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2(x − δx, y − δy) − 2 Z Z =⇒ arg min J(δx, δy) = arg max δx,δy δx,δy D E1(x − δx, y − δy)E2(x, y)dxdy + D Z Z E1(x − δx, y − δy )E2(x, y)dxdy Z Z D 2(x, y)dxdy E2 Since the first and third terms are constant, and since we are minimizing the negative of a scaled correlation objective, this is equivalent to m...
https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/63bf90743360a41d70eda346816d8304_MIT6_801F20_lec13.pdf
above looks similar to our BCCE constraint from optical flow! • Gradient-based methods are cheaper to compute but only function well for small deviations δx, δy . • Correlation methods are advantageous over least-squares methods when we have scaling between the images (e.g. due to optical setting differences): E1(x, y...
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match. Are there any issues with this approach? If parts/whole images of objects are obscured, this will greatly affect correlation computations at these points, even with proper normalization and offsetting. With these preliminaries set up, we are now ready to move into a case study: a patent for object detection and...
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training/template image to obtain boundary points. 3. Connect neighboring boundary points that have consistent directions. 4. Organize connected boundary points into chains. 5. Remove short or weak chains. 6. Divide chains into segments of low curvature separated by conrner of high curvature. 7. Create evenly-spac...
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and construct features for feature matching using a Histogram of Oriented Gradients (HoG) [1]. 4 • For running this framework at multiple scales/resolutions, we want to use different probes at different scales. • For multiscale, there is a need for running fast low-pass filtering. Can do so with rapid convolutions, w...
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ly • At different levels of resolution • Hexagonally, rather than on a square grid - there is a 4 π detection is used, and to break ties, we arbitrarily set 3 of the 6 inequalities as ≥, and the other 3 as >. advantage of work done vs. resolution. Here, hexagonal peak What is pose? Pose is short for position and ...
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_______________________ 6 MIT OpenCourseWare https://ocw.mit.edu 6.801 / 6.866 Machine Vision Fall 2020 For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms
https://ocw.mit.edu/courses/6-801-machine-vision-fall-2020/63bf90743360a41d70eda346816d8304_MIT6_801F20_lec13.pdf
Summary from last week • Linear systems f1(t) x1(t) f2(t) x2(t) a1f1(t)+ a2f2(t) a1x1(t)+ a2x2(t) • Translational & rotational mechanical elements & systems x(t) M f(t) K fv M x¨ + fvx˙ + Kx = f T (t) T(t) θ(t) q(t) J D K Figures by MIT OpenCourseWare. • Solving 1st order linear ODEs with constant coefficients Jω˙ + bω...
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s) input, output expressed as functions of new variable s Benefits: • Simplifies solution • • particularly useful in control s-domain offers additional insights 2.004 Fall ’07 Lecture 03 – Monday, Sept. 10 Laplace transform: definition Given a function f (t) in the time domain we define its Laplace transform F (s) as ...
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ω) = σ − jω σ2 + ω2 . Alternatively, we can represent the complex number s in polar form s = |s| ejφ, jω σ2 + ω2 where |s| = φ ≡ 6 s = atan (ω/σ) the phase of s. is the magnitude and 1/2 ¡ ¢ It is straightforward to derive 1 s = 1 |s| e− jφ ⇒ 2.004 Fall ’07 = 1 |s| 1 s ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ and 1 s = −6 s. Lecture 03 – Mond...
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a = 1 sec− 1 Nise Table 2.1 2.004 Fall ’07 Lecture 03 – Monday, Sept. 10 Laplace transforms of commonly used functions f(t) d(t) u(t) tu(t) tnu(t) e-atu(t) sin wtu(t) cos wtu(t) F(s) 1 1 s 1 s2 n! sn + 1 1 s + a w s2 + w2 s s2 + w2 Figure by MIT OpenCourseWare. Nise Table 2.1 2.004 Fall ’07 Lecture 03 – Monday, Sept...
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� δ(t) = 1; (unit energy) and + ∞ Z −∞ δ(t)f (t) = f (0) (sifting.) Properties of the Laplace transform Let F (s), F1(s), F2(s) denote the Laplace transforms of f (t), f1(t), f2(t), respectively. We denote L [f (t)] = F (s), etc. • Linearity L [K1f1(t) + K2f2(t)] = K1F1(s) + K2F2(s), where K1, K2 are complex constan...
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) = K1 s + 3 + K2 s + 5 . (2) That would be convenient because we know the inverse Laplace transform of the 1/(s + a) function (it’s a decaying exponential) and we can also use the linearity theorem to finally find f (t). All that’d be left to do would be to find the coefficients K1, K2. This is done as follows: first multip...
