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} a T yi + b < 0, y1, . . . , yM } { by a hyperplane: i = 1, . . . , M homogeneous in a, b, hence equivalent to a T xi + b ≥ 1, i = 1, . . . , N, a T yi + b 1, ≤ − i = 1, . . . , M a set of linear inequalities in a, b Geometric problems 8–8 Robust linear discrimination (Euclidean) distance between hyperplan...
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margin of separation interpretation change variables to θi = λi/1T λ, γi = µi/1T µ, t = 1/(1T λ + 1T µ) invert objective to minimize 1/(1T λ + 1T µ) = t • • minimize t subject to � �� � θ � N i=1 θixi 0, � − 1T θ = 1, M i=1 γiyi γ � � � 2 ≤ 0, � t 1T γ = 1 optimal value is distance between convex ...
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0 v � � i = 1, . . . , N i = 1, . . . , M produces point on trade-off curve between inverse of margin 2/ classification error, measured by total slack 1T u + 1T v �2 and a � same example as previous page, with γ = 0.1: Geometric problems 8–12 Nonlinear discrimination separate two sets of points by a nonlinear ...
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4th degree polynomial Geometric problems 8–14 Placement and facility location N points with coordinates xi R2 (or R3) ∈ some positions xi are given; the other xi’s are variables for each pair of points, a cost function fij(xi, xj) • • • placement problem minimize � i6 =j fij(xi, xj) variables are position...
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MIT OpenCourseWare http://ocw.mit.edu 18.727 Topics in Algebraic Geometry: Algebraic Surfaces Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ALGEBRAIC SURFACES, LECTURE 9 LECTURES: ABHINAV KUMAR 1. Castelnuovo’s Criterion for Rationality Theorem 1...
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is C, but C moves in the pencil): after blowing up finitely many base points, we P1 with a fiber isomorphic to C ∼ P1 . Therefore, by the get a morphism X˜ ˜ Noether-Enriques theorem, X is ruled over P1 and X is rational (as is X). � → = ˜ Reduction 2: Let X be a minimal surface with q = p2 = 0. It is enough to sh...
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) = h0(−K) + h2(−K) (2) ≥ 1 + 1 2 K · 2K = 1 + K 2 = 1 · so |−K = ∅. Take a hyperplane section H of X. Then there is an n ≥ 0 | � nK| =� ∅ but |H + (n + 1)K| = ∅. Since −K ∼ an effective nonzero s.t. |H + · divisor, H K < 0 and H (H + nK) is eventually negative and H + nK is not effective. Let D ∈ |H + nK|: th...
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but |C + (n + 1)K| = ∅. Choosing D ∈ |C + nK| gives the desired divisor. · · · We now find the claimed E. Again, let H be a hyperplane section: if K H < 0, we can take E = H; if K H = 0, we can take K + nH for n >> 0; so assume K H > 0. Let γ = −K· > 0 so that (H + γK) K = 0. Also, · · · H · K2 (3) (H + γK)2...
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− m(D H). Thus, · mD is effective for large m, and we can take E ∈ |mD|. · 2 · Case 3 (K 2 > 0): Assume that there is no such D as in reduction 2, i.e. K · D ≥ 0 for every effective divisor D s.t. |K + D| = ∅. We will obtain a contradiction. Lemma 1. If X is a minimal surface with p2 = q = 0, K 2 > 0 and K D ≥ 0 ...
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nd, since K · D = −K 2 < 0). If D = C + C �, |K + C| = |−D + C| = |−C �| = ∅ since C � is effective. Also, C K < 0, contradicting the hypothesis. So D is irreducible, and similarly D is not a multiple. Furthermore, pa(D) = 2 D(D + K) + 1 = 1, showing (2). Next, we claim that the only effective divisor s.t. D + K = ∅...
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for all n ≥ 1. ⇒ We claim that adjuction terminates: if D is any divisor on X, then there is an integer nD s.t. |D + nK| = ∅ for n ≥ nD. To see this, note that (D +nK)·(−K) will eventually become negative. −K is represented by an irreducible curve of positive self-intersection, so by the useful lemma D +nK is not eff...
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of C. This would be a contradiction, since pa(C) = 1 = absurd. So K 2 ≤ 5. To see the existence of this C, let ⇒ (7) Ix = Ker (OX → OX,x/m 2 x), Iy = Ker (OX → OX,y/m 2 y) Then we get, by the Chinese Remainder theorem, (8) 0 → OX (−K) ⊗ Ix ⊗ Iy → OX (−K) → k6 → 0 2 have dimension 3 over k. Taking the long exac...
