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IV.E Perturbative RG (First Order) The last section demonstrated how various expectation values associated with the Landau–Ginzburg Hamiltonian can be calculated perturbatively in powers of u. However, the perturbative series is inherently divergent close to the critical point and cannot be used to characterize cri...
https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/051af10c926b5c8b2ceafc7257669a65_MIT8_334S14_Lec10.pdf
Dm˜ (q) exp ~ − Z = � exp − � nV 2 � 0 � Λ ddq Λ/b (2π)d � Λ/b ddq (2π)d t + Kq2 2 � m(q)|2 | ˜ × � � (IV.30) ln t + Kq2 −U [m,~~˜ σ] e � � � � ≡ � σ � Dm˜ (q)e ~ −βH˜ [ ~˜ m] . Here we have defined the partial averages hOiσ ≡ � q) σ( D~ Z σ O exp − � Λ d Λ/b (2 � dq )d π t + Kq2 ...
https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/051af10c926b5c8b2ceafc7257669a65_MIT8_334S14_Lec10.pdf
(−1)ℓ ℓ! ×ℓth cumulant of U +· · · . (IV.33) The cumulants can be computed using the rules set in the previous sections. For example, at the first order we need to compute = u U m, ~˜ σ ~ �� � � ~˜ m(q1) + ~ σ � · m˜ (q2) + ~ ~ σ(q2) σ(q1) ddq1ddq2ddq3ddq4 (2π)4d (2π)dδd(q1 + q2 + q3 + q4) m˜ (q3) + ~ ~ σ(...
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) · ~σ(q2) ~σ(q3) · ~σ(q4)iσ σ � σ � σ (IV.35) . σ σ � � The second element in each line is the number of terms with the a given ‘symmetry’. The total of these coefficients is 24 = 16. Since the averages hOiσ, involve only the short wavelength fluctuations, only contractions with ~σ appear. The resulting int...
https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/051af10c926b5c8b2ceafc7257669a65_MIT8_334S14_Lec10.pdf
no dependence on ~˜ We shall denote the sum of these terms by uV δfb 1 . terms, the coarse grained Hamiltonian at order of u is given by m. Summing up all β˜H[ ~˜ =V m] δf 0 1 b + uδfb Λ/b + d (2 dq )d π t˜+ Kq2 2 m(q)|2 | ˜ 0 � � Λ/b ddq1 dd q 2 )3 π (2 d � + u 0 � � � ddq3 m~˜ (q1) · m~˜ (q2)m~˜...
https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/051af10c926b5c8b2ceafc7257669a65_MIT8_334S14_Lec10.pdf
dd q d b (2 ) π 0 � � ′ ′ ddq3 ′ ddq2 Λ ddq1 (2π)3d −d z 2 � ′ (q 1) ′ m ~ t˜+ Kb 2 −2q ′2 ′ |m (q ′ )|2 � . ′ m · ~ (q ′ ′ 2) ~ m ′ (q 3) ′ m · ~ (−q ′ ′ 1 − q2 − q3) ′ (IV.40) The renormalized Hamiltonian is characterized by the triplet of interactions (t ′, K ′ , u ′ ), such that ′ ...
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for t is different. It is common to convert the discrete recursion relations to continuous differential equations by setting b = eℓ, such that for an infinitesimal δℓ, ′ b ≡ t(b) = t(1 + δℓ) = t + δℓ t dt dℓ + O(δℓ2) , ′ b ≡ u(b) = u + δℓ u du dℓ + O(δℓ2). 61 Expanding eqs.(IV.42) to order of δℓ, gives t ...
https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/051af10c926b5c8b2ceafc7257669a65_MIT8_334S14_Lec10.pdf
  The recursion relations can be linearized in the vicinity of the fixed point t ∗ = u ∗ = 0, by setting t = t ∗ + δt and u = u ∗ + δu, as d dℓ δt δu = � � � 4(n+2)KdΛ d−2 K 4 − d 2 0 δt δu � � � (IV.45) In the differential form of the recursion relations, the eigenvalues of the matrix determine the r...
https://ocw.mit.edu/courses/8-334-statistical-mechanics-ii-statistical-physics-of-fields-spring-2014/051af10c926b5c8b2ceafc7257669a65_MIT8_334S14_Lec10.pdf
Unfortunately, the recursion relations have no other fixed point at this order and it appears that we have learned little from the perturbative RG. However, since we are dealing with an alternating series we can anticipate that the recursion relations at the next order are modified to dt dℓ du dℓ      ...
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D-Lab Spring 2010 Development through Dialogue, Design and Dissemination 1 Today’s Class • Logistics • Design Box Presentations • Design, Innovation, Invention and the Design Process • Discussion – Readings • Case Studies 2 Some Logistics • Turning in Homework • Course website • Textbooks 3 Techno...
https://ocw.mit.edu/courses/ec-720j-d-lab-ii-design-spring-2010/0544f53d0d8501e7ea1e16d5dba634ba_MITEC_720JS10_lec02.pdf
15 The Design Process • Information Gathering • Problem Definition • Design Specifications • Idea Generation • Analysis & Experimentation • Concept Evaluation • Detail Design • Fabrication • Testing & Evaluation 16 Design Specifications • Translate customer needs into quantitative design performance targets • Defi...
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The Design Process Get feedback Test Problem Gather Information Think of ideas Solution Build Work out details Experiment Choose the best idea 28 Design for Developing Countries 29 criteria “Brute force engineering options often meet but the somewhere there is a profound solution, which is simple, ch...
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6.897: Selected Topics in Cryptography Lectures 9 and 10 Lecturer: Ran Canetti Highlights of past lectures Presented two frameworks for analyzing protocols: • A basic framework: – Only function evaluation – Synchronous – Non-adaptive corruptions – Modular composition (only non-concurrent) • A stronger framework ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
oldwasser-Wigderson88, Two-party functionalities • Known protocols do not work. (“black-box simulation with rewinding” cannot be used). • Many interesting functionalities (commitment, ZK, • coin tossing, etc.) cannot be realized in plain model. In the “common random string model” can do: – UC Commitment [Canett...
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about password-based key exchange? • What about modeling symmetric encryption and message authentication as ideal functionalities? UC commitments The commitment functionality, Fcom 1. Upon receiving (sid,C,V,“commit”,x) from (sid,C), do: 1. Record x 2. Output (sid,C,V, “receipt”) to (sid,V) 3. Send (sid,C,V, “re...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
: (I) Consider the following environment Zc and real-life adversary Ac that controls the committer C: – Ac is the dummy adversary: It reports to Zc any message received from the verifier V, and sends to V any message provided by Zc. – Zc chooses a random bit b, and runs the code of the honest C by instructing Ac ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
crs (with prescribed distribution D) 1. Choose a value r from distribution D, and send r to the adversary. 2. Upon receiving (“CRS”,sid) from party P, send r to P. Note: The Fcrs-hybrid model is essentially the “common reference string model”, as usually defined in the literacture (cf., Blum-Feldman-Micali89). I...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
easy to compute f -1(x). i i – Commitment Scheme: • CRS: f0,f1 • To commit to bit b, choose random x in the domain of f and send fb(x). To open, send b,x. – Simulator chooses the CRS so that it knows the trapdoors f0 Now can equivocate: find x0,x1 s.t. f0(x0)=f1(x1)=y, send y. -1,f1 . -1 • But: Not extractable… ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
and send (sid,C,V,y,c0,c1) to V. • When receiving (sid,C,V,y,c0,c1) from C, V outputs (sid,C,“receipt”,C). • On input (sid,“open”), C does: – Send b,x,rb to V. • Having received b,x,r, V verifies that Fb(x)=y and cb=Ee(r,x). If verification succeeds then output (“Open”,sid,cid,C,b). Else output nothing. Proof ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
then send (sid,C,V,“commit”,0) to Fcom. -1(y), then send (sid,C,V,“commit”,1) to Fcom. If c0 decrypts to x0 where x0=f0 If c1 decrypts to x1 where x1=f1 • -1(y) and cb’=Ee(r,x)), If C sends a valid opening message (b’,x,r) (I.e., x=fb’ then S checks whether b’ equals the bit sent to Fcom. If yes, then S sends (sid...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
bits committed to be the uncorrupted parties, and can simulate the interaction perfectly. Furthermore, whenever S aborts then D finds a claw in (f0, f1): S aborts if A provides a valid commitment to a bit b and then a valid opening to 1-b. But in this case A generated a claw! Step 2: Show that HYB ~ IDEALFcom S,Z...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
,c0,c1) from C, V outputs (sid,C,“receipt”,C). • On input (sid,“open”), C does: – Send b,x,rb to V. • Having received b,x,r, verifies that Fb(x)=y and cb=Ee(r,x). If verification succeeds then output (“Open”,sid,cid,C,b). Else output nothing. Problem: When the committer is corrupted, it needs to present the ran...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
id,x) to (sid,V) 2. Send (sid,cid,x) to S How to realize Fmcom? • Trivial solution: Run multiple copies of protocol ucc, where each copy uses its own copy of Fcrs… • But, can we do it with a single copy of Fcrs? • Does protocol ucc do the job? Attempt 1: Run as is. Bad: Adversary can copy commitments. Attempt...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
The proof that EXECucc,A,Z ~ HYB remains essentally the same, except that here there are many commitments and decommitments. – The proof that HYB ~ IDEALFmcom is similar in S,Z structure to the proof for the single commitment case, except that here the reduction is to the CCA security of the encryption: Simula...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
”,sid,cid) to Fmcom. If b’ != b, then S aborts the simulation -1(y), then let -1(y), Analysis of S: Let Z be an environment. define first the following hybrid interaction HYB: Interaction HYB is identical to IDEALFmcom S,Z, except that when S generates commitments by uncorrupted parties, it “magically learns” t...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
Fmcom the commitments generated by S are real, whereas in IDEALFmcom these commitments are fake. S,Z is that in HYB S,Z Assume a env. Z that distinguishes between the two interactions. Construct a CCA-adversary B that breaks the security of (E,D). (In fact, B will interact in a Left-or-Right CCA interaction): Give...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
standard encryption based on hard-core bits of tradoor permutations). • E” is CCA-secure. Note: E is not CCA-secure, but is good enough… UC Zero-Knowledge from UC commitments • Recall the ZKPoK ideal functionality, Fzk, and the version with weak soundness, Fwzk. • Recall the Blum Hamiltonicity protocol • Show t...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
: (sid.i,P,V,“Commit”,b ) . i • V (cid:198) P: When getting “receipt”, send a random bit c. • P (cid:198) V: If c=0 then send Fcom: (sid.i,“Open”) for all i. If c=1 then open only commitments of edges in h. • V accepts if all the commitment openings are received from and in addition: Fcom – If c=0 then the opened ...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
A’s opening of the commitments in step 3 of the protocol. If c=0, all openings are obtained and are consistent with G, then send b’=1 to Fwzk . If c=0 and some openings are bad or inconsistent with G then send b’=0 (I.e., no cheating, and V should not accept.) If c=1 then obtain A’s openings of the commitments to t...
https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/054e4a5552a39239b46a1e5a5c09cf13_lecture9_10.pdf
times, where (x,b) is the output value. This is easy to simulate: S obtains (x,b) from TP, gives it to A k times, and outputs whatever A outputs. A controls the prover: Here, A should provide k inputs x1 …xk to the k R , and copies of Fwzk should give witnesses w1 …wk in return. S runs A, obtains x1 …xk, gives it...
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Welcome back to 8.033! Emmy Noether 1882-1935 Image courtesy of Wikipedia. MIT Course 8.033, Fall 2006, Lecture 2 Max Tegmark PRACTICAL STUFF: • PS1 due Friday 4PM (cid:129) Symmetry notes posted TODAY’S TOPIC: SYMMETRY IN PHYSICS (cid:129) Key concepts: frame, inertial frame, transformation, invariant, invari...
https://ocw.mit.edu/courses/8-033-relativity-fall-2006/055852a54ae99cb8b6a6386f84e71bef_lecture2_symme1.pdf
ESD.86 Markov Processes and their Application to Queueing Richard C. Larson March 5, 2007 Photo courtesy of Johnathan Boeke. http://www.flickr.com/photos/boeke/134030512/ Outline (cid:139) Spatial Poisson Processes, one more time (cid:139) Introduction to Queueing Systems (cid:139) Little’s Law (cid:139) Markov Proce...
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Random Point Nearest Neighbors: Euclidean Define D2= distance from a random point to nearest Poisson entity Want to derive fD2 1 γ E[D2] = (1/2) (r). "Square Root Law" 2 = (2 −π/2) σD2 1 2πγ f D2 (r) = d dr FD2 (r) = 2rγπe−γπr 2 r ≥ 0 Rayleigh pdf with parameter 2γπ r Random Point Nearest Neighbor: Taxi Metric FD...
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(t) − d(t) L(t) = number of customers in the system (in queue and in service) at time t Little’s Law for Queues t∫ γ(t) = [a(τ) − d(τ)]dτ t∫ γ(t) = total number of customer minutes spent in the system L(τ)dτ = 0 0 a(t) = cumulative # arrivals to system in (0,t] d(t) = cumulative # departures from system in (0,t] L(t...
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139) Look at role of γ(t). Can change queue statistics by changing queue discipline. Cumulative # of Arrivals FCFS=First Come, First Served SJF=Shortest Job First FCFS SJF L(t) LSJF(t) 0 What about LJF, Longest Job 1st? t = time “System” is General L = λW (cid:139) Our results apply to entire queue system, queue p...
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3.37 (Class 7) Review • Book on explosive welds • Wire Bonding o Diagram on board (squeeze wire-thermal compression weld) o Perimeter Bonding o Up to 200-400 I/O (approx 50 per side, can make double rows but lose real-estate on semiconductor) • TAB Bonding o Diagram on board o Solder connection (gold and tin-...
https://ocw.mit.edu/courses/3-37-welding-and-joining-processes-fall-2002/058777e333c43dcdd2ab4faec95745b6_33707.pdf
Bonding • Unique among welding and joining processes • Only one that just buries the contamination+ • For copy machines, essentially bonding toner to paper o Company X, book starts with “gluons” o Similar to starting with diagram of binding energy between atoms o But don’t need to start at this level in ...
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MIT 3.016 Fall 2005 c � W.C Carter Lecture 5 25 Sept. 16 2005: Lecture 5: Introduction to Mathematica IV Graphics Graphics are an important part of exploring mathematics and conveying its results. An infor­ mative plot or graphic that conveys a complex idea succinctly and naturally to an educated observer is a ...
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dimensional pro­ jection. You get some depth information in a projection by the perspective (i.e, the trick that artists use of making parallel lines converge at a non­infinite point. (e.g. 15th cen­ tury Italian School, Donetello)). You also get information by changing your viewpoint. In Mathematica r� you need to sp...
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import your own drawing and images into Mathematica r � .
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Electricity and Magnetism • More on – Electric Flux – Gauss’ Law Feb 20 2002 More on Electric Flux and Gauss’ Law Maxwell Equations (1873) Feb 20 2002 Electric Flux Note absence of ‘ ‘ Electric Flux ΦE = E A ‘ΦE’ is a Scalar: How much? I.e. how much field passes through surface A? Feb 20 2002 A? • Direction – Norm...
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+ + + + + Last example + + + + + + + + + + + + + + Feb 20 2002 Faraday Cage Hollow Metal Sphere + + + + + + + + + + - - - -- HV Feb 20 2002 Van der Graaf Generator Figure by MIT OCW. Faraday Cage Hollow Metal Sphere + ++ + + +++ + Large E; E~1/r2 Feb 20 2002 Van der Graaf Generator Figure by MIT OCW. ‘Challenge’ ...
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Lecture 2 8.251 Spring 2007 Lecture 2 - Topics • Energy and momentum • Compact dimensions, orbifolds • Quantum mechanics and the square well Reading: Zwiebach, Sections: 2.4 - 2.9 1 x± = √ 2 (x 0 ± x 1) x+ l.c. time Leave x2 and x3 untouched. −ds2 = −(dx0)2 + (dx1)2 + (dx2)2 + (dx3)2 = ηµvdxµdxv u, v = 0, 1...
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dimensions (d5, d6, ...). Why don’t we live with 4 space dimensions? If we lived with 4 space dimesnions, planetary orbits wouldn’t be stable (which would be a problem!) Maybe there’s an extra dimension where we can unify gravity and ... Maybe if so, then the extra dimensions would have to be very small – too small to ...
https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/061e1f35aad94b88fee585e928db8f61_lec2.pdf
. Say the same (P1 ≈ P2) if and only if x(P1) = x(P2) + (2πR)n (n ∈ Z) Write as: x ≈ x + (2πR)n Define: Fundamental Domain = a region sit. 1. No two points in it are identified 2. Every point in the full space is either in the fundamental domain or has a representation in the fundamental domain. So on our x line, we...
