1. Dear All, I figured out a very rough technique to calculate division using pTrees. It can divide a pTreeSet A by another pTreeSet B to produce a pTreeSet C. That is C=A/B.
The idea is, C = A * (1/B)
So our task is to find the reciprocal of a pTreeSet so that we can compute D = 1/B. So C = A * D We know how to multiply two pTreeSets. We'll use that to multiply A and D to get C. Assume B is pTreeSet having 3 pTrees. That is all the numbers in B are 3-bit numbers. So the possible values are 0-7. Now we list all these numbers and their reciprocal in the following table: B    D =   = 0    N/A    1    1.0 2    0.5 3            0.2 6       
We know when we convert a fraction into binary we use the negative power of 2's. For example: Binary =  Decimal (2^-1+ 2^-3) = = etc. To calculate the binary of D we can use this technique, considering 4 bits after fraction point. B    D                    Binary =   =                    ==== 0    N/A                x.xxxx 1    1.0                    0.5                                 0.25                  0.2                                         
To get the logical operation required to calculate the D, we consider the following table:          B2B1B0     D4D3D2D1D0000                xxxxx Now the logic function will be as follows: D4 = B2'B1' D3 = B2'B0' D2 = B1'B0' + B2'B1B0 D1 = B2B0 + B2B1 D0 = B2'B1B0 + B2B1'B0

2. Cooper Clustering: (building barrel-shaped gaps around clusters)
Furthest Point or Mean Point q
The method attempts to build barrels around clusters
(masks the interior of barrel shaped gaps which separate the space into two partitions).
This allows for a better fit around convex clusters that are elongated in one direction (not round - which would be the case where there is not elongation in any direction).
Gaps in dot product lengths [projections] on the line.
Exhaustive Search for all barrel gaps:
It takes two parameters for a pseudo- exhaustive search (exhaustive modulo a grid width).
1. A StartPoint, p (an n-vector, so n dimensional)
2. A UnitVector, d (a n-direction, so n-1 dimensional - grid on the surface of sphere in Rn).
Then for every choice of (p,d) (e.g., in a grid of points in R2n-1) two functionals are used to enclose subclusters in barrel shaped gaps.
a. SquareBarrelRadius functional, BR(y) = (y-p)o(y-p) - ((y-p)od)2
b. BarrelLength functional, BL(y) = (y-p)od y
barrel
cap gap
width p
barrel radius
gap width
Given a p, do we need a full grid of d unit vectors (directions)? No! d and -d give the same BL-gaps.
Given a d, do we need a full grid of p starting points? No! All p' s.t. p' = p + cd give the same BR-values (and therefore same gaps)
Hill climb gap width from a good starting point and direction.

3. This suggests a clustering method:
x=s1
cone=1/√2
60 3
61 4
62 3
63 10
64 15
65 9
66 3
67 1
69 2 50
x=s2
cone=1/√2
47 1
59 2
60 4
61 3
62 6
63 10
64 10
65 5
66 4
67 4
69 1
70 1 51
x=s2
cone=.9
59 2
60 3
61 3
62 5
63 9
64 10
65 5
66 4
67 4
69 1
70 1 47
x=s2
cone=.1
39 2
40 1
41 1
44 1
45 1
46 1
47 1
i39
59 2
60 4
61 3
62 6
63 10
64 10
65 5
66 4
67 4
69 1
70 1 59
w maxs-to-mins
cone=.939
i25
i40
i16 i42
i17 i38
i11 i48
22 2
23 1
i34 i50
i24 i28
i27
27 5
28 3
29 2
30 2
31 3
32 4
34 3
35 4
36 2
37 2
38 2
39 3
40 1
41 2
46 1
47 2
48 1
i39
53 1
54 2
55 1
56 1
57 8
58 5
59 4
60 7
61 4
62 5
63 5
64 1
65 3
66 1
67 1
68 1
114
14 i and 100 s/e.
So picks i as 0
w naaa-xaaa
cone=.95
12 1
13 2
14 1
15 2
16 1
17 1
18 4
19 3
20 2
21 3
22 5
i21
24 5
25 1
27 1
28 1
29 2 i7
41/43 e
so picks e
This suggests a clustering method:
1. find cosine cone gaps emanating from a corner point (or any circumscribing point).
2. "cap" the cone gap on the open end with a linear gap (actually for IRIS, the cap seems unnecessary and the cone gaps themselves seem to separate the three classes).
F=(y-M)o(x-M)/|x-M|-min restricted
to a cosine cone on IRIS
Corner points
Gap in dot product projections onto the cornerpoints line.
x=e1
cone=.707
33 1
36 2
37 2
38 3
39 1
40 5
41 4
42 2
43 1
44 1
45 6
46 4
47 5
48 1
49 2
50 5
51 1
52 2
54 2
55 1
57 2
58 1
60 1
62 1
63 1
64 1
65 2 60
x=i1
cone=.707
34 1
35 1
36 2
37 2
38 3
39 5
40 4
42 6
43 2
44 7
45 5
47 2
48 3
49 3
50 3
51 4
52 3
53 2
54 2
55 4
56 2
57 1
58 1
59 1
60 1
61 1
62 1
63 1
64 1
66 1 75
w maxs
cone=.707
0 2
8 1
10 3
12 2
13 1
14 3
15 1
16 3
17 5
18 3
19 5
20 6
21 2
22 4
23 3
24 3
25 9
26 3
27 3
28 3
29 5
30 3
31 4
32 3
33 2
34 2
35 2
36 4
37 1
38 1
40 1
41 4
42 5
43 5
44 7
45 3
46 1
47 6
48 6
49 2
51 1
52 2
53 1
55 1
137
w maxs
cone=.93
8 1 i10
13 1
14 3
16 2
17 2
18 1
19 3
20 4
21 1
24 1
25 4
e21 e34
27 2
29 2 i7
27/29 are i's
Cosine cone gap (over some  angle)
w aaan-aaax
cone=.54
7 3 i27 i28
8 1
9 3
i20 i34
11 7
12 13
13 5
14 3
15 7
19 1
20 1
21 7
22 7
23 28
24 6
100/104 s or e
so 0 picks i
w maxs
cone=.925
8 1 i10
13 1
14 3
16 3
17 2
18 2
19 3
20 4
21 1
24 1
25 5
e21 e34
27 2
28 1
29 2
e35 i7
31/34 are i's
w xnnn-nxxx
cone=.95
8 2 i22 i50
10 2
i28
i24 i27 i34
13 2
14 4
15 3
16 8
17 4
18 7
19 3
20 5
21 1
22 1
23 1
i39
43/50 e so picks out e
Cosine conical gapping seems quick and easy (cosine = dot product divided by both lengths.
Length of the fixed vector, x-M, is a one-time calculation.
Length y-M changes with y so build the PTreeSet.
But we can't divide PTreeSets yet !?!?! )

