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again in the asynchronous setting. Spanning tree o Leader • Given any spanning tree of an undirected graph, elect a leader: – Convergecast from the leaves, until messages meet at a node (which can become the leader) or cross on an edge (choose endpoint with the larger UID). – Complexity: Time O(n); Messages O(n) ...
https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf
’s MIS Algorithm (sketch) • Each process chooses a random val in {1,2,…,n4}. – Large enough set so it’s very likely that all numbers are distinct. • Neighbors exchange vals. • If node i’s val > all neighbors’ vals, then process i declares itself a winner and notifies its neighbors. • Any neighbor of a winner declar...
https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf
Filter Banks: time domain Filter Banks: time domain ((HaarHaar example) and frequency domain; example) and frequency domain; conditions for alias cancellation conditions for alias cancellation and no distortion and no distortion trivial) example of a two channel FIR Simplest (non--trivial) example of a two channe...
https://ocw.mit.edu/courses/18-327-wavelets-filter-banks-and-applications-spring-2003/204a826459bc8236e14f4787e1575cc2_Slides3.pdf
downsampler -----------------jj ----------------- Similarly Similarly yy11[n] = [n] = 11 � 22 (x[2n] –– x[2nx[2n –– 1])1]) (x[2n] ------------------kk ------------------ 33 � � � � � � � Matrix form Matrix form MM yy00[0][0] yy00[1][1] :: :: yy11[0][0] yy11[1][1] M == 11 � 22 1 1 LL 1 1 0 0 LL ...
https://ocw.mit.edu/courses/18-327-wavelets-filter-banks-and-applications-spring-2003/204a826459bc8236e14f4787e1575cc2_Slides3.pdf
Perfect reconstruction! Perfect reconstruction! In general, we will make all filters causal, so we will In general, we will make all filters causal, so we will havehave ^^ x[n] = x[n –– nn00]] x[n] = x[n � PR with delay PR with delay 77 � � � � Matrix form Matrix form MM ^^ x[x[--1]1] ^^ x[0]x[0] ^^ x[1]...
https://ocw.mit.edu/courses/18-327-wavelets-filter-banks-and-applications-spring-2003/204a826459bc8236e14f4787e1575cc2_Slides3.pdf
So we have an Synthesis bank = Transpose of Analysis bank Synthesis bank = Transpose of Analysis bank filter bank, where orthogonal filter bank, where ff00[n] [n] ff11[n] [n] = = = = h00[[-- n]n] h h11[[-- n]n] h 99 ł � � Ł � � fi Perfect Reconstruction Filter Banks Perfect Reconstruction Filter Banks cha...
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p --p p --p 22 00 p 22 p w � Downsampling operation in each channel can Downsampling operation in each channel can aliasing produce aliasing produce 1111 � p p w w w w Let’s see why: Let’s see why: channel has Lowpass channel has Lowpass YY00(z) = (z) = ½½{R{R00(z(z½½) + R ) + R00((--zz½½)})} downsamplin...
https://ocw.mit.edu/courses/18-327-wavelets-filter-banks-and-applications-spring-2003/204a826459bc8236e14f4787e1575cc2_Slides3.pdf
) } X( z) } X(--z)z) 1414 Compare terms in X(z) and X(--z):z): Compare terms in X(z) and X( Condition for no distortion (terms in X (z) amount 1)1) Condition for no distortion (terms in X (z) amount to a delay) to a delay) FF00(z) H(z) H00(z) + F (z) + F11(z) H(z) H11(z) = 2z (z) = 2z-- ll --------------jj ----...
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( Example Example alternating signs alternating signs rule rule hh00[n] = { a } [n] = { a00, a, a11, a, a22} f00[n] = { b f [n] = { b00,, --bb11, b, b2 2 }} hh11[n] = { b } [n] = { b00, b, b11, b, b22} f11[n] = { f [n] = {--aa00, a, a11,, --aa22}} 1616 w Product Filter Product Filter Define Define PP00(z) ...
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Equation nn asas So we can rewrite Equation zz-- ll P(z) + z P(z) + z-- ll P(P(--z) = 2z z) = 2z-- ll i.e. i.e. z) = 2 P(z) + P(--z) = 2 P(z) + P( ---------------------------pp --------------------------- This is the condition on the normalized product filter This is the condition on the normalized product filte...
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6.867 Machine learning, lecture 9 (Jaakkola) 1 Lecture topics: • Kernel optimization • Model (kernel) selection Kernel optimization Whether we are interested in (linear) classification or regression we are faced with the problem of selecting an appropriate kernel function. A step in this direction might be to ta...
https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/2051efc0159bf145f2050469b7589fc5_lec9.pdf
� K(x, x�) K(x, x)K(x�, x�) (1) Another approach to optimizing the kernel function is kernel alignment. In other words, we would adjust the kernel parameters so as to make it, or its Gram matrix, more towards an ideal target kernel. For example, in a classification setting, we could use as the Gram matrix of the t...
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vectors and define their the target kernel, Kij inner product in the usual way i=1 i=1 �K ∗, Kθ� = n � i,j=1 Kij ∗ Kij (θ) (5) The parameters θ can be now set so as to maximize the cosine of the angle between the Gram matrices: � �K ∗, Kθ� �K ∗, K ∗��Kθ, Kθ� (6) Model (kernel) selection Optimizing the ...
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Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10) 6.867 Machine learning, lecture 9 (Jaakkola) 3 Classifiers making use of the quadratic polynomial kernel can in princi...
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1, K2, . . ., where the models associated with the kernels are nested F1 ⊆ F2 ⊆ . . .. This is a model selection problem in a standard nested form. From here on we will be referring to discriminant functions rather than kernels so as to emphasize the point that the discussion applies to other types of classifiers as ...
https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/2051efc0159bf145f2050469b7589fc5_lec9.pdf
Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10) 6.867 Machine learning, lecture 9 (Jaakkola) 4 where the loss could be the hinge loss (SVM), logistic, or other. The regularization param­ eter λn would in general depend on the ...
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zero-one loss and it behaves much better in terms of the resulting optimization problem we have to solve during training (quadratic rather than integer programming problem). The quantity of interest to us is the generalization error R(θ, ˆ θˆ 0), or R(fˆ i) for short, corre­ sponding to the classifier or discriminant ...
https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/2051efc0159bf145f2050469b7589fc5_lec9.pdf
a lower probability of error. Our task is Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10) 6.867 Machine learning, lecture 9 (Jaakkola) 5 much more difficult ...
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the model, some measure of size or power of Fi, against their fit to the training data. There are a number of answers to this question depending on your perspective. We will briefly go over a few possibilities and return to them later on. Figure 1: Training and test errors as a function of model order (e.g., degree of...
https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/2051efc0159bf145f2050469b7589fc5_lec9.pdf
∈ Fi could be any estimate derived from the training set that approximately tries to minimizing the empirical risk. We are interested in quantifying how much R(fˆ i) can deviate from Rn(fˆ i). The larger the deviation the less representative the training error is about the generalization error. This happens with mo...
https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/2051efc0159bf145f2050469b7589fc5_lec9.pdf
course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10) 6.867 Machine learning, lecture 9 (Jaakkola) 7 Figure 2: Bound on the generalization error as a function of model order (e.g., degree...
https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/2051efc0159bf145f2050469b7589fc5_lec9.pdf
6.863J Natural Language Processing Lecture 16: the boundaries of syntax & semantics – towards constraint-based systems Robert C. Berwick The Menu Bar • Administrivia: • Lab 4 due April 9? (what about Friday) • Start w/ final projects, unless there are objections • Agenda: • Shallow instead of ‘deep’ semantic...
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NP : Apply(lambda (x) (DEF/SING x), N) Lexicon: V:kissed = lambda(o) lambda(x) (kiss past [agent x] [theme o]) N:guy = person N:dog = DOG Det:the = DEF/SING : Top-down parse sentence The guy kissed the dog S � N P V P VP � V NP Lexicon look-u p Apply(Apply(lambda(o) lambda(x) (kiss past [agent x] [theme o]),...
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and semantic processing is collapsed in a single framework • Like a regular grammar but terminal symbols are replaced by semantic categories • Example: • [VP read [NP a book]] or [write [a book]] VP � V NP � READ-VP � READ-VERB READ-STUFF WRITE-VP � WRITE-VERB WRITE-STUFF 6.863J/9.611J Lecture 16 Sp03 Example ...
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J/9.611J Lecture 16 Sp03 Solution (1) • FLIGHT-MOD � from SOURCE- LOCATION Book a flight [FLIGHT-MOD from SOURCE- LOCATION Boston] to Chicago for me • FLIGHT-MOD � to DEST-LOCATION Book a flight [FLIGHT-MOD from SOURCE- LOCATION Boston] [FLIGHT-MOD to DEST-LOCATION Chicago] for me • FLIGHT-MODS � FLIGHT-MOD Book...
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FLIGHT-NOUN flight [FLIGHT-MODS [FLIGHT-MOD from SOURCE-LOCATION Boston] [FLIGHT-MODS [FLIGHT-MOD to DEST-LOCATION Chicago]]]]] [RES-MOD for PERSON me]] 6.863J/9.611J Lecture 16 Sp03 Useful? • Semantic grammars are useful in a limited domain • Dialogue system to book flights through the telephone • For general...
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Location El Salvador: San Salvador Incident: Type Bombing Perpetrator: Individual ID urban guerrillas Perpetrator: Organization ID FMLN Perpetrator: Organization conf suspected or accused Physical target: description vehicle Physical target: effect some damage Human target: name Roberto Garcia Alvarado Human ta...
