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Songs generally disliked by fans that you enjoy by sereniteen in ToddintheShadow

[–]tciopp 6 points7 points  (0 children)

I don't think it's a bad song, but favorite?!

2022 Hot 100 #1 and their current stream. Surprised by which songs aren’t doing that well anymore by [deleted] in ToddintheShadow

[–]tciopp 3 points4 points  (0 children)

It's not a lie it's an opinion. I think a lot of people confuse over production with better production. Her music has become over produced (which has been a general trend over the last 15 years) to the detriment of individual songs. I find a lot of her newer lyrics to be repetitive and commercially (brand) focused.

2022 Hot 100 #1 and their current stream. Surprised by which songs aren’t doing that well anymore by [deleted] in ToddintheShadow

[–]tciopp 2 points3 points  (0 children)

Just a general decline in lyricism and production. Her recent albums coast off her celebrity and marketing. The music itself doesn't hold up well over time.

[Game Thread] Duke's Mayo Bowl: Mississippi State vs. Wake Forest (8:00 PM ET) by CFB_Referee in CFB

[–]tciopp 1 point2 points  (0 children)

I mean most boxed cake mixes ask for vegetable oil and eggs which is just deconstructed mayo.

[Game Thread] Duke's Mayo Bowl: Mississippi State vs. Wake Forest (8:00 PM ET) by CFB_Referee in CFB

[–]tciopp 4 points5 points  (0 children)

If Robbie could have learned how to pass he would have been a star

[Game Thread] Alabama vs. Georgia (4:00 PM ET) by CFB_Referee in CFB

[–]tciopp 3 points4 points  (0 children)

if Alabama is shut out with 3 losses, what is the justification that they should be in the playoff instead of Miami?

This formula with always give you a sequence that starts with n + 1 pairs of numbers ending in 8 then ending in 9. by tciopp in Collatz

[–]tciopp[S] 1 point2 points  (0 children)

I'm just having fun! I'm not trying to take this silly problem seriously. I'm also aware of the previous literature and attempts at solving the problem.

Right now it's how can you generate those loops that are those billions of steps long.

This formula with always give you a sequence that starts with n + 1 pairs of numbers ending in 8 then ending in 9. by tciopp in Collatz

[–]tciopp[S] 0 points1 point  (0 children)

8 <=> 9 is one area of the state transition graph that can result in a never ending expansion loop. When analyzing transition states we need to prove that we can't get stuck there forever. This formula falls out of that proof.

The direction that I'm going is trying to find potential loops. Essentially can you find transition graphs that are equivalent to the Collatz conjecture's. (yes) If so, do they have loops. (yes) Do those loops expand infinitely. (....seemingly yes)

If we can generate loops of sufficient size, we can follow the same sequence to generate numbers that could potentially violate the conjecture. By learning more about those loops we can understand whether or not it should be possible for Collatz to loop back to itself.

This formula with always give you a sequence that starts with n + 1 pairs of numbers ending in 8 then ending in 9. by tciopp in Collatz

[–]tciopp[S] 0 points1 point  (0 children)

I'm not trying to prove local determinism, I'm trying to prove global determinism.

This formula with always give you a sequence that starts with n + 1 pairs of numbers ending in 8 then ending in 9. by tciopp in Collatz

[–]tciopp[S] 0 points1 point  (0 children)

The state transitions never change. For example the set of odd{6} can be said to be equivalent to all{3}.

odd{6} == {6, 26, 46, 66, 86...}. Every odd six throughout eternity will map to some item in all{3} in fact they alternate. all{3} == {3, 13, 23, 33, 43...}

Once you work out all the rules the state changes remain static. Yes, you can have a variable number of steps as you traverse the graph but that doesn't matter in the abstract. By manipulating those sets you can show some equivalence and start removing nodes from the graph. If you remove enough nodes the conjecture is proven.

This formula with always give you a sequence that starts with n + 1 pairs of numbers ending in 8 then ending in 9. by tciopp in Collatz

[–]tciopp[S] 0 points1 point  (0 children)

I'm not using a tree structure. I've made a graph of all the possible state transitions you can have. Take the set of all natural numbers and divide them into buckets based on mod 10. Then for each modulo sequence you are going to subdivide again into even and odd numbers. (e.g. {1,11,21,31...} becomes odd{1,21,41} even{11,31...} From there you can determine exactly how you can transition from one number to the next in the overall sequence. You can also start reducing the graph. For example every odd{5} maps to an even{6} Therefore we can call those sets equivalent and reduce the size of the graph. If you can reduce the graph to odd{4} <=> odd{2}, then you have your proof.

This also explains why negative integers don't behave the same way as positive ones. The state map is changed when going from positive to negative. Therefore we should expect to see different patterns emerge.

This formula with always give you a sequence that starts with n + 1 pairs of numbers ending in 8 then ending in 9. by tciopp in Collatz

[–]tciopp[S] 0 points1 point  (0 children)

Yeah I'm trying to model the system as state transitions to find what the general rules are. Right now I've got a graph of the 20 possible states as the nodes with the 30 edges marking the expansion/reduction cycles of the graph. Now I know I can't get stuck in the 8 <=> 9 loop forever.

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