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## Tbilisi Mathematical Journal

### A note on the new set operator $\psi_{r}$

#### Abstract

Recently many published works made on local function used in ideal topological spaces can be found in related literature. "Semi Local Functions in Ideal Topological Spaces", "Closure Local Functions", and "$()^p$ and $\psi_p$-Operator" can be mentioned among such works those aim to define such functions. In general, the researchers prefer using the generalized open sets instead of topology in ideal topological spaces. Obtaining a Kuratowski closure operator with the help of local functions is an important detail in ideal topological space. However, it is not possible to obtain a Kuratowski closure operator from many of these local functions proposed by the above mentioned works. In order to address the lack of such an operator, the goal of this paper is to introduce another local function to give possibility of obtaining a Kuratowski closure operator. On the other hand, regular local functions defined for ideal topological spaces have not been found in the current literature. Regular local functions for the ideal topological spaces has been described within this work. Moreover, with the help of regular local functions Kuratowski closure operators $cl_I^{*r}$ and $\tau^{*r}$ topology are obtained. Many theorems in the literature have been revised according to the definition of regular local functions.

#### Article information

Source
Tbilisi Math. J., Volume 11, Issue 4 (2018), 43-52.

Dates
Accepted: 10 July 2018
First available in Project Euclid: 4 January 2019

https://projecteuclid.org/euclid.tbilisi/1546570884

Digital Object Identifier
doi:10.32513/tbilisi/1546570884

Mathematical Reviews number (MathSciNet)
MR3954206

#### Citation

Atay, Arife; İlhan Tutalar, Hasan. A note on the new set operator $\psi_{r}$. Tbilisi Math. J. 11 (2018), no. 4, 43--52. doi:10.32513/tbilisi/1546570884. https://projecteuclid.org/euclid.tbilisi/1546570884

#### References

• A. Vadivel and Mohanarao Navuluri, Regular semi local functions in ideal topological spaces, Journal of Advenced Research in Scientific Computing 5, (2013), 1-6.
• A. Vadivel and K. Vairamanickam, $rg\alpha$-Closed Sets and $rg\alpha$-Open Sets in Topological Spaces, Int. Journal of Math. Analysis 3, (2009), 1803 - 1819.
• Ahmad Al Omari and Takashi Noiri, Local closure functions in ideal topological spaces, Novi Sad J. Math. 43, (2013), 139-149.
• D. Jankovic and T.R. Hamlett, New topologies from old via ideals, Amer. Math. Monthly 97, (1990), 295 - 310.
• J. Dontchev, M. Ganster and D. Rose, Ideal resolvability, Topology Appl. 93(1), (1999), 1 - 16.
• K. Kuratowski, Topology, Academic Press, NewYork 1, (1966).
• M. Khan and T. Noiri, Semi local functions in ideal topological spaces, Journal of Advenced Research in Pure Mathematics (2010), 36-42.
• Nirmala Rebecca Paul, RgI-closed sets in ideal topological spaces, International Journal of Computer Applications 69(2013), no.4.
• N. Levine, Semi-open sets and semi-continuity in topological spaces, Am. Math. Mon. 70(1963), 36-41.
• O. Njastad, Remarks on topologies defined by local properties, Anh. Norske Vid. -Akad. Oslo (N.S.) 8(1966), 1-6.
• Sukalyan Mistry and Shyamapada Modak, $()^{*p}$ and $\psi_{p}$ operator, International Mathematical Forum 7(2012), no.2, 89-96.
• S. Modak and C. Bandyopadhyay, A note on $\psi$ - operator, Bull. Malyas. Math. Sci. Soc. 30(2007), no. (2), 43-48.
• T.R. Hamlett and D. Jankovic, Ideals in topological spaces and the set operator $\psi$, Bull. U.M.I., 7, (1990), no. 4-B, 863-874.
• R. Vaidyanathaswamy, The localization theory in set topology, Proc. Indian Acad. Sci 20, (1945), 51-61.


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# A unified approach for real and complex zeros of zero-dimensional ideals.

 Author(s):Ph. Rostalski, J.B. Lasserre, M. Laurent Conference/Journal:vol. The IMA Volumes in Mathematics and its Applications, accepted as a book chapter in: "Emerging Applications of Algebraic Geometry","The IMA Volumes in Mathematics and its Applications", Seth Sullivant and Mihai Putinar ed. Abstract:Abstract. In this paper we propose a unified methodology for computing the set VK(I) of complex (K = C) or real (K = R) roots of an ideal I \in R[x], assuming V_K(I) is finite. We show how moment matrices, defined in terms of a given set of generators of the ideal I, can be used to (numerically) find not only the real variety V_R(I), as shown in the authors’ previous work, but also the complex variety V_C(I), thus leading to a unified treatment of the algebraic and real algebraic problems. In contrast to the real algebraic version of the algorithm, the complex analogue only uses basic numerical linear algebra because it does not require positive semidefiniteness of the moment matrix and so avoids semidefinite programming techniques. The links between these algorithms and other numerical algebraic methods are outlined and their stopping criteria are also related. Year:2008 Type of Publication: Book Chapter Supervisor: File Download: Request a copy of this publication. (Uses JavaScript) % No recipe for automatically generating a BibTex entry for Book Chapter Permanent link

 © 1999-2014 by ETH Zurich | Webmaster | Tuesday, May 22, 2018

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# Can Jimmy get around the world?

Yet another Jimmy challenge, but this time... we expand into the third dimension.

Jimmy has conquered the platform, performed in the circus, fought inequality and even battled against monsters. In fact, he's gone around the whole world on his travels. However, we need to look at the big picture: Jimmy was successful in this world, on this planet. But can he do the same on, let's say, a different planet? Or, is something as simple as the shape and terrain enough to stop him?

## Input

In this challenge, the input will be a 'planet' - (roughly) an ASCII circle made out of = characters. The circle can be any size, and be shaped like the outputs in this challenge (except hollow), like the following:

    =====
=     =
=       =
=       =
=       =
=       =
=       =
=       =
=     =
=====

These circles are made with the Midpoint Circle algorithm, which computes the coordinates of each point of the discrete circle. You can find examples on how to implement this algorithm on this Wikipedia page. A slightly less technical much simpler and clearer explanation, with code examples in common languages, is also available on GeeksForGeeks' website. You can assume all circles are made using this algorithm. See the 'Notes' section for a little more info on this, if you'd like.

The planet may also be (horizontally) thicker or thinner in places, but will always be justified by spaces to be circular. There can also be holes (where the wall is infinitely thin), in which case justification obviously doesn't make sense, but you can require that as part of your input if you wish (see near bottom of this question). Two opposite walls will never both extend so much that one wall will touch the other.

The top/bottom of the circle count as individual sections, so they can also be made 0-width.

An example circle:

    ==  =
==    =
========
=     ===
=
=      ==
=====
==
=== ===
====

But here's the thing: there will also be a Jimmy in the image, somewhere. He might be on the circle, inside it, hanging off it, or floating around in space somewhere:

    ==  =
==    =
========
=     ===
=
= /o\  ==    // Jimmy is standing safely
=====
==
=== ===
====

or

    ==  =
==    =
========
=     ===
=
=      ==
=====
== /o\      // Jimmy is balancing between two platforms
=== ===
====

or

    ==  =
==    =
========
=     ===
=
=   /o\==      // Jimmy will fall (only one body part on the ground)
=====
==
=== ===
====

or

    ==  =
==    =
========      /o\     // Jimmy's very lost. :(
=     ===
=
=      ==
=====
==
=== ===
====

## Output

The output should be a truthy/falsey value representing whether Jimmy can safely make it around the circle (not just whether he is safe in his starting position).

To clarify, 'get around the circle' means to get back to exactly your starting position, as in, your legs and head both need to be above the same blocks as they were when you started. This does also mean you cannot shift blocks that are below your feet. Further, to 'get around' means you've visited every 'column' (with a width of one character each, i.e he's visited each point along the X axis) within the bounds of the starting, unaltered circle at least once. For instance, if the circle looks like:

    ==  =
==    =
========
=     ===
=
=      ==
=====
==
=== ===
====

at least one body part of Jimmy has to, at some point, touch each of these columns:

  vvvvvvvvv
==  =
==    =
========
=     ===
=
=      ==
=====
==
=== ===
====

To do this, Jimmy has to be able to move. He can move in the following ways:

• Move up to (can move less if you wish) up to one block vertically and two blocks across, but he cannot go end up inside any blocks in either direction. He can go through blocks though:
Start (a _cropped_ section of a circle, obviously not a valid input):

=
=
===/o\
===

Step 2:
=
= /o\
===
===
// Move up 1 and across two, and not colliding with anything in start or end position

• "Bump" the block directly beside Jimmy 1 horizontal layer, but:
• Jimmy has to be moving exclusively in that direction (e.g if he is trying to shift a block to the left, he has to be moving just left: not right or diagonally in any way).
• You can't bump blocks diagonally or vertically.
• You can't bump a block with another block immediately beside it (i.e you can only move one block at a time)
• Jimmy cannot move a block beyond the bounds of the current circle - see below
Start:

== =
=/o\
= ==

Step 2:

== =
=/o\     // Jimmy has moved himself, as well as the block beside him, to the left.
= ==

Regarding the last bumping rule:

== =. // Jimmy cannot bump this block to the left as it would
=/o\  // go out of the bounds of the original circle.
= == // (also, Jimmy is stuck in this case)

Obviously, Jimmy can at no point be in such a position that less than two body parts are touching the ground - otherwise, he will tip over.

