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FAUST T op KO utliers : Use D 2 NN=SqDist(x, X')=rank 2 S x for TopKOutlier -slider .

FAUST Oblique Analytics : X(X 1 ..X n )R n |X|=N, Classes={C 1 ..C K }, d=(d 1 ..d n ) |d|=1, p=(p 1 ..p n )R n , L, R:. Then T op KO utliers uses SPTS, R p,d , which measures Square Radial Reach of each xX from the d-line thru p. x. (X-p) o (X-p) -. [(X-p) o d] 2 =.

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FAUST T op KO utliers : Use D 2 NN=SqDist(x, X')=rank 2 S x for TopKOutlier -slider .

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  1. FAUST Oblique Analytics:X(X1..Xn)Rn |X|=N, Classes={C1..CK}, d=(d1..dn) |d|=1, p=(p1..pn)Rn, L, R: Then TopKOutliers uses SPTS, Rp,d, which measures Square Radial Reach of each xX from the d-line thru p. x (X-p)o(X-p) - [(X-p)od]2 = XoX+pop-2Xop - [Xod-pod]2  (x-p)o(x-p) - (x-p)od2 (x-p)o(x-p) d (x-p)od = |x-p| cos p XoX, -2Xop, Xod pre-computed, then 2 scalar adds, 1 mult., 2 adds Lp,d  (X-p)od, llp,d,k=minLp,d&Ck, hlp,d,k=maxLp,d&Ck, Sp  (X-p)o(X-p) lsp,d,k=minSp&Ck, hsp,d,k=maxSp&Ck. Rp,d  Sp, - Lp,d2 lrp,d,k=minRp,d&Ck, hrp,d,k=maxRp,d&Ck. FAUST CountChangeClusterer: If DensThres not reached, cut C at PCCsLp,d&C w next (p,d)pdSet FAUST TopKOutliers: Use D2NN=SqDist(x, X')=rank2Sx for TopKOutlier-slider. FAUST PiecewiseLinearClassifier: y is Ck iff yLHk  { z | llp.d,k  (z-p)od  hlp,d,k}  (p,d)pdSet LHk is Linear Hull of Class=k, pdSet is chosen set of (p,d) pairs, e.g., (DiagonalStartPt, Diagonal). XoD is a central computation for FAUST. e.g., Xod is the only SPTS needed in FAUST CCClusterer and PLClassifier. xX, D2(X,x)=(X-x)o(X-x)=XoX+xox-2Xox. XoX is pre-computed 1 time, then xox is read from XoX, leaving Xox. Then the RankiPTR(x,ptr-to-RankiD2(X,x)) SPTS and the RankiSD(x,RankiD2(X,x))) valueTree (ordered descending on RankiD2(X,x), i=2..q) are constructed. If X is a high-value classification training set (eg, Enron emails), pre-compute what? 1. column statistics(min, avg, max, std,...) ; 2. XoX; Xop, p=class_Avg/Median); 3. Xod, d=interclass_Avg/Median_UnitVector; 4. Xox, d2(X,x), Rankid2(X,x), xX, i=2,3...; 5. Lp,d and Rp,d for all p's and d's above FAUST LinearAndRadialClassifiery is Ck iff yLRHk {z | llp.d,k  (z-p)od  hlp,d,k AND lrp.d,k  (z-p)o(z-p) - (z-p)od2 hrp,d,k(p,d)pdSet }

  2. 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] 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] LARC on IRIS150 y isa O if yoD (-,-184)(382,590)(2725,) y isa O or S(50) if yoD  C1,1 [-184 , 123] Dse 9 -6 27 10; xoDes: -184 123 S 590 1331 E 381 2046 I y isa O or I(1) if yoD  C1,2  [ 381 , 590] 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] 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 C1,1 0 154 S SRR(AVGs,dse) on C1,2 only one such I y isa O if y isa C1,3 AND SRR(AVGs,Dse)(-,2)U(392,) 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. SRR(AVGs,dse) onC1,3 2 137 E 7 143 I Dei 1 .7 -7 -4; xoDei on C2,1: 1.4 19 E -2 3 I SRR(AVGe,dei) onC3,1 2 370 E 8 106 I 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.

