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cube router „ CM-2: 2048B/PE, plus 2,048 32-bit floating-point units ● Maspar MP-1 (1989) „ 16K 4-bit processors, 16-64KB/PE, 2D + Xnet router „ MP-2: 16K 32-bit processors, 64KB/PE Prof. Saman Amarasinghe, MIT. 25 6.189 IAP 2007 MIT Shared Instruction: SIMD Architecture ● Central controller broadcasts instruc...
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V4 V5 V6 V7 S0 S1 S2 S3 S4 S5 S6 S7 A0 A1 A2 A3 A4 A5 A6 A7 Vi Vj Vk Sj Sk Si Aj Ak Ai V. Mask V. Length FP Add FP Mul FP Recip Int Add Int Logic Int Shift Pop Cnt Addr Add Addr Mul 64-bitx16 4 Instruction Buffers NIP LIP CIP memory bank cycle 50 ns processor cycle 12.5 ns (8...
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A[20] B[20] A[21] B[21] A[22] B[22] A[23] B[23] A[16] B[16] A[17] B[17] A[18] B[18] A[19] B[19] A[12] B[12] A[13] B[13] A[14] B[14] A[15] B[15] C[2] C[1] C[0] C[8] C[4] C[9] C[5] C[10] C[6] C[11] C[7] C[0] C[1] C[2] C[3] Prof. Saman Amarasinghe, MIT. 30 6.189 IAP 2007 MIT Vector Unit Structure Func...
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Compiler schedules parallel execution ● Multiple parallel operations packed into one long instruction word ● Compiler must avoid data hazards (no interlocks) Prof. Saman Amarasinghe, MIT. 33 6.189 IAP 2007 MIT VLIW Instruction Execution 1 IF 2 ID IF Successive instructions Cycles 3 4 5 6 EX WB EX EX ...
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: Superscalar Processors ● Explicit Parallelism ● Shared Instruction Processors ● Shared Sequencer Processors ● Shared Network Processors ● Shared Memory Processors ● Multicore Processors Prof. Saman Amarasinghe, MIT. 37 6.189 IAP 2007 MIT Shared Network: Message Passing MPPs (Massively Parallel Processors) ...
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messages is expensive, need to poll or interrupt Prof. Saman Amarasinghe, MIT. 39 6.189 IAP 2007 MIT Outline ● Implicit Parallelism: Superscalar Processors ● Explicit Parallelism ● Shared Instruction Processors ● Shared Sequencer Processors ● Shared Network Processors ● Shared Memory Processors ● Multicore Pr...
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Each node has 256MB-2GB local DRAM memory ● Load and stores access global memory over network ● Only local memory cached by on-chip caches ● Alpha microprocessor surrounded by custom “shell” circuitry to make it into Y X Z Image by MIT OpenCourseWare. effective MPP node. Shell provides: „ multiple stream buffers i...
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● Bus provides broadcast and serialization point for simple snooping cache coherence protocol ● Modern microprocessors integrate support for this protocol Prof. Saman Amarasinghe, MIT. 45 6.189 IAP 2007 MIT Sun Starfire (UE10000) • Up to 64-way SMP using bus-based snooping protocol μP μP μP μP μP μP μP μP ...
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008 Picochip Ambric PC102 AM2045 Cisco CSR-1 Intel Tflops Raw Raza Cavium XLR Octeon Niagara Cell Boardcom 1480 Xbox360 PA-8800 Opteron Power4 Opteron 4P Xeon MP Tanglewood PExtreme Power6 Yonah P2 P3 Itanium P4 Athlon Itanium 2 1970 1975 1980 1985 1990 1995 2000 2005 20?? Prof. Saman Ama...
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latencies of chip-to-chip communication are drastically reduced „ ARM multi-chip core Private IRQ Configurable # of hardware intr Per-CPU aliased peripherals Configurable between 1 & 4 symmetric CPUs Private peripheral bus Interrupt Distributor CPU Interface CPU Interface CPU Interface CPU Inter...
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Processor Elements „ Or “Streaming Processor Elements” „ Co-processors with dedicated 256kB of memory (not cache) ● IO „ Dual Rambus XDR memory controllers (on chip) – 25.6 GB/sec of memory bandwidth 76.8 GB/s chip-to-chip bandwidth (to off-chip GPU) „ PP P P U U M I C M I C Image removed due to copyright rest...
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or migrate „ back to the era of assembly programming Prof. Saman Amarasinghe, MIT. 58 6.189 IAP 2007 MIT
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18.409 An Algorithmist’s Toolkit September 24, 2009 Lecturer: Jonathan Kelner Scribe: Shaunak Kishore Lecture 5 1 Administrivia Two additional resources on approximating the permanent • Jerrum and Sinclair’s original paper on the algorithm • An excerpt from Motwani and Raghavan’s Randomized Algorithms 2 Review of Monte...
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subset of V . In this case, it would take exponentially many samples from V to get O( log 1/δ ) successes. Second, it could be difficult to sample uniformly from a large and complicated set V . We will see ways to solve both these problems in two examples today. (cid:15)2 3 DNF Counting and an Exponentially Small Target ...
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solution out of 2n assignments. 3.1 Reducing the Sample Space Instead of picking assignments uniformly at random and testing each clause, we will instead sample from the set of assignments that satisfy at least one clause (but not uniformly). This algorithm illustrates the general strategy of sampling only the impo...
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n−ki . 1To see this, understand that the negation of a DNF formula is just a CNF formula by application of De Morgan’s laws. Therefore counting solutions to a DNF formula is equivalent to counting (non-)solutions of a CNF formula, which is the canonical example of a #P -Hard problem. 5-2 Our probability of pickin...
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Matrix and Perfect Matchings Given an n × n 0-1 matrix M , we construct a subgraph G of Kn,n, as follows. Let the vertices on the left be v1, v2, . . . vn and let the vertices on the right be w1, w2, . . . wn. There is an edge between vi and wj if and only if Mij is 1. Suppose σ is a permutation of {1, 2, . . . n}....
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2001: Jerrum, Sinclair and Vigoda showed how to approximate the permanent of an arbitrary graph (and therefore for any matrix with nonnegative entries). We will show today the result of 1989 for approximating the permanent of a dense graph. 4.2.2 General Strategy We can’t do the na¨ıve Monte Carlo here, since the p...
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2. Then every partial matching in Mk−1 has an augmenting path of length ≤ 3. Proof Let m ∈ Mk−1 be a partial matching. Let u be an unmatched node of m. Now suppose that there are no augmenting paths of length 1 starting from u in this matching m (i.e. there is no unmatched node v such that there is an edge connecti...
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f −1(m�)| ≤ (n − k + 1)2 ≤ n2 . Thus |Mk| ≤ n2|Mk−1|. Now we show that 1/n2 ≤ rk. Fix some m ∈ Mk. By Lemma 6, every partial matching in Mk−1 has an augmenting path of length ≤ 3. There are at most k partial matchings in Mk−1 that can by augmented by a path of length 1 to equal m. In addition, there are at most k(k ...
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canonical paths. Lemma 8 Let G = (V, E) be a graph for which we wish to bound Φ(G). For every v, w ∈ V , we specify a canonical path pv,w from v to w. Suppose that for some constant b and for all e ∈ E, we have � v,w∈V I[e ∈ pv,w] ≤ b|V | that is, at most b|V | of the canonical paths run through any given edge e. ...
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walk on G will converge in polynomial time. 4.2.5 The Graph Ck We will only do Cn. It should be clear later how to extend this construction for all k. Recall that our vertices correspond to matchings in Mn ∪ Mn−1. We show how to connect our vertices with 4 different types of directed edges: • Reduce (Mn −→ Mn−1): I...
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,2. 4.2.6 Canonical Paths We still need to define the canonical paths pv,w for our graph Cn. For each node s ∈ Mn ∪Mn−1, we associate with it a “partner” s� ∈ Mn, as follows: • If s ∈ Mn, s� = s. • If s ∈ Mn−1 and has an augmenting path of length 1, augment to get s�. • If s ∈ Mn−1 and has a shortest augmenting pa...
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are perfect matchings, d consists of a collection of disjoint, even-length, alternating (from s� or from t�) cycles of length at least 4. Our canonical path from s� to t� will in a sense “unwind” each cycle of d individually. Now, in order for the path to be canonical, we need to provide some ordering on the cycles...
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of transitions. There is only partner for the first, O(n) for the second, and O(n2) for the third. Therefore there are at most O(n2) nodes s� that can count s� as their partner. Now we wish to count the number of canonical paths for type B. Lemma 11 Let T be a transition (i.e. an edge of Cn). Then the number of pair...
