text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
idefinite, denoted X (cid:23) 0. The set of PSD matrices is a convex set. Also, the constraint
121 (cid:107)ui(cid:107) = 1 can be expressed as Xii = 1. This means that the relaxed problem is equivalent to the
following semidefinite program (SDP):
max
s.t.
(cid:80)
i<j wij(1 − ij)
1
X
2
X (cid:23) 0 and Xii = 1, i = 1, .... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
the procedure described above. Note that W is a
random variable, whose expectation is easily seen (see Figure 20) to be given by
E[W ] =
(cid:88)
(cid:9)
wij Pr sign(rT ui) = sign(rT uj)
(cid:8)
i<j
(cid:88) 1
wij
π
i<j
=
arccos(uT
i uj).
122(cid:54)
If we define αGW as
it can be shown that αGW > 0.87.
It is then clear... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
) is the following: Given a graph and a set
of k colors, and, for each edge, a matching between the colors, the goal in the unique games problem
is to color the vertices as to agree with as high of a fraction of the edge matchings as possible. For
example, in Figure 21 the coloring agrees with 3 of the edge constraints... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
Sums-of-Squares interpretation
We now give a different interpretation to the approximation ratio obtained above. Let us first slightly
reformulate the problem (recall that wii = 0).
1
max
yi=±1 2
(cid:88)
i<j
wij(1 − yiyj) = max
yi=±1
1
4
= max
1
yi=±1 4
= max
1
yi=±1 4
= max
1
yi=±1 4
= max
1
yi=±1 4
(cid:88)
i,j
(cid:8... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
±1, i = 1, . . . , n.
Similarly, (68) can be written (by taking X = yyT ) as
max 1 Tr (LGX)
4
s.t. X (cid:23) 0
Xii = 1, i = 1, . . . , n.
(72)
(73)
Indeed, given
Next lecture we derive the formulation of the dual program to (73) in the context of recovery in
the Stochastic Block Model. Here we will simply show weak du... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
19)T
LG
+
1
4
n
(cid:88)
i=1
D
ii
(cid:0)y2
i − 1(cid:1) .
(75)
Note that (75) certifies that no cut of G is larger than RMaxCut. Indeed, if y ∈ {±1}2 then y2
i = 1
and so
RMaxCut −
1
4
yT LGy = yT
(cid:18)
D(cid:92) −
(cid:19)T
LG
.
1
4
125 Since D(cid:92) − 1
v1, . . . , vn. This means that meaning that yT (cid:0)D(ci... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
but for details and actual formulations, please see the references.)
The remarkable fact is that, if one bounds the degree of the sum-of-squares certificate, it can be
found using Semidefinite programming [Par00, Las01]. In fact, SDPs (74) and (74) are finding the
smallest real number Λ such that Λ − 1 yT LGy is a sum-of-... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
yiyj + yjyk + ykyi ≥ −1,
Xij + Xjk + Xik ≥ −1,
however it is not difficult to see that the SDP (73) of degree 2 is only able to constraint
33This is related with Hilbert’s 17th problem [Sch12] and Stengle’s Positivstellensatz [Ste74]
Xij + Xjk + Xik ≥ −
3
2
,
126 which is considerably weaker. There are a few different way... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
1 X
ij
1T
ij Xjk Xki
1 Xik Xjk
1 Xij
1
X
Xjk Xik
Xki Xjk Xij
1
≥
0.
Also, note that the inequality yiyj + yjyk + ykyi ≥ −1 can inde
with degree 4 (recall that y2
i = 1):
ed be proven using sum-of-squares proof
(1 + yiyj + yjyk + ykyi)2 ≥ 0 ⇒ yiyj + yjyk + ykyi ≥ −1.
Interestingly, it is known [KV13] tha... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
max (cid:80)
s.t.
ij AijuT
i vj
ui ∈ Rn+m,
vj ∈ R +m, (cid:107)vj(cid:107) = 1
n
(cid:107)ui(cid:107) = 1,
(77)
and 76, and its exact value is still unknown, the best known bounds are available here [] and are 1.676 <
KG <
√ . See also page 21 here [F+14]. There is also a complex valued analogue [Haa87].
