text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
R I'(t) + (1/C) I(t) = V'(t)
The circuit is the system and it is represented by the left hand side.
The input signal is V , the voltage increase across the power source.
I(0)
|
|
|
V
______________
| |
--------------> | Circuit | -------------->
V(t) |______________| I(t)
The *derivative* of the in... | https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/17828cb4e899b98aa351a2d6d6f8da2e_MIT18_03S10_c03.pdf |
Lecture 14: Cluster States
Scribed by: Ilia Mirkin
Department of Mathematics, MIT
October 21, 2003
A cluster state is a highly entangled rectangular array of qubits. We measure qubits
one at a time. The wiring diagram tells us which basis to measure each qubit in, and which
order to measure them in. A wiring diagr... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
K (a) commutes with K (b) when a = b. To show that this is true, we
can look at the following cases:
neighborhood(a) ∩ neighborhood(b) = ∅
(2)
When this is true, then there is absolutely no overlap between K (a) and K (b) and thus
the two commute.
neighborhood(a) �� b
(3)
This means that neighborhoods overlap, ... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
a), a ∈ C (cluster)
is a cluster state. Each K (a) has eigenvalue ±1, making for 2n vectors of eigenvalues {Ka}
.
φ
�
�
0 if {κ
{κa}
C
�
�
a} �
= {κ
a
}
.
For example,
is a cluster state with eigenvalue κa on qubit a, {κa} =
{±1}
. Thus φ
{κa }| {κ
φ
=
�
a
}
�
C
�
(8)
(9)
(10)
(11)
(12)
(13)
κ
=
b
a =
κ
{... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
�
�
|+�
a, where
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 −1
⎞
⎟
⎟
⎠
Sab
=
=
⎜
⎜
⎝
1
2
I + σ(a) + σ(b) − σ(a) ⊗ σ(b)
z
z
�
z
z
�
|
+� =
√1
|
0� + 1�). We
(
2
|
(14)
(15)
Here are a few examples of gates that can be made using wiring diagrams:
CNOT Gate
p
p
p
p
p
p
p
σp x
p
σp x σp x σp y
p σx
p σy
p σy
p σx
p
p ... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
Demonstration of a transmission line effect:
Sab |+� |+�
(|00� + 01� + 10� + 11�)
|
|
|
1
= Sab 2
1
|
|
( 00� + 01� +
=
2
1
= √
2
(|+� |0� + |−� |
1�)
|10� − |
11�)
(16)
(17)
(18)
With this, we apply Sab and measure both a and b in the +�, |−� basis. This is equivalent
Sab, �−−| Sab basis, which is also equi... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
Introduction to Simulation - Lecture 5
QR Factorization
Jacob White
Thanks to Deepak Ramaswamy, Michal Rewienski,
and Karen Veroy
QR Factorization
Singular Example
LU Factorization Fails
Strut
Joint
Load force
The resulting nodal matrix is SINGULAR, but a solution exists!
SMA-HPC ©2003 MIT
QR Factorization
Singular... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
equation by ith column:
=
≠
j
i
j
(cid:71)
(cid:71)
(cid:71)
M x M x M
+
•
1
1
2
i
(
+
(cid:34)
+
2
(cid:71)
x M
N
)
=
N
(cid:71)
M b
•
i
Simplifying using orthogonality:
(cid:71)
(cid:71)
(
)
x M M
i
i
(cid:71)
M b
i
=
•
i
• ⇒ =
x
i
(cid:71)
M b
•
i
(cid:71)
(cid:71)
M M
•
i
(
)
i
SMA-HPC ©2003 MIT
QR Factorization
O... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)
x
N
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
b
1
b
2
(cid:35)
b
N
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
y
1
y
2
(cid:35)
y
N
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
b
1
b
2
(cid:35)
b
N
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
(cid:34)
(cid:34)
(cid:34)
↑
↑
(cid:71)
(cid:71)
Q Q
1
2
↓
↓
⎡
⎤
⎡
↑ ⎢
(cid:71)
⎥
⎢
⎢
Q
⎥
⎢
N
⎢
⎥
⎢
↓ ⎢
⎥
⎢
⎦ ⎢
⎣
(cid:8)(cid:11)(cid:11)(cid:11)(cid:9)(cid:11)(ci... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
