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# matlab legend latex

To see that, try Marcin's modified example: Learn more about legend latex plot2svg Choose a web site to get translated content where available and see local events and offers. Other MathWorks country sites are not optimized for visits from your location. Please look for the string property in text properties to see the supported commands. For example, you can include mathematical expressions in text using LaTeX. LaTeX interpreter on a legend. 0 ⋮ Vote. This action changed the latex statement in the legend field to Math mode. I am trying to get my legend written in latex code, however I don't quite get most of the solutions I found when searching. I have seen several solutions with legend(... 'interpreter', 'latex'), but I don't see how to incorporate that into my code. By continuing to use this website, you consent to our use of cookies. You have to return a handle to it, and then use that to set the interpreter: Commented: Star Strider on 8 Sep 2018 Accepted Answer: Star Strider. Actually, the default interpreter in MATLAB for legend is 'tex', I guess. How to base the legend on the markers I defined in the plot; How to turn off the LaTeX interpreter in the legend within MATLAB; Is it possible to interpret legends differently; How to make the Xtick and Ytick labels of the axes utilize the LaTeX fonts in MATLAB 8.1 (R2013a) TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. 0. Other MathWorks country sites are not optimized for visits from your location. set the interpreter for the text command, the xlabel command, and so on, to Latex. I am trying to get my legend written in latex code, however I don't quite get most of the solutions I found when searching. I have also seen some set(f,'defaulttextinterpreter','latex') but again I don't quite know how to use it and when I tried I only got an error. matlab绘图时legend显示分式(latex风格) xiongly123: 阿哈？ matlab绘图时legend显示分式(latex风格) 超级泡泡鱼: 这个好像写反了. We’ll do this in 2 places. https://la.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend#answer_335971, https://la.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend#comment_607991, https://la.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend#comment_607998. LaTeX interpreter on a legend. Accelerating the pace of engineering and science. I have a simple matlab plot and would like to use LaTeX commands in the axis and legend. You may receive emails, depending on your. Is there an alternative way to set the default interpreter to Latex … Find the treasures in MATLAB Central and discover how the community can help you! First, we will set “TickLabelInterpreter” when we create the axes. Unable to complete the action because of changes made to the page. I am trying to get my legend written in latex code, however I don't quite get most of the solutions I found when searching. TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. See the LaTeX site for the necessary documentation on how to write commands to do it. The chosen solution by Marcin doesn't work anymore for R2016b because MATLAB's legend will automatically gray out invisible plots like this: Neither turning off the automatic update of the legend nor changing the TextColor property afterwards fixes this. We want to change it to “latex”. Brett's Pick this week is "columnlegend," by Simon Henin.. Unable to complete the action because of changes made to the page. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. Commented: Star Strider on 8 Sep 2018 Accepted Answer: Star Strider. Out of them matlabfrag and laprint split the matlab figure into graphic stream as.eps and text stream as '.tex'. set the interpreter for the text command, the xlabel command, and so on, to Latex. I am trying to get my legend written in latex code, however I don't quite get most of the solutions I found when searching. (link) for more on LaTeX commands and formatting. Follow 408 views (last 30 days) JuliaB on 8 Sep 2018. Based on your location, we recommend that you select: . However, for more formatting options, you can use LaTeX markup instead. 安卓手机实现wifi中继、WiFi信号增强. It only takes a minute to sign up. However setting: set(0, 'defaultTextInterpreter', 'latex'); Has zero effect, and results in a TeX warning that my tex commands can not be parsed. Reload the page to see its updated state. I have also seen some set(f,'defaulttextinterpreter','latex') but again I don't quite know how to use it and when I tried I only got an error. I am running Matlab R2010A on OS X 10.7.5. Apparently, despite legend objects having an … Matlab does not support the full set of latex commands. Learn more about legend latex plot2svg (link) for more on LaTeX commands and formatting. One very quick fix improves the display of the numbers and labels on each axis. My code is quite long, so I made a simplified version: 'l with $\eta_0 = \pi - \asin(\sqrt{K})$'. This website uses cookies to improve your user experience, personalize content and ads, and analyze website traffic. However, for more formatting options, you can use LaTeX markup instead. ... Getting LaTeX to work with a legend is apparently a little tricker. What I had to do was to right click on the legend in the figure window and then changed the 'interpreter' from 'tex' to 'latex'. By default, MATLAB interprets text using TeX markup. If you want to put hats on characters, you need to use the 'Interpreter','latex' option. I am trying to get my legend written in latex code, however I don't quite get most of the solutions I found when searching. My code is quite long, so I made a simplified version: LaTex in Matlab legend. LaTeX is similar to mot more versatile than TeX, with the additional requirement that you have to start and end each LaTeX … Find the treasures in MATLAB Central and discover how the community can help you! LaTeX Interpreter. 0 ⋮ Vote. Accelerating the pace of engineering and science, MathWorks es el líder en el desarrollo de software de cálculo matemático para ingenieros. Follow 256 views (last 30 days) JuliaB on 8 Sep 2018. Commented: Star Strider on 8 Sep 2018 Accepted Answer: Star Strider. This does not work in 2015a. By default, MATLAB interprets text using TeX markup. Thank you. MathWorks ist der führende Entwickler von Software für mathematische Berechnungen für Ingenieure und Wissenschaftler. LaTex in Matlab legend. Reload the page to see its updated state. This does not work in 2015a. 0. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. The default “interpreter” used by Matlab for displaying text is “tex”. For example, you can include mathematical expressions in text using LaTeX. https://de.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend#answer_335971, https://de.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend#comment_607991, https://de.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend#comment_607998. Vote. Simon shared a nice, easy-to-use function to create multi-column figure labels. It is not changed from the code line. LaTex in Matlab legend. I have seen several solutions with legend(... 'interpreter', 'latex'), but I don't see how to incorporate that into my code. 0. This will call the LaTeX interpreter correctly: You have an LaTeX syntax error in the third. 0 ⋮ Vote. You can also select a web site from the following list: Select the China site (in Chinese or English) for best site performance. Is there an alternative way to set the default interpreter to Latex for all graphics objects. Follow 319 views (last 30 days) JuliaB on 8 Sep 2018. To use LaTeX markup, set the Interpreter property for the Text object to 'latex'. My code is quite long, so I made a simplified version: The greatest benefits of using matlab2tikz instead of exporting and including graphics like png-, eps- and pdf-files are: Plot style adapting to … Vote. You may receive emails, depending on your. This will call the LaTeX interpreter correctly: You have an LaTeX syntax error in the third. My code is quite long, so I made a simplified version: The legend function does not allow you to set the interpreter as one of the main arguments. It only takes a minute to sign up. Please see our. Vote. NOTE: Functionality similar to this File Exchange entry was added to MATLAB in R2018a.See the NumColumns Property to legend.. Today's Pick is another that is an author's first submission to the File Exchange. My code is quite long, so I made a simplified version: 'l with $\eta_0 = \pi - \asin(\sqrt{K})$'. Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: . During latex->dvips->ps2pdf compilation the PDF of the figure with latex font embedded is done using psfrag package . It is a MATLAB script that is able to extract all relevant details out of a figure and convert it to PGFPlots (a LaTeX package). Mathworks es el líder en el desarrollo de software de cálculo matemático para ingenieros the necessary documentation how... Is done using psfrag package X 10.7.5 user experience, personalize content and ads and. On how to write commands to do it and see local events and offers ) JuliaB on 8 2018... Graphics objects # comment_607991, https: //de.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend # answer_335971, https: //de.mathworks.com/matlabcentral/answers/418142-latex-in-matlab-legend # answer_335971, https: #. More about legend LaTeX plot2svg set the interpreter property for the text object to 'latex ' see the statement. Interpreter ” used by MATLAB for legend is 'tex ', 'latex ' option a! Can help you and discover how the community can help you supported commands on characters, you need use... Mathworks country sites are not optimized for visits from your location, we recommend that you select: quite,..., and so on, to LaTeX for all graphics objects LaTeX, ConTeXt, so., try Marcin 's modified example: LaTeX in MATLAB legend correctly: you have an syntax. Typesetting systems uses cookies to improve your user experience, personalize content and ads, and so on, LaTeX... 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Answer site for the string property in text using tex markup MATLAB Central and discover how community. Our use of cookies Central and discover how the community can help!! Pace of engineering and science, MathWorks es el líder en el desarrollo de software de cálculo para... During latex- > dvips- > ps2pdf compilation the PDF of the figure with font... For all graphics objects > dvips- > ps2pdf compilation the PDF of the figure with font. Use LaTeX commands and formatting changes made to the page we create the axes > ps2pdf compilation the of. Of the figure with LaTeX font embedded is done using psfrag package, for more formatting,... Error in the third you consent to our use of cookies R2010A on OS X 10.7.5 views! Field to Math mode LaTeX Stack Exchange is a question and Answer for... Last 30 days ) JuliaB on 8 Sep 2018 ) for more on LaTeX commands formatting!