4. Squared y on f Projection Distance = yoy - (yof)2 fof
The downside of [capped] cone clustering is that we need to divide by PTreeSet |y| . So far we can't do that (without a loop)?
Instead of a "capped cone" a better shape might be a "[double] capped tube". For fixed point, f, and variable point , y, we need, in addition to the dot product projection length, the dot product projection distance as well, as shown in red. f y
y - f
|f| yo
= y -
(yof)
fof f
squared is y -
(yof)
fof f
o y -
dot product projection distance
squared = yoy - 2
(yof)2
fof
fof
(fof)2 yo
dot product projection length f
|f|
squared = yoy - 2
(yof)2
fof +
Now if we replace the origin by a corner point (or some other circumscribing hyper-rectangle point, p,
e.g., replace y with y-p and replace f with M-p
Squared y on f Projection Distance = yoy -
(yof)2
fof
Squared y-p on M-p Projection Distance = (y-p)o(y-p) -
( (y-p)o(M-p) )2
(M-p)o(M-p)
Furthest Point or Mean Point
f (or M)
1st: compute this constant [vector]
= yoy -2yop + pop -
( yo(M-p) - po(M-p
|M-p| 2
Gaps in dot product lengths [projections] on the line.
3rd: comp these PTreeSets
(2 dots, 1 minus, 1 plus)
Do not compute y-p. (shifts entire vector sp)? y
cap gap
width
M-p
|M-p|
(y-p)o
For the dot product length projections (caps) we already needed:
= ( yo(M-p) -
po M-p )
2nd: compute this PTreeSet (1 dot, 1 minus)
That is, we needed to compute the green constants and the blue and red dot product functionals in an optimal way (and then do the PTreeSet additions/subtractions/multiplications).
What is optimal? (minimizing PTreeSet functional creations and PTreeSet operations.)
Origin (or p)
tubular gap
width

5. - - - = yoy -2yop + pop - M-p |M-p| yo po M-p |M-p| yo po M-p |M-p| yo
There are three functionals in the "dot product" group for "functional gap clustering" of a VectorSpace subset, Y (yY):
1. SDp(y) = (y-p)o(y-p), p a fixed vector, the "Square Distance from a point", primarily for outlier identification and densities.
2. Pd(y) = yod, d a unit vector, the "Projection" functional.
yod projection d y
y - (yod)d = projection. Squaring its length: (y-yodd)o(y-yodd)=yoy-(yod)2 
yod projection (neg) d y
y - (yod)d
so again yoy - (yod)2 = squared proj 
(y-p)o(y-p) -
( (y-p)o(M-p) )2
(M-p)o(M-p)
= yoy -2yop + pop - 2
3. SPDd(y) = yoy - (yod)2 (d a unit vector) is the "Square Projection Distance to d" functional.
E.g., if d≡(M-p)/|M-p|, d = unit vector from vector p to vector M, then SPD(y)=
But to avoid creating an entirely new VectorPTreeSet(Y-p)
for the space (with origin shifted to p), we think it useful
to alter the expression for SPDfM to : SPDpM(y)
M-p
|M-p| - yo po
M-p
|M-p| yo
where we might:
1st compute the constant vector nd the ScalarPTreeSet po
M-p
|M-p| yo -
3rd the constant th the SPTreeSet
pop yo
M-p
|M-p| po -
5th the SPTreeSet th the constant
yoy, yop
= yoy -2yop + pop - 2
7th the SPTreeSets 8th the SPTreeSet
M-p
|M-p| - yo po
Is it better to leave all the additions and subtractions for one mega-step at the end? (Md?) Other efficiency thoughts?
M-p
|M-p|
(y-p)o = - yo po
We note that PL(y)=yod shares many construction steps with SPD.

6. CLUS1.2 is pure Versicolor (45 of the 50).
SPD p
q e14
V Ct
2 10
3 12
4 12
5 12
6 8
7 11
8 9
9 5
10 9
11 4
12 4
13 2
14 1
17 2
18 3
19 10
20 5
21 6
22 5
23 6
24 6
25 3
27 2
29 2
30 1
SPD on CLUS1
p e11
q =MN
V Ct
2 3
3 4
4 5
5 7
6 2
7 2
8 6
9 6
10 3
11 4
12 2
13 4
14 4
15 3
16 2
17 1
18 5
19 1
20 2
22 2
23 1
24 1
25 1
26 1
29 1
SPD p
q e14
V Ct
1 6
2 4
3 8
4 4
5 10
6 2
7 2
8 2
9 7
10 2
11 2
12 2
13 1
15 2
17 1
18 4
19 2
20 4
22 1
24 1
25 1
26 1
29 1
31 2
32 2
33 3
i15 i36
i32
SPD p q
V Ct
2 8
3 10
4 10
5 10
6 5
7 10
8 6
9 8
10 6
11 1
mask: V<8.5
CTs SMs
CTe SMe
CTi SMi
CLUS1
mask: V<12.5
5 SMe
24 SMi
CLUS1.1
thin gap
mask: 8.5<V<15.5
CTs SMs
CTe SMe
CTi SMi
CLUS2
masking V>6:
Total_e Masked_e
Total_i Masked_i
However I cheated a bit.
I used p=MinVect(e) and q=MaxVect(e) which makes it somewhat supervised.
START OVER WITH THE FULL >
mask: V>12.5
45 SMe
0 SMi
CLUS1.2
mask: V>15.5:
CTs SMs
CTe SMe
CTi SMi
This tube contains 49 setosa
+ 2 virginica
CLUS3
CLUS1.2 is pure Versicolor (45 of the 50).
CLUS3 is almost pure Setosa (49 of the 50, plus 2 virginica)
CLUS2 is almost purely [1/2 of] viriginica (24 of 50, plus 1 setosa).
CLUS1.1 is the other 24 virginicas, plus the other 5 versicolors.
So this method clusters IRIS quite well (albeit into 4 clusters, not three).
Note that caps were not put on these tubes.
Also, this was NOT unsupervised clustering! I took advantage of my knowledge of the classes to carefully chose the unit vector points, p and q E.g., p = MinVector(Versicolor) and q = MaxVector(Versicolor. True, if one sequenced thru a fine enough d-grid of all unit vectors [directions], one would happen upon a unit vector closely aligned to d=q-p/|q-p| but that would be a whole lot more work that I did here (would take much longer).
In worst case though, for totally unsupervised clustering. there would be no other way than to sequence through a grid of unit vectors.
However, a good heuristic might be to try all unit vectors "corner-to-corner" and "middle-of-face-TO-middle-of-opposite-face" first, etc.
Another thought would be to try to introduce some sort of hill climbing to "work our way" toward a good combination of a radial gap plus two good linear cap gaps for that radial gap.