71)
, find
,
M M
1
(cid:71)
(cid:71)
(
M Q M M r M
•
1
(cid:71)
(cid:71)
M M
•
(cid:71)
(cid:71)
1
M M
•
1
r
12
=
=
=
−
0
•
12
1
1
1
1
2
2
2
(cid:71)
2Q
SMA-HPC ©2003 MIT
12r
(cid:71)
2M
(cid:71)
1M
QR Factorization
Orthogonalization
Normalization
Formulas simplify if we normalize
(cid:71)
(cid:71)
Q Q
⇒ •
1
1
(ci... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)
(cid:71)
⎢
⎥
Q Q
⎢
⎥
1
2
⎢
⎥
↓
↓
⎢
⎥
⎦
⎣
(cid:11)
(cid:11)(cid:9) (cid:10)
(cid:8)
Orthonormal
r
r
⎤
⎡
12
11
⎢
⎥
r
0
⎦
⎣
22
(cid:10)
(cid:8)(cid:11)(cid:9)
(cid:11)
Upper
Triangular
x
1
x
2
⎡
⎢
⎣
⎤
⎥
⎦
=
b
1
b
2
⎡
⎢
⎣
⎤
⎥
⎦
Two Step Solve Given QR
= (cid:4)
T
Rx Q b b
QRx b
Step 1)
= ⇒ =
Step 2) Backsolve Rx b= (cid... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
column is orthogonal
(cid:71)
M
1
(cid:71)
M
(cid:71)
(cid:71)
(
M M
•
1
3
(cid:71)
(cid:71)
(
M M
•
(cid:71)
M
2
(cid:71)
M
)
=
)
r
2
3
r
2
3
r
13
r
13
=
−
−
−
−
0
0
2
3
1
2
SMA-HPC ©2003 MIT
QR Factorization
Orthogonalization
Must Solve Equations for
Coefficients in 3x3 Case
1
(cid:71)
(cid:71)
(
M M
•
(cid:71)
(ci... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)
M
•
N
1
−
(cid:71)
M
(cid:34)
(cid:37)
(cid:34)
1
(cid:71)
(cid:71)
M M
•
1
N
1
−
(cid:35)
•
(cid:71)
M
(cid:71)
M
N
1
−
N
1
−
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
r
1,
N
(cid:35)
r
N
1
,
−
N
=
⎤
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎣
(cid:71)
(cid:71)
M M
•
1
(cid:35)
(cid:71)
M
•
N
1
−
N
(cid:71)
M
⎤
⎥
⎥
⎥
⎦
N
⎡
⎢
⎢
⎢
⎣
2
N
inner products requires
N
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
r
2
3
2
−
⎤
⎥
⎥
⎥
⎥
⎦
To Insure the third column is orthogonal
(cid:71)
(cid:71)
(cid:71)
(cid:71)
(
Q M Q
Q
1
1
1
= ⇒ =
(cid:71)
Q
2
r
23
r
13
r
13
)
−
−
0
•
3
(cid:71)
M
•
3
(cid:71)
(cid:71)
(cid:71)
(
Q M Q
1
−
•
3
2
(cid:71)
Q
2
r
23
)
r
13
−
= ⇒ =
0
r
23
(cid:71)
Q
2
(cid:71)
M
•
3
SMA-HPC ©2003 MIT
QR Factoriza... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
13
0
0
r
12
0
0
r
11
0
0
(cid:34)
(cid:34)
⎡
⎢
⎢
⎢
⎣
0
0
(cid:34)
(cid:34)
(cid:34)
r
1
N
0
0
⎤
⎥
⎥
⎥
⎦
1) Do not try to normalize the column.
2) Do not use the column as a source for orthogonalization.
3) Perform backward substitution as well as possible
SMA-HPC ©2003 MIT
QR Factorization
Basic Algorithm
Zero Column ... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
(cid:34)
(cid:34)
(cid:34)
⎡
⎤
↑ ⎢
(cid:71)
⎥ ⎢
M
⎥ ⎢
⎥
↓ ⎢
⎦
⎣
N
x
1
x
2
(cid:35)
x
N
⎤
⎥
⎥
⎥
⎥
⎦
=
b
1
b
2
(cid:35)
b
N
⎡
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎦
(cid:71)
(cid:71)
x M x M
+
1
2
1
+
(cid:34)
+
2
(cid:71)
x M
N
N
=
b
Two Cases when M is singular
Case 1)
Case 2)
(cid:71)
∈
(cid:71)
b span M M
{
(cid:71)
b span M M
{
,.... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
imization
Suppose
x
=
(cid:71)
x e
1 1
and therefo
re
Mx
=
(cid:71)
x Me
1
1
=
T
( )
R x R x
( )
One dimensional Minimization
(
(cid:71)
b x Me
1
1
−
T
=
=
T
b b
−
−
(cid:71)
) (
b x Me
1
1
(cid:71)
T
x b Me
2
1
1
(cid:71)
T
b Me
2
1
+
(cid:71)
(
e
x M
2
1
1
(cid:71)
T
b Me
1
(cid:71)
(cid:71)
T
T
e M Me
1
1
+
)
(cid:... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
(cid:71)
b x Me
1
1
(cid:71)
Tx b Me
1
1
+
(
2
x
1
(cid:71)
Tx b Me
2
2
−
2
(cid:71)
b x Me
1
1
(cid:71)
x Me
2
2
)
−
(cid:71)
x Me
2
2
(cid:71)
Me
1
T
T
) (
−
(cid:71)
) (
)
e
M
1
(cid:71)
(
) (
2
x Me
2
2
(cid:71)
(
x x Me
1
1 2
) (
T
T
+
(cid:71)
Me
2
(cid:71)
Me
2
)
)
Coupling
Term
+
2
SMA-HPC ©2003 MIT
QR F... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
2003 MIT
0
Minimization
s D
ecouple!!
