text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
u :=
(T − k)/
(cid:19)(cid:25)
(cid:18)
k + 1
2
gives a contradition (note that this choice of u is smaller than T as long as k > 2). This implies that
|A2| ≤ k which means that
2
n = |A| = |A0| + |A1| + |A2| ≤ k +
(cid:18) (cid:19)
T
u
(cid:18)
= k +
(cid:108)
T
(T − k)/(cid:0)k+1
2
(cid:19)
(cid:1)(cid:109) .
This me... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
7.4 Given n and k, there exists a family A satisfying the k-disjunct property for a number
of tests
T = O
(cid:18) k2 log n
log k
(cid:26)
min
log k,
(cid:27)(cid:19)
.
log n
log k
While the upper bound in Corollary 7.4 and the lower bound in Theorem 7.2 are quite close, there
is still a gap. This gap was recently clos... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
number of elements of the codeword gets erased or replaced
there is no risk for the codeword sent to be confused with another codeword. The set C of codewords
(which is a subset of ΣN ) is called the codebook and N is the blocklenght.
If every two codewords in the codebook differs in at least d coordinates, then there i... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
.
• Hamming bound follows essentially by noting that if a code has distance d then balls of radius
(cid:98) d−1
2 (cid:99) centered at codewords cannot intersect. It says that
R ≤ 1 − Hq
(cid:19)
(cid:18) 1
2
d
N
+
o(1)
• Another particularly simple bound is Singleton bound (it can be easily proven by noting that
the fi... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
d − 1} the pinched Hamming ball of radius d (recall that ∆(d) is the Hamming
weight of x, meaning the number of non-zero entries). In the Boolean Classification problem one is
willing to confuse two codewords as long as they are sufficiently close (as this is likely to mean they are
116 in the same group, and so they are ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
will construct a family A of sets achieving the upper bound in Theorem 7.3. We will do this
by using a Reed-Solomon code [q, m, q − m + 1]q. This code has qm codewords. To each codework c we
will correspond a binary vector a of length q2 where the i-th q-block of a is the indicator of the value
of c(i). This means that... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
(cid:54)
This would give us a family (for large enough parameters) that is k-disjunct for
(cid:23)
(cid:22) q − 1
m − 1
≥
(cid:37)
(cid:36) 2k log n
log k − 1
log n
log q + 1 − 1
(cid:22)
=
2k
(cid:23)
log q
log k
−
log q
log n
≥ k.
T ≈
(cid:18) log n
log k
2k
(cid:19)
2
.
Noting that
concludes the proof.
(cid:50)
7.3... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
underdetermined. Note that the unknown vector, 1i:n has only k non-zero components, meaning it
is k−sparse. Interestingly, despite the similarities with the setting of sparse recovery discussed in a
previous lecture, in this case, O(k2) measurements are needed, instead of O(k) as in the setting of
Compressed Sensing.
˜... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
can then be written as α C⊕2
(cid:1)
= 5. This motivates the definition of
∈ E hold.
(cid:0)
5
Shannon Capacity [Sha56]
(cid:16)
G⊕k(cid:17) 1
k .
θS (G) sup
k
Lovasz, in a remarkable paper [Lov79], showed that θS (C5) =
an open problem for many graphs of interested [AL06], including C7.
√
5, but determining this quanti... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
has the number of times the receiver needs to receive the message
2
so that she can decode the message exactly, with high probability.
(with independent corruptions)
It is easy to see that D n; 1 ≤ 2n, since roughly once in every 2n times the whole message will go
(cid:1) ≤ 2 n but it is not
through the channel unharme... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
set S ⊂ V for which cut(S) is maximal. Goemans and Williamson [GW95] introduced
an approximation algorithm that runs in polynomial time and has a randomized component to it, and
is able to obtain a cut whose expected value is guaranteed to be no smaller than a particular constant
αGW times the optimum cut. The constant... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
exploit the fact that X having a decomposition of the form X = Y T Y is equivalent to being
positive semidefinite, 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
followi... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
u1, . . . , un by a random hyperplane (perpendicular to r). We
will show that the cut given by the set S(cid:48) is comparable to the optimal one.