π
2 log(1+
2)
... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
suggest that this approach has the potential to improve on the upper bound ω(p)
√
≤ p
Open Problem 8.4 What are the asymptotics of the Paley Graph clique number ω(p) ? Can the the
SOS degree 4 analogue of the theta number help upper bound it? 34
Interestingly, a polynomial improvement on Open Problem 6.4. is known to i... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
that are meaningful already for d = 3.
129 9 Community detection and the Stochastic Block Model
9.1 Community Detection
Community detection in a network is a central problem in data science. A few lectures ago we discussed
clustering and gave a performance guarantee for spectral clustering (based on Cheeger’s Inequalit... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
of spectral
clustering to attempt to partition the graph.
Let A be the adjacency matrix of G, meaning that
Aij =
(cid:26) 1
0
if (i, j) ∈ E(G)
otherwise.
Note that in our model, A is a random matrix. We would like to solve
(cid:88)
max
Aijxixj
s.t. x
i,j
i = ±1, ∀i
(cid:88)
xj = 0,
The intended solution x takes the val... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
:0)A − E[A](cid:1)
. In previous lectures we saw that for large
enough λ, the eigenvalue associated with λ pops outside
the distribution of eigenvalues of W and
whenever this happens, the leading eigenvector has a non-trivial correlation with g (the eigenvector
associated with λ).
q
= −
p
2 n
λvvT
and
n
n
A = W + λvvT
... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
partition vector g as long as
(cid:114)
p − q
2
>
1
√
n
p(1 − p) + q(1 − q)
2
,
(84)
However, this argument is not necessarily valid because the matrix is not a Wigner matrix. For the
special case q = 1 − p, all entries of A − E[A] have the same variance and they can be made to
be identically distributed by conjugating... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
1 Consider the balanced Stochastic Block Model for k > 3 (constant) communities
with inner probability p = a
n and outer probability q = b , what is the threshold at which it becomes
possible to make an estimate that correlates with the original partition is open (known as the par-
tial recovery or detection threshold)... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
5 The algorithm
Note that if we remove the constraint that
Let us define B = 2A − (11T − I), meaning
j xj = 0 in (79) then the optimal solution becomes x = 1.
that
(cid:80)
Bij =
0
1
−1
if i = j
if (i, j) ∈ E(G)
otherwise
(88)
It is clear that the problem
133 (cid:88)
max
Bijxixj
s.t. x
i,j
i = ±1, ∀i
(cid:88)
xj ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
∀i
X (cid:23) 0
rank(X) = 1.
We now relax the problem by removing the non-convex rank constraint
max
Tr(BX)
s.t. Xii = 1, ∀i
X (cid:23) 0.
(92)
(93)
134 This is an SDP that can be solved (up to arbitrary precision) in polynomial time [VB96].
Since we removed the rank constraint, the solution to (93) is no longer guaran... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
= D+
G − D−
G − A
A standard technique to show that a candidate solution is the optimal one for a convex problem is to
use convex duality.
We will describe duality with a game theoretical intuition in mind. The idea will be to rewrite (93)
without imposing constraints on X but rather have the constraints be implicitly ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
this game theoretical intuition in mind, it is clear that if we change the “rules of the game” and
have the dual player decide their variables before the primal player (meaning that the primal player
can pick X knowing the values of Z and Q) then it is clear that the objective can only increase, which
means that:
max T... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
Z − B (cid:23) 0
(95)
is called the dual problem of (93).