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y + L2) Blue: Fundamental domain for first identification Red: Fundamental domain for second identification 6 Lecture 2 8.251 Spring 2007 � � pµ = mγ dx0 d�x , dt dt = (mcγ, mγ�v) � � E c , �p = E: relativistic energy = µc√ 2 1−β2 � p: relativistic momentum Scalar: µp = − + �p 2 pµ = (p 0)2 + (p�)2...
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2007 Cone! We focus on these two since quite solvable in string theory. SE: p�ˆ = h�/i ih ∂Ψ E = Ψ ∂x0 c ih ∂ c ∂t Ψ = EΨ So for our x+, want ih ∂Ψ ∂x+ = ElccΨ � Et − �p �x = − − ct + p� �x · � · E c = −p · x = −(p+x + p−x− + . . .) Now have isolated dependence on x+, so can take derivative: So: ...
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get different E � �2 l value from R contribution. Only noticeable at very high temperatures. 10 �
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6.852: Distributed Algorithms Fall, 2009 Class 6 Today’s plan f+1-round lower bound for stopping agreement, cont’d. • • Various other kinds of consensus problems in synchronous networks: – k-agreement – Approximate agreement (skip) – Distributed commit • Reading: – [Aguilera, Toueg] – [Keidar, Rajsbaum] – Chapter 7 ...
https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0628b22125dc0598ad8b1f27422a4f2c_MIT6_852JF09_lec06.pdf
if α is either 0-valent or 1-valent (essentially decided). – Bivalent, if both decisions occur in some extensions (undecided). Univalence and Bivalence α α α 0 0 0 1 1 1 0 1 1 0-valent 1-valent bivalent α univalent Initial bivalence • Lemma 1: There is some 0-round execution (vector of initial values) that is bivale...
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a chain of (k+1)-round executions, α0, α1, α2,…, αm. • Each αl in this sequence is the same as α0 except that i also sends messages to j1, j2,…jl. – Adding in messages from i, one at a time. α α* α0 round k+1 1-valent 0-valent • Each αl is univalent, by assumption. • Since α0 is 0-valent, either: – At least one of t...
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reement after f rounds • Lemma 3: There is an f-round execution in which two nonfaulty processes decide differently. • Proof: – Use Lemma 2 to get a bivalent (f-1)-round execution α with ≤ f-1 failures. – In every 1-round extension of α, everyone who hasn’t failed must decide (and agree). – Let α* be the 1-round exte...
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′ + 2 ≤ f, decide “early”, within f′ + 2 rounds; in any case decide within f+1 rounds. [Keidar, Rajsbaum 02] Lower bound of f′ + 2 for early- stopping agreement. – Not just f′ + 1. Early stopping requires an extra round. • Theorem 2: Assume 0 ≤ f′ ≤ f – 2 and f < n. Every early- stopping agreement algorithm tolerating...
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1, in which process i fails, sends only to j. • Then: – β0 looks to j like ff extension of α0, so j decides 0 in β0 after 1 round. – β1 looks to j like ff extension of α1, so j decides 1 in β1 after 1 round. • β0 and β1 are indistinguishable to all processes except i, j. • Define: – γ 0, infinite extension of β0, in wh...
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How many rounds are enough? • 1-agreement: f+1 rounds – Argument like those for previous stopping agreement algorithms. • k-agreement: ⎣f/k⎦ + 1 rounds. • Allowing k values divides the runtime by k. FloodMin correctness • Theorem 1: FloodMin, for ⎣f/k⎦ + 1 rounds, solves k- agreement. • Proof: • Define M(r) = set of ...
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That is, |M(⎣f/k⎦ + 1)| > k. – Then by Lemma 1, |M(r)| > k for every r, 0 ≤ r ≤ ⎣f/k⎦+1. – So by Lemma 2, at least k processes fail in each round. – That’s at least (⎣f/k⎦+1) k total failures, which is > f failures. – Contradiction! Lower Bound (sketch) • Theorem 2: Any algorithm for k-agreement requires ≥ ⎣f/k⎦ + 1 r...
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algorithm vanishes somewhere in the interior. • Label nodes with executions: – Corner: No failures, all have same initial value. – Boundary edge: Initial values chosen from those of the two endpoints – For k > 2, generalize to boundary faces. – Interior: Mixture of inputs • Label so executions “morph gradually” i...
