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Posted by JohnSmith on Apr-16-2002 23:42:

DJ's are good at math right?

i have an interesting problem, i am trying to solve it.

observe the following set of numbers:

PHP:

there is a formula, that will allow you, to take the number on the left (x), and derive from it, the number on the right (y). the numbers above, are simply counts of how many numbers are in each set. this number is essential in the calculation, as the relationship changes each time you add a number..

i will not rest until i find this formula.

Posted by skywarp on Apr-17-2002 00:58:

And WHAT exactly does this have to do with playing records ?!

Posted by DJ LIQUID on Apr-17-2002 01:11:

hey...im really good at math....it must say something bout my djing skillz

Posted by JohnSmith on Apr-17-2002 01:29:

quote:
Originally posted by skywarp
And WHAT exactly does this have to do with playing records ?!

has nothing to do with playing records.

but, like i said, DJs are good at math right?

Posted by Special_K on Apr-17-2002 01:42:

im not good at math.....

Posted by Dmatrox on Apr-17-2002 01:48:

im pretty good at math, but im better at biology

Yeah, what does math have to do with playing records?

Posted by patticus on Apr-17-2002 01:50:

wtf? are those derivatives of some sort??
being asian, i used to be good at math, but then my overall capacity for the arts (in other words, stupidity) came to the forefront..
i got 78 x 150 wrong today. i said like 340 or something stupid

Posted by sym on Apr-17-2002 01:55:

Well what looks like to me is the following...

Find the distance between the first x and the first y. This number we will call z.

Than take the next x value you have and need the y for. Find the distance betweent his 2nd x and the first x you used to find z.

Than subtract that number from z and add the result to x2 to get your y2.

At least thats how I saw it... You'd have to be more specific with what you want... You can AIM me @ SyM2oo1

Posted by IronDragon on Apr-17-2002 02:21:

Re: DJ's are good at math right?

quote:
Originally posted by JohnSmith
i have an interesting problem, i am trying to solve it.

observe the following set of numbers:

PHP:

there is a formula, that will allow you, to take the number on the left (x), and derive from it, the number on the right (y). the numbers above, are simply counts of how many numbers are in each set. this number is essential in the calculation, as the relationship changes each time you add a number..

i will not rest until i find this formula.

::giggles like a bimbo:: math is hard.
But really, I am only so-so at best in math. I assume you'll find that everyone else in here will range from friggin' genius to friggin' idiot. Maybe you'll get lucky and some'll know what the hell you're trying to do.

Suerte,
Dave

Posted by uberclkgtr on Apr-17-2002 03:01:

ah, this looks kinda fun.

i can get it to work for all except the 1st set. here's how i set it up and solved it:

y - x = a + b

where a = the set number (ie 1, 2, 3, 4 or 5)
and b we have to figure out

if you make table of y and b values for a given a such as the following:

code:
for a = 5
 y   b
 -----
 10  4
 11  3
 12  2
 13  1
 14  0

for a = 4
 y   b
 -----
 8   3
 9   2
 10  1
 11  0

you can see that b can be expressed as

b = 3a - y - 1

so we have the two equations

x - y = a + b
b = 3a - y -1

which we can solve by substitution to get

y = (4a + x - 1) / 2

this only works if a > 1.

i suppose you could use an iterative approach, where you specify y1, and based on the value of a and y1, you get y2. but i've done enough of your homework for tonight.

i'm better at math than spinning records...

Posted by hapamoto on Apr-17-2002 04:34:

haha.. fuck this.. im asian too but that doesn't mean im good at math because of it.. im just good in math because I am.. anyways if i wanted to be doing math problems i'd be taking math classes.. sorry, i could probably help you but seems like other people already figured it out.. yay less work for me

Posted by Xquisite on Apr-17-2002 06:56:

heh i got a messy equation. I dont think you want to see it. It has a ton of mods in it.

Posted by DJ Bazz on Apr-17-2002 08:36:

--> Chill out room post

you got the formule already, you use PHP, so come on with the formule

Posted by JohnSmith on Apr-17-2002 15:35:

thanks guys!

sheesh.. asians are good at math? who knew? i'm white, i love math.. it's the language of the universe. and DJing, has a lot of math to it, the 4 beat structures, the different speeds..

anyway.. thanks for your answers and time. i actually found a simpler solution though.. now, i just need to optimize it.

this is not PHP, i just used that tag so it would come out formatted properly. now, i will use the code one. what i am doing is a javascript mouseover. it has to swap images.

I did it one way, but it was inefficient, and tough to update. i wanted a more elegant solution, and i saw the relations between the numbers.

this is what i started with:

code:

and this is how i [i]optimized it:

code:

some people would say i am crazy, but i think i can make it even tighter. these 6 lines here:

code:
	if ((aryImages.length / 2) == (Math.round(aryImages.length / 2))) {
		document['imgDisplay'].src = aryImages[imgNum + aryImages.length / 2];
	}
	else {
		document['imgDisplay'].src = aryImages[imgNum + (aryImages.length+1) / 2];
	}

can be shrunk down to 1, they are so similar. all i am doing is adding one to the length, if the array has an odd number of elements.. there is a way of doing that inline, in know there is, it's a simple thing.

i just can't FIND it.. grr.. I am INSANE!

Posted by Great Outdoors on Apr-18-2002 09:39:

SHEESH.

Posted by Xquisite on Apr-18-2002 10:11:

oh lol.. doing it by Java/C++ would be the easiest. I thought you wanted a math equation. If you did it by java it can be done with for loops and such.

Posted by SpykeChyld on Apr-18-2002 15:33:

Ummm....This is the guy that has failed 4 math courses in high school/college, lmao. No, I don't have a fucking clue what any of that means. I like music, lmao.

I was in straight college english courses, but never liked math. Good luck though. :shrugs: