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FAUST Analytics X=(X 1 ..X n )R n , |X|=N. X C =(X 1 ..X n ,C}. x.C{C 1 ..C c }=

Today’s main takeaway: We can attack both the curse of cardinality (CC) and the curse of dimensionality (CD) with pTree structuring. We datamine Big Cardinality DataSets (many rows) by Horizontal Processing of Vertical Data (HPVD).

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FAUST Analytics X=(X 1 ..X n )R n , |X|=N. X C =(X 1 ..X n ,C}. x.C{C 1 ..C c }=

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  1. Today’s main takeaway: We can attack both the curse of cardinality (CC) and the curse of dimensionality (CD) with pTree structuring. We datamine Big Cardinality DataSets (many rows) by Horizontal Processing of Vertical Data (HPVD). We datamine Big Dimensionality DataSets (many columns) with HPVD because it reduces the cost of attribute relevance analysis (eliminating attributes irrelevant to the classification or clustering or ARM or???.) by orders of magnitude. Wrt to Attribute Relevance Analysis, we may be wise to pre-compute, not only the bitslice pTrees of all numeric columns and bitmap pTrees of all categorical columns (the basic pTrees), but to also pre-compute (one time) all attribute-value bitmap pTrees (a bitmap for each value in the extent domain (the values that actually occur there). This would be a matter of treating numeric columns, not only as binary numbers to be bitsliced but also as categorical columns (treating each actual numerical value as a category and bit mapping it). We may, in fact, find it worth the trouble to pre-compute the value bit map of all composite attributes as well. (See the next few slides) We need a Genetic Algorithm implementation!!!! (See next slide.) • TODO: • Attribute selection prior to FAUST Oblique (for speed/accuracy; use Treeminers, use hi pTree correlation with class?, info_gain, gini_index, other?, … • Replicate the graph below (and expand that accuracy performance study to a full experimental design). • Do a comprehensive speed performance study. • Implement FAUST Hull Classifier. Enhance/compare it using a good exp. design and real datasets (Treeminer hasn’t gotten to that yet. So we can help a lot there!). • Research pTree density based clustering (https://bb.ndsu.nodak.edu/bbcswebdav/pid-2819939-dt-content-rid-13579458_2/courses/153-NDSU-20309/dmcluster.htm + 1stppt_set) Related papers: • DAYVD: Iterative Density-Based Approach for Clusters with VarYing Density, International Journal of Computers and Their Applications, V17:1, ISSN 1076-5204, B. Wang and W. Perrizo, March, 2010.52. . • A Hierarchical Approach for Clusters in Different Densities”, Proceedings of the International Conference on Software Engineering and Data Engineering, Los Angeles, B. Wang, W. Perrizo, July, 2006. • A Comprehensive Hierarchical Clustering Method for Gene Expression Data Association of Computing Machinery, Symposium on Applied Computing, ACM SAC 2005, Mar., Santa Fe, NM, B. Wang, W. Perrizo. • A P-tree-based Outlier Detection Alg”, International Society of Computer Applications Conference on Applications. in Industry and Engineering., ISCA CAINE 2004, Orlando, FL, Nov., 2004 (with B. Wang, D. Ren) • A Cluster-based Outlier Detection Method with Efficient Pruning”, International Society of Computer Applications Conf. on Applics. in Industry and Eng., ISCA CAINE, Nov., 2004 (with B. Wang, D. Ren) • A Density-based Outlier Detection Alg using Pruning Techniques”, Intl Society of Computer Applications Conf. on Applics. in Industry and Eng., ISCA CAINE 2004, Nov., 2004 (with B. Wang, K. Scott, D. Ren) • Parameter Reduction for Density-based Clustering on large Data Sets”, Intl Society of Computer Applications Conference on Applications in Industry and Engineering, ISCA CAINE 2004, Nov., 2004 (with B. Wang) • Outlier Detection with Local Pruning”, Association of Computing Machinery Conference on Information and Knowledge Management, ACM CIKM 2004, Nov., 2004, Washington, D.C., (with D. Ren). • RDF: A Density-based Outlier Detection Method using Vertical Data Representation”, IEEE International Conference On Data Mining, IEEE ICDM 2004, Nov., 2004, Brighton, U.K., (with D. Ren, B. Wang). • A Vertical Outlier Detection Method with Clusters as a By-Product”, IEEE International Conf. On Tools in Artificial Intelligence, IEEE ICTAI 2004, Nov., 2004, Boca Raton, FL, (with D. Ren). • AGNES Clusterer: Start w singles clusters.  C, find closest nbr, C’. If (C,C’) Qualifies, CCC’. Qualifies::always; d(C,C’)<T; dens(CC’)<dens(C)-; minFdis(F(C),F(C’))< (easy computation by comparing cutpoints under each F?) • Fuzzy Classifier: fully replicate pTreeSet. Start w all singles classes. Each processor assigned a class, C, do One-classification on C result in C’. If C’ qualifies, CC’. • F’s are distance dominated (actual separation ≥ F separation). If a hull is close to the convex hull, maxFSingleLinkSeperation ~= SL_seperation? • MiniMax approaches: If it’s true that Max(F-sep) approximates actual separation (of sets – distance between sets generalizes standard distance between points by SingleLink, CompleteLink, AvgLink, …), then Minimizing the maxFsep should be a good classifier? Etc.? Maximizing the maxFsep should tease out potential outliers? FAUST Hull Classifying: Classify yCk iff yHullk{z|minFCk-F(z)maxFCk+}. FAUST Clustering:Start C=X. Uuntil STOP (ClusDens> thres), recursively cut C at F-gaps (mdpt or adjust w variance). Use PCC gaps instead for suspected aberrations. SPRING PLAN: Develop Treeminer platform and datasets (+ Matt Piehl’s datasets) on pTree1 incl Hadoop (+Ingest procs for new datasets and convert them to pTreeSets. Each pick a VDMp TopicSetPPT or book / produce enhanced/up-to-date version / embed audio / Blackboard based AssignmentSets (+solutions) + (TestSets +answers. Md: Ph.D. CS, VDMp operations. Proposal defended (X) Rajat: MS CS, finished coursework. SE Quality Metrics (Dr. Magel) imp api Maninder: Ph.D. CS, (book needs structure) Plan: develop in Treeminer env., Spoorthy: MS SE Damian: Ph.D. in SE Arjun: Ph.D. CS. (VDMp algs, 2 tasks s15, proposal, journal paper, Arijit: Ph.D. in CS, VDMp DM of fanacials, proposal defense this term Bryan: Multilevel pTrees Mark11/25 FAUST text classification capable of accuracy as high as anything out there. Stanford_newsgroup dataset (7.5Kdocs) FAUST got 80% boost by eliminating terms in <= 2 docs. Chi-squared to reduce attrs, 20% (pick 80% best attr.). Vertical allows toss atts easily before we expend CPU. Tossing intelligently improvse accuracy and reduces time!  We eliminated ~70% attribs from TestSet and achieved better accuracy than the classifiers referenced on Stanford NLP site! About to turn this loose on datasets approaching 1TB in size. FAUST Analytics X=(X1..Xn)Rn, |X|=N. XC=(X1..Xn,C}. x.C{C1..Cc}= TrainSetCl. d=(d1..dn), p=(p1..pn)Rn. F:RnR, F=L, S, R : PTSSPTS. Mark 11/26 Adjusting midpt as well based on cluster deviation. Gives extra 4% accuracy. The hull is interesting case, as we’re looking at situations. We are already able to predict which members are poor matches to a class. Sp (X-p)o(X-p)=XoX+Xo(-2p)+pop=L-2p+XoX+pop Ld,p (X-p)od=Xod-pod LdXod Rd,p Sp-L2d,p =XoX+L-2p+pop-(Ld)2-2pod*Xod+(pod)d2=L-2p-(2pod)d -(Ld)2+XoX+pop+(pod)2 TEST MINING COMMMENTS: For text corpus (d docs (rows) and t terms (cols), so far recorded only a very tiny part of the total info. Wealth of other info, not captured . 2010-2015 notes tried to capture more information than just term existence or wtf info (~2012_07-09). FAUST Top K OutlierDetector : rankn-1Sx

  2. From Mark 1/23/2015: So how would you use GA in conjunction with faust?  Use the weightings to adjust the splits? WP: See below, but we used GAs for attribute selection (weighting). For attribute selection (or weighting-selection can be viewed as setting most weights =0) we think three tools (at least) will help: correlation with the class column; Information Gain wrt to the class column (see later slides); A GA tool (yet to be developed, but it would be for attribute selection from very wide datasets). MS: Also, check this out...  we are doing clustering of data in hadoop, building a classification model directly from the clustering results, and applying it real time to streaming data (real time, multi-class, classification of text documents) It's all about attribute relevance (well, maybe not "all", but it's huge.... especially with 100,000 attributes), I've wanted to play with GA for a while!!! https://www.youtube.com/watch?v=5X65WV0n4rU (everyone should watch this impressive video!!!) Training Data Genetic Algorithm Attribute Relevance Param. Fitness Top M Attributes Training Data Nearest Neighbor Vote + Boundary Based Classification Algorithm Optimized Classification Algorithm Classified Results Test Data First we describe and reference the uses of Genetic Algorithms in the two ACM KDD Cup wins: ACM KDD Cub 2002: Full paper at in the ACM SIGMOD Explorations Journal: http:///home/perrizo/public_html/saturday/papers/paper Due to the diversity of the experimental class label and the nature of the attribute data, we need a classifier that would not require a fixed importance measure on its features, i.e., we needed an adaptable mechanism to arrive at the best possible classifier. In our approach we optimize (column based scaling) the weight space, W, with a standard Genetic Algorithm (GA). The set of weights on the features represented the solution space that was encoded for the GA. The AROC value of the classified list was used as the GA fitness evaluator in the search for an optimal solution. RESULTS AND LESSONS: This work resulted in both biological and data mining related insights. The systematic analysis of the relevance of different attributes is essential for a successful classification. We found that the function of a protein did not help to classify the hidden system. Sub-cellular localization, which is a sub-hierarchy of the function hierarchy, on the other hand, contributed significantly. Furthermore, it was interesting to note that quantitative information, such as the number of interactions, played a significant role. The fact that a protein has many interactions may suggest that the deletion of the corresponding gene would cause changes to many biological processes. Alternatively it could be that a high number of listed interactions is an indication of the fact that previous researchers have considered the gene important and that it therefore is more likely to be involved in further experiments. For the purpose of classification we didn’t have to distinguish between these alternatives. ACM KDD Cub 2006: Full paper at in the ACM SIGMOD Explorations Journal: http:///home/perrizo/public_html/saturday/papers/paper The multiple parameters involved in the two classification algorithms were optimized via an evolutionary algorithm. The most important aspect of an evolutionary algorithm is the evaluation or fitness function, to guide the evolutionary algorithm towards a optimized solution. Negative Predictive Value NPV and True Negatives (TN) are calculated based on the task specification for KDDCup06. NPV is calculated by TN/(TN+FN). The above fitness function encourages solutions with high TN, provided that NPV was within a threshold value. Although the task specified threshold for NPV was 1.0, with the very low number of negative cases it was hard to expect multiple solutions that meet the actual NPV threshold and also maintained a high TN level. In a GA, collections of quality solutions in each generation potentially influence the set of solutions in the next generation. Since the training data set was small, patient level bootstrapping was used to validate solutions. Bootstrap implies taking out one sample from the training set for testing and repeating until all samples were used for testing. In this specific task which involves multiple candidates for the same patient we removed all candidates from the particular training set for a given patient when used as a test sample. The overall approach of the solution is summarized in the following diagram. As described in the above figure, attribute relevance is carried out on the training data to identify the top M attributes. Those attributes are subsequently used to optimize the parameter space, with a standard Genetic Algorithm (GA) [8]. The set of weights on the parameters represented the solution space that was encoded for the GA. Multiple iterations with the feedback from the solution evaluation is used to arrive at the best solution. Finally the optimized parameters are used on the combined classification algorithm to classify the unknown samples.

  3. ANDs of Green, Class value pairs (Roll Up Red by pTree ORing Value pTrees in a Data Cube A data warehouse is usually based on a multidimensional data model which views data in the form of a data cube describing the subject of interest A data cube allows data to be modeled and viewed in multiple dimensions Auxiliary dimension tables are added to the central cube for additional information Fact cube contains measurement(s) and keys (references) to each of the related dimension tables. Class avg bad good great hi Red hi avg This is a RSI dataset with 2 feature bands (Red {lo,avg,hi} Green {lo,avg,hi} Class (bad,avg,good,great} crop yield In each cell we store the AND(R-value,G-value,C-value) The rollup for all others? That is, do we want to pre-compute this rollup and store it using datacube software? Or do we want to store all the cuboids ANDs separately (that is, store all the rollups as pre-computed pTrees?)? lo avg Green lo

  4. Rollup yield by Oring R and G Yield avg lo good great hi Red hi avg lo avg Green lo

  5. Rollup Green by Oring Red,Class Date 2Qtr 1Qtr 3Qtr 4Qtr TV Product U.S.A PC VCR Canada Country Mexico

  6. Rollup Class = Red& Green RolluR=G RollupC=G Rollup Red = Class&Green RollupC=R RollupC=R RollupG=C ??? ??? ??? RollupG=C RollupG=C Class avg bad good great hi Red hi avg lo avg Green lo It’s probably better to drill down!!! Start with the R={lo,avg,hi} and G={lo,avg,hi} feature value pTrees and the C={bad,avg,good.great} class value pTrees Then drill down by ANDing all value pairs (producing the 2D “sides” first, then the 3D cube. Or it’s proabably better still to do all that as pre-computation and store everything ahead of time (or have a background process drilling down to create all of these combo value pTrees in the background while users are using the bitslice pTrees???

  7. EXPECTED INFO to classify: I= -i=1..m((rcPci)/|X|) log2((rcPci)/|X|), m=5 I = -(3/16*log3/16+1/4*log1/4+1/4*log1/4+1/8*log1/8+3/16*log3/16)= 2.8 (If basic pTree rc’s are pre-computed (actually just value pTrees), this is arithmetic!) 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 rcPci = 3 4 4 2 3 pj=rcPci/16 = 3/16 1/4 1/4 1/8 3/16 S1,j= 0 0 0 0 3 0 0 3 0 3 rc(Pc=2^PBk=aj) ENTROPY: E(Aj)=j=1..v[(s1j+..+smj)*I(s1j..smj)/s] I(s1j..smj)=-i=1..m[pij*log2(pij)] pij=sij/|Aj| sij=s1,j+..+s5,j= 2 4 2 4 4 6 2 8 115 |Aj| where Aj's are the rootcounts of Pk(aj)'s 2 4 2 4 4 6 2 8 11 5 S2,j= 0 2 2 0 0 0 0 4 3 1 rc(Pc=3^PBk=aj) S3,j= 2 2 0 0 0 2 2 0 4 0 rc(Pc=7^PBk=aj) S4,j= 0 0 0 1 1 1 0 1 1 1 rc(Pc=10^PBk=aj) P1j = 0 0 0 0 .75 0 0 .375 0 .6 P2j = 0 .5 .5 0 0 0 0 .5 .273 .2 P3j = 1 .5 0 0 0 .33 1 0 .363 0 P4j = 0 0 0 .25 .25 .17 0 .125 .091 .2 P5j = 0 0 0 .75 0 .5 0 0 .273 0 rc(Pc=15^PBk=aj) S5,j= 0 0 0 3 0 3 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 .75 0 0 .53 0 .44 0 .5 .5 0 0 0 0 .5 .51 .46 0 .5 0 0 0 .52 0 0 .53 0 0 0 0 .5 .5 .43 0 .37 .31 .46 0 0 0 .31 0 .5 0 0 .51 0 -p1j*log2(p1j) -p2j*log2(p2j) -p3j*log2(p3j) -p4j*log2(p4j) -p5j*log2(p5j) (s1j+..+s5j)*I(s1j..s5j)/16 0 .25 .13 .2 .31 .54 0 .7 .127 .43 GAIN(B2)=2.81-.89 =1.92 GAIN(B3)=2.81-1.24 =1.57 GAIN(B4)=2.81-.557 =2.53 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 P2,0 rc=10 P3,3 rc=8 P3,0 rc=2 P4,3 rc=16 Pc=10 rc=2 P3,1 rc=0 P4,1 rc=16 P2,2 rc=2 Pc=15 rc=3 P3,2 rc=8 P1,0 rc=11 P4,0 rc=16 P1,1 rc=16 Pc=7 rc=4 P2,1 rc=16 P1,2 rc=7 P4,2 rc=5 Pc=3 rc=4 Pc=2 rc=3 P1,3 rc=5 P2,3 rc=8 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 0000000000110111 0 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 1 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 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 1 1 1 1 1 1 1 1 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 0 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 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 1 1 0 0 1 1 0 0 0 1 0 0 1 1 • So it’s all just arithmetic, except for the #Classes * #FeatureValues ANDs and RootCounts, Should these be pre-computed at capture? • Are they part of the correlation calculation? Other often-used calculations? • Other speedups include: • Use approx. Value pTrees. Intervalize feature values and use the IntervalBitMaps (which can be calculated either from BitSlice or ValueMappTrees. • Using bitslice intervals, we’d pre-calculate all then use HiOrderBitIntervalBitMaps, Pi,j,Kj where Kj is the HiOrderBit of attrib, j. For 2ndHiOrderBit Intervals, just take Pi,j,Kj & P i,j,Kj-1 etc. • Aside note: There must be mistakes in the arithmetic above since I get different GAIN values than on the previous slide. Who can correct? rc(PC=ci^PBk=aj). 3 4 4 2 3 rcPci 0.1875 0.25 0.25 0.125 0.187 pi=rcPci/16 -2.4150 -2 -2 -3 -2.41 log2(pi) -0.4528 -0.5 -0.5 -0.37 -0.45 pilog2(pi) 2.280 -SUM(pilog2(pi)=I(c1..cm) Pi,j,k = PC=ci^Pj,k An RSI Dataset 16 pixels (4 rows 4 cols): 4 bitslices Band B1: Band B2: Band B3: Band B4: 3 3 7 7 7 3 3 2 8 8 4 5 11 15 11 11 3 3 7 7 7 3 3 2 8 8 4 5 11 11 11 11 2 2 10 15 11 11 10 10 8 8 4 4 15 15 11 11 2 10 15 15 11 11 10 10 8 8 4 4 15 15 11 11 S: X-Y B1 B2 B3 B4 0,0 0011 0111 1000 1011 0,1 0011 0011 1000 1111 0,2 0111 0011 0100 1011 0,3 0111 0010 0101 1011 1,0 0011 0111 1000 1011 1,1 0011 0011 1000 1011 1,2 0111 0011 0100 1011 1,3 0111 0010 0101 1011 2,0 0010 1011 1000 1111 2,1 0010 1011 1000 1111 2,2 1010 1010 0100 1011 2,3 1111 1010 0100 1011 3,0 0010 1011 1000 1111 3,1 1010 1011 1000 1111 3,2 1111 1010 0100 1011 3,3 1111 1010 0100 1011 . Value BitMap pTrees BitSlice pTrees PB3=4 PB3=4 rc=6 PB2=3 rc=4 PB2=3 PB2=7 PB2=7 rc=2 PB2=10 PB2=10 rc=4 PB2=11 rc=4 PB2=11 PB3=5 PB3=5 rc=2 PB3=8 PB3=8 rc=8 PB4=11 rc=11 PB4=11 PB4=15 PB4=15 rc=5 PB2=2 PB2=2 rc=2 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0