Figure 20: θ = arccos(uT
i uj)
Let W be the value of the cut produced by the procedure described above. Note that W is a
random variable, whose expectation is easily seen (s... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
≥ αGW MaxCut(G)
8.2 Can αGW be improved?
A natural question is to ask whether there exists a polynomial time algorithm that has an approxi-
mation ratio better than αGW .
Figure 21: The Unique Games Problem
© Thore Husfeldt. All rights reserved. This content is excluded from our Creative Commons
license. For more infor... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
, if the Unique
Games Conjecture is true, the semidefinite programming approach described above produces optimal
approximation ratios for a large class of problems [Rag08].
Not depending on the Unique Games Conjecture, there is a NP-hardness of approximation of 16
17
for Max-Cut [Has02].
Remark 8.3 Note that a simple gr... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
1
2
(cid:34)
(cid:88)
(cid:88)
(cid:35)
wij y2
j
j
i
2
deg(j)yj
w
ijyiyj +
2
deg(i)yi
(cid:88)
i
(cid:88)
i,j
(cid:88)
i,j
(cid:88)
i,j
124 where LG = DG − W is the Laplacian matrix, DG is a diagonal matrix with (DG)ii = deg(i) = (cid:80)
and Wij = wij.
j wij
This means that we rewrite (66) as
max 1 yT LGy
... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
) −
1
4
Tr (LGX) = Tr(D) −
1
4
Tr (LGX) ,
since D is diagonal and Xii = 1. This shows weak duality, the fact that the value of (74) is larger
than the one of (73).
If certain conditions, the so called Slater conditions [VB04, VB96], are satisfied then the optimal
values of both programs are known to coincide, this is kn... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
. This means
4 LG
(cid:13)V T y(cid:13)
2 = (cid:80)n
(cid:13)
(cid:1)T = (cid:13)
k=1(vT
RMaxCut −
yT LGy =
(vT
k y)2.
n
(cid:88)
1
4
k=1
1
In other words, RMaxCut − yT LGy is, in the hypercube (y
1 2) a sum-of-squares of degree 2.
4
This is known as a sum-of-squares certificate [BS14, Bar14, Par00, Las01, Sho87, Nes00... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
certificate of
degree 4 (which corresponds to another, larger, SDP that involves all monomials of degree ≤ 4 [Bar14])
one can improve the approximation of αGW to Max-Cut. Remarkably this is open.
4
Open Problem 8.2
1. What is the approximation ratio achieved by (or the integrality gap of ) the
Sum-of-squares degree 4 re... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
this is that the inequality yiyj + y y
j k + ykyi ≥ − 2 can be proven using sum-of-squares
proof with degree 2:
3
(yi + yj + yk)2 ≥ 0 ⇒ yiyj + yjyk + ykyi ≥ −
3
2
However, the stronger constraint cannot.
On the other hand, if degree 4 monomials are involved, (let’s say XS = (cid:81)
and XijXik = Xjk) then the constrain... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
2 ≥ 0 ⇒ yiyj + yjyk + ykyi ≥ −1.
Interestingly, it is known [KV13] that these extra inequalities alone will not increase the approximation
power (in the worst case) of (68).
8.4 The Grothendieck Constant
There is a somewhat similar remarkable problem, known as the Grothendieck problem [AN04, AMMN05].
Given a matrix A ∈... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
<
√ . See also page 21 here [F+14]. There is also a complex valued analogue [Haa87].
π
2 log(1+
2)
Open Problem 8.3 What is the real Grothendieck constant KG?
