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18.409 An Algorithmist’s Toolkit September 22, 2009 Lecturer: Jonathan Kelner Scribe: Dan Iancu (2009) Lecture 4 1 Random walks Let G = (V, E) be an undirected graph. Consider the random process that starts from some vertex v ∈ V (G), and repeatedly moves to a neighbor of the current vertex chosen uniformly at random. ...
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8) π(i) = (cid:88) u∈ V (G) u∈ V (G) d(u ) (cid:80) v∈V (G) d(v) (cid:80) = u∈V (G) d(u) v∈V (G) d(v) (cid:80) = 1. We next show that, if the random walk follows the distribution π at time t, then it has the same distribution at time t + 1. This is expressed using matrix notation in the following claim. 4-1 Claim 2 W ...
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6}. Assume without loss of generality that the random walk starts at time t0 = 1 at vertex 6. Then, at time t, the current vertex is odd if and only if t is odd. Therefore, the walk does not converge to any distribution. 3 Lazy Random Walks There is an easy way to fix the above periodicity problem. We introduce a m...
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is an eigenvector of N , with eigenvalue λ. Let q = D1/2 · v. Then, N · v = λ · v = D−1/2 · W · D1/2 · v = D−1/2 · W · q. Multiplying by D1/2 on the left we obtain Therefore, q is an eigenvector of W with eigenvalue λ. W · q = λ · D1/2 · v = λ · q. Observe that, by Claim 2, W has eigenvector D · 1, with eigenvalue...
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�2 Convergence Define the spectral gap to be λ := 1 − μ(cid:3) 2. For probability distributions p, q, we define their (cid:4)2 distance to be (cid:4)p − q(cid:4)2 = � � (p(i) − q(i))2 . i The following theorem gives a bound on the rate of convergence of the lazy random walk to the stationary distribution π. The...
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μk i · vi � n i=2 (cid:4)pk − π(cid:4)2 = (cid:4) ci · μk i n � i=2 � n � � � · vi(cid:4)2 = � n � 2 · μ2 ci i � � � � k ≤ μk 2 n � 2 ci i=2 i=2 ≤ μk 2 (vi T p0)2 ≤ μk 2 = (1 − λ)k . i=1 Using a similar argument, we can also show an analogous bound for (cid:4)∞ convergence. Theorem 6 F...
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of the connection between conductance and the rate of convergence is as follows. If a graph has high conductance, it is well-connected. Therefore, a large amount of probability mass can very quickly move from one part of the graph to another. 7 Introduction to Monte Carlo methods Assume that we want to estimate π ...
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Theorem 8 (One version of the Chernoff bound) The probability that R fails to (cid:5)-approximate E[R] is Pr [|R − E[R]| ≥ (cid:5)E[R]] ≤ 2e −np(cid:2)2/12 = 2e −E[R](cid:2)2/12 . Some notes on the above bound: • The bound is near tight. • It is necessary for the trials to be independent, in order for the bound to ...
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if we have a big dartboard, and a small circle. In particular, if p is exponentially small, then we need exponentially many trials to expect a constant number of successes. We can also run into trouble if it is hard to throw darts at all. That is, if it is hard to draw samples uniformly at random from the ambient ...
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DIVISOR CLASSES ON THE MODULI SPACE OF CURVES 1. The cohomology of the moduli space of pointed genus zero curves In this section we discuss the Chow rings of the moduli spaces of n-pointed genus zero curve M0,n. Recall that we are working over the complex numbers C. The cohomology and Chow groups of M0,n turn out t...
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λ is the class of the exceptional divisor. Now we introduce the generators of the Chow ring. Let S be a subset of {1, . . . , n} with the property that both S and its complement have at least two elements. We will denote the number of elements of S by #S. Given such a set we can define the class ζS on M0,n as the cl...
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≥S i,l≥S,j,k / ⎪ ≥S 1 (3) For two subsets S and T ζS ζT = 0 unless S ⊗ T, T ⊗ S, S ⊗ T c or T c ⊗ S. � Example 1.3. Since M0,4 = P1, the classes of the three boundary divisors �{1,2}, �{1,3} and �{1,4} are linearly equivalent. If we specialize the statement of the theorem to n = 4 we recover the cohomology of ...
