Here's my current idea:
Imagine a building with various cubicles in columnn and rows. Let the amount of total desks (the amount of rows multiplied by the number of columns) be one number and the perimeter of the cubicles be another. Is there another number of cubicles whose numbers match up with the originals?
Mathematically:
A = {a_0, a_1, a_2, ..., a_n | a_i > 1 }
B = {b_0, b_1, b_3, ..., b_n | b_i > 1 }
?
Σ a_i = Σ b_i
Π a_i = Π b_i
Pigeon Holing:
n is the amount of objects
m is the amount of holes
If n>m, then at least one object will occupy the same hole.
Reapplied, we have:
n be the total amount of numbers
m be the total amount of outcomes
n = inf.
m tends toward infinity
Is n>m?
If so, then there must be two sets whose multiplicational and additional values equal each other.
If the two numbers are unique to each set, then a large array of numbers can be reduced to only two terms.
(This also has some relation to P vs. NP, but I'd rather not get into that)
My second idea:
Scatter Plot Accurate Lining
Both of these ideas I developed when I was in the car on a long trip. I didn't have any mathematical tools handy, so, I don't know if this was already done or not.
--Shorthand Method
The shorthand method involves finding the slope between every two consecutive points.
[m 1 b]
[-Σ m_i/n 1 Σ b_i/n]
Where:
Code
n
Σ m_i/n = (y_2 - y_1 / x_2 - x_1 + y_3 - y_2 / x_3 - x_2 + ... + y_n - y_n-1 / x_n - x_n-1 )/n
i=1[/code[
[code]n
Σ b_i/n = (y_1 - m_1 * x_1 + y_2 - m_2 * x_2 + ... + y_n-1 - m_n * x_n-1) / n
i=1
Σ m_i/n = (y_2 - y_1 / x_2 - x_1 + y_3 - y_2 / x_3 - x_2 + ... + y_n - y_n-1 / x_n - x_n-1 )/n
i=1[/code[
[code]n
Σ b_i/n = (y_1 - m_1 * x_1 + y_2 - m_2 * x_2 + ... + y_n-1 - m_n * x_n-1) / n
i=1
So:
Code
n n
(Σ m_i/n)x + (Σ b_i/n) = y
i=1 i=1
(Σ m_i/n)x + (Σ b_i/n) = y
i=1 i=1
--Full Method
The Full Method involves finding the slope between EVERY two point combinations.
Code
m = (y_2 - y_1 / x_2 - x_1 + y_3 - y_2 / x_3 - x_2 + ... + y_n - y_1 / x_n - x_1 + ... + y_n - y_n-1 / x_n - x_n-1 )/ (n!/n)
Code
n
Σ b_i/n = (y_1 - m_1 * x_1 + y_2 - m_2 * x_2 + ... + y_n-1 - m_n * x_n-1) / n
i=1
Σ b_i/n = (y_1 - m_1 * x_1 + y_2 - m_2 * x_2 + ... + y_n-1 - m_n * x_n-1) / n
i=1
y = mx + b
Now, I'm not sure if my sigma notation is right since I've never had formal training in sigma and pi notation. Also, I'm not sure if this has been done before, nor if it can be reduced.
Post has been edited 2 time(s), last time on Apr 29 2008, 5:00 am by A_of-s_t.