8.5 The Paley Graph
Let p be a prime such that p ∼= 1 mod 4. The Paley graph of order p is a graph on p nodes (each
node associated with an element of Zp) where ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
a polynomial improvement on Open Problem 6.4. is known to imply an improvement
on this problem [BMM14].
8.6 An interesting conjecture regarding cuts and bisections
Given d and n let Greg(n, d) be a random d-regular graph on n nodes, drawn from the uniform
distribution on all such graphs. An interesting question is to u... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
. A few lectures ago we discussed
clustering and gave a performance guarantee for spectral clustering (based on Cheeger’s Inequality)
that was guaranteed to hold for any graph. While these guarantees are remarkable, they are worst-case
guarantees and hence pessimistic in nature. In what follows we analyze the performan... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
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 value +1 in one cluster and −1 in the other.
j
(78)
(79)
130 Relaxing the condition xi = ±1, ∀i to (cid:107)x(... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
value is zero and a rank-1
matrix, i.e.
where W = (cid: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
asso... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
is q(1
q).
If we w
ere
to take σ =
2
p(1
−p)+q(1
2
)
−q and use (83) it would suggest that the leading eigenvector of
A correlates with the true 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.... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
probability p if i and j are in the same set, and with probability q if
they are in different sets. Each edge is drawn independently and p > q. In the sparse regime, p = a
n
and q = b , the threshold at which it is possible to make an estimate that correlates with the original
n
partition is open.
132 Open Problem 9.1 C... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
(cid:112)β >
√
2,
(87)
then, with high probability, (79) recovers the true partition. Moreover, if
√
α − (cid:112)β <
√
2,
no algorithm (efficient or not) can, with high probability, recover the true partition.
We’ll consider a semidefinite programming relaxation algorithm for SBM and derive conditions for
exact recovery.... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
T Bx = Tr(xT Bx) = Tr(BxxT ) = Tr(BX)
(cid:88)
i,j
Also, the condition xi = ±1 implies Xii = x2
i = 1. This means that (90) is equivalent to
max
s.t.
Tr(BX)
Xii = 1, ∀i
X = xxT for some x ∈ Rn.
(91)
The fact that X = xxT for some x ∈ Rn is equivalent to rank(X) = 1 and X (cid:23) 0.This means that
(90) is equivalent to... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
definitions
Recall that the degree matrix D of a graph G is a diagonal matrix where each diagonal coefficient Dii
corresponds to the number of neighbours of vertex i and that λ2(M ) is the second smallest eigenvalue
of a symmetric matrix M .
Definition 9.1 Let G+ (resp. G ) be the subgraph of G that includes the edges that... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
ii
135 dual player can take Q = cvvT where v is such that vT Xv < 0. If, on the other hand, X satisfy the
constraints of (93) then
Tr(BX) ≤
min
Z, Q
Z is diagonal
Q(cid:23)0
Tr(BX) + Tr(QX) + Tr (Z (In
×n − X)) ,
since equality can be achieve if, for example, the dual player picks Q = 0n n, then it is clear that the
va... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
to chose Q = Z − B and so we can write
min
Z, Q
Z is diagonal
Q(cid:23)0
max Tr ((B + Q − Z) X) + Tr(Z) =
X
min
Z,
Z is diagonal
Z−B(cid:23)0
max Tr(Z),
X
which clearly does not depend on the choices of the primal player. This means that
max Tr(BX)
X,
≤
Xii
X
∀
i
(cid:23)0
Tr(Z).
min
Z,
Z is diagonal
Z−B(cid:23)0
This ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
,
(this condition is known as complementary slackness)
then X = ggT must be an optimal solution of (93). To ensure that ggT is the unique solution we
just have to ensure that the nullspace of Z − B only has dimension 1 (which corresponds to multiples
of g). Essentially, if this is the case, then for any other possible ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
Lemma 9.2 If
λ2(2LSBM + 11T ) > 0,
(96)
then the relaxation recovers the true partition.