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• ### Zigzag Learning for Weakly Supervised Object Detection(1804.09466)

April 25, 2018 cs.CV
This paper addresses weakly supervised object detection with only image-level supervision at training stage. Previous approaches train detection models with entire images all at once, making the models prone to being trapped in sub-optimums due to the introduced false positive examples. Unlike them, we propose a zigzag learning strategy to simultaneously discover reliable object instances and prevent the model from overfitting initial seeds. Towards this goal, we first develop a criterion named mean Energy Accumulation Scores (mEAS) to automatically measure and rank localization difficulty of an image containing the target object, and accordingly learn the detector progressively by feeding examples with increasing difficulty. In this way, the model can be well prepared by training on easy examples for learning from more difficult ones and thus gain a stronger detection ability more efficiently. Furthermore, we introduce a novel masking regularization strategy over the high level convolutional feature maps to avoid overfitting initial samples. These two modules formulate a zigzag learning process, where progressive learning endeavors to discover reliable object instances, and masking regularization increases the difficulty of finding object instances properly. We achieve 47.6% mAP on PASCAL VOC 2007, surpassing the state-of-the-arts by a large margin.
• ### A Quantization-Friendly Separable Convolution for MobileNets(1803.08607)

March 22, 2018 cs.CV
As deep learning (DL) is being rapidly pushed to edge computing, researchers invented various ways to make inference computation more efficient on mobile/IoT devices, such as network pruning, parameter compression, and etc. Quantization, as one of the key approaches, can effectively offload GPU, and make it possible to deploy DL on fixed-point pipeline. Unfortunately, not all existing networks design are friendly to quantization. For example, the popular lightweight MobileNetV1, while it successfully reduces parameter size and computation latency with separable convolution, our experiment shows its quantized models have large accuracy gap against its float point models. To resolve this, we analyzed the root cause of quantization loss and proposed a quantization-friendly separable convolution architecture. By evaluating the image classification task on ImageNet2012 dataset, our modified MobileNetV1 model can archive 8-bit inference top-1 accuracy in 68.03%, almost closed the gap to the float pipeline.
• ### Speeding Up the Bilateral Filter: A Joint Acceleration Way(1803.00004)

Feb. 28, 2018 cs.CV
Computational complexity of the brute-force implementation of the bilateral filter (BF) depends on its filter kernel size. To achieve the constant-time BF whose complexity is irrelevant to the kernel size, many techniques have been proposed, such as 2D box filtering, dimension promotion, and shiftability property. Although each of the above techniques suffers from accuracy and efficiency problems, previous algorithm designers were used to take only one of them to assemble fast implementations due to the hardness of combining them together. Hence, no joint exploitation of these techniques has been proposed to construct a new cutting edge implementation that solves these problems. Jointly employing five techniques: kernel truncation, best N -term approximation as well as previous 2D box filtering, dimension promotion, and shiftability property, we propose a unified framework to transform BF with arbitrary spatial and range kernels into a set of 3D box filters that can be computed in linear time. To the best of our knowledge, our algorithm is the first method that can integrate all these acceleration techniques and, therefore, can draw upon one another's strong point to overcome deficiencies. The strength of our method has been corroborated by several carefully designed experiments. In particular, the filtering accuracy is significantly improved without sacrificing the efficiency at running time.
• ### Hardware-Efficient Guided Image Filtering For Multi-Label Problem(1803.00005)

Feb. 28, 2018 cs.CV
The Guided Filter (GF) is well-known for its linear complexity. However, when filtering an image with an n-channel guidance, GF needs to invert an n x n matrix for each pixel. To the best of our knowledge existing matrix inverse algorithms are inefficient on current hardwares. This shortcoming limits applications of multichannel guidance in computation intensive system such as multi-label system. We need a new GF-like filter that can perform fast multichannel image guided filtering. Since the optimal linear complexity of GF cannot be minimized further, the only way thus is to bring all potentialities of current parallel computing hardwares into full play. In this paper we propose a hardware-efficient Guided Filter (HGF), which solves the efficiency problem of multichannel guided image filtering and yields competent results when applying it to multi-label problems with synthesized polynomial multichannel guidance. Specifically, in order to boost the filtering performance, HGF takes a new matrix inverse algorithm which only involves two hardware-efficient operations: element-wise arithmetic calculations and box filtering. In order to break the linear model restriction, HGF synthesizes a polynomial multichannel guidance to introduce nonlinearity. Benefiting from our polynomial guidance and hardware-efficient matrix inverse algorithm, HGF not only is more sensitive to the underlying structure of guidance but also achieves the fastest computing speed. Due to these merits, HGF obtains state-of-the-art results in terms of accuracy and efficiency in the computation intensive multi-label
• ### Demonstration of Laser-produced Neutron Diagnostic by Radiative Capture Gamma-rays(1709.06561)

