text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
− f (x�
i))2 .
n
�
i=1
Use the Symmetrization Lemma with ξ1 = Z(x), ξ2 = Z(x�), and
ξ3 = sup
f ∈F
n
�
i=1
(f (xi) − f (x� ))2 .
i
It is enough to prove that
⎛
P ⎝Z(x) ≥ Z(x�) + 2 �t sup
(f (xi) − f (x� ))2⎠ ≤ 4e−t/4 ,
i
n
�
�
� �
f ∈F i=1
⎞
i.e.
⎛
P ⎝sup
n
�
f ∈F i=1
n
�
f ∈F i=1
n
�
�
� �
f ∈F i=1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
= 0) = P (εi = 1) = 1/2.
⎛
Px,x� ⎝sup
f ∈F
n
�
i=1
f (xi) ≥ sup
f ∈F
n
�
i=1
f (xi
�
�
� �
� ) + 2 �t sup
f ∈F
n
i=1
⎞
(f (xi) − f (x� ))2⎠
i
= Px,x�,ε
Define
and
�
n
�
i=1
sup
f ∈F
(f (x� ) + εi(f (xi) − f (x� ))) ≥ sup
f ∈F
i
i
(f (xi) − εi(f (xi) − f (x� )))
i
n
�
i=1
�
�
� �
+2�t sup
f ∈... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Moreover, M edian(Φ1) =
i=1
i
and
�
Φ1 ≤ M (Φ1) + L
√
�
t
Pε
≥ 1 − 2e−t/4
�
Φ2 ≤ M (Φ2) − L
√
�
t
Pε
≥ 1 − 2e−t/4 .
With probability at least 1 − 4e−t/4 both above inequalities hold:
Φ1 ≤ M (Φ1) + L
√
t = M (Φ2) + L
√
t ≤ Φ2 + 2L
√
t.
Thus,
and
�
Φ1 ≥ Φ2 + 2L
√
�
t
Pε
≤ 4e−t/4
�
Px,x�,ε Φ1 ≥ Φ2 + ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
iance
�
�
�
�
97
Lecture 36
Talagrand’s two-point inequality.
18.465
Let x ∈ X n . Suppose A1, A2 ⊆ X n . We want to define d(A1, A2, x).
Definition 36.1.
d(A1, A2, x) = inf{card {i ≤ n : xi =�
1 and xi =�
yi
2}, y 1 ∈ A1, y 2 ∈ A2}
yi
Theorem 36.1.
and
E2d(A1 ,A2 ,x)
=
�
2d(A1,A2,x)dP n(x) ≤
1
P n(A1)P... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
3
�
min 2,
�
1
1
,
g1 g2
+ g1 + g2 ≤ 3.
If min is equal to 2, then g1, g2 ≤ 1 and the sum is less than 3.
If min is equal to 1 , then g1 ≥ 1 and g1 ≥ g2, so min +g1 + g2 ≤ 1 + 2g1 ≤ 3.
2
g1
2
g1
We now prove the Theorem:
Proof. Proof by induction on n.
n = 1 :
d(A1, A2, x) = 0 if x ∈ A1 ∪ A2 and d(A1, A2, ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
=
A2(xn+1)
xn+1
xn+1
d(A1, A2, x) = d(A1, A2, (z, xn+1)) ≤ 1 + d(B1, B2, z),
d(A1, A2, (z, xn+1)) ≤ d(A1(xn+1), B2, z),
d(A1, A2, (z, xn+1)) ≤ d(B1, A2(xn+1), z).
�
2d(A1,A2,x)dP n+1(z, xn+1) =
� �
�
2d(A1,A2,(z,xn+1 ))dP n(z) dP (xn+1)
��
I(xn+1)
�
The inner integral ca ne bounded by induction as follow... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
n(A1(xn+1)/P n(B1) P n(A2(xn+1)/P n(B2) ⎠
�
� �
�
1
1
��
1/g2(xn+1)
��
1/g1(xn+1)
99
Lecture 36
Talagrand’s two-point inequality.
18.465
So,
�
I(xn+1)dP (xn+1) ≤
≤
=
1
P n(B1)P n(B2)
1
P n(B1)P n(B2)
1
P n(B1)P n(B2)
�
�
min 2,
1
,
g1
1
g2
�
dP
· �
1
g1dP �
·
g2dP
·
1
P n+1(A1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Let A = {y ∈ X n, V (y) ≤ M } ⊆
X n . Hence, A consists of points with typical behavior. We will use control by 2 points to show that any
other point is close to these two points.
By control by 2 points,
P (d(A, A, x) ≥ t) ≤ P (A) P (A)
1
· 2−t ≤ 4 · 2−t
Take any x ∈ X n . With probability at least 1 − 4 2−t , d(A, A... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(x� ))2 + Ex� sup
i
(f (xi) − f (x� ))2
i
�
f ∈F i∈I3
≤ Ex� sup
f ∈F
(f (yi
1) − f (xi
� ))2 + Ex� sup
f ∈F
i=1
(f (yi
2) − f (x�
i))2 + (b − a)2t
�
f ∈F i∈I2
n
�
i=1
= V (y 1) + V (y 2) + (b − a)2t
≤ M + M + (b − a)2t
because y1, y2 ∈ A. Hence,
Finally, M ≤ 2EV because
�
P V (x) ≤ 2M + (b − a)2t ≥ 1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1. Assume a ≤ f ≤ b for all f ∈ F. Let Z = supf ∈F | n f (xi)| and V = supf ∈F
f (x�
i=1
i))2 . Then
�
�
Z ≤ EZ + 4
P
√
�
EV t + 2(b − a)t
≥ 1 − (4e)e−t/4 − 4 2−t .
