text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
• Airlock / Dust Mitigation
• Habitat / Laboratory
• Mobility
• EVA Suits
• Subsurface Access / Sample
Acquisition
• Science Instrumentation
• Resource Extraction, Utilization
– Ascent Systems
• Earth Return
– Entry system
– Planetary protection
• Cross-Cutting
– Human Health and Performance
– Space and Surface
• Nuc... | https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1eb80d3e0594fe46d8a6cc8a1321d6dc_architecture_mit.pdf |
i
/O /T l
d T h i
39
Near Term Decisions and Forward Work
•
Earth to Orbit (ETO) Transportation
– Understand the options for Shuttle heritage hardware for heavy lift and/or crewed
launch systems
• Moon as a Testbed
– Identify system, subsystem, and operational capabilities we need to test on the
Moon to enable Mars... | https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1eb80d3e0594fe46d8a6cc8a1321d6dc_architecture_mit.pdf |
Space Shuttle to new exploration focused launch systems
– Explore the origin, evolution, structure and destiny of the Universe
– Determine the behavior of the dynamic Earth system effected by natural and
human induced processes and understand the consequences for life on Earth and
beyond
– Explore the Sun-Earth syste... | https://ocw.mit.edu/courses/16-892j-space-system-architecture-and-design-fall-2004/1eb80d3e0594fe46d8a6cc8a1321d6dc_architecture_mit.pdf |
Lab 3 Revisited
• Zener diodes
R
C
6.091 IAP 2008 Lecture 4
1
Lab 3 Revisited
ready
• Voltage regulators
• 555 timers
Vs = 5 V
Vin
Vc
V
c
=
V
s
−
t
RC
1
−
e
⎛
⎜
⎜
⎝
⎞
⎟
⎟
⎠
+15
270
1N758
0.1uf
5K
pot
VCC
8
Threshold
Control Voltage
Trigger
6
5
2
5k
5k
5k
V+
V-
.
+
Comp
A
_
+
Comp
B
_
2N2222
Vo
0.1uf
R
S
Flip
Flop
Q
I... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
>2.0V
HCMOS 0 (low)
– Output low: <0.4v
– Input voltage
low: 0.0 – 0.7v
+5V
+3.98V
input high
range
+2.0V
0.7V
Forbidden Zone
0.4V
input low
rage
output high
range
noise
margin
noise
margin
output low
range
6.091 IAP 2008 Lecture 4
.
5
Power Requirements
• The following power supplies are common for analog
and digit... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
GND
This device contains four independent gates each
of which performs the logic NAND function.
Figure by MIT OpenCourseWare, adapted from the National Semiconductor 54LS00 datasheet.
6.091 IAP 2008 Lecture 4
11
74LS02 NOR Gate
Dual-In-Line Package
VCC
Y4
B4
A4
Y3
B3
A3
14
13
12
11
10
9
8
1
2
3
4
5
6
7
Y1
A1
B1
Y2
A2
... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
H
L
L
L
L
L
L
L
L
Outputs
Y
L
D0
D1
D2
D3
D4
D5
D6
D7
W
H
D0
D1
D2
D3
D4
D5
D6
D7
H = High Level, L = Low Level, X = Don’t Care
D0, D1_D7 = Level of the Respective D Input
6.091 IAP 2008 Lecture 4
14
Figures by MIT OpenCourseWare.
Building Logic
• From basic gates, we
can build other
functions: Exclusive
OR Gate
X
... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
SET
Q
Q
CLR
B1
D
SET
Q
Q
CLR
B1
SET
D
Q
Q
CLR
18
Building a Synchronous Counter
• All bits clock on
the same clock
signal.
• Next count based
on current count.
CLK
Power connections
not shown
6.091 IAP 2008 Lecture 4
B1 B0
0 0
0 1
1 0
1 1
B0
D
SET
Q
Q
CLR
B1 Next
0
1
1
0
B1
SET
D
Q
Q
CLR
B1
Next
19
Shift Registe... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
Hex
Dec Binary Hex
0
1
2
3
4
5
6
7
0000
0001
0010
0011
0100
0101
0110
0111
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
1000
1001
1010
1011
1100
1101
1110
1111
8
9
A
B
C
D
E
F
6.091 IAP 2008 Lecture 4
22
Binary Adder – mth bit
Cin A
0
0
0
0
1
0
1
0
0
1
0
1
1
1
1
1
B
0
1
0
1
0
1
0
1
Sum Cout
0
0
0
1
0
1
1
0
0
1
1
0
1
0
1
1
A
... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
-3
-2
-1
6.091 IAP 2008 Lecture 4
24
Propagation Delays
• All digital logic have
propagation delay
• Typical discrete logic gate
propagation delay ~10ns
X
Z
X
Z
6.091 IAP 2008 Lecture 4
25
Lab Exercise Ring Oscillator
1
0
1
0
1
0
.
