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We analyze S: when the verifier is corrupted, the views of Z in the hybrid and ideal interactions are identically distributed. When the prover is corrupted, the only difference is that S may fail with probability 2−k; conditioned on non­failure, the views are identical. Therefore the views are statistically indistingui...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/40bda230e290076f112388b0d14497e7_l11.pdf
. Receive (sid, T, R, v1, . . . , vm) from (sid, T ). 2. Receive (sid, R, T, i ∈ {1, . . . , m} from sid, R). 3. Output (sid, vi) to (sid, R). 4. Halt. To realize F 2 ot, we can use the protocol from [EGL85]: let F be a family of trapdoor permutations and let B be a hard­core predicate for F . Then the protocol i...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/40bda230e290076f112388b0d14497e7_l11.pdf
(t0, t1). S works as follows: obtain vi from F 2 ot, give the ideal­process input i to A, select (f, f −1) from F and give the pair to A, receive (y0, y1) from A where yi = f (xi) and y1−i = x1−i for random x0, x1 due to semi­honesty, and send (t0, t1) to A where ti = vi ⊕ B(xi) and t1−i is random. We now analyze ...
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standard functionality;” i.e. it has a “shell” and a “core,” where the core does not know who is corrupted. Our protocol evaluates the core only. • F is written as two­input ⊕ (addition mod 2) and ∧ (multiplication mod 2) gates. • The circuit has 5 types of input lines: inputs of P0, inputs of P1, inputs of S, rando...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/40bda230e290076f112388b0d14497e7_l11.pdf
gate, P0 and P1 use F 4 as follows: P0 chooses c0 at random, and acts as the sender with input ot v00 = a0b0 + c0 v01 = a0(1 − b0) + c0 v10 = (1 − a0)b0 + c0 v11 = (1 − a0)(1 − b0) + c0 while P1 plays the receiver with input (a1, b1) and sets the output to be c1. It is easy to verify that c0 ⊕ c1 = (a0 ⊕ a1)(b0 ...
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P1 receives the corresponding information, plus random shares of all intermediate values from Fot. This is also easy to simulate. • Upon corruption, it’s easy to generate local state that is consistent with the protocol. � Some remarks: there is a protocol by Yao that works in a constant number of rounds, which can...
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zero­ knowledge, statements about those values. Here is the formal definition of the F R cp functionality for a given poly­time relation R: 1. Upon receiving (sid, C, V, commit, w) from (sid, C), add w to the list W of committed values, and output (sid, C, V, receipt) to (sid, V ) and S. 2. Upon receiving (sid, C, ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/40bda230e290076f112388b0d14497e7_l11.pdf
F Rp zk , where Rp = {((x, A), (W, R)) : W = w1 · · · wn, A = a1 · · · an, R = r1 · · · rn, R(x, W ) = 1 and ai = Com(wi, ri) for all i.} 4. Upon receiving (sid, C, V, (x, A), 1) from Fzk , V verifies that A agrees with its local list A, and if so, Rp outputs (sid, C, V, x). Theorem 1 The above protocol realizes...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/40bda230e290076f112388b0d14497e7_l11.pdf
) for random r. In the proof phase, S obtains from Fcp a (sid, C, V, x) message, and simulates for A the message (sid, C, V, (x, A)) from Fzk , where A is the list of commitments that S has generated so far. Rp Let’s analyze S: for a corrupted committer, the simulation is perfect. For a corrupted verifier, the only ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/40bda230e290076f112388b0d14497e7_l11.pdf
, commit, r) to Fcp. Q0 receives r1 from Q1, and sets r = r0 ⊕ r1. 2. Commit to Q1’s randomness. Q0 receives (sid.1, Q1, Q0, receipt) from Fcp and sends a random value s0 to Q1. 3. Receive the input x in the ith invocation. Q0 sends (sid.0, Q0, Q1, commit, x) to Fcp. Let M be the list of messages seen so far. Q0 r...
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A there exists a semi­honest adversary S such that for any envi­ ronment Z we have: EXECP,S,Z ≈ EXECQ,A,Z . Fcp Corollary 1 If protocol P securely realizes F for semi­honest adversaries then Q = C(P ) securely realizes F in the Fcp­hybrid model for malicious adversaries. Proof Sketch: We will skip the details of th...
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composable two­ party and multi­party secure computation. STOC 2002, pages 494–503, 2002. [EGL85] Shimon Even, Oded Goldreich, and Abraham Lempel. A randomized protocol for signing con­ tracts. Communications of the ACM, 28(6):637–647, June 1985. [GMW87] O. Goldreich, S. Micali, and A. Wigderson. How to play any men...
