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# American Institute of Mathematical Sciences

January  2015, 2(1): 103-115. doi: 10.3934/jdg.2015.2.103

## Reversibility and oscillations in zero-sum discounted stochastic games

 1 Sorbonne Universités, UPMC Univ Paris 06, Institut de Mathématiques de Jussieu-Paris Rive Gauche, UMR 7586, CNRS, Univ Paris Diderot, Sorbonne Paris Cité, F-75005, Paris, France 2 Université Paris-Dauphine, CEREMADE, Place du Maréchal De Lattre de Tassigny, 75775 Paris cedex 16, France

Received  November 2014 Revised  January 2015 Published  June 2015

We show that by coupling two well-behaved exit-time problems one can construct two-person zero-sum dynamic games having oscillating discounted values. This unifies and generalizes recent examples of stochastic games with finite state space, due to Vigeral (2013) and Ziliotto (2013).
Citation: Sylvain Sorin, Guillaume Vigeral. Reversibility and oscillations in zero-sum discounted stochastic games. Journal of Dynamics & Games, 2015, 2 (1) : 103-115. doi: 10.3934/jdg.2015.2.103
##### References:
 [1] T. Bewley and E. Kohlberg, On stochastic games with stationary optimal strategies,, Mathematics of Operations Research, 3 (1978), 104.  doi: 10.1287/moor.3.2.104.  Google Scholar [2] J. Bolte, S. Gaubert and G. Vigeral, Definable zero-sum stochastic games,, Mathematics of Operation Research, 40 (2015), 171.  doi: 10.1287/moor.2014.0666.  Google Scholar [3] G. Grimmett and D. Stirzaker, Probability and Random Processes,, Oxford University Press, (2001).   Google Scholar [4] R. Laraki, Explicit formulas for repeated games with absorbing states,, International Journal of Game Theory, 39 (2010), 53.  doi: 10.1007/s00182-009-0193-2.  Google Scholar [5] J.-F. Mertens, A. Neyman and D. Rosenberg, Absorbing games with compact action spaces,, Mathematics of Operation Research, 34 (2009), 257.  doi: 10.1287/moor.1080.0372.  Google Scholar [6] J.-F. Mertens, S. Sorin and S. Zamir, Repeated Games,, Cambridge University Press, (2015).   Google Scholar [7] A. Neyman, Stochastic games and nonexpansive maps,, in Stochastic Games and Applications (eds. A. Neyman and S. Sorin), (2003), 397.  doi: 10.1007/978-94-010-0189-2_26.  Google Scholar [8] D. Rosenberg and S. Sorin, An operator approach to zero-sum repeated games,, Israel Journal of Mathematics, 121 (2001), 221.  doi: 10.1007/BF02802505.  Google Scholar [9] S. Sorin, A First Course on Zero-SumRepeated Games,, Springer-Verlag, (2002).   Google Scholar [10] S. Sorin, The operator approach to zero-sum stochastic games,, in Stochastic Games and Applications (eds. A. Neyman and S. Sorin), (2003), 417.   Google Scholar [11] S. Sorin and G. Vigeral, Existence of the limit value of two person zero-sum discounted repeated games via comparison theorems,, Journal of Opimization Theory and Applications, 157 (2013), 564.  doi: 10.1007/s10957-012-0193-4.  Google Scholar [12] G. Vigeral, A zero-sum stochastic game with compact action sets and no asymptotic value,, Dynamic Games and Applications, 3 (2013), 172.  doi: 10.1007/s13235-013-0073-z.  Google Scholar [13] B. Ziliotto, Zero-sum repeated games: Counterexamples to the existence of the asymptotic value and the conjecture maxmin=$\lim v_n$,, to appear in Annals of Probability, (2013).   Google Scholar

show all references

##### References:
 [1] T. Bewley and E. Kohlberg, On stochastic games with stationary optimal strategies,, Mathematics of Operations Research, 3 (1978), 104.  doi: 10.1287/moor.3.2.104.  Google Scholar [2] J. Bolte, S. Gaubert and G. Vigeral, Definable zero-sum stochastic games,, Mathematics of Operation Research, 40 (2015), 171.  doi: 10.1287/moor.2014.0666.  Google Scholar [3] G. Grimmett and D. Stirzaker, Probability and Random Processes,, Oxford University Press, (2001).   Google Scholar [4] R. Laraki, Explicit formulas for repeated games with absorbing states,, International Journal of Game Theory, 39 (2010), 53.  doi: 10.1007/s00182-009-0193-2.  Google Scholar [5] J.-F. Mertens, A. Neyman and D. Rosenberg, Absorbing games with compact action spaces,, Mathematics of Operation Research, 34 (2009), 257.  doi: 10.1287/moor.1080.0372.  Google Scholar [6] J.-F. Mertens, S. Sorin and S. Zamir, Repeated Games,, Cambridge University Press, (2015).   Google Scholar [7] A. Neyman, Stochastic games and nonexpansive maps,, in Stochastic Games and Applications (eds. A. Neyman and S. Sorin), (2003), 397.  doi: 10.1007/978-94-010-0189-2_26.  Google Scholar [8] D. Rosenberg and S. Sorin, An operator approach to zero-sum repeated games,, Israel Journal of Mathematics, 121 (2001), 221.  doi: 10.1007/BF02802505.  Google Scholar [9] S. Sorin, A First Course on Zero-SumRepeated Games,, Springer-Verlag, (2002).   Google Scholar [10] S. Sorin, The operator approach to zero-sum stochastic games,, in Stochastic Games and Applications (eds. A. Neyman and S. Sorin), (2003), 417.   Google Scholar [11] S. Sorin and G. Vigeral, Existence of the limit value of two person zero-sum discounted repeated games via comparison theorems,, Journal of Opimization Theory and Applications, 157 (2013), 564.  doi: 10.1007/s10957-012-0193-4.  Google Scholar [12] G. Vigeral, A zero-sum stochastic game with compact action sets and no asymptotic value,, Dynamic Games and Applications, 3 (2013), 172.  doi: 10.1007/s13235-013-0073-z.  Google Scholar [13] B. Ziliotto, Zero-sum repeated games: Counterexamples to the existence of the asymptotic value and the conjecture maxmin=$\lim v_n$,, to appear in Annals of Probability, (2013).   Google Scholar
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# Solving a radical equation.

Begin with an equation you know to be true such as $$3+5 = 8$$.

Try the following:

1. Square each of the three numbers in the equation. Is the equation still true?
2. Square each side of the equation (square the sum of three and five). Is the equation still true?

Take what you learned about keeping the numerical equation true to solve the equations below.

Solve the following two equations by isolating the radical on one side and squaring both sides. Check to be sure the solutions actually work by substituting the solution into the equation.

i. $$\sqrt{2x+1}-5 = -2$$

ii. $$\sqrt{2x+1}+5 = 2$$

format: plain


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# Shooting method - Matlab ODE

I'm trying to solve these equations of hypersonic adiabatic flow over a flat plate. I did all the simplifications and got these equations for the stagnation point flow. $$\left(Cf''\right)' + f f'' = \left(f'\right)^2-g$$ $$\left(\frac{C}{p_r} g'\right) + fg'=0$$

For a calorically perfect gas, $g=h/he$, and $he = \rm stagnation ~enthalpy$

$p_r$ is a constant and $C=g^{-1/3}$

$f(0)=0$ (over the streamline of the body)

$f'(0)=0$ (this is $u$, the velocity over the body which is null)

$f''(0)=?$ white suggests to try as a first value $0.664/\sqrt{C_w}$

I'm doing this for the adiabatic case so

$g(0)=g_{wall}$

$g'(0)=0$ (no thermic flux)

UPDATE

Control conditions:
$f'(\delta)=1$ and $g(\delta)=1$ I want to integrate from $\delta=0$ to $\delta=6$, in the end, if $f'(\delta)$ is different than $1$ I'll have to change the value assumed for $f''(0)$.