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and after T we agree with t. But what we can do at this point is notice that one of these two edges (which we denote by es,t) always has the start vertex of the current cycle as one of its end-points. Therefore by removing it we end up with a matching again. This is further illustrated in Figure 6. Theorem 12 The c...
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5-8 Figure by MIT OpenCourseWare. Figure 5: Unwinding a collection of cycles (type B path). Figure by MIT OpenCourseWare. Figure 6: The encoding σT (s, t). 5-9 MIT OpenCourseWare http://ocw.mit.edu 18.409 Topics in Theoretical Computer Science: An Algorithmist's Toolkit Fall 2009 For information about citing the...
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MIT OpenCourseWare http://ocw.mit.edu 18.726 Algebraic Geometry Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.726: Algebraic Geometry (K.S. Kedlaya, MIT, Spring 2009) Cohen-Macaulay schemes and Serre duality In this lecture, we extend Serre dual...
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X is equidimensional (each irreducible component has dimension n) and Cohen-Macaulay. (b) The maps θi are isomorphisms for all i ≥ 0 and all coherent sheaves F on X. This is of course meaningless if I don’t tell you what a Cohen-Macaulay scheme is. For the moment, suffice to say that a scheme is Cohen-Macaulay if and...
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sufficiently large, we have H i(X, F (−q)) = 0 for all i < n. (c′) For q sufficiently large, we have H i(X, OX (−q)) = 0 for all i < n. Note that condition (c) is a sort of opposite to Serre’s vanishing theorem, which gives the vanishing of H i(X, F (q)) for i > 0 and q sufficiently large. Proof. Given (b), for any loca...
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, Ext P (j∗OX , ωP ) = 0. N −i Remember that no matter what X is, we have Ext P proved this in the course of constructing the dualizing sheaf ω◦ X . N −i (j∗OX , ωP ) = 0 for i > n: we Proof. By Serre duality on P (and choosing an isomorphism H n(P, ωP ) = k), we may identify ∼ H i(X, OX (−q)) = H i(P, j∗OX (−q)...
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is the ideal of A defining X at x, then for all i < n, ExtN −i A (A/I, A) = 0. Proof. This translates directly from (d) once we remember that ωP is locally free of rank 1 on P . This is almost the local condition we are seeking, except that it still refers to the position of X within P . 2 3 The Cohen-Macaulay...
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is projective if and only if pdA(M) = 0. For M a module over a ring A, a regular sequence is a sequence x1, . . . , xn of elements of A such that for i = 1, . . . , n, xi is not a zerodivisor on M/(x1, . . . , xi−1)M. For A a local ring, the depth of M is the maximal length of a regular sequence with all xi in the m...
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. On the other hand, we always have depthB(B) ≤ dim(B) ≤ n, so it is equivalent to require depthB(B) = dim(B) = n. This condition depthB (B) = dim(B) is in fact the definition of a Cohen-Macaulay lo­ cal ring B. Any regular local ring is Cohen-Macaulay, since we can use generators of the cotangent space as a regular...
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MEDIA AND BOUNDARY CONDITIONS Media: conductivity σ, permittivity ε, permeability μ Media are the only tools we have to create or sense EM fields Conductivity (σ): ⎯J (current density, A/m2) = nq v = σ E n = #q’s/m3, q = charge (Coulombs), =v average velocity (m/s) Semiconductors: P{escape} ∝ e-W/kT + - ∞ ...
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nucleus +q Atom - d ⎯E polarized atoms Metal plates + + + + + - - - - - E P ρv = 0 ρp + + + + + - - - - - =P “Polarization Vector” L2-2 MAGNETIC MATERIALS Basic Equations: B n • ˆda = 0 (cid:119)∫∫S B = μoH in vacuum B = μH = μo (H + M) M = “Magnetization Vector” μ = permeability e + + e...
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Teflon, petroleum 2.1 Vaseline Paper Polystyrene 0.99983 0.99998 0.999991 0.999991 1.000000 1.0000004 1.00002 250 600 2000 5000 100,000 Supermalloy 1,000,000 2.2 Water 2-3 Vacuum 2.6 Air 2.6 Aluminum 3.8 Fused quartz Cobalt 4.15 Nickel Ice Pyrex glass 5.1 Aluminum oxide 8.8 Ethyl alcohol...
