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Article Contents
Article Contents

# Homoclinic orbits of first-order superquadratic Hamiltonian systems

• In this article, we study the existence of homoclinic orbits for the first-order Hamiltonian system \begin{equation*} J\dot{u}(t)+\nabla H(t,u(t))=0,\quad t\in\mathbb{R}. \end{equation*} Under the Ambrosetti-Rabinowitz's superquadraticy condition, or no Ambrosetti-Rabinowitz's superquadracity condition, we present two results on the existence of infinitely many large energy homoclinic orbits when $H$ is even in $u$. We apply the generalized (variant) fountain theorems established recently by the author and Colin. Under no Ambrosetti-Rabinowitz's superquadracity condition, we also obtain the existence of a ground state homoclinic orbit by using the method of the generalized Nehari manifold for strongly indefinite functionals developed by Szulkin and Weth.
Mathematics Subject Classification: Primary: 37J45; Secondary: 35B38, 70H05.

 Citation:

•  [1] R. A. Adams, Sobolev Spaces, Academic Press, New York, 1975. [2] T. Bartsch and A. Szulkin, Hamiltonian Systems: Periodic and Homoclinic Solutions by Variational Methods, Elsevier B. V., Amsterdam, 2005. [3] C. J. Batkam and F. Colin, Generalized fountain theorem and applications to strongly indefinite semilinear problems, J. Math. Anal. Appl., 405 (2013), 438-452.doi: 10.1016/j.jmaa.2013.04.018. [4] C. J. Batkam and F. Colin, On multiple solutions of a semilinear Schrödinger equation with periodic potential, Nonlinear Anal., 84 (2013), 39-49.doi: 10.1016/j.na.2013.02.006. [5] H. Brezis and E. Lieb, A relation between pointwise convergence of functions and convergence of functionals, Proc. Amer. Math. Soc., 88 (1983), 486-490.doi: 10.2307/2044999. [6] G. Chen and S. Ma, Homoclinic orbits of superlinear Hamiltonian systems, Proc. Amer. Math. Soc., 139 (2011), 3973-3983.doi: 10.1090/S0002-9939-2011-11185-7. [7] V. Coti-Zelati, I. Ekeland and E. Séré, A variational approach to homoclinic orbits in Hamiltonian systems, Math. Ann., 228 (1990), 133-160.doi: 10.1007/BF01444526. [8] Y. H. Ding, Multiple homoclinics in a Hamiltonian system with asymptotically or super linear terms, Commun. Contemp. Math., 8 (2006), 453-480.doi: 10.1142/S0219199706002192. [9] Y. H. Ding and M. Girardi, Infinitely many homoclinic orbits of a Hamiltonian system with symmetry, Nonlinear Anal., 38 (1999), 391-415.doi: 10.1016/S0362-546X(98)00204-1. [10] L. Jeanjean, On the existence of bounded Palais-Smale sequences and application to a Landesman-Lazer type problem set on $\mathbbR^N$, Proc. Roy. Soc. Edinburgh A, 129 (1999), 787-809.doi: 10.1017/S0308210500013147. [11] W. Kryszewski and A. Szulkin, Generalized linking theorem with an application to a semilinear Schrödinger equation, Adv. Differ. Equat., 3 (1998), 441-472. [12] P.-L. Lions, The concentration-compactness principle in the calculus of variations. The locally compact case. II, Ann. Inst. H. Poincaré, 1 (1984), 223-283. [13] A. Mao and S. Luan, Critical points theorems concerning strongly indefinite functionals and infinite many periodic solutions for a class of Hamiltonian systems, Appl. Math. Comput., 214 (2009), 187-200.doi: 10.1016/j.amc.2009.03.084. [14] A. Pankov, Periodic nonlinear Schrödinger equation with application to photonic crystals, Milan J. Math., 73 (2005), 259-287.doi: 10.1007/s00032-005-0047-8. [15] P. H. Rabinowitz and K. Tanaka, Some results on connecting orbits for a class of Hamiltonian systems, Math. Z., 206 (1991), 473-499.doi: 10.1007/BF02571356. [16] E. Séré, Existence of infinitely many homoclinic orbits in Hamiltonian systems, Math. Z., 209 (1992), 27-42.doi: 10.1007/BF02570817. [17] M. Schechter and W. Zou, Weak linking theorems and Schrödinger equations with critical Sobolev exponent, ESAIM Control Optim. Calc. Var., 9 (2003), 601-619.doi: 10.1051/cocv:2003029. [18] M. Struwe, Variational Methods. Applications to Nonlinear Partial Differential Equations and Hamiltonian Systems, Second Edition, Springer-Verlag, Berlin, 1996. [19] C. A. Stuart, Bifurcation into spectral gaps, Bull. Belg. Math. Soc., Supplement, (1995). [20] A. Szulkin and T. Weth, Ground state solutions for some indefinite variational problems, J. Funct. Anal., 257 (2009), 3802-3822.doi: 10.1016/j.jfa.2009.09.013. [21] A. Szulkin and T. Weth, The Method of Nehari Manifold, Int. Press, Somerville, MA, 2010. [22] A. Szulkin and W. Zou, Homoclinic orbits for asymptotically linear Hamiltonian systems, J. Funct. Anal., 187 (2001), 25-41.doi: 10.1006/jfan.2001.3798. [23] K. Tanaka, Homoclinic orbits in a first order super-quadratic Hamiltonian system: Convergence of subharmonic orbits, J. Differential Equations, 94 (1991), 315-339.doi: 10.1016/0022-0396(91)90095-Q. [24] J. Wang, J. X. Xu and F. B. Zhang, Homoclinic orbits for superlinear Hamiltonian systems without Ambrosetti-Rabinowitz growth condition, Discrete Contin. Dyn. Syst. Ser. A, 27 (2010), 1241-1257.doi: 10.3934/dcds.2010.27.1241. [25] M. Willem, Minimax Theorems, Birkhauser, Boston, 1996.doi: 10.1007/978-1-4612-4146-1. [26] M. Yang, Ground state solutions for a periodic Schrödinger equation with superlinear nonlinearities, Nonlinear Anal., 72 (2010), 2620-2627.doi: 10.1016/j.na.2009.11.009. [27] W. Zou, Variant fountain theorems and their applications, Manuscripta Math., 104 (2001), 343-358.doi: 10.1007/s002290170032.

• on this site

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# EKF SLAM (prediction of new landmarks)

Prediction of new landmarks are commonly expressed as:

        Xm = Xr + r*cos(phi + theta_r),
Ym = Yr + r*sin(phi + theta_r)

However this is only true for point landmarks. What if I am extracting line feature?

format: plain


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# ▸ Neural Networks: Learning :

1. You are training a three layer neural network and would like to use backpropagation to compute the gradient of the cost function. In the backpropagation algorithm, one of the steps is to update$\inline&space;\Delta_{ij}^{(2)}&space;:=&space;\Delta_{ij}^{(2)}&space;+&space;\delta_i^{(3)}&space;*&space;(a^{(2)})_j$
for every i,j. Which of the following is a correct vectorization of this step?

1. Suppose Theta1 is a 5x3 matrix, and Theta2 is a 4x6 matrix. You set thetaVec = [Theta1(:), Theta2(:)]. Which of the following correctly recovers ?
• reshape(thetaVec(16 : 39), 4, 6)
This choice is correct, since Theta1 has 15 elements, so Theta2 begins at
index 16 and ends at index 16 + 24 - 1 = 39.

• reshape(thetaVec(15 : 38), 4, 6)

• reshape(thetaVec(16 : 24), 4, 6)

• reshape(thetaVec(15 : 39), 4, 6)

• reshape(thetaVec(16 : 39), 6, 4)

1. Which of the following statements are true? Check all that apply.
• For computational efficiency, after we have performed gradient checking to verify that our backpropagation code is correct, we usually disable gradient checking before using backpropagation to train the network.
Checking the gradient numerically is a debugging tool: it helps ensure a correct implementation, but it is too slow to use as a method for actually computing gradients.

• Computing the gradient of the cost function in a neural network has the same efficiency when we use backpropagation or when we numerically compute it using the method of gradient checking.