6.091 IAP 2008 Lecture 4
26
Lab Exercise
4 Bit Counter – Logic Analyzer
+5
Power c... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
R ohms
R
R
R
R
R
2R
2R
2R
2R
+5V
Vo
30
DA Summary
• Output from digital to analog conversion
are discrete levels.
• More bits means better resolution.
• An example of DA conversion
– Current audio CD’s have 16 bit resolution or
65,536 possible output levels
– New DVD audio samples at 192 khz with 24
bit resolution... | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
based on
sequence of 0, 1.
6.091 IAP 2008 Lecture 4
38 | https://ocw.mit.edu/courses/6-091-hands-on-introduction-to-electrical-engineering-lab-skills-january-iap-2008/1ebc7a3c1f59bab488ae4a8b41c31296_lec4a.pdf |
6.867 Machine learning, lecture 6 (Jaakkola)
1
Lecture topics:
• Active learning
• Non-linear predictions, kernels
Active learning
We can use the expressions for the mean squared error to actively select input points
x1, . . . , xn, when possible, so as to reduce the resulting estimation error. This is an active... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
should choose the inputs. When
the choice of inputs is indeed up to us (e.g., which experiments to carry out) we can select
them so as to minimize T r (XT X)−1 . One caveat of this approach is that it relies on the
underlying relationship between the inputs and the responses to be linear. When this is no
longer the... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
T )−1
�
The matrix inverse can actually be carried out in closed form (easy enough to check)
(A−1 + vv T )−1 = A −
1
(1 + vT Av)
AvvT A
so that the trace becomes
�
T r (A−1 + vv T )−1 = T r [A] −
�
= T r [A] −
= T r [A] −
�
T r AvvT A
�
�
T r v T AAv
�
1
(1 + vT Av)
1
(1 + vT Av)
vT AAv
(1 + vT Av)... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
If we constrain �v� ≤ c,
then the maximizing v is the normalized eigenvector of A with the largest eigenvalue,
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... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
1 =
�
�
1 1 0
0 1
2
(9)
v = [x, 1]T and therefore vT Av = (x2 + 1)/2 and vT AAv = (x2 + 1)/4. The criterion to
be maximized becomes
vT AAv
(1 + vT Av)
=
(x2 + 1)/4
1 + (x2 + 1)/2
(10)
Since z/(1 + z) is an increasing function of z, it follows that the criterion is maximized
when (x2 + 1)/2 is maximized. G... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
�
�
−
ˆθ
ˆθ0
�
�T
=
x
1
= σ∗2 · v T Av
�
σ∗2(XT X)−1 x
1
θ∗
θ∗
0
�
�
�2
0
|x, X
�� ��
�
�
−
θ∗
θ∗
0
��T �
x
1
ˆθ
ˆθ0
�
|x, X
(11)
�
(12)
(13)
(14)
where the expectation is over responses for existing training examples, again assuming that
there is a correct underlying linear mo... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
T
(15)
(16)
(17)
√
The role of
sion model is then given by
2 and other constants will become clear shortly. The new polynomial regres
y = θT φ(x) + θ0 + �, � ∼ N (0, σ2)
(18)
Cite as: Tommi Jaakkola, course materials for 6.867 Machine Learning, Fall 2006. MIT OpenCourseWare
(http://ocw.mit.edu/), Massachusetts... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
the degree of polynomial expansion, especially when the dimension of the
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)
−2−1012−505xy−2−1012−505xy−2−1012−505... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
vectors. The effect is more striking with higher dimensional inputs and higher polynomial
degrees. (We did have to specify the constants appropriately in the feature vectors to make
this work). To shift the modeling from explicit feature vectors to inner products (kernels)
we obviously have to first turn the estimatio... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
estimation (see previous lecture
notes). The effect of the regularization penalty is to pull all the parameters towards zero.
So any linear dimensions in the parameters that the training feature vectors do not pertain
to are set explicitly to zero. We would therefore expect the optimal parameters to lie in
the span ... | https://ocw.mit.edu/courses/6-867-machine-learning-fall-2006/1eedc5b3427ca3eef198d707f016f295_lec6.pdf |
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 8-1
Lecture 8 - Carrier Drift and Diffusion (cont.),
Carrier Flow
February 21, 2007
Contents:
1. Quasi-Fermi levels
2. Continuity equations
3. Surface continuity equations
Reading assignment:
del Alamo, Ch. 4, §4.6; Ch. 5, §§5.1, 5.2
Cite as... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
F1/2(
Ev − EF
kT
)
Outside TE, EF cannot be used.
Define two ”quasi-Fermi levels” such that:
n = NcF1/2(
Efe − Ec
)
kT
p = NvF1/2(
Ev − Efh
)
kT
Under Maxwell-Boltzmann statistics (n (cid:3) Nc, p (cid:3) Nv):
n = Nc exp
Efe − Ec
kT
p = Nv exp
Ev − Efh
kT
What are quasi-Fermi levels good for?
Cite as: J... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
meaning of ∇Ef
For electrons,
Then:
Je = μen
dEfe
dx
= −qnve
dEfe
dx
q
= − ve
μe
∇Efe linearly proportional to electron velocity!
Similarly for holes:
dEfh
dx
=
q
μh
vh
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
MIT OpenCourseWare (http://ocw.... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
0
• ∇Efe (cid:5)
• if n high, ∇Efe small to maintain a certain current level
• if n low, ∇Efe large to maintain a certain current level
Examples:
thermal equilibrium
under bias
n-type
p-type
Ec
EF
Ev
Ec
EF
Ev
Ec
Efe
Ev
Ec
Efh
Ev
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelec... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
(p − n + ND
(cid:2)
(cid:2)
drift
+ qDe∇n
(cid:2)
−)
+ − NA
Hole current equation:
(cid:2)
Jh = qp (cid:2)vh
drift
− qDh∇p
(cid:2)
Total current equation:
J(cid:2)
t = J(cid:2)
e + J(cid:2)
h
Carrier dynamics:
dp
dn
dt = dt = G − R
Still, can’t solve problems like this:
hυ
S
n
S
-L/2
0
L/2
x
Equat... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.720J/3.43J - Integrated Microelectronic Devices - Spring 2007
Lecture 8-11
In terms of current density:
For holes:
∂n
∂t
(cid:2) (cid:2)
= G − R + ∇.Je
1
q
∂p
∂t
1 (cid:2) (cid:2)
= G − R − ∇.Jh
q
Cite as: Jesús del Alamo, c... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
to flow out of device.
For n-type:
contact area: Ac
I
Jes
Jhs
Kirchoff’s law demands current continuity at metal-semiconductor
interface:
|I| = Ac|Jes + Jhs| = qAc|Fes − Fhs|
Equation is sign sensitive:
• IEEE convention: I entering into device is positive
• sign of Jes and Jhs depend of choice of axis in semic... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
semiconductor interface:
(cid:6) = ps
ns
(cid:6) = 0 or S = ∞
Cite as: Jesús del Alamo, course materials for 6.720J Integrated Microelectronic Devices, Spring 2007.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.720J/3.43J - Integrated Microelectr... | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
7.
MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. | https://ocw.mit.edu/courses/6-720j-integrated-microelectronic-devices-spring-2007/1f1fc5af273b0f02a04022ef0899c178_lecture8.pdf |
3.37 (Class8)
Review
C4 (Area Array) 1000-2000 I/O
Cold welding
• Aluminum is the second easiest metal to cold weld
• Make near perfect welds in aluminum wire
Adhesive Bonding
• Unique in that it does not remove surface contamination
• Type I Adhesive Bonding results from attractive force of wetted liquid at th... | https://ocw.mit.edu/courses/3-37-welding-and-joining-processes-fall-2002/1f455cedc7837bb4390f13e61e2c94b0_33708.pdf |
product = see equation on board
o Viscosity
o Initial and final separations
o Radius for a circular disc
• Looking at different forces, viscosities, radii, and separations
o Water at given parameters 7.5ms
o As the joint gets thinner, time gets longer
o Also works in reverse, how long will it take the joint to s... | https://ocw.mit.edu/courses/3-37-welding-and-joining-processes-fall-2002/1f455cedc7837bb4390f13e61e2c94b0_33708.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ALGEBRAIC SURFACES, LECTURE 4
LECTURES: ABHINAV KUMAR
We recall the theorem we stated and lemma we proved f... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
S�
φ
The projections q, q� are birational morphisms and the diagonal morphism com
mutes. Since φ−1(q) is not defined, (q�)−1(p) is not defined either, so ∃C1 ⊂ S an
irreducible curve s.t. q�(C1) = {p}. Moreover, q(C1) = C is a curve in S: if not,
since S1 ⊂ S × S�, q(C1) a point = C1 ⊂ {x} × S� for some x ∈ S; but su... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
that there is a local coordinate y on
S at p s.t. f ∗y ∈ m2
q . To see this, let (x, t) be a local system of coordinates at
p. If f ∗t ∈ m2
q , then f ∗t vanishes on
f −1(p) with multiplicity 1, so it defines a local equation for f −1(p) in OX,q. So
f ∗(x) = u f ∗t for some u ∈ Ox,q. Let y = x − u(q)t; then
q then... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
from S˜ to a variety X that contracts E to a
point must factor through S.
Proof. We can reduce to X affine, then to X = An, then to X = A1 . Then f
defines a function on ˜
�
Theorem 2. Let f : S
→
sequence of blowups πk : Sk →
s.t. f = π1 ◦ · · · ◦ πn ◦ u, i.e. f factors through blowups and an isomorphism.
Proof. If ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
ow-up maps, i.e.