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Coloring the nodes • Now color each node v with a “color” in {0,1,…,k}: – If v is labeled with (α,i) then color(v) = i’s decision value in α. • Properties: – Colors of the major corners are all different. – Color of each boundary edge node is the same as one of the endpoint corners. – For k > 2, generalize to bo...
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Agreement on real number values: – Readings of several altimeters on an aircraft. – Values of approximately-synchronized clocks. • Consider with Byzantine participants, e.g., faulty hardware. • Abstract problem: – Inputs, outputs are reals – Agreement: Within ε. – Validity: Within range of initial values of nonfaulty p...
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, 2), mean = 1.5. One-round guarantees • Lemma 1: Any nonfaulty process’ val after the round is in the range of nonfaulty processes’ vals before the round. • Proof: All elements of reduce(W) are in this range, because there are at most f faults, and we discard the top and bottom f values. • Lemma 2: Let d be the ra...
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manager (TM) at each site decides whether it would like to “commit” or “abort” the transaction. • Based on whether the transaction’s work has been successfully completed at that site, and results made stable. – All TMs must agree on whether to commit or abort. • Assume: – Process stopping failures only. – n-node, com...
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the coordinator decides. • Weak termination: – If no one fails, everyone terminates by end of round 2. • Strong termination? – No: If coordinator fails before sending its round 2 messages, then others with initial value 1 will never terminate. Add a termination protocol? • We might try to add a termination protoco...
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0. – Anyone else who receives “ready” becomes ready. – Now p1 decides 1 if it hasn’t already decided. • Round 3: – If p1 has decided 1, bcasts “decide 1”. – Anyone else who receives “decide 1” decides 1. 3-Phase Commit • Key invariants (after 0, 1, 2, or 3 rounds): – If any process is in ready or dec-1, then all proce...
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2 becomes ready, but doesn’t decide yet. • Round 5 (like round 2): – If p1 has (ever) decided 0, broadcasts “decide 0”, and similarly for 1. – Else broadcasts “ready”. – Any undecided process who receives “decide()” decides accordingly. – Any process who receives “ready” becomes ready. – Now p2 decides 1 if it hasn’t a...
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a LAN, how to reliably distinguish failure of a process from lost communication? • Other consensus algorithms can be used for commit, including some that don’t depend on such strong timing and reliability assumptions. Paxos consensus algorithm • A more robust consensus algorithm, could be used for commit. • Tolerat...
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MIT OpenCourseWare http://ocw.mit.edu 6.334 Power Electronics Spring 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . Chapter 2 Introduction to Rectifiers Read Chapter 3 of “Principles of Power Electronics” (KSV) by J. G. Kassakian, M. F. Schlecht, and G. ...
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(so ripple current is small). Therefore, we can approximate load as a “special” current source. “Special” since < vL >= 0 in P.S.S. Id =< ⇒ vx R > (2.3) Assume we start with D2 conducting, D1 off (V sin(ωt) < 0). What happens when V sin(ωt) crosses zero? 12 CHAPTER 2. INTRODUCTION TO RECTIFIERS Lc i1 D1 ...
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1 − ωLcId Vs (2.5) − As compared to the case of no commutating inductance, we lose a piece of output voltage during commutation. We can calculate the average output voltage in P.S.S. from < Vx >: < Vx > = = Vs sin(Φ)dΦ π 1 2π u Z Vs [cos(u) + 1] 2π f rom bef ore cos(u) = 1 − ωLcId Vs XcId Vs = 1 − ...
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in Model All due to non-zero commutation time because of ac-side reactance. This effect occurs in most rectifier types (full-wave, multi-phase, thyristor, etc.). Full-bridge rectifier has similar problem (similar analysis). Read Chapter 4 of KSV. Lc D1 D2 D3 D4 VsSin( t) ω <Vx> 2Vs π Vs π + Vx Id − F...
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Chapter 4 State Machines 6.01— Spring 2011— April 25, 2011 117 Chapter 4 State Machines State machines are a method of modeling systems whose output depends on the entire history of their inputs, and not just on the most recent input. Compared to purely functional systems, in which the output is purely determined ...