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�T contain the point represeting a curve C in their intersection if and only if there are two nodes on C that divide C into C1, C2 and ∅ , C ∅ where the labeling on C1 is S and the labeling on C ∅ is T . Observe that C1 unless the conditions S ⊗ T, T ⊗ S, S ⊗ T c or T c ⊗ S are satisfied �S and �T are disjoint, henc...
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that one can give a very simple presentation of the coho­ mology ring of D5 realizing it as the blow-up of P2 in four points. Sending the divisors ζi,5 to the classes of the four exceptional divisors E1, . . . , E4 and ζi,j to H − Ek − El (where {k, l} is disjoint from {i, j, 5}) for the remaining i, j gives a ring ...
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of n − 3 blow-ups along explicit smooth centers. Set X1 = M0,n × M0,4. If S is a subset of {1, . . . , n}, we can embed the divisors �S into X1 by first mapping �S by the universal section corresponding to the i-th point to M0,n+1, then following it with the map to X1. Let X2 be the blow-up of X1 along �S where #Sc ...
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homology. Remark 1.9. Observe that M0,n is an affine variety. Fixing three of the points at 0, 1 and → we can view this space as the complement of hyperplanes in Cn−3 . Hence, M0,n is affine of dimension n − 3. Recall that the homology of an affine manifold vanishes above half its real dimension. Theorem 1.10. Let X be ...
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the canonical divisor of M0,n. In fact, we can do better than the previous corollary. Proposition 1.13. Let n ∼ 4. Fix three distinct indeces i, j, k. The second coho­ mology group of M0,n has basis ζ{j,k}, ζS where i ⊕ S and #S � n − 3. Proof. We can give an elementary proof of this result that does not depend on t...
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forgetting a point other than i, j, k. It immediately follows that all the coefficients of the relation have to be zero. � Remark 1.14. Note that the following proposition implies that the rank of the second cohomology group is 2n−1 − n2 − n + 2 . 2 2. The second homology group of the moduli space of curves Orig...
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isors. Let ζirr be the class of the divisor of curves �irr that contain a non-separating node. Let 0 � h � g be an integer and let S be a subset of {1, . . . , n}. Let ζh,S be the class of the divisor �h,S of curves that contain a node which separates the curve into two components of genus h with marked points pi f...
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among these classes are generated by (2) If g = 2, there is the additional relation ζh,S = ζg−h,Sc . (3) If g = 1, there are the following two additional relations 5� = 5ξ + ζirr − 5ζ0 + 7ζ1. � = ξ − ζ0, 12ξp = ζirr + 12 ζ0,S . p≥S,#S�2 ⎪ Since Theorem 1.2 already determines the genus zero case we will omit it ...
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´e duality and the long exact sequence of cohomology Hc k (Mg,n, Q) � H k (Mg,n, Q) � H k(ζMg,n, Q) � Hc k+1(Mg,n, Q) we conclude the following proposition. Proposition 2.2. The map H k(Mg,n, Q) � H k (ζMg,n, Q) is an isomorphism when k < d(g, n) and injective when k = d(g, n), where d(g, n) is defined by n − 4 ...
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boundary of Mg,n. The map H k (Mg,n, Q) � ≥i≥I H k(Xi, Q) is injective if k � d(g, n). Sketch. This proposition follows from the fact that the map H k (Mg,n, Q) � H k(ζMg,n, Q) is a morphism of Hodge structures. Since the map is an injection in the claimed range and H k(Mg,n, Q) is pure of weight k, the cohomolog...
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eth decomposition, Proposition 2.3 and the fact that H 1(Mg,n, Q) = 0 for every g and n. There are many ways of proving the last statement. It follows, for example, from the fact that Mg,n is simply connected. We will see an elementary proof in the next section. 2.1. The relations among tautological classes. In thi...
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,S≤{x,y} otherwise if g = 2h, n = 0 ⎨ Finally, we need to know the pull-backs of tautological classes by the morphism ath,S : Mg−h,n−S≤{x} � Mg,n obtained by attaching a fixed curve of genus h and marking S ∞ {y} to curves in Mg−h,n−S≤{x} by identifying x and y. Exercise 2.8. Show that the following relations hold: (1...
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obtained the claimed relations in The­ orem 2.1. Recall that the Hodge class ω is the first chern class of the Hodge bundle. ⎧ ⎧ ⎧ � Lemma 2.9 (Mumford’s relation). On any Mg,n there is the following relation � = 12ω − ζ + ξ. Proof. It suffices to prove the formula when n = 0. The general case follows by pulling-back ...