How can I solve these equations, someone suggested me using the shooting method but how could I code this in matlab? I would like just a help on how to implement the equations on matlab, not a final solution.

• Similar kind of question is found in scicomp.stackexchange.com/questions/21498/…. Here you should assume different values of $f''(0)$ to get $g'(0)=0$. Still having problem pls. comment here.
– AGN
Jan 4 '16 at 1:15
• Are there any wrong/missing boundary conditions? You can't apply the shooting method unless you have a boundary condition at a location not at x=0. The whole premise of the shooting method is you treat a boundary value problem like an initial value problem and "shoot" from the one boundary, say the left, using the left conditions as initial conditions, and then compare the results to whatever the right boundary conditions are. If the results don't match up with the right BCs, you modify the unknown left BCs until you get there. Jan 4 '16 at 2:19
• @ArunGovindNeelanA Yes that's exactly what I need todo, different values of f''(0) to get g'(0)=0. But I have 2 equations, since h is unknown. In the other question and materials you suggested there was only one function, I have 2, f and g Jan 4 '16 at 10:36
• Right now I'm slightly busy, I will provide some details later. I think you are clear in procedure and only problem is computation. First split the third order ode into three first order ode then you can use RK methods. Then feed four ode with IC to ode45, for that you shall refer www3.nd.edu/~nancy/Math20750/Demos/3dplots/dim3system.html
– AGN
Jan 5 '16 at 6:36
• I'm guessing $\delta$ is fixed? I'm not sure about specifying both $f'$ and $g$ at $\delta$; looks like too many boundary conditions. Also if you set $f''(0)=0$ then you wont have a shooting parameter to use. Jan 7 '16 at 16:49

To change this into a system of first order ODEs, you can define

$$Y(x;s) := \left( \begin{split} f''\\ f'\\ f\\ g'\\ g \end{split} \right) = \left(\begin{split} Y_1\\ Y_2\\ Y_3\\ Y_4\\ Y_5 \end{split} \right)\,.$$ Your first equation then gives $$CY_1' + C'Y_1 + Y_3Y_5= (Y_2)^2 - Y_5\,,$$ or, with $C=g^{-1/3}$, $C' = -\frac13g^{-4/3}g'$, $$Y_5^{-1/3}Y_1' - \tfrac13Y_5^{-4/3} Y_4Y_1 + Y_3Y_5= (Y_2)^2 - Y_5\,,$$ and the second differentiated is $$\left(\frac{Y_5^{-1/3}}{p_r}+Y_3\right)Y_4' + \left(\frac{-\frac13Y_4Y_5^{-4/3}}{p_r}+Y_2\right)Y_4 = 0\,,$$ with the additional equations $$Y_2' = Y_1, \quad Y_3' = Y_2, \quad Y_5' = Y_4\,.$$

Bringing these together, gives $$Y'(x;s) = \left(\begin{split} Y_5^{1/3}\left(\tfrac13Y_5^{-4/3} Y_4Y_1 - Y_3Y_5 + (Y_2)^2 - Y_5\right)\\ Y_1\\ Y_2\\ - \left(\frac{Y_5^{-1/3}}{p_r}+Y_3\right)^{-1} \left(\frac{-\frac13Y_4Y_5^{-4/3}}{p_r}+Y_2\right)Y_4\\ Y_4 \end{split}\right)$$ and initial conditions $$Y(0;s) := \left( \begin{split} s\\ 0\\ 0\\ 0\\ g_{\text{wall}} \end{split} \right) \,.$$

You can use an ODE solver to integrate this the find $Y(\delta)$ for some $\delta>0$, and change $s$ to match your additional condition $$F(s) := Y_2(\delta;s) - 1 = 0\,,$$ or $$F(s) := Y_5(\delta;s) - 1 = 0\,.$$

This could be by bisection (find 2 values $s_1$ and $s_2$ with $F(s_1)<0$, $F(s_2)>0$, and then home in by testing $\tilde{s}=0.5(s_1+s_2)$ and replace $s_1$ or $s_2$ as appropriate) or even Newton's method (more work is required to get $F_s(s)$, but it's possible).

• It might be cleaner to define $Y_4 = C$, not sure. Jan 6 '16 at 13:43
• My bad, I didn't give the control conditions. n=f'(delta)=1 and g(delta)=1 I want to integrate from n=0 to n=6, in the end, if f'(delta) is different than 1 I'll have to change the value assumed for f''(0) Jan 7 '16 at 15:12
• I'm still struggling with the second equation $(C/p_r +f)g'=0$. Is this correct? Could you do something like saying in general $g' \neq 0$ so $C/p_r+f = 0$? In which case you could differentiate it to get an expression for $g'$ in terms of $Y_2$ and eliminate it from the equation for $Y_1'$ and carry on with the shooting method from there. Jan 7 '16 at 16:22
• Maybe you could try to minimise both your conditions with $F(s) := (Y_2(\delta;s)-1)^2+(Y_5(\delta;s)-1)^2$, but this is a different problem Jan 8 '16 at 11:43
• Does this make sense now? Hopefully it's now in a form where you can use the previous answers. Note you have to have $\delta >0$. Jan 12 '16 at 10:11


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# Combinatorics: $2n$ sets with $n$ entries, disjointness.

The following combinatorial problem came up that seems easy but has turned out to be very intricate.

Assume that you are given $2n$ sets, say $S_1, \ldots, S_{2n}$ such that

• each $S_i$ has exactly $n$ elements,
• each element is in at most $n$ sets, $\forall x \colon | \{ S_i | x \in S_i \}| \leq n$,
• for each $S_i$, there is an $S_j$ such that they are disjoint, $S_i \cap S_j = \emptyset$.

The final goal is to prove that it is possible to pick in each $S_i$ a different element $x_i \in S_i$ (i.e. $i \neq j \Rightarrow x_i \neq x_j$).

Assume we could prove that there is always a pair of elements $x,y$ such that none of the $S_i$ contains both $x$ and $y$. Then the statement above follows easily by induction.

But how to prove this? Is it actually true?

Okay, the original problem can be solved using the marriage theorem. Using the marriage theorem, it is sufficient to prove that each collection of $S_i$ of size $k$ contains at least $k$ many different elements.
For $k \leq n$, there is nothing to do since already one single $S_i$ contains $n$ elements. For an arbitrary $k$, note that a collection of $k$ many $S_i$ contains $n \cdot k$ elements (counted with multiplicity). As each element occurs in at most $n$ many $S_i$, this means we have at least $\frac{n \cdot k}{n} = k$ many different elements.
The problem whether it is possible to pick a pair of elements such that none of the $S_i$ contains both remains unanswered.


text:
Opuscula Math. 30, no. 4 (2010), 457-464
http://dx.doi.org/10.7494/OpMath.2010.30.4.457

Opuscula Mathematica

# Minimal and co-minimal projections in spaces of continuous functions

Dominik Mielczarek

Abstract. Minimal and co-minimal projections in the space $$C[0,1]$$ are studied. We construct a minimal and co-minimal projection from $$C[0,1]$$ onto a subspace $$Y$$ defined in the introduction.

Keywords: minimal projection, co-minimal projection.

Mathematics Subject Classification: 46A22, 47A30, 47A58.

Full text (pdf)

Dominik Mielczarek, Minimal and co-minimal projections in spaces of continuous functions, Opuscula Math. 30, no. 4 (2010), 457-464, http://dx.doi.org/10.7494/OpMath.2010.30.4.457

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text:
# How do you simplify -50-4((-10-4(-3+1)^2)/ (-2)) ?

