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Sp (X-p)o(X-p) = XoX + Xo(-2p) + pop = L-2p + XoX+pop
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:RnR, 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 Fd,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 PCCsLd,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,PtrRankiSx). RkiSD(x,RankiSx) ordered as constructing desc on rankiSx. LIN: Linear Classifier yCk iff yLHk {z | minLd,p,k Ld,p,k(z) maxLd,pd,k} (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 yCk iff yLSRHk{z | minFd,p,k Fd,p,k(z) maxFd,p,k d,pdpSet, 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=AvCiAvCj 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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boundary pieces boundary plate pairs.
FAUST LSR classifier on IRIS X=X(X1,...,Xn,C); oultiers are O=O(O1,...,On,OC)X initially empty; OT=Outlier_Threshold; Carve off outliers from X into O (O:=O{x|Rankn-1DNN(x,X)OT; O:=O{x|D2NN(X,x)>OT )... DkNN=distance to the kth Nearest Neighbor, meaning, in the distribution, UDR(dis(x,X)) it is the k+1st 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{zRn | minFd,p,k Fd,p,k(z) maxFd,p,k (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 the Ck that minimizes dis(y,Ck), kMH (note dis(y,Ck)=dis(y,X)&Ck). Else (y is in no hulls), if dis(y,O)min{dis(y,o)|oO}=dis(y,oi)<OT, declare y to be in the class of oi, else declare y to be other. 1. This algorithm deals with singleton outliers but ignores doubleton and tripleton outliers etc. 2. In Elseif, rather than compute dis(y,Ck) (single link distance) one could use dis(y,meanCk) for the pre-computed class means. We set OT=5 and carve off outliers: i7, i9, i10, i35, i36, i39, s42 Ld d=e4 S 1 6 E I Ld d=e1 S E I Ld d=e3 S 10 19 E I Ld d=e2 S E I Ld d=e1+e2 S E I Ld d=e1-e2 S7 15 E I Ld d=e1+e3 S38 53 E I Ld d=e1-e3 S E I Ld d=e1+e4 S E I Ld d=e1-e4 S E I Ld d=e2+e3 S E I Ld d=e2-e3 S E I Ld d=e2+e4 S E I Ld d=e2-e4 S E I Ld d=e3+e4 S8 16 E I Ld d=e3-e4 S5 12 E I 3 5 e e e i i i i i 48.497 66.972 60.621 86.025 76.210 105.07 Le1+e2+e3 42.723 46.188 56.002 79.096 Le1+e2+e4 31.754 44.455 53.116 77.364 72.168 97.572 Le1+e3+e4 24.248 36.373 37.527 56.580 50.229 73.323 Le2+e3+e4 11.547 21.361 32.331 50.806 69.282 Le1-e2+e3 6.9282 13.856 20.207 27.135 Le1-e2+e4 17.897 27.712 13.279 14.433 23.671 Le1-e3+e4 9.2376 19.052 -5.196 3.4641 -11.54 1.7320 Le2-e3+e4 34.063 49.652 18.475 30.599 Le1+e2-e3 40.991 34.641 35.795 Le1+e2-e4 40.414 59.467 49.074 71.591 Le1+e3-e4 23.094 25.403 47.920 Le2+e3-e4 -3.464 4.0414 -12.12 -1.154 -17.32 -6.928 Le1-e2-e3 4.618 10.39 5.196 16.16 2.886 16.74 Le1-e2-e4 15.588 -4.041 -12.70 Le1-e3-e4 6.3508 15.011 -23.09 -9.237 -38.10 -21.36 Le2-e3-e4 42.5 60 57.5 82 73.5 102 e1+e2+e3+e4 41.5 56 47.5 67.5 80.5 e1+e2+e3-e4 30.5 45 22.5 34.5 23.5 36.5 e1+e2-e3+e4 11 20 33.5 52 48.5 71.5 e1-e2+e3+e4 28.5 42 10 20.5 5.5 16.5 e1+e2-e3-e4 9 17.5 22.5 37 48.5 e1-e2+e3-e4 -1.5 4.5 -3 5 e1-e2-e3+e4 -5 2.5 -19.5 -6 -27.5 -13.5 e1-e2-e3-e4 I didn't check every pair of boundary plates but many of them. It appears that the 6 samples are so tight that we do not separate them with Linear plates along! So FAUST-L-TP-accuracy = 96% Looking at the numbers, I don't expect S or R plates to help much.
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At any node in the Hull Tree, create children by:
FAUST LSR Hull classifier, recursive X=X(X1,...,Xn,C); oultiers are O=O(O1,...,On,OC)X initially ; OT=Outlier_Thres; Carve off outliers from X into O (O:=O{x|Rankn-1DNN(x,X)OT; O:=O{x|D2NN(X,x)>OT ).... We set OT=5 and carve off outliers: i7, i9, i10, i35, i36, i39, s42 59 69 56 d=e1 6 9 25 32 22 d=e2 13 14 18 15 24 d=e4 5 6 14 10 19 30 51 48 69 d=e3 There is TP purity except in Le2[48,51] so restricting to that interval: ( i.e., X = L-1[48,51] ) Count reductions are insufficient for e1 and e3, so use e4, 61.5 70.7 57.9 64.3 e1+e2 3 5 75.6 79.1 77.7 80.6 e1+e3 2 4 53.7 55.1 54.4 57.2 e1+e4 1 so restricting to Le4-1[15,18]: restrict to Le1+e2-1[61,64.3]: restrict to Le1+e3-1[77.7,79.1]: 79.0 80.8 e1+e2+e3 1 restrict Le1+e4-1[55.1,54.4]: 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)X initially ; OT=Outlier_Thres; Carve off outliers from X into O ( O:=O{x|Rankn-1DNN(x,X)OT; O:=O{x|D2NN(X,x)>OT .... ) At any node in the Hull Tree, create children by: Moving to next 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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Thanksgiving clustering (carve off clusters as one would carve a thanksgiving turkey)
2.62 AvgDNN 2.44 MedDNN Let m be a furthest point from aAvgX (i.e., pt in X that maximizes SPTS, Sa=(X-a)o(X-a) ) DNNS (top) 16.0 i39 7.34 i7 6.32 i10 6.24 s42 5.56 i9 5.38 i36 5.38 i35 4.89 i15 4.89 e13 4.58 s23 4.35 i20 4.24 e15 4.24 i1 4.12 i32 4.12 i19 4.12 i18 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 ) SLL-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 UDR[Le4(X)] L Ct Gp 25 2 49s 49e 3i 1e 41i
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Thanksgiving clustering (carve off clusters as one would carve a thanksgiving turkey)
2.62 AvgDNN 2.44 MedDNN 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 33 1 no gap or PCCs Let m be a furthest point from aAvgX (i.e., pt in X that maximizes SPTS, Sa=(X-a)o(X-a) ) DNNS (top) 16.0 i39 7.34 i7 6.32 i10 6.24 s42 5.56 i9 5.38 i36 5.38 i35 4.89 i15 4.89 e13 4.58 s23 4.35 i20 4.24 e15 4.24 i1 4.12 i32 4.12 i19 4.12 i18 If m is an outlier, carve {m} off from X. Repeat until m is a non-outlier. Construct L=Ld where d=am/|am| 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 ) SLL-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., mimj using pillars), before going to pkmeans. IRIS: Carve off outliers i39 i7 i10 s42 i9 i36 i35 i15 e13 s23 i20 e15 i1 i32 i19 i18 m = i23 = is furthest point from AvgX = ) Construct L=Ld where d=am/|am|= UDR[L(X)] gap count L+3_val Carve off cluster C1=L-1[0,7] (s=48) No PCCs remain (except 1st and last) so add Avg(X) to CCS={( )} Construct L=Ld where d=am/|am| = UDR(LX) gap count value