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6.241 Dynamic Systems and Control Lecture 8: Solutions of State-space Models Readings: DDV, Chapters 10, 11, 12 (skip the parts on transform methods) Emilio Frazzoli Aeronautics and Astronautics Massachusetts Institute of Technology February 28, 2011 E. Frazzoli (MIT) Lecture 8: Solutions of State-space Models F...
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0] = C [0]x0 y [1] = C [1] A[0] x[0] x[0] = x0 x[1] = A[0] x[0] x[2] = A[1] A[0] x[0] y [2] = C [2] A[1] A[0] x[0] . . . . . . y [k] = C [k] Φ[k, 0] x[0] x[k] = Φ[k, 0] x[0] where we defined the state transition matrix Φ[k, �] as � Φ[k, �] = A[k − 1] A[k − 2] . . . A[l], k > � ≥ 0 I , k = � E. Frazzoli (M...
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� . . . B[k − 1] , ⎡ ⎤ u[0] u[1] ⎥ U = ⎢ ⎥ ⎢ ⎣ . . . ⎦ u[k − 1] . The output is y [k] = C [k]Γ[k, 0]U[k, 0]. E. Frazzoli (MIT) Lecture 8: Solutions of State-space Models Feb 28, 2011 4 / 19 Summary (DT) In general, state/output trajectories of a DT state-space model can be computed as: x[k] = Φ[k, 0]x...
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t Φ(t, τ ) = A(t)Φ(t, τ ), Φ(t, t) = I . The function Φ can in general be computed numerically, integrating a differential equation in n unknown functions, with n initial conditions (assuming x ∈ Rn). Then, x(t) = Φ(t, t0)x0, and y (t) = C (t)Φ(t, t0)x0. E. Frazzoli (MIT) Lecture 8: Solutions of State-space Model...
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Φ(t, τ )B(τ )u(τ )]τ =t = A(t) Φ(t, τ )B(τ )u(τ ) dτ + B(t)u(t) = A(t)x(t) + B(t)u(t). t0 Similarly for the output. E. Frazzoli (MIT) Lecture 8: Solutions of State-space Models Feb 28, 2011 8 / 19 Further properties of the state transition function Φ(t2, t0) = Φ(t2, t1)Φ(t1, t0). Look up on the lecture notes...
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i.e., r = T −1x. This is called a similarity transformation. The standard state-space model can be written as Tr + = ATr + Bu y = CTr + Du i.e., (T −1AT )r + (T −1B)u = ˆ r + = y = (CT )r + Du = ˆ C r + ˆ Du Ar + ˆ Bu E. Frazzoli (MIT) Lecture 8: Solutions of State-space Models Feb 28, 2011 11 / 19 Modal Coor...
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shows that αi = wi i-th mode. �x0 is the contribution of the initial condition to the i=1 E. Frazzoli (MIT) Lecture 8: Solutions of State-space Models Feb 28, 2011 13 / 19 Diagonalization of the system If T = V = matrix of eigenvectors, then V −1AV = Λ (prove by AV = V Λ). Decoupled system for each mode. E. F...
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44 1.21. Bialgebras. Let C be a finite monoidal category, and (F, J) : C → Vec be a fiber functor. Consider the algebra H := End(F ). This algebra has two additional structures: the comultiplication Δ : H → H ⊗ H and the counit ε : H k. Namely, the comultiplication is defined by the formula → Δ(a) = α−1 Δ(a)), F,F (...
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.1 is called a bialgebra. Thus, Theorem 1.21.1 claims that the algebra H = End(F ) has a natural structure of a bialgebra. Now let H be any bialgebra (not necessarily finite dimensional). Then the category Rep(H) of representations (i.e., left modules) of H and its subcategory Rep(H) of finite dimensional represent...
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Proof. Straightforward from the above. � Theorem 1.21.3 is called the reconstruction theorem for finite dimen­ sional bialgebras (as it reconstructs the bialgebra H from the category of its modules using a fiber functor). Exercise 1.21.4. Show that the axioms of a bialgebra are self-dual if H is a finite dimensional ...
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G (not necessarily finite). · Exercise 1.21.6. Let H be a k-algebra, C = H −mod be the category of H-modules, and F : C → Vec be the forgetful functor (we don’t assume finite dimensionality). Assume that C is monoidal, and F is given a monoidal structure J. Show that this endows H with the structure of a bialgebra,...
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additional structure on the bialgebra H = End(F ) from the previous subsection in the case when the category C has right duals. In this case, one can define a linear map S : H H by the formula → S(a)X = a∗ X ∗ , where we use the natural identification of F (X)∗ with F (X ∗). Proposition 1.22.1. (“the antipode axio...