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Oan → 1 gives X )∗ X H 1(O ) → H 1((O )∗) → H 2(X, Z) → H 2(O ) → · · · (9) By Serre’s GAGA, H i(X, F) ∼ H i(X an , F an) for an OX -module F. pg = 0, h1(Oan) = h2(Oan) = 0, and an X an X an X = X Since q = an X )∗) ∼ = H 1(O∗ X ) = Pic X ∼ (10) This implies that b2 = rank H 2(X, Z) = rank Pic X = 1 contr...
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Bluespec Tutorial: Rule Scheduling and Synthesis Michael Pellauer Computer Science & Artificial Intelligence Lab Massachusetts Institute of Technology Based on material prepared by Bluespec Inc, January 2005 March 4, 2005 BST-1 Improving performance via scheduling Latency and bandwidth can be improved by performing...
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same clock cycle as this method March 4, 2005 BST-5 Rule scheduling info For each rule, there is an entry like this: name of the rule expression for the rule’s condition Rule: fetch Predicate: the_bf.i_notFull_ && the_started.get Blocking rules: imem_put, start more urgent rules which can block the execution of this ...
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will fire Which one? (cid:132) The compiler chooses (and informs you, during compilation) (cid:132) The “descending_urgency” attribute allows the designer to control the choice March 4, 2005 BST-10 5 Demo Example 2: Concurrent Updates Process 0 increments register x; Process 1 transfers a unit from register x to re...
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s g BST-14 7 Example3.bsv Demo Compiling Examining FIFO signals, enables Examining conservative conditions (cid:132) What are the predicates for R1, R2? -aggressive-conditions (cid:132) What are the predicates now? -expand-if (cid:132) Why can certain generated rules never March ...
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particularly when the interface is "always_enabled“ (cid:132) To handle transient situations e.g., interrupts March 4, 2005 BST-18 9 no_implicit_conditions Asserts that rule actions do not introduce any implicit conditions (cid:132) That the rule’s condition is exactly as the user has written, and nothing more Can b...
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removed March 4, 2005 BST-23 Interface attributes These attributes are used to match externally-specified port lists which do not have RDY and EN wires Or for a synchronous module which should receive input on every cycle March 4, 2005 BST-24 12 Synchronous Binary Multiplier Interface interface Design_IFC; method...
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busy); m.start (a[i], b[i]); i <= i+1; busy <= True; endrule rule data_out (busy); Tout x = m.result(); $display (“%0.h X %0.h = %0.h Status: %0.d”, a[j], b[j], x, x==ab[j] ); j <= j+1; busy <= False; endrule endmodule: mkTest March 4, 2005 BST-27 Example1.bsv Demo Compiling with -u The (* synthesize *) pragma Method R...
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Lecture 23: Fault­Tolerant Quantum Computation Scribed by: Jonathan Hodges Department of Nuclear Engineering, MIT December 4, 2003 1 Introduction Before von Neumann proposed classical fault­tolerance in the 1940’s, it was assumed that a compu­ tational device comprised of more than 106 components could not perfor...
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put them through the AND gate. Ideally one should get 1111111. Instead, suppose the strings received are 1110101 on a and 0111111 on b. If one performs a bit­wise AND on each successive bit of the bit strings a and b, the result is 0110101. Taking “triples” of bits of this resulting addition, one performs a majority...
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these will prove feasible. Consistency checks, like the parity of a bit string, work classically, but in the quantum world are simply not powerful enough. Checkpoints require stoping the computation at a 1 P. Shor – 18.435/2.111 Quantum Computation – Lecture 23 2 specific point, checking the result, then starting ...
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the CNOT and single­qubit gates are required, but formulating fault­ tolerant operations becomes easier with a finite gate set. CNOT, Hadamard, σx, σz, T, Toffoli, and π 8 will be proven useful. Theorem 2 (Kitaev­Solovay Theorem). Given a set of gates on SU(2) ( or SU(k) generally) that generates a dense set in SU(2)...
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two classical codes; v ∈ C1 − C2. Since the encoded 0� and 1� are orthogonal, a σx should just interchange the two encodings. These two states are separated by v, which amounts to peforming a σx on each individual qubit. Now suppose an error is made in performing σx on one of the qubits, where the errors on each qub...
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3 1 � � |C2| x∈C2 � 1 |x + v� −→ � |C2| x∈C2 (−1)x.w (−1)v.w x + v� = | − � 1 � |C2| x∈C2 | x + v� (5) since at least 1 vector in C1 gives v.w = 1. 3.3 Fault Tolerance of Hadamard Gate The fault tolerance of the Hadamard gate under this CSS encoding can be seen under the additional ⊥. If the function E(x) ...
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( 0�) is composed of all codeword in C2 and E( 1�) is everything in C1, but not in C2 by definition. A Hadamard transformation on E( 1�), simply adds in the phase factor of (−1)y.v , which obviously follows from above. This phase factor will be unity if y ∈ C2, but ­1 if y ∈ (C1 − C2), thus appropriately adding a pha...
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| x + y + va + vb� |C2| 1 |C2| y∈C2 x∈C2 � � |x� ⊗ x∈C2 y∈C2 |y + va + vb� P. Shor – 18.435/2.111 Quantum Computation – Lecture 23 4 If va = vb, then va + vb = 0 in binary addition and the vector is unchanged. Otherwise, the resulting state will be a string of 1’s added to all states y�, which is just the e...
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flipped depending on the states of the controlled qubits. Measurement of this ancilla qubit would unveil the syndrome, but not with fault­tolerant precision. This can easily be seen under a Hadamard transformation, H ⊗k+1, which reverses the direction of the CNOT gates and gives the dual CSS code in the H⊥ space. (H...
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. | | H ⊗k ψ� = | 1 √ 2 ( 0� + 1�) | | (9) (The state x� represents a k length string of x’s.) | Now suppose a maximally entangled state can be created and verified by performing a few CNOT gates between the bits of the cat state and an ancilla. Fault tolerance is not an issue here; one only wants to know if the s...
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18.445 Introduction to Stochastic Processes Lecture 13: Countable state space chains 2 Hao Wu MIT 1 April 2015 Hao Wu (MIT) 18.445 1 April 2015 1 / 5 Recall Suppose that P is irreducible. The Markov chain is recurrent if and only if Px [τ + x < ∞] = 1, for some x. The Markov chain is positive recurrent if and only if ...
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, x) > 0}. n Then gcd(T (x)) = gcd(T (y )), for all x, y . We say the chain is aperiodic if gcd(T (x)) = 1. Theorem Suppose that the Markov chain is irreducible and aperiodic. If the chain is positive recurrent, then lim ||Pn(x, ·) − π||TV = 0. n Hao Wu (MIT) 18.445 1 April 2015 5 / 5 MIT OpenCourseWare http://ocw.mit...
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MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 32: Exam 4 Review 18.01 Fall 2006 Exam 4 Review 1. Trig substitution and trig integrals. 2. Partial fractions. 3. Integrati...
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3 (x → 1) 1 Lecture 32: Exam 4 Review 18.01 Fall 2006 To find C, C = (−2)2 − 2 + 1 (−2 − 1)2 = 1 3 (x → −2) To find A, one method is to plug in the easiest value of x other than the ones we already used (x = 1, −2). Usually, we use x = 0. 1 A (−1)2(2) −1 = + 1 (−1)2 + 1/3 2 and then solve to find A. ...
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2 + 9) can be integrated using the trigonometric substitution x = 3 tan u. This method can be used to evaluate the integral of any rational function. In practice, the hard part turns out to be factoring the denominator! In recitation you encountered two other steps required to cover every case systematically, namely...
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(Pay attention to the range of θ to be sure that you are not double-counting regions or missing them.) 1For example, we rewrite the denominator x2 + 4x + 13 = (x + 2)2 + 9 = u2 + a2 with u = x + 2 and a = 3. 2Long division is used when the degree of P is greater than or equal to the degree of Q. It expresses P (x)/Q...
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4 Lecture 32: Exam 4 Review 18.01 Fall 2006 Postscript: Systematic integration of rational functions For a general rational function P/Q, the first step is to express P/Q as the sum of a polynomial and a ratio in which the numerator has smaller degree than the denominator. For example, x3 x2 − 2x + 1 = x + 2 + ...
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� dx (Ax2 + Bx + C)n are handled by first completing the square: Ax2 + Bx + C = A(x − B/2A)2 + C − � � B2 4A Using the variable u = √ A(x − B/2A) yields combinations of integrals of the form � � udu (u2 + k2)n and du (u2 + k2)n The first integral is handled by the substitution w = u2 + k2 , dw = 2udu. The ...
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2.160 Identification, Estimation, and Learning Lecture Notes No. 3 February 15, 2006 2.3 Physical Meaning of Matrix P The Recursive Least Squares (RLS) algorithm updates the parameter vector t y ) in such a way that the overall squared error may ( ˆ θ(t − 1) based on new data ϕT (t ), be minimal. This is done by...
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real numbers, it has all real eigenvalues. The eigen vectors associated with the individual eigenvalues are also real. Therefore, the matrix ΦΦT can be reduced to a diagonal matrix using a coordinate transformation, i.e. using the eigen vectors as the bases. λ 0 L 0   1   0 M  ∈ R mxm ΦΦT ⇒ D =   M  ...
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by the data set. max In the direction of λ , there are plenty of input data: ϕ(i)L. This direction has been well explored, well excited. Although new data are obtained, the correction to the parameter vector θ(t − 1) is small, if the new input data ϕ(t) is in the same direction as that of λ . See the second figure a...
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. For this, Po must be a positive definite matrix, such as the identity matrix I. (21) Depending on initial values of θ (0) and Po , the (best) estimation thereafter will be different. ˆ Question: How do the initial conditions influence the estimate? The following theorem shows exactly how the RLS algorithm work...
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) θ (t ) = P ∑ i y ) ϕ ϕ (i ) T  i = 1 ( t ) − 1 (i ) + P θ (0) 0    t − 1  t y )ϕ (t ) + ∑ i y ) ϕ ϕ (i ) T = Pt  (  i = 1 ( )− 1 (i ) + P θ (0) 0    (23) (24) 3 − 1 ˆ − 1) Pt θ(t − 1 − 1 = ϕ ϕ T (t ) − 1 (t ) + P t − 1 Recall Pt ) t [ − 1 ) 1) θ (t ) = P P θ (t + − P ϕ )[ ( t...
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in (25) yields (18), the RLS algorithm, ˆ θ(t ) =θ (t ˆ − 1) + Pt − 1ϕ (t ) ( 1 +ϕ ( P T ) t − 1 t ϕ (t ))( t y ) −ϕ (t )θ (t − 1)) ( ˆ T (25) (26) (18) Q.E.D. Discussion on the Theorem of RLS ) θ (t ) = arg min θ t     1 ∑ ( i y ) − ϕ (i )θ) (  2 i = 1  1 4 4 42 4 4 43  Squared estimati...
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the eigenvalues of P0 in response to the prediction error, T t y ) −ϕ (t )θ . ( − 1 are small, θ tends to change more quickly 4) The initial matrix P0 represents the level of confidence for the initial parameter ) valueθ (0) . Note: The P matrix involved in RLS with an initial condition Po has been extended in t...
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i t − J t (θ) = ∑α e 2 (i ) ˆθ (t ) = i = 1 arg min θ J (θ ) t θ ˆ(t ) is given by the following recursive algorithm. ˆ 1) + − ˆ θ(t ) = θ (t Pt − 1ϕ (t ) ( ϕ α + T ) t − 1 ( P (t ) T 1   t ) ( P Pt − 1 ϕ ϕ t − 1 α Pt − 1 − ( ϕ α  t ϕ (t )) + T ) t − 1 ( P  Exercise: Obtain (34) and (35...
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) The Covariance Re-Setting method is to solve these shortcomings by occasionally re-setting the P matrix to: P ∗ = kI t 0 < k < ∞ (37) This re-vitalizes the algorithm. 2.6 Orthogonal Projection The RLS algorithm provides an iterative procedure to converge to its final parameter value. This may take more than m...
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R (  y l   l T (t )θ Ψ ∈ R l × m (40) M M For each output ˆ i ( ty ) = ϕ T (t )θ i T ϕ 1    vˆ(ty ) =  M θ Ψ=  T   ϕ   v( ty ) Ψ− T (t )θ Error  e 1    r te ) =  M  = (  e    Consider that each squared error is weighted differently, or Weighted Multi-Output Squared Erro...
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1 Ρ − t − 1Ψ (t )[ W Ρ =Ρ t − 1 T (t )W ( (t )θ (t − 1)) r( ˆ ty ) Ψ− T (t )Ρ Ψ (t )]− 1 Ψ T (t )Ρ Ψ+ t − 1 t − 1 (41) (42) (43) (44) 8
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2.160 Identification, Estimation, and Learning Lecture Notes No. 1 February 8, 2006 Mathematical models of real-world systems are often too difficult to build based on first principles alone. Figure by MIT OCW. System Ident cation; “Let the data speak about the system”. ifi Figure by MIT OCW. Image removed fo...
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model Cons 1. No direct connection to physical parameters 2. No solid ground to support a model structure 3. Not available until an actual system has been built 2 Introduction: System Identification in a Nutshell b 3 b 2 b 1 u(t ) y(t ) FIR Finite Impulse Response Model t y ) = ( ( t u b 1 − 1) + ( t...
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∑ ( N t = 1 N t = 1 t y ) −ϕ (t )θ )(−ϕ ) = 0 ( T ∑ t y )ϕ (t ) = ∑ (ϕ (t )θ )ϕ (t ) ( T N t = 1 3 N  ∑ (ϕ(t )ϕ (t )  t =1 T N  θ = ∑ t y )ϕ(t ) (   t =1 = RN ∴ θN = RN ∑ t y )ϕ(t ) −1 ( ˆ N t =1 Question1 What will happen if we repeat the experiment and obtain θˆ again? N Consider the expecta...
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0 N ) t e t ) ( RN ∴ θN −θ0 = RN ∑ϕ( −1 ˆ ) ( t e ) t N t =1 Taking expectation E [θ −θ ] = E RN ∑ϕ( −1 0 ˆ N   N t =1  ) t e t ) (  −1 = RN ∑ϕ(t ) ⋅ E [ N t =1 ( t e )] = 0 Question2 Since the true parameter θ is unknown, how do we know how close 0 ˆ N will be toθ ? How many data points, ...
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� t = 1 s = 1 N N [ = RN ∑∑ϕ ( E t e ( ) ( s e t )  t = 1 s = 1 T )]ϕ (s )  R − 1  N Assume that {e(t)} is stochastically independent [ ( ) ( t e E s e )] =  )] = 0 s e t e ( ) ( 2 t ≠ s  E [  E [e (t )] = λ t = s Then PN = RN ∑ϕ (t )λϕ (t )RN = λRN − 1 T − 1 N − 1   t = 1   As...
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   rij = N ∑ t 1= t u ( − i ) ( − j) t u III. The convergence of ˆ θN to θ0 may be accelerated if we design inputs such that R is large. IV. The covariance does not depend on the average of the input signal. Only the second moment What will be addressed in 2.160? A) How to best estimate the parameters W...
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Introduction to C++ Massachusetts Institute of Technology January 12, 2011 6.096 Lecture 5 Notes: Pointers 1 Background 1.1 Variables and Memory When you declare a variable, the computer associates the variable name with a particular location in memory and stores a value there. When you refer to the variable b...
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structures efficiently, even if their data is scattered in differ­ ent memory locations • Use polymorphism – calling functions on data without knowing exactly what kind of data it is (more on this in Lectures 7-8) 2 Pointers and their Behavior 2.1 The Nature of Pointers Pointers are just variables storing integers –...
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x. We can have pointers to values of any type. The general scheme for declaring pointers is: data_type * pointer_name ; // Add "= initial_value " if applicable pointer name is then a variable of type data type * – a “pointer to a data type value.” 2.2.2 Using Pointer Values Once a pointer is declared, we can derefer...
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{ int x = 5; 6 squareByPtr (& x ) ; 7 cout << x ; // Prints 25 8 9 } Note the varied uses of the * operator on line 2. 2.2.3 const Pointers There are two places the const keyword can be placed within a pointer variable declaration. This is because there are two different variables whose values you might want ...
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often set to 0 to signal that they are not currently valid. Dereferencing pointers to data that has been erased from memory also usually causes runtime errors. Example: 1 2 3 4 int * myFunc () { int phantom = 4; return & phantom ; } phantom is deallocated when myFunc exits, so the pointer the function return...
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whereas you can change the location to which a pointer points. Because of this, references must always be initialized when they are declared. • When writing the value that you want to make a reference to, you do not put an & before it to take its address, whereas you do need to do this for pointers. 4 3.1 The Ma...
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element that is 3 away from the starting el­ ement of myArray. This explains why arrays are always passed by reference: passing an array is really passing a pointer. This also explains why array indices start at 0: the first element of an array is the element that is 0 away from the start of the array. 4.1 Pointer ...
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. Similarly, we can add/subtract two pointers: ptr2 - ptr gives the number of array elements between ptr2 and ptr (2). All addition and subtraction operations on pointers use the appropriate step size. 4.1.2 Array Access Notations Because of the interchangeability of pointers and array names, array-subscript notat...
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{ ’6 ’ , ’. ’ , ’0 ’ , ’9 ’ , ’6 ’ , ’ \0 ’ }; char * courseName2 = " 6.096 " ; Attempting to modify one of the elements courseName1 is permitted, but attempting to modify one of the characters in courseName2 will generate a runtime error, causing the program to crash. 6 MIT OpenCourseWare http://ocw.mit.edu 6.0...
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2.092/2.093 — Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 5 - The Finite Element Formulation Prof. K. J. Bathe MIT OpenCourseWare In this system, (X, Y, Z) is the global coordinate system, and (x, y, z) is the local coordinate system for the element i. We want to satisfy the following equations...