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NG] in [Prep] Taiwan [location NG] with [Prep] a local concern [NG] and [Conj] a Japanese trading house [NG] to produce [VG] golf clubs [NG] to be shipped [VG] to [Prep] Japan [location NG] 6.863J/9.611J Lecture 16 Sp03 Architecture – Step 4 • Construction of complex nominal and verbal groups • Apposition • ...
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A leak • “By the time their son was born, though, Honus Whiting was beginning to understand and privately share his wife’s opinion, at least as it pertained to Empire Falls” at least as 6.863J/9.611J Lecture 16 Sp03 Subcategorization: what we have been doing • Eat vs. devour: John ate the meal/John ate Bill d...
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– doesn’t support the linguists Consider as: • The boys consider her as family and she participates in everything we do. • Greenspan said, “I don't consider it as something that gives me great concern. • “We consider that as part of the job,” Keep said. • Although the Raiders missed the playoffs for the second ti...
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syntax (“hidden structure”) for the frequency information to interact productively with a formal theory • Goal: to get productive mutual feedback 6.863J/9.611J Lecture 16 Sp03 Incorporating knowledge • Do density estimation P(form | meaning context) 6.863J/9.611J Lecture 16 Sp03 Application: retire • Step 1: ...
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We can recalculate entire frame 0.25 = • P(NP[subj]___|V=retire) • P(NP[subj]___NP[obj]|V=retire) = 0.50 • P(NP[subj]___PP[from]|V=retire) = 0.04 • P(NP[subj]___PP[from]PP[after]|V=retire) = 0.003 … (Sum of pr’s of all frames adds to 1) 6.863J/9.611J Lecture 16 Sp03 Then we can do things like this • Integrat...
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• A pattern that can be matched against unrestricted text • NP NP � (OBJ|SUBJ_OBJ|CAP) (PUNC|CC) • […] greet Peter, […] • […] see him. […] • […] love it, if […] • […] came Thursday, […] � error! • Hypothesis Testing • Initial or null hypothesis (H0) � frame is not appropriate for verb • If cue indicates with ...
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• Verb = greet � occurs 80 times (n = 80) • Cue = (OBJ|SUBJ_OBJ|CAP) (PUNC|CC) � has e = 0.25 • Frame = NP__ NP • C(greet,(OBJ|SUBJ_OBJ|CAP) (PUNC|CC)) = 11 (m = 11 and r = 11) 6.863J/9.611J Lecture 16 Sp03 Hypothesis Testing – Example ( greet P = ( NP NP ) p E = ( C greet,((OBJ|SUBJ_OBJ|CAP)(PUN 0 C|...
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�� food) • Selectional preference (eatfi 6.863J/9.611J Lecture 16 Sp03
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Foundations of Program Analysis Node find (Node prev, Node cur, int key) { while (cur.key < key) { prev = cur; cur = cur.next; } return cur; } 6.820 Armando Solar-Lezama Course Staff • Armando Solar-Lezama – Instructor L01-2 What this course is about The top N good ideas in programming langu...
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01-8 Grading o 6 homework assignments - Each is 15-20% of your grade - start on them early! L01-9 6 Homework Assignments o Pset 1 (out now, due in about 2 weeks!) - Practice functional programming - Build some Lambda Calculus interpreters o Pset 2 - Practice more functional programming - Implement a ty...
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is true for all functional programs regardless whether they are right or wrong A formal definition will be given later September 9, 2015 L01-14 Blocks let x = a * a y = b * b in (x - y)/(x + y) • a variable can have at most one definition in a block • ordering of bindings does not matter September...
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plus' a b = a + b plus and plus' are the same because plus' can be obtained by systematic renaming of bounded identifiers of plus September 9, 2015 L01-19 Capture of Free Variables f x = . . . g x = . . . foo f x = f (g x) Suppose we rename the bound identifier f to g in the definition of foo fo...
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+dx/2) 0) * dx sum dx b f x tot = if x > b then tot else sum dx b f (x+dx) (tot+(f x)) Any function definition can be “closed” and “lifted” September 9, 2015 L01-23 Types All expressions in Haskell have a type 23 :: Int "23 belongs to the set of integers" "The type of 23 is Int" true :: Bool ...
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“->” is right associative September 9, 2015 L01-28 Type of a Block (let x1 = e1 . . . xn = en in e ) :: t provided e :: t September 9, 2015 L01-29 Type of a Conditional (if e then e1 else e2 ) :: t provided Bool :: e e1 :: t t e2 :: T...
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= t2 = t3  twice :: (t0 -> t0) -> t0 -> t0 September 9, 2015 L01-32 Another Example: Compose compose f g x = f (g x) What is the type of compose ? 1. Assign types to every subexpression x :: t0 f :: t1 g :: t2 g x :: t3 f (g x) :: t4  compose :: 2. Set up the constraints t1 -> t2 -> t0 -> t4 ...
