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# Lecture 19: Joint, Conditional and Marginal Distributions; 2D LOTUS; Expected Distance between Uniforms; Chicken-egg Problem

## Stat 110, Prof. Joe Blitzstein, Harvard University

### Joint, Conditional and Marginal Distributions

#### Joint CDF

A joint CDF is simply where we are dealing with multiple random variables. As an example, a case where we have two random variables $X, Y$, the joint CDF of two random variables $X, Y$ can be expressed as:

\begin{align} F(x,y) &= P(X {\le} x, Y {\le} y) \end{align}

Note that the random variables may be discrete, continuous, or a mixture of both.

#### Joint PDF

The joint PDF, in the case of continuous random variables, is what you would integrate to get the joint CDF. Continuing with our example of two (continous) random variables, we have:

\begin{align} f(x,y) &= \frac{\partial^2}{\partial{x}\partial{y}} F(x,y) \end{align}

Conversely, if we want to know the probability of $X,Y$ in some set $A$, we integrate the density to get that probability.

\begin{align} P\left((X,Y) \in A\right) &= \iint\limits_{A} f(x,y) \, dxdy \end{align}

Integrate by holding one variable constant, and then do the other. The key thing is to be sure to get the limits of integration correct.

#### Marginal PDF

The marginal PDF of $X$ is obtained by integrating out the $Y$. Recall the $X,Y$ contigency table and the definition of marginal probability.

\begin{align} \int_{-\infty}^{\infty} f(x,y) \, dy \end{align}

Notice that by keeping $X$ constant and integrating over all $Y$, the marginal PDF of $X$ no longer depends on $Y$.

And we can do vice-versa for the marginal PDF of $Y$, but keeping $Y$ constant and integrating over all $X$.

Do not forget that taking the marginal PDF of $X$ and then integrating over all $X$, we should get 1.0.

\begin{align} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} f(x,y) \, dx dy &= 1.0 \end{align}

#### Conditional PDF

Given that we know $X$, what is the appropriate PDF for $Y$?

Well, we can apply what we know about conditional probability to get a conditional PDF.

\begin{align} f_{Y|X} (y|x) &= \frac{f_{XY}(x,y)}{f_{X}(x)} \\ &= \frac{f_{X|Y}(x|y) \, f_{Y}(y)}{f_{X}(x)} \end{align}

This is completely analogous to conditional probability.

#### Independence

$X,Y$ are independent if

\begin{align} f_{X,Y}(x,y) &= f_{X}(x) \, f_{Y}(y) &\quad \text{for all }x, y \text{ from PDF p.o.v.} \\ \\ F(x,y) &= F(x) \, F(y) &\quad \text{for all }x, y \text{ from CDF p.o.v.} \end{align}

These statements are equivalent, but in most cases it might be easier to work the PDFs.

### A Uniform example

Let's revisit that distribution that is uniform on the unit disc $x^2 + y^2 \le 1$.

In [1]:

import matplotlib
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

plt.xkcd()

fig = plt.figure(figsize = (5.0,5.0))
ax.set_xlim([-1.5,1.5])
ax.set_ylim([-1.5,1.5])
circ = plt.Circle((0, 0), radius=1.0, color="b", alpha=0.2, lw=5)
plt.axhline(0, color="black", alpha=0.4)
plt.axvline(0, color="black", alpha=0.4)
plt.show()

#### Joint PDF

A valid PDF is required to integrate to 1.0, and since we are looking at a Uniform distribution over the unit circle with area equal to $\pi$, we have

\begin{align} f_{XY}(x,y) &= \begin{cases} \frac{1}{\pi} & \quad \text{if } x^2 + y^2 \le 1 \\ 0 & \quad \text{otherwise}\\ \end{cases} \end{align}

Understand that while this is uniformly distributed, $x$ and $y$ are closely related with $x^2 + y^2 \le 1$, and so $X,Y$ cannot be independent. More on that coming up.

#### Marginal PDF of $X$

\begin{align} f_{X}(x) &= \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} dy \\ \\ &= \frac{2}{\pi} \sqrt{1-x^2} &\quad \text{where } -1 \le x \le 1 \end{align}

Notice that while the joint PDF is Uniform, the marginal PDF is not Uniform: it grows as $x$ approaches 0.