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2.093, Fall ‘09 The Finite Element Formulation ⎡ ⎤ uˆ = ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ u1 v1 w1 . . . uN vN wN ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ N is the number of nodes (3N = n) and H is the displacement interpolation matrix. For the moment, let’s assume Su = 0. We use � uˆT = u1 u2 u3 . . . un � Then, we obtain We also as...
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� H Si(m)T f Si(m) f f dSi(m) f � m i Si(m) f uˆ is the unknown to be found. When evaluated on Sf i(m) , i(m) u¯Sf = H Sf i(m) u ¯ˆ With the transformed equation above, we can insert the following identity matrices: H Si(m) f = H (m) � � Sf i(m) � Let u ¯ˆ T = Then u ¯ˆ T = Then u¯ˆ = T 1 0 � � 0 ...
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system, we can define U T = [ u1 u2 u3 ]. We want to find: u(1)(x) = H (1) ⎢ ⎡ ⎡ ⎤ ⎤ u1 u1 ⎦ ; u(2)(x) = H (2) ⎢ ⎣ u2 ⎥ ⎣ u2 ⎥ ⎦ u3 u3 3 MIT OpenCourseWare http://ocw.mit.edu 2.092 / 2.093 Finite Element Analysis of Solids and Fluids I Fall 2009 For information about citing these materials or our Terms of Us...
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Lecture 1 8.821/8.871 Holographic duality Fall 2014 8.821/8.871 Holographic duality MIT OpenCourseWare Lecture Notes Hong Liu, Fall 2014 Lecture 1 1: HINTS FOR HOLOGRAPHY In this chapter, we will get a favor of the holographic duality. We first study gravity system and derive black hole thermodynamics where holography p...
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1997, Juan Maldacena discovered the famous duality: Quantum gravity in Anti de Sitter spacetime = Field theories (many-body system) in a fixed spacetime (2) The two sides should be be considered as different descriptions of the same quantum system. This duality provides a ”unification”, which has far-reaching implications...
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gravity can be emergent. For example, in Quantum Chromodyanmics (QCD) , there are indeed massive spin-2 excitations. Could one tweak such a theory that massless spin-2 particles emerge? Such hopes were however dashed by a powerful theorem of Weinberg and Witten [1]. 2 Theorem 1 : A theory that allows the construction o...
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spacetime as the original theory. This loophole was utilized by holographic duality. Proof. Suppose we have such a theory that allows Lorentz-covariant conserved current and stress tensor, and there exist massless particles of spin-J. One-particle state are denoted as We have |k, σ(cid:105), kµ = (k0, k), σ = ±j(helici...
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104)k, σ|T µν|k(cid:48), σ(cid:105) k→k(cid:48) −−−→ µ qk k0 kµkν k0 1 (2π)3 1 (2π)3 (7) (8) where (cid:104)k, σ|k(cid:48), σ(cid:105) = δσ,σ(cid:48) δ(3)(k − k(cid:48)). You need to prove the claim in pest 1, one self-consistency check would be when looking at 0-component of Eq. (7), we have (cid:104)k, σ|J 0|k(cid:48...
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8), j|J µ|k, j(cid:105) = Λµ ν (θ)(cid:104)k(cid:48), j|J ν|k, j(cid:105) here Λµ ν (θ) is defined by the rotational transformation acting on a vector by a angle θ around z-axis, i.e. Λµ ν (θ) = 1 0 0 cos θ   0 − sin θ  0 0 0 sin θ cos θ 0  0 0   0  1 Similarly: e2ijθ(cid:104)k(cid:48), j|T µν|k, j(cid:105) = Λµ...
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fferent spacetime, as in a holographic duality. We are not yet ready to go there without some preparations: 1. Black hole thermodynamics ⇒ holographic principle. 2. Large N gauge theories ⇒ gauge/string duality. 3. A bit of string theory which would be useful for building intuitions and perspectives. References [1] S. W...
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Consequence of Electrons as Waves on Free Electron Model • Boundary conditions will produce quantized energies for all free electrons in the material • Two electrons with same spin can not occupy same electron energy (Pauli exclusion principle) Imagine 1-D crystal for now Traveling wave picture Standing wave pic...
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Lk F π g(E) = 2 m dN dk 1 dk dE L π h = 2m = 2k πh dE dk 1 − 2 E 2mE h g(E)=density of states=number of electrons per energy per length n = 2k N = L π F = 2 2mEF hπ or kF = nπ 2 •n=is the number of electrons per unit length, and is determined by the crystal structure and valence •The electron dens...
