to crisp version (and more) for a,b ∈ { 0, 1 } a b a ∧ b t(a,b) 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 a b a ∨ b s(a,b) 0 0 0 0 0 1 1 1 1 0 1 1 1 1 1 1 a a c(a) 0 1 1 1 0 0 Fuzzy Logic Lecture 03 G. Rudolph: C [...] X is A THEN Y is B x1 x2 x3 0.5 1.0 0.6 y1 y2 1.0 0.4 A: B: R‘ x1 x2 x3 y1 1.0 1.0 1.0 y2 0.9 0.4 0.8 ⇒ given fact y1 y2 0.9 0.7 B‘: using Imp(a,b) = min{1, 1-a+b } thus: B‘ ◦ R-1 = A‘ with max-min-composition [...] rule: IF X is A THEN Y is B using x1 x2 x3 0.5 1.0 0.6 y1 y2 1.0 0.4 A: B: R y1 y2 x1 1.0 0.9 x2 1.0 0.4 x3 1.0 0.8 ⇒ given fact x1 x2 x3 0.6 0.9 0.7 A‘: Imp(a,b) = min{1, 1-a+b } thus: A‘ ◦ R = B‘ with m …

Microsoft PowerPoint - lec01.ppt

end“: 1. Multilayer Perceptrons: 2. Nonlinear separating functions: x1 x2 2 x1 x2 2 1 ⇒ realizes XOR XOR 0 1 1 0 g(x1, x2) = 2x1 + 2x2 – 4x1x2 -1 with  = 0 g(0,0) = –1 g(0,1) = +1 g(1,0) = +1 g(1,1) = –1 [...] Lecture 01 OR NAND NOR = 0 = 1 AND 0 1 1 0 XOR 0 1 1 0 ? 101 011 110 000 xorx2x1 ⇒ 0 <  ⇒ w2 ≥  ⇒ w1 ≥  ⇒ w1 + w2 <  w1, w2 ≥  > 0 ⇒ w1 + w2 ≥ 2 contradiction! w1 x1 + w2 x2 ≥  Introduction to [...] 2009/10 28 Lecture 01 Example threshold as a weight: w = (, w1, w2)‘ ⇒ ≥0x2 x1 1 w2 w1 - suppose initial vector of weights is w(0) = (1, -1, 1)‘ Introduction to Artificial Neural Networks 8 G. Rudolph: …

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Vorlesung Maschinelles Lernen - SVM -- optimale Hyperebene

= N∑ i=1 αiyi~xi führt zu 1 2 ||~β||2 − N∑ i=1 αiyi〈~xi, ~β〉 + N∑ i=1 αi = 1 2 N∑ i=1 N∑ j=1 αiαjyiyj〈~xi, ~xj〉 − N∑ i=1 N∑ j=1 αiαjyiyj〈~xi, ~xj〉 + N∑ i=1 αi = + N∑ i=1 αi − 1 2 N∑ i=1 N∑ j=1 αiαjyiyj〈~xi [...] Problem 1 2 ||~β||2 − N∑ i=1 αi [ yi ( 〈~xi, ~β〉+ β0 ) − 1 ] = 1 2 ||~β||2 − N∑ i=1 αi yi ( 〈~xi, ~β〉+ β0 ) + N∑ i=1 αi = 1 2 ||~β||2 − N∑ i=1 αiyi〈~xi, ~β〉 − N∑ i=1 αiyiβ0 + N∑ i=1 αi (6) = 1 2 ||~β||2 [...] kein Beispiel zwischen H1 und H2, d.h. 〈~x, ~β〉+ β0 ≥ +1∀~x ∈ C+ (1) 〈~x, ~β〉+ β0 ≤ −1∀~x ∈ C− (2) Der Abstand d(H1, H2) = 2 ||~β|| heißt Margin und soll maximiert werden! H2 H H1 d− d+ + + + + + − − − − …

