1. A Spatial Data and Sensor Network Application:
CubE for Active Situation Replication (CEASR)
Nano-sensors dropped
into the Situation space
Situation space
Wherever threshold level is sensed (chem, bio, thermal...)
a ping is registered in 1 compressed Ptree for that location.
Using Alien Technology’s Fluidic Self-assembly (FSA) technology, clear layers are laminated into a cube, with a embedded nano-LED at each voxel.
.:.:.:.:..::….:. : …:…:: ..:
. . :: :.:…: :..:..::. .:: ..:.::..
The Ptree is transmitted to the cube, where the pattern is reconstructed (uncompress Ptree, display on the cube).
Each energized nano-sensor transmits a ping (location is triangulated from the ping). These locations are then translated to 3-dimensional coordinates at the display. The corresponding voxel on the display lights up. This is the expendable, one-time, cheap sensor version.
A more sophisticated CEASR device could sense and transmit the intensity levels, lighting up the display voxel with the same intensity.
Soldier sees replica of sensed
situation prior to entering space
==================================
\ CARRIER /

2. Spatial Data Pixel – a point in a space
Band – feature attribute of the pixels
Value – usually one byte (0~255)
Images have different numbers of bands
TM4/5: 7 bands (B, G, R, NIR, MIR, TIR, MIR2)
TM7: 8 bands (B, G, R, NIR, MIR, TIR, MIR2, PC)
TIFF: 3 bands (B, G, R)
Ground data: individual bands (Yield, Moisture, Nitrate level, Temperature, elevation…)
These notes contain NDSU confidential &
Proprietary material.
Patents pending on Ptree technology

3. RSI dataset example
RSI data can be viewed as collection of pixels. Each pixel has a value for each feature attribute
TIFF image
Yield Map
For example, the RSI dataset above has 320 rows and 320 columns of pixels (102,400 pixels) and 4 feature attributes (B,G,R,Y). The (B,G,R) feature bands are in the TIFF image and the Y feature is color coded in the Yield Map.

4. Spatial Data Formats Existing formats BSQ (Band Sequential)
BIL (Band Interleaved by Line)
BIP (Band Interleaved by Pixel)
New format
bSQ (bit Sequential)
BAND-1
( ) ( )
( ) ( )
BAND-2
( ) ( )
( ) ( )
BSQ format (2 files)
Band 1:
Band 2:

5. Spatial Data Formats (Cont.)
BAND-1
( ) ( )
( ) ( )
BAND-2
( ) ( )
( ) ( )
BSQ format (2 files)
Band 1:
Band 2:
BIL format (1 file)
BAND-1
( ) ( )
( ) ( )
BAND-2
( ) ( )
( ) ( )
BSQ format (2 files)
Band 1:
Band 2:
BIL format (1 file)
BIP format (1 file)

6. Spatial Data Formats (Cont.)
BAND-1
( ) ( )
( ) ( )
BAND-2
( ) ( )
( ) ( )
BSQ format (2 files)
Band 1:
Band 2:
BIL format (1 file)
BIP format (1 file)
bSQ format (16 files)
B11 B12 B13 B14 B15 B16 B17 B18 B21 B22 B23 B24 B25 B26 B27 B28

7. Spatial Formats
Split each band into eight separate files, one for each bit position.
Reasons of using bSQ format
Different bits contribute to the value differently.
bSQ format facilitates representation of a precision hierarchy (from 1 to 8 bit precision).
bSQ format facilitates creation of an efficient data structure, the P-tree, algebra and cube.
BSQ and bSQ are “tabular” formats
BSQ consist of a separate table for each feature band
bSQ consist of a separate table for each bit of each band
One can view it this way:
The data set is initially 1 relation or table, R(K1,..,Kk, A1, …, An) where K1,..,Kk are structure attributes and Ai are feature attributes.
Structure attributes of a 2-D image are X,Y coordinates of the pixels (rows).
Feature attributes are the bands, B,G,R, NIR, …
BSQ we separate each feature into a separate file and suppress the structure attributes altogether (assuming pixels are always arranged in raster order. (aka: Decomposition Storage Model (DSM), Copeland et al, SIGMOD85, )
bSQ, separate each bit of each feature into separate file (raster order assumption) (aka: Bit Transpose File (BTF) model, Wong et al, VLDB85, pp )