• Using gradient checking can help verify if one’s implementation of backpropagation is bug-free.
If the gradient computed by backpropagation is the same as one computed numerically with gradient checking, this is very strong evidence that you have a correct implementation of backpropagation.

• Gradient checking is useful if we are using one of the advanced optimization methods (such as in fminunc) as our optimization algorithm. However, it serves little purpose if we are using gradient descent.

### Check-out our free tutorials on IOT (Internet of Things):

1. Which of the following statements are true? Check all that apply.
• If we are training a neural network using gradient descent, one reasonable “debugging” step to make sure it is working is to plot J(Θ) as a function of the number of iterations, and make sure it is decreasing (or at least non-increasing) after each iteration.
Since gradient descent uses the gradient to take a step toward parameters
with lower cost (ie, lower J(Θ)), the value of J(Θ) should be equal or less at each iteration if the gradient computation is correct and the learning rate is set properly.

• Suppose you have a three layer network with parameters $\inline&space;\theta^{(1)}$ (controlling the function mapping from the inputs to the hidden units) and $\inline&space;\theta^{(2)}$ (controlling the mapping from the hidden units to the outputs). If we set all the elements of $\inline&space;\theta^{(1)}$ to be 0, and all the elements of $\inline&space;\theta^{(2)}$ to be 1, then this suffices for symmetry breaking, since the neurons are no longer all computing the same function of the input.

• Suppose you are training a neural network using gradient descent. Depending on your random initialization, your algorithm may converge to different local optima (i.e., if you run the algorithm twice with different random initializations, gradient descent may converge to two different solutions).
The cost function for a neural network is non-convex, so it may have multiple minima. Which minimum you find with gradient descent depends on the initialization.

• If we initialize all the parameters of a neural network to ones instead of zeros, this will suffice for the purpose of “symmetry breaking” because the parameters are no longer symmetrically equal to zero.

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If you find this helpful by any mean like, comment and share the post.
This is the simplest way to encourage me to keep doing such work.

Thanks & Regards,
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# openlearn

## September 22, 2021

Essay and report writing skills

This is a repost

Hopefully reposting this here, will inspire people.

I completed Essay and report writing skills [3] with OpenLearn [2], earlier this year. This is part of the Open University [1]. In fact, this course mirrors a student tool kit book I have here, also from the OU. It is nice to have a printed and bound version.

Very useful course, lots of useful ideas, pointers to writing good essays and reports. Topics included structure, referencing, quotations,drafting, polishing and letting go. So probably just as applicable in the workplace as it is in the Academic world

I am looking forward to further study.

The UK government want all adults to spend time learning. To this end have created the skills toolkit [4] to help adults find learning and development courses.

This is a great idea, provided that employers:-

• Support employees in learning
• Value any learning an employee undertakes
• Value the commitment and work put in in order to learn new skills
• Recognise employee achievements on their CV
• See potential in their employee(s)

REFERENCES

TAGS

## May 7, 2021

OpenLearn Courses

Education & Development course

Am I ready to be a distance learner? Introductory level Education & Development course Last accessed: 04 February 2021

Essay and report writing skills Introductory level Education & Development course Last accessed: 27 February 2021

Encouraging book talk in the school library Introductory level Education & Development course Last accessed: 14 August 2018

Science, Maths & Technology course

Galaxies, stars and planet Introductory level Science, Maths & Technology course Last accessed: 27 October 2020

Health and safety in the laboratory and field Introductory level Science, Maths & Technology course Last accessed: 18 August 2018

In the night sky: Orion Introductory level Science, Maths & Technology cours Last accessed: 30 June 2020

The Moon Introductory level Science, Maths & Technology course Last accessed: 10 February 2019

The Sun Introductory level Science, Maths & Technology course Last accessed: 31 December 2019

The evolving Universe Introductory level Science, Maths & Technology course Last accessed: 27 December 2019

Particle physics Introductory level Science, Maths & Technology course Last accessed: 10 November 2020

Simple coding Introductory level Science, Maths & Technology course Last accessed: 28 October 2020

Sociology course

Children’s rights Intermediate level Sociology course Last accessed: 16 November 2018

Environmental Studies course

Energy resources: Tidal energy Intermediate level Environmental Studies course Last accessed: 28 October 2020

Facilitating group discussions Introductory level Money & Business course Last accessed: 16 August 2018

Mathematics & Statistics course

Numbers, units and arithmetic Introductory level Mathematics & Statistics course Last accessed: 27 January 2021

Education course

Teaching assistants: support in action Introductory level Education course Last accessed: 11 September 2018

REFERENCES

TAGS

## May 5, 2021

OpenLearn : Discovering Chemistry

Just signed upto Discovering Chemistry with Open Learn [1]. This is a 24 hour course from the Open University [2] via their OpenLearn platform.

This is partly for personal interest, partly for career development and will help me develop useful knowledge going forward. I have set up a thread on Discourse [3] to facilitate discussion.

I would like to learn more about Isotope shift , which is mentioned in a arXiv article [4] from 2017. This topic is probably beyond the scope of the chemistry course, but is a good starting point. Hopefully however this Chemistry course can build on what I already know about isotopes at least.

REFERENCES

TAGS

## March 14, 2021

OpenLearn : Childhood in the digital age

I have just started a 12 hour OpenLearn course, Childhood in the digital age. However I really need to be working in a school to do this course as the first activity asks

What do you think? How might technology hinder children’s physical, social and cognitive development, and how might it provide exciting opportunities for learning?

Make a note of you thoughts at the start of this course. Use a notebook or digital notepad that you can refer back to at a later date.

So while I can answer the question. I would really like to answer this and other course activities, following discussion with children / young people and teachers, in order to help me answer this with an informed and evidence based approach.

I may therefore defer going any further until I am able to do this. Adds value to working as a TA if I can also use this as a way to help with OU Open Learn courses.

In the meantime I am open to offers of employment.

REFERENCES

TAGS

## February 27, 2021

Essay and report writing skills

I have just completed a course on Essay and report writing skills [3] with OpenLearn [2], which is part of the Open University [1]. In fact, this course mirrors a student tool kit book I have here, also from the OU. It is nice to have a printed and bound version.

Very useful course, lots of useful ideas, pointers to writing good essays and reports. Topics included structure, referencing, quotations,drafting, polishing and letting go. So probably just as applicable in the workplace as it is in the Academic world

I am looking forward to further study.

The UK government want all adults to spend time learning. To this end have created the skills toolkit [4] to help adults find learning and development courses.

This is a great idea, provided that employers:-

• Support employees in learning
• Value any learning an employee undertakes
• Value the commitment and work put in in order to learn new skills
• Recognise employee achievements on their CV
• See potential in their employee(s)

REFERENCES

TAGS

## January 27, 2021

Numbers, units and arithmetic

I have just completed the course 'Numbers, units and arithmetic' with Open Learn. This is an excellent 5 hour course and will be really useful going forward.

Well worth doing if you need a refresher as there are lots of worked examples and activities to have a go at to re-enforce learning.

You can of course, as with any OpenLearn course use this as a reference in the future if you need to revise or revisit how to do something. Full PDF of the course is also .

REFERENCES

TAGS

## June 25, 2020

In the night sky : Orion Completed

I have finally, after months of going back and forth to this, partly due to other study and things to do. Completed the 24 hour OU / OpenLearn Course. I was just doing this casually anyway.

Note: This is a Level 1 Open University course.

After studying this course, you should be able to:

* Understand facts, concepts, principles, theories, classifications and language used in astronomy * Understand the range of sizes, distances and motions of objects in the night sky * Understand the structure, evolution and the main processes operating in stars * Understand the properties of planets in our Solar System and exoplanetary systems * Understand the history of the universe.

Really interesting course, lots to think about and learn just from doing some of the research for the questions posed during the course.

Astronomy uses the greek alphabet for star names for example so this post may help.

## In the night sky : Orion

I am currently working through another short OpenLearn course. This time I am exploring the constellation of Orion.

Having previously studied other Space science courses, some of the concepts are easier to understand, such as the lifecycle of stars.

