## Slide #1.

FAUST Analytics X(X1..Xn)Rn, |X|=N. If X is a classified training set with classes=C={C1..CK}, X((X1..Xn,C}. d=(d1..dn), |d|=1. p=(p1..pn)Rn. We have functionals, F:RnR, F=L, S, R (as well as others, but these are the focus here). Ld,p  (X-p)od = Xod - pod = Ld - pod, (where LD=XoD for any vector, D) Sp  (X-p)o(X-p) = XoX + Xo(-2p) + pop = L-2p + XoX+pop Rd,p  Sp - L2d,p = XoX+L-2p+pop-(Ld)2-2pod*Xod+(pod)d2 = L-2p-(2pod)d - (Ld)2 +pop+(pod)2+XoX Assuming XoX is pre-calculated, for all 3, calculate Ld, L-2p and do pTree arithmetic (if just L and R, calculate Ld, L-2p-(2pod)d). Fmind,p,k= min(Fd,p&Ck), Fmaxd,p,k= max(Fd,p&Ck) FPCCd,p,k,j = jth precipitous count change (from left-to-right) of F d,p,k. Same notation for PCIs and PCDs (incr/decr) GAP: Gap Clusterer If DensityThreshold, DT, isn't reached, cut C mid-gap of Ld,p&C using the next (d,p) from dpSet PCC: Precipitous Count Change Clusterer If DT isn't reached, cut C at PCCsLd,p&C using the next (d,p) from dpSet Fusion step may be required? Use density, proximity, or use Pillar pkMeans (next slide). TKO: Top K Outlier Detector Use D2NN=rank2Sx for TopKOutlier-slider. or use RkiPtr(x,Ptr Ptr(x,PtrRankiSx). RkiSD(x,RankiSx) ordered as constructing desc on rankiSx. LIN: Linear Classifier y yCk iff y yLHk  {z | minLd,p,k  Ld,p,k(z)  maxLd,pd,k}  (d,p) (d,p)dpSet LHk is a Linear hull around Ck. dpSet is a set of (d,p) pairs, e.g., (Diag,DiagStartPt). LSR: Linear Spherical Radial Classifier yCk iff y yLSRHk{z | minFd,p,k Fd,p,k(z)  maxFd,p,k d,p d,pdpSet, F=L,S,R} (Examine and remove outliers first, then use first PCI instead of min and last PCD instead of max?) Express the Hulls as decision trees, one for every d. Then y isa k iff y isa k in every d-tree. Build each d-tree using Ld at the root and then from any multi-class inode use F=L,R,S with d=AvCiAvCj and p=AvCi distinct pair Ci, Cj, where Ci,Cj have nonempty restrictions at that node, using every F=L,S,R except the parent. This assumes convex classes. If it's known/suspected there are non-convex classes, judicious use of PCCs may provide tighter hulls. What should we pre-compute besides XoX? stats(min/avg/max/std); Xop; p=class_Avg/Med; Xod; Xox; d2(X,x); Rkid2(X,x); Ld,p, Rd,p We need a "Basic pTree Operations Timing Manual" to show users the cost of various pTree computations.
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## Slide #2.

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## Slide #3.

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## Slide #4.

FAUST L Hull classifier, recursive on MG44d60w d=q-p, q=0, p=avCk Ld,avC1 Ld,avC2 Ld,avC3 Ld,avC4 Ld,avC5 Ld,avC6 Ld,avC7 -0.2 0.08 1.73 1.73 0.61 0.86 0.99 1.19 0.75 0.75 0.57 0.97 0.58 0.94 1.45 1.45 -1.1 0.76 -0.0 0.99 0.78 0.99 0.5 1 0.16 0.97 0.58 0.94 0.77 1.45 1.15 1.73 -0.5 0.74 0.78 1.19 0.25 1 0.57 0.97 0.58 0.94 1.28 1.45 1.34 1.73 0.48 0.86 -0.2 0.78 0.75 0.75 0.36 0.97 0.23 0.94 0.77 1.45 1.34 1.73 0.74 0.99 0.78 1.19 -0.2 0.5 0.57 0.97 0.58 0.94 1.11 1.45 1.34 1.53 0.61 0.99 0.78 1.19 0.5 1 -0.6 0.36 0.23 0.94 1.11 1.45 1.53 1.73 0.74 0.99 0.99 1.19 0.75 1 0.77 0.77 -0.1 0.23 C1 0 overlpas C2 0 overlaps C3 overlaps C4 overlaps C5 overlaps C6 overlaps C7 overlaps w 1,2,4,5,6,7 w 2,3,5,6 w 2,3,6 w 2,4 w 4,6 d=q-p, q=0, p=maxCk C1:L3 C2:L34 C3:L47 C4:L49 C5:L8 C6:L52 Ld,avC7 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 d=q-p, q=0, p=avCk recursive Class sizes: 4 3 8 3 4 5 3 Ld,avC1 -.2, 1.45, 0.77, 1.28, 0.77, 1.11, 1.11, Since all points are outlierish, we should be able to carve off one class at a time by connecting averages? (Note, here I'm not even bothering to chop into intervals other than to carve off the Ck interval when it is non=overlapping with the others). Note: avCk' estimated as av(avCh | hk) .08 1.45 1.45 1.45 1.45 1.45 1.45 I1=[-.2 , .08] I2=[.77 ,1.12) I3=[1.12,1.28) I4=[1.29,1.46] Ld,avC2 on 0.9 , 2.9 0 , .57 0 , .38 0 , .38 0.19, .38 0 , .19 4000000 0010111 0030011 0343331 I4 C2 C3 C4 C5 C6 C7 I41=[ 0 , .19) I42=[.19, .38] I43=(.38, .57] I44=[.9 , 3] Ld,avC4 on 0.78,1.19 -0.2,0.78 0.78,1.19 0.78,1.19 FAUST L Recursing On Averages: 001201 042130 300000 I42 C3 C4 C5 C6 I421=[-.2, .78) I422=[.78,1.19] Lp=avC1,q=avC1' -0.2, 1.52, 1.14, 1.42, 0.85, 1.35, 1.13, 0.10 2.04 1.67 1.64 1.54 1.54 1.44 C1 Carve off C1: Lp=avC2,q=avC2' -1.1, 1.35, 1.48, 1.41, 1.37, 1.59, 0.70 1.82 2.00 1.88 1.66 1.80 C2 Carve off C2: Lp=avC3,q=avC3' 0200 4013 Ld,avC3 on I422 -0.5, 0.74 C3 0.74, 0.99 C5 0.61, 0.99 C6 I4221=[-.5, .61) 3000 I4222=[.61, .74) 1001 I4223=[.72, .99] 0012 0 C1 0 C2 5 C3 2 C4 2 C5 2 C6 2 C7 overlap overlap overlap overlap overlaps overlaps overlaps Consider the following interesting facts about the unit sphere and unit cube in high dimensions: pTrees lie in the nonnegative polytant of the unit sphere, npus,. So, they all lie on the sphere centered at (½,...,½)½ with radius=n/2. So it seems likely that ½ is a better centroid for pTree analysis than the origin. Other interesting observations: npus has hypervolume=1 no matter the dimension so lim ninfHypVol(npuc)=1. But we know that the volume of the nsphere goes to zero as n goes to infinity (albeit the descent to zero kicks in later the larger the radius). This means that the as n goes to infinity, the npus radius increases just fast enough to keep the hypervolume stable at 1. I put down these remarks before developing FAUST L Recursing On Averages, which seems to be great for classification of text!! It may save some time to use ½ instead of av(avCk') ? , which is a lazy computation anyway. we should keep sumC k and countCk then we can calculate avCk' precisely at almost no additional expense. But the main use of the ½ centroid may be in clustering. -0.4, 0.71, 0.91, 0.63, 0.84, 0.60 1.25 1.16 1.28 1.15 C3 Carve off C3: Lp=avC4,q=avC4' -0.1, 0.88, 0.98, 1.13, 0.67 1.58 1.58 1.36 C4 Carve off C4: Lp=avC5,q=avC5' -0.2, 0.34 0.97, 1.59 0.92, 1.28 C5 Carve off C5: Lp=avC6,q=avC6' -0.4, 0.52 1.01, 1.29 C6 Carve off C6:
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## Slide #5.

FAUST Thanksgiving q=½ clusterer (always use q=1/2, p=longest left in the table) 17FEC L Gap R Gap 12OWF 0.00 1.67 0.00 10.1 05HDS 1.68 0 10.18 0 28BBB 1.68 0 10.18 0 44HLH 1.68 0.12 10.18 0.55 49WLG 1.81 0 10.73 1 08JSC 1.81 0 11.73 -1 39LCS 1.81 0 10.73 0 37MBB 1.81 0 10.73 0 04LMM 1.81 0 10.73 -1 48OTB 1.81 0 9.73 1 06SPP 1.81 0.12 10.73 0.51 09HBD 1.94 0 11.25 -1 35SSS 1.94 0 10.25 1 47CCM 1.94 0 11.25 0 03DDD 1.94 0 11.25 -1 22HLH 1.94 0 10.25 1 14ASO 1.94 0 11.25 0 23MTB 1.94 0 11.25 0 46TTP 1.94 0 11.25 0 07OMH 1.94 0 11.25 0 29LFW 1.94 0 11.25 0 38YLS 1.94 0.12 11.25 0.48 32JGF 2.07 0 11.73 0 17FEC 2.07 0 11.73 0 45BBB 2.07 0 11.73 0 50LJH 2.07 0 11.73 0 36LTT 2.07 0 11.73 0 11OMM 2.07 0 11.73 0 25WOW 2.07 0 11.73 0 18HTP 2.07 0 11.73 0 10JAJ 2.07 0.12 11.73 0.45 33BFP 2.19 0 12.18 0 41OKC 2.19 0 12.18 0 02TLP 2.19 0 12.18 0 26SBS 2.19 0 12.18 1 16PPG 2.19 0 13.18 -1 27CBC 2.19 0 12.18 0 30HDD 2.19 0.12 12.18 -0.5 01TBM 2.32 0 11.60 0 43HHD 2.32 0 11.60 1 13RRS 2.32 0 12.60 -1 21LAU 2.32 0 11.60 1 42BBC 2.32 0.25 12.60 0.73 15PCD 2.58 -2.5 13.33 carve off 12OWF outlier. L Gap R Gap 39LCS 0.00 0.77 0.00 5.4 01TBM 0.77 0.12 5.40 0.78 38YLS 0.90 0 6.18 0 16PPG 0.90 0 6.18 0 27CBC 0.90 0 6.18 1 35SSS 0.90 0 7.18 -1 29LFW 0.90 0.12 6.18 0.75 17FEC 1.03 0 6.93 0 04LMM 1.03 0 6.93 1 28BBB 1.03 0 7.93 -1 06SPP 1.03 0 6.93 0 25WOW 1.03 0 6.93 1 44HLH 1.03 0.12 7.93 -0.2 50LJH 1.16 0 7.65 0 07OMH 1.16 0 7.65 0 42BBC 1.16 0 7.65 0 05HDS 1.16 0 7.65 0 10JAJ 1.16 0 7.65 0 08JSC 1.16 0 7.65 0 23MTB 1.16 0 7.65 1 37MBB 1.16 0 8.65 -1 22HLH 1.16 0.12 7.65 0.68 48OTB 1.29 0 8.33 0 03DDD 1.29 0 8.33 0 26SBS 1.29 0 8.33 0 21LAU 1.29 0 8.33 0 11OMM 1.29 0 8.33 0 49WLG 1.29 0 8.33 0 41OKC 1.29 0.12 8.33 0.65 47CCM 1.42 0 8.98 0 30HDD 1.42 0 8.98 0 33BFP 1.42 0 8.98 0 13RRS 1.42 0 8.98 1 15PCD 1.42 0 9.98 -1 46TTP 1.42 0.12 8.98 -1.3 36LTT 1.55 0 7.60 1 12OWF 1.55 0 8.60 0 14ASO 1.55 0 8.60 2 43HHD 1.55 0.12 10.60 -0.4 45BBB 1.68 0.12 10.18 0.55 32JGF 1.81 0 10.73 0 02TLP 1.81 0.25 10.73 1 18HTP 2.07 11.73 carve off 39LCS outlier L 50LJH 0.00 38YLS 0.90 06SPP 1.03 45BBB 1.03 14ASO 1.03 46TTP 1.03 15PCD 1.03 48OTB 1.03 26SBS 1.16 36LTT 1.16 29LFW 1.16 23MTB 1.16 39LCS 1.16 32JGF 1.16 28BBB 1.16 33BFP 1.16 11OMM 1.16 17FEC 1.16 37MBB 1.16 18HTP 1.16 35SSS 1.29 21LAU 1.29 03DDD 1.29 44HLH 1.29 01TBM 1.29 04LMM 1.29 10JAJ 1.42 30HDD 1.42 05HDS 1.42 13RRS 1.42 27CBC 1.42 02TLP 1.42 41OKC 1.42 22HLH 1.55 47CCM 1.55 08JSC 1.55 42BBC 1.55 43HHD 1.55 07OMH 1.68 49WLG 1.81 16PPG 1.81 25WOW 2.58 carve 50LJH Gap R Gap 0.90 0.00 7.18 0.12 7.18 -0.2 0 6.93 0 0 6.93 1 0 7.93 -1 0 6.93 0 0 6.93 1 0.12 7.93 -0.2 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0.12 7.65 0.68 0 8.33 0 0 8.33 0 0 8.33 0 0 8.33 0 0 8.33 0 0.12 8.33 0.65 0 8.98 0 0 8.98 1 0 9.98 -1 0 8.98 0 0 8.98 0 0 8.98 1 0.12 9.98 -1.3 0 8.60 0 0 8.60 1 0 9.60 0 0 9.60 1 0.12 10.60 -0.4 0.12 10.18 -1.4 0 8.73 2 0.77 10.73 2.6 13.33 as outlier mistakes mistakes!!! Anyway it just appears to be carving outliers and everything is eventually an outlier! L 0.00 22HLH 0.77 47CCM 0.77 23MTB 0.77 06SPP 0.90 45BBB 0.90 36LTT 0.90 35SSS 1.03 37MBB 1.03 18HTP 1.03 28BBS 1.03 13RRS 1.03 48OTB 1.03 38YLS 1.03 25WOW 1.03 46TTP 1.03 11OMM 1.03 30HDD 1.16 02TLP 1.16 42BBC 1.16 26SBS 1.16 49WLG 1.16 29LFW 1.16 03DDD 1.16 14ASO 1.16 32JGF 1.16 10JAJ 1.16 01TBM 1.29 27CBC 1.29 39LCS 1.29 08JSC 1.29 07OMH 1.29 50LJH 1.42 43HHD 1.42 33BFP 1.42 41OKC 1.42 15PCD 1.42 44HLH 1.55 21LAU 1.68 04LMM 1.68 05HDS 2.19 carve 17FEC Gap R Gap 0.77 0.00 5.4 0 5.40 0 0 5.40 1 0.12 6.40 0.78 0 7.18 0 0 7.18 -1 0.12 6.18 0.75 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 1 0 7.93 -1 0 6.93 0 0 6.93 0 0 6.93 0 0.12 6.93 0.71 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0.12 7.65 0.68 0 8.33 0 0 8.33 0 0 8.33 0 0 8.33 0 0.12 8.33 -0.3 0 7.98 2 0 9.98 -2 0 7.98 1 0 8.98 0 0.12 8.98 0.61 0.12 9.60 0.58 0 10.18 -2 0.51 8.18 4 12.18 05HDS outliers 25WOW 0.13 30HDD 0.65 17FEC 0.77 41OKC 0.77 35SSS 0.77 44HLH 0.90 33BFP 0.90 47CCM 0.90 04LMM 0.90 02TLP 0.90 32JGF 0.90 21LAU 0.90 07OMH 0.90 01TBM 0.90 28BBB 0.90 03DDD 0.90 36LTT 1.03 43HHD 1.03 26SBS 1.03 37MBB 1.03 23MTB 1.03 10JAJ 1.03 38YLS 1.03 14ASO 1.03 05HDS 1.03 08JSC 1.03 13RRS 1.16 50LJH 1.16 39LCS 1.16 27CBC 1.16 49WLG 1.29 42BBC 1.29 48OTB 1.29 29LFW 1.42 45BBB 2.32 carve 45BBB 0.51 -0.02 4.6 0.12 4.58 0.81 0 5.40 0 0 5.40 -1 0.12 4.40 1.78 0 6.18 0 0 6.18 0 0 6.18 0 0 6.18 0 0 6.18 0 0 6.18 0 0 6.18 1 0 7.18 -1 0 6.18 0 0 6.18 0 0.12 6.18 0.75 0 6.93 0 0 6.93 1 0 7.93 -1 0 6.93 0 0 6.93 1 0 7.93 -1 0 6.93 0 0 6.93 0 0 6.93 0 0.12 6.93 0.71 0 7.65 0 0 7.65 0 0 7.65 0 0.12 7.65 0.68 0 8.33 0 0 8.33 0 0.12 8.33 0.65 0.90 8.98 3.61 12.60 25WOW royalty? L 44HLH 0.13 23MTB 0.65 32JGF 0.65 22HLH 0.77 07OMH 0.90 46TTP 0.90 47CCM 0.90 29LFW 0.90 26SBS 1.03 35SSS 1.03 43HHD 1.03 02TLP 1.03 45BBB 1.03 42BBC 1.03 25WOW 1.03 05HDS 1.03 14ASO 1.03 11OMM 1.03 33BFP 1.03 39LCS 1.03 28BBB 1.16 10JAJ 1.16 01TBM 1.16 37MBB 1.16 38YLS 1.16 15PCD 1.16 04LMM 1.16 21LAU 1.29 13RRS 1.29 41OKC 1.29 48OTB 1.29 03DDD 1.29 08JSC 1.29 18HTP 1.42 50LJH 1.42 30HDD 1.55 49WLG 1.68 36LTT 1.68 27CBC 2.19 carve 44HLH Gap R Gap 0.51 -0.02 3.6 0 3.58 3 0.12 6.58 -1.1 0.12 5.40 0.78 0 6.18 1 0 7.18 -2 0 5.18 1 0.12 6.18 0.75 0 6.93 -1 0 5.93 1 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0 6.93 0 0.12 6.93 0.71 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0 7.65 0 0.12 7.65 0.68 0 8.33 -1 0 7.33 1 0 8.33 0 0 8.33 0 0 8.33 0 0.12 8.33 0.65 0 8.98 0 0.12 8.98 0.61 0.12 9.60 0.58 0 10.18 0 0.51 10.18 2 12.18 23MTB 26SBS 0.00 33BFP 0.65 35SSS 0.77 22HLH 0.77 32JGF 0.77 14ASO 0.90 07OMH 0.90 08JSC 0.90 42BBC 0.90 23MTB 0.90 25WOW 0.90 27CBC 0.90 05HDS 0.90 47CCM 0.90 02TLP 0.90 21LAU 0.90 03DDD 0.90 28BBB 1.03 15PCD 1.03 36LTT 1.03 29LFW 1.03 50LJH 1.03 48OTB 1.03 38YLS 1.03 41OKC 1.03 45BBB 1.03 01TBM 1.16 13RRS 1.16 43HHD 1.16 49WLG 1.16 44HLH 1.29 30HDD 1.29 17FEC 1.29 10JAJ 1.42 39LCS 1.55 37MBB 1.55 04LMM 2.32 carve 26SBS 0.64 0.12 0 0 0.12 0 0 0 0 0 0 0 0 0 0 0 0.12 0 0 0 0 0 0 0 0 0.12 0 0 0 0.12 0 0 0.12 0.12 0 0.77 04LMM 0.00 4.58 5.40 6.40 6.40 6.18 6.18 6.18 6.18 6.18 6.18 6.18 6.18 7.18 6.18 6.18 6.18 6.93 6.93 6.93 6.93 6.93 6.93 6.93 6.93 6.93 7.65 7.65 8.65 7.65 7.33 9.33 8.33 8.98 9.60 7.60 12.60 4.58 0.81 1 0 -0.2 0 0 0 0 0 0 0 1 -1 0 0 0.75 0 0 0 0 0 0 0 0 0.71 0 1 -1 -0.3 2 -1 0.65 0.61 -2 5
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## Slide #6.

