluelight.org/xf/BL_Images/Orig_Emoji/wrygrin.gif "wry grin :\")
RedLeader
Bluelight Crew
You should bump the complain about professors thread and tell a detailed story!
Rogue Robot
Bluelight Crew
Nah. 
That's about all my gripe is. So I'm staying home until my German class and getting overly caffeinated.
That's about all my gripe is. So I'm staying home until my German class and getting overly caffeinated.
Binge Artist
Bluelighter
^gefaellt dir der Deutschunterricht?
RedLeader
Bluelight Crew
Nicht Deutsch! Reminds me of that alte Frau I got the cannabis for.
Binge Artist
Bluelighter
And she was German TOO???
Geeeesh. Now I DEFINITELY can't believe you didn't hit that. Are you GAY?!?!?
Geeeesh. Now I DEFINITELY can't believe you didn't hit that. Are you GAY?!?!?
RedLeader
Bluelight Crew
Dude. I told you I was sleeping with an Irish vixen at the time...
That German woman was so bad at math that I think she avoided even teaching us eins, zwei, drei,...
That German woman was so bad at math that I think she avoided even teaching us eins, zwei, drei,...
Binge Artist
Bluelighter
Yeah, thats all well and good. But when you boink a German chick, you're GUARANTEED anal. In fact, they get pissey if you don't.
Especially true of the older ones.
Especially true of the older ones.
RedLeader
Bluelight Crew
^ I see you being more into Swedes.
Binge Artist
Bluelighter
Even though I'm laughing my ass off, I'm not sure if I get the joke.
Are swedes supposed to be anal freaks?
Are swedes supposed to be anal freaks?
RedLeader
Bluelight Crew
All Europeans are.
But I just see you as liking long legs and blonde hair.
But I just see you as liking long legs and blonde hair.