→
(4)
S
q
� �
S
��������
q�
� �
� S�
φ
Proof. First resolve the indeterminacy of φ using S and then note that q� is a
�
birational morphism, i.e. a composition of blowups by the above.
�
ALGEBRAIC SURFACES, LECTURE 4
3
1. Minimal Surfaces
We say that a surface S1 dominates S2 if there is ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
exceptional curve (classically called an “exceptional curve of the first kind”).
Then ∃ a morphism S →
−1.
S�
Proof. We will find S� as the image of a particular morphism from S to a pro
jective space: informally, we need a “nearly ample” divisor which will contract
E and nothing else. Let H be very ample on S s.t. H... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
(H + (i − 1)E)) → H 0(S, OS (H + iE)) → H 0(E, OE (k − i)) →
H 1(S, OS (H + (i − 1)E)) → H 1(S, OS (H + iE)) → 0
Thus, the latter map is surjective for i = 1, . . . , k+1: for i = 0, H 1(S, OS (H)) = 0
so all those H 1(S, OS (H + iE)) are zero.
Next, we claim that M is generated by global sections. Since M is local... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
the image. Since
→
(f �)∗O(1) = M and deg M |E is 0, we see that f � maps E to a point p�. On the
other hand, since H is very ample, H + kE separates points and tangent vectors
away from E as well as separates points of E from points outside E. So f � is an
isomorphism from S − E
So M defines a morphism S
S� � {p�}.... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
�
→
= lim H 0(En, OEn ). Now, it is enough to show that
ˆ
Op
= k[[x, y]]. Let’s show for every n,
∼
H 0(En, OEn = k[[x, y]]/(x, y)n = k[[x, y]]/(x, y)n
) ∼
(8)
For n = 1, H 0(E, OE ) = k. For n > 1, we have
0 → IE
(9)
where E ∼ P1 = ⇒ IE /I 2
obtain
=
n /IE
E ≡ OP1 (1), IE
n+1 → OEn+1 → OEn → 0
n /IE =
n+1 ... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
x, y)n+1 . The truncations are compatible, so ˆ =
�
k[[x, y]] =
p is nonsingular.
Op
⇒
= | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/1f4f36355341a0e235b980608b418882_lect4.pdf |
Kinematics of Rigid Bodies
1
2.003J/1.053J Dynamics and Control I, Spring 2007
Professor Thomas Peacock
2/28/2007
Lecture 7
2-D Motion of Rigid Bodies - Kinematics
Kinematics of Rigid Bodies
Williams 3-3 (No method of instant centers)
”Kinematics” - Description and analysis of the motions of objects without con
s... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [DD Month YYYY].
Kinematics of Rigid Bodies
3
dR
dt
(r2
d
=
dt
= ω × (r2
= ω × R
− r1)
− r1)
B is moving on a circular path relative to A → although neithe... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
combination of translation of a
point fixed on the body and rotation about an axis passing through this point
→ need (x, y, θ).
(In a rigid body, particles are constrained to be the same distance apart.)
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and
Control I, ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
for the angular momentum principle.
dr
dt
+
vP = vG + ω × r
We can express the motion of any point on a rigid body in terms of translation
of another point on the body and a rotation about that point.
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and
Control I... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
YYYY].
Kinematics of Rigid Bodies
7
Figure 9: Kinematic Diagram of Rod AB. Figure by MIT OCW.
ω = θ˙eˆz
vAB = ˙ ex + (θ˙eˆz
xˆ
rAB
) × (L sin θˆ
= (L sin θ)ˆex − (L cos θ)ˆey
x + ˙
ex − L cos θeˆy) = ( ˙
θL cos θ)ˆex
+ ( ˙
θL sin θ)ˆey
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
Rolling hoop. Hoop rolls without slipping. Mass m is attached to
the hoop. Figure by MIT OCW.
Want to specify the position of the mass.
Pick c as the reference point.
xm = xc − r sin θ
ym = yc − r cos θ
xc, yc, θ form a complete coordinate system.
But there are 2 constraints on the surface.
1. Rolls on surface ... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
��nes the system.
Note: If slipping is allowed, need xc and θ to describe state of the system. This
is because the hoop could slide (translate without rotation).
Cite as: Thomas Peacock and Nicolas Hadjiconstantinou, course materials for 2.003J/1.053J Dynamics and
Control I, Spring 2007. MIT OpenCourseWare (http://... | https://ocw.mit.edu/courses/2-003j-dynamics-and-control-i-spring-2007/1f67510b4e0e8758705be60bc6957982_lec07.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.006 Introduction to Algorithms
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 4
Balanced Binary Search Trees
6.006 Spring 2008
Lecture 4: Balanced Binary Search Trees
Lecture Overview
• The importance ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/1f940ab45156183a323edd0011317452_lec4.pdf |
-balancing binary search trees. These trees are named after their two
inventors G.M. Adel’son-Vel’skii and E.M. Landis 1
An AVL tree is one that requires heights of left and right children of every node to differ
by at most ±1. This is illustrated in Fig. 4)
k-1
k
Figure 4: AVL Tree Concept
In order to implement an... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/1f940ab45156183a323edd0011317452_lec4.pdf |
5
1 +
2
√
5
where φ =
(rounded to nearest integer)
≈ 1.618
(golden ratio)
= ⇒ maxh ≈
logφ(n) ≈ 1.440 lg(n)
AVL Insert:
1. insert as in simple BST
2. work your way up tree, restoring AVL property (and updating heights as you go).