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0. � Now all the relations follow when we observe that on M2 we have the relation 10ω = ζ0 + 2ζ1. To prove this relation, for instance, consider the following test families. (1) To a fixed genus 1 curve attach a fixed point of a genus 1 curve at a variable point. (2) On a genus 1 curve identify a variable point wit...
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,n is the morphism that identifies the two points x, y. Since by induction H 2(Mg−1,n≤{x,y}, Q) is tautological ��d may be expressed as a linear combination of tautological classes. Moreover, since the morphism is symmetric under exchanging x and y, the expressions of divisors involving x and y need to be symmetric....
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the points x and y. The classes of these two pull-backs have to coincide. This gives a relation that shows that ��(d − dt) must be identically zero. Since by Proposition 2.4, the map �� is injective, we conclude that d is tautological. To conclude the proof then one needs to analyze the cases of genus 1 and 2 in gr...
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section we have the following Reduction Lemma. ⎧ � Lemma 3.2 (Reduction Lemma). Let k be an odd integer. Suppose that H q (M g,n, Q) = 0 for all odd q � k, and for all g and n such that q > d(g, n), then for all odd q � k and all g and n. H q (M g,n, Q) = 0 In other words, as long as all the odd cohomology for j...
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m or k = l. In this case either the genus is smaller than g or if the genus is equal to g the number of marked points is smaller than n. A � double induction concludes the proof. Proof of vanishing of the first cohomology. By the Reduction Lemma to prove that the first cohomology groups of M g,n vanish we need to ch...
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cohomology group of P1−{0, 1, →} has rank 2. In contrast we saw above that all odd cohomology groups of M 0,n vanish. To emphasize the point, observe that the Euler characteristic of M0,n is given by the formula α(M0,n) = (−1)(n−3)(n − 3)!. To prove this formula consider the map M0,n � M0,n−1 given by forgetting ...
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omorphic to P1, hence their third cohomology clearly vanishes. The moduli spaces M0,5 and M1,2 both have complex dimension 2 or real dimen­ sion 4. By Poincar´e duality we conclude that the dimension of H 3 is equal to the dimension of H 1 . Since H 1 vanishes we conclude that H 3 vanishes. To show the vanishing of ...
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i Y = Y d ⊗ Y d−1 ⊗ · · · ⊗ Y 1 ⊗ Y 0 11 so that Yi = Y i\Y i−1 is empty or of pure dimension i for every i, then by the exact sequence of cohomology with compact supports the Euler characteristic of Y with cohomology with compact supports is the sum of those of Yd and Y d−1. Repeating the process and using Poin...
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of M1,2, M1,3, M0 ,5 denote the space obtained by taking the quotients of M0,4 and M0,5 under the operation of interchanging the labeling of two marked points. To calculate the Euler characteristics of the latter two we note that we have morphisms from M0,4 and M0,5 to these spaces. Both morphisms have degree 2 sinc...
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ramified over the points 0, 1, −1, → of the sphere and it has the extra automorphism coming from rotating the sphere by ψ along the 0 − → axis (multiplication by −1). In the other case the elliptic curve can be realized as ramified over the cube roots of unity and →. Its automorphism group has order 6 and it can be g...
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p2 is not a 2-torsion point and not the special point considered in the previous case. In this case the fiber is an elliptic curve with two points removed. Adding up the various Euler characteristics we conclude that α(M1,3) = 0. This information together with an enumeration of the strata of M 1,3 suffices to calculate...
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group of the moduli functor In this section we will determine the Picard group of the moduli functor following [AC1]. A very good introduction to Picard groups of moduli functors is contained in [Mum]. Let Mg,n denote the moduli functor of genus g stable curves with n marked points. Let (C � S, π1, . . . , πn) den...
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node. The class ζi is the class of the divisor of curves that contain a node that separates the curve to a subcurve of genus i and genus g − i. Similarly ω, ξ1, . . . , ξn, ζirr, ζh,S are elements of P ic(Mg,n). Recall that ω is the Hodge class. The class ξi is the class of the cotangent line at the i-th marked poi...
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ω, ξ1, . . . , ξn and the classes of boundary divisors. The Picard group P ic(Mg,n) is freely generated by ω and ξ1, . . . , ξn. Sketch of the proof of Theorem 4.1. We first remark that P ic(Mg) is torsion free and contains P ic(Mg) as a finite index subgroup. To see that P ic(Mg) is torsion free one uses Teichm¨ulle...