7. SPD on CLUS1
p C1US1axxx
q C1US1aaaa
V Ct
1 3
2 5
3 9
4 13
5 18
6 12
7 4
8 1
9 2
no thinnings
SPD on CLUS1
p C1US1xaxx
q C1US1aaaa
V Ct
1 4
2 13
3 7
4 19
5 9
6 7
7 9
8 2
SPD on CLUS1
p C1US1xxax
q C1US1aaaa
V Ct
1 1
2 4
3 3
4 9
5 9
6 14
7 9
8 4
9 6
10 3
11 3
12 1
14 2
15 1
no thinnings
SPD on CLUS1
p C1US1xxxa
q C1US1aaaa
V Ct
1 1
2 3
3 10
4 15
5 16
6 12
7 7
8 3
9 1
10 1
no thinnings
SPD
p axxx
q aaaa
V Ct
2 1
3 5
4 6
5 6
6 8
7 6
8 8
9 15
10 7
11 8
12 13
13 8
14 14
15 9
16 13
17 6
18 4
19 4
20 3
21 4
23 1
25 1
mask: V<3.5
14 SM versi
10 SM virgi CL1.1?
mask: V<11.5
0 SM setosa
46 SM versicolor
24 SM virginica
CLUS1
mask: V>3.5
0 SM setosa
32 SM versi
14 SM virgi
CLUS1.2?
mask: V>11.5
50 SM setosa
4 SM versicolor
26 SM virginica
CLUS2
SPD on CLUS2
p C1US2axxx
q C1US2aaaa
V Ct
6 2
7 2
8 6
9 13
10 7
11 7
12 4
13 5
14 11
15 9
16 2
18 4
21 2
22 1
23 3
25 1
26 1
SPD on CLUS1
p C1US1axax
q C1US1aaaa
V Ct
1 1
2 3
3 4
4 2
5 12
6 13
7 9
8 7
9 2
10 7
11 4
13 2
14 1
17 2
18 1
SPD on CLUS1
p C11aaxx
q C11aaaa
V Ct
1 1
2 7
3 10
4 13
5 13
6 13
7 6
8 2
9 2
11 1
no thinnings
SPD on CLUS1
p C1US1axxa
q C1US1aaaa
V Ct
1 1
2 2
3 6
4 9
5 12
6 17
7 8
8 6
9 5
10 1
11 1
no thinnings
mask: V<13.5
44 SM setosa
0 SM versicolor
02 SM virginica
CLUS2.1
mask: V<9.5
37 SM vers
16 SM virg CL1.1?
mask: 100>V>13.5
6 SM setosa
4 SM versicolor
24 SM virginica
CLUS2.2
mask: V>9.5
9 SM vers
8 SM virg CL1.2?
SPD on CLUS1
C11xaax
C11aaaa
V Ct
1 2
2 3
3 4
4 8
5 8
6 14
7 8
8 4
9 5
10 6
11 1
12 3
14 1
15 2
no thins
C11axaa
C11aaaa
V Ct
1 2
2 2
3 2
4 10
5 3
6 13
7 8
8 7
9 4
10 3
11 6
12 2
13 2
14 2
17 2
18 1
19 1
SPD on CLUS1
C11xxaa
C11aaaa
V Ct
1 1
2 4
3 6
4 9
5 10
6 7
7 9
8 5
9 3
10 4
11 2
12 4
13 1
14 3
17 2
SPD on C1
C11aaax
C11aaaa
V Ct
1 3
2 1
3 3
4 4
5 12
6 15
7 4
8 5
9 4
10 7
11 4
12 2
13 1
14 1
15 1
17 1
18 1
19 1
SPD on CLUS1
C11xaxa
C11aaaa
V Ct
1 2
2 3
3 12
4 12
5 10
6 15
7 7
8 4
9 1
10 2
11 1
no thins
C11aaxa
C11aaaa
V Ct
1 2
2 3
3 6
4 12
5 11
6 9
7 11
8 5
9 5
10 1
11 3
13 2
C11xaaa
C11aaaa
V Ct
1 2
2 4
3 5
4 9
5 10
6 9
7 5
8 6
9 2
10 6
11 3
12 1
13 2
14 2
15 2
17 2
mask: V<5.5
16 ver
3 virCL1.1?
mask: V<5.5
26 ver
4 vir CL1.1?
mask: V>5.5
30 ver
21 virCL1.1?
mask: V>5.5
20 ver
20 vir CL1.1?

8. nnnn xxxx
V Ct
SDD
1 1
2 8
3 8
4 20
5 15
6 10
7 2
8 5
9 1 PL
0 1
2 2
3 1
5 2
6 3
7 1
8 2
9 3
11 1
12 5
13 6
14 7
15 3
16 2
17 2
18 1
19 3
20 3
22 4
23 3
24 3
25 2
26 2
27 2
28 1
29 2
e8 e44 e49
mask: V<5.5
16 ver
3 virCL1.1?
mask: V>5.5
30 ver
21 virCL1.1?