QR Factorization
Minimization View
Forming MTM orthogonal
Minimization Directions
ith search direction equals MTM orthogonalized unit vector
(cid:71)
p
i
(cid:71)
e
= −
i
i
1
−
∑
j
1
=
(cid:71)
r p
ji
j
T
(cid:71)
(cid:71)
T
p M Mp
i
j
=
0
Use previous orthogonalized
Se... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)
(cid:71)
Mp•
Mp
i
i
(cid:71)
1
p⇐
i
ir
(cid:71)
i
v p
x
= +
i
i
Orthogonalize Search Direction
Normalize search direction
r
ii
(cid:71)
p
i
=
x
j
SMA-HPC ©2003 MIT
QR Factorization
(cid:71)
1Q
(cid:71)
2Q
Minimization and QR
Comparison
(cid:71)
NQ
Orthonormal
M
M
M
(cid:71)
1
e
1
r
(cid:78)
11
(cid:71)
p
1
(
(cid:71... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
:11)(cid:11)(cid:10)
Krylov-Subspace
MTM
Orthogonalization
,
2
(cid:71)
p
(cid:71)
1,
p
}
{
…
,
(cid:8)(cid:11)(cid:9)(cid:11)(cid:10)
Search Directions
N
SMA-HPC ©2003 MIT
Why?
Summary
• QR Algorithm
– Projection Formulas
– Orthonormalizing the columns as you go
– Modified Gram-Schmidt Algorithm
• QR and ... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
6.825 Techniques in Artificial Intelligence
Resolution Theorem Proving:
First Order Logic
Resolution with variables
Clausal form
Lecture 8 • 1
We’ve been doing first-order logic and thinking about how to do proofs. Last time
we looked at how to do resolution in the propositional case, and we looked at
how to do... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
,ψ) = θ
P(x) v Q(x,y)
P(A) v R(B,z)
¬
(Q(x,y) v R(B,z))θ
θ = {x/A}
Lecture 8 • 4
So, we get rid of the P literals, and end up with Q(x,y) v R(B,z), but then we have to
apply our substitution to the result.
4
Resolution with Variables
α v φ [rename]
ψ v β [rename]
¬
(α v β)θ
MGU(φ,ψ) = θ
P(x) v Q(x,y)
P... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
xy.
P(x) v Q(x,y)
P(A) v R(B,z)
z.
¬
∀
∀
(Q(x,y) v R(B,z))θ
Q(A,y) v R(B,z)
θ = {x/A}
Lecture 8 • 7
The x’s in the two sentences are actually different. There is an implicit universal
quantifier on the outside of each of these sentences (we’ll see exactly how we
get sentences ready for resolution in the nex... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
]
ψ v β [rename]
¬
(α v β)θ
MGU(φ,ψ) = θ
xy. P(x) v Q(x,y)
∀
x.
∀
¬
P(A) v R(B,x)
All vars implicitly
univ. quantified
xy.
P(x) v Q(x,y)
P(A) v R(B,z)
z.
¬
∀
∀
(Q(x,y) v R(B,z))θ
Q(A,y) v R(B,z)
Scope of var is local to a clause.
Use renaming to keep vars distinct
∀
x2.
∀
x1y. P(x1) v Q(x1,y) ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
quantifiers (implicit universal
quantifiers).
How do we go from the full range of sentences in
FOL, with the full range of quantifiers, to sentences
that enable us to use resolution as our single
inference rule?
Lecture 8 • 11
So the question is: how do we go from sentences with the whole rich set of
quantifier... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
α v β
¬
(α v β)
(α Æ β)
α
¬¬
x. P(x)
x. P(x)
¬
¬
¬∃
¬∀
¬
¬
β
¬
β
¬
α Æ
α v
α
P(x)
P(x)
x.
x.
¬
¬
∀
∃
Lecture 8 • 15
The next thing you do is to drive in negation. And you already basically know how
to do that. We have deMorgan’s laws to deal with conjunction and disjunction,
and we can eliminate dou... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
doesn’t change the semantics at all).
In this example, we have two quantifications involving the variable x. It’s
especially confusing in this case, because they’re nested. The rules are like
those for a programming language: a variable is captured by the enclosing
quantifier. So the x in Q(x,y) is really a differe... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
instead of writing exists an X such that P of X, you say P of Fred. The trick is
that it absolutely must be a new name. It can't be any other name of any other
thing that you know about. If you're in the process of inferring things about
John and Mary, then it's not good to say, oh, there's a unicorn and it's John -... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
20
Converting to Clausal Form, II
4. Skolemize
• Substitute brand new name for each existentially quantified
variable
•
•
•
•
•
P(X11, Y13)
P(Fred)
x. P(x)
⇒
x. P(x,y)
x. P(x) Æ Q(x)
y.
x.
⇒
x. Loves(x,y)
y. Loves(x,y)
⇒
∀
∃
∃
∃
∃
∃
∀
P(Blue) Æ Q(Blue)
Lecture 8 • 21
All right. If that's ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
lose that information. We’d
get the same result as above, which would be wrong.
So, when you are skolemizing an existential variable, you have to look at the other
quantifiers that contain the one you’re skolemizing, and instead of substituting
in a new constant, you substitute in a brand new function symbol, appli... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
x. Loves(x, Beloved(x))
⇒
x. Loves(x,y)
y. Loves(x,y)
P(Blue) Æ Q(Blue)
P(X11, Y13)
P(Fred)
⇒
∃
∃
∃
∃
∀
∀
∃
⇒ ∀
⇒ ∀
5. Drop universal quantifiers
Lecture 8 • 24
Now we can drop the universal quantifiers because we just replaced all the
existential quantifiers with these skolem constants or functions and s... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
function of all universally quantified variables
in enclosing scopes for each existentially quantified variable.
