.9 1 0 3 .8 6 1 4 8 1 .9 8 8 5 4 0 .2 0 1 .7 6 7 4 1 0 6 9 0 0 .4 4 8 2 9 .6 5 1 1 5 .2 4 3 0 1 .9 5 1 2 .9 9 E st o n ia 1 .3 1 1 .6 1 .1 1 0 .1 0 1 2 9 .4 1 1 .2 1 7 3 .6 5 1 5 .0 4 1 .4 3 8 9 1 6 8 [...] 0 1 .4 2 1 1 .1 1 P o la n d 3 8 .1 8 .2 0 .5 7 0 .2 0 4 6 .3 1 6 .4 1 7 6 3 .6 2 6 2 4 .9 2 0 .7 5 1 3 3 1 4 3 4 9 .4 5 1 3 2 .4 6 4 6 .9 4 3 .6 5 3 .0 4 P o rt u g a l 1 0 .6 1 5 .4 1 .2 1 0 .1 1 1 8 [...] a n 1 2 7 .4 2 5 .0 3 .4 4 0 .2 7 8 6 1 .9 6 7 .0 1 1 0 1 1 5 .9 7 8 5 5 4 .2 3 1 .8 8 7 2 4 1 1 5 6 8 .3 8 1 5 2 0 .7 1 1 1 8 .1 3 1 6 6 .5 0 1 7 .9 6 K o re a ( R e p u b li c o f) (S o u th ) 4 9 .0 …

: Class PlannerTypeC

PNode  p0, PNode  p1)           Markiert zwei Propositionen p0, p1 als "wechselseitig exklusiv", falls in A_{i-1} alle p0-erzeugenden Aktionen "wechselseitig exklusiv" mit allen p1-erzeugenden Aktionen [...] ty ( PNode  p0, PNode  p1) Markiert zwei Propositionen p0, p1 als "wechselseitig exklusiv", falls in A_{i-1} alle p0-erzeugenden Aktionen "wechselseitig exklusiv" mit allen p1-erzeugenden Aktionen sind [...] ANode [] Since: version 1.0 interferenceActions private boolean interferenceActions ( ANode  a0, ANode  a1, boolean oneDirectionChecked) "Interference"-Test fuer zwei Aktionen a0 und a1. Returns: boolean true …

Training Support Vector Machines – Lecture Notes

f(x_1)-y_1)-(f(x_2)-y_2)}{\left\lVert \smash {x_1-x_2}\mathstrut \right\rVert ^2} = \lambda _2 + y_2\frac{(f(x_1)-y_1)-(f(x_2)-y_2)}{\left\langle x_1,x_1\mathstrut \right\rangle -2\left\langle x_1,x_2 [...] by \eqref{eqn:svm-lambda0}}} b & +\sum \nolimits _{i=1}^{N} \lambda _i \\ & = \tfrac {1}{2} & \hfill \sum \nolimits _{i=1}^{N} \sum \nolimits _{j=1}^{N} \lambda _i\lambda _j y_i y_j \left\langle \vec{x_i} [...] Equation K3 becomes (for any two \(i\) , we simply use 1, 2 here): \[\begin{align*} \sum \nolimits _{i=1}^{N} \lambda _i y_i & = 0 \\ \lambda _1 y_1 + \lambda _2 y_2 & = - \sum \nolimits _{i=3}^{N} \lambda …

Training Support Vector Machines – Lecture Notes

f(x_1)-y_1)-(f(x_2)-y_2)}{\left\lVert \smash {x_1-x_2}\mathstrut \right\rVert ^2} = \lambda _2 + y_2\frac{(f(x_1)-y_1)-(f(x_2)-y_2)}{\left\langle x_1,x_1\mathstrut \right\rangle -2\left\langle x_1,x_2 [...] by \eqref{eqn:svm-lambda0}}} b & +\sum \nolimits _{i=1}^{N} \lambda _i \\ & = \tfrac {1}{2} & \hfill \sum \nolimits _{i=1}^{N} \sum \nolimits _{j=1}^{N} \lambda _i\lambda _j y_i y_j \left\langle \vec{x_i} [...] Equation K3 becomes (for any two \(i\) , we simply use 1, 2 here): \[\begin{align*} \sum \nolimits _{i=1}^{N} \lambda _i y_i & = 0 \\ \lambda _1 y_1 + \lambda _2 y_2 & = - \sum \nolimits _{i=3}^{N} \lambda …

LNAI 5212 - Hierarchical, Parameter-Free Community Discovery

intuition. Let A1 := {0, 1, 0, 1, · · · , 0, 1}, with alternating values. We have C(A1|M (1) 1 ) = 16H(1/2) = 16 and C(M (1) 1 ) = log!16 + #log(16 + 1)$ = 10 + 5 = 15. However, for M (2) 1 the best choice [...] Figures 2a and 3 follows such a structure, consisting of G1 = (I1, J1, A1,1), G2 = (I1, J2, A1,2), G3 = (I2, J1, A2,1), and G4 = (I2, J2, A2,2), where I1 is the set of “computer science researchers” and I2 the [...] choice is n1 = 15, with C(A1|M (2) 1 ) ' 15 and C(M (2) 1 ) ' 19. The total code lengths are C(A1, M (1) 1 ) ' 16 + 15 = 31 and C(A1, M (2) 1 ) ' 15 + 19 = 34. Thus, based on total code length, the simpler …

Microsoft PowerPoint - 2020-01-23 Jugendarbeit AKJStat_Dokumentation

Angaben in 1.000 EUR sowie pro 6- bis unter 27-Jährigen) 1. 41 1. 45 9 1. 43 2. 06 0 1. 45 9. 09 9 1. 38 7. 14 2 1. 34 9. 77 6 1. 37 7. 59 1 1. 40 0. 84 6 1. 45 1. 26 0 1. 54 4. 48 8 1. 55 9. 52 5 1. 56 5. [...] 07 8 1. 62 7. 10 5 1. 62 5. 80 3 1. 69 3. 55 1 1. 70 9. 14 2 1. 76 8. 27 1 1. 83 9. 86 6 1. 90 4. 17 3 1. 97 2. 59 9 74 75 76 72 71 73 74 78 84 86 87 92 92 97 99 101 106 110 115 0,00 20,00 40,00 60,00 [...] tätigen Personen in Arbeitsbereichen der KJA1 1998 44.560 33.292 23,1 17,3 2002 39.137 27.541 20,4 14,4 2006 33.631 19.814 17,9 10,5 2010 35.959 20.734 20,1 11,6 2014 29.126 17.565 16,9 10,2 2016 30.302 …

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