choose bit i with equal probability from 1st or 2nd parent 1 0 0 1 0 1 1 1 1 1 1 1a) ⇒ 1 0 0 1 0 1 1 1 0 0 0 1c) ⇒ 1 0 0 1 0 1 1 1 1 1 0 1b) ⇒ Individuals ∈ { 0, 1 }nVariation in n Lecture 05 G. Rudolph: [...] index ke at random invert order of bits between start and end index 1 0 0 1 1 1 1 0 1 1a) k=2 0 0 1 0 1b) 1 1 0 0 1 ks ke d) 0 0 0 0 1c) K=2 → → Variation in n Lecture 05 G. Rudolph: Computational Intelligence [...] offspring (skip values already contained) x x x 7 1 6 x 5 3 2 7 1 6 4 2 3 5 7 1 6 4 6 4 5 3 7 2 1 2 3 5 7 1 6 4 6 4 5 3 7 2 1 x x x 7 1 6 x x 4 5 7 1 6 x 3 4 5 7 1 6 2 Variation in n Lecture 05 G. Rudolph: C …

choose bit i with equal probability from 1st or 2nd parent 1 0 0 1 0 1 1 1 1 1 1 1a) ⇒ 1 0 0 1 0 1 1 1 0 0 0 1c) ⇒ 1 0 0 1 0 1 1 1 1 1 0 1b) ⇒ Individuals ∈ { 0, 1 }nVariation in n Lecture 05 G. Rudolph: [...] index ke at random invert order of bits between start and end index 1 0 0 1 1 1 1 0 1 1a) k=2 0 0 1 0 1b) 1 1 0 0 1 ks ke d) 0 0 0 0 1c) K=2 → → Variation in n Lecture 05 G. Rudolph: Computational Intelligence [...] offspring (skip values already contained) x x x 7 1 6 x 5 3 2 7 1 6 4 2 3 5 7 1 6 4 6 4 5 3 7 2 1 2 3 5 7 1 6 4 6 4 5 3 7 2 1 x x x 7 1 6 x x 4 5 7 1 6 x 3 4 5 7 1 6 2 Variation in n Lecture 05 G. Rudolph: C …

choose bit i with equal probability from 1st or 2nd parent 1 0 0 1 0 1 1 1 1 1 1 1a) ⇒ 1 0 0 1 0 1 1 1 0 0 0 1c) ⇒ 1 0 0 1 0 1 1 1 1 1 0 1b) ⇒ Individuals ∈ { 0, 1 }nVariation in n Lecture 05 G. Rudolph: [...] index ke at random invert order of bits between start and end index 1 0 0 1 1 1 1 0 1 1a) k=2 0 0 1 0 1b) 1 1 0 0 1 ks ke d) 0 0 0 0 1c) K=2 → → Variation in n Lecture 05 G. Rudolph: Computational Intelligence [...] offspring (skip values already contained) x x x 7 1 6 x 5 3 2 7 1 6 4 2 3 5 7 1 6 4 6 4 5 3 7 2 1 2 3 5 7 1 6 4 6 4 5 3 7 2 1 x x x 7 1 6 x x 4 5 7 1 6 x 3 4 5 7 1 6 2 Variation in n Lecture 05 G. Rudolph: C …

choose bit i with equal probability from 1st or 2nd parent 1 0 0 1 0 1 1 1 1 1 1 1a) ⇒ 1 0 0 1 0 1 1 1 0 0 0 1c) ⇒ 1 0 0 1 0 1 1 1 1 1 0 1b) ⇒ Individuals ∈ { 0, 1 }nVariation in n Lecture 06 G. Rudolph: [...] index ke at random invert order of bits between start and end index 1 0 0 1 1 1 1 0 1 1a) k=2 0 0 1 0 1b) 1 1 0 0 1 ks ke d) 0 0 0 0 1c) K=2 → → Variation in n Lecture 06 G. Rudolph: Computational Intelligence [...] offspring (skip values already contained) x x x 7 1 6 x 5 3 2 7 1 6 4 2 3 5 7 1 6 4 6 4 5 3 7 2 1 2 3 5 7 1 6 4 6 4 5 3 7 2 1 x x x 7 1 6 x x 4 5 7 1 6 x 3 4 5 7 1 6 2 Variation in n Lecture 06 G. Rudolph: C …

diss_bohem.dvi

2 1 1 1 9 1 1 3 1 4 1 1 1 1 10 1 4 2 1 1 17 1 3 2 1 1 1 1 1 11 1 2 1 2 1 1 10 1 1 1 1 1 3 1 12 2 1 1 1 3 1 8 1 1 1 1 1 13 1 1 1 2 1 10 1 3 1 1 1 1 1 14 1 1 1 1 4 1 1 15 1 1 1 1 1 7 2 1 1 16 1 1 1 1 1 1 [...] 100 110 001 011 101 111 +++ 1 1 1 1 1 1 1 1 ++ 1 1 1 1 1 1 1 1 ++ 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 ++ 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Tabelle 2.3: Die Charaktertafel [...] 1 3 5 1 1 1 1 17 1 1 1 1 18 1 1 3 1 2 1 1 8 1 2 1 19 1 1 1 1 1 3 1 20 1 1 1 1 1 5 1 1 1 21 1 1 1 2 1 3 1 1 22 1 1 1 1 1 2 1 23 1 1 1 1 2 24 1 1 1 Tabelle 2.6: Der Nachbarschaftsgraph des Geschlechtes des …

