text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
field (instructive)
∂αgσµ + ���������
�
∂µgασ − ∂σgµα
µ Aα
µα
Γµ = gµσ
�µAµ = ∂µAµ + Γµα
1
2
1
2
1
= √
g
= gµσ∂αgσµ
∂α
√
g
[To prove ∂αg = ggµσ∂αgµσ , use expansion by minors and expansion for inverse matrix. Check
on diagonal matrices!]
�√
√
√
d4x ∂µ(
gaµ) is semi-trivial: it is a
d4
x
g �µ
Aµ
gAµ �
= � ... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
gµν ∂ν φ) = −V �(φ)
δΛ
δφ
√
∂µ (
√
gV �(φ)
2. Transverse vector field
�
d4
x
S = −
1
4
√
αγ
gg g βδ (∂αAβ − ∂β Aα) (∂γ Aδ − ∂δAγ ) −
�
√
d4
gjµAµ
x
�
��
�
coupling to current
√
gL
δ
∂µ
δ∂µAν
√
gL
δ
δ∂µAν
�√
�
ggµγ g νδ (∂γ Aδ − ∂δAγ )
= −∂µ
gF µν )
·
=
exercise!
−
√
g �µ
F µν
√
= −∂µ (
√
= −
... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
we have that √1
So � √
g �µAµ�ν Aν → λ2 � √
g ∂σpσ = λ, which is constant.
g gives a “dynamical” cosmological term
4. Field equation for gravity
7
a. The hard part of finding the field equation for gravity is varying
√
ggαβδRαβ is a total derivative.
we can use the trick that
To prove this relatively painlessly, ... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
cosmological term
1 √
2
ggαβ δgαβ R +
√
gRαβ δgαβ
�
g Rαβ −
�
gαβ R δgαβ
1
2
�
√
δ
g
δS = −Λ
�
= Λ
√
�
g
�
1
gαβ δgαβ
2
c. Matter
S =
δS =
�
√
gΛ
� �
√
δ
gΛ
δgαβ − ∂µ δ∂µgαβ
gΛ
√
δ
�
δgαβ
We define the energy-momentum tensor by
√
gΛ
√
δ
gΛ
δgαβ − ∂µ δ∂µgαβ
δ
√
g
2
=
Tαβ
We will now see t... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
− (�φ)2
m 2φ2 −
1 �
1
2
2
1 2φ2 �
(�φ)2 + m = W SB
2
•
Maxwell field: (now we use
δ )
δgµν
�√
δ
αγ
gg g βδFαβ Fγδ
�
= −
1
2
2
Tµν = √
g
√
αγ
gg g βδFαβ Fγδ + 2
gµν
� �
�
1
√
−
δ
4
gg g βδFαβ Fγδ
αγ
√
gg βδFµβ Fµδ
�
= −Fµβ Fνδ g βδ +
1
gµν g αγ g βδFαβ Fγδ ⇒
4
gµν Tµν = 0 (�)
In flat space we ha... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
gµν �α
g
Thus δgµν = g�µν (x) = ∂α�µgαν + ∂α�ν gµα − ∂αgµν �α
↑
not
x�
9
Now notice that
δgµν = gαµ�α�ν + g αν �α�µ
(Killing equations)
αρ
Γν
��
�
�
1
g νβ
= gαµ∂α�ν + gαµ
·
2
gµβ �
+ g αν ∂α�µ + g αν 1
·
2
= gαµ∂α�ν + g αν ∂α�µ + gαµg ρν ∂ρgαβ �ρ
= gαµ∂α�ν + g αν ∂α�µ − ∂ρg αβ �ρ
�
�
�ρ
∂αgβρ + ... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
µ
g Tµν �µ �
Tµν �µ �
�
√
g �ν Tµν +
�
√
gTµν �µ;ν
��
�
0
Since this holds for an arbitrary �µ, we conclude that �ν Tµν = 0.
1.8 Newtonian Limit
We should be able to identify Newtonian gravity (and fix the coupling constant) by looking at the
situation with nearly flat space and only T00 = ρ significant. (For ... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
2φ = ρ
This fixes κ =
.