Jun 14, 2018

$- 102$

#### Explanation:

$- 50 - 4 \left(\frac{- 10 - 4 {\left(- 3 + 1\right)}^{2}}{- 2}\right)$

To simplify this, we will use PEMDAS, shown here:

This is a common method to simplify expressions, and you can remember it using:
Please Excuse My Dear Aunt Sandy

We also start from inside to outside.

First thing we do is simplify the parenthesis:
$- 50 - 4 \left(\frac{- 10 - 4 {\left(- 2\right)}^{2}}{- 2}\right)$

Exponent:
$- 50 - 4 \left(\frac{- 10 - 4 \left(4\right)}{- 2}\right)$

Multiplication:
$- 50 - 4 \left(\frac{- 10 - 16}{- 2}\right)$

Subtract on numerator:
$- 50 - 4 \left(\frac{- 26}{- 2}\right)$

Division:
$- 50 - 4 \left(13\right)$

Multiplication:
$- 50 - 52$

And lastly, subtraction:
$- 102$

Hope this helps!


text:
Algebra

# Exponential Inequalities - Different Base

Solve the inequality $2^{5x}\ge 5^{22-5x}.$

If the solution to the inequality $2 ^{1-5x} \le 50 ^{5x}$ is $x\ge k,$ what is $10^{10k}?$

Solve for $x:$ $3^{-3x+12}\leq11.$

Given that $\log2=0.301,$ what is the largest integer $x$ that satisfies $2^{x-3}<10^{2}?$

Solve for $x:$ $16^{x+2}\geq27^{-2x+3}.$

×


text:
Vol. 12, No. 4, 2019

 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editors’ Interests Scientific Advantages Submission Guidelines Submission Form Ethics Statement Editorial Login Author Index Coming Soon Contacts ISSN: 1944-4184 (e-only) ISSN: 1944-4176 (print) Other MSP Journals
Euler's formula for the zeta function at the positive even integers

Samyukta Krishnamurthy and Micah B. Milinovich

Vol. 12 (2019), No. 4, 541–548
DOI: 10.2140/involve.2019.12.541
Abstract

We give a new proof of Euler’s formula for the values of the Riemann zeta function at the positive even integers. The proof involves estimating a certain integral of elementary functions two different ways and using a recurrence relation for the Bernoulli polynomials evaluated at $\frac{1}{2}$.

However, your active subscription may be available on Project Euclid at
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Keywords
Riemann zeta function, Euler, Basel problem, Bernoulli numbers, Bernoulli polynomials
Mathematical Subject Classification 2010
Primary: 11M06
Secondary: 11B68, 11B37


text:
## Monte Carlo simulations of the three-dimensional XY spin glass focusing on the chiral and the spin order    [PDF]

Tomoyuki Obuchi, Hikaru Kawamura
The ordering of the three-dimensional isotropic XY spin glass with the nearest-neighbor random Gaussian coupling is studied by extensive Monte Carlo simulations. To investigate the ordering of the spin and the chirality, we compute several independent physical quantities including the glass order parameter, the Binder parameter, the correlation-length ratio, the overlap distribution and the non-self-averageness parameter, etc, for both the spin-glass (SG) and the chiral-glass (CG) degrees of freedom. Evidence of the spin-chirality decoupling, i.e., the CG and the SG order occurring at two separated temperatures, $T_{CG}>T_{SG}>0$, is obtained from the glass order parameter, which is fully corroborated by the Binder parameter. By contrast, the CG correlation-length ratio yields a rather pathological and inconsistent result in the range of sizes we studied, which may originate from the finite-size effect associated with a significant deviation of the spatial CG correlations from the standard Ornstein-Zernike form. Finite-size-scaling analysis yields the CG exponents $\nu_{CG}=1.4 \pm 0.1$ and $\eta_{CG}=0.29 \pm 0.12$, and the SG exponents $\nu_{SG}=1.23^{+0.17}_{-0.06}$ and $\eta_{SG}=-0.42^{+0.12}_{-0.27}$. The obtained exponents are close to those of the Heisenberg SG, but are largely different from those of the Ising SG. The chiral overlap distribution and the chiral Binder parameter exhibit the feature of a continuous one-step replica-symmetry breaking (1RSB), consistently with the previous reports. Such a 1RSB feature is again in common with that of the Heisenberg SG, but is different from the Ising one, which may be the cause of the difference in the CG critical properties from the Ising SG ones despite of a common $Z_2$ symmetry.
View original: http://arxiv.org/abs/1212.3804


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnlilne Submission ㆍMy Manuscript - For Reviewers - For Editors
 On the Tate-Shafarevich group of elliptic curves over $\mathbb Q$ Bull. Korean Math. Soc. 2012 Vol. 49, No. 1, 155-163 https://doi.org/10.4134/BKMS.2012.49.1.155Published online January 1, 2012 Dohyeong Kim Pohang University of Science and Technology Abstract : Let $E$ be an elliptic curve over $\mathbb Q$. Using Iwasawa theory, we give what seems to be the first general upper bound for the order of vanishing of the $p$-adic $L$-function at $s=0$, and the $\mathbb Z_p$-corank of the Tate-Shafarevich group for all sufficiently large good ordinary primes $p$. Keywords : good ordinary reduction, Tate-Shafarevich group, elliptic curves, Iwasawa theory MSC numbers : 11G05 Downloads: Full-text PDF


text:
Lemma 77.6.2. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $G$ be a group algebraic space over $B$. Assume $G \to B$ is locally of finite type. Then $G \to B$ is unramified (resp. locally quasi-finite) if and only if $G \to B$ is unramified (resp. quasi-finite) at $e(b)$ for all $b \in |B|$.

Proof. By Morphisms of Spaces, Lemma 66.38.10 (resp. Morphisms of Spaces, Lemma 66.27.2) there is a maximal open subspace $U \subset G$ such that $U \to B$ is unramified (resp. locally quasi-finite) and formation of $U$ commutes with base change. Thus we reduce to the case where $B = \mathop{\mathrm{Spec}}(k)$ is the spectrum of a field. Let $g \in G(K)$ be a point with values in an extension $K/k$. Then to check whether or not $g$ is in $U$, we may base change to $K$. Hence it suffices to show

$G \to \mathop{\mathrm{Spec}}(k)\text{ is unramified at }e \Leftrightarrow G \to \mathop{\mathrm{Spec}}(k)\text{ is unramified at }g$

for a $k$-rational point $g$ (resp. similarly for quasi-finite at $g$ and $e$). Since translation by $g$ is an automorphism of $G$ over $k$ this is clear. $\square$

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text:
# Did the spacecrafts Galileo or Juno use the Galilean moons for a gravity assist before entering Jovian orbit?

Galileo and Juno are the only two spacecraft to have entered orbit around Jupiter.

Did either of them use one of the Galilean moons for a gravity brake before in order to enter a stable orbit around Jupiter?

Or did they just directly insert into Jupiter's orbit using deep space delta-v maneuvers?

• Answers are still coming in! Thanks for the accept but perhaps you can un-accept and hold off for another few days at least? There could be even more information added, and a quick accept can discourage some folks from writing up additional answer. Thanks! – uhoh Mar 23 at 18:33

I checked JPL's Horizons with the following setup. It looks like Juno was Galilean-agnostic but Galileo had some definite close encounters, especially with Io.

Gravity has no specific range, it just falls of as $$1/r^2$$ so it definitely had a significant interaction with Io but whether it can classified as an assist or just an effect will need further investigation, but I think whatever effect there was must have been extremely small compared to all the velocity gained by falling into Jupiter's huge gravity well.