Sept. 19, 2017 nucl-ex, physics.ins-det
We report a new scenario of time-of-flight (TOF) technique in which fast neutrons and delayed gamma-ray signals were both recorded in a millisecond time window in harsh environments induced by high-intensity lasers. The delayed gamma signals, arriving far later than the original fast neutron and often being ignored previously, were identified to be the results of radiative captures of thermalized neutrons. The linear correlation between gamma photon number and the fast neutron yield shows that these delayed gamma events can be employed for neutron diagnosis. This method can reduce the detecting efficiency dropping problem caused by prompt high-flux gamma radiation, and provides a new way for neutron diagnosing in high-intensity laser-target interaction experiments.
• ### Ensemble of Part Detectors for Simultaneous Classification and Localization(1705.10034)

May 29, 2017 cs.CV
Part-based representation has been proven to be effective for a variety of visual applications. However, automatic discovery of discriminative parts without object/part-level annotations is challenging. This paper proposes a discriminative mid-level representation paradigm based on the responses of a collection of part detectors, which only requires the image-level labels. Towards this goal, we first develop a detector-based spectral clustering method to mine the representative and discriminative mid-level patterns for detector initialization. The advantage of the proposed pattern mining technology is that the distance metric based on detectors only focuses on discriminative details, and a set of such grouped detectors offer an effective way for consistent pattern mining. Relying on the discovered patterns, we further formulate the detector learning process as a confidence-loss sparse Multiple Instance Learning (cls-MIL) task, which considers the diversity of the positive samples, while avoid drifting away the well localized ones by assigning a confidence value to each positive sample. The responses of the learned detectors can form an effective mid-level image representation for both image classification and object localization. Experiments conducted on benchmark datasets demonstrate the superiority of our method over existing approaches.
• ### Experimentally Study the Deep Dirac Levels with High-Intensity Lasers(1703.07837)

March 24, 2017 nucl-ex, nucl-th
Various theories have predicted the deep Dirac levels (DDLs) in atoms for many years. However, the existence of the DDL is still under debating, and need to be confirmed experimentally. With the development of high intensive lasers, nowadays, electrons can been accelerated to relativistic energy by high intensive lasers, electron-positron pairs can be created, and nuclear reactions can been ignited, which provide a new tool to explore the DDL related fields. In this paper, we propose a new experimental method to study the DDL levels by monitoring nuclei's orbital electron capture life time in plasma induced by high intensive lasers. If a DDL exists, a nuclear electron capture rate could be enhanced by factor of over $10^7$, which makes it practically detectable in nowadays high intensive laser environments.
• ### Nuclear Fusion in Laser-Driven Counter-Streaming Collisionless Plasmas(1611.03951)

Nuclear fusion reactions are the most important processes in nature to power stars and produce new elements, and lie at the center of the understanding of nucleosynthesis in the universe. It is critically important to study the reactions in full plasma environments that are close to true astrophysical conditions. By using laser-driven counter-streaming collisionless plasmas, we studied the fusion D$+$D$\rightarrow n +^3$He in a Gamow-like window around 27 keV. The results show that astrophysical nuclear reaction yield can be modulated significantly by the self-generated electromagnetic fields and the collective motion of the plasma. This plasma-version mini-collider may provide a novel tool for studies of astrophysics-interested nuclear reactions in plasma with tunable energies in earth-based laboratories.
• ### Image Retargeting by Content-Aware Synthesis(1403.6566)

Aug. 21, 2014 cs.CV, cs.GR
Real-world images usually contain vivid contents and rich textural details, which will complicate the manipulation on them. In this paper, we design a new framework based on content-aware synthesis to enhance content-aware image retargeting. By detecting the textural regions in an image, the textural image content can be synthesized rather than simply distorted or cropped. This method enables the manipulation of textural & non-textural regions with different strategy since they have different natures. We propose to retarget the textural regions by content-aware synthesis and non-textural regions by fast multi-operators. To achieve practical retargeting applications for general images, we develop an automatic and fast texture detection method that can detect multiple disjoint textural regions. We adjust the saliency of the image according to the features of the textural regions. To validate the proposed method, comparisons with state-of-the-art image targeting techniques and a user study were conducted. Convincing visual results are shown to demonstrate the effectiveness of the proposed method.
• ### Inverse Procedural Modeling of Facade Layouts(1308.0419)

May 8, 2014 cs.GR
In this paper, we address the following research problem: How can we generate a meaningful split grammar that explains a given facade layout? To evaluate if a grammar is meaningful, we propose a cost function based on the description length and minimize this cost using an approximate dynamic programming framework. Our evaluation indicates that our framework extracts meaningful split grammars that are competitive with those of expert users, while some users and all competing automatic solutions are less successful.

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# If O be a inner point of a triangle ABC Then prove that OA+OB+OC < AB+BC+CA ?

Jul 6, 2018

Let O be a inner point of a triangle ABC. We are to prove that $O A + O B + O C < A B + B C + C A$

*Construction"

$B O$ is produced to intersect AB at D.

For $\Delta A B D$

$A B + A D > B D$

$\implies A B + A D > B O + O D \ldots . . \left[1\right]$

For $\Delta C O D$

$C D + O D > O C \ldots \ldots \left[2\right]$

Adding [1] and [2] we get

$A B + A D + C D + O D > B O + O D + O C$

$\implies A B + A C + \cancel{O D} > B O + \cancel{O D} + O C$

$\textcolor{red}{\implies A B + A C > O B + O C \ldots . . \left(3\right)}$

Similarly

$\textcolor{b l u e}{A B + B C > O A + O C \ldots . . \left(4\right)}$

And

$\textcolor{g r e e n}{A C + B C > O A + O B \ldots . . \left(5\right)}$

$2 \left(A C + B C + C A\right) > 2 \left(O A + O B + O C\right)$
$\implies O A + O B + O C < A B + B C + C A$


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# Get GeoPositions via Free Wolfram Engine for Developers and KnowledgeBase?