1
16πGN
�
φ = −
GN M
r
; �φ =
GN
M
2
r
ˆr
Gauss
−−−→
�
dV
�
2φ = 4πGN M
�
����
M = 4κ
Note on the appendices and scholia:
These are not necessarily self-contained, specifically, they refer to facts about quantum field
theory and the standard model that are not assumed elsewhere... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
postulate local Lorentz invariance under transformations of this form with Λ(x) a function
of x, Λ(x) ∈ SO(3, 1). To connect this structure to the metric we introduce a vierbein e (x) such
that
a
µ
ηab eµ
gµν e a
a (x)eν
µ(x)e b
b (x) = gµν (x)
ν (x) = ηab
(“square root” of the metric)
(“moving frame”)
Now, for ... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
+ ω a
µ
b e b = 0
ν
ν
(1)
leading to a unique determination
ω a
µ c = −ec
ν ∂µeν
a + e ν e a Γα
c α µν
Appendix 2: Moving Frames Method and Recipe
The discussion of Appendix 1 is not as profound as it should be: ωµ
than Γ. This is worth pursuing, since it leads to a beautiful analogy and useful formulae.
We ca... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
a
ρ = 2ωµ
ac eaρecν
using ωac = −ωca . Multiplying both sides by eeρef ν , we get
µ
µ
A
� �� �
ef = e f ν ∂µe e
B
� �� �
ν − e f ν ∂ν e a
B
�
�
� �� �
ν + e ρe∂ρe f
ρ + eaµe eρ e f ν ∂ρe a
µ − eaµe ρe e νf ∂ν e a
C
��
C
��
A
� �� �
µ − e ρe∂µe f
ρ
�
�
2ωµ
or
ef =
ωµ
�
ef ν
2
∂µe e
ν − ∂ν e e
µ + eaµe eρ∂ρe... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
locally flat” was developed by E. Cartan and is called the moving
frame method. It is usually presented in very obscure ways.
Scholium 1: Structure and Redundancy
Allowing a very general framework and demanding symmetry is an alternative to finding a canon
ical form that “solves” the symmetry. Thus we consider genera... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
allowed if
φ1(x), φ2(x) are).
In General Relativity there is still symmetry, and minimal coupling leads to weak cutoff de
pendence below the Planck scale. However, the fields manifold is not linear (gµν + g may not
be invertible, hence not allowed) and there is a dimensional fundamental coupling. It looks like an
eff... | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
a scale symmetry retains considerable appeal.
A closely related variant is
�αgµν = ∂αφ gµν
�
= λgµν
g
µν
φ� = φ + λ
φ-fields of this sort are called “dilatons”.
14 | https://ocw.mit.edu/courses/8-952-particle-physics-of-the-early-universe-fall-2004/1723f9a7383f25b91c34216de12ab80e_89521.pdf |
Bluetooth Tutorial
Larry Rudolph
1
Pervasive Computing MIT 6.883 SMA 5508 Spring 2006 Larry Rudolph
from bluetooth import *target_name = "My
Phone"target_address = Nonenearby_devices =
discover_devices()for address in nearby_devices:
if target_name == lookup_name( address ):
target_address = address breakif
targe... | https://ocw.mit.edu/courses/6-883-pervasive-human-centric-computing-sma-5508-spring-2006/174043c6531c12d2db27fb724842c42f_r2_bluetooth_tut.pdf |
port
from bluetooth import *socket =
BluetoothSocket( RFCOMM )while True:
free_port = get_available_port( RFCOMM )
try: socket.bind( ( "", free_port ) ) break
except BluetoothError: print "couldn't bind
to ", free_port# listen, accept, and the rest of
the program...
Asynchronous
from bluetooth import *from selec... | https://ocw.mit.edu/courses/6-883-pervasive-human-centric-computing-sma-5508-spring-2006/174043c6531c12d2db27fb724842c42f_r2_bluetooth_tut.pdf |
Introduction to Simulation - Lecture 8
1-D Nonlinear Solution Methods
Jacob White
Thanks to Deepak Ramaswamy Jaime Peraire, Michal
Rewienski, and Karen Veroy
Outline
• Nonlinear Problems
– Struts and Circuit Example
• Richardson and Linear Convergence
– Simple Linear Example
• Newton’s Method
– Derivation of Newton
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
2
(cid:14)
(cid:11)
y
2
(cid:16)
y
0
(cid:12)
2
(cid:11)
x
2
L
2
(cid:32)
f
1
x
(cid:32)
x
1
2
(cid:12)
(cid:14)
(cid:11)
y
2
(cid:16)
2
y
1
(cid:12)
(
(cid:72)
L
o
(cid:16)
L
1
)
x
2
f
2
x
(cid:32)
x
1
(
(cid:72)
L
o
(cid:16)
L
2
)
(cid:16)
x
0
x
2
(cid:16)
L
1
(cid:16)
L
2
(cid:166)
f
1
x
f(cid:14)
2
x
(cid:32)
0
(ci... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
:16)
1)
(cid:16) (cid:32)
0
Need to Solve
I
d
I
srcv
I(cid:14)
r
I(cid:16)
(cid:32)
0
(cid:32)
0
r
Nonlinear
problems
Solve Iteratively
Hard to find analytical solution for
f x (cid:32)
( )
0
Solve iteratively
x(cid:32)
guess at a solution
0
repeat for k = 0, 1, 2, ….