Your challenge is to state whether there is any possible set of moves in which Jimmy can get around the circle either clockwise or anticlockwise. You can use any valid truthy/falsey values, and take input in any valid format, such as a string with \n (or even \\n, if your language interprets them as literal characters) separators, or as an array of lines.

A circle should return a falsey value if and only if in every possible combination of moves, Jimmy eventually either:

• gets stuck, i.e cannot make any legal move before getting around the circle
• falls (i.e the only valid move is one that results in an invalid position)

This is , so shortest answer in bytes wins.

## Notes

• If Jimmy starts in an invalid place, the answer is necessarily falsey. Such inputs do count as valid, but as no possible move can be made without falling, it will fail to condition 2.
• Outside the circle is not necessarily invalid, as long as his feet are touching the ground - e.g, this is valid:
     /o\
==  =
==    =
========
=     ===
=
=      ==
=====
==
=== ===
====

• You can assume the input will always be based on a valid circle, with no more than half (round down) of the wall segments being thickness-0.
• If you wish, you can define how you want inputs to be space-justified in the case that there are 0-width walls - e.g:
('@' is used to represent space)

=====
==@@@@=
====@@===
=@@@@@===
=@@@@@@@=
=@@@@@@==
=====@@@=  // This is pretty straightforward
==@@@@@@=
===@===
=====

==  =
==    =
========
=     ===
=
= /o\   ==
=====
==
=== ===
====

You can choose between:

==@@=
==@@@@=
========
=@@@@@===
@@@@@@@@=
=@/o\@@==
=====
==
===@===
====

or

==@@=
==@@@@=
========@
=@@@@@===
@@@@@@@@=
=@/o\@@==
=====@@@@
==@@@@@@@
===@===
====

• Additional spaces in the input may also be requested, to ensure equal line length:
@@==@@=@@
@==@@@@=@
========@
=@@@@@===
@@@@@@@@=
=@/o\@@==  // While this _is_ an option, it's likely less-than-ideal
=====@@@@  // as you have to manually work out/guess the dimensions of
==@@@@@@@  // the circle in the case of 0-width edges (I'm not even sure if
@===@===@  // this is always possible)
@@====@@@

vs

==@@=
==@@@@=
========@
=@@@@@===
@@@@@@@@=
=@/o\@@==
=====@@@@
==@@@@@@@
===@===
====

• Once again, remember that planets can not only vary in geology, but also radius! I've just arbitrarily used size-5 circles as they are not too big to fit on one line/make by hand, but the size can theoretically be anything between 2 and infinity. As to what the radius means, see this question for some examples, or look at a pseudocode writeup of the circle generation algorithm here.
• If you wish, you can take the radius as an additional argument/input.
• Default loopholes are forbidden!
• All your circles seem to be 9 wide but 10 high. – Jonathan Allan Aug 6 at 18:31
• @JonathanAllan I know - there's no way I know of to do it with 10 width and 10 height ̶w̶i̶t̶h̶o̶u̶t̶ ̶t̶w̶o̶ ̶h̶o̶r̶i̶z̶o̶n̶t̶a̶l̶ ̶c̶e̶l̶l̶s̶ ̶n̶e̶x̶t̶ ̶t̶o̶ ̶e̶a̶c̶h̶ ̶o̶t̶h̶e̶r̶.̶ ̶I̶'̶l̶l̶ ̶h̶a̶v̶e̶ ̶t̶o̶ ̶t̶h̶i̶n̶k̶ ̶a̶b̶o̶u̶t̶ ̶w̶h̶e̶t̶h̶e̶r̶ ̶I̶ ̶c̶o̶u̶l̶d̶ ̶a̶l̶l̶o̶w̶ ̶t̶h̶a̶t̶.̶ while keeping the circle symmetrical. It works with even numbered radii though. I'm simply following the convention in the Pie Charts post. – Geza Kerecsenyi Aug 6 at 19:01
• Why the unupvote? It's a hard challenge, sure, but nothing that's crazy/radical about it. It's just a combination of a number of different pre-existing formats. If you're going to blame me for the irregular circles, why does the Pie Chart question have 16 upvotes, when my circles are based off of theirs (it's literally the only source for ASCII circles I found on the internet)? – Geza Kerecsenyi Aug 10 at 19:57
• I have not downvoted, but revisiting it there is actually no specification of your input circle - either here or in the linked challenge. I'm also not sure what "get around the circle" actually means. – Jonathan Allan Aug 10 at 20:14
• @JonathanAllan I've added a clarification about what 'getting around' means - I realise this was unclear. I'll try to add a spec of what an ASCII circle is meant to look like, but I just assumed it was obvious, based on how the (very well received) pie chart question didn't need a description either. – Geza Kerecsenyi Aug 10 at 20:19


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# Trying to prove that the sequence: 3, 3, sin(1), 3, 3, sin(2), 3, 3, sin(3), 3, 3, sin(4), $\ldots$ does not converge to 3

$\textbf{Proof}$:

I need to show that $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$

Let $\epsilon = 2$, then if you take $n = 30$, we have that $a_{30} = sin(10) \approx -0.544 \ldots$

So we have that $\left | sin(10) - 3 \right | \ge 2$

Therefore I have showed that, $\exists \epsilon > 0$, such that, $\forall N \in \mathbb{N},\ \exists n \ge N$, such that, $\ \left | a_n - 3 \right | \ge \epsilon$, where $\epsilon = 2$ and $n = 30$.

I am wondering if this proof is right.

• You don't need to calculate (even approximately) values of sine. Just use the fact that $\sin x \le 1$ for all $x$. – njguliyev Oct 31 '13 at 21:02
• I will remove the approximations, they do look rather ugly. I was wondering if general argument I've made is correct though? – Michael Oct 31 '13 at 21:03
• Ok Andre, I see my mistake. – Michael Oct 31 '13 at 21:05
• You didn't make a general argument yet. You chose correct $\varepsilon$, but for any $N$ you must show that the inequality is true for some $n > N$. – njguliyev Oct 31 '13 at 21:06

You have to find $\varepsilon>0$ such that, for all $N\in\mathbb N$, there exists $n\geq N$ such that $|a_n-3|\geq \varepsilon$. What you did will only work for $N\leq 30$, but, if for example $N=31$, you have to find an $n\geq 31$ such that $|a_n-3|\geq\varepsilon$.
Instead, you could do the following: given $N\in\mathbb N$, there exists $n\geq N$ such that $a_n=\sin k$ for some $k\in\mathbb N$. So, for that $n$, $|a_n-3|=|\sin k-3|\geq\left||3|-|\sin k|\right|\geq 3-|\sin k|\geq 2$.
It suffices you show $\langle \sin n\rangle$ cannot converge to $3$. That is, if the sequence converged to $3$, then the subsequence $\sin 1,\sin 2,\ldots$ would converge to $3$. But for any $n$, we have that $\sin n \leqslant 1$. Thus, this is impossible. Note that if we replaced $3$ with $1$; things'd get much more delicate!


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# Power of signal in specific frequency in FM and AM modulation

My professor gave us the following problem:

Given a sinusoidal signal with $$f_m = 1KHz$$ which is put into an AM and FM modulation find the modulation index of FM and AM if you know that (amongst other things) that the amplitude of the spectrum at $$f_c = \pm 1000Hz$$ is same for both systems.

When solving it, he computed the output of AM as \begin{align} X_{AM}(f) &= \frac{A_c \mu}{2} \delta(f-f_c) + \frac{A_c \mu}{2} \delta(f+f_c) \\ &\phantom{\,}+ \frac{A_c\mu}{4} \delta(f+f_c + 1000)\\ &\phantom{\,} + \frac{A_c\mu}{4} \delta(f+f_c-1000) \frac{A_c\mu}{4} \delta(f-f_c+1000)\\ &\phantom{\,} + \frac{A_c\mu}{4} \delta(f-f_c-1000) \end{align}

and then said that the power at $$f_c \pm 1000Hz$$ is same, and finally conluded that for AM: $$P_1 = 2 \left( \frac{A_c \mu}{4} \right )^2$$.

I have two questions on the final part:

Firstly, why exactly is the power same for both? (I think it has to do with the amplitudes, but he doesn't actually say it, so I have to make sure)

Secondly, what is the formula used to compute the power for just one point? I suppose the 2 in front is because of the two frequencies ($$\pm 1000Hz$$) and the inside is the amplitude at these two points, but how is that derived exactly? It seems like to take advantage of the spectrum he uses Parseval's Theorem, but we have been taught that $$\int_{-\infty}^{\infty} |X(f)|^2 df$$ gives as the Energy, and not the Power. What am I missing? Thanks for any help

## 1 Answer

This is a common homework or quiz/test problem so I won’t bypass the learning experience by providing the full solution, but I will provide further details that will give intuitive insight into the conditions in which the FM and AM sideband levels for a sinusoidal modulation will be the same and how the modulation index for FM relates to the sideband level.

This is not applicable to general AM and FM modulations, so I will first clarify what types of AM and FM modulations are being considered, and then show the very intuitive way to see what the modulation index would be for the sideband levels to be the same.

To be clear, the AM referenced by the OP is specifically "Large Carrier AM Modulation" where the sidebands are significantly lower than the carrier which is not suppressed (in contrast to suppressed carrier AM modulations where the carrier is completely removed).

And for the FM case, the FM referenced by the OP is specifically "Narrowband FM Modulation". Narrowband FM Modulation is when the modulation index (which is the peak angle of the carrier as modulated by a sinusoidal modulation) is small enough such that there are only one pair of significant sidebands on the carrier, spaced at the modulation rate similar to the Large Carrier AM Modulation.