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APPENDIX FAUST LSR classifier on IRIS X=X(X1,...,Xn,C); oultiers are O=O(O1,...,On,OC)X initially empty; OT=Outlier_Threshold; Carve off outliers from X into O (O:=O{x|Rankn-1Dis(x,X)OT; O:=O{x|Rankn-2Dis(X,x)>OT );...; Define class hulls: Hk{zRn | minFd,p,k Fd,p,k(z) maxFd,p,k (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 the Ck that minimizes dis(y,Ck), kMH (note dis(y,Ck)=dis(y,X)&Ck). Else (y is in no hulls), if dis(y,O)min{dis(y,o)|oO}=dis(y,oi)<OT, declare y to be in the class of oi else declare y to be other. Notes: 1. This algorithm deals with singleton outliers but ignores doubleton and tripleton outliers etc. 2. In Elseif, rather than compute dis(y,Ck) (single link distance) one could use dis(y,meanCk) for the pre-computed class means. We create hull boundaries for the d=ek=(0..1at k..0) standard basis and check for overlaps. Then the goal is to reduce False Positives. We set OT=5 and carve off outliers: i7, i9, i10, i35, i36, i39, s42 Ld d=e4=SW p=origin S 1 6 E I 49 15 12 32 35 Ld d=e1=PL p=origin S E I 15 3 13 5 31 11 26 29 Ld d=e3=SL p=origin S 10 19 E I 6 14 30 44 Ld d=e2=PW p=origin S E I 1 15 2 26 16 6 10 5 6 5 6 Start here 5 6 e e e e e i i i i i i Ld d=e1+e2 p=origin E I 3 5 Ld d=e1-e2 p=origin E I 1 5 Note that these last 3 are applied recursively. We get 100% TP accuracy quickly with recursion (building hull trees). This isn't what the algorithm says to do so we need to make this "LSR Hull Tree" algorithm precise. First, let's see how FAUST-L along as described does wrt TPs. Ld d=-e1+e2 p=origin E I 5 1
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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 aAvgX If m1 is an outlier (Check using Sm1or better using D2NN?), repeat until m1 is a non-outlier. AvX1 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). (Instead of using Smi or D2NN to eliminate outliers each round, one might get better pillars by constructing Lmi-1mi:XR, eliminating outliers that show up on L, then picking the pillar to be the mean (or vector of medians) of the slice L-1[(3PCC1+PCC2)/4 , PCC2) ? ) m3 m2 A PCC Pillar pkmeans clusterer: Assign each (object, class) a ClassWeightReals (all CW init at 0) Classes numbered as they are revealed. As we are identifying pillar mj's, compute Lmj = Xo(mj-mj-1) and 1. For the next larger PCI in Ld(C), left-to-right. 1.1a If followed by PCD, CkAvg(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 d minL = pci1L A potential advantage of the classifier: FAUST Linear-Spherical-Radial (LSR) pcd1L 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 On each PCC interval (ill-defined but here [pci1L,pcd1L] (pcd1L,pci2L) [pci2L,pcd2L] Build hull segments on each interval and OR them? pci2L Whereas the convex hull in orange (lots of false positives) maxL = pcd2L
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DNN or D2NN or D2NNS are powerful constructs
2.62 AvgDNN 2.44 MedianDNN 3.87 e10 3.60 e11 3 e12 4.89 e13 1.41 e14 4.24 e15 1.41 e16 2 e17 2.44 e18 2.64 e19 1.73 e20 3 e21 3.31 e22 3.60 e23 2.23 e24 2 e25 1.41 e26 3.16 e27 3.16 e28 2 e29 2 e40 2.64 e41 1.41 e42 1.41 e43 1.41 e44 1.73 e45 1.41 e46 1.41 e47 2 e48 3.87 e49 1.41 e50 4.24 i1 2.64 i2 3.87 i3 2.44 i4 3 i5 2.64 i6 7.34 i7 2.64 i8 5.56 i9 6.32 i10 2.23 i11 3.46 i12 1.73 i13 2.64 i14 4.89 i15 3 i16 1.41 i17 4.12 i18 4.12 i19 4.35 i20 2.23 i21 3.16 i22 2.64 i23 1.73 i24 3 i25 3.46 i26 1.73 i27 2.44 i28 1 i29 3.46 i30 2.64 i31 4.12 i32 1 i33 3.31 i34 5.38 i36 2.44 i37 1.41 i38 16.0 i39 1.73 i40 2.44 i41 2.44 i42 2.64 i43 2.23 i44 2.44 i45 2.44 i46 2.44 i47 2.23 i48 2.44 i49 2.82 i50 DNN = 1 s1 1.41 s2 1.41 s3 1.41 s4 1.41 s5 3.31 s6 2.23 s7 1 s8 1.41 s9 1.73 s10 1 s11 2.23 s12 1.41 s13 2.44 s14 4.12 s15 3.60 s16 3.46 s17 1 s18 3.31 s19 1.41 s20 2.82 s21 1.41 s22 4.58 s23 2 s24 3 s25 2 s26 2 s27 1.41 s28 1.41 s29 1.41 s30 1.41 s31 2.82 s32 3.46 s33 3.46 s34 1.73 s35 2.23 s36 3 s37 1.73 s38 1.41 s39 1 s40 1.41 s41 6.24 s42 2 s43 2.23 s44 3.60 s45 1.41 s46 1.41 s47 1.41 s48 1 s49 1.41 s50 2.64 e1 2.64 e2 2.64 e3 2 e4 2.44 e5 3 e6 2.64 e7 1.41 e8 2.44 e9 DNNS = Distance to Nearest Neighbor Sorted 16.0 i39 7.34 i7 6.32 i10 6.24 s42 5.56 i9 5.38 i36 5.38 i35 4.89 i15 4.89 e13 4.58 s23 4.35 i20 4.24 e15 4.24 i1 4.12 i32 4.12 i19 4.12 i18 4.12 s15 3.87 e49 3.87 e10 3.87 i3 3.74 e36 3.60 s16 3.60 s45 3.60 e11 3.60 e23 3.46 e30 3.46 s33 3.46 s34 3.46 i12 3.46 i26 3.46 i30 3.46 s17 3.31 s19 3.31 i34 3.31 e22 3.31 s6 3.31 e34 3.16 e27 3.16 i22 3.16 e28 3 e6 3 s37 3 i5 3 e21 3 i16 3 s25 3 e12 3 i25 2.82 e37 2.82 s32 2.82 i50 2.82 s21 2.64 e41 2.64 i31 2.64 i43 2.64 e2 2.64 i2 2.64 i8 2.64 e38 2.64 i23 2.64 e3 2.64 i14 2.64 e7 2.64 e19 2.64 i6 2.64 e1 2.44 i37 2.44 e5 2.44 i4 2.44 i45 2.44 i28 2.44 i42 2.44 i49 2.44 i47 2.44 i41 2.44 i46 2.44 e18 2.44 e9 2.44 s14 2.23 i21 2.23 i48 2.23 i11 2.23 s7 2.23 e24 2.23 i44 2.23 s12 2.23 s36 2.23 s44 2 s24 2 e35 2 e29 2 s43 2 s27 2 e17 2 e48 2 e40 2 s26 2 e4 2 e25 1.73 i13 1.73 i27 1.73 i24 1.73 s38 1.73 e45 1.73 i40 1.73 e39 1.73 e20 1.73 s35 1.73 s10 1.41 e26 1.41 s41 1.41 e50 1.41 s47 1.41 s4 1.41 e44 1.41 s46 1.41 e42 1.41 e47 1.41 e33 1.41 e46 1.41 e31 1.41 s5 1.41 e16 1.41 s39 1.41 e8 1.41 s31 1.41 s3 1.41 s30 1.41 s13 1.41 s29 1.41 i17 1.41 i38 1.41 s2 1.41 s28 1.41 s50 1.41 s22 1.41 e43 1.41 s20 1.41 e14 1.41 e32 1.41 s48 1.41 s9 1 s40 1 i33 1 i29 1 s18 1 s11 1 s8 1 s49 1 s1 REMEMBER! 1. The pTree Rule: Never throw a pTree away! 2. In the process of creating D2NN we create, for each xX, 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 (top portion) 16.0 i39 GAP 7.34 i 6.32 i 6.24 s 5.56 i 5.38 i 5.38 i 4.89 i 4.89 e 4.58 s 4.35 i 4.24 e 4.24 i 4.12 i 4.12 i 4.12 i outlier slider 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. 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.