  8. Access to Ptree1 and Ptree2 William Perrizo To all my students (plus 3 former students who may be interested in an opportunity to use a pTree environment replete with massive, real-life big datasets that was developed by Treeminer Inc (the company that licensed pTree patents and is becoming a strong player in the vertical data mining area).  The only requirement from this side is your willingness to share and willingness to sign a Treeminer NDA.I particularly am interested in having available for everyone to use a Genetic Algorithm tool.  As we all know, Dr. Amal Shehan Perera was the lead scientist implementing a GA tool in pTree data mining software and it helped to win two ACM KDD Cups!!!  Since that time my suggestion to anyone has been "When your done getting all you can out of your algorithm, GA it!!!"Baoying (Elizabeth)Wang did some amazing work on pTree clustering (and classification and ARM) and Dr. Imad Rahal did amazing pTree work on classification (and ARM and Clustering).  That's why I invite Amal, Elizabeth and Imad to use our system once it gets up and going.  My hope is that they will be willing to let us use any tools they develop in that system? ;-)  (synergism).  There may be other former DataSURG members who would be interested as well (please advise).Just to update Amal, Elizabeth and Imad, Treeminer is a startup that has implemented a pTree data mining system in Java (about 70.000 lines of code).  It is wonderful and is commercially successful.  Treeminer has used several BigData real-life dataset (captured in pTrees).  They have demonstrated both orders of magnitude improvement in speed (which is the traditional focus of pTree technology - speedup through Horizontal Processing of Vertical Data or HPVD) but also improvements in accuracy at the same time!  This is unheard of and wonderful.  Maybe the main reason we are able to get accuracy improvements while getting great speedup is that vertical structuring facilitates fast and effective attribute selection (which might take forever with horizontal data and therefore is not attempted in the horizontal data world).  E.g., when datamining a text corpus of 100,000 vocab words (the columns) and 1oo,000,000 documents (the rows). it is important to find out the important words (attribute selection) before running an algorithm (solve the curse of dimensionality first).  Amal, Elizabeth and Imad, if you would like to get involved, let us know.  We still have Saturday meetings at 10 CST and we have two people who skype in every week (the more the merrier!).  Arjun Roy or Arijit Chatterjee can tell how to skype in.Nathan Olson is our department technician who has helped us a lot.At this point we are nearly done getting all of Treeminer's stuff to the two pTree1 and pTree2 servers two servers, so that you can remote login to add anything to the GIT repository there (and run and test it against a very impressive suite of datasets.)Please note I have posted all recent Saturday notes and other good materials on my web site: http://www.cs.ndsu.nodak.edu/~perrizo You can get these materials by clicking on dot or period to the right of the bullet "media coverage"  William Perrizo:I have put all the Saturday notes for the past few years and many other useful files (e.g., topic synopses) on my web site so you all can access anything you want.  I'm not worried at all about security since, if someone outside our group were to think they had stolen something important from there, then they probably would study it and become a pTree data mining expert.  How could that be bad?  You can get these materials by clicking on dot or period to the right of the bullet "media coverage"  My homepage is:   http://www.cs.ndsu.nodak.edu/~perrizoThere was a time in the past when I contemplated a "pTree Data Mining" monograph.  I will put the very rough preliminary version of that book project on the website also.  Anyone at all that every wants to co-author such a thing with me is welcome to work on it using the material at the site. I just wanted to ask if the 4 experience students (Damian, Mohammad, Arjun and Arijit) would be willing to help the students who will/may join our group (manider, Rajat, Spoorthy) and who will then be tasked to do coding, implementation and testing using Treeminer Software Environment and using Treeminer big vertical data sets? That would be great and would also help each of you, I believe.  I'm sure Mark, Puneet will help also but they are, no doubt, buried under the weight of other needs right now.From: Bryan Mesich bryan.mesich@ndsu.edu Sent: Friday, January 23, 2015 12:41 PM> I had the same question. We can make changes to code on ptree1/2 directly but it will be difficult via VI editor. 2 relevant options. Take a copy in your local system, make changes and move the java file to ptree1/2. Compile and run.> 2. Take a copy in your local system, make changes and compile. Move class file to ptree1/2 and run. I do all of my development (Perl, C/C++. Assembly, Java, Bash) using VIM. This is my preference.  I'm not advocating to use one IDE/Editorover another as its a personal choice. The best way to accomplish this would be to use a version system. That way you can do development on a remote machine and commit your changes.  You could then log into ptree[12].ndsu, checkout thelatest version and run the code.  Version control works well in collaborations efforts when merging code together. Also, I went ahead and installed both Java7 and Java8 JDKs on both servers.  They are located in /usr/java.  By default, Java8 will be used when running/compiling. Bryan So we have any other alternatives I would suggest running GIT or SVN.Rajat> From: Arjun Roy> My IP address is 192.168.0.4> If for some reason I am not able to copy, Damian can give a try on campus?> I think the way we are going to proceed is that we will have Eclipse software on our client machine (pointing to code on ptree1) but its going to be compiled on ptree1.> I don't know if we can directly make changes to code stored on ptree1 or if we would have to make changes locally and push it to ptree1.> I dont have much knowledge on Java environment/Eclipse but does this sound logical (Rajat/Maninder)?> Thanks, Arjun>> From: Bryan Mesich <bryan.mesich@ndsu.edu>> > > Please give me a remote access. Once I have access, will put everything on Share.> > I have used it with Jdk 7 without any compiler complaints.> I need your IP address in order to make the exception in the firewall.> Also, did you use OpenJDK, or the official JDK from Oracle?> > > > Subject: Re: Access to Ptree1 and Ptree2> > Logins are ready for use.  I will be installing a firewall shortly> > that will only allow on-campus access.  If you need remote access,> > please let me know and I'll make an exception for you in the firewall.> > Otherwise I'll assume everyone has access unless contacted.> > >> > >    Bryan, I have a few months old version of TreeMiner. Its about 2.5 GB. Is> > >    it ok if I put the entire thing in my space on ptree1 (might take a while  to transfer)?> > Lets have you put the TreeMiner code under the Shared directory> > (/stggroups/perrizo/Shared).  That way everyone will have access to the code.> > >    For latest version, we will have to contact Mark and go through some layers of security.> > We'll want to investigate a way to pull the code in an automated> > fashion in the future.  The current version(s) you have should work for now.> > Does anyone know what JDK version TreeMiner is using?

  9. Damian Lampl Fri 1/23/2015 2:12 PM I'll need remote access, too.  My IP is: 208.107.126.159 One reason we wanted to get things centralized on the pTree servers was so we wouldn't have to configure everyone's development environment separately (Java, Eclipse dependencies, etc) since that's the primary hang-up to getting started with the Treeminer code. I was thinking we could just remote desktop into the ptree servers using xrdp or something similar if that works?  I really don't know the best way of setting it up so we all have access to the same preconfigured Java/Eclipse environment (.NET and Visual Studio would be a different story: install Visual Studio, done). You mentioned maven, Bryan, but I'm not familiar with how that works?  Or if there's a configuration file we can get set up that makes sure our local Java, Eclipse, and any other dependencies are configured properly? I agree on a git repository so we'll have one main trunk and then separate branches for everyone, and when we need to merge with Mark's code, we can use the main trunk for that since he's also using git. Mark also has things working with Hadoop and some of his datasets will be stored in that. Bryan Mesich <bryan.mesich@ndsu.edu> Fri 1/23/2015 12:41 PM Rajat Singh wrote: had the same question.> We can make changes to code on ptree1/2 directly but it will be difficult via VI editor. So we have two relevant options> 1. Take a copy in your local system, make changes and move the java file to ptree1/2. Compile and run.> 2. Take a copy in your local system, make changes and compile. Move class file to ptree1/2 and run. I do all of my development (Perl, C/C++. Assembly, Java, Bash) using VIM. This is my preference.  I'm not advocating to use one IDE/Editor over another as its a personal choice.The best way to accomplish this would be to use a version control system.  That way you can do development on a remote machine and commit your changes.  You could then log into ptree[12].ndsu, checkout thel atest version and run code.  Version control works well in collaborative efforts when merging code together. Also, I went ahead and installed both Java7 and Java8 JDKs on both servers. They are located in /usr/java.  By default, Java8 will be used when running/compiling. Bryan Maninder Singh Fri 1/23/2015 12:21 PM Just a thought, can we think of some way to write a batch file in VI editor to do some of our work directly on ptree? Rajat Singh Fri 1/23/2015 12:15 PM I had the same question. We can make changes to code on ptree1/2 directly but it will be difficult via VI editor. So we have two relevant options 1. Take a copy in your local system, make changes and move the java file to ptree1/2. Compile and run. 2. Take Arjun Roy Fri 1/23/2015 12:08 PM My IP address is 192.168.0.4 If for some reason I am not able to copy, Damian can give a try on campus? I think the way we are going to proceed is that we will have Eclipse software on our client machine (pointing to code on ptree1) but its going to be compiled on ptree1. I don't know if we can directly make changes to code stored on ptree1 or if we would have to make changes locally and push it to ptree1. I dont have much knowledge on Java environment/Eclipse but does this sound logical (Rajat/Maninder)? Arjun In my opinion, you can stay back at home and relax this Saturday :)We students will collectively make it operational now that Bryan has  created login for us.Logins are ready for use.  I will be installing a firewall shortly thatwill only allow on-campus access.  If you need remote access, please letme know and I'll make an exception for you in the firewall.  OtherwiseI'll assume everyone has access unless contacted.> >    Bryan, I have a few months old version of TreeMiner. Its about 2.5 GB. Is>    it ok if I put the entire thing in my space on ptree1 (might take a while>    to transfer)?Lets have you put the TreeMiner code under the Shared directory(/stggroups/perrizo/Shared).  That way everyone will have access to thecode.>    For latest version, we will have to contact Mark and go through some>    layers of security.We'll want to investigate a way to pull the code in an automated fashionin the future.  The current version(s) you have should work for now.Does anyone know what JDK version TreeMiner is using?

  10. Current Treeminer methods, hi pTree correlation with class?; hi information_gain; Hi gini_index?… http://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmlearn.htm Attribute Selection; In information theory and machine learning, information gain is a synonym for Kullback–Leibler divergence. The expected value of information gain is the mutual information I(X; A) of X and A – i.e. reduction in the entropy of X achieved by learning the state of the random variable A. In machine learning, this concept can be used to define a preferred sequence of attributes to investigate to most rapidly narrow down the state of X. Such a sequence (depends on outcome of investigation of previous attribs at each stage) is a decision tree. Usually an attribute with hi mutual info is preferred to others. In general terms, the expected information gain is the change in information entropy from a prior state to a state that takes some information as given: Drawbacks: Altho info gain is a good measure of attrib relevance, it is not perfect. E.g., when applied to attributes that can take on many distinct values. E.g., building a decision tree for some data describing the customers of a business. Info gain is often used to decide which attribs are most relevant, so they can be tested near the tree root. Credit card number uniquely identifies customer, but deciding how to treat a customer based on credit card # will not generalize to customers we haven't seen yet (overfitting). Information gain ratio instead? This biases the decision tree against attributes with many distinct values. But, attributes with low info values then receive an unfair advantage. In statistics, dependence is any relationship between two random variables or two sets of data. Correlation refers to any of a class of statistical relationships involving dependence. E.g., correlation between physical statures of parents and offspring,; between demand for a product and its price. Correlations can indicate a predictive relationship to exploited in practice. E.g., an electrical utility may produce less power on a mild day based on the correlation between electricity demand and weather. In this example there is a causal relationship, because extreme weather causes the use of more electricity; however, statistical dependence is not sufficient to demonstrate the presence of such a causal relationship (i.e., correlation does not imply causation). Formally, dependence refers to random variables not satisfy inga math condition of probabilistic independence. Loosely correlation is any departure of two or more random variables from independence, but technically it refers to any of several more specialized types of relationship between mean values. There are several correlation coefficients, ρ or r, measuring the degree of correlation., e.g., Pearson correlation coefficient, sensitive only to a linear relationship between 2 variables. Other correlation coeffs have been developed to be more robust than Pearson correlation. Here are several sets of (x, y) points, with the Pearson correlation coefficient of x and y for each set. Note , correlation reflects noisiness /direction of a linear relationship (top row), but not slope (middle), nor many aspects of nonlinear relationships (bottom). N.B.: figure in center has a slope of 0 but correlation coefficient is undefined because the variance of Y=0 Contents: Pearson's product-moment coefficient: Main article: Pearson product-moment correlation coefficient :The most familiar measure of dependence between two quantities is the Pearson product-moment correlation coefficient, or "Pearson's correlation coefficient", commonly called simply "the correlation coefficient". It is obtained by dividing the covariance of the two variables by the product of their standard deviations. Karl Pearson developed the coefficient from a similar but slightly different idea by Francis Galton.[4] The population correlation coefficient ρX,Y between two random variablesX and Y with expected values μX and μY and standard deviations σX and σY is defined as: where E is expected value operator, cov means covariance, and, corr an alternative notation for the correlation coefficient. Pearson correlation is defined only if both standard deviations are finite and nonzero. It is a corollary of the Cauchy–Schwarz inequality that the correlation cannot exceed 1 in absolute value. The correlation coefficient is symmetric: corr(X,Y) = corr(Y,X). The Pearson correlation is +1 in the case of a perfect direct (increasing) linear relationship (correlation), −1 in the case of a perfect decreasing (inverse) linear relationship (anticorrelation),[5] and some value between −1 and 1 in all other cases, indicating the degree of linear dependence between variables. As it approaches 0 there is less of a relationship (closer to uncorrelated). The closer coefficient is to either −1 or 1, stronger the correlation between variables. If variables are indep Pearson's correlation coeff is 0, but the converse is not true because the correlation coefficient detects only linear dependencies between two variables. E.g., random variable X is symmetrically distributed about 0, Y = X2. Then Y is completely determined by X, so that X and Y are perfectly dependent, but their correlation is zero; uncorrelated. Special case X and Y are jointly normal, uncorrelatedness = independence. If a series of n measmnts of X ,Y written xi and yii = 1...n, sample correlation coefficient can be used to estimate pop Pearson correlation r between X and Y. Sample correlation coeff is written where x and y are the sample means of X and Y, and sx and sy are the sample standard deviations of X and Y. This can also be written as: If x,y are results of meamnts containing error, realistic limits on correlation coef are not −1 to +1 but a smaller range..Rank correlation coefficients Main articles: Spearman's rank correlation coefficient and Kendall tau rank correlation coefficientRank correlation coefficients, such as Spearman's rank correlation coefficient and Kendall's rank correlation coefficient (τ) measure one variable increases to the other variable increase, w/o requiring that increase to be represented by a linear relationship (alternatives to Pearson's coefficient). To illustrate rank correlation, and its difference from linear correlation, consider 4 pairs (x, y): (0, 1), (10, 100), (101, 500), (102, 2000). As we go from each pair to the next pair x increases, and so does y. This relationship is perfect, in the sense that an increase in x is always accompanied by an increase in y. This means that we have a perfect rank correlation, and both Spearman's and Kendall's correlation coefficients are 1, whereas in this example Pearson product-moment correlation coefficient is 0.7544, indicating that the points are far from lying on a straight line. In the same way if y always decreases when xincreases, the rank correlation coefficients will be −1, while the Pearson product-moment correlation coefficient may or may not be close to −1, depending on how close the points are to a straight line. Although in the extreme cases of perfect rank correlation the two coefficients are both equal (being both +1 or both −1) this is not in general so, and values of the two coefficients cannot meaningfully be compared.[7] For example, for the three pairs (1, 1) (2, 3) (3, 2) Spearman's coefficient is 1/2, while Kendall's coefficient is 1/3. Other measures of dependence: The info given by a correlation coef is not enough to define the dependence structure between random variables.[9] Correlation coef completely defines dependence structure only in particular cases, for ex when distrib is a multivariate normal distribution. In the case of elliptical distributions it characterizes (hyper-) ellipses of equal density, but, it does not completely characterize dependence structure (for ex, a multivariate t-distribution's degrees of freedom determine level of tail dep). Distance correlation and Brownian covariance / Brownian correlation[10][11] were introduced to address deficiency of Pearson's corr that it can be zero for dependent random variables; 0 distance correlation and 0 Brownian correlation imply indep SEE: Association (statistics)AutocorrelationCanonical correlationCoefficient of determinationCointegrationConcordance correlation coefficientCophenetic correlationCopulaCorrelation functionCovariance and correlationCross-correlationEcological correlationFraction of variance unexplainedGenetic correlationGoodman and Kruskal's lambdaIllusory correlationInterclass correlationIntraclass correlationModifiable areal unit problemMultiple correlationPoint-biserial correlation coefficientQuadrant count ratioStatistical arbitrageSubindependence