The derivation above explains why the objective value of the dual is always larger or equal to
the primal. Nevertheless, there is a much simpler proof (although not as enlightening): let X, Z be
respectively a feasible point of (93) and (95). Since Z is diagonal... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
to (Z − B)g = 0. This means that we need to
B[i, :]g. Since B = 2A − (11T − I) we have
construct Z such that Zii = 1
gi
Zii =
1
gi
(2A
− (11T − I))[i, :]g = 2
1
gi
(Ag)i + 1,
137 meaning that
Z = 2(D+
G − D−) + I
G
is our guess for the dual witness. As a result
Z − B = 2(D+
G − D−)G − I − (cid:2)2A − (11T − I)(cid:3) =... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
n) I + 1 − (α + β)
T (cid:3)
(cid:18)
(cid:19)
log n
n
11T − (α − β)
log n
n
ggT .
Since 2LSBM g = 0 we can ignore what happens in the span of g and it is not hard to see that
(cid:19)
(cid:18)
(cid:20)
(cid:21)
λ2
((α − β) log n) I +
1 − (α + β)
11T − (α − β)
= (α − β) log n.
log n
n
ggT
log n
n
This means that it is ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
:= max
∞
ij
(cid:107)Xij(cid:107) .∞
Using these techniques one can show (this result was independently shown in [Ban15c] and [HWX14],
with a slightly different approach)
Theorem 9.4 Let G be a random graph with n nodes drawn accordingly to the stochastic block model
on two communities with edge probabilities p and q. L... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
can be similarly
defined for any k ≥ 2 communities: G is a graph on n = km nodes divided on k groups of m nodes
each. Similarly to the k = 2 case, for each pair (i, j) of nodes, (i, j) is an edge of G with probability p
if i and j are in the same set, and with probability q if they are in different sets. Each edge is dra... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
(see [ABKK15]).
9.11 Euclidean Clustering
The stochastic block model, although having fascinating phenomena, is not always an accurate model
for clustering. The independence assumption assumed on the connections between pairs of vertices
may sometimes be too unrealistic. Also, the minimum bisection of multisection obje... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
the set of points, in order to formulate the k-medians LP we use variables yp indicat-
ing whether p is a center of its cluster or not and zpq indicating whether q is assigned to p or not
(see [ABC+15] for details), the LP then reads:
35When the points are in Euclidean space there is an equivalent more common formulati... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
-means SDP and the k-medians
LP) are integral in instances that have clustering structure and not necessarily arising from generative
It is unclear however how to define what is meant by “clustering structure”. A
random models.
particularly interesting approach is through stability conditions (see, for example [AJP13]),... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
above) [IMPV15b].
9.13 Another conjectured instance of tightness
The following problem is posed, by Andrea Montanari, in [Mon14], a description also appears in [Ban15a].
We briefly describe it here as well:
Given a symmetric matrix W ∈ Rn×n the positive principal component analysis problem can be
written as
max
s. t.
xT... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
Remark 9.6 The dual of this SDP motivates a particularly interesting statement which is implied by
the conjecture. By duality, the value of the SDP is the same as the value of
which is thus conjectured to be
Λ = 0) is known.
√
2 + o(1), although no bound better than 2 (obtained by simply taking
min λmax (W + Λ) ,
Λ≥0
1... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
imator. Naturally, the measure of “consistency” is application specific. While there is a general
way of describing these problems and algorithmic approaches to them [BCS15, Ban15a], for the sake
of simplicity we will illustrate the ideas through some important examples.
10.2 Angular Synchronization
The angular synchron... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
spectral relaxation of (103) into
max x∗Y x.
(cid:107)x(cid:107)2=n
(104)
Notice that, since the solution to (104) will not necessarily be a vector with unit-modulus entries,
a rounding step will, in general, be needed. Also, to compute the leading eigenvector of A one would
likely use the power method. An interesting ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
SDP relaxation similar to the one for Max-Cut and minimum
bisection.
max
Tr(Y X)
s.t. Xii = 1, ∀i
X (cid:23) 0.
(105)
In [BBS14] it is shown that, in the model of Y = zz∗zz∗ + σW , as long as σ = O(n1/4) then (105)
is tight, meaning that the optimal solution is rank 1 and thus it corresponds to the optimal solution
of ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
random variables.