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ω(t) + bω(t) = Ts(t), where Ts(t) = T0u(t) (step function) and ω(t = 0) = 0 (no spin—down). Ts(t) ω(t) Taking the Laplace transform of both sides, J sΩ(s) + bΩ(s) = T0 s ⇒ Ω(s) = T0 b 1 s (J/b)s + 1 = T0 b 1 s(τ s + 1) , ´ where τ ≡ J/b is the time constant (see also Lecture 2). ³ We can now apply the partial fraction ...
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(t), where now Ts(t) is an arbitrary function, but still ω(t = 0) = 0 (no spin—down). Ts(t) ω(t) Proceeding as before, we can write Ω(s) = Ts(s) Js + b ⇔ Ω(s) Ts(s) = 1 Js + b . Generally, we define the ratio L output h L input i = Transfer Function; in this case, TF(s) = 1 Js + b . h We refer to the (TF)− i 1 of a sing...
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= f (t). x(t) M f(t) K fv Figures by MIT OpenCourseWare. F(s) 1 Ms2 + fvs + K X(s) Impedances X(s) = Forces . £ P x(t) ¤ £ P x(t) ¤ Kx(t) fv dx dt d2x dt2M M f(t) KX(s) fvsX(s) Ms2X(s) Figures by MIT OpenCourseWare. M f(t) 2.004 Fall ’07 Lecture 03 – Monday, Sept. 10 Summary • Laplace transform L [f (t)] ≡ F (s) = + ...
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Computational Ocean Acoustics • Ray Tracing • Wavenumber Integration • Normal Modes • Parabolic Equation 13.853 COMPUTATIONAL OCEAN ACOUSTICS 1 Lecture 7 Wavenumber Integration • Range-independent – Integral Transform solution • Exact depth-dependent solution – Global Matrix Approach – Propagator Matrix Approach – In...
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USTICS 6 Lecture 7 Blows up – k ,z large r < 1 inside layer m 13.853 COMPUTATIONAL OCEAN ACOUSTICS 7 Lecture 7 h zm-1 zm z e-g (Z-Z ) m-1 e-g (Zm-Z) e-g (Z-Z ) m-1 e-g (Zm-Z) gh ~ 1 gh à1 Layer m-1 m m+1 Layer m-1 m m+1 Interface m-1 m Interface m-1 m – Φm +Φm Diagonal Unstable 13.853 COMPUTATIONAL OCEAN ACOUSTICS – ...
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Layer 3 Lower halfsp. 2 ‘separate’ systems: No coupling between upper and lower matrices => No error propagation from bottom to top => Numerically stable Intfc 1 Intfc 2 Intfc 3 layer 2 x x x x x 0 x 0 00 0 0 0 0 x x xx x x xx 0 0 00 0 0 x x xx x x x x A A + 2 - 2 + 3 - 3 + 4 + 4 A A A B = x x x 0 0 0 13.853 Bloc...
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7 Evanescent Wave Tunneling See Fig. 4.14 and 4.15 in Jensen, Kuperman, Porter and Schmidt. Computational Ocean Acoustics. New York: Springer-Verlag, 2000. 13.853 COMPUTATIONAL OCEAN ACOUSTICS 13 Lecture 7 Evanescent Wave Tunneling See Fig. 4.14 in Jensen, Kuperman, Porter and Schmidt. Computational Ocean Acoustics...
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8.022 (E&M) - Lecture 2 Topics: Energy stored in a system of charges (cid:132) (cid:132) Electric field: concept and problems Gauss’s law and its applications (cid:132) Feedback: Thanks for the feedback! (cid:132) Scared by Pset 0? Almost all of the math used in the course is in it… (cid:132) (cid:132) Math...
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Qqr 2 r = − Coulomb . F = − 1 to r2? q (cid:71) d s ( W r 1 ) r → 2 = (cid:71) ∫ F I (cid:71) • ∫ ds = − r 1 r 2 ˆ Qqr ˆ i dr 2 r = Qq Qq − r r 1 2 (cid:132) Does this result depend on the path chosen? (cid:132) No! You can decompose any path in segments // to the radial direct...
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Q ( ) 0 no other charges: F=0 P 1 = (cid:71) (cid:71) ∫ F ds • = ∞ (cid:71) ∫ F I W = (cid:71) ds W + • = 1 2 + W + = 1 2 W 1 2 3 + + q q 1 2 r 12 + W 1 3 + 2 3 + q q 1 2 + r 12 q q 1 3 + r 13 q q 2 3 r 23 U i N = j N = = ∑ ∑ 1 2 i 1 = j q q i r ij j j 1 = i ...
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: the denser the lines, the stronger the field. Faucet Sink Demo + - Properties: (cid:132) (cid:132) eld lines never cross (if so, that’s where E=0) Fi They are orthogona to equipotential surfaces (wil see this ater). l l l September 8, 2004 8.022 – Lecture 1 8 4 Electric field of a ring of charge Pro...