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. absorb energy and contribute to properties TF~104K (Troom~102K), EF~100Eclass, vF 2 2~100vclass Element rs /ao Li Na K Rb Cs Cu Ag Au Be Mg Ca Sr Ba Nb Fe Mn Zn Cd Hg 3.25 3.93 4.86 5.20 5.62 2.67 3.02 3.01 1.87 2.66 3.27 3.57 3.71 3.07 2.12 2.14 2.30 2.59 2.65 eF 4.74 eV 3.24 2.12 1.85 1.59 7.00 5.49 5.53 14....
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1 1.02 0.98 1.18 1.71 1.70 1.58 1.40 1.37 nF 1.29 x 108 cm/sec 1.07 0.86 0.81 0.75 1.57 1.39 1.40 2.25 1.58 1.28 1.18 1.13 1.37 1.98 1.96 1.83 1.62 1.58 8.63 8.15 11.7 10.4 Al Ga In Tl Sn Pb Bi Sb 2.07 2.19 2.41 2.48 2.22 2.30 2.25 2.14 13.6 12.1 10.0 9.46 11.8 11.0 11.5 12.7 Fermi energies, fermi temperatures, fermi w...
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thermal properties in conductors (i.e. cv) • Electrons at the Fermi surface are able to increase energy: responsible for properties Fermi-Dirac distribution • • NOT Bolltzmann distribution, in which any number of particles can occupy each energy state/level Originates from: EF ...N possible configurations T=0 ...
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6.087 Lecture 8 – January 21, 2010 Review Pointers Void pointers Function pointers Hash table 1 Review:Pointers • pointers: int x; int∗ p=&x; • pointers to pointer: int x; int∗ p=&x;int∗∗ pp=&p; • Array of pointers: char∗ names[]={"abba","u2"}; • Multidimensional arrays: int x [20][20]; 1 Review: Stacks • ...
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enced. The pointers should always be cast before dereferencing. void∗ p; printf ("%d",∗p); /∗ invalid ∗/ void∗ p; int ∗px=(int∗)p; printf ("%d",∗px); /∗valid ∗/ 5 Function pointers • In some programming languages, functions are first class variables (can be passed to functions, returned from functions etc.). • ...
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∗ ) pb ) ; } / ∗ c a l l b a c k ∗ / i n t desc ( void ∗ pa , void ∗ pb ) { r e t u r n ( ∗ ( i n t ∗ ) pb − ∗ ( i n t ∗ ) pa ) ; i n ascending o r d e r ∗ / } / ∗ s o r t q s o r t ( a r r , s i z e o f ( a r r ) / s i z e o f ( i n t ) , s i z e o f ( i n t ) , asc ) ; / ∗ s o r t q s o r t ( a r r , s i z...
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s t r u c t node∗ phead ; / ∗ p o p u l a t e somewhere ∗ / void p r i n t ( void ∗ p , void ∗ arg ) { s t r u c t node∗ np =( s t r u c t node ∗ ) p ; p r i n t f ( "%d " , np−>data ) ; } a p p l y ( phead , p r i n t , NULL ) ; 10 Callback (cont.) Counting nodes: void d o t o t a l ( void ∗ p , void ∗ arg...
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r u c t shape∗ ps )= {& draw_square ,& draw_rec ,& d r a w _ c i r c l e ,& draw_poly } ; typedef void ( ∗ f p ) ( s t r u c t shape∗ ps ) drawfn ; drawfn f p [ 4 ] = {& draw_square ,& draw_rec ,& d r a w _ c i r c l e ,& draw_poly } ; void draw ( s t r u c t shape∗ ps ) { ( ∗ f p [ ps−>t y p e ] ) ( ps ) ; / ∗ c a l ...
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. 15 Hash tables Hash functions: • A hash function maps its input into a finite range: hash value, hash code. • The hash value should ideally have uniform distribution. why? • Other uses of hash functions: cryptography, caches (computers/internet), bloom filters etc. • Hash function types: • Division type • M...
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= c u r r −>n e x t ) / ∗ search ∗ / ; n o t f ou n d : i f ( c r e a t e ) / ∗ add t o r e t u r n c u r r ; } f r o n t ∗ / 19 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MIT OpenCourseWare http://ocw.mit.edu 6.087 Practical Programming in C January (...
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7. Rational Cherednik algebras and Hecke algebras for varieties with group actions 7.1. Twisted differential operators. Let us recall the theory of twisted differential oper­ ators (see [BB], section 2). Let X be a smooth affine algebraic variety over C. Given a closed 2-form ω on X, the algebra Dω(X) of differential op...