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order to specify the behavior of a system whose output depends on the history of its inputs mathematically, you could think of trying to specify a mapping from i1, . . . , it (sequences of previous inputs) to ot (current output), but that could become very complicated to specify or execute as the history gets longe...
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world to reach desirable states, by considering different courses of action and using the model to predict their results. We will develop a single formalism, and an encoding of that formalism in Python classes, that will serve all three of these purposes. Our strategy for building very complex state machines will b...
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combinations of segments illuminated in the display). • The controller for a bank of elevators in a large office building: it transduces the current set of buttons being pressed and sensors in the elevators (for position, open doors, etc.) into commands to the elevators to move up or down, and open or close their do...
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.. It uses the states 0, 1, and 2 to stand for the situations in which it is expecting an a, b, and c, respectively; and it uses state 3 for the situation in which it has seen an input that was not the one that was expected. Once the machine goes to state 3 (sometimes called a rejecting state), it never exits that ...
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4.1 State transition diagram for language acceptor. We will use tables like the following one to examine the evolution of a state machine: time input state output 0 i0 s0 o1 1 i1 s1 o2 2 i2 s2 o3 ... ... ... ... For each column in the table, given the current input value and state we can use the out...
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time input state 0 ’a’ 0 1 ’b’ 1 2 ’c’ 2 3 ’a’ 0 4 ’c’ 1 5 ’a’ 3 6 ’b’ 3 7 3 output True True True True False False False The output sequence is (True, True, True, True, False, False, False). Clearly we don’t want to analyze a system by considering all input sequences, but this table ...
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s0 = 0 Here is a table showing what the up and down counter does with input sequence u, u, u, d, d, u: time input state output 0 u 0 1 1 u 1 2 2 u 2 3 3 d 3 2 4 d 2 1 5 u 1 2 6 2 The output sequence is 1, 2, 3, 2, 1, 2. 4.1.1.3 Delay An even simpler machine just takes the input and pas...
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3, 1, 2, 5. 4.1.1.4 Accumulator Here is a machine whose output is the sum of all the inputs it has ever seen. S = numbers I = numbers O = numbers n(s, i) = s + i o(s, i) = n(s, i) s0 = 0 Here is a table showing what the accumulator does with input sequence 100, −3, 4, −123, 10: time 0 input 100 state 0 outpu...
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-oriented facilities to make this convenient. We have an abstract class, SM, which will be the superclass for all of the particular state machine classes we define. It does not make sense to make an instance of SM, because it does not actually specify the behavior of the machine; it just provides some utility methods...
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To run a state machine, we make an instance of the appropriate state-machine class, call its start method (a built in method we will see shortly) to set the state to the starting state, and then ask it to take steps; each step consists of generating an output value (which is printed) and updating the state to the ne...
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() >>> b.step(10) 10 >>> b.state 10 >>> a.state 5 Now, we have two accumulators, a, and b, which remember, individually, in what states they exist. Figure 4.2 shows the class and instance structures that will be in place after creating these two accumulators. Figure 4.2 Classes and instances for an Accumulato...
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We do not promise anything about how many times this method will be called and in what circumstances. Sometimes, it is convenient to arrange it so that the class really defines a range of machines with slightly different behavior, which depends on some parameters that we set at the time we create an instance. So, fo...
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methods allows us to run a state machine. To run a machine is to provide it with a sequence of inputs and then sequentially go forward, computing the next state and generating the next output, as if we were filling in a state table. To run a machine, we have to start by calling the start method. All it does is create...
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, -3, 4, -123, 10], verbose = True, compact = True) See the Infrastructure Guide for details on different debugging options. To simplify the code examples we show in these notes, we have omitted parts of the code that are responsible for debugging printouts. Chapter 4 State Machines 6.01— Spring 2011— April 25, 20...
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that its output is the same as its next state. startState = None def getNextValues(self, state, inp): nextState = self.getNextState(state, inp) return (nextState, nextState) Because these methods are provided in SM, we can define, for example, a state machine whose out­ put is always k times its input, with this simpl...
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’c’, ....]. class ABC(SM): startState = 0 def getNextValues(self, state, inp): if state == 0 and inp == ’a’: return (1, True) elif state == 1 and inp == ’b’: return (2, True) elif state == 2 and inp == ’c’: return (0, True) else: return (3, False) It behaves as we would expect. As soon as it sees a character that devi...