These spacecraft were "flying fuel tanks", mostly because of the huge deceleration necessary to fall into orbits around these massive bodies to compensate the gravitational acceleration.

above: Galileo capture at Jupiter, below: same for Juno

note: "relative speed" is really the rate of change of range (distance)

             Io      Europa    Ganymede    Callisto
-------   -------   ---------   ---------
Juno      348,720   675,054   1,022,210   1,897,202
Galileo     2,719    34,560   1,191,685     792,388

import numpy as np
import matplotlib.pyplot as plt

names = 'Io', 'Europa', 'Ganymede', 'Callisto'

datas = []
for name in names:
fname = 'Juno ' + name + ' horizons_results.txt'
with open(fname, 'r') as infile:
print(len(lines))
a = [i for (i, line) in enumerate(lines) if 'SOE' in line][0]
b = [i for (i, line) in enumerate(lines) if 'EOE' in line][0]
lines = lines[a+1: b]
data = [[float(x) for x in line.split(',')[-3:-1]] for line in lines]
datas.append(data)
datas = np.swapaxes(np.array(datas), 2, 1)
print(datas.shape)

fig, (axr, axv) = plt.subplots(2, 1, figsize=[7, 7])
for (r, v), name in zip(datas, names):
t = np.linspace(0, 2, len(r))
axr.plot(t, r*1E-06, label=name)
axv.plot(t, v, label=name)
print('Juno closest to ', name, np.round(r.min(), 1), ' km')
axr.set_ylabel('distance (million km)')
axv.set_ylabel('relative speed (km/sec)')
axv.set_xlabel('time since 2016-Jul-04 00:00 (days)')
axr.legend()
axv.legend()
axr.set_ylim(0, None)
axv.plot(t, np.zeros_like(t), '-k', linewidth=0.5)
plt.show()

datas = []
for name in names:
fname = 'Galileo ' + name + ' horizons_results.txt'
with open(fname, 'r') as infile:
print(len(lines))
a = [i for (i, line) in enumerate(lines) if 'SOE' in line][0]
b = [i for (i, line) in enumerate(lines) if 'EOE' in line][0]
lines = lines[a+1: b]
data = [[float(x) for x in line.split(',')[-3:-1]] for line in lines]
datas.append(data)
datas = np.swapaxes(np.array(datas), 2, 1)
print(datas.shape)

fig, (axr, axv) = plt.subplots(2, 1, figsize=[7, 7])
for (r, v), name in zip(datas, names):
t = np.linspace(0, 2, len(r))
axr.plot(t, r*1E-06, label=name)
axv.plot(t, v, label=name)
print('Galileo closest to ', name, np.round(r.min(), 1), ' km')
axr.set_ylabel('distance (million km)')
axv.set_ylabel('relative speed (km/sec)')
axv.set_xlabel('time since 1995-Dec-06 12:00 (days)')
axr.legend()
axv.legend()
axr.set_ylim(0, None)
axv.plot(t, np.zeros_like(t), '-k', linewidth=0.5)
plt.show()

UPDATED with final trajectory prediction prior to atmospheric entry and
end of mission. Schedule of events for final day:

2017-Sep-15

Event time at Saturn                          Signal receipt time at Earth

5:08 UTC (10:08 pm PDT - Sept. 14)

High above Saturn, Cassini crosses the orbital distance of Enceladus
for the last time

7:14 UTC (12:14 am PDT)      8:37 UTC (1:37 am PDT)
Spacecraft begins a 5-minute roll to point instrument (INMS) that will
sample Saturn's atmosphere and reconfigures systems for real-time data
transmission at 27 kilobits per second (3.4 kilobytes per second).
Final, real-time relay of data begins

7:22 UTC (12:22 am PDT)      High above Saturn, Cassini crosses the orbital distance of the F ring
(outermost of the main rings) for the last time

10:31 UTC (3:31 am PDT)                        11:54 UTC (4:54 am PDT)
Atmospheric entry begins; thrusters firing at 10% of capacity

10:32 UTC (3:32 am PDT)                        11:55 UTC (4:55 am PDT)
Thrusters at 100% of capacity; high-gain antenna begins to point away
from Earth, leading to loss of signal.

• This is amazing! – Jonathan L. Mar 23 at 14:00
• Can you plot gravitational acceleration induced vs time? Ie, Gm/r^2. – Yakk Mar 24 at 2:27
• @Yakk it is certainly possible and one could do it, but in what frame? Everything is always accelerating in everything else's frame all the time. I suppose one could plot $GM/r^2$ for Jupiter and Io to get some idea, or do a numerical integration with Io's mass turned on and turned off, but that's more work than I'd like to do for this answer. Please feel free to try it yourself (the first one only takes a few more lines of Python, what better way to learn it!) or to post a new question and let others give it a try. Thanks! – uhoh Mar 24 at 2:40
• @uhoh It is mostly the annoyance of getting the data from that interface, not adding a k/r^2 term in python, but ok, no is a reasonable response. – Yakk Mar 24 at 3:09
• @uhoh No, I mean, I can edit python quite easily, I just have no idea how to get the data files the python script uses from that website. I tried to duplicate your screenshot with their web controls and checkbox forest, but failed (QUANTITIES=20 has no obvious correspondence in the full page of widgets I could see, for example). The script is useless without the data. My point is, graphing induced acceleration (maybe logarithmic) gives an idea of how big the use of the moons gravity is compared to (say) Jupiter, visually. But I don't want to bother you further. – Yakk Mar 24 at 15:36

https://history.nasa.gov/sp4231.pdf

Galileo definitely did gravity assist at Io. See the link (3.7 Mb pdf), pages 202-203.

quote:

Two of the first events of Arrival Day were the Orbiter’s flybys of the moons Europa (at 3:09 a.m. PST) and Io (at 7:46 a.m. PST). The Orbiter passed 32,500 kilometers (20,200 miles) from Europa, but only 890 kilometers (550 miles) from Io, closer than originally planned (see figure 8.4). The Orbiter’s trajectory had been altered to make maximum use of Io’s gravitational field for decelerating the spacecraft, thereby conserving propellant for later in the mission.

The saving of propellant because of the gravity assist was rather significant. Galileo insertion burn has reduced the velocity by ~600 m/s (ibid, page 208). Without the assist delta-v would be closer to 900 m/s.

(Deduced from similar numbers for future ESA Juice mission, 7.5 Mbytes, page 85).

EDIT: ok, 175 m/s from another answer by @kwan3217, but still significant.

Juno mission could not use gravity assist from the moons. It was inserted on polar orbit around Jupiter. Galilean moons orbits are coplanar to equator of Jupiter, so Juno has flow too far from them to use gravity assist.

Other answers discuss how close Galileo got. Here is a source identifying the intent, Galileo Messenger v37:

https://planetary.s3.amazonaws.com/galileo_messenger/GM037.pdf

"The Io flyby is also critical for JOI: Passing in front of Io in its orbital path delivers a gravity-assised brake on Galileo's trajectory. The reduction in the Orbiter's speed (its ΔV) by 175m/s accouts for about a fifth of the total change needed for the capture orbit, the first in the orbital tour."

I remember watching this on the early World Wide Web, when I was growing up -- back in the day, NASA had one of the stronger presences on the web. I remember thinking that the Io pass, probe relay, and Jupiter orbit insertion was way too much to expect to work within just one day. As it happened, the tape recorder got stuck and they had to cancel all of the Io observations.


text:
# Homework Help: Using the formula for a rigid pendulum.

1. Sep 6, 2012

### ardnassac.95

How to use the formula for a rigid pendulum?

Given the formula T = 2π√(2L/3g)
(1) how do you change to get T^2 = ?
(2) And then how (1) to get g = ?

Here's what I've tried, but I'm not sure if they are correct:
(1) T^2 = 4π^2 x (2L/3g)
(2) g = [4π^2 x (2L/T^2)] ÷ 3

(3) The experiment is to find the relationship between length of rod, L and period, T, for a rigid pendulum (i.e rod) hooked onto a stand. When we plot the transformed data onto a graph it should be L vs T^2, but the formula in (2) as I've tried above give 2L/T^2 not L/T^2 as given as the gradient of my graph. So how would I use the gradient to put into the formula (2) to get g (gravity) which should be 9.81ms^-2? Do I double my gradient and substitute that as 2L/T^2?