10.2.1 Orientation estimation in Cryo-EM
A particularly challenging application of this framework is the orientation estimation problem in
Cryo-Electron Microscopy [SS11].
Cryo-EM is a technique used to determine the three-dimensional structure of biological macro-
molecules. The molecules are rapidly... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
fferent com-
munities, then this problem essentially reduces to the Max-Cut problem.
10.3 Signal Alignment
In signal processing, the multireference alignment problem [BCSZ14] consists of recovering an unknown
signal u ∈ RL from n observations of the form
yi = Rliu + σξi,
(106)
where Rli is a circulant permutation matrix... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
information between observations, do not
suffer from model bias as they do not use any information besides the data itself.
In order to recover the shifts li from the shifted noisy signals (106) we will consider the following
estimator
argminl1,...,ln
∈Z
L
(cid:88)
(cid:13)R liyi −
(cid:13)
−
R y
−lj j
(cid:13)
2
(cid:1... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
)
we will refer to this estimator as the quasi-MLE.
It is not surprising that solving this problem is NP-hard in general (the search space for this
optimization problem has exponential size and is nonconvex). In fact, one can show [BCSZ14] that,
conditioned on the Unique Games Conjecture, it is hard to approximate up t... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
RT
l RT
l
1
2
· · ·
T
Rl
n
(cid:3)
∈
R × ,
nL nL
(109)
and can rewrite (108) as
max Tr(CX)
s. t. Xii = IL×L
Xij is a circulant permutation matrix
X (cid:23) 0
rank(X) ≤ L,
(110)
where C is the rank 1 matrix given by
C
1
y
y
2
=
..
.
yn
T
with blocks Cij = yiyj .
(cid:2)
T
y1
T
y2
· · ·
T (cid:3)
nL ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
n the
SDP is tight with high probability?
10.3.3 Sample complexity for multireference alignment
Another important question related to this problem is to understand its sample complexity. Since
the objective is to recover the underlying signal u, a larger number of observations n should yield
a better recovery (consider... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
,
no need to round: integrality of clustering formulations. 6th Innovations in Theoretical
Computer Science (ITCS 2015), 2015.
[ABFM12] B. Alexeev, A. S. Bandeira, M. Fickus, and D. G. Mixon. Phase retrieval with polarization.
available online, 2012.
[ABG12]
L. Addario-Berry and S. Griffiths. The spectrum of random lifts... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
D. Amelunxen, M. Lotz, M. B. McCoy, and J. A. Tropp. Living on the edge: phase
transitions in convex programs with random data. 2014.
[Alo86]
N. Alon. Eigenvalues and expanders. Combinatorica, 6:83–96, 1986.
152 [Alo03]
[AM85]
N. Alon. Problems and results in extremal combinatorics i. Discrete Mathematics, 273(1–
3):31... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
. Bai. Methodologies in spectral analysis of large dimensional random matrices, a
review. Statistics Sinica, 9:611–677, 1999.
[Ban15a]
A. S. Bandeira. Convex relaxations for certain inverse problems on graphs. PhD thesis,
Program in Applied and Computational Mathematics, Princeton University, 2015.
[Ban15b]
A. S. Bande... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
.OC], 2014.
[BCS15]
A. S. Bandeira, Y. Chen, and A. Singer. Non-unique games over compact groups and
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3.044 MATERIALS PROCESSING
LECTURE 8
Radiation:
∂T−k
∂x
(cid:2)
=
−εσ
surf − T 4
T 4
source
(cid:3)
M = εσ
L
k
T 3
surf
∂T
∂t
= α
∇2T
q = εσ(T 4
surf − T 4
source)
⇒ Very few analytical solutions, some charts
Date: March 5th, 2012.