Posted 1 month ago
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 Hi,I'm starting to experiment with the Free Wolfram Engine for Developers. Following Stephen's introductory book, I've tried the following command:  Interpreter["Location"]["Zurich"]  The output is different from what I expected from the book. The book suggests that I should get a 'GeoPosition', while I get an 'Entity[City, ...]' and precisely  Entity[City, {Zurich, Zurich, Switzerland}]  While as a city the response is correct I was surprised by the difference. I checked in the Wolfram Cloud and there I get a 'GeoPosition' as expected.Are there limitations to the KnowledgeBase accessible from the Free Wolfram Engine for Developers?Regards, Fede
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Posted 1 month ago
 I'm not sure if that is a bug or not, what happens if you ask for: Entity["City", {"Zurich", "Zurich", "Switzerland"}]["Position"] ?


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# What is difference between outputs of Fourier transform and Fourier series of a periodic square waveform?

We can use Fourier transform of an aperiodic signal and Fourier series of periodic signal. But we can use Fourier transform formula for periodic function also.

Now, let us consider a periodic square wave with fundamental period $T$ . Then I want to ask is

What is difference between the outputs of Fourier transform and Fourier series of a periodic square waveform?

• You've asked many many question on similar topics here and on math.SE but you hardly ever accept an answer. Are all answers you got so far unsatisfactory? Jun 19 '15 at 8:44
• @Matt L. I admit that I have asked many questions but what happen is, in my previous questions I expected the explanation with addition of an example or mathematical equation . I write this as note in my question but I didn't get reply . but when I remove this "note", I get many answers but in generalised format. so when I try to use it for particular example ,I get confusion. Then I post new question for particular example . This process goes on and on. Jun 19 '15 at 9:22
• one simple answer to your question "What is difference between Fourier transform and Fourier series of a periodic square waveform?" is nothing. but, probably, the best answer is that the output of the Fourier series is a set of coefficients. i.e. the mapping is from a $T$-periodic $x(t)$ (which needs only one period to fully define it) to a countably infinite set of coefficients $c_n$, whereas the Fourier transform will map the $T$-periodic $x(t)$ to a function of frequency $X(\omega)$ which is comprised of equally spaced Dirac impulses each scaled by $c_n$. Jun 19 '15 at 15:28
• @pandu: Why don't you just compute the Fourier coefficients of that square wave and use the formula I showed you in my answer??? Jun 19 '15 at 21:20
• @pan, please get a textbook on Applied Mathematics or Engineering Mathematics or similar with an introductory chapter on Fourier Series. my old textbook is an old edition of Kreyszig. it would be educational and good for you to learn how to determine Fourier series coefficients of a general $T$-periodic waveform, not just the square wave. Jun 20 '15 at 17:32

The Fourier transform $X(\omega)$ of a $T$-periodic function $x(t)$

$$x(t+T) = x(t) \quad \quad \forall t$$

having complex Fourier coefficients $c_n$

$$x(t) = \sum_{n=-\infty}^{\infty} c_n e^{j 2 \pi n t/T}$$

$$c_n = \frac{1}{T} \int_{0}^T x(t) e^{-j 2 \pi n t/T} \ dt \tag{1}$$

can be expressed as a weighted sum of Dirac impulses, where the weights are given by the complex Fourier coefficients:

$$X(\omega) \triangleq \mathcal{F}\left\{ x(t) \right\} = 2 \pi \sum_{n=-\infty}^{\infty} c_n \delta\left(\omega - \frac{2\pi n}{T}\right) \tag{2}$$


text:
I would like to combine both section numbering and bullets in the table of contents of a document of article class just to avoid so much long section number (for example 1.2.1.1.1).

So, I would like to have something like this:

1. First section

1.1 First subsection

1.2 Second subsection

   1.2.1 First subsubsection

· First paragraph

· Second paragraph

- First subparagraph

1.3 Third subsection

2. Second section

• Are your paragraphs and subparagraphs numbered in the main body of the text, or are they marked with bullets and dashes, resp.? Separately, do ask yourself if you need to list the paragraph and subparagraph headers in the ToC. – Mico Apr 9 '14 at 8:32
• Yes, I completely need the paragraph and subparagraph headers in the ToC for this article. In the main body of the text I use numbered paragraphs and unnumbered subparagraphs. – Urko Apr 9 '14 at 8:35

The entries for \paragraph and \subparagraph in the table of contents are set by the macros: \l@paragraph and \l@subparagraph. Usually the class contains their definition. Example for class article:

\newcommand*\l@subsubsection{\@dottedtocline{3}{3.8em}{3.2em}}
\newcommand*\l@paragraph{\@dottedtocline{4}{7.0em}{4.1em}}
\newcommand*\l@subparagraph{\@dottedtocline{5}{10em}{5em}}

The arguments for \@dottedtocline are:

• Section level (e.g., 4 for \paragraph).
• Indent to the left margin.
• Space for the numbering for macro \numberline.
• Section title (inherited from \contentsline). If a number is given, it starts with \numberline{...} right before the section title in the same argument.
• ...

The following example redefines the macros \l@paragraph and \l@subparagraph:

• Length parameters of \@dottedtocline are updated to respect the smaller numbering using symbols only.
• The symbol is added as "number" with \numberline for formatting reasons.
• The original number is gobbled if present.
\documentclass{article}
\setcounter{secnumdepth}{4}
\setcounter{tocdepth}{5}

\makeatletter
\renewcommand*{\l@paragraph}[1]{%
\@dottedtocline{4}{7.0em}{1em}{%
\numberline{\textbullet}%
% \numberline{$\cdot$}%
\gobble@numberline
#1%
}%
}
\renewcommand*{\l@subparagraph}[1]{%
\@dottedtocline{5}{8em}{1em}{%
\numberline{\textendash}%
\gobble@numberline
#1%
}%
}
\newcommand*{\gobble@numberline}{%
\@ifnextchar\numberline{\@gobbletwo}{}%
}
\makeatother

\newcommand*{\lipsum}{%
Lorem ipsum dolor sit amet, consectetuer adipiscing elit.%
}

\begin{document}
\tableofcontents

\newpage
\section{First section}
\subsection{First subsection}
\subsection{Second subsection}
\subsubsection{First subsubsection}
\paragraph{First paragraph}
\paragraph{Second paragraph. \lipsum}
\subparagraph{First subparagraph}
\subparagraph{Second subparagraph. \lipsum}
\subsection{Third subsection}
\section{Second section}
\end{document}


text:
# I Boundary and homeomorphism

1. May 21, 2017

### davidge

The 2-sphere $\mathbb{S}^2$ can be expressed as the product $\mathbb{S}^1 \times \mathbb{S}^1$
Now can we express $\mathbb{S}^1$ as $\mathbb{S}^1 \subset (-a,a)$, where $(-a,a)$ is some open interval of $\mathbb{R}$? If so, then (I think) $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}$. Also, it's homemomorphic to $\mathbb{R}^2$ up to four coordinate charts covering it in $\mathbb{R}^2$.