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5 L01-35 MIT OpenCourseWare http://ocw.mit.edu 6.820 Fundamentals of Program Analysis Fall 2015 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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Optimality Conditions for Constrained Optimization Problems Robert M. Freund February, 2004 2004 Massachusetts Institute of Technology. 1 1 Introduction Recall that a constrained optimization problem is a problem of the form (P) minx f (x) s.t. g(x) ≤ 0 h(x) = 0 x ∈ X, where X is an open set and g(x)) = (...
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. 2 2 Necessary Optimality Conditions 2.1 Geometric Necessary Conditions A set C ⊆ (cid:4)n is a cone if for every x ∈ C, αx ∈ C for any α > 0. A set C is a convex cone if C is a cone and C is a convex set. Suppose ¯x ∈ S. We have the following definitions: • F0 := {d : ∇f (¯x)td < 0} is the cone of “improving” ...
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x) = 0 for all i ∈ I (for i (cid:10) x + λd) = x + λd) < 0), and h(¯ x + λd ∈ S for all λ > 0 sufficiently small. (A¯ On the other hand, for all sufficiently small λ > 0, f (¯ x). This contradicts the assumption that ¯x is a local minimum of (P). x − b) + λAd = 0. Therefore ¯ ∈ I, since λ is small, gi(¯ x + λd) ≤ gi(¯ ...
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= 0 for i ∈ I and h(¯ x + λd) ≤ gi(¯ The proof of Theorem 2 is rather awkward and involved, and relies on the Implicit Function Theorem. We present this proof at the end of this note, in Section 6. 2.2 Separation of Convex Sets We will shortly attempt to restate the geometric necessary local opti- mality conditions...
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said to strictly separate S and T if ptx > α for all x ∈ S and ptx < α for all x ∈ T . • H is said to strongly separate S and T if for some (cid:1) > 0, p x ≥ α + (cid:1) t for all x ∈ S and p x ≤ α − (cid:1) for all x ∈ T . t 4 Theorem 3 Let S be a nonempty closed convex set in (cid:4)n, and suppose that y (cid...
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uniqueness, suppose that there is some x(cid:2) ∈ S for which (cid:15)y − 2 (¯x + x(cid:2)) ∈ S. But by the triangle . By convexity of S, 1 (cid:2)(cid:15) ¯x(cid:15) = (cid:15)y − x inequality, we have: (cid:7) (cid:7) (cid:7)y − (cid:7) 1 2 (cid:7) (cid:7) (cid:7) 1 2 x + x (cid:2))(cid:7) ≤ (cid:15)y − ¯...
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(cid:15)¯ 2 x−y(cid:15) . Conversely, assume that x is the minimizing point. For any x ∈ S, λx + ¯ 5 (1 − λ)¯ x ∈ S for any λ ∈ [0, 1]. Also, (cid:15)λx + (1 − λ)¯ x − y(cid:15) ≥ (cid:15)x − y(cid:15). Thus, ¯ (cid:15)x − y(cid:15)2 ≤ (cid:15)λx + (1 − λ)¯ ¯ x − y(cid:15)2 = (cid:15)λ(x − ¯ x) + (¯ 2 x − y)(cid:...
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≤ ¯ x t(y−x) = ¯ ¯ 1 x t(y−x)+ (cid:15)y−x(cid:15)2−(cid:1) = 2 ¯ ¯ 1 t 2 1 ¯ 2 y y− x x−(cid:1) = α−(cid:1). t ¯ Therefore p x ≤ α − (cid:1) for all x ∈ S. On the other hand, p y = (y − ¯ α + (cid:1), establishing the result. t t x)ty = Corollary 5 If S is a closed convex set in (cid:4)n, then S is the intersecti...
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Weierstrass Theorem, some i=1 subsequence of {yi} converges to a point ¯ y ∈ S1. Then zi = yi − xi → y¯ − ¯ x y − ¯ x is a limit point of {zi}. Since S2 (over this subsequence), so that ¯ x ∈ T , proving that T is a closed y − ¯ z ∈ S2, and then ¯ is also closed, ¯ set. z = ¯ x = ¯ i=1 By hypothesis, S1 ∩ S2 = ∅...
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2) ≥ α + α = α + (cid:1) 2 1 ¯ 2 and p t z ≤ α2 = 1 2 (α1 + α2) − 1 (α1 − α2) ≤ α − α = α − (cid:1). 2 1 ¯ 2 Theorem 8 (Farkas’ Lemma) Given an m × n matrix A and an n-vector c, exactly one of the following two systems has a solution: (i) Ax ≤ 0, c x > 0 t (ii) A y = c, y ≥ 0. t Proof: First note that both ...