Despite understanding the maths behind huge numbers. Distances such as 1 light year are massive : (ly) $$9.5 \times 10^{12} km.$$

The subsequent distance to the Orion Nebula which is 1600 light years. Therefore $$1600 \times 9.5 \times 10^{12} km$$.

I think, that this should be expressed as $$1.6 \times 9.5 \times 10^{15} km$$

Understandably it is difficult to imagine the sort of distances involved.

If you intend to study this course. I would recommend other courses first. Courses such as 'The Sun' may be a good starting point.

I am about ½ way through the second week of the course. Currently looking at the life cycle of a star beyond the main sequence phase, so this includes Red Giants, Brown / White dwarf, Supernova and black holes, or at least how and under what conditions these are formed.

You do need an astronomy package, to help with the course. This comes in more useful, if you don't have clear skies to observe Orion. I have discussed how to use the kstars package to find Orion in post yesterday (30/12/2019).

What I can take from this course, is some inspiration to write some posts about other constellations of the night sky.

References :

It may be beneficial to have a look at these courses before embarking on this longer course:

The Galaxies stars and planets course has a section on the scale of the universe, which is helpful to help you understand very large numbers.

You can find me on Friendica at zleap@social.isurf.ca

Licenced under Attribution 4.0 International (CC BY 4.0)

## OpenLearn course Update: 27/12/2019

I have just completed another OU / OpenLearn course, this time on The Evolving Universe. Despite its rather complex subject matter it was very interesting and expanded a little further on the 'How the universe works' course I undertook a few years ago.

The course looked at the early universe, the big bang, how particles, energy, time and temperature all changed over time and how all these are inter-related. How the matter we know today formed and why the universe is mostly matter and not antimatter. The fact there was more matter and what would have happened if the amount of antimatter and matter had been equal. Plus other interesting topics.

How sub-atomic particles such as Bosons, Leptons, Quarks etc interact. How research finds new particles and the huge energy levels required for this to take place.

This is a level 1 introductory course, and like the more formal course mentioned. The course also asks other questions which gets the reader thinking about different scenarios that could have occurred.

## References

You can find me on Friendica at zleap@social.isurf.ca

Licenced under Attribution 4.0 International (CC BY 4.0)


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# Solving Equations by Using Addition and Subtraction

Solving Equations Using Addition and Subtraction

## Practice Questions

Question #1:

$$2x+13=41$$

$$x=27$$
$$x=7$$
$$x=14$$
$$x=28$$

The correct answer is $$x=14$$.
$$2x+13=41$$
The first thing to do with this equation is subtract 13 from both sides.
$$2x=28$$
Then, divide both sides by 2.
$$x=14$$

Question #2:

$$8x-29=14$$

$$x=-1$$
$$x=5$$
$$x=-1\frac{7}{8}$$
$$x=5\frac{3}{8}$$

The correct answer is $$x=5\frac{3}{8}$$.
$$8x-29=14$$
The first thing to do with this equation is add 29 to both sides.
$$8x=43$$
Then, divide both sides by 8.
$$x=\frac{43}{8}$$
Finally, turn $$\frac{43}{8}$$ into a mixed number.
$$x=5\frac{3}{8}$$

Question #3:

$$41-2x=19$$

$$x=30$$
$$x=11$$
$$x=-30$$
$$x=-11$$

The correct answer is $$x=11$$.
$$41-2x=19$$
The first thing to do with this equation is subtract 41 from both sides.
$$-2x=-22$$
Then, divide both sides by -2.
$$x=11$$

Question #4:

$$21+13-x=3x-6$$

$$x=10$$
$$x=11$$
$$x=14$$
$$x=9$$

The correct answer is $$x=10$$.
$$21+13-x=3x-6$$
The first thing to do with this equation is combine like terms.
$$34-x=3x-6$$
Next, add 6 to both sides.
$$40=4x$$
Finally, divide both sides by 4.
$$10=x$$

Question #5:

$$x-17+11=21-5$$

$$x=22$$
$$x=10$$
$$x=14$$
$$x=27$$
The correct answer is $$x=22$$.
$$x-17+11=21-5$$
$$x-6=16$$
$$x=22$$


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# Normal distribution with multiple variance components

#1

I’m trying to model a Generalised Least Squares panel data model with a diagonal variance-covariance matrix, i.e. y_i|X_i,\beta,\Sigma \sim N(X_i\beta,\Sigma),\ \Sigma=diag(\sigma^2_1,\dots,\sigma^2_T). I’ve been able to model a Panel Data model with a common variance, i.e. y_i|X_i,\beta,\sigma^2_\varepsilon\sim N(X_i\beta,\sigma^2_\varepsilon I_T), but I fail with multiple components. I have, however, been able to write the code for an arbitrary \Sigma, but that code failed due to memory limitations (N=2376) and I know that the matrix is diagonal anyway. Could anyone give a hint on how to make this code work for multiple variance components?

data {
int<lower=0> N; \\ Number of cross-sectional units
int<lower=0> K; \\ Number of regressors
real bval[N]; \\ y-value
real btreat[N]; \\ Regressor 1
real bcountrd[N]; \\ Regressor 2
}

parameters {
real beta[K];
real sigma[N];
}

transformed parameters {
real mu[N];
for (i in 1:N) {
mu[i] = beta[1]+beta[2]*btreat[i]+beta[3]*bcountrd[i];
}
}

model {
// Likelihood
bval ~ normal(mu,sigma);
}

#2

I think you’ve coded the model you intended. If you move mu to the model block it’ll use less (there’s less output) but you still have O(N) sigmas:

model {
real mu[N];
for (i in 1:N) {
mu[i] = beta[1]+beta[2]*btreat[i]+beta[3]*bcountrd[i];
}

bval ~ normal(mu,sigma);
}

I think it’s really unlikely this model fits well though honestly. There’s no structure on the sigmas. Do you have multiple samples of each y?


text:
Browse Questions

# Which of the following obeys the octet rule?

$\begin{array}{1 1}(a)\;CCl_4\\(b)\;O_2\\(c)\;NF_3\\(d)\;\text{ All of the above}\end{array}$

As per octet rule, all the atoms contain 8 electrons in the valence shell.
Hence (d) is the correct answer.


text:
It is currently 20 Mar 2019, 19:18

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# 2^3 * 17 * 5^2

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2^3 * 17 * 5^2 [#permalink]  27 Jul 2018, 09:28
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Question Stats:

100% (01:17) correct 0% (00:00) wrong based on 6 sessions
 Quantity A Quantity B $$\frac{2^3 * 17 * 5^2}{60}$$ $$\frac{255}{2}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E
[Reveal] Spoiler: OA

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Re: 2^3 * 17 * 5^2 [#permalink]  27 Jul 2018, 09:44
1
KUDOS
Carcass wrote:
 Quantity A Quantity B $$\frac{2^3 * 17 * 5^2}{60}$$ $$\frac{255}{2}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

QTY A = $$\frac{2^3 * 17 * 5^2}{60} = \frac{8 * 17 * 25}{60} = \frac{170}{3}$$

As we compare we find the numerator of QTY A is less than the numerator of QTY B and

the denominator of QTY A is greater than the denominator of QTY B

Hence QTY A < QTY B

Option B
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Re: 2^3 * 17 * 5^2   [#permalink] 27 Jul 2018, 09:44
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text:
# Identify the kind of friction (static or kinetic)

I have a block that is on a surface. The coefficient of static friction between the two is $0.3$, while the coefficient of kinetic friction is $0.2$. If I apply a force of $18\,\mathrm{N}$ to the left, is my block moving? How can I tell? Please show work and explain the logic.

My attempt:

$$N = mg$$

Force of friction static $= \mu N = 29\,\mathrm{N}$ -> I need a force greater than $29\,\mathrm{N}$ for the block to move. I have a resultant force of $18\,\mathrm{N}$ to the left. Therefore the block won't move. BUT this is wrong; the solution says that the block will move. What am I doing wrong?

Either way, can you explain it to me in your words with logic that I can stick to?