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TO BOUNDARIES Using Gauss’s Law: ( (cid:119) ∫∫ S ( ) D D A → 2 ⊥ ˆD nda • 1 ⊥ − ∫∫(cid:119) S ˆ D • n da = ∫∫∫ v ρ dv A = ρ s ˆn A (Lim A → 0, δ2 <<A) ) : 1D δ 2D surface charge density ρs surface S Therefore: D 1 ⊥ − D 2 ⊥ = ρ s yields: ) nˆ D D − 2 1 ( • = ρ s ∫∫(cid:119) A ...
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/ t) • da ∂ ∂ = ∫∫ A H • ds ∫(cid:118) c c∫(cid:118) i H ds ( H → 1// − H ) 2 // Therefore: ( × ˆn H H J − s 1 = ) 2 L = A∫∫(cid:119) ( J nˆ • s → a i J da ) ∂ A∫∫(cid:119) i D da t ∂ → 0 − L L2-8 PERFECT CONDUCTORS Electric Fields: {if σ→∞ and⎯E ≠ 0} ⇒ { (J D/ t) da + ∂ ∂ } ⇒ ...
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0 inside Cooper pairs of electrons disassociate and superconductivity fails when the external B(T) is above a critical threshold L2-9 SUMMARY: BOUNDARY CONDITIONS General Boundary Conditions: ) ˆn D D − 2 1 ) 2 ˆn B B 0 − ( • ( • = 1 = ρ s 1 1E ,D 1 1H ,B ρs ρs ⎯Js ( × ( × ˆn E E 0...
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6.012 - Microelectronic Devices and Circuits – Fall 2009 Inverter Analysis and Design The inverter stage is a basic building block for digital logic circuits and memory cells. A generic inverter stage is illustrated below on the left. It consists of two devices, a pull-up device, which is typically either a bipolar...
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To analyze this circuit we not first that with a MOSFET pull- down, the static input current is zero and if the stage output is connected to the input of a similar stage, the static stage load current will also be zero, and the equation above is simply iPU = iPD. With a resistor pull-up, the pull-up current, iPU, is (...
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- vOUT/2)vOUT, and the desired relationship is found by solving the quadratic (VDD - vOUT)/R = K(vIN - VT - vOUT/2)vOUT The resulting transfer characteristic is plotted in the course text in Figure 15.8. Curves for other pull-up devices are also shown in this section of the text. We can identify six features of a g...
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smallest values are valid; the middle solution is unstable and will not be realized in practice. VHI and VLO are often most easily found with the aid of the graphical technique shown in Figure 15.5 of the course text. To determine the switching times we must first recognize that the reason an inverter output does not ...
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the switching time analytically. There are two cases in which we can calculate the switching time, 3 +-vIN+-PullUpPullDownStage Load+ VqN(vOUT) vOUT however, and to the extent that we can approximate a given situation as one or the other of these to cases, we can estimate the switching time analytically. The first s...
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a differential equation for vOUT(t). We should be able to solve this equation numerically, if we can not do so analytically. If the charge store can be modeled as a linear capacitor and the charging and dischargind currents can be modeled as constant, then τLO→HI ≈ CL(VHI - VLO)/ICH, and τHI→LO ≈ CL(VHI - VLO)/IDCH...
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to design the circuit to have equal magnitude charging and discharging currents, if possible. An example to the switching transient situation in the MOSFET inverter with a resistive pull-up that we illustrated earlier is shown in the figures below. The gate capacitance of the loading stage(s) has been modeled as a li...
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PU = (V - vOUT)/ROFFCLVLO !VHI The dynamic power dissipation is due the fact that each low-high-low cycle the output node is charged to roughly VDD, and then discharged to roughly zero volts. If we can approximate the node charge store with a linear capacitor, CL, the energy 2/2, and the amount of energy stored on t...
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T. Speller January 11, 2007 Scenarios as a creative practice Scenarios are another creative practice increasingly being used by companies and gaining favor from the success of Shell Oil with their future scenarios process (see references below). We intend our system architectures to not just satisfy current requir...
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design as much as you judge is appropriate. In this viewpoint one might consider the selected present SA to be the result of all plausibly considered futures. By modularity and platforming along with other SA strategies, you can achieve flexibility and its close relative, extensibility, with minimal or modest added ...
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MIT OpenCourseWare http://ocw.mit.edu 6.854J / 18.415J Advanced Algorithms Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. � � 18.415/6.854 Advanced Algorithms November 19, 2008 Approximaion Algorithms: MAXCUT Lecturer: Michel X. Goemans 1 MAX-CUT pro...
