0 0 0 0 2 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1  [...] RM(1, 4). Wählt man als Erzeugermatrix von RM(1, 4) die Matrix 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 [...]  0 1 0 0 1 0 1 1 1/2 0 0 1 1 1 0 0 1 −3/2 0 1 0 1 0 1 0 1 −1/2 0 0 0 0 0 0 0 0 −1/2 0 0 0 0 0 0 0 0 −1/2 0 0 0 1 0 1 1 1 −1/2 0 1 0 1 1 0 0 1 1/2 0 1 1 1 1 1 1 0 −1/2 0 1 1 0 0 0 1 1 1/2 0 0 1 1 0 0 1 …

0, -1, 0, -1; 0, 1, 1, 1; εaba; abaa • 0, 0, 1, 1; 0, -1, 0, 1; εaba; abaa • 0, 1, 1, 1; 0, 1, 1, 1; εaba; εaaa • 0, 1, 1, 1; 0, 1, 1, -1; abaa; εaaa • 0, 1, 1, 1; 0, -1, -1, 0; abaa; εaaa • 0, 1, 0, 1; [...] 1; 0, 1, 0, 1; abaa; abaa • 0, -1, 1, 1; 0, 0, 0, 1; abaa; abaa Aufgabe 12.2 Erstellen Sie einen Aho-Corasick-Automaten für die Texte {an, bibi, iban, anna, banbi}. Aufgabe 12.3 Erstellen Sie einen S [...] Algorithmen auf Sequenzen Blatt 12 Ausgabe: Donnerstag, 09. Juli Besprechung: Dienstag, 14. Juli Aufgabe 12.1 Erstellen Sie mit dem Four-Russians-Trick ein globales Alignment für die Texte r = ababaa, s = aaabaa …

Algorithmen auf Sequenzen - Fehlertolerantes Pattern-Matching

, P = MAOAM, k = 1: A M O A M A M A O M 0 0 0 0 0 0 0 0 0 M 1 1 1 1 0 1 1 1 0 A 2 1 1 1 1 0 1 1 1 O 3 2 2 2 2 1 1 1 1 A 4 3 3 2 2 2 2 1 1 M 5 4 3 3 2 2 2 2 2 x 0 0 0 0 1 0 1 0 1 1 1 1 D. Kopczynski | [...] folgt: di ,j ≥ di−1,j−1. Laut Rekursionsgleichung kann nur gelten: di ,j = di−1,j−1 oder di ,j = di−1,j−1 + 1. di−1,j−1 di,j di−2,j−1 min 0||1 Beweis für minimalen Pfad aus di,j−1 analog. D. Kopczynski [...] Anzahl der Alignments N(m, n) für 0 ≤ m, n ≤ 4. m\n 0 1 2 3 4 . . . 0 1 1 1 1 1 . . . 1 1 3 5 7 9 . . . 2 1 5 13 25 41 . . . 3 1 7 25 63 129 . . . 4 1 9 41 129 321 . . . ... ... ... ... ... ... . . . D. …

∑ m1∈N 1 3m2<m1≤m2−1 m1≡m2 mod 2 1 = 1 2 ∑ m1∈N 1 3m2<m1≤m2−1 1 +O(1) = 1 2 ( m2 − 1− ⌊ 1 3 m2 ⌋) +O(1) = 1 2 ( m2 − 1 3 m2 +O(1) ) +O(1) = m2 3 +O(1) . Similar steps yield ∑ m1∈N 1 3m2<m1≤ D m2 m1≡m2 [...] (s) = 1 4s ∞∑ e=1 1 es 2e−1∑ d=1 1 (d+ e)s lim k→∞ ( (d+ e)1−s − k1−s s− 1 + 1 2ks − 1 2(d+ e)s − χ1,0 d+e,k(s) ) = 1 4s(s− 1) ∞∑ e=1 1 es 2e−1∑ d=1 1 (d+ e)2s−1 − 1 2 · 4s ∞∑ e=1 1 es 2e−1∑ d=1 1 (d+ e)2s [...] −d<f<0<d 1 (4d(d− f)2)s = 1 4s ∞∑ d=1 1 ds ∑ f∈Z −d<f<0 1 (d− f)2s = 1 4s ∞∑ d=1 1 ds 2d−1∑ f=d+1 1 f 2s = 1 4s ∞∑ d=1 1 ds 2d∑ f=d+1 1 f 2s − ζ(3s) 16s = 1 4s ∞∑ d=1 1 ds ( d1−2s − (2d)1−2s 2s− 1 − ϕ1,0 d …