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) 2 2 (cid:198) Integrating on r: 0(cid:198)R: (cid:71) r R = E 2 z σ π (cid:71) dE r R = = = ˆ z = ∫ r = 0 2 ( r + z dr 3 ) 2 2 ∫ r = 0 ˆ z 2 πσ ˆ zz ⎛ ⎜ | ⎝ 1 z | − 1 2 + R 2 z ⎞ ⎟ ⎠ P z R r September 8, 2004 8.022 – Lecture 1 11 Special case 1: R(cid:198)infinit...
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⎟ ⎠ What happens when h>>R? (cid:132) Physicist’s approach: P z R (cid:132) The disk will look like a point charge with Q= 2 σπ r2 (cid:198) E=Q/z (cid:132) Mathematician's approach: (cid:132) Calculate from the previous result for z>>R (Taylor expansion): (cid:71) E = 2 πσ ˆ zz ⎛ ⎜ | ⎝ 1 z ...
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a river The water velocity is described by (cid:71) v x y z v x v y v z v v v ( , ) + y z (cid:71) ˆ A An≡ Immerse a squared wire loop of area A in the water (surface S) , y + ≡ ˆ ˆ ) ( ≡ ˆ , , x x z (cid:132) Define the loop area vector as Q: how much water will flow through the loop? E.g.: What ...
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(cid:132) At any point in space, dA is perpendicular to the surface It points towards the “outside” of the surface (cid:132) (cid:132) Examples: . . . ˆn Intuitively: (cid:132) “dA i s oriented in such a way that if we have a hose inside the surface the flux through the surface will be positive” September 8, ...
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θ (cid:71) i E d S (cid:198) The total flux through the cylinder is zero! September 8, 2004 8.022 – Lecture 1 18 9 ΦE through closed empty surface Q1: Is this a coincidence due to shape/orientation of the cylinder? (cid:132) Clue: (cid:132) Think about interpretat on of through the surface… i (cid:132) ...
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i d A E = ∫ S Q 2 R d A = Q 2 R ∫ S d A = Q 2 R 4 R π 2 = 4 Q π September 8, 2004 8.022 – Lecture 1 21 ΦE through a generic surface What if the surface is not spherical S? Impossible integral? Use intuition and interpretation of flux! (cid:132) Version 1: (cid:132) (cid:132) Consider ...
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(cid:132) No, it’s useful only when the problem has symmetries September 8, 2004 8.022 – Lecture 1 23 Applications of Gauss’s law: Electric field of spherical distribution of charges Problem: Calculate the electric field (everywhere in space) due to a spherical distribution of positive charges or radius R. (NB...
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(cid:71) S ym m etry: E is constant on S and (cid:38) to dA . (cid:118)∫ S1 (cid:71) E dA E 4πr = 4πQ enclosed = i (cid:71) 1 2 → E = Q r 2 For r>R, sphere looks like a point charge! 2) Inside the sphere (r<R) S2 r + R S1 r Apply Gauss on a sphere S2 of radius r: (cid:71) A g ain : Φ = (cid:118)∫ S ...
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8.022 – Lecture 1 r + r R S1 R r 26 13 Another application of Gauss’s law: Electric field of spherical shell Problem: Calculate the electric field (everywhere in space) due to a positively charged spherical shell or radius R (surface charge density σ) Physicist’s solution:apply Gauss 1) Outside the sp...
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(cid:132) (cid:132) Trick #2: apply Gauss’s theorem (cid:132) Φtot = Φ + Φ + Φbottom Symmetry: E // z axis (cid:132) side top (cid:198) Φ =0 and Φtop = Φbottom side (cid:118) ∫ (cid:71) (cid:71) i E dA cylinder (cid:71) (cid:71) i E dA = 2 Φ = (cid:118) ∫ cylinder ∫ top = 4 Q π enclosed E...
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expansions are more than limits… ) and “massage” the result , or ln(1+x) or e x<<1 you until x September 8, 2004 8.022 – Lecture 1 29 Summary and outlook (cid:132) What have we learned so far: (cid:132) Energy of a system of charges Concept of electric field E (cid:132) (cid:132) To describe the effect ...
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34 1.13. Exactness of the tensor product. Proposition 1.13.1. (see [BaKi, 2.1.8]) Let C be a multitensor cate­ gory. Then the bifunctor ⊗ : C × C → C is exact in both factors (i.e., biexact). Proof. The proposition follows from the fact that by Proposition 1.10.9, the functors V ⊗ and ⊗V have left and right adjoin...
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the tensor product of bimodules. But the tensor product functor is not C-bilinear on morphisms (it is only R-bilinear). Definition 1.13.3. A multiring category over k is a locally finite k- linear abelian monoidal category C with biexact tensor product. If in addition End(1) = k, we will call C a ring category. Thus,...