  3. LARC on IRIS150-2 We use the diagonals. Also we set a MinGapThreshold=2 which will mean we stay 2 units away from any cut y isa O if yoD(-,43)(79,) y isa O or S( 9) if yoD[43,47] y isa O if yoD[43,47]&SRR(-,52)(60,) y isa O or S(41) or E(26) or I( 7) if yoD(47,60) (yC1,2) d=e1=1000; The xod limits: 43 58 S 49 70 E 49 79 I y isa O or E(24) or I(32) if yoD[60,72] (yC1,3) y isa O or I(11) if yoD(72,79] y isa O if yoD[72,79]&SRR(-,49)(78,) 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) d=e2=0100 on C1,2 xod lims: 30 44 S 20 32 E 25 30 I 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,) y isa O or E(17) if yoD[60,72]&SRR[1.2,20] d=e2=0100 on C1,3 xod lims: 22 34 E 22 34 I zero differentiation! 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,) d=e3=0010 on C2,2 xod lims: 30 33 S 28 32 E 28 30 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 if yoD(-,1)(5,12)(24,) y isa O or S(13) if yoD[1,5] d=e3=0001 xod lims: 12 18 E 18 24 I y isa O or E( 9) if yoD[12,16) y isa O if yoD[12,16)&SRR[0,208)(558,) 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]

  4. LARC IRIS150. d=AvgEAvgI p=AvgE, L=(X-p)od -36 -25 S -14 11 E -17 33I d=AvgSAvgE p=AvgS, L=(X-p)od -6 4 S 18 42 E 11 64I d=AvgSAvgI p=AvgS, L=(X-p)od -6 5 S 17.5 42 E 12 65I [-14,11) (50, 13) 0 2.8 76 134 [11,33] I(36) [-17,-14)] I(1) [17.5,42) (50,12) 4.7 6 192 205 [18,42) (50,11) 2 6.92 133 137 [11,33] I(37) [42,64] 38 [12,17.5)] I(1) [11,18)] I(1) R(p,d,X) SEI .3 .9 4.7 150 204 213 R(p,d,X) SEI 0 2 6 137 154 393 R(p,d,X) SEI 0 2 32 76 357 514 30ambigs, 5 errs 38ambigs 16errs d=e3 p=AvgS, L=(X-p)od -5 5 S&L 15 37 E&L 4 55I&L d=e2 p=AvgS, L=(X-p)od -11 10 S&L -14 0 E&L -13 4I&L d=e4 p=AvgE, L=(X-p)od -13 -7 S&L -3 5 E&L 1 12I&L d=e4 p=AvgS, L=(X-p)od -2 4 S&L 7 16 E&L 11 23I&L d=e1 p=AvgS, L=(X-p)od -8 8 S&L -2 20 E&L -2 29I&L -5,4) 47 [4,15) 3 1 [15,37) 50, 15 157 297 536 792 [37,55] I=34 ,-13) 1 -13,-11 0, 2, 1 all=-11 -11,0 29,47,46 0 66 310 352 1749 4104 [0,4) [4, 15 3 6 -2,4) 50 -7] 50 [-3,1) 21 [7,11) 28 [1,5) 22, 16 .7 .7 4.8 4.8 [11,16) 22, 16 11 16 11 16 [16,23] I=34 [5,12] 34 -8,-2 16 [-2,8) 34, 24, 6 0 99 393 1096 1217 1825 [8,20) 26, 32 270 792 1558 2567 [20,29] 12 3, 1 E=32 I=14 E=18 I=12 E=22 I=16 E=22 I=16 E=32 I=14 E=26 I=5 1, 1 46,11 9, 3 2, 1 d=e3 p=AvgE, L=(X-p)od -32 -24 S&L -12 9 E&L -25 27I&L d=e1 p=AvgE, L=(X-p)od -17 -1 S&L -11 11 E&L -11 20I&L d=e2 p=AvgE, L=(X-p)od -5 `17 S&L -8 7 E&L -6 11I&L ,-25) 48 -25,-12 2 11 -17-11 16 [-11,-1) 33, 21, 3 0 27 107 172 748 1150 [-12,9) 49, 15 2(17) 16 158 199 [9,27] I=34 [-1,11) 26, 32 1 51 79 633 [11,20] I12 ,-6) 1 [-6, -5) 0, 2, 1 15 18 58 59 [-5,7) 29,47, 46 3 58 234 793 1103 1417 [7,11) [11, 15 3 6 1 err E=5 I=3 E=26 I=11 E=45 I=12 E=22 I=16 E=47 I=12 E=47 I=22 E=39 I=11 E=7 I=4 13, 21 21, 3 E=46 I=14 d=e4 p=AvgI, L=(X-p)od -19 -14 S&L -10 -3 E&L -6 5I&L d=e3 p=AvgI, L=(X-p)od -44 -36 S&L -25 -4 E&L -37 14I&L d=e1 p=AvgI, L=(X-p)od -22 -8 S&L -17 4 E&L -17 14I&L d=e2 p=AvgI, L=(X-p)od -7 `15 S&L -10 4 E&L -8 9I&L -7] 50 [-3,1) 21 ,-25) 48 -25,-12 2 1 1 [-17,-8) 33, 21, 3 38 126 132 730 1622 2181 [-6,-3) 22, 16 same range [5,12] 34 [-25,-4) 50, 15 5 11 318 453 [9,27] I=34 [-8,4) 26, 32 0 34 1368 730 [-8, -7) 2, 1 allsame [5, 9] 9, 2, 1 allsame ,-6) 1 [-7, 4) 29,46,46 5 36 929 1403 1893 2823 [6,11) [11, 15 3 6 S=9 E=2 I=1 E=2 I=1 E=2 I=1