Vorlesung Wissensentdeckung - Stützvektormethode

αiyixi führt zu 1 2 ||β||2 − N i=1 αiyi xi , β + N i=1 αi = 1 2 N i=1 N j=1 αiαjyiyj xi , xj − N i=1 N j=1 αiαjyiyj xi , xj + N i=1 αi =+ N i=1 αi − 1 2 N i=1 N j=1 αiαjyiyj xi , xj unter den Nebenbedingungen [...] Problem 1 2 ||β||2 − N i=1 αi yi xi , β + β0 − 1 = 1 2 ||β||2 − N i=1 αi yi xi , β + β0 + N i=1 αi = 1 2 ||β||2 − N i=1 αiyi xi , β − N i=1 αiyiβ0 + N i=1 αi (6) = 1 2 ||β||2 − N i=1 αiyi xi [...] liegt kein Beispiel zwischen H1 und H2, d.h. x, β+ β0 ≥ +1∀x ∈ C+ (1) x, β+ β0 ≤ −1∀x ∈ C− (2) Der Abstand d(H1, H2) = 2 ||β|| heißt Margin und soll maximiert werden! H2 H H1 d− d+ + + + + + − − − − − − …

6)} ) 1.) 1.8.) 1.8.9.) 1.8.9.7 4 1.2.) 1.2.3.) 1.2.3.4 2 1.2.5.) 1.2.5.6 6 l = 2 " h HHH2 = {(1.2.5.6)} ) 1.) 1.8.) 1.8.9.) 1.8.9.7 4 1.2.) 1.2.3.) 1.2.3.4 2 1.2.5.) 1.2.5.6 6 l = 1 " h HHH1 = {(1.2.5.6) [...] 5 ) 1.) 1.8.) 1.8.9.) 1.8.9.7 4 1.2.) 1.2.3.) 1.2.3.4 2 1.2.5.) 1.2.5.6 6 l = h = 4 HHH4 = {(1.2.5.6)} ) 1.) 1.8.) 1.8.9.) 1.8.9.7 4 1.2.) 1.2.3.) 1.2.3.4 2 1.2.5.) 1.2.5.6 6 l = 3 " h HHH3 = {(1.2.5.6)} [...] 5.) 1.) ) ) 5.6.) 1.) 5.) 1.2.) ) ) 5.6.7.) 1.) 5.6.) 1.2.) 5.) 1.2.3.) ) ) 5.6.7.8 1.) 5.6.7.) 1.2.) 5.6.) 1.2.3.) 5.) 1.2.3.4 ) 1.) 5.6.7.8 1.2.) 5.6.7.) 1.2.3.) 5.6.) 1.2.3.4 5.) 1.2.) 5.6.7.8 1.2.3 …

1−θ2 − 1−θ2 1−θ2 = θ 1 1−θ2 1 · (1 − θ2) 1 1−θ2 · (1 − θ2)− 1−θ2 1−θ2 · (t − α0) 1 1−θ2 − 1−θ2 1−θ2 = θ 1−θ2 1−θ2 1 · θ θ2 1−θ2 1 · (1 − θ2) θ2 1−θ2 · (t − α0) θ2 1−θ2 = θ1 · ( (θ1 · (1 − θ2)) 1 1−θ2 · (t [...] t)− 1 θ2−1−1 · (−1) = θ −1 θ2−1 1 · (θ2 − 1) −1 θ2−1 · (θ2 − 1)− θ2−1 θ2−1 · (α0 − t)− 1 θ2−1− θ2−1 θ2−1 = θ θ2−1 θ2−1 1 · θ −θ2 θ2−1 1 · (θ2 − 1)− θ2 θ2−1 · (α0 − t)− θ2 θ2−1 = θ1 · ( (θ1 · (θ2 − 1))− 1 [...] ∂t [α1 · (t − α0)α2] = ∂ ∂t [ (θ1 · (1 − θ2)) 1 1−θ2 · (t − α0) 1 1−θ2 ] = (θ1 · (1 − θ2)) 1 1−θ2 · (1 − θ2)−1 · (t − α0) 1 1−θ2 −1 = θ 1 1−θ2 1 · (1 − θ2) 1 1−θ2 · (1 − θ2)− 1−θ2 1−θ2 · (t − α0) 1 1−θ2 …