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4 State Machines 6.01— Spring 2011— April 25, 2011 130 Count up and down This is a direct translation of the machine defined in section 4.1.1.2. class UpDown(SM): startState = 0 def getNextState(self, state, inp): if inp == ’u’: return state + 1 else: return state - 1 We take advantage of the default getNextValu...
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7) >>> d.transduce([3, 1, 2, 5, 9], verbose = True) Start state: 7 In: 3 Out: 7 Next State: 3 In: 1 Out: 3 Next State: 1 In: 2 Out: 1 Next State: 2 In: 5 Out: 2 Next State: 5 In: 9 Out: 5 Next State: 9 [7, 3, 1, 2, 5] >>> d100 = Delay(100) >>> d100.transduce([3, 1, 2, 5, 9], verbose = True) Start state: 100 In: 3 Out: ...
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state machine whose output at time t is the average of the input values from times t − 1 and t. class Average2(SM): startState = 0 def getNextValues(self, state, inp): return (inp, (inp + state) / 2.0) It needs to remember the previous input, so the next state is equal to the input. The output is the average of ...
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the sum of it−2, it−1 and it; that is, of the last three inputs. In order to do this, it has to remember the values of two previous inputs; so the state is a pair of numbers. We have defined it so that the initial state is (0, 0). The getNextValues Chapter 4 State Machines 6.01— Spring 2011— April 25, 2011 132 me...
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(10, 1) In: 2 Out: 13 Next State: (1, 2) In: 1 Out: 4 Next State: (2, 1) In: 5 Out: 8 Next State: (1, 5) [2, 3, 6, 8, 17, 15, 13, 4, 8] Selector A simple functional machine that is very useful is the Select machine. You can make many different versions of this, but the simplest one takes an input that is a stream of ...
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are there? A: 12. A: 12. The gate has three possible outputs (think of them as controls to the motor for the gate arm): ’raise’, ’lower’, and ’nop’. (Nop means “no operation.”) Roughly, here is what the gate needs to do: • If a car wants to come through, the gate needs to raise the arm until it is at the top positio...
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parking gate. For com­ pactness, the getNextValues method starts by determining the next state of the gate. Then, depending on the next state, the generateOutput method selects the appropriate output. class SimpleParkingGate (SM): startState = ’waiting’ def generateOutput(self, state): if state == ’raising’: return...
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State: raising In: (’bottom’, True, False) Out: raise Next State: raising In: (’middle’, True, False) Out: raise Next State: raising In: (’middle’, True, False) Out: raise Next State: raising In: (’middle’, True, False) Out: raise Next State: raising In: (’top’, True, False) Out: nop Next State: raised In: (’top’...
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’, ’lower’, ’lower’, ’nop’, ’raise’] Exercise 4.1. What would the code for this machine look like if it were written without using the generateOutput method? 4.2 Basic combination and abstraction of state machines In the previous section, we studied the definition of a primitive state machine, and saw a number of e...
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Then Cascade(m1, m2) is a new state machine, constructed by making the output of m1 be the input of m2. Now, imagine we feed a sequence of values, 3, 8, 2, 4, 6, 5, into the composite machine, m. What will come out? Let’s try to understand this by making a table of the states and values at different times: m1m2o1 =...
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[0] = init2 Now, since we have connected the output of m1 to the input of m2, we also have that i2[t] = o1[t] for all values of t. This lets us make the following derivation: o2[t] = i2[t − 1] = o1[t − 1] = i1[t − 2] This makes it clear that we have built a “delay by two” machine, by cascading two single delay m...
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the individual machines. The result is a new composite machine, whose input vocabulary is the same as the input vocabulary of the component machines (which is the same for both machines) and whose output vocabulary is pairs of elements, the first from the output vocabulary of the first machine and the second from the...
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having the input be a single item which is fed to both machines, the input is a pair of items, the first of which is fed to the first machine and the second to the second machine. This composition is shown in the second part of figure 4.5. Here is a Python class that implements this two-input parallel composition. It ...
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, we would check and generate error messages for all sorts of potential problems (in this case, for instance, if v is neither None nor a two-element list or tuple.) Chapter 4 State Machines 6.01— Spring 2011— April 25, 2011 139 class ParallelAdd (Parallel): def getNextValues(self, state, inp): (s1, s2) = state (...
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