x
0
k
(cid:12)
x
(cid:11)
1k
(cid:14) (cid:32)
W x... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
(
) 10
(cid:32)
1
f x
(
) 13
(cid:32)
2
3
f x
(
f x
(
) 13.9
(cid:32)
) 14.17
(cid:32)
5
x
6
x
7
x
8
x
(cid:32)
14.25
(cid:32)
14.27
(cid:32)
14.28
(cid:32)
14.28
Converged
Richardson
Iteration
Example 1
f x
( )
(cid:32) (cid:16)
0.7
x
(cid:14)
10
kx
*
x(cid:16)
Richardson
Iteration
Example 2
f x
( )
x(cid:32)
2
(cid... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
11) (cid:12)
f x
(cid:4)
(cid:119)
x
(cid:119)
Convergence
Richardson Theorem
(cid:74)
(cid:100) (cid:31)
1 for all
x
(cid:4)
s.t.
x
(cid:4)
(cid:16)
*
x
(cid:31)
(cid:71)
And
0
x
(cid:16)
*
x (cid:71)
(cid:31)
Then
x
k
1
(cid:14) (cid:16)
*
x
(cid:100)
(cid:74)
k
x
(cid:16)
*
x
Or
*
k
1
(cid:14)
k
x
x
(cid:16)
(cid:1... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
.
(cid:170)
1
(cid:14)
(cid:16) (cid:171)
(cid:172)
(cid:32)
x
x
k
k
df
dx
k
(
x
)
1
(cid:16)
(cid:186)
(cid:187)
(cid:188)
k
f x
(
)
if
df
dx
(cid:170)
(cid:171)
(cid:172)
k
(
x
)
1
(cid:16)
(cid:186)
(cid:187)
(cid:188)
exists
until convergence
Newton’s Method
Graphically
Newton’s Method
Example
Newton’s Method
Ex... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
(cid:12)
(cid:12)
or x
(
k
1
(cid:14)
(cid:16)
*
x
)
(cid:32)
1
x
2
k
k
(
x
(cid:16)
2
*
x
)
Convergence is quadratic
Newton’s Method
Convergence
Example 2
(cid:32)
2
x
(cid:32)
0,
*
x
(cid:32)
0
k
)
(cid:32)
2
x
k
1
(cid:16)
Note : not bounded
df
dx
(cid:167)
(cid:168)
(cid:169)
away from zero
(cid:183)
(cid:184)
(ci... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
all
x
if
L x
0
(cid:16)
*
x
(cid:100)
(cid:74)
(cid:31)
1
x
then
k
Proof
converges to
*
x
x
1
(cid:16)
*
x
(cid:100)
L x
(
0
(cid:16)
*
x
)
x
0
(cid:16)
*
x
x
(cid:159) (cid:16)
1
*
x
(cid:100)
(cid:74)
x
0
(cid:16)
*
x
x
(cid:159) (cid:16)
2
*
x
(cid:100)
L x
(cid:74)
0
(cid:16)
*
x
x
1
(cid:16)
*
x
or x
2
(cid:16)
*... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
false convergence
f(x)
k
1
(cid:14)
x
(cid:16)
k
x
(cid:33)
(cid:72)
x
a
(cid:14)
(cid:72)
x
r
k
1
(cid:14)
x
1kx (cid:14)
kx
*x
X
k
(cid:11)
f x
(cid:14)
(cid:12)1
(cid:31)
(cid:72)
f
a
SMA-HPC ©2003 MIT
Newton’s Method
Convergence
Convergence Checks
Also need an "
(cid:11) (cid:12)
f x
" check to avoid false converg... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/176914834032f0c2438eb23cf185a65f_lec8a.pdf |
TUPLES, LISTS,
ALIASING,
MUTABILITY, CLONING
(download slides and .py files and follow along!)
6.0001 LECTURE 5
6.0001 LECTURE 5
1
LAST TIME
functions
decomposition – create structure
abstraction – suppress details
from now on will be using functions a lot
6.0001 LECTURE 5
2
TODAY
have seen variable types... | https://ocw.mit.edu/courses/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/1776670e271578eeb99fc25975f20586_MIT6_0001F16_Lec5.pdf |
IPULATING TUPLES
aTuple:(( ),( ),( ))
can iterate over tuples
def get_data(aTuple):
nums = ()
words = ()
for t in aTuple:
nums( )
words( )
?
?
if not already in words
i.e. unique strings from aTuple
?
nums = nums + (t[0],)
if t[1] not in words:
words = words + (t[1],)
min_n = min(nums)
max_n = max(nums)
unique_wor... | https://ocw.mit.edu/courses/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/1776670e271578eeb99fc25975f20586_MIT6_0001F16_Lec5.pdf |
� compute the sum of elements of a list
common pattern, iterate over list elements
total = 0
total = 0
for i in range(len(L)):
for i in L:
total += L[i]
total += i
print total
print total
notice
• list elements are indexed 0 to len(L)-1
• range(n) goes from 0 to n-1
6.0001 LECTURE 5
10
OPERATIONS ON LISTS - A... | https://ocw.mit.edu/courses/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/1776670e271578eeb99fc25975f20586_MIT6_0001F16_Lec5.pdf |
,6,3,7,0] # do below in order
L.remove(2) mutates L = [1,3,6,3,7,0]
L.remove(3) mutates L = [1,6,3,7,0]
del(L[1]) mutates L = [1,3,7,0]
L.pop()
returns 0 and mutates L = [1,3,7]
6.0001 LECTURE 5
13
CONVERT LISTS TO STRINGS
AND BACK
convert string to list with list(s), returns a list with every
character f... | https://ocw.mit.edu/courses/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/1776670e271578eeb99fc25975f20586_MIT6_0001F16_Lec5.pdf |
Again, Python Tutor is your best friend
to help sort this out!
http://www.pythontutor.com/
6.0001 LECTURE 5
16
LISTS IN MEMORY
lists are mutable
behave differently than immutable types
is an object in memory
variable name points to object
any variable pointing to that object is affected
key phrase to keep... | https://ocw.mit.edu/courses/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/1776670e271578eeb99fc25975f20586_MIT6_0001F16_Lec5.pdf |
ups(L1, L2):
L1_copy = L1[:]
for e in L1_copy:
if e in L2:
L1.remove(e)
for e in L1:
if e in L2:
L1.remove(e)
L1 = [1, 2, 3, 4]
L2 = [1, 2, 5, 6]
remove_dups(L1, L2)
L1 is [2,3,4] not [3,4] Why?
• Python uses an internal counter to keep track of index it is in the loop
• mutating changes the list length but Python d... | https://ocw.mit.edu/courses/6-0001-introduction-to-computer-science-and-programming-in-python-fall-2016/1776670e271578eeb99fc25975f20586_MIT6_0001F16_Lec5.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.438 Algorithms for Inference
Fall 2014
1 Course Overview
This course is about performing inference in complex engineering settings, providing a
mathematical take on an engineering subject. While driven by applicati... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
ob
servations of a hidden state are obtained. Accurately inferring the hidden state
allows us to control and achieve a desired trajectory for spacecraft. Formally,
such scenarios are well modeled by an undirected Guassian graphical model
shown in Figure 2. An efficient inference algorithm for this graphical model is ... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
7-bit codeword before transmission over the
noisy channel. Note that there are 24 = 16 different 7-bit codewords (among
27 = 128 possible 7-bit sequences), each codeword corresponding to one of 16
possible 4-bit messages. The 16 possible codewords can be described by means
of constraints on the codeword bits. These ... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
viduals based on an audio signal. One way to do this is to take the audio input,
segment it into 10ms (or some small) time interval, and then capture appro
priate ‘signature’ or ‘structure’ (in the form of frequency response) of each of
these time sgements (the so-called cepstral coefficient vector or the “features”... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
as an inference
problem. The key observation here is that images have structural properties.
Specifically, in many real-world images, nearby pixels look similar. Such a
similarity constraint can be captured by a nearest-neighbor graphical model of
image. This is also known as a Markov random field (MRF). An example o... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
partition function is expensive. This is also called the
“marginalization” operation. For example, suppose X is a discrete set. Then,
px1 (x1) =
px1,x2 (x1, x2).
x2∈X
(2)
But this requires |X| number of operations for a fixed x1. There are |X| many
values of x1, so overall, the number of operations needed scales a... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
xN ) = px1 (x1)px2 (x2) · · · pxN (xN ).