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by the De Rham differential acting from 1-forms to closed plex of sheaves Ω1 X 2-forms (sitting in degrees 1 and 2, respectively). If X is projective then this space is isomorphic to H2,0(X, C) ⊕ H1,1(X, C). We refer the reader to [BB], Section 2, for details. ΩX → X X Remark 7.1. One can show that Dω(X) is the u...
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X|Z → N . OX (Z) ⊗OX → We have an exact sequence of OX -modules: 0 → OX → OX (Z) → i∗N φ − → 0 Thus we have a natural surjective map of OX -modules ξZ : T X → OX (Z)/OX . 7.3. The Cherednik algebra of a variety with a finite group action. We will now generalize the definition of Ht,c(G, h) to the global case. Let...
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]) of twisted (by ω/t) differential operators on X with rational coefficients. Definition 7.2. A Dunkl-Opdam operator for (X, G) is an element of Dω/t(X)r[c] given by the formula (7.1) D := tLv − · fY (x) · (1 − g), � 2c(Y, g) 1 − λY,g (Y,g)∈S where λY,g is the eigenvalue of g on the conormal bundle to Y , v ∈ Γ(X...
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,λω(G, X) = Ht,c,ω(G, X). Example 7.6. X = h is a vector space and G is a subgroup in GL(h). Let v be a constant vector field, and let fY (x) = (αY , v)/αY (x), where αY ∈ h∗ is a nonzero functional vanishing on Y . Then the operator D is just the usual Dunkl-Opdam operator Dv in the complex reflection case (see Sec...
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true. Now consider arbitrary X. We have a homomorphism of graded algebras ψ : grF (Ht,c,ω(G, X)) → G � O(T ∗X)[t, c, ω] (the principal symbol homomorphism). The homomorphism ψ is clearly surjective, and our job is to show that it is injective (this is the nontrivial part of the proof). In each degree, ψ is a morph...
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ology group H1(G, AutU (D)), which is trivial since G is � finite and AutU (D) is prounipotent over C. It follows from Lemma 7.8 that it suffices to prove the theorem in the linear case, which � has been accomplished already. We are done. Remark 7.9. The following remark is meant to clarify the proof of Theorem 7.7....
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a Weyl group and X = H the corresponding torus. Then H1,c,0(G, H) is called the trigonometric Cherednik algebra. 7.4. Globalization. Let X be any smooth algebraic variety, and G ⊂ Aut(X). Assume that X admits a cover by affine G-invariant open sets. Then the quotient variety X/G exists. For any affine open set U in X/...
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the case G = 1). X Remark 7.11. (1) The construction of Ht,c,ω(G, X) and the PBW theorem extend in a straightforward manner to the case when the ground field is not C but an algebraically closed field k of positive characteristic, provided that the order of the group G is relatively prime to the characteristic. 5...
https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/02ec876b48928aad62c32c782ce96699_MIT18_735F09_ch07.pdf
12. One has an isomorphism Ht,c,η,ψ,G,X → Ht,c,ψ+ P Y η(Y )ψY ,G,X . Proof. Let y ∈ Y and z be a function on the formal neighborhood of y such that z|Y = 0 and dzy = 0. Extend it to a system of local formal coordinates z1 = z, z2, . . . , zd near y. A Dunkl-Opdam operator near y for the vector field ∂z can be writte...
https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/02ec876b48928aad62c32c782ce96699_MIT18_735F09_ch07.pdf
⊂ X be the set of points with trivial stabilizer. Fix a base point x0 ∈ X �. Then the braid group of X/G is defined to be BG = π1(X �/G, x0). We have an exact sequence 1 K BG → G → → Now let S be the set of pairs (Y, g) such that Y is a component of X g of codimension 1 in X (such Y will be called a reflection hyper...
https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/02ec876b48928aad62c32c782ce96699_MIT18_735F09_ch07.pdf
G, X, x0), to be the quotient of the group algebra of the braid group, C[BG][[τ ]], by the relations (7.3) nY � (T − e 2πji/nY e τjY ) = 0, T ∈ CY j=1 (i.e., by the closed ideal in the formal series topology generated by these relations). Thus, A is a deformation of A0. It is clear that up to an isomorphism thi...
https://ocw.mit.edu/courses/18-735-double-affine-hecke-algebras-in-representation-theory-combinatorics-geometry-and-mathematical-physics-fall-2009/02ec876b48928aad62c32c782ce96699_MIT18_735F09_ch07.pdf