8. An example of PC-tree Peano or Z-ordering1
Given a bSQ file, Bij, (shown in spatial positions also) we create its basic PC-tree, Pij as follows. 55 16 8 15 3 4 1 55 1 3 1 16 15 16 8 1 4 1 4 3 4 4 1 1
Peano or Z-ordering
Pure (Pure-1/Pure-0) quadrant
Root Count
Level
Fan-out
QID (Quadrant ID)

9. Our example of PC-tree (again)
001 55 16 8 15 3 4 1
Level-3 1 2 3
Level-2 2 3
Level-1
111
Level-0
Peano or Z-ordering
Pure (Pure-1/Pure-0) quadrant
Root Count
Level
Fan-out
QID (Quadrant ID)
( 7, 1 )
( 111, 001 )

10. P-tree variation – PM-tree1
Peano Mask tree (PM-tree) uses mask instead of count.
1 denotes pure-1, 0 denotes pure-0 and m denotes mixed.
It provides an efficient way for ANDing.
Most compact form (all lossless)
Predicate Tree (1 iff predicate is true for quadrant)
E.g., Pure1-Tree (predicate: quad is all 1’s)

11. Depth-first Pure 1 path code
Ptree Algebra
And Or
Complement
Other
Ptree:
____________/ / \ \___________
/ ___ / \___ \
/ / \ \
____8__ _15__
/ / | \ / | \ \
//|\ //|\ //|\
Complement:
____8__ __1__
PM-tree1: m
______/ / \ \______
/ / \ \
/ / \ \
m m
/ / \ \ / / \ \
m m m 1
//|\ //|\ //|\
PM-tree2: m m
/ / \ \ m
//|\
0100
AND Result: m
________ / / \ \___
/ ____ / \ \
/ / \ \ m
/ | \ \
1 1 m m
//|\ //|\
Depth-first Pure 1 path code
&  RESULT
 0
 20
 21 
 231

12. Basic, Value and Tuple Ptrees
Basic Ptrees (a Pure1-Trees predicate-tree for target bit of target attribute)
e.g., P11, P12, …, P18, P21, …, P28, …, P71, …, P78
Target Attribute
Target Bit Position
Value Ptrees (predicate: quad is purely target value in target attribute)
e.g., P1, 5 = P1, = P11 AND P12’ AND P13
AND
Target Attribute
Target Value
Tuple Ptrees (predicate: quad is purely target tuple)
e.g., P(1, 2, 3) = P(001, 010, 111) = P1, AND P2, AND P3, 111
AND
Cube Ptrees (predicate: quad is purely in target cube (product of intervals)
e.g., P([13],, [0.2]) = (P1,1 OR P1,2 OR P1,3) AND (P3,0 OR P3,1 OR P3,2)
AND/OR

13. Creating Peano-Count-trees (PC-trees) from Spatial Relations
Take any spatial relation, R(K1,..,Kk, A1, A2, …, An) (Ki=structure, Ai=feature attributes).
Eg, Structure attributes of a 2-D image = X-Y coords, feature attribs = bands (e.g., B,G,R)
We create BSQ files from it by projection, Bi = R[Ai].
We create bSQ files from each of these BSQ files, Bi1, Bi2 , …, Bin
We create a Peano Tree, Pij, from each bSQ file, Bij
Peano trees (P-trees):
P-tree represents bSQ, BSQ, relational data in a recursive quadrant-by-quadrant,
lossless, compressed, datamining-ready format.
P-trees come in many forms
Count-trees (PC-trees);
Predicate-trees (P1, P0, PN1, PNZ, value-P-trees, tuple-P-trees, cube-P-trees)