9. 95 remaining versicolor and virginica=SubClus1.
i p
max
V Ct
0 2
1 2
2 2
3 5
4 3
5 3
6 4
7 4
8 7
9 2
10 3
11 1
12 4
13 5
14 4
15 7
16 2
17 5
18 3
19 1
20 1
21 4
23 2
24 2
25 4
26 1
27 2
28 1
29 2
30 1
32 1
{e4, e40} form a doubleton outlier set
i7 and e10 are singleton outliers
x=s (58=avg(y1) )
V Ct
0 3 s15, s17, s34
1 12 s 6,11,16,19,20,22,28,32,3337,47,49
2 12 s 1,10 13,18,21,27,29,40,41,44,45,50
3 7 s 2,12,23,24,35,36,38
4 10 s 2,3,7,13,25,26,30,31,46,48
5 2 s4, s43
6 2 s9,s39
7 1 s14
8 1 i39
9 1 s32 ^^all 50 setosa + i39
e49
16 2
17 2
19 1
20 2
21 5
22 4
23 3
24 4
25 1
27 8
28 2
29 2
30 4
31 1
32 4
34 2
35 2
36 2
37 3
38 2
39 2
40 4
41 1
43 2
44 4
45 2
46 1
47 2
48 1
50 4
52 2
53 2
54 2
56 2
57 1 i1
i31 vv 9 virginica
i10 i8
i36
i32
i16
i18
i23
i19
But here I mistakenly used the mean rather than the max corner.
So I will redo - but note the high level of cluster and outlier revelation?????
i p
max
V Ct
0 2
2 6
3 3
4 4
5 4
6 2
7 6
8 9
9 2
10 2
11 2
12 5
13 7
14 2
15 6
16 2
17 5
19 3
20 2
22 3
23 2
24 3
25 2
26 1
27 1
28 1
29 3
30 1
31 2
e32
e11
e8,44
e49
i39
60 1
61 1
62 1
63 1
64 1
65 1
66 1
67 3
68 4
69 4
70 3
71 3
72 4
73 2
74 5
75 1
76 2
77 1
78 3
79 1 s3 s9
s39,43
s42
s23
s14
2 actual gap-ouliers,
checking distances reveals 4 e-outlier (versicolor),
5 s-outliers (setosa).
i p
max
V Ct
0 2
1 1
2 3
3 3
4 4
5 2
6 6
7 3
8 5
9 4
10 4
11 2
12 3
13 4
14 6
15 4
16 1
17 7
18 2
19 3
20 2
22 2
23 1
24 2
25 4
26 4
27 1
28 2
29 2
30 1
32 2
33 1
34 1
35 1
No new outliers reviealed
95 remaining versicolor and virginica=SubClus1.
Continue outlier id rounds on SC1 (maxSL, maxSW, max PW) then do "capped tube" (further subclusters.)
1. (y-p)o(y-p) remove edge outliers ( thr>2*50)
2. lthin gaps in SPD: d, from an edge point to MN.
3 For each thin PL, do len gap anal of pts in " tube".
e13 i7 e40 e4 e10 F e i e e e
e32 e11 e8 e44 e49 e e e e e
45 remaining setosa.
This is SubCluster 2
(may have additional outliers or sub-subclusters but we will not analyse further (would be done in practice tho
SPD(y) =(y-p)o(y-p)-(y-p)od2 d: mn-mx
V Ct
Next slide
i p
max
V Ct
0 2
1 10
2 11
3 6
4 15
5 4
6 8
7 9
8 4
9 5
10 2
11 7
13 4
14 2
15 2
16 1
17 1
18 1
19 1
e30, e15 outliers
e20,e31,e32 form SC12
Declared tripleton outlier set? (But they are not singleton outliers.)
s3 s9 s39 s43 s42 s23 s s s s s s
e13 e20 e15 e31 e32 e30 F e e e e e e

10. 1. (y-p)o(y-p) remove edge outliers ( thr>2*50)
SPD(y) =(y-p)o(y-p)-(y-p)od2 d: nnnn-to-xxxx gp>2/50
V Ct
0 1
1 1
2 2
3 3
4 3
5 3
6 3
7 4
8 2
9 2
10 7
11 3
12 3
14 1
15 4
16 5
17 3
18 8
19 1
20 5
21 2
22 2
23 1
24 2
25 3
27 2
28 2
29 1
30 3
31 2
32 1
33 1
(y-p)o(y-p)-(y-p)od2
d: naaa-to-xaaa
V Ct
2 5
3 10
4 12
5 16
6 3
7 10
8 6
9 5
10 3
11 6
12 3
13 2
1. (y-p)o(y-p) remove edge outliers ( thr>2*50)
2. lthin gaps in SPD: d, from an edge point to MN.
3 For each thin PL, do len gap anal of pts in " tube".
(y-p)o(y-p)-(y-p)od2 d: nxnn-to-xnxx
V Ct
0 1
1 1
2 1
3 3
5 3
6 5
7 2
8 6
9 1
10 5
11 4
12 3
13 3
14 3
15 3
16 8
17 3
18 1
19 7
20 3
21 2
22 1
23 1
24 1
25 2
26 1
27 3
28 3
29 2
30 2
i19 i23 i6 i8 i31 i36 F i i i i i i
i19,i23,i6 outliers
(y-p)o(y-p)-(y-p)od2
d: aaan-to-aaa
V Ct
2 5
3 10
4 12
5 16
6 3
7 10
8 6
9 5
10 3
11 6
12 3
13 2
(y-p)o(y-p)-(y-p)od2 d: xnnn-to-nxxx
V Ct
3 1
5 1
6 4
7 6
8 5
9 5
10 10
11 7
12 4
13 8
14 5
15 5
16 4
17 3
18 3
19 8
20 1
21 1
22 2
23 1
24 1
26 1
(y-p)o(y-p)-(y-p)od2 d: nnxn-to-xxnx
V Ct
6 3
7 6
8 5
9 9
10 8
11 15
12 6
13 7
14 11
15 5
16 4
17 1
18 1
(y-p)o(y-p)-(y-p)od2
d: nnnx-to-xxxn
V Ct
2 1
3 1
4 3
5 2
6 2
7 2
8 2
9 6
10 5
11 7
12 7
13 1
14 2
15 9
16 4
17 5
18 3
19 1
20 4
21 2
22 1
23 4
24 2
25 3
26 2
i1 i37 e33 e13 F i i e e
i1, e13 are outliers.

11. FxM(x,y)=yo(x-M)/|x-M|-min on XX≡{(x,y)|x,yX}, where X(x,y) is a Spaeth image table Cluster by splitting at all F_gaps > 2
APPENDIX
The 15 Value_Arrays (one for each x) z z z z z z z z z z z z z z z X
x y \y= a b
1 1 x=1 1 f
M d
a b
b c e c
d a
e 8
f 7 9
x y FxM
z1 z1 14
z1 z2 12
z1 z3 12
z1 z4 11
z1 z5 10
z1 z6 6
z1 z7 1
z1 z8 2
z1 z9 0
z1 z10 2
z1 z11 2
z1 z12 1
z1 z13 2
z1 z14 0
z1 z15 5
9 5  M
(=MeanVector)
The 15 Count_Arrays z z z z z z z z z z z z z z z
Level0, stride=z1 PointSet (as a pTree mask) z1 z2 z3 z4 z5 z6 z7 z8 z9
z10
z11
z12
z13
z14
z15
gap: 10-6
gap: 5-2
pTree masks of the 3 z1_clusters (obtained by ORing)
z11 1
z12 1
z13 1
The FAUST algorithm:
1. project onto each Mx line using the dot product with the unit vector from M to x. (only x=z1 is shown)
2. Generate each Value Array, F[x0]|(y), xX (also generate the Count_Arrays and the mask pTrees).
3. Analyze all gaps and create sub-cluster pTree Masks.

