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The Age of Infinite Storage or the age of data mining

The Age of Infinite Storage or the age of data mining. William Perrizo Dept of Computer Science North Dakota State Univ. How much information is there?. Yotta Zetta Exa Peta Tera Giga Mega Kilo. Everything! Recorded. Soon everything can be recorded and indexed.

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The Age of Infinite Storage or the age of data mining

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  1. The Age ofInfinite Storageor the age of data mining William Perrizo Dept of Computer Science North Dakota State Univ.

  2. How much information is there? Yotta Zetta Exa Peta Tera Giga Mega Kilo Everything! Recorded • Soon everything can be recorded and indexed. • Most bytes will never be seen by humans. • Data summarization, trend detection, anomaly detection, data mining, are key technologies All Books MultiMedia All books (words) .Movie A Photo A Book 10-24 Yocto, 10-21 zepto, 10-18 atto, 10-15 femto, 10-12 pico, 10-9 nano, 10-6 micro, 10-3 milli

  3. E.g., A recent Purchase Order Company: NDSU Date: 8/7/03 System Board: Intel D865 GBFL system board w/LAN 800mhz FSB Processor: Intel Pentium 4 2.6 GHz Hard Drives: 4 x 250 GB IDE (total = 1 TB) Controller: Onboard IDE Controller 2nd IDE Controller: Video: Integrated Diskette Drive: 1.44 MB Memory: 4 GB 400 mhz memory CD/DVD Drive: DVD/CDRW Sound: Integrated AC97 Audio w/Soundmax Case: Performance Minitower ATX w/300 Watt PS Keyboard: Microsoft 104 Internet keyboard Mouse: Microsoft Intellimouse Optical Operating System: none Network Cards: Integrated Intel 10/100 Ethernet w/D845GEBV2L board Price:$2,899.00

  4. Trying to fill a terabyte in a year

  5. The TerabyteHow Will We Find Anything? • 80% of data is personal/individual • 20% is Corporate, Governmental SQL ++DBMS

  6. TIFF image Yield Map What are the relationships between the color intensities and yield? An intuitive hypothesis: hi_green^low_redhi_yield. We confirm it, through data analysis. The stronger rule, hi_NIR^low_redhi_yieldis un-intuitive. The hypothesis itself came from Data Mining. RSI Data Mining

  7. Grasshopper Infestation Prediction • Grasshopper caused significant economic loss each year. • Early infestation prediction is key to damage control. Association rule mining on remotely sensed imagery holds promise to achieve early detection.

  8. Gene1 Gene2, Gene3 Gene4, Gene 5, Gene6 Gene7, Gene8 Gene9 Clustering ARM Gene4 Gene7 Gene1 Gene3 Gene5 Gene2 Gene9 Gene6 Gene8 Gene Regulation Pathway Discovery • Association Rule Mining (ARM) on that cluster may discover the relationships among the genes • Clustering can isolate genes in a metabolic pathway.

  9. Sensor Network Data Mining • Micro and Nano scale sensors are being developed for sensing • Biological agents • Chemical agents • Motion detection • coatings deterioration • RF-tagging of inventory • Structural materials fatigue • The data must be mined for it’s information.

  10. Network Security Application(Network security through Vertical Structured data) • Network layers do their own partitioning • Packets, frames,… (independent of data structure – e.g., record structure) • Data privacy is compromised when the horizontal (stream) message content is eavesdropped upon at the reassembled level (in network) • A standard solution is to host-encrypt horizontal structure so that any network reassembled message is meaningless. • Alternative: Vertically structure data. • Send one bit slice per packet • Send intra-message packets separately • The message is only meaningful after destination demux-ing • Only bit-slice holding info is high-order one, so encrypt it!

  11. Situation space ================================== \ CARRIER / Sensor Network Application: CubE for Active Situation Replication (CEASR) Nano-sensors dropped into the Situation space Drop or mortar “smart dust” sensors into the situation space to detect armour, chemical, biological, thermal…. Wherever a threshold level is senseda ping is sent for that location to a 3-D viewing device, which is constructed using .:.:.:.:..::….:. : …:…:: ..: . . :: :.:…: :..:..::. .:: ..:.::.. .:.:.:.:..::….:. : …:…:: ..: . . :: :.:…: :..:..::. .:: ..:.::.. .:.:.:.:..::….:. : …:…:: ..: . . :: :.:…: :..:..::. .:: ..:.::.. Alien Technology’s Fluidic Self-assembly (FSA) process (clear plastic layers with embedded nano-LEDs, laminated together into a viewing cube. In all these applications, standard horizontal data structures will be inadequate for the speed and accuracy required. Soldier sees replica of sensed situation prior to entering space