14. PM= P1 xor NP0 PCT: .--- 55 ---. / / \ \ 16 8 15 16 // \ \ // \\
Other forms: Predicate Ptrees (1 if condition is true thruout the quadrant, else 0) (P1 and P0 are lossless)
PCT:
/ / \ \
// \ \ // \\
//|\ //|\ //|\
Pure1Tree (P1T)
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
Pure0Tree (P0T)
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
NotPure0(NP0T)
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
NotPure1(NP1T)
/ / \ \
// \ \ // \\
//|\ //|\ //|\
Vector Implemented Ptrees (Vector Ptrees have 1 row for each mixed quadrant, with that quadrant’s (qid, P-vector)
P1V
Qid PgVc []
[1]
[1.0]
[1.3]
[2]
[2.2]
P0V []
[1]
[1.0]
[1.3]
[2]
[2.2]
NP0V []
[1]
[2]
NP1V []
[1]
[2]
PeanoMixed (PM)
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
PMV
Qid PgVc []
[1]
[2]
Leaf-vectors always Can be omitted.
We may need Peano Mixed (PM) trees (e.g., distributed P-trees).
Note:
PM= P1 xor NP0

15. The Peano Cube of a relation (P-cube)
Suppose we have
R(K, A1, A2, A3 ) with each Ai a 2-bit number
Construct the cube of all tuple-P-trees for R
Form the cube of all RootCountP(t)
P-Cube of R
P-Cube(A1, A2, A3, rcP(A1,A2,A3))
(rootcounts form the feature attributes
and Ai’s form the structure attributes)
We can intervalize the RCs,
(eg, 4 intervals, [0,0], [1,8], [9,63], [64,),
labelled, 00, 01, 10 ,11 respectively).
Meta-P-trees of R, by forming basic Ptrees over the P-Cube of R
(1 feature attribute and, if we intervalize as above, 4 basic Ptrees).
- |HR|  |R| and = iff (A1, A2, A3 ) candidate key for R
- what is the relationship to the Haar wavelet low-pass tree? rc
P(0,0,3) rc
P(1,0,3) rc
P(2,0,3) rc
P(3,0,3) rc
P(0,0,2) rc
P(1,0,2) rc
P(2,0,2) rc
P(3,0,2) 1 5 rc
P(0,0,1) 11 rc
P(1,0,1) rc
P(2,0,1) rc
P(3,0,1) 14 5 3 11 rc
P313 5 5 17 11 10 rc
P312 rc
P(0,0,0) rc
P(1,0,0) rc
P(2,0,0) 00 rc
P(3,0,0) 10 rc
P311 rc
P323 10 01 rc
P322 rc
P(0,0,0) rc
P(1,1,0) 1 rc
P(2,1,0) rc
P(3,1,0) 01 01 rc
P321 rc
P333 1 A2 01 00 rc
P332 rc
P(0,2,0) rc
P(1,2,0) rc
P(2,2,0) rc
P(3,2,0) 1 11 10 00 00 01 10 rc
P331 10 00 00 01 10 rc
P(0,3,0) rc
P(1,3,0) rc
P(2,3,0) rc
P(3,3,0) 01 11 A3 00 00 01 10 11 A1

16. The P-tree Algebra (Complement, AND, OR, …)
Complement Tree = the Ptree for the bit-complement of the bSQ file) (‘)
We will use the “prime” notation.
PC-tree of a complement formed by purity-complementing each count.
Truth-tree of a complement: by bit-complementing leaves only.
Tree Complement = Complement of the tree - each tree entry is complemented. (“)
Not the same as the Complement Tree!
We will use”double prime” notation.
P1 = P0’
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
P = P1’
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
NP0 = NP1’
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
NP1=NP0’=P1”
/ / \ \
// \ \ // \\
//|\ //|\ //|\
P1V
Qid PgVc
[] [1] [1.0] [1.3] [2] [2.2] 1101
P0V
Qid PgVc
[] [1] [1.0] 0001 [1.3] 1101 [2] [2.2] 0010
NP0V
Qid PgVc []
[1]
[1.0] 1110
[1.3] 0010
[2]
[2.2] 1101
NP1V
Qid PgVc []
[1]
[1.0] 0001
[1.3] 1101
[2]
[2.2] 0010
P1”
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
P0”
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
NP0” = P0
/ / \ \
// \ \ // \ \
//|\ //|\ //|\
NP1” = P1
/ / \ \
// \ \ // \\
//|\ //|\ //|\
P1V”
Qid PgVc
[] [1] [1.0] [1.3] [2] [2.2] 0010
P0V”
Qid PgVc
[] [1] [1.0] 1110
[1.3] 1101 [2] [2.2] 1101
NP0V”
Qid PgVc []
[1]
[1.0] 0001
[1.3] 1101
[2]
[2.2] 1101
NP1V”
Qid PgVc []
[1]
[1.0] 0001
[1.3] 0010
[2]
[2.2] 1101