• What is the mass of the block? – Anthonny May 24 '15 at 23:57
• 10kg, how did JoDraX even answer me without that? – Edward Newgate May 25 '15 at 0:06
• Could it be possible that the surface is tilted at an angle? – fibonatic May 25 '15 at 1:22

When you have a block with a mass of 10 kg, the normal force is approximately 98 N.

If the block starts out stationary, it will continue to be stationary until you apply sufficient force to exceed the static friction. At that point it will start to slide, and will continue sliding until the force drops below the force of dynamic friction.

For the parameters you gave, minimum force needed to start moving the block is $0.3\cdot 98 = 29.4 N$. And the force needed to keep the block moving is $0.2 \cdot 98 = 19.6 N$.

So if the block is on a horizontal surface (no other forces) it will not move.

Unless you left something else out... Clearly the mass of the block is critical.

What about the car when there is no force applied to the block? The block will experience no frictional force, so when dealing with static friction, we have to say $f_s \geq \mu_s n$, since there are instances when the frictional force may not act at all or may at at a reduced capacity. So here, $f_s = 18N$, since that is the only force necessary to keep the block from moving.

• Does the block move or not? – Edward Newgate May 25 '15 at 0:01
• The block would not move. Because the net force is zero. – JoDraX May 25 '15 at 0:02


text:
# Tag Info

10

Thomas is correct; there's no attack on CFB mode if you can predict the IV; NIST is just being cautious. With CBC, the value of the first encrypted block $C_0 = E_k( IV \oplus P_0)$, where $IV$ is the IV used for that packet, $P_0$ is the value of the first plaintext block, and $E_k$ is the evaluation of the block cipher. If an attacker can predict the ...

8

One obvious thing that it is vulnerable to a known plaintext attack that truncates the known message. This attack is quite simple; suppose the attacker knows a message $(P_1, P_2, ..., P_n)$ and the corresponding ciphertext $(C_1, C_2, ..., C_n, T)$ (using some IV; we don't care what it is). Here is how the attacker can generate a ciphertext that would ...

8

I recommend that you prepend a random 16-byte prefix. Prepending a random 16-byte prefix, before encrypting with your CFB mode, will be just as good as using a random IV. The argument is pretty similar to Using CBC with fixed IV. If we use CFB with an all-zeros IV and a random 16-byte value prefixed to the message before encryption, as you suggested, we ...

8

I found a little more info on Google, so let me provide a partial answer to my own question. In particular, I found a post by David Wagner to sci.crypt in 2004, titled "IND-CPA for CFB mode", which in turn led me to a paper titled "Practical symmetric on-line encryption", published in FSE 2003 by Fouque, Martinet and Poupard. In this paper, the authors ...

7

CFB with a fixed IV? Yikes! That is completely insecure: for the first 16 bytes of plaintext, it is even worse than ECB mode, and that's saying something. Please go enlighten whoever thought it was a good idea to expose this as the only mode of encipherment available (or even one among multiple options). Let me elaborate. It sounds like the baseline is ...

6

The modes you are referencing are specifically modes of operations for block ciphers, and therefore are not directly applicable to hash functions. Block cipher operations take 2 inputs, the key and a block-sized input value, and output a block-sized keyed permutation of the input. Hash functions take a variable length input, and output a fixed length value. ...

6

Actually, for CFB mode, the IV is the same size as the block size, 16 bytes. As for your question "does keeping the IV secret help security", the answer is "not really". CFB mode processes the message in blocks, and for each block of plaintext, combines that with the previous block of ciphertext to generate the next block of ciphertext. What the IV is ...

4

Actually, s is in CFB mode to handle transmission channels for the encrypted data that can add or drop individual bytes. In the olden times (say, the 70's), it was common to transmit data over serial channels, for example, RS-232. These channels were not perfect, and one common error we see is that if the transmitter sent 7 bytes, the receiver might get ...

3

Yes, it is correct. Just follow the bits in the decryption pictures on the Wikipedia page about modes of operation. Modes of operation don't have to have a meaning compared to other modes of operation. I don't see CFB or OFB used too much anymore. OFB with partial feedback has been shown to be less secure, so that shouldn't be used anymore. Currently the ...

3

Forget OFB mode. You should use CTR (counter) mode. It has the best bounds, and is parallelizable. This means that when you are using the AES-NI instruction set, encrypt with CTR is about 7 times faster than CBC, OFB etc. If you encrypt in OpenSSL you will get this performance. For a good thorough analysis and comparison of modes of operation, see ...

3

According to Handbook of Applied Cryptography (15.3.2, ii), ANSI X9.9 (which SEJPM mentioned in the comments but I have no access to) defined CFB-MAC only as a compatible alternative to CBC-MAC: The X9.9 MAC algorithm may be implemented using either the cipher-block chaining (CBC) or 64-bit cipher feedback (CFB-64) mode, initialized to produce the same ...

3

No, because CFB isn't commutative. You can see this by looking at the decryption of double-CFB encrypted ciphertext. Even assuming a constant IV (so single-use keys), if you decrypt in the wrong order it cannot work, since the ciphertext is used as input into the block cipher and will differ from what was used with that key when encrypting. The exception, ...

3

For CFB mode: NEVER make the IV constant, it must be unique for every message. The IV does not need to be secret or impossible to predict, only unique. It can be a simple counter, for example. The IV may not be chosen by the attacker. I can not emphasis UNIQUE enough, if your IV is not unique you've basically lost all security.

2

At least one thing goes very wrong: if an adversary can obtain the ciphertext for a few short chosen plaintexts and the same reused key/initial state, that allows reconstructing the state because the indexes vary in a controlled way. That certainly works if we obtain the ciphertext for the $2^{24}$ plaintexts consisting of the 3 bytes $u$ $v$ $w$ $0$ ...

2

I have a problem with OFB mode, because I have heard that it is stronger than CFB. On the contrary I would say that CFB is stronger. OFB means encrypting the IV again and again to produce the keystream. If you end up in a cycle, the keystream will start repeating itself. (This should not be a practical weakness, but why chance it?) CFB is more like ...

2

There is no real advantage, other than the fact that it allows you to convert a block cipher into a stream cipher securely. Since there has been a large amount of research put into block ciphers and ciphers such as AES are commonly implemented in hardware (such as AES-NI), it allows for reuse of the primitives. Side note: the nonce generally does not need ...

2

I've been thinking a little bit about it, and now I think it is possible, but you have to consider the generalization of CFB in ISO 10116 (I don't have access to the ISO 10116 standard, so I will assume that the description by Rogaway is correct). The generalization of CFB from the ISO standard seems to have two main changes: The feedback block (FB), of ...

1

A predictable nonce that cannot be controlled by the adversary is safe as a CFB IV (with some assumptions), as shown in the other answers. However, a nonce that can be chosen by an adversary is not safe against chosen plaintext attacks, as shown in Evaluation of Some Blockcipher Modes of Operation (page 36): Assume s = n. The adversary asks its oracle to ...

1

Cipher Feedback mode turns the block cipher (AES) into a self-synchronizing stream cipher which feeds back the full ciphertext block as the next IV. If you encrypt something smaller than a multiple of the block size, it will not use all of the block cipher output to create the ciphertext, just the amount it needs. Therefore there is not a padding ...

1

Initialization Key hash function hashstring Configuration: E is Rijndael with a block size and key size of 256 bits Input: keystring Output: hash The keystring is padded up with bytes valued zero and split up in blocks of 256 bits $S_0$ to $S_n$; The key blocks $K_0$ to $K_n$ are generated, where $K_0$ consist of $S_0$. The following blocks - if any, ...

1

It just leaks information about the blocks with the same IV. Specifically, if the two messages encrypted with the same IV started with blocks $M_0$ and $M'_0$, then the attacker learns the value $M_0 \oplus M'_0$ (and if those where the same, then the attacker also learns the corresponding xor with the second block, etc) However, the attacker learns ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Which linearisation technique is correct?

I have the objective function (Maximally Diverse Grouping Problem) as

$$\max\sum_{g=1}^G\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}d_{ij}x_{ig}x_{jg}$$

Here, $$d_{ij}$$ are known parameters, and $$x_{ig}$$ and $$x_{jg}$$ are binary variables.