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, ζ∪g/2∈ . The point is to show that it may be expressed as an integral linear combination. The strategy is to construct two different sets of one-parameter families of curves F1, . . . , F∪g/2∈+2 and G1, . . . , G∪g/2∈+2 such that their intersection matrices with respect to ω, ζirr, ζ1, . . . , ζ∪g/2∈ are non-sin...
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zero. All the intersections with the boundary divisors vanish unless i = 1, h or h + 1. The degree of ζ1 on Fh is 1 if h > 1, 0 if g − h − 1 > h = 1 and −1 if g = 3 and g − h − 1 = h = 1. The degree of ζh on Fh is −1 if g − h − 1 > h = 1 or if g − h − 1 = h = 1, 0 if g − h − 1 > h = 1 and −2 if g − h − 1 = h > 1. ...
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genus is 2m + 2, the intersection matrix for the families Kh, C, CE, F1, . . . , Fm has determinant (−1)m+1(h + 1) if m ∼ h ∼ 2. Again taking h = 2 and h = 3 gives � two relatively prime determinants. 5. The Tautological ring of Mg In this course we will not have time to discuss the tautological ring. In this se...
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a rank g locally free sheaf called the Hodge bundle E. The Hodge bundle is defined by E = ψ1��Mg,1 /Mg . The chern classes ωl = cl (E) also define classes in Al (Mg ). Ths subring of the Chow ring generated by these classes is called the tautological ring. = ψ1�K l+1 One of the first things to observe is that the coho...
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2] E. Arbarello and M. Cornalba. Calculating cohomology groups of moduli spaces of curves via algebraic geometry. Inst. Hautes Etudes Sci. Publ. Math. (1998), 97–127 (1999). ´ [Fab] C. Faber. A conjectural description of the tautological ring of the moduli space of curves. In Moduli of curves and abelian varieties,...
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J. Math. 25(2001), 237–243. [GV2] T. Graber and R. Vakil. Relative virtual localization and vanishing of tautological classes on moduli spaces of curves. Duke Math. J. 130(2005), 1–37. [Har1] J. Harer. The second homology group of the mapping class group of an orientable surface. Invent. Math. 72(1983), 221–239. ...
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Picard groups of moduli problems. In Arithmetical Algebraic Geometry (Proc. Conf. Purdue Univ., 1963), pages 33–81. Harper & Row, New York, 1965. R. Vakil. The moduli space of curves and its tautological ring. Notices Amer. Math. Soc. 50(2003), 647–658. [V] 17
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18.409 An Algorithmist’s Toolkit 10/8/2009 Lecturer: Jonathan Kelner Lecture 9 At the end of the previous lecture, we began to motivate a technique called Sparsification. In this lecture, we describe sparsifiers and their use, and give an overview of Combinatorial and Spectral Sparsifiers. We also define Spectral Spa...
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in G(cid:2). Say the answer is a cut with value S(cid:2); then our algorithm will output the estimate S = S(cid:2)/p for the original graph G. Denoting the number of edges between S and S ¯ by e(S) = pc, we have the following concentration result due to Chernoff’s inequality: So our result will be close to the corre...
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a slightly modified version of the Chernoff bound: 9-1 Theorem 2 (Chernoff Bound) Let X1, . . . , Xn be random variables so that Xi ∈ [0, 1], and let X = Xi. Then, (cid:2) P r[|X − E[X]| ≥ (cid:2)X] ≤ 2e −Θ(1)(cid:2)2E[X] Proof The only difference here is that the random variables Xi are no longer discrete variables...
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to be small. In this way, we can approximate cut problems while throwing away more edges which are present in only cuts of high size. Thus, a natural choice for we would be the size of the smallest cut containing e. Unfortunately, we do not know we; however, it is possible to approximate it quickly. The final result ...
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the instances x ∈ [−1, 1]n by normalization. We now have a good definition for a spectral version of sparsification: Definition 3 A Spectral Sparsifier G(cid:2) of a graph G is one for which the relation (1 − (cid:2))x T LG� ≤ x T LGx ≤ (1 + (cid:2))x T LG� x for all x ∈ [0, 1]n 9-2 It is clear from this definition t...