  5. C13 APPENDIX C8,1: D=0110 Ch,1: D=10-10 Cg,1: D=1-100 Cf,1: D=1111 Ce,1: D=0111 C1,1: D=1000 Cc,1: D=1101 C2,3: D=0100 Cd,1: D=1011 C3,3: D=0010 C4,1: D=0001 Ca,1: D=0011 C5,1: D=1100 C6,1: D=1010 C9,1: D=0101 C7,1: D=1001 Cb,1: D=1110 54 146 y isa O if yoD(-,54)(146,) L H y isa O|S if yoD C7,1  [54,146] 44 100 y isa O if yoD(-,44)(100,) L H y isa O|S if yoD C8,1  [44,100] 43 58 y isa O if yoD(-,43)(58,) L H y isa O|S if yoD C2,3  [43,58] 23 44 y isa O if yoD(-,23)(44,) L H y isa O|S if yoD C3,3  [23,44] 10 19 y isa O if yoD(-,10)(19,) L H y isa O|S if yoD C4,1  [10,19] 1 6 y isa O if yoD(-,1)(6,) L H y isa O|S if yoD C5,1  [1,6] 68 117 y isa O if yoD(-,68)(117,) L H y isa O|S if yoD C6,1  [68,117] 71 137 y isa O if yoD(-,71)(137,) L H y isa O|S if yoD Cd,1  [71,137] 26 61 y isa O if yoD(-,26)(61,) L H y isa O|S if yoD Ca,1  [26,61] 10 22 y isa O if yoD(-,10)(22,) L H y isa O|S if yoD Ch,1  [10,22] 3 46 y isa O if yoD(-,3)(46,) L H y isa O|S if yoD Ci,1  [3,46] 84 204 y isa O if yoD(-,84)(204,) L H y isa O|S if yoD Cg,1  [84,204] 12 91 y isa O if yoD(-,12)(91,) L H y isa O|S if yoD Cb,1  [12,91] 39 127 y isa O if yoD(-,39)(127,) L H y isa O|S if yoD Cf,1  [39,127] 55 169 y isa O if yoD(-,55)(169,) L H y isa O|S if yoD Ce,1  [55,169] 81 182 y isa O if yoD(-,81)(182,) L H y isa O|S if yoD Cc,1  [81,182] 36 105 y isa O if yoD(-,36)(105,) L H y isa O|S if yoD C9,1  [36,105] 400 1000 1500 2000 2500 3000 LARC on IRIS150 y isa OTHER if yoDse (-,495)(802,1061)(2725,) Dse 9 -6 27 10 495 802 S 1270 2010 E 1061 2725 I L H 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 -3 -2 3 3 -117 -44 E y isa O if yoDei (-,-117)(-3,) -62 -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 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] 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? 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 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??