Last edited: Sep 6, 2012


text:
Photoelectric effect and wavelengths

1. Jan 2, 2005

punjabi_monster

Here is a quesiton I am having trouble with :uhh: , thanks for your help.

Electrons are ejected from a photoelectric surface with a maximum speed of 4.20 * 10^5 m/s. If the work function of this surface is 2.55 eV, what is the wavelength of the incident light?

This is how i tried to solve this question:
Ek=(1/2)mv^2
Ek=(1/2)(9.11*10^-31 kg)(4.20*10^5 m/s)^2
Ek=8.04*10^-20 J

E=hc/λ
λ=[(6.63*10^-34 Js)(3.00*10^8 m/s)]/(8.04*10^-20 J)
λ=(2.48*10^-6 m) :yuck:

The actual answer is (4.07*10^-7 m).

Can you please tell me the correct way to do this question. Thanks.

Last edited: Jan 2, 2005
2. Jan 2, 2005

apchemstudent

Your Ek was correct, but you forgot to include the energy needed to eject an electron from the surface 2.55 eV.

3. Jan 2, 2005

punjabi_monster

Would you use this formula:
Ekmax = hf - W

4. Jan 2, 2005

quasar987

That's right.

$$E_{k_{max}}=E_{photons}-W$$

Because the incident photons give away all of their energy to the electrons. So all the electrons have to do to break free from the metal is to escape its "potential well" of energy 2.55 eV.

Last edited: Jan 2, 2005
5. Jan 2, 2005

punjabi_monster

oooo thanks..........i got the right answer!!!!!


text:
# One time pad, get plaintext using ciphers encoded with the same key

I'm trying to solve a problem:

I have 11 ciphers encoded with the same key. My aim is to decode target cipher.

If I do xor C1, C2 (ciphers encoded with the same key) I do get M1 xor M2 (where M1, M2 are plain text messages).

Please say what to do next. I don't understand how to get plain texts from M1 xor M2

UPD: it's a two time pad (i.e. the same key has been use more than one time)

• What cipher? I'm guessing a one-time-pad that's been reused for several messages, is that right? See Taking advantage of one-time pad key reuse? – Gilles 'SO- stop being evil' Nov 18 '12 at 13:21
• Is it correct: here are two methods, named statistical analysis or Frequency analysis and pattern matching. So I have to implement one of them and apply to xor results. – Sergey Nov 18 '12 at 14:02
• You have more information since you have a depth of 11 rather than 2. So your task is much easier, though you probably still have to apply the same kind of analysis. – Yuval Filmus Nov 18 '12 at 19:00

So I've passed coursera Scala course and deciced to apply my fresh knowledge. Here is my algo:

1. Xor ciphers with each other and collect xor result for each position. As a result I got a Map[Int, List[String]] where Int is a position of char in chiper and List[String] are all xor results for current position.

2. I have an assumption that if xor in List[String] equals to UPPER ASCII alpha, it means that source of this computation was: *space xor lower_alpha = UPPER_alpha*.

3. I can try to guess key value: c1 (char from 1st cipher), c2(char from 2nd cipher), xorRes (c1 xor c2 = xorRes) => c2 xor(c1 xor xorRes) or c1 xor(c2 xor xorRes) = *lower_alpha*.

Here is a part of code:

  /**
* hex1 is a hex from 1st cipher
* hex2 is a hex from second cipher
* xor is ahex1 xor hex2
* @return List with pos of possible key and possible key value else return empty list
* */
def getPossibleKey(hex1: String, hex2: String, xor: String, pos: Int): List[(Int, String)] = {
if(isUpperAphpa(xor)){
if( isLowerAphpa(xorHexStrings(xorWithSpace(hex1), hex2)) ){
List( (pos, xorWithSpace(hex1)) )
}else if( isLowerAphpa(xorHexStrings(xorWithSpace(hex2), hex1)) ){
List( (pos, xorWithSpace(hex2)) )
}else{
List.empty
}
}else{
List.empty
}

}

As a result i got: List[Int, String] where String is the most popular lower_alpha got from getPossibleKey function.

Then i just applied collected key to each cipher and started to fix the key. I know that c1 xor m1 = k1 where c1 is a cipher fragment, k1 is a key fragment and m1 is a plain text.

• This is not going to work. If $X$ is an XOR of two uppercase letters, then there are lots of ways to represent it as an XOR of two letters, for example by flipping the LSB of both, or by switching the letters (unless they're equal). As an extreme example, if $X = 0$ then there are $26$ different ways. When considering $11$ different messages together you get more information, but I'm not sure whether that's enough. In general, you need to look at more than one letter at the same time, say using dynamic programming to make it computationally feasible. – Yuval Filmus Dec 23 '12 at 21:02
• Sorry, I did solve the problem and got 100% score on this task. As I mentioned before, I did have to spend some time picking right values for the key. I can agree that for just two encrupted messages I would have to apply much more coplicated algorithm. – Sergey Dec 23 '12 at 21:09


text:
What is the final temperature after 840 J is absorbed by 10.0 g of water at 25°C?

May 15, 2016

The final temperature is equal to ${45}^{\circ}$ $C$

Explanation:

We have the amount of energy gain
$Q = m \cdot c \cdot \Delta T = m c \Delta t$

where $c$ = $4.184$ $J$/$g . C$ is the specific heat of water, $m$ is the mass of water

$\implies$$840 = 10$ x $4.184 \cdot \left(t - 25\right)$
$t = \frac{840}{10 \text{ x } 4.184} + 25 = 45$ i.e. ${45}^{\circ}$ $C$


text:
ISSN 1683-3414 (Print)   •   ISSN 1814-0807 (Online)

## Contacts

362027, RNO-A, Russia
Phone: (8672)50-18-06
E-mail: rio@smath.ru

DOI: 10.23671/VNC.2017.1.5827

# Contractive Projections in Variable Lebesgue Spaces

Tasoev B. B.
Vladikakazian Mathematical Journal 2017. Vol. 19. Issue 1.
Abstract:
In this article we describe the structure of positive contractive projections in variable Lebesgue spaces $$L_{p(\cdot)}$$ with $$\sigma$$-finite measure and essentially bounded exponent function $$p(\cdot)$$. It is shown that every positive contractive projection $$P:L_{p(\cdot)}\rightarrow L_{p(\cdot)}$$ admits a matrix representation, and the restriction of $$P$$ on the band, generated by a weak order unite of its image, is weighted conditional expectation operator. Simultaneously we get a description of the image $$\mathcal{R}(P)$$ of the positive contractive projection $$P$$. Note that if measure is finite and exponent function $$p(\cdot)$$ is constant, then the existence of a weak order unit in $$\mathcal{R}(P)$$ is obvious. In our case, the existence of the weak order unit in $$\mathcal{R}(P)$$ is not evident and we build it in a constructive manner. The weak order unit in the image of positive contractive projection plays a key role in its representation.
Keywords: conditional expectation operator, contractive projection, variable Lebesgue space, Nakano space, $$\sigma$$-finite measure.


text:
# Cross-flow filtration

(Redirected from Crossflow filtration)
Diagram of cross-flow filtration

In chemical engineering, biochemical engineering and protein purification, crossflow filtration[1] (also known as tangential flow filtration[2]) is a type of filtration (a particular unit operation). Crossflow filtration is different from dead-end filtration in which the feed is passed through a membrane or bed, the solids being trapped in the filter and the filtrate being released at the other end. Cross-flow filtration gets its name because the majority of the feed flow travels tangentially across the surface of the filter, rather than into the filter.[1] The principal advantage of this is that the filter cake (which can blind the filter) is substantially washed away during the filtration process, increasing the length of time that a filter unit can be operational. It can be a continuous process, unlike batch-wise dead-end filtration.

