studis:lernraeume [Fachschaft Informatik]

02 N 1 1 1 - 4 OH 14 U 03 N 2 1 4 - 16 OH 14 E 33 J 1 1 2 - 8 OH 14 E 34 J 1 1 2 - 8 OH 14 E 37 J 1 8 6 - 16 OH 14 E 38 J 1 7 6 - 16 OH 12 3.026 N 1 1 1 - 4 OH 12 3.028 N 1 1 1 - 4 OH 12 3.029 N 1 1 1 - [...] - 4 OH 12 3.044 J 1 1 2 - 6 OH 12 4.026 N 1 1 2 - 8 OH 12 4.027 J 1 1 2 - 8 OH 12 4.029 J 1 2 4 - 16 OH 12 4.033 J 1 2 4 - 16 OH 12 4.037 J 1 2 8 - 16 OH 12 4.042 N 1 1 2 - 6 Poolräume der Fakultät für …

PHYSICS AND PHILOSOPHY

mixed states of the subsystems S1 and S2 are W1 = 1 2 P [ ϕ(1) n ] + 1 2 P [ ϕ (1) −n ] = 1 2 1l1 , (22) W2 = 1 2 P [ ϕ(2) n ] + 1 2 P [ ϕ (2) −n ] = 1 2 1l2 , (23) where by P [ϕ] we denote the projection [...] subsystems S1 and S2 are in the mixed states W1 = 1 2 P [ ϕ(1) n ] + 1 2 P [ ϕ (1) −n ] , W2 = 1 2 P [ ϕ(2) n ] + 1 2 P [ ϕ (2) −n ] . (13) A Lüders measurement (cf. Appendix B) of A1(n) = σ1(n) ⊗ 1l2 transforms [...] observables σ1(n) and σ2(n) can also be written as observables A1(n) and A2(n) of the compound system S = S1 + S2 A1(n) := σ1(n)⊗ 1l2 and A2(n) := 1l1 ⊗ σ2(n) . (3) A measurement of the observable A1(n), say …

Vorlesung Wissensentdeckung - Strukturelle Modelle – Conditional Random Fields

Modelle allgemein CRF vs. HMM HMM: p(y,x) = ∏3 t=1 p(yt|yt−1)p(xt|yt) y1 y2 x1 x2 y3 x3 CRF: p(y|x) = Z(x)−1 ∏3 t=1 exp [ ayt−1,yt,xt + byt,xt ] y1 y2 x1 x2 y3 x3 Bemerkung: y0 = start, p(y|start) = π(y) [...] (yT+1 = stop, yT = yi, xT ) Rekursion für 1 < t < T: αt(x) = αt−1 ·Mt(x) = αt−1(x) · exp[ ∑K k λkfk(yt−1, yt, xt) + ∑K k λkfk(yt, xt)] βt(x) = βt+1(x) ·Mt+1(x) = βt+1(x) · exp[ ∑K k λkfk(yt, yt+1t, xt) [...] αt(i) = P (x1, ..., xT , Yt = yi|λ) Anfang: π mal Wahrscheinlichkeit für erste Beobachtung α1(i) = P (x1, yi|λ) = πibi(x1) Rekursion für 1 < t ≤ T : αt(j) = P (x1, ..., xt, yj |λ) = |Y |∑ i=1 αt−1(i)aijbj(xt) …

schlechtsehr gut n=210 mw=1,3 md=1 s=0,65 77,1% 1 18,1% 2 3,3% 3 0,5% 4 1% 5 0% 6 Wie war Ihr Lernerfolg im Tutorium?3.6) sehr schlechtsehr gut n=211 mw=1,6 md=1 s=0,85 57,8% 1 29,4% 2 9,5% 3 2,4% 4 0,5% [...] 73% 1 21,3% 2 4,7% 3 0,5% 4 0,5% 5 0% 6 War Ihre Tutorin / Ihr Tutor gut vorbereitet?3.4) sehr schlechtsehr gut n=211 mw=1,22 md=1 s=0,55 81% 1 17,1% 2 1,4% 3 0% 4 0% 5 0,5% 6 Wie bewerten Sie das Engagement [...] statt.3.1) n=211vollständig 98,1% fast immer 0,9% überwiegend 0,9% manchmal 0% selten 0% gar nicht 0% Wie hat Ihnen Ihr Tutorium insgesamt gefallen?3.3) sehr schlechtsehr gut n=211 mw=1,34 md=1 s=0,64 73% …

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: …

Threshold Logic Units

Conjunction \(x_1\wedge x_2\) : \(x_1\) \(x_2\) \(3x_1 + 2x_2\) \(y\) 0 0 0 0 0 1 2 0 1 0 3 0 1 1 5 1 Implication \(x_1\rightarrow x_2 \; \equiv \; \neg x_1 \vee x_2\) : \(x_1\) \(x_2\) \(-2 x_1 + 2x_2\) \(y\) [...] 0 0 0 1 0 1 2 1 1 0 -2 0 1 1 0 1 Geometric Interpretation of TLUs We can interpret these functions as hyperplanes: Conjunction \(x_1\wedge x_2\) : Hyperplane:   \(3x_1+2x_2=4\) Implication \(x_1\rightarrow [...] \(x_2\) \(y\) 0 0 0 0 1 1 1 0 1 1 1 0 Proof that a TLU cannot compute XOR: \[\begin{alignat}{3} (0,0) & \mapsto 0 & \quad & \Rightarrow \quad & 0 & \lt{} \theta \tag {1} \\ (1,0) & \mapsto 1 & \quad & \Rightarrow …