Note that 2LSBM + 11T is a random matrix and so this boils down to “an exercise” in random matrix
theory.
9.9 Matrix Concentration
Clearly,
E (cid:2)2LSBM + 11T (cid:3) = 2ELSBM + 11T = 2ED+
G − 2ED−
G − 2EA + 11T ,
α log(n)
n
I, E... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
to show that
(cid:107)LSBM − E [LSBM ](cid:107) <
α
− β
2
log n,
(97)
which is a large deviations inequality. ((cid:107) · (cid:107) denotes operator norm)
We will skip the details here (and refer the reader to [Ban15c] for the details), but the main idea is
to use an inequality similar to the ones presented in the lec... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
β are
constants. Then, as long as
and q = β log n
α − (cid:112)β >
the semidefinite program considered above coincides with the true partition with high probability.
(99)
√
√
2,
n
n
Note that, if
√
√
2,
α − (cid:112)β <
then exact recovery of the communities is impossible, meaning that the SDP algorithm is optimal.
Furt... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
q if they are in different sets. Each edge is drawn
independently and p > q. In the logarithmic degree regime, we’ll define the parameters in a slightly
different way: p = α(cid:48) log m
and q = β(cid:48) log m . Note that, for k = 2, we roughly have α = 2α(cid:48) and β = 2β
(cid:48),
which means that the exact recovery... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
be too unrealistic. Also, the minimum bisection of multisection objective may not be
the most relevant in some applications.
One particularly popular form of clustering is k-means clustering. Given n points x1, . . . , xn
and pairwise distances d(xi, xj), the k-means objective attempts to partition the points in k clus... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
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 formulation in which each cluster is
assign a mean and the objective function is the sum of the ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
show that these relaxations (both the k-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... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
Ban15b]). More recently, a PCC algorithm was also analyzed for k-means
clustering (based on the SDP described 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:
Give... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
1
X ≥ 0
X (cid:23) 0.
(102)
Prove or disprove the following conjectures.
1. The expected value of this program is
√
2 + o(1).
2. With high probability, the solution of this SDP is rank 1.
Remark 9.6 The dual of this SDP motivates a particularly interesting statement which is implied by
the conjecture. By duality, the v... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
i, j) ∈ E, reveals information about the ratios gi(gj)−1.
j)− and
1
In its simplest form, for each edge (i, j) ∈ E of the graph, we have a noisy estimate of gi(g
the synchronization problem consists of estimating the individual group elements g : V → G that are
the most consistent with the edge estimates, often corresp... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
6And it will follow somewhat the structure in Chapter 1 of [Ban15a]
144 Figure 22: Given a graph G = (V, E) and a group G, the goal in synchronization-type problems is to
estimate node labels g : V → G from noisy edge measurements of offsets gig−1
.
j
There are several approaches to try to solve (103). Using techniques ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
well with z), with high probability?37
37We thank Nicolas Boumal for suggesting this problem.
-x ixjggji~ggij-1145 Figure 23: An example of an instance of a synchronization-type problem. Given noisy rotated copies
of an image (corresponding to vertices of a graph), the goal is to recover the rotations. By comparing
pai... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
probability (cid:107)W x(cid:92)(cid:107) = O(n1/2),
where x(cid:92) is the optimal solution to (103).
∞
˜
˜
38Note that this makes (in this regime) the SDP relaxation a Probably Certifiably Correct algorithm [Ban15b]
146 Image courtesy of Prof. Amit Singer, Princeton University. Used with permission.
Figure 24: Illustr... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
not go into details
here, there is a mechanism that, from two such projections, obtains information between their ori-
entation. The problem of finding the orientation of each projection from such pairwise information
naturally fits in the framework of synchronization and some of the techniques described here can be
adap... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
the multireference alignment of signals (or the orien-
tation estimation problem in Cryo-EM), the alignment step is only a subprocedure of the estimation
of the underlying signal (or the 3d density of the molecule).