Diagram of cross-flow filtration

This type of filtration is typically selected for feeds containing a high proportion of small particle size solids (where the permeate is of most value) because solid material can quickly block (blind) the filter surface with dead-end filtration. Industrial examples of this include the extraction of soluble antibiotics from fermentation liquors.

## Operation

Ceramic membrane for industrial cross-flow filtration

In crossflow filtration, the feed is passed across the filter membrane (tangentially) at positive pressure relative to the permeate side. A proportion of the material which is smaller than the membrane pore size passes through the membrane as permeate or filtrate; everything else is retained on the feed side of the membrane as retentate.

With crossflow filtration the tangential motion of the bulk of the fluid across the membrane causes trapped particles on the filter surface to be rubbed off. This means that a crossflow filter can operate continuously at relatively high solids loads without blinding.

## Benefits over conventional filtration

• A higher overall liquid removal rate is achieved by the prevention of filter cake formation
• Process feed remains in the form of a mobile slurry, suitable for further processing
• Solids content of the product slurry may be varied over a wide range
• It is possible to fractionate particles by size [3]
• Tubular pinch effect

## Industrial applications

Filtration unit for industrial cross-flow filtration

Cross flow membrane filtration technology has been used widely in industry globally. Filtration membranes can be polymeric or ceramic, depending upon the application. The principles of cross-flow filtration are used in reverse osmosis, nanofiltration, ultrafiltration and microfiltration. When purifying water, it can be very cost effective in comparison to the traditional evaporation methods.

In protein purification, the term Tangential Flow Filtration (TFF) is used to describe cross-flow filtration with membranes. The process can be used at different stages during purification, depending on the type of membrane selected.[2]

In the photograph of an industrial filtration unit (right), it is possible to see that the recycle pipework is considerably larger than either the feed pipework (vertical pipe on the right hand side) or the permeate pipework (small manifolds near to the rows of white clamps). These pipe sizes are directly related to the proportion of liquid flows through the unit. A dedicated pump is used to recycle the feed several times around the unit before the solids-rich retentate is transferred to the next part of the process.

## Techniques to improve performance of cross flow filtration

### Backwashing

In backwashing, the transmembrane pressure is periodically inverted by the use of a secondary pump, so that permeate flows back into the feed, lifting the fouling layer from the surface of the membrane. Backwashing is not applicable to spirally wound membranes and is not a general practice in most applications. (See CIP) [4]

### Clean-in-place (CIP)

Clean-in-place systems are typically used to remove fouling from membranes after extensive use. The CIP process may use detergents, reactive agents such as sodium hypochlorite and acids and alkalis such as citric acid and sodium hydroxide. Sodium hypochlorite (bleach) must be removed from the feed in some membrane plants. Bleach oxidizes thin-film membranes. Oxidation will degrade the membranes to a point where they will no longer perform at rated rejection levels and have to be replaced. Bleach can be added to a NaOH sodium hydroxide CIP during an initial system start-up before spirally-wound membranes are loaded into the plant to help disinfect the system. Bleach is also used to CIP perforated stainless steel (Graver) membranes, as their tolerance for sodium hypochlorite is much higher than a spirally-wound membrane. Caustics and acids are most often used as primary CIP chemicals. Caustic removes organic fouling and acid removes minerals. Enzyme solutions are also used in some systems for helping remove organic fouling material from the membrane plant. The pH and temperature are important to a CIP program. If pH and temperature are too high the membrane will degrade and flux performance will suffer. If pH and temperature are too low, the system simply will not be cleaned properly. Every application has different CIP requirements. e.g. a dairy RO reverse osmosis plant most likely will require a more rigorous CIP program than a water purification RO plant. Each membrane manufacturer has their own guidelines for CIP procedures for their product.

### Concentration

The volume of the fluid is reduced by allowing permeate flow to occur. Solvent, solutes, and particles smaller than the membrane pore size pass through the membrane, while particles larger than the pore size are retained, and thereby concentrated. In bioprocessing applications, concentration may be followed by diafiltration.

### Diafiltration

In order to effectively remove permeate components from the slurry, fresh solvent may be added to the feed to replace the permeate volume, at the same rate as the permeate flow rate, such that the volume in the system remains constant. This is analogous to the washing of filter cake to remove soluble components.[4] Dilution and re-concentration is sometimes also referred to as "diafiltration."

### Process flow disruption (PFD)

A technically simpler approach than backwashing is to set the transmembrane pressure to zero by temporarily closing off the permeate outlet, which increases the attrition of the fouling layer without the need for a second pump. PFD is not as effective as backwashing in removing fouling, but can be advantageous.

## Flow rate calculation

The flux or flow rate in cross-flow filtration systems is given by the equation:

${\displaystyle J={\frac {\Delta P}{(R_{m}+R_{c})\mu }}}$ [4]

in which:

• ${\displaystyle J}$: liquid flux
• ${\displaystyle \Delta P}$: transmembrane pressure (should also include effects of osmotic pressure for reverse osmosis membranes)
• ${\displaystyle R_{m}}$: Resistance of the membrane (related to overall porosity)
• ${\displaystyle R_{c}}$: Resistance of the cake (variable; related to membrane fouling)
• ${\displaystyle \mu }$: liquid viscosity

Note: ${\displaystyle R_{m}}$ and ${\displaystyle R_{c}}$ include the inverse of the membrane surface area in their derivation; thus, flux increases with increasing membrane area.


text:
# Velocity Vector Transformation for Point Ahead Angle and Doppler Shift Calculation for Inter-Satellite Link

So, I have the Radial and Tangential Velocity Vector (like in Figure 1) and also the Cartesian coordinate of 3D velocity vector. For reference, you can see in this link. Where, $$o_{dot}$$ vector matrix have 3 components (only X,Y component), first element $$V_R$$ and second element $$V_T$$ like in the Figure 1.

But to calculate Point Ahead Angle and Doppler Effect I should follow, this velocity vector like in Figure 2. Where Vp is the perpendicular or the tangential (I guess so!) one and Vr is aligned with Line-of-Sight, where radius is the Range between 2 satellites, as describe here by @uhoh

But how to transform the velocity vector as in orbital mechanics point of view (lets take we all are in same inertial frame) from derived in Figure 1 to Figure 2 formation. Is it simply rotation of the angle (angle between Earth-Satellite1-Satellite2) where radial velocity must align with line-of-sight, how?

Thank you

• I mentioned you should ask a new question and you did right away, that's great! +1 I did add a little bit more to this answer but I don't think that addresses this. – uhoh Sep 26 '19 at 1:55
• Careful with your definitions. In a circular orbit, the tangential and radial velocities of the spacecraft are orthogonal. But in any real orbit, the velocity and the position are not usually orthogonal. Still, the modulus of the tangential velocity is the modulus of the velocity. – Mefitico Oct 16 '19 at 15:37


text:
08-03-2014, 01:07 PM (This post was last modified: 08-04-2014 01:20 PM by Dave Britten.)
Post: #1
 Dave Britten Senior Member Posts: 1,182 Joined: Dec 2013
EDIT (Aug 4): Fixed a really stupid bug. Thanks Joe.

I needed to generate a WEP key to set up an access point for legacy gear, but didn't have a tool handy that would allow me to restrict the output to a 13-digit hexadecimal number. But I've got a 48, so that problem didn't last long!