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Computing the Rank values and Rank pTrees, one at a time, using our pTree code.
(n=3) c=Count(P&P4,3)= < 6 p=6–3=3; P=P&P’4,3 masks off highest (val 8) {0} X P4, P4, P4, P4,0 10 5 6 7 11 9 3 1 1 1 1 (n=2) c=Count(P&P4,2)= >= 3 P=P&P4,2 masks off lowest (val 4) {1} (n=1) c=Count(P&P4,1)= < 3 p=3-2=1; P=P&P'4,1 masks off highest (val8-2=6 ) {0} (n=0) c=Count(P&P4,0 )= >= 1 P=P&P4,0 {1} 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) */ 23 * * * * = RankKval= P=MapRankKPts= ListRankKPts={2} 1 {0} {1} {0} {1}
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(n=3) c=Count(P&P4,3)= 3 < 6 p=6–3=3; P=P&P’4,3 masks off 1s {0}
What if there ar duplicates? (n=3) c=Count(P&P4,3)= < 6 p=6–3=3; P=P&P’4,3 masks off 1s {0} X P4, P4, P4, P4,0 10 3 6 7 11 9 1 1 1 1 (n=2) c=Count(P&P4,2)= < 3 p=3-2= P=P&P'4,2 masks off 1s {0} (n=1) c=Count(P&P4,1)= >= 1 P=P&P 4,1 masks off 0s (none) {1} (n=0) c=Count(P&P4,0 )= >= 1 P=P&P4,0 {1} (n=3) c=Count(P&P4,3)= < 5 p=5–3=2; P=P&P’4,3 masks off 1s {0} X P4, P4, P4, P4,0 10 3 7 11 9 1 1 1 1 (n=2) c=Count(P&P4,2)= >= 2 P=P&P4,2 masks off 0s {1} (n=1) c=Count(P&P4,1)= >= 2 P=P&P 4,1 masks off 0s (none) {1} (n=0) c=Count(P&P4,0 )= >= 2 P=P&P4,0 {1}
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(n=3) c=Count(P&P4,3)= 3 < 4 p=4–3=1; P=P&P’4,3 masks off 1s {0}
X P4, P4, P4, P4,0 10 3 7 11 9 1 1 1 1 (n=2) c=Count(P&P4,2)= >= 1 P=P&P4,2 masks off 0s {1} (n=1) c=Count(P&P4,1)= >= 1 P=P&P 4,1 masks off 0s (none) {1} (n=0) c=Count(P&P4,0 )= >= 1 P=P&P4,0 {1} (n=3) c=Count(P&P4,3)= >= 3 P=P&P4,3 masks off 0s {1} X P4, P4, P4, P4,0 10 3 7 11 9 1 1 1 1 (n=2) c=Count(P&P4,2)= < 3 p-3-0= P=P&P'4,2 masks off 1s (none ) {0} (n=1) c=Count(P&P4,1)= < 3 p=3-2= P=P&P' 4,1 masks off 1s {0} (n=0) c=Count(P&P4,0 )= >= 1 P=P&P4,0 {1}
12
(n=3) c=Count(P&P4,3)= 3 >= 2 P=P&P4,3 masks off 0s {1}
X P4, P4, P4, P4,0 10 3 7 11 9 1 1 1 1 (n=2) c=Count(P&P4,2)= < 2 p-2-0= P=P&P'4,2 masks off 1s (none ) {0} (n=1) c=Count(P&P4,1)= >=2 P=P&P 4,1 masks off 0s {1} (n=0) c=Count(P&P4,0 )= < 2 P=P&P4,0 mask off 1s {0} (n=3) c=Count(P&P4,3)= >= 1 P=P&P4,3 masks off 0s {1} X P4, P4, P4, P4,0 10 3 7 11 9 1 1 1 1 (n=2) c=Count(P&P4,2)= < 1 p-1-0= P=P&P'4,2 masks off 1s (none ) {0} (n=1) c=Count(P&P4,1)= >=1 P=P&P 4,1 masks off 0s {1} (n=0) c=Count(P&P4,0 )= <= 1 P=P&P'4,0 mask off 0s {1} 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.
13
UDR Univariate Distribution Revealer (on Spaeth:)
15 UDR Univariate Distribution Revealer (on Spaeth:) applied to S, a column of numbers in bistlice format (an SpTS), will produce the DistributionTree of S DT(S) depth=h=0 depth=h=1 Y y1 y2 y y y y y y y y y ya 13 4 pb 10 9 yc 11 10 yd 9 11 ye 11 11 yf 7 8 yofM 11 27 23 34 53 80 118 114 125 110 121 109 83 p6 1 p5 1 p4 1 p3 1 p2 1 p1 1 p0 1 p6' 1 p5' 1 p4' 1 p3' 1 p2' 1 p1' 1 p0' 1 node2,3 [96.128) f= depthDT(S)b≡BitWidth(S) h=depth of a node k=node offset Nodeh,k has a ptr to pTree{xS | F(x)[k2b-h+1, (k+1)2b-h+1)} and its 1count p6' 1 5/64 [0,64) p6 10/64 [64,128) p5' 1 3/32[0,32) 2/32[64,96) p5 2/32[32,64) ¼[96,128) p3' 1 0[0,8) p3 1[8,16) 1[16,24) 1[24,32) 1[32,40) 0[40,48) 1[48,56) 0[56,64) 2[80,88) 0[88,96) 0[96,104) 2[194,112) 3[112,120) 3[120,128) p4' 1 1/16[0,16) p4 2/16[16,32) 1[32,48) 1[48,64) 0[64,80) 2[80,96) 2[96,112) 6[112,128) 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 fact are often all we need to know about the pTree to get the answers we are after.).
14
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=0001 p=origin S 1 6 E I 50 22 16 34 28 Ld d=1000 p=origin S E I 16 26 32 12 Ld d=0010 p=origin S E I 48 50 15 2 1 Ld d=0100 p=origin S E I 1 15 3 2 1 6 i i i i i i 1 of the 6 occurs, i7. i i i i i i i i i i i i 5 of the 6 occur. 6 of the 6 occur. i i i i i i i i i i i i i i i 6 of the 16 occur. i i i i i i i i i i i i i i i What does this tell us about FAUST LSR on IRIS? 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 e i The sum of the distances to class corner vectors is 38.1, average is 6.4.