17. ANDing (for all Truth-trees, just AND bit-wise)
Pure1-quad-list method: For each operand, list the qids of the pure1 quad’s in depth-first order. Do one multi-cursor scan across the operand lists , for every pure1 quad common to all operands, install it in the result.
AND 
P1operand1
// \ \ // \\
//|\ //|\ //|\
P0operand1
// \ \ // \ \
//|\ //|\ //|\
NP0operand1 1
// \ \ // \\
//|\ //|\ //|\
NP1operand1 NP0’ 1
// \ \ // \\
//|\ //|\ //|\
bitwise
Depth first traversal using
1^1=1, 1^0=0, 0^0=0.
AND
P1operand2
/ / \ \
//|\
0100
P0op2 = P1’op2
/ / \ \
//|\
1011
NP0operand2 1
/ / \ \
1 11 1
//|\
0100
NP1operand2 NP0’ 1
/ / \ \
//|\
1011 =
P1op1^P1op2
// | \
//|\ //|\
P1op1^P0op2 = P1op1^P1’op2 0
// \ \ //\ \
//|\ //|\ //|\
NP0op1^NP0op2 1
// | \
//|\ //|\
NP0op1^NP0’op2 1
// \ \ /// \
//|\ //|\ //|\

18. Example1: One band, B1, with 3-bit precision
PNP0V11 P1V11 (combined into 1 table)
qid NP0 P1
[ ]
[01]
[10]
[01.00]
[01.11]
[10.10]
P12
qid NP0 P1
[ ]
[10]
[10.11] 0111
P13
qid NP0 P1
[ ]
[01]
[10]
[01.11] 0110
[10.00] 1000
Redundant! Since, at leaf, NP0=P

19. Data Mining in Genomics
There is (will be?) an explosion of gene expression data.
Current emphasis is on extracting meaningful information from huge raw data sets.
Methods employed are Clustering and Classification
Microarray data is most often represented as a relation G(Gid, T1, T2, ., Tn) where Gid is the gene identifier; T1…. Tn are the various treatments (or conditions) and the data values are gene expression levels. We will call this the " Gene Table”.
Currently, data-mining techniques concentrate on the Gene table, G(Gid, T1, T2, ., Tn) - specifically, on finding clusters of genes that exhibit similar expression patterns under selected treatments (clustering the gene table).

20. Gene Table …. G4 G3 G2 G1 T4 T3 T2 T1 Treatmt-ID Gene-ID .
P13 [ ]
qid NP0 P1
[ ]
[01]
[10]
[01.11] 0110
[10.00] 1000
Using the Universal Relation approach to mining across different Microarray datasets, one can use a consistent Gene-id. Each Microarray will be embedded in a subquadrant. Therefore the data will be sparse and can be handled by Vector Implemented P-trees in which the prefix of the subquadrant can be listed only once:

21. Example1: ANDing to get rc P1(6)
BpQid NP P1
11[ ]
12[ ]
13[ ]
11[01]
13[01]
11[01.00]
11[01.11]
13[01.11]
11[10]
12[10]
13[10]
13[10.00]
11[10.10]
12[10.11]
P1(6) = P1(110) = P111^P112^P013 = P11^P12^NP0”13
PM1(110)= P1(110) xor NP01(110) = P11^P12^NP0”13 xor NP011^NP012^P1”13
At [ ]: CNT[ ]=1-cnt*4level =1*42=16 since P1(110)[ ] = 1001^1000^1000=1000
PM1(110)[ ] = P11 ^ P12 ^NP0”13 xor NP011^NP012^P1”13
=1001^1000^ xor ^ 1010 ^1110 = 0010
At [10]: CNT[10]= 1-cnt*4level=0*41=0 since P1(110)[10]= 1101^1110^0001=0000
PM1(110)[10] = P11^P 12 ^NP0”13 xor NP011^NP012^P1”13
=1101^1110^0001 xor 1111^1111^1001= 0000 xor 1001=1001
At [10.00]: CNT=[10.00]1-cnt*4level=3*40=3 since P1(110)[10.00]= 1111^1111^0111=0111
At [10.11]: CNT=[10.11]1-cnt*4level=3*40=3 since P1(110)[10.11]= 1111^0111^1111=0111
Thus, rcP1(6) = = 22
[10] only mixed child
[10.00], [10.11] mixed children
For P(p)= P( , … , ): At each [..]
1. swap and take bit comp of each [..]NP0V [..]P1V pair corresponding to 0-bits.
2. AND the resulting vector-pairs Result: [..]NP0V(p)[..]P1V(p). To get PMV(p) for the next level, xor the two vectors.