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F (X) � F (X) ⊗ F (X)∗ ⊗ F (X) F (X) Id � � F (X) η1 � � F (X) JX,X∗ F (coevX ) � � � F (X ⊗ X ∗) ⊗ F (X) ηX⊗X∗ F (coevX ) � � � F (X ⊗ X ∗) ⊗ F (X) Id � � F (X) � evF (X) J −1 X,X∗ � � F (X) ⊗ F (X)∗ ⊗ F (X), for any η ∈ End(F ). Namely, the commutativity of the upper and the lower square fol­ lows from...
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sense: if H is a finite dimensional bialgebra with antipode SH , then the bialgebra H ∗ also admits an antipode SH ∗ = S∗ H . The following is a “linear algebra” analog of the fact that the right dual, when it exists, is unique up to a unique isomorphism. Proposition 1.22.4. An antipode on a bialgebra H is unique i...
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� b b 3 j . j Then using the definition of the antipode, we have � � S(ab) = S(a 2 1 i b)ai S(ai ) = 3 S(a 11 3 i bj )ai bj S(bj )S(ai ) = S(b)S(a). 22 3 i i,j Thus S is an antihomomorphism of algebras (which is obviously unital). The fact that it is an antihomomorphism of coalgebras then follows using the...
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. Proof. Part (i) follows from the antipode axiom and Proposition 1.22.5. Part (ii) follows from part (i) and the fact that the operation of taking � the left dual is inverse to the operation of taking the right dual. Remark 1.22.7. A similar statement holds for finite dimensional co­ modules. Namely, if X is a fini...
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(Rep(H), Forget) are mutually inverse bijections between 1) finite tensor categories C with a fiber functor F , up to monoidal equivalence and isomorphism of monoidal functors; 2) finite dimensional Hopf algebras over k up to isomorphism. Proof. Straightforward from the above. 49 � Exercise 1.22.12. The algebra of...
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ike elements is grouplike. In particular, grouplike elements of any Hopf algebra H form a group, denoted G(H). Show that this group can also be defined as the group of isomorphism classes of 1­ dimensional H-comodules under tensor multiplication. Proposition 1.22.15. If H is a finite dimensional bialgebra with an ant...
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collapsing the first two factors using the antipode axiom, we get b = a. Thus a = S�(S(a)) and thus a ∈ H1, so H = H1, a contradiction. � Exercise 1.22.16. Let µop and Δop be obtained from µ, Δ by permu­ tation of components. (i) Show that if (H, µ, i, Δ, ε, S) is a Hopf algebra, then Hop := (H, µop, i, Δ, ε, S−1), ...
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to generalize the reconstruction theory to the situation when the category C is not assumed to be finite. Let C be any essentially small k-linear abelian category, and F : C → Vec an exact, faithful functor. In this case one can define the space Coend(F ) as follows: Coend(F ) := (⊕X∈CF (X)∗ ⊗ F (X))/E where E is s...
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�CF (X)∗ ⊗ X. For X, Y ∈ C and f ∈ Hom(X, Y ), let jf : F (Y )∗ ⊗ X → F (X)∗ ⊗ X ⊕ F (Y )∗ ⊗ Y ⊂ Q 51 be the morphism defined by the formula jf = Id ⊗ f − F (f )∗ ⊗ Id. −→ Let I be the quotient of Q by the image of the direct sum of all jf . In other words, I = lim (F (X)∗ ⊗ X). The following statements are easy ...
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since End(F ) may now be infinite dimensional, the algebra End(F ⊗ F ) is in general isomorphic not to the usual tensor product End(F )⊗End(F ), but rather to its completion End(F )⊗� End(F ) with respect to the inverse limit topology. Thus the comultiplication of End(F ) is a continuous linear map Δ : End(F ) → En...
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functors, and Hopf algebras over k, up to isomorphism. Remark 1.23.3. This theorem allows one to give a categorical proof of Proposition 1.22.4, deducing it from the fact that the right dual, when it exists, is unique up to a unique isomorphism. Remark 1.23.4. Corollary 1.22.15 is not true, in general, in the infi­...
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fine the coproduct on H by setting Δ(x) = x ⊗ 1 + 1 ⊗ x for all x ∈ g. It is easy to show that this extends to the whole H, and that H equipped with this Δ is a Hopf algebra. Moreover, it is easy to see that the tensor category Rep(H) is equivalent to the tensor category Rep(g). This example motivates the following...
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Lie algebra over a field of characteristic zero. Show that Prim(U (g)) = g. Hint. Identify U (g) with Sg as coalgebras by using the symmetriza­ tion map. Example 1.24.5. (Taft algebras) Let q be a primitive n-th root of unity. Let H be the algebra (of dimension n2) generated over k by g and x satisfying the followi...
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of type h, g in a Hopf algebra H. (i) Show that ε(x) = 0, S(x) = −h−1xg−1 . (ii) Show that if a, b ∈ H are grouplike elements, then axb is a skew- primitive element of type (ahb, agb). Example 1.24.9. (Nichols Hopf algebras) Let H = C[Z/2Z]�∧(x1, ..., xn), where the generator g of Z/2Z acts on xi by gxig−1 = −xi. ...