  11. Clustering: Partition; Hierarchical; Density; Grid; Model-based Information Gain as an Attribute Selection Measure Minimizes expected number of tests needed to classify an object and guarantees simple tree (not necessarily the simplest) S = {s1,...,sm} be a TRAINING SUBSET. S[C] = {C1,...,Cc} be the distinct classes in S. EXPECTED INFORMATION needed to classify a sample given S is: I{s1,...,sm} = -∑i=1..mpi*log2(pi) pi= |S∩Ci|/|S|. Choosing A as decision attribute, the Expected Info gained is E(A) = ∑j=1..v; i=1..m ( si,j/|S| * I{sij..smj} ) where Skh = SA=ak∩Ch. Gain(A) = I(s1..sm) - E(A) - expected reduction of info required to classify after splitting via A-values.. Alg computes the information gain of each attribute and selects the one with the highest information gain as the test attribute. Branches are created for each value of that attribute and samples are partitioned accordingly. When a decision tree is built, many branches will reflect anomalies in the training data due to noise or outliers. Tree pruning addresess "overfitting" data (classifying situations that are erroneous or accidental). http://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmcluster.htm Agnes(Agglomerative Nesting) Kaufmann, Rousseeuw (90). Use Single-Link (distance between two sets is the minimum pairwise dist) meth. Merge nodes most similarity. Eventually all nodes belong to the same cluster Diana (Divisive Analysis) Inverse order of AGNES (start: all objects in 1 cluster; split by some criteria (e.g., max some aggregate or pairwise dissimilarity. Agglomerative clustering doesn’t scale well (imecomplexity≥O(n2), n=#objs. Can never undo what was done previously (greedy alg). Integration w distance-based: BIRCH 96: uses Cluster Feature tree (CF-tree).Incr adjusts quality of sub-clusters CURE98: selects well-scattered pts from cluster, shrinks to cluster ctr by fraction CHAMELEON 99: hierarchical clustering using dynamic modeling Density Clustering, Discover clusters of arbitrary shape, Handle noise; One scan; Need density parameters as stop condition. Several interesting studies: DBSCAN: Ester, et al. (KDD’96); OPTICS: Ankerst, et al (SIGMOD’99).; DENCLUE: Hinneburg & D. Keim (KDD’98); CLIQUE: Agrawal, et al. (SIGMOD’98) Decision Tree Classification (A flow-chart-like tree structure) Internal node denotes test on an attrib. Branch represents an outcome of test. Leaf nodes represent class labels or class distribution. Tree pruning (Identify and remove branches that reflect noise or outliers). http://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmlearn.htm CLASSIFICATIONTRAINING SETT(A1..An,C), CLASS C, FEATURES (A1..An) unclassified sample, (a1;;an) SELECT Max (Count (T.Ci));FROM T;WHERE T.A1=a1 AND T.A2=a2 ... T.An=anGRP BY T.C; i.e., just a SELECTION C-Classificatn is assigning to (a1..an) most frequent C-val in RA=(a1..an). Nearest Neighbor Classification (NNC) selecting a set of R-tuples with similar features (to the unclassified sample)and then letting corresponding class values vote. NNC won't work well if vote inconclusive or if similar (near) is not well defined, then we build MODEL of TRAINSET(at, possibly, great 1-time expense?) Eager Classifiers models decision trees, Bayesian, Neural Nets,SVM... Preparing Data: Data CleaningRemove Noise (or reduce noise) by "smoothing", Fill in missing values (with most common or statistical val). Noise/Missing Val mgnt done by a NN Vote! (interpolation) Feature Extraction eliminate irrelevant attrs Compare Different Methods Predictive Accuracy (predicting the class label of new data) Speed (computation costs for generating and using the model) Robustness (~same predictions when Training Set are almost the same?) Scalability (Model construction efficiency - massive datasets) Decision Trees: each inode is a test on a feature attrib. Each test outcome is assigned a link to next level (outcome=a val / range of vals or?). Leaf = distribution of classes) Some branches may represent noise or outliers (and should be pruned?) ID3 algorithm for inducing a decision tree from training tuples is: 1. The tree starts as a single node containing the entire TRAINING SET. 2. If all TRAINING TUPLES have the same class, this node is a leaf. DONE. 3. else, use info gain, for selecting the best decision attribute for that node 4. Branch created for each val [interval of vals] of test attr and TrainSet partitioned. 5. Recurses 2,3,4, til STOP? All samples same class. ∃ no candidate attribs Info Gain (ID3/C4.5) Select attrib with highest info gain. Assume two classes, P and N (positive/negative). Let the set of examples S contain p elements of class P and n elements of class N. Amount of info, needed to decide if arbitrary example in S belongs to P or N is defined: Class P: buys_computer = “yes” Class N: buys_computer = “no” I(p, n) = I(9, 5) =0.940 Compute the entropy for age: Hence Similarly Bayesian: thm:X inclassified sample. H be the hypothesis that X belongs to class, H. P(H|X)=condprobof H given X. P(H) is prob of H, P(H|X) = P(X|H)P(H)/P(X) Assume using attribute A, set S will be partitioned into sets {S1, S2 , …, Sv}. If Si contains piexamples of P and ni examples of N, the entropy, or the expected info needed to classify objects in all subtrees Si is info gained by branching on A Naïve Bayesian: Given training set, R(A1..An, C) where C={C1..Cm} is the class label attribute. The naive Bayesian Classifier will predict the class of unknown data sample, X, to be the class, Cj having the highest conditional probability, conditioned on X P(Cj|X) ≥ P(Ci|X), i  j. From the Bayes theorem: P(Cj|X) = P(X|Cj)P(Cj)/P(X) P(X) is constant for all classes so we maximize P(X|Cj)P(Cj).. Max P(X|Cj)P(Cj). To reduce comp complexity of calculating all P(X|Cj)'s the naive assumption: class conditional indep

  12. EXPECTED INFO to classify: I= -i=1..m((rcPci)/|X|) log2((rcPci)/|X|), m=5 I = -(3/16*log3/16+1/4*log1/4+1/4*log1/4+1/8*log1/8+3/16*log3/16)= 2.8 (If basic pTree rc’s are pre-computed (actually just value pTrees), this is arithmetic!) 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 rcPci = 3 4 4 2 3 pj=rcPci/16 = 3/16 1/4 1/4 1/8 3/16 S1,j= 0 0 0 0 3 0 0 3 0 3 rc(Pc=2^PBk=aj) ENTROPY: E(Aj)=j=1..v[(s1j+..+smj)*I(s1j..smj)/s] I(s1j..smj)=-i=1..m[pij*log2(pij)] pij=sij/|Aj| sij=s1,j+..+s5,j= 2 4 2 4 4 6 2 8 115 |Aj| where Aj's are the rootcounts of Pk(aj)'s 2 4 2 4 4 6 2 8 11 5 S2,j= 0 2 2 0 0 0 0 4 3 1 rc(Pc=3^PBk=aj) S3,j= 2 2 0 0 0 2 2 0 4 0 rc(Pc=7^PBk=aj) S4,j= 0 0 0 1 1 1 0 1 1 1 rc(Pc=10^PBk=aj) P1j = 0 0 0 0 .75 0 0 .375 0 .6 P2j = 0 .5 .5 0 0 0 0 .5 .273 .2 P3j = 1 .5 0 0 0 .33 1 0 .363 0 P4j = 0 0 0 .25 .25 .17 0 .125 .091 .2 P5j = 0 0 0 .75 0 .5 0 0 .273 0 rc(Pc=15^PBk=aj) S5,j= 0 0 0 3 0 3 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 .75 0 0 .53 0 .44 0 .5 .5 0 0 0 0 .5 .51 .46 0 .5 0 0 0 .52 0 0 .53 0 0 0 0 .5 .5 .43 0 .37 .31 .46 0 0 0 .31 0 .5 0 0 .51 0 -p1j*log2(p1j) -p2j*log2(p2j) -p3j*log2(p3j) -p4j*log2(p4j) -p5j*log2(p5j) (s1j+..+s5j)*I(s1j..s5j)/16 0 .25 .13 .2 .31 .54 0 .7 .127 .43 GAIN(B2)=2.81-.89 =1.92 GAIN(B3)=2.81-1.24 =1.57 GAIN(B4)=2.81-.557 =2.53 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 Pc=2 P2,0 rc=10 P3,3 rc=8 P3,0 rc=2 P4,3 rc=16 Pc=10 rc=2 P3,1 rc=0 P4,1 rc=16 P2,2 rc=2 Pc=15 rc=3 P3,2 rc=8 P1,0 rc=11 P4,0 rc=16 P1,1 rc=16 Pc=7 rc=4 P2,1 rc=16 P1,2 rc=7 P4,2 rc=5 Pc=3 rc=4 Pc=2 rc=3 P1,3 rc=5 P2,3 rc=8 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=15 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=7 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=3 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 Pc=10 0000000000110111 0 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 1 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 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 1 1 1 1 1 1 1 1 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 0 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 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 1 1 0 0 1 1 0 0 0 1 0 0 1 1 • So it’s all just arithmetic, except for the #Classes * #FeatureValues ANDs and RootCounts, Should these be pre-computed at capture? • Are they part of the correlation calculation? Other often-used calculations? • Other speedups include: • Use approx. Value pTrees. Intervalize feature values and use the IntervalBitMaps (which can be calculated either from BitSlice or ValueMappTrees. • Using bitslice intervals, we’d pre-calculate all then use HiOrderBitIntervalBitMaps, Pi,j,Kj where Kj is the HiOrderBit of attrib, j. For 2ndHiOrderBit Intervals, just take Pi,j,Kj & P i,j,Kj-1 etc. • Aside note: There must be mistakes in the arithmetic above since I get different GAIN values than on the previous slide. Who can correct? rc(PC=ci^PBk=aj). 3 4 4 2 3 rcPci 0.1875 0.25 0.25 0.125 0.187 pi=rcPci/16 -2.4150 -2 -2 -3 -2.41 log2(pi) -0.4528 -0.5 -0.5 -0.37 -0.45 pilog2(pi) 2.280 -SUM(pilog2(pi)=I(c1..cm) Pi,j,k = PC=ci^Pj,k An RSI Dataset 16 pixels (4 rows 4 cols): 4 bitslices Band B1: Band B2: Band B3: Band B4: 3 3 7 7 7 3 3 2 8 8 4 5 11 15 11 11 3 3 7 7 7 3 3 2 8 8 4 5 11 11 11 11 2 2 10 15 11 11 10 10 8 8 4 4 15 15 11 11 2 10 15 15 11 11 10 10 8 8 4 4 15 15 11 11 S: X-Y B1 B2 B3 B4 0,0 0011 0111 1000 1011 0,1 0011 0011 1000 1111 0,2 0111 0011 0100 1011 0,3 0111 0010 0101 1011 1,0 0011 0111 1000 1011 1,1 0011 0011 1000 1011 1,2 0111 0011 0100 1011 1,3 0111 0010 0101 1011 2,0 0010 1011 1000 1111 2,1 0010 1011 1000 1111 2,2 1010 1010 0100 1011 2,3 1111 1010 0100 1011 3,0 0010 1011 1000 1111 3,1 1010 1011 1000 1111 3,2 1111 1010 0100 1011 3,3 1111 1010 0100 1011 . Value BitMap pTrees BitSlice pTrees PB3=4 PB3=4 rc=6 PB2=3 rc=4 PB2=3 PB2=7 PB2=7 rc=2 PB2=10 PB2=10 rc=4 PB2=11 rc=4 PB2=11 PB3=5 PB3=5 rc=2 PB3=8 PB3=8 rc=8 PB4=11 rc=11 PB4=11 PB4=15 PB4=15 rc=5 PB2=2 PB2=2 rc=2 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0

  13. APPENDIX: level-1 TermFreqPTrees (E.g., the predicate of tfP0: mod(sum(mdl-stride),2)=1) <--dfP0 ... Term (Vocab) ..doc freq <--dfP2 0 0 1 1 2 0 0 0 0 0 df (cnt) 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 3 1 1 1 8 2 0 1 0 2 1 1 8 1 1 2 3 8 1 1 3 0 0 1 1 0 0 0 1 0 0 1 0 1 1 0 0 1 3 3 0 3 0 . . . 0 0 0 0 0 . . . 0 . . . 0 . . . 0 . . . ... tf0 0 . . . ... tf1 0 . . . ... tf 0 . . . 0 . . . 0 0 0 0 0 d=3 d=3 d=3 t=a t=again t=all ... doc=3 doc=3 doc=3 term=a trm=again term=all ... doc=1 d=1 d=1 term=a t=again t=all doc=1 doc=1 doc=1 term=a trm=again term=all doc=2 doc=2 doc=2 term=a trm=again term=all d=2 d=2 d=2 t=a t=again t=all 0 0 0 0 0 0 0 0 0 0 5 6 ...doc ...Term Freq ... tf2 ... tf0 ... tf1 ... Term Ex 1 0 0 JSE 0 0 1 0 . . . HHS LMM 0 0 0 0 0 0 0 . . . 0 0 0 0 0 0 0 . . . a apple 1 3 always. 1 1 1 1 April an again all and 1 2 are 3 7 1 2 3 4 Length of this level-1 TermExistencePTree =VocabLen*DocCount pred is NOTpure0 Length of this level-0 pTree= mdl*VocabLen*DocCount . . . 0 0 0 1 2 3 4 5 6 7 mdl reading-positions for doc=1, term=a (mdl = max doc length) 1 2 3 4 5 6 7 mdl reading-positions: doc=1, term=again 1 2 3 4 5 6 7 mdl reading-positions for doc=1, term=all DocTrmPos pTreeSet dfk isn't a level-2 pTree since it's not a predicate on level-1 te strides. Next slides shows how to do it differently so that even the dfk's come out as level-2 pTrees. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 . . . 0 0 0 0 0 0 0 . . . 0 0 0 0 0 0 0 . . . pTree Text Mining (from 2012_08_04 Data Cube Text Mining ... Position

  14. level-2 PTree, hdfP?? (Hi Doc Feq): pred=NOTpure0 applied to tfP1 <--dfP0 Vocab Terms ..doc freq hdfP <--dfP3 1 0 2 0 1 doc1 doc2 doc3 0 0 0 0 0 df count 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 doc1 doc2 doc3 1 0 1 1 3 0 1 1 1 0 0 0 0 8 8 0 2 1 8 1 0 1 0 1 1 0 1 0 0 0 . . . 0 0 . . . 2 . . . 0 . . . 3 3 3 . . . . . . ... tfP1 . . . ... tf 0 . . . . . . ... tfP0 . . . 0 . . . . . . 0 0 0 0 0 tePt=all d=1 d=2 d=3 t=all t=all t=all ... doc=1 d=2 d=3 term=a t=a t=a d=1 d=2 d=3 t=again t=again t=again 0 0 0 0 0 0 0 0 0 0 tePt=again tePt=a tr=all t=all t=all doc1 doc2 doc3 ... t=again t=again t=again doc1 doc2 doc3 trm=a trm=a term=a doc1 doc2 doc3 5 6 ...doc ... tf2 ... tf0 ...Term Freq ... tf1 ... Term Ex 0 0 0 JSE 0 0 1 0 . . . HHS LMM 0 0 0 0 0 0 0 . . . 0 0 0 0 0 0 0 . . . 1 always. 1 apple a 1 again 1 an 1 April 1 3 are 1 2 and all 7 1 2 3 4 These level-2 pTrees, dfPk have len= VocabLength level-1 PTrees, tfPk e.g., pred of tfP0: mod(sum(mdl-stride),2)=1 This one, overall, level-1 pTree, teP, has length = DocCount*VocabLength term=a doc2 term=a doc3 term=a doc1 term=again doc1 ... This one, overall, level-0 pTree, corpusP, has length = MaxDocLen*DocCount*VocabLen Corpus pTreeSet 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 . . . pTree Text Mining data Cube layout: ... Pos