That said, observe the plots below comparing phasor diagrams in time for both cases with the resulting magnitude spectrums in frequency:

Large Carrier AM:

A normalized 1 KHz sinusoid in time has a Fourier Transform that results in two impulses at $$\pm 1$$ KHz each with a magnitude of $$1/2$$ consistent with Euler's formula:

$$\cos(2\pi 1000t) = \frac{1}{2}e^{j2\pi 1000 t} + \frac{1}{2}e^{-j2\pi 1000t}$$

From this we should get our first "Aha!" that each $$e^{j\omega t}$$ term is a spinning phasor in time time domain since $$Ke^{j\theta}$$ is the same as $$K\angle\theta$$. The Fourier Transform is just the magnitude and initial phase for each spinning phasor.

So for the AM case, the blue phasors shown are these components of the sinusoidal modulation, that when applied to Large Carrier AM as a modulation signal, end up adding to the carrier phasor in phase. The above diagrams are the "baseband equivalent" of the modulated signal such that we can "freeze" the carrier as shown (consider if we had a strobe light on it flashing at the carrier frequency), such that we see just the sidebands rotating relative to the carrier. With the carrier added to the sidebands in phase, the net result will always be on the real axis, resulting in a change in the amplitude only (the phase will always be zero). In this case after summing everything we would see in the time domain on the complex plot shown, that the resulting phasor would go stay on the real axis and change in amplitude sinusoidally but due to being Large Carrier AM (small sidebands) the amplitude will always stay positive. If you can visualize all that, then the FM case will make complete sense.

Narrow Band FM:

With narrow band FM the difference is the two phasors from the sinusoidal modulation are added in quadrature, as shown in the lower part of the diagram.

With FM in general with the carrier depicted as a phasor, we would see the carrier phase change only and always remain at a constant amplitude (frequency is the derivative of phase, so if the phase is changing sinusoidally, the frequency is also changing sinusoidally). So we would see the phasor changing its phase only, sinusoidally and thus staying on a constant radius. We note that the sum of the carrier with the two sidebands if the sidebands are large, will deviate from a constant radius (AM will be induced if there are only two sidebands if the sidebands are large enough). But if the sidebands are small (resulting in a small angle), the deviation in amplitude will be insignificant, and this is the condition for Narrowband FM.

Under this case specifically there will only be one pair of significant sidebands, consistent with the drawings given above. Due to small angle approximations ($$\sin(\theta) \approx \theta$$) we can see that the magnitude for each of the phasors (which we saw is directly related to their magnitude in the frequency domain) will be directly related to the peak angle (when the two phasors both sum in quadrature to the carrier, their sum will be equal to the peak angle $$\beta$$ if we normalized the carrier). For FM modulation that peak angle $$\beta$$ IS the modulation index! So from this we see the relationship between the modulation index and the sideband level for Narrowband FM, and we see the equivalence to Large Carrier AM.

Note that for general FM with large angles, many sidebands are needed such that when they all add together the sum will stay on the constant radius circle as depicted in the graphic below. In this case the carrier phasor and all the sideband phasors necessary to stay on the constant radius circle are predicted with Bessel functions.

• Thank you for the detailed answer. I now understand it a bit better. However, my questions remain, as I also care for the way he gets there, and for that I had the questions I stated above. When saying that this is not applicable, you mean that the power thing is wrong to be assumed as true? thanks again =) May 1 at 7:46
• I wanted to give you an intuitive insight that would allow you to work out the detailed answer - We tend to not provide the complete solutions since it is a typical homework/quiz problem. If you follow my explanation carefully the answer will be quite obvious; try to work it out from that and I will help you where you are stuck. It may help you to know with modulation that it really doesn’t matter where the carrier is when you work with baseband equivalent signals the “carrier” is DC which is what these diagrams depict- what happens there as a complex signal with mag and phase is what happens May 1 at 13:07
• … at any carrier frequency. So please try to work it out and show your progress by updating your question which will highlight where you are stuck rather than asking for what the complete solution is :) May 1 at 13:08


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# Control system in Matlab with RSA encryption

I want to implement an encrypted controller in Matlab using RSA. I am trying to replicate what is written in this scheme from 'Cyber-Security Enhancement of Networked Control Systems Using Homomorphic Encryption' written by Kiminao Kogiso and Takahiro Fujita:

The input of the controller has to be the output of the plant, and the input of the plant has to be the output of the controller. I think the problem happens when I try to decrypt, but I am not sure.

I am using the state space representation of a system and the following code:

A_p = [9.9998*(10)^-1 1.9700*(10)^-2; -1.970*(10)^-2 9.7025*(10)^-1];
B_p = [9.9900*(10)^-5; 9.8508*(10)^-3];
C_p = [1 0];
D_p = 0;

X0 = [1;0]; %initial conditions for the plant

t = [0:0.01:5];  %sampling time

u = ones(1,length(t));        %encrypted input
y_p = zeros(1, length(t));    %output of the plant
x_p = zeros(2, length(t));    %state of the plant
x_c = zeros(2, length(t));    %state of the controller

x_p(:, 1) = X0;               %assign the initial values to the satet of the
%plant

%parameters of the controller

A_c = [1 0.0063; 0 0.3678];
B_c = [0; 0.0063];
C_c = [10 -99.90];
D_c = 3;

end_t = length(t);

N = 94399927;    % N=p*q
e = 11;          %public key
d = 85800451;    %private key

%encrypted parameters for the controller

A_enc = mod(A_c.^e, N);
B_enc = mod(B_c.^e, N);
C_enc = mod(C_c.^e, N);
D_enc = mod(D_c.^e, N);

xc_enc = zeros(2,length(t));   %x_c encrypted
y_enc = zeros(1,length(t));    %y_c encrypted

u_dec = ones(1, length(t));    %u decripted

y_quant = zeros(1, length(t));

for ind=1:end_t-1

%plant

x_p(:, ind+1) = A_p*x_p(: ,ind) + B_p*u_dec(ind);
y_p(ind) = C_p*x_p(: ,ind);

%encryption of x_c and y_p

y_quant(ind) = round(y_p(ind));  %quantization

y_enc(ind) = modpow(y_quant(ind), e, N);

%controller with encrypted parameter
x_c(:, ind+1) = A_enc*x_c(:, ind) - B_enc*y_enc(ind);
u(ind+1) = C_enc*x_c(:, ind) - D_enc*y_enc(ind);

%decrypting u
u_dec(ind+1) = modpow(u(ind+1),d,N);

function result = modpow(base,exp,m)
result = 1;
while (exp > 0)
if bitand(exp,1) > 0
result = mod((result * base),m);
end
exp = bitshift(exp,-1);
base = mod(base^2,m);
end
end

end

figure
plot(t,y_enc);

figure
plot(t,u);

Can somebody help me? Thank's in advance.

EDIT:

sorry, I had to delete some pictures because I can't post all of them because I don't have enough reputation.

Now I am just trying to simply encrypt the output of the plant and then encrypt it before it enters in the controller, and then I encrypt the output of the controller and before it enters in the plant I decrypt it. The idea is the following:

The problem is that if I don't use encryption the outputs I get when I plot

hold on plot(t, y_p, 'r') figure plot(t, u_dec)

are

but when i use encryption, the outputs are

my code is:

A_p = [9.9998*(10)^-1 1.9700*(10)^-2; -1.970*(10)^-2 9.7025*(10)^-1];
B_p = [9.9900*(10)^-5; 9.8508*(10)^-3];
C_p = [1 0];
D_p = 0;

X0 = [1;0];

t = [0:0.01:5];

u = zeros(1,length(t));
y_p = zeros(1, length(t));
x_p = zeros(2, length(t));
x_c = zeros(2, length(t));

x_p(:, 1) = X0;

A_c = [1 0.0063; 0 0.3678];
B_c = [0; 0.0063];
C_c = [10 -99.90];
D_c = 3;

end_t = length(t);

N = 94399927;
e = 11;
d = 85800451;
xc_enc = zeros(2,length(t));
y_enc = zeros(1,length(t));

u_dec = ones(1, length(t));

xc_dec = zeros(2, length(t));

x_dec = zeros(2,length(t));
y_dec = zeros(1, length(t));

u_enc = zeros(1,length(t));

y_quant = zeros(1, length(t));

u_quant = zeros(1, length(t));

for ind=1:end_t-1

x_p(:, ind+1) = A_p*x_p(: ,ind) + B_p*u_dec(ind);
y_p(ind) = C_p*x_p(: ,ind);

y_quant(ind) = round(y_p(ind));

y_enc(ind) = modpow(y_quant(ind),e,N);

y_dec(ind) = modpow(y_enc(ind),d,N);

x_c(:, ind+1) = A_c*x_c(:, ind) - B_c*y_dec(ind);
u(ind+1) = C_c*x_c(:, ind) - D_c*y_dec(ind);

u_quant(ind+1) = round(u(ind+1));        %quantization of u(ind+1)

u_enc(ind+1) = modpow(u_quant(ind+1),e,N);
u_dec(ind+1) = modpow(u_enc(ind+1),d,N);

disp(u(ind +1) + " " + u_dec(ind+1))

end

hold on
plot(t, y_p, 'r')
figure
plot(t, u_dec)

function result = modpow(base,exp,m)
result = 1;
while (exp > 0)
if bitand(exp,1) > 0
result = mod((result * base),m);
end
exp = bitshift(exp,-1);
base = mod(base^2,m);
end
end

function y = powermod_matrix(a,z,n)
% This function calculates y = a^z mod n
%If a is a matrix, it calculates a(j,k)^z mod for every element in a
[ax,ay]=size(a);

% If a is negative, put it back to between 0 and n-1
a=mod(a,n);

% Take care of any cases where the exponent is negative
if (z<0)
z=-z;
for j=1:ax
for k=1:ay
a(j,k)=invmodn(a(j,k),n);
end
end
end

for j=1:ax
for k=1:ay
x=1;
a1=a(j,k);
z1=z;
while (z1 ~= 0)
while (mod(z1,2) ==0)
z1=(z1/2);
a1=mod((a1*a1), n);
end  %end while
z1=z1-1;
x=x*a1;
x=mod(x,n);
end
y(j,k)=x;
end %end for k
end %end for j
end

Shouldn't I get almost the same result as the ones without encryption? I think that something is wrong with the input u and how I initialize and increment it, but I don' t know how to fix this. Can somebody help me? Thank's in advance.