•
•
•
•
•
x. P(x)
⇒
x. P(x,y)
x. P(x) Æ Q(x)
y.
x.
x. Loves(x, Englebert)
x. Loves(x, Beloved(x))
⇒
x. Loves(x,y)
y. Loves(x,y)
P(Blue) Æ Q(Blue)
P(X11, Y13)
P(Fred)
⇒
∃
∃
∃
∃
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
for
owns and J stand for John.
28
Example: Converting to clausal form
a. John owns a dog
x. D(x) Æ O(J,x)
∃(cid:32)
D(Fido) Æ O(J, Fido)
Lecture 8 • 29
Okay. To convert this to clausal form, we can start at step 4, skolemization,
because the previous three steps are unnecessary for this sentence. Since we just... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
∀(cid:32)
¬∃(cid:32)
First, we get rid of the arrow. Note that the parentheses are such that the existential
quantifier is part of the antecedent, but the universal quantifier is not.
Lecture 8 • 31
31
Example: Converting to clausal form
a. John owns a dog
x. D(x) Æ O(J,x)
∃(cid:32)
D(Fido) Æ O(J, Fido)
b. Anyo... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
cid:32)
y.
∀(cid:32)
¬
¬
(D(y) Æ O(x,y)) v L(x)
D(y) v
¬
O(x,y) v L(x)
D(y) v
¬
¬
O(x,y) v L(x)
There’s no skolemization to do, since there aren’t any existential quantifiers. So,
we can just drop the universal quantifiers, and we’re left with a single clause.
Lecture 8 • 33
33
Example: Converting to clau... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
A to stand for “is an animal” and the predicate symbol K to stand for x kills y.
34
Example: Converting to clausal form
a. John owns a dog
x. D(x) Æ O(J,x)
∃(cid:32)
D(Fido) Æ O(J, Fido)
c. Lovers-of-animals do not kill
animals
x. L(x)
∀(cid:32)
(
→(cid:32)
∀(cid:32)
y. A(y)
K(x,y))
→(cid:32)¬
b. Anyone who... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
O(J, Fido)
c. Lovers-of-animals do not kill
animals
x. L(x)
y. A(y)
K(x,y))
∀(cid:32)
(
→(cid:32)
∀(cid:32)
→(cid:32)¬
b. Anyone who owns a dog is a
lover-of-animals
x.
∀(cid:32)
¬
L(x) v (
y. A(y)
∀(cid:32)
K(x,y))
→(cid:32)¬
x. (
∀(cid:32)
∃(cid:32)
y. D(y) Æ O(x,y))
L(x)
→(cid:32)
x. (
y. (D(y) Æ O(x... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
More converting to clausal form
d. Either Jack killed Tuna
or curiosity killed Tuna
K(J,T) v K(C,T)
e. Tuna is a cat
C(T)
“Tuna is a cat” just turns into C(T).
Lecture 8 • 38
38
More converting to clausal form
d. Either Jack killed Tuna
or curiosity killed Tuna
K(J,T) v K(C,T)
e. Tuna is a cat
C(T)
f. Al... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
of the conclusion, which is not K(C,T), and start doing
the proof.
Lecture 8 • 41
41
Curiosity Killed the Cat
1
2
3
4
5
6
7
8
9
D(Fido)
O(J,Fido)
D(y) v
O(x,y) v L(x)
L(x) v
K(x,y)
¬
A(y) v
¬
¬
¬
¬
K(J,T) v K(C,T)
C(T)
C(x) v A(x)
¬
K(C,T)
¬
K(J,T)
a
a
b
c
d
e
f
Neg
5,8
Lecture 8 • 42
We can ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
)
O(J,Fido)
D(y) v
O(x,y) v L(x)
L(x) v
K(x,y)
¬
A(y) v
¬
¬
¬
¬
K(J,T) v K(C,T)
C(T)
C(x) v A(x)
¬
K(C,T)
¬
K(J,T)
A(T)
L(J) v
A(T)
¬
¬
a
a
b
c
d
e
f
Neg
5,8
6,7 {x/T}
4,9 {x/J, y/T}
Using lines 4 and 9, and substituting J for x and T for Y, we get not L(J) or not
A(T).
Lecture 8 • 44
44
C... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(y) v
¬
¬
¬
¬
K(J,T) v K(C,T)
C(T)
C(x) v A(x)
¬
K(C,T)
¬
K(J,T)
A(T)
L(J) v
A(T)
¬
¬
L(J)
¬
D(y) v
¬
¬
O(J,y)
a
a
b
c
d
e
f
Neg
5,8
6,7 {x/T}
4,9 {x/J, y/T}
10,11
3,12 {x/J}
From 3 and 12, substituting J for X, we get not D(y) or not O(J,y).