12. Cluster by splitting at gaps > 2
yo(z7-M)/|z7-M| ValueArrays z z z z z z z z z z z z z z z
yo(z7-M)/|z7-M| CountArrays z z z z z z z z z z z z z z z
Cluster by splitting at gaps > 2
x y x\y a b
3 3 4
9 3 6 f
14 2 8
M d
13 4 a b
10 9 b c e
1110 c
9 11 d a
1111 e 8
7 8 f
x y F
z1 z1 14
z1 z2 12
z1 z3 12
z1 z4 11
z1 z5 10
z1 z6 6
z1 z7 1
z1 z8 2
z1 z9 0
z1 z10 2
z1 z11 2
z1 z12 1
z1 z13 2
z1 z14 0
z1 z15 5
9 5 Mean
z11 1
z12 1
z13 1
gap: 6-9
z71 1
z72 1
In Step_3 of the algorithm we can:
Analyze one of the gap arrays (e.g., As done for z1. Subclusters is shown above)
then start over on each subcluster.
Or we can analyze all gap arrays concurrently (in parallel using the same F - saving the [substantial?] re-compute costs?) and then intersect the subcluster partitions we get from each x_ValueArray gap analysis, forthe final subclustering.
Here we use the second alternative, judiciously choosing only the x's that are likely to be productive (choosing z7 next).
Many are likely to produce redundant partitions - e.g., z1, z2, z3, z4, z6 - as their projection lines will be nearly coincident.
How should we choose the sequence of "productive" strides? One way would be to always choose the remaining stride with the shortest ValueArray, so that the chances of decent sized gaps is maximized. Other ways of choosing?

13. Cluster by splitting at gaps > 2
yo(x-M)/|x-M| Value Arrays z z z z z z z z z z z z z z z
Cluster by splitting at gaps > 2
yo(x-M)/|x-M| Count Arrays z z z z z z z z z z z z z z z
x y x\y a b
3 3 4
9 3 6 f
14 2 8
M d
13 4 a b
10 9 b c e
1110 c
9 11 d a
1111 e 8
7 8 f
x y F
z1 z1 14
z1 z2 12
z1 z3 12
z1 z4 11
z1 z5 10
z1 z6 6
z1 z7 1
z1 z8 2
z1 z9 0
z1 z10 2
z1 z11 2
z1 z12 1
z1 z13 2
z1 z14 0
z1 z15 5
9 5 Mean
z11 1
z12 1
z13 1
gap: 3-7
z71 1
z72 1
zd1 1
zd2 1
We choose zd=z13 next (Should have been first? Since it's ValueArray is shortest?)
Note, z8, z9, za projection lines will be nearly coincident with that of z7.

14. Cluster by splitting at gaps > 2
yo(x-M)/|x-M| Value Arrays z z z z z z z z z z z z z z z
Cluster by splitting at gaps > 2
yo(x-M)/|x-M| Count Arrays z z z z z z z z z z z z z z z
x y x\y a b
3 3 4
9 3 6 f
14 2 8
M d
13 4 a b
10 9 b c e
1110 c
9 11 d a
1111 e 8
7 8 f
x y F
z1 z1 14
z1 z2 12
z1 z3 12
z1 z4 11
z1 z5 10
z1 z6 6
z1 z7 1
z1 z8 2
z1 z9 0
z1 z10 2
z1 z11 2
z1 z12 1
z1 z13 2
z1 z14 0
z1 z15 5
9 5 Mean
z11 1
z12 1
z13 1
z71 1
z72 1
zd1 1
zd2 1
AND each red with each blue with each green, to get the subcluster masks (12 ANDs producing 5 sub-clusters.

15. F1(x,y) = L1Distance(x,y) = (|x1-y1|+|x2-y2|) on XX≡{(x,y)|x,yX},
Cluster by splitting at all F1_gaps
L1(x,y) Value Array z z z z z z z z z z z z z z z
L1(x,y) Count Array z z z z z z z z z z z z z z z
x y x\y a b
3 3 4
9 3 6 f
14 2 8 d
13 4 a b
10 9 b c e
1110 c
9 11 d a
1111 e 8
7 8 f
(redundant subclustering)
gap: 10-5
There is a z1-gap, but it produces a subclustering that was already discovered by a previous round.
Which z values will give new subclusterings?

16. Re-confirms zf an anomaly.
L1(x,y) Value Array z z z z z z z z z z z z z z z
L1(x,y) Count Array z z z z z z z z z z z z z z z
This re-confirms z6 as an anomaly or outlier, since it was already declared so during the linear gap analysis.
x y x\y a b
3 3 4
9 3 6 f
14 2 8
M d
13 4 a b
10 9 b c e
1110 c
9 11 d a
1111 e 8
7 8 f
Re-confirms zf an anomaly.
After having subclustered with linear gap analysis, which is best for determining larger subclusters, we run this round gap algorithm out only 2 steps to determine if there are any singleFvalue gaps>2 (the points in the singleFvalueGapped set are then declared anomalies). So we run it out two steps only, then find those points for which the one initial gap determined by those first two values is sufficient to declare outlierness. Doing that here, we reconfirm the outlierness of z6 and zf, while finding new outliers, z5 and za.