Input:
2: Length of password
1: Allowable characters (enter as a single string, e.g. "0123456789ABCDEF")

Output:

Code:
\<< DUP SIZE \-> L C S   \<< "" 1 L     START C RAND S * CEIL DUP SUB +     NEXT   \>> \>>

Disclaimer: This is only as cryptographically secure as the quality of the random number generator built in to the 48. Don't use it for securing warheads or anything like that.
08-04-2014, 03:35 AM (This post was last modified: 08-04-2014 03:41 AM by Joe Horn.)
Post: #2
 Joe Horn Senior Member Posts: 1,505 Joined: Dec 2013
RE: (48) Password Generator
(08-03-2014 01:07 PM)Dave Britten Wrote:
Code:
\<< DUP SIZE \-> L C S   \<< "" 1 L     START C RAND S * 1 + DUP SUB +     NEXT   \>> \>>

Please note that the first character in the input string will have a greater probability of getting picked than any other, and the final one will have less probability, and sometimes the output will be shorter than L, since SUB rounds to the nearest integer. I therefore suggest replacing 1 + with CEIL. This evens out the probabilities, fixes the length bug, and is a tad shorter and faster too.

<0|ɸ|0>
-Joe-
08-04-2014, 11:28 AM
Post: #3
 Dave Britten Senior Member Posts: 1,182 Joined: Dec 2013
RE: (48) Password Generator
(08-04-2014 03:35 AM)Joe Horn Wrote:
(08-03-2014 01:07 PM)Dave Britten Wrote:
Code:
\<< DUP SIZE \-> L C S   \<< "" 1 L     START C RAND S * 1 + DUP SUB +     NEXT   \>> \>>

Please note that the first character in the input string will have a greater probability of getting picked than any other, and the final one will have less probability, and sometimes the output will be shorter than L, since SUB rounds to the nearest integer. I therefore suggest replacing 1 + with CEIL. This evens out the probabilities, fixes the length bug, and is a tad shorter and faster too.

Thanks Joe, can't believe I missed that. I blame the fact that I'm in the hospital for a couple days, and was on pain medication when I wrote that.
03-22-2017, 04:38 PM
Post: #4
 pier4r Senior Member Posts: 2,016 Joined: Nov 2014
RE: (48) Password Generator
This is a nice idea!

I will try to expand it in phonetic passwords, since a crude random password may be too difficult to type.

PS: a thing is that all the programs are according to a calculator but actually many calculators are compatible to each other, so I discard the non-50g threads while actually I miss many useful threads.

Wikis are great, Contribute :)
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User(s) browsing this thread: 1 Guest(s)


text:
Friday, June 6, 2008

Malware Trend in 2007

I read the report IBM Internet Security System X-Force 2007 TrendStatistics. This is a report describing trends for various threats in 2007.This team has been tracking trends since 2000. I found the reportto be quite interesting. In the rest of this post, I highlight someof the interesting points from the report and what they mean in thecontext of malware detection.(I) The X-Force team reports continued growth in Web browser exploitation. Thisclearly shows that the infection vector is changing to the Web. Earlierthe primary infection vectors were email and the network. Therefore,for detecting malware, drive-by-downloads (DBD) and other threats targeted at hacking through the Web browser need a lot of attention.(II) X-Force also reports a marked increase in obfuscated exploits, i.e.,exploits that use various code obfuscation techiques (such as encryption).Here is a quote, "X-Force estimated that nearly 80 percent of Web exploitsused obfuscation and/or self decryption ... By the end of 2007, X-Forcebelieved this rate had reached 100 percent, ...". This means that goingforward, Web exploits will increasingly harbor indiscernible code rending signature-based techniques less effective. Advancedtechniques (such as behavior-based detection) are clearly needed to detectsuch malware. To exacerbate the situation the X-Force report stated thatthere was a 30% increase in new malware samples in 2007 over 2006. Thisfurther drives home the point that signature-based detectors will have troublein keeping up with the number of malware as they cannot detect new threats.(III) There was another very interesting point made by the report. Modernmalware use features from various types of classic malware (such as viruses, worms,and spyware) by pulling the successful features of each into new strains. To quote the report, "Modern malware is now the digital equivalentof the Swiss Army knife, and 2007 data continues to support this." This trendalso indicates that the behavior of malware is becoming more sophisticated, whichagain supports my claim that detection techniques based on analyzing behavior arebetter suited to handle malware of the future. Another interesting tidbit from thereport: "Trojans make up the largest class of malware in 2007 as opposed to downloaders,which were the largest category in 2006." Recall that a Trojan appears to be alegitimate file with some hidden functionality (for example, that of a rootkit).Trojans are historically a problematic class of malware for signature-baseddetection.Overall, I found the report to be very interesting. Read it for yourself.You can find the report here.

Wednesday, April 23, 2008

Zero Day Threat by Acohido and Swartz

I read the book Zero Day Threat (ZDT) by Byron Acohido and Jon Swartz. I really liked the book! Zero Day Threat is about the underground cyber-economy. It makes some surprising points grounded in real truths. I liked that the book paints a complete picture, i.e., how malware,
identity theft, and "drop off" gangs collaborate to facilitate
a well oiled cyber-economy. Since my research area is security,
I was very familiar with the different types of malware brought up in Zero Day Threat. However, this book gave me a complete picture of the problem.

I particularly appreciated two features of the book:

Structure: Each chapter is broken into three sections: exploiters,
enablers, and expeditors. Exploiter sections focus on crooks (such
as scam artists and drug addicts) and how they benefit from the
underground economy. The Enablers sections focus on credit card
companies, banks, and credit bureaus, and how their current practices
enable the underground cyber-economy. Expediters
are guys (good and bad) that allow the cybercrooks to exploit
vulnerabilities in an expeditious manner. I thought this structure
was just brilliant! It really brings out the correlation between
various factors and actors that enable the underground cyber-economy.

Narrative Style: I really enjoyed various anecdotes in the book.
There are several stories about people being scammed or getting
lured into the profitable cyber-underground. For example, there is a story of
a "drop off" gang in Edmonton which is narrated throughout the
book. These anecdotes make the book very interesting and provide
a "human side" to the cyber-underground.

I highly recommend this book.

Wednesday, March 19, 2008

Botnets in USA Today

I got a call from Byron Acohido over at the USA Today last weekend,
and we had an interesting talk about botnets. Byron and Jon Swartz ended
up writing an article about botnets which appeared as the cover story
in the Money section of the USA Today on March 17, 2008. Here's a link to the full
story (link). I found the entire article to be a fascinating read
on the nature of botnets. Here are some of the highlights, but
definitely go and read the entire article.
• On a typical day, 40% of the 800 million computers connected to the Internet are bots engaged in various nefarious activities, such as spamming, stealing sensitive data, and engaging in denial-of-service attacks. Think about it. Approximately 320 million computers are engaged these illicit actiivities!
• Later on in the article they describe various features of Storm, the state-of-the-art for botnets. Storm introduced various innovations into the bot landscape, such as using P2P style communication to converse with the bots and encrypting the command-and-control (C&C) traffic. Command-and-control is the traffic from the bot-herder to the bots instructing them to perform various nefarious activities. Note that this means that various network-based botnet solutions that simply look for centralized C&C communication will not work. Moreover, encrypted traffic is a major problem for the network-based solutions. See my earlier blog where I argue that we should move to a cooperative solution. This is looking like a very good idea. Storm also has a self-defense mechanism, i.e., anyone trying to probe the botnet is punished with a denial-of-service attack. I found this self-defense mechanism of Storm to be very interesting.
Overall a fascinating article!
I plan to drop by Byron's book signing at the RSA Conference in San
Francisco on April 7th. Byron also has an interesting blog which is related to the
material in the book.

Wednesday, March 5, 2008

Model Checking and Security

Model checking is a technique of verifying temporal properties of finite-state systems. One of
the attractive features of model checking over other techniques (such as theorem proving)
is that if a property is not true, a model checker provides a counter-example which
explains why the property is not true. Inventors of model checking, Edmund Clarke,
Allen Emerson, and Joseph Sifakis, won the 2008 ACM Turing award (see the announcement here). I have a personal connection to two of the recipients. Edmund Clarke was my adviser
at Carnegie Mellon, and Allen Emerson and I have collaborated on few projects and he
has supported me through out my career.