In fact, if the underlying signal was
known, finding the shifts would be nearly trivial: for the case of ... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
example of one such measurements is the third image in the first row. We
then proceeded to align these images to a reference consisting of a famous image of Albert Einstein
(often used in the model bias discussions). After alignment, an estimator of the original image was
constructed by averaging the aligned measurement... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
�
Rl
u
1
.
.
.
uL
=
.
u1−l
.
.
.
uL−l
This corresponds to an L-dimensional represen
tation
of the cyclic
group. Then, (108) can be rewritten:
(cid:88)
i,j∈[n]
(cid:104)R−liyi, R−lj yj(cid:105) =
=
=
=
(cid:88)
i,j∈[n]
(cid:88)
i,j
∈[n]
(cid:88)
i,j∈[n]
(cid:88)
i,j∈[n]
T
(R−liyi) R−lj yj
(cid:104)
T... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
�� y
y
2
=
..
.
yn
T
with blocks Cij = yiyj .
(cid:2)
T
y1
T
y2
· · ·
T (cid:3)
nL nL
yn ∈ R × ,
(111)
150 The constraints Xii = IL L and rank(X) ≤ L imply that rank(X) = L and X
only doubly stochastic matrices in O(L) are permutations, (110) can be rewritten as
×
ij ∈ O(L). Since the
max Tr(CX)
s. t.... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
the model in (??)). Another open question is the consistency of the
quasi-MLE estimator, it is known that there is some bias on the power spectrum of the recovered
signal (that can be easily fixed) but the estimates for phases of the Fourier transform are conjecture
to be consistent [BCSZ14].
Open Problem 10.4
1. Is the... | https://ocw.mit.edu/courses/18-s096-topics-in-mathematics-of-data-science-fall-2015/5f0f7205d1cf274e80d77345a7edbf2a_MIT18_S096F15_TenLec.pdf |
.
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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 |
Tm
2. heat balance:
qin
(cid:5)(cid: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)
... | https://ocw.mit.edu/courses/3-044-materials-processing-spring-2013/5f208d00cad80cf13ccba92887fe7aa3_MIT3_044S13_Lec08.pdf |
(cid:8)
(cid:8)
(cid:8)
(cid:8)
l
(within factor of two)
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 |
yt
∗
yu
∗
yv
∗
ys
Proof:
⎡
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,... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5f5e9ce28082dce02c359a8b43a40b62_MIT15_083JF09_lec06.pdf |
:2) (cid:2) (cid:2) (cid:2) (cid:2)
n
n
n
n
n
αi +
βj −
γij =
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 ... | https://ocw.mit.edu/courses/15-083j-integer-programming-and-combinatorial-optimization-fall-2009/5f5e9ce28082dce02c359a8b43a40b62_MIT15_083JF09_lec06.pdf |
row corresponding to wk the random number U lies
strictly to the left of the interval xik.
U
• xij = 1 if mi and wj are matched.
U
E[xij ] = P(U lies in the interval spanned by xij ) = xij .
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
... | 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 |
(x) if
�
|fj (x)| < ∞.
j=1
j
g1(x) = |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.
... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
but we can simply make the series converge less rapidly by adding
a ‘pointless’ subseries. Namely replace gk by
�
k
(6.8)
hn(x) =
⎧
⎪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 convergen... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
a null function in the sense that
�
|f |
(6.11)
Conversely, if (6.11) holds then f ∈ L1(R) and
�
f (x) = 0 ∀ x ∈ R \ E where 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 conver... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
n
�
n
because of (6.12). Thus the null function f ∈ L1(R), and so is |f | and from (6.15)
(6.16)
�
|f | =
�
�
k
�
gk = lim
k→∞
fk = 0
where the last statement follows from the absolute summability.