22. ANDing in the NP0V-P1V Vector-Pair Format
For P(p)= P( , … , ) (previous example, P1(6) at qid[ ] )
At each [..]
1. swap and complement each [..]NP0V [..]P1V pair corresponding to 0-bits. Result denoted with *
2. AND the resulting vector-pairs Result: [..]NP0V(p)[..]P1V(p). To get PMV(p) for the next level,
3. xor the two vectors to get [..]PMV(p)
pos NP0V P1V - …
bit NP0V* P1V* - …
_____________________
NP0V P1V p
PMV(p) =

23. Striping P-trees? BpQid NP0 P1 00 11[01.00] 1110 13[10.00] 1000
11[01.00]
13[10.00]
BpQid NP0 P1 01
11[01]
13[01]
BpQid NP P1
11[ ]
12[ ]
13[ ]
11[01]
13[01]
11[01.00]
11[01.11]
13[01.11]
11[10]
12[10]
13[10]
13[10.00]
11[10.10]
12[10.11]
BpQid NP0 P1 C
11[ ]
12[ ]
13[ ]
BpQid NP0 P1 10
11[10]
11[10.10]
12[10]
13[10]
BpQid NP0 P1 11
11[01.11]
12[10.11]
13[01.11]
Assume 5-computer cluster; NodeC, Node00, Node01, Node10, Node11
Send to Nij if qid ends in ij:
P11(110) = P111^P112^P013 = P11^P12^NP0” PM1(110) = P11^P12^NP0”13 xor NP011^NP012^P1”13
At NC: CNT[ ]=1-cnt*4level =1*42= since P1(110)[ ]= 1001^1000^1000=1000
PM1(110)[ ] =1001^1000^1000 xor 1111^1010^1110= 0010
At N10: CNT[10]= 1-cnt*4level=0*41=0 since P1(110)[10]= 1101^1110^0001=0000
PM1(110)[10] = 1101^1110^0001 xor 1111^1111^1001= 0000 xor 1001=1001
At N00: CNT=[10.00]1-cnt*4level=3*40=3 since P1(110)[10.00]= 1111^1111^0111=0111
At N11: CNT=[10.11]1-cnt*4level=3*40=3 since P1(110)[10.11]= 1111^0111^1111=0111
Every node sends accumulated CNT to C, where rcP1(6) = = 22 calculated.

24. Striping P-trees? P11 P12 P13 qid NP0 P1 [ ] 1111 1001 [01] 1011 0010
[ ]
[01]
[10]
[01.00] 1110
[01.11] 0010
[10.10] 1101
qid NP0 P1
[ ]
[10]
[10.11] 0111
qid NP0 P1
[ ]
[01]
[10]
[01.11] 0110
[10.00] 1000
Striping P-trees?
Bp qid NP0 P
Bp qid NP0 P
11[01]
11[01.00] 1110
11[01.11] 0010
13[01]
13[01.11] 0110
Bp qid NP0 P C
11[ ]
12[ ]
13[ ]
Bp qid NP0 P
11[10]
11[10.10] 1101
12[10]
12[10.11] 0111
13[10]
13[10.00] 1000
Bp qid NP0 P
Alternatively, Send to Nodeij if qid starts with qid segment, ij. Is this better? How would the AND code be revised? AND performance?
OR: Send to Nodeij if the largest qid segment divisible by p is ij eg if p=4: [0]->0; [0.3]->0; [0.3.2]->0; [ ]->2; [ ]->2; [ ]->2; [ ]->2; [ ]->2; [ ]->1 etc.
Similar to fanout  4. Implement by multicasting externally only every 4th segment. More generally, choose any increasing sequence, p=(p1..pL), define x p = {max pi  x},
then multicast [s1.s2…sk] --> Node k p