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has a solution: (i) Ax < 0, Bx ≤ 0, Hx = 0 ¯ (ii) Atu + Btv + H tw = 0, u ≥ 0, v ≥ 0, e u = 1. t ¯ Proof: It is easy to show that both (i) and (ii) cannot have a solution. Suppose (i) does not have a solution. Then the system ¯ Ax + eθ ≤ 0, θ > 0 Bx ≤ 0 Hx ≤ 0 −Hx ≤ 0 has no solution. This system can be re-wri...
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This can be rewritten as ¯ At u + Bt v + H t(w 1 − w 2) = 0, t e u = 1 . 8 Letting w = w1 − w2 completes the proof of the lemma. 2.3 Algebraic Necessary Conditions Theorem 10 (Fritz John Necessary Conditions) Let x be a feasible solution of (P). If x is a local minimum of (P), then there exists (u0, u, v) such ...
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⎥ ⎢ ⎢ ∇g1(¯ ⎥ ⎢ ⎥ , H = ⎣ A = ⎢ .. ⎥ ⎢ ⎦ ⎣ . ∇gp(¯ x)t ⎤ ⎥ ⎦ . x)t ∇h1(¯ . .. ∇hl(¯ x)t Then there is no d that satisfies Ad < 0, Hd = 0. From the Key Lemma there exists (u0, u1, . . . , up) and (v1, . . . , vl) such that p (cid:16) (cid:16) l u0∇f (¯ x) + ui∇gi(¯ x) + vi∇hi(¯ x) = 0, i=1 i=1 9 ...
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x) for i = 1, . . . , l and ∇gi(¯ x) for i ∈ I are linearly inde- pendent. If ¯x is a local minimum, there exists (u, v) such that ∇f (¯ x) + ∇g(¯ x)t u + ∇h(¯ x)t v = 0, u ≥ 0, uigi(¯x) = 0 , i = 1, . . . , m. Proof: x must satisfy the Fritz John conditions. If u0 > 0, we can redefine u ← u/u0 and v ← v/u0. If u0 ...
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(x) = ⎛ ⎝ 12(x1 − 10) ⎞ ⎠ 8(x2 − 12.5) ∇g1(x) = ⎛ ⎝ 2x1 ⎞ ⎠ 2(x2 − 5) ∇g2(x) = ⎛ ⎝ ⎞ ⎠ 2x1 6x2 ⎛ ⎝ ∇g3(x) = 2(x1 − 6) ⎞ ⎠ 2x2 Let us determine whether or not the point x = (¯ x2) = (7, 6) is a x1, ¯ ¯ candidate to be an optimal solution to this problem. 11 We first check for feasibility: g1(¯x) = 0 ≤ 0 ...
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⎠ + ⎝ ⎠ u1 + ⎝ ⎠ u2 + ⎝ ⎛ ⎞ 14 14 ⎛ ⎞ 2 ⎠ u3 = 2 36 12 ⎛ ⎞ 0 ⎝ ⎠ 0 u ≥ 0 Notice that ¯ u2 = 0. Therefore x is a candidate to be an optimal solution of this u1, ¯ u3) = (2, 0, 4) solves this system, and that ¯ u = (¯ u2, ¯ ¯ and ¯ problem. 12 Example 2 Consider the problem (P): (P) : maxx xT Qx s.t. (c...
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(P) with the largest objective function value is x = 0 if the largest eigenvalue of Q is nonpositive. If the largest eigenvalue of Q is positive, then the optimal objective value of (P) is the largest eigenvalue, and the optimal solution is any eigenvector x corresponding to this eigenvalue, normalized so that (cid...
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⎝ ⎛ ⎝ −20 ⎞ ⎠ 14 7 ⎞ ⎠ ∇f (x) = ∇g1(x) = −2.5 ⎛ ⎝ −14 ⎞ ⎠ ∇g2(x) = 2 ⎛ ⎞ 8 ⎝ ⎠ 4 ∇h1(x) = We next check to see if the gradients “line up”, by trying to solve for u1 ≥ 0, u2 = 0, v1 in the following system: ⎛ ⎝ −20 ⎞ ⎛ ⎠ + ⎝ 7 ⎛ ⎞ ⎠ u1 + ⎝ 14 −2.5 −14 2 ⎞ ⎛ ⎞ 8 ⎠ u2 + ⎝ ⎠ v1 = 4 ⎛ ⎞ 0 ⎝ ⎠ ...
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)}. f (x) is quasiconcave if for all x, y ∈ X and for all λ ∈ [0, 1], f (λx + (1 − λ)y) ≥ min{f (x), f (y)}. 15 If f (x) : X → (cid:4), then the level sets of f (x) are the sets Sα = {x ∈ X : f (x) ≤ α} for each α ∈ (cid:4). Proposition 12 If f (x) is convex, then f (x) is quasiconvex. Proof: If f (x) is conve...