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software studio an overview of Rails Daniel Jackson 1 what is Rails? an application framework › full stack: web server, actions, database a programming environment › eg, rake (like make), unit testing an open-source community › many plugins 2 history of Rails genesis in Basecamp › project management tool...
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? › next slide... 11 Pull request by jeyb on GitHub. 12 in summary... rich environment many libraries code generation helpful community friendly online guides invisible magic quirky conventions no static checking masking of failures ? 13 an easy life? or a deadly cocktail? 14 actually, neither every too...
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MATH 18.152 COURSE NOTES - CLASS MEETING # 9 18.152 Introduction to PDEs, Fall 2011 Professor: Jared Speck Class Meeting # 9: Poisson’s Formula, Harnack’s Inequality, and Liouville’s Theorem 1. Representation Formula for Solutions to Poisson’s Equation We now derive our main representation formula for solution’s to Poi...
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− ) ˆN σ Φ x σ dσ. Recall also that (1.0.4) where (1.0.5) and (1.0.6) G(x, y) = Φ(x − y) − φ(x, y), ∆yφ(x, y ) = 0, x ∈ Ω, ( G x, σ ) = 0 when x Ω and σ ∂Ω. ∈ ∈ The expression (1.0.3) is not very useful since don’t know the value of fix this, we will use Green’s identit and recalling that ∆yφ x, y ) = ( y. Applying Gree...
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P x, σ) = ∇ ˆN G(x, σ from (1.0.2) a technique works for special domains. ( def Warning 2.0.1. Brace yourself for a bunch of tedious computations that at the end of the day will lead to a very nice expression. ( ) The basic idea is to hope that φ x, y from the decomposition G x, y ( ) φ x, y , where the Newtonian that ...
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which implies that (2.0.9) Simple algebra then leads to 1 4π∣x − y ∣ = q ∗ − ∣ 4π∣x y . (2.0.10) ∗ ∣ x − y 2 ∣ = q2 x y 2. ∣ − ∣ When ∣y∣ = R we , use x∗ 2 ∣ ∣ − (2.0.10) to compute 2x∗ ⋅ y + R2 = ∣x∗ − y that ∣2 = q2∣ − y∣2 = q2 x (∣ x∣2 − 2x ⋅ y + R , ) 2 the Euclidean dot product. Then performing simple algeb ra, it...
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2 x. φ(x, y ) = 1 4π x R ∣ ∣ R 2 − ∣ x∣2 x y ∣∣ , ( φ 0, y ) = 1 4πR , where we took a limit as x → 0 Next, using (2.0.8), we have in (2.0.16) to derive (2.0.17). (2.0.18) G(x, y) = − (2.0.19) ( G 0, y ) = − 1 4π x − y∣ ∣ + 1 4π ∣x∣∣ R 2 R x 2 x − y ∣ ∣ ∣ , 1 ∣ 4π y ∣ + 1 4πR . For future use, we also compute that (2.0...
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∣ ∣ x 2 R2 ) . ( − 1 = = − − x σ ∣ − ∣3 σ π 4 x + ∣ ∣ 1 x 2 4π R2 R2 − x 2 x σ ∣ ∣ − σ∣3 ∣x Using (2.0.22) and the fact that (2.0.23) ∇ ( ) ( ˆN σ G x, σ ˆ σ ( ) = 1 N R σ, w e deduce def ) = ∇ ( σG x, σ N σ ) ⋅ ( ) = ˆ R 2 − 4π ∣ ∣ x 2 R 1 ∣ − ∣ x σ 3 . 4 MATH 18.152 COURSE NOTES - CLASS MEETING # 9 Remark 2.0.2. If ...
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. R2 x−p R − (x ∣2 p ) − (y − p )∣ , ≠ x p, Furthermore, if x ∈ BR(p) and σ ∈ ∂BR(p), then (2.0.25c) ∇ ( ˆN (σ)G x, σ ) = ∣ − 2 R2 − ∣x p 4πR 1 ∣ − ∣ 3 x σ . We can now easily derive a representation formula for solutions to the Laplace equation on a ball. Theorem 2.1 (Poisson’s formula). Let BR(p) ⊂ = ( ) ( x3 , x2, x...
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g σ ∣ − σ dσ, n Proof. The identity (2.0.27) follows immediately from Theorem 1.1 and Lemma 2.0.1. (cid:3) 3. Harnack’s inequality We will now use some of our tools to prove a famous inequality for Harmonic functions. The theorem provides some estimates that place limitations on how slow/fast harmonic functions are all...