25. Node-C Node-00 Node-01 Node-10 Node-11 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0
B11
Example 1 (bottom-up)
Bp qid NP P1
11[00.00]
Band, B1, with 3-bit values
Bp qid NP0 P1
11[00.00]
11[00.01]
Bp qid NP0 P1
11[00.00]
11[00.01]
11[00.10]
Bp qid NP P1
11[00.00]
11[00.01]
11[00.10]
11[00.11]
Bp qid NP P1
11[00]
Node-C
Bp qid NP P1
11[] __ 10__
This ends the possibility
of a larger pure1 quad.
So 00 can be installed in
parent as a pure1.
Bp qid NP P1
11[00]
11[01.00]
Node-00
Bp qid NP P1
11[01.00]
Bp qid NP P1
11[01.00]
11[01.01]
Mixed leaf quad sent.
Also ends possibility
parent is pure so it &
all siblings are installed
as bits in parent.
Node-01
Bp qid NP P1
11[01]
11[01.10]
Node-10
Bp qid NP P1
11[01.11]
Mixed leaf quad sent.
Ends parent so install bits in grandparent also
Node-11
Bp qid NP P1
11[01.11]

26. Node-C Node-00 Node-01 Node-10 Node-11 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0
B11
Example 1 (bottom-up)
Bp qid NP P1
11[10.00]
Band, B1, with 3-bit values
Bp qid NP P1
11[10.00]
11[10.01]
Bp qid NP P1
11[10.00]
11[10.01]
11[10.10]
11[10.11]
Ends the possibility
of a larger pure1 quad.
All can be installed in
parent/grandparent
as a 1-bit.
10.10 can be installed.
Bp qid NP P1
11[11.00]
11[11.01]
11[11.10]
11[11.11]
Node-C
Bp qid NP P1
11[]
Bp qid NP P1
11[11]
Node-00
Bp qid NP P1
11[01.00]
Node-01
Bp qid NP P1
11[01]
Ends quad-11.
All can be installed in
Parent as a 1-bit.
Node-10
Bp qid NP P1
11[10.10]
11[10]
Node-11
Bp qid NP P1
11[01.11]
Bottom-up bottom-line: Since it is better to use 2-D than 3-D (higher compression), it should be better to use 1-D than 2-D? This should be investigated.

27. X, Y, B1, B2
Example2 B1
B11
B12
B13 B2
B21
B22
B23
Example2

28. Example2: Striping Raster order Peano order __PNP0V_ __P1V__
Band
bit-pos
[ ] === === === ===
Raster order
Peano order
X, Y, B1, B2
X, Y, B11B12B13B21B22B23
x1y1x2y2x3y3 B11B12B13B21B22B23
OR for PNP0
AND for P1
00_PNP0V__ __P1V__
Send B21B22B23 to Node00
01_PNP0V__ __P1V__
Send B11B13 B22B23 to Node01
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Purity Template
[ ]
10_PNP0V__ __P1V__
Send B12B13 B21B22B23 to Node10
11_PNP0V__ __P1V__
Send nothing to Node11

29. Example2: striping at Node 00
Bp qid NP0 P
21[ ]
22[ ]
23[ ]
PurityTemplate [00]
11[ ]
23[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
_PNP0V__ __P1___
Band
bit-pos
[00 ] === === === ===
x1y1x2y2x3y3B11B12B13 B21B22B23
_PNP0V__ __P1V__
Send nothing to Node00
_PNP0V__ __P1V__
Send [ ]B21B22 to Node01
_PNP0V__ __P1V__
Pages on disk
Send nothing to Node10
Bp qid NP0 P
12[ ]
13[ ]
21[ ]
21[ ]
22[ ]
22[ ]
23[ ]
23[ ]
23[ ]
11[ ]
_PNP0V__ __P1V__
Send nothing to Node11 P1
Band
bit-pos 13
[01.00 ] == 11 10 00 P1
Band
bit-pos
[10.00 ] == ===
11 011
10 110
x1y1x2y2x3y3 B11 B23
x1y1x2y2x3y3 B12B12 B23B23B23
From [01 ]
From [10 ]
To [01 ]