1
2
LECTURE 8
CVD: Chemical Vapor Deposition
At steady state the thermocouple outputs a... | https://ocw.mit.edu/courses/3-044-materials-processing-spring-2013/5f208d00cad80cf13ccba92887fe7aa3_MIT3_044S13_Lec08.pdf |
6)
∂T
∂x
(cid:7)
(cid:8)
(cid:8)
(cid:8)
(cid:8)
−ks
=
(cid:4)
s
(cid:4)
qout
(cid:5)(cid:6)
∂T
∂s
(cid:7)
(cid:8)
(cid:8)
(cid:8)
(cid:8)
l
−kl
3. heat of fusion
Look closely:
qin = −kl
(cid:8)
(cid:8)
(cid:8)
(cid:8)
∂T
∂x
x=s,l
4
LECTURE 8
qout = −ks
Fusion: − Hf
(cid:8)
(cid:8)
(cid:8)
(cid:8)
∂T
∂x
(cid:9)
kJ
kg
... | https://ocw.mit.edu/courses/3-044-materials-processing-spring-2013/5f208d00cad80cf13ccba92887fe7aa3_MIT3_044S13_Lec08.pdf |
)
MIT OpenCourseWare
http://ocw.mit.edu
3.044 Materials Processing
Spring 2013
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/3-044-materials-processing-spring-2013/5f208d00cad80cf13ccba92887fe7aa3_MIT3_044S13_Lec08.pdf |
15.083J/6.859J Integer Optimization
Lecture 6: Ideal formulations II
1 Outline
• Randomized rounding methods
2 Randomized rounding
• Solve c(cid:2)x subject to x ∈ P for arbitrary c.
• x be optimal solution.
∗
• From x ∗ create a new random integer solution x, feasible in P : E[c(cid:2)x] =
ZLP = c(cid:2)x ∗ .
... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5f5e9ce28082dce02c359a8b43a40b62_MIT15_083JF09_lec06.pdf |
⎡
ZIP ≤ E[ZH] = E ⎣
(cid:2)
⎤
⎦
cuv xuv
(cid:2)
{u,v}∈E
(cid:2)
=
=
{u,v}∈E
(cid:7)
(cid:8)
(cid:10)
(cid:9)
(cid:9)
∗
∗
cuv P min yu, yv ≤ U < max yu, yv
(cid:8)
∗
∗
∗
∗
cuv |yu − yv |
{u,v}∈E
= ZLP ≤ ZIP
2.3 Stable matching
• n men {m1, . . . , mn} and n women {w1, . . . , wn}, with each person h... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5f5e9ce28082dce02c359a8b43a40b62_MIT15_083JF09_lec06.pdf |
=
xij .
• Complementary slackness of optimal primal and dual solutions.
i=1
j=1
i=1 j=1
i=1 j=1
3
0
mi
(cid:2)
xij
0011 01
01
x0011
ik
j }
w
{k|wk <mi
1
(cid:2)
{k|mk >wj
xkj
mi }
wj
01
01 01
2.4 Key Theorem
PSM = conv(S).
2.4.1 Randomization
• Generate a random number U uniformly in [0,1].
... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5f5e9ce28082dce02c359a8b43a40b62_MIT15_083JF09_lec06.pdf |
Slide 10
Slide 11
Slide 12
(cid:12)
• xij = 0
1 u
xij du: x can be written as a convex combination of stable matchings
x u as u varies over the interval [0, 1].
4
MIT OpenCourseWare
http://ocw.mit.edu
15.083J / 6.859J Integer Programming and Combinatorial Optimization
Fall 2009
For information about citing ... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5f5e9ce28082dce02c359a8b43a40b62_MIT15_083JF09_lec06.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.102 Introduction to Functional Analysis
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
LECTURE NOTES FOR 18.102, SPRING 2009
27
Lecture 6. Tuesday, Feb 24
By now the structure of the proofs should be getting... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
f1(x)|, gk(x) = |
fj (x)| − |
fj (x)| ∀ x ∈ R.
k
�
k−1
�
j=1
j=1
Then, for sure,
(6.4)
N
�
N
�
gk(x) = |
fj (x)| → |f (x)| if
�
|fj (x)| < ∞.
k=1
j=1
j
So, what we need to check, for a start, is that {gj } is an absolutely summable series
of step functions.