455 7,1 8,6 -1,4 Bayern 27 96 28,1 337 5,3 11,4 -6,1 Berlin 1 1 100,0 1.217 19,1 9,5 9,6 Brandenburg 3 18 16,7 124 1,9 4,1 -2,2 Bremen 1 2 50,0 125 2,0 1,2 0,7 Hamburg 1 1 100,0 151 2,4 1,2 1,1 Hessen [...] 6 0,7 Bayern 26 96 27,1 323 5,6 11,4 -5,8 Berlin 1 1 100,0 1.093 19,0 9,5 9,5 Brandenburg 3 18 16,7 61 1,1 4,1 -3,1 Bremen 1 2 50,0 85 1,5 1,2 0,2 Hamburg 1 1 100,0 76 1,3 1,2 0,1 Hessen 11 33 33,3 320 [...] 3 578 9,6 8,6 1,1 Bayern 26 96 27,1 334 5,6 11,4 -5,9 Berlin 1 1 100,0 1.083 18,0 9,5 8,5 Brandenburg 4 18 22,2 132 2,2 4,1 -1,9 Bremen 1 2 50,0 83 1,4 1,2 0,1 Hamburg 1 1 100,0 119 2,0 1,2 0,7 Hessen 11 …

[ (1− γG) + γGSt 1−η ]− 1 1−η P ∗Ft PG∗t = [ (1− γG∗)St−(1−η) + γG∗ ]− 1 1−η xi πHt/πt = [ (1− γ) + γS1−η t ]− 1 1−η [ (1− γ) + γS1−η t−1 ]− 1 1−η π∗Ft/π ∗ t = [ (1− γ∗)S−(1−η) t + γ∗ ]− 1 1−η [ (1− γ∗)S−(1−η) [...] γ∗)S−(1−η) t−1 + γ∗ ]− 1 1−η πHt/πGt = [ (1− γG) + γGS1−η t ]− 1 1−η [ (1− γG) + γGS1−η t−1 ]− 1 1−η π∗Ft/π ∗ Gt = [ (1− γG∗)S−(1−η) t + γG∗ ]− 1 1−η [ (1− γG∗)S−(1−η) t−1 + γG∗ ]− 1 1−η Consumption and [...] t + γ∗ ]− 1 1−η πHt/πt = [ (1− γ) + γS1−η t ]− 1 1−η [ (1− γ) + γS1−η t−1 ]− 1 1−η π∗Ft/π ∗ t = [ (1− γ∗)S−(1−η) t + γ∗ ]− 1 1−η [ (1− γ∗)S−(1−η) t−1 + γ∗ ]− 1 1−η Consumption and government demand: CHt …

choose bit i with equal probability from 1st or 2nd parent 1 0 0 1 0 1 1 1 1 1 1 1 a) ) 1 0 0 1 0 1 1 1 0 0 0 1 c) ) 1 0 0 1 0 1 1 1 1 1 0 1 b) ) Individuals 2 { 0, 1 }n Lecture 10 G. Rudolph: Computational [...] index ke at random invert order of bits between start and and index 1 0 0 1 1 1 1 0 1 1 a) k=2 0 0 1 0 1 b) 1 1 0 0 1 ks ke d) 0 0 0 0 1 c) K=2 → → Lecture 10 G. Rudolph: Computational Intelligence ▪ Winter [...] Algorithm Basics Specific example: (1+1)-EA in Bn for minimizing some f: Bn → R population size = 1, number of offspring = 1, selects best from 1+1 individuals parent offspring 1. initialize X(0) 2 Bn uniformly …