  6. Dot Product SPTS computation:XoD = k=1..nXkDk D2,0 D2,1 D1,0 D1,1 D 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 1 1 3 3 1 1 pXoD,1 pXoD,0 pXoD,3 pXoD,2 X X1 X2 p11 p10 p21 p20 XoD 0 1 1 0 1 0 0 1 1 1 1 1 1 1 0 1 1 1 0 1 0 1 1 0 1 0 0 1 0 0 6 9 9 0 1 1 0 1 1 1 3 2 1 0 1 0 1 1 1 1 0 0 0 0 1 0 1 & & & & 0 1 1 0 1 1 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 0 1 0 pX1*X2,0 0 1 0 pX1*X2,1 0 1 0 0 1 0 X X1 X2 pX1*X2,2 pX1*X2,3 p11 p10 p21 p20 X1*X2 D2,0 D2,1 D1,0 D1,1 D ( ( = 22 = 22 1 p1,1 1 p1,1 + 1 p2,1 ) + 1 p2,1 )   + 1 p2,0 ) + 1 p2,0 ) + 21 (1 p1,0 + 21 (1 p1,0 + 1 p11 + 1 p11 + 20 (1 p1,0 + 20 (1 p1,0 + 1 p2,0 + 1 p2,0 + 1 p2,1 ) + 1 p2,1 ) 1 3 2 1 3 1 0 1 1 1 1 0 0 1 0 1 1 1 1 9 2 1 1 0 0 0 0 0 1 0 1 0 1 1 0 3 3 1 1 & & 0 1 0 0 0 0 0 0 1 CAR12,3 1 1 0 0 1 0 0 0 1 0 1 0 1 0 1 CAR11,2 0 0 0 1 0 0 CAR10,1  CAR22,3    & pX1*X2,1 pX1*X2,2 pX1*X2,3 pX1*X2,0 & & & CAR21,2 0 1 1 0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 0 0 CAR13,4 PXoD,0 PXoD,3 PXoD,2 PXoD,1 0 1 0 0 0 0 1 0 1 0 1 0 1 1 0 PXoD,4 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 Different data. CAR10,1 pTrees XoD 0 0 1 X 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 0 1 1 1 0 0 1 1 1 1 0 1 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 1 1 3 2 1 3 1 0 1 1 1 1 0 0 1 0 1 1 1 6 18 9 PXoD,0 PXoD,2 PXoD,1        PXoD,3        1 1 0 1 1 0 1 1 1 1 0 1 & & & & & & & & & & & & & & /*Calc PXoD,i after PXoD,i-1 CarrySet=CARi-1,i RawSet=RSi */ INPUT: CARi-1,i, RSi ROUTINE: PXoD,i=RSiCARi-1,i CARi,i+1=RSi&CARi-1,i OUTPUT: PXoD,i, CARi,i+1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 1 1 0 1 1 0 1 0 1 1 1 0 1 0 We have extended the Galois field, GF(2)={0,1}, XOR=add, AND=mult to pTrees. SPTS multiplication: (Note, pTree multiplication = &)