EDIT:

moreover, if i add the following line of code:

disp(y_p(ind) + " " + y_dec(ind))

i can see that y_p and y_dec, which should be the same, are different at each iteration.

EDIT:

the problem is that even if i add a quantization ,and so i get integers, i get the same output.

disp(u(ind +1) + " " + u_dec(ind+1))

after the decryption of u_enc, so after u_dec, i get :

so u(ind+1) and u_dec(ind+1) are different numbers. And another fact that leaves me confused is that if i quantize u(ind+1), this result doesn't become an integer.

EDIT:

I think that the problem of this is that u(ind+1) becomes a negative number and too big. Infact, if i for example try to do:

x = -19495615862231.37;

z = round(x);

x_enc = modpow(x,11,94399927);

x_dec = modpow(x_enc,85800451,94399927)

as result, after the decryption i get 31779346, and not the original number. I think modpow calculates the modular exponential for large numbers, but it looks like it has a limit.

• Hi Flavio, I would suggest editing the question to first explain what you are trying, then the code (with maybe some variable clarification) and then clarify what you observe that "doesn't work properly" – N. Staub Dec 4 '18 at 18:30
• Also, there is no control action in your controller and your first control input is arbitrarily equal to 1. I also not that you fill the variable xc_enc(:, ind+1) with two different expressions, one after the other. – N. Staub Dec 4 '18 at 18:36
• you are still updating twice the value xc_enc(:, ind+1) which seems suspicious. I would suggest you to try to see if without encryption your "controller" really controls something. – N. Staub Dec 5 '18 at 8:55
• without encryption i get the correct output i should get if i don't use RSA. The problems come when i encrypt and i can't understand why. – J.D. Dec 5 '18 at 8:58
• You simulated your plant with u_dec but only update its value at the end of the loop (so the plant is always simulated with u_dec[ind]=1). y_quant is never defined. xc_enc  is never used. – N. Staub Dec 5 '18 at 9:11

# :EDIT:

This is probably the root of the trouble you're having. Interesting that the paper didn't mention anything about this. I'm not sure how they arrived at the encrypted control parameters but, as I mentioned below, it's not my field so I haven't "digested" the paper. In any event, RSA encryption doesn't work with non-integer or negative integer numbers so you can't pass fractional numbers back and forth and you can't pass negative numbers back and forth.

I've removed the original answer here, but it's in the edit history if anyone is curious.

Consider the following example. I used much smaller numbers to reduce my own search times, and I'm sure there are better ways of finding these numbers but, as I mentioned previously, crypto is not my field. Here's some checks:

p = 7
q = 11
N = p*q
lambda = lcm(p-1,q-1)
e = 17
d = 1;
while true
if(mod(d*e,lambda)==1)
break;
else
d = d+1;
end
end

>>d

d =

23

So now I've got e = 17, d = 23, and N = 77. Now, check that it works:

>> modpow(29,e,N)

ans =

50

>> modpow(50,d,N)

ans =

29

Okay, seems to work fine. Now, try it with a negative number, then see if there's a positive number that gives the same encrypted result:

i=1;
while true
if(modpow(i,e,N)==modpow(-3,e,N))
break;
else
i=i+1;
end
end

>>i

i =

74

Try it again, with a different negative number:

i=1;
while true
if(modpow(i,e,N)==modpow(-29,e,N))
break;
else
i=i+1;
end
end

>>i

i =

48

So, as you can see, a negative integer can wind up giving you the same encryption result as (a different) positive integer.

Since N is always positive (because p and q are prime and thus positive), Matlab documentation states that the modulus operation will always return a positive result. This again confirms that modpow of some negative number, with N>0, will always result in some positive result which could be duplicated by a positive number.

It also means that modpow(~, d, N) will never return a negative number.

# Solution

As I mentioned at the top, I'm surprised this wasn't mentioned in the paper you linked, but my approach would be to bias your numbers by some "sufficiently large" positive number and un-bias them before use. Also, since decimal numbers can't be used, you could "inflate" your numbers by multiplying by some scale factor before rounding, then "deflate" the numbers before use by dividing by the same scale factor. Consider the following edits to your code (the powermod_matrix function wasn't being used, so I removed it for brevity):

scaleFactor = 1000; % <-- New
biasValue = 100; % <-- New

A_p = [9.9998*(10)^-1 1.9700*(10)^-2; -1.970*(10)^-2 9.7025*(10)^-1];
B_p = [9.9900*(10)^-5; 9.8508*(10)^-3];
C_p = [1 0];
D_p = 0;

X0 = [1;0];

t = [0:0.01:5];

u = zeros(1,length(t));
y_p = zeros(1, length(t));
x_p = zeros(2, length(t));
x_c = zeros(2, length(t));

x_p(:, 1) = X0;

A_c = [1 0.0063; 0 0.3678];
B_c = [0; 0.0063];
C_c = [10 -99.90];
D_c = 3;

end_t = length(t);

N = 94399927;
e = 11;
d = 85800451;
xc_enc = zeros(2,length(t));
y_enc = zeros(1,length(t));

u_dec = ones(1, length(t));

xc_dec = zeros(2, length(t));

x_dec = zeros(2,length(t));
y_dec = zeros(1, length(t));

u_enc = zeros(1,length(t));

y_quant = zeros(1, length(t));

u_quant = zeros(1, length(t));

for ind=1:end_t-1

x_p(:, ind+1) = A_p*x_p(: ,ind) + B_p*u_dec(ind);
y_p(ind) = C_p*x_p(: ,ind);

%y_quant(ind) = round(y_p(ind));
y_quant(ind) = round(scaleFactor*(y_p(ind) + biasValue)); % <-- New

y_enc(ind) = modpow(y_quant(ind),e,N);

y_dec(ind) = modpow(y_enc(ind),d,N);

y_dec(ind) = y_dec(ind)/scaleFactor - biasValue; % <-- New

x_c(:, ind+1) = A_c*x_c(:, ind) - B_c*y_dec(ind);
u(ind+1) = C_c*x_c(:, ind) - D_c*y_dec(ind);

%u_quant(ind+1) = round(u(ind+1));        %quantization of u(ind+1)
u_quant(ind+1) = round(scaleFactor*(u(ind+1) + biasValue)); % <-- New

u_enc(ind+1) = modpow(u_quant(ind+1),e,N);
u_dec(ind+1) = modpow(u_enc(ind+1),d,N);

u_dec(ind+1) = u_dec(ind+1)/scaleFactor - biasValue; % <-- New

disp(u(ind +1) + " " + u_dec(ind+1))

end

hold on
plot(t, y_p, 'r')
figure
plot(t, u_dec)

function result = modpow(base,exp,m)
result = 1;
while (exp > 0)
if bitand(exp,1) > 0
result = mod((result * base),m);
end
exp = bitshift(exp,-1);
base = mod(base^2,m);
end
end

I commented all the lines (six of them) with % <-- New to highlight the changes; again all I did was shift your values by +100 (biasValue) and then multiply by 1000 (scaleFactor) before rounding, and then doing the inverse after decrypting. Seems like that's got your controller working!

• Thank's for your answer, i have edited the post with the name of the paper i am trying to replicate and at the beginning of the post there is a link to it. – J.D. Dec 6 '18 at 8:03


text:
# The Infinite square well

I am studying Quantum Mechanics by David J. Griffiths. I am kind of stuck on the part of "The infinite square well" where griffith says

"negative solutions give nothing new, since $\sin(-\omega) = - \sin \omega$ and we can absorb the minus sign into $A$."

To be more specific, he uses boundary conditions on the solution $y(x) = A \sin(kx) + B \cos(kx)$, to show that $B$ is $0$ and $K$ can only take values $\frac{n\pi}{a}$ where $n$ can take values $1,2,3,4$...

My question is: Since $A$ is $\sqrt{\frac{2}{a}}$, how can you "absorb" minus sign into it.

Ps: i couldn't find symbols for psi and theta, so i have used $y$ and $\omega$ in their stead.