  15. level-2 PTree, hdfP?? (Hi Doc Feq): pred=NOTpure0 applied to tfP1 <--dfP0 Vocab Terms ..doc freq hdfP <--dfP3 1 2 1 0 0 doc1 doc2 doc3 0 0 0 0 0 df count 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 1 3 1 0 1 2 0 1 0 0 0 0 1 1 1 8 8 8 0 0 0 0 0 1 1 1 0 1 1 0 . . . 3 0 3 2 . . . . . . ... tfP0 . . . ... tf 3 0 . . . 0 . . . . . . ... tfP1 . . . . . . . . . 0 . . . 2 . . . 0 0 0 0 0 0 0 0 0 0 This overall, level-1 pTree, teP, has length = DocCount*VocabLength 0 0 0 0 0 tePt=again tePt=all tePt=a tr=all t=all t=all doc1 doc2 doc3 ... t=again t=again t=again doc1 doc2 doc3 trm=a trm=a term=a doc1 doc2 doc3 0 0 0 1 0 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 1 0 0 Pt=a,d=3 0 Pt=a,d=2 5 6 Verb pTree Refrncs pTree EndofSentence Preface pTree LastChpt pTree ...doc ... tf1 ... tf2 ... tf ... tf0 ... te 0 0 0 JSE 0 0 0 0 Pt=a,d=1. . . HHS LMM 0 0 0 0 0 0 0 Pt=again,d=1 0 0 0 0 0 0 0 . . . 1 1 1 a apple always. 1 1 April 1 again an all 2 and 3 1 are 7 1 2 3 4 These level-2 pTrees, dfPk have len= VocabLength doc1 doc2 doc3 level-1 PTrees, tfPk e.g., pred of tfP0: mod(sum(mdl-stride),2)=1 d=1 d=2 d=3 t=all t=all t=all ... doc=1 d=2 d=3 term=a t=a t=a d=1 d=2 d=3 t=again t=again t=again This overall level-0 pTree corpusP length MaxDocLen*DocCount*VocabLen term=again doc1 ... term=a doc3 term=a doc2 term=a doc1 Any of these masks can be ANDed into the Pt= , d= pTrees before they are concatenated as above (or repetitions of the mask can be ANDED after they are concatenated). 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 . . . pTree Text Mining data Cube layout: ... Pos

  16. APPENDIX I have put together a pBase of 75 Mother Goose Rhymes or Stories. Created a pBase of the 15 documents with  30 words (Universal Document Length, UDL) using as vocabulary, all white-space separated strings. Little Miss Muffet Lev1 (term freq/exist) Lev-0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20... pos 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 . . . 182 te tf tf1 tf0 VOCAB Little Miss Muffet sat on a tuffet eating 1 2 1 0 a 0 0 0 0 0 1 0 0 0 0 0 0 again. 0 0 0 0 0 0 0 0 0 0 0 0 all 0 0 0 0 0 0 0 0 0 0 0 0 always 0 0 0 0 0 0 0 0 0 0 0 0 an 0 0 0 0 0 0 0 0 1 3 1 1 and 0 0 0 0 0 0 0 0 0 0 0 0 apple 0 0 0 0 0 0 0 0 0 0 0 0 April 0 0 0 0 0 0 0 0 0 0 0 0 are 0 0 0 0 0 0 0 0 0 0 0 0 around 0 0 0 0 0 0 0 0 0 0 0 0 ashes, 0 0 0 0 0 0 0 0 0 0 0 0 away 0 0 0 0 0 0 0 0 0 0 0 0 away 0 0 0 0 0 0 0 0 1 1 0 1 away. 0 0 0 0 0 0 0 0 0 0 0 0 baby 0 0 0 0 0 0 0 0 0 0 0 0 baby. 0 0 0 0 0 0 0 0 0 0 0 0 bark! 0 0 0 0 0 0 0 0 0 0 0 0 beans 0 0 0 0 0 0 0 0 0 0 0 0 beat 0 0 0 0 0 0 0 0 0 0 0 0 bed, 0 0 0 0 0 0 0 0 0 0 0 0 Beggars 0 0 0 0 0 0 0 0 0 0 0 0 begins. 0 0 0 0 0 0 0 0 1 1 0 1 beside 0 0 0 0 0 0 0 0 0 0 0 0 between 0 0 0 0 0 0 0 0 . . . 0 0 0 0 your 0 0 0 0 0 0 0 0 of curds and whey. There came a big spider and sat down... 0 0 0 0 0 0 1 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 1 0 0 0 0 0 0 1 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 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 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 Level-2 pTrees (document frequency) df3 df2 df1 df0 df VOCAB te04 te05 te08 te09 te27 te29 te34 1 0 0 0 8 a 1 1 0 1 0 0 0 0 0 0 1 1 again. 0 1 0 0 0 0 0 0 0 1 1 3 all 0 1 0 0 0 0 0 0 0 0 1 1 always 0 0 0 0 0 1 0 0 0 0 1 1 an 0 0 0 0 0 0 0 1 1 0 1 13 and 1 1 1 1 1 1 1 0 0 0 1 1 apple 0 0 0 0 0 0 0 0 0 0 1 1 April 0 0 0 0 0 0 0 0 0 0 1 1 are 0 0 0 0 0 0 0 0 0 0 1 1 around 0 0 0 0 0 0 0 0 0 0 1 1 ashes, 0 0 0 0 0 0 0 0 0 1 0 2 away 0 0 0 0 0 1 0 0 0 1 0 2 away 0 0 0 0 0 1 0 0 0 0 1 1 away. 1 0 0 0 0 0 0 0 0 0 1 1 baby 0 0 0 0 1 0 0 0 0 0 1 1 baby. 0 0 0 1 0 0 0 0 0 0 1 1 bark! 0 0 0 0 0 0 0 0 0 0 1 1 beans 0 0 0 0 0 0 1 0 0 0 1 1 beat 0 0 0 0 0 0 0 0 0 0 1 1 bed, 0 0 0 0 0 1 0 0 0 0 1 1 Beggars 0 0 0 0 0 0 0 0 0 0 1 1 begins. 0 0 0 0 0 0 0 0 0 0 1 1 beside 1 0 0 0 0 0 0 0 0 0 1 1 between 0 0 1 0 0 0 0 Humpty Dumpty Lev1 (term freq/exist) Lev-0 1 2 3 4 5 6 7 8... pos 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 . . . 182 te tf tf1 tf0 05HDS Humpty Dumpty sat on a wall. Humpt yDumpty 1 2 1 0 a 0 0 0 0 1 0 0 0 1 1 0 1 again. 0 0 0 0 0 0 0 0 1 2 1 0 all 0 0 0 0 0 0 0 0 0 0 0 0 always 0 0 0 0 0 0 0 0 0 0 0 0 an 0 0 0 0 0 0 0 0 1 1 0 1 and 0 0 0 0 0 0 0 0 0 0 0 0 apple 0 0 0 0 0 0 0 0 0 0 0 0 April 0 0 0 0 0 0 0 0 0 0 0 0 are 0 0 0 0 0 0 0 0 0 0 0 0 around 0 0 0 0 0 0 0 0 0 0 0 0 ashes, 0 0 0 0 0 0 0 0 0 0 0 0 away 0 0 0 0 0 0 0 0 0 0 0 0 away 0 0 0 0 0 0 0 0 0 0 0 0 away. 0 0 0 0 0 0 0 0 0 0 0 0 baby 0 0 0 0 0 0 0 0 0 0 0 0 baby. 0 0 0 0 0 0 0 0 0 0 0 0 bark! 0 0 0 0 0 0 0 0 0 0 0 0 beans 0 0 0 0 0 0 0 0 0 0 0 0 beat 0 0 0 0 0 0 0 0 0 0 0 0 bed, 0 0 0 0 0 0 0 0 0 0 0 0 Beggars 0 0 0 0 0 0 0 0 0 0 0 0 begins. 0 0 0 0 0 0 0 0 0 0 0 0 beside 0 0 0 0 0 0 0 0 0 0 0 0 between 0 0 0 0 0 0 0 0 . . . 0 0 0 0 your 0 0 0 0 0 0 0 0

  17. FAUST Clustering1 L-GapClustererCut, C, mid-gap (of F&C) using next (d,p) from dpSet, where F=L|S|R 2-1 separates 7,50 2-2 separates.27s 2^?1 0 -1 -2 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 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 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.27 0 0 0 1 0.55 0 0 1 0 0.55 0 0 1 0 3.60 1 1 1 0 D=d35 0 d26 0 d1 0 d27 0 d3 0 d44 0 d16 0 d6 0 d17 0 d47 0 d18 0 d10 0 d43 0 d12 0 d33 0 d14 0 d23 0 d49 0 d25 0 d45 0 d2 0 d29 0 d13 0 d9 0 d32 0.27 d28 0.27 d41 0.27 d42 0.27 d30 0.27 d21 0.27 d22 0.27 d15 0.27 d36 0.27 d11 0.27 d38 0.27 d46 0.27 d5 0.27 d8 0.27 d37 0.27 d48 0.27 d39 0.27 d4 0.55 d50 0.55 d7 3.60 d35 35, 7, 50 outliers D=.27s 0 d9 0 d49 0 d45 0.09 d6 0.09 d3 0.09 d33 0.09 d18 0.09 d44 0.18 d43 0.18 d25 0.18 d22 0.18 d12 0.18 d16 0.18 d2 0.27 d27 0.27 d23 0.27 d42 0.27 d15 0.27 d13 0.27 d47 0.36 d26 0.36 d29 0.36 d36 0.46 d38 0.46 d14 0.46 d48 0.46 d8 0.46 d10 0.46 d37 0.55 d32 0.55 d1 0.55 d5 0.64 d21 0.64 d4 0.64 d11 0.64 d17 0.92 d30 1.01 d41 1.01 d28 1.10 d39 1.29 d46 {28,30,39,41,46} Cluster D=.64s 0 d26 0 d33 0 d3 0 d27 0 d45 0 d2 0 d44 0 d23 0 d9 0 d15 0 d49 0 d16 0 d38 0 d6 0 d18 0 d22 0.25 d1 0.25 d37 0.25 d43 0.25 d8 0.25 d29 0.25 d25 0.25 d42 0.25 d12 0.25 d47 0.25 d48 0.51 d32 0.51 d14 0.51 d4 0.51 d36 0.51 d13 0.51 d5 0.77 d10 1.03 d11 1.29 d17 1.54 d21 the 0's, .25s, .51s are clusters. d10, d11, d17, d21 outliers Going back to D=d35, how close does HOB come? 21, 20 separate 35 C1 (.17  xod  .25)={2,3,6,16,18,22,42,43,49} D=sum of all C31docs 0.63 d17 0.63 d29 0.63 d11 0.84 d50 0.84 d13 0.84 d30 0.95 d26 0.95 d28 0.95 d10 0.95 d41 1.16 d21 C311(..63) ={11,17,29} C312(.84) ={13,30,50} C313(.95) ={10,26,28,41} 21 outlier C2 (.34  xod  .56)={1,4,5,8,9,12,14,15,23,25,27,32,33,36,37,38,44,45,47,48} C3 (.64xod.86)={10,11,13,17,21,26,28,29,30,39,41,50} Single: 46 (xod=.99); 7 (=1.16); 35 (=1.47) D=sum of allC2docs 0.27 d23 0.36 d25 0.36 d4 0.36 d38 0.45 d15 0.45 d33 0.45 d12 0.45 d36 0.54 d8 0.54 d44 0.54 d47 0.63 d1 0.63 d37 0.63 d5 0.63 d32 0.63 d50 0.72 d27 0.72 d45 0.72 d9 0.81 d14 Next, on each Ck try D=Ck, Thres=.2 D=sum of all C1docs 0.42 d16 0.42 d2 0.42 d3 0.42 d42 0.42 d43 0.42 d22 0.63 d18 0.63 d49 0.85 d6 C11(xod=.42)={231622,42,43} 6,18,49 outliers D=sum of all C11docs 0.57 d2 0.57 d3 0.57 d16 0.57 d22 0.57 d42 0.57 d43 D=sum of all C3docs 0.56 d11 0.66 d17 0.66 d29 0.75 d13 0.85 d30 0.85 d10 0.94 d28 0.94 d26 0.94 d41 0.94 d50 1.03 d21 1.41 d39 C31(.56xod1.03) ={10,11,13,17,21,26,28,29,30,41,50} 39 outlier Other Clustering methods later D=44docs GT=.08 0.17 d22 0.17 d49 0.21 d42 0.21 d2 0.21 d16 0.25 d18 0.25 d3 0.25 d43 0.25 d6 0.34 d23 0.34 d15 0.34 d44 0.34 d38 0.34 d25 0.34 d36 0.38 d33 0.38 d48 0.38 d8 0.43 d4 0.43 d12 0.47 d47 0.47 d9 0.47 d37 0.51 d5 0.56 d1 0.56 d32 0.56 d45 0.56 d14 0.56 d27 0.64 d10 0.64 d17 0.64 d21 0.64 d29 0.64 d11 0.69 d26 0.69 d50 0.69 d13 0.73 d30 0.77 d28 0.82 d41 0.86 d39 0.99 d46 1.16 d7 1.47 d35 C11: 2. This little pig went to market. This little pig stayed at home. This little pig had roast beef. This little pig had none. This little pig said Wee, wee. I can't find my way home. 3. Diddle diddle dumpling, my son John. Went to bed with his breeches on, one stocking off, and one stocking on. Diddle diddle dumpling, my son John. 16. Flour of England, fruit of Spain, met together in a shower of rain. Put in a bag tied round with a string. If you'll tell me this riddle, I will give you a ring. 22. Had a little husband no bigger than my thumb. I put him in a pint pot, and there I bid him drum. I bought a little handkerchief to wipe his little nose and a little garters to tie his little hose. 42. Bat bat, come under my hat and I will give you a slice of bacon. And when I bake I will give you a cake, if I am not mistaken. 43. Hark hark, the dogs do bark! Beggars are coming to town. Some in jags and some in rags and some in velvet gowns. C2: 1. Three blind mice! See how they run! They all ran after the farmer's wife, who cut off their tails with a carving knife. Did you ever see such a thing in your life as three blind mice? 4. Little Miss Muffet sat on a tuffet, eating of curds and whey. There came a big spider and sat down beside her and frightened Miss Muffet away. 5. Humpty Dumpty sat on a wall. Humpty Dumpty had a great fall. All the Kings horses, and all the Kings men cannot put Humpty Dumpty together again. 8. Jack Sprat could eat no fat. His wife could eat no lean. And so between them both they licked the platter clean. 9. Hush baby. Daddy is near. Mamma is a lady and that is very clear. 12. There came an old woman from France who taught grown-up children to dance. But they were so stiff she sent them home in a sniff. This sprightly old woman from France. 14. If all seas were one sea, what a great sea that would be! And if all the trees were one tree, what a great tree that would be! And if all the axes were one axe, what a great axe that would be! And if all the men were one man what a great man he would be! And if the great man took the great axe and cut down the great tree and let it fall into great sea, what a splish splash it would be! 15. Great A. little a. This is pancake day. Toss the ball high. Throw the ball low. Those that come after may sing heigh ho! 23. How many miles is it to Babylon? Three score miles and ten. Can I get there by candle light? Yes, and back again. If your heels are nimble and light, you may get there by candle light. 36. Little Tommy Tittlemouse lived in a little house. He caught fishes in other mens ditches. 37. Here we go round mulberry bush, mulberry bush, mulberry bush. Here we go round mulberry bush, on a cold and frosty morning. This is way we wash our hands, wash our hands, wash our hands. This is way we wash our hands, on a cold and frosty morning. This is way we wash our clothes, wash clothes, wash our clothes. This is way we wash our clothes, on a cold and frosty morning. This is way we go to school, go to school, go to school. This is the way we go to school, on a cold and frosty morning. This is the way we come out of school, come out of school, come out of school. This is the way we come out of school, on a cold and frosty morning. 38. If I had as much money as I could tell, I never would cry young lambs to sell. Young lambs to sell, young lambs to sell. I never would cry young lambs to sell. 44. The hart he loves the high wood. The hare she loves the hill. The Knight he loves his bright sword. The Lady loves her will. 47. Cocks crow in the morn to tell us to rise and he who lies late will never be wise. For early to bed and early to rise, is the way to be healthy and wealthy and wise. 48. One two, buckle my shoe. Three four, knock at the door. Five six, ick up sticks. Seven eight, lay them straight. Nine ten. a good fat hen. Eleven twelve, dig and delve. Thirteen fourteen, maids a courting. Fifteen sixteen, maids in the kitchen. Seventeen eighteen. maids a waiting. Nineteen twenty, my plate is empty. C311: 11. One misty moisty morning when cloudy was weather, I met an old man clothed all in leather. He began to compliment and I began to grin. How do And how do? And how do again 17. Here sits the Lord Mayor. Here sit his two men. Here sits the cock. Here sits the hen. Here sit the little chickens. Here they run in. Chin chopper, chin chopper, chin chopper, chin! 29. When little Fred went to bed, he always said his prayers. He kissed his mamma and then his papa, and straight away went upstairs. C312: 13. A robin and a robins son once went to town to buy a bun. They could not decide on plum or plain. And so they went back home again. 30. Hey diddle diddle! The cat and the fiddle. The cow jumped over the moon. The little dog laughed to see such sport, and the dish ran away with the spoon. 50. Little Jack Horner sat in the corner, eating of Christmas pie. He put in his thumb and pulled out a plum and said What a good boy am I! C313: 10. Jack and Jill went up the hill to fetch a pail of water. Jack fell down, and broke his crown and Jill came tumbling after. When up Jack got and off did trot as fast as he could caper, to old Dame Dob who patched his nob with vinegar and brown paper. 26. Sleep baby sleep. Our cottage valley is deep. The little lamb is on the green with woolly fleece so soft and clean. Sleep baby sleep. Sleep baby sleep, down where the woodbines creep. Be always like the lamb so mild, a kind and sweet and gentle child. Sleep baby sleep. 28. Baa baa black sheep, have you any wool? Yes sir yes sir, three bags full. One for my master and one for my dame, but none for the little boy who cries in the lane. 41. Old King Cole was a merry old soul. And a merry old soul was he. He called for his pipe and he called for his bowl and he called for his fiddlers three. And every fiddler, he had a fine fiddle and a very fine fiddle had he. There is none so rare as can compare with King Cole and his fiddlers three.

