Hessen 1.2 1.2 0.9 0.7 1.2 1.0 1.5 Rheinland-Pfalz 1.0 1.0 0.9 0.7 1.2 1.9 1.7 Baden-Württemberg 1.8 1.8 1.7 1.5 2.2 2.0 2.1 Bayern 1.4 1.5 1.1 1.2 1.6 1.5 1.6 Saarland 1.4 1.9 0.8 1.1 0.8 1.8 1.8 Berlin 1 [...] 1.3 0.7 1.3 0.6 1.1 0.6 1.3 Brandenburg 1.3 1.1 1.0 1.1 1.1 1.2 1.7 Mecklenburg-Vorpommern 1.5 1.3 1.8 0.4 1.0 0.9 1.5 Sachsen 0.9 0.8 0.9 0.4 1.2 1.0 1.0 Sachsen-Anhalt 1.0 0.8 1.0 1.7 1.0 0.3 1.3 Thüringen [...] lstein 1.8 1.8 1.7 1.2 1.7 1.4 1.9 Hamburg 1.8 2.4 1.5 1.8 1.4 1.6 1.8 Niedersachsen 2.2 2.2 2.0 1.6 2.2 1.8 2.3 Bremen 1.6 1.5 1.2 1.4 0.7 1.6 1.6 Nordrhein-Westfalen 1.1 0.8 0.9 0.8 1.0 0.8 1.6 Hessen …

0, q1(s1) = 0, q1(s∗) = 1, qk(s1) = 1, qk(s∗) = 0, and `k(s1) = `k(s∗) = 0 and `1(s1) = `1(s∗) = 1. 13 This implies that c11(s1)− c11(s∗) = 2 k , c12(s1)− c12(s∗) + c13(s1)− c13(s∗) = 2 k , c22(s1)− c22(s∗) [...] c11(U) = k − 1 k t∑ j=1 nj(s)2, c12(U) = t∑ j=1 mjj(s)− 1 k t∑ j=1 nj(s)`j(s), c13(U) = t∑ j=1 mjj(s)− 1 k t∑ j=1 nj(s)qj(s), c23(U) = t∑ j=1 m̃jj(s)− 1 k t∑ j=1 `j(s)qj(s)− 1 t (k − 2) + 1 tk (k − 1)2, c22(U) [...] c22(U) = (k − 1)− 1 k t∑ j=1 `j(s)2 − 1 t (k − 1) + 1 tk (k − 1)2, c33(U) = (k − 1)− 1 k t∑ j=1 qj(s)2 − 1 t (k − 1) + 1 tk (k − 1)2. Based on these formulae, the reduced sequence-specific function H̃U for …

1989 0 0 0 0 0 0 1 0 1 1 Wayne’s World 1992 1 1 0 1 0 1 0 1 0 1 Bang Boom Bang 1999 1 1 0 1 1 0 0 0 1 1 Bridget Jones 2001 1 0 0 1 0 0 0 1 0 0 Simpsons (Film) 2007 0 0 0 1 1 0 0 0 0 1 Aufgabe 2.2 (10 Punkte) [...] erfüllen) an. Titel Jahr z1 z2 z3 z4 z5 z6 z7 z8 z9 z10 Star Wars 1977 1 1 0 0 1 0 1 0 1 1 E.T. der Ausserirdische 1982 1 1 0 1 1 0 1 0 1 1 Indiana Jones 1989 1 1 1 0 0 0 1 0 1 1 Otto - der Ausserfriesische [...] gefunden. Gegeben sei die nachfolgende Aufstellung von Filmen, die von Zuschauern z1, . . . , z10 besucht worden sind. 1. (2 Punkte) Formen Sie die Tabelle in eine Transaktionsdatenbank um! 2. (8 Punkte) …

(II) ∀i ∈ {0, 1, . . . , n− 1} : Li := { x ∈ {0, 1}n \ ( i−1 ⋃ j=0 Lj ) | BinVal(x) < i ∑ j=0 2n−1−j } obvious si ≥ 1 n ( 1− 1 n )n−1 ≥ 1 en Theorem E ( T(1+1) EA,BinVal ) ≤ en2 365 Introduction Fitness-Based [...] partition. To leave Li, flip exactly 1 out of n− i 0-bits. si ≥ (n−i 1 ) · 1 n · ( 1− 1 n )n−1 ≥ n−ien E ( T(1+1) EA,OneMax ) ≤ n−1 ∑ i=0 en n− i = en · n ∑ i=1 1 i < en ln(n) + en = O(n log n) 362 In [...] ⇒ E ( T(1+1) EA,f ) = O(n2) Proof f(x) = n ∑ i=1 wix[i] mit w1 ≥ w2 ≥ · · · ≥ wn Definition fitness levels for i ∈ {0, 1, . . . , n− 1} Li := { x ∈ {0, 1}n \ ( i−1 ⋃ j=0 Lj ) | f(x) < i+1 ∑ j=1 wj } Ln …

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generator f(-1)(x) pseudo-inverse Examples a) f(x) = k – k · x (k ≥ 1) f(-1)(x) = 1 – x/k c(a) = b) f(x) = 1 – xw f(-1)(x) = (1 – x)1/w c(a) = f-1(aw) = (1 – aw)1/w (Yager) Theorem c: [0,1] → [0,1] is involutive [...] Examples: a) standard fuzzy complement c(a) = 1 – a ad (A1): c(0) = 1 – 0 = 1 and c(1) = 1 – 1 = 0 ad (A2): c‘(a) = –1 < 0 (monotone decreasing) ad (A3):  ad (A4): 1 – (1 – a) = a Lecture 02 G. Rudolph: Computational [...] Examples ad (A1): c(0) = 1 and c(1) = 0 ad (A2):  ad (A3): is continuous as a composition of continuous functions ad (A4): e) c(a) = ( 1 – aw )1/w for w > 0 Yager class  (1 – aw)1/w ≥ (1 – bw)1/w ⇔ 1 – aw ≥ …