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not a spectral sparsifier of G. Let (cid:4) x = 0 1 . . . n/2 − 1 n/2 − 1 . . . 1 0 (cid:5) . Then, we have that since each vertex contributes Θ( x T LG� x = Θ(nk3) (cid:2) k =1 k2) to the sum. On the other hand, i x T LGx = Θ(nk3) + ( n − 1)2 2 If k is constant, we get that we need (cid:2) = Θ(1/n) for ...
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possibly affect the graph orders. We try again with another definition: M (cid:5) N if the ith eigenvalue of M is ≥ the ith eigenvalue of N for all indices i This is better in that it is basis independent - but it is too basis independent. Under this definition, we have both 9-3 as well as (cid:6) 1 0 0 −1 (cid...
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:5) LH . Claim 6 Let G = (V, EG, wG) and H = (V, EH , wH ) be weighted graphs on the same vertex set such that wG(i, j) ≥ wH (i, j) for all edges (i, j) ∈ E. Then, G (cid:5) H 2.2 Towards Spectral Sparsification With this order relation on graphs, we can now restate the goal of spectral sparsification: Given a dense ...
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probabilities are based on a linear algebra sense of importance, and have a nice interpretation in terms of effective resistance of circuits. To proceed with our analysis, however, we need to develop the ideas of pseudoinverses, calculating effective resistances, and a matrix version of the Chernoff Bound. 9-4 2.3 Ps...
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ent probability. It turns out that the correct way to do this is to sample each edge with probability proportional to its “effective resistance.” The basic idea is to treat each edge as a resistor with resistance 1. If the edge had a capacity of c, we give it a resistance of 1/c. After calculating these values, we s...
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ext = Lv, and v = L+iext We define U (e, v) to be the adjacency matrix with ±1 values. Let ue be the eth row, and v = L+iext. We have and as a result, T Ref f (e) = ueL+ ue Ref f (e) = (U L+U T )e,e Thus, calculating the effective resistance of an edge is as simple as calculating the pseudoinverse of the Laplacian...
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z zT M T LGM z for all vectors z, provided that x ⊥ (LG) ⇒ x ∈ range(M ). Choose M so that M T LGM is a projection. Then, it suffices to show that ≤ 1 + (cid:2) 1 − (cid:2) ≤ (cid:10) M T LH M − M T LGM (cid:10)2≤ (cid:2) From before, we have that LG = U T CU . Choose M = L+U T C 1/2 . Then, we have G Π = M T LGM =...
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)C 1/2) (cid:12) G √ We then set τe = Recall that n−1 ceRef f (e) πe with (cid:10) τe (cid:10)= n − 1. Choose edges with probability pe = c2Ref f (e) n−1 . (cid:3) ceRef f (e) = (cid:3) Πe,e = n − 1 e e Then, we find that E[τeτe T ] = (cid:3) peτeτe T = (cid:3) πeπe T = Π Sample q times with replacem...
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by solving logarithmically many linear systems. This uses the Johnson-Lindenstrauss Lemma. References [1] “Randomized Approximation Schemes for Cuts and Flows in Capacitated Graphs, ” A. Benczur, D. Karger, manuscript. 9-7 MIT OpenCourseWare http://ocw.mit.edu 18.409 Topics in Theoretical Computer Science: An A...
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6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007 Lecture 2-1 Lecture 2 - Carrier Statistics in Equilibrium February 8, 2007 Contents: 1. Conduction and valence bands, bandgap, holes 2. Intrinsic semiconductor 3. Extrinsic semiconductor 4. Conduction and valence band density of states Reading ass...
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bands, bandgap, holes y g r e n e Ec Ev conduction band bandgap ↓ valence band Ec Ev Eg ↓ space coordinate Conduction and valence bands: • bonding electrons occupy states in valence band • ”free” electrons occupy states in conduction band • holes: empty states in valence band • CB electrons and VB holes ...
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e n n o o r r t t c c e e e e l l -- ++ a) υ > Eg hhυ > Eg -- ++ hhυ >υ > EgEg b) Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.72...
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J/3.43J - Integrated Microelectronic Devices - Spring 2007 Lecture 2-6 Key dependencies of ni: • Temperature: • Bandgap: T ↑⇒ ni Eg ↑⇒ ni What is detailed form of dependencies? Use analogy of chemical reactions. Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2...