·
�
n (f (xi)−
i=1
This is an analog of Bernstein’s inequality:
√
4
EV t −→ Gaussian behavior
2(b − a)t −→ Poisson behavior
Now, consider the fo... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
sup
f ∈F
n
�
i=1
εi(f (xi) − f (x�
i))2 + 2nσ2
�
εi(f (xi) − f (xi
� ))2 + 2nσ2
+
102
Lecture 37
Talagrand’s concentration inequality for empirical processes.
18.465
Note that the square function [−(b−a), (b−a)] �→ R is a contraction. Its largest derivative on [−(b−a), (b−a)]
is at most 2(b − a). Note that... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
| n f (xi)| and σ2 = supf ∈F Var(f ).
Then
i=1
�
�
�
P Z ≤ EZ + 4 (8(b − a)EZ + 2nσ2)t + 2(b − a)t ≥ 1 − (4e)e−t/4 − 4 2−t .
·
�
Using other approaches, one can get better constants:
�
�
P Z ≤ EZ +
(4(b − a)EZ + 2nσ2)t + (b − a)
�
t
3
≥ 1 − e−t .
103
Lecture 38
Applications of Talagrand’s concentration ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
E sup |. . .| +
f ∈F
4(b − a)E sup |. . .| + 2σ2
f ∈F
�
t
n
+ (b − a)
t
3n
Compare this result to the Martingale-difference method (McDiarmid):
�
�
n1
�
�
sup
�
�
f ∈F n
i=1
f (xi) − Ef ≤ E sup |. . .| +
f ∈F
�
2(b − a)2t
n
The term 2(b − a)2 is worse than 4(b − a)E supf ∈F |. . .| + 2σ2 .
An algorit... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Ef −
�
�
�
≤ sup
f ∈Fε
n
�
1
n
1
n
is too conservative.
−0
i=1
n
�
i=1
�
�
�
f (xi)
�
�
Pretend we know Ef0 ≤ ε, f0 ∈ Fε.
Then with probability at least 1 − e−t ,
≤ E sup Ef −
f ∈Fε
�
�
�
�
�
1
n
n
�
i=1
�
�
�
+
f (xi)
�
�
�
�
�
�
�
4E sup . . . + 2σε
2
|
|
f ∈Fε
�
t
n
+
t
3n
where σε
2 = su... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
Then, for a fixed k, with probability at least
1 − e−t
1
(k+2)2 ,
�
�
Ef0 −
�
�
�
1
n
n
�
i=1
�
�
�
�
�
�
f0(xi) ≤ ϕ(ε) +
(4ϕ(ε) + 2ε)
t + 2 log(k + 2)
n
+
t + 2 log(k + 2)
3n
For all k ≥ 0, the statement holds with probability at least
∞
�
1 −
k=1
�
e−t ≥ 1 − e−t
1
(k + 2)2
�
��
π2
6 −1
For f0, fi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
n
�
n f0(xi) + Φ(Ef0). Denote x = Ef0. Then x ≤
i=1
f ¯ + Φ(x).
Theorem 38.1. Let 0 ≤ f ≤ 1 for all f ∈ F. Define Fε = {f ∈ F, Ef ≤ ε} and ϕ(ε) = E supf ∈Fε
Then, with probability at least 1 − e−t, for any f0 ∈ F, Ef0 ≤ x∗, where x∗ is the largest solution of
n
�
�
�
Ef − n
1
�
n
i=1
�
f (xi) .
�
x∗ =
1
n
i=1
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, yn) ∈ A},
and
U (A, x) = conv V (A, x)
d(A, x) = min{|s 2 =
|
n
�
2 , s ∈ U (A, x)}
si
i=1
In the previous lectures, we proved
Theorem 39.1.
Today, we prove
P (d(A, x) ≥ t) ≤ P (A)
e−t/4 .
1
Theorem 39.2. The following are equivalent:
(1) d(A, x) ≤ t
(2) ∀α = (α1, . . . , αn), ∃y ∈ A, s.t.
�
n
i=1
αi... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
�
� �
α2
i · t
i=1
i=1
Note that
�
0 is constant on L because s0 is perpendicular to the face.
αis
i
�
��
�
αisi
0 ≤
αiI(xi =�
yi) ≤
Hence,
�
0)2
(s
i
≤
��
(s0)2t and
i
��
(s0)2 ≤
i
√
t.
Therefore, d(A, x) ≤
0)2
i ≤ t.
2t
αi
�
(s
We now turn to an application of the above results: Bin Pack... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
one. Hence, B − 1 ≤
�
Theorem 39.3.
�
P B(x1, . . . , xn) ≤ M + 2
i t + 1 ≥ 1 − 2e−t/4 .
x2
·
��
�
Proof. Let A = {y : B(y1, . . . , yn) ≤ M }, where P (B ≥ M ) ≥ 1/2, P (B ≤ M ) ≥ 1/2. We proved that
P (d(A, x) ≥ t) ≤ P (A)
e−t/4 .
1
Take x such that d(A, x) ≤ t. Take α = (x1, . . . , xn). Since d(A, x) ≤ ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
≤ nEx + nEx · t + t
3
2
1
2
1
≥ 1 − e−t .
Hence,
B(x1, . . . , xn) � M + 2 nEx2
1
�
t·
�
108
Lecture 40
Entropy tensorization inequality. Tensorization of Laplace transform.
18.465
In this lecture, we expose the technique of deriving concentration inequalities with the entropy tensorization
inequality. The... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
− (u − x)) dP : x ∈ R+
�
��
= sup
(u · g) dP :
�
exp(g)dP ≤ 1
�
.
Proof. For the first formulation, we define x pointsizely by ∂
∂x
�
get x = udP > 0.