In this note I try to summarize various applications of model checking to security.

Protocol verification: Protocols in the realm of security (henceforth referred to
as security protocols) are very tricky to get correct. For
example, flaws in authentication protocols have been discovered several years after they have been published. Techniques based on model checking have been extensively used to verify these protocols. The tricky part in applying these techniques for verifying security protocols is
modeling the capabilities of the attacker. Gavin Lowe used the FDR model checker to find
a subtle attack on the Needham-Schroeder authentication protocol (this
publication can found here). Following Lowe's work there
was a flurry of activity on this topic. Interested readers can look at the proceedings of
the Computer Security Foundations Symposium (CSF).

Vulnerability assessment: Imagine you are given an enterprise network with various components (firewalls, routers, and Intrusion Prevention Systems (IPSs)).
Vulnerability assessment tries to ascertain how an attacker can penetrate the specified network. Vulnerability assesment is crucial in updating policies of various security appliances (such as firewalls and IPSs) and ascertaining the risk of various decisions. Traditionally, vulnerability assessment has been performed by red teams. Red teaming is a very valuable activity but can provide no guarantees that the entire state space of vulnerabilities has been explored. I along with (Oleg Sheyner and Jeannette Wing) explored techniques based on model checking for vulnerability assessment. We formally specify the network and express the negation of the attackers goal (e.g., attacker gets root access on a critical server) as a property to be verified. If
the specified network is vulnerable, then the model checker will output a counter-example (which is an attack on the network). The innovation we devised was to output the set of all
counter-examples or attacks as an attack graph, which is succinct representation of all attacks on the network. Analysis of the attack graph can provide a basis for vulnerability assessment. This paper can be downloaded here.

Other applications: There are several problems in security that can be addressed using model checking. For example, I and Tom Reps have used model checking properties to analyze properties of security policies in trust-management systems. Ninghui Li and his collaborators have used techniques based on model checking to analyze several classes of security properties.
In the context of security, the advantage model checking has over other techniques (such as
testing) is that it exhaustively covers the state-space. After all, if you just have one
vulnerability, an attacker will exploit that vulnerability, i.e., an attacker just needs one
door to get through your system. Thus the completeness guarantee that a model checker provides is very valuable in the context of security.

Wednesday, February 13, 2008

Cooperating Detectors

A malware detector tries to determine whether a program is malicious (examples
Malware detection is primarily performed at two vantage points: host and
network. This post explains why cooperation between host-based and network-
based detectors is a good thing.

Traditionally, detection has been performed either at the network or host level, but
not both. First, let me examine both approaches separately.

A network-based detector monitors events by examining a session or
network flow and tries to determine whether it is malicious. The
advantage of a network-based detector is ease of deployment -- there
are not that many points of deployment for a network-based detector
(typically they are deployed behind border routers).

Unfortunately, network-based detectors have a limited view of each
network session. In fact, if a session happens to be
encrypted such as is common with VPNs, Skype, and some bots, a
network-based detector is essentially blind. For example, a botmaster
can hide its communication with the bots by simply encrypting the session.

By contrast, host-based detectors have a more comprehensive view of system activities, i.e.,
they have the potential to observe every event at the host, including malicious ones. However, the major drawback of a host-based detector is that it has to be widely deployed. Typically in a managed
network (such as in an enterprise), a host-based detector has to be deployed at
every host.

Cooperation between host-based and network-based detectors can potentially
address the shortcomings of each detector. I've come up with three possible scenarios.

1) Host-based detector helping the network-based detector.
A network-based detector can pull alerts from a host-based
detector and a host-based detector can push alerts to a network-based
detector. This is a simple solution and I suspect the easiest
scenario for cooperation.

2) Queue up suspicious activity on a virtual machine.
If a network-based detector determines that a session is
"suspicious," it can divert the suspicious traffic to a virtual machine
with a host-based detector for more in-depth analysis. The trick here
is figuring out what events are indeed "suspicious" (you do not want
too much traffic to go through the "slow path" corresponding to a
host-based detector). There is already a
startup called Fireeye adapting this solution. I find this line of work quite intriguing.

3) Pushing signatures.
This third scenario has been explored quite thoroughly in academic
literature. It involves the cooperation of host-based and network-based detectors to
push signatures for malware in real-time. For example, if a host-
based detector recognizes an attack, it pushes out a signature to a
network-based detector. The advantage of this
approach is that, by updating a network-based detector, an entire
enterprise can be protected against that particular threat. However, in my
view this is not a good approach in the long run. Hackers are creating malware variants at
an alarming rate and signatures won't be able to keep up.

Wednesday, January 30, 2008

Case for kernel-level detection

Why kernel-level detection?
These are my thoughts on why malware detection should performed at the
kernel level. In general, the lower in the system hierarchy your
detector resides, the harder it is for an attacker to evade your detector.
For example, if a detector uses system-call interposition, an attacker can
evade this system by directly using kernel calls. For example,
system-call interposition can be done on Windows using the following
package. In my conversations with
a guy from NSA (name withheld for obvious reasons:-)) he confirmed that
new malware they are observing in their lab are using kernel calls directly.
Also, look at the following article

The semantic-gap problem:
A natural question that comes to mind is: why not perform detection at even a lower layer
in the heirarchy? Say the VM layer or even better at hardware. As you move
down in the system hierarchy, you lose some high-level semantics. Let me explain.
Lets say you are doing detection at the VM layer. A high-level event (such as
opening a file) manifests itself as a sequence of events (such as writing to
memory page or an interrupt). In other words, there is a gap between
the events you observe at the VM level and the corresponding high-level event. To my knowledge
the "semantic gap" issue was first articulated in the following paper:

Peter M. Chen, Brian D. Noble, "When virtual is better than real",
Proceedings of the 2001 Workshop on Hot Topics in Operating Systems (HotOS),
May 2001.


text:
# Homework Help: Help with trig identities

1. Feb 18, 2015

### Marcus27

1. The problem statement, all variables and given/known data
Show that (sin^4 x + (sin^2 x * cos^2 x)) / (cos^2 x - 1) == -1
2. Relevant equations
Sin^2 x + cos^2 x == 1

3. The attempt at a solution
(sin ^4 x + (sin^2 x * cos^2 x)) / (cos^2 x - 1)
= ((sin^2 x)(sin^2 x) + (sin^2 x * cos^2 x)) / (cos^2 x - 1)
=((sin^2 x)(1 - cos^2 x) + (sin^2x * cos^2 x)) / (cos^2 x -1 )
= (sin^2 x - (sin^2 x * cos^2 x) + (sin^2 x * cos^2 x)) / (cos^2 x - 1 )
= (sin^2 x) / (cos^2 x - 1 )
= (1 - cos^2 x ) / (cos^2 x -1)
= -1

I think this is correct, but when I looked up the answer in the back of the textbook it showed completely different working using different substitutions. Did I make any mistakes? or are there two or more solutions to this problem?, if this is the case would I be marked down in an exam for using this method?.

2. Feb 18, 2015

### SteamKing

Staff Emeritus
In working trig identity problems, there may be more than one valid substitution which can be used to obtain a simplification, especially with complicated or lengthy expressions.

If this were an exam exercise, no, you should not be penalized for using a valid method of simplification, even if it differs from a method preferred by the instructor.

3. Feb 18, 2015

### Marcus27

Thanks, that puts my mind at ease.

4. Feb 22, 2015

### HallsofIvy

I, for example, would, seeing that "$sin^4(x)$" change everything else to "sin(x)".
Since $cos^2(x)= 1- sin^2(x)$, the numerator is $sin^4(x)+ sin^2(x)(1- sin^2(x))= sin^4()+ sin^2(x)- sin^4(x)= sin^2(x)$. Is that what your text book does?