It is nonlinear as we have the bilinear term, $$x_{ig}x_{jg}$$.

### Linearization Technique 1

We can linearise it by introducing an binary variable $$z_{ijg}=x_{ig}x_{jg}$$

\begin{align}z_{ijg} &\le x_{ig}\\z_{ijg} &\le x_{jg}\\z_{ijg}&\ge x_{ig}+x_{jg}-1\end{align}

Now, the new objection function becomes

$$\max\sum_{g=1}^G\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}d_{ij}z_{ijg}$$

### Linearization Technique 2

I do not introduce any extra variables. Instead I rewrite the objective function as

$$\max\sum_{g=1}^G\sum_{i=1}^{N-1}\sum_{j=i+1}^{N}d_{ij}(x_{ig}+x_{jg}-1)$$

They are equivalent except when $$x_{i,g}=x_{j,g}=0$$, in which case the second linearization incorrectly contributes $$-d_{ij}$$ to the objective.
Assuming $$d_{ij} \ge 0$$, I recommend a third linearization (relaxing $$z$$ and omitting two constraints from linearization 1): \begin{align} z_{ijg}&\ge x_{ig}+x_{jg}-1 \\ z_{ijg}&\ge 0 \end{align}
• Yes, if $d>0$ and $x$ is declared binary, $z$ will automatically be binary-valued even if you relax to $z \ge 0$. Relaxing $z$ might or might not speed things up, but that is the motivation. A best practice is to try it both ways on realistic instances. May 11 at 12:32
• I see now that you have changed the objective sense from minimization to maximization, and that invalidates my answer. For maximization, you should instead keep the other two ($\le$) constraints. May 26 at 12:40


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# Native Language Support¶

## Overview¶

Native Language Support (NLS) allows the application to facilitate internationalization. It provides support to manipulate messages and translate them in different languages. Each message to be internationalized is referenced by a key, which can be used in the application code instead of using the message directly.

Messages must be defined in PO files, located in the Classpath of the application (for example in the src/main/resources folder). Here is an example:

msgid ""
msgstr ""
"Language: en_US\n"
"Language-Team: English\n"
"MIME-Version: 1.0\n"
"Content-Type: text/plain; charset=UTF-8\n"

msgid "Label1"
msgstr "My label 1"

msgid "Label2"
msgstr "My label 2"

PO files are declared in Classpath *.nls.list files (and to *.externresources.list for an external resource, see Application Resources).

The file format is a standard Java properties file, each line represents the Full Qualified Name of a Java interface that will be generated and used in the application. Example:

com.mycompany.myapp.Labels
com.mycompany.myapp.Messages

The message can be accessed with a call to ej.nls.NLS.getMessage(). Example:

import com.mycompany.myapp.Labels;

public class MyClass {

String label = Labels.NLS.getMessage(Labels.Label1);

...

## Usage¶

For each line, PO files whose name starts with the interface name (Messages and Labels in the example) are retrieved from the Classpath and used to generate:

• a Java interface with the given FQN, containing a field for each msgid of the PO files
• a NLS binary file containing the translations

So, in the example, the generated interface com.mycompany.myapp.Labels will gather all the translations from files named Labels*.po and located in the Classpath. PO files are generally suffixed by their locale (Labels_en_US.po) but it is only for convenience since the suffix is not used, the locale is extracted from the PO file’s metadata.

Once the generation is done, the application can use the Java interfaces to get internationalized messages, for example:

import com.mycompany.myapp.Labels;

public class MyClass {

String label = Labels.NLS.getMessage(Labels.Label1);

...

The generation is triggered when building the application or after a change done in any PO or *.nls.list files. This allows to always have the Java interfaces up-to-date with the translations and to use them immediately.

The NLS-PO module must be added to the module.ivy of the Application project.

<dependency org="com.microej.library.runtime" name="nls-po" rev="2.2.0"/>

The NLS External Loader allows to update the PO files of an application executed on a Virtual Device without rebuilding it. PO files can be dropped in a given location in the Virtual Device folders to dynamically replace the language strings packaged in the application.

This is typically useful when testing or translating an application in order to have a quick feedback when changing the PO files. Once the PO files are updated, a simple restart of the Virtual Device allows to immediately see the result.

### Installation¶

To enable the NLS External Loader in the Virtual Device, add the following dependency to the module.ivy file of the Firmware project:

<dependency org="com.microej.tool" name="nls-po-external-loader" rev="2.3.0" transitive="false"/>

Then rebuild the Firmware project to produce the Virtual Device.

### Usage¶

Once the project built:

• unzip the Virtual Device and create a folder named translations in the root folder.

• copy all the PO files from the project into the translations folder. All PO files found in this folder are processed, no matter their folder level.

• start the Virtual Device with the launcher. The following logs should be printed if the NLS External Loader has been executed and has found the PO files:

externalPoLoaderInit:init:

[mkdir] Created dir: <PATH>\tmp\microejlaunch1307817858\resourcebuffer
[po-to-nls] *.nls files found in <PATH>\output\<FIRMWARE>\resourceBuffer :
[po-to-nls]   - com.mycompany.Messages1
[po-to-nls]   - com.mycompany.Messages2
[po-to-nls]   => loaded locales : fr_FR,de_DE,ja_JP,en_US
[po-to-nls]   => loaded locales : fr_FR,de_DE,ja_JP,en_US

• update the languages strings in the PO files of the Virtual Device (the files in the translations/ folder).

• restart the Virtual Device and check the changes.

It is important to know the following rules about the NLS External Loader:

• the external PO files names must match with the default PO files names of the application to be processed.
• when PO files with a given name are loaded, the default translations for these PO files are replaced, there is no merge. It means that:
• if messages are missing in the new PO files, they are not available anymore for the application and may very probably make it crash.
• if languages are missing (the application has 3 PO files for English, French and Spanish, and only PO files for English and French are available in the translations folder), the messages of the missing languages are not available anymore for the application and may very probably make it crash.
• if new messages are added in the PO files, it has no impact, they are ignored by the application.
• External PO files are loaded at Virtual Device startup, so any change requires a restart of the Virtual Device to be considered

### Troubleshooting¶

#### java.io.IOException: NLS-PO:S=4¶

The following error occurs when at least 1 PO file is missing for a language:

[parallel2] NLS-PO:I=6
[parallel2] Exception in thread "main" java.io.IOException: NLS-PO:S=4 323463627 -1948548092
[parallel2]     at java.lang.Throwable.fillInStackTrace(Throwable.java:79)
[parallel2]     at java.lang.Throwable.<init>(Throwable.java:30)
[parallel2]     at java.lang.Exception.<init>(Exception.java:10)
[parallel2]     at java.io.IOException.<init>(IOException.java:16)
[parallel2]     at com.microej.nls.BinaryNLS.<init>(BinaryNLS.java:157)
[parallel2]     at com.microej.nls.BinaryNLS.newBinaryNLS(BinaryNLS.java:118)

Make sure that all PO files are copied in the translations folder.

### Crowdin¶

Crowdin is a cloud-based localization platform which allows to manage multilingual content. The NLS External Loader can fetch translations directly from Crowdin to make the translation process even easier. Translators can then contribute and validate their translations in Crowdin and apply them automatically in the Virtual Device.

A new dependency must be added to the module.ivy file of the Firmware project to enable this integration:

<dependency org="com.microej.tool" name="nls-po-crowdin" rev="1.0.0" transitive="false"/>

Once the module has been built, edit the file platform/tools/crowdin/crowdin.properties to configure the Crowdin connection:

• set crowdin.token to the Crowdin API token. A token can be generated in the Crowdin in Settings > API > click on New Token.
• set crowdin.projectsIds to the id of the Crowdin project. The project id can be found in the Details section on a project page. Multiple projects can be set by separating their id with a comma (for example crowdin.projectsIds=12,586,874).