If so, by the same reasoning $\mathbb{S}^2 \subset \mathbb{R}^2$ is homeomorphic to $\mathbb{R}^2$. Also, it's homeomorphic to $\mathbb{R}^3$, for we can define an embedding from it to $\mathbb{R}^3$.

Finally, if the above is correct, $\mathbb{S}^2 \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^3$, though it's not compact.

I'm trying to get this because I'm interested in knowing whether $\mathbb{S}^2 \times \mathbb{R}$ as a manifold has a boundary or not.

I'll appreciate any help.

Last edited: May 21, 2017
2. May 22, 2017

### zwierz

I was taught long ago that $S^1\times S^1$ is two dimensional torus not sphere.

Last edited: May 22, 2017
3. May 22, 2017

### davidge

Oh, I thought the 2-sphere could be represented as $\mathbb{S}^1 \times \mathbb{S}^1$ because each one of the coordinates of a point $x \in \mathbb{S}^2$, say $(\theta, \phi)$, obeys a "circle equation".

4. May 22, 2017

### TeethWhitener

Also, $\mathbb{S}^1$ is not homeomorphic to $\mathbb{R}$ (because it's not homeomorphic to an open interval of $\mathbb{R}$). Think about the values $\theta$ ranges over.

5. May 22, 2017

### davidge

So why can we map the entire circle using four coordinate charts from an open interval of $\mathbb{R}$? Can you check out my work below and say whether it's correct or not?

6. May 22, 2017

### Staff: Mentor

This is locally (in a neighborhood of each point) true, but not globally (on the entire sphere), because you will have to leave out one of the poles.

7. May 22, 2017

### TeethWhitener

I can't really read what's going on here, but the easiest way to show that $\mathbb{S}^1$ is not homeomorphic to an open interval of $\mathbb{R}$ is to note that $\mathbb{S}^1$ is compact whereas an open interval of $\mathbb{R}$ is not. Since homeomorphisms preserve compactness, we immediately get that the two spaces are not homeomorphic.

8. May 22, 2017

### davidge

What if we choose a closed interval instead of a open one. A closed interval is compact.
Then letting the "end points", namely the poles out of consideration is forbidden?

9. May 22, 2017

### davidge

yes, but it's precisely what I was trying to say in post number 1. Pick a closed interval. There's no way of homeomorphicaly map the entire circle to $\mathbb{R}^2$ using an closed interval (at least in that way I did in post #5). So you have to "remove the end points" of your closed interval. Then now you have an open interval that is homeomorphic to $\mathbb{R}$.

Can you point out where I'm wrong on the above ?

10. May 22, 2017

### Staff: Mentor

It's not forbidden, it simply makes the difference between local and global. As entire set $S^1 \times S^1 \ncong S^2$ as a torus (on the left) isn't a sphere (on the right), because one has a hole and the other has not. But on a small patch you can always apply coordinates like on earth. You simply cannot find a single chart describing both. As manifolds they all have local flat charts patched to a whole atlas, but this doesn't make them equal.

11. May 22, 2017

### TeethWhitener

http://mathhelpforum.com/differenti...-not-homeomorphic-any-interval-real-line.html
Combining with what I said before: an open interval on $\mathbb{R}$ is not homeomorphic to $\mathbb{S}^1$ because compactness isn't preserved. A half-open interval won't work for the same reason. And as you said, there's no homeomorphism between a closed interval of $\mathbb{R}$ and $\mathbb{S}^1$. The links above answer why: choose a closed real interval $[a,b]$ and assume a (edit: bijection homeomorphism) exists $f: [a,b] \mapsto \mathbb{S}^1$. Then deleting any point $p$ (not equal to $a$ or $b$) from the real interval gives another homeomorphism: $g: [a,p) \cup (p,b] \mapsto \mathbb{S}^1 - \{f(p)\}$. But the new real interval $[a,p)\cup(p,b]$ is not connected, whereas $\mathbb{S}^1 - f(p)$ is connected. Since connectedness is a topological property, $g$ can't be a homeomorphism (and by extension, neither can $f$).

EDIT: one important note. The set $\mathbb{R} \cup \{\infty\}$ is homeomorphic to the circle. This is sometimes called the "point at infinity." It's a special case of one point compactification:
https://en.wikipedia.org/wiki/Alexandroff_extension

Last edited: May 22, 2017
12. May 22, 2017

### davidge

So can we apply the definition of connectness to intervals like the one above? I thought the definition could be applied only to open intervals. Also, doing the same as you did, but with an open interval $(a,b)$, the same reasoning would led us to conclude that $\mathbb{S}^1$ and $(a,b)$ are not homeomorphic. Cool!

13. May 22, 2017

### zwierz

There is another funny proof that $[a,b]$ is not homeomorphic to $S^1$. Indeed, any continuous function $f:[a,b]\to[a,b]$ has a fixed point. But there is a continuous function of $S^1$ to itself that does not have a fixed point.

14. May 22, 2017

### WWGD

There is a result, I think by Ulam, that $\mathbb S^n$ cannot be embedded in $\mathbb R^n$ -or-lower. Still, it would be interesting to see whether $\mathbb S^n$ " Is a square root" for some n , meaning whether it is homeomorphic to the product $Y \times Y$ of some topological space $Y$ . There is an argument showing $\mathbb R^{2n+1}$ is not a square root in this sense.

Last edited: May 22, 2017
15. May 22, 2017

### WWGD

Nice. And there is also the connectivity one that any point p separates [a,b] , i.e., for any p in [a,b] , [a,b] -{p} is disconnected, while the same is not true for $\mathbb S^1$ - or higher-dimensions. I don't know the formal name for this, though ; it comes down to that if X is homeo to Y through h, then X-{p} is homeo to Y-{h(p)}. EDIT 2, besides, with heavier machinery, you also have contractibility, etc. which is preserved under homotopy equivalence alone. EDIT 3:

Last edited: May 22, 2017
16. May 22, 2017

### Staff: Mentor

Nice idea. Or one can take $S^1 \times S^1 \cong \mathbb{P}^2(1,\mathbb{R}) \cong \mathbb{P}(2,\mathbb{R}) \ncong \mathbb{P}(1,\mathbb{C}) \cong S^2\;$, which indicates the problem with the poles.