FAUST Thanksgiving clusterer (ffa to ffffa (outliers in red)) p=ffa=35SSS q=ffffa=26SBS F Gap R Gap 0.00 2.4596 0.00 9.95 35SSS 2.46 0.2236 9.95 -0.1 07OMH 2.68 0 9.80 -1 21LAU 2.68 0 8.80 -2 41OKC 2.68 0 6.80 0 48OTB 2.68 0 6.80 0 05HDS 2.68 0 6.80 1 50LJH 2.68 0 7.80 -2 37MBB 2.68 0.2236 5.80 2.75 42BBC 2.91 0 8.55 0 01TBM 2.91 0 8.55 0 14ASO 2.91 0 8.55 -2 46TTP 2.91 0 6.55 0 23MTB 2.91 0 6.55 2 06SPP 2.91 0 8.55 -1 28BBB 2.91 0 7.55 -1 33BFP 2.91 0 6.55 3 02TLP 2.91 0 9.55 0 13RRS 2.91 0 9.55 -5 10JAJ 2.91 0 4.55 4 04LMM 2.91 0 8.55 -2 47CCM 2.91 0 6.55 -1 25WOW 2.91 0 5.55 4 38YLS 2.91 0.2236 9.55 -2.3 39LCS 3.13 0 7.20 -1 29LFW 3.13 0 6.20 0 09HBD 3.13 0 6.20 -2 12OWF 3.13 0 4.20 2 08JSC 3.13 0 6.20 0 27CBC 3.13 1.3416 6.20 -6.2 45BBB 4.47 -4.472 -0.00 0 26SBS p=ffa=39LCS q=ffffa=07OMH F Gap R Gap 0.00 1.3363 0.00 6.21 23MTB 1.34 0.2672 6.21 -1.7 28BBB 1.60 0 4.43 4 02TLP 1.60 0 8.43 -2 10JAJ 1.60 0 6.43 2 47CCM 1.60 0 8.43 -2 09HBD 1.60 0.2672 6.43 0.07 38YLS 1.87 0 6.50 -1 48OTB 1.87 0 5.50 2 04LMM 1.87 0 7.50 -1 33BFP 1.87 0 6.50 0 42BBC 1.87 0 6.50 -1 26SBS 1.87 0 5.50 1 25WOW 1.87 0 6.50 1 12OWF 1.87 0 7.50 -2 08JSC 1.87 0 5.50 1 27CBC 1.87 0.2672 6.50 1.92 50LJH 2.14 0 8.43 -4 14ASO 2.14 0 4.43 1 29LFW 2.14 0 5.43 2 45BBB 2.14 0 7.43 -2 13RRS 2.14 0 5.43 0 21LAU 2.14 0 5.43 0 39LCS 2.14 0 5.43 -1 35SSS 2.14 0 4.43 0 46TTP 2.14 0.2672 4.43 1.78 06SPP 2.41 0 6.21 -2 01TBM 2.41 1.3363 4.21 -4.2 41OKC 3.74 -0.00 37MBB p=ffa=21LAU q=ffffa=39LCS F Gap R Gap 0.00 1.1094 0.00 5.76 04LMM 1.11 0 5.77 0 27CBC 1.11 0.2773 5.77 3.30 48OTB 1.39 0 9.08 -5 09HBD 1.39 0.2773 4.08 3.15 26SBS 1.66 0 7.23 -2 12OWF 1.66 0 5.23 1 01TBM 1.66 0 6.23 0 23MTB 1.66 0 6.23 0 37MBB 1.66 0 6.23 -1 46TTP 1.66 0 5.23 1 33BFP 1.66 0 6.23 0 28BBB 1.66 0 6.23 0 02TLP 1.66 0 6.23 0 38YLS 1.66 0 6.23 0 10JAJ 1.66 0 6.23 0 21LAU 1.66 0 6.23 -1 41OKC 1.66 0.2773 5.23 3 14ASO 1.94 0 8.23 -4 50LJH 1.94 0 4.23 2 08JSC 1.94 0 6.23 -1 45BBB 1.94 0 5.23 1 25WOW 1.94 0 6.23 0 29LFW 1.94 0 6.23 2 13RRS 1.94 0 8.23 -1 35SSS 1.94 1.6641 7.23 -7.2 47CCM 3.61 -0.00 *****42BBC There came an old woman from France who taught grown children to Three blind mice! See how they run! They all ran after the farme How many miles is it to Babylon? Three score miles and ten. Can I Here we go round the mulberry bush, the mulberry bush, the mulber Tom Tom the pipers son, stole a pig and away he run. The pig was Buttons, a farthing a pair! Come, who will buy them of me? They a Baa baa black sheep, have you any wool? Yes sir yes sir, three ba This little pig went to market. This little pig stayed at home. If I had as much money as I could tell, I never would cry young l Jack and Jill went up the hill to fetch a pail of water. Jack fe The Lion and the Unicorn were fighting for the crown. The Lion be Old King Cole was a merry old soul. And a merry old soul was he. If all the seas were one sea, what a great sea that would be! And Little Jack Horner sat in the corner, eating of Christmas pie. He Jack Sprat could eat no fat. His wife could eat no lean. And so b Bye baby bunting. Father has gone hunting. Mother has gone milkin There was an old woman, and what do you think? She lived upon not When little Fred went to bed, he always said his prayers. He kiss A robin and a robins son once went to town to buy a bun. They cou Sing a song of sixpence a pocket full of rye. Four and twenty bla
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## Slide #7.

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## Slide #8.

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## Slide #9.

FAUST LSR classifier, recursive LR on IRIS X=X(X1..Xn,C); oultiers are O=O(O1...On,OC) ,OC)X init ; OT=Outlier_Thres; Carve off outliers from X into O (O:=O (O:=O{x|Rankn-1DNN(x,X) DNN(x,X)OT; O:=O O:=O{x|D2NN(X,x)>OT ).... We set OT=5 and carve off outliers: i7, i9, i10, i35, i36, i39, s42 There is TP purity except in C1=Le4-1[15,18]. We will apply R on each interval, then restrict to X = C1 and use R: d=e R on [15,18], p=avE then on [2.5,8.1], p=avI, then SAvI Note: using Le4=L(0001)=X4 only, the method gives 100% True Positive accuracy. 4 1 6 2.53 15 4 8.03 2.51 1 9.10 0 0 10 18 15 2.13 19.8 1.67 2.96 2 14 17 15 12 25 d=e3 10 19 R on [48,51], p=avE 30 51 6 48 14 69 Ld d=e1 p=orig S43 58 E 49 70 I 56 79 15 31 3 11 13 26 5 29 10 Ld d=e2 p=orig S 29 44 E20 34 I 22 38 28 15 6 1 26 23 16 26 2 then SAvI Using Le3=L(0010)=X3 only, the method gives 100% True Positive accuracy also. 3.2 3 Next we check the other two for TP accuracy, Le1 and Le2, then the question 9.6 will be "How should one choose the low number of columns to use in the text mining case, when there are 10,000, not just 4?" 13.7 1.5 Interestingly, each column, by itself, gives 13.7 6.0 2 100% TP accuracy using L,R and S (L once, -1 -1 -1 -1 -1 R on L [49,56) w R on L [56,58] w R on L (58,70] w R on R (4.38,9.97] R on R (1.6,2.2] then R mostly - S was used only when it got p=avS down to just 1 or 2 in a class). p=avS 0.8 p=avE .92 p=avI 2.2 p=avI 2.27 2.9 3 1 31 3.5 Next let's see if it works for text. I will take 7.8 9.97 15 2.27 10 10 10 10 26 13 MG44d60w, classify as many docs as I can 34 as an expert, then use the others as a test set. 19 4.38 1.6 1.66 40 5 2 11 33 18.2 6.6 2.18 43 Once I have a LSR Hull model built from the training set, I will classify each of the R on L-1[22,29) R R-1[9.69,12.8) R Le2-1[29,34] R on R-1(39.82,42.56] R Le2-1(34,38] test set documents using the FAUST p=avE .87 p=avI 2.3 p=avE 0.57 p=avS p=avI 55.83 15 3 Multiclass LSR hull Classifier which I 0.948 12.8 8 2.70 66.82 built. 69 22. 6.647 I will then look at the test doc classifications 9.69 9 8.12 2 2.061 19 32 2 and assess how good they are (basically to 35.7 10.2 2.061 25.07 see if the method reveals affinities that make 2.4does it now SAvE sense and that I hadn't notice when I put 42.56 5 together the "expert training set" in the 3.13 6 6 6.57 1.77 10.3 then on [3.13,6.57], p=avI, That would seem to be good news for text classification, for example, where there are thousands of columns and a need to reduce that number significantly. Of course, eliminating False Positives is also important but we think R will be effective in doing that (more effective than L) even in high dimensions. These are conjectures only, of course, and require careful further study. Next, we check to see if Le3 is as efficient wrt TP accuracy. We also note that the R numbers here are Sqrt(R) (i.e., actual radial reach, not radial reach squared). In practice, using pTree algebra, one would probably use radial reach squared since square root is a difficult pTree computation (would have to estimate using truncated Taylor series? Very ugly!). 30.7 45.7
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## Slide #10.