  7. Example: FAUST Oblique: XoD used in CCC, TKO, PLC and LARC) and (x-X)o(x-X) p1 p1 p1 p,0 p,0 p,0 p3 p3 p3 p2 p2 p2 X X1 X2 p11 p10 p21 p20 XoD XoD XoD = -2Xox+xox+XoX is used in TKO. 0 0 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 3 9 2 2 3 3 3 6 5 0 1 0 0 0 0 0 0 0 1 0 1 1 1 0 0 1 1 1 3 2 1 0 1 0 1 1 1 1 0 0 0 0 1 0 1 n=1 p=2 n=0 p=2 P &p0 P=p0&P P p1 P=P&p1 D2,0 D2,0 D2,0 D2,1 D2,1 D2,1 D1,0 D1,0 D1,0 D1,1 D1,1 D1,1 D=x2 D=x1 D=x3 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 32 1*21+ 22 1*21+1*20=3 so -2x1oX = -6 0 1 0 1 0 0 2 1 1 1 3 0 0 1 1 1 1 0 n=3 p=2 n=2 p=1 n=1 p=1 n=0 p=1 P &p2 P=p'2&P P p3 P &p1 P &p0 P=P&p'3 P=p1&P P=p0&P RankN-1(XoD)=Rank2(XoD) 0 0 0 1 0 1 1 0 1 1 0 1 1 0 1 1 0 0 0 1 0 1 0 1 1 1 0 1 1 1 1 0 1 1 0 1 1<2 2-1=1 0*23+ 0<1 1-0=1 0*23+0*22 21 0*23+0*22+1*21+ 11 0*23+0*22+1*21+1*20=3 so -2x2oX= -6 RankN-1(XoD)=Rank2(XoD) n=2 p=2 n=1 p=2 n=0 p=1 P=p'1&P P &p1 P=P&p2 P=p0&P P p2 P &p0 0 0 1 1 1 0 0 1 1 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 22 1*22+ 1<2 2-1=1 1*22+0*21 11 1*22+0*21+1*20=5 so -2x3oX= -10 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 he__ 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)

  8. pTree Rank(K) computation: (Rank(N-1) gives 2nd smallest which is very useful in outlier analysis?) p1 p,0 p3 p2 X X1 X2 p11 p10 p21 p20 XoD 1 0 0 1 1 1 2 3 3 0 0 0 0 1 1 1 3 2 1 0 1 0 1 1 1 1 0 0 0 0 1 0 1 D2,0 D2,1 D1,0 D1,1 D 0 1 3 3 0 1 RankKval= 0 1 0 0 0 0 0 23 * + 22 * + 21 * + 20 * = 5P=MapRankKPts= ListRankKPts={2} RankN-1(XoD)=Rank2(XoD) n=3 p=2 n=2 p=2 n=1 p=2 n=0 p=2 P &p2 P p3 P &p1 P &p0 1 0 0 0 1 1 0 1 1 1 0 0 0 1 1 1 1 1 0 1 1 0 1 1 22 1*23+ 0<2 2-0=2 1*23+0*22+ 0<2 2-0=2 1*23+0*22+0*21+ 22 1*23+0*22+0*21+1*20=9 0 1 1 0 1 1 0 1 1 0 1 1 P=P&p3 P=p'2&P P=p'1&P P=p0&P Cross out the 0-positions of P each step. (n=3) c=Count(P&P4,3)= 3 < 6 p=6–3=3; P=P&P’4,3 masks off highest 3 (val 8) {0} X P4,3P4,2P4,1 P4,0 0 1 1 1 0 0 0 0 1 0 1 1 1 1 10 5 6 7 11 9 3 1 0 0 0 1 1 0 1 0 1 1 1 0 1 (n=2) c=Count(P&P4,2)= 3 >= 3 P=P&P4,2 masks off lowest 1 (val 4) {1} (n=1) c=Count(P&P4,1)=2 < 3 p=3-2=1; P=P&P'4,1 masks off highest 2 (val8-2=6 ) {0} {1} (n=0) c=Count(P&P4,0 )=1 >= 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) */ {0} {1} {0} {1}