�
(u (log u − log x) − (u − x)) dP = 0, and
·
��
(ug) dP − λ
For the second formulation, the Laplacian corresponding to sup
�
is L(g, λ) =
exp(g)dP − ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
.2. [Tensorization of entropy] X = (X1, · · · , Xn), Pn = P1 × · · · × Pn, u = u(x1, · · · , xn),
EntPn (u) ≤
� �
( n EntPi (u)) dPn .
i=1
Proof. Proof by induction. When n = 1, the above inequality is trivially true. Suppose
�
u log udPn ≤
�
�
udPn log udPn +
� n
�
EntPi (u)dPn .
i=1
109
Lecture 40
Entrop... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
theorem
�
� �
� �
udPn+1 ·
log udPn+1 + EntPn+1 (u)dPn +
EntPi (u)dPn+1
convexity of entropy
udPn+1
log udPn+1 + EntPn+1 (u)dPn+1 +
EntPi (u)dPn+1
� �
·
� �
� �
� � n+1
�
udPn+1 ·
log udPn+1 +
EntPi (u)dPn+1 .
=
����
=
����
≤
����
�
�
=
≤
� n
�
i=1
� n
�
i=1
i=1
By definition of entropy, EntPn... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
·
��
.
i=1
Moreover,
n
� �
�
φ −λ Z − Z i
·
��
Ee λZ
i=1
= E
n
�
λZ φ �
e
�
−λ Z − Z i
·
i=1
⎛
�
�� ⎜ �
⎝I Z ≥ Z i + I
Z i
·
�
��
�
I
�
�
⎞
�⎟
≥ Z ⎠
�
��
II
n
⎛
� ⎜ λZi �
= E ⎝e
�
i=1
��
�
�
· I Z ≥ Z i
φ −λ · Z i −
Z
��
switch Z and Zi in II
⎛
� ⎜ λ(Zi
⎝e
·
�
�
I Z ≥ Z i
·
�
��... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, xn be independent random
1,
· · ·
, x� be their independent copies, Z = Z(x1,
variables, x�
· · ·
and φ(x) = ex − x − 1. We have EeλZ − EeλZ log EeλZ ≤ EeλZ
, xn), Z i = (x1,
, xi−1, x�
, xn),
�
n
=1 φ(−λ(Z − Z i)). We will use the ten
i
sorization of entropy technique to prove the following Hoeffding-type in... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
follows that
0 ≤ Z − Z i ≤
� �
◦ −
fi
◦ − ai = fi
fj
◦ − ai ≤ bi(f ◦) − ai(f ◦).
i
j=i
Since φ(x) = e x−x−1 is increasing in R and limx
→
x2
x2
φ(x)
0 x2
→
1 , it follows that ∀x < 0, φ(x) ≤ 1 x2, and
2
2
Ee λZ λZ − Ee λZ log Ee λZ ≤ Ee λZ
�
�
�
φ −λ(Z − Z i)
i
≤
≤
Center Z, and we get
1 Ee λZ �
2
... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
1
t
0
��
log F (t) dt
t
F (λ) ≤ exp(
1
2
Lλ2).
By Chebychev inequality, and minimize over λ, we get
P(Z ≥ EZ + t)
≤
e−λtEe λ(Z−EZ)
≤ e−λt 2
e
1 Lλ2
minimize over λ
P(Z ≥ EZ + t) ≤ e−t /(2L)
2
Let fi above be Rademacher random variables and apply Hoeffding’s inequality, we get P(Z ≥ EZ+ Lt/2) ≤
e−t . As a... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
i
fj .
It follows that 0 ≤ Z − Z i ≤ fi
◦ ≤ 1. Since φ = e − x − 1 is a convex function of x,
x
φ(−λ(Z − Z i)) = φ(−λ (Z − Z i) + 0 (1 − (Z − Z i))) ≤ (Z − Z i)φ(−λ)
·
·
113
Lecture 41
Application of the entropy tensorization technique.
18.465
and
�
E λZeλZ − Ee λZ log Ee λZ ≤ E e λZ
�
�
n
�
�
�
φ −λ(... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
(
F
(λ − φ(−λ)) (log F (λ))�
− log F (λ) ≤ vφ(−λ)
λ − log F (λ) ≤ vφ(−λ).
Solving the differential equation
⎛
⎞
�
(41.2)
(λ − φ(−λ)) ⎝log F (λ)⎠ − log F (λ) = vφ(−λ),
⎜
�
��
Ψ0
⎟
�
λ
�
��
Ψ0
�
yields Ψ0 = v φ(λ). We will proceed to show that Ψ satisfying 41.1 has the property Ψ ≤ Ψ0:
·
1
eλ −1
λ
(e −1)(1−... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, we get P (Z ≥ EZ + t) ≤ e−v·h(t/v) where h(x) = (1 + x) log(1 + x) − x.
·
The following sub-additive increments bound can be found as Theorem 2.5 in
Olivier Bousquet. Concentration Inequalities and Empirical Processes Theory Applied to the
Analysis of Learning Algorithms. PhD thesis, Ecole Polytechnique, 2002.
1... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
λ)) (Z − Zk) e λZ + e λZk − e λZ
= f (λ) (Z − Zk) e λZ + g(Z − Zk)e λZk .
In the above, g(x) = 1 − eλx + (λ − f (λ)) xeλx, and we define f (λ) = 1 − eλ + λeλ / eλ + α − 1 where
�
� �
�
α = 1/ (1 + u). We will need the following lemma to make use of the bound on the variance.
Lemma 41.4. For all x ≤ 1, λ ≥ 0 and α... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
�
(λ − f (λ))
F �(λ)
F (λ)
� �� �
Ψ�(λ)
− log F (λ)
� �� �
Ψ(λ)
f (λ) �
αnσ2 + EZ �
.
≤
Solve this inequality, we get F (λ) ≤ evψ(−λ) where v = nσ2 + (1 + u)EZ.
�
116
Lecture 42
Stein’s method for concentration inequalities.