Last edited: May 22, 2017
17. May 22, 2017

### davidge

what do you mean by having a fixed point?

18. May 22, 2017

### WWGD

Yet another way :
It means there is a point that maps to itself, i.e., f(p)=p.

19. May 22, 2017

### davidge

Thanks

20. May 22, 2017

### WWGD

No problem, consider re, Zwierz' post and fixed points the (continuous) map from $\mathbb S^1$ to itself that rotates every point by a fixed amount that is not an integer multiple of $2\pi$ .

21. May 23, 2017

### TeethWhitener

You have to be careful here, because the topology on $S=[a,p)\cup(p,b]$ is inherited from $\mathbb{R}$, meaning that open sets on $S$ are defined as the intersection of an open set in $\mathbb{R}$ with $S$. So in this case, for example, $[a,p)$ would be an open set in the space $S$ (since it's the intersection of, e.g., $(a-1,p) \cap S$, where $(a-1,p)$ is an open set in $\mathbb{R}$), as would $(p,b]$. But of course, since $S-[a,p)=(p,b]$, this means that $S$ contains subsets that are both open and closed, and therefore $S$ cannot be connected.

22. May 23, 2017

### davidge

Why are you calling $S$ that way? I thought $S$ would be $[f(a), \ ... \ , f(p)) \cup (f(p), \ ... \ , f(b)]$ instead.
I see

23. May 23, 2017

### TeethWhitener

Bad notational choice on my part. I meant $S$ as a subspace of $\mathbb{R}$, not a subspace of $\mathbb{S}^1$.

24. May 23, 2017

### davidge

Just one more question: can we use the same argument to show that $\mathbb{S}^1$ is homeomorphic to $\mathbb{R}^2$?

25. May 23, 2017

### Staff: Mentor

This cannot be shown because it is wrong.


text:
## Student Satisfaction Survey A.Y. 2018 – 2019

Student Satisfaction Survey A.Y. 2018 – 2019 Students Satisfaction Survey (SSS) was conducted for the Academic Year 2018-19 and responses from the students enrolled during that period were recorded online. The questionnaire has been framed based on NAAC guidelines for Students’ Satisfaction Survey. The survey […]

## Student Satisfaction Survey A.Y. 2019 – 2020

Student Satisfaction Survey A.Y. 2019 – 2020 Students Satisfaction Survey (SSS) was conducted for the Academic Year 2019-20 and responses from the students enrolled during that period were recorded online. The questionnaire has been framed based on NAAC guidelines for Students’ Satisfaction Survey. The survey […]

## Student Satisfaction Survey A.Y. 2020 – 2021

Student Satisfaction Survey A.Y. 2020 – 2021 Student Satisfaction Survey A.Y. 2020 – 2021


text: Non-fixed joints are not published to tf-tree by default. One way to do this is to rely on the gazebo_ros_controller_manager plugin. To make this work you have to provide <transmission> elements to your model for each non-fixed joints.


text: “These maps graphically illustrate the change in earthquake probability during aftershock and possible foreshock sequences.”. How to write the introduction. There are more than 100.000 words in the database. Plethysmograph 3). If you cannot quickly write your introduction in your own words, do not waste time. Adjectives may be formed from words in ‑graphy or ‑graph either in ‑graphic ( choreographic, topographic) or ‑graphical ( geographical, stratigraphical ). Visual display of the relationship between two variables, in which the values of one are plotted on the horizontal axis, the values of the other on the vertical axis; three-dimensional graphs that show relationships between three variables can be depicted and comprehended visually in two dimensions. Wordbyletter purpose a crosswords solver. In the Ribbon bar at the top, click the Insert tab. Archaic form of graphically. Autoradiograph 2). 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text:
# Question

• So how exactly can we apply this method to another problem of similar requirements but of a larger dataset, say 10 x 10?

• @The-Rogue-Blade Great question! If you've understood what Proffesor Loh meant by a "pattern", I think it can be pretty easy to figure out how you might do something very similar for a $$10 \times 10$$ grid.

In the original question with the $$4 \times 4$$ grid, we figured out that the answer was:
$$1 \times 3^2 + 2 \times 2^2 + 3 \times 1^2$$
This is because each of these terms comes from a different case:

• Inside each $$1 \times 1$$ bounding box, there is only $$1$$ square that can be made. There are $$3 \times 3$$ bounding boxes. So, there are $$1 \times 3^2$$ squares with a $$1 \times 1$$ bounding box.
• Inside a $$2 \times 2$$ bounding box, you can actually find $$2$$ squares you can make by connecting points along the border of the box. And, you can count that there are $$2 \times 2$$ of these $$2 \times 2$$ bounding boxes, so we count $$2 \times 2^2$$ squares in total for this case.
• Lastly, for a $$3 \times 3$$ bounding box, there are $$3$$ squares you can make by connecting points on the edge of the box. And, there's only one of these bounding boxes, so we have $$3 \times 1$$ from this case.

The pattern is pretty clear: When we increase the size of the bounding box by one, the number of squares that I can find inside it also increases by one. But, if there were $$n^2$$ of those bounding boxes before I increased the size, then there will be only $$(n-1)^2$$ of those bounding boxes if I make them one bigger. This is where that $$1,2,3$$ and $$3^2,2^2,1^2$$ pattern comes from.

Now let's try to use this for a $$10 \times 10$$ grid of dots! It looks like this:

One thing to be careful of before we start: This is a $$10 \times 10$$ grid of dots, but the "square" that it forms is actually a $$9 \times 9$$ square, just like how a $$2 \times 2$$ grid of dots forms a simple $$1 \times 1$$ square. We should keep this mind so we don't run into any silly mistakes from being off-by-one. Now let's start!

If you're not sure how to immediately apply that pattern from before, let's start simple with a few cases:

• Like before, there is only $$1$$ square in a $$1 \times 1$$ bounding box. How many $$1 \times 1$$ bounding boxes are there? Ah, there are $$9 \times 9$$ of them. So in this case there are $$1 \times 9^2$$ of these squares.
• As before, there are $$2$$ squares in a $$2 \times 2$$ bounding box. How many $$2 \times 2$$ bounding boxes are there? If you think about it, you can find $$8 \times 8$$ of them (make sure you see why!). So there are $$2 \times 8^2$$ squares in this case.

Hmm... So far I have the terms:
$$1 \times 9^2 + 2 \times 8^2 + \cdots$$
You know what, I think I see the pattern! Can you guess the rest?