15
FAUST LSR Classification on IRIS150, a new version
Ld d=1000 p=origin MinL, MaxL for classes S,E,I S E I 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). We have used just d=1000 so many more edges could be placed on these hulls to eliminate false positives. 2. If there is a whole table of unclassified samples to be classified (e.g., millions 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 accomplish the classification as one batch process (no loop required). 16 26 32 12 99 393 1096 1217 1826 p=AvE 270 1558 2568 Ld d=avgI-avgE p=origin E I 24,0 2,4 0,1 p=AvgS This is the {y isa EI)2 recursive step pseudo code: if R1000,AvgS(y) < 792 {y isa I} elseif 792 R1000,AvgS(y) 1558 {y isa EI}3 elseif 1558 R1000,AvgS(y) 2568 {y isa I} else {y isa O} p=AvgE 22.69 35.51 54.32 Ld d=avgI-avgE p=origin E 1.78 I 1,0 0,1 This is the {y isa EI}3 recursive step: if LAvE-AvI(y) < 13.6 {y isa E } elseif 13.6 LAvE-AvI(y) 15.9 {y isa EI}4 elseif 15.9 < LAvE-AvI(y) 16.6 {y isa I} else {y isa O } if L1000(y)=y1 < 49 {y isa S } elseif 49 L1000(y)=y1 58 {y isa SEI}1 elseif 59 < L1000(y)=y1 70 {y isa EI}2 elseif 70 < L1000(y)=y1 79 {y isa I} else {y isa O } This is the {y isa EI)4 recursive step pseudo code: if RAvE-AvI,AvgE(y)<31.02 {y isa E } elseif RAvE-AvI,AvgE(y)35.51 {y isa EI}5 elseif RAvE-AvI,AvgE(y)54.32 {y isa I} else {y isa O } This is the {y isa SEI)1 recursive step pseudo code: if R1000,AvgS(y) 99 {y isa S } elseif 99 < R1000,AvgS(y) < 393 {y isa O } elseif 393 < R1000,AvgS(y) 1096 {y isa E } elseif 1096 < R1000,AvgS(y) < 1217 {y isa O } elseif 1217 R1000,AvgS(y) 1826 {y isa I} else {y isa O } This is the {y isa EI}5 recursive step: if =LAvgE-AvgI,origin(y) {y isa E } elseif 6.26=LAvgE-AvgI,origin(y) {y isa I} else {y isa O } L1000,origin(y) [43,49)[49,58](58,70](70,79]else OTHER yS yI R1000,AvgE(y) [0,99][399,1096][1217,1826]else OTHER yS yE yI R1000,AvgE(y) [270,792)[792,1558](1558,2568]else OTHER yE yI LAvEAvI,origin(y) [5.7,13.6)[13.6,15.9](15.9,16.6]else OTHER yE yI LSR Decision Tree algorithm is, Build decision tree for each ek (also for some ek combos?). Build branches to 100% TP (no class duplication exiting). Then y isa C iff y isa C in every tree else y isa Other. node build a branch for each pair of classes in each interval. RAvEAvI,AvgE(y) [22.7,31)[31,35.52](35.52,54.32]else OTHER yE yI LAvEAvI,origin(y)= 1.78 yE yI else OTHER
16
FAUST LSR DT Classification on IRIS150, d= 0100
Instead of calculating R's wrt a freshly calculated Avg in each slice, we calculate R0100,AvgS R0100,AvgE R0100,AvgI once then & w mask, P20L0100,0rigin<22 and later & with masks, P22L0100,0rigin<23 , P23L0100,0rigin34 , P34<L0100,0rigin38 and P38<L0100,0rigin44 L 0100.Origin(y) S E I 1 2 29 47 46 15 3 6 On 22L0100,O<23 R0100,AvgE 15 18 58 59 On 23L0100,O34 R0100,AvgS 0 66 46,12 3 234 R0100,AvgI 36 929 On 34<L0100,O38 0 55 96 273 On 23L0100,O34 & 352R0100,AvgS1750 LAvgEAvgI,Origin 44,11 It takes 7 recursive rounds to separate E and I (build this branch to 100% TP) in this branch of the e2=0100 tree 0100 (Pedal Width). It seems clear we are mostly pealing off outliers a few at a time. Is it because we are not revising the Avg Vectors as we go (to get the best angle)? On the next slide we make a fresh calculation of Avg for each subcluster. It also appears to be unnecessary to position the starting point of the AvgEAvgI vector to both AvgE and AvgI On 23L0100,O34 & 352R0100,AvgS1750 & 53LAvgEAvgI,Origin77 RAvgEAvgI,AvgE 40,10 RAvgEAvgI,AvgI On 23L0100,O34 & 352R0100,AvgS1750 & 53LAvgEAvgI,Origin77 & 2.8RAvgEAvgI,AvgE75.2 LAvgEAvgI,Origin 7,7 On 23L0100,O34 & 352R0100,AvgS1750 & 53LAvgEAvgI,Origin77 & 2.8RAvgEAvgI,AvgE75.2 & 74.1LAvgEAvgI,Origin76.2 RAvgEAvgI,AvgE 6,4 On 23L0100,O34 & 352R0100,AvgS1750 & 53LAvgEAvgI,Origin77 & 2.8RAvgEAvgI,AvgE75.2 & 74.1LAvgEAvgI,Origin76.2 & 15.4RAvgEAvgI,AvgE57.3 & 74.2LAvgEAvgI,Origin75.6 RAvgEAvgI,AvgE 15 37 57 6,0 0,1 On 23L0100,O34 & 352R0100,AvgS1750 & 53LAvgEAvgI,Origin77 & 2.8RAvgEAvgI,AvgE75.2 & 74.1LAvgEAvgI,Origin76.2 & 15.4RAvgEAvgI,AvgE57.3 LAvgEAvgI,Origin 6,1
17
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. L 0100.Origin(y) S E I 1 2 29 47 46 15 3 6 L 0100.Origin(y) S E I 1 2 29 47 46 15 3 6 L 0010.Origin(y) S10 19 E I 2 1 50 15 On 23L0100,O34 R0100,AvgS 0 43 45,12 On 23L0100,O34 R0100,AvgE 3 234 13,21 On 30L0010,O51 R0010,AvgE 33,14 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. We pause the algorithm and try SBarrelAvgE and SBarrelAvgI in addition to LAvEAvI,O Next try inserting SLinearAvgE and SLinearAvgI in serial w LAvEAvI,O instead of parallel. On 23L0100,O34 & 320R0100,AvgS1820 LAvgEAvgI,Origin 24,9 On 23L0100,O34 & 58R0100,AvgS234 SBarrelAvgE 13,18 23L0100,O34 & 58R0100,AvgS234 SBarrelAvgI 7,14 On 23L0100,O34 & 58R0100,AvgS234 LAvgEAvgI,Origin 7,13 On 30L0010,O51 & 16.3R0100,AvgS157.6 LAvEAvI,O 19,13 On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 RAvgEAvgI,AvgE 24,6 On 23L0100,O34 & 58R0100,AvgS234 & 25LAvEAvI,O32 SLinearAvgE 1,5 On 23L0100,O34 & 58R0100,AvgS234 & 25LAvEAvI,O32 SLinearAvgI 6,11 On 30L0100,O51 & 16.3R0100,AvgS157.6 & 66.3LAvEAvI,O78.4 RAvgEAcI,AvE 5,6 Seems very beneficial! Use only LinearAvg with the smallest count, in this case LinearAvgE? On 23L0100,O34 & 58R0100,AvgS234 & 25LAvEAvI,O32 & 27SLinearAvgE66.1 Sp 11 23 1, ,5 On 30L0100,O51 & 16.3R0100,AvgS157.6 & 66.3LAvEAvI,O78.4 & 1416RAvgEAcI,AvE1449 & L 5,6 On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 & 4RAvgEAvgI,AvgE88 LAvgEAvgI,Origin 18,5 On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 & 4RAvgEAvgI,AvgE88 & 3411LAvgEAvgI,Origin4397 RAvgEAvgI,AvgE 10,5 On 23L0100,O34 & 320R0100,AvgS1820 & 25LAvgEAvgI,Origin30 & 4RAvgEAvgI,AvgE88 & 3411LAvgEAvgI,Origin4397 & 5.9RAvgEAvgI,AvgE20.5 LAvgEAvgI,Origin 1,1
18
FAUST Oblique LSR Classification IRIS150
Ld d=1000 p=origin S E I Ld d=0100 p=origin S E I Ld d=0010 p=origin S E I Ld d=0001 p=origin S 1 6 E I 16 26 32 12 1 15 3 2 1 6 48 50 15 2 1 34 50 22 16 34 28 99 393 1096 1217 1826 270 1558 2568 66 310 1750 4104 55 3139 3850 4954 8134 9809 3000 6120 6251 p=AvgS 279 5 171 186 748 998 1 517, 4 79 633 3 234 793 1417 712 636 9, 3 983 1369 2.8 158 199 5 3617 7 152 611 p=AvgE 24 132 730 1622 2281 388 1369 5 1403 929 1892 2824 96 273 1776 2747 5.9 319 453 454 1397 p=AvgI In pTree psuedo-code: Py<43=PO P43y<49=PS P49y58=PSEI P59<y70=PEI P70<y79=PI PO:= PO or Py>70
19
This first example suggests that recursion can be important.