Each Step:
• suppose x is lowest node violating AVL
• assume x is right-heavy ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/1f940ab45156183a323edd0011317452_lec4.pdf |
left-right case55155412011501φφ22623129φφ65Done3φ Lecture 4
Balanced Binary Search Trees
6.006 Spring 2008
Balanced Search Trees:
There are many balanced search trees.
Adel’son-Velsii and Landis 1962
AVL Trees
B-Trees/2-3-4 Trees Bayer and McCreight 1972 (see CLRS 18)
BB[α] Trees
Red-black Trees
Splay-Trees
Ski... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/1f940ab45156183a323edd0011317452_lec4.pdf |
u [Van Ernde
Boas; see 6.854 or 6.851 (Advanced Data Structures)]
Big Picture:
Abstract Data Type(ADT): interface spec.
e.g. Priority Queue:
• Q = new-empty-queue()
• Q.insert(x)
• x = Q.deletemin()
vs.
Data Structure (DS): algorithm for each op.
There are many possible DSs for one ADT. One example that we will ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2008/1f940ab45156183a323edd0011317452_lec4.pdf |
Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.438 Algorithms for Inference
Fall 2014
8
Inferences on trees: sum-product algorithm
Recall the two fundamental inference problems in graphical models:
1. marginalization, i.e. computing the marginal distribution p... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
Elimination algorithm on trees
Recall that a graph G is a tree if any two nodes are connected by exactly one path.
1 edges and no cycles.
This definition implies that all tree graphs have exactly N
Throughout this lecture, we will use the recurring example of the tree graphical
model shown in Figure 1. Suppose we ... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
x1x3x5x1x3x1x1x6x2x3x4x5x6x2x3x4x5 Figure 4: The sequence of graph structures and messages obtained from the elimina
tion algorithm on the graph from Figure 1 using an optimal ordering (4, 5, 3, 2, 1).
Fortunately, for tree graphs, it is easy to find an ordering which adds no edges.
Recall that, in last lecture, we s... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
xj ).
i∈V
(i,j)∈E
(1)
The messages produced in the course of the algorithm are:
m4(x2) =
φ4(x4)ψ24(x2, x4)
m5(x2) =
m3(x1) =
m2(x1) =
x4
x5
x3
x2
φ5(x5)ψ25(x2, x5)
φ3(x3)ψ13(x1, x3)
φ2(x2)ψ12(x1, x2)m4(x2)m5(x2)
(2)
Finally, we obtain the marginal distribution over x1 by multiplying the incoming
messages wit... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
marginal for another variable
x3. If we use the elimination ordering (5, 4, 2, 1, 3), the resulting messages are:
m5(x2) =
m4(x2) =
m2(x1) =
m1(x3) =
(cid:88)
x5
(cid:88)
x4
(cid:88)
x2
(cid:88)
x1
φ5(x5)ψ25(x2, x5)
φ4(x4)ψ24(x2, x4)
φ2(x2)ψ12(x1, x2)m4(x2)m5(x2)
φ1(x1)ψ13(x1, x3)m2(x1)
(4)
Notice that t... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
would, in fact, have higher computational complexity. However,
with the proper bookkeeping, we can reuse computations between messages to ensure that the total
complexity is O(N
X
2).
|
|
4
x1x2x3x4x5
Figure 6: The message mi(xi) depends on each of the incoming messages mk(xi) for
xi’s other neighbors N (... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
��erent variables, which suggests that it might be par
allelizable. This intuition turns out to be correct: if the updates (6) are repeatedly
applied in parallel, it is possible to show that the messages will eventually converge
t(xj ) denote the messages at time
to their correct values. More precisely, letting mi
... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
that this requires O (N
|
Parallel sum-product is unlikely to pay off in practice unless the diameter of the tree
is small. However, in a later lecture we will see that it naturally leads to loopy belief
propagation, where the update rule (7) is applied to a graph which isn’t a tree.