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egativity of (i.e. σ we have that ∣x R x σ ∣ ≤ e g, w − ∣ ≤ deduce R ), ∈ ∣ + ∣ x R. that (3.0.31) ) ∂BR(0 Now recall that by the mean value property, we have that u(x) ≤ R R2 + ∣x − ∣ x ∣ 1 ∣2 4πR ∫ ( ) g σ dσ. (3.0.32) u( ) = 0 1 4πR2 Thus, combining (3.0.31) and (3.0.32), we have ( ) g σ dσ. (0) ∫ ∂BR that (3.0.33) ...
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� ∈ ∣ def = (3.0.34) Rn−2(R − ∣ ∣) x − ( + ∣ ∣) R x n 1 v(0) ≤ v ) (x ≤ − ( + ∣ n 2 R R ∣) ∣ − ( x R ∣) x n−1 ) v(0 . Allo wing R (and therefore u is to → ∞ o). in (3.0.34), we conclude that v x ( ) = ( ) 0 v . Th us, v is a constant-valued function 6 MATH 18.152 COURSE NOTES - CLASS MEETING # 9 To handle the case u(x...
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18.413: Error­Correcting Codes Lab February 24, 2004 Lecturer: Daniel A. Spielman Lecture 6 6.1 Introduction Begin by describing LDPC codes, and how they are described by many local constraints. Point out that random graphs locally look like trees (from the birthday paradox), and so we will learn to do belief p...
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| Ppost [X1 = a1|Y1Y2 = b1b2] = cb1,b2Pprior [X1 = a1] Pext [X1 = a1 Y1Y2 = b1b2] , | so it suffices to prove Lemma 6.2.2. Pext [X1 = a1|Y1Y2 = b1b2] = cb1,b2Pext [X1 = a1 Y1 = b1] Pext [X1 = a1 Y2 = b2] . | | 6­1 Lecture 6: February 24, 2004 6­2 Proof. We begin by examining the right­hand­sides. We have Pext [X1 ...
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2 = b1b2 X1 = a1] = cb1,b2 � | P [Y1Y2 = b1b2|X1X2 = a1a2] P [X2 = a2 X1 = a1] | = cb1,b2 a2:(a1,a2)∈C � a2:(a1,a2)∈C = cb1,b2P [Y1 = b1|X1 = a1] | P [Y1 = b1|X1 = a1] P [Y2 = b2 X2 = a2] P [X2 = a2 X1 = a1] | � a2:(a1,a2)∈C P [Y2 = b2|X2 = a2] P [X2 = a2 X1 = a1] . | To conclude, we observe that this last te...
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1X3 = a1a3|X2 = a2] = P [X1 = a1 X2 = a2] P [X3 = a3 X2 = a2] . | | In this case, we can say that all the information that X3 contains about X1 is transmitted through X2. This fact can be used to simplify the belief computation. Lemma 6.4.1. Pext [X1 = a1|Y2Y3 = b2b3] = � a2:(a1,a2)∈C P [X2 = a2|X1 = a1] Pext [X2...
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3 = b3] | | 6.5 Trees A hypergraph is given by a collection of vertices x1, . . . , xn and a collection of edges e1, . . . , em, where each ei ⊆ {x1, . . . , xn}. A path in a hypergraph is a sequence of vertices xi1, . . . , xik such that each consecutive pair in the sequence lie in some edge. That is, for each 2 ≤ ...
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6.763 Applied Superconductivity Lecture 1 Terry P. Orlando Dept. of Electrical Engineering MIT September 8, 2005 Outline • What is a Superconductor? • Discovery of Superconductivity • Meissner Effect • Type I Superconductors • Type II Superconductors • Theory of Superconductivity • Tunneling and the Josephs...
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physics/laureates 5 Discovery of Superconductivity “As has been said, the experiment left no doubt that, as far as accuracy of measurement went, the resistance disappeared. At the same time, however, something unexpected occurred. The disappearance did not take place gradually but (compare Fig. Please see: Figu...
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the magnetic vortices (blue cylinders feel a force Lorentz force that pushes the ( vortices at right angles to the current flow. This movement dissipates energy and produces resistance from D. J. Bishop et al., Scientific American, 48 (Feb. 1993)]. [ ) i Upper Critical Fields of Type II Superconductors Image rem...
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Interaction Image removed for copyright reasons. The origin of superconductivity in conventional superconductors http://www.physics.carleton.ca/courses/75.364/mp-2html/node16.html 13 14 •7 Cooper Pairs & Energy Gap Images removed for copyright reasons. Please see: Figure 5 and figure 6 from http://nobelprize.o...
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Tunneling between a normal metal and another normal metal or a superconductor Images removed for copyright reasons. Please see: Figures 3 and 4 from http://nobelprize.org/physics/laureates/ 1973/giaever-lecture.html http://www.nobel.se/physics/laureates/1973/giaever-lecture.pdf 18 •9 Tunneling between two superco...