• For formatting math, use MathJax Apr 14 '18 at 16:29
• It means that for negative solutions you have a different value of $A$. But that "new" solution is identical to the original solution. Apr 14 '18 at 16:31
• Thanks garyp. I think i understand it now. Since, A^2 = 2/a, it will give 2 values for A, -√(2/a) and +√(2/a). Apr 14 '18 at 17:03
• not sure what you mean by boundary conditions on the solution $y(x)=A\sin(kx)+B\sin(kx)$. Is this not just $(A+B)\sin(kx)$? Apr 14 '18 at 17:13
• Ow i am so sorry. It should be Bcos(kx). I will edit it right away. Thanks for pointing it out. Apr 14 '18 at 17:22

Griffiths finds that the nontrivial solutions of the stationary Schroedinger equation in the quantum well with infinite barriers at $x=0$ and $x=a$ (where $\psi =0$) are $$\psi(x)=A\sin(k_nx) \tag 1$$ with $$k_n=±n\frac {\pi}{a} \tag 2$$ where $n=1,2,...$ However only the solutions with positive signs for $k_n$ in (2) have to be considered because the solutions with negative $k_n$ are equivalent to the ones with positive sign as the negative sign for $k_n$ in eq. (1) can be taken out of the argument of the sinus and considered to be part of a new arbitrary constant $A'$ so that the solution is the same as the solution with positive $k_n$.
The variable $k,$ which later in the book will be interpreted as momentum, can be either positive or negative. However, $A\sin(\pm kx)$ is the same as $\pm A\sin(kx)$, which is the same as just $A\sin(kx)$ as long as we don't make any restrictions on $A$'s sign.


text:
## October 23, 2011

### 5.4 Pressure

Recall and use the relationship between pressure, force and area:
NB: 1N/m2 = 1Pa

Using the Formula

Calculate the pressure generated by an ordinary shoe heel (person of mass 40kg, heel 5cm x 5cm), an elephant (of mass 500kg, foot of 20cm diameter) and a high-heeled shoe (person of mass 40kg, heel area 0.5cm2). Which ones will damage a wooden floor that starts to yield at a pressure of 4000 kPa?

1. Ordinary shoe heel

2. Elephant

$A=0.5cm^{2}=0.00005m^{2}$
$p=400/0.00005$
$p=8,000,000Pa$
$p=8,000kPa$
$p=8MPa$

Clearly one can see that the high-heeled show will do the most damage to the floor.


text:
# [Video] How I do my U perm

#### yurivish

##### Member
I'm not that fast at most things, but my current specialty is the counterclockwise U perm.

It's the same alg as Erik uses, but with different execution. The annoying thing about holding it my way is that I can't do U' turns without regripping, so when I do a clockwise U perm I end up cycling the pieces twice counterclockwise, which is noticeably slower; I can get high 1.xes on clockwise U perms.

I don't have Stackmat so I can't time those three attempts there, but I'm pretty sure they were all comfortably sub-1. I can get sub-1 pretty easily, and my best is .80.

I hope this helps/converts someone with/to the MU U perms!

Edit: Tip - Do the M2 U M as one motion, and the U2 M' U M2 as another. That should get you to around 1.2. At some point, I realized I had a pause between the two phases, and trying to remove it is what brought me down to below 1.

Last edited by a moderator:

#### fcwy1

##### Member
fast but i think it is around 1.02sec. get a stackmat.


text:
• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The terminal zeros of a number are the zeros to the right of tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 0 member replies ### GMAT/MBA Expert ## The terminal zeros of a number are the zeros to the right of ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [Math Revolution GMAT math practice question] The terminal zeros of a number are the zeros to the right of its last nonzero digit. For example, 30,500 has two terminal zeros because there are two zeros to the right of its last nonzero digit, 5. How many terminal zeros does n! have? 1) n^2 - 15n + 50 < 0 2) n > 5 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$99 for 3 month Online Course
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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

The terminal zeros of an integer are the zeros to the right of its last nonzero digit. For example, 30,500 has two terminal zeros because there are two zeros to the right of its last nonzero digit, 5. How many terminal zeros does n! have?

1) n^2 - 15n + 50 < 0
2) n > 5
$?\,\,\,:\,\,\,{\text{number}}\,\,{\text{of}}\,\,{\text{terminal}}\,\,{\text{zeros}}\,\,{\text{of}}\,\,n\,{\text{!}}$
$\left( 1 \right)\,\,\,{n^2} - 15n + 50 < 0\,\,\,\, \Leftrightarrow \,\,\,\,5 < n < 10$
$\left. \begin{gathered} n! = 6! = 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{6!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{6!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ n! = 7! = 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{7!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{7!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ n! = 8! = 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{8!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{8!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\, \hfill \\ n! = 9! = 9 \cdot 8 \cdot 7 \cdot 6 \cdot \boxed5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{9!}}{{10}} = \operatorname{int} \,\,\,\,\,{\text{but}}\,\,\,\,\,\frac{{9!}}{{{{10}^2}}} \ne \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1\,\, \hfill \\ \end{gathered} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1$
$\left( 2 \right)\,\,n > 5\,\,\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,n = 6\,\,\,\, \Rightarrow \,\,\,? = 1\,\, \hfill \\ \,{\text{Take}}\,\,n = 10\,\, \Rightarrow \,\,{\text{?}}\,\,{\text{ = }}\,\,{\text{2}}\,\, \hfill \\ \end{gathered} \right.$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Consider 1)
n^2 - 15n + 60 < 0
=> (n-5)(n-10) < 0
=> 5 < n < 10

The number of terminal zeros of a number is determined by the number of 5s in its prime factorization.

The integers satisfying 5 < n < 10 are 6, 7, 8 and 9. We count the 5s in the prime factorizations of 6!, 7!, 8! and 9!:
6! has one 5 in its prime factorization.
7! has one 5 in its prime factorization.
8! has one 5 in its prime factorization.
9! has one 5 in its prime factorization.

Thus, for 5 < n < 10, n! has one terminal zero.
As it gives us a unique answer, condition 1) is sufficient.

Condition 2)

If n = 6, then 6 > 5 and 6! = 720 has one terminal 0.
If n = 10, then 10 > 5 and 10! = 3,628,800 has two terminal 0s.

Condition 2) is not sufficient since it does not give a unique solution.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

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text:
# Search for four-top-quark production in the single-lepton and opposite-sign dilepton final states in proton-proton collisions at $\sqrt{s}=13$ TeV with the ATLAS detector.

Faraj M.

Comunicazione
I - Fisica nucleare e subnucleare
Aula Palazzo dell'Emiciclo - Sala Ipogea - Mercoledì 25 h 16:30 - 19:00

A search for the production of $t\bar{t}t\bar{t}$ using proton-proton collisions data at center of mass energy $\sqrt{s} = 13$ TeV with an integrated luminosity of 36 fb${}^{-1}$ recorded by the ATLAS experiment at LHC is presented. The considered final states are events with multiple jets, $b$-jets and either one lepton or two opposite sign leptons with high-transverse momentum. The result is combined with the previous same-sign dilepton and multilepton searches carried out by ATLAS. No significant excess above the Standard Model expectation is observed.

Società Italiana di Fisica - Via Saragozza 12 40123 Bologna P.IVA 00308310374 | credits


text:
# Step Change Detection

I am using a non linear least squares method to fit an analytical function to some experimental data. I have to provide some initial guess values to the algorithm, so I am trying to figure out how to do this automatically (rather than by eye, which is what I have been doing).

This is some simulated data, created by adding normally distributed random noise to the analytical function

I am trying reliably detect the position of this step change in the data. I have had some limited success by calculating the mean variance in the data points and looking for points in the data that differ significantly from this value, but this approach seems very limited by the signal to noise ratio.

I am hoping for some direction on what I need to look into to solve my problem, as I don't know much statistics at all.

Thank you!

-Edit paste bin link to xy data

http://pastebin.com/QTawFex3

• Maybe it would be useful to consider change-point detection techniques. Circular binary segmentation might be just what you need. – Mur1lo Aug 15 '16 at 18:22
• Chen and Liu 1993 describe a method of detecting level shift in a time series. Their algorithm is implemented in the tsoutliers R package. – pionpi_ Aug 16 '16 at 13:34
• A new powerful change detection technique is impulse-indicator saturation and step-indicator saturation in the works of Castle, Doornik, Hendry & Co. (Google these keywords to find more.) – Richard Hardy Aug 16 '16 at 18:33
• The best reply I could find on that is there, works very well in my case and looking at your data it should work as well. stackoverflow.com/questions/48000663/… – francois-xavier sireta Oct 4 '18 at 16:19

There may be some more sophisticated methods for this but here is my first thought.

You basically want to take the derivative of the function and find where it is the largest. Numerically, you can just take the difference between data points and find which two points have the biggest difference. Then the midpoint of the x-values for these two points is your location of biggest change.

This simple method is susceptible to the noise. So you can first filter the data using a filter that does not shift the data to the right or left. If you use a simple FIR filter, then filter front to back and then filter the result from back to front. The result is a doubly filtered and NON shifted data set. Then follow the procedure above to find the point with the largest difference between values.

You can also use more sophisticated numerical differential calculations that use more then the difference of two points.

• Thank you for the answer. I was hung up on the fact that this was discrete data, and didn't think to just differentiate it! I'm going to look into using the FIR filter on some of my smaller peaks, to see if this works – jm22b Aug 15 '16 at 14:13
• I have implemented your idea and it works like a charm, thanks a lot! – jm22b Aug 16 '16 at 9:40
• That is great to hear. Thank you for letting me know. – jason Aug 16 '16 at 21:10

Your data visually suggests an asymptotic (gradual) change to the new level. Time series methods can often be used to detect these kinds of structures even if the data is not time series. Please post your data and I may be able to demonstrate this with "toys" at my disposal. If your data is time series then as @jason reflected one needs to deal effectively with the noise model to correctly "see" the structure.