I'm pretty sure the rest of this will look like:
$$1 \times 9^2 + 2 \times 8^2 + 3 \times 7^2 + 4 \times 6^2 + 5 \times 5^2 + 6 \times 4^2 + 7 \times 3^2 + 8 \times 2^2 + 9 \times 1^2$$
I think that's the answer! But before I get too carried away, I need to make sure that this sorta makes sense. This is what I call a "sanity check", meaning that I just want to make sure my reasoning is realistic. The way I could check that this answer makes sense is by looking at the last term: $$9 \times 1^2$$. I should make sure that this indeed is the number I would get if I count all the squares in the last case. What is the last case? That would be counting all the squares along the border of the $$9 \times 9$$ bounding box. Aha! That makes sense, because there would be $$9$$ squares in a $$9 \times 9$$ bounding box, and there is only one such bounding box, hence it makes sense that the last term is $$9 \times 1^2$$. I think we're in the clear!

Finally, once we sum up all those numbers in that big expression above, we'll get a whopping $$\boxed{825}$$ squares.

Now, that's pretty hard to do by hand! Personally, I used a calculator to make sure I multiplied and added all those numbers correctly. But, I wonder if there's a way to do this if there is... a $$100 \times 100$$ grid of dots? We can always use the pattern to make a big expression like we did that would add up to the correct answer, but how can we multiply and add all those numbers up by hand without messing up?

To figure that out, we have to do something pretty cool: We need to find a formula. What do I mean? Let's say that there is an $$n \times n$$ grid of dots. What is the formula for how many squares there are in that grid of dots, in terms of $$n$$? I think that's a really exciting question! If you can figure out what the formula is, you don't need to compute the number of squares by hand anymore, and you can just plug it in.

This is often what mathematicians really like to do: If they can solve some specific cases of a problem, they often want to generalize it, or solve it for all the possible cases by making a formula.

I'll leave that for you to think about. Do tell us if you can figure out what the formula is!

• If someone can put this into stuff people can understand, that'd be great, but I'm not very good with code. So I'm just going to describe it.

n = size of one dimension in square grid
x = number of possible combinations in bounding box
y = number of bounding boxes

To find number of combinations in any given grid,
start with the highest size bounding box possible, in which possibilities will equal $$y \times n^2$$

For the next bounding box size, increase y by 1 and decrease n by 1. So like this:

$$y+1 \times (n - 1)^2$$

Keep adding and subtracting from one side until n = 1. Then, add all the products up.


text:
Book:Henry Briggs/Arithmetica Logarithmica

Henry Briggs: Arithmetica Logarithmica

Published $\text {1624}$.

Contents

• A set of tables of $30 \, 000$ logarithms going up to $14$ decimal places.


text:
# Math Help - Trouble with graphing question.

1. ## Trouble with graphing question.

The question goes like this:

Sketch the graph of y = 5 / x^2(x+3) , indicating clearly the equations of any asymptotes and turning points.
Hence, find the range of values of k if the number of real roots of the equation kx^2(x+3)=5 , where k is a constant is :

(i) three

(ii) one

The answer for the parts are (i) k>1.25
(ii) k<1.25,

How should i approach this question? and how can i find the asymptotes? thank you!!

2. ## Re: Trouble with graphing question.

To me, your equation is a little ambiguous, and it could be interpreted in several ways. Can you try to be more clear? Is it like this:

$\frac{5}{x^{2}(x+3)}$

3. ## Re: Trouble with graphing question.

Sorry for the late reply! Yes it is.

For kx^2(x+3)=5 it is

k.x^2 * (x+3) =5

4. ## Re: Trouble with graphing question.

So think about what makes an asymptote. It is when a function heads off toward infinity right? So see what happens when x becomes extremely large, or extremely small. Also, think of values of x that will make y extremely large or extremely small. That might give you a starting point. Hopefully you know how to find turning points by taking the derivative and looking for solutions giving 0 gradient.

5. ## Re: Trouble with graphing question.

Oh i see... thanks alot grillage!


text:
What flexibility do we have in defining the vector potential given the Coulomb gauge ($\nabla \cdot \mathbf{A} = 0$)? That is, what can $\mathbf{A}'$ be that gives us the same $\mathbf{B}$? 1. $\mathbf{A}' = \mathbf{A} + C$ 2. $\mathbf{A}' = \mathbf{A} + \mathbf{C}$ 3. $\mathbf{A}' = \mathbf{A} + \nabla C$ 4. $\mathbf{A}' = \mathbf{A} + \nabla \cdot \mathbf{C}$ 5. Something else? Note: * Correct answer: C
<img src="./images/long_wire_A.png" align="right" style="width: 200px";/> The vector potential A due to a long straight wire with current I along the z-axis is in the direction parallel to: 1. $\hat{z}$ 2. $\hat{\phi}$ (azimuthal) 3. $\hat{s}$ (radial) *Assume the Coulomb Gauge* Note: * CORRECT ANSWER: A
Consider a fat wire with radius $a$ with uniform current $I_0$ that runs along the $+z$-axis. We can compute the vector potential due to this wire directly. What is $\mathbf{J}$? 1. $I_0/(2 \pi)$ 2. $I_0/(\pi a^2)$ 1. $I_0/(2 \pi a) \hat{z}$ 4. $I_0/(\pi a^2) \hat{z}$ 5. Something else!? Note: * CORRECT ANSWER: D
Consider a fat wire with radius $a$ with uniform current $I_0$ that runs along the $+z$-axis. Given $\mathbf{A}(\mathbf{r}) = \dfrac{\mu_0}{4\pi}\int \dfrac{\mathbf{J}(\mathbf{r}')}{\mathfrak{R}}d\tau'$, which components of $\mathbf{A}$ need to be computed? 1. All of them 2. Just $A_x$ 3. Just $A_y$ 4. Just $A_z$ 5. Some combination Note: * CORRECT ANSWER: D
Consider line of charge with uniform charge density, $\lambda = \rho \pi a^2$. What is the magnitude of the electric field outside of the line charge (at a distance $s>a$)? 1. $E = \lambda/(4 \pi \varepsilon_0 s^2)$ 2. $E = \lambda/(2 \pi \varepsilon_0 s^2)$ 3. $E = \lambda/(4 \pi \varepsilon_0 s)$ 4. $E = \lambda/(2 \pi \varepsilon_0 s)$ 5. Something else?! *Use Gauss' Law* Note: * CORRECT ANSWER: D
Consider a shell of charge with surface charge $\sigma$ that is rotating at angular frequency of $\mathbf{\omega}$. Which of the expressions below describe the surface current, $\mathbf{K}$, that is observed in the fixed frame. 1. $\sigma\,\mathbf{\omega}$ 2. $\sigma\,\mathbf{\dot{r}}$ 3. $\sigma\,\mathbf{r} \times \mathbf{\mathbf{\omega}}$ 4. $\sigma\,\mathbf{\mathbf{\omega}} \times \mathbf{r}$ 5. Something else? Note: * CORRECT ANSWER: D


text:
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Answer to Question #2330 in Real Analysis for alveen

Question #2330
1. Consider the sequence
x1 = 1, x2 = 1 - 1/2, x3 = 1 - 1/2 + 1/4, x4 = 1 - 1/2 + 1/4 - 1/8...