  12. Our Approach • Vertical, compressed data structures, processed horizontally • Use Predicate-trees (Ptrees1) to address the curses of scalability and curse of dimensionality. • Next slides illustrates the construction of a set of BASIC P-TREE representation of data. 1 Ptree Technology is patent pending by North Dakota State University

  13. Current practice: Sets of horizontal records Ptrees: vertically project each attribute; R( A1 A2 A3 A4) R[A1] R[A2] R[A3] R[A4] 010 111 110 001 011 111 110 000 010 110 101 001 010 111 101 111 101 010 001 100 010 010 001 101 111 000 001 100 111 000 001 100 010 111 110 001 011 111 110 000 010 110 101 001 010 111 101 111 101 010 001 100 010 010 001 101 111 000 001 100 111 000 001 100 Horizontal structures (records) Scanned vertically R11 R12 R13 R21 R22 R23 R31 R32 R33 R41 R42 R43 0 1 0 1 1 1 1 1 0 0 0 1 0 1 1 1 1 1 1 1 0 0 0 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 1 0 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 1 0 1 0 0 0 0 1 0 01 0 1 0 0 1 01 1. Whole is pure1? false 0 2. 1st half pure1? false  0 0 0 0 0 1 01 P11 P12 P13 P21 P22 P23 P31 P32 P33 P41 P42 P43 3. 2nd half pure1? false  0 0 0 0 0 1 0 0 10 01 0 0 0 1 0 0 0 0 0 0 0 1 01 10 0 0 0 0 1 10 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 10 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 4. 1st half of 2nd ? false  0 0 0 1 0 1 01 5. 2nd half of 2nd half? true1 0 1 0 6. 1sthalf of 1st of 2nd? true1 Eg, to count, 111 000 001 100s, use “pure111000001100”: 0 23-level P11^P12^P13^P’21^P’22^P’23^P’31^P’32^P33^P41^P’42^P’43 = 0 0 22-level=2 01 21-level 7. 2ndhalf of 1st of 2nd false0 vertically project each bit pos of each attribute; processed vertically (vertical scans) compress each bit slice into a basic Ptree; Horizontally AND basic Ptrees R11 0 0 0 0 1 0 1 1 The 1-Dimensional Ptree, P11, of R11 built by recording the truth of predicate “pure 1” recursively on halves, until purity is reached. But it is pure (pure0) so this branch ends

  14. 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0 1 1 2-D Ptrees Node = 1 iff that quadrant is pure 1-bits An image bit-slice. Compress it into a quadrant tree using Peano order.

  15. 1=001 55 level-3 (pure=43) 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 2 3 1 16 0 0 0 15 1 16 level-2 2 0 3 0 0 4 1 1 0 0 4 0 4 0 3 1 4 level-1 3 7=111 1 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0 1 1 level-0 2 . 2 . 3 Quad-id ( 7, 1 ) Coordinates 10.10.11 ( 111, 001 ) Count-trees: (counts are needed in data miningPredicate-trees are very compressed and can be used to produce counts quickly. However, Count-trees are an alternative - each inode counts 1s in that quadrant):

  16. Hardware ANDer for Ptrees? P-4, Itanium, Opteron, G5 card? Active network switch? Ptree 1 Ptree 2 AND result OR result Ptree AND is faster than bit-by-bit AND due to shortcuts (any pure0 operand node means result node is pure0.) (any pure1, copy subtree of the other operand to the result) More operands means greater benefit (more pure0 nodes).

  17. DATA MINING DATA Data Integration Language DIL Ptree (Predicates) Query Language PQL Data Integration Interface Data Mining Interface DataMIME™and theDIL PQL InterfaceData Mining for Information Materialization and Extraction - no noise Internet Ptree Repository lossless, compressed, distributed, vertical

  18. Basic Ptrees (a Pure1-Trees predicate-tree for target bit of target attribute) e.g., P11, P12, …, P18, P21, …, P28, …, P71, …, P78 AND Target Attribute Target Bit Position Value Ptrees (predicate: quad is purely target value in target attribute) e.g., P1, 5 = P1, 101 = 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, 001 AND P2, 010 AND P3, 111 AND/OR 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) Basic, Value and Tuple Ptrees

  19. 3-Dimensional Ptrees(e.g., for the CEASR sensor network

  20. Generalizing Peano compression to any table with numeric attributes. Raster Sorting: Attributes 1st Bit position 2nd Peano Sorting: Bit position 1st Attributes 2nd Same relation, showing values in binary Unsorted relation

  21. Unsorted Generalized Raster Generalized Peano crop adult spam function mushroom Generalize Peano Sorting can increase classification speed Classification speed improvement (sample-based KNN classifier) Using 5 UCI Machine Learning Repository data sets 120 100 80 Time in Seconds 60 40 20 0