2.62 AvgDNN 2.44 MedDNN ENDIX Thanksgiving clustering (carve off clusters as one would carve a thanksgiving turkey) Let m be a furthest point from aAvgX (i.e., pt in X that maximizes SPTS, Sa=(X-a)o(X-a) ) If m is an outlier, carve {m} off from X. Repeat until m is a non-outlier. Construct L=Ld where d=am/|am| Carve off L-gapped cluster(s). Pick centroid, cc=mean of slice, SL: A. If (PCC2=PCD1) declare L-1[PCC1,PCC2] to be a cluster and carve it off (mask it off) of X; else (PCC2=PCI2 ) SLL-1[(3PCC1+PCC2)/4 ,PCC2) and look for a Scc or Rd,cc gap. If one is found, declare it to be a cluster and carve it off of X; Else add cc to the Cluster Centroid Set, CCS. B. Do A. from the high side of L also. Repeat until no new clusters carve off. If X (not completely carved up) use pkmeans on the remains with initial centroids, CCS One can also continue to carve using other vectors (e.g., mimj using pillars), before going to pkmeans. DNNS 16.0 7.34 6.32 6.24 5.56 5.38 5.38 4.89 4.89 4.58 4.35 4.24 4.24 4.12 4.12 4.12 IRIS: Carve off outliers i39 i7 i10 s42 i9 i36 i35 i15 e13 s23 i20 e15 i1 i32 i19 i18 m = i23 = 77 28 67 20 is furthest point from AvgX = 57.89 3 0.70 36.18 11.38) UDR[L(X)] Construct L=Ld where d=am/|am|= -0.12 -0.17 0.811 0.545 gap 1 count 1 L+3_val 0 1 1 1 1 1 1 1 6 10 14 10 2 3 4 5 1 15 3 2 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 3 1 2 4 1 3 2 3 3 5 6 2 6 7 5 1 4 3 2 3 5 5 4 1 4 4 4 1 1 1 6 7 22 25 27 29 30 31 32 33 34 35 36 37 39 4041 42 43 44 45 46 47 48 49 50 53 54 Carve off cluster C1=L-1[0,7] (s=48) No PCCs remain (except 1st and last) so add Avg(X) to CCS={(57.8 30.7 36.1 11.3)} Construct L=Ld where d=am/|am| = 0.58 0.01 0.80 0.08 UDR(LX) gap count value 1 1 0 2 1 1 4 1 3 1 3 7 2 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 3 2 4 4 2 2 3 5 3 4 7 4 2 2 3 4 1 5 3 2 3 1 1 1 2 1 2 8 10 11 12 13 14 15 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 37 39 (top) i39 i7 i10 s42 i9 i36 i35 i15 e13 s23 i20 e15 i1 i32 i19 i18 UDR(C3) L CT GP 0 2 2 2 1 1 3 1 1 4 2 1 5 2 1 6 1 4 10 1 1 11 2 2 13 1 1 14 1 UDR(C2) L CT GP 17 1 1 18 4 1 19 4 1 20 4 1 21 2 1 22 3 1 23 2 1 24 3 2 26 4 1 27 4 1 28 2 1 29 2 1 30 1 1 31 2 1 32 1 1 33 1 no gap or PCCs
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## Slide #11.

FAUST LSR classifier on IRIS X=X(X1,...,Xn,C); oultiers are O=O(O1,...,On,OC) ,OC)X initially empty; OT=Outlier_Threshold; Carve off outliers from X into O (O:=O (O:=O{x|Rankn-1Dis(x,X) Dis(x,X)OT; O:=O O:=O{x|Rankn-2Dis(X,x)>OT );...; Define class hulls: Hk{z {zRn | minFd,p,k  Fd,p,k(z)  maxFd,p,k (d,p) (d,p)dpSet, F=L,S,R} If y is in just one hull, declare y to be in that class. Elseif y is in multiple hulls, MH, declare y y the Ck that minimizes dis(y,Ck), k kMH (note dis(y,Ck)=dis(y,X)&Ck). Else (y is in no hulls), if dis(y,O) dis(y,O)min{dis(y,o)|o min{dis(y,o)|oO}=dis(y,oi)
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## Slide #12.

m1 m4 Finding the Pillars of X (So, e.g., the k can be chosen intelligently in k-means) :Let m1 be a point in X that maximizes the SPTS, dis2(X,a)=(X-a)o(X-a) where aAvgX If m1 is an outlier (Check using Sm1or better using D2NN?), repeat until m1 is a non-outlier. A point, m1, found in this manner is called a non-outlier pillar of X wrt a, or nop(X,a) ) Let m2  nop(X,m1) In general, if non-outlier pillars m1..mi-1 have been chosen, choose mi from nop(X,{m1,...,mi-1}) (i.e., mi maximizes k=1..i-1dis2(X,mk) and is a non-outlier). AvX1 m3 m2 (Instead of using Smi or D2NN to eliminate outliers each round, one might get better pillars by constructing Lmi-1mi:XR, eliminating outliers that show up on L, then picking the pillar to be the mean (or vector of medians) of (object, the slice L-1[(3PCC , PCC ? )at 0) 1+PCC2)/4 (all A PCC Pillar pkmeans clusterer: Assign each class) a ClassWeightReals CW2)init As we are identifying pillar mj's, compute Lmj = Xo(mj-mj-1) and Classes numbered as they are revealed. 1. For the next larger PCI in Ld(C), left-to-right. 1.1a If followed by PCD, CkAvg(Ld-1[PCI,PCD]) (or VoM). If Ck is center of a sphere-gap (or barrel gap), declare Classk and mask off. 1.1b If followed by another PCI, declare next Classk=the sphere-gapped set around Ck=Avg( Ld-1[ (3PCI1+PCI2)/4,PCI2) ). Mask it off. 2. For the next smaller PCD in Ld from the left side. 2.1a If preceded by a PCI, declare next Classk= subset of Ld-1[PCI, PCD] sphere-gapped around Ck=Avg. Mask off. 2.1b If preceded by another PCD declare next Classk=subset of same, sphere-gapped around Ck=Avg(Ld-1( [PCD2,(PCD1+PCD2)/4] ). Mask off @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ A potential advantage of the classifier: @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ FAUST Linear-Spherical-Radial (LSR) @ @ The parallel part lets us build a pair of L,S,R hull segments @ for every pTree computation (the more the merrier) @ Serial part allows possibility of better hull than ConvexHull @ @ E.g., in a linear step, if we not only use min and max but also PCIs @ and PCDs, potentially we could do the following on class=@: @ On each PCC interval (ill-defined but here [pci1L,pcd1L] (pcd1L,pci2L) [pci2L,pcd2L] @ Build hull segments on each interval and OR them? @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ Whereas the convex hull in orange (lots of false positives) @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ d @ @ @ @ @ @ @ @ @ @ @ @ @ @ minL = pci1L pcd1L pci2L maxL = pcd L
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## Slide #13.

2.62 AvgDNN 2.44 MedianDNN DNN or D2NN or D2NNS are powerful constructs REMEMBER! 1. The pTree Rule: Never throw a pTree away! outlier slider 2. In the process of creating D2NN we create, for each xX, the mask pTree of all nearest neighbors of x (all those points that tie as being nearest to x), which BTW, in high dimension is likely to be a large number. This is useful information (reason #1: no ties, maybe that one point is also an outlier? or?) In RANKk(x) pTree code, you may be able to see how we can compute all RANKk(x)s (all k) in parallel with efficiency (sharing sub-procedures). DNNS 16.0 7.34 6.32 6.24 5.56 5.38 5.38 4.89 4.89 4.58 4.35 4.24 4.24 4.12 4.12 4.12 (top portion) i39 GAP i7 8.68 i10 1.02 s42 0.07 i9 0.67 i36 0.18 i35 0 i15 0.48 e13 0 s23 0.31 i20 0.22 e15 0.11 i1 0 i32 0.11 i19 0 i18 0 If not, we can (serially) mask off the ties and apply RANKn-1 again to get RANKn-2 ( those points that are next nearest neighbors to x. I believe this has value too, e.g., if DNN(x)=1 and y is the only point in that mask of points distance=1 from x, and DNN(y)=1 and x is the only point distance=1 from y, then if RANKn-2(x)>outlier threshold+1, {x,y} is a doubleton outlier. With a little more work, tripleton and quadrupleton outliers can be identified, etc. At some point we have to stop and call the set a "small cluster" rather than an outlier polyton. If we construct tables, RANKk(x, Rkn-1Dis(x), PtrToRkn-1Mask(x),...,Rkn-kDis(x), PtrToRkn-kMask(x) ), we have a lot of global information about our dataset. It is a version of the "neighbor" network that is studied so actively for social networks, etc. (i.e., Rankn-1Mask(X) is a bit map of the edges emanating from x in the "nearest neighbors" network. DNN = 1 1.41 1.41 1.41 1.41 3.31 2.23 1 1.41 1.73 1 2.23 1.41 2.44 4.12 3.60 3.46 1 3.31 1.41 2.82 1.41 4.58 2 3 2 2 1.41 1.41 1.41 1.41 2.82 3.46 3.46 1.73 2.23 3 1.73 1.41 1 1.41 6.24 2 2.23 3.60 1.41 1.41 1.41 1 1.41 2.64 2.64 2.64 2 2.44 3 2.64 1.41 2.44 s1 s2 s3 s4 s5 s6 s7 s8 s9 s10 s11 s12 s13 s14 s15 s16 s17 s18 s19 s20 s21 s22 s23 s24 s25 s26 s27 s28 s29 s30 s31 s32 s33 s34 s35 s36 s37 s38 s39 s40 s41 s42 s43 s44 s45 s46 s47 s48 s49 s50 e1 e2 e3 e4 e5 e6 e7 e8 e9 Task: Construct a theory of large networks (or engineering handbook) using pTrees to identify edges (nearest nbrs). Rkn-2Mask(x) gives all pts "straight line distance" second closest to x, which we don't get in standard network theory. If y is 2 hops from x, we know y is a nearest nbr of a nearest nbr of x . We don't know how far away it is. Next we suggest that the Rkk calculations may be more efficiently done using UDR in one fell swoop. Why? 1. the UDR provides all of them. 2. UDR takes care of the duplicate problem (e.g., if looking for Nearest Nbr, it may not be Rankn-1 due to duplicates). 3. In the process of building UDR we get the Distribution Tree, which has lots of useful approximation information. We note that we still have to build DNN, D2NN, D2NNS one row at a time. 3.87 3.60 3 4.89 1.41 4.24 1.41 2 2.44 2.64 1.73 3 3.31 3.60 2.23 2 1.41 3.16 3.16 2 2 2.64 1.41 1.41 1.41 1.73 1.41 1.41 2 3.87 1.41 4.24 2.64 3.87 2.44 3 2.64 7.34 2.64 5.56 6.32 2.23 3.46 1.73 2.64 4.89 3 1.41 4.12 4.12 4.35 2.23 3.16 2.64 1.73 3 3.46 1.73 2.44 1 3.46 2.64 4.12 1 3.31 5.38 2.44 1.41 16.0 1.73 2.44 2.44 2.64 2.23 2.44 2.44 2.44 2.23 2.44 2.82 e10 e11 e12 e13 e14 e15 e16 e17 e18 e19 e20 e21 e22 e23 e24 e25 e26 e27 e28 e29 e40 e41 e42 e43 e44 e45 e46 e47 e48 e49 e50 i1 i2 i3 i4 i5 i6 i7 i8 i9 i10 i11 i12 i13 i14 i15 i16 i17 i18 i19 i20 i21 i22 i23 i24 i25 i26 i27 i28 i29 i30 i31 i32 i33 i34 i36 i37 i38 i39 i40 i41 i42 i43 i44 i45 i46 i47 i48 i49 i50 DNNS = Distance to Nearest Neighbor Sorted 16.0 7.34 6.32 6.24 5.56 5.38 5.38 4.89 4.89 4.58 4.35 4.24 4.24 4.12 4.12 4.12 4.12 3.87 3.87 3.87 3.74 3.60 3.60 3.60 3.60 3.46 3.46 3.46 3.46 3.46 3.46 3.46 3.31 3.31 3.31 3.31 3.31 3.16 3.16 3.16 3 3 3 3 3 3 3 3 2.82 2.82 2.82 2.82 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.64 2.44 2.44 2.44 2.44 i39 i7 i10 s42 i9 i36 i35 i15 e13 s23 i20 e15 i1 i32 i19 i18 s15 e49 e10 i3 e36 s16 s45 e11 e23 e30 s33 s34 i12 i26 i30 s17 s19 i34 e22 s6 e34 e27 i22 e28 e6 s37 i5 e21 i16 s25 e12 i25 e37 s32 i50 s21 e41 i31 i43 e2 i2 i8 e38 i23 e3 i14 e7 e19 i6 e1 i37 e5 i4 i45 2.44 2.44 2.44 2.44 2.44 2.44 2.44 2.44 2.44 2.23 2.23 2.23 2.23 2.23 2.23 2.23 2.23 2.23 2 2 2 2 2 2 2 2 2 2 2 1.73 1.73 1.73 1.73 1.73 1.73 1.73 1.73 1.73 1.73 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1.41 1 1 1 1 1 1 1 1 i28 i42 i49 i47 i41 i46 e18 e9 s14 i21 i48 i11 s7 e24 i44 s12 s36 s44 s24 e35 e29 s43 s27 e17 e48 e40 s26 e4 e25 i13 i27 i24 s38 e45 i40 e39 e20 s35 s10 e26 s41 e50 s47 s4 e44 s46 e42 e47 e33 e46 e31 s5 e16 s39 e8 s31 s3 s30 s13 s29 i17 i38 s2 s28 s50 s22 e43 s20 e14 e32 s48 s9 s40 i33 i29 s18 s11 s8 s49 s1
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## Slide #14.

Computing the Rank values and Rank pTrees, one at a time, using our pTree code. X P4,3 P4,2 P4,1 P4,0 10 1 0 1 0 5 0 1 0 1 6 0 1 1 0 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 (n=3) c=Count(P&P4,3)= 3 < 6 {0} p=6–3=3; P=P&P’4,3 masks off highest 3 (n=2) c=Count(P&P4,2)= 3 >= 3 (val 8) {1} P=P&P4,2 masks off lowest 1 (n=1) c=Count(P&P4,1)=2 < 3 p=3-2=1; P=P&P'4,1 masks off highest 2 (n=0) c=Count(P&P4,0 )=1 >= 1 (val 4) {0 (val}8-2=6 ) {1} P=P&P4,0 RankKval=0; p=K; c=0; P=Pure1; /*Note: n=bitwidth-1. The RankK Points are returned as the resulting pTree, P*/ For i=n to 0 {c=Count(P&Pi); If (c>=p) {RankVal=RankVal+2i; P=P&Pi }; else {p=p-c; P=P&P'i }; return RankKval, P; /* Above K=7-1=6 (looking for the Rank6 or 6th highest vaue (which is also the 2nd lowest value) */ RankKval= 23 * {0}+ 22 * {1}+ 21 * {0}+ 20 * {1}= 5 P=MapRankKPts= 0 1 0 0 0 0 0 ListRankKPts={2}
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## Slide #15.

What if there ar duplicates? X P4,3 P4,2 P4,1 P4,0 10 1 0 1 0 3 0 0 1 1 6 0 1 1 0 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 (n=3) c=Count(P&P4,3)= 3 < 6 {0} p=6–3=3; P=P&P’4,3 masks off 1s (n=2) c=Count(P&P4,2)= 2 < 3 {0} p=3-2=1 P=P&P'4,2 masks off 1s (n=1) c=Count(P&P4,1)=2 >= 1 {1} P=P&P 4,1 masks off 0s (none) (n=0) c=Count(P&P4,0 )=2 >= 1 {1} P=P&P4,0 X P4,3 P4,2 P4,1 P4,0 10 1 0 1 0 3 0 0 1 1 7 0 1 1 1 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 (n=3) c=Count(P&P4,3)= 3 < 5 {0} p=5–3=2; P=P&P’4,3 masks off 1s (n=2) c=Count(P&P4,2)= 2 >= 2 {1} P=P&P4,2 masks off 0s (n=1) c=Count(P&P4,1)=2 >= 2 {1} P=P&P 4,1 masks off 0s (none) (n=0) c=Count(P&P4,0 )=2 >= 2 P=P&P4,0 {1}
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## Slide #16.