Lecture 8 • 46
46
Curiosity Killed the Cat
1
2
3
4
5... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(Fido)
O(J,Fido)
D(y) v
O(x,y) v L(x)
L(x) v
K(x,y)
¬
A(y) v
¬
¬
¬
¬
K(J,T) v K(C,T)
C(T)
C(x) v A(x)
¬
K(C,T)
¬
K(J,T)
A(T)
L(J) v
A(T)
¬
¬
L(J)
¬
D(y) v
¬
¬
O(J,y)
D(Fido)
¬
•
a
a
b
c
d
e
f
Neg
5,8
6,7 {x/T}
4,9 {x/J, y/T}
10,11
3,12 {x/J}
13,2 {x/Fido}
14,1
Lecture 8 • 48
And fina... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
set of sentences proves φ]
• { }
`
Lecture 8 • 51
What does it mean for a sentence to be valid, in the language of entailment? That
it's true in all interpretations. What that means really is that it should be
derivable from nothing. A valid sentence is entailed by the empty set of
sentences. The valid sentence i... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(B))
x.
(
¬
¬
∀
P(x) v P(A)) v
¬
P(x) v P(B))
(
¬
x. (P(x) Æ
∀
¬
P(A)) v (P(x) Æ
P(B))
¬
Lecture 8 • 54
We start by negating it and converting to clausal form. It takes quite a few steps to
drive in all the negations, but eventually we end up with this universally
quantified statement.
54
Proving ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
v P(A)) Æ (
¬
P(x) v P(B))
¬
∃
¬
∃
x.
((
¬
¬
∀
P(x) v P(A)) Æ (
¬
P(x) v P(B))
x.
(
¬
¬
∀
P(x) v P(A)) v
¬
P(x) v P(B))
(
¬
x. (P(x) Æ
∀
¬
P(A)) v (P(x) Æ
P(B))
¬
(P(x) Æ
¬
P(A)) v (P(x) Æ
P(B))
¬
(P(x) v P(x)) Æ (P(x) v
Æ (
P(A) v P(x)) Æ (
P(B))
¬
P(A) v
¬
¬
¬
P(B))
Lecture 8 • 56
And... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
P(B))
¬
1
2
3
4
5
6
P(x)
P(x) v
¬
P(B)
P(A) v P(x)
¬
P(A) v
¬
¬
P(B)
(P(x) v P(x)) Æ (P(x) v
Æ (
P(A) v P(x)) Æ (
P(B))
¬
P(A) v
¬
¬
¬
We enter the clauses into our proof.
P(B))
Lecture 8 • 57
57
Proving validity: example
Prove validity of:
x. (P(x)
P(A)) Æ (P(x)
∃
→
P(B))
→
x. (P(x) ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
P(A) v
¬
¬
¬
P(B))
Lecture 8 • 58
Now, we can resolve lines 1 and 4, substituting A for x, to get not P(B).
58
Proving validity: example
Prove validity of:
x. (P(x)
P(A)) Æ (P(x)
∃
→
P(B))
→
x. (P(x)
→
¬
∃
P(A)) Æ (P(x)
P(B))
→
x. ((
¬
¬
∃
P(x) v P(A)) Æ (
¬
P(x) v P(B))
x.
((
¬
¬
∀
P(x) ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
others using resolution refutation.
xy.F(x,y)
xy.F(y,x)
∀
∀
x.F(x)
(G(x) Ç H(x))
∀
G(A)
→
(H(A)Æ
↔
G(A))
¬
F(A)
¬
∀
x.F(x)Ç G(x)
∀
x.
G(x)
∃
¬
x.H(x)
F(x)
x.F(x)
y.F(y)
∀
xyz.F(x,y)Æ F(y,z)
→
F(x,z)
∃
∃
∃
→
¬
H(x)
x.
¬
F(x,x)
¬
xy.F(x,y)
→
∀
F(y,x)
¬
∀
x.
y.L(x,y)
∀
∃
xy.L(x,y)
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
wl at night
• No cat fails to kill mice
• No animals ever like me, except those that are in this
house
• Kangaroos are not suitable for pets
• None but carnivorous animals kill mice
• I detest animals that do not like me
• Animals that prowl at night always love to gaze at the
moon
• Therefore, I always avoid ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
6.858 Lecture 13
Kerberos
Administrivia
Quiz
Post
review
today
(Actual
idea
your
project
final
quiz
by
next
tomorrow.
Wednesday.
)
Kerberos setting:
• Distributed architecture, evolved from a single time-‐sharing system.
• Many servers providing
services: remote login, mail, printing, file server.
• ... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
set
• Users
don't
need
u
accounts,
passwords,
etc
on
each
server.
p
1
Overall architecture diagram
+-----------------------+
|
c, tgs
[ User: Kc ] <--------> [ Kerberos ]
^
|
V
\
\
\
s
|
|
|
Database... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
store
K_c
and
forget the
user
password? Password-‐
equivalent.
Naming.
• Critical to
Kerberos:
mapping between keys and principal names.
• Each principal name consists of ( name, instance, realm )
o Typically written name.instance@realm
• What
entities have principals?
o Users: name is username... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
‐ willing
to respond to any request.
• How does the
client authenticate
the
Kerberos
server?
o Decrypt the
response
and
check if the
ticket looks
valid.
o Only the Kerberos server would
know K_c.
•
In what ways is this better/worse
than sending
password to server?
o Password
d... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
the design, packet format.
• Difficult to switch to another cryptosystem when DES became too weak.