When the configuration is done, the fetch of the Crowdin translations can be done by executing the script crowdin.bat or crowdin.sh located in the folder platform/tools/crowdin/. The PO files retrieved from Crowdin are automatically pasted in the folder translations, therefore the new translations are applied after the next Virtual Device restart.


text:
# Reversal of limits of integration in the derivation of probability current density

by hnicholls
Tags: current, density, derivation, integration, limits, probability, reversal
 P: 25 In working out the derivation of the probability current density, I see (based on the definition of j(x,t)) that the limits of integration are changed from d/dt∫(b to a) P(x.t) dx = iħ/2m[ψ*(x.t)∂/∂xψ(x.t) - ψ(x.t)∂/∂xψ*(x.t)](b to a) to d/dt∫(b to a) P(x.t) dx = -iħ/2m[ψ*(x.t)∂/∂xψ(x.t) - ψ(x.t)∂/∂xψ*(x.t)](a to b) Thus, the prefactor becomes -iħ/2m as a result of reversing the limits of integration. Is there a reason that the prefactor must be in terms of -i? Thanks
 Mentor P: 10,853 The changed limits just changed the sign from + to -, the ##\frac{i\hbar}{2m}##-part is quantum mechanics.
 P: 25 I understand that the sign change is a result of reversing the limits and that the iħ/2m part is what quantizes the result, my question is why does the result need to me negative, rather than positive. Thanks again.
P: 123

## Reversal of limits of integration in the derivation of probability current density

Let me give you an alternative way to derive the current's expression, which maybe has more sense. What we are looking for is a function that satisfies the “continuity equation”:

$${{\partial }_{t}}\rho +{{\partial }_{x}}j=0$$

which comes from the requirement of local conservation of probability. In the above equation $\rho ={{\Psi }^{*}}\Psi$ is the probability density, so when you calculate ${\partial \rho }/{\partial t}\;$ using S.E. and its complex conjugate, you find:

$${{\partial }_{t}}\rho =-\frac{i\hbar }{2m}{{\partial }_{x}}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$

So, when you compare this with the continuity equation, you have to set:

$$j=\frac{i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$
 Mentor P: 10,853 A minus sign does not mean that the result is negative, the integral itself can be negative as well. I don't know the context of that equation, but I think it is just a complex phase anyway. In other words, the sign (together with i) has no physical significance on its own.
P: 25
 Quote by cosmic dust Let me give you an alternative way to derive the current's expression, which maybe has more sense. What we are looking for is a function that satisfies the “continuity equation”: $${{\partial }_{t}}\rho +{{\partial }_{x}}j=0$$ which comes from the requirement of local conservation of probability. In the above equation $\rho ={{\Psi }^{*}}\Psi$ is the probability density, so when you calculate ${\partial \rho }/{\partial t}\;$ using S.E. and its complex conjugate, you find: $${{\partial }_{t}}\rho =-\frac{i\hbar }{2m}{{\partial }_{x}}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$ So, when you compare this with the continuity equation, you have to set: $$j=\frac{i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$
But isn't j (x,t) defined as,

$$j=\frac{-i\hbar }{2m}\left( \Psi {{\partial }_{x}}{{\Psi }^{*}}-{{\Psi }^{*}}{{\partial }_{x}}\Psi \right)$$
 Sci Advisor Thanks P: 2,152 The sign is uniquely defined by the Schrödinger equation, which reads (setting $\hbar=1$) $$\mathrm{i} \partial_t \psi(t,x)=-\frac{\Delta}{2m} \psi(t,x)+ V(x) \psi(t,x).$$ Multiplying with $\psi^*$ leads to $$\psi^*(t,x) \mathrm{i} \partial_t \psi(t,x)=\psi^*(t,x) \left [-\frac{\Delta}{2m} \psi(t,x)+ V(x) \psi(t,x) \right].$$ Then subtracting the conjugate complex of this equation and multiplying with (-i) leads to $$\partial_t |\psi(t,x)|^2=\vec{\nabla} \cdot \frac{\mathrm{i}}{2m} [\psi^*(t,x) \vec{\nabla} \psi(t,x)-\psi(t,x) \vec{\nabla} \psi^*(t,x)].$$ Comparing this with the continuity equation leads to $$\rho(t,x)=|\psi(t,x)|^2, \quad \vec{j}=-\frac{\mathrm{i}}{2m} [\psi^*(t,x) \vec{\nabla} \psi(t,x)-\psi(t,x) \vec{\nabla} \psi^*(t,x)],$$ as already given by cosmic dust.
 P: 25 So, proceeding with the assumption that ψ(t,x) satisfies the TDSE, (−ℏ2/2m)∂x2ψ(t,x) = (iℏ)∂tψ(t,x) with V(x)ψ(t,x) = 0 Dividing by iℏ (−ℏ/2mi)∂x2ψ(t,x) = ∂tψ(t,x) Multiply by -i (−iℏ/2m)∂x2ψ(t,x) = ∂tψ(t,x) Multiplying by ψ(t,x)∗ (−iℏ/2m)∂x2ψ(t,x)∗ψ(t,x) = ∂tψ(t,x)ψ(t,x)∗ P(t,x) = ψ(t,x)ψ(t,x)∗ So, (−iℏ/2m)∂x2ψ(t,x)∗ψ(t,x) = ∂tP(t,x) and the left side of this equation can be rewritten as (−iℏ/2m)∂x[ψ(t,x)∗∂xψ(t,x) - ψ(t,x)∂xψ(t,x)∗] but, this is ∂xj(t,x) where j(t,x) is defined as (−iℏ/2m)[ψ(t,x)∗∂xψ(t,x) - ψ(t,x)∂xψ(t,x)∗] and ∂xj(t,x) = ∂tP(t,x) because the "continuity equation" requires that ∂xj(t,x) - ∂tP(t,x) = 0 So, the reversal of the limits of integration is necessary so that the TDSE and the "continuity equation" are both satisfied. That seems right.

 Related Discussions Quantum Physics 8 Quantum Physics 9 Calculus & Beyond Homework 0 Advanced Physics Homework 4 Calculus & Beyond Homework 0


text:
# 11

(This is a (possibly perpetual) draft of some work that we (I) did at the Vancouver meetup. Thanks to my meetup buddies for letting me use their brains as supplementary computational substrate. Sorry about how ugly the LaTeX is; is there a way to make this all look a bit nicer?)

(Large swaths of this are obsolete. Thanks for the input, LW!)

## The Problem of Decision Under Preference Uncertainty

Suppose you are uncertain whether it is good to eat meat or not. It could be OK, or it could be very bad, but having not done the thinking, you are uncertain. And yet you have to decide what to eat now; is it going to be the tasty hamburger or the morally pure vegetarian salad?

You have multiple theories about your preferences that contradict in their assessment, and you want to make the best decision. How would you decide, even in principle, when you have such uncertainty? This is the problem of Preference Uncertainty.

Preference Uncertainty is a daily fact of life for humans; we simply don't have introspective access to our raw preferences in many cases, but we still want to make the best decisions we can. Just going with our intuitions about what seems most awesome is usually sufficient, but on higher-stakes decisions and theoretical reasoning, we want formal methods with more transparent reasoning processes. We especially like transparent formal methods if we want to create a Friendly AI.

There is unfortunately very little formal analysis of the preference uncertainty problem, and what has been done is incomplete and more philosophical than formal. Nonetheless, there has been some good work in the last few years. I'll refer you to Crouch's thesis if you're interested in that.

## Using VNM

I'm going to assume VNM. That is, that rational preferences imply a utility function, and we decide between lotteries, choosing the one with highest expected utility.

The implications here are that the possible moral theories ($m \in M$) each have an associated utility function ($U_m$) that represents their preferences. Also by VNM, our solution to preference uncertainty is a utility function $U'$.

We are uncertain between moral theories, so we have a probability distribution over moral theories $\sum\nolimits_{m \in M}P(m) = 1$.

To make decisions, we need a way to compute the expected value of some lottery $l$. Each lottery is essentially a probability distribution over the set of possible outcomes $\sum\nolimits_{o \in O}P_l(o) = 1$.

Since we have uncertainty over multiple things ($m,o$), the domain of the final preference structure is both moral theories and outcomes: $U' : (M \times O) \rightarrow \mathbb{R}$.