(5)
5
Then posterior belief calculation can be done separately for each variable. Comput
ing the posterior belief of a particular variable has complexity |X|. Similarly, MAP
estimation can be done by finding each variable’s assignment that maximizes its own
probability.... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/177720360c220f677921a76b6cc33174_MIT6_438F14_Lec1.pdf |
18.03 Class 3, Feb 8, 2010
First order linear equations; systems and signals perspective
[1] First order linear ODEs
[2] Bank Accounts; rate and cumulative total
[3] Systems and signals language
[4] RC circuits
[1] If I had to name the most important general class of differential
equations it would be "linear equ... | https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/17828cb4e899b98aa351a2d6d6f8da2e_MIT18_03S10_c03.pdf |
a bank would pay interest at the end of the month on
the balance at the beginning of the month. We can model this
mathematically:
With Delta t = 1/12 , the statement at the end of the month will read:
x( t + Delta t ) = x(t) + I x(t) Delta t
+ [deposits - withdrawals between t and t+Delta t]
... | https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/17828cb4e899b98aa351a2d6d6f8da2e_MIT18_03S10_c03.pdf |
negative too, from time to time,
because withdrawals are merely negative deposits.
Q3.1. At the indicated point of the graph of Q(t) am I making
(1) deposits
(2) withdrawals
|
| __
| / \ /
| / \ /
_____|/______\______/_________________________________
| \ /<---- here
\__/
... | https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/17828cb4e899b98aa351a2d6d6f8da2e_MIT18_03S10_c03.pdf |
signal and yields the function x(t),
the "output signal" or SYSTEM RESPONSE. Here's a picture:
initial condition
|
|
|
V
______________
| |
--------------> | System | -------------->
|______________|
input output
For example,
x(0)
|
|
|
V
____... | https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/17828cb4e899b98aa351a2d6d6f8da2e_MIT18_03S10_c03.pdf |
electrons flow to the left, because they are negatively
charged. It's confusing. Sorry, I didn't invent this.) The current is measured
in "amperes" and is denoted by I . (I don't know what language has a word
for current starts with an I!) In this "series" circuit, the current is the
... | https://ocw.mit.edu/courses/18-03-differential-equations-spring-2010/17828cb4e899b98aa351a2d6d6f8da2e_MIT18_03S10_c03.pdf |
capacitor is proportional to the *integral*
of the current; it results from a buildup of charge on the two plates of
the capacitor. High capacitance means lots of space for the charge.
A very large capacitor is like no capacitor at all.
To relate these, differentiate KVL:
V'(t) = V'_R(t) + V'_C(t) = R I'(t) + (1/... | 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 |
�
b
σ(b) b ∈ neighborhood(a)
z
|
(1)
The claim is that 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.
neighbor... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
three cases K (a) and K (b) both commute, so the claim holds. This means that K (a)
are all simultaneously diagonalizable. Any simultaneous eigenvector of K (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
}
.
... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
� σ(b)
κ
a
K φ{κa}
a
z
�
�
�
z κ φ
a
�
�
C
{κa}
{κa}
�
C
C
z φ
�
�
C =
a
�
�
|+�
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 diagra... | https://ocw.mit.edu/courses/18-435j-quantum-computation-fall-2003/17936fff47adc2117ae819eeec8cdb1c_qc_lec14.pdf |
with Sa�b . In the 0�, 1� basis, Sab can be represented
is an
by a diagonal matrix, which means that they have to commute.
eigenvector of K (a).