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Consider a locally maximum cut for this neighborhood. Lemma 1 If (S : S¯) is a local maximum for the MOVE neighborhood, then w(S : S¯) ≥ 2 w(E) ≥ 1 OP T . 2 1 Proof of lemma 1: Look at a vertex i ∈ V . Let Ci be the set of all edges (i, j) ∈ E that are part of the cut (S : S¯) (that is if i ∈ S then j ∈ S ¯ and vice...
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solution with only k/2 edges in the cut. (b) The local search algorithm based on the MOVE neighborhood for MAX-CUT takes expo­ nentially many steps in the worst-case. This is true even for graphs that are 4-regular (each vertex has exactly 4 neighbors) (Haken and Luby [1]). For 3-regular graphs the algorithm is poly...
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In particular, MAX-CUT with the MOVE neighborhood is PLS-complete [5]. Furthermore, it follows from Johnson et al. [3] that the obvious local search algorithm is not an efficient way of finding a local optimum for a PLS-complete problem; indeed, for any PLS-complete problem, there exist instances for which the local s...
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= E[f |A]P r(A) + E[f |A¯](1 − P r(A)) | | {E[f |A], E[f |A¯] }. ≤ max In our setting, we consider the vertices in a specific order, say v1, v2, already decided/conditioned on the position (i.e. whether or not they are in S) of v1, Now, condition on whether vi ∈ S. Letting f = w(S : S¯), we get: , and suppose we ha...
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, we notice that we will place vi on the side , vi−1}. of the cut that maximizes the total weight between vi and the previous vertices {v1, v2, This is therefore a simple greedy algorithm. · · · Remarks: (a) The performance guarantee of the randomized algorithm is no better than 0.5; just consider the complete gr...
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(e)xe s.t. e∈E 0 ≤ xe ≤ 1 ∀e ∈ E � � xe − xe ≤ |F | − 1 ∀cycle C ⊆ E ∀F ⊆ C, |F | odd. e∈F e∈C\F This isa relaxation of the maximum cut problem, and thus provides an upper bound on the value of the optimum cut. We could try to solve this linear program and devise a scheme to “round” the possibly fractional so...
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The idea is to use semidefinite programming to get a more useful relaxation of the maximum cut problem. This is due to Goemans and Williamson [2]. Instead of defining variables on the edges as we did in the previous section, let’s use variables on the vertices to denote which side of the cut a given vertex is. This le...
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T Ly = n � n � yiyj lij = n � � y 2 i w(i, k) − � yiyj w(i, j) i=1 j=1 i=1 k�=i (i,j)∈E = 2w(E) − � yiyj w(i, j) = 4 ⎝ � (i,j)∈E (i,j)∈E ⎛ w(i, j) 1 − yiyj 2 ⎞ ⎠ , and thus � (i,j)∈E w(i, j) 1 − yiyj 2 = 1 4 y T Ly. Lec18-4 � � Thus the maximum cut value is thus equal to max{ y T LY...
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equal to 1. It is easy to see that the coverse is also true: if Y � 0, rank(Y ) = 1 and Yii = 1 for all i then Y = yyT where y ∈ {−1, 1}n . Thus we can reformulate the problem as: max s.t. • L Y 1 4 rank(Y ) = 1, ∀i ∈ V : Yii = 1, Y � 0. This is almost a semidefinite program except that the rank condition is ...
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strong duality). Our semidefinite programming relaxation of MAXCUT is particularly simple and indeed satisfies both the primal and dual regularity conditions: (a) Primal regularity conditions ∃Y � 0 s.t. Yii = 1 ∀i. This condition is obviously satisfied (consider Y = I). Lec18-5 (b) Dual regularity condition: First ...
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, we have that OP T ≥ 0.87856. SDP In order to prove this theorem, we will propose an algorithm which derives a cut from the solution to the semidefinite program. To describe this algorithm, we first need some preliminaries. From the Cholesky’s decomposition, we know that: Y � 0 ⇔ ∃V ∈ Rk×n , k = rank(Y ) ≤ n, s.t. Y...
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pp. 822-839. [5] A.A. Sch¨affer and M. Yannakakis, “Simple local search problems that are hard to solve”, SIAM Journal on Computing, 20, 56–87, 1991. Lec18-7
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6.858 Lecture 6 Capabilities and  other Protection Mechanisms   What's the problem the authors of "confused deputy" encountered? • Their system had a Fortran compiler, /sysx/fort (in Unix filename syntax) • They wanted the Fortran compiler to record usage statistics,  but where? o Created  a special statistics ...