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), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007 Lecture 2-8 In general, relatively few bonds are broken. Hence: and Then: Two important results: • First, [bonds] � no, po [bonds] � constant nopo ∼ exp(− Eg kT ) ni ∼ exp...
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ni Equilibrium np product in a semiconductor at a certain temper­ ature is a constant specific to the semiconductor. Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD...
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electronic Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. 6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007 Lecture 2-12 Representation of donor and acceptor states in energy band diagram: Ed Ec Ev Ea ED EA E...
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electronic Devices - Spring 2007 Lecture 2-14 4. Conduction and valence band density of states Image removed due to copyright restrictions. Figure 1b) on p. 468 in Laux, S. E., M. V. Fischetti, and D. J. Frank. "Monte Carlo Analysis of Semiconductor Devices: The DAMOCLES Program." IBM Journal of Research and Develo...
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E Common expressions for DOS: gc(E) = 4π ⎛ ⎝ 2m∗ de h2 ⎞ ⎠ 3/2 √ E − Ec E ≥ Ec gv(E) = 4π ⎛ ⎝ 2m∗ dh h2 ⎞ ⎠ 3/2 √ Ev − E E ≤ Ev m∗ m∗ de ≡ density of states electron effective mass dh ≡ density of states hole effective mass Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devi...
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in semiconductors feature DOS ∼ E. √ • Order of magnitude of key parameters for Si at 300 K: – intrinsic carrier concentration: ni ∼ 1010 cm−3 – typical doping level range: ND, NA ∼ 1015 − 1020 cm−3 Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007. MIT OpenCour...
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Lecture 2 8.821/8.871 Holographic duality Fall 2014 8.821/8.871 Holographic duality MIT OpenCourseWare Lecture Notes Hong Liu, Fall 2014 Lecture 2 1.2: BLACK HOLE THERMODYNAMICS 1.2.1: IMPORTANT SCALES Planck scale We can construct physical units using fundamental constants l (reduced Planck constant), GN (...
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137 λG = VG(rc) GN m 1 = 2 2 m l/mc mc2 m2 p = = l2 p r2 c mc2 Then λG « 1, for m « mp. For example, in the case of electron, me = 5 × 10−4GeV /c2, we have λG ∼ 10−43 λEM The gravity effect is quite weak in this case. But if the mass is at Planck mass scale mp, then λG ∼ O(1), which means quantum gravity eff...
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important); 2. m « mp, rs « rc: rs is not relevant, gravity effect is weak and not important; 3. m ∼ mp, rc ∼ rs, quantum gravity effects are important. If this were the whole story, life would be much simpler, but much less interesting. However, black holes can make quantum gravity effects manifest at macroscopic lev...
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object of mass M. If we consider the object to be spherically symmetric, non-rotational, neutral, we have the Schwarzschild metric solution: ds2 = −f dt2 + dr2 + r 2(dθ2 + sin θ2dφ2), 1 f f = 1 − 2GN M r = 1 − rs r (1) Note that from now on, we have adapted the convention to take c = 1. The event horizon is de...
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· · = −dτ 2 + · · · . We have dτh = f 1/2(rh)dt, with τh to be the proper time for Oh. Then 2, t is the proper time for O∞. On the other hand, at r = rh: h dτh dt = (1 − 1 2 rs ) rh As rh → rs, Consider some event of energy Eh happening at r = rh, to O∞ this event has energy → 0, i.e. compared to the time at...
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6.252 NONLINEAR PROGRAMMING LECTURE 3: GRADIENT METHODS LECTURE OUTLINE • Quadratic Unconstrained Problems • Existence of Optimal Solutions • Iterative Computational Methods • Gradient Methods - Motivation • Principal Gradient Methods • Gradient Methods - Choices of Direction QUADRATIC UNCONSTRAINED PROBLEMS ...
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): (cid:1)2 (cid:2)→ (cid:1) given by f (x, y) = 1 αx2 + βy2 − x (cid:1) 2 (cid:2) for various values of α(cid:160)and β. EXISTENCE OF OPTIMAL SOLUTIONS• Consider min f (x)  x∈X Two possibilities: (cid:3) • The set f (x) | x ∈ X and there is no optimal solution • The set f (x) | x ∈ X is bounded below (cid:...