X
|
8.4 Efficient implementation
I... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
xi∈X
ψi(xi, xj )µt(xi)
i
tm (xi)
j→i
(8)
(9)
a
Using this algorithm, each update (8) can be computed in O(
di) time, so the cost
X
2) per
per iteration is O (
di) = O (
|
|
X
2N d) total.
node, so the overall running time is O (
|
(Recall that d is the diameter of the graph.) A similar strategy can be applied to ... | https://ocw.mit.edu/courses/6-438-algorithms-for-inference-fall-2014/1faaccb44f78c4f4e99c6814842082a0_MIT6_438F14_Lec8.pdf |
Corn Sheller
This project is a low-cost device for removing corn kernels from the cob. It is made from a piece
of sheet metal which is formed and then joined together. This device was developed by Marco
Villagarcia, an engineer from Cusco, Peru, based on an injection molded plastic corn sheller from
Malawi. It is ma... | https://ocw.mit.edu/courses/ec-720j-d-lab-ii-design-spring-2010/1fab1f92881a2a0aa9ee50b0154a1960_MITEC_720JS10_bldit_csm.pdf |
the ridges on your part and then use the die to
bend the ridges.
Forming the Cone
You can form the cone by bending it over an anvil or a piece of pipe, by
hand or with a hammer or you can bend the whole things by hand. Be
careful not to damage the ridges as you form the cone.
Fig 1 The sheet metal sheller, made
at ... | https://ocw.mit.edu/courses/ec-720j-d-lab-ii-design-spring-2010/1fab1f92881a2a0aa9ee50b0154a1960_MITEC_720JS10_bldit_csm.pdf |
deliver enough current
through the sheets in a concentrated area that the sheets melt together
at that spot. Spot welding is not appropriate for all materials, however the
sheet metal that we are using can easily be welded using the spot welder
in the Pappalardo Lab.
Finishing the sheller
In order to finish your sh... | https://ocw.mit.edu/courses/ec-720j-d-lab-ii-design-spring-2010/1fab1f92881a2a0aa9ee50b0154a1960_MITEC_720JS10_bldit_csm.pdf |
Introduction to Engineering
Introduction to Engineering
Systems, ESD.00
Networks III/Stakeholders
Lecture 9
Lecturers:
Professor Joseph Sussman
Dr. Afreen Siddiqi
TA: Regina Clewlow
f
Outline
Outline
Introduction to networks (Lecture 8)
Infrastructure networks ((Lecture 8))
Institutional networks (Lecture 8)... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1fc7ee99f40b5f81cfa1765ed1304033_MITESD_00S11_lec09.pdf |
structure and function of complex networks”, SIAM
[1]M E J Newman The structure and function of complex networks SIAM
review, 2003
[2] Duncan J. Watts & Steven H. Strogatz, “Collective dynamics of ‘small
[2] Duncan J. Watts & Steven H. Strogatz, Collective dynamics of small
world’ networks’, Nature, Vol. 393, 4 June... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1fc7ee99f40b5f81cfa1765ed1304033_MITESD_00S11_lec09.pdf |
PProposiitiion 1
related
the cumulative number of stakeholder
attributes –power, legitimacy, and urgency –perceived by
(Mitchell)
present. (Mitchell)
managers toto bebe present
managers
will b
to
i i
li
l
St eholder
s
Stakeholders
ak
• Represent in three‐circled Venn diagram
• Power, legitimacy, urgency
... | https://ocw.mit.edu/courses/esd-00-introduction-to-engineering-systems-spring-2011/1fc7ee99f40b5f81cfa1765ed1304033_MITESD_00S11_lec09.pdf |
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
Physics 8.07: Electromagnetism II
Prof. Alan Guth
September 21, 2012
LECTURE NOTES 4
CONDUCTORS: SURFACE FORCES AND CAPACITANCE
These notes are an addendum to Lecture 7, Wednesday September 19, 2012. The
notes will not repeat what I said in class, but rather will... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/1ffda29d6ddadef8e37e4be9c071b4e7_MIT8_07F12_ln4.pdf |
6)E · d(cid:6)(cid:13)
∞
(cid:4)
= +
r
∞
(cid:6)E · d(cid:6)(cid:13) =
Q
4π(cid:8)0
(cid:4) ∞ dr
r2
r
=
Q
4π(cid:8)0r
.
(4.2)
(4.3)
8.07 LECTURE NOTES 4, FALL 2012
p. 2
The potential on the surface is therefore Q/(4π(cid:8)0R, and the potential inside the sphere
is constant, with this value:
V (r) = V (R) −
(cid:4) r
... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/1ffda29d6ddadef8e37e4be9c071b4e7_MIT8_07F12_ln4.pdf |
(cid:4)
dWmech = P dR
da = 4πR2P dR .
(4.7)
(4.8)
Even though we have integrated, I am still calling the work dWmech, since it is an in-
finitesimal quantity proportional to dR. From Eq. (4.5), the change in the total potential
energy is given by
dW =
dR = −
dR .