The triangle inequality in the form ||v| − ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
�
⎪gk(x)
⎨
fk(x)
−fk(x)
⎩
⎪
if n = 3k − 2
if n = 3k − 1
if n = 3k.
This series converges absolutely if and only if both the |gk(x)| and |fk(x)| series
converge – the convergence of the latter implies the convergence of the former so
(6.9)
�
|hn(x)| < ∞ ⇐⇒
|fk(x)|.
�
n
k
On the other hand when this hold... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
E is of measure zero.
|f | = 0.
Proof. The main part of this is the first part, that the vanishing of
|f | implies
that f is null. This I will prove using the next Proposition. The converse is the
easier direction.
�
Namely, if f is null in the sense of (6.11) then, by the definition of a set of
measure zero, the... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
also closely
related to the completeness of L1(R).
Proposition 11. If fn ∈ L1(R) is an absolutely summable series, in the sense that
� �
|fn| < ∞ then
n
(6.17)
E = {x ∈ R;
�
|fn(x)| = ∞} has measure zero
and if f : R −→ C and
(6.18)
then f ∈ L1(R) and
(6.19)
n
f (x) =
�
fn(x) ∀ x ∈ R \ E
n
�
f =
� ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
summability – so the tail of the series is
small. Having done this, we replace the series fn,j by
(6.22)
f � =
n,1
fn,j (x), n,j
f �
(x) = fn,Nn
+k−1(x) ∀ j ≥ 2.
�
j≤Nn
30
LECTURE NOTES FOR 18.102, SPRING 2009
This still converges to fn on the same set as in (6.20). So in fact we can simply
replace fn,j by ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
convergent series we see that
(6.27)
|fn,j (x)| < ∞ = ⇒ f (x) =
|gk(x)| =
fn,j (x).
�
�
� �
n,j
k
n
j
From a result last week, we know that the set on the left here is of the form R \ E
|fn,j (x)| = ∞.
where E is of measure zero; E is the union of the sets En on which
�
j
So, take another absolutel summa... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
0
} has measure zero
which is what we wanted to show.
Finally this allows us to define the standard Lebesgue space
(6.29)
and to check that
L1(R) = L1(R)/N , N = {null functions}
|f | is indeed a norm on this space.
�
�
�
LECTURE NOTES FOR 18.102, SPRING 2009
31
Problem set 3, Due 11AM Tuesday 3 Mar.
This pro... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
(5) Show that the integral is a continuous linear functional
�
(6.31)
: L1(R) −→ C.
Problem 3.2 If I ⊂ R is an interval, including possibly (−∞, a) or (a, ∞), we
define Lebesgue integrability of a function f : I −→ C to mean the Lebesgue
integrability of
(6.32)
f˜ : R −→ C,
f˜(x) =
�
f (x) x ∈ I
0
x ∈ R \ I... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
102, SPRING 2009
(7) Show that the preceeding construction gives a surjective and continuous
linear map ‘restriction to I’
(6.35)
L1(R) −→ L1(I).
(Notice that these are the quotient spaces of integrable functions modulo
equality a.e.)
Problem 3.3 Really continuing the previous one.
(1) Show that if I = [a, b) and... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
to be zero outside, is Lebesgue integrable. Using this,
and the fact that step functions are dense in L1(R) show that the linear space of
continuous functions on R each of which vanishes outside a compact set (which
depends on the function) form a dense subset of L1(R).
Problem 3.6
(1) If g : R −→ C is bounded and... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
of contin
uous functions, with supremum norm, on [0, 1].
34
LECTURE NOTES FOR 18.102, SPRING 2009
Solutions to Problem set 2
I was originally going to make this problem set longer, since there is a missing
Tuesday. However, I would prefer you to concentrate on getting all four of these
questions really right!
... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
(6.45)
�
n
�bn�B < ∞.
We have to show that it converges in B. The first task is to guess what the limit
should be. The idea is that it should be a series which adds up to ‘the sum of the
(n) in V. So, we
bn’s’. Each bn is represented by an absolutely summable series vk
can just look for the usual diagonal sum of... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
∞ �
uk� and
�uk�V < ∞. Choosing M large it follows that
�
N
k=1
(6.50)
�uk�V ≤ 2−n−1 .
With this choice of M set v(n) =
1
uk and v(n) = uM +k−1 for all k ≥ 2. This does
k
still represent bn since the difference of the sums,
(6.51)
N
� �
(n) −
vk
N
uk = −
N +M −1
�
uk
k=1
k=1
k=N
for all N. The sum... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
�
�b −
n=1
�
n
�
N
n=1
with nth term
(6.55)
and the norm is then
(6.56)
N
�
(n)
vk
wk −
n=1
lim
p→∞
p
� �
(wk −
�
N
k=1
n=1
v(n)
k )�V .
The norm here is itself a limit – b −
bn is represented by the summable series
Then we need to understand what happens as N → ∞! Now, wk is the diagonal
(n... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
2 Let’s consider an example of an absolutely summable sequence
of step functions. For the interval [0, 1) (remember there is a strong preference
for left-closed but right-open intervals for the moment) consider a variant of the
construction of the standard Cantor subset based on 3 proceeding in steps. Thus,
remove ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
to one on the union of all
the subintervals of [0, 1) which are removed in the construction and zero
elsewhere. Show that g is Lebesgue integrable and compute its integral.
Solution.
(1) The total length of the intervals is being reduced by a factor of
1/3 each time. Thus l(Ck) = 3k . Thus the integral of f, which... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
k+1, which is not an
empty set, f (x) = k.
(5) The set F, which is the union of the intervals removed is [0, 1) \ E. Taking
step functions equal to 1 on each of the intervals removed gives an absolutely
summable series, since they are non-negative and the kth one has integral
1/3×(2/3)k−1 for k = 1, . . . . This s... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
of the areas of the rectangles is at least as large of
that of the rectangle contained in the union.
(5) Prove the extension of the preceeding result to a countable collection of
rectangles with union containing a given rectangle.
Solution.
(1) For subdivision of one rectangle this is clear enough. Namely
we eith... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
number of
points in C1 and the finite number of points in C2. The total area remains
the same and now the rectangles covering [a1, b1) × [A2, b2) are precisely
the Ai × Bj where the Ai are a set of disjoint intervals covering [a1, b1)
and the Bj are a similar set covering [a2, b2). Applying the one-dimensional
resu... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
rectangles cover D. Now look at
the next one, DN −k+1. Subdivide it using all the endpoints of the intervals
for the earlier rectangles D1, . . . , Dk and D. After subdivision of DN −k+1
each resulting rectangle is either contained in one of the Dj , j ≤ N − k
or is not contained in D. All these can be discarded an... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
9
39
[a2, b1 − δ). Applying the preceeding finite result we see that
(6.66)
Sum of areas + 2Cδ ≥ Area D − 2Cδ.
Since this is true for all δ > 0 the result follows.
�
I encourage you to go through the discussion of integrals of step functions – now
based on rectangles instead of intervals – and see that everythin... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
function f and then
add afterwards. By the uniform continuity of a continuous function on a
compact set, in this case [0, 1], given n there exists N such that |x − y| ≤
2−N = ⇒ |f (x)−f (y)| ≤ 2−n . So, if we divide into 2N equal intervals, where
N depends on n and we insist that it be non-decreasing as a function ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
� 1
0
f (x)dx
(6.71)
(6.72)
where on the right is the Riemann integral. However this follows from the
fact that
�
�
f = lim
n→∞
Fn
and the integral of the step function is between the Riemann upper and
lower sums for the corresponding partition of [0, 1].
� | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
OpencourseWare
2.11 The p diacritic (and the %r tone label)
9 August 2006
Note for this 2006 ToBI tutorial: Prosodic labelling of disfluencies has turned out to be one of
the biggest challenges to the ToBI system, and because these events happen often in spontaneous
speech, extending the system to label them appro... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
isi-Kelm and Sun-Ah Jun (2005) “A comparison of disfluency patterns in normal and stuttered
speech” in Proceedings of Disfluency in Spontaneous Speech Workshop, Aix-en Provence, France.
1 1 1
1 3-
2p 0
4
which leave after 4:00 p.m.
2p 2p 3p 2p 4
The p diacritic is used in conjunction with a break index 1, 2, o... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
I mean the stuff he knows is kind of
0 0 1p 3 1 4 1
1 1
4 1 1 1
amazing 'coz he does a lot of um environmental
3 1p 1 1 X 0 1 4
1
impact stuff
2p 4
EXAMPLE <cheapest>: I want to see the cheapest flight from Atlanta
1
1 1 1 1 3p
1 1
3
to Baltimore
1
4
In general the p diacritic should be used conse... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
In these cases there is clearly no disfluency and 1p should not be used. Conversely,
an abrupt cutoff that needs a 1p can occur when no phonological material is missing, as when a
final vowel is abruptly interrupted, e.g. in an utterance like “I-I don’t know”. In these cases
there is a perceived disfluency i.e. a su... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
H*
H-L%
EXAMPLE <connections>: What are the plane sizes for these flights and
4 1
4 1
1
1 1 1
H*
1
L* H-H% H* H* H-L%
do they ha(ve)- do are there any other flights
2p 1
1p 1p 1
%r
1 1
H*
1
1
!H*
that have s-
connections
1 1 1p
4
%r H* L-L%
As with the use of the p diacritic, one should be cons... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
index for the intermediate phrase.
Summary of ToBI labels introduced so far:
Tones:
H*: high pitch accent
L*: low pitch accent
L+H*: bitonal pitch accent with low tone followed by high tone prominence
L*+H: bitonal pitch accent with low tone prominence followed by high tone
!H*: downstepped high pitch accent
L+... | https://ocw.mit.edu/courses/6-911-transcribing-prosodic-structure-of-spoken-utterances-with-tobi-january-iap-2006/5f9c5ac91b6352207d27efe39712e480_chap2_11.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
ESD.70J / 1.145J Engineering Economy Module
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ESD.70J Engineering Economy
Fall 2009
Session One
Michel-Alexandre Cardin
Prof. Richard de Neufville
ESD.70J Engineering Eco... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
Analysis
– today
2. Monte Carlo simulations
3. Modeling uncertainty using different
stochastic models and probability
distributions
4. Analyzing the system in the context of
flexibility
ESD.70J Engineering Economy Module - Session 1
6
Course Materials
• Excel spreadsheets
– ESD70session# –1.xls : setup before the... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
movie of THE BEST MAN, 1964
… weather … oil supply … stock market …
ESD.70J Engineering Economy Module - Session 1
10
Modeling with dynamic mentality
• Cannot ignore the intrinsic uncertainty of
the future (cid:198) DYNAMIC mentality in
decision-making to the rescue
• Excel is a decent tool for decision analysis
• W... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
linked to
entries => NEVER HARD CODE INPUTS
ESD.70J Engineering Economy Module - Session 1
16
Net Present Value (NPV) primer
• NPV = PV (Cash Inflows) - PV (Cash Outflows)
=NPV
T
∑
0=t
CF
t
r)+(1
t
NPV ≥ 0 ⇒ valuable project
‘r’ reflects risk of project, expressed as required
rate of return. Also called discount ra... | https://ocw.mit.edu/courses/esd-70j-engineering-economy-module-fall-2009/5fa0b84793362aca263cb900be59ba34_MITESD_70Jf09_lec01.pdf |
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