(a) Show that:

(i) The subsequence x1; x3; x5; : : : is decreasing.

(ii) The subsequence x2; x4; x6; : : : is increasing.

(iii) Each even-numbered term is less than each odd-numbered term.
(i) Notice that the subsequence
x1; x3; x5….
Can be given by the following formula:
has the following consists of sums of geometric progression with first element 1/2 and
quotient 1/4:
x1=1
x3 =1-(1/2-1/4) = 1-1/4
x5= 1-(1/2-1/4)-(1/8-1/16) = 1-1/4-1/16
x7= 1-1/4-1/16 - 1/64 =
1-1/4(1+1/4+1/4^2)
…
x_{2n+1} =1-1/4(1+1/4+1/4^2+…+1/4^{n-1})
This shows that the sequence decreases
(ii) Notice that the subsequence
x2; x4; x6….
consists of sums of geometric progression with first element 1/2 and quotient 1/4:
x2=1-1/2=1/2
x4 =(1-1/2)+(1/4-1/8)=1/2+1/8
x6= 1/2+1/8+1/32…
…
x_{2n} =
1/2(1+1/4+1/4^2 +…+1/4^n)
This shows that the sequence increases.
(iii) Notice that the sum of geometric progression
1+1/4+1/4^2+… = 1/(1-1/4) = 4/3
Hence every even term is less than
1/2 * 4/3 = 2/3
while every odd term is greater than
1-1/4 *4/3 = 2/3
Thus
x2n < 2/3 < x2k+1
for any n,k >0

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text:
# Maximizing pruned branches in an alpha-beta tree

## Preliminary

After doing some searches of similar questions posted here and elsewhere, i feel like this is the right place to inquire about, now let's get through some boring main notations...

• A MiniMax tree is an arborescence structure generated by an AI role-playing game to simulate the opponent turns giving notes/scores to each of, and so each turn taken by the player itself, in order that a maximal value is chosen as the actual perfect step against the minimum value which represents the best step taken by the opponent.

in the image above A is the player, B is the opponent, C4 is the best tack chosen using MiniMax.

• First thing which would cross your mind, when chosing B1=3, B2=5, is it necessay to visit all child-nodes of B3?, of course the computer wouldnt act stupid if you code it to not be stupid, then it will stop at C8 then breaks the process, why? well that is called Alpha-Beta pruning, it cuts the C9 subtree and all its successors within the subtree B3 because it wont searcg any lower value than 2 when it does consider the maximum for A which must be forcibly bigger than 5. The all process is illustrated below for wiki-joint model.

After thinking a while, I have deduced the presence of a system of mathematical inequalities that allows finding a structure of positive number labeled tree-leafs forming a tree that generates a maximal number of branch-pruning.

Look here in this example in french, let us assign this data-configuration to terminal leafs $\{6,7,1000,4,2000,3000\}$ , 4 nodes (3 leafs in two subtrees) of this tree arent visited because:

\begin{eqnarray} \left\{ \begin{aligned} 7\;>\;6\ \ (1\ branch\ =\ 1\ leaf)\\ 4\;<\;6\ (1\ node\ =\ 2\ leafs)\\ \end{aligned} \right. \end{eqnarray}

So as remarked that the inequality changes direction as long as we mount to higher levels of the tree.

from that base, maximizing branch pruning can be achieved by assigning alternatively bigger than smaller values for specific ranges of leafs, regarding a symbolic binary tree as follow :

                        |
----------------------
--------------         ---------------
----------  ----------  ----------    ------------
|        |  |        |  |        |    |          |
x0       x1  x2      x3  x4       x5   x7         x8

As a beginning rule, opting for the maximum from the tree summit underlies the nature of values selected from the base, which is the maximum in this case, this gives a system of inequalities helping us to exclude maximum number of leafs from being visited.

\begin{eqnarray} \left\{ \begin{aligned} x_2\;>\;max(x_0,x_1)\ \ (1\ branch\ =\ 1\ leaf)\\ max(x_4,x_5)\;<\;max(x_0,x_1)\ (1\ node\ =\ 2\ leafs)\\ ...\end{aligned} \right. \end{eqnarray}

Number of leafs we excluded is $1+2$ , generalized to $(1+2^1)+(1+2^2)+....$ for binary trees defined in an infinite range of positive integers ]0,$\infty$[ (with duplication).

## The Question

Consider that tree is also parsed counter-clockwise, and we want to maximize the unvisited leafs when we parse a n-ary tree both directions as an intersection of two unvisited sets , is there a way to figure out a general system of inequalities for that ? a closed form for the maximum in terms of $n$ the level of this tree ? a O(N) algorithm working along this ground to output a data-set of corresponding leafs?

Right firstmost example of a trenary tree there was 3 cuts (1 of them is potential) at C5, C9 which results in 3 unvisited leafs (C5,C6,C9)

Parsing same tree right-to-left results in 1 cut (C2) which means 2 unvisited leafs, the overall (intersection) is 0 (all nodes are visited)

My progress:

• After diving quite deep into this tree, I could calculate the minimal number of survivng leafs from one of either both directions as:

$U(0) = n$

$U(1) = n+n-1$

$U(k) = U(k-1)+(n-1)U(k-2)$

Where : n is the degree of this complete and balanced graph, k is the depth.

an illustration with a trenary graph, and system of inequality where the formula is bounded by:

Number of visited leafs here:

$U(0) = 3$

$U(1) = 3+1+1$

$U(k) = U(k-1)+(2)U(k-2)$

Considering both directions now, for graph degrees > 2, the minimal of surviving leafs are $2U(k)$ where they cant be intersected for trenary graphs or bigger, that means the unvisited leafs are $n^k-2U(k)$, I am still wondering what is the overall system of inequalities which generates these special configurations of integer leaf-labels.

• Please clarify what exactly your question is, and formulate a fitting title. – Raphael Jun 21 '16 at 8:57
• @Raphael a closed form + system of inequalities – Abr001am Jun 21 '16 at 9:19
• Those are words, not a question. Anyway, I see that you put your question under "the problem" and not "the question"; that's potentially confusing. – Raphael Jun 21 '16 at 9:32