18.465
This lecture reviews the method for proving concentration inequaliti... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
)
E (h(X)f (X)) = E (h(X) · E (F (X, X �)|X)) = E (h(X) · F (X, X �))
X,X � are exchangeable
= E (h(X �) F (X �, X))
·
F (X,X �) is anti-symmetric
= −E (h(X �) F (X, X �))
·
=
1
2
E ((h(X) − h(X �)) · F (X, X �)) .
Take h(X)=1, we get Ef (x) = 0. Take h(X) = f (X), we get Ef 2 = 1 E ((f (X) − f (X �)) F (X, X ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
R
1
e0
=
b+t(a−b)
� �
1
λf (X) + e λf (X �)
e
2
�
|λ| · E e
⎛
�
1
�
·
�
�
2
λf (X)
b
b
)+e
a+(1−t)e )dt=
1 a
2 (e
(f (X) − f (X �)) F (X, X �)
�
�
�
�
�
·
·
�
�
�
�
�
(f (X) − f (X �)) F (X, X �)
|λ| · E
e λf (X) ·
E
(f (X) − f (X �)) F (X, X �)
·
��
1
�
�
�
2
�
⎜
⎜
⎜
⎜
⎝
�
��
Δ(X)
�
≤ |λ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
) ≤
λ · C
(1 − λB)
� λ
0 1 −
s C
·
·
s B
�
ds ≤
C
� λ
·
1 − λ B 0
sds =
1 C λ2
·
·
λ B
1 −
2
·
.
By Chebyshev’s inequality P(f (x) ≥ t) ≤ exp −λt + 1
·
2 1
�
.
, and P(f (x) ≥ t) ≤ exp − ·
2
t
2 (C+Bt)
we get λ =
t
C+Bt
�
λ2
·
C
·
λ
B
−
�
1 ,
. Minimize the inequality over 0 ≤ λ < B
�
We ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
method for concentration inequalities.
18.465
Example 42.3. Let (aij )i,j=1,
uniformly distributed over the permutations of 1,
i,j ai,j , and
our goal is to bound |X − EX| probabilistically. To apply the above theorem, we define exchangeable pairs
i=1 ai,π(i), then EX = 1
, n. Let X =
,n be a real matrix where ai... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
, X �)|X)
n �
2
n 1 �
2
n
�
=
·
I
=
ai,π(i) −
aI,π(I) +
1 �
n
ai,j
Δ(X)
def. =
·
i,j
=
i
X − EX
�
�
n
1
(X − X �)2 |π
E
2
2
��
n
E
aI,π(I) + aJ,π(J) − aI,π(J) − aJ,π(I)
4
E �
aI,π(I) + aJ,π(J) − aI,π(J) − aJ,π(I)|π �
n
2
= X + EX
=
≤
�
�2 |π
=
f (X) + 2EX.
Apply the theore... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
around tanh (β m(σ) + β h) where tanh(x) =
, n}, and generate σ� according
· · ·
·
·
I
�
119
�
Lecture 42
Stein’s method for concentration inequalities.
18.465
P(σ� = +1|{σj : j = i}) =
i
2 (exp( β
n
·
�
j=i
�
exp( β
n
j=i σj + β h)
�
·
β
·
σj + β h) + exp(−
n
�
σj − β h))
·
j=i
P(σ�
i = −1|{σj : ... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/336b787dc798473a4fa55da69591b190_toc.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.080 / 6.089 Great Ideas in Theoretical Computer Science
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
6.080/6.089 GITCS
Feb 07, 2008
Lecturer: Scott Aaronson
Scribe: Mergen Nachin
Lecture 2
Administrative... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
down a set
of simple, clear rules that one can repeatedly apply to construct complicated objects. We also
talked about Euclid’s GCD algorithm, which was one of the first non-trivial algorithms known to
humankind.
Here is some digression. Area of a circle is A = πr2 . It’s obvious that the area of a circle should
go... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
it to logical scrutiny itself?
The credit for being the first logician is usually given to Aristotle, who formalized the concept
of the syllogism.
All men are mortal, Socrates is man, therefore Socrates is a mortal.
This is a syllogism. In more modern language, we call it transitivity of implications. In general,
a... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
J and the 2, not the 5. 80-90 percent of college students get this wrong.
On the other hand, suppose you ask people the following: you say, you’re a bouncer in a bar,
and you want to make sure the rule ”If you are under 21 the you are not drinking”. Who do you
have to check to test this rule: someone who is drinking... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
also the sentence that’s
talking about them. These three sentences you can think of as meaningless pieces of code, the
sentence is addressed to us; it’s telling us one of the rules of the code.
Was Aristotle the first person in history to apply such an inference? Obviously he wasn’t. As
we all see in our everyday li... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
and uncertain, and people don’t agree how much weight different facts entered into evidence should
be assigned. On top of that, the laws themselves are necessarily vague, and people disagree in their
preferred interpretation of the law.
Nevertheless, this idea of Leibniz that we could automate even part of human thou... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
do this sort of thinking every morning: ”My socks go on my feet, these are my
socks, therefore these go on my feet.”
Yet suppose you strung together hundreds or thousands of these baby steps. Then maybe you’d
end up with the most profound thought in the history of the world! Conversely, if you consider the
most pro... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
then
A ⇒ ¬A is valid. But ¬A ⇒ A is not valid. Similarly if we assign A=true then A ⇒ ¬A is not
valid. Now consider the following example.
• A ⇒ B
2-3
• ¬C ⇒ A
• ¬A ⇒ ¬C
• B ⇒ ¬A
Can these sentences simultaneously be satisfied? I.e. is there some way of setting A,B,C,D to
”true” or ”false” that satisfies all fo... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
A ⇒ C actually means C is reachable from A. Start with
A=true and if we reach ¬ A then it means A ⇒ ¬A. If we also end up connecting ¬A and A, in
other words if we have cycle, then we have discovered a contradiction.