EDITED UPON RECEIPT OF DATA:

Modelling is often an iterative approach with interim steps providing valuable clues to a useful model. I took your data and introduced it to AUTOBOX (one of my toys which I have helped develop). An initial graph strongly suggested a longitudinal (chronological) data set where the X series is reported at fixed intervals. AUTOBOX automatically suggested a standard ARIMA model (with Intervention Detection) replacing the non-stationary X with a differencing operator. Here is the actual/fit/forecast graph and the suggested model.

Upon examination another possible model incorporating a lag structure for an indicator variable suggested itself. I introduced a Pulse at time period 76 (a Dynamic Predictor expressly allowing up to a possible lag effect of 50 periods) (the beginning of the transition) to deal with the relationship between the original Y and the user-suggested X in order to more fully investigate the effect of X than accept the total setting-aside of X.

Following is the actual-fit-forecast graph for that approach and the identified robust transfer function model . with residual plot and residual acf here

The final model captures the dynamics in certain lags of the Dynamic Predictor and a few pulses and a reasonable memory structure.

Even the most powerful analysis packages often need some guidance when dealing with complex real world data sets like this one as nothing compares to the creative human mind.

• I am just curious ...1) What was you analytic function and 2) what was the variance of the error series (your normally distributed random noise) : AUTOBOX delivered : Variance =SOS/(n) .402675E-04 Adjusted Variance =SOS/(n-m) .443625E-04 Standard Deviation RMSE =SQRT(Adj Var) .666052E-02 – IrishStat Aug 16 '16 at 10:23
• My analytical function is $T(\lambda) = C_1 + C_2 [\mathrm{erfc(\frac{\lambda_0 - \lambda}{\sqrt{2} \sigma})} - \mathrm{Exp}(\frac{\lambda_0 - \lambda}{\tau} + \frac{\sigma^2}{2 \tau^2}) * erfc(\frac{\lambda_0 - \lambda}{\sqrt{2} \sigma} + \frac{\sigma}{\sqrt{2} \tau})]$ Where, $C_1, C_2, \lambda_0, \sigma, \tau$ are parameters to be fitted The image of the plot in my original post was simply this function + normally distributed random noise with a stdev of 0.1 (I think). The data I shared with you was real data, and I haven't the foggiest what it's properties are! – jm22b Aug 16 '16 at 10:30

One technique is to test all values of the x variable for the standard deviation of the data before and after it. For a true step function, the sum of those two will be minimal at the step location, and the minimum should be a good starting parameter for your nonlinear function.

Here is a plot of your original data (black), the standard deviation before x (blue), after x (red) and the sum of the last two (green).

I recognize that this question is old. But I wanted to throw another method out there. Canny wrote a paper (A Computational Approach to Edge Detection) where he solved this problem in the two dimensional case for edge detection in images. You can read the paper if you like, but to cut to the chase, you can get a very good approximation to the change point by doing the following:

1. Perform the convolution of your signal with the derivative of the Gaussian.

$$f(x) = \frac{-x}{\chi^2} e^{-\frac{x^2}{\sigma^2}}$$

where $\chi$ is a scaling factor.

1. The peak of the response is where the change point occurs.

In my experience with this method, I have found that choosing the correct scaling parameters is difficult. But there may be more work on this that I am not aware of.

I have summarized Canny's paper and provided an example here.

• Hi, I appreciate the response and interesting method presented. Interesting blog post too! The project that generated this question involved some image processing, so I had read a bit about canny edge detection then. – jm22b Mar 24 '17 at 12:56

You may investigate Wavelet tranformed time-series using short period types haar/db4. I have no pointers but just some search terms, try 'wavelet change point detection'.

There are several R packages on Wavelets, see time-series task view: https://cran.r-project.org/web/views/TimeSeries.html

For theory look for Mallat et Hwang paper: "Singularity detection and processing with wavelets"

See related answer: Application of wavelets to time-series-based anomaly detection algorithms


text:
1. ## Greatest Common Divisor

Which of the following CANNOT be the greatest common divisor of the two positive integers x and y?

A. 1
B. x
C. y
D. x - y
E. x + y

I do not understand exactly what this question is asking?
The answer is choice E. I do not understand why E is the answer. Can someone state this question another way?

2. ## Re: Greatest Common Divisor

Originally Posted by harpazo
Which of the following CANNOT be the greatest common divisor of the two positive integers x and y?
A. 1
B. x
C. y
D. x - y
E. x + y
I do not understand exactly what this question is asking?
The answer is choice E. I do not understand why E is the answer. Can someone state this question another way?
To find the $GCF(x,y)$ write down the prime factorization of each. Then list the common factors with the least power present.
EX: $x=2^5\cdot3^5\cdot 5^3\cdot 7^4\cdot 11^5~\& ~y=2^3\cdot 7^5\cdot 11^8\cdot 13^5\cdot$ $GCF(x,y)=2^3\cdot7^4\cdot 11^5$

NOTE $GCD(x,y)\le\min\{x,y\}$

3. ## Re: Greatest Common Divisor

It's much simpler than that. Because x and y are "two positive integers", x+ y is larger than either x or y so cannot be a divisor of either one, much less a "common" divisor!

4. ## Re: Greatest Common Divisor

Originally Posted by Plato
To find the $GCF(x,y)$ write down the prime factorization of each. Then list the common factors with the least power present.
EX: $x=2^5\cdot3^5\cdot 5^3\cdot 7^4\cdot 11^5~\& ~y=2^3\cdot 7^5\cdot 11^8\cdot 13^5\cdot$ $GCF(x,y)=2^3\cdot7^4\cdot 11^5$

NOTE $GCD(x,y)\le\min\{x,y\}$


text:
## Monday, January 28, 2008

### Loose Change

This is an incredible documentary. I would request you watch it and then see if your suspicions still hold true. Mine never have changed because I know my system of government. This is a capitalist country and capital gains have been won.

### Pentagon Strike

Out of all of the vids I like this one the best. It was my first.

I can't embed it so you will have to click on the title above if you want to see it. It explains all of the cameras that the plane would have had to pass. Even if they photoshopped the plane in by now it would have been better than leaving it out of public arena. It wasn't even released after the 9/11 commission report.

## Saturday, January 26, 2008

### 9/11 falsehoods.

I said that this was a set up from day 1. Flying planes into buildings to start a war really isn't that big of a motivator. However flying planes into buildings to effectively take complete control of the world's most powerful nation is. This is exactly what the patriot act does for us.

### 9/11 It is all a fake

Regardless of what you may think of this General "Expert" what he says is correct there is no way a plane hit that building.

### United 93 was a scam.

I've always thought from the beginning that 9/11 was a farce of smoke and mirrors. Both the pentagon footage and the flight 93 footage confirm my suspicions. There is no way a 100 ft plane is going to leave a 20' x 10' hole in the ground. That is scientifically impossible. Period. It was all a ploy for the patriot act which strips Americans of all of their rights and gives complete power to the government IF anything else were to happen.

## Saturday, January 19, 2008

### Synthesis of Hydrogen technology HELP!!!

I know I haven't posted in weeks but I have been beating my head over this hydrogen thing. Basically I want to get something for next to nothing. Hydrogen is the almost easiest way to do this. but it naturally isn't turning out so easy. I think personally a lot of the problem is that there are a lot of people looking at pieces of the puzzle from very different angles and not synthesizing all known facts into a nice little bundle and coming out with the solution. I am by no means a genius all I want is to power my house and my car off of some hydrogen.

So the first thing I do which was a giant pain in the buttocks is figure out exactly how much hydrogen I would need for an Internal Combustion Engine vehicle ICE for short. I ran around trying to get the numbers which was very frustrating as I like many people hate math, but I am coming to learn that I need it, especially when trying to figure out how to snatch up some stinking hydrogen. So I start with my car:

2001 Nissan Quest Performance

-
3,275 cc 3.3 liters 6 V engine with 91.5 mm bore, 83 mm stroke, 9 compression ratio, overhead cam and two valves per cylinder
-
-
Fuel economy EPA highway (l/100km): 10.2
-
Multi-point injection fuel system
-
Main 76 liter unleaded fuel tank
-
Power: SAE and 127 kW , 170 HP @ 4,800 rpm; 200 ft lb , 271 Nm @ 2,800 rpm

Then I try to go through all that math and figure out how much fuel this thing burns a minute. So I start guestimating. I first figure that I run through a tank at my normal driving speed on long trips to be about 80mph or 130 km/h for you Euros. I figure I probably leave a liter or two in the tank before I underestimate and have to get out and walk with my gas can. So let's guess 75 liters. I can travel on average 310 miles or roughly 500km. So the math starts 500km / 130 km/h = 3.85 hours which in turn is 230.77 minutes which means I burn 75L / 230.77 minutes = .325 Liters per minute I went around the house looking for something that had roughly the same specs and I found this a bottle of Heinz Malt Vinegar that was 355mL which looking at it I thought it was quite a bit of gas to be going through in one minute.

That means even if I believe this John Heyward guy I am giving up 15 miles if my fuel situation is in fact achieving a 97% burn. I don't believe that it is, but hey what do I know right. So then there is more math, now this math I think is extremely sketchy because it involves liquid to gas conversions and moles and joules and it gets really confusing but here goes. So I figure that 1 liter of water is 1kg so I then transpose that onto fuel. 1kg for a liter of gasoline this would mean that guessing on the high side I would have to make .400 kg per minute to be safe to use in my ICE to drive down the street that translates into 24kg per hour. Ok what does that really mean? Well under normal room temperature electrolysis I would need 60kwh to get 1kg of hydrogen. yeah but what does that really mean. Well 1kg of hydrogen holds 12oMegaJoules per kg, gasoline in comparison holds 44 MegaJoules but to get 1kg of hydrogen I would have to expend 216 Megajoules of regular electricity. So I would be at a energy deficit. This in effect is the energy conundrum.