X P4,3 P4,2 P4,1 P4,0 10 1 0 1 0 3 0 0 1 1 7 0 1 1 1 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 (n=3) c=Count(P&P4,3)= 3 < 4 {0} p=4–3=1; P=P&P’4,3 masks off 1s (n=2) c=Count(P&P4,2)= 2 >= 1 {1} P=P&P4,2 masks off 0s (n=1) c=Count(P&P4,1)=2 >= 1 {1} P=P&P 4,1 masks off 0s (none) (n=0) c=Count(P&P4,0 )=2 >= 1 {1} P=P&P4,0 X P4,3 P4,2 P4,1 P4,0 10 1 0 1 0 3 0 0 1 1 7 0 1 1 1 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 (n=3) c=Count(P&P4,3)= 3 >= 3 {1} P=P&P4,3 masks off 0s (n=2) c=Count(P&P4,2)= 0 < 3 p-3-0=3 P=P&P'4,2 masks off 1s (n=1) c=Count(P&P4,1)=2 < 3 p=3-2=1 P=P&P' 4,1 masks off 1s (n=0) c=Count(P&P4,0 )=1 >= 1 P=P&P4,0 {0} (none ) {0} {1}
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## Slide #17.

X P4,3 P4,2 P4,1 P4,0 10 1 0 1 0 3 0 0 1 1 7 0 1 1 1 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 (n=3) c=Count(P&P4,3)= 3 P4,2 P4,1 P4,0 10 1 0 1 0 3 0 0 1 1 7 0 1 1 1 7 0 1 1 1 11 1 0 1 1 9 1 0 0 1 3 0 0 1 1 {1} P=P&P4,3 masks off 0s (n=2) c=Count(P&P4,2)= 0 {0} < 2 p-2-0=2 P=P&P'4,2 masks off 1s (n=1) c=Count(P&P4,1)=2 >=2 (none ) {1} P=P&P 4,1 masks off 0s (n=0) c=Count(P&P4,0 )=1 < 2 P=P&P4,0 X P4,3 >= 2 {0} mask off 1s (n=3) c=Count(P&P4,3)= 3 >= 1 {1} P=P&P4,3 masks off 0s (n=2) c=Count(P&P4,2)= 0 p-1-0=1 P=P&P'4,2 masks off 1s (n=1) c=Count(P&P4,1)=2 >=1 P=P&P 4,1 masks off 0s (n=0) c=Count(P&P4,0 )=1 <= 1 P=P&P'4,0 {0} < 1 (none ) {1} {1} mask off 0s So what we get is really the same output as the UDR but it seems more expensive to calculate. Unless all we need is Rank(n-1), but then we won't know for sure that there are no duplicates. We have to check Rank(n-2), Rank(n-3), ... until we see a non-duplicate.
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## Slide #18.

UDR Univariate Distribution Revealer (on Spaeth:) Y y2 y3 y4 y5 y6 y7 y8 y9 ya pb yc yd ye yf f= y1 y1 1 3 2 3 6 9 15 14 15 13 10 11 9 11 7 y2 1 1 2 3 2 3 1 2 3 4 9 10 11 11 8 yofM 11 27 23 34 53 80 118 114 125 114 110 121 109 125 83 applied to S, a column of numbers in bistlice format (an SpTS), will 15 produce the DistributionTree of S DT(S) 5 p6 p5 p4 p3 p2 p1 p0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 0 1 1 0 1 0 1 1 0 1 0 0 0 0 1 1 1 0 1 1 0 1 1 1 0 0 1 0 1 1 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 0 1 1 0 1 0 0 1 1 depth=h=1 node p6'p5'p4'p3 1 0 1 0 1 1 0 1 0 0 1 0 0 0 1 0 1 0 0 1 1 0 0 1 0 1 p6'p5'p4 p3' 0 1 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 0 0 1 1[8,16) 2/16[16,32) 1[16,24) 1[24,32) p6 p5'p4'p3' 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 10/64 [64,128) p6 p5'p4'p3 0 1 0 1 0 1 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 p6 p5'p4 p3' 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 2[80,96) 2[80,88) 0[64,80) 10 node2,3 p6' p5' p4' p3' p2' p1' p0' 3 2 2 8 [96.128) 1 1 1 0 1 0 0 1 1 0 0 1 0 0 1 1 0 1 0 0 0 2 1 1 0 2 2 6 1 0 1 1 1 0 11 1 0 0 1 0 1 0 0 1 0 1 1 1 1 0 0 0 1 0 0 1 0 0 0 1 1 0 01 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0 2 0 0 2 3 3 0 0 0 1 1 0 1 depthDT(S)b≡BitWidth(S) h=depth of a node k=node offset 0 0 1 0 0 0 1 depthDT(S) 0 0 0 0 1 1 0 0 0 1 0 0 1 0 Nodeh,k has a ptr to pTree{x pTree{xS | F(x) F(x)[k2b-h+1, (k+1)2b-h+1)} and 0 0 0 0 0 1 0 0 1 0 1 1 0 0 its 1count p6'p5'p4'p3' 1 0 0 1 1 0 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 0 0 0 1 5/64 [0,64) 3/32[0,32) 1/16[0,16) 0[0,8) 2/32[64,96) depth=h=0 p6'p5'p4 p3 0 1 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 1 0 0 1 1 0 1 0 p6'p5 p4'p3' 1 0 1 0 0 1 1 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 p6'p5 p4'p3 1 0 0 1 1 0 1 0 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 p6'p5 p4 p3' 1 0 1 0 0 1 0 1 1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 p6'p5 p4 p3 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 1 1 0 0 1 0 1 0 1 0 1 1 0 1[32,48) 1[32,40) 0[40,48) 1[48,64) 1[48,56) 0[56,64) p6 p5'p4 p3 0 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 p6 p5 p4'p3' 0 1 0 1 0 0 1 1 0 1 0 0 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 p6 p5 p4'p3 1 0 1 0 0 0 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 1 1 0 1 0 p6 p5 p4 p3' 0 1 0 1 0 0 1 0 1 0 1 1 1 0 1 1 0 1 0 1 0 1 0 1 1 0 p6 p5 p4 p3 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 1 1 0 1 1 0 0[88,96) ¼[96,128) 2[96,112) 0[96,104) 2[194,112) 6[112,128) 3[112,120) 3[120,128) 2/32[32,64) Pre-compute and enter into the ToC, all DT(Yk) plus those for selected Linear Functionals (e.g., d=main diagonals, ModeVector . Suggestion: In our pTree-base, every pTree (basic, mask,...) should be referenced in ToC( pTree, pTreeLocationPointer, pTreeOneCount ).and these OneCts should be repeated everywhere (e.g., in every DT). The reason is that these OneCts help us in selecting the pertinent pTrees to access - and in
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## Slide #19.

FAUST Oblique LSR Classification on IRIS150 We create hull boundaries for the d=ek standard basis vectors and check for overlaps. Then the only goals is to reduce False Positives. Ld d=1000 p=origin S43 E 49 I 49 59 S 23 E 20 I 22 71 80 16 34 24 6 26 32 12 i1 i7 i14 i15 i22 i43 63 49 57 58 56 58 33 25 25 28 28 27 1 of the 6 occurs, i7. 60 45 50 51 49 51 Ld d=0010 p=origin Ld d=0100 p=origin 25 17 20 24 20 19 44 34.1 38.1 1 2 1 29 47 46 15 3 6 of the 6 occur. i4 i9 i17 i20 i24 i27 i28 i34 i35 i38 i39 i50 63 67 65 60 63 62 61 63 61 64 60 59 29 25 30 22 27 28 30 28 26 31 30 30 56 58 55 50 49 48 49 51 56 55 18 51 18 18 18 15 18 18 18 15 14 18 18 18 S 10 19 E 30 I 18 Ld d=0001 p=origin 48 2 1 0 50 15 6 i2 i7 i11 i14 i15 i20 i22 i24 i27 i28 i34 i42 i43 i47 i50 58 49 65 57 58 60 56 63 62 61 63 69 58 63 59 27 25 32 25 28 22 28 27 28 30 28 31 27 25 30 S 1 6 E 10 18.1 I 14 26 51.1 69 34 51 45 51 50 51 50 49 49 48 49 51 51 51 50 51 50 19 17 20 20 24 15 20 18 18 18 15 23 19 19 18 i4 i7 i8 i9 i17 i20 i24 i26 i27 i28 i30 i34 i35 i38 i39 28 22 16 63 49 73 67 65 60 63 72 62 61 72 63 61 64 60 29 25 29 25 30 22 27 32 28 30 30 28 26 31 30 34 56 45 63 58 55 50 49 60 48 49 58 51 56 55 18 18 17 18 18 18 15 18 18 18 18 16 15 14 18 18 What does this tell us 6 of the 16 occur. about FAUST LSR on IRIS? 5 of the 6 occur. The LSR hulls are 96% True Positive accurate on IRIS using only the pre-computed min and max of each given column, PL, PW, SL, SW as cut points (no further pTree calculations beyond the attribute min and max pre-calculations). That's pretty good! Note i7 and i20 are prominent outlies (see IRIS_DNNS on slide 4) so if we had eliminated outliers first using DNNS, the TPaccuracy is 97.3% Next we address False Positives. How does one measure FP accuracy? One way would be to measure the area of Hull-Class for each Class. That would give us a FP accuracy for each Class. The sum of those would give us an FP accuracy for the model. These areas are difficult numbers to calculate however, for many reasons. First, what do we mean by Class? The mathematical convex hull of the Class? How do we calculate area? An easier way would be to measure up a large set of IRIS samples, none of which are Setosa, Versicolor or Virginica. The problem with this approach is that other varieties may well share some or all measurements with S, E and I, so we would not expect to be able to separate them into an "other" class using the data we have. So a vector space based FP assessment might be preferable. Since area of the symmetric difference is hard to calculate, how about measuring the maximum distance from any hull corner to it's closest class point (or the sum of those distances?)? Easier, use max distance to the main corners only. That's easy enough for strictly linear hulls but what about hulls that have S and R components? Since the above is a linear hull, we use it. The main corners are: MIN VECTOR MAX VECTOR MnVecDis MxVecDis s 43 23 10 1 59 44 19 6 4.1 4.9 The sum of the distances to class corner vectors is 38.1, average is 6.4. e 49 20 30 10 71 34 51 18 4.4 5.1 i 49 22 18 14 80 38 69 26 14.2 5.4
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## Slide #20.

FAUST LSR Classification on IRIS150, a new version 1. If you're classifying individual unclassified samples one at a time, applying these formulas gives 100% accuracy in terms of true positives (assuming the given training set fully characterizes the classes). S43 58 We have used just d=1000 so many more edges could be placed on these hulls to eliminate false positives. E 49 70 I 49 79 2. If there is a whole table of unclassified samples to be classified (e.g., millions 16 34 24 6 26 32 12 or billions) then it might be time-cost effective to convert that table to a pTreeSet and then convert these inequalities to pTree inequalities (EIN Ring technology) to L d=avgI-avgE p=origin 0 p=AvE p=AvgS 99 accomplish the classification as one batch process (no loop required). E 5.74 15.9 270 50 34 15 2 393 792 I 13.6 16.6 26 5 This is the {y isa EI)2 recursive step pseudo code: 1096 1217 1558 if 270  R1000,AvgS(y) < 792 {y isa I} 1826 2568 24,0 2,4 0,1 elseif 792  R1000,AvgS(y)  1558 {y isa EI}3 L d=avgI-avgE p=origin elseif 1558  R (y)  2568 {y isa I} p=AvgE 1000,AvgS This is the {y isa EI} recursive step: E 1.78 22.69 3 else {y isa O} I 6.26 31.021 1 if 5.7  LAvE-AvI(y) < 13.6 {y isa E } 35.51 elseif 13.6  L (y)  15.9 {y isa EI}4 1,0 0,1 54.32 AvE-AvI elseif 15.9 < L (y)  16.6 {y isa I} if 43  L1000(y)=y1 < 49 {y isa S } AvE-AvI else {y isa O } elseif 49  L1000(y)=y1  58 {y isa SEI}1 This is the {y isa EI) recursive step pseudo code: 4 elseif 59 < L1000(y)=y1  70 {y isa EI}2 if 22.69 RAvE-AvI,AvgE(y)<31.02 {y isa E } elseif 70 < L1000(y)=y1  79 {y isa I} elseif 31.02 RAvE-AvI,AvgE(y)35.51 {y isa EI}5 else {y isa O } elseif 35.51 RAvE-AvI,AvgE(y)54.32 {y isa I} This is the {y isa SEI)1 recursive step pseudo code: {y isa O } if 0  R1000,AvgS(y)  99 {y isa S } else This is the {y isa EI}5 recursive step: elseif 99 < R1000,AvgS(y) < 393 {y isa O } if 1.78=LAvgE-AvgI,origin(y) {y isa E } elseif 393 < R1000,AvgS(y)  1096 {y isa E } elseif 6.26=LAvgE-AvgI,origin(y) {y isa I} elseif 1096 < R1000,AvgS(y) < 1217 {y isa O } else {y isa O } elseif 1217  R1000,AvgS(y)  1826 {y isa I} else {y isa O } L1000,origin(y) [43,49)[49,58](58,70](70,79]else OTHER yS yI Ld d=1000 p=origin MinL, MaxL for classes S,E,I d d R1000,AvgE(y) [0,99][399,1096][1217,1826]else OTHER R1000,AvgE(y) yI [270,792)[792,1558](1558,2568]else OTHER yE yS yE yI LAvEAvI,origin(y) yE [5.7,13.6)[13.6,15.9](15.9,16.6]else OTHER yI LSR Decision Tree algorithm is, Build decision tree for each ek (also for some ek combos?). RAvEAvI,AvgE(y) Build branches to 100% TP (no class duplication exiting). yE [22.7,31)[31,35.52](35.52,54.32]else OTHER yI Then y isa C iff y isa C in every tree else y isa Other. LAvEAvI,origin(y)= node build a branch for each pair of classes in each interval.
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## Slide #21.

FAUST LSR DT Classification on IRIS150, d= 0100 L 0100.Origin(y) S 23 44 E20 34 I 22 38 1 2 1 15 18 and P38
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## Slide #22.

FAUST LSR DT Classification on IRIS150 On this slide we do the same as on the last but make a fresh calculation of Avg for each recursive steps. It takes 7 recursive rounds again to separate E and I in this branch of the e2=0100 tree 0100 (Pedal Width). From this incomplete testing, it seems not to be beneficial to make expensive fresh Avg calculations. L 0100.Origin(y) S 23 44 E20 34 I 22 38 1 2 1 L 0100.Origin(y) S 23 44 E20 34 I 22 38 29 15 6 47 46 3 On 23L0100,O34 R0100,AvgS 1 45,12 2 50 15 1 On 23L0100,O34 R0100,AvgE 3 234 58 793 1417 1103 13,21 On 30L0010,O51 R0010,AvgE 2.8 16.3 157.6 33,14 199 On 23L0100,O34 & 320R0100,AvgS1820 LAvgEAvgI,Origin On 23L0100,O34 & 23L0100,O34 & 58R0100,AvgS234 58R0100,AvgS234 SBarrelAvgE SBarrelAvgI On 23L0100,O34 & 58R0100,AvgS234 LAvgEAvgI,Origin On 30L0010,O51 & 16.3R0100,AvgS157.6 LAvEAvI,O 23 25 68 241.1 36.1 951 58.6 272 5.9 417 52 52.7 24,9 30 34 13,18 Seems very beneficial! Use only LinearAvg with the smallest count, in this case LinearAvgE? On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 RAvgEAvgI,AvgE 0 88 4 24,6 108 On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 & 4RAvgEAvgI,AvgE88 LAvgEAvgI,Origin 3133 4961 3411 4397 On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 & 4RAvgEAvgI,AvgE88 & 3411LAvgEAvgI,Origin4397 RAvgEAvgI,AvgE On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 & 4RAvgEAvgI,AvgE88 & 3411LAvgEAvgI,Origin4397 & 5.9RAvgEAvgI,AvgE20.5 LAvgEAvgI,Origin 1 38 18,5 5.9 29 15 6 47 46 3 2 1 We pause the algorithm and try S BarrelAvgE and SBarrelAvgI in addition to LAvEAvI,O Next try inserting SLinearAvgE and SLinearAvgI in serial w LAvEAvI,O instead of parallel. 0 43 320 1820 399 4213 L 0010.Origin(y) S10 19 E 30 51 I 18 69 20.5 26 54.4 66 7,14 79 7,13 On 23L0100,O34 & 58R0100,AvgS234 & 25LAvEAvI,O32 SLinearAvgE 2 27.8 1,5 66.1 357 83 78.4 19,13 On 23L0100,O34 & 58R0100,AvgS234 & 25LAvEAvI,O32 SLinearAvgI 2.4 66.3 7.4 135 6,11 171 80 On 30L0100,O51 & 16.3R0100,AvgS157.6 & 66.3LAvEAvI,O78.4 RAvgEAcI,AvE 1416 2522.2 936.4 1748.2 5,6 On 23L0100,O34 & 58R0100,AvgS234 & 25LAvEAvI,O32 & 27SLinearAvgE66.1 Sp On 30L0100,O51 & 16.3R0100,AvgS157.6 & 66.3LAvEAvI,O78.4 & 1416RAvgEAcI,AvE1449 & L 0 1416 2522.2 936.4 1748.2 1,0 11 23 0,5 5,6
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## Slide #23.