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important break-through in the discovery of superconductivity in ceramic materials" Images removed for copyright reasons. ____________________ Please see: http://nobelprize.org/physics/ __________________ laureates/1987/index.html Image removed for copyright reasons. ______________ Please see: Figure 1.5 from ht...
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the Earth’s magnetic field) • Quantum Computing Picture source: http://www.superconductors.org Massachusetts Institute of Technology 25 26 •13 Large-Scale Applications Technical Points Application Power cables Current Limiters Transformers Motors/Generators Energy Storage Magnets NMR magnets (MRI) Cavitie...
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, low cost of the raw materials, and absence of weak-link limitations that allows the use of mature powder-in-tube technology to fabricate long wires. The inclusion of MgB2 presentations in the symposium will bring together both communities and will encourage the discussion of problems that are common to all superco...
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over existing computers • Capable of reversible computation • e.g. Can factorize a 250-digit number in seconds while an ordinary computer will take 800 000 years! Current Research in my group focuses on Quantum Computation using Superconductors Massachusetts Institute of Technology 34 •17 The “Magic” of Quantum...
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Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Lecture 3: Continuous Dependence On Parameters1 Arguments based on continuity of functions are common in dynamical system analysis. They rarely apply to...
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) − a(¯ x2)| ∃ M |x1 − ¯ x2| ¯ x1, ¯ [t0, tf ] ∈� Rn of (3.1). The proof of both for all ¯ x2 from a neigborhood of a solution x : existence and uniqueness is so simple in this case that we will formulate the statement for a much more general class of integral equations. Theorem 3.1 Let X be a subset of Rn containin...
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. When a does not depend on the third argument, we have the standard ODE case x˙ (t) = a(x(t), t). In general, Theorem 3.1 covers a variety of nonlinear systems with an infinite dimensional state space, such as feedback interconnections of convolution operators and memoryless nonlinear transformations. For example,...
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, t)|d� ∃ K|xk (� ) − xk−1(� )|d� t0 ∃ 0.5 max {|xk (t) − xk−1(t)|}. t�[t0,tf ] Therefore one can conclude that max {|xk+1(t) − xk (t)|} ∃ 0.5 max {|xk (t) − xk−1(t)|}. t�[t0,tf ] t�[t0,tf ] Hence xk (t) converges exponentially to a limit x(t) which, due to continuity of a with respoect to the first argument, ...
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≤ Br (¯ x1, ¯ x0), t0 ∃ � ∃ t ∃ t1, and x, �, t)| ∃ m(t) � ¯ where the functions K(·) and M (·) are integrable over [t0, t1]. x ≤ Br (¯ x0), t0 ∃ � ∃ t ∃ t1, |a(¯ 4 3.2 Continuous Dependence On Parameters In this section our main objective is to establish sufficient conditions under which solutions of ODE depend con...
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K ≤ R such that |a(¯ x, �, t, q)| ∃ M � ¯ x ≤ X d, t0 ∃ � ∃ t ∃ tf , q ≤ (q0 − d, q0 + d); (c) for every � > 0 there exists � > 0 such that |x0(q1) − ¯ ¯ x0(q2)| ∃ � � q1, q2 ≤ (q0 − d, q0 + d) : |q1 − q2| < �, (3.6) (3.7) |a(¯ x, �, t, q1) − a(¯ x, �, t, q2)| ∃ � � q1, q2 ≤ (q0 − d, q0 + d) : |q1 − q2| < �, ¯ x ...
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to derive qualitative statements about nonlinear systems. 5 3.3.1 Differential flow Consider a time-invariant autonomous ODE where a : Rn ∈� Rm is satisfies the Lipschitz constraint x˙ (t) = a(x(t)), |a(¯ x1) − a(¯ x2)| ∃ M |x1 − ¯ x2| ¯ (3.8) (3.9) on every bounded subset of Rn . According to Theorem 3.1, this ...
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3.2, x : � ∈� Rn is a continuous function defined on an open subset � ∀ R × Rn . With x considered a parameter, t ∈� x(t, ¯ x) defines a family of smooth curves in Rn . When t is fixed, ¯ x) defines a continuous map form an open subset of Rn and with values in Rn . Note that x(t1, x(t2, ¯ x)) = x(t1 + t2, ¯ x) whenever ...
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< �. In other words, all solutions starting sufficiently close to an asymptotically stable equilibrium ¯x0 converge to it as t � ⊂, and none of such solutions can escape far away before finally converging to ¯x0. 6 Theorem 3.3 Let ¯x0 ≤ Rn be an asymptotically stable equilibrium of (3.8). The set x0 as t � ⊂ is an o...
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 4 DENIS AUROUX 1. Pseudoholomorphic Curves For (X 2n, ω) symplectic, J a compatible a.c.s. ∈ J ...