What in the world is a Joule you ask?

One joule is the work done, or energy expended, by a force of one newton moving one meter along the direction of the force. This quantity is also denoted as a newton-meter with the symbol N·m. Note that torque also has the same units as work, but the quantities are not identical. In elementary units:

$\, 1\, \mathrm{J}=1\, \mathrm{N} \cdot \mathrm{m}$

$\, 1\, \mathrm{J}=1\, \mathrm{kg} \cdot \frac{\mathrm{m}^{2}}{\mathrm{s}^{2}}$

One joule is also:

• The work required to move an electric charge of one coulomb through an electrical potential difference of one volt; or one coulomb volt, with the symbol C·V.
• The work done to produce power of one watt continuously for one second; or one watt second (compare kilowatt-hour), with the symbol W·s. Thus a kilowatt-hour is 3,600,000 joules or 3.6 megajoules
I desperately searched for the conversion formula that would sustain my vehicle flat out but I really couldn't figure out a formula so I am posting it here in hopes that some science major with a love of volume math post me a response that I have seen so wittingly posted on other forums. So I guess on if I need to produce 24kg per hour then I need to sustain a constant pressure guessing here of 17.7psi. I guess that much because hydrogen is 3 times a powerful as gasoline so that is a 3rd of what I need to produce per hour. The consumption of which gets into flow rates and other stuff that is over my head really. Oh I estimated my RPMs to be around 2500 - 3000 I know that info is needed.

So then I realize that ultimately the problem with hydrogen is really a Rate of Production problem more than it is anything else. Rate of production/ Rate of consumption = Unity or viability to do work. What is aimed at is Over Unity or a ratio above 1:2 this is the ultimate hydrogen goal. If you could do that on a consistent basis you could really go anywhere and do anything for free. You could drive all over the US for pennies literally. So in come the innovators those who have claimed overunity all of which have met with untimely deaths after being visited by the oil companies and government per se. You know how I love Conspiracy theories. I'm convinced that United flight 93 was shot out of the sky somewhere else because that golf divot that they showed on the news was no gigantic plane. I'm sorry there would have been much more than that and that was unofficially confirmed by someone in the intelligence community.
I however digress, so don't get me started on the pentagon. ANYWAY Muhammad if you will please.

Ok so collectively what do we know about hydrogen production and water? I'm not going to get into all of the beneficial ways to get hydrogen out of hydrocarbons or oxides or zinc or a million other chemicals because frankly I don't understand it. I will stick to the simple electromechanical ways of producing hydrogen.

First the Laws as we know them:

The mass of a substance produced at an electrode during electrolysis is directly proportional to the number of electrons (the quantity of electricity) transferred at that electrode.
The number of faradays of electric charge required to discharge one mole of substance at an electrode is equal to the number of "excess" elementary charges on that ion.
More expertly expressed by:

In modern form, Faraday's laws are summarised by:

$m \ = \ { QM \over zF }$

where

m is the mass of the substance produced at the electrode
Q is the total electric charge passed through the solution
z is the valence number of ions of the substance (electrons transferred per ion)
F = 96 485 C mol-1 is the Faraday constant
M is the molar mass of the substance

The total charge Q is the integral electric current I(t) over time t:

$Q = \int_0^T I(t) \ dt$

where T is the total amount of time of the electrolysis.

In the simple case of constant current electrolysis this reduces to:

Q = It

$n \ = \ { I t\over zF }$

where n is the amount of substance ("number of moles") produced.

Uggh math breathe breathe. Anyway that little snippet of math will tell us some neat things later. However it doesn't include something that I thought is of dire importance temperature.

So these are the building blocks of getting Hydrogen from electricity which I actually know a little something about, and so here starts my quest at getting hydrogen for my ICE. So I'm researching and researching and finding more and more neat little clues of how to get more hydrogen faster.

First off I will list everything by numbers and then give my hypothesis at the end of how to get on demand hydrogen for my ICE. I feel it is rather simple actually and some key elements have been overlooked. However people who don't know much actually think that way. I think it is a blessing because people who know it all or a little bit of something overlook the obvious, OFTEN.

1. We find that passing a relatively small amount of current can produce electrolysis. The separating of hydrogen from oxygen in water. (Yes there are other kinds of electrolysis but we are talking about this one).
2. We know that molecules are made up of two or more elements and are bonded together covalently or ionically (I suppose).
3. We know that mass at certain temperatures has certain characteristics. At higher temperatures it moves around more.
4. We know that lines of magnetism has its effect on matter some more than others.
5. We know that ICE engines are about 18-25% efficient the rest of their efficiency is lost primarily in heat or energy transfer from one direction to another.
6. We know that at 2500 C electrolysis happens instantaneously in a process known as thermolysis.
7. We know that ridiculously high voltage and current can atomize water.
8. We know that salt water can burn when exposed to radiation.
9. We know that hydroxy (HHO or Brown's Gas) can increase fuel efficiency.
10. We know that John Keely made claims that oscillating water at 42,800khz causes it to dissociate. This is unconfirmed but ultrasonic vibrations do cause water to vaporize commonly used in Fog Machines in concerts.
11. We know that the resonant frequency for hydrogen is 1,420,405,752 Hz. Presumably to effectively agitate hydrogen should probably be attempted at one of its harmonics. One of which would bring us close to John Keely's frequency -15th 43347.344 hz. If working on a harmonics theory it would be safe to say that the length of the conductor be brought into account. It is also said that John Keely's frequency was done in a cord. Therefore it would be safe to say that other frequencies should be taken into account which would be the resonant frequency of Oxygen (currently unknown to me) and either the resonance of the container of water or the resonance of the cathode/anode used to induce the frequency during electrolysis.
12. We know that transformers induce an emf on coils nearby and if the primary winding has less turns than the secondary the voltage will be increased on the secondary.
13. We know a man named Stan Meyer claims to have come up with a way to get 300% efficiency from cold hydrolysis.
14. We know that his design has been improved on by several people Dave Lawton being one of them.
15. We know that water molecules have certain ways of acting in water.
16. We know that HTE is more efficient than normal room temperature Electrolysis.
Now some of this stuff maybe hoax, some of it maybe ignorant guessing but either way it is the asking of the question "What if" that is at the core of true science. In science there are laws and laws are meant to be broken, so they say. Or at least they remain the same when all of the variables that made them laws are in place.

So here is my hypothesis: Since lighting off your vehicle on Hydroxy will undoubtedly not yield enough to run your car you can do the second best thing boost your octane with it. So I say you start your car like normal sucking down cups of gas, then you flip on your hydroxy unit. Your primary hydroxy unit first bubbles through a water bubbler to clear it and then through a fuel bubbler to mix with your engines fuel and effectively giving you a higher octane rating. Possibly having a second hydrolysis machine that performs hydrolysis on the fuel itself who knows but safely just water. That hydrogenized fuel is then fed into the car giving it an extra boost. The secondary hydrolysis unit is fed into the air intake to provide a richer mixture and allowing you to use less fuel. Then after the engine gets hot run pressurized water around the exhaust to collect the heat from there creating superheated water. Run this super heated water through an electrolysis process getting even more hydrogen and as hydrogen production increases with the heat from the engine hydroxy production becomes more and more efficient using less energy to create. The fuel hydroxy mixture can then be leaned out further and further so that less and less fuel is used because hydroxy gas will be so high and effectively increasing fuel efficiency to extremely acceptable levels and contributing to the overall efficiency of the engine.

This way you aren't really getting something for nothing you are just using that which is wasted, so you are in effect recycling energy rather than pulling it out of thin air. If you could reclaim 50% of the energy that is wasted it would effectively increase miles per gallon and my financial happiness.

I know it is a long post but hopefully a brainiac will come along and either agree with me or give me the necessary formulas so I can figure out what will possibly work and what won't.

## Sunday, January 13, 2008

### First Rule of Science

The Quran sums it up best where it says "And of knowledge, you (mankind) have been given only a little." Surah Bani Isra'il 17:85

I have been searching for a solution to run vehicles off of water for a couple of days now. Ideastorming and thinking and storming some more, losing sleep and having a lot of hyper brain activity. I keep running into the same answers over and over and over again. They've already done that. WHAT!! If this stuff has already been done over and over and over again then why the hell are we still using fossil fuels? The answer consistently "It's too expensive, It's not cost effective." To fricking who? If you can give me a motor for $100,000 that will provide all of my energy needs and the only thing that I have to worry about is maintenance? Are you fricking serious? Let's see the average light bill these days is about$150 to $400 depending on the size of house for normal people size houses. At$400 a month that would translate into 20 years worth of free energy and I could run all of my fricking appliances all year round. So please really give me a break. If it was in a car which are already reaching astronomical proportions I could drive for 20 years to and from work. When I was living in VA I commuted 120+ miles a day creating a $300 dollar a month gas bill. So if all of my vehicles and home was equipped with these engines I would pocket over$1000 a month. Increasing my standard of living. So really you must ask yourself Cost inefficient for whom? If everyone did this and they made engines so they wouldn't break after 20 miles then the biggest companies in the world would theoretically go out of business tomorrow. So yeah I guess that wouldn't be cost effective.