Now for some conditions. In the degenerate case of full confidence in one moral theory $m$, the overall preferences should agree with that theory:

$P(m) = 1 \Rightarrow EU(o) = U'(m,o) = k_m*U_m(o) + c_m$

For some $k_m > 0 \in \mathbb{R}$ and $c_m \in \mathbb{R}$ representing the degrees of freedom in utility function equality. That condition actually already contains most of the specification of $U'$.

$U'(m,o) = k_m *U_m(o) + c_m$.

So we have a utility function, except for those unknown scaling and offset constants, which undo the arbitrariness in the basis and scale used to define each individual utility function.

Thus overall expectation looks like this:

$EU(l) = \sum\limits_{o \in O}\sum\limits_{m \in M}P_l(o)*P(m)*(k_m *U_m(o) + c_m)$.

This is still incomplete, though. If we want to get actual decisions, we need to pin down each $k_m$ and $c_m$.

## Offsets and Scales

You'll see above that the probability distribution over $M$ is not dependent on the particular lottery, while $P_l(o)$ is a function of lottery. This is because I assumed that actions can't change what is right.

With this assumption, the contribution of the $c_m$'s can be entirely factored out:

$EU(l) = (\sum\limits_{m \in M}P(m)*c_m) + (\sum\limits_{o \in O}\sum\limits_{m \in M}P_l(o)*P(m)*k_m *U_m(o))$.

This makes it obvious that the effect of the $c_m$'s is an additive constant that affects all lotteries the same way and thus never affects preferences. Thus we can set them to any value that is convenient; for this article, all $c_m = 0$.

A similar process allows us to arbitrarily set exactly one of the $k_m$.

The remaining values of $k_m$ actually affect decisions, so setting them arbitrarily has real consequences. To illustrate, consider the opening example of choosing lunch between a $Burger$ and $Salad$ when unsure about the moral status of meat.

Making up some details, we might have $U_{veg}(Burger) < U_{veg}(Salad)$ and $U_{meat}(Burger) > U_{meat}(Salad)$ and $P(meat) = 0.7$. Importing this into the framework described thus far, we might have the following payoff table:

Meat OK (meat) 1 0 (0.7)
Meat Bad (veg) 0 k_veg (0.3)
(expectation) 0.7 0.3*k_veg (1)

We can see that with those probabilities, the expected value of $Salad$ exceeds that of $Burger$ when $k_{veg} > 7/3$ (when $0.3*k_{veg} > 0.7$), so the decision hinges on the value of that parameter.

The value of $k_m$ can be interpreted as the "intertheoretic weight" of a utility function candidate for the purposes of intertheoretic value comparisons.

In general, if $|M| = n$ then you have exactly $n-1$ missing intertheoretic weights that determine how you respond to situations with preference uncertainty. These could be pinned down if you had $n-1$ independent equations representing indifference scenarios.

For example, if we had $EU(Burger) = EU(Salad)$ when $P(meat) = 0.8$, then we would have $k_{veg} = 4$, and the above decision would be determined in favor of the $Salad$.

## Expressing Arbitrary Preferences

Preferences are arbitrary, in the sense that we should be able to want whatever we want to want, so our mathematical constructions should not dictate or limit our preferences. If they do, we should just decide to disagree.

What that means here is that because the values of $k_m$ drive important preferences (like at what probability you feel it is safe to eat meat), the math must leave them unconstrained, to be selected by whatever moral reasoning process it is that selected the candidate utility functions and gave them probabilities in the first place.

We could ignore this idea and attempt to use a "normalization" scheme to pin down the intertheoretic weights from the object level preferences without having to use additional moral reasoning. For example, we could dictate that the "variance" of each candidate utility function equals 1 (with some measure assignment over outcomes), which would divide out the arbitrary scales used to define the candidate utility functions, preventing dominance by arbitrary factors that shouldn't matter.

Consider that any given assignment of intertheoretic weights is equivalent to some set of indifference scenarios (like the one we used above for vegetarianism). For example, the above normalization scheme gives us the indifference scenario $EU(Burger) = EU(Salad)$ when $P(meat) = 0.5$.

If I find that I am actually indifferent at $P(meat) = 0.8$ like above, then I'm out of luck, unable to express this very reasonable preference. On the other hand, I can simply reject the normalization scheme and keep my preferences intact, which I much prefer.

(Notice that the normalization scheme was an unjustifiably privileged hypothesis from the beginning; we didn't argue that it was necessary, we simply pulled it out of thin air for no reason, so its failure was predictable.)

Thus I reassert that the $k_m$'s are free parameters to be set accordance with our actual intertheoretic preferences, on pain of stupidity. Consider an analogy to the move from ordinal to cardinal utilities; when you add risk, you need more degrees of freedom in your preferences to express how you might respond to that risk, and you need to actually think about what you want those values to be.

## Uncertainty Over Intertheoretic Weights

(This section is less solid than the others. Watch your step.)

A weakness in the constructions described so far is that they assume that we have access to perfect knowledge of intertheoretic preferences, even though the whole problem is that we are unable to find perfect knowledge of our preferences.

It seems intuitively that we could have a probability distribution over each $k_m$. If we do this, making decisions is not much complicated, I think; a simple expectation should still work.

If expectation is the way, the expectation over $k_m$ can be factored out (by linearity or something). Thus in any given decision with fixed preference uncertainties, we can pretend to have perfect knowledge of $k_m$.

Despite the seeming triviality of the above idea for dealing with uncertainty over $k_m$, I haven't formalized it much. We'll see if I figure it out soon, but for now, it would be foolish to make too many assumptions about this. Thus the rest of this article still assumes perfect knowledge of $k_m$, on the expectation that we can extend it later.

## Learning Values, Among Other Things

Strictly speaking, inference across the is-ought gap is not valid, but we do it every time we act on our moral intuitions, which are just physical facts about our minds. Strictly speaking, inferring future events from past observations (induction) is not valid either, but it doesn't bother us much:

We deal with induction by defining an arbitrary (but good-seeming, on reflection) prior joint probability distribution over observations and events. We can handle the is-ought gap the same way: instead of separate probability distributions over events $O$ and moral facts $M$, we define a joint prior over $M \times O$. Then learning value is just Bayesian updates on partial observations of $O$. Note that this prior subsumes induction.

Making decisions is still just maximizing expected utility with our constructions from above, though we will have to be careful to make sure that $P(m)$ remains independent of the particular lottery.

The problem of how to define such a prior is beyond the scope of this article. I will note that this "moral prior" idea is the solid foundation on which to base Indirect Normativity schemes like Yudkowsky's CEV and Christiano's boxed philosopher. I will hopefully discuss this further in the future.

## Recap

The problem was how to make decisions when you are uncertain about what your object-level preferences should be. To solve it, I assumed VNM, in particular that we have a set of possible utility functions, and we want to construct an overall utility function that does the right thing by those utility functions and their probabilities. The simple condition that the overall utility function should make the common sense choices in cases of moral certainty was sufficient to construct a utility function with a precise set of remaining degrees of freedom. The degrees of freedom being the intertheoretic weight and offset of each utility function candidate.

I showed that the offsets and an overall scale factor are superfluous, in the sense that they never affect the decision if we assume that actions don't affect what is desirable. The remaining intertheoretic weights do affect the decision, and I argued that they are critical to expressing whatever intertheoretic preferences we might want to have.

Uncertainty over intertheoretic weight seems tractable, but the details are still open.

I also mentioned that we can construct a joint distribution that allows us to embed value learning in normal Bayesian learning and induction. This "moral prior" would subsume induction and define how facts about the desirability of things could be inferred from physical observations like the opinions of moral philosophers. In particular, it would provide a solid foundation for Indirect Normativity schemes like CEV. The nature of this distribution is still open.

## Open Questions

What are the details of how to deal with uncertainty over the intertheoretic weights? I am looking in particular for construction from an explicit set of reasonable assumptions like the above work, rather than simply pulling a method out of thin air unsupported.

What are the details of the Moral Prior? What is its nature? What implications does it have? What assumptions do we have to make to make it behave reasonably? How do we construct one that could be safely given to a superintelligence. This is going to be a lot of work.