• DES key
space
is too
small:
keys are only 56 bits, 2^56 is not that big.
• Cheap
to
days
• How could
an adversary
break Kerberos
give
this
weakness?
($20--$200
break
these
DES
via
https://w... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
need to send an authenticator,
in
addition
to the ticket?
o Prove to the server that an adversary is not replaying an old message.
o Server must keep last few authenticators in memory, to detect replays.
• How
does Kerberos use time? What happens if the clock is wrong?
o Prevent stolen tickets from being us... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
appens if the KDC is down?
• Cannot log in.
• Cannot obtain new tickets.
• Can keep using existing tickets.
Authenticating to a Unix system.
4
• No Kerberos
protocol involved
when
accessing
local files,
processes.
•
... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
? Why?
o Special flag in ticket indicates which interface
was used to obtain it.
o Password-‐changing
service only
accepts
tickets
obtained
by
using K_c.
o Ensure that
client
knows old password,
doesn't
just
have the ticket.
• How does the
client change
the
user's
passwo... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
master/slave server means all passwords/keys have to change.
• Must
be physically secure,
no bugs in
Kerberos server software,
o no bugs in any other network service on server machines, etc.
• Can we
do better? SSL
CA
infrastructure slightly better, but not much.
detail
browser
about
when
more
look
o ... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
secrecy (avoiding the
password-‐change
problem).
• Abstract problem: establish a shared secret between two parties.
• Kerberos approach: someone picks the secret, encrypts it, and sends it.
• Weakness: if the encryption
key is stolen,
can
get
the secret later.
• Diffie-‐Hellman
key exchange proto... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
.
• Proxy problem: still no great solution in Kerberos, but ssh-‐agent
is nice.
• Workstation
security (can
trojan
login,
and did happen
in
practice).
o Smartcard-‐based approach hasn't
taken
off.
o Two-‐step
authentication (time-‐based OTP) used by Google Authenticator.
o Shared
preva... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
LECTURE 15
LECTURE OUTLINE
Subgradient methods
Calculation of subgradients
•
•
Convergence
•
***********************************************
Steepest descent at a point requires knowledge
•
of the entire subdifferential at a point
Convergence failure of steepest descent
•
3
2
1
0
-1
-2
2
x
-3
-3
-2
-1
0
x1
1
2
3
60
40
z... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
L(x, µ) = inf
X
x
⌦
⌦
X
x
f (x) +µ �g(x)
⇤
⌅
or minµ
⇧
0 F (µ), where F (
µ)
−
q(µ).
⌃ −
2 ALGORITHMS: SUBGRADIENT METHOD
Problem: Minimize convex function f :
over a closed convex set X.
n
�
→
Subgradient method:
•
�
•
xk+1 = PX (xk −
αkgk),
where gk is any subgradient of f at xk, αk is a
positive stepsize, and PX ( )... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
k),
�
⇥
xk+1 −
for all αk such that
�
y
�
<
xk −
�
y
,
�
0 < αk <
2
�
f (y)
f (xk)
−
2
gk�
�
⇥
.
4 PROOF
Proof of nonexpansive property
•
PX (x)
�
PX (y)
x
� ⌥ �
y
,
�
−
−
x, y
n.
⌘ �
Use the projection theorem to write
PX (x)
�
x
z
−
PX (x)
−
0,
⌥
z
⌘
X
�
�
⇥
PX (y)
from which
Similarly, PX (x)
�
Adding and using ... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
(x
y
−
�
2αk
⌥ �
⌘
αkgk)
2
k −
f (xk)
−
−
−
�
y
2
y
⌃
⌃
) +α 2
gk�
k�
+ α2
k�
2
2,
gk�
where the last inequality follows from the subgra-
�
dient inequality. Q.E.D.
f (y)
⇥
5 CONVERGENCE MECHANISM
Assume constant stepsize: αk ⌃
If
c for some constant c and all k,
α
•
•
�
gk� ⌥
x⇤�
2
�
x
−
k+1
x⇤�
so the distance to the... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
−
If fk = f ⇤, makes progress at every iteration.
If fk < f ⇤ it tends to oscillate around the
If fk > f ⇤ it tends towards the
optimum.
fk}
level set
fk can be adjusted based on the progress of
the method.
f (x)
⌥
x
{
|
.
Example of dynamic stepsize rule:
•
fk = min f (xj)
j
⌅
⌅
k
0
⌅k,
−
and ⌅k (the “aspiration level... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/1814d845fc72f1ad285be44cf8cb3663_MIT6_253S12_lec15.pdf |
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
6.001 Notes: Section 3.1
Slide 3.1.1
In this lecture, we are going to put together the basic pieces of
Scheme that we introduced in the previous lecture, in order to
start capturing computational pro... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
to determine its value. If the expression is
a simple expression there are several possibilities. If it is a self-
evaluating expression, like a number, we just return that value.
If the expression is a name (something we created with a define
expression) then we replace the name with the corresponding
value associ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
the formal
parameter x in the body of the procedure and replace the original expression with this instantiated body expression,
as shown.
This reduces the evaluation of (square 4) to the evaluation of the expression (* 4 4). This is also a compound
expression, so again we get the values of the subexpressions. In th... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
the procedure associated with
average, now substituting 5 and 9 for x and y in that body.