a,b Sab |+�⊗n
K (a)
�
|
|
�
Demonstration of a transmission line effect:
Sab |+� |+�
(|00� + 01� + 10� + 11�)
|
|
|
1
= Sab 2
1
|
|
( 00� + 01� +
=
2
1
= √
2
(|... | 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 |
)
(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
Orthogonalization
Orthonormal M - Picture
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
:34)
N
x
1
x
2
(cid:35)
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... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
(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
(cid:71)
Q
1
(cid:71)
M
(cid:71)
M
=
=
1
1
1
=
1
(cid:71)
(cid:71)
M M
•
1
1
(cid:7... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
⎢
⎥
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:4)
SMA-HPC ©2003 MIT
QR Factorization
Orthogonalization
The General Case
↑
↑
↑
(cid:71)
(cid:71)
(cid:71)
M M... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
ization
Orthogonalization
Must Solve Equations for
Coefficients in 3x3 Case
1
(cid:71)
(cid:71)
(
M M
•
(cid:71)
(cid:71)
(
M M
•
2
(cid:71)
M
(cid:71)
M
−
r
13
3
−
r
2
3
1
−
3
r
13
−
1
r
2
3
(cid:71)
M
(cid:71)
M
•
(cid:71)
(cid:71)
(cid:71)
(cid:71)
r
⎤ ⎡
M M M M
1
3
2
1
(cid:71)
(cid:71)
(cid:71)
(cid:71)
⎥ ⎢
M r
M... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
⎡
⎢
⎢
⎢
⎣
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
3
work
SMA-HPC ©2003 MIT
QR Factorization
Orthogonalization
Use previously
orthogonalized vectors
↑
↑
↑
(cid:71)
(cid:71)
(cid:71)
M M ... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
71)
Q
2
r
23
)
r
13
−
= ⇒ =
0
r
23
(cid:71)
Q
2
(cid:71)
M
•
3
SMA-HPC ©2003 MIT
QR Factorization
Basic Algorithm
“Modified Gram-Schmidt”
For i = 1 to N “For each Source Column”
r
ii
(cid:71)
Q
i
=
(cid:71)
(cid:71)
M•
M
i
(cid:71)
1
M=
ir
i
i
i
Normalize
N
≈∑
2
N
i
1
=
2
2
N
operations
For j = i+1 to N {
(cid:71)
Q
i... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
2
0
r
3
N
(cid:35) (cid:37) (cid:37) (cid:35)
(cid:34)
(cid:34)
N
0
(cid:34)
0
r
NN
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
QR Factorization
Basic Algorithm
“By Picture”
↑
↑
(cid:71)
(cid:71)
1M
1Q
↓
↓
↑
↑
(cid:71)
(cid:71)
2M
2Q
↓
↓
↑
↑
(cid:71)
(cid:71)
3Q
3M
↓
↓
↑
↑
(cid:71)
(cid:71)
4Q
4M
↓
↓
11r
r
12
22r
r
13
r
23
33r
r
14
r
24
34r
44r
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
possible
SMA-HPC ©2003 MIT
QR Factorization
Basic Algorithm
Zero Column Continued
Resulting QR Factorization
↑
(cid:71)
Q
1
↓
⎡
⎢
⎢
⎢
⎢
⎣
0
0
0
↑
(cid:71)
Q
3
↓
(cid:34)
(cid:34)
(cid:34)
↑
(cid:71)
Q
N
↓
⎤
⎥
⎥
⎥
⎥
⎦
r
11
0
0
0
0
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
r
12
0
0
0
0
r
13
0
r
33
0
0
(cid:34)
(cid:34)
r
N
1
0
(cid:34)
r
3
N
(ci... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)
+
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
{
,..,
,..,
∉
(cid:71)
1
1
(cid:71)
b span Q Q
N
(cid:71)
{ ,..,
1
⇒ ∈
}
}
N
}, How accurate is ?
x
N
SMA-HPC ©2003 MIT
QR Factorization
Minimization View
Alternative Formulations
( )
Definiti... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)
(cid:71)
T
T
e M Me
1
1
+
)
(cid:71)
(
2
x Me
1
1
(cid:71)
e
M
1
) (
T
T
) (
)
=
0
Normalization
d
d
x
T
( )
R x R
( )
x
= −
x
1
=
SMA-HPC ©2003 MIT
(cid:71)
x M
1
1
(cid:71)
Me
1
)
QR Factorization
Minimization View
One-dimensional
Minimization, Picture
(cid:71)
(cid:71)
Me M=
1
1
x
1
=
(cid:71)
T
b Me
1
(cid:... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
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 Factorization
Minimization View
Two-dimensional
Minimization Continued
x
=
T
( )
R x R x
( )
2
(cid:71)
and
v p
+
2
2
(cid:71)
(cid:71)
{
p p
,
span
1
More General Search Direc... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
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
Search directions
r
⇒ =
ji
(cid:71)
(
Mp
(cid:71)
(
Mp
j
j
SMA-HPC ©2003 MIT
T
) (
) (
T
(cid:71)
Me
i
(cid:71)
Mp
j
)
)
QR Factorization
Minimization View
Minimizing in the Search
Direction
Decoupled minimization... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
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)
(cid:71)
1
)
e
e
r−
2
12
... | https://ocw.mit.edu/courses/6-336j-introduction-to-numerical-simulation-sma-5211-fall-2003/17c121310705fdd3f5652441104719a4_lec5.pdf |
)(cid:10)
Search Directions
N
MTM
Orthogonalization
Could use other sets of starting vectors
}
{
b Mb M b …
,
,
(cid:8)(cid:11)(cid:11)(cid:9)(cid: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
W... | 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 |
β [rename]
¬
(α v β)θ
MGU(φ,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
MG... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(B,x)
Scope of var is local to a clause.
