Latrinsorm
03-22-2014, 10:08 PM
Suppose we want to measure things. We devise a system where each dimension of measurement has 4 bins, but even if it turns out that every bin has exactly the same population, we will still have fringes of varying degrees. In one dimension, we have...
http://img.photobucket.com/albums/v456/johnnyoldschool/DimensionsMan2a_zps13393b46.png
...2 bins with 2 neighbors (green), 2 bins with 1 neighbor (black). Black will always be the fringiest color, but in this dimension it is also the only fringe color. Thus we have:
absolute fringe %: 2 / 4
total fringe %: 2 / 4
.
In two dimensions, we have...
http://img.photobucket.com/albums/v456/johnnyoldschool/DimensionsMan2b_zps29005610.png
...4 bins with 4 neighbors (brown), 8 bins with 3 neighbors (green), 4 bins with 2 neighbors (black). Now we have two fringe types: black corners and green edges. Now our figures are:
af%: 4 / 16
tf%: 12 / 16
.
The last dimension we can draw with any legibility is three dimensions, which looks like...
http://img.photobucket.com/albums/v456/johnnyoldschool/DimensionsMan2c_zps03c47ff5.png
...8 bins with 6 neighbors (red), 24 with 5 (brown), 24 with 4 (green), 8 with 3 (black). The dots kind of overlap, so I drew them in both shades where that happens. Our new dimension creates a new fringe from what once was normal: brown faces.
af%: 8 / 64
tf%: 56 / 64
.
Three dimensions and four bins per dimension are all well and good, but wouldn't it be nice to generalize to any number of dimensions D and bins L? Let's look at our figures geometrically:
dimens. 0 1 2 3
corners 1 2 4 8
edges 0 1 4 12
faces 0 0 1 6
Corners is easy to see, it just doubles every time. If we call Corners a function C of dimensions D then...
C[D] = 2^D
Edges is a little trickier because it starts at 0 and doesn't quite double; instead, each step after the first doubles and adds increasing multiples of 2. Put symbolically...
E[D] = 2 * E[D-1] + 2^(D-1)
...but we can get a little recursive on its ass like so...
E[D-1] = 2 * E[D-2] + 2^(D-2)
E[D] = 2 * (2 * E[D-2] + 2^(D-2)) + 2^(D-1)
E[D] = 4 * E[D-2] + 2^(D-1) + 2^(D-1)
E[D] = 4 * E[D-2] + 2^D
...and if we turn the crank again, we get...
E[D] = 8 * E[D-3] + (3/2) * 2^D
...so we can see a pattern here that goes...
E[D] = 2^a * E[D-a] + (a/2) * 2^D
...when a = 0, we get E[D] = 1 * E[D], when a = 1 we get our original equation, and so on. Let's set a = D-1...
E[D] = 2 ^ (D-1) * E[1] + (D-1) * 2 ^ (D-1)
...but we know E[1] = 1, so now we have a formula for Edges...
E[D] = 2 ^ (D-1) + D * 2 ^ (D-1) - 2 ^ (D-1)
E[D] = D * 2 ^ (D-1)
.
Faces we know have to be zero for 0 and 1 dimensions, so we can take a tip from our last equations and guess that it's going to be a function of D and (D-1), and it turns out that gives us...
F[D] = D * (D-1) * 2 ^ (D-3)
Altogether now, and rearranged more seductively...
C[D] = 2 ^ D
E[D] = 2 ^ D * D/2
F[D] = 2 ^ D * D/2 * (D-1)/4
And this pattern continues for any arbitrary dimension.
.
So now we know how many of each geometric form we have, but how many of those are fringe elements?
1. Corners are always fringe after the 0th dimension: C[D] = fringe
2. Edges are always fringe after the 1st dimension, but we don't want to double count the corners so we'll do E[D] * (L - 2) = fringe: just the green dots.