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if D∂ is onto. Theorem 1. The set J reg(X, β) of J ∈ J (X, ω) s.t. every simple J-holomorphic curve in class β is regular is a Baire subset. For J ∈ J reg(X, β), the subset of simple maps M∗ (X, J, β) ⊂ Mg,k(X, J, β) is smooth and oriented of dimension 2d. g,k Let g(·, ) = ω( , J ) be the associated Riemannian met...
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uniformly to x �→ (x, 0). But if you reparameterize to ˜x = nx, 1 ) and away from x = ∞, it converges uniformly to ˜ → (0, 1 x ). x˜ �→ ( n ˜ 1 ˜ x x, ˜ x nx The general idea is: • Identify bubbling regions where sup |dun| → ∞. • Away from those, ∃ convergent subsequences. • Near them, we can rescale the doim...
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defined as follows: given α1, . . . , αk ∈ H ∗(X), deg (αi) = 2d, � (8) �α1, . . . , αk�g,β = ev∗ 1α1 ∧ · · · ∧ evk ∗αk ∈ Q � [M g,k(X,J,β)] i ev−1(Ci)) (or rather #(ev X (choose Ci trans­ Equivalently, if we represent P D(αi) by a cycle Ci verse to the evaluation map), then the pairing is simply #([M g,k(X, ...
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β MIRROR SYMMETRY: LECTURE 4 3 is a universal curve C → M0,k (the fiber over a point is the corresponding curve), and J is now given by a map C → J (X, ω). The holomorphic curve equation becomes u : (Σ, j) X, du + J(u(z), z)du j = 0. We choose a superposition of a finite number perturbations, which break the symme...
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xed j, we have a ∂-operator on sections of u∗T X, and the cokernel of this operator is precisely H 1(Σ, u∗T X). Where du = 0, we have u∗T X = T Σ ⊕ u∗N and H 1(Σ, T Σ) is simply the deformations of j. There is also an obstruction bundle Obsu = H 1(Σ, u∗NΣ) if u is an immersion. We claim that we can define an obstruc...
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MIT 6.035 Semantic Analysis Martin Rinard Laboratory for Computer Science Massachusetts Institute of Technology Error Issue • Have assumed no problems in building IR • But are many static checks that need to be done as part of translation • Called Semantic Analysis Goal of Semantic Analysis • Ensure that program obe...
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parameter descriptor for x while (i < v.length) v[i] = v[i]+x; while < ldl len ldf sta ldf ldl + lda ldp ldf ldl field descriptor for v local descriptor for i parameter descriptor for x while (i < v.length) v[i] = v[i]+x; while < ldl len ldf sta ldf ldl + lda ldp ldf ldl field descriptor for v local descriptor for ...
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and return type declarations of superclass Load Instruction • What does compiler have? Variable name. • What does it do? Look up variable name. – If in local symbol table, reference local descriptor – If in parameter symbol table, reference parameter descriptor – If in field symbol table, reference field descriptor ...
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Int • Class C compatible with Class D if C inherits from D (but not vice-versa) Store Instruction • What does compiler have? – Variable name – Expression • What does it do? – Look up variable name. • If in local symbol table, reference local descriptor • If in parameter symbol table, error • If in field symbol tabl...
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MIT OpenCourseWare http://ocw.mit.edu 18.969 Topics in Geometry: Mirror Symmetry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIRROR SYMMETRY: LECTURE 2 DENIS AUROUX Reference for today: M. Gross, D. Huybrechts, D. Joyce, “Calabi-Yau Mani­ folds ...
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· } ∂ q(X, E) = ker∂/im∂ H ∂ ∂ ∂ Deforming J to a “nearby” J � gives (4) Ω1,0 ⊆ T ∗C = Ω1,0 ⊕ Ω0,1 J J � J is a graph of a linear map (−s) : Ω1 0 (acted , ,1 . J � is determined by Ω1 ,0 → Ω0 � J J J on by i) and Ω0,1 (acted on by i�). s is a section of (Ω1,0)∗ ⊗ Ω0,1 = T1,0 ⊗ Ω0,1 J 1,0X. If z1, . . . , z...
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, v�] giving it the structure of a differential graded Lie algebra. Proposition 1. J � is integrable ∂s + 2 Proof. We want to check that the bracket of two 0, 1 tangent vectors is still 0, 1, i.e. that 1 [s, s] = 0. ⇔ (6) [ ∂ � + ∂zk � s�k ∂ � + ∂ , ∂z� ∂zk s�k ∂ ∂z� � 1 0, ] ∈ T XJ � � 0,q⊗ q ΩX...