  9. applied to S, a column of numbers in bistlice format (an SpTS), will produce the DistributionTree of S DT(S) depth=h=0 15 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 5/64 [0,64) p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 2/32[64,96) p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1[32,48) p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 3/32[0,32) p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 2[96,112) p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0[64,80) p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 1/16[0,16) p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 6[112,128) p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 1[48,64) p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 1[16,24) p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 2/16[16,32) p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 2[80,96) p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 2/32[32,64) p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 ¼[96,128) p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 1[48,56) p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 1[24,32) p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 0[56,64) p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 0[0,8) p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 1[32,40) p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 1[8,16) p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 0[40,48) p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 10/64 [64,128) p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 2[80,88) p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 3[112,120) p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 0[88,96) p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 3[120,128) p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 0[96,104) p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 2[194,112) UDR Univariate Distribution Revealer (on Spaeth:) 5 10 depth=h=1 node2,3 [96.128) yofM 11 27 23 34 53 80 118 114 125 114 110 121 109 125 83 p6 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 p5 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 p4 0 1 1 0 1 1 1 1 1 1 0 1 0 1 1 p3 1 1 0 0 0 0 0 0 1 0 1 1 1 1 0 p2 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 p1 1 1 1 1 0 0 1 1 0 1 1 0 0 0 1 p0 1 1 1 0 1 0 0 0 1 0 0 1 1 1 1 p6' 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 p5' 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 p4' 1 0 0 1 0 0 0 0 0 0 1 0 1 0 0 p3' 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 p2' 1 1 0 1 0 1 0 1 0 1 0 1 0 0 1 p1' 0 0 0 0 1 1 0 0 1 0 0 1 1 1 0 p0' 0 0 0 1 0 1 1 1 0 1 1 0 0 0 0 Y y1 y2 y1 1 1 y2 3 1 y3 2 2 y4 3 3 y5 6 2 y6 9 3 y7 15 1 y8 14 2 y9 15 3 ya 13 4 pb 10 9 yc 11 10 yd 9 11 ye 11 11 yf 7 8 3 2 2 8 f= 1 2 1 1 0 2 2 6 0 1 1 1 1 0 1 000 2 0 0 2 3 3 depthDT(S)b≡BitWidth(S) h=depth of a node k=node offset Nodeh,k has a ptr to pTree{xS | F(x)[k2b-h+1, (k+1)2b-h+1)} and its 1count 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.).

  10. D2,0 D2,1 D1,0 D1,1 D 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 1 1 3 3 1 1 ( ( = 22 = 22 1 p1,1 1 p1,1 + 1 p2,1 ) + 1 p2,1 ) ( ( ( ( + 1 p2,0 ) + 1 p2,0 ) 1 p1,0 1 p1,0 + 1 p11 + 1 p11 1 p1,0 1 p1,0 + 21 + 21 + 1 p2,0 + 1 p2,0 + 1 p2,1 ) + 1 p2,1 ) + 20 + 20 pTrees XoD X 1 0 0 1 0 0 0 1 1 0 1 1 1 3 2 1 0 1 0 1 1 1 1 0 0 0 0 1 0 1 6 9 9 0 1 1 0 1 1 1 1 0 1 1 0 1 0 1 1 0 1 0 1 1 1 1 1 1 0 0 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=RSiCSPi-1,iCSNi-1,i CSNi,i+1=CSNi-1,iPXoD,i; CSPi,i+1=CSPi-1,iRSi-1; OUTPUT: PXoD,i, CSNi,i+1 CSPi,i+1 0 1 1 0 1 1 1 1 0 0 0 0 1 0 1 1 1 0 1 0 1 CSN-1.0PXoD,0 CSP-1,0RS0 RS1 CSN0,1= CSP0,1= CSP-1,0=CSN-1,0= RS0 PXoD,0 PXoD,1 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 0 1  =      = 1 0 1 0 0 0 0 1 1 1 0 1 0 0 0 