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) is a convex function, its level sets are convex sets. Suppose X is a convex set in (cid:4)n . The differentiable function f (x) : X → (cid:4) is a pseudoconvex function if for every x, y ∈ X the following holds: ∇f (x)t(y − x) ≥ 0 ⇒ f (y) ≥ f (x) . Theorem 15 (i) A differentiable convex function is pseudoconvex. ...
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(z)t(−(1 − λ)(y − x)) = −(1 − λ)∇f (z)td ≥ 0, so f (z) ≤ f (x) ≤ max{f (x), f (y)}. Thus f (x) is quasiconvex. Incidentally, we define a differentiable function f (x) : X → (cid:4) to be pseudoconcave if for every x, y ∈ X the following holds: ∇f (x)t(y − x) ≤ 0 ⇒ f (y) ≤ f (x) . 4 Sufficient Conditions for Optimality...
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is also a convex set, the feasible region S = {x ∈ X : g(x) ≤ 0, h(x) = 0} is a convex set. Let I = {i | gi(¯ x) = 0} denote the index of active constraints at ¯ x. Then λx + (1 − λ)¯ x. Let x is feasible for all x ∈ S be any point different from ¯ λ ∈ (0, 1). Thus for i ∈ I we have gi(λx + (1 − λ)¯ x) = gi(¯ x +...
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. , m are convex functions, hi(x), i = 1 . . . , l are linear functions, and X is an open convex set. Corollary 17 The KKT conditions are sufficient for optimality of a convex program. Example 4 Continuing Example 1, note that f (x), g1(x), g2(x), and g3(x) are all convex functions. Therefore the problem is a convex...
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important definition: Definition 5.1 A point x is called a Slater point if x satisfies g(x) < 0 and h(x) = 0, that is, x is feasible and satisfies all inequalities strictly. Theorem 18 (Slater condition) Suppose that gi(x), i = 1, . . . , m are pseudoconvex, hi(x), i = 1, . . . , l are linear, and ∇hi(x), i = 1, . . . ...
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td < 0, unless ui = 0 for all i ∈ I. But if this is true, then we would have v (cid:10)= 0 and ∇h(¯x)tv = 0, violating the linear independence assumption. This is a contradiction, and so u0 > 0. Theorem 19 (Linear constraints) If all constraints are linear, the KKT conditions are necessary to characterize an optim...
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second order conditions for optimality, we will define the following function, known as the Lagrangian function, or simply the Lagrangian: L(x, u, v) = f (x) + uigi(x) + vihi(x) = f (x) + u g(x) + v th(x). t (cid:16) m (cid:16) l i=1 i=1 Using the Lagrangian, we can, for example, re-write the gradient conditi...
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∇hi(¯x)td = 0, i ∈ I, i = 1 . . . , l dt∇xxL(¯x, u, v)d ≥ 0 . Theorem 21 (KKT second order sufficient conditions) Suppose the point ¯x ∈ S together with multipliers (u, v) satisfies the KKT conditions. Let I + = {i ∈ I : ui > 0} and I 0 = {i ∈ I : ui = 0}. Additionally, suppose that every d (cid:10)= 0 that satisfies...
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N z − b. Let s(z) = B−1b−B−1N z. Then for any z, h(s(z), z) = Bs(z)+N z −b = 0, i.e., x = (s(z), z) solves h(x) = 0. This idea of “invertability” of a system of equations is generalized (although only locally) by the following version of the Implicit Function Theorem, where we will preserve the notation used above...
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and j = 1, . . . , n − l we have: l (cid:16) ∂hi(y, z) ∂sk(z) · ∂zj + ∂hi(y, z) ∂zj = 0 . k=1 ∂yk Proof of Theorem 2: Let A = ∇h(¯x) ∈ (cid:4)l×n . Then A has full row rank, and its columns (along with corresponding elements of ¯x) can be re-arranged so that A = [B | N ] and ¯ x = (¯ z), where B is non-singu...
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we have: 0 = ∂hi(x(θ)) ∂θ = l (cid:16) ∂hi(s(z(θ)), z(θ)) ∂sk(z(θ)) n−l ∂hi(s(z(θ)), z(θ)) ∂zk(θ) ∂θ (cid:16) ∂θ + · · ∂yk ∂zk . k=1 k=1 Let rk = ∂sk(z(θ)) , and recall that ∂zk(θ) = pk. The above equation system ∂θ can then be re-written as 0 = Br + N p, or r = −B−1N p = q. Therefore, ∂xk(θ) = dk ...
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) + θ∇f (¯ x)td + |θ|α(θ) < f (¯ x) for θ > 0 sufficiently small, which contradicts the local optimality of x.¯ Therefore no such d can exist, and the theorem is proved. 7 Constrained Optimization Exercises 1. Suppose that f (x) and gi(x), i = 1, . . . , m are convex real-valued functions over (cid:4)n, and that X ⊂ ...