By symmetry we can just replace $x$ with $y$ to get the marginal PDF of $Y$ $f_{Y}(y)$

#### Conditional PDF of $Y|X$

\begin{align} f_{Y|X}(y|x) &= \frac{f_{XY}(x,y)}{f_{X}(x)} \\ &= \frac{\frac{1}{\pi}}{\frac{2}{\pi} \sqrt{1-x^2}} \\ &= \frac{1}{2 \sqrt{1-x^2}} &\quad \text{if } -\sqrt{1-x^2} \le y \le \sqrt{1-x^2} \end{align}

But since we are treating $x$ as a constant, and the above equation does not depend on $y$, the conditional PDF $f_{Y|X}(y|x)$ is actually Uniform.

\begin{align} Y|X &\sim \operatorname{Unif}(-\sqrt{1-x^2}, \sqrt{1-x^2}) &\quad \text{or also } \\ \\ Y|X=x \, &\sim \operatorname{Unif}(-\sqrt{1-x^2}, \sqrt{1-x^2}) \end{align}

#### Non-independence

Since in our example above

\begin{align} f_{XY}(x,y) \neq f_{X}(x) \, f_{Y}(y) \end{align}

we can see that $X,Y$ are not independent.

Or, because the unconditional distribution of $Y$ is not the same as the conditional distribution of $Y|X$, it necessarily follows that $X,Y$ are not independent as learning $X$ gives us information about $Y$ (and vice-versa).

### 2D LOTUS

Let $X,Y$ have joint PDF $f(x,y)$, and let $g(x,y)$ be a real-valued function of $x,y$.

Then, without trying to find the PDF of $g(x,y)$, LOTUS lets us use joint PDF:

\begin{align} \mathbb{E} g(x,y) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x,y) \, f(x,y) \, dx dy \end{align}

We will illustrate this by proving the following theorem.

#### Theorem

If $X,Y$ are independent, then $\mathbb{E}(XY) = \mathbb{E}(X)\mathbb{E}(Y)$.

Or, in other words, the independence of r.v. $X$ and $Y$ implies that they are uncorrelated.

#### Proof

Consider the continuous case

\begin{align} \mathbb{E}(XY) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy \, f(x,y) \, dx dy &\quad \text{2D LOTUS} \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xy \, f_{X}(x) \, f_{Y}(y) \, dx dy &\quad \text{since } X,Y \text{ are independent} \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} xy \, f_{X}(x) \, f_{Y}(y) \, dx \right) dy &\quad \text{this is a double integral, so we can group thusly} \\ &= \int_{-\infty}^{\infty} y \, f_{Y}(y) \underbrace{ \left( \int_{-\infty}^{\infty} x \, f_{X}(x) \, dx \right)}_{\text{actually }\mathbb{E}(X) \text{, which is constant}} dy &\quad \text{since we are holding } y \text{ constant} \\ &= \mathbb{E}(X) \int_{-\infty}^{\infty} y \, f_{Y}(y) \, dy \\ &= \mathbb{E}(X) \, \mathbb{E}(Y) &\quad \blacksquare \end{align}

### Example: expected distance between 2 Uniformly distributed random points

Given i.i.d. $X,Y \sim Unif(0,1)$, find the expected distance between $X$ and $Y$, $\mathbb{E}|X-Y|$.

In [2]:

plt.xkcd()

fig,ax = plt.subplots(1, figsize=(6.0,1.0))

plt.plot([1/3, 2/3],[0,0], 'ro')
ax.axhline(y=0, color='k', alpha=0.2)

ax.set_xticks([0.0, (1.0/3.0), (2.0/3.0), 1.0])
ax.set_xticklabels(['0', '1/3', '2/3', '1'])

ax.get_xaxis().set_ticks_position('bottom')
ax.get_yaxis().set_visible(False)
ax.spines['right'].set_visible(False)
ax.spines['left'].set_visible(False)
ax.spines['top'].set_visible(False)
ax.spines['bottom'].set_visible(False)

plt.show()

We don't need to directly find the distribution of $|X-Y|$, because we are only interested in the average distance between points $x,y$. Using LOTUS, we have:

\begin{align} \mathbb{E}|X-Y| &= \int_{0}^{1} \int_{0}^{1} |x-y| \, dx dy &\quad \text{since }X,Y \text{ are i.i.d., the joint PDF }=1 \\ &= \iint_\limits{x \gt y} (x-y) \, dx dy + \iint_\limits{x \le y} (y-x) \, dx dy &\quad \text{split into 2 integrals to deal with abs. value} \\ &= 2 \int_{0}^{1} \int_{y}^{1} (x-y) \, dx dy &\quad \text{by symmetry; and since } x \gt y \text{, inner integral starts from }y \\ &= 2 \int_{0}^{1} \left( \frac{x^2}{2} - yx \right) \bigg|_{y}^{1} \, dy \\ &= 2 \int_{0}^{1} \frac{1}{2} - y + \frac{y^2}{2} \, dy \\ &= 2 \left( \frac{1}{6} \right) \\ &= \boxed{ \frac{1}{3} } \end{align}

But looking at that line-segment illustration above suggests another way of looking at this problem.