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2 → X1 ⊗ I2 → 0. X1 ⊗ X2 → I1 ⊗ I2 → 0. Arguing similarly, we show that the sequence 0 → I1 ⊗ I2 → Y1 ⊗ Y2 is exact. This implies the statement. � Proposition 1.13.5. If C is a multiring category with right duals, then the right dualization functor is exact. The same applies to left duals. Proof. Let 0 X show ...
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the sequence 0 X ⊗ T → → Y ⊗ T → Z ⊗ T is exact, by the exactness of the functor ⊗T . This implies that the � sequence Z ∗ 0 is exact. Y ∗ X ∗ → → → Proposition 1.13.6. Let P be a projective object in a multiring cate­ gory C. If X ∈ C has a right dual, then the object P ⊗ X is projective. Similarly, if X ∈ C h...
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⊗ F ( ) • → • F (• ⊗ •), and F (1) = 1. (ii) A quasi-tensor functor (F, J) is said to be a tensor functor if J is a monoidal structure (i.e., satisfies the monoidal structure axiom). Example 1.14.2. The functors of Examples 1.6.1,1.6.2 and Subsection 1.7 (for the categories Vecω G) are tensor functors. The identity...
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, which is zero. So K ⊗ J = 0. Now tensoring the exact sequence 0 K → → → → 1 J and applying Proposition 1.13.1, we get that J = 0, so a = 0. 0 with J, � Let {pi}i∈I be the primitive idempotents of the algebra End(1). Let 1i be the image of pi. Then we have 1 = ⊕i∈I 1i. Corollary 1.15.2. In any multiring catego...
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ned, map Cij to Cji. Exercise 1.15.6. Prove Proposition 1.15.5. Proposition 1.15.5 motivates the terms “multiring category” and “multitensor category”, as such a category gives us multiple ring cate­ gories, respectively tensor categories Cii. Remark 1.15.7. Thus, a multiring category may be considered as a 2-cat...
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X is simple and X ⊗X ∗ = 0 (because the coevaluation morphism is nonzero) we obtain that X ⊗ X ∗ = X. So we have a surjective composition morphism 1 → X ⊗ X ∗ → X. From this and (1.15.1) we have a nonzero composition morphism 1 � X � 1. Since End(1) = k, → � this morphism is a nonzero scalar, whence X = 1. ∼ Cor...
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A) ⊗ (V �, W �, A�) = (V ⊗ V �, W ⊗ W �, A ⊗ A�), with obvious associativity isomorphisms, and the unit object (k, k, Id). Of course, this category has neither right nor left duals. 1.16. Grothendieck rings. Let C be a locally finite abelian category over k. If X and Y are objects in C such that Y is simple then we ...
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][Xk ⊗ Xp : Xl]. k � [Xj ⊗ Xp : Xk][Xi ⊗ Xk : Xl]. k Thus the associativity of the multiplication follows from the isomor­ � phism (Xi ⊗ Xj ) ⊗ Xp ∼= Xi ⊗ (Xj ⊗ Xp). Thus Gr(C) is an associative ring with the unit 1. It is called the Grothendieck ring of C. The following proposition is obvious. Proposition 1...
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G more details. → → → Here are some examples of groupoids. (1) Any group G is a groupoid G with a single object whose set of morphisms to itself is G. 40 (2) Let X be a set and let G = X × X. Then the product groupoid G(X) := (X, G) is a groupoid in which s is the first projection, t is the second projection, u...
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The unit object is 1 = ⊕x∈X 1x, where 1x is a 1-dimensional vector space which sits in degree idx in G. The left and right duals are defined by (V ∗)g = (∗V )g = Vg−1 . We invite the reader to check that the component subcategories C(G)xy are the categories of vector spaces graded by Mor(y, x). We see that C(G) is ...
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��nitely many non-zero diagonals, and morphisms are matrices of linear maps. The tensor product in this category is defined by the formula (1.17.2) (V ⊗ W )il = � Vij ⊗ Wjl, j and the unit object 1 is defined by the condition 1ij = kδij . The left and right duality functors coincide and are given by the formula...
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elian category C is said to be finite if it is equivalent to the category A − mod of finite dimensional modules over a finite dimensional k-algebra A. Of course, the algebra A is not canonically attached to the category C; rather, C determines the Morita equivalence class of A. For this reason, it is often better to ...
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the exactness property of F , and the condition that P is a generator (i.e., covers any simple object) translates into the property that F is faithful (does not kill nonzero objects or morphisms). Moreover, the algebra A = End(P )op can be alternatively defined as End(F ), the algebra of functorial endomorphisms of ...
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-vector spaces, such that F (1) = k, equipped with an isomorphism J : F ( ) ⊗ F ( ) F (• ⊗ •). If in addition J is a monoidal structure (i.e. satisfies the monoidal structure axiom), one says that F is a fiber functor. • → C • → G → Example 1.19.2. The forgetful functors VecG → Vec, Rep(G) Vec are naturally fiber f...