FAUST Oblique LSR Classification IRIS150 Ld d=1000 p=origin S43 E 49 I 49 59 71 80 16 34 24 6 26 32 12 p=AvgS 50 34 15 2 p=AvgE 59 28 43 13 p=AvgI 66 30 55 20 Ld d=0010 p=origin Ld d=0100 p=origin 0 270 99 792 26 5 393 1558 1096 2568 1217 1826 1 517, 4 279 5 171 79 186 633 748 998 0 24 126 2 1 3426 10 132 388 730 1369 1622 2281 0 In pTree psuedo-code: Py<43=PO P43y<49=PS P49y58=PSEI P5970 S 23 E 20 I 22 S 10 19 E 30 I 18 44 34.1 38 1 2 1 29 47 46 15 3 51 69 48 2 1 0 50 15 34 6 0 0 66 55 310 3139 35246 12 3850 1750 4104 712 5813 21 636 234 983 793 1369 21 3 1103 1417 5 96 273 36 47 23 1776 1403 2747 929 1892 2824 Ld d=0001 p=origin 3000 331547 6120 6251 3 9, 3 2.8 1633 158 199 5.9 1146 319 453 14 S 1 6 E 10 18 I 14 26 50 28 22 16 34 4954 482422 8134 9809 11 5 14 3617 152 611 7 0 14 2522 454 1397 12
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## Slide #24.

Ld,p=(X-p)od (if p=origin we use Ld=Xod) is a distance dominated functional, meaning dis(Ld,p(x),Ld,p(y))  dis(x, y) x,yX. 7 6 5 2 Therefore there is no conflict between Ld,p gap enclosed clusters for different d's. I.e., consecutive Ld,p gaps a separate cluster always (but not necessarily vice versa). A PCI followed by a PCD  a separate cluster (with nesting issues to be resolved!). 4 Recursion solves problems, e.g., gap isolating point4 is revealed by a Le (X)=Attr1 gap. 1 Recursively restricting to {123 5678} and applying Le2(X)=Attr2 reveals the 2 other gaps 1, 3, 8 This first example suggests that recursion can be important. A different example suggests that recursion order can also be important: 7 6 5 25 Using ordering, d=e2, e1 recursively, Le =Attr2 reveals no gaps, so 2 Le1=Attr1 is applied to all of X and reveals only the gap around point4. 100 Using ordering d=e1, e2 instead: Le1=Attr1 on X reveals a gap of at least 100 around point4 (actual gap: 103.078) 2 Le2=Attr2 is applied to X-{4} reveals a gap of 50 between {123} and {567} also. 1 3 103.07 8 4 X Row 1 2 3 4 5 6 7 8 Row 1 2 3 4 5 6 7 Attr1 0 0 0 110 0 0 0 0 Attr1 0 0 0 75 0 0 0 Attr2 0 100 0 110 114 123 145 0 Attr2 0 25 50 75 100 125 150 StD: ~30 ~55 Note StD doesn't always reveal best order! What about the other functionals? Sp=(X-p)o(X-p) and Rd,p=Sp-L2d,p In an attempt to be more careful, we can only say that Sp (and therefore also Rd,p) is eventually distance dominated meaning dis (Sp(x), Sp(y))dis(x, y) provided 1dis(p,x)+dis(p,y) Letting r=dis(p,x)=Sp(x), s=dis(p,y)=Sp(y) and r>s, then r-s  dis(x,y) and dis(Sp(x),Sp(y)) = r2-s2 = (r-s)*(r+s)  dis(x,y)*[dis(p,x)+dis(p,y)] When does FAUST Gap suffice for clustering? For text mining?
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## Slide #25.

LSR IRIS150-. Consider all 3 functionals, L, S and R. What's the most efficient way to calculate all 3?\ o=origin; pRn; dRn, |d|=1; {Ck}k=1..K are the classes; An operation enclosed in a parallelogram, , means it is a pTree op, not a scalar operation (on just numeric operands) Lp,d  (X - p) o d = Lo,d - [pod] minLp,d,k = min[Lp,d & Ck] maxLp,d,k = max[Lp,d & Ck[ = [minLo,d,k] - pod = min(Xod & Ck) - pod = min(X&Ck) o d - pod = [maxLo,d,k] - pod = max(Xod & Ck) - pod OR = max(X&Ck) o d - pod Sp = (X - p)o(X - p) = -2Xop+So+pop = Lo,-2p + (So+pop) minSp,k=minSp&Ck OR Rp,d  Sp, - Lp,d2 maxSp,k = maxSp&Ck = min[(X o (-2p) &Ck)] + (XoX+pop) =max[(X o (-2p) &Ck)] + (XoX+pop) = min[(X&Ck)o-2p] + (XoX+pop) =max[(X&Ck)o-2p] + (XoX+pop) minRp,d,k=min[Rp,d&Ck] maxRp,d,k=max[Rp,d&Ck] I suggest that we use each of the functionals with each of the pairs, (p,d) that we select for application (since, to get R we need to compute L and S anyway). So it would make sense to develop an optimal (minimum work and time) procedure to create L, S and R for any (p,d) in the set.
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## Slide #26.

LSR on IRIS150 400 1000 1500 2000 C13 2500 3000 y isa OTHER if yoDse (-,495)(802,1061)(2725,) Dse 9 -6 27 10 495 802 S y isa OTHER or S if yoDse  C1,1  [ 495 , 802] 1270 2010 E y isa OTHER or I if yoDse  C1,2  [1061 ,1270] 1061 2725 I y isa OTHER or E or I if yoDse  C1,3  [1270 ,2010 L H C1,3: y isa OTHER or I if yoDse  C1,4  [2010 ,2725] 0 s Dei -3 -2 3 3 -117 -44 E y isa O if yoDei (-,-117)(-3,) 49 e -62 -3 I y isa O or E or I if yoDei  C2,1  [-62 ,-44] 11 i L H y isa O or I if yoDei  C2,2  [-44 , -3] C2,1: Dei 6 -2 3 1 420 459 E y isa O if yoDei (-,420)(459,480)(501,) 480 501 I y isa O or E if yoDei  C3,1  [420 ,459] 2 e L H y isa O or I if yoDei  C3,2  [480 ,501] 4 i Continue this on clusters with OTHER + one class, so the hull fits tightely (reducing false positives), using diagonals? C1,1: C2,3: 43 58 y isa O if yoD(-,43)(58,) 23 44 y isa O if yoD(-,23)(44,) L H y isa O|S if yoD C2,3  [43,58] L H y isa O|S if yoD C3,3  [23,44] D=1000 D=0100 C3,3: C4,1: 10 19 y isa O if yoD(-,10)(19,) 1 6 y isa O if yoD(-,1)(6,) L H y isa O|S if yoD C4,1  [10,19] L H y isa O|S if yoD C5,1  [1,6] D=0010 D=0001 C5,1: C6,1: 68 117 y isa O if yoD(-,68)(117,) 54 146 y isa O if yoD(-,54)(146,) L H y isa O|S if yoD C6,1  [68,117] L H y isa O|S if yoD C7,1  [54,146] D=1100 D=1010 C7,1: C8,1: 44 100 y isa O if yoD(-,44)(100,) 36 105 y isa O if yoD(-,36)(105,) L H y isa O|S if yoD C8,1  [44,100] L H y isa O|S if yoD C9,1  [36,105] D=1001 D=0110 Ca,1: C9,1: 12 91 y isa O if yoD(-,12)(91,) 26 61 y isa O if yoD(-,26)(61,) L H y isa O|S if yoD Cb,1  [12,91] D=0011 L H y isa O|S if yoD Ca,1  [26,61] D=0101 Cb,1: 81 182 y isa O if yoD(-,81)(182,) Cc,1: 71 137 y isa O if yoD(-,71)(137,) L H y isa O|S if yoD Cc,1  [81,182] L H y isa O|S if yoD Cd,1  [71,137] D=1110 D=1101 Cd,1: Ce,1: 55 169 y isa O if yoD(-,55)(169,) 39 127 y isa O if yoD(-,39)(127,) L H y isa O|S if yoD Ce,1  [55,169] L H y isa O|S if yoD Cf,1  [39,127] D=1011 D=0111 Cg,1: Cf,1: 10 22 y isa O if yoD(-,10)(22,) 84 204 y isa O if yoD(-,84)(204,) L H y isa O|S if yoD Ch,1  [10,22] D=1-100 L H y isa O|S if yoD Cg,1  [84,204] D=1111 Ch,1: 3 46 y isa O if yoD(-,3)(46,) The amount of work yet to be done., even for only 4 attributes, is immense.. L H y isa O|S if yoD Ci,1  [3,46] For each D, we should fit boundaries for each class, not just one class. D=10-10 For 4 attributes, I count 77 diagonals*3 classes = 231 cases. How many in the Enron email case with 10,000 columns? Too many for sure!! D, not only cut at minCoD, maxCoD but also limit the radial reach for each class (barrel analytics)? Note, limiting the radial reach limits all other directions [other than the D direction] in one step and therefore by the same amount. I.e., it limits all directions assuming perfectly round clusters). Think about Enron, some words (columns) have high count and others have low count. Our radial reach threshold would be based on the highest count and therefore admit many false positives. We can cluster directions (words) by count and limit radial reach differently for different clusters??
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## Slide #27.

Dot Product SPTS computation: XoD = k=1..nXkDk p11 p10 p21 p20 0 1 1 1 1 0 0 0 0 p p p p XoD XoD,3 XoD,2 XoD,1 XoD,0 6 9 9 1 0 1 1 0 0 0 1 1 1 0 0 D1,1 D1,0 D2,1 D2,0 1 1 1 1 0 1 0 13,4 CAR1 2,3 0  01 & 0 0 0 CAR22,3 0 1 1 0 1 1 1 1 0 0 1 0 0 1 1  & 0 0 1 0 1 1 0 1 0  &0 1 0 1 OUTPUT: PXoD,i, CARi,i+1 0 1 0 1 0 1 CAR11,2 0 0 0  & 0 1 CAR21,2 0 0 1 1 0  & PXoD,2 D2,1 D2,0 0 1 1 0 1 0  &  0& 0 1 PXoD,3 0 1 1 1 0 1 1 0 0 0 1 0 0 0 0 1 1 0 CAR10,1   & &1 1 0  & 1 0 1 0 1 1 PXoD,1 PXoD,0 = 22 ( 1 p1,1 + 1 p2,1 ) + 21 (1 p1,0 + 1 p11 + 1 p2,0 + 1 p2,1 ) 0 1 1 0 0 0 0 1 1 1 0 1 0 1 ROUTINE: PXoD,i=RSiCARi-1,i CARi,i+1=RSi&CARi-1,i 1 1 1 1 XoD 0 0 1 1 0 6 18 1 0 0 1 0 0 1 0 0 1 9 0 1 0 0 1 1 1 1 0 + 20 (1 p1,0 + 1 p2,0 ) INPUT: CARi-1,i, RSi 0 1 D1,1 D1,0 3 0 1 1 1 0 1 0 0 0 PXoD,4 0 1 1 /*Calc PXoD,i after PXoD,i-1 CarrySet=CARi-1,i RawSet=RSi */  & PXoD,3 D Different data. 3 pTrees 0 1 1 = 22 ( 1 p1,1 + 1 p2,1 ) + 21 (1 p1,0 + 1 p11 + 1 p2,0 + 1 p2,1 ) 0 1 0 1 0 1 0 1 0 0 1 0   & &1 1 0 1 0 1 1 0 0 0 1 1 1 1 0 0 0 0 1 1 0  & 0 1 1 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 0 1 1 0 1 1 0    & &0&1 1 0 0 1 PXoD,2 1 1 1 1 0 + 20 (1 p1,0 + 1 p2,0 ) CAR10,1 PXoD,1 1 1 1  & 1 1 1 0 0 1 PXoD,0 We have extended the Galois field, GF(2)={0,1}, XOR=add, AND=mult to pTrees. SPTS multiplication: X1*X2= (21 p1,1 +20 p1,0) (21 p2,1 +20 p2,0) = 22 p1,1 p2,1 +21( p1,1 p2,0+ p2,1 p1,0) + 20 p1,0 p2,0 (Note, pTree multiplication = &) 0 0 1 &1 1 0 0 1 0 X X1 X2 1 3 2 1 3 1 p11 p10 0 1 1 1 1 0 p21 p20 X1*X2 pX1*X2,3 pX1*X2,2 pX1*X2,1 pX1*X2,0 0 1 1 1 0 1 1 9 2 0 1 0 0 0 0 0 0 1 1 1 0 0 1 0 pX1*X2,3 0 1 0 0 1 1 &1 1 1 0 1 0 0 0 0  & pX1*X2,2 0 1 1 1 1 1 &1 0 1 0 1 1 &1 0 0 0 1 0 0 0 1  & pX1*X2,1 1 1 0 pX1*X2,0
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## Slide #28.

mple: ST Oblique: XoD used in CCC, TKO, PLC and LARC) and (x-X)o(x-X) So in FAUST, we need to construct lots of SPTSs of the type, X dotted with a fixed vector, a costly pTree calculation (Note that XoX is costly too, but it is a 1-time calculation (a pre-calculation?). x ox is calculated for each individual x but it's a scalar calculation and just a read-off of a row of XoX, once XoX is calculated.. Thus, we should optimize the living be__ out of the XoD calculation!!! The methods on the previous seem efficient. Is there a better method? Then for TKO we need to computer ranks: RankK: p is what's left of K yet to be counted, initially p=K V is the RankKvalue, initially 0. For i=bitwidth+1 to 0 if Count(P&Pi)  p { KVal=KVal+2i; else /* < p */ { p=p-Count(P&Pi); D=x RankN-1(XoD)=Rank2(X oD)1 1 1 p11 p10 0 1 1 1 1 0 p XoD 3 p21 p20 0 0 0 2 3 3 1 0 1 0 0 0 D1,1 D1,0 D2,1 D2,0 0 1 p2 p1 1 0 0 1 1 1 0 1 p,0 0 1 1 RankN-1(XoD)=Rank2(XoD) n=3 p=2 x2 D1,1 D1,0 D2,1 D2,0 3 = -2Xox+xox+XoX is used in TKO. 0 1 1 p2 p1 0 0 0 1 0 1 P=P&p'3 1 1 0 1<2 1<2 0 1 1 2-1=1 1 1 0 3 0*2 + p,0 1 1 0 D=x RankN-1(XoD)=Rank2(X oD)3 2 1 p XoD 3 3 6 5 0 0 0 p2 p1 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 P=P&p1 1 1 32 1 1 0 0 0 1 0 D1,1 D1,0 D2,1 D2,0 1 0 n=0 p=2 P &p0 1*21+ n=2 p=1 P &p2 P p3 0 0 n=1 p=2 P p1 0 1 p,0 1 0 1 1 1 P=p'2&P 1 0 1 0<1 0<1 1-0=1 0*23+0*22 n=2 p=2 P p2 P=P&p2 1 0 22 0 1 1 1 1 1 1 1*22+ P=P&Pi P=P&P'i }; }; P=p0&P 0 1 1 22 1*21+1*20=3 so -2x1oX = -6 n=1 p=1 P &p1 P=p1&P n=0 p=1 P &p0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 0 21 0*23+0*22+1*21+ n=1 p=2 P &p1 0 1 1 1 1 0 P=p'1&P 0 1<2 1<2 0 2-1=1 1 1*22+0*21 n=0 p=1 P &p0 0 0 1 1 0 1 P=p0&P 1 0 0 11 0*23+0*22+1*21+1*20=3 so -2x2oX= -6 P=p0&P 11 0 0 1 1*22+0*21+1*20=5 so -2x3oX= -10
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## Slide #29.