Let $M = max(X,Y)$

Let $L = min(X,Y)$

And so

\begin{align} |X-Y| &= M - L \\ \\ \mathbb{E}(M-L) &= \frac{1}{3} &\quad \text{from our previous proof} \\ \mathbb{E}(M) - \mathbb{E}(L) &= \frac{1}{3} &\quad \text{since } X,Y \text{ are i.i.d.} \\ \\ \mathbb{E}(M+L)&= \mathbb{E}(M) + \mathbb{E}(L) &\quad \text{by linearity} \\ &= \frac{1}{2} + \frac{1}{2} &\quad \text{since} X \sim \operatorname{Unif}(0,1) \text{, } Y \sim \operatorname{Unif}(0,1) \\ &= 1 \\ \\ \mathbb{E}(M) &= \frac{2}{3} \\ \mathbb{E}(L) &= \frac{1}{3} \end{align}

### Chicken-egg Problem

Say we have $N$ eggs; let $N \sim \operatorname{Pois}(\lambda)$

Some of the eggs hatch, while some don't. Each egg hatches with probability $p$. The event of an egg hatching is independent of the others.

Let $X$ be the number of eggs out of $N$ that do hatch, so $X|N \sim \operatorname{Bin}(N,p)$.

Let $Y$ be the number of eggs that don't hatch, so $X+Y = N$.

Find the joint PMF of $X,Y$. Are they independent?

\begin{align} P(X{=}i, Y{=}j) &= \sum_{n=0}^{\infty} P(X{=}i, Y{=}j |N{=}n) \, P(N{=}n) \\ \\ &= P(X{=}i, Y{=}j |N{=i+j}) \, P(N{=i+j})　&\quad \text{since we know that } i + j = n \\ \\ &= P(X{=}i | N{=i+j}) \, P(N{=i+j})　&\quad Y{=}j \text{ is redundant} \\ \\ &= \binom{i+j}{i} p^i \, q^j \cdot \frac{e^{-\lambda} \lambda^{i+j}}{(i+j)!} &\quad \text{from definition of binomial and Poisson} \\ \\ &= \frac{(i+j)!}{i! \, j!} p^i \, q^j \cdot \frac{e^{-\lambda} \lambda^{i+j}}{(i+j)!} \\ \\ &= \frac{(\lambda p)^i}{i!} \, \frac{(\lambda q)^j}{j!} \, e^{-\lambda} \\ \\ &= \left(e^{-\lambda p} \frac{(\lambda p)^i}{i!} \right) \left(e^{-\lambda q} \frac{(\lambda q)^j}{j!} \right) &\quad \text{since } p + q = 1 \\ \\ &\Rightarrow X,Y \text{ are independent, } X \sim \operatorname{Pois}(\lambda p), Y \sim \operatorname{Pois}(\lambda q) \end{align}


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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Substituting $x= 4$ to the given equation, $6-x=-2$, then \begin{array}{l} 6-4=-2 \\\\ 2=-2 \text{ (FALSE)} .\end{array} Since the solution above ended with a $\text{ FALSE }$ statement, then the given value of the variable is $\text{ NOT a solution }$ to the given equation.

format: plain


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# Chrome extension

I just made a Chrome extension. I followed this tutorial to get a nice button in my toolbar. Now if I load a page, the javascript in my extension gets executed.

Unfortunately, if I click the extension's button in my toolbar, nothing happens.

Question: How do I make the js get executed when I click the extension's button in my toolbar?

Edit I added background.html and it still doesn't work:

<html>
<script>
// Called when the user clicks on the browser action.
var action_url = "javascript:mathjax.js";
chrome.tabs.update(tab.id, {url: action_url});
});
</script>
</html>

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You have two options. Either, you can specify a "popup": "popup.html" in your manifest.json, then put the code in popup.html. In this case, the code will get executed in the popup (though of course you can communicate with your background page or anything else).

Probably better would be to put a callback in background.html by attaching a listener to chrome.browserAction.onClicked.