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angle with ∇f (x) that is greater than 90 degrees, ∇f(cid:160)(x) � d < (cid:160)0,(cid:160) x + δd d there is an interval (0, δ) of stepsizes such that f(cid:160)(x+ αd) < f (cid:160)(x) for all α(cid:160) ∈ (0, δ). PRINCIPAL GRADIENT METHODS• xk+1 = xk + αkdk , k = 0, 1, . . . where, if ∇f (xk) (cid:9)= 0,...
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) = 1. Given xk(cid:160), the method ob­ tains xk+1 as the minimum of a quadratic approxima­ tion of f(cid:160) based on a sec­ ond order Taylor expansion around xk(cid:160). OTHER CHOICES OF DIRECTION • Diagonally Scaled Steepest Descent Dk = Diagonal approximation to (cid:1) ∇2f (xk) (cid:2)−1 • Modified New...
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More Symple Types Progress And Preservation Armando Solar-Lezama Computer Science and Artificial Intelligence Laboratory M.I.T. September 28, 2015 September 28, 2015 L06-1 Formalizing a Type System Recap September 28, 2015 Static Semantics • Typing rules – Typing rules tell us how to derive typing j...
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06-10 Preservation • Using Big-Step semantics we can argue global preservation Γ ⊢ 𝑒1: 𝜏 ∧ 𝑒1 → 𝑒2 ⇒ Γ ⊢ 𝑒2: 𝜏 • Prove by induction on the structure of derivation of 𝑒1 → 𝑒2 September 28, 2015 L06-11 Proof by induction on Structure of Evaluation • Base cases: trivial • Inductive cas...
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September 28, 2015 L06-14 Small Step Example • Contexts H ::= o | H e1 | H + e | n + H | if H then e1 else e2 | H == e1 | n == H • Local Reduction Rules – n1 + n2  n (where n = plus n1 n2) – n1 == n2  b (where b =(equals n1 n2)) – if true then e1 else e2  e1 – if false then e1 else e2  e2 – (𝜆x:...
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valid context. L06-18 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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3.15 Photoconductors, Photovoltaics and Photodetectors C.A. Ross, Department of Materials Science and Engineering Reference: Pierret, chapter 9.2 and 9.3 Photoconductors – conductivity a function of light Photovoltaics – generate power from light Photodetectors – use a pn junction to detect light Photoconducting materi...
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orphous Si: uncertainly principle DxDp ≥ h -the localization of carriers gives them an uncertain momentum, so direct absorption of light can occur. Use PIN design because mobility is low. Handout 5 † 1 Scanned article removed due to copyright restrictions. Please See "This Month in Physics History, October 22, ...
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Lecture 7 8.821/8.871 Holographic duality Fall 2014 8.821/8.871 Holographic duality MIT OpenCourseWare Lecture Notes Hong Liu, Fall 2014 Lecture 7 In fact, any orientable two dimensional surface is classified topologically by an integer h, called the genus. The genus is equal to the number of “holes” that the surface ha...
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Theorem: Given a surface composed of polygons with F faces, E edges and V vertices, the Euler character satisfy χ = F + V − E = 2 − 2h Since each Feynman diagram can be considered as a partition of the surface separating it into polygons, then the above theorem also works for our counting in N. Thus in this limit, to t...
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of Feynman diagrams. General observables Now we have introduced two theories: (a) L = − 1 g2 Tr (cid:20) 1 g2 Y M (cid:20) 1 2 1 4 − (∂Φ)2 + (cid:21) Φ4 1 4 Tr FµνF µν (cid:21) − iΨ(D/ − m)Ψ (b) L = (a) is invariant under the global U (N ) transformation: Φ→U ΦU † with U constant U (N ) matrix, i.e. the theory has a gl...
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n)(cid:105)c (1) Note that it is enough to focus on single-trace operators since multiple-trace ones are products of them. Since we are working in the t’ hooft limit, we want to know how correlation (Eq. 1) scales in the large N limit. There is a trick, consider ˆ ˆ  Z [J1, · · · , Jn] = DAµDΦ · · · exp (iSef f ) = DA...
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∼ O(N ) + O(N −1) + · · · (cid:104)O1O2(cid:105)c ∼ O(N 0) + O(N −2) + · · · 2O3(cid:105)c ∼ O(N − ) + O(N − ) + · · · 1 3 (cid:104)O1O All leading order contributions come from planar diagrams. Physical implications: 1. In the large N limit, O(x)|0(cid:105) can be interpreted as creating a single-particle state (”glue...