I assumed that it is meaningful to assign probabilities over moral theories. Probability is closely tied up with utility, and probability over epiphenomena like preferences is especially difficult. It remains to be seen how much the framing here actually helps us, or if it effectively just disguises pulling a utility function out of a hat.

Is this at all correct? I should build it and see if it type-checks and does what it's supposed to do.


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# Upslope area – Recursive calculation procedure

So far in this series I've talked about calculating pixel flow directions and handling plateaus, but I haven't yet discussed the actual upslope area calculation. The Tarboton paper presents a recursive formulation of the upslope area calculation (from page 313):

  Procedure DPAREA(i, j)
if AREA(i, j) is known
then
no action
else
AREA(i, j) = 1 (the area of a single pixel)
for each neighbor (location in, jn)
p = proportion of neighbor (in, jn) that drains to
pixel (i, j) based on angle
if (p > 0) then
call DPAREA(in, jn) (this is the recursive call to
calculate the area for the neighbor)
AREA(i, j) = AREA(i, j) + p x AREA(in, jn)
Return

I had to read the surrounding text a couple of times to figure out exactly how the author intends to calculate p. Consider the flow direction for a particular neighbor, (in, jn). If the flow direction points directly at pixel (i, j), then the corresponding weight p is 1. In other words, pixel (i, j) gets all of the flow from (in, jn). If the flow direction is pi/4 or greater away from the direction toward (i, j), then the weight p is 0. Pixel (i, j) gets none of the flow from (in, jn). In between an angular difference of 0 and pi/4, the weight p varies proportionally between 1 and 0.

For example, consider the computation of upslope area for pixel number 5 in this set of 9 pixels:

   1 4 7
2 5 8
3 6 9

If the flow direction for pixel 2 is zero radians, its flow is pointing directly at pixel 5, so the corresponding weight is 1.0. If the flow direction for pixel 8 is pi/4 radians, then its flow is pointing directly at pixel 4. The corresponding weight for pixel 5 is 0.0. If the flow direction is pi/8 radians, its flow is pointing between pixels 5 and 4, and the weight for the pixel 5 computation is 0.5.

It also took me a few minutes to figure out how to compute the radial difference between two angles, properly taking into account the 2*pi periodicity of angles. (For example, the radial difference between pi/8 and 2*pi - pi/8 should be pi/4.) Here's what I came up with:

   radial_difference = abs(mod(theta1 - theta2 + pi), 2*pi) - pi)

(Does anyone else have a better way to compute this quantity?) To calculate the weight for a given neighbor of pixel (i,j), first determine the angle that points directly from that neighbor to the pixel (i, j). Call that angle theta_c. Then find the flow direction for that neighbor; call it theta_f. Now compute the radial difference:

   radial_difference = abs(mod(theta_c - theta_f + pi), 2*pi) - pi)

And finally compute the weight:

   p = max(1 - (radial_difference * 4 / pi), 0)

So, are we ready to start writing a recursive MATLAB function based on DPAREA above? Nope! I don't actually want to use a recursive formulation at all.

Here's a different way to think about it:

  AREA(i,j) = 1 + p1*AREA(i1,j1) + p2*AREA(i2,j2) + ... + p8*AREA(i8,j8)

If we write down this equation for each pixel in the image, we'll end up with a system of N linear equations in N unknowns. Although the system is very large (N-by-N, where N is the number of pixels), it is also very sparse, because each equation involves no more than nine of the unknowns.

So we are getting very close to calculating upslope area. We "just" have to set up an extremely large sparse system of equations and then solve it.

Next time I'll tackle that.

Published with MATLAB® 7.4

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text:
zbMATH — the first resource for mathematics

Local properties of accessible injective operator ideals. (English) Zbl 0926.46054
In contrast to Pisier’s example of a non-accessible maximal Banach ideal, the author exhibits a large class of maximal Banach ideals which are accessible. The construction is based on a characterization of total accessibility of injective hulls of maximal Banach ideals (if $$(\mathcal A,\mathbf A)$$ is a maximal Banach ideal, then $$(\mathcal A^{inj},\mathbf A^{inj})$$ is totally accessible if and only if $$(\mathcal A\circ \mathcal A^*,\mathbf A\circ \mathbf A^\ast)\subseteq (\mathcal P_1,\mathbf P_1)$$, the ideal of absolutely summing operators) and an operator version of Grothendieck’s inequality. As a consequence, it is obtained, for instance, that if $$(\mathcal A,\mathbf A)$$ is a maximal Banach ideal such that $$(\mathcal D_2,\mathbf D_2)\subseteq (\mathcal A,\mathbf A)\subseteq (\mathcal L_1,\mathbf L_1)$$ and both $$(\mathcal A,\mathbf A)$$ and $$(\mathcal A^*, \mathbf A^\ast)$$ are metrically $$\varepsilon$$-tensorstable, then $$(\mathcal A^{inj}, \mathbf A^{inj})$$ is totally accessible. Several other criteria for right- and total accessibility are established, and applications to normed products of operator ideals are also given.

MSC:
 46M05 Tensor products in functional analysis 47L20 Operator ideals 47A80 Tensor products of linear operators
Full Text:
References:
 [1] B. Carl, A. Defant, and M. S. Ramanujan: On tensor stable operator ideals. Michigan Math. J. 36 (1989), 63-75. · Zbl 0669.47025 [2] A. Defant: Produkte von Tensornormen. Habilitationsschrift. Oldenburg 1986. [3] A. Defant and K. Floret: Tensor Norms and Operator Ideals. North-Holland Amsterdam, London, New York, Tokio, 1993. · Zbl 0774.46018 [4] J. E. Gilbert and T. Leih: Factorization, tensor products and bilinear forms in Banach space theory. Notes in Banach spaces, Univ. of Texas Press, Austin, 1980, pp. 182-305. · Zbl 0471.46053 [5] Y. Gordon, D. R. Lewis, and J. R Retherford: Banach ideals of operators with applications. J. Funct. Analysis 14 (1973), 85-129. · Zbl 0272.47024 [6] A. Grothendieck: Résumé de la théorie métrique des produits tensoriels topologiques. Bol. Soc. Mat. São Paulo 8 (1956), 1-79. · Zbl 0074.32303 [7] H. Jarchow: Locally convex spaces. Teubner, 1981. · Zbl 0466.46001 [8] H. Jarchow and R. Ott: On trace ideals. Math. Nachr. 108 (1982), 23-37. · Zbl 0523.47030 [9] H. P. Lotz: Grothendieck ideals of operators in Banach spaces. Lecture notes, Univ. Illinois, Urbana, 1973. [10] J. Lindenstrauss and H. P. Rosenthal: The Lp-spaces. Israel J. Math. 7 (1969), 325-349. · Zbl 0205.12602 [11] F. Oertel: Konjugierte Operatorenideale und das $${\mathcal A}$$-lokale Reflexivitätsprinzip. Dissertation. Kaiserslautern, 1990. [12] F. Oertel: Operator ideals and the principle of local reflexivity. Acta Universitatis Carolinae-Mathematica et Physica 33 (1992), no. 2, 115-120. · Zbl 0803.47038 [13] A. Pietsch: Operator Ideals. North-Holland Amsterdam, London, New York, Tokio, 1980. · Zbl 0455.47032 [14] A. Pietsch: Eigenvalues and s-numbers. Cambridge Studies in Advanced Mathematics 13 (1987). · Zbl 0615.47019
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Probability

# Discrete Mathematics Warmups: Level 2 Challenges

If two fair coins are tossed simultaneously, what is the probability that exactly one head appears?

In a business meeting, a total of 45 handshakes occurred. How many people were at the meeting?

Assume that each person shakes hands with each other person exactly once.

How many triangles are there in the above image?

A fair coin is tossed until it lands heads up. What is the probability that heads first appears on an odd numbered toss?

If someone painted the outside of a $5 \times 5 \times 5$ cube made out of $1 \times 1 \times 1$ unit cubes, how many of the unit cubes would have paint on exactly 3 sides?

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