What we’re talking about are two properties of logical systems called ”soundness” and ”com
pletenes... | https://ocw.mit.edu/courses/6-080-great-ideas-in-theoretical-computer-science-spring-2008/33742c58e0e02d6ae058fc87c4b981d7_lec2.pdf |
Introduction to Simulation - Lecture 10
Modified Newton Methods
Jacob White
Thanks to Deepak Ramaswamy, Jaime Peraire, Michal
Rewienski, and Karen Veroy
Outline
• Damped Newton Schemes
– Globally Convergent if Jacobian is Nonsingular
– Difficulty with Singular Jacobians
• Introduce Continuation Schemes
– Problem with... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
be used to find the zero
of the function provided you all ready know the
answer.
Need a way to develop Newton methods which
converge regardless of initial guess!
SMA-HPC ©2003 MIT
Non-converging
Case
f(x)
1-D Picture
1x
X
0x
Limiting the changes in X might improve convergence
SMA-HPC ©2003 MIT
Newton Method
wit... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
x
k
1
+
)
Method Performs a one-dimensional search in
Newton Direction
SMA-HPC ©2003 MIT
Newton Method
with Limiting
Damped Newton
Convergence Theorem
If
a
)
b
)
J
−
F
J
F
Then
1
k
x
(
( )
x
)
−
≤
β
(
Inverse is bounded
)
J
F
(
y
)
≤
(cid:65)
x
−
y
(
Derivative is Lipschitz Cont
)
There exists a set of
(
)
F x
(
F... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
I
r
−
1
10
V
r
=
0
V
d
I e
(
s
V
t
I
d
−
1)
− =
0
Nodal Equations with Numerical Values
(
f v
2
)
=
)
1
(
−
v
2
10
(
2 0
v
−
)
+
10
−
16
(
e
0.025
− =
1) 0
Newton Method
with Limiting
Damped Newton
Example cont.
(
f v
2
)
=
)
1
(
−
v
2
10
(
2 0
v
−
)
+
10
−
16
(
e
0.025
− =
1) 0
Newton Method
with Limiting
0
x
Repe... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
k
J
(cid:8)(cid:11)(cid:11)(cid:9)(cid:11)(cid:11)(cid:10)
Newton Direction
Multidimensional Mean Value Lemma
(
F y
( )
F x
)(
−
−
−
J
y
x
)
(
F
y
)
≤
(cid:65)
2
2
x
−
y
Combining
k
(
F x
)
1
+ −
F
(
k
x
)
+
J
F
(
)
k
k
x α
⎡
⎢
⎣
1
−
J
F
(
k
x
)
k
(
F x
)
≤⎤
⎥
⎦
(cid:65)
2
k
α
J
F
1
−
(
k
x
)
F
(
k
x
)
2
SMA-HPC ©2003... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
)
(
)
)
x
(cid:65)
1
+
2
k
k
2
β
2
k
) 2
⎤
⎥
⎦
⎡
⎢
⎣
Yields a quadratic in the damping coefficient
SMA-HPC ©2003 MIT
Newton Method
with Limiting
Damped Newton
Theorem Proof-Cont-II
Simplifying quadratic from previous slide
⎡
(
kF x
1
⎢
⎣
(
k
+
α α
(
F x
2
β
2
≤
−
)
)
(cid:65)
1
+
2
k
k
)
⎤
⎥
⎦
k
(
F x
)
Two C... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
(
F x
≤
)
0
not yet a convergence theorem
γ
independent from
k
(
kF x
(cid:65)
For the case where
2
β
2
1
(
kF x
Note the proof technique
2
2
β
−
1
(cid:65)
)
)
>
1
2
≤
1
−
1
2
2
β
(cid:65)
0
(
F x
)
≤
0
γ
First – Show that the iterates do not increase
Second – Use the non-increasing fact to prove convergence
SMA-HPC... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
initial guess
– Generate a sequence of problems
– Make sure previous problem generates guess for next problem
• Heat-conducting bar example
1. Start with heat off, T= 0 is a very close initial guess
Increase the heat slightly, T=0 is a good initial guess
2.
Increase heat again
3.
SMA-HPC ©2003 MIT
Continuation Schemes... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
λ
s
)
=
0
)
λ
,
(cid:4)
(
f v
∂
v
∂
=
i
∂
diode
v
∂
( )
v
1
+ ←
R
Not
λ
dependent!
f
(cid:71)(cid:4)
(
,F x λ =
)
xf
f
y
(
(
Lf
)
x y =
,
)
x y
,
0
fλ+
=
0
l
Source/Load Stepping Does Not Alter Jacobian
SMA-HPC ©2003 MIT
Continuation Schemes
Jacobian Altering Scheme
Description
(
(cid:4)
F x
)
(
λ λ λ
=
)
,
)
(
(
)
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
(
λ
prev
=
x
)
(
λ λ λ
=
,
x
Else
1
2
,
δλ
λ
=
λ δ
+
λ
prev
+
δλ δλ δ
=
2
λ
,
δλ
=
}
SMA-HPC ©2003 MIT
Continuation Schemes
Jacobian Altering Scheme
Initial Guess for each step.
(
)
x λ
)
(
x λ δλ+
Initial Guess Error
(
)
0x
+
λ δλ
(
)
λ
=
x
0
λ δλ+λ
SMA-HPC ©2003 MIT
λ
1
Continuation Schemes
Jacobian Altering Schem... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
puted
SMA-HPC ©2003 MIT
Continuation Schemes
Jacobian Altering Scheme
Update Improvement Cont. II.