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What are the necessary opti- mality conditions for this problem? Use these conditions to show that ∗ z = |(cid:15)c(cid:15) −α |. What is the optimal solution x ∗? 4. Let S1 and S2 be convex sets in (cid:4) . Recall the definition of strong separation of convex sets in the notes, and show that there exists a hyperpl...
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. . , λk satisfy λ1, . . . , λk ≥ 0 and (cid:19) k j=1 λj = 1. 8. Let f1(·), . . . , fk (·) : (cid:4)n → (cid:4) be convex functions, and consider the function f (·) defined by: f (x) := max{f1(x), . . . , fk (x)} . Prove that f (·) is a convex function. State and prove a similar result for concave functions. 26...
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if and only if ∇f (¯ x) ≥ 0 for every x ∈ S. x)t(x − ¯ 12. Consider the following problem: maximizex 2 3x1 − x2 + x3 s.t. x1 + x2 + x3 ≤ 0 −x1 + 2x2 + x2 = 0 3 x ∈ (cid:4)3 . • Write down the KKT optimality conditions. • Argue why this problem is unbounded. 27 13. Consider the following problem: minimizex ...
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solves the problem to minimize f (x) subject to gi(x) ≤ 0 for i ∈ M . Let V := {i | gi(ˆx) > 0}. If ∗ z > f (ˆx), show that gi(x ) = 0 for some i ∈ V . (This shows that if an unconstrained minimum of f (·) is infeasible and has an objective value that is less than z , then any constrained minimum lies on the bound...
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(cid:4)n , b ∈ (cid:4)m , A ∈ (cid:4)m×n, and H ∈ (cid:4)n×n. Consider the following two problems: P1 : minimizex c x + 1 xT Hx t 2 s.t. Ax ≤ b x ∈ (cid:4) n and P2 : minimizeu htu + 1 uT Gu 2 s.t. u ≥ 0 u ∈ (cid:4) , m where G := AH −1AT and h := AH −1c+b. Investigate the relationship between the KKT condi...
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a positive multiple of the optimal solution of the following problem: minimized ∇f (¯x)T d s.t. Aβ d dT d d ∈ (cid:4)n . 0 0 ≤ 1 • Suppose that A = −I and b = 0, that is, the constraints are of the form “x ≥ 0”. Develop a simple way to construct d ¯ in this case. x be a feasible point, and let I := {i | gi(...
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TWO HYPOTHETICAL EXAMPLES 1. A furniture manufacturer produces two types of desks: Standard and Executive. These desks are sold to an office furniture wholesaler, and for all practical purposes, there is an unlimited market for any mix of these desks, at least within the manufacturer’s production capacity. Each des...
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1/10 ton of B, and 1/12 ton of C. Compound X costs $250 per ton, compound Y $400 per ton. The cost of processing is $250 per ton of X and $200 per ton of Y. Amounts produced in excess of daily requirements have no value as the products undergo chemical changes if not used immediately. The problem is to find the min...
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066J 27 Summer 2003 (0, 20) Y Y (0, 12) 15.066J X/4 + Y/10 >= 2 (8, 0) X X/12 + Y/12 >= 1 (12, 0) 28 X Summer 2003 Y (0, 20) Z = 9 K Z = 6.4K (16/3, 20/3) (8, 4) Z = 4 K (16, 0) X 15.066J 29 Summer 2003 OBSERVATIONS ON THE GRAPHICAL METHOD FOR SOLVING A LINEAR PROGRAM 1. The set of f...
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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 4: Analysis Based On Continuity 1 This lecture presents several techniques of qualitative systems analysis based on what is frequently called to...
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2) ¯ x ⊂ Rn . Then with x(0) = x exist and are unique on the time interval t ⊂ [0, 1] for all ¯ discrete time system (4.1) with f (¯) = x(1, ¯) describes the evolution of continuous time x system (4.2) at discrete time samples. In particular, if a is continuous then so is f . x 2 Let us call a point in the closur...
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boundary of a subset Y � X � Rn in X is he set of all x ⊂ X such that for every r > 0 there exist y ⊂ Y and z ⊂ X/Y such that |y − x| < r and z − x| < r. For example, the half-open interval Y = (0, 1] is a relatively closed subset of X = (0, 2), and its boundary in X consists of a single point x = 1. Example 4.1 A...
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. ¯ 4.1.2 Proof of Theorem 4.1 x0). Let x1 = x1(t) be the solution of (4.1) with x(0) = ¯ According to the definition of local attractiveness, there exists d > 0 such that x(t) � ¯x0 as t � → for every x = x(t) satisfying (4.1) with |x(0) − ¯x0| < d. Take an arbitrary x1. Then x1(t) � ¯0 as x1 ⊂ A(¯ ¯ t � →, and hen...
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A, this implies f (¯ x) ⊂ d(A). Let us show that the opposite is impossible. Indeed, if f (¯) ⊂ A then, since A is proven open, there exists � > 0 such that z ⊂ A for every z ⊂ X such that |z − f (¯)| < �. Since f is continuous, there exists � > 0 such that |f (y) − f (¯ x)| < � whenever y ⊂ X is such that |y − ¯| <...