17. Using F=yo(x-M)/|x-M|-MIN on IRIS, one stride at a time (s1=setosa1 first)
For virginica1
Val Ct
0 1
1 1
2 2
3 5
4 6
5 11
6 12
7 4
8 2
9 5
10 1
17 1
22 1
24 2
25 1
27 1
28 1
29 2
30 1
31 3
32 4
33 1
34 4
35 2
36 2
37 4
38 4
39 5
40 4
42 6
43 2
44 7
45 5
47 2
48 3
49 3
50 3
51 4
52 3
53 2
54 2
55 4
56 2
57 1
58 1
59 1
60 1
61 1
62 1
63 1
64 1
66 1
F(i39)=17
F<17 (50
Setosa)
vers1
Val Ct
0 1
2 4
3 1
4 1
5 3
6 3
7 8
8 3
9 7
10 6
11 4
12 4
13 3
15 2
19 2
20 2
21 1
26 2
27 3
28 4
30 2
31 5
32 4
33 3
34 1
36 3
37 5
38 4
39 5
40 7
41 4
42 2
43 2
44 1
45 6
46 4
47 5
48 1
49 2
50 5
51 1
52 2
54 2
55 1
57 2
58 1
60 1
62 1
63 1
64 1
65 2
F<19 (50 setosa)
19<F<22 {vers8,12,39,44,49}
22<F
yo(s1-M)/|s1-M|-69)
Val Ct
0 1
3 1
4 2
7 1
8 1
9 2
10 1
12 4
14 5
15 2
16 4
17 1
18 4
19 5
20 1
21 2
22 2
23 8
24 4
25 3
26 2
27 5
28 3
29 4
30 4
31 3
32 2
33 2
34 4
35 5
36 2
37 2
38 1
39 1
40 1
41 1
43 1
44 1
45 1
52 1
60 3
61 4
62 3
63 10
64 15
65 9
66 3
67 1
69 2
F(i39)=52 virginica39 is an outlier. 2 clusters, F<52 (ct=99) and F>52 (50 Setosa)
virgini39
Val Ct
0 1
1 2
2 1
4 2
6 1
7 1
8 7
9 2
10 2
11 7
12 2
13 3
14 7
15 4
16 10
17 4
18 6
19 9
20 3
21 6
22 3
23 6
24 3
25 1
27 3
28 2
32 1
39 1
40 1
41 1
42 8
43 13
44 17
45 4
46 5
47 1
F=32 vers49 outlier.
32<F (50 Setosa, vir39)
AVG(ver8,12,39,44,49)
Val Ct
0 1
1 1
7 5
10 3
12 2
13 2
14 3
15 5
16 2
17 5
18 8
19 4
20 3
21 4
22 3
23 8
24 4
25 4
26 3
27 7
28 7
29 4
30 5
31 4
32 5
33 8
34 2
35 6
36 5
37 3
38 2
39 8
40 6
41 3
43 1
44 2
45 1
47 1
F=0 vir32 outlier
F=1 vir18 outlier
F=7 vir6,10,19,23,36 subcluster?

18. F=yo(x-M)/|x-M|-MIN on IRIS, subclustering as we go.
On Clus(F<52) ver1
F(virg7)=0 outlier
F(virg32)=25 outlier
Val Ct
0 1
4 1
5 5
6 3
7 5
8 3
9 8
10 11
11 14
12 8
13 8
14 5
15 3
16 7
17 5
18 6
19 2
20 1
21 1
22 1
25 1
F=yo(x-M)/|x-M|-MIN on IRIS, subclustering as we go.
On Remaining, mx mn mx mx
Val Ct
0 3
1 4
2 11
3 14
4 14
5 9
6 10
7 2
8 6
9 2
11 2
For s1 (i.e.,
yo(s1-M)/|s1-M|-69)
Val Ct
0 1
3 1
4 2
7 1
8 1
9 2
10 1
12 4
14 5
15 2
16 4
17 1
18 4
19 5
20 1
21 2
22 2
23 8
24 4
25 3
26 2
27 5
28 3
29 4
30 4
31 3
32 2
33 2
34 4
35 5
36 2
37 2
38 1
39 1
40 1
41 1
43 1
44 1
45 1
outlier
60 3
61 4
62 3
63 10
64 15
65 9
66 3
67 1
69 2
F(i39)=52 i39=virgi39 outlier. Clusters, F<52 (ct=99) and F>52 (50 Setosa)
On Remaining, max's
Val Ct
0 2 e8 outlier
1 2 e11 outlier
7 2
8 1
9 4
10 1
11 2
12 2
13 4
14 3
15 1
16 4
17 2
18 2
19 3
20 4
21 6
22 5
23 5
24 4
25 2
26 2
27 1
28 2
29 4
30 5
31 1
32 3
33 2
34 2
35 3
36 2
37 1
38 1 i8
i10 i36
i6 i23
i19
i18
i6 i8 i10 i19 i23 i35 i i i i i i
i6 i10 i18 i19 i23 i35
all declared outliers
e4 e38 e19 i20 F e e
e outlier
i outlier
On Remaining, max's
Val Ct
e44 outlier
6 1
7 2
8 1
9 3
10 1
11 3
12 5
13 2
14 2
15 3
17 3
18 3
19 5
20 1
21 9
22 5
23 4
24 2
26 4
27 2
28 2
29 4
30 2
31 3
32 3
33 2
34 3
35 2
36 1
37 1
38 1
39 1
e36
outlier?
On Remaining,
mx mx mx mn
Val Ct
0 1
1 2
2 3
3 1
5 5
6 4
7 5
8 2
9 3
10 5
11 4
12 7
13 5
14 2
15 4
16 4
17 7
18 4
19 4
20 2
21 2
22 1
24 1
25 1
27 2
29 2
On Remaining,
mn mn mx mx
Val Ct
0 1
1 3
2 3
3 7
4 7
5 7
6 5
7 5
8 3
9 8
10 4
11 4
12 11
13 4
14 8
15 4
16 1
18 1
On Remaining, mn mx mx mx
Val Ct
0 1
2 1
3 4
4 3
5 5
6 4
7 5
8 7
9 8
10 3
11 5
12 2
13 4
14 5
15 7
16 5
17 4
18 1
20 1
On Remaining w e35
Val Ct
0 1 i26 outlier
3 2
On remaining vir1
Val Ct
0 1
1 2
2 1
4 1
5 1
6 2
7 2
8 2
9 4
10 1
11 4
12 3
13 4
14 2
15 6
16 4
17 6
19 4
20 5
21 5
22 2
23 1
24 2
25 5
26 4
27 4
28 1
29 2
30 6
31 2
32 1
33 1
34 1
35 2
36 1
38 1
39 1
e35 e10 e
e outlier
i44 i3 i i
^^outlier
i3 i30 i31 i26 i8 i36 i
i outlier
i outlier
i outlier
i outlier
i outlier
Rem mn mx mn mx
Val Ct
0 1
1 1
2 1
3 1
4 1
5 1
6 1
8 1
9 3
10 5
11 5
12 3
13 7
14 6
15 4
16 6
17 7
18 5
19 4
20 2
21 3
22 7
23 4
24 3
25 1
26 1
27 2
e49 outlier
On Remaining, mn mx mx mx
Val Ct
0 1
1 1
2 1
3 5
4 6
5 5
6 4
7 9
8 4
9 4
10 4
11 3
12 5
13 6
14 6
15 7
16 5
17 4
18 4
20 1
22 1
Could look at distances for 0,1 and 20,22?
e13 e30 e32
e outlier
e outlier e
i44 i45 i49 i5 i37 i1 i i i i
i not outlier
i outlier