m+1 ∏ r=1 1 lr! n −lr) g(∑ m+1 r=1 lr) (i − (m + 1) n ) + k−1 ∑ l1=1 ⋯ k−1−∑m−1 r=1 lr ∑ lm=1 ( m ∏ r=1 1 lr! n −lr)R (g(k), k − 1 − m ∑ r=1 lr, i −m n , 1 n) + rm(g, i, n) − rm(g, i − 1, n) = k−1 ∑ l1=1 [...] l1=1 ⋯ k−1−∑m r=1 lr ∑ lm+1=1 ( m+1 ∏ r=1 1 lr! n −lr) g(∑ m+1 r=1 lr) (i − (m + 1) n ) + rm+1(g, i, n). (4.5.2) Hence, by (4.5.1) and (4.5.2), we have on the one hand △ k i,ng = rk(g, i, n) = k−1 ∑ l,r=0 [...] Lemma 1.5.2. By combining (3.1.5) to (3.1.7), we get An ≤ K M ∑ d=1 ( M ∏ j=d+1 ( 1 n n ∑ i=k ∣Y n i ∣Qjpj) 1 Qjr )( d−1 ∏ j=1 ( 1 n n ∑ i=k ∣Y ∞ i ∣Qjpj) 1 Qjr ) ( 1 n n ∑ i=k ∣Y n i − Y ∞ i ∣Qdpd) 1 Qdr …

d−1ik ) =1− P ( T−1ij ≥ j∑ k=1 d−1ik ∣∣∣∣∣xi(t), w−1i(t), T−1ij > j−1∑ k=1 d−1ik ) =1− exp ( − ∫ ∑j k=1 d−1ik∑j−1 k=1 d−1ik λ(0)∗ −1i(t|xi(t), w−1i(t))dt ) =1− exp ( −v−1ij ∫ ∑j k=1 d−1ik∑j−1 k=1 d−1ik [...] d̃−1i; η−1) × f̄−1i(d−1i, d̃−1i; η−1)dd̃−1i . . . dd−1i1 + ∫ ∞ 0 . . . ∫ ∞ 0 ∂ ∂ ( θ(0)> −1 , η−1 )> log f̄−1i(d−1i, d̃−1i; η−1) × ∂ ∂ ( θ(0)> −1 , η−1 ) f̄−1i(d−1i, d̃−1i; η−1)dd̃−1i . . . dd−1i1. (4.43) [...] θ(0)> −1 , η−1 ) f̄−1i(d−1i, d̃−1i; η−1) = f̄−1i(d−1i, d̃−1i; η−1) × ∂ ∂ ( θ(0)> −1 , η−1 ) log f̄−1i(d−1i, d̃−1i; η−1), Gleichung (4.38) und ∂2 ∂θ∗−1∂θ ∗> −1 log L̄−1i(θ∗−1) ∣∣∣∣∣( θ (0)> −1 ,η−1 )> = ∂2 …

{MPLTEXT 1 0 48 " T2:=linalg[matrix](8,8,[1, 1, 1, 1, 1, 1, 1, 1, " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " 1,-1, 1,-1, 1,-1, 1,-1," }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " 1, 1,-1,-1, 1, 1,-1,-1," }}{PARA [...] "> " 0 "" {MPLTEXT 1 0 47 " \+ 1,-1,-1, 1, 1,-1,-1, 1," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " \+ 1, 1, 1, 1,-1,-1,-1,-1," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " 1,-1, 1,-1,-1, 1,-1, 1," }} {PARA 0 "> " 0 [...] {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "T2:=linalg[matrix](4,4,[1,1, 1,1, 1,-1,1,-1, 1,1,-1,-1, 1,-1,-1,1]);" }}{PARA 11 "" 1 "" {XPPMATH …