Row Attr1 Attr2 1 0 0 2 0 100 3 0 0 4 110 110 5 0 114 6 0 123 7 0 145 8 0 0 1, 3, 8 4 2 6 5 7 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,yX. 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!). Recursion solves problems, e.g., gap isolating point4 is revealed by a Le1(X)=Attr1 gap. Recursively restricting to { } and applying Le2(X)=Attr2 reveals the 2 other gaps This first example suggests that recursion can be important. A different example suggests that recursion order can also be important: Row Attr1 Attr2 1 0 0 2 0 25 3 0 50 4 75 75 5 0 100 6 0 125 7 0 150 1 2 6 4 7 3 5 Using ordering, d=e2, e1 recursively, Le2=Attr2 reveals no gaps, so Le1=Attr1 is applied to all of X and reveals only the gap around point4. 100 25 Using ordering d=e1, e2 instead: Le1=Attr1 on X reveals a gap of at least 100 around point4 (actual gap: ) StD: ~ ~55 Note StD doesn't always reveal best order! Le2=Attr2 is applied to X-{4} reveals a gap of 50 between {123} and {567} also. 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 1dis(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?
20
LSR IRIS150-. Consider all 3 functionals, L, S and R. What's the most efficient way to calculate all 3?\ o=origin; pRn; dRn, |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 = [maxLo,d,k] - pod = min(Xod & Ck) - pod = max(Xod & Ck) - pod OR = min(X&Ck) o d - pod = max(X&Ck) o d - pod Sp = (X - p)o(X - p) = -2Xop+So+pop = Lo,-2p + (So+pop) minSp,k=minSp&Ck maxSp,k = maxSp&Ck = min[(X o (-2p) &Ck)] + (XoX+pop) =max[(X o (-2p) &Ck)] + (XoX+pop) OR = min[(X&Ck)o-2p] + (XoX+pop) =max[(X&Ck)o-2p] + (XoX+pop) Rp,d Sp, - Lp,d2 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.
21
LSR on IRIS150 C13 Dse S E I L H y isa OTHER if yoDse (-,495)(802,1061)(2725,) y isa OTHER or S if yoDse C1,1 [ 495 , 802] y isa OTHER or I if yoDse C1,2 [1061 ,1270] y isa OTHER or E or I if yoDse C1,3 [1270 ,2010 C1,3: 0 s 49 e 11 i y isa OTHER or I if yoDse C1,4 [2010 ,2725] Dei E y isa O if yoDei (-,-117)(-3,) I y isa O or E or I if yoDei C2,1 [-62 ,-44] L H y isa O or I if yoDei C2,2 [-44 , -3] C2,1: 2 e 4 i Dei E y isa O if yoDei (-,420)(459,480)(501,) I y isa O or E if yoDei C3,1 [420 ,459] L H y isa O or I if yoDei C3,2 [480 ,501] Continue this on clusters with OTHER + one class, so the hull fits tightely (reducing false positives), using diagonals? C1,1: D=1000 y isa O if yoD(-,43)(58,) L H y isa O|S if yoD C2,3 [43,58] C2,3: D=0100 y isa O if yoD(-,23)(44,) L H y isa O|S if yoD C3,3 [23,44] C3,3: D=0010 y isa O if yoD(-,10)(19,) L H y isa O|S if yoD C4,1 [10,19] C4,1: D=0001 1 6 y isa O if yoD(-,1)(6,) L H y isa O|S if yoD C5,1 [1,6] C5,1: D=1100 y isa O if yoD(-,68)(117,) L H y isa O|S if yoD C6,1 [68,117] C6,1: D=1010 y isa O if yoD(-,54)(146,) L H y isa O|S if yoD C7,1 [54,146] C7,1: D=1001 y isa O if yoD(-,44)(100,) L H y isa O|S if yoD C8,1 [44,100] C8,1: D=0110 y isa O if yoD(-,36)(105,) L H y isa O|S if yoD C9,1 [36,105] C9,1: D=0101 y isa O if yoD(-,26)(61,) L H y isa O|S if yoD Ca,1 [26,61] Ca,1: D=0011 y isa O if yoD(-,12)(91,) L H y isa O|S if yoD Cb,1 [12,91] Cb,1: D=1110 y isa O if yoD(-,81)(182,) L H y isa O|S if yoD Cc,1 [81,182] Cc,1: D=1101 y isa O if yoD(-,71)(137,) L H y isa O|S if yoD Cd,1 [71,137] Cd,1: D=1011 y isa O if yoD(-,55)(169,) L H y isa O|S if yoD Ce,1 [55,169] Ce,1: D=0111 y isa O if yoD(-,39)(127,) L H y isa O|S if yoD Cf,1 [39,127] Cf,1: D=1111 y isa O if yoD(-,84)(204,) L H y isa O|S if yoD Cg,1 [84,204] Cg,1: D=1-100 y isa O if yoD(-,10)(22,) L H y isa O|S if yoD Ch,1 [10,22] Ch,1: D=10-10 y isa O if yoD(-,3)(46,) L H y isa O|S if yoD Ci,1 [3,46] The amount of work yet to be done., even for only 4 attributes, is immense.. For each D, we should fit boundaries for each class, not just one class. For 4 attributes, I count 77 diagonals*3 classes = 231 cases. How many in the Enron 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??