So let us look at ways of doing the work to calculate XoD = k=1..nXk*Dk As we recall from the below, the task is to ADD bitslices giving a result bitslice and a set of carry bitslices to carry forward X 1 1 3 0 2 1 =2 ( 1 2 pTrees 0 1 1 1 1 0 0 1 0 0 0 1 p1,1 + 1 0 1 10 1 1 XoD D 3 3 1 1 1 D2,1 D2,0 p1,0 + 1 p11 + 1 p2,0 1 p2,1 ) + 21 ( 1 6 9 9 D1,1 D1,0 1 1 0 1 1 1 1 0 1 + 1 p2,1 ) 1 0 1 0 1 1 0 1 0 0 1 1 0 0 1 + 20 ( 1 +1 p1,0 1 1 0 1 0 0 p2,0 ) 1 0 1 I believe we add by successive XORs and the carry set is the raw set with one 1-bit turned off iff the sum at that bit is a 1-bit Or we can characterize the carry as the raw set minus the result (always carry forward a set of pTrees plus one negative one). We want a routine that constructs the result pTree from a positive set of pTrees plus a negative set always consisting of 1 pTree. The routine is: successive XORs across the positive set then XOR with the negative set pTree (because the successive pset XOR gives us the odd values and if you subtract one pTree, the 1-bits of it change odd to even and vice versa.): /*For PXoD,i (after PXoD,i-1). CarrySetPos=CSPi-1,i CarrySetNeg=CSNi-1,i INPUT: CSPi-1, CSNi-1, RSi ROUTINE: PXoD,i=RSiCSPi-1,iCSNi-1,i CSNi,i+1=CSNi-1,iPXoD,i; OUTPUT: PXoD,i, CSNi,i+1 = 22 ( 1 p1,1 + 1 0 1 1 p2,1 ) + 21 ( 1 p1,0 + 1 0 1 1 1 1 0 RS1 p11 + 1 p2,0 + 1 p2,1 )   0 1 1 1 0 1  1 0 1 0 0 0  0 1 1 0 1 1  1 1 0 1 0 1  1 0 1 0 0 0 = 0 0 0 1 1 0 p2,0 ) 1 0 1 CSP-1,0RS0 PXoD,1 1 1 0 +1 p1,0 1 1 0 CSN-1.0P CSP XoD,0 0,1= CSN0,1= CSPi,i+1=CSPi-1,iRSi-1; CSPi,i+1 + 20 ( 1 0 0 0 1 0 1 RawSet=RSi CSP-1=CSN-1=*/  CSP-1,0=CSN-1,0= PXoD,0 RS0 1 0 1 = 0 1 1
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## Slide #30.

CCC Clusterer If DT (and/or DUT) not exceeded at C, partition C further by cutting at each gap and PCC in C oD For a table X(X1...Xn), the SPTS, Xk*Dk is the column of numbers, xk*Dk. XoD is the sum of those SPTSs, k=1..nXk*Dk XoD = k=1..nXk*Dk Xk*Dk = Dkb2bpk,b = 2BDkpk,B +..+ 20Dkpk,0 2B = Dk(2Bpk,B +..+20pk,0) = (2BDk,B+..+20Dk,0) (2Bpk,B +..+20pk,0) = 2 + 22B-1 2B D p ) 2B-1 0 = 2 ( k,B k,B + 2 (Dk,B-1pk,B +Dk,Bpk,B-1) +..+2 Dk,0pk,0 + 22B-2 + 22B-3 pTrees D D D D D 2,0 2,1 1,0 1,1 ... B=1 0 1 0 1 1 0 1 2 0 1 1 1 1 0 0 0 0 1 XoD=k=1,2Xk*Dk with pTrees: qN..q0, N=22B+roof(log2n)+2B+1  ( D p = 22 k=1..2 k,1 k,1 + 21 k=1..2 ( Dk,1 pk,0 + 20 k=1..2 ( Dk,0 pk,0 + 23 + 22 + 21 + 20 k=1..n ( k=1..n ( k=1..n ( k=1..n ( Dk,B pk,B Dk,B pk,B-1 Dk,B pk,B-2 Dk,B pk,B-3 k=1..n ( k=1..n ( k=1..n ( k=1..n ( Dk,B pk,0 Dk,2 pk,0 Dk,1 pk,0 Dk,0 pk,0 = 22 ( D1,1 p1,1 + D2,1 p2,1 ) + 21 ( D1,1 p1,0 + D1,0 p11 + D2,1 p2,0 + D2,0 p2,1 ) + 20 ( D1,0 p1,0 0 0 So, DotProduct pTree 0 involves just multi-operand 1 addition. (no SPTSs and no multiplications) 0 1 1 0 1 Engineering shortcut tricka would be huge!!! D 3 3 =2 ( 1 0 1 1 p1,1 + 1 + Dk,2 pk,1 + Dk,1 pk,1 + Dk,0 pk,1 p2,1 ) + 2 ( 1 1 1 1 0 D2,1 D2,0 p1,0 + 1 p11 + 1 p2,0 1 1 0 1 1 1 + D2,0 p2,0 ) + D2,0 p2,0 ) 1 1 0 1 + 1 p2,1 ) 0 0 0 +2 ( 1 0 1 1 0 p1,0 q0 =11 p1,0 = 1 1 0 1 1 0 D1,1 D1,0 + Dk,1 pk,2 + Dk,0 pk,2 +1 +Dk,0 pk,3 no carry 0 + Dk,0 pk,1 = 22 ( D1,1 p1,1 + D2,1 p2,1 ) + 21 ( D1,1 p1,0 + D1,0 p11 + D2,1 p2,0 + D2,0 p2,1 ) + 20 ( D1,0 p1,0 2 + Dk,B-1 pk,B + Dk,B-1 pk,B-1 + Dk,B-2 pk,B + Dk,B-1 pk,B-2 + Dk,B-2 pk,B-1 +Dk,B-3 pk,B p2,0 ) 1 0 1 A carryTree is a valueTree or vTree, as is the rawTree at each level (rawTree = valueTree before carry is incl.). In what form is it best to carry the carryTree over? (for speediest of processing?) 1. multiple pTrees added at next level? (since the pTrees at the next level are in that form and need to be added) 2. carryTree as a SPTS, s1? (next level rawTree=SPTS, s2, then s10& s20 = qnext_level and carrynext_level ? q 1= q2=carry 0 1= carry 0 1= 0 1 no carry 0 1 0 1 1 q0 = 10carry0= 1 2 1 1 0 1 q1=carry0+raw 1= 1carry1= 1 1 q2=carry1+raw 2= 1 1 q3=carry 2 = 1 1 1 1carry = 2 1 carry3=
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## Slide #31.

X(X1...Xn) D2NN yields a 1.a-type outlier detector (top k objects, x, dissimilarity from X-{x}). Question: (x-X)o(x-X)= k=1..n(xk-Xk)(xk-Xk)= )=k=1..n(b=B..02bxk,b-2bpk,b)( (b=B..02bxk,b-2bpk,b) ----ak,b--=k=1..n( b=B..02b(xk,b-pk,b) ) (  b b=B..02 (xk,b-pk,b) ) B B-1 1 =k (2 ak,B+ 2 ak,B-1+..+ 2 ak, 1+ 20ak, 0) (2Bak,B+ 2B-1ak,B-1+..+ 21ak, 1+ 20ak, 0) } Which primitives are needed and how do we compute them? ( 22Bak,Bak,B + 22B-1( ak,Bak,B-1 + ak,B-1ak,B ) + { 22Bak,Bak,B-1 22B-2( ak,Bak,B-2 + ak,B-1ak,B-1 + ak,B-2ak,B ) + {2B-1ak,Bak,B-2 + 22B-2ak,B-12 22B-3( ak,Bak,B-3 + ak,B-1ak,B-2 + ak,B-2ak,B-1 + ak,B-3ak,B ) + { 22B-2( ak,Bak,B-3 + ak,B-1ak,B-2 ) } 22B-4(ak,Bak,B-4+ak,B-1ak,B-3+ak,B-2ak,B-2+ak,B-3ak,B-1+ak,B-4ak,B)... {22B-3( ak,Bak,B-4+ak,B-1ak,B-3)+22B-4ak,B-22} =22B ( ak,B2 + ak,Bak,B-1 ) + 22B-1( ak,Bak,B-2 ) + 22B-2( ak,B-12 + ak,Bak,B-3 + ak,B-1ak,B-2 ) + 22B-3( ak,Bak,B-4+ak,B-1ak,B-3) + 22B-4ak,B-22 ... Should we pre-compute all pk,i*pk,j ANOTHER TRY! D2NN = each min[D2NN(x)] p'k,i*p'k,j D2NN=multi-op pTree adds? When xk,b=1, ak,b=p'k,b and when xk,b=0, ak,b= -pk.b So D2NN just multi-op pTree mults/adds/subtrs? Each D2NN row (each xX) is separate calc. pk,i*p'k,j X(X1...Xn) RKN (Rank K Nbr), K=|X|-1, yields1.a_outlier_detector (top y dissimilarity from X-{x}). Install in RKN, each RankK(D2NN(x)) (1-time construct but for. e.g., 1 trillion xs? |X|=N=1T, slow. Parallelization?) xX, the square distance from x to its neighbors (near and far) is the column of number (vTree or SPTS) d2(x,X)= (x-X)o(x-X)= k=1..n|xk-Xk|2= k=1..n(xk-Xk)(xk-Xk)= k=1..n(xk2-2xkXk+Xk2) = -2 kxkXk + kxk2 + kXk2 = -2xoX 5. Add 3 to this 3. Pick this from XoX for each x and add to 2. + xox + XoX k=1..n i=B..0,j=B..02i+jpk,ipk,j 1. precompute pTree products within each k i,j 2i+j kpk,ipk,j 2. Calculate this sum one time (independent of the x) -2xoX cost is linear in |X|=N. xox cost is ~zero. XoX is 1-time -amortized over xX (i.e., =1/N) or precomputed The addition cost, -2xoX + xox + XoX, is linear in |X|=N So, overall, the cost is linear in |X|=n. Data parallelization? No! (Need all of X at each site.) Code parallelization? Yes! (After replicating X to all sites, Each site creates/saves D2NN for its partition of X, then sends requested number(s) (e.g., RKN(x) ) back.
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## Slide #32.

LSR on IRIS150-3 Here we use the diagonals. Here we try using other p points for the R step (other than the one used for the L step). d=e1 p=AVGs, L=(X-p)od 43 49 49 58 70 R&L [43,49) [49,58) [58,70) S(16) E(24)I(6) E(26) I(32) 0 0 S(34) 99 128 270 393 792 1096 1217 1558 1825 2567 S E 79 I [70,79] I(12) d=e1 p=AvgS, L=Xod 43 R(p,d,X) S 0 58 70 79 S&L E&L I&L d=e1 p=AS L=(X-p)od (-pod=-50.06) E -7.06 7.94 S&L -1;06 19.94 E&L -1.06 28.94 I&L I 128 270 I(1) 393 E(50) I(7) 49 1558 I(42) 70 2081 3444 49 49 3444 Only overlap L=[58,70), R[792,1557] (E(26), I(5)) With just d=e1, we get good hulls using LARC: While  Ip,d containing >1class, for next (d,p) create L(p,d)X Xod-pod, R(p,d)X XoX+pop-2X p-2Xop-L2 1.  MnCls(L), MxCls(L), create a linear boundary. 2.  MnCls(R), MxCls(R).create a radial boundary. 3. Use R&Ck to create intra-Ck radial boundaries Hk = {I | Lp,d includes Ck} 49 (36,7) 63 (11) -8,-2 [-2,8) 16 34, 24, 6 0 99 393 1096 1217 1825 [8,20) 26, 32 270 792 1558 2567 [20,29] 12 E=26 I=5 p=AvgS d=e4 p=AvgS, L=(X-p)od -2 4 7 11 16 S&L E&L 23 I&L -2,4) [7,11) [11,16) 50 28 22, 16 127.5 648.7 1554.7 2892 [16,23] I=34 E=22 I=7 p=AvgS 30ambigs, 5 errs d=e1 p=AS L=(X-p)od (-pod=-50.06) -7.06 7.94 S&L -1;06 19.94 E&L -1.06 28.94 I&L d=e4 p=AvgS, L=(X-p)od -2 4 7 11 16 S&L E&L 23 I&L -8,-2 [-2,8) [8,20) w [20,29] 16 34, 24, 6 p=AvgE 12 0 26, 32 99 1.9 393 51.8 1096 78.6 <--E=6 1217 633 I=4 p=AvgE 1825 -2,4) [7,11) [11,16) 50 28 22, 16 5.7 36.2 151.06 611 d=e1 p=AS L=(X-p)od (-pod=-50.06) d=e4 p=AvgS, L=(X-p)od -7.06 7.94 S&L -1;06 19.94 E&L -1.06 28.94 I&L -8,-2 [-2,8) 16 34, 24, 6 0 99 393 1096 1217 1825 [8,20) w [20,29] p=AvgI 12 26, 32 0.62 34.9 E=25 387.8 <-I=10 1369 p=AvgI There is a best choice of p for the R step (p=AvgE) but how would -2 4 7 11 16 [16,23] I=34 E=17 I=7 p=AvgE S&L E&L 23 I&L [16,23] -2,4) [7,11) [11,16) I=34 50 28 22, 16 127.5 E=22 1555 I=8 2892 p=AvgI For e4, the best choice of p for the R step is also p=AvgE. (There are mistakes in this column
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## Slide #33.