Folie 1

Setze i = 2 1 E 1 Test i ≤ n 1 E n Test ai < x 1 E n - 1 Setze j = i, x = aj 2 E A Erhöhe i 1 E n - 1 Insgesamt also: T(n) = 2 + 1 + n + (n-1) + 2A + (n-1) E = (3n + 2A + 1) E Rudolph: EINI (WS 2006/07) [...] Altersangaben a1, a2, …, an in Jahren, n ≥ 1 Gesucht: aj = min(a1, a2, …, an), wobei j die erste Stelle in der Folge sei, an der das Minimum auftritt (1) [ Wähle 1. Kandidat ] Setze j = 1 und x = aj. (2) [...] Raum“ (3) Beispiel Gegeben: Folge von 8 Altersangaben 20, 21, 20, 19, 18, 19, 18, 20 1 2 3 4 5 6 7 8 j (1) 20 1 (2) 1 1 4 5 5 5 5 (2) (2) (2) (2) (2) (2) 20 21 20 21 20 20 21 20 19 20 21 20 19 18 20 21 20 …

1, 1, 0, 0, 0) 9 (1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0) 9 (1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0) 9 (1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1) 10 (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0) 10 (1, 1, 1, 1, 0, 1, 1, 1, [...] 1, 1, 1, 0, 0, 1, 0) 8 (1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0) 8 (1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0) 8 (1, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0) 8 (1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0) 9 (1, 1, 1, 1, 1, 1, 1, [...] 0, 0, 1, 1, 1, 1, 0, 1, 0) 7 (0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0) 7 (1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0) 8 (1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0) 8 (1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0) 8 (1, 1, 1, 1, 0, 1 …

Slide 1

2019/20 15 OR NAND NOR = 0 = 1 AND 0 1 1 0 XOR 0 1 1 0 ? x1 x2 xor 0 0 0 0 1 1 1 0 1 1 1 0 ⇒ 0 < θ ⇒ w2 ≥ θ ⇒ w1 ≥ θ ⇒ w1 + w2 < θ w1, w2 ≥ θ > 0 ⇒ w1 + w2 ≥ 2θ contradiction! w1 x1 + w2 x2 ≥ θ Introduction [...] leave the „dead end“: 1. Multilayer Perceptrons: x1 x2 2 x1 x2 2 1 ⇒ realizes XOR XOR 0 1 1 0 g(x1, x2) = 2x1 + 2x2 – 4x1x2 -1 with θ = 0 g(0,0) = –1 g(0,1) = +1 g(1,0) = +1 g(1,1) = –1 Introduction to Artificial [...] yet: by construction x1 x2 NAND 0 0 1 0 1 1 1 0 1 1 1 0 example: NAND-gate ⇒ 0 ≥ θ ⇒ w2 ≥ θ ⇒ w1 ≥ θ ⇒ w1 + w2 < θ requires solution of a system of linear inequalities (∈ P) (e.g.: w1 = w2 = -2, θ = -3) now: …

r1 ≤ E ( Td− d 2k ) d− d 2k ≤ √ nπ Vn(r1) r1 √ 2 ⇔ 2dVn(r1) r1 ( 1− 1 2k ) ≤ E ( Td− d 2k ) ≤ d √ nπ Vn(r1) r1 √ 2 ( 1− 1 2k ) or, under the common assumption 1 À 1/dk, after removing the parentheses and [...] normal c.d.f.. We have next 1 σt ≤ 1 σ1 Φ ( bt − µt σt ) ≥ Φ(0) = 1 2 Φ ( α− µt σt ) = 1− Φ ( µt − α σt ) and hence Cα,bt ≤ 1/σ1 1 2 − 1 + Φ ( µt−α σt ) ≤ 1/σ1 Φ ( µ1−α σ2 ) − 1 2 < ∞ which yields the constant [...] E(Xt)E(Yt) µ1 ∞∑ t=1 E(Yt) ≤E ( T∑ t=1 Xt ) ≤ µ2 ∞∑ t=1 E(Yt) µ1 ∞∑ t=1 P{T ≥ t} ≤E ( T∑ t=1 Xt ) ≤ µ2 ∞∑ t=1 P{T ≥ t} µ1E(T ) ≤E ( T∑ t=1 Xt ) ≤ µ2E(T ). That E(T ) = ∑∞ t=1 P{T ≥ t} can be seen - for any …

R(t) = 1 R(t) R(vk) = 1 R(γk(v1)) R(γk(0)) if t = γk(u), 0 ≤ u < v1 = γ(0). Thus by assumption, LIM( R(vk) R(t) ) ⊆ [1, lim R(γk(0)) R(γk(v1)) ] = [1, 1 S(v1) ] or LIM(Rt(Γ(t)(x))) ⊆ S(x) · [1, 1 T (γ(0)−) [...] 2.4.6 : R(x1, x2) = (1 + x1 + x2) −1 , x1, x2 > 0. We define T (t) ∈ CAT(R, d) by T (t) := γ et·I,(et−1)· 1 2 ⇀ e . Then we have R(T (t)( ⇀ x)) = (1 + et〈⇀x,⇀ e 〉+ et − 1 2 · 〈⇀e ,⇀ e 〉)−1 = (et〈⇀x,⇀ e [...] definition 2.8.1(a) that (1) and (2) are satisfied. Then R(γn+1( ⇀ x)) R(γn+1( ⇀ 0)) = R(γn(γ−1 n γn+1)( ⇀ x)) R(γn( ⇀ 0)) · R(γn( ⇀ 0)) R(γn+1( ⇀ 0)) n→∞−→ Λ( ⇀ x) (2.8.1) Hence we have R(γn(γ−1 n γn+1)( ⇀ x)) …