  18. FAUST Cluster 1.2 OUTLIER: 46. Tom Tom the piper's son, stole a pig and away he run. The pig was eat and Tom was beat and Tom ran crying down the street. WS0= 2 3 13 20 22 25 38 42 44 49 50 DS1= | WS1= 2 20 25 46 49 51 46 | DS2 46 DS0=|WS1= 7 10 17 23 25 28 33 34 37 40 43 45 50 35 |---| |DS2| |35 | OUTLIER: 35. Sing a song of sixpence, a pocket full of rye. 4 and 20 blackbirds, baked in a pie. When the pie was opened, the birds began to sing. Was not that a dainty dish to set before the king? The king was in his counting house, counting out his money. Queen was in the parlor, eating bread and honey. The maid was in the garden, hanging out the clothes. When down came a blackbird and snapped off her nose. WS0= 2 3 13 32 38 42 44 52 DS1 |WS1= 42(Mother) 7 9 |DS2|WS2=WS1 11 |7 27 |9 27 29 45 29 32 29 41 45 C1: Mother theme 7. Old Mother Hubbard went to the cupboard to give her poor dog a bone. When she got there cupboard was bare and so the poor dog had none. She went to baker to buy him some bread. When she came back dog was dead. 9. Hush baby. Daddy is near. Mamma is a lady and that is very clear. 27. Cry baby cry. Put your finger in your eye and tell your mother it was not I. 29. When little Fred went to bed, he always said his prayers. He kissed his mamma and then his papa, and straight away went upstairs. 45. Bye baby bunting. Father has gone hunting. Mother has gone milking. Sister has gone silking. And brother has gone to buy a skin to wrap the baby bunting in. WS0 22 38 44 52 DS1 WS1= 27 38 44 {fiddle(32 41) man(11 32) old(11 44) 11 DS2 32 11 41 22 44 C2 fiddle old man theme 11. One misty moisty morning when cloudy was weather, I chanced to meet an old man clothed all in leather. He began to compliment and I began to grin. How do you do How do you do? How do you do again 32. Jack come and give me your fiddle, if ever you mean to thrive. No I will not give my fiddle to any man alive. If I'd give my fiddle they will think I've gone mad. For many a joyous day my fiddle and I have had 41. Old King Cole was a merry old soul. And a merry old soul was he. He called for his pipe and he called for his bowl and he called for his fiddlers three. And every fiddler, he had a fine fiddle and a very fine fiddle had he. There is none so rare as can compare with King Cole and his fiddlers three. DS0|WS1 2 9 12 18 19 21 26 27 30 32 38 39 42 44 45 47 49 52 54 55 57 60 1 |DS1| WS2 12 19 26 39 44 10 |10 | DS2| WS3 13 | | 10 | DS3 10 17 37 14 39 21 41 26 44 28 30 47 50 OUTLIER: 10. Jack and Jill went up hill to fetch a pail of water. Jack fell down, and broke his crown and Jill came tumbling after. When up Jack got and off did trot as fast as he could caper, to old Dame Dob who patched his nob with vinegar and brown paper. DS0| WS1=2 9 18 21 30 38 41 45 47 49 52 54 55 57 60 1 | DS1|WS2=2 9 18 30 39 45 55 13 | 39 |DS2 14 | |39 17 21 39 28 41 30 47 37 50 OUTLIER: 39. A little cock sparrow sat on a green tree. He chirped and chirped, so merry was he. A naughty boy with his bow and arrow, determined to shoot this little cock sparrow. This little cock sparrow shall make me a stew, and his giblets shall make me a little pie. Oh no, says the sparrow I will not make a stew. So he flapped his wings\,away he flew C3: men three 1. Three blind mice! See how they run! They all ran after the farmer's wife, who cut off their tails with a carving knife. Did you ever see such a thing in your life as three blind mice? 5. Humpty Dumpty sat on a wall. Humpty Dumpty had a great fall. All the Kings horses, and all the Kings men cannot put Humpty Dumpty together again. 14. If all the seas were one sea, what a great sea that would be! And if all the trees were one tree, what a great tree that would be! And if all the axes were one axe, what a great axe that would be! And if all the men were one man what a great man he would be! And if the great man took the great axe and cut down the great tree and let it fall into the great sea, what a splish splash that would be! 17. Here sits the Lord Mayor. Here sit his two men. Here sits the cock. Here sits the hen. Here sit the little chickens. Here they run in. Chin chopper, chin chopper, chin chopper, chin! 23. How many miles is it to Babylon? Three score miles and ten. Can I get there by candle light? Yes, and back again. If your heels are nimble and light, you may get there by candle light. 28. Baa baa black sheep, have you any wool? Yes sir yes sir, three bags full. One for my master and one for my dame, but none for the little boy who cries in the lane. 36. Little Tommy Tittlemouse lived in a little house. He caught fishes in other mens ditches. 48. One two, buckle my shoe. Three four, knock at the door. Five six, pick up sticks. Seven eight, lay them straight. Nine ten. a good fat hen. Eleven twelve, dig and delve. Thirteen fourteen, maids a courting. Fifteen sixteen, maids in the kitchen. Seventeen eighteen. maids a waiting. Nineteen twenty, my plate is empty. WS0 38 52 DS1 WS1= 38 52 1 ---------- 5 17 23 28 36 48 C4: 4. Little Miss Muffet sat on a tuffet, eating of curds and whey. There came a big spider and sat down beside her and frightened Miss Muffet away. 6. See a pin and pick it up. All the day you will have good luck. See a pin and let it lay. Bad luck you will have all the day. 8. Jack Sprat could eat no fat. Wife could eat no lean. Between them both they licked platter clean. 12. There came an old woman from France who taught grown-up children to dance. But they were so stiff she sent them home in a sniff. This sprightly old woman from France. 15. Great A. little a. This is pancake day. Toss the ball high. Throw the ball low. Those that come after may sing heigh ho! 18. I had two pigeons bright and gay. They flew from me the other day. What was the reason they did go? I can not tell, for I do not k 21. Lion and Unicorn were fighting for crown. Lion beat Unicorn all around town. Some gave them white bread and some gave them brown. Some gave them plum cake, and sent them out of town. 25. There was an old woman, and what do you think? She lived upon nothing but victuals, and drink. Victuals and drink were the chief of her diet, and yet this old woman could never be quiet. 26. Sleep baby sleep. Our cottage valley is deep.Little lamb is on green with woolly fleece so soft, clean. Sleep baby sleep. Sleep baby sleep, down where woodbines creep. Be always like lamb so mild, a kind and sweet and gentle child. Sleep baby sleep. 30. Hey diddle diddle! The cat and the fiddle. The cow jumped over the moon. The little dog laughed to see such sport, and the dish ran away with the spoon. 33. Buttons, a farthing a pair! Come, who will buy them of me? They are round and sound and pretty and fit for girls of the city. Come, who will buy them of me? Buttons, a farthing a pair! 37. Here we go round mulberry bush, mulberry bush, mulberry bush. Here we go round mulberry bush, on a cold and frosty morning. This is way we wash our hands, wash our hands, wash our hands. This is way we wash our hands, on a cold and frosty morning. This is way we wash our clothes, wash our clothes, wash our clothes. This is way we wash our clothes, on a cold and frosty morning. This is way we go to school, go to school, go to school. This is the way we go to school, on a cold and frosty morning. This is the way we come out of school, come out of school, come out of school. This is the way we come out of school, on a cold and frosty morning. 43. Hark hark, the dogs do bark! Beggars are coming to town. Some in jags and some in rags and some in velvet gowns. 44. The hart he loves the high wood. The hare she loves the hill. The Knight he loves his bright sword. The Lady loves her will. 47. Cocks crow in the morn to tell us to rise and he who lies late will never be wise. For early to bed and early to rise, is the way to be healthy and wealthy and wise. 49. There was a little girl who had a little curl right in the middle of her forehead. When she was good she was very very good and when she was bad she was horrid. 50. Little Jack Horner sat in the corner, eating of Christmas pie. He put in his thumb and pulled out a plum and said What a good boy am I! WS0=2 5 8 11 14 15 16 22 24 25 29 31 36 41 44 47 48 53 54 57 59 DS1|WS1(17wds)=2 5 11 15 16 22 24 25 29 31 41 44 47 48 54 57 59 4 6 8|DS2=DS1 12 15 18 21 25 26 30 33 37 43 44 47 49 50 DS0|WS1 2 5 8 13 14 15 16 22 24 25 29 36 41 44 47 48 51 54 57 59 13 |DS2|WS2 4 13 47 51 54 14 |13 |DS3 13 21 26 30 37 47 50 OUTLIER: 13. A robin and a robins son once went to town to buy a bun. They could not decide on plum or plain. And so they went back home again. OUTLIERS: 2. This little pig went to market. This little pig stayed home. This little pig had roast beef. This little pig had none. This little pig said Wee, wee. I can't find my way home 3. Diddle diddle dumpling, my son John. Went to bed with his breeches on, one stocking off, and one stocking on. Diddle diddle dumpling, my son John. 16. Flour of England, fruit of Spain, met together in a shower of rain. Put in a bag tied round with a string. If you'll tell me this riddle, I will give you a ring. 22. Had little husband no bigger than my thumb. Put him in a pint pot, there I bid him drum. Bought a little handkerchief to wipe his little nose, pair of little garters to tie little hose 42. Bat bat, come under my hat and I will give you a slice of bacon. And when I bake I will give you a cake, if I am not mistaken. DS0|WS1=6 7 8 14 43 46 48 51 53 57 2 3|DS2=DS1 16 22 42 Each of the 10 words occur in 1 doc, so all 5 docs are outliers real HOB Alternate WS0, DS0 OUTLIER:38. If I had as much money as I could tell, I never would cry young lambs to sell. Young lambs to sell, young lambs to sell. I never would cry young lambs to sell. Notes Using HOB, the final WordSet is the document cluster theme! When the theme is too long to be meaningful (C4) we can recurse on those (using the opposite DS)|WS0?). The other thing we can note is that DS) almost always gave us an outliers (except for C5) and only WS) almost always gave us clusters (excpt for the first one, 46). What happens if we reverse it? What happens if we just use WS0?

  19. real HOB Alternate WS0, DS0 recuring on C3 and C4 FAUST Cluster 1.2.1 DS0|WS1=41 47 57 (on C4) 21|DS2 WS2=41(morn) 57(way) 26| 37 DS3=DS2 30| 47 . 37 47 50 C4.1 37. Here we go round mulberry bush, mulberry bush, mulberry bush. Here we go round mulberry bush, on a cold and frosty morning. This is way we wash our hands, wash our hands, wash our hands. This is way we wash our hands, on a cold and frosty morning. This is way we wash our clothes, wash our clothes, wash our clothes. This is way we wash our clothes, on a cold and frosty morning. This is way we go to school, go to school, go to school. This is the way we go to school, on a cold and frosty morning. This is the way we come out of school, come out of school, come out of school. This is the way we come out of school, on a cold and frosty morning. 47. Cocks crow in the morn to tell us to rise and he who lies late will never be wise. For early to bed and early to rise, is the way to be healthy and wealthy and wise. WS0=2 5 11 15 16 22 24 25 29 31 44 47 54 59 DS1|WS1=2 5 15 16 22 24 25 44 47 54 59 4 DS2 WS2=2 15 16 24 25 44 47 54 59 6 4 DS3 WS3=WS2 8 6 4 12 8 8 15 12 12 18 21 21 21 25 25 25 26 26 26 30 30 30 43 43 43 50 50 49 50 DS0|WS1= 47 (plum) 21 DS2 WS2=WS1 26 21 30 50 50 C4.2.1 word47(plum) 21. Lion &Unicorn were fighting for crown. Lion beat Unicorn all around town. Some gave them white bread and some gave them brown. Some gave them plum cake sent them out of town. 50. Little Jack Horner sat in corner, eating of Christmas pie. He put in his thumb and pulled out a plum and said What a good boy am I! WS0= 2 15 16 23 24 27 36 DS1|WS1 = 2 15 16 25 44 59 4 |DS2 WS2=15 16 25 44 59 8 |4 DS3 WS3=15 16 44 59 12 |8 8 DS4 WS4=15 44 59 25 |12 12 12 DS5 WS544 59 26 |25 25 25 12 DS6=DS5 30 |26 26 26 25 C4.2.2 word44(old) word59(woman) 12. There came an old woman from France who taught grown-up children to dance. But they were so stiff she sent them home in a sniff. This sprightly old woman from France. 25. There was old woman. What do you think? She lived upon nothing but victuals, and drink. Victuals and drink were the chief of her diet, and yet this old woman could never be quiet. Final WordSet is too long. Recurse 4.2 OUTLIER: 6. See a pin and pick it up. All the day you will have good luck. See a pin and let it lay. Bad luck you will have all the day. WS0= 5 11 22 25 29 31 DS1 WS1=5 22 6 1518 49 DS2 6 C4.2.3 (day eat girl) 4. Little Miss Muffet sat on tuffet, eating curd, whey. Came big spider, sat down beside her, frightened Miss Muffet away 8. Jack Sprat could eat no fat. Wife could eat no lean. Between them both they licked platter clean. 15. Great A. little a. This is pancake day. Toss the ball high. Throw the ball low. Those that come after may sing heigh ho! 18. I had 2 pigeons bright and gay. They flew from me other day. What was the reason they did go? I can not tell, for I do not know. 33. Buttons, farthing pair! Come who will buy them? They are round, sound, pretty, fit for girls of city. Come, who will buy ? Buttons, farthing a pair 49. There was little girl had little curl right in the middle of her forehead. When she was good she was very good and when she was bad she was horrid. DS0|WS1=22 25 29 4 DS2 =WS1 8 |4 8 15|15 18Recursing 18|33 49 no change 33 43 49 DS0|WS1=1 2 3 15 16 23 24 27 30 36 49 60 26 |DS1=DS0 30 Doc26 and doc30 have none of the 12 words in commong so these two will come out outliers on the next recursion! OUTLIERS: 26. Sleep baby sleep. Cottage valley is deep.Little lamb is on green with woolly fleece soft, clean. Sleep baby sleep. Sleep baby sleep, down where woodbines creep. Be always like lamb so mild, a kind and sweet and gentle child. Sleep baby sleep. 30. Hey diddle diddle! Cat and the fiddle. Cow jumped over moon.Little dog laughed to see such sport, and dish ran away with spoon. DS0=|WS1=21 38 49 52 1 |DS1 |WS2=21 38 49 14 |1 |DS3=DS2 17 |14 28 |17 C31 [21]cut [38]men [49]run 1. Three blind mice! See how run! All ran after farmer's wife, cut off tails with carving knife. Ever see such thing in life as 3 blind mice? 14. If all seas were 1 sea, what a great sea that would be! And if all trees were 1 tree, what a great tree that would be! And if all axes were 1 axe, what a great axe that would be! if all men were 1 man what a great man he would be! And if great man took great axe and cut down great tree and let it fall into great sea, what a splish splash that would be! 17. Here sits Lord Mayor. Here sit his 2 men. Here sits the cock. Here sits hen. Here sit the little chickens. Here they run in. Chin chopper, chin chopper, chin chopper, chin! C32: [38]men [52] three 5. Humpty Dumpty sat on wall. Humpty Dumpty had great fall. All Kings horses, all Kings men cannot put Humpty Dumpty together again. 23. How many miles to Babylon? 3 score miles and 10. Can I get there by candle light? Yes, back again. If your heels are nimble, light, you may get there by candle light. 28. Baa baa black sheep, have you any wool? Yes sir yes sir, three bags full. One for my master and one for my dame, but none for the little boy who cries in the lane. 36. Little Tommy Tittlemouse lived in a little house. He caught fishes in other mens ditches. 48. One two, buckle my shoe. Three four, knock at the door. Five six, pick up sticks. Seven eight, lay them straight. Nine ten. a good fat hen. Eleven twelve, dig and delve. Thirteen fourteen, maids a courting. Fifteen sixteen, maids in the kitchen. Seventeen eighteen. maids a waiting. Nineteen twenty, my plate is empty. WS0=38 52 DS1|WS1=WS0 5 | . 23 28 36 48 Doc43 and doc44 have none of the 6 words in commong so these two will come out outliers on the next recursion! OUTLIERS: 43. Hark hark, the dogs do bark! Beggars are coming to town. Some in jags and some in rags and some in velvet gowns. 44. The hart he loves the high wood. The hare she loves the hill. The Knight he loves his bright sword. The Lady loves her will. recurse on C3:

  20. HOB Alternate WS0, DS0 FAUST Cluster 1.2.2 eat girl day men 33 4 15 5 49 18 36 8 32 men 11 fiddle old 41 1 run cut 17 men 14 28 three 23 three 48 three morn old 37 12 47 25 way woman 16 OUTLIERS: 2 3 6 10 13 16 22 26 30 35 38 39 42 43 44 46 Categorize clusters (hub-spoke, cyclic, chain, disjoint...)? Separate disjoint sub-clusters? Each of the 3 C423 words gives a disjoint cluster!Each of the 2 C32 work gives a disjoint sub-clusters also. C4231 day 15. Great A. little a. This is pancake day. Toss ball high. Throw ball low. Those come after sing heigh ho! 18. I had 2 pigeons bright and gay. They flew from me other day. What was reason they go? I can not tell, I do not know. C4232 eat 4. Little Miss Muffet sat on tuffet, eat curd, whey. Came big spider, sat down beside her, frightened away 8. Jack Sprat could eat no fat. Wife could eat no lean. Between them both they licked platter clean. C4233 girl 33. Buttons, farthing pair! Come who will buy them? They are round, sound, pretty, fit for girls of city. Come, who will buy ? Buttons, farthing a pair 49. There was little girl had little curl right in the middle of her forehead. When she was good she was very good and when she was bad she was horrid. C1: mother 7. Old Mother Hubbard went to cupboard to give her poor dog a bone. When she got there cupboard was bare, so poor dog had none. She went to baker to buy some bread. When she came back dog was dead. 9. Hush baby. Daddy is near. Mamma is a lady and that is very clear. 27. Cry baby cry. Put your finger in your eye and tell your mother it was not I. 29. When little Fred went to bed, he always said his prayers. He kissed his mamma and then his papa, and straight away went upstairs. 45. Bye baby bunting. Father has gone hunting. Mother has gone milking. Sister has gone silking. And brother has gone to buy a skin to wrap the baby bunting in. C2: fiddle old men {cyclic} 11. 1 misty moisty morning when cloudy was weather, Chanced to meet old man clothed all leather. He began to compliment,I began to grin. How do you do How do? How do again 32. Jack come give me your fiddle, if ever you mean to thrive. No I'll not give fiddle to any man alive. If I'd give my fiddle they will think I've gone mad. For many joyous day fiddle and I've had 41. Old King Cole was merry old soul. Merry old soul was he. He called for his pipe, he called for his bowl, he called for his fiddlers 3. And every fiddler, had a fine fiddle, a very fine fiddle had he. There is none so rare as can compare with King Cole and his fiddlers three. C11 cut men run {cyclic} 1. Three blind mice! See how run! All ran after farmer's wife, cut off tails with carving knife. Ever see such thing in life as 3 blind mice? 14. If all seas were 1 sea, what a great sea that would be! And if all trees were 1 tree, what a great tree that would be! And if all axes were 1 axe, what a great axe that would be! if all men were 1 man what a great man he would be! And if great man took great axe and cut down great tree and let it fall into great sea, what a splish splash that would be! 17. Here sits Lord Mayor. Here sit his 2 men. Here sits the cock. Here sits hen. Here sit the little chickens. Here they run in. Chin chopper, chin chopper, chin chopper, chin! C321 men 5. Humpty Dumpty sat on wall. Humpty Dumpty had great fall. All Kings horses, all Kings men can't put Humpty together again. 36. Little Tommy Tittlemouse lived in little house. He caught fishes in other mens ditches. C322 three 23. How many miles to Babylon? 3 score 10. Can I get there by candle light? Yes, back again. If your heels are nimble, light, you may get there by candle light. 28. Baa baa black sheep, have any wool? Yes sir yes sir, 3 bags full. One for my master and one for my dame, but none for the little boy who cries in the lane. 48. One two, buckle my shoe. Three four, knock at the door. Five six, pick up sticks. Seven eight, lay them straight. Nine ten. a good fat hen. Eleven twelve, dig and delve. Thirteen fourteen, maids a courting. Fifteen sixteen, maids in the kitchen. Seventeen eighteen. maids a waiting. Nineteen twenty, my plate is empty. C4.1 morn way 37. Here we go round mulberry bush, mulberry bush, mulberry bush. Here we go round mulberry bush, on cold and frosty morn. This is way wash our hands, wash our hands, wash our hands. This is way wash our hands, on a cold and frosty morning. This is way we wash our clothes, wash our clothes, wash our clothes. This is way we wash r clothes, on a cold and frosty morning. This is way we go to school, go to school, go to school. This is the way we go to school, on a cold and frosty morning. This is the way we come out of school, come out of school, come out of school. This is the way we come out of school, on a cold and frosty morning. 47. Cocks crow in the morn to tell us to rise and he who lies late will never be wise. For early to bed and early to rise, is the way to be healthy and wealthy and wise. C421 plum 21. Lion &Unicorn were fighting for crown. Lion beat Unicorn all around town. Some gave them white bread and some gave them brown. Some gave them plum cake sent them out of town. 50. Little Jack Horner sat in corner, eating of Christmas pie. He put in his thumb and pulled out a plum and said What a good boy am I! C422 old woman 12. There came an old woman from France who taught grown-up children to dance. But they were so stiff she sent them home in a sniff. This sprightly old woman from France. 25. There was old woman. What do you think? She lived upon nothing but victuals, and drink. Victuals and drink were the chief of her diet, and yet this old woman could never be quiet. Let's pause and ask "What are we after?" Of course it depends upon the client. 3 main categories for relatioinship mining? text corpuses, market baskets (includes recommenders), bioinformatics? Others? What do we want from text mining? (anomalies detection, cliques, bicliques?) What do we want from market basket mining? (future purchase predictions, recommendations...) What do we want in bioinformatics? (cliques, strong clusters, ...???)

  21. FAUST Cluster 1.2.3 run 1 30 three 1 cut two old plum cut 23 three 10 fall brown 5 14 21 33 33 21 48 48 men three fall crown buy 48 three girl girl 5 14 king buy plum town 28 old three bread old 32 bake men buy back back 7 13 42 6 23 23 49 49 men 11 fiddle tree maid bad 7 bake sing men men old 41 36 day bake bread house town run 1 36 35 15 43 run dog cloth run 37 17 cloth 11 11 hill cut 37 morn way 14 dog men day old nose morn way day 32 22 eat dish 32 three day high fiddle round old son plum thumb king fiddle 18 44 47 bright wife 47 2 bed cock way old 41 41 pig 3 8 8 41 son round men fiddle pie eat merry mother 17 three clean mother 17 bed run cock always child away 4 26 29 4 old woman 12 away away 30 39 25 away 46 mother 29 old woman 12 25 eat cry mother green baby pie eat boy 9 50 50 boy 28 28 16 bag 9 lamb cry money mother baby 27 mother lamb 38 cry 27 baby full mother mother 45 lady 45 eat men word-labeled document graph We have captured only a few of the salient sub-graphs. Can we capture more of them? Of course we can capture a sub-graph for each word, but that might be 100,000. Let's stare at what we got and try to see what we might wish we had gotten in addition. A bake-bread sub-corpus would have been strong. (docs{7 21 35 42) A bake-bread sub-corpus would have been strong. (docs{7 21 35 42) There are many others. Using AVG+1 2 9 10 25 45 47 d21 0 0 1 0 0 1 d35 0 0 1 1 1 0 d39 1 1 0 0 1 0 d46 1 0 0 1 0 0 d50 0 1 0 1 1 1

  22. HOB2 Alt (use other HOBs) FAUST Cluster 1.2.4 run 1 30 three cut two old 10 fall brown 5 14 21 33 21 48 men three fall crown buy girl king buy plum town old bread bread old bake buy back back 7 13 42 6 23 49 men tree maid bad bake sing men 36 day bake bread house town run 35 35 15 43 dog cloth 37 cloth 11 hill 35 morn way dog day old nose 32 22 eat dish three day high fiddle round old son plum plum thumb king old 18 44 47 bright wife 2 bed cock way 41 pig 3 8 son round men fiddle pie pie eat merry mother 17 pie three clean 12. There came an old woman from France who taught grown-up children to dance. But they were so stiff she sent them home in a sniff. This sprightly old woman from France. bed run eat cock always child away 26 29 4 old woman 39 12 away away 30 39 25 away away 46 46 eat child 26 mother old woman 12 39 eat eat eat 46 cry pie boy pie boy green eat baby boy pie baby c o w h l o i d m l a d n 15 44 59 d12 1 1 1 9 50 50 boy 28 16 bag 9 50 lamb cry money mother baby 27 baby lamb 38 cry 27 baby full mother 45 39 lady boy boy 50 boy 28 wAvg+1, dAvg+1 a b b e p p w o r a i l a y e t e u y a m d 2 9 10 25 45 47 d21 0 0 1 0 0 1 d35 0 0 1 1 1 0 d39 1 1 0 0 1 0 d46 1 0 0 1 0 0 d50 0 1 0 1 1 1 recurse: wAv+2,dAvg-1 e p a i t e 2 9 10 25 45 d35 0 0 1 1 1 d39 1 1 0 0 1 d46 1 0 0 1 0 d50 0 1 0 1 1 And if we want to pull out a particular word cluster, just turn the word-pTree into a list.: w=baby a b w a b a y 2 3 d9 0 1 d26 1 1 d27 0 1 d45 1 w=boy a b w o a y 2 9 d28 0 1 d39 1 1 d50 0 1 For a particular doc cluster, just turn the doc-pTree into a list:

  23. FAUST HULL Classification 1 Using the clustering of FAUST Clustering1 as classes, we extract 80% from each class as TrainingSet (w class=cluster#). How accurate is FAUST Hull Classification on the remaining 20% plus the outliers (which should be "Other"). C11={2,3,16,22,42,43} C2 ={1,4,5,8,9,12,14,15,23,25,27,32,33,36,37,38,44,45,47,48} C11={3} C11={2,16,22,42,43} C311= {11,17,29} C312={13,30,50} C313={10,26,28,41} C2 ={4,14,23,45} C2 ={1,5,8,9,12,15,25,27,32,33,36,37,38,44,47,48} Full classes from slide: FAUST Clustering1 20% Test Set C311= {11,17} C312={30,50} C313={10,28,41} C311= {29} C312={13} C313={26} 80% Training Set OUTLIERS {18,49} {6} {39} {21} {46} {7} {35} O={18 49 6 39 21 46 7 35} .305 .439 C312 D11=C11 p=avC11 L MIN MAX CLASS .63 .63 C11 0 .63 C2 0 0 C311 .31 .31 C312 0 .31 C313 0 C311 .31 C312 0 .22 C11 .44 .66 C313 D2=C2 p=avC2 L MIN MAX CLS 0 .22 C11 .44 .77 C2 .66 .66 C311 .11 .22 C312 .44 .66 C313 -.09 .106 C11 D1=TS p=avTS Lpd MIN MAX CLASS -0.09 .106 C11 0.106 .439 C2 0.572 .572 C311 0.305 .439 C312 0.505 .771 C313 .572 C311 .11 .22 C312 .44 .77 C2 0 .31 C313 .63 C11 .106 .439 C2 .505 .771 C313 .66 C311 0 .63 C2 0 C11 .31 C11 .31 C2 .31 C311 .31 C313 D312=C312 p=avC312 L MN MX CLAS 0 .31 C11 0 .31 C2 0 .31 C311 1.58 1.58 C312 0 .31 C313 0 .22 C11 0 .44 C2 0 .44 C311 D311=C311 p=avC311 L MN MX CLAS 0 0 C11 0 0.66 C2 1.33 1.66 C311 0 0.33 C312 0 0.33 C313 D313=C313 p=avC313 L MN MX CLAS 0 .22 C11 0 .44 C2 0 .44 C311 0.22 .22 C312 1.34 1.56 C313 1.3 1.6 C311 0 .33 C312 0 .33 C313 1.34 1.56 C313 .22 C312 1.58 C312 0 .66 C2 D1=TS p=avTS Sp 4.2 C313 5.4 1.9 C11 2.1 2.4 C311 3.4 4.6 C312 4.7 1.8 C2 3.8 Use Lpd, Sp, Rpd with p=ClassAvg and d=unitized ClassSum. All 6 class hulls separated using Lpd, p=CLavg, D=CLsum. D311 separates C311, D312 separates C312 and D313 separates C313 from all others. D2 separates C11 and C2. Now, remove some false positives with S and R using the same p's and d's: D11=C11 p=avC11 Sp [1.6]C11 [3.4 4 4]C311 [5.4 6]C313 [2.4 4.4]C2 [5]C312 D2=C2 p=avC2 Sp [2 2.3]C11 [4.5 5.8]C313 [1.8 3.5]C2 [5 5.1]C312 [2.5 3.5]C311 D313=C313 p=avC313 Sp [3.5 4.2]C11 [6.5]C312 [2.8 6.2]C2 [3.8 6.2]C311 [2.5 3.5]C313 D311=C311 p=avC311 Sp [1.2]C311 [4.2]C11 [6.2 7.2]C312 [2.2 6.2]C2 [6.2 8.2]C313 D312=C312 p=avC312 Sp [3.5 4.5]C11 [6.5 7.5]C313 [4.5 6.5]C2 [2.5]C312 [5.5]C311 Sp removes a lot of the potential for false positives. (Many of the classes lie a single distance from p.) D11=C11 p=avC11 Rpd [1.2]C11 [1.4 2]C2 [1.7 2]C311 [2.2 2.]]C312 [2.2 2.4]C313 D1=TS p=avTS Rpd [1.3 1.4]C11 [1.3 1.9]C2 [1.5 1.8]C311 [2.1]C312 [2.0 2.2]C313 D2=C2 p=avC2 Rpd [1.3 1.4]C11 [1.3 1.8]C2 [1.6 1.8]C311 [2.2]C312 [2.1 2.4]C313 D313=C313 p=avC313 Rpd [1.3 1.4]C11 [1.3 2]C2 [1.6 2]C311 [2.2]C312 [1.5 1.8]C313 D312=C312 p=avC312 Rpd [1.3 1.4]C11 [1.4 2]C2 [1.7 1.9]C311 [1.5]C312 [2.2 2.4]C313 D311=C311 p=avC311 Rpd [1.4]C11 [1.2 2]C2 [1.1]C311 [2.2]C312 [2.2 2.4]C313 Rpd removes even more of the potential for false positives.

  24. Test Set FAUST Hull Classification 2 (TESTING) D1=TS p=avTS Rpd [1.3 1.4]C11 [1.3 1.9]C2 [1.5 1.8]C311 [2.1]C312 [2.0 2.2]C313 D1=TS p=avTS Sp C11={3} [1.9 2.1]C11 [2.4 3.4]C311 [4.2 5.4]C313 D1=TS p=avTS Lpd [4.6 4.7]C312 [1.8 3.8]C2 C2 ={4,14,23, 45} [.57]C311 [.31 .44]C312 [-.09 .11]C11 [.11 .44]C2 [.51 .77]C313 C311= {29} C312={13} C313={26} D11=C11 p=avC11 Rpd [1.2]C11 [1.4 2]C2 [1.7 2]C311 [2.2 2.]]C312 [2.2 2.4]C313 D11=C11 p=avC11 Sp [1.6]C11 [3.4 4 4]C311 [5.4 6]C313 [2.4 4.4]C2 [5]C312 D11=C11 p=avC11 Lpd [0]C311 [.31]C312 O={18 49 6 39 21 46 7 35} [.63]C11 [0 .31]C313 [0 .63]C2 D2=C2 p=avC2 Lpd D2=C2 p=avC2 Sp [2 2.3]C11 [4.5 5.8]C313 [1.8 3.5]C2 [5 5.1]C312 [2.5 3.5]C311 D2=C2 p=avC2 Rpd [1.3 1.4]C11 [2.1 2.4]C313 [1.3 1.8]C2 [1.6 1.8]C311 [2.2]C312 .[44 .66]C313 [0 .22]C11 [.44 .77]C2 [.66] C311 [.11 .22]C312 D311=C311 p=avC311 Sp [1.2]C311 [4.2]C11 [6.2 7.2]C312 [2.2 6.2]C2 [6.2 8.2]C313 D311=C311 p=avC311 Rpd [1.4]C11 [1.2 2]C2 [1.1]C311 [2.2]C312 [2.2 2.4]C313 D311=C311 p=avC311 Lpd [0]C11 [1.3 1.6]C311 [0 .33]C312 [0 .33]C313 [0 .66]C2 D312=C312 p=avC312 Lpd D312=C312 p=avC312 Sp [3.5 4.5]C11 [6.5 7.5]C313 [4.5 6.5]C2 [2.5]C312 [5.5]C311 D312=C312 p=avC312 Rpd [1.3 1.4]C11 [2.2 2.4]C313 [1.4 2]C2 [1.7 1.9]C311 [1.5]C312 .31 C11 .31 C2 .31 C311 .31 C313 1.58 C312 D313=C313 p=avC313 Rpd [1.3 1.4]C11 [1.3 2]C2 [1.6 2]C311 [2.2]C312 [1.5 1.8]C313 D313=C313 p=avC313 Sp [3.5 4.2]C11 [6.5]C312 [2.8 6.2]C2 [3.8 6.2]C311 [2.5 3.5]C313 D313=C313 p=avC313 Lpd [0 .22]C11 [0 .44]C2 [0 .44]C311 [1.3 1.6]C313 [.22]C312 ε=.8 predicted Class 11 2 2 2 311(all 311|2 all) 312(all 312|313 a Other . . . . . . . Other D=TS Rpd Sp Lpd trueCL Predicted____CLASS Final R S L predicted 1.41 2.19 -0.4 11 d3 2 Oth 11 Other 1.40 2.06 -0.3 2 d4 2 2 11 Other 1.92 3.71 0.01 2 d14 Oth 2 11 Other 1.38 1.99 -0.2 2 d23 2|11 2|11 Oth Other 1.97 3.92 -0.1 311 d29 Oth 312|313 11 Other 2.22 4.99 -0.2 312 d13 313 313 11 Other 2.60 6.78 -0.0 313 d26 Oth Oth 11 Other 1.40 2.13 -0.3 d6 2|11 2 11 Other 2.50 6.37 0.34 d7 313 Oth 2 Other 1.40 2.06 -0.3 d18 2|11 2|11 Oth Other 2.42 5.92 -0.1 d21 313 Oth Oth Other 3.46 12.2 0.47 d35 Oth Oth Oth Other 2.60 6.78 -0.0 d39 Oth Oth 11 Other 2.35 5.57 0.14 d46 Oth Oth 2 Other 1.41 2.19 -0.4 d49 2 2 Oth Other 8/15 = 53% correct just with D=TS p=AvgTS Note: It's likely to get worse as we consider more D's. Let's think about TrainingSet quality resulting from clustering. This a poor quality TrainingSet (from clustering Mother Goose Rythmes. MGR is a difficult corpus to cluster since: 1., in MGR, almost every document is isolated (an outlier), so the clustering is vague (no 2 MGRs deal with the same topic so their word use is quite different.). Instead of tightening the class hulls by replacing CLASSmin and CLASSmax by CLASSfpci (fpci=first percipitous count increase) and CLASSlpcd, we might loosen class hulls (since we know the classes somewhat arbitrary) by expanding the [CLASSmin, CLASSmax] interval as follows: Let A = Avg{ClASSmin, CLASSmax} and R (for radius) = A-CLASSmin (=CLASSmax-A also). Use [A-R-ε, A+R+ε]. Let ε=.8 increases accuracy to 100% (assuming all Other stay Other.). Finally, it occurs to me that Clustering to produce a TrainingSet, then setting aside a TestSet gives a good way to measure the quality of the clustering. If the TestSet part classifies well under the TrainingSet part, the clustering must have been high quality (produced a good TrainingSet for classification). This clustering quality test method is probably not new (check the literature?). If it is new, we might have a paper here? (discuss this quality measure and assess using different ε's?)