(4.9)
dW
dR
Q2
8π(cid:8)0R2
By conservation of energy, th... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/1ffda29d6ddadef8e37e4be9c071b4e7_MIT8_07F12_ln4.pdf |
(We are treating the
surface as if it is a plane, since the thickness of the surface charge layer is far smaller
than R.) Thus, the average piece of charge in the surface experiences an electric field
that is midway between the value outside and the vanishing value inside, as we found
by the method of virtual work. The ... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/1ffda29d6ddadef8e37e4be9c071b4e7_MIT8_07F12_ln4.pdf |
,
and nˆ is a unit outward normal vector to the surface. The total charges on each conductor
i would be specified by
(cid:3)
(cid:4)
σ da = Qi .
Si
(4.14)
The charges on each conductor will distribute themselves so that the field inside the
conductor is zero, and we will prove later that this condition determines the dis... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/1ffda29d6ddadef8e37e4be9c071b4e7_MIT8_07F12_ln4.pdf |
used.
The most commonly discussed situation involves just
two conductors, with charges that are equal in magnitude
but opposite in sign. Here V is used to denote the poten-
tial difference between the two conductors. This pair of
conductors is called a capacitor. In this case we define the
capacitance C by
Q = CV .
(4.18... | https://ocw.mit.edu/courses/8-07-electromagnetism-ii-fall-2012/1ffda29d6ddadef8e37e4be9c071b4e7_MIT8_07F12_ln4.pdf |
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https://ocw.... | https://ocw.mit.edu/courses/2-062j-wave-propagation-spring-2017/201818c21f2a69ccbd7afc473b5aa143_MIT2_062J_S17_Chap5.pdf |
Machine Learning for Healthcare
HST.956, 6.S897
Lecture 1: What makes healthcare unique?
Prof. David Sontag & Pete Szolovits
1
The Problem
• Cost of health care expenditures in the US are
over $3 trillion, and rising
• Despite having some of the best clinicians in
the world, chronic conditi... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
indicate
information flow.
© Addison-Wesley Publishing Company, Inc. This content is excluded from our Creative Commons license. For more information, see https://ocw.mit.edu/help/faq-fair-use/
FIGURE 33-1 Short sample dialogue. The physician’s
appear in capital letters after the double asterisks.
inputs
4
... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
L<CC<.0:/)/@)76<;/6<;()F<CC<.0:/)/@)K-;-16:<f-
cP<::-1)-/)6:$()d*'()"5M- -/)6:$()d%2e
5 1980’s: automating medical discovery
Discovers that prednisone
elevates cholesterol
(Annals of Internal Medicine, ‘86)
[Robert Blum, “Discovery, Confirmation and Incorporation of Causal Relationships
from a Large Time-Oriented... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
8
96%
85.2%*
00000
%*
94
96.9%*
Certi(cid:1)ed EHR
83.8%*
Basic EHR
Percentage
of hospitals
in the US
71.9%
75.5%*
59.4%*
44.4%*
27.6%*
9.4%
12.2%
15.6%
2008 2009 2010 2011 2012 2013 2014 2015
Courtesy of Health and Human Services. Image is in the public domain.
[Henry et al., ONC Data Brief, May 2016]
9... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
lab tests
imaging
social media
phone
vital signs
(cid:47)mages are (cid:104)(cid:94) (cid:39)overnment wor(cid:364). (cid:47)mages are in the public domain.
devices
genomics
12
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5//E93``-;$M<8<E-F... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
All rights reserved. This content is excluded from our Creative Commons
license. For more information, see https://ocw.mit.edu/help/faq-fair-use/
14
"/6;F61F<f6/<@;
(cid:47)mage is in the public domain.
15 Standardization
OMOP
Common
Data
Model v5.0
Image is in the public domain.
16
]1-68/51@0K59)<;)76.... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
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b0/@76/<.6::O)-D/16./-F)C1@7)
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25 456/)M<::)/5-)S^)@C)/5-)C0/01-)A-):<8->
?1@E6K6/<;K)A-9/)E... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
Output
Pneumonia Positive (85%)
b115O/57<6>
© (cid:51)a(cid:75)pur(cid:76)ar et al. All rights reserved. This content is excluded from our
Creative Commons license. For more information,
see https://ocw.mit.edu/help/faq-fair-use/
l<K01-) 9@01.-93)^6iE01861 -/)6:$()61h<H32+22$U&RR&)Q2+
^6iE01861 -/)6:$()61h<H32+U... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
cid:14)(cid:71)(cid:66)(cid:74)(cid:83)(cid:14)(cid:86)(cid:84)(cid:70)(cid:16)
30 456/)<9)/5-)C0/01-)@C)5@M)M-)/1-6/)
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R%@'
Disease burden
Undiagnosed
condition
#<7-
Courtesy of the CDC. Image is in the public domain.
l<K01-).1-F</3)5//E93``MM... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
of diabetes, Alzheimer's,
cancer
• Continuous monitoring and coaching, e.g. for the
elderly, diabetes, psychiatric disease
• Discovery of new disease subtypes; design of
new drugs; better targeted clinical trials
33
Outline for today’s class
1. Brief ... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
• How to set up as machine learning problems
• Understand which learning algorithms are
likely to be useful and when
• Appreciate subtleties in safely & robustly
applying ML in healthcare
• Set the research agenda for the next decade
38
6... | https://ocw.mit.edu/courses/6-s897-machine-learning-for-healthcare-spring-2019/2020812c6b3be9e2c27873595d1a9aed_MIT6_S897S19_lec1.pdf |
6.852: Distributed Algorithms
Fall, 2009
Class 3
Today’s plan
• Algorithms in general synchronous networks (continued):
Shortest paths spanning tree
Minimum-weight spanning tree
Maximal independent set
• Reading: Sections 4.3-4.5
• Next:
– Distributed consensus
– Reading: Sections 5.1, 6.1-6.3
Last time... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
for termination).