22
Dot Product SPTS computation: XoD = k=1..nXkDk
( = 22 + 21 (1 p1,0 + 1 p11 + 20 (1 p1,0 1 p1,1 + 1 p2,1 ) + 1 p2,0 + 1 p2,0 ) 1 3 3 D D1,1 D1,0 1 1 D2,1 D2,0 1 3 2 X X1 X2 p11 p10 p21 p20 6 9 XoD 1 pXoD,3 pXoD,2 pXoD,1 pXoD,0 /*Calc PXoD,i after PXoD,i-1 CarrySet=CARi-1,i RawSet=RSi */ INPUT: CARi-1,i, RSi ROUTINE: PXoD,i=RSiCARi-1,i CARi,i+1=RSi&CARi-1,i OUTPUT: PXoD,i, CARi,i+1 1 CAR12,3 & 1 & 1 CAR22,3 PXoD,2 1 1 CAR11,2 & & 1 CAR21,2 PXoD,1 1 & 1 1 1 1 PXoD,0 CAR10,1 & & 1 PXoD,3 CAR13,4 Different data. 3 3 D D1,1 D1,0 1 1 D2,1 D2,0 1 1 ( = 22 + 21 (1 p1,0 + 1 p11 + 20 (1 p1,0 1 p1,1 + 1 p2,1 ) + 1 p2,0 + 1 p2,0 ) 1 3 2 X pTrees 6 18 9 XoD 1 1 & 1 1 1 & 1 & 1 1 1 1 & 1 1 & 1 & 1 & PXoD,2 1 & PXoD,1 1 & 1 PXoD,0 CAR10,1 & 1 & PXoD,3 PXoD,4 We have extended the Galois field, GF(2)={0,1}, XOR=add, AND=mult to pTrees. SPTS multiplication: (Note, pTree multiplication = &) = (21 p1,1 +20 p1,0) (21 p2,1 p2,0) = 22 +21( p2,0+ + 20 p1,0 p2,0 X1*X2 1 & 1 & 1 & 1 1 & pX1*X2,0 1 & pX1*X2,2 1 & pX1*X2,1 1 3 2 X X1 X2 p11 p10 p21 p20 9 X1*X2 pX1*X2,3 pX1*X2,2 pX1*X2,1 pX1*X2,0 1 pX1*X2,3
23
FAUST Oblique: XoD used in CCC, TKO, PLC and LARC) and (x-X)o(x-X)
Example: FAUST Oblique: XoD used in CCC, TKO, PLC and LARC) and (x-X)o(x-X) = -2Xox+xox+XoX is used in TKO. 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?). xox 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; P=P&Pi }; else /* < p */ { p=p-Count(P&Pi); P=P&P'i }; RankN-1(XoD)=Rank2(XoD) 1 1 D=x1 D1,1 D1,0 0 1 D2,1 D2,0 1 P=P&p1 32 1*21+ P p1 n=1 p=2 1 P=p0&P 22 1*21+1*20=3 so -2x1oX = -6 P &p0 n=0 p=2 1 3 2 X X1 X2 p11 p10 p21 p20 2 3 XoD 1 p3 p2 p1 p,0 RankN-1(XoD)=Rank2(XoD) 3 0 D=x2 D1,1 D1,0 1 1 D2,1 D2,0 0 0 3 9 2 XoD 1 p3 p2 p1 p,0 1 P=P&p'3 1<2 2-1=1 0*23+ P p3 n=3 p=2 1 P=p'2&P 0<1 1-0=1 0*23+0*22 P &p2 n=2 p=1 1 P=p1&P 21 0*23+0*22+1*21+ P &p1 n=1 p=1 1 P=p0&P 11 0*23+0*22+1*21+1*20=3 so -2x2oX= -6 P &p0 n=0 p=1 RankN-1(XoD)=Rank2(XoD) 2 1 D=x3 D1,1 D1,0 1 0 D2,1 D2,0 0 1 3 6 5 XoD 1 p3 p2 p1 p,0 1 P=P&p2 22 1*22+ P p2 n=2 p=2 1 P=p'1&P 1<2 2-1=1 1*22+0*21 P &p1 n=1 p=2 1 P=p0&P 11 1*22+0*21+1*20=5 so -2x3oX= -10 P &p0 n=0 p=1
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So let us look at ways of doing the work to calculate 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 XoD = k=1..nXk*Dk X pTrees XoD 1 3 2 1 1 1 1 6 9 1 1 1 1 3 3 D D1,1 D1,0 1 1 D2,1 D2,0 1 1 ( = 22 + 21 1 p1,0 p11 + 20 1 p1,1 p2,1 ) p2,0 + 1 p2,1 ) p2,0 ) 1 1 1 1 1 1 1 1 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 RawSet=RSi CSP-1=CSN-1=*/ INPUT: CSPi-1, CSNi-1, RSi ROUTINE: PXoD,i=RSiCSPi-1,iCSNi-1,i CSNi,i+1=CSNi-1,iPXoD,i; CSPi,i+1=CSPi-1,iRSi-1; OUTPUT: PXoD,i, CSNi,i CSPi,i+1 ( = 22 + 21 1 p1,0 p11 + 20 1 p1,1 p2,1 ) p2,0 + 1 p2,1 ) p2,0 ) 1 1 1 1 1 1 RS1 CSN0,1= CSN-1.0PXoD,0 CSP0,1= CSP-1,0RS0 RS0 CSP-1,0=CSN-1,0= PXoD,0 PXoD,1 1 1 1 1 1 1 1 1 1 = = 1 1 1
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XoD=k=1,2Xk*Dk with pTrees: qN..q0, N=22B+roof(log2n)+2B+1
CCC Clusterer If DT (and/or DUT) not exceeded at C, partition C further by cutting at each gap and PCC in CoD 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 Xk*Dk = Dkb2bpk,b = 2BDkpk,B Dkpk,0 = Dk(2Bpk,B +..+20pk,0) = (2Bpk,B +..+20pk,0) (2BDk,B+..+20Dk,0) + 22B-1(Dk,B-1pk,B +..+20Dk,0pk,0 = 22B( Dk,Bpk,B) +Dk,Bpk,B-1) XoD = k=1..nXk*Dk k=1..n ( = 22B + 22B-1 Dk,B pk,B-1 + Dk,B-1 pk,B + 22B-2 Dk,B pk,B-2 + Dk,B-1 pk,B-1 + Dk,B-2 pk,B + 22B-3 Dk,B pk,B-3 + Dk,B-1 pk,B-2 + Dk,B-2 pk,B-1 +Dk,B-3 pk,B + 23 Dk,B pk,0 + Dk,2 pk,1 + Dk,1 pk,2 +Dk,0 pk,3 + 22 Dk,2 pk,0 + Dk,1 pk,1 + Dk,0 pk,2 + 21 Dk,1 pk,0 + Dk,0 pk,1 + 20 Dk,0 pk,0 Dk,B pk,B . . . 1 3 2 X pTrees 1 2 D D1,1 D1,0 0 1 D2,1 D2,0 1 0 B=1 XoD=k=1,2Xk*Dk with pTrees: qN..q0, N=22B+roof(log2n)+2B+1 k=1..2 ( = 22 + 21 Dk,1 pk,0 + Dk,0 pk,1 + 20 Dk,0 pk,0 Dk,1 pk,1 q0 = p1,0 = no carry 1 q1= carry1= 1 ( = 22 + 21 D1,1 p1,0 + D1,0 p11 + 20 D1,0 p1,0 D1,1 p1,1 + D2,1 p2,1 ) + D2,1 p2,0 + D2,0 p2,1 ) + D2,0 p2,0 ) q2=carry1= no carry 1 ( = 22 + 21 D1,1 p1,0 + D1,0 p11 + 20 D1,0 p1,0 D1,1 p1,1 + D2,1 p2,1 ) + D2,1 p2,0 + D2,0 p2,1 ) + D2,0 p2,0 ) 1 So, DotProduct involves just multi-operand pTree addition. (no SPTSs and no multiplications) Engineering shortcut tricka would be huge!!! q0 = carry0= 1 3 3 D D1,1 D1,0 1 1 D2,1 D2,0 1 1 q1=carry0+raw1= carry1= 1 2 ( = 22 + 21 1 p1,0 p11 + 20 1 p1,1 p2,1 ) p2,0 + 1 p2,1 ) p2,0 ) 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 ? q2=carry1+raw2= carry2= 1 q3=carry2 = carry3= 1
26
Should we pre-compute all pk,i*pk,j p'k,i*p'k,j pk,i*p'k,j
Question: Which primitives are needed and how do we compute them? X(X1...Xn) D2NN yields a 1.a-type outlier detector (top k objects, x, dissimilarity from X-{x}). D2NN = each min[D2NN(x)] (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) =k=1..n( b=B..02b(xk,b-pk,b) ) ( ----ak,b--- b=B..02b(xk,b-pk,b) ) ( 22Bak,Bak,B + 22B-1( ak,Bak,B-1 + ak,B-1ak,B ) + { 22Bak,Bak,B } 22B-2( ak,Bak,B-2 + ak,B-1ak,B-1 + ak,B-2ak,B ) + {2B-1ak,Bak,B B-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} (2Bak,B+ 2B-1ak,B-1+..+ 21ak, 1+ 20ak, 0) =k 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 xX) is separate calc. =22B ( ak,B2 + ak,Bak,B-1 ) + 22B-1( ak,Bak,B-2 ) + 22B-2( ak,B-12 22B-3( ak,Bak,B-4+ak,B-1ak,B-3) + 22B-4ak,B + ak,Bak,B-3 + ak,B-1ak,B-2 ) + Should we pre-compute all pk,i*pk,j p'k,i*p'k,j pk,i*p'k,j ANOTHER TRY! 