Is there a mathematical function for this? 
Mr_Stabby 
Im way off the deep end with this analytical algorithm im building, my gut told me the problem is solvable so i just followed the seemingly random pile of numbers until i found some order. My crazies aside is there a math function that would turn 5 into 5+4+3+2+1=15, 4 into 4+3+2+1=10? Sorta like ! except with + instead of *
read 668 times 8/31/2012 2:48:58 AM (last edit: 8/31/2012 2:48:58 AM)

jareu 
Yes there is, I can't remember if it is dot product or a variation of dot product which does what you want but I'll look up summation in a min and see what I can find for ya
He who says it cannot be done is interrupting the man doing it.
read 659 times 8/31/2012 3:24:59 AM (last edit: 8/31/2012 3:53:29 AM)

gogodr 
n(n+1)/2 5*6/2 = 15 4*5/2 = 10
Hello therebeautiful ;3
read 654 times 8/31/2012 3:38:18 AM (last edit: 8/31/2012 3:40:06 AM)

jareu 
ok forget dot product, I was confusing vector math for a minute.
edit: yeah use gogodr's function, I forgot about that one :)
He who says it cannot be done is interrupting the man doing it.
read 645 times 8/31/2012 3:46:47 AM (last edit: 8/31/2012 3:53:17 AM)

Mr_Stabby 
gogodr  oh ya! *flashback to high school* that'll cut down the execution time, thanks.
jaeru  int lvl(int n) {int v = 0; for (int i = 0; i < n; i++) {v += i;} return v;} that was my current one, even though it looks nice on paper not very efficient :
read 643 times 8/31/2012 3:48:00 AM (last edit: 8/31/2012 3:49:10 AM)

herfst1 
Hey, didn't some French kid work that out when he was asked to do it from 100 as a way for the teacher to get 30 mins peace? Think he was sent to detention for being a smart ass before they worked out that he was a genius.
Or I could be mistaken.
read 605 times 8/31/2012 10:52:25 AM (last edit: 8/31/2012 10:52:25 AM)

Garp 
Stabby, care to share a little about what it is you're doing?
read 595 times 8/31/2012 11:35:31 AM (last edit: 8/31/2012 11:35:31 AM)

Mr_Stabby 
trying to extrapolate an array of values that follow unknown amount of unknown but mathematically definable trajectories
read 576 times 8/31/2012 8:37:11 PM (last edit: 8/31/2012 8:37:11 PM)

advancesoftware 
I reckon stabby's working on a new way of preventing side fumbling.
read 571 times 8/31/2012 8:45:40 PM (last edit: 8/31/2012 8:48:54 PM)

Garp 
So if you walk on a landmine, you want to know where all your bits are going to fall? You need to factor in the wind.
read 558 times 8/31/2012 9:10:50 PM (last edit: 8/31/2012 9:10:50 PM)

Mr_Stabby 
Its not for anything specifically, just curiosity mostly. The theory is that if i could observe the coordinates of a tea leaf in relation to the center of the galaxy cluster over X samples then a) when the algorithm starts producing 0's at the top of its hierarchial tree then i have all the data i need and the level where it produces the 0's will be X which is the number of mathematically definable trajectories that are affecting it b) the first member of each array in the tree will be the velocity/energy or whatever im measuring representation of the effect on the tea leaf. In short it creates a series of numbers that when extrapolated define the entire galaxy cluster in relation to the tea leaf.
read 529 times 8/31/2012 11:22:31 PM (last edit: 8/31/2012 11:22:31 PM)

advancesoftware 
read 522 times 8/31/2012 11:32:23 PM (last edit: 8/31/2012 11:32:23 PM)

Garp 
Hmm. Aren't you afraid that X might be larger than the total number of particles in the Universe?
read 517 times 8/31/2012 11:40:00 PM (last edit: 8/31/2012 11:40:00 PM)

Mr_Stabby 
quite possibly is, a more practical example would be using it as a smoothing function  you draw a line, it places x number of points on the line at equal distance and recreates the line with x precision. The cool thing about it is that after the sampling process which could be multicast each point can work independently while holding the pattern of the whole within itself  perfect for gpu computing stuff!
read 512 times 8/31/2012 11:48:14 PM (last edit: 8/31/2012 11:48:14 PM)

Mr_Stabby 
Anybody like retarded math puzzles?
5 6 4 > 5 4 2 2 4 6 > 2 1 3 5 13 10 > 5 8 5 8 10 6 > 8 7 3 3 7 8 > 3 3 4 3 8 8 > 3 4 4 3 9 8 > 3 5 4
i know how to get from the right side to the left side, the question is how do i get from the left to the right
EDIT: fixed typo in values.. which was also on my set.. which suddenly makes it simple
a' = a c' = c/2 b' = b + c'  c
read 492 times 9/1/2012 2:27:53 AM (last edit: 9/1/2012 2:58:03 AM)