  11. D2,0 D2,0 D2,1 D2,1 D1,0 D1,0 D1,1 D1,1 D D XoD = k=1..nXk*Dk 1 1 1 0 3 3 1 2 1 1 0 1 k=1..n ( = 22B Dk,B pk,B k=1..n ( Dk,B pk,B-1 + Dk,B-1 pk,B + 22B-1 k=1..n ( Dk,B pk,B-2 + Dk,B-1 pk,B-1 + Dk,B-2 pk,B + 22B-2 Xk*Dk = Dkb2bpk,b XoD=k=1,2Xk*Dk with pTrees: qN..q0, N=22B+roof(log2n)+2B+1 k=1..n ( +Dk,B-3 pk,B Dk,B pk,B-3 + Dk,B-1 pk,B-2 + Dk,B-2 pk,B-1 + 22B-3 = Dk(2Bpk,B +..+20pk,0) = (2BDk,B+..+20Dk,0) (2Bpk,B +..+20pk,0) . . . k=1..2 ( = 2BDkpk,B +..+ 20Dkpk,0 = 22 Dk,1 pk,1 k=1..n ( Dk,Bpk,B) = 22B( +Dk,Bpk,B-1) + 22B-1(Dk,B-1pk,B Dk,B pk,0 + Dk,2 pk,1 + Dk,1 pk,2 +Dk,0 pk,3 + 23 +..+20Dk,0pk,0 k=1..2 ( Dk,1 pk,0 + Dk,0 pk,1 + 21 pTrees k=1..n ( X Dk,2 pk,0 + Dk,1 pk,1 + Dk,0 pk,2 + 22 B=1 1 3 2 1 0 1 0 1 1 1 1 0 0 0 0 1 0 1 k=1..2 ( k=1..n ( Dk,0 pk,0 Dk,1 pk,0 + Dk,0 pk,1 + 20 + 21 q0 = p1,0 = no carry 1 1 0 k=1..n ( Dk,0 pk,0 + 20 ( ( = 22 = 22 1 p1,1 D1,1p1,1 + 1 p2,1 ) + D2,1p2,1 ) ( ( ( ( + 1 p2,0 ) + D2,0p2,0) D1,1p1,0 1 p1,0 + 1 p11 + D1,0p11 1 p1,0 D1,0p1,0 + 21 + 21 + 1 p2,0 + D2,1p2,0 + 1 p2,1 ) + D2,0p2,1) + 20 + 20 q1= carry1= 1 1 0 0 0 1 ( = 22 D1,1 p1,1 + D2,1 p2,1 ) ( ( + D2,0 p2,0) D1,1 p1,0 +D1,0 p11 D1,0 p1,0 + 21 + D2,1 p2,0 +D2,0 p2,1) + 20 0 0 0 q2=carry1= no carry 0 1 1 1 0 1 1 1 0 0 0 1 q0 = carry0= 0 1 1 1 0 0 0 1 1 0 1 1 1 1 0 0 0 0 1 1 0 1 0 1 1 0 1 1 1 1 2 1 1 q1=carry0+raw1= carry1= 1 1 1 1 1 1 q2=carry1+raw2= carry2= 1 1 1 q3=carry2 = carry3= 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 ? CCC ClustererIf 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 So, DotProduct involves just multi-operand pTree addition. (no SPTSs and no multiplications) Engineering shortcut tricka would be huge!!!

  12. 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) ----ak,b--- b=B..02b(xk,b-pk,b) ) ( 22Bak,Bak,B + =k=1..n( b=B..02b(xk,b-pk,b) )( 22B-1( ak,Bak,B-1 + ak,B-1ak,B ) + { 22Bak,Bak,B-1 } =k (2Bak,B+ 2B-1ak,B-1+..+ 21ak, 1+ 20ak, 0) (2Bak,B+ 2B-1ak,B-1+..+ 21ak, 1+ 20ak, 0) 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 ... X(X1...Xn) RKN (Rank K Nbr), K=|X|-1, yields1.a_outlier_detector (top y dissimilarity from X-{x}). ANOTHER TRY! 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) Should we pre-compute all pk,i*pk,j p'k,i*p'k,j pk,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. = -2 kxkXk + kxk2 + kXk2 3. Pick this from XoX for each x and add to 2. = -2xoX + 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 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.

  13. LARC on IRIS150-3 Here we use the diagonals. d=e1 p=AVGs, L=(X-p)od 43 58 S 49 70 E 49 79I R(p,d,X) SEI 0 128 270 393 1558 3444 [43,49) S(16) 0 128 [49,58) E(24)I(6) 0 S(34) 99 393 1096 1217 1825 [70,79] I(12) 2081 3444 [58,70) E(26) I(32) 270 792 1558 2567 E(50) I(7) 49 49 (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} R & L I(1) I(42)

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