3. Same story for faces, we only want the brown dots so we subtract off the edges and get F[D] * (L - 2) ^ 2 = fringe.
4. etc.
So let's calculate the total fringe elements for every dimension!
D=1
We have 2^1 corners and 0 everything else on the fringe.
No dependence on L.
2 for all L.
D=2
2^2 * (1 corners, 1 edge)
= 4 * (1 + (L-2))
= 4 * (L - 1)
D=3
2^3 * (1 + (3/2) * (L-2) + (3/4) * (L-2)^2)
= 8 * (1 + (3/2)L - 3 + (3/4)L^2 - 3L + 3)
= 8 * (1 - (3/2)L + (3/4)L^2)
= 2 * (4 - 6L + 3L^2)
That doesn't factor neatly, but let's try subtracting off 3 * (L - 1)^2! We get...
(4 - 6L + 3L^2)
-(3 - 6L + 3L^2)
=
1
Sweet! So our total fringe population is...
= 2 * (3 * (L - 1)^2 + 1)
.
The algebra gets more intense for higher dimensions, here's the list out to 6 dimensions, and while we're at it let's define a quantity W ≡ L - 1...
1D - 2
2D - 2 * 2W
3D - 2 * (3W^2 + 1)
4D - 2 * (4W^3 + 4W)
5D - 2 * (5W^4 + 10W^2 + 1)
6D - 2 * (6W^5 + 20W^3 + 6W)
7D - 2 * (7W^6 + 35W^4 + 21W^2 + 1)
The first and last terms in every row are really obvious, but how to get at those middle coefficients? I propose a pyramid schema...
1
2 0
3 0 1
4 0 4 0
5 0 10 0 1
6 0 20 0 6 0
7 0 35 0 20 0 1
Aha! Look at the southwesterly diagonals, which we shall call X!
X1
Is just D.
X2
Is a little more mysterious, but we're ready for its tricks. The pyramid swims in a sea of zeros, so X2 has to be zero for D=1 and D=2, so it must look like...
X2 = (D-1) * (D-2) * a mystery
So if we divide through by those D terms, we go from...
1, 4, 10, 20, 35
...to...
1/2, 2/3, 5/6, 1, 7/6
That's just 1/6th of D! And we are also wise to that little game, so...
X2 = D * (D-1)/2 * (D-2)/3
...is the likely form.
X3
Only three terms to work with? Nope. Three nonzero terms and four zeros to prime the pump, and we end up with...
X3 = D * (D-1)/2 * (D-2)/3 * (D-3)/4 * (D-4)/5
.
So now we can calculate the total fringe elements for any D and L in three easy steps!
1. Calculate X up to X[D/2].
2. Multiply each X by powers of W beginning with D-1 and decreasing by 2.
3. Multiply by 2.
For instance, the total fringe elements for a 9 dimensional matrix with 7 bins per dimension would be...
X = 9,
(9*8*7) / (2*3) = 84,
(9*8*7*6*5) / (2*3*4*5) = 126
(9*8*7*6*5*4*3) / (2*3*4*5*6*7) = 36
9! / 9! = 1
9 * 6 ^ 8 + 84 * 6 ^ 6 + 126 * 6 ^ 4 + 36 * 6 ^ 2 + 1 = 19,200,241
19,200,241 * 2 = 38,400,482
Which out of 7^9 = 40,353,607 total members gives us 95.2%! But alas, the absolute fringe is only 2^9 / 7^9 = (2/7) ^ 9 = 0.001%.
.
What does it all mean? It means that Tisket is right (which I have never disputed) and pretty much everyone is weird in some way. But it also means that I am right (which I have never disputed) and some people are weirder than others, and are in fact so weird that they are weird even to weird people. This does not cancel out and make them normal, in the same way that if you drive a car over your foot twice it doesn't doubly break and therefore become unbroken.
.