Use renaming to keep vars distinct
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)
θ = {x/A}
Lecture 8 • 7
The x’s in the two sentences are actually different. There is an implicit universal
quantif... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
or if you find yourself with the same variable in both sentences
and it's getting confusing, then you should rename the sentences apart.
8
Resolution with Variables
α v φ [rename]
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... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
anything about quantifiers. I told
you that clauses have kind of an implicit quantifier in them. But now we've
been looking at languages that have quantifiers.
10
Resolution
Input are sentences in conjunctive normal form with
no apparent quantifiers (implicit universal
quantifiers).
How do we go from the full r... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
14
The first step you guys know very well is to eliminate implications. So you know
how to do that. Anywhere you see a A right arrow B, you just change it into
not A or B.
14
Converting to Clausal Form
1. Eliminate
,
→
↔
2. Drive in
¬
β
α
→
α v β
¬
(α v β)
(α Æ β)
α
¬¬
x. P(x)
x. P(x)
¬
¬
¬∃
¬∀
¬
¬
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
. (P(x1)
∀
∃
→ ∀
→ ∀
x. Q(x,y))
x3.Q(x3,y2))
Lecture 8 • 16
The next step is to rename variables apart. The idea here is that if every quantifier
in your sentence should be over a different variable. So, if you had two
different quantifications over x, you should rename one of them to use a
different variable (w... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
•
∃
x. P(x)
⇒
P(Fred)
Lecture 8 • 18
So, the Skolem insight is that when you have an existential quantification like this,
you're saying there is such a thing as a unicorn, let's say that P is a unicorn.
There exists a thing such that it's a unicorn. You can just say, all right, well, if
there is one, let's ca... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
P(x,y)
x. P(x) Æ Q(x)
⇒
P(X11, Y13)
∃
∃
∃
P(Blue) Æ Q(Blue)
⇒
Lecture 8 • 20
But the names also have to persist so that if you have exists an X such that P of X
and Q of X, then if you skolemize that expression you should get P of Blue and
Q of Blue. You make up a name and you put it in there, but every occur... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
Loves(x,y)
x. Loves(x, Englebert)
P(Blue) Æ Q(Blue)
P(X11, Y13)
P(Fred)
⇒ ∀
⇒
∃
∃
∃
∃
∀
∀
∃
Lecture 8 • 22
In the first case, there is a single y that everyone loves. So we do ordinary
skolemization and decide to call that person Englebert.
In the second case, there is a different y, potentially, for each x... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
call it Beloved of x. Now it’s clear that the person who is loved by x
depends on the particular x you’re talking about.
23
Converting to Clausal Form, II
4. Skolemize
• Substitute brand new name for each existentially quantified
variable
• Substitute a new function of all universally quantified variables
in en... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
Y13)
P(Fred)
⇒
∃
∃
∃
∃
∀
∀
∃
⇒ ∀
⇒ ∀
5. Drop universal quantifiers
6. Convert to CNF
Lecture 8 • 25
And then we convert to conjunctive normal form. At this point, converting to
conjunctive normal form just means multiplying out the and's and the or's,
because we already eliminated the arrows and pushed in the n... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
Converting to clausal form
Lecture 8 • 27
So, let’s do an example from the book, starting with English sentences, writing them
down in first-order logic, converting to clausal form, and then finally doing a
resolution proof.
27
Example: Converting to clausal form
a. John owns a dog
x. D(x) Æ O(J,x)
∃(cid:32)
L... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
write that in FOL as “For all
x, if there exists a y such that D(y) and O(x,y), then L(x).” We’ve added a new
predicate symbol L to stand for “is a lover of animals”.
30
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. Anyone who owns a dog is a
lover-of... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
¬
O(x,y) v L(x)
Next, we drive in the negations. We’ll do it in two steps.