  22. Association of Computing Machinery KDD-Cup-022-yeast gene deletion problem – Ptree solution NDSU Team

  23. Ptree for classification(AKA: regression, prediction, case-based reasoning,… • Whenever predictions or classifications need to be made based on past behavior (training data) • Precision Ag (Predict yield, id grasshoppers…) • Homeland Security (predict ill-intent) • Materials (metal fatigue, chemical reaction) • Gene annotation (Genetic disease prediction) • Cancers • Alzeheimers • Diabetes • Other • Mineral exploration • Etc. • Current practice: scan training data (horizontal records) • Our approach AND Ptrees (vertical, compressed, flexible

  24. Unclassified sample ( a5 a6 a1’ a2’ a3’ a4’ ) = Relevant attributes 0 0 0 0 0 0 a5 a6 C a1’a2’a3’a4’ dis 3NNs C=1 wins! d=2, don’t replace d=4, don’t replace d=4, don’t replace d=2, don’t replace d=3, don’t replace d=3, don’t replace d=3, don’t replace d=2, don’t replace d=2, don’t replace d=2, don’t replace d=3, don’t replace d=2, don’t replace d=2, don’t replace 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d=1, replace 0 0 0 0 0 0 Class label=C 3 nearest nbrs prediction using Hamming distance = # of mismatches t1 t2 0 0 1 0 1 1 0 2 t1 t3 0 0 1 0 1 0 0 1 t1 t5 0 0 1 0 1 0 1 2 t5 t3 0 0 0 0 1 0 0 1 0 1 d1 d2 a1 a2 a3 a4 a5 a6 a7 a8 a9 C a1'a2'a3'a4'a5'a6'a7'a8'a9‘ C' t1 t2 1 0 1 0 0 0 1 1 0 1 0 1 1 0 1 1 0 0 0 1 t1 t3 1 0 1 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 1 t1 t5 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1 1 0 0 t1 t6 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 0 0 1 0 t2 t1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 1 0 1 t2 t7 0 1 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 0 0 t3 t1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 1 t3 t2 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 1 0 0 0 1 t3 t3 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 0 0 0 1 1 t3 t5 0 1 0 0 1 0 0 0 1 1 0 1 0 1 0 0 1 1 0 0 t5 t1 0 1 0 1 0 0 1 1 0 0 1 0 1 0 0 0 1 1 0 1 t5 t3 0 1 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1 t5 t5 0 1 0 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 t5 t7 0 1 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 t6 t1 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 1 0 1 t7 t2 0 0 1 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 t7 t5 0 0 1 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0

  25. To find all training pts within dis=2 (fairer prediction?) requires another scan. d=2, include it also d=4, don’t include d=4, don’t include d=2, include it also d=3, don’t include d=2, include it also d=3, don’t include d=2, include it also d=2, include it also d=3, don’t replace d=2, include it also d=2, include it also d=2, include it also d=3, don’t include 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d=1, already have d=1, already have d=2, already have 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 a5 a6 C a1’a2’a3’a4’ d t1 t2 0 0 1 0 1 1 0 2 t1 t3 0 0 1 0 1 0 0 1 t5 t3 0 0 0 0 1 0 0 1 0 1 d1 d2 a1 a2 a3 a4 a5 a6 a7 a8 a9 C a1'a2'a3'a4'a5'a6'a7'a8'a9‘ C' t1 t2 1 0 1 0 0 0 1 1 0 1 0 1 1 0 1 1 0 0 0 1 t1 t3 1 0 1 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 1 t1 t5 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 1 1 0 0 t1 t6 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 0 0 1 0 t2 t1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 0 0 1 1 0 1 t2 t7 0 1 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 0 0 t3 t1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 1 0 1 t3 t2 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 1 0 0 0 1 t3 t3 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 0 0 0 1 1 t3 t5 0 1 0 0 1 0 0 0 1 1 0 1 0 1 0 0 1 1 0 0 t5 t1 0 1 0 1 0 0 1 1 0 0 1 0 1 0 0 0 1 1 0 1 t5 t3 0 1 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 1 t5 t5 0 1 0 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 t5 t7 0 1 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 t6 t1 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 1 0 1 t7 t2 0 0 1 1 0 0 1 1 0 0 0 1 1 0 1 1 0 0 0 1 t7 t5 0 0 1 1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0