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Not so convenient:  can't just open  the  file  like  any  other. • What if we make gcc setuid to some non-­‐root user  (owner  of stats  file)? o Hard  to  access  user's  original files. • What  if gcc is setuid-­‐root? (Bad  idea, but let's  figure  out why..) o Lots  of potential...
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be  also  the  privileges  for accessing  it. 2. Complex  permission checks: hard for privileged app to replicate. With simpler checks, privileged apps might be able to correctly check if another  user should have access to some object. What are examples of ambient authority? • Unix UIDs, GIDs. • Fire...
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in various  applications. • Overall  plan: o Break up an application into smaller components. o Reduce privileges of components that are most vulnerable to attack. o Carefully  design interfaces so one component can't compromise another. • Why is this difficult? o Hard  to reduce privileges of code ("sand...
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project directory. o Risk: Bob could replace the file with a symlink to Alice's private file. o Alice's process will implicitly use Alice's ambient privileges to open. o Can think of this  as  sandboxing an individual file  operation. What sandboxing plans (mechanisms) are out there (advantages, limitations)? • O...
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isolation. § Finer-­‐grained  isolation is often hard  to get right  (Javascript,  NaCl). § E.g., Native  Client  uses both a fine-­‐grained  sandbox + OS-­‐level   sandbox. o Will look at these in more detail in later lectures. Plan 0: Virtualize everything  (e.g., VMs). • Run untrustworthy code in...
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V Object -> Permissions -> Allow? How  would you sandbox a program on a DAC system (e.g., Unix)? • Must  allocate a new  principal  (user ID): o Otherwise,  existing principal's privileges will be used implicitly! • Prevent process from reading/writing other files: o Change  permissions on every file syst...
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"deputy" for multiple principals. • • One solution: check if group permissions allow access (manual, error-­‐prone). o Alternative solution: explicitly specify  privileges  for each operation. § Capabilities  can help: capability (e.g., fd) combines object + privileges. Unix § Some by name) features incompat. w...
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o When  process reads low-­‐integrity  data, it becomes low integrity too. o Transitive, prevents adversary from indirectly tampering with files. • Not immediately useful for sandboxing: only a fixed number of levels. SElinux. Idea: system administrator specifies a system-­‐wide security policy. • • Policy  file ...
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some server? • Note: quite is seccomp_filter about talks Capsicum  paper different the from regular/old regular/old seccomp. seccomp, and the ] Is it a good idea to separate policy from application code? • Depends  on overall goal. • Potentially good if user/admin wants to look at or change policy. • Problematic i...
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sandbox's  access to objects in global namespaces • Hard  to  control Kernel  changes. • Just to  double-­‐check:  why  do we  need kernel changes? . o Can  we implement everything in a library (and LD_PRELOAD it)? • Represent more things as file descriptors: processes (pdfork). o Good idea in general...
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up the first "..", which goes to "/foo" Look up the second "..", which goes to "/" ## should return a cap • Do Unix permissions still apply? can't o Yes o But intent is that sandbox shouldn't rely on Unix permissions. because  you files  in have acce just cap dir all a s for dir. • For file  descriptors, add ...
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pipe, lease  file. (Even a sandbox can  create  a sandbox.) Advantages: any process can create a new sandbox. • Advantages: fine-­‐grained control of access to resources (if they map to FDs). • Files, network  sockets,  processes. Disadvantage:  weak story  for keeping track of access  to  persi...
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open lots of files all over the place may be cumbersome. • tcpdump. o 2-­‐line version: just  cap_enter() after opening  all FDs. o Used procstat to  look at resulting  capabilities. o 8-­‐line version: also restrict  stdin/stdout/stderr. o Why?   E.g., avoid reading  stderr log,  changing termina...
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, lazy initialization seems to be a problem. § No general-­‐purpose  solution  -­‐-­‐ either  change  code or initialize   early. o Suggested plan: sandbox and see what breaks. § Might  be subtle: gzip  compression level bug. • What  are the security guarantees it  provides? o Guarantees  prov...
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icum_and_Casper.pdf   § § o There's a port of Capsicum  to Linux (but not in upstream kernel repo). What applications wouldn't be a good fit for Capsicum?   • Apps that need to control access to non-­‐kernel-­‐managed  objects. o E.g.: X server state,  DBus,  HTTP  origins in a web browser,  etc. o...