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it ε : C k such that → (i) Δ is coassociative, i.e., (Δ ⊗ Id) Δ = (Id ⊗ Δ) Δ ◦ ◦ as maps C C ⊗3;→ (ii) one has (ε ⊗ Id) Δ = (Id ⊗ ε) Δ = Id ◦ ◦ as maps C → C (the “counit axiom”). Definition 1.20.2. A left comodule over a coalgebra C is a vector → C ⊗ M (called the space M together with a linear map π : M coacti...
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Exercise 1.20.4. (i) Show that any coalgebra C is a sum of finite dimensional subcoalgebras. Hint. Let c ∈ C, and let (Δ ⊗ Id) ◦ Δ(c) = (Id ⊗ Δ) ◦ Δ(c) = � i ⊗ c 2 c 1 i ⊗ c 3 i . Show that span(c2 i ) is a subcoalgebra of C containing c. (ii) Show that any C-comodule is a sum of finite dimensional subco­ modu...
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1 2 3 4 5 6 7 8 MIT OpenCourseWare http://ocw.mit.edu 6.851 Advanced Data Structures Spring 2012 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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Statistical Models Parameter Estimation Fitting Probability Distributions MIT 18.443 Dr. Kempthorne Spring 2015 MIT 18.443 Parameter EstimationFitting Probability Distributions 1 Statistical Models Definitions General Examples Classic One-Sample Distribution Models Outline 1 Statistical Models Definitions...
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5 Statistical Models Definitions General Examples Classic One-Sample Distribution Models Statistical Models: General Examples Example 1. One-Sample Model X1, X2, . . . , Xn i.i.d. with distribution function F (·). E.g., Sample n members of a large population at random and measure attribute X E.g., n independent...
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measurement errors {Ej } are i.i.d. N(0, σ2), with σ2 > 0, unknown. Semi-Parametric Model: Symmetric measurement-error distributions with mean µ {Ej } are i.i.d. with distribution function G (·), where G ∈ G, the class of symmetric distributions with mean 0. Non-Parametric Model: X1, . . . , Xn are i.i.d. w...
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n cases i = 1, 2, . . . , n 1 Response (dependent) variable yi , i = 1, 2, . . . , n p Explanatory (independent) variables xi = (xi,1, xi,2, . . . , xi,p)T , i = 1, 2, . . . , n Goal of Regression Analysis: Extract/exploit relationship between yi and xi . Examples Prediction Causal Inference Approximation ...
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Fitting Probability Distributions 11 Statistical Models Definitions General Examples Classic One-Sample Distribution Models Outline 1 Statistical Models Definitions General Examples Classic One-Sample Distribution Models MIT 18.443 Parameter EstimationFitting Probability Distributions 12 Statistical Models D...
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t e = e E [X k ] = −λ λe x! |t=0 , k = 0, 1, 2, . . . tx e MIT 18.443 Parameter EstimationFitting Probability Distributions 14 Statistical Models Definitions General Examples Classic One-Sample Distribution Models Berkson (1966) Data: National Bureau of Standards experiment measruing 10, 220 times between succ...
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EstimationFitting Probability Distributions 16 Statistical Models Definitions General Examples Classic One-Sample Distribution Models Classic Probability Models Normal Distribution Two parameters: µ : mean σ2 : variance Probability density function: 1 (x − µ)2 σ2 , −∞ < x < ∞. f (x | µ, σ2) = √ 1 − 2πσ...
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stand and apply optimality principles in parameter estimation. Important Methodologies Method-of-Moments Maximum Likelihood Bayesian Approach MIT 18.443 Parameter EstimationFitting Probability Distributions 19 MIT OpenCourseWare http://ocw.mit.edu 18.443 Statistics for Applications Spring 2015 For information a...
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3.052 Nanomechanics of Materials and Biomaterials Thursday 03/08/07 I Prof. C. Ortiz, MIT-DMSE LECTURE 9: QUANTITATIVE ASPECTS INTRA- AND INTERMOLECULAR FORCES Outline : LAST LECTURE : INTRODUCTION TO INTRA- and INTERMOLECULAR FORCES.............................. 2 BRIDGING THE GAP BETWEEN LENGTH SCALES ..........
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- and intermolecular forces are electrostatic in origin → key to life on earth (e.g. water, cell membranes, protein folding, etc.), also materials science (what holds matter together?). -strength measured relative to the thermal energy (room temperature) : kBT= 4.1 ● 10-21 J : "ruler" -Classifications; primary or ch...