24,5 4,0 175039 54,5 1,0 178485 52,5 1,3 175064 45,0 1,7 178487 23,0 5,0 175073 14,0 5,0 178659 30,5 3,3 175107 48,0 1,3 178671 41,5 2,3 175113 45,0 1,7 178679 47,0 1,7 175114 46,0 1,7 178684 43,0 2,0 175116 [...] 147309 46,5 1,7 174413 25,5 4,0 147329 52,5 1,3 174633 32,0 3,3 147557 41,5 2,3 174654 38,0 2,7 147718 48,5 1,3 174657 49,0 1,3 147785 46,5 1,7 174679 39,5 2,3 147792 21,0 5,0 174714 52,0 1,3 147945 43 [...] 28,0 3,7 174932 52,0 1,3 167734 30,5 3,3 174937 36,0 2,7 167956 40,0 2,3 174939 55,5 1,0 168372 23,0 5,0 174942 54,5 1,0 168392 46,0 1,7 174943 46,0 1,7 168458 10,0 5,0 174944 50,5 1,3 168967 32,5 3,3 174946 …

Threshold Logic Units

Conjunction \(x_1\wedge x_2\) : \(x_1\) \(x_2\) \(3x_1 + 2x_2\) \(y\) 0 0 0 0 0 1 2 0 1 0 3 0 1 1 5 1 Implication \(x_1\rightarrow x_2 \; \equiv \; \neg x_1 \vee x_2\) : \(x_1\) \(x_2\) \(-2 x_1 + 2x_2\) \(y\) [...] 0 0 0 1 0 1 2 1 1 0 -2 0 1 1 0 1 Geometric Interpretation of TLUs We can interpret these functions as hyperplanes: Conjunction \(x_1\wedge x_2\) : Hyperplane:   \(3x_1+2x_2=4\) Implication \(x_1\rightarrow [...] \(x_2\) \(y\) 0 0 0 0 1 1 1 0 1 1 1 0 Proof that a TLU cannot compute XOR: \[\begin{alignat}{3} (0,0) & \mapsto 0 & \quad & \Rightarrow \quad & 0 & \lt{} \theta \tag {1} \\ (1,0) & \mapsto 1 & \quad & \Rightarrow …

24,5 4,0 175039 54,5 1,0 178485 52,5 1,3 175064 45,0 1,7 178487 23,0 5,0 175073 14,0 5,0 178659 30,5 3,3 175107 48,0 1,3 178671 41,5 2,3 175113 45,0 1,7 178679 47,0 1,7 175114 46,0 1,7 178684 43,0 2,0 175116 [...] 147309 46,5 1,7 174413 25,5 4,0 147329 52,5 1,3 174633 32,0 3,3 147557 41,5 2,3 174654 38,0 2,7 147718 48,5 1,3 174657 49,0 1,3 147785 46,5 1,7 174679 39,5 2,3 147792 21,0 5,0 174714 52,0 1,3 147945 43 [...] 28,0 3,7 174932 52,0 1,3 167734 30,5 3,3 174937 36,0 2,7 167956 40,0 2,3 174939 55,5 1,0 168372 23,0 5,0 174942 54,5 1,0 168392 46,0 1,7 174943 46,0 1,7 168458 10,0 5,0 174944 50,5 1,3 168967 32,5 3,3 174946 …

E[ Dk+1 ‖Xk = x ] = E[ Dk+1 ‖Xk = x, J = 0 ]× P{ J = 0 } + E[ Dk+1 ‖Xk = x, J = 1 ]× P{ J = 1 } . (1) Analyse Weitere Vereinfachung durch Ausnutzung der Symmetrien: Da für alle x ∈ IR2 und z1 > 0 E[ [...] E[ Dt+1 ‖Xt = (x1, x2), J = 1 ] = E[ Dt+1 ‖Xt = (z1 − x1, x2), J = 0 ] (2) Analyse beschränkbar auf Selektion bzgl. Ziel f0. Hinreichend: Betrachte Fall x2 ≥ 0 da für alle x ∈ IR2 gilt E[ Dt+1 ‖Xt = (x1 [...] (x1, x2) ] = E[ Dt+1 ‖Xt = (x1,−x2) ] . Also bestimme nur: E[ Dk+1 ‖Xk = x, J = 0 ] = E0[ Dk+1 ] für alle x mit x1 ∈ IR und x2 > 0. Analyse Berechne q(x) = E0[ d(x + sU) ] d(x) = ∫ · · · Bis jetzt: Mo …