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15.083J/6.859J Integer Optimization Lecture 9: Duality II 1 Outline • Solution of Lagrangean dual • Geometry and strength of the Lagrangean dual 2 The TSP � xe = 2, i ∈ V, e∈δ({i}) � xe ≤ |S| − 1, e∈E(S) S ⊂ V, S =(cid:5) ∅, V, min s.t. xe ∈ {0, 1}. � cexe e∈E � xe = 2, i ∈ V \ {1}, e∈δ({i}) � xe =...
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gradients • Prop: f : (cid:8)n (cid:9) → (cid:8) is concave if and only if for any x ∗ ∈ (cid:8)n s ∈ (cid:8)n such that f (x) ≤ f (x ∗ ) + s (cid:3) (x − x ∗ ). 1 , there exists a vector Slide 4 • Def: f concave. A vector s such that for all x ∈ (cid:8)n: f (x) ≤ f (x ∗ ) + s (cid:3) (x − x ∗ ), ∗ is ca...
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s is a subgradient of the function Z(·) at λ ∗ if and only if Z(λ ∗ ) is a convex combination of the vectors f k , k ∈ E(λ ∗ ). � � 3.2 The subgradient algorithm Input: A nondifferentiable concave function Z(λ). Output: A maximizer of Z(λ) subject to λ ≥ 0. Algorithm: 1. Choose a starting point λ1 ≥ 0; let t = 1. ...
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�Z(λt) is rarely met. Typically, the algorithm is stopped after a fixed number of iterations. 3.3 Example � � • Z(λ) = min 3 − 2λ, 6 − 3λ, 2 − λ, 5 − 2λ, − 2 + λ, 1, 4 − λ, λ, 3 , • θt = 0.8t . Slide 8 2 • λt s t Z(λt) 1.5.00 −3 −9.00 2.2.60 −2 −2.20 3.1.32 −1 −0.68 2 −0.66 4.1.83 1 −0.99 5.1.01 1 −0.6...
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y + λ(cid:3) z, • Geometrically, the hyperplane Z(λ) = y + λ(cid:3) z lies below the set Y . • Theorem: ∀(y, z) ∈ Y. ZD = min y s.t. (y, 0) ∈ conv(Y ). 4.1 Figure 4.2 Example again X = {(1, 0)(cid:3) , (2, 0)(cid:3) , (1, 1)(cid:3) , (2, 1)(cid:3) , (0, 2)(cid:3) , (1, 2)(cid:3) , (2, 2)(cid:3) , (1, 3)(cid:3) ,...
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gradient s t; λt = max{λj + j θtsj t , 0}, where θt = g Zˆ − ZLP(λ) ||st||2 with Zˆ an upper bound on ZD, and 0 < g < 2. With λ = λt solve minx∈X f (x)+ λ(cid:3) g(x) to obtain the optimal value Zt and an optimal solution x t . � � 4. (Solution update) Update x ← αx t + (1 − α)x where 0 < α <1. 5. (Improving st...
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Lecture Notes for LG’s Diff. Analysis trans. Paul Gallagher Feb. 23, 2015 1 The Sobolev Inequality Suppose that u 2 C 1 ∇u is \small", does this imply that u is small? c (Rn). Clearly, if ∇u = 0, then u = 0. So, we can ask if Question 1. If u 2 C 1 c (Rn) and ∫ j∇uj = 1, is there a bound for sup juj? Answer 1. If n = 1,...
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∥∇u∥L1. As in the scaling example, pick (cid:17) a smooth bump function, and de(cid:12)ne (cid:17)(cid:21) as before. Then n(cid:0)1 1 ̸ ∫ j(cid:17)(cid:21)jp = (cid:21)n ∫ ( (cid:17)pdx (cid:20) (cid:21)n ∫ (∫ ) p j∇(cid:17)jdx ) p = (cid:21)n (cid:21)(cid:0)n ( = (cid:21)n (cid:21)(cid:0)n+1 j(∇(cid:17))(x=(cid:21))...
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(cid:1) (cid:1) (cid:1) dxn(cid:0)1 j @nu(x1; (cid:1) (cid:1) (cid:1) ; xn) jdxndx1 (cid:1) (cid:1) (cid:1) dxn(cid:0)1 (cid:20) (cid:20) Rn(cid:0)1 ∫ R j∇uj Rn Then we can use the Loomis-Whitney theorem which we proved in the homework: Theorem 1.2 (Loomis-Whitney). If U (cid:26) Rn is open, and j(cid:25)j(U ) all j, t...