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.4 0.6 0.8 1 → -0.04 Interatomic or Intermolecular Separation Distance, r(nm) w(r) or U(r) → f(r) (one atom, ion, or molecule) -1st step is to assume a mathematical form of the potential rr repulsive regime attractive regime kc 0 ) N n ( F , e c r o F Tip-Sample Separation Distance, D (nm) W(D) → F(D) (net ...
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⎡ ⎢ ⎢ ⎣ ⎛ ⎜ ⎝ σ r ⎞ ⎟ ⎠ - σ r ⎛ ⎜ ⎝ 6 ⎞ ⎟ ⎠ ⎤ ⎥ ⎥ ⎦ (4) F ( LJ m = 6, n = 12 ) = -6A 12B + 13 r r 7 (5) B E = "binding energy" or "bond dissociation energy"; or depth of potential well r = distance at which U(r ) s F(r )= minimum = F r = equilibrium bond length e = distance at which U(r )= minimum, F(r )= 0 e r = σ =...
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ipole-dipole H-H, Schematic - - - - - - r r u θ r Q r Q u θ - - Cu2+ Cu2+ Cu2+ Cu2+ - - - - Q2 Q2 Q1 Q1 fixed dipole u r Q r Q u freely rotating dipole φ φ u1 θ1 u1 θ1 r r θ2 θ2 u2 u2 r r u2 u2 fixed dipole w(r)= - charge-induced dipole freely rotating dipoles rQ rQ α α dipole-induced dipole u θ u θ r r α...
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(r)= - o ⎞ Q Q 1 2 ⎟ 4 ⎠πε Qucos 4 πε o 2 2 2 2 -2 r ( 3 θ ⎞ ⎟ ⎠ k T B 2 u u 1 )2 4 πε o ( 6 4 πε o [ ⎛ ⎜ ⎝ ⎛ 2 Q u ⎜ ⎜ ) ⎝ ⎛ u u 2cos cos 1 2 ⎜ ⎝ ⎛ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎝ ⎛ ⎜ ⎜ ⎝ )2 4 πε o ( 2 1+ 3cos α )2 ( 2 2 α )2 3h ( 4 πε o 4 πε o 4 πε o 2 να 2 α ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ Q u 2 4 u ( ( ) r r r -6 -4 -6 2 ) 2...
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electric polarizability (C2 m2 J-1), r = distance between interacting atoms or molecules (m), kB= Boltzmann's constant = 1.381●10-23 J K-1, T = absolute temperature (K), h = Planck's constant = 6.626●10-34 J s, ν = electronic absorption (ionization) frequency (s-1), εo=dielectric permittivity of free space = 8.854 ●1...
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� ⎞ ⎟ ⎠ θ -2 r (1) w(r)= - Qucos 4 πε o r= charge-dipole separation distance (nm) u= electric dipole moment = ql (Cm) q= charge of dipole (C) l = separation distance between dipole charges(m) Q= charge of the ion (C) θ = dipole angle relative to horizontal Charge Charge Q=ze Q=ze dipole moment = ql dipole momen...
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(4πε ) o ⎞ ⎟ ⎠ -6 r = C induced -6 r (3) Dispersion Energy : Induced dipole-induced dipole : w(r) dispersion = ⎛ ⎜ ⎝ 2 -3h α ν 2 4(4πε ) o ⎞ ⎟ ⎠ -6 r = C -6 r (4) dispersion Biomolecular Adhesion : -controlled by bonds between molecular “ligands” and cell surface “receptors” which exhibit the “lock-n-key principle...
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, kBT 0.05-40 0.4 2-10 0.02-16 0.17 0.8-4 2.5 5 13 10-40 25 62 125 380 630 840 1 2 5 4-16 10 25 50 150 250 340 Interaction dispersion hydrophobic H-bonding ion-ion covalent Material metals ceramics and glasses semiconductors diamond water inert gases solid salt crystals alkanes...
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Coordinate Systems and Separation of Variables Revisiting the homogeneous wave equation… 2 +∇ ψ 1 2 c 2 ∂ ψ 2 t ∂ = 0 where previously in Cartesian coordinates, the Laplacian was given by 2 =∇ 2 ∂ x ∂ + 2 2 ∂ y ∂ + 2 2 ∂ z ∂ 2 We are now faced with a spherical polar coordinate system, with the motivation that we might...
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of r only… 1 2 r ⎡ ⎢ ⎣ ∂ r ⎛ ⎜ ⎝ 2 r ∂ r ∂ ⎞ +⎟ ⎠ 2 k ⎤ ψ ⎥ ⎦ 0)( r = Which is known to have the two solutions… rψ )( = rA ( )/ ⎧ ⎨ rB )/ ( ⎩ exp( exp( ikr ) ikr ) − (Makes sense as the field decreases with r as we expect) As we want to consider the sphere as the only source in the medium (radiation condition), we ca...