{MPLTEXT 1 0 48 " T2:=linalg[matrix](8,8,[1, 1, 1, 1, 1, 1, 1, 1, " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " 1,-1, 1,-1, 1,-1, 1,-1," }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " 1, 1,-1,-1, 1, 1,-1,-1," }}{PARA [...] "> " 0 "" {MPLTEXT 1 0 47 " \+ 1,-1,-1, 1, 1,-1,-1, 1," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " \+ 1, 1, 1, 1,-1,-1,-1,-1," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 " 1,-1, 1,-1,-1, 1,-1, 1," }} {PARA 0 "> " 0 [...] {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "T2:=linalg[matrix](4,4,[1,1, 1,1, 1,-1,1,-1, 1,1,-1,-1, 1,-1,-1,1]);" }}{PARA 11 "" 1 "" {XPPMATH …

r a1 a2 b1 b2 c1 c2 1 1 -1 1 0 -1 0 0 0 2 1 -1 0 1 0 -1 0 0 3 1 -1 1 0 0 0 -1 0 4 1 -1 0 1 0 0 0 -1 5 1 -1 0 0 1 0 -1 0 6 1 -1 0 0 0 1 0 -1 7 1 -1 -1 0 1 0 0 0 8 1 -1 0 -1 0 1 0 0 9 1 -1 -1 0 0 0 1 0 10 [...] 0 0 0 2 1 -1 1 0 -1 0 0 0 3 1 -1 0 1 -1 0 0 0 4 1 -1 -1 1 0 0 0 0 5 1 -1 -1 0 1 0 0 0 6 1 -1 0 -1 1 0 0 0 7 1 -1 0 0 0 1 -1 0 8 1 -1 0 0 0 1 0 -1 9 1 -1 0 0 0 0 1 -1 10 1 -1 0 0 0 -1 1 0 11 1 -1 0 0 0 -1 [...]  1 −1 0 1 0 −1 0 1 −1 ⊗ [ 1 −1 ] =  1 −1 −1 1 0 0 1 −1 0 0 −1 1 0 0 1 −1 −1 1  . Damit hat die Kontrastmatrix C4 die Form C4 = [ 03 03 P 3 ⊗ P 2 ]T =  0 0 1 −1 −1 1 0 0 0 0 1 −1 0 0 …

1 -0 1 2 0 0 8 -0 1 -0 1 2 0 0 9 -0 1 -0 1 2 0 1 0 -0 1 -0 1 2 0 1 1 -0 1 -0 1 2 0 1 2 -0 1 -0 1 2 0 1 3 -0 1 -0 1 2 0 1 4 -0 1 -0 1 2 0 1 5 -0 1 -0 1 2 0 1 6 -0 1 -0 1 2 0 1 7 -0 1 -0 1 2 0 1 8 -0 1 -0 [...] 1 2 0 0 5 -0 1 -0 1 2 0 0 6 -0 1 -0 1 2 0 0 7 -0 1 -0 1 2 0 0 8 -0 1 -0 1 2 0 0 9 -0 1 -0 1 2 0 1 0 -0 1 -0 1 2 0 1 1 -0 1 -0 1 2 0 1 2 -0 1 -0 1 2 0 1 3 -0 1 -0 1 2 0 1 4 -0 1 -0 1 2 0 1 5 -0 1 -0 1 2 [...] 1 2 0 0 5 -0 1 -0 1 2 0 0 6 -0 1 -0 1 2 0 0 7 -0 1 -0 1 2 0 0 8 -0 1 -0 1 2 0 0 9 -0 1 -0 1 2 0 1 0 -0 1 -0 1 2 0 1 1 -0 1 -0 1 2 0 1 2 -0 1 -0 1 2 0 1 3 -0 1 -0 1 2 0 1 4 -0 1 -0 1 2 0 1 5 -0 1 -0 1 2 …