Note that there is no confusion about which x I am referring to,
it's the one in the procedure associated with average, not the
one in square.
As before, I substitute into the body of this procedure, and then
continue. I rec... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
given this common pattern, we can write a procedure to
capture this idea. Here it is. The first part says to give the name
fact to the procedure created by the lambda, which has a single
argument n and a particular body. Now, what does the lambda
say to do? It has two different parts, which we need to look at
care... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
some argument, we first test to see if that value is 1. The if
expression does this by first evaluating the predicate, before ever considering the other expressions, thus changing
the order of evaluation. If the predicate is true, then we simply return the value 1. If it is false, then we will
evaluate the last expr... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
simpler version of fact. Once we get to a
simple case, we can start accumulating those stacked up
operations. We will come back to this idea next time.
6.001 Notes: Section 3.3
Slide 3.3.1
Now that we have seen our first more interesting procedure,
fact, let's step back and generalize the ideas we used to capture... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
of the smaller version of factorial, and that will continue
to unwind to another version, ad infinitum.
The problem is that I didn't use my third step; I didn't consider
the smallest sized problem that I can solve directly, without
using wishful thinking. In the case of factorial, that is just
knowing that factori... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
our factorial procedure from
last time. Remember that this was a nice little procedure that
first checked to see if we were in the base case, and if we were,
returned the answer 1. If we were not, the procedure reduced
the problem to multiplying n by a recursive call to factorial of
n-1. One of the properties of t... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
6
Notice that in this version, we only need to keep track of 2
things, the current product, and the next thing to do. This is
different from the recursive case, where we were blindly
keeping track of each pending multiplication. Here, we replace
the previous product by the new one, and the next multiplier by
the ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
, and that becomes
the next entry in the product column.
Slide 3.4.11
Similarly, I need a rule for getting the next value for counter.
That, we know, is simply given by adding 1 to the current row's
value for counter, thus generating the next row's value for
counter.
6.001 Structure and Interpretation of Comput... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
I use a define
to give it a name. Notice that the body of the procedure is just a
call to another procedure, a helper procedure that I am also
going to define.
The reason I need a helper procedure is clear from my previous
discussion. I need to keep track of three things during the
computation, so I need a proced... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
this works, we use our substitution model. Calling
ifact on some argument, say 4, reduces to calling ifact-
helper on arguments 1, 1, and 4, since that is what the
body of ifact contains, where we have substituted 4 for
the parameter n.
Slide 3.4.22
The substitution model then says to substitute the arguments for ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
evolution to be different. It grew with each new
evaluation, because that extra pending operation had to be
tracked, including information relevant to each operation.
6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 3.4.26
Compare that to th... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology.
Slide 3.5.2
First, we need to talk about what constitutes a formal proof.
Technically, a proof of a mathematical or logical proposition is
a chain or sequence of logical deductions that starts with a base
set of axioms ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
does equivalence, since
the combination can only be true when both simpler elements
are.
Implication is a bit more puzzling. The proposition is defined to
be true if either P is false or Q is true. To see this, take a simple
example. Suppose we consider the proposition: "If n is greater
than 2, then n squared is ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
prove statements
of fact. Unfortunately, the things we want to prove about
programs (e.g. does this code run for all correct inputs) require
potentially infinite propositions, since we would have to take a
conjunction of propositions, one for each input value. So we
need to extend our propositional logic to predic... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
true.
Slide 3.5.7
Let's make this a bit less murky with an example. Suppose I
want to prove the predicate shown here for all non-negative
values of n, that is, that the sum of the powers of 2 can be
captured by the simple expression shown on the right.
To prove this by induction, I should first really specify the... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
assumption here, namely that we give it a correct input. Note
that here, we have made the unstated assumption that factorial
only applies to integers greater than or equal to one. If we
provide some other value, all bets are off. In other words, our
universe here is positive integers.
So here is our code, and our ... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
new problem to solve, it is valuable to identify the base case
and find a solution for it. Then, one turns to the issue of
breaking the problem down into a simpler version of the same
problem, assuming that the code will solve that version, and
using that to construct the inductive step: how to solve the full
vers... | https://ocw.mit.edu/courses/6-001-structure-and-interpretation-of-computer-programs-spring-2005/182629e35d886325280dbc1bb4b5643c_lecture3webhand.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.014 Calculus with Theory
Fall 2010
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-014-calculus-with-theory-fall-2010/182970ab73d4377688a774a8db9db7f3_MIT18_014F10_ChDnotes.pdf |
Queueing Systems: Lecture 6
Amedeo R. Odoni
November 6, 2006
Lecture Outline
• Congestion pricing in transportation: the
fundamental ideas
• Congestion pricing and queueing theory
• Numerical examples
• A real example from LaGuardia Airport
• Practical complications
Reference: Handout on “Congestion Pricing
an... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
”) user is given by:
dW q
dC
dλ
dλ
= c W + cλ
q
MC =
Marginal
cost
Internal
cost
External
cost
Numerical Example
• Three types of aircraft; Poisson; FIFO service
_ Non-jets: λ1 = 40 per hour; c1 = $600 per hour
_ Narrow-body jets: λ2 = 40 per hour; c2 = $1,800 per hour
_ Wide-body jets: λ3 = 10 ... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
E 2[S] +σ S
2 ⋅ (1 − ρ)
+
2 ]
λ⋅[E 2[S] +σ S
⋅
1
2 ⋅ (1 − ρ)2 μ
≈ 5.1556 ×10−6 hours ≈ 18.6 sec
Numerical Example [3]
dC
dλ1
= c1 ⋅Wq + c ⋅
dWq
dλ
≈ $28 + $711 = $739
internal
cost
external cost=
congestion toll
dC
dλ2
dC
dλ3
= c2 ⋅Wq + c ⋅
dWq
dλ
= c3 ⋅Wq + c ⋅
dWq
dλ
≈ $85 + $711 = $796
≈ $1... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
• When we have explicit expressions for Wq, we
can also compute explicitly the total marginal
delay cost MC(i), the internal (or private) cost
and the external cost associated with each
additional user of type i
Example
For an M/G/1 system:
MC(i) =
dC
dλi
2
λ⋅ E[S ]
= ci 2(1 − ρ)
+ cλ
(1 − ρ)E[Si
E[S 2 ]... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
⋅λ (
)x
i
+ K
i
∀i
The missing piece: Demand functions can
only be roughly estimated, at best!