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4)O(cid:105) =(cid:54) (cid:104)O1O2(cid:105) = (cid:104)O1(cid:105)(cid:104)O2(cid:105) + (cid:104)O the large N limit, it is more like a classical theory. 0 ∼ O(N ), the variance of (cid:104)O(cid:105) is (cid:104)O2(cid:105) − (cid:104)O(cid:105)2 = (cid:104)O2(cid:105)c ∼ O(1), i.e. (cid:104)O2(cid:105)c N (cid:104...
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Lecture Five: The Cacciopolli Inequality 1 The Cacciopolli Inequality The Cacciopolli (or Reverse Poincare) Inequality bounds similar terms to the Poincare inequalities studied last time, but the other way around. The statement is this. Theorem 1.1 Let u : B2r → R satisfy u�u ≥ 0. Then � Br 2 |�u| ≤ 4 2 r � ...
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||�u| . (5) (6) B2r 1 �� Recall the inequality f g ≤ of the Cauchy­Schwarz inequality), and apply it above to get f 2 �1/2 �� � g2 �1/2 for any functions f and g (this is one form � B2r �� φ2|�u|2 ≤ 2 �1/2 �� φ2|�u|2 �1/2 |u|2|�φ|2 . B2r B2r Dividing and squaring then gives � B2r � φ2|�u|2 ≤ 4 ...
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be a test function as before, and consider � B2r |�(φu)| 2 = = � B2r � B2r 2 |φ�u + u�φ| φ2|�u|2 + u |�φ|2 + 2uφ�φ · �u. 2 Apply Cauchy­Schwarz and lemma 1.2 to get 2 � B2r |�(φu)| ≤ 2 � φ2 � |�u|2 + �� u |�φ|2 + 2 2 �1/2 �� 2 |�u| φ2 �1/2 |�φ|2 2 u B2r � ≤ 2 B2r � φ2|�u|2 + 2 2 |�φ|2 . u ...
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Br � (1 + k(n)) B2r \Br � 2 u ≤ 2 u . Br B2r and This completes the proof. (11) (12) (13) (14) (15) 2.2 Bounding the growth of the energy of a harmonic function We will now prove a similar inequality for the Dirichlet energy of a harmonic function. Proposition 2.2 There are dimensional constants c(n) suc...
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Lecture 2 Acoustics of Speech and Hearing 6.551J / HST.714J LECTURE 2: One-Dimensional ‘Traveling Waves’ Main Points - Exponential and sine-wave solutions to the one-dimensional wave equation. - The distributed compressibility and mass in acoustic plane waves are analogous with the distributed capacitance and in...
https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf
(the sound pressure) at any position (x) and time (t) in a one dimensional system like Figure 2.1. Newtons 2nd Law: ∂p(x, t) ∂x = −ρ0 ∂vx (x,t) ∂t Conservation of Mass-Compressibility Relationship: ∂v x (x,t) ∂x = − 1 BA ∂p(x,t) ∂t . The Wave-Equation for Sound Pressure in a Plane Wave: 14-Sept-2004 (1.2.7) (...
https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf
C C C C C C C A A A A A A A A D D D D D D D E D D E E E E E E E E 14-Sept-2004 page 2 Lecture 2 Acoustics of Speech and Hearing 6.551J / HST.714J In the top row the balls and springs are at rest In the second row, ball A is displaced to the left, stretching the AB spring In the third row, ball B h...
https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf
pressure). The inductance/length: The compliance per length: ∂e(x,t) ∂x = −Lx ∂i(x,t) ∂t ∂i(x,t) ∂x = −C x ∂e(x,t) ∂t . (2.2) (2.3) The variations in voltage and current in time and space can be described by wave equations: ∂2e(x,t) ∂x 2 = 1 c 2 ∂2e(x,t) ∂t 2 ∂2i(x,t) ∂x 2 = 1 c 2 ∂2i(x,t) ∂t 2 , and c = 1 LxC...
https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf
) determined by the absolute time t and the time needed to travel to x, i.e. x/c. You should note: –vx and p in each wave are related by the characteristic impedance of the medium z0. – The two scalar pressure terms add. – Because of a difference in direction, the two velocity terms subtract. – Because of the dif...
https://ocw.mit.edu/courses/6-551j-acoustics-of-speech-and-hearing-fall-2004/1149b68a00a45f7c67d4a5c24ed08b51_lec_2_2004.pdf