Please note! It might appear from the 4s case that we can just say that absolute normal is equal to absolute fringe, but that is a coincidence because 4 - 2 = 2. The relationship does not hold for any other L, and anyone who says it does is a dead man.
http://img.photobucket.com/albums/v456/johnnyoldschool/DimensionsMan2a_zps13393b46.png
...2 bins with 2 neighbors (green), 2 bins with 1 neighbor (black). Black will always be the fringiest color, but in this dimension it is also the only fringe color. Thus we have:
absolute fringe %: 2 / 4
total fringe %: 2 / 4
.
In two dimensions, we have...
http://img.photobucket.com/albums/v456/johnnyoldschool/DimensionsMan2b_zps29005610.png
...4 bins with 4 neighbors (brown), 8 bins with 3 neighbors (green), 4 bins with 2 neighbors (black). Now we have two fringe types: black corners and green edges. Now our figures are:
af%: 4 / 16
tf%: 12 / 16
.
The last dimension we can draw with any legibility is three dimensions, which looks like...
http://img.photobucket.com/albums/v456/johnnyoldschool/DimensionsMan2c_zps03c47ff5.png
...8 bins with 6 neighbors (red), 24 with 5 (brown), 24 with 4 (green), 8 with 3 (black). The dots kind of overlap, so I drew them in both shades where that happens. Our new dimension creates a new fringe from what once was normal: brown faces.
af%: 8 / 64
tf%: 56 / 64
.
Three dimensions and four bins per dimension are all well and good, but wouldn't it be nice to generalize to any number of dimensions D and bins L? Let's look at our figures geometrically:
dimens. 0 1 2 3
corners 1 2 4 8
edges 0 1 4 12
faces 0 0 1 6
Corners is easy to see, it just doubles every time. If we call Corners a function C of dimensions D then...
C[D] = 2^D
Edges is a little trickier because it starts at 0 and doesn't quite double; instead, each step after the first doubles and adds increasing multiples of 2. Put symbolically...
E[D] = 2 * E[D-1] + 2^(D-1)
...but we can get a little recursive on its ass like so...
E[D-1] = 2 * E[D-2] + 2^(D-2)
E[D] = 2 * (2 * E[D-2] + 2^(D-2)) + 2^(D-1)
E[D] = 4 * E[D-2] + 2^(D-1) + 2^(D-1)
E[D] = 4 * E[D-2] + 2^D
...and if we turn the crank again, we get...
E[D] = 8 * E[D-3] + (3/2) * 2^D
...so we can see a pattern here that goes...
E[D] = 2^a * E[D-a] + (a/2) * 2^D
...when a = 0, we get E[D] = 1 * E[D], when a = 1 we get our original equation, and so on. Let's set a = D-1...
E[D] = 2 ^ (D-1) * E[1] + (D-1) * 2 ^ (D-1)
...but we know E[1] = 1, so now we have a formula for Edges...
E[D] = 2 ^ (D-1) + D * 2 ^ (D-1) - 2 ^ (D-1)
E[D] = D * 2 ^ (D-1)
.
Faces we know have to be zero for 0 and 1 dimensions, so we can take a tip from our last equations and guess that it's going to be a function of D and (D-1), and it turns out that gives us...
F[D] = D * (D-1) * 2 ^ (D-3)
Altogether now, and rearranged more seductively...
C[D] = 2 ^ D
E[D] = 2 ^ D * D/2
F[D] = 2 ^ D * D/2 * (D-1)/4
And this pattern continues for any arbitrary dimension.
.
So now we know how many of each geometric form we have, but how many of those are fringe elements?
1. Corners are always fringe after the 0th dimension: C[D] = fringe
2. Edges are always fringe after the 1st dimension, but we don't want to double count the corners so we'll do E[D] * (L - 2) = fringe: just the green dots.
3. Same story for faces, we only want the brown dots so we subtract off the edges and get F[D] * (L - 2) ^ 2 = fringe.
4. etc.