  25. WP Wed 11/26 Yes, we have discovered also that one has to think about the quality of the training set.   If it is very high quality (expected to fully encompass all borderline cases of all classes) then using exact gap endpoints is probably wise, but if there is reason to worry about the comprehensiveness of the training set (e.g., when there are very few training samples - which is often the case in medical expert systems where getting a sufficient number of training samples is difficult and expensive), then it is probably better to move the cutpoints toward the midpoint (reflecting the vagueness of training set class boundaries).  What does one use to decide how much to move away from the endpoints?  That's not an easy question.  Cluster deviation seems like a useful measure to employ. One last though on how to decide whether to cut at gap midpoint, endpoints, or to move the cut-points away from the endpoints toward the midpoint, If one has a time-stamp on training samples, one might assess the "class endpoint" change rate over time. As the training set gets larger and larger, if an endpoint stops moving much and isn't an outlier, then cutting at the endpoint seems wise.   If an endpt is still changing a lot, then moving away from that endpoint seems wise (maybe based on the rate of change of that endpoint as well as other measures?). A complete subgraph is a clique. A maximal clique is not a proper subset of any other clique. In G=(X,Y,E), a bipartite graph, a clique (Sx, Sy) is a complete bipartite subgraph induced by bipartite vertex set (Sx, Sy). The Consensus Set or clique of Sx, CLQ(Sx) = xSxNy(x), i.e., the set of all y's that are adjacent (edge connected) to every x in Sx. Clearly, (Sx, CLQ(Sx)) is a clique. Thm2:  SyY s.t. CLQ(Sy) ( CLQ(Sy), CLQ(CLQ(Sy)) ) is maximal. Thm1: (Sx, Sy) is a maximal clique iff Sy=CLQ(Sx) and Sx=CLQ(Sy) Find all cliques starting with Sy=singletons. Then examine Sy1y2-doubletons s.t. Px(Sy1y2) Then tripletons etc. Examining MGRs, (x=docs, y=words) all singleton wordsets, Sy, form a nonempty clique. AND pairwise to find all nonempty doubleton wordset cliques, Sy1y2. AND those nonempty doubleton wordset with each other singleton wordset to find all nonempty tripleton wordset cliques, Sy1y2y3... Start w singleton docs, incl another... until . The last nonempty set is a max-clique and all subsets are cliques. Remove them. Iterate. 7 13 w4 7 35 42 w7 7 35 w10 7 13 33 45 w13 7 30 43 w24 7 9 23 29 45 w42 7 10 11 12 25 41 w44 7 13 w4 w13 7 35 w7 w10 7 45 w13 w42 10 11 12 25 41 w44 10 14 w42 10 10 44 w32 10 21 w12 w19 2 37 w57 2 46 w45 2 47 w57 2 37 47 w57 #CLQs #docs #words 13 2 2 1 6 1 7 5 1 9 4 1 23 3 1 48 2 1 4 8 w25 4 29 w2 4 30 w2 4 35 w25 4 39 w2 4 46 w2 w25 4 50 w25 4 8 35 46 50 w25 4 29 30 39 46 w2 1 8 w58 1 14 w21 1 17 w49 1 23 w52 1 28 w52 1 30 w49 1 41 w52 1 46 w49 1 48 w52 1 8 none 1 14 none 1 17 30 46 w49 1 23 28 41 48 w52 1 28 23 41 48 1 30 17 46 1 41 23 28 48 1 46 17 30 1 48 23 28 41 11 12 25 41 w44 11 14 17 32 36 w38 11 35 37 w17 3 13 w51 3 29 w8 3 46 w51 3 47 w8 3 13 46 w51 3 29 47 w8 12 25 41 w44 12 25 w59 12 26 w15 12 25 w44 w59 5 10 14 w26 5 11 17 32 36 w38 5 36 41 w34 5 36 w34 w38 8 26 w16 8 35 46 50 w25 9 26 27 45 w3 9 27 29 45 w42 9 44 w35 9 45 w3 w42 6 15 18 32 w22 6 49 w5 • There is something wrong here. • This does not find all maximal cliques. • Next I try the following logic: • Find all 1WdC (1 Word Cliques). • A kWdC contains each of k (k-1)WdCs, so of a (k-1) wordset is not the wordset of a clique than none of its supersets are either (downward closure property). • Thus, the wordset of any 2WdCs can be composed by unioning the wordsets of two 1WdCs and any k WdCwordset is the union o f a (k-1)WdCwordset with a 1WdCwordset. 18 32 w22 18 44 w10 13 21 50 w47 13 23 w4 13 33 45 w13 13 21 43 w54 13 46 w51 13 21 w47 w54 14 17 32 36 w47 14 39 w55 16 33 37 w48 16 28 w6 23 28 41 48 w52 27 35 w43 27 50 w53 21 35 w10 21 42 w4 21 43 w54 21 50 w47 25 41 w44 17 30 46 w42 17 32 36 w38 17 39 47 w18 17 48 w56 28 35 w28 28 38 46 w20 28 39 50 w9 28 41 48 w52 15 18 32 w22 15 35 w50 15 44 w31 26 27 45 w3 26 28 w60 26 29 w1 26 38 w36 26 39 w30 22 35 w43 22 50 w53 29 45 w28 29 47 w8 29 30 39 46 w2 33 37 w48 33 45 w13 33 49 w29 37 47 w41 w57 45 38 46 w57 46 39 41 w39 39 46 w2 39 47 w18 39 50 w9 w45 47 50 w25 30 32 41 w27 30 35 w23 30 43 w24 30 46 w2 w49 35 36 w33 35 37 w17 35 38 w40 35 39 w45 35 41 w34 35 42 w7 48 49 41 48 w52 32 36 w22 32 41 w27 50 44 42 43 36

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0 1 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 1 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 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 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 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 & w1 1 1 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 00 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 & w2 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 1 1 2 0 1 0 0 1 0 0 00 0 0 0 0 1 0 0 1 0 0 1 1 0 0 2 0 1 0 0 0 1 0 0 0 0 0 & w3 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 • A first goal of our team might be to implement (in optimized Treeminer parallel code ), that is, a maximal complete subgraph finder (maximal clique finder). The benefits would be substantial! • This would be an exercise in parallel programming (e.g., in a TreeminerHadoop environment). • This is a typical exponential growth case. If you can find an engineering breakthru here, it will be a breakthru for a massive collection of similar existing big data parallel programming problems . & w4 0 0 1 0 0 1 0 0 2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 0 & w5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 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 & w7 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 & w8 0 0 2 0 0 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w9 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 & w10 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 & w11 0 1 1 1 0 0 1 0 1 0 0 0 1 1 1 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 & w12 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 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 & w13 0 1 0 0 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 & w14 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 0 0 0 What is the next step? We would have to have all the wh&wk, k>h results (not just the 1 counts) but that would have taken about 60 more slides to display ;-( Just looking at the pTee results of w1&wk k>1 above, we see that even though w1&w2 and w1&w3 both have counts=1, their AND (which is w1&w2&w3) has 0 count and therefore we need not consider any combinations of the type w1&w2&w3&… by the downward closure). In fact, the only wh for which we need to look further is h=42. Note that ct(w1&w2&w42)=1 but all other ct(w1&w2&wh)=0 The only maximal clique involving w1 and w2 is {DocSet={29}, WordsSet={1,2,42}, right? Next we would look at the pTrees of w2&wk, k>2. Clearly we only need to consider the 16 WDpTrees {8,9,18,20,23,24,25,27,30,39,42,45,46,49,51,55}, not all 58 of them. And going down to w30 the WDpTreeSet is (30,48} only To appreciate that we need engineering breakthroughs here, recall that a typical vocabulary might be 100,000 words, not just 60. & w15 0 0 0 0 1 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 1 0 0 0 0 0 0 1 0 0 0 0 0 0 & w16 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 & w17 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 & w18 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 1 1 0 0 1 1 0 1 1 0 1 1 1 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 & w19 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 & w20 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 & w21 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 0 1 & w22 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 & w23 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w24 1 1 0 1 1 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 & w25 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 & w26 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 2 1 1 0 1 1 1 0 1 0 0 0 0 1 0 0 & w27 0 0 0 0 0 1 0 1 0 0 0 2 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 & w28 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 & w29 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 & w30 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 & w31 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 & w32 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w33 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 & w34 1 0 0 1 1 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w35 0 0 1 1 1 1 0 0 1 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 & w36 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 & w37 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 & w38 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 & w39 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 & w40 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 & w41 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w42 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 2 0 0 0 & w43 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w43 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 & w44 0 0 0 0 0 0 0 1 0 0 0 0 0 0 2 0 & w45 0 1 0 0 1 0 0 1 0 1 0 0 0 0 0 & w46 0 0 1 0 1 0 0 0 0 0 1 0 0 0 & w47 0 0 0 1 0 1 2 0 0 0 0 0 0 & w48 0 0 0 0 0 0 0 0 1 0 0 0 & w49 0 1 1 0 0 0 1 0 1 0 0 & w50 0 0 0 0 0 0 0 0 0 0 & w51 0 0 1 0 0 0 0 0 0 & w52 0 0 0 1 0 1 0 1 & w53 0 0 0 0 0 0 0 & w54 0 0 0 0 0 0 & w55 0 0 0 0 0 & w56 0 0 0 0 & w57 0 0 0 & w58 0 0 & w59 0

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0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 & w4 0 0 1 0 0 1 0 0 2 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 0 & w5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 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 & w7 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 & w8 0 0 2 0 0 1 1 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 1 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w9 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 & w10 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 & w11 0 1 1 1 0 0 1 0 1 0 0 0 1 1 1 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 & w12 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 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 & w13 0 1 0 0 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 & w14 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 0 0 0 We can discard those subWordSets that have the same DocSet & w15 0 0 0 0 1 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 1 0 0 0 0 0 0 1 0 0 0 0 0 0 & w16 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 & w17 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 & w18 0 0 0 0 0 1 0 1 0 0 1 0 0 0 0 1 1 0 0 1 1 0 1 1 0 1 1 1 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 & w19 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 W1&wW 2 3 4 5 6 7 8 910 1 2 3 4 5 6 7 8 920 1 2 3 4 5 6 7 8 930 1 2 3 4 5 6 7 8 940 1 2 3 4 5 6 7 8 950 1 2 3 4 5 6 7 8 960 & w20 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 & w21 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 0 1 & w22 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 21 22 23 25 26 27 28 29 30 32 33 35 36 37 38 39 41 42 43 44 45 46 47 48 49 50 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 & w23 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w24 1 1 0 1 1 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 & w25 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 & w26 0 0 1 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 2 1 1 0 1 1 1 0 1 0 0 0 0 1 0 0 & w27 0 0 0 0 0 1 0 1 0 0 0 2 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 & w28 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 & w29 0 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 & w30 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 & w31 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 & w32 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w33 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 & w34 1 0 0 1 1 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w35 0 0 1 1 1 1 0 0 1 1 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 & w36 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 & w37 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 & w38 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 & w39 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 & w40 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 & w41 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 & w42 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 2 0 0 0 & w43 1 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 One possible way to process is to start with w1CountVector: & w43 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 & w44 0 0 0 0 0 0 0 1 0 0 0 0 0 0 2 0 & w45 0 1 0 0 1 0 0 1 0 1 0 0 0 0 0 & w46 0 0 1 0 1 0 0 0 0 0 1 0 0 0 & w47 0 0 0 1 0 1 2 0 0 0 0 0 0 & w48 0 0 0 0 0 0 0 0 1 0 0 0 & w49 0 1 1 0 0 0 1 0 1 0 0 & w50 0 0 0 0 0 0 0 0 0 0 & w51 0 0 1 0 0 0 0 0 0 & w52 0 0 0 1 0 1 0 1 & w53 0 0 0 0 0 0 0 & w54 0 0 0 0 0 0 & w55 0 0 0 0 0 & w56 0 0 0 0 & w57 0 0 0 & w58 0 0 & w59 0

  28. FAUST Hull Classification, F(x)=D1ox: Scalar pTreeSet (column of reals) , SPF(X) pTree calculated: mod( int(SP F(X)/(2exp) , 2 ) = SPD1oX = SPD1,iXi SPF(X)-min pD1,0 pD1,-1 pD1,-2 pD1,-3 pe1,0 pe1,-1 pe1,-2 pe1,-3 pe2,0 pe2,-1 pe2,-2 pe2,-3 F(a)= F(b)= F(c)= F(d)= F(e)= F(f)= F(g)= F(h)= 1*1.0 -½*3.0 = 1*1.5 -½*3.0 = 1*1.2 -½*2.4 = 1*0.6 -½*2.4 = 1*2.2 -½*2.1 = 1*2.3 -½*3.0 = 1*2.0 -½*2.4 = 1*2.5 -½*2.4 = 0.1 0.6 0.6 0 1.75 1.4 1.4 1.9 -.5 0 0 -.6 1.15 .8 .8 1.3 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1 1 0 1 1 0 1 0 0 1 1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 1 1 1 1 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 SPe1oX h1 [.6,1.5] h2 [2,2. 5] SPe2oX h1 [2.4,3] h2 [2.1,3] SPD1oX -mn h1 [0,.6] h2 [1.4,1.9] Idea: Incrementally build clusters one at a time using all F values. E.g., start with one pt, x. Recall F dis dominated, which means actual distance ≥ F difference. If the hull is close to convex hull, max Fdiff approximates distance? Then 1st gap in maxFdiss isolates x-cluster? (1.5,3) b (1,3) a 2.3,3) f c (1.2,2.4) g (2,2.4) d (.6,2.4) h (2.5,2.4) e (2.2,2.1) F(b=F(c)) F(a) F(f)=F(g) F(h) F(d) F(e) D1=(1 , -½)