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(k ≥ 1) f(-1)(x) = 1 – x/k c(a) = b) f(x) = 1 – xw f(-1)(x) = (1 – x)1/w c(a) = f-1(aw) = (1 – aw)1/w (Yager) Theorem c: [0,1] → [0,1] is involutive fuzzy complement iff continuous function f: [0,1] → with [...] Examples: standard fuzzy complement c(a) = 1 – a ad (A1): c(0) = 1 – 0 = 1 and c(1) = 1 – 1 = 0 ad (A2): c‘(a) = –1 < 0 (monotone decreasing) ad (A3):  ad (A4): 1 – (1 – a) = a Lecture 02 4 G. Rudolph: Com [...] Examples ad (A1): c(0) = 1 and c(1) = 0 ad (A2):  ad (A3): is continuous as a composition of continuous functions ad (A4): e) c(a) = ( 1 – aw )1/w for w > 0 Yager class  (1 – aw)1/w ≥ (1 – bw)1/w  1 – aw ≥ …

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generator f(-1)(x) pseudo-inverse Examples a) f(x) = k – k · x (k ≥ 1) f(-1)(x) = 1 – x/k c(a) = b) f(x) = 1 – xw f(-1)(x) = (1 – x)1/w c(a) = f-1(aw) = (1 – aw)1/w (Yager) Theorem c: [0,1] → [0,1] is involutive [...] Examples: a) standard fuzzy complement c(a) = 1 – a ad (A1): c(0) = 1 – 0 = 1 and c(1) = 1 – 1 = 0 ad (A2): c‘(a) = –1 < 0 (monotone decreasing) ad (A3):  ad (A4): 1 – (1 – a) = a Lecture 02 G. Rudolph: Computational [...] Examples ad (A1): c(0) = 1 and c(1) = 0 ad (A2):  ad (A3): is continuous as a composition of continuous functions ad (A4): e) c(a) = ( 1 – aw )1/w for w > 0 Yager class  (1 – aw)1/w ≥ (1 – bw)1/w ⇔ 1 – aw ≥ …

Algorithmen auf Sequenzen - Paarweises Sequenzalignment

Alignment angewendet, a b b a b 0 1 0 -1 a 1 1 1 0 b 1 1 1 1 a 2 2 2 1 VB = 0, 1, 0, 1 VC = 0, 1,−1,−1 und in einem Array abgespeichert: 1, 0, 1; 1,−1,−1; aba; bba → 2, 2, 1, 1, 0 D. Kopczynski | Algorithmen [...] Alignment angewendet, a b b a b 0 1 0 -1 a 1 1 1 0 b 1 1 1 1 a 2 2 2 1 VB = 0, 1, 0, 1 VC = 0, 1,−1,−1 und in einem Array abgespeichert: 1, 0, 1; 1,−1,−1; aba; bba → 2, 2, 1, 1, 0 D. Kopczynski | Algorithmen [...] ch beträgt O(n log3σ(n)). F = { ... 1, 0, 0; 1,−1,−1; aba; bba → 1, 2, 1, 1, 0 1, 0, 1; 1,−1,−1; aba; bba → 2, 2, 1, 1, 0 1, 1, 0; 1,−1,−1; aba; bba → 2, 2, 1, 1, 0 ... } D. Kopczynski | Algorithmen auf …

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x ) g-1(x) = x ) c(a) = 1 – a (Standard) b) g(x) = xw ) g-1(x) = x1/w ) c(a) = (1 – aw)1/w (Yager class, w > 0) c) g(x) = log(x+1) ) g-1(x) = ex – 1 ) c(a) = exp( log(2) – log(a+1) ) – 1 1 – a 1 + a = [...] Examples: standard fuzzy complement c(a) = 1 – a ad (A1): c(0) = 1 – 0 = 1 and c(1) = 1 – 1 = 0 ad (A2): c‘(a) = –1 < 0 (monotone decreasing) ad (A3):  ad (A4): 1 – (1 – a) = a Lecture 06 G. Rudolph: Computational [...] Examples ad (A1): c(0) = 1 and c(1) = 0 ad (A2):  ad (A3): is continuous as a composition of continuous functions ad (A4): e) c(a) = ( 1 – aw )1/w for w > 0 Yager class  (1 – aw)1/w ≥ (1 – bw)1/w , 1 – aw ≥ …

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0) f(-1)(x) = 1 – x/k c(a) = b) f(x) = 1 – xw f(-1)(x) = (1 – x)1/w c(a) = f-1(aw) = (1 – aw)1/w (Yager) Theorem c: [0,1] → [0,1] is involutive fuzzy complement iff  continuous function f: [0,1] → with [...] Examples: standard fuzzy complement c(a) = 1 – a ad (A1): c(0) = 1 – 0 = 1 and c(1) = 1 – 1 = 0 ad (A2): c‘(a) = –1 < 0 (monotone decreasing) ad (A3):  ad (A4): 1 – (1 – a) = a Lecture 06 G. Rudolph: Computational [...] Examples ad (A1): c(0) = 1 and c(1) = 0 ad (A2):  ad (A3): is continuous as a composition of continuous functions ad (A4): e) c(a) = ( 1 – aw )1/w for w > 0 Yager class  (1 – aw)1/w ≥ (1 – bw)1/w  1 – aw ≥ …