  26. (a5 a6a1’a2’a3’a4’) a5^a6^a1’^a2’^a3’^a4’ a5^a6^a1’^a2’^a3’^a4’ a5^a6^a1’^a2’^a3’^a4’ a5^a6^a1’^a2’^a3’^a4’ a5^a6^a1’^a2’^a3’^a4’ a5^a6^a1’^a2’^a3’^a4’ 0 1 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 OR Using Ptree:Find all training pts  dis 1 from a=a5 a6 a1’a2’a3’a4’ = (000000) The 1-ring: given by 1-bits in the Ptree, P, constructed: C=1 vote count = root count of P^C. a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 C=0 vote count = root count of P^C. (never need to know which tuples voted) a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 C 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 P 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 a4 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a7 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a8 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a9 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 0 0 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 a5‘ 1 1 0 1 0 0 0 1 1 0 0 1 0 0 0 1 0 a6‘ 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 a7‘ 0 0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 a8‘ 0 0 1 0 1 1 1 0 0 1 1 0 1 1 1 0 1 a9‘ 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 d1 d2 t1 t2 t1 t3 t1 t5 t1 t6 t2 t1 t2 t7 t3 t1 t3 t2 t3 t3 t3 t5 t5 t1 t5 t3 t5 t5 t5 t7 t6 t1 t7 t2 t7 t5 C' 1 1 0 0 1 0 1 1 1 0 1 1 0 0 1 1 0 a2 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 a3 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1

  27. a’s 2-ring? a=a5 a6 a1’a2’a3’a4’ = (000000) 0 1 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 For each of the following Ptrees, a 1-bit corresponds to a training point in a’s 2-ring: Pa5a6a1’a2‘a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6a1‘a2’a3‘a4‘Pa5a6a1‘a2‘a3’a4‘Pa5a6a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3‘a4’ 1st line first: a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 d1 d2 t1 t2 t1 t3 t1 t5 t1 t6 t2 t1 t2 t7 t3 t1 t3 t2 t3 t3 t3 t5 t5 t1 t5 t3 t5 t5 t5 t7 t6 t1 t7 t2 t7 t5

  28. 0 1 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 For each of the following Ptrees, a 1-bit corresponds to a training point in a’s 2-ring: Pa5a6a1’a2‘a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6a1‘a2’a3‘a4‘Pa5a6a1‘a2‘a3’a4‘Pa5a6a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3‘a4’ 2nd line: a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 d1 d2 t1 t2 t1 t3 t1 t5 t1 t6 t2 t1 t2 t7 t3 t1 t3 t2 t3 t3 t3 t5 t5 t1 t5 t3 t5 t5 t5 t7 t6 t1 t7 t2 t7 t5

  29. 0 1 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 For each of the following Ptrees, a 1-bit corresponds to a training point in a’s 2-ring: Pa5a6a1’a2‘a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6a1‘a2’a3‘a4‘Pa5a6a1‘a2‘a3’a4‘Pa5a6a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3‘a4’ 3rd line: a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 d1 d2 t1 t2 t1 t3 t1 t5 t1 t6 t2 t1 t2 t7 t3 t1 t3 t2 t3 t3 t3 t5 t5 t1 t5 t3 t5 t5 t5 t7 t6 t1 t7 t2 t7 t5

  30. 0 1 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 a4‘ 1 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 0 a3‘ 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 0 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 For each of the following Ptrees, a 1-bit corresponds to a training point in a’s 2-ring: Pa5a6a1’a2‘a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6a1‘a2’a3‘a4‘Pa5a6a1‘a2‘a3’a4‘Pa5a6a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3‘a4’ 4th line: a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 d1 d2 t1 t2 t1 t3 t1 t5 t1 t6 t2 t1 t2 t7 t3 t1 t3 t2 t3 t3 t3 t5 t5 t1 t5 t3 t5 t5 t5 t7 t6 t1 t7 t2 t7 t5

  31. 0 1 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a2‘ 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 0 a1‘ 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 1 1 a6 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 a5 1 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 For each of the following Ptrees, a 1-bit corresponds to a training point in a’s 2-ring: Pa5a6a1’a2‘a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘ Pa5a6a1‘a2’a3‘a4‘Pa5a6a1‘a2‘a3’a4‘Pa5a6a1‘a2‘a3‘a4’ Pa5a6 a1‘a2’a3‘a4‘Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3’a4‘Pa5a6 a1‘a2‘a3‘a4’ Pa5a6 a1‘a2‘a3‘a4’ 5th line: a5 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 0 0 a6 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 C 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 a1‘ 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 a2‘ 1 1 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 a3‘ 1 0 0 1 1 1 1 1 0 0 1 0 0 1 1 1 0 a4‘ 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 1 d1 d2 t1 t2 t1 t3 t1 t5 t1 t6 t2 t1 t2 t7 t3 t1 t3 t2 t3 t3 t3 t5 t5 t1 t5 t3 t5 t5 t5 t7 t6 t1 t7 t2 t7 t5

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