This then in fact means it is much cheaper to tell the world it can't be done than having fifty million Einstiens in their back yard discovering the wonders of cold-fusion. Knowing the money at stake would make it in your best interest to infest the education centers of the world and give them bogus information and systematically force everyone into focus on working than on thinking to maximize efforts of enslaving a planet. Paycheck to Paycheck is probably the reality of the worlds brightest individuals.

Question everything is definitely the answer. Non-Religious folk would probably chime in including God, that however is not a wise position to take. I can go into more than that if you wish to comment on it. However Question everything in this universe. The Laws of Science definitely should be one of the things questioned. For the establishment lack of questioning only leads to predictability and higher profits.

## Thursday, January 10, 2008

### 300 mile per Gallon Vehicle and the Big Oil Conspiracy

Whilst contemplating my latest ideaquest I was seriously pondering starting a lottery. A lottery for renewable energy, mainly for my car. I was going to name it FreedomFromOilLottery.com and on such site I would collect monies that would eventually lead to the invention that would free us from such a massive dependency on Oil. Then I found this.

While I don't want to belittle blue-collar ingenuity, the vapor carb's inventors are trying to solve a nonexistent problem. According to John Heywood, a professor of mechanical engineering at MIT and an authority on internal-combustion engines, incomplete burning of fuel is insignificant in modern cars. Fuel combustion today typically exceeds 97 percent. While it's true cars aren't very efficient--only 20-35 percent of the fuel energy is converted to useful work--that's mostly due to heat loss (through the engine block, out the exhaust pipe) and unavoidable energy loss during burning itself.

The theoretical (and unobtainable) maximum efficiency for a small car like a Honda Civic is around 200 mpg; for your big beaters it's much lower. Claims to the contrary are fraudulent, and I gather Professor Heywood said as much in a report he wrote for the Postal Service, which was investigating high-mileage carb vendors for fraud.

So I'm sitting here thinking the maximum efficiency for a small car like a Honda Civic is around 200mpg? Well hell most people would take half of that. Half is better than all of the hybrids combined just about. Then I found this article:

Eric's Autos: 300 mpg or Conspiracy Theory? - The 2008 Aptera
Eric Peters
Maybe the conspiracy theorists were right after all. That was the first thought to pop into my head as I read about an engineer named Steve Fambro - and his 300 mpg hybrid Aptera two-seater. Yes, you read that right. Three hundred miles per gallon. Really. And even if he's off by half, the mileage of the snarky little gullwing coupe would be at least double that of the best hybrid a major automaker ever delivered - the 70 mpg Honda Insight- and two to three times the best-case mileage of an '08 Toyota Prius. Wow.

I dug further and found this little tidbit from the Shell corporation:

Some folks at Shell Oil Co. wrote “Fuel Economy of the Gasoline Engine” (ISBN 0-470-99132-1); it was published by John Wiley & Sons, New York, in 1977. On page 42 Shell Oil quotes the President of General Motors, he, in 1929, predicted 80 MPG by 1939. Between pages 221 and 223 Shell writes of their achievements: 49.73 MPG around 1939; 149.95 MPG with a 1947 Studebaker in 1949; 244.35 MPG with a 1959 Fiat 600 in 1968; 376.59 MPG with a 1959 Opel in 1973. The Library of Congress (LOC), in September 1990, did not have a copy of this book. It was missing from the files. I bought my copy from Maryland Book Exchange around 1980 after a professor informed me that it was used as an engineering text at the University of West Virginia. VPI published a paper, March 1979, concerning maximum achievable fuel economy. This paper has several charts illustrating achievable and impossible fuel economy. About 1980 I contacted the author concerning conflicts between the paper and documented achieved “impossible” mpg. The author said, “I will get back to you.” I am still waiting for his response.

So then I start to get pissed and think that this company who has a great idea but is really thinking too small. Automotive X Prize has a measly 10 million dollar prize for someone to make a 100 mpg vehicle with various and sundry stipulations. So I write them a letter:

Sent: Thursday, January 10, 2008 4:47 PM
To: 'autoprize@xprize.org'
Subject: Not thinking big enough

Hello all,

I think you have an outstanding concept here. I was just thinking of doing it myself and naming it Freedom from oil lottery or something to that effect. I think that there is one significant problem with your concept and that is that it simply isn’t big enough. I think you need or should shift it from a prize to a lottery. One where Me and every other consumer can say you know what I’m sick and tired of paying $4.00 at the pump because people are buying patents and locking them away in a drawer. That’s BS. The only way we the consumer can compete is if our purse is bigger than the purse of the oil empire.$10 mil is no doubt a nice chunk of change but it is petty cash to big oil business and for that price they can buy up 50 of your winners a week every week and pass the savings back down to us. I don’t think anyone will really take you serious until you get into the $100 million dollar range. I’m being honest here. There literally needs to be enough money to win and then start manufacturing the next day, and or possibly take over a car company. Seriously. You have to go bigger, much bigger. If you start a lottery how many people would pay a buck just to make the pot ridiculous? The pot needs to be huge. Large enough so that someone who knows the simple alteration of getting the results you ask for to say hey this is definitely better than retirement. If the pot gets large enough greed will take over and someone will roll over off of the solution. Think about it, I have my$5 dollars a month standing by and I’m sure the entire state of California isn’t too far behind me. I know that Oprah’s audience will chip in a buck or two.

Sincerely,

P.S. That Virgin Millionaire just put up $25 Million for anyone coming up with a solution to reducing greenhouse gasses. His$25 and your $10 that’s$35 and now you are talking enough money to make people listen.

Oh yeah and you might need to think about sponsoring it in a neutral country, fearing that our government will claim the winning solution as some national defense secret or some nonsense.

Because really the only way to get past this is to be able to pay more than the Big Oil Companies and possibly the US government and World Banks. Realistically you need to be in the neighborhood of \$400 million. Honestly before greedy people will begin to renege on promises they made to colleagues and friends.

I went on to say that to be really in the ball park you need to say that the vehicle is a SUV or something that everybody is currently driving.

So then I go back to my first love. Hydrogen. It is clean and readily available all over the planet. However common scientific knowledge dictates by repeatedly drilling it into our heads that getting more out of something than what you put into it is scientifically impossible. This however is a lie. The only two circumstances that the Science establishment says that you get more out than what you put in is in terms of Fusion which our sun is a ready example, but the problem is that you can't control it, or presumably Cold Fusion which is a figment of all of our collective imaginations.

Well I would have to look around me to find out that is a lie. My family started with 2 people now there are 6. Cells replicate multiplying themselves numerous amounts of times. 1 kernel of corn goes on to produce hundreds of kernels of corn and if replanted will reproduce several times over. These are biological examples however properties in one science can spread to other sciences. In Physics this must be true as well. If Energy is Neither created nor destroyed then it stands to reason that it can be recycled indefinitely. There is a perpetual machine out there. It has to exist, I would suppose that it is only limited by piss poor management.

Free energy is available on every coast around the world with a decent tide flow. Completely capable of producing millions of kilowatts of electricity. If I had a beach house I would have a small building with a bank of batteries and a Ocean Generator. Free Power.

Over and over Physicists talk about heat loss, heat loss and the inability to control heat loss however the heat isn't used. If it was used then I suppose that it could be put to work.

The only thing that I could really establish is that if Conspiracy Theorists question the establishment then the entire establishment must be questioned. The Laws of science are no different. Everything must be verified to assess the veracity of the truth. If we accept Science as given by the establishment then we have surrendered the creative spirit that makes us human, and are thus doomed to become subservient mindless gnomes PHD's and all.

Be sure to check out the Quasiturbine. Been watching it for a while now.

## Tuesday, January 08, 2008

### At first I wasn't concerned!

Then I realized that if something else stupid conveniently happens between now and the time the next president is inaugurated then we are STUCK with BUSH forever. Now that is scary. Check Mate!!

Furthermore, on May 9 2007, Bush signed the
National Security and Homeland Security Presidential Directive,
which gives Bush near full dictatorial powers in the event of a "National Emergency". Bush, and the people he works for, will be able to do ANYTHING they want if a "National Emergency" is deemed. All they need is a "New 9/11".
The US Constitution will no longer be applicable.

In regards to this Directive, Paul Craig Roberts, a former Assistant
Secretary of the Treasury during the Reagan administration, has come
out and said he feels the White House is preparing to STAGE A NEW 9/11 and the executive order is designed to capitalize on such an attack, specifically to impose Martial Law and Invade Iran.

IN ORDER TO STOP MARTIAL LAW AND THE TOTAL DESTRUCTION
OF THE US CONSTITUTION, WE MUST STOP ANOTHER FALSE
FLAG ATTACK. THE FIRST STEP IS TO EXPOSE THE 9/11 FRAUD.
THIS IS THE EVENT THAT IS BEING USED TO JUSTIFY THE DEVELOPING POLICE STATE, NOT TO MENTION WORLDWIDE OPPRESSION.

I read Zietgiest and looked at the full 1:58 hour movie and it was pretty convincing. Especially since I don't believe most of the 9.11 nonsense anyway. Some of the lost news clips prove that United 93 was shot out of the air though. Chances also are that the Pentagon got hit by a Missile. I believe that. Now if you go to zeitgeistmovie you can watch the whole thing for yourself. What I have posted is only the end. Also the first part is pretty anti-Christian which I think is a stupid move on his part. But hey it's his movie.

## Tuesday, January 01, 2008

### First Sunrise of 2008

This is not where I wanted to go to get the sunrise but when I set out for my spot I had a flat so oh well. It was a lovely sunrise anyway. I hope you like it.