An illustrative example from airports
Service rate
(movements per hour)
Standard deviation of
service time (seconds)
Cost of delay time
($ per hour)
Type 1
(Big)
80
Type 2
(Medium)
90
10
10
$2,500
$1,000... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
5
17,6
12,3
6,6
0,5
-
6
-
12,9
20,2
-
27,9
-
-
36
40
39,8
39,4
38,8
38
37
35,8
34,4
32,8
31
29
26,8
24,4
21,8
19
16
12,8
9,4
5,8
2
1200
1600
1000
1400
1800
2000
600
800
400
200
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
0
1800
1900
2000
Ty pe 1
Ty pe 2
Ty pe 3
Case 1: No... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
135
$853
$988
29.2
$54
$750
$804
34.6
78.7
$22
$670
$692
14.9
3 minutes 15 seconds
89.9%
Demand Functions for three types of users
o
)
e
m
i
t
t
i
n
u
/
s
r
e
s
U
(
e
t
a
r
l
a
v
i
r
r
A
70
60
50
40
30
20
10
0
o
+
+
+
0
200
400
600
800
1000
1200
1400
1600
1800
Total cost ($)
Type 1
Ty... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
delays
• June 2001: Notice for Public Comment posted with regards to
longer-term policy that would use “market-based” mechanisms
• Process stopped after September 11, 2001; re-opened in 2004
Scheduled aircraft movements at LGA
before and after slot lottery
120
Scheduled
movements
100
per hour
80
60
40
20... | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
Time of day
23
1
3
Issues that arise in practice
-- Toll may vary in time and by location
-- Facility users may be driven by “network”
considerations
-- “Social benefit” considerations
-- Political issues
-- What to do with the money? | https://ocw.mit.edu/courses/1-203j-logistical-and-transportation-planning-methods-fall-2006/18342098bf20f7b29102666916eba761_lec10.pdf |
PDE Examples Sheet
Problem 1. Prove that a Harmonic function with an interior maximum is constant.
Problem 2. Write out the laplacian in planepolar coordinates.
Problem 3. A Green’s function on Rn is a harmonic function on Rn \ {0} which
depends only on the radius (for example log r on R2). Find nontrivial Green’s f... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2005/184a3abfb02fb6d9a8d27996023feb00_pset.pdf |
Lecture 3
8.251 Spring 2007
Lecture 3 - Topics
• Relativistic electrodynamics.
•
Gauss’ law
• Gravitation and Planck’s length
Reading: Zwiebach, Sections: 3.1 - 3.6
Electromagnetism and Relativity
Maxwell’s Equations
Source-Free Equations:
� × E� = −
1 ∂B�
c ∂t
� · B� = 0
With Sources (Charge, Current):
... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
c ∂t
( �
�
E, B� ) encoded as (Φ, A)
Φ, A are the fundamental quantities we’ll use
Gauge Transformations
A� → A�� = A� + �
B� � = � × A� = � × (A + �) = B�
function of �x,t. � function = vector.
Φ → Φ� = Φ −
1 ∂
c ∂t
�
E� � = −�(Φ�) = −� Φ −
�
1 ∂
c ∂t
−
1 ∂
c ∂t
(A + �) = E�
So under gauge transformati... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
Φ) = −Ei
F12 = ∂xAy − ∂yAx = Bz
⎛
Fµν = ⎜
⎜
⎝
0
0 −Ex −Ey −Ez
Bz −By
Ex
Bx
0
Ey −Bz
0
−Bx
Ez
By
⎞
⎟
⎟
⎠
What happens under gauge transformation?
Aµ → Aµ
� = Aµ + ∂µ
Then get:
F � = ∂µA�
µν
ν − ∂ν A�
µ
= ∂µ(Aν + ∂ν ) − ∂ν (Aµ + ∂µ)
= Fµν + ∂µ∂ν − ∂ν ∂µ
= Fµν
Define:
Tλµν = ∂xFµν + ∂µFνλ + ∂ν Fλµ
Note ... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
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