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?) xX, 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 kxk kXk2 3. Pick this from XoX for each x and add to 2. = xoX xox XoX 5. Add 3 to this 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 xX (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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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 S E I d=e1 p=AvgS, L=Xod S&L E&L I&L R & L [43,49) S(16) 128 [49,58) E(24)I(6) 0 S(34) 99 393 1096 1217 1825 [58,70) E(26) I(32) 270 792 1558 2567 [70,79] I(12) 2081 3444 R(p,d,X) S E I 128 270 393 1558 3444 d=e1 p=AS L=(X-p)od (-pod=-50.06) S&L -1; E&L I&L -8,-2 16 [-2,8) 34, 24, 6 99 393 1096 1217 1825 [20,29] 12 [8,20) 26, 32 270 792 1558 2567 E=26 I=5 p=AvgS 30ambigs, 5 errs d=e4 p=AvgS, L=(X-p)od S&L E&L I&L -2,4) 50 [7,11) 28 [16,23] I=34 [11,16) 22, 127.5 648.7 1554.7 2892 E=22 I=7 p=AvgS I(1) E(50) I(7) (36,7) 63 70 (11) 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)Xod-pod, R(p,d)XoX+pop-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} I(42) d=e1 p=AS L=(X-p)od (-pod=-50.06) S&L -1; E&L I&L -8,-2 16 [-2,8) 34, 24, 6 99 393 1096 1217 1825 [20,29] 12 [8,20) w p=AvgE 26, 32 1.9 51.8 78.6 633 <--E=6 I=4 d=e4 p=AvgS, L=(X-p)od S&L E&L I&L -2,4) 50 [7,11) 28 [16,23] I=34 [11,16) 22, 5.7 36.2 151.06 611 E=17 I=7 p=AvgE d=e1 p=AS L=(X-p)od (-pod=-50.06) S&L -1; E&L I&L -8,-2 16 [-2,8) 34, 24, 6 99 393 1096 1217 1825 [20,29] 12 [8,20) w p=AvgI 26, 32 0.62 34.9 387.8 1369 <--E=25 I=10 d=e4 p=AvgS, L=(X-p)od S&L E&L I&L -2,4) 50 [7,11) 28 [16,23] I=34 [11,16) 22, 127.5 1555 2892 E=22 I=8 p=AvgI For e4, the best choice of p for the R step is also p=AvgE. (There are mistakes in this column on the previous slide!) There is a best choice of p for the R step (p=AvgE) but how would we decide that ahead of time?
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LSR on IRIS150 SRR(AVGs,dse) on C1,1 S y isa O if yoD (-,-184)(123,381)(2046,) y isa O or S(50) if yoD C1,1 [-184 , 123] 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] y isa O or I(1) if yoD C1,2 [ 381 , 590] Dse ; xoDes: S E I y isa O or E(50) or I(11) if yoD C1,3 [ 590 ,1331] y isa O or I(38) if yoD C1,4 [1331 ,2046] SRR(AVGs,dse) on C1,2only one such I SRR(AVGs,dse) onC1,3 E I y isa O if y isa C1,3 AND SRR(AVGs,Dse)(-,2)U(143,) y isa O or E(10) if y isa C1,3 AND SRR in [2,7) y isa O or E(40) or I(10) if y isa C1,3 AND SRR in [7,137) = C2,1 y isa O or I(1) if y isa C1,3 AND SRR in [137,143] etc. y isa O if yoD (-,-2) (19,) y isa O or I(8) if yoD [ -2 , 1.4] y isa O or E(40) or I(2) if yoD C3,1 [ 1.4 ,19] Dei ; xoDei on C2,1: E I SRR(AVGe,dei) onC3,1 E I y isa O if y isa C3,1 AND SRR(AVGs,Dei)[0,2)(370,) y isa O or E(4) if y isa C3,1 AND SRR(AVGs,Dei)[2,8) y isa O or E(27) or I(2) if y isa C3,1 AND SRR(AVGs,Dei)[8,106) y isa O or E(9) 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 Dline (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 case the dimensions would be words that have about the same count, reducing false positives. d=e1=1000; The xod limits: S E I y isa O if yoD(-,43)(79,) y isa O or S( 9) if yoD[43,47] y isa O or S(41) or E(26) or I( 7) if yoD(47,60) (yC1,2) y isa O or E(24) or I(32) if yoD[60,72] (yC1,3) y isa O if yoD[43,47]&SRR(-,52)(60,) y isa O or I(11) if yoD(72,79] y isa O if yoD[72,79]&SRR(-,49)(78,) LSR on IRIS150-2 We use the diagonals. Also we set a MinGapThres=2 which will mean we stay 2 units away from any cut y isa O or E( 3) if yoD[18,23) y isa O if yoD(-,18)(46,) y isa O or E(13) or I( 4) if yoD[23,28) (yC2,1) y isa O or S(13) or E(10) or I( 3) if yoD[28,34) (yC2,2) y isa O or S(28) if yoD[34,46] y isa O if yoD[18,23)&SRR[0,21) y isa O if yoD[34,46]&SRR[0,32][46,) d=e2=0100 on C1,2 xod lims: S E I d=e2=0100 on C1,3 xod lims: E I zero differentiation! y isa O or E(17) if yoD[60,72]&SRR[1.2,20] y isa O or I(25)if yoD[60,72]&SRR[66,799] y isa O or E( 7) or I( 7)if yoD[60,72]&SRR[20, 66] y isa O if yoD[0,1.2)(799,) d=e3=0010 on C2,2 xod lims: 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] d=e3=0001 xod lims: E I y isa O or S(13) if yoD[1,5] y isa O if yoD(-,1)(5,12)(24,) y isa O or E( 9) if yoD[12,16) y isa O or E( 1) or I( 3) if yoD[16,24) y isa O if yoD[12,16)&SRR[0,208)(558,) 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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LSR IRIS150. Next, we examine:
d=AvgEAvgI p=AvgE, L=(X-p)od S E I R(p,d,X) S E I 2 32 76 357 514 [-17,-14)] I(1) [-14,11) (50, 13) 2.8 134 [11,33] I(36) E=47 I=12 R(p,d,X) S E I .3 .9 4.7 150 204 213 [12,17.5)] I(1) d=AvgSAvgI p=AvgS, L=(X-p)od S E I [17.5,42) (50,12) 6 192 205 [11,33] I(37) E=45 I=12 d=AvgSAvgE p=AvgS, L=(X-p)od S E I R(p,d,X) S E I 2 6 137 154 393 [11,18)] I(1) [18,42) (50,11) 6.92 133 [42,64] 38 E=39 I=11 d=e1 p=AvgS, L=Xod S&L E&L I&L Note that each L=(X-p)od is just a shift of Xod by -pod (for a given d). Next, we examine: For a fixed d, the SPTS, Lp,d. is just a shift of LdLorigin,d by -pod we get the same intervals to apply R to, independent of p (shifted by -pod). Thus, we calculate once, lld=minXod hld=maxXod, then for each different p we shift these interval limit numbers by -pod since these numbers are really all we need for our hulls (Rather than going thru the SPTS calculation of (X-p)od anew new p). There is no reason we have to use the same p on each of those intervals either. d=e1 p=AS L=(X-p)od (-pod=-50.06) S&L -1; E&L I&L -8,-2 16 [-2,8) 34, 24, 6 99 393 1096 1217 1825 [20,29] 12 [8,20) 26, 32 270 792 1558 2567 E=26 