LSR on IRIS150 Dse 9 -6 27 10; -184 123 590 381 xoDes: 1331 2046 y isa O y isa O or S(50) y isa O if yoD (-,-184)(123,381)(2046,) if yoD  C1,1  [-184 , 123] or I(1) if yoD  C1,2  [ 381 , 590] y isa O or I(11) if yoD  C1,3  [ 590 ,1331] S E I or E(50) y isa O or I(38) if yoD  C1,4  [1331 ,2046] y isa O y isa O or E(10) 137 E 143 y isaI O or E(40) or I(10) y isa O or I(1) etc. y Dei 1 .7 -7 -4; xoDei on C2,1: y 1.4 19 E if if AND AND SRR(AVGs,dse) onC1,3 2 7 -2 3 I SRR(AVGe,dei) onC3,1 2 8 106 370 if y isa C1,3 if y isa C1,3 isa O isa O or I(8) 0 154 S y isa O if y isa C1,1 AND SRR(AVGs,Dse)(154,) y isa O or S(50) if y isa C1,1 AND SRR(AVGs,DSE)[0,154] SRR(AVGs,dse) on C1,2only one such I y isa C1,3 AND SRR(AVGs,Dse)(-,2)U(143,) y isa C1,3 AND SRR in [2,7) SRR in [7,137) = C2,1 SRR in [137,143] if yoD (-,-2) (19,) if yoD  [ -2 , 1.4] y isa O or E(40) or I(2) y isa O SRR(AVGs,Dei)[0,2)(370,) y isa isa O O or or E(4) E(27) or I(2) y y isa O or E(9) E I SRR(AVGs,dse) on C1,1 if yoD  C3,1 [ 1.4 ,19] if y isa C3,1 AND if y y isa isa C C3,1 AND AND SRR(AVGs,Dei)[2,8) SRR(AVGs,Dei)[8,106) if 3,1 if y isa C3,1 AND SRR(AVGs,Dei)[106,370] We use the Radial steps to remove false positives from gaps and ends. We are effectively projecting onto a 2-dim range, generated by the Dline and the D line (which measures the perpendicular radial reach from the D-line). In the D projections, we can attempt to cluster directions into "similar" clusters in some way and limit the domain of our projections to one of these clusters at a time, accommodating "oval" shaped or elongated clusters giving a better hull fit. E.g., in the Enron email case the dimensions would be words that have about the same count, reducing false positives. LSR on IRIS150-2 d=e1=1000; The xod limits: 43 We use the diagonals. Also we set a MinGapThres=2 which will mean we stay 2 units away from any cut d=e2=0100 on C1,2 xod lims: 20 30 25 30 32 49 49 44 58 S E I 70 79 I y isa y Sisa y Eisa y isa y isa y isa y isa y isa O if yoD(-,18)(46,) y isa O or E( 3) if yoD[18,23) y isa O if yoD[18,23)&SRR[0,21) y isa O or E(13) or I( 4) if yoD[23,28) (yC2,1) y isa O or S(13) or E(10) or I( 3) if yoD[28,34) (yC2,2) y isa O or S(28) if yoD[34,46] y isa O if yoD[34,46]&SRR[0,32][46,) d=e3=0001 xod lims: 12 18 18 24 I E y isa O if yoD(-,1)(5,12)(24,) y isa O or S(13) if yoD[1,5] y isa O or E( 9) if yoD[12,16) y isa O if yoD[12,16)&SRR[0,208)(558,) O if yoD(-,43)(79,) O or S( 9) if yoD[43,47] O if yoD[43,47]&SRR(-,52)(60,) O or S(41) or E(26) or I( 7) if yoD(47,60) (yC1,2) O or E(24) or I(32) if yoD[60,72] (yC1,3) O or I(11) if yoD(72,79] O if yoD[72,79]&SRR(-,49)(78,) d=e2=0100 on C1,3 xod lims: 22 22 34 34 d=e3=0010 on C2,2 xod lims: E I zero differentiation! y isa O or E(17) if yoD[60,72]&SRR[1.2,20] y isa O or E( 7) or I( 7)if yoD[60,72]&SRR[20, 66] y isa O or I(25)if yoD[60,72]&SRR[66,799] y isa O if yoD[0,1.2)(799,) 28 28 30 30 32 33 S E I y isa O if yoD(-,28)(33,) y isa O or S(13) or E(10) or I(3) if yoD[28,33] y isa O or E( 1) or I( 3) if yoD[16,24) y isa O if yoD[16,24)&SRR[0,1198)(1199,1254)1424,) y isa O or E(1) if yoD[16,24)&SRR[1198,1199] y isa O or I(3) if yoD[16,24)&SRR[1254,1424]
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## Slide #34.

d=AvgEAvgI p=AvgE, L=(X-p)od LSR IRIS150. d=e1 p=AvgS, L=Xod 43 49 49 58 70 -36 -25 -14 11 -17 R(p,d,X) S&L SE&L E 79 I&L 0 I 2 S E 33 I [-17,-14)] [-14,11) (50, 13) I(1) 0 2.8 E=47 76 I=12 134 [11,33] I(36) Note that each L=(X-p)32 od is76 just a shift of Xod by -pod (for a given d).357 514 d=e1 p=AS L=(X-p)od (-pod=-50.06) -7.06 7.94 S&L -1;06 19.94 E&L -1.06 28.94 I&L -8,-2 [-2,8) 16 34, 24, 6 0 99 393 1096 1217 1825 [8,20) 26, 32 270 792 1558 2567 [20,29] 12 E=26 I=5 30ambigs, 5 errs d=e1 p=AE L=(X-p)od (-pod=-59.36) -17 -1 -11 11 -11 20 -17-11[-11,-1) 16 33, 21, 3 0 E=5 27 I=3 107 172 748 1150 S&L E&L I&L [-1,11) 26, 32 1 51 79 633 [11,20] I12 E=7 I=4 d=e1 p=AI L=(X-p)od (-pod=-65.88) -22 -8 -17 4 -17 14 S&L E&L I&L [-17,-8) [-8,4) 33, 21, 3 26, 32 38 0 126 34 E=2 132 1368 I=1 730 730 1622 2181 E=26 I=11 d=AvgSAvgI p=AvgS, L=(X-p)od d=AvgSAvgE p=AvgS, L=(X-p)od -6 -6 5 12 R(p,d,X) S .3 150 E I .9 4.7 204 17.5 42 [12,17.5)] [17.5,42) (50,12) I(1) 4.7 6 192 205 S E 65 I 4 11 [11,33] R(p,d,X) I(37) S E E=45 0 I=12 2 154 137 [11,18)] I(1) I 6 18 S E 64 I 42 [18,42) (50,11) 2 6.92 133 137 [42,64] 38 E=39 I=11 393 213 Next, we examine: For a fixed d, the SPTS, -5 5 S&L -2 4 S&L -11 10 S&L 15 37 E&L 7 16 E&L Lp,d. is just a shift of -14 0 E&L 4 55 I&L 11 23 I&L -13 4 I&L LdLorigin,d by -pod we get [15,37) [37,55] -2,4) [7,11) [11,16) [16,23] ,-13) -13,-11 -11,0 [0,4) [4, -5,4) [4,15) 1 50 28 50, 15 I=34 22, 16 I=34 1 0, 2, 1 29,47,46 15 3 6 47 3 the same intervals to apply 157 11 16 all=-11 0 9, 3 3, 1 R to, independent of p 297 E=18 11 16 E=22 2, 1 66 I=12 I=16 536 310 1, 1 (shifted by -pod). 792 352 46,11 1749 38ambigs 16errs 4104 Thus, we calculate once, d=e2 p=AvgE, L=(X-p)od d=e3 p=AvgE, L=(X-p)od d=e4 p=AvgE, L=(X-p)od ll =minXod hl =maxXod, d d S&L -32 -24 S&L -13 -7 -5 `17 S&L -3 5 E&L -12 9 E&L then for each different p we -8 7 E&L 1 12 I&L -25 27 I&L -6 11 I&L shift these interval limit [1,5) [5,12] ,-6) [-6, -5) [-5,7) [7,11) [11,,-25) -25,-12 [-12,9) [9,27] -7] [-3,1) numbers by -pod since 50 21 22, 16 34 1 0, 2, 1 29,47, 46 15 3 6 48 2 1 1 49, 15 I=34 .7 .7 2(17) 15 3 1 err these numbers are really all E=22 4.8 4.8 16 E=32 18 58 13, 21 I=16 I=14 we need for our hulls 158 58 234 199 59 793 21, 3 (Rather than going thru the 1103 1417 SPTS calculation of (Xd=e p=Avg , L=(X-p) o d d=e p=Avg , L=(X-p) o d p)od anew  new p). d=e2 p=AvgI, L=(X-p)od 4 I 3 I d=e2 p=AvgS, L=(X-p)od -10 -7 -8 4 `15 9 S&L E&L I&L d=e3 p=AvgS, L=(X-p)od d=e4 p=AvgS, L=(X-p)od -44 -36 -25 -4 -37 [-8, -7) [5, 9] [11,,-25) -25,-12 ,-6) [-7, 4) [6,11) 9, 32, 16 48 2 1 1 1 2, 1 29,46,46 15 allsame allsame S=9 5 E=2 36 E=47 E=2 I=1 929 I=22 I=1 1403 1893 2823 [-25,-4) 50, 15 5 11 318 453 S&L S&L -19 -14 -10 -3 E&L E&L -6 5 I&L 14 I&L [9,27] I=34 E=32 E=46 I=14 I=14 [-6,-3) 22, 16 same range [5,12] E=22 I=1634 There is no reason we have to use the same p on each of those intervals either. So on the next slide, we consider all 3 functionals, L, S and R. E.g., Why not apply S first to limit the spherical reach (eliminate FPs). S is calc'ed anyway?
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## Slide #35.

FAUST Oblique, LSR Linear, Spherical, Radial classifier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 XoX 4026 3501 3406 3306 3996 4742 3477 3885 2977 3588 4514 3720 3401 2871 5112 5426 4622 4031 4991 4279 4365 4211 3516 4004 3825 3660 3928 4158 4060 3493 3525 4313 4611 4989 3588 3672 4423 3588 3009 3986 3903 2732 3133 4017 4422 3409 4305 3340 4407 3789 8329 7370 8348 5323 7350 6227 7523 4166 7482 5150 4225 6370 5784 6967 5442 7582 6286 5874 6578 5403 7133 6274 7220 6858 6955 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 146 148 149 150 7366 7908 8178 6691 5250 5166 5070 5758 7186 6066 7037 7884 6603 5886 5419 5781 6933 5784 4218 5798 6057 6023 6703 4247 5883 9283 7055 9863 8270 8973 11473 5340 10463 8802 10826 8250 7995 8990 6774 7325 8458 8474 12346 11895 6809 9563 6721 11602 7423 9268 10132 7256 7346 8457 9704 10342 12181 8500 7579 7729 11079 8837 8406 5148 9079 9162 8852 7055 9658 9452 8622 7455 8229 8445 7306 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 Ld 51 49 47 46 50 54 46 50 44 49 54 48 48 43 58 57 54 51 57 51 54 51 46 51 48 50 50 52 52 47 48 54 52 55 49 50 55 49 44 51 50 45 44 50 51 48 51 46 53 50 70 64 69 55 65 57 63 49 66 52 50 59 60 61 56 67 56 58 62 56 59 61 63 61 64 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 146 148 149 150 Form Class Hulls using linear d boundaries thru min and max of Lk.d,p=(Ck&(X-p))od  On every I {[ep ,ep ) | ep =minL or maxL for some k,p,d} interval add spherical and barrel k,p,d i i+1 j k,p,d k,p,d d=1000 66 boundaries with Sk,p and Rk,p,d similarly 68 67 60 57 55 all p,d k,p,d 55 58 60 54 o o d k,L,d k d 60 (pre-ccompute?) pre-ccompute?) 67 k,L,d k d 63 56 o o o o d,p d k,L,d,p k d,p k,L,d k,L,d.p k d,p k,L,d 55 55 61 58 p,d 50 S E I 56 d=1000 p=0000 =L -p od 57 d nk,L,d xk,L,d d=1000 p=AS=(50 34 15 2) d=1000 p=AE=(59 28 43 13) d=1000 p=AI=(66 30 55 20) 57 62 S 43 58 -7 8 -16 -1 -23 -8 51 57 E 49 70 -1 20 -10 11 -17 4 63 I 49 79 -1 29 -10 20 -17 13 58 1 71 63 d=0100 p=0000 65 nk,L,d xk,L,d d=0100 p=AS=(50 34 15 2) d=0100 p=AE=(59 28 43 13) d=0100 p=AI=(66 30 55 20) 76 49 S 23 44 -5 16 -7 14 -11 10 73 E 20 34 67 -8 6 -10 4 -14 0 72 I 22 38 -6 10 -8 8 -12 4 65 2 64 68 d=0010 p=0000 57 nk,L,d xk,L,d d=0010 p=AS=(50 34 15 2) d=0010 p=AE=(59 28 43 13) d=0010 p=AI=(66 30 55 20) 58 64 S 10 19 -33 -24 -45 -36 -5 4 65 E 30 51 77 -13 8 -25 -4 15 36 77 I 18 69 -25 26 -37 14 3 54 60 69 3 56 d=0001 p=0000 77 nk,L,d xk,L,d d=0001 p=AS=(50 34 15 2) d=0001 p=AE=(59 28 43 13) d=0001 p=AI=(66 30 55 20) 63 67 -1 4 -12 -7 -25 -20 S 1 6 72 8 16 -3 5 -16 -8 E 10 18 62 61 12 23 1 12 -12 -1 I 14 25 64 72 74 4 79 We have introduce 36 linear bookends to the class hulls, 1 pair for each of 4 ds, 3 ps , 3 class. For fixed d, C , the k 64 63 pTree mask is the same over the 3 p's. However we need to differentiate anyway to calculate R correctly. 61 77 63 64 That is, for each d-line we get the same set of intervals for every p (just shifted by -pod). The only reason we need to 60 69 have them all is to accurately compute R on each min-max interval. In fact, we computer R on all intervals (even 67 69 those where a single class has been isolated) to eliminate False Positives (if FPs are possible - sometimes they are 58 68 not, e.g., if we are to classify IRIS samples known to be Setosa, vErsicolor or vIriginica, then there is no "other"). 67 67 63 Assuming L , n d k,L,d and xk,L,d have been pre-computed and stored, the cut-pt pairs of (nk,L,d,p; xk,L,d,p) are computed 65 62 59 without further pTree processing, by the scalar computations: o o (use enough (p,d) pairs so that no 2 class hulls overlap) Points outside all hulls are declared as "other".  dis(y,I ) = unfitness of y being classed in k. Fitness of y in k is f(y,k) = 1/(1-uf(y,k)) On IRIS150 d, precompute! p, L (X-p) d=L -p d L n X X, L =X d min(C &L )=n p=Avg -p d n x  Lmin(C &L ) x  max(C &L ) max(C &L )=x p=Avg p=Avg d=e d=e d=e d=e n =n -p d x =x -p d. -p d
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## Slide #36.