�
For the converse argument we will use the following result, which is also closely
related to the complet... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
) =
�
j
�
fn,j (x).
j
We can expect f (x) to be given by the sum of the fn,j (x) over both n and j, but
in general, this double series is not absolutely summable. However we can make it
so. Simply choose Nn for each n so that
(6.21)
� �
j>Nn
|fn,j | < 2−n .
This is possible by the assumed absolute summabili... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
23).
�
j≥2
|fn| by (6.1) as N → ∞. Using (6.21) again gives
So, now dropping the primes from the notation and using the new series as fn,j
we can set
(6.25)
gk(x) =
�
n+j=k
fn,j .
This gives a new series of step functions which is absolutely summable since
(6.26)
N �
�
k=1
|gk| ≤
�
�
n,j
|fn,j | ≤
�... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
pointwise series
converges off a set of measure zero – can only diverge on a set of measure zero. It
is rather shocking but this allows us to prove the rest of (10)! Namely, suppose
f ∈ L1(R) and
|f | = 0. Then Proposition 11 applies to the series with each term
being |f |. This is absolutely summable since all the... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
the line
show that
�
(1) If f (x) ≥ 0 a.e. then f ≥ 0.
(2) If f (x) ≤ g(x) a.e. then f ≤ g.
(3) If f is complex valued then its real part, Re f, is Lebesgue integrable and
�
�
�
| Re f | ≤
�
|f |.
(4) For a general complex-valued Lebesgue integrable function
�
|
f | ≤
|f |.
�
(6.30)
Hint: You can loo... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
it
L1(I).
(2) Show that is f is integrable on I then so is |f |.
�
(3) Show that if f is integrable on I and I |f | = 0 then f = 0 a.e. in the sense
that f (x) = 0 for all x ∈ I \ E where E ⊂ I is of measure zero as a subset
of R.
(4) Show that the set of null functions as in the preceeding question is a linear ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
) Prove that the function x−1 cos(1/x) is not Lebesgue integrable on the
interval (0, 1]. Hint: Think about it a bit and use what you have shown
above.
Problem 3.4 [Harder but still doable] Suppose f ∈ L1(R).
(1) Show that for each t ∈ R the translates
(6.37)
ft(x) = f (x − t) : R −→ C
are elements of L1(R).
(2... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
|
|
f .
|
R
(2) Suppose now that G ∈ C([0, 1] × [0, 1]) is a continuous function (I use C(K)
to denote the continuous functions on a compact metric space). Recall from
the preceeding discussion that we have defined L1([0, 1]). Now, using the
first part show that if f ∈ L1([0, 1]) then
(6.41)
F (x) =
G(x, )f ( ) ∈... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
February 10. The description of that construction can be found in the
notes to Lecture 3 as well as an indication of one way to proceed.
Solution. The proof could be shorter than this, I have tried to be fairly complete.
To recap. We start of with a normed space V. From this normed space we
construct the new linear... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
for the usual diagonal sum of the double series and set
(6.46)
wj =
�
(n) .
vk
n+k=j
The problem is that this will not in generall be absolutely summable as a series in
V. What we want is the estimate
(6.47)
�
�wj � =
� �
�
(n)� < ∞.
vk
j
j
j=n+k
The only way we can really estimate this is to use the ... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
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 on the right tends to 0 in V (since it is a fixed number of terms).
Moreover, because of (6.50),
(6.52)
�
N
�
N
�
M
�
�
�
(n)
uj �V +
�uk� ≤ �
uj � + 2
�uk� ≤ �
uj �... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
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) over the
(n)’s so sum over k gives the difference of the sum of the vj
sum of the vj
first p anti-diagonals minus t... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
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 the ‘central interval [1/3, 2/3). This leave C1 = [0, 1/3) ∪ [2/3, 1). Then
remove the central interval from each of... | https://ocw.mit.edu/courses/18-102-introduction-to-functional-analysis-spring-2009/5f60894efb394c09e45df3ba7120c74b_MIT18_102s09_lec06.pdf |
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