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(or, assuming that Aut(X, J) is discrete, we want that near J, ∃ a universal family X → U ⊂ MCX (complex manifolds, holomorphic fibers ∼= X) s.t. any family of integrable complex structures X � → S induces a map S → U s.t. X pulls back to X �). We have an action of the diffeomorphisms of X: for φ ∈ Diff(X) close to id...
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, this should satisfy (12) ∂s(t) + 1 2 [s(t), s(t)] = 0 In particular, s1 = dt t=0 solves ∂s1 = 0. We obtain an infinitesimal action of Diff(X): for (φt), φ0 = id , dt |t=0 = v a vector field, dφ | ds (13) d dt |t=0(−(∂φt)−1 ◦ ∂φt) = − d dt |t=0(∂φt) = −∂v This implies that first-order deformations are given as (...
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vij = −vji and vij + vjk = vik on Uijk. Cech 1-cocycle conditions in the sheaf of holomorphic tangent vector fields. Modding out by holomorphic functions ψi : Ui → Ui (which act by φij �→ ψj φij ψ−1) is precisely modding by the ˇ Cech coboundaries. Thus, Def 1(X, J) = ˇ This is precisely the ˇ i H 1(X, T X 1,0). = φi...
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). If it is zero, then there is an s2 s.t. ∂s2 + 1 2 [s1, s1] = 0, and the next obstructure is the class of [s1, s2] ∈ H 2(X, T X 1,0). We are basically attempting to apply by brute force the implicit function theorem. If it happens that H 2(X, T X) = 0, then the deformations are unobstructed and the moduli space o...
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} � X ⇒ Jt (17) αt = i � I,J||I|=p,|J|=q ( αIJ (t)dzi 1 t) ∧ · · · ∧ dzi ) t ) ∧ · · · ∧ dz( t t) ∧ dz( ( j j p q 1 d | Taking dt t=0, the result follows from the product rule. We mostly get (p, q) terms (t) � and a few (p + 1, q − 1), (p − 1, q + 1) forms (the latter from dt |t=0(dzik ). d
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Massachusetts Institute of Technology 18.413: Error­Correcting Codes Laboratory Professor Daniel A. Spielman Handout 0 February 3, 2004 Signing Up If too many people sign up for the course, I will perform a lottery among those who have signed up, and announce the results by email on Wednesday, Feb 4th. First Re...
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System Identification 6.435 SET 9 – Asymptotic distribution of PEM Munther A. Dahleh Lecture 9 6.435, System Identification 1 Prof. Munther A. Dahleh Central Limit Theorem (Generalization) • Basic Theorem II: Consider are both ARMA processes, possibly correlated, with underlying white noise (bounded 4th moment) Lect...
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leh Examples ARX: Lecture 9 6.435, System Identification 11 Prof. Munther A. Dahleh MA: Lecture 9 6.435, System Identification 12 Prof. Munther A. Dahleh ARMA: Lecture 9 6.435, System Identification 13 Prof. Munther A. Dahleh Comments: As , then . If , then the model structure is over parametrized so ( has pol...
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20.110/5.60 Fall 2005 Lecture #1 page 1 Introduction to Thermodynamics Thermodynamics: → Describes macroscopic properties of equilibrium systems → Entirely Empirical → Built on 4 Laws and “simple” mathematics 0th Law ⇒ Defines Temperature (T) 1st Law ⇒ Defines Energy (U) 2nd Law ⇒ Defines Entropy (S) 3rd La...
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• Intensive: Independent of the size of the system (T,p, V = V n ,…) The State of a System at Equilibrium: • Defined by the collection of all macroscopic properties that are described by State variables (p,n,T,V,…) [INDEPENDENT of the HISTORY of the SYSTEM] • For a one-component System, all that is required is “n...
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(g, 1 bar, 150 °C) initial state final state 20.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field 20.110/5.60 Fall 2005 Lecture #1 page 5 • Path: Seque...
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are in thermal equilibrium. Consequence of the zero’th law: B acts like a thermometer, and , , and are all C A B at the same “temperature”. 20.110J / 2.772J / 5.601JThermodynamics of Biomolecular SystemsInstructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field ...
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(Kelvin) ) = T K t C ( ( ° ) + 273.15 T = 0K corresponds to absolute zero ( t = − 273.15 ) C ° Better reference points used for the Kelvin scale today are T = 0K (absolute zero) and Ttp = 273.16K (triple point of H2O) Ideal Gases Boyle’s Law and the Kelvin scale ( lim → 0 p pV ) T = ⎡ ⎢ ⎢ ⎢ ⎣ pV ) ( lim → 0 p 273...
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work to surroundings ( the system. If the system does work on the w )0 then , that 0 if < . 0 ∆ > V w ∆ < V 0 > convention: • Heat: “q” That quantity flowing between the system and the surroundings that can be used to change the temperature of the system and/or the surroundings. Sign convention: If heat en...
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