0
x
(
)
+
λ δλ
=
x
(
)
λ
(cid:4)
(
F x
∂
)
(
)
,
λ λ
x
∂
⎛
− ⎜
⎜
⎝
⎞
⎟
⎟
⎠
1
−
(cid:4)
F
∂
(
x
)
(
)
,
λ λ
λ
∂
δλ
Graphically
(
)
x λ
)
0x λ δλ+
(
0
λ δλ+λ
λ
1
SMA-HPC ©2003 MIT
Continuation Schemes
(
)
x λ
Jacobian A... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
1
+
x
x
−
k
1
+
λ λ
−
k
⎤
⎥ ⎡
⎥ ⎢
⎥ ⎣
⎥
⎦
=
⎤
⎥
⎦
(
(cid:4)
F x
,k
k
λ
)
(
λ
prev
⎤
⎥
⎥
⎥
⎦
) 2
2
−
arc
2
k
(
−
λ λ
prev
2
)
+
k
x
−
x
−
⎡
⎢
⎢
⎢
⎣
SMA-HPC ©2003 MIT
Continuation Schemes
)x λ
(
0
Jacobian Altering Scheme
Arc-length Turning point
What happens here?
λ
1
Upper left-hand
Block is singular
SMA-HPC ©2003 MIT... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/33912d2b3446ff62fcd51b7238469f10_lec10.pdf |
6.867 Machine learning, lecture 8 (Jaakkola)
1
Lecture topics:
• Support vector machine and kernels
• Kernel optimization, selection
Support vector machine revisited
Our task here is to first turn the support vector machine into its dual form where the exam
ples only appear in inner products. To this end, assume ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
0 means that all the components αt are non-negative. Let’s try to see first that
J(θ, θ0) really is equivalent to the original problem. Suppose we set θ and θ0 such that at
least one of the constraints, say the one corresponding to (xi, yi), is violated. In that case
�
�
− αi yi(θT φ(xi) + θ0) − 1 > 0
(4)
for any... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
≥0 θ,θ0
(6)
The left hand side, equivalent to minimizing Eq.(5), is known as the primal form, while the
right hand side is the dual form. Let’s solve the right hand side by first obtaining θ and θ0
as a function of the Lagrange multipliers (and the data). To this end
d
dθ0
d
dθ
J(θ, θ0; α) = −
αtyt = 0
J(θ, θ0... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
quadratic
optimization problem. The constraints are simpler, however. Moreover, the dimension of
t=1
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10)
6.867 ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
t)T φ(x)] + θˆ
0
t∈SV
(14)
(15)
(16)
where SV is the set of support vectors corresponding to non-zero values of αt. We don’t
know which examples (feature vectors) become as support vectors until we have solved the
optimization problem. Moreover, the identity of the support vectors will depend on the
feature map... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
�
γˆgeom =
n
n
� �
�
αˆiαˆjyiyj K(xi, xj )
−1/2
4
(18)
i=1 j=1
Would it make sense to compare geometric margins we attain with different kernels? We
could perhaps use it as a criterion for selecting the best kernel function. Unfortunately
this won’t work without some care. For example, if we multiply all the f... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
larger values for constraints
that are harder to satisfy. Without any upper limit, they would simply reach ∞ for any
constraint that cannot be satisfied. The limit C specifies the point when we should stop
from trying to satisfy such constraints. More formally, the dual form is
n
�
n
n
� �
αt − (1/2)
αiαj yiyj [φ(... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
we are faced with the
problem of selecting an appropriate kernel function. A step in this direction might be to
tailor a particular kernel a bit better to the available data. We could, for example, introduce
additional parameters in the kernel and optimize those parameters so as to improve the
performance. These pa... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].(cid:13)(cid:10) | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/33a6c8e66c62602f9f03ab6a2c632eed_lec8.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.969 Topics in Geometry: Mirror Symmetry
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
MIRROR SYMMETRY: LECTURE 9
DENIS AUROUX
1. The Quintic (contd.)
To recall where we were, we had
(1)
Xψ = {(x0 : · · ·... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
5
ψ)5n
(n!)
T0
In terms of z = (5ψ)−5, the period is proportional to
(5)
φ0(z) =
∞ (5n)! n
�
z
(n!)5
n=0
�
�
cnzn) =
ncnzn, we obtained the Picard-Fuchs equation
d
Setting Θ = z dz : Θ(
(6)
θ4φ0 = 5z(5Θ + 1)(5Θ + 2)(5Θ + 3)(5Θ + 4)φ0
Proposition 1. All periods Ωˇ
ψ satisfy this equation.
�
Note that a... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
C).
on X which is ideally a 3-form on X, but is at least a class in H 3(X,
�
1
πi S1 φ(z)dz.
Recall from complex analysis, if φ(z) has a pole at 0, res0(φ) = 2
Now, let’s say that we have a 3-cycle C in X: we can associate a “tube” 4-cycle
in P4 which is the preimage of C in the boundary of a tubular neighborhood ... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
dxj ∧ · · · ∧ dx4
with deg (g0 · · · g4) = 5� − 4, then
(13)
dφ =
�
�
1 � ∂fψ � ∂gj
gj ∂xj
f �+1
∂xj
ψ
− fψ
j
j
�
Ω
MIRROR SYMMETRY: LECTURE 9
3
In particular, if we have something of the form (
�
gj
∂fψ
∂xj
)
Ω
+1 (the Jacobian ideal
f �
ψ
∂fψ
is the span of { ∂xi
}), it can be written as somethi... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
(z)
Θf (z)
. . .
Θs−1f (z)
⎞
⎟
⎟
⎠
1
−Bs−1(z)
.
.
.
0
· · ·
The fundamental theorem of these differential equations states that there exists
a constant s × s matrix R and an s × s matrix of holomorphic functions S(z) s.t.
(16)
Φ(z) = S(z) exp((log z)R) = S(z)(id + (log z)R +
log2 z
2
R2 +
· · ·
)
is a fundamenta... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
= e
⎛
⎜
2πiR ⎜ 0
=
⎝
0
0
1 2πi
1
0
0
(2πi)2
2
2πi
1
0
⎞
(2πi)3
6
(2πi)2 ⎟
⎟
2
2πi
⎠
1
�
If ω(z) = β Ωˇ
ψ is a period, then it is a solution of the Picard-Fuchs equation, and
thus a linear combination of Φ(z)1i’s. There exists a basis b1, . . . , b4 of H3( ˇX, C)
�
s.t.