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Spherical Harmonics Re[ 1 1Y ] Im[ 1 1Y ] 0 1Y [ ] Various quadrupole orientations in terms of Spherical Harmonics Im[ 1 2Y ] Im[ 2 2Y ] Re[ 1 2Y ] Propagating fields from spherical boundaries rp ,( , ) φθ = ∞ ∑ n = 0 h h n n ( ( kr ka ) ) n ∑ m −= Y n m n ,( ) φθ ∫ ap ,( ′ ′ , φθ Y ) m n ′ ′ ( , φθ *) d Ω′ Where th...
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Spring 2009 Spring 2009 Lecture 9: Introduction to Lecture 9: Introduction to Program Analysis and Optimization Optimization Outline Introduction • Introduction • Basic Blocks • Common Subexpression Elimination • Copy Propagation C ti P • Dead Code Elimination ead Code at o • Algebraic Simplification • Sum...
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• Nodes Represent Computation – Each Node is a Basic Block – Basic Block is a Sequence of Instructions with o p a o a q u u o • No Branches Out Of Middle of Basic Block • No Branches Into Middle of Basic Block • Basic Blocks should be maximal – Execution of basic block starts with first instruction – Includes a...
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1; return s; Saman Amarasinghe 17 6.035 ©MIT Fall 2006 s = 0; s = 0; a = 4; i = 0; k == 0 k == 0 s = 0; a = 4; i = 0; k == 0 k == 0 b = 2; b = 1; b = 2; b = 1; i < n s = s + a*b; i + 1; i = i + 1; i i < n return s; s = s + a*b; i = i + 1; return s; Saman Amarasinghe 18 6.035 ©MIT Fall 2006 s = 0; s = 0; a = 4; i = 0;...
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Split point has multiple successors – conditional branch statements only split points • Merge point has multiple predecessors • Each basic block – Either starts with a merge point or its predecessor ends with a split point – Either ends with a split point or its successor i starts with a merge point t t i h t t...
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Introduced as part of instruction flattening a o – Introduced by optimizations/transformations Typically assigned to only once – Typically assigned to only once odu d a u o p p a • Program Variables – Declared in original program Declared in original program – May be assigned to multiple times – May transfer val...
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+v2  v3 v3+v4  v5 v5+v2  v6 v1+v2  t1 v1+v2  t1 v3+v4  t2 v5+v2  t6 Saman Amarasinghe 27 6.035 ©MIT Fall 1998 y Value Numbering Summary g • Forward symbolic execution of basic block • Each new value assigned to temporary – a=x+y; becomes a=x+y; t=a; – Temporary preserves value for use later in pro...
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y=a+b; – y=a+b; t=y; x=b; z=t y a+b; t y; x b; z t – Why? Because computes with symbolic values x=b; z=a+x becomes i • Finds common subexpressions even if variable • Finds common subexpressions even if variable that originally held the value was overwritten y a+b; – y=a+b; – y=a+b; t=y; y=1; z=t – Why? Because...
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are unnecessary S dead code elimination Saman Amarasinghe 32 6.035 ©MIT Fall 1998 Problems II • Expressions have to be identical – a=x+y+z; b=y+z+x; c=x*2+y+2*z–(x+z) • We use canonicalization • We use algebraic simplification Saman Amarasinghe 33 6.035 ©MIT Fall 1998 Copy Propagation p g py • Once ag...
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x+y a b = a+z c = x+y a = b • After CSE a = x+y t1 = a b = a+z t2 = b c = t1 c t1 a = b Saman Amarasinghe 37 6.035 ©MIT Fall 2006 Copy Propagation Example p g py p Basic Block After CSE Aft CSE a = x+y t1t1 = a Basic Block After CSECSE and C d Copy PProp a = x+y t1 t1 = a tmp to var tmp to var ...
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�{{t2}t2} Saman Amarasinghe 40 6.035 ©MIT Fall 2006 Copy Propagation Example p g py p Basic Block After CSE Aft CSE Basic Block After CSECSE and C d Copy PProp a = x+y t1t1 = a b = a+z t2 = b c = t1 tmp to var tmp to var t1  a t2  b t2  b a = x+y t1 t1 = a b = a+z t2 = b c = a var to set va...
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b = a+z t2 = b c = a a = b var to set var to set a {} bb {{t2}t2} Saman Amarasinghe 43 6.035 ©MIT Fall 2006 Outline Introduction • Introduction • Basic Blocks • Common Subexpression Elimination • Copy Propagation C ti P • Dead Code Elimination ead Code at o • Algebraic Simplification • Summary ...
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Needed Set { , } {a, b} Saman Amarasinghe 48 6.035 ©MIT Fall 2006 Basic Block After CSE and Copy Prop a = x+y t1 = a t1 b = a+z t2 = b c = a a = b Needed Set { , } {a, b} Saman Amarasinghe 49 6.035 ©MIT Fall 2006 Basic Block After CSE and Copy Prop a = x+y t1 = a t1 b = a+z c = a a = b Need...