  29. FAUST Hull Classification, F(x)=D1ox: Scalar pTreeSet (column of reals) , SPF(X) pTree calculated: mod( int(SP F(X)/(2exp) , 2 ) = SPD1oX = SPD1,iXi SPF(X)-min mxFdf(h) 1.8 1.3 1.3 1.9 .3 .6 .5 0 {e,f,g,h} h-core Gap=.7 Density= 4/.6^2= 11.1 mxFdf(e) 1.65 1.15 1.15 1.75 0 .9 .35 .3 {e,g,h} e-core Gap=.55 Density= 3/.35^2= 24.5 mxFdf(g) 1.3 .8 .8 1.4 .35 .6 0 .5 {b,c,e,f,g,h} g-core Gap=.5 Density= 6/.8^2= 9.4 mxFdf(f) 1.3 .8 1.1 1.7 .9 0 .6 .6 No f-core mxFdf(a) 0 .5 .6 .6 1.65 1.3 1.3 1.8 {a,b,c,d} a-core. Gap=.7 Density= 4/.6^2= 11.1 mxFdf(b) .5 0 .6 .9 1.15 .8 .8 1.3 No b-core mxFdf(c) .6 .6 0 .6 1.15 1.1 .8 1.3 No c-core mxFdf(d) .6 .9 .6 0 1.75 1.7 1.4 1.9 {a,b,c,d} d-core Gap=.5 Density= 4/.9^2= 4.9 F(a)= F(b)= F(c)= F(d)= F(e)= F(f)= F(g)= F(h)= 1*1.0 -½*3.0 = 1*1.5 -½*3.0 = 1*1.2 -½*2.4 = 1*0.6 -½*2.4 = 1*2.2 -½*2.1 = 1*2.3 -½*3.0 = 1*2.0 -½*2.4 = 1*2.5 -½*2.4 = 0.1 0.6 0.6 0 1.75 1.4 1.4 1.9 -.5 0 0 -.6 1.15 .8 .8 1.3 SPe1oX h1 [.6,1.5] h2 [2,2. 5] SPe2oX h1 [2.4,3] h2 [2.1,3] SPD1oX -mn h1 [0,.6] h2 [1.4,1.9] Incrementally build clusters 1 at a time with F values. E.g., start with 1 pt, x. Recall F dis dominated, which means actual separation ≥ F separation. If the hull is well developed (close to convex hull) max Fdiff approximates distance? Then 1st gap in maxFdis isolates x-cluster? (1.5,3) b (1,3) a 2.3,3) f g (2,2.4) c (1.2,2.4) d (.6,2.4) h (2.5,2.4) e (2.2,2.1) F(b=F(c)) F(a) F(f)=F(g) F(h) F(d) F(e) D1=(1 , -½)

  30. FAUST Oblique, F(x)=D1ox: Scalar pTreeSet (column of reals) , SPF(X) pTree calculated: mod( int(SP F(X)/(2exp) , 2 ) = SPD1oX = SPD1,iXi SPF(X)-min F(a)= F(b)= F(c)= F(d)= F(e)= F(f)= F(g)= F(h)= 1*3.0 -½*1.5 = 1*1.5 -½*3.0 = 1*1.2 -½*2.4 = 1*0.6 -½*2.4 = 1*2.2 -½*2.1 = 1*2.3 -½*3.0 = 1*2.0 -½*2.4 = 1*2.5 -½*2.4 = 2.85 0.6 0.6 0 1.75 1.4 1.4 1.9 2.25 0 0 -.6 1.15 .8 .8 1.3 (1.5,3) b 2.3,3) f c (1.2,2.4) g (2,2.4) d (.6,2.4) h (2.5,2.4) e (2.2,2.1) (3,1.5) a F(b=F(c)) F(a) F(f)=F(g) F(h) F(d) F(e) D1=(1 , -½)

  31. http://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmcluster.htmhttp://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmlearn.htmhttp://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmcluster.htmhttp://www.cs.ndsu.nodak.edu/~perrizo/saturday/teach/879s15/dmlearn.htm Clustering: Partition; • TODO: • Attribute selection prior to FAUST (for speed/accuracy, clustering/classification; Treeminer methods, hi pTree correlation with class?, hi info_gai/gini_index/other?,… In information theory and machine learning, information gain is a synonym for Kullback–Leibler divergence. The expected value of the information gain is the mutual information I(X; A) of X and A – i.e. the reduction in the entropy of X achieved by learning the state of the random variable A. In machine learning, this concept can be used to define a preferred sequence of attributes to investigate to most rapidly narrow down the state of X. Such a sequence (which depends on the outcome of the investigation of previous attributes at each stage) is called a decision tree. Usually an attribute with hi mutual info should be preferred to others. In general terms, the expected information gain is the change in information entropy from a prior state to a state that takes some information as given: Formal definition: Let denote a set of training examples, each of the form where is the value of the th attribute of example and is the corresponding class label. The information gain for an attribute is defined in terms of entropy as follows: The mutual information is equal to the total entropy for an attribute if for each of the attribute values a unique classification can be made for the result attribute. In this case, the relative entropies subtracted from the total entropy are 0. Drawbacks: Although information gain is usually a good measure for deciding the relevance of an attribute, it is not perfect. A notable problem occurs when information gain is applied to attributes that can take on a large number of distinct values. For example, suppose that one is building a decision tree for some data describing the customers of a business. Information gain is often used to decide which of the attributes are the most relevant, so they can be tested near the root of the tree. One of the input attributes might be the customer's credit card number. This attribute has a high mutual information, because it uniquely identifies each customer, but we do not want to include it in the decision tree: deciding how to treat a customer based on their credit card number is unlikely to generalize to customers we haven't seen before (overfitting). Information gain ratio is sometimes used instead. This biases the decision tree against considering attributes with a large number of distinct values. However, attributes with very low information values then appeared to receive an unfair advantage. In statistics, dependence is any statistical relationship between two random variables or two sets of data. Correlation refers to any of a class of statistical relationships involving dependence. Familiar examples of dependent phenomena include the correlation between the physical statures of parents and their offspring, and the correlation between the demand for a product and its price. Correlations are useful because they can indicate a predictive relationship that can be exploited in practice. For example, an electrical utility may produce less power on a mild day based on the correlation between electricity demand and weather. In this example there is a causal relationship, because extreme weather causes people to use more electricity for heating or cooling; however, statistical dependence is not sufficient to demonstrate the presence of such a causal relationship (i.e., correlation does not imply causation). Formally, dependence refers to any situation in which random variables do not satisfy a mathematical condition of probabilistic independence. In loose usage, correlation can refer to any departure of two or more random variables from independence, but technically it refers to any of several more specialized types of relationship between mean values. There are several correlation coefficients, often ρ or r, measuring the degree of correlation. Most common of these is Pearson correlation coefficient, sensitive only to a linear relationship between 2 variables. Other correlation coeffs have been developed to be more robust than Pearson correlation. Several sets of (x, y) points, with the Pearson correlation coefficient of x and y for each set. Note that the correlation reflects the noisiness and direction of a linear relationship (top row), but not the slope of that relationship (middle), nor many aspects of nonlinear relationships (bottom). N.B.: the figure in the center has a slope of 0 but in that case correlation coefficient is undefined because the variance of Y=0 Contents: Pearson's product-moment coefficient: Main article: Pearson product-moment correlation coefficient :The most familiar measure of dependence between two quantities is the Pearson product-moment correlation coefficient, or "Pearson's correlation coefficient", commonly called simply "the correlation coefficient". It is obtained by dividing the covariance of the two variables by the product of their standard deviations. Karl Pearson developed the coefficient from a similar but slightly different idea by Francis Galton.[4] The population correlation coefficient ρX,Y between two random variablesX and Y with expected values μX and μY and standard deviations σX and σY is defined as: where E is expected value operator, cov means covariance, and, corr an alternative notation for the correlation coefficient. Pearson correlation is defined only if both standard deviations are finite and nonzero. It is a corollary of the Cauchy–Schwarz inequality that the correlation cannot exceed 1 in absolute value. The correlation coefficient is symmetric: corr(X,Y) = corr(Y,X). The Pearson correlation is +1 in the case of a perfect direct (increasing) linear relationship (correlation), −1 in the case of a perfect decreasing (inverse) linear relationship (anticorrelation),[5] and some value between −1 and 1 in all other cases, indicating the degree of linear dependence between variables. As it approaches 0 there is less of a relationship (closer to uncorrelated). The closer coefficient is to either −1 or 1, stronger the correlation between variables. If variables are indep Pearson's correlation coeff is 0, but the converse is not true because the correlation coefficient detects only linear dependencies between two variables. E.g., random variable X is symmetrically distributed about 0, Y = X2. Then Y is completely determined by X, so that X and Y are perfectly dependent, but their correlation is zero; uncorrelated. Special case X and Y are jointly normal, uncorrelatedness = independence. If a series of n measmnts of X ,Y written xi and yii = 1...n, sample correlation coefficient can be used to estimate pop Pearson correlation r between X and Y. Sample correlation coeff is written where x and y are the sample means of X and Y, and sx and sy are the sample standard deviations of X and Y.

  32. Info Gain as an Attribute Selection Measure Minimizes expected number of tests needed to classify an object and guarantees simple tree (not the simplest) At any stage, let S = {s1,...,sm} be a TRAINING SUBSET. S[C] = {C1,...,Cc} be the distinct classes in S. EXPECTED INFORMATION needed to classify a sample given S as TRAINING SET is: I{s1,...,sm} = -∑i=1..mpi*log2(pi) pi= |S∩Ci|/|S| Choosing A as decision attribute, Expected Info gained is E(A) = ∑j=1..v; i=1..m ( si,j/|S| * I{sij..smj} ) where Skh = SA=ak∩Ch Gain(A) = I(s1..sm) - E(A) - expected reduction of info required to classify after splitting via A-values. The algorithm above computes the information gain of each attribute and selects the one with the highest information gain as the test attribute. The example: Training Data: Band B1: Band B2: Band B3: Band B4: 3 3 7 7 7 3 3 2 8 8 4 5 11 15 11 11 3 3 7 7 7 3 3 2 8 8 4 5 11 11 11 11 2 2 10 15 11 11 10 10 8 8 4 4 15 15 11 11 2 10 15 15 11 11 10 10 8 8 4 4 15 15 11 11 S: X-Y B1 B2 B3 B4 0,0 0011 0111 1000 1011 0,1 0011 0011 1000 1111 0,2 0111 0011 0100 1011 0,3 0111 0010 0101 1011 1,0 0011 0111 1000 1011 1,1 0011 0011 1000 1011 1,2 0111 0011 0100 1011 1,3 0111 0010 0101 1011 2,0 0010 1011 1000 1111 2,1 0010 1011 1000 1111 2,2 1010 1010 0100 1011 2,3 1111 1010 0100 1011 3,0 0010 1011 1000 1111 3,1 1010 1011 1000 1111 3,2 1111 1010 0100 1011 3,3 1111 1010 0100 1011 B1 is the class label (2, 3, 7, 10, 15 are C1,..,C5). We need to know the count of the number of pixels (rows in the table above) that contain each value in each attribute. We also need to know the count of pixels that contain pairs of values, one from a descriptive attribute and the other from the class label attribute. B2=(a1,a2,a3,a4,a5}={ 2, 3, 7,10,11 } as 1st candidate attribute. Aj={t: :t(B2)=aj}, a1=0010, a2=0011, a3=0111, a4=1010, a5=1011. sij is number of samples of class, Ci, in a subset, Aj. so sij=rc( P1(ci)^P2(aj) ), where ci{2,3,7,10,15}, aj{2,3,7,10,11}. • Therefore, • j=1 j=2 j=3 j=4 j=5 • --- --- --- --- --- • 0 0 0 0 .75 <- p1j • 0 .5 .5 0 0 <- p2j • 1 .5 0 0 0 <- p3j • 0 0 0 .25 .25 <- p4j • 0 0 0 .75 0 <- p5j • and • 0* 0 0 0 -.31 <- p1j*log2(p1j) • 0 -.5 -.5 0 0 <- p2j*log2(p2j) • 0 -.5 0 0 0 <- p3j*log2(p3j) • 0 0 0 -.5 -.5 <- p4j*log2(p4j) • 0 0 0 -.31 0 <- p5j*log2(p5j) • -- --- --- ---- ---- • 0 1 .5 .81 .81 <- I(s1j..s5j) • 2 4 4 4 4 <- s1j+..+s5j so that, • 0 .25 .125 .203 .203 (s1j+..+s5j)*I(s1j..s5j)/16 • E(B2)=.781 • I(s1..sm)=2.28 • GAIN(B2) -> 1.750 I(s1..sm) - E(B2) • GAIN(B3) -> 1.331 I(s1..sm) - E(B3) • GAIN(B4) -> .568 I(s1..sm) - E(B4) • For Info Gain, IG, we can (for that matter, we can do this for correlation too) • eliminate attributes based on an IG threshold or • use 1/IG as a weighting in the classification vote or • both! ENTROPY based on partition into subsets by B2 is E(B2)=j=1..v[ (s1j+..+smj)*I(s1j..smj)/s ] where Ij=I(s1j..smj)=- i=1..m [pij*log2(pij)], pij=sij/|Aj| Since m=5, the sij's are: j=1 j=2 j=3 j=4 j=5 --- --- --- --- --- 0 0 0 0 3 <-- s1j 0 2 2 0 0 <-- s2j 2 2 0 0 0 <-- s3j 0 0 0 1 1 <-- s4j 0 0 0 3 0 <-- s5j --- --- --- --- --- 2 4 2 4 4 <- s1j+..+s5j j=1 j=2 j=3 j=4 j=5 --- --- --- --- --- 2 4 2 4 4 <- |Aj| where Aj's are the rootcounts of P2(aj)'s. EXPECTED INFO needed to classify the sample is: I = I(s1..sm) = -SUM(i=1..m)[ pi * log2(pi) ], m=5 s=16 si=3,4,4,2,3 (rootcounts for class labels, rc(P1(ci))'s) pi = s1/s = 3/16, 1/4, 1/4, 1/8, 3/16 I = -(3/16*log(3/16)+4/16*log(4/16)+4/16*log(4/16) +2/16*log(2/16)+3/16*log(3/16)) = -( -.453 -.5 -.5 -.375 -.453 ) = -( -2.281) = 2.281

  33. From: Arjun Roy January 14, 2015 2:03 AM • For starting 'd', we have been using the vector from mean to median which has worked quite well. But I am curious to know if there is any other starting 'd' which would give even better separation between the classes. I could randomly choose some d's but also wanted to avoid too many computations. • WP: For clustering, we use the vector from mean to median for recursive clustering (to break apart a cluster into subclusters), but those vectors may not be very helpful for classification (of course, they are not hurtful either - the more d's the merrier for classification, right?  Computation cost is the only limiting issue.) • For classification, the vectors connecting pairs of class means (or medians or...) would be good vectors to pick (together with the freebees, e1=(1.0...0), ... , en=(0...0,1) ). In the picture below, d seems to connect the two class means? We might also take the "single link" class pair vectors (connecting the pair of points that give the single link distance between the two classes (=the min pairwise distance)).  Do we know how to find single link distance using pTrees?  (seems like we may have done that????).   I apologize if I am forgetting an idea already mentioned in MY lecture slides. ;-) Then of course we can look at the "complete link" vectors (max instead of min).   The "average link" vectors are those connecting the means. Of course the single link and complete link pairs can be approximate.  • For convex hull algorithm, I was looking for the minimum number of d required to build the class model (kind of an end condition). For e.g. in last Saturday slide, one d was sufficient enough (the red line) and black line computation was redundant. • WP: Remember the black ( using d=e2=(0,1) )and gold lines ( using d=e1=(1,0) ) are very helpful in eliminating false positives in the case where "none of the classes" is a possibility, right? (e.g., one-class classification).. So I recommend (as a minimal set of d's): 1. Use all ei (the freebees).2. Use all pairwise class mean connectors. • Then if more are desired, add the following in the order given:3. Main diagonals (connecting each corner of the (0,x2,...,xn) face with the opposite corner on the (1,x2,...,xn) face). The alg for main diagonal endpts:  Let x2,...,xn be any sequence of 0's and 1's.  D is a main diagonal iff D connects (0,x2,x3,...,xn) to (1,x2',x3',...,xn') where ' =bit complement (flip each bit).  So exactly 2^(n-1) main diagonals??4. All non-main diagonals (basically fix a set of coordinates at some fixed sequence of 0's and 1's then from the other coordinates, pick one, say k.  Run a vector from xk=0 with the fixed sequence and all others 0 or 1, to xk=1 with the fixed sequence and all others flipped.5.  Do 3 using xh=1/26. Do 4 using xh=1/2 • So the classifier could build initial hulls using 1 first (Use only the freebees , for a quick and dirty FAUST HULL classifier.).   • Then as the user is using those results, reclassify (add to the class hulls) in the background using 1 and 2. • Then as the user is using those results, reclassify in the background using 1 and 2 and 3. • Then as the user is using those results, reclassify in the background using 1 and 2 and 3 and 4. • Then as the user is using those results, reclassify in the background using 1 and 2 and 3 and 4 and 5. • Then as the user is using those results, reclassify in the background using 1 and 2 and 3 and 4 and 5 and 6 • Is computing variance the only way to choose the next d? I need to check with Mohammad how expensive it is.   Graphically, its intuitive to choose d just by looking at the points but algorithmically its quite a computational pain. WP:  Maybe you can re-phrase this question?  I'm not catching it. • WP: Alternatively, we could use the freebees, then use the PTM partitioning of the surface of the unit sphere (vectors from the origin to a vertex of a great circle triangle under PTM or equivalently, HTM), starting with the partition into 8 triangles (which gives just the feebees, e1,e2,e3), then descend one layer (subdivide all triangles by connecting all side midpoints, etc. Basically, again, at each level in PTM, we would run a vector from the northern hemisphere to the point in the southern hemisphere directly opposite it (or more simply, from the origin to each northern hemisphere vertex. That way, the vectors, D, are coincident with the vertexes of the triangles at that level (there are formulas in the literatire – see HTM – but it seems like we could get the same thing by: • use the freebees {e1..en}, which is the top level PTM triangulation. • use all V1={ ei ej | i=1,2,3; ij=2,3}, which I believe is the 2nd level PTM triangulation. • use all V2={ v + w | v ,w in V1}, • use all V3={ v + w | v ,w in V2}, …

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