ψ = Φ(z)1i. The monodromy action... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
j (f (z) log z) = (Θj f ) log z + j(Θj−1f )
If we write F (x) = x4 − 5z �
4
=1(5x + j), then
j
(21)
Dφ1(z) = F (Θ)(φ0(z) log z + φ˜(z))
= (F (Θ)φ0) log z + F �(Θ)φ0 + F (Θ)φ˜
Since 0 = Dφ0 = Dφ1, we find Dφ˜(z) = −F �(Θ)φ0(z). This gives a recurrence
relation on the coefficients of φ˜(z), and one obtains:
�
1
n
z ... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/33dfafe4343fb216e890aa96617797e3_MIT18_969s09_lec09.pdf |
Substitution of Power Series
We can find the power series of e−t by starting with the power series for ex
and making the substitution x = −t2 .
2
e x = 1 + x +
2
x
2!
+
e−t2
= 1 + (−t2) +
+
· · ·
(R = ∞)
+
(−t2)3
3!
+
· · ·
3
x
3!
(−t2)2
2!
t6
3!
= 1 − t2 +
t4
2!
−
+ · · ·
The signs of the terms a... | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/33f76c6f2abf4190beb0a0cf29065926_MIT18_01SCF10_Ses100d.pdf |
L2: Combinational Logic Design
L2: Combinational Logic Design
(Construction and Boolean Algebra)
(Construction and Boolean Algebra)
Acknowledgements:
Materials in this lecture are courtesy of the following sources and are used with permission.
Prof. Randy Katz (Unified Microelectronics Corporation Distinguished Prof... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
Introductory Digital Systems Laboratory
3
NMOS Device Characteristics
NMOS Device Characteristics
body
source
polysilicon gate
drain
gate
n+
n+
n+
p
inversion layer�
channel
gate oxide
p+
n
D
ID
-4
x 10
6
5
4
)
A
(
3
I
D
2
1
0
0
VGS= 2.5 V
Resistive
Saturation
VGS= 2.0 V
VGS= 1.5 V
VGS= 1.0 V
0.5
1
1.5
2... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
G
G
IN
ID n
Vin = 0
Vin = 2.5
PMOS
Vin = 0.5
Vin = 2
NMOS
Vin = 1
Vin = 1.5
Vin = 1.5
Vin = 1
Vin = 1.5
Vin = 2
Vin = 2.5
Vin = 1
Vin = 0.5
Vin = 0
Vout
CMOS gates have:
(cid:131) Rail-to-rail swing (0V to VDD)
(cid:131) Large noise margins
(cid:131) “zero” static power dissipation
2.5
2
1.5
)
V
(
t
u
o
V
1
0.5
0
0
0.5... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
0
X AND Y
0
0
1
0
X
In general, there are 2 (2^n) functions of n inputs
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
8
Common Logic Gates
Common Logic Gates
Gate
Symbol
Truth-Table Expression
NAND
AND
NOR
OR
X
Y
X
Y
X
Y
X
Y
Z
Z
Z
Z
X
0
0
1
1
X
0
0
1
1
X
0
0
1
1
X
0
0
1
1
Y
0
1
0
1
Y
0
1
0
1
Y
0
1
0
1
... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
Generic CMOS Recipe
Generic CMOS Recipe
Vdd
A1
.
.
.
An
.
.
.
.
.
.
pullup: make this connection
when we want F(A1,…,An) = 1
F(A1,…,An)
pulldown: make this connection
when we want F(A1,…,An) = 0
Note: CMOS gates
result in inverting
functions!
(easier to build NAND
vs. AND)
A
B
A
B
A
PDN
B
PUN
O
CL
A B PDN PUN O
1
0 ... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
(X • Z)
8D. X + (Y • Z) = (X + Y) • (X + Z)
(cid:132) Uniting:
9. X • Y + X • Y = X
9D. (X + Y) • (X + Y) = X
(cid:132) Absorption:
10. X + X • Y = X
11. (X + Y) • Y = X • Y
10D. X • (X + Y) = X
11D. (X • Y) + Y = X + Y
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
12
Theorems of Boolean Algebra (II)
T... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
(X1,X2,...,Xn,0,1,+,•) ⇔ f(X1,X2,...,Xn,1,0,•,+)
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
13
Simple Example: One Bit Adder
Simple Example: One Bit Adder
(cid:132) 1-bit binary adder
(cid:134) inputs: A, B, Carry-in
(cid:134) outputs: Sum, Carry-out
A
B
Cin
S
Cout
A B
Cin S Cout
Sum-of-Products Can... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
B
S = A B Cin + A B Cin + A B Cin + A B Cin
=( A B + A B )Cin + (A B + A B) Cin
=(A ⊕ B) Cin + (A ⊕ B) Cin
= A ⊕ B ⊕ Cin
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
15
SumSum--ofof--Products & Product
Sum
Products & Product--ofof--Sum
(cid:132) Product term (or minterm): ANDed product of literals –... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
maxterms
A + B + C M0
A + B + C M1
A + B + C M2
A + B + C M3
A + B + C M4
A + B+ C M5
A + B +C
M6
A +B + C M7
short-hand notation for maxterms of 3 variables
F in canonical form:
F(A, B, C)
= ΠM(0,2,4)
= M0 • M2 • M4
= (A + B + C) (A + B + C) (A + B + C)
canonical form ... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
list the indices not already used in F
E.g., F(A,B,C) = Σm(3,4,5,6,7) F'(A,B,C) = Σm(0,1,2)
= ΠM(0,1,2) = ΠM(3,4,5,6,7)
4. Minterm expansion of F to Maxterm expansion of F':
rewrite in Maxterm form, using the same indices as F
E.g., F(A,B,C) = Σm(3,4,5,6,7) F'(A,B,C) = ΠM(3,4,5,6,7)
= ΠM(0,1,2) = Σm(0,1,2)
L2: 6.111 Sp... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
ube
X
XYZ
111
01
Y
00
X
0111
XY
11
10
2-cube
1011
WXYZ
1111
011
010
Y
Z
001
000
X
100
3-cube
0011
1010
110
0010
101
Y
Z
0001
0000
W
X
0110
0101
1000
0100
4-cube
1110
1001
1101
1100
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
19
Mapping Truth Tables onto Boolean Cube... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
A