Lecture 8 • 32
32
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. Anyone who owns a dog is a
lover-of-animals
x. (
∀(cid:32)
∃(cid:32)
y. D(y) Æ O(x,y))
L(x)
→(cid:32)
x. (
y... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
:32)¬
x. (
∀(cid:32)
∃(cid:32)
y. D(y) Æ O(x,y))
L(x)
→(cid:32)
x. (
y. (D(y) Æ O(x,y)) v L(x)
∀(cid:32)
¬∃(cid:32)
x.
∀(cid:32)
y.
∀(cid:32)
x.
∀(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)
Lecture 8 • 34
Lovers of animals do not kill animals. We can... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(x)
∀(cid:32)
¬∃(cid:32)
x.
∀(cid:32)
y.
∀(cid:32)
x.
∀(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)
x.
∀(cid:32)
¬
L(x) v (
y.
∀(cid:32)
A(y) v
¬
¬
K(x,y))
L(x) v
¬
¬
A(y) v
¬
K(x,y)
Lecture 8 • 36
Then we’re left with only universal quanti... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
’m sure
you can fill them in.
Lecture 8 • 39
39
Curiosity Killed the Cat
1
2
3
4
5
6
7
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)
¬
a
a
b
c
d
e
f
Lecture 8 • 40
Given all these premises, we’re interested in proving that curiosity killed ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(x)
¬
K(C,T)
¬
K(J,T)
a
a
b
c
d
e
f
Neg
5,8
Lecture 8 • 42
We can apply the resolution rule to any pair of lines that contain unifiable literals.
Here’s one way to do the proof. We’ll use the “set-of-support” heuristic (which
says we should involve the negation of the conclusion in the proof), and resolve
awa... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
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
Curiosity Killed the Cat
1
2
3
4
5
6
7
8
9
10
11
12
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) ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
)
¬
¬
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
6
7
8
9
10
11
12
13
14
D(Fido)
O(J,Fido)
D(y) v
O(x,y) v L(x)
L(x) v
K(x,y)
¬
A(... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
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 finally, from lines 13 and 2, we ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
φ [empty set of sentences entails φ]
φ [empty 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 t... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
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))
¬
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 ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
(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) 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 ... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
¬
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))
¬
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
Pr... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
v
P(B)
¬
P(A) v P(x)
¬
¬
¬
P(A) v
¬
P(B)
P(B)
1,4
{x/A}
(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 • 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)
∃
... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
v P(x)) Æ (P(x) v
Æ (
P(A) v P(x)) Æ (
P(B))
¬
P(A) v
¬
¬
¬
P(B))
Lecture 8 • 59
And we can resolve 1 and 5, substituting B for x, to get a contradiction.
59
Recitation Problems
In each group, derive the last sentence from the
others using resolution refutation.
xy.F(x,y)
xy.F(y,x)
∀
∀
x.F(x)
(G... | https://ocw.mit.edu/courses/6-825-techniques-in-artificial-intelligence-sma-5504-fall-2002/17e67a940531979f0e93adb8a5281bf1_Lecture8FinalPart1.pdf |
ian and
author of Alice in Wonderland.)
Lecture 8 • 61
61
Another, Sillier Problem
You don’t have to do this one. It’s just for fun. Same type as
the previous one. Also from Lewis Carroll.
• The only animals in this house are cates
• Every animal that loves to gaze at the moon is suitable for
a pet
• When I d... | 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 |
application's job to decide
this.
Why do we need this trusted Kerberos server?
to
set
• Users
don't
need
u
accounts,
passwords,
etc
on
each
server.
p
1
Overall architecture diagram
+-----------------------+ ... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
just
use "Kerberos"
protocol?
o Client
machine can forget user password after it gets TGS ticket.
o Can we
just 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 ( n... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
{ T_{c,s}
}_{K_s}
}_{K_c}
• How does the
Kerberos
server authenticate
the
client?
o Doesn't need
to
-‐-‐ 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 t... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
for a ticket encrypted with user's password.
• Then try
to
brute-‐force
the
user's
password
offline:
easy
to parallelize.
• Better design: require client to interact with server for each login attempt.
General weakness:
DES hard-‐coded
into the design, packet format.
• Difficult to switc... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
's principal name from ticket.
o Addr tries to prevent stolen ticket from being used on another machine.
o Lifetime similarly tries to limit damage from stolen ticket.
• How does a network protocol use
Kerberos?
o Encrypt/authenticate all messages with K_{c,s}
o Mail server commands, documents sent to printer, shell... | https://ocw.mit.edu/courses/6-858-computer-systems-security-fall-2014/1812c0fe15b67c5b73b983475243e19d_MIT6_858F14_lec13.pdf |
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