• Required:
– Shortest-paths tree, giving shortest paths from i0 to every other node.
– Shortest path = path with minimum total weight.
– Each node should output parent, “distance” from root (by weight).
Shortest paths
10
1
5
1
8
6
3
3
2
5
4
2
4
7
11
9
6
Shortest paths
3
2
5
10
1
2
4 ... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
10
11 5 1
11
1
6
3
8 0
4
0
9
6
3
5
2
0
2
4
Round 1 (trans)
Shortest paths
2
5
10
1
4
2
4
5
2
11 1
8
6
3
3
7
11
0
9
6
Round 2 (start)
Shortest paths
7
2
10
11 5 2 1
11
1
6
3
11
8 0
4
0
9
6
3
5
2
0
2
4 2
Round 2 (msgs)
... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
4
0
9 6
6
6
5
2
0
2
4 2
Round 3 (trans)
Shortest paths
3
2
5
10
1
4
2
4
5
2
10
1
8
6
3
9
3
7
11
0
9
6
6
Round 4 (start)
Shortest paths
10 1
3
7
2
3
11 3 5 2 1
310
10
9 8 0
6
3
9
6
3
4
0
9 6
6
6
5
2
0
2
4 2
Round 4 (msgs)
Shortest paths
10
1
3
7
2
3
11 3 5 2 1
310
10
9 8 0
6
3
9
6
3
4
0
9... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
10
1
4
2
4
5
2
10 1
8
6
3
9
3
7
11
0
9
6
6
End configuration
Correctness
z Need to show that, after round n-1, for each
process i:
disti = shortest distance from i0
parenti = predecessor on shortest path from i0
z Proof:
z Induction on the number r of rounds.
z But, what statement should w... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
path.
If disti is finite, then it really is the distance on some at-most-r-hop path to i,
and parent is its parent on such a path.
Claim that disti and parenti correspond to a shortest at-most-r-hop path.
Any shortest at-most-r-hop path from i0 to i, when cut off at i’s predecessor
j on the path, yields a s... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
edges.
Processes have UIDs.
Nodes know (a good upper bound on) n.
z Required:
Each process should decide which of its incident edges are in MST
and which are not.
Minimum spanning tree theory
• Graph theory definitions (for undirected graphs)
– Tree: Connected acyclic graph
– Forest: An acyclic graph (not... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
Contradicts assumed properties of T.
Minimum spanning tree algorithms
• General strategy:
– Start with n isolated nodes.
– Repeat (n-1 times):
• Choose some component i.
• Add the minimum-weight outgoing edge (MWOE) of component i.
• Sequential MST algorithms follow (special cases of) this
strategy:
– Dijkstr... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
component know which incident edges are in the tree.
• Each level k component has at least 2k nodes.
• Every level k+1 component is constructed from two or more level k
components.
• Level 0 components: Single nodes.
• Level k o level k+1:
Level k o Level k+1
• Each level-k component leader finds MWOE of its
com... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
• In particular, test messages are supposed to compare
leader UIDs to determine whether endpoints are in the
same component.
• Requires that the node being queried has up-to-date UID
information.
Minimum spanning tree
4
g
a
d
12
5
c
1
2
f
9
i
10
0
b
8
3
7
e
11
j
6
h
k
13
Minimum spanning tr... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
11
j
k
13
Minimum spanning tree
4
g
a
d
12
5
c
1
2
f
9
i
10
0
b
8
3
7
e
11
j
6
h
k
13
Minimum spanning tree
a
d
4
g
12
5
2
c
1
ok
f
8
3
e
0
b
9
i
10
ok
7
k
13
11
j
6
h
Minimum spanning tree
4
g
a
d
12
5
c
1
2
f
9
i
10
0
b
8
3
7
e
11
j
6
h ... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
i
10
0
b
8
3
7
e
11
j
6
h
k
13
Simplified GHS MST Algorithm
z Proof?
z Use invariants; but this is complicated because the
algorithm is complicated.
z Complexity:
Time: O(n log n)
n rounds for each level
log n levels, because there are t 2k nodes in each level k component.
Messages: O( (n + |E|)... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/203e61d32dfad9ba4279f269f1e4595b_MIT6_852JF09_lec03.pdf |
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