The on-set is completely covered by the combination (OR) of the subcubes of
lower dimensionality - note that “111” is covered three times
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
20
Higher Dimension Cubes
Higher Dimension Cubes
011
111
010
B C
110
001
101
000
A
100
F(A,B,C) = Σm(4,5,6,7)
on-set... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
0
1
1
0
2
3
A
0
0
1
1
B
0
1
0
1
F
1
0
1
0
(cid:132) Numbering scheme based on Gray–code
(cid:134) e.g., 00, 01, 11, 10 (only a single bit changes in code for adjacent map cells)
2-variable
K-map
0
1
0
1
2
3
B
A
0
1
AB
A
C
00
01
11
10
3-variable
K-map
0
1
0
1
6
7
4
5
2
3
B
AB
CD
00
01
11 ... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
) = Σm(0,4,5,7)
F' simply replace 1's with 0's and vice versa
F =
F'(A,B,C) = Σm(1,2,3,6)
F' =
L2: 6.111 Spring 2006
Introductory Digital Systems Laboratory
23
Four Variable Karnaugh
Four Variable
Karnaugh MapMap
AB
CD
A
00
01
11
10
00
1
01
0
C
11
1
10
1
1
0
1
1
D
0
1
1
1
0
0
1
1
B
F(A... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
1
1
11
1
0
1
1
C
10
0
X
X
X
0
0
B
0
1
0
0
D
In PoS form: F = D (A + C)
Equivalent answer as above,
but fewer literals
F(A,B,C,D) = Σm(1,3,5,7,9) + Σd(6,12,13)
F = A D + B C D w/o don't cares
F = C D + A D w/ don't cares
By treating this DC as a "1", a 2-cube
can be formed rather than one 0-cube
AB... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
fix it, cover it up with
another grouping or product term!
A
C
B
AB
C
00 01
11 10
0 0
1
0
0
1
1
1
1
0
F
F = A * C + B * C + A * B
Figure by MIT OpenCourseWare.
(cid:131) In general, it is difficult to avoid hazards – need a robust
design methodology to deal with hazards.
L2: 6.111 Spring 2006
Introductory Digital Syst... | https://ocw.mit.edu/courses/6-111-introductory-digital-systems-laboratory-spring-2006/340fab3bd54502c96ef177100c99b0f8_l2_combi_logic.pdf |
2.3.2 Swollen (coil) polymers in good solvents
Most of the terms in the trial free energy of Eq. (2.49) have definite sign. The exception
is the term proportional to N 2(a/R)3 which has opposing contributions from the repulsive
1/2). The sign of this term
and attractive parts of the potential, and is proportional to (χ
... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
treatment is not trivial, and one of the triumphs
of renormalization group theory is to estimate the exact value of ν = 0.591 . . . , remarkably
close to the Flory approximation of 3/5. While not directly relevant to real polymers, it is
possible to inquire about the exponent ν for self-avoding walks in d-spatial dimen... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
)3. The leading terms in the expansion of
the variational free energy can now be recast as
−
ln Z(ρ)
N
−
ln g +
=
−
2χ
1
−
2
ρ +
ρ2
6
+ higher order terms.
(2.54)
The optimal density for T < θ is obtained by minimizing the above free energy,
leading to
1
N
d ln Z
dρ
−
=
1
2 −
(cid:18)
χ
+
(cid:19)
ρ
3
+
,
· · ·
ρ = 3
χ... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
cooling of liquids typically leads to frozen states with even lower entropy. We may thus
inquire if such a freezing transition also exists for polymers.
→
43
2.3.4 The Random Energy Model (REM) for compact heteropolymers
Deep in the globular phase, the states of the compact polymer can be visualized as the
collection ... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
and as long as the
number of terms NB in Eq. (2.58) is large, taken from a Gaussian distribution. The mean
and variance of the distribution are given by
Eαi
h
= NBh
Nε0
Vabi ≡
1)N (of the z per each site of the lattice, one is polymeric), we
E2
αic = NBh
V 2
abic ≡
Nσ2 ,
(2.60)
h
,
where noting that NB = (z
have folded... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
S(Ec) = 0 =
⇒
Ec
N
= ε0 −
σ
2 ln g′ .
(2.63)
(Note the connection to the extreme value problem studied earlier: Ec is also the mean value
of the lowest of g′N energies randomly selected from p(E).) The singularity of entropy at Ec
signifies a phase transition into a glassy state, at a temperature Tc given by
p
1
Tc
=
dS... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
energy (En < Ec) representing the native configuration.
With the added state at En, the system makes a transition to the native state (i.e. folds)
at a temperature Tf , high enough that there are still many equivalent states to explore. The
location of Tf , and the corresponding energy Ef , can be obtained by equating f... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
at
Ef for β
N.
≥
≤
which after taking the logarithm leads to the tangent rule in Eq. (2.65).
e−βf En = Ω(Ef )e−βf Ef ,
(2.67)
We can eliminate Ef in terms of βf by noting that E = Nε0−
Using these expressions and defining a quantity βn = (En −
reduces to
Nσ2β, and ln g′ = (βcσ)2/2.
Nε0)/(Nσ2), the above equation
β2
f
β2... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/341e304796bde8c040895e83d45706ea_MIT8_592JS11_lec11.pdf |
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