30. Example2: striping at Node 01
To [00 ]
_PNP0V__ __P1___
Band
bit-pos
[01 ] === === === ===
Bp qid NP0 P
11[ ]
13[ ]
22[ ]
23[ ]
PurityTemplate [01]
21[ ]
22[ ]
23[ ]
x1y1x2y2x3y3 B11 B13 B22B23
_PNP0V__ __P1V__
Send [01]B11B23 to Node00
_PNP0V__ __P1V__
Send nothing to Node01
_PNP0V__ __P1V__
Send nothing to Node10
Pages on disk
_PNP0V__ __P1V__
Bp qid NP0 P
11[ ]
Send nothing to Node11
Bp qid NP0 P
13[ ]
Bp qid NP0 P
21[ ] P1
Band
bit-pos 12
[00.01 ] == 10 01 P1
Band 2
bit-pos 3
[10.01 ] == 1
x1y1x2y2x3y B21B22
x1y1x2y2x3y B23
Bp qid NP0 P
22[ ]
22[ ]
Bp qid NP0 P
23[ ]
23[ ]
From [00 ]
From [10 ]

31. Example2: striping at Node 10
To [00 ]
_PNP0V__ __P1___
Band
bit-pos
[10 ] === === === ===
To[01 ]
Bp qid NP0 P
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
PurityTemplate [10]
x1y1x2y2x3y3 B12B13B21B22B23
_PNP0V__ __P1V__
Send [10]B13B21B23 to Node00
_PNP0V__ __P1V__
Send [10] B23 to Node01
_PNP0V__ __P1V__
Send nothing to Node10
Pages on disk
_PNP0V__ __P1V__
Bp qid NP0 P
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Send [10]B12B21B22 to Node11
To [11 ]

32. Example2: striping at Node11
Bp qid NP0 P
12[ ] 01
22[ ] 10
23[ ] 01
Pages on disk
Bp qid NP0 P
12[ ] 01 P1
Band
bit-pos 223
[10.11 ] ===
010
101
x1y1x2y2x3y3 B B21B22
Bp qid NP0 P
22[ ] 10
Bp qid NP0 P
23[ ] 01
From [10 ]

33. Example2.1 AND at NodeC or [ ]
NP0-pattern
NP0 P1
xxxx
prime
xxxx
prime
xxxx
prime
Example2.1 AND at NodeC or [ ]
RC(P 101,010) = P11^ P’12^ P13^ P’21^ P22^ P’23
[]NP0
1111
0111
------AND
Sum= 8 so far.
Invocation= [ ] 101,010 send to Nodes 01, 10
P1-pattern
NP0 P1
xxxx
prime
xxxx
prime
xxxx
prime
[]P1
1011
0101
0001
------AND
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

34. Example2.1 AND at Node01 [ ] 101,010 received Sent to Node00
NP0-pattern
NP0 P1
xxxx
prime
xxxx
prime
xxxx
prime
[01] NP0 12 21
AND------
1000
Invocation= [01] 101,010
Sent to Node00
[ ] 101,010 received
P1-pattern
NP0 P1
xxxx
prime
xxxx
prime
xxxx
prime
[01] P1 12 21
AND------
0000
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

35. Example2.1 AND at Node10 [ ] 101,010 received Sent nowhere (no mixed)
NP0-pattern
NP0 P1
11 xxxx
prime
13 xxxx
prime
22 xxxx
prime
[10] NP0 11
AND------
0000
[ ] 101,010 received
Invocation= [10] 101,010
Sent nowhere (no mixed)
P1-pattern
NP0 P1
xxxx
12 prime
xxxx
21 prime
xxxx
23 prime
[10] P1 11 12 13 21 22 23
AND------
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

36. Example2.1 AND at Node00 [01] 101,010 received
Sum=1, sent to NodeC gives a sum total of = 9
P1-pattern P1
11 xxxx
12 prime
13 xxxx
21 prime
22 xxxx
23 prime
[01.00] P1 12 13 21 22
AND------
0100
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

37. Example2.2 AND at NodeC or [ ]
P1-pattern
NP0 P1
xxxx
prime
prime
xxxx
prime
xxxx
NP0-pattern
xxxx
prime
prime
xxxx
prime
xxxx
Example2.2 AND at NodeC or [ ]
RC(P 100,101) = P11^ P’12^ P’13^ P21^ P’22^ P23
[]NP0
------AND
0010
Sum= 0 so far.
Invocation= [ ] 100, 101 send to Node 10
[]P1
------AND
0000
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