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+y b = a+z c = a a = b Needed Set { , , {a, b, z}} Saman Amarasinghe 55 6.035 ©MIT Fall 2006 Outline Introduction • Introduction • Basic Blocks • Common Subexpression Elimination • Copy Propagation C ti P • Dead Code Elimination ead Code at o • Algebraic Simplification • Summary Algebraic Simplifi...
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– a * 2 – a * 8  a*a  a + a  a << 3 Saman Amarasinghe 60 6.035 ©MIT Fall 1998 Opportunities for Algebraic Simplification Algebraic Simplification • In the code • In the code – Programmers are lazy to simplify expressions – Programs are more readable with full expressions Programs are more readable with ful...
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Saman Amarasinghe 65 6.035 ©MIT Fall 1998 Canonical Format • Put expression trees into a canonical p format – Sum of multiplicands – Variables/terms in a canonical order Example – Example (a+3)*(a+8)*4  4*a*a+44*a+96 p – Section 12.3.1 of whale book talks about this Saman Amarasinghe 66 6.035 ©MIT Fall 19...
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Si S b li Symbolically Simulate Execution of Program f P – CSE and Copy Propagation go forward Dead Code Elimination goes backwards – Dead Code Elimination goes backwards l t E ti • Transformations stacked – Group of basic transformations work together – Often, one transformation creates inefficient code that ...
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6.772/SMA5111 - Compound Semiconductors Lecture 4 - Carrier flow in heterojunctions - Outline • A look at current models for m-s junctions (old business) Thermionic emission vs. drift-diffusion vs. p-n junction • Conduction normal to heterojunctions (current across HJs) Current flow: 1. Drift-diffusion 2. Balli...
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A*T 2 -qf e bm / kT (eqvAB / kT -1) * A : Thermionic emission coefficient bm : Barrier in metal, = f f b + (kT /q) ln(NC / N D ) Substituting for f id / A = q A*T 2 qNC N D e -qf bm and rearranging : -1) b / kT (eqvAB / kT = A*T 2 /qNC from which we see RTE C. G. Fonstad, 2/03 Lecture 4 - Slide 3 Drift-di...
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Ê kT ˆ Ë q ¯ = N D e -qf b / kT (eqvAB / kT -1). For this diode R + = De / w p , the diffusion velocity. pn C. G. Fonstad, 2/03 Lecture 4 - Slide 5 Comparing the results - Thermionic emission RTE = A*T 2 /qNC , modeling from metal side Drift-diffusion = m RDD p-n+ diode E pk , the drift velocity at x = 0. ...
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+ Egp - EgN| x Lecture 4 - Slide 8 -xp 0 xn Conduction parallel to heterojunctions Modulation doped structures N-n heterojunctions and accumulated electrons Modulation doped structure with surface pinning and HJ WHITE BOARD Back to foils Carrier scattering and mobilities in accumulated two-dimensional electr...
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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.02 Lecture 1. – Thu, Sept 6, 2007 Handouts: syllabus; PS1; flashcards. Goal of multivariable calculus: tools to handle prob...
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displayed example. Dot product. Definition: A� B� = a1b1 + a2b2 + a3b3 (a scalar, not a vector). Theorem: geometrically, A� · B� = |A�||B� | cos θ. Explained the theorem as follows: first, A� A� = A� 2 cos 0 = |A� 2 is consistent with the definition. | Next, consider a triangle with sides A�, B� , C� = A� − B� . Then th...
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than 90◦, zero if perpendicular. 2) detecting orthogonality. Example: what is the set of points where x + 2y + 3z = 0? (possible answers: empty set, a point, a line, a plane, a sphere, none of the above, I don’t know). Answer: plane; can see “by hand”, but more geometrically use dot product: call A� = �1, 2, 3�, ...
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� | sin θ (= 1/2 area of parallelogram). Could get sin θ using dot product to compute cos θ and 2 sin2 + cos2 = 1, but it gives an ugly formula. Instead, reduce to complementary angle θ� = π/2 − θ by considering A�� = A� rotated 90◦ counterclockwise (drew a picture). Then area of parallelogram = |A�||B� | sin θ = |A...
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b1 b3 c1 c3 � � +a3 � � � � � � b1 b2 c1 c2 � � . � � Geometrically: det( � B, � A, � C) = ± volume of parallelepiped. Referred to the notes for more about determinants. Cross-product. (only for 2 vectors in space); gives a vector, not a scalar (unlike dot-product). Definition: A� × B� = � � � � � � ˆj kˆ ˆı a...
https://ocw.mit.edu/courses/18-02-multivariable-calculus-fall-2007/649253ba60d11b0598cc58e9dcf58142_lec_week1.pdf