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First, if tk,q � → and x(tk,q , x(0)) � ¯q as k � → for every q, and ¯ xq � ¯ q � → then one can select q = q(k) such that tk,q(k) � → and x(tk,q(k), x(0) � ¯ k � →. This proves the closedness (continuity of solutions was not used yet). x x� x� as as Second, by assumption ¯x0 = lim x(tk , x(0)). k�� Hence, by the...
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�� R2 of (4.2), each of which has a limit (possibly infinite) as t � t1 or t � t2. The proof of Theorem 4.3 is based on the more specific topological arguments, to be discussed in the next section. 4.2 Map index in system analysis The notion of index of a continuous function is a remarkably powerful tool for proving...
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� x�Sn det(Jx(Fˆ))dm(x), where Jx(Fˆ) is the Jacobian of Fˆ at x, and m(x) is the normalized Lebesque measure on Sn (i.e. m is invariant with respect to unitary coordinate transformations, and the total measure of Sn equals 1). Once it is proven that the integral in (b) is always an integer (uses standard vol- ume...
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the point of Sn−1 which is the (unique) intersection of the open ray starting from G(x) and passing through x with S n−1 . Then H : Sn−1 × [0, 1] ∞� Sn−1 defined by ˆ H(x, t) = ˆG(tx) is a homotopy between the identity map H(·, 1) and the constant map H(·, 0). Due to existence of the index function, such a homotopy ...
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MIT OpenCourseWare http://ocw.mit.edu 18.306 Advanced Partial Differential Equations with Applications Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Lecture 01 2009 09 09 WED TOPICS: Mechanics of the course. Example pde. Initial and boundary value pr...
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14)(cid:27)(cid:8)(cid:6)(cid:8)(cid:28)(cid:6)(cid:13)"(cid:14)(cid:16)(cid:8)(cid:9)(cid:6)(cid:28)(cid:3)(cid:6))(cid:7)(cid:25)(cid:8)(cid:3)(cid:10)(cid:8)(cid:5) (cid:25)(cid:8)(cid:31)(cid:28)(cid:14))(cid:6))(cid:3)(cid:7)(cid:3)(cid:5)%(cid:24)(cid:26)(cid:25)(cid:3)(cid:4) (cid:5)(cid:25)"(cid:8)(cid:6)(cid:9...
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/1) = 5.7 bits. For an event comprising M such (mutually exclusive) outcomes, the probability is M/52. Q. If I tell you the card is a spade (cid:7), how many bits of information have you received? A. Out of N=52 equally probable cards, M=13 are spades (cid:7), so probability of drawing a spade is 13/52, and the a...
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, Slide #18 Expected Information as Uncertainty or Entropy ete random variable , which may represent Consider a discr the set of possible symbols to be transmitted at a particular time, taking possible values , with respective pr s1, s2,..., sN pS (s1), pS (s2 ),..., pS (sN ) obabilities . S The entropy H (S) ...
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1024 binary digits (C=0 or 1) to code the results of 1024 tosses of this particular coin seems inordinately wasteful, i.e., 1 binary digit per trial. Can we get closer to an average of .0112 binary digits/trial? • Yes! Confusingly, a binary digit is also referred to as a bit! • Binary coding: Mapping source symbols t...
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(cid:1)B(cid:2) (cid:1)C(cid:2) (cid:1)D(cid:2) 1/3 1.58 bits 1/2 1 bit 1/12 3.58 bits 1/12 3.58 bits The expected information content in a choice is given by the entropy: = (.333)(1.58) + (.5)(1) + (2)(.083)(3.58) = 1.626 bits Can we find an encoding where transmitting 1000 choices requires 1626 binary ...
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in S. Repeat the following steps until there is only 1 symbol left in S: –  Choose the two members of S having lowest probabilities. Choose arbitrarily to resolve ties. –  Remove the selected symbols from S, and create a new node of the decoding tree whose children (sub-nodes) are the symbols you've removed. Labe...
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)(cid:25)(cid:16)(cid:15)(cid:3) (cid:26)(cid:13)(cid:11)(cid:8)(cid:17)(cid:10)(cid:13)(cid:21)(cid:3) (cid:12)(cid:8) 0(cid:8) (cid:15)(cid:8) (cid:2)(cid:8) ,(cid:8) -(cid:8) ,,(cid:8) ,-(cid:8) Why isn’t this a workable code? The expected length of an encoded message is (.333+.5)(1) + (.083 + .083)(2) = 1.22 bit...
https://ocw.mit.edu/courses/6-02-introduction-to-eecs-ii-digital-communication-systems-fall-2012/21637440fda8bd0e28d06a9eac518b03_MIT6_02F12_lec01.pdf