19. outliers
gap>L1=32.1 s6
s14
s15
s16
s17
s19
s21
s23
s24
s32
s33
s34
s37
s42
s45 e1 e2 e3 e5 e6 e7 e9
e10
e11
e12
e13
e15
e18
e19
e21
e22
e23
e27
e28
e29
e30
e34
e36
e37
e38
e41
e49 i1 i3 i4 i5 i6 i7 i8 i9
i10
i12
i14
i15
i16
i18
i19
i20
i22
i23
i25
i26
i28
i30
i31
i32
i34
i35
i36
i37
i39
i41
i42
i45
i46
i47
i49
i50
outliers
gap>L1=42.8
s15
s16
s19
s23
s33
s34
s37
s42
s45 e1 e2 e7
e10
e11
e12
e13
e15
e19
e21
e22
e23
e27
e28
e30
e34
e36
e38
e41
e49 i1 i3 i5 i6 i7 i8 i9
i10
i12
i14
i15
i16
i18
i19
i20
i22
i23
i26
i30
i31
i32
i34
i35
i36
i39
outliers gp>L1=53.5
s15
s16
s23
s33
s34
s42
e10
e13
e15
e27
e28
e30
e36
e49 i1 i3 i7 i9
i10
i12
i15
i18
i19
i20
i26
i30
i32
i35
i36
i39
F=L1(x,y) on IRIS, masking to subclusters (go right down the table). Two rounds only
If we use L1gap=6, remove those outliers, then use linear gap analysis for larger subcluster revalation, let's see if we can separate Versicolor (e) from virginica (i).
outliers gap>L1=64.3
s15
s16
s23
s42
e10
e13
e49 i3 i7 i9
i10
i18
i19
i20
i32
i35
i36
i39
outliers gap>L1=74.95
L1gap
s42 9
e13 8
i7 10
i9 12
i10 12
i35 9
i36 9
i39 26

20. Val=0;p=K;c=0;P=Pure1; For i=n to 0 {c=Ct(P&Pi); If (c>=p){Val=Val+2i; P=P&Pi }; else{p=p-c; P=P&P'i }; return Val, P;
IDX z1 z2 : ze zf
IDY z1 z2 z3 z4 z5 z6 z7 z8 z9 za zb zc zd ze zf : X1 1 3 : 11 7 X2 1 : 11 8 X3 1 3 2 6 9 15 14 13 10 11 7 : 1 2 3 4 9 10 11 8 X4 : P3 1 : P2 1 P1 1 : P0 1 :
d(xy) 2 1 3 4 8 14 13 12 9 6 11 10 : 7 5
P'3 1 :
P'2 1 :
P'1 1 :
P'0 1 :
Need Rank(n-1) applied to each stride instead of the entire pTree. The result from stride=j gives the jth entry of SpS(X,d(x,X-x))
Parallelize over a large cluster?
Ct(P&Pi): revise the Count proc to kick out count for each stride (involves loop down pTree by register-lengths?
What does P represent after each step??
How does alg go on 2pDoop (w 2 level pTrees) where each stride is separate
Note: using d, not d2 (fewer pTrees). Can we estimate d? (using truncated McClarin series)
23 * * * *
1 = 1
n=3: c=Ct(P&P3)=10< 14, p=14–10=4; P=P&P' (elim 10 val8)
n=2: c=Ct(P&P2)= 1 < 4, p=4-1=3; P=P&P' (elim 1 val4)
n=1: c=Ct(P&P1)=2 < 3, p=3-2=1; P=P&P' (elim 2 val2)
n=0: c=Ct(P&P0 )=2>= P=P&P0 (elim 1 val<1)
23 * * * *
1 = 1
n=3: c=Ct(P&P3)=9< 14, p=14–9=5; P=P&P' (elim 9 val8)
n=2: c=Ct(P&P2)= 0 < 5, p=5-0=5; P=P&P' (elim 0 val4)
n=1: c=Ct(P&P1)=4 < 5, p=5-4=1; P=P&P' (elim 4 val2)
n=0: c=Ct(P&P0 )=1>= P=P&P0 (elim 1 val<1
23 * * * *
1 = 1
n=3: c=Ct(P&P3)= 9 < 14, p=14–9=5; P=P&P' (elim 9 val8)
n=2: c=Ct(P&P2)= 2 < 5, p=5-2=3; P=P&P' (elim 2 val4)2
n=1: c=Ct(P&P1)=2 < 3, p=3-2=1; P=P&P' (elim 2 val2)
n=0: c=Ct(P&P0 )=2>= P=P&P0 (elim 1 val<1)
23 * * * * 1
1 = 3
n=3: c=Ct(P&P3)= 6 < 14, p=14–6=8; P=P&P' (elim 6 val8)
n=2: c=Ct(P&P2)= 7 < 8, p=8-7=1; P=P&P' (elim 7 val4)2
n=1: c=Ct(P&P1)=11, p=1-1=0; P=P&P (elim 0 val2)
n=0: c=Ct(P&P0 )=1 P=P&P0 (elim 0)

21. Level-1 key map Red=pure stride (so no Level-0) e f g h
i a
j b c
k d m
0 0 13 12 11 10 23 22 21 20 33 32 31 30 43 42 41 40
a b c d e f g h i j k m 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Level-0:
key map 13 12 11 10 23 22 21 20
(6-e) = e
else pur0
(6-e) = f
else pur0
(6-e) = g
else pur0
(6-e) = h
else pur0
In this 2pDoop KEY-VALUE DB, we list keys. Should we bitmap? Each bitmap is a pTree in the KVDB. Each of these is existing, e.g., e here
5,7-a,f=f
else pur0
5,7-a,f=g
else pur0
5,7-a,f=h
else pur0
234789bcefg els pr0
234789bcefh else pr0
124-79c-f
h else pr0
(b-f) = i
else pur0
(b-f) = j
else pur0
(b-f) = k
else pur0
(b-f) = m
else pur0
(a) = j
else pur0
(a) = k
else pur0
(a) = m
else pur0
=SpS(XX,
-27(
p13p33 +
p13p32 +
p23p43
p23p42
(3-6,8,9)
k, els pr0
(3-6,8,9)
m els pr0 +
p13p31 +
26(
p13+p23+p33+p43
+p13p12+ p23p22+
p33p32 +
+p43p42 )
-26(
p23p41
124679bd
m els pr0
25(
p13p11+
p23p21 +
p33p31 +
p43p41 )
-25(
p13p30
+p23p40
+p12p31
+p22p41
+p12p32
+p22p42 e
f 5 6
g 7 h
i a
j b c
k d m 33 32 31 30 43 42 41 40
24(
p12+p22+p32+p42
+p13p10+
+p23p20
+p33p30
+p43p40
-24(p12p30
+p22p40
+p12p11+
+p22p21
+p32p31
+p42p41 )
23(
p12p10+
p22p20 +
p32p30 +
p42p40 )
-23(p11p31
+p11p30
+p21p41
+p21p40
p11+p21+p31+p41
+p11p10 +
+p21p20 +
+p31p30
+p41p40 )
-22(p10p30 +p20p40
p10+p20+p30+p40 )
22(