38. Example2.2 AND at Node10 [ ] 100,101 received Sent to Node 11
P1-pattern
NP0 P1
xxxx
prime
prime
xxxx
prime
xxxx
NP0-pattern
xxxx
prime
prime
xxxx
prime
xxxx
[10] NP0 11 12 13 21 22 23
AND------
0001
[ ] 100,101 received
Invocation= [10] 100, 101
Sent to Node 11
[10] P1 11 12 13 21 22 23
AND------
0000
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

39. Example2.2 AND at Node11 [10] 100,101 received
Sum=1, sent to NodeC gives a sum total of 1
[10] P1 12 13 21
AND------ 01
Disk 10 PT[10]
Disk 01 PT[01]
Disk 00 PT[00]
Disk C PT[ ]
Bp qid NP0 P1
12[ ]
23[ ]
22[ ]
Disk 11
Bp qid NP0 P1
11[ ]
Bp qid NP P1
11[ ]
Bp qid NP0 P1 C
11[ ]
12[ ]
13[ ]
21[ ]
22[ ]
23[ ]
Bp qid NP P1
12[ ]
Bp qid NP0 P1
12[ ]
Bp qid NP P1
13[ ]
Bp qid NP P1
13[ ]
Bp qid NP0 P1
13[ ]
Bp qid NP P1
21[ ]
Bp qid NP P1
21[ ]
Bp qid NP0 P1
21[ ]
21[ ]
Bp qid NP P1
22[ ]
22[ ]
Bp qid NP P1
22[ ]
Bp qid NP0 P1
23[ ]
23[01.00]
23[10.00]
Bp qid NP0 P1
22[ ]
22[ ]
Bp qid NP P1
23[ ]
23[ ]
Bp qid NP P1
23[ ]

40. Example2, bottom-up Peano order Bp qid NP0 P1 11[00.00] 1111
11[00.00]
12[00.00]
13[00.00]
21[00.00]
22[00.00]
23[00.00]
Example2, bottom-up
Peano order
x1y1x2y2x3y3 B11B12B13B21B22B23

41. Mixed quads (can be sent to node01)
Bp qid NP0 P1
11[00.00]
11[00.01]
12[00.00]
12[00.01]
13[00.00]
13[00.01]
21[00.00]
21[00.01]
22[00.00]
22[00.01]
23[00.00]
23[00.01]
Example2, bottom-up
Peano order
x1y1x2y2x3y3 B11B12B13B21B22B23
Mixed quads (can be sent to node01)
Bp qid NP0 P1
21[00.01]
22[00.01]

42. Mixed quads (sent to node00)
Bp qid NP0 P1
11[00.00]
11[00.01]
11[00.10]
12[00.00]
12[00.01]
12[00.10]
13[00.00]
13[00.01]
13[00.10]
21[00.00]
21[00.01]
21[00.10]
22[00.00]
22[00.01]
22[00.10]
23[00.00]
23[00.01]
23[00.10]
Example2, bottom-up
Peano order
x1y1x2y2x3y3 B11B12B13B21B22B23
Mixed quads (sent to node00)
Bp qid NP0 P at 00
23[00]
Bp qid NP0 P at 01
21[00.01]
22[00.01]

43. 00 quads that are pure are:
Bp qid NP0 P1
11[00.00]
11[00.01]
11[00.10]
11[00.11]
12[00.00]
12[00.01]
12[00.10]
12[00.11]
13[00.00]
13[00.01]
13[00.10]
13[00.11]
21[00.00]
21[00.01]
21[00.10]
21[00.11]
22[00.00]
22[00.01]
22[00.10]
22[00.11]
23[00.00]
23[00.01]
23[00.10]
23[00.11]
Example2, bottom-up
Peano order
x1y1x2y2x3y3 B11B12B13B21B22B23
00 quads that are pure are:
Bp qid NP0 P1
11[00]
12[00]
13[00]
At 00
Bp qid NP0 P1
23[00]
At 01
Bp qid NP0 P1
21[00.01]
22[00.01]