So let's calculate the total fringe elements for every dimension!
D=1
We have 2^1 corners and 0 everything else on the fringe.
No dependence on L.
2 for all L.
D=2
2^2 * (1 corners, 1 edge)
= 4 * (1 + (L-2))
= 4 * (L - 1)
D=3
2^3 * (1 + (3/2) * (L-2) + (3/4) * (L-2)^2)
= 8 * (1 + (3/2)L - 3 + (3/4)L^2 - 3L + 3)
= 8 * (1 - (3/2)L + (3/4)L^2)
= 2 * (4 - 6L + 3L^2)
That doesn't factor neatly, but let's try subtracting off 3 * (L - 1)^2! We get...
(4 - 6L + 3L^2)
-(3 - 6L + 3L^2)
=
1
Sweet! So our total fringe population is...
= 2 * (3 * (L - 1)^2 + 1)
.
The algebra gets more intense for higher dimensions, here's the list out to 6 dimensions, and while we're at it let's define a quantity W ≡ L - 1...
1D - 2
2D - 2 * 2W
3D - 2 * (3W^2 + 1)
4D - 2 * (4W^3 + 4W)
5D - 2 * (5W^4 + 10W^2 + 1)
6D - 2 * (6W^5 + 20W^3 + 6W)
7D - 2 * (7W^6 + 35W^4 + 21W^2 + 1)
The first and last terms in every row are really obvious, but how to get at those middle coefficients? I propose a pyramid schema...
1
2 0
3 0 1
4 0 4 0
5 0 10 0 1
6 0 20 0 6 0
7 0 35 0 20 0 1
Aha! Look at the southwesterly diagonals, which we shall call X!
X1
Is just D.
X2
Is a little more mysterious, but we're ready for its tricks. The pyramid swims in a sea of zeros, so X2 has to be zero for D=1 and D=2, so it must look like...
X2 = (D-1) * (D-2) * a mystery
So if we divide through by those D terms, we go from...
1, 4, 10, 20, 35
...to...
1/2, 2/3, 5/6, 1, 7/6
That's just 1/6th of D! And we are also wise to that little game, so...
X2 = D * (D-1)/2 * (D-2)/3
...is the likely form.
X3
Only three terms to work with? Nope. Three nonzero terms and four zeros to prime the pump, and we end up with...
X3 = D * (D-1)/2 * (D-2)/3 * (D-3)/4 * (D-4)/5
.
So now we can calculate the total fringe elements for any D and L in three easy steps!
1. Calculate X up to X[D/2].
2. Multiply each X by powers of W beginning with D-1 and decreasing by 2.
3. Multiply by 2.
For instance, the total fringe elements for a 9 dimensional matrix with 7 bins per dimension would be...
X = 9,
(9*8*7) / (2*3) = 84,
(9*8*7*6*5) / (2*3*4*5) = 126
(9*8*7*6*5*4*3) / (2*3*4*5*6*7) = 36
9! / 9! = 1
9 * 6 ^ 8 + 84 * 6 ^ 6 + 126 * 6 ^ 4 + 36 * 6 ^ 2 + 1 = 19,200,241
19,200,241 * 2 = 38,400,482
Which out of 7^9 = 40,353,607 total members gives us 95.2%! But alas, the absolute fringe is only 2^9 / 7^9 = (2/7) ^ 9 = 0.001%.
.
What does it all mean? It means that Tisket is right (which I have never disputed) and pretty much everyone is weird in some way. But it also means that I am right (which I have never disputed) and some people are weirder than others, and are in fact so weird that they are weird even to weird people. This does not cancel out and make them normal, in the same way that if you drive a car over your foot twice it doesn't doubly break and therefore become unbroken.
.
Please note! It might appear from the 4s case that we can just say that absolute normal is equal to absolute fringe, but that is a coincidence because 4 - 2 = 2. The relationship does not hold for any other L, and anyone who says it does is a dead man.