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∫ Rn u(cid:0)1 n (cid:20) ∫ Rn j∇uj. We will proceed by induction. ∫ Rn jujn=(n(cid:0)1) (cid:20) (cid:20) (cid:20) (cid:20) ∫ (∫ R ∫ [∫ Rn(cid:0) 1 R ∫ ∫ Rn(cid:0)1 ) dxn 1=(n 1) (cid:0) juju1=(n(cid:0)1) n ] dx1 [∫ n n (cid:0)2 (cid:0)1 (cid:0) n 1 j j u n(cid:0)2 (cid:1) (cid:1) (cid:1) dx n (cid:0)1 ] j (∫ j un Rn(...
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Chapter 9 Notes, Part 1 - Inference for Proportion and Count Data We want to estimate the proportion p of a population that have a specific attribute, like “what percent of houses in Cambridge have a mouse in the house?” We are given X1, . . . , Xp where Xi’s are Bernoulli, and P (Xi = 1) = p. Xi is 1 if house i ...
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width 2E: pˆ − E ≤ p ≤ pˆ + E so (cid:114) E = zα/2 pˆqˆ n which means n = zα/2 (cid:16) E 2 (cid:17) pˆq. ˆ Question: We’ll take ˆpqˆ to be its largest possible value, (1/2) × (1/2). Why do we do this? Why don’t we just use the ˆp and ˆq that we measure from the data? So, we need: n = (cid:16) zα/2 E (cid...
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(cid:17) p2 1−p2 (cid:16) p1 1−p1 → we’ll use this one “relative risk” “odds ratio” 2 For large samples, we’ll use the CLT: pˆ1 − pˆ2 − (p1 − p2) Z = - pˆ1qˆ1 n1 pˆ2qˆ2+ n2 ≈ N (0, 1) where ˆp1 = X/n1 and ˆp2 = Y /n2. To test H0 : p1 − p2 = δ0, H1 : p1 − p2 = δ0, we can just compute zscore...
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Data Networks Lecture 1 Introduction Eytan Modiano Eytan Modiano Slide 1 6.263: Data Networks • Fundamental aspects of network Design and Analysis: – Architecture Layering Topology design – Protocols Pt.-to-Pt. Multiple access End-to-end – Algorithms Error recovery Routing Flow Control – Analysis tool...
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istributed routing algorithms, optimal routing Flow Control - Window/Credit Schemes Flow Control - Rate Based Schemes Transport layer and TCP/IP ATM Networks Special topic: Optical Networks, Wireless networks Final Exam during final exam week. Date and time to be announced. 13 14 15 16 17 18 19 20 21 22...
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bursty – E.g., Interactive sessions, file transfers, email • Connection oriented services – Long sustained session – Orderly and timely delivery of packets – E.g., Telnet, FTP • Connectionless services – One time transaction (e.g., email) • QoS Eytan Modiano Slide 8 Switching Techniques • Circuit Switching ...
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lengths λ = arrival rate of messages R = channel rate in bits per second X = message transmission delay = L/R – R must be large enough to keep X small – Bursty traffic => λx << 1 => low utilization • Example λ = 1 message per second – L = 1000 bytes (8000 bits) – – X < 0.1 seconds (delay requirement) – => R ...
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to set-up virtual circuit – Once established, local virtual circuit numbers can then be used to represent the virtual circuits on a given link: VC number changes from link to link • Merits of virtual circuits – Save on route computation Need only be done once at start of session – Save on header size – Facilitate ...
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Virtual link for reliable packets Virtual bit pipe DLC DLC DLC DLC phys. int. phys. int. phys. int. phys. int. Data link Control physical interface Physical link subnet node subnet node External site Data link Control physical interface External Site Eytan Modiano Slide 18 Layers • Presentation layer – Provid...
https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf
error-free transmission of packets across a single link – Framing Determine the start and end of packets – Error detection Determine which packets contain transmission errors – Error correction Retransmission schemes (Automatic Repeat Request (ARQ)) Eytan Modiano Slide 22 Physical Layer • Responsible for transmiss...
https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf
driver token ring Protocol token ring driver Ethernet token ring Eytan Modiano Slide 25 Encapsulation user data Appl header user data TCP header application data TCP segment IP header TCP header IP datagram application data Ethernet header 14 IP header 20 TCP header 20 application data Ethernet trailer 4 Ethernet fr...
https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/0feb60d4d98f77a9c08991b4960139f3_Lecture1.pdf