LSR IRIS150 e1 only Analyze R:RnR1 (and S:RnR1?) projections on each interval formed by consecutive L:RnR1 cut-pts. Sp  (X-p)o(X-p) = XoX + L-2p + pop Rp,d Sp-L2p,d = L-2p-(2pod)d + pop + pod2 + XoX - L2d Ld d=1000 p=origin Setosa 43 vErsicolor 49 vIrginica 49 58 70 79 nk,S,p = min(Ck&Sp) nk,R,p,d = min(Ck&Rp,d) xk,S,p  max(Ck&Sp) xk,R,p,d  max(Ck&Rp,d) d=1000 p=AS=(50 34 15 2) d=1000 p=AE=(59 28 43 13) d=1000 p=AI=(66 30 55 20) -7 -1 -1 -16 -10 -10 -23 -17 -17 8 20 29 -1 11 20 -8 4 13 34 24 6 26 32 34 24 6 26 32 12 16 12 16 16 34 24 6 26 32 12 24 0 0 1 1641 17 723 249 0 0 270 2081 2391 517,4 794 128 99 126 2 1 3426 10 220 279 5 792 26 5 3445 1258 393 79 1558 132 388 171 1096 633 2568 730 1369 186 1217 1622 748 1826 2281 998 with AE with AI 17 1 What is the cost for these additional cuts (at new p-values in an 517,4 220 eliminates 78 It looks like: make the one additional calculation: L-2p-(2po -2p-(2pod)d 633 FPs better? L-interval)? then AND the interval masks, then AND the class masks? If we have computed, S:RnR1, how can we utilize it?. (Or if we already have all interval-class mask, only one mask AND step.) We can, of course simply put spherical hulls boundaries by centering on the class Avgs, e.g., S p=AvgS p Recursion works wonderfully on IRIS: The only hull overlaps after only d=1000 are And the 4 i's common to both are {i24 i27 i28 i34}. We could call those "errors". If on the L 1000,avgE interval, [-1, 11) we recurse using SavgI we get Setosa 0 154 E=50 I=11 vErsicolor 394 1767 vIrginica 369 4171 7 4 36 540,4 72 170 Thus, for IRIS at least, with only d=e1=(1000), with only the 3 ps avgS, avgE, avgI, using full linear rounds, 1 R round on each resulting interval and 1 S, the hulls end up completely disjoint. That's pretty good news! There is a lot of interesting and potentially productive (career building) engineering to do here. What is precisely the best way to intermingle p, d, L, R, S? (minimizing time and False Positives)?
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## Slide #37.

FAUST LSR Hull classifier, recursive LR on MG44d60w We won't carve off outliers since they're all sort of outliers! We first hull the classes. To do the hulling we choose the keyword from each class for our d that occurs in the maximum number of class docs and use the Avg for p. Then we will look at the test doc classifications and assess how good they are (basically to see if the method reveals affinities that make sense and that I hadn't notice when I put together the "expert training set" in the first place - therefore answering the question "Has new info been uncovered thru SML?") First attempt: Try to isolate one class at a time. Take the column(word), e k, with maximum class count and use Lek C1: k=3 (baby) isolates C1 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (4/4 so 0% FN). 0 test cases are in class1. C2: k=34 (King) isolates C2 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (3/3 so 0% FN). 0 test cases are in class2 C3: k=47 (plum) isolates C3 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (3/8 so 62.5% FN). 0 test cases are in class3 C4: k=47 (plum) isolates C4 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (2/4 so 50% FN). 2 test cases are in class4 test 17FEC Here sits the Lord Mayor. Here sit his two men... test 30HDD Hey diddle diddle! The cat and the fiddle. The cow jumped... C5: k=8 (bed) isolates C5 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (2/3 so 33.3% FN). 1 test cases are in class5 test 03DDD Diddle diddle dumpling my son John. Went to bed with... C6: k=528 (plum) isolates C6 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (3/5 so 40% FN). 0 test cases are in class6 FPs 41OKC Old King Cole... 01TBM Three Blind Mice... C7: k=29 (girl) isolates C7 from the other classes and 0 test cases are in that hull, so 100% TP accuracy (1/3 so 66.7% FN). 1 test cases are in class7 test 49WLG There was a little girl who had a little curl right in the...
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## Slide #38.

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## Slide #39.

Thanksgiving Clustering MG44d60w using min to max L 0.13 0.13 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.26 0.39 0.39 0.39 0.39 0.39 0.39 0.39 0.39 0.39 0.39 0.39 0.52 0.52 0.52 0.52 0.52 0.52 0.52 0.65 0.65 0.65 0.65 0.65 0.77 0.77 0.90 0.90 0.90 0.90 1.68 Gap 0 0.12 0 0 0 0 0 0 0 0 0 0 0 0.12 0 0 0 0 0 0 0 0 0 0 0.12 0 0 0 0 0 0 0.12 0 0 0 0 0.12 0 0.12 0 0 0 0.77 03DDD 04LMM Diddle diddle dumpling my son John. Went to be Little Miss Muffet sat on a tuffet, eating of 02TLP 06SPP 08JSC 18HTP 22HLH 23MTB 25WOW 36LTT 42BBC 43HHD 49WLG This little pig went to market. See a pin and pick it up. All the day you Jack Sprat could eat no fat. His wife could ea I had two pigeons bright and gay. They flew fr I had a little husband no bigger than my thumb How many miles is it to Babylon? Three score m There was an old woman, and what do you think? Little Tommy Tittlemouse lived in a little hou Bat bat, come under my hat and I will give you Hark hark, the dogs do bark! Beggars are comin There was a little girl who had a little curl 05HDS 09HBD 11OMM 12OWF 15PCD 27CBC 32JGF 33BFP 38YLS 45BBB 48OTB Humpty Dumpty sat on a wall. Humpty Dumpty ha Hush baby. Daddy is near. Mamma is a lady and One misty moisty morning when cloudy was the w There came an old woman from France who taught Great A. little a. This is pancake day. Toss t Cry baby cry. Put your finger in your eye and Jack, come give me your fiddle, if ever you me Buttons, a farthing a pair! Come, who will buy If I had as much money as I could tell, I neve Bye baby bunting. Father has gone hunting. Mot One two, buckle my shoe. Three, four, knock at 13RRS 14ASO 29LFW 37MBB 44HLH 47CCM 50LJH 01TBM 10JAJ 17FEC 41OKC 46TTP 21LAU 28BBB 07OMH 26SBS 30HDD 39LCS 35SSS A robin and a robins son once went to town to If all the seas were one sea, what a great sea When little Fred went to bed, he always said h Here we go round the mulberry bush, the mulber The hart he loves the high wood. The hare she Cocks crow in the morn to tell us to rise and Little Jack Horner sat in the corner, eating o Three blind mice! See how they run! They all Jack and Jill went up the hill to fetch a pail Here sits the Lord Mayor. Here sit his two me Old King Cole was a merry old soul. And a merr Tom Tom the pipers son, stole a pig and away h The Lion and the Unicorn were fighting for the Baa baa black sheep, have you any wool? Yes si Old Mother Hubbard went to the cupboard to giv Sleep baby sleep. Our cottage valley is deep. Hey diddle diddle! The cat and the fiddle. The A little cock sparrow sat on a green tree. And Sing a song of sixpence a pocket full of rye.
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## Slide #40.

Thanksgiving clustering (carve off clusters as one would carve a thanksgiving turkey) Let m be a furthest point from aAvgX (i.e., pt in X that maximizes SPTS, Sa=(X-a)o(X-a) ) If m is an outlier, carve {m} off from X. Repeat until m is a non-outlier. Construct L=Ld where d is next in dSet Carve off L-gapped cluster(s). Pick centroid, cc=mean of slice, SL: A. If (PCC2=PCD1) declare L-1[PCC1,PCC2] to be a cluster and carve it off (mask it off) of X; else (PCC2=PCI2 ) SLL-1[(3PCC1+PCC2)/4 ,PCC2) and look for a Scc or Rd,cc gap. If one is found, declare it to be a cluster and carve it off of X; Else add cc to the Cluster Centroid Set, CCS. B. Do A. from the high side of L also. Repeat until no new clusters carve off. If X (not completely carved up) use pkmeans on the remains with initial centroids, CCS IRIS: Carve off outliers with DNN>=5: i39 i7 i10 s42 i9 i36 i35. Le4 PCC=90% L 1 2 3 4 5 6 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Ct 6 28 6 7 1 1 7 3 5 13 7 12 4 1 10 5 6 6 3 7 3 2 Gp 1 1 1 1 1 4 49s 1 1 1 1 1 1 1 1 49e 3i 1 1 1 1 1 1 1 1e 41i i20 i30 i34 60 22 50 15 72 30 58 16 63 28 51 15 e21 59 32 48 18 Sp Ct Gp 2 1 1 3 1 1 4 3 1 5 7 1 6 4 1 7 3 1 8 2 1 9 4 1 10 3 1 11 4 1 12 4 1 13 2 1 14 1 1 15 2 1 16 3 1 17 2 1 18 2 6 24 3 1 where p=Av(¾,1)= 25 1 64.4 30.8 50.2 16.2 e8 e11 e44 e49 49 50 50 51 24 20 23 25 emn emx eav 49 70 59.36 20 34 27.7 33 35 33 30 10 10 10 11 30 51 42.6 10 18 13.26 2.62 AvgDNN 2.44 MedDNN DNNS 16.0 7.34 6.32 6.24 5.56 5.38 5.38 4.89 4.89 4.58 4.35 4.24 4.24 4.12 4.12 4.12 (top) i39 i7 i10 s42 i9 i36 i35 i15 e13 s23 i20 e15 i1 i32 i19 i18 GT=4 Discovered there is a Versicolor plume led by {e8 e11 e44 e49} (I checked S centered at their average and found they are gapped away from all other by 3 and a scatter of other versicolor connect them to versicolor central). taking Sp1=S(Av(last 4)= 50 23 32.75 10.25 the 4 are gapped away from the other 47 by 3 Sp1 Gp 1 3 6 7 8 9 11 12 13 14 15 16 17 18 19 20 21 22 23 24 26 Ct 1 2 1 2 2 3 2 6 2 1 3 1 4 3 1 2 7 2 1 2 3 2 3 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 8
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## Slide #41.

Thanksgiving clustering-2 Let m be a furthest point from aAvgX (i.e., pt in X that maximizes SPTS, Sa=(X-a)o(X-a) ) If m is an outlier, carve {m} off from X. Repeat until m is a non-outlier. Construct L=Ld where d is next in dSet If a PCI is followed by another PCI, skip the second one. Same for a PCD. Therefore PCI1 will be followed by a PCD. A. declare each L-1[PCC,PCD] to be a cluster and carve it off (mask it off) of X; B. Do A. from the high side of L also. Repeat until no new clusters carve off. Recurse using a different d on each slice where a PCC was skipped. If X (not completely carved up) use pkmeans starting with a pillar set. IRIS: Carve off outliers with DNN>=5: i39 i7 i10 s42 i9 i36 i35. GT=4 Le4 PCC=75% Le1 PCC=75% L Ct Gp PCI1 L Ct Gp 1 2 3 4 5 PCI1 106 11 12 13 14 PCD1 15 PCD2 16 17 PCI2 18 19 20 21 22 23 24 PCD  25 3 6 28 6 7 1 1 7 3 5 13 7 12 4 1 10 5 6 6 3 7 3 2 i30 1 1 1 1 1 4 49s 1 1 1 1 1 1 C1 1 49e 3i 1 1 1 1 1 1 1 1 1e 41i PCD1 PCI2 i20 i34 i30 60 22 50 15 63 28 51 15 72 30 58 16 e21 59 32 48 18 Le3 30 33 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 72 30 58 16 58 on C1 1 3 2 2 2 1 1 1 1 1 1 1 3 1 5 1 3 1 4 1 2 1 4 1 7 1 3 1 5 1 1 1 2 1 2 1 2 7 1 PCD2 43 1 44 3 46 4 47 2 48 5 49 5 50 10 51 9 52 4 53 1 54 6 55 7 56 6 57 8 58 7 59 3 60 5 61 5 62 4 63 9 64 7 65 5 66 2 67 7 68 3 69 4 70 1 71 1 72 2 73 1 74 1 76 1 77 3 79 1 1 2 1 1 1 1 1 1 1 38s 5e 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 C1 1 11s 44e 34i 1 1 1 1 2 1 2 2.62 AvgDNN 2.44 MedDNN DNNS 16.0 7.34 6.32 6.24 5.56 5.38 5.38 4.89 4.89 4.58 4.35 4.24 4.24 4.12 4.12 4.12 (top) i39 i7 i10 s42 i9 i36 i35 i15 e13 s23 i20 e15 i1 i32 i19 i18
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## Slide #42.

FAUST LSR classifier on IRIS X=X(X1,...,Xn,C); oultiers are O=O(O1,...,On,OC) ,OC)X initially empty; OT=Outlier_Threshold; Carve off outliers from X into O (O:=O (O:=O{x|Rankn-1DNN(x,X) DNN(x,X)OT; O:=O O:=O{x|D2NN(X,x)>OT )... th DkNN=distance to the k Nearest Neighbor, meaning, in the distribution, UDR(dis(x,X)) it is the k+1 st value, since 0 is always the first value). In the future, we'll use SQDNN...SQDkNN, for the SPTS of squares of such distances. Define class hulls: Hk{z {zRn | minFd,p,k  Fd,p,k(z)  maxFd,p,k (d,p) (d,p)dpSet, F=L,S,R}. In the future, we'll call each pair of boundary pieces boundary plate pairs. If y is in just one hull, declare y to be in that class. Elseif y is in multiple hulls, MH, declare y y the Ck that minimizes dis(y,Ck), k kMH (note dis(y,Ck)=dis(y,X)&Ck). Else (y is in no hulls), if dis(y,O) dis(y,O)min{dis(y,o)|o min{dis(y,o)|oO}=dis(y,oi)
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## Slide #43.

FAUST LSR Hull classifier, recursive L X=X(X1,...,Xn,C); oultiers are O=O(O1...On,OC) ,OC)X initially ; OT=Outlier_Thres; Carve off outliers from X into O (O:=O (O:=O{x|Rankn-1DNN(x,X) DNN(x,X)OT; O:=O O:=O{x|D2NN(X,x)>OT ).... We set OT=5 and carve off outliers: i7, i9, i10, i35, i36, i39, s42 d=e3 10 19 There is TP purity except in Le2[48,51] so restricting to that interval: ( i.e., X = L-1[48,51] ) 30 51 6 48 14 69 so restricting to Le4-1[15,18]: 5 61.5 70.7 57.9 64.3 d=e2 d=e4 59 69 6 25 32 6 14 18 5 56 9 69 22 13 32 15 6 24 Count reductions are insufficient for e1 and e3, so use e4, e1+e3 e1+e2 3 d=e1 2 restrict to Le1+e2-1[61,64.3]: 4 75.6 79.1 e1+e4 1 53.7 55.1 1 54.4 57.2 restrict to Le1+e3-1[77.7,79.1]: 77.7 80.6 e1+e2+e3 restrict Le1+e4-1[55.1,54.4]: 1 79.0 79.0 1 80.8 80.8 No overlap! So 100% TP accuracy already. This doesn't address FP rates, however. A first cut description of the FAUST Linear Hull - Recursive algorithm: X=X(X1,...,Xn,C); oultiers are O=O(O1,...,On,OC) ,OC)X initially ; OT=Outlier_Thres; Carve off outliers from X into O ( O:=O O:=O{x|Rankn-1DNN(x,X) DNN(x,X)OT; O:=O O:=O{x|D2NN(X,x)>OT .... ) At any node in the Hull Tree, create children by: Moving to next d, d,dSet in each non-pure region, L-1[opti,optj], until the sum of the counts decreases by at least THRESHOLD, then restrict to that region and recurse, until purity is reach.
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