I=5 30ambigs, 5 errs d=e2 p=AvgS, L=(X-p)od S&L E&L I&L ,-13) 1 -13,-11 0, 2, 1 all=-11 [0,4) [4, -11,0 29,47,46 66 310 352 1749 4104 1, 1 46,11 2, 1 9, 3 d=e3 p=AvgS, L=(X-p)od S&L E&L I&L -5,4) 47 [4,15) [37,55] I=34 [15,37) 50, 15 157 297 536 792 E=18 I=12 3, 1 d=e4 p=AvgS, L=(X-p)od S&L E&L I&L -2,4) 50 [7,11) 28 [16,23] I=34 [11,16) 22, 16 11 16 E=22 I=16 38ambigs 16errs d=e1 p=AE L=(X-p)od (-pod=-59.36) S&L E&L I&L -17-11 16 [-11,-1) 33, 21, 3 27 107 172 748 1150 [11,20] I12 [-1,11) 26, 32 1 51 79 633 E=7 I=4 E=5 I=3 d=e2 p=AvgE, L=(X-p)od -5 `17 S&L E&L I&L ,-6) 1 [-6, -5) 0, 2, 1 15 18 58 59 [7,11) [11, 1 err [-5,7) 29,47, 46 3 234 793 1103 1417 13, 21 21, d=e3 p=AvgE, L=(X-p)od S&L E&L I&L ,-25) 48 -25,-12 [9,27] I=34 [-12,9) 49, 15 2(17) 16 158 199 E=32 I=14 d=e4 p=AvgE, L=(X-p)od S&L E&L I&L -7] 50 [-3,1) 21 [5,12] 34 [1,5) 22, 16 E=22 I=16 d=e1 p=AI L=(X-p)od (-pod=-65.88) S&L E&L I&L [-17,-8) 33, 21, 3 38 126 132 730 1622 2181 [-8,4) 26, 32 34 1368 E=26 I=11 E=2 I=1 d=e2 p=AvgI, L=(X-p)od -7 `15 S&L E&L I&L ,-6) 1 [6,11) [11, [-7, 4) 29,46,46 5 36 929 1403 1893 2823 [-8, -7) 2, 1 allsame E=2 I=1 E=47 I=22 [5, 9] 9, 2, 1 S=9 d=e3 p=AvgI, L=(X-p)od S&L E&L I&L ,-25) 48 -25,-12 [9,27] I=34 [-25,-4) 50, 15 5 11 318 453 E=32 I=14 E=46 d=e4 p=AvgI, L=(X-p)od S&L E&L I&L [5,12] 34 [-6,-3) 22, 16 same range E=22 I=16 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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p=AvgS p=AvgE p=AvgI d=e1 d=e2 d=e3 d=e4 FAUST Oblique, LSR Lp,d
Linear, Spherical, Radial classifier Form Class Hulls using linear d boundaries thru min and max of Lk.d,p=(Ck&(X-p))od On every Ik,p,d{[epi,epi+1) | epj=minLk,p,d or maxLk,p,d for some k,p,d} interval add spherical and barrel boundaries with Sk,p and Rk,p,d similarly (use enough (p,d) pairs so that no 2 class hulls overlap) Points outside all hulls are declared as "other". all p,ddis(y,Ik,p,d) = unfitness of y being classed in k. Fitness of y in k is f(y,k) = 1/(1-uf(y,k)) 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 3672 4423 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 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 7366 7908 8178 6691 5250 5166 5070 5758 7186 6066 7037 7884 6603 5886 5419 5781 6933 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 9658 9452 8622 7455 8229 8445 7306 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 148 149 150 Ld 51 49 47 46 50 54 44 48 43 58 57 52 55 45 53 70 64 69 65 63 66 59 60 61 56 67 62 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 68 71 72 73 74 75 d=1000 76 77 79 78 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 148 149 150 On IRIS150 d, precompute! XoX, Ld=Xod nk,L,d Lmin(Ck&Ld) xk,L,d max(Ck&Ld) p, (pre-ccompute?) Ld,p(X-p)od=Ld-pod nk,L,d,pmin(Ck&Ld,p)=nk,L,d-pod xk,L,d.pmax(Ck&Ld,p)=xk,L,d-pod Lp,d =Ld-pod d=e1 d=e2 d=e3 d=e4 p=AvgS p=AvgE p=AvgI d=1000 p=0000 nk,L,d xk,L,d S E I d=1000 p=AS=( ) -7 8 -1 20 d=1000 p=AE=( ) d=1000 p=AI=( ) S E I d=0100 p=0000 nk,L,d xk,L,d d=0100 p=AS=( ) d=0100 p=AE=( ) d=0100 p=AI=( ) S 10 19 E I d=0010 p=0000 nk,L,d xk,L,d d=0010 p=AS=( ) -5 4 15 36 d=0010 p=AE=( ) -13 8 d=0010 p=AI=( ) -25 -4 S 1 6 E I d=0001 p=0000 nk,L,d xk,L,d d=0001 p=AS=( ) -1 4 8 16 12 23 d=0001 p=AE=( ) -12 -7 -3 5 1 12 d=0001 p=AI=( ) We have introduce 36 linear bookends to the class hulls, 1 pair for each of 4 ds, 3 ps , 3 class. For fixed d, Ck, the pTree mask is the same over the 3 p's. However we need to differentiate anyway to calculate R correctly. 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 have them all is to accurately compute R on each min-max interval. In fact, we computer R on all intervals (even those where a single class has been isolated) to eliminate False Positives (if FPs are possible - sometimes they are not, e.g., if we are to classify IRIS samples known to be Setosa, vErsicolor or vIriginica, then there is no "other"). Assuming Ld, nk,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 without further pTree processing, by the scalar computations: nk,L,d,p = nk,L,d-pod xk,L,d.p = xk,L,d-pod.
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LSR IRIS150 e1 only Analyze R:RnR1 (and S:RnR1?) projections on each interval formed by consecutive L:RnR1 cut-pts. Sp (X-p)o(X-p) = XoX + L-2p + pop nk,S,p = min(Ck&Sp) xk,S,p max(Ck&Sp) Rp,d Sp-L2p,d = L-2p-(2pod)d + pop + pod2 + XoX - L2d nk,R,p,d = min(Ck&Rp,d) xk,R,p,d max(Ck&Rp,d) Ld d=1000 p=origin Setosa vErsicolor vIrginica d=1000 p=AS=( ) d=1000 p=AE=( ) d=1000 p=AI=( ) 270 1558 2568 16 128 24 6 34 99 393 1096 1217 1826 12 2081 3445 26 32 1 517,4 79 633 16 723 1258 279 5 171 186 748 998 12 249 794 16 1641 2391 24 132 730 1622 2281 12 17 220 26 32 388 1369 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". with AI 17 220 with AE 1 517,4 78 633 eliminates FPs better? What is the cost for these additional cuts (at new p-values in an L-interval)? It looks like: make the one additional calculation: L-2p-(2pod)d then AND the interval masks, then AND the class masks? (Or if we already have all interval-class mask, only one mask AND step.) If we have computed, S:RnR1, how can we utilize it?. We can, of course simply put spherical hulls boundaries by centering on the class Avgs, e.g., Sp p=AvgS Setosa E=50 I=11 vErsicolor vIrginica If on the L 1000,avgE interval, [-1, 11) we recurse using SavgI we get 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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