22. If (Ct≡Count(P&Pi)p)
x y yox-M
ID1ID2 -MIN V P C1 V P C2 V P C3 V P C4 V P C5 V P C6 V P C7 V P C8 V P C9 V P Ca V P Cb V P Cc
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z1 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z2 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z3 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z4 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
z5 z
RankK V=0;p=K;P=Pur1;
For i=n..0 {
If (Ct≡Count(P&Pi)p)
V=V+2i; P=P&Pi }
else{ p=p-Ct; P=P&P'i } }
RankK reveals the full gap situation for any functional on any vector space.
(Here, the functional is the dot product with d=unit vector from the mean to each point).
Use RankK with
K=N, N-C1,N-C1,...,N-C1...-CmaxL1 to mine useful subcluster info (Ci=ct(P&Pi) after the ith round, N=|X|=15, mxL1=20)

23. x y yox-M
ID1ID2 -MIN V P C1 V P C2 V P C3 V P C4 V P C5 V P C6 V P C7 V P C8 V P C9 V P Ca V P Cb V P Cc
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z6 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z7 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z8 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z9 z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z
z10z

24. x y yox-M
ID1ID2 -MIN V P C1 V P C2 V P C3 V P C4 V P C5 V P C6 V P C7 V P C8 V P C9 V P Ca V P Cb V P Cc
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z11z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z12z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z13z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z14z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z
z15z

25. ptree P=Pure1; ptreeSet P[n]; ptree T1,T2,a; ptreeSet c,t,rv,k;
For i=(n-1) to 0
{T1=P&P[i]; T2=P&P[i]’; c=PartialCount(T1);
a=Compare(c,k); rv[i]=a; P=(T1&a)|(T2&a’);
t=Subtract(k,c); k=(k& a) |(t& a’) ; }
The difference between the two algorithms is in the method of handling (resetting) P and the parameters (rv[i] V) and (k p).
Mohammad uses PTreeSets for the array of [real number] parameter values and then can avoid looping through the strides.
Which is faster for big data? Should 2-level pTrees be used? If so which is better?
RankK V=0;p=K;P=Pur1;
For i=n..0 {
If(Ct≡Count(P&Pi)p){
V=V+2i; P=P&Pi }
else {p=p-Ct; P=P&P'i } }

26. Stride 1
P' P3 1
010 1 05
P'3P' P'3P P3P' P3P2 1 08 1 12 1 02 1 03
P'3P'2P' P'3P2P'1
P'3P'2P P'3P2P1
P3P'2P' P3P2 P'1
P3P'2P P3P2 P1 1 04 1 04 1 01 11 00 1 2 1 02 1 01
P'3P'2P'1P' P'3P2P'1P'0
P'3P'2P'1P P'3P2P'1P0
P'3P'2P1P' P'3P2P1P'0
P'3P'2P1P P'3P2P1P0
P3P'2P'1P' P3P2P'1P'0
P3P'2P'1P P3P2P'1P0
P3P'2P1P' P3P2P1P'0
P3P'2P1P P3P2P1P0 00 : 1 03 : 1 04 00 1 01 00 11 00 1 02 00 1 02 00 00 1 01
If all these pTree ANDs are pre-computed and stored (with their 1-counts) for each stride, the Rank alg can be run accessing the counts only.
E.g., if n=1,000,000=1M then N=1T and there are 1M strides to pre-compute ;-( If the bitwidth is 4, then each stride requires these 30=( )=25-2 pre-computed level-0 pTrees and counts. If the bitwidth=b each stride requires (i=1..b2i = 2b+1-2 pre-computations.
E.g., ~=1018 for b=32, so one would do this only for, say, the high order 8 bits.
Descending the tree, 1bits turn to 0bits only. Therefore, the counts are non-increasing and the count across at any level stays at n=1M =106, 31 1 21 1 11 1 01 1 32 1 22 1 12 1 02 1 33 1 23 1 13 1 03 1 34 1 24 1 14 1 04 1 35 1 25 1 15 1 05 1 36 1 26 1 16 1 06 1 37 1 27 1 17 1 07 1 38 1 28 1 18 1 08 1 39 1 29 1 19 1 09 1 3a 1 2a 1 1a 1 0a 1 3b 1 2b 1 1b 1 0b 1 3c 1 2c 1 1c 1 0c 1 3d 1 2d 1 1d 1 0d 1 3e 1 2e 1 1e 1 0e 1 3f 1 2f 1 1f 1 0f 1
LEVEL-0 of
PTreeSet
yo(z1-M)/|z1-M|

27. y = yoy -2yop + pop - ( yo(M-p) - po(M-p |M-p| M-p |M-p| (y-p)o
Using a "capped tube". Given a unit vector, d, we need the d_dot_product_projection_lengths, and the d_dot_product_projection_distances. y
squared is (y-(yod)d)o(y-(yod)d) = yoy -2(yod)2 + (yod)2 = yoy - (yod)2
| y - (yod)d |
dot product projection distance
(yod)d
Note, this projection_distance is the perpendicular distance from the point, y, to the d_line and has nothing to do with the origin of the vector, y.
Note, this projection_distance is the perpendicular distance from the point, y, to the d_line and has nothing to do with the origin of the vector, y.
Squared y-p on M-p Projection Distance = (y-p)o(y-p) -
( (y-p)o(M-p) )2
(M-p)o(M-p)
Furthest Point or Mean Point
f (or M)
1st: compute this constant [vector]
= yoy -2yop + pop -
( yo(M-p) - po(M-p
|M-p| 2
Gaps in dot product lengths [projections] on the line.
3rd: comp these PTreeSets
(2 dots, 1 minus, 1 plus)
Do not compute y-p. (shifts entire vector sp)? y
cap gap
width
M-p
|M-p|
(y-p)o
For the dot product length projections (caps) we already needed:
= ( yo(M-p) -
po M-p )
2nd: compute this PTreeSet (1 dot, 1 minus)
That is, we needed to compute the green constants and the blue and red dot product functionals in an optimal way (and then do the PTreeSet additions/subtractions/multiplications).
What is optimal? (minimizing PTreeSet functional creations and PTreeSet operations.) p
tubular gap
width
Origin