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Lovely, thank you, can you tell me how to attach a listener to onClicked? – Matt N. Dec 6 '11 at 7:56
I put this into background.html, but it still won't trigger: <!doctype html> <html> <head> <title>Background Page</title> <script chrome.browserAction.onClicked.addListener(function(tab) { chrome.tabs.executeScript(null, {code:"mathjax.js"}); });></script> </head> <body> </body> </html> – Matt N. Dec 6 '11 at 8:13
Odd... I'll play with it later today and see what's up. What happens if you just try to surface an alert onClicked? – Alex Churchill Dec 6 '11 at 23:56
Thank you! I think the problem is with the mathjax queue. I posted a separate question: stackoverflow.com/questions/8397331/chrome-extension-revisited – Matt N. Dec 7 '11 at 11:51

Chrome extension doesn't support inline javascript code in background.html.You must specify external javascript file.Also instead of specifying background.html you can directly specify external file in manifest file in background property.

I have answered similar question here.

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Maybe add some of the details from the other answer into this one, rather than providing only a link / Review – Ms Yvette Jan 1 at 6:25

## protected by Community♦Dec 7 '15 at 5:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.


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# MiniKanren Second Commandment

To transform a function whose value is not a Boolean into a function whose value is a goal, add an extra argument to hold its value, replace cond with conde, and unnest each question and answer.


text:
# Gauss Theorem and Weil Conjecuters for elliptic curves

Hi there!

It is known (by Gauss) that for a prime $p \equiv 1 \ (mod \ 3)$ there is a "unique" writing of $4p=A^2+27B^2$ where $A=1+p-M_p$ and $M_p$ is the number of solutions of $X^3+Y^3+Z^3=0$ in the projective plane $\mathbb{P^2(\mathbb F_p)}$.

It's also known that the points on the cubic $y^3=x^3+1$ are in 1-1 correspondance with the points on the elliptic curve $y^2=x^3-432$.

I noted that $A$ is the same (up to its sign) of the Frobenius trace in the Weil conjectures for elliptic curves; and the number $A^2-4p$ is actually the discriminant of the caracteristic polinomyal of the map the Frobenius automorphism induces on the Tate module.

So, I've the following question to you:

Is there a proof of $p \equiv 1 \ (mod \ 3) \Rightarrow 4p=A^2+27B^2, \ A=1+p-M_p$ using Weil conjectures?

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What does "using Weil conjectures" mean? That statement follows from identifying Frobenius in the ring of endomorphisms of the elliptic curve, which is an order in the field of cube roots of one. –  Felipe Voloch Sep 23 '11 at 11:49
The problem is that I do not know how to show that Frobenius is in $\mathbb Z[\zeta_3]$. –  user18019 Sep 23 '11 at 12:00
The elliptic curve has an automorphism of order 3, $x \mapsto \zeta_3 x$, so the ring of endomorphisms contains $\zeta_3$. The condition $p \equiv 1 \mod 3$ implies that the curve is ordinary so the endomorphism ring is an order in a quadratic field. So the quadratic fiels has to be generated by $\zeta_3$ and you are done. But the "Weil conjectures" are not used anywhere. –  Felipe Voloch Sep 23 '11 at 12:32
I've read of this. I found this proof in many books, but the point it that I would like to know if or not a prove by WCs exists. Thank you anyway!! –  user18019 Sep 23 '11 at 13:03
Just becouse I would like -if possible of course!- to put in my thesis this fact as an application of the Weil conjectures. –  user18019 Sep 23 '11 at 14:00

I am going to interpret "the Weil conjectures" as a synechdoche for "thinking about counting points in terms of eigenvalues of Forbenius operators." In this case, the answer is a resounding yes! However, the specific facts about eigenvalues of Frobenius which which are being used here are easier than the ones discussed in the Weil conjectures. Silverman's The Arithmetic of Elliptic Curves might be a good reference for you.

Let $E$ be the genus one curve $X^3+Y^3+Z^3$ over $\mathbb{F}_p$. This curve has an automorphism $\Phi: (x:y:z) \mapsto (z^p:y^p:z^p)$. The map $\Phi$ sends every point with values in $\mathbb{F}_p$ to itself. However, points with values in $\mathbb{F}_{p^n}$ for higher values of $n$ are permuted. Our goal is to find the number of fixed points of $\Phi$.

The Weil conjecture way to do this is to associate to $E$ certain vector spaces $H^0(E)$, $H^1(E)$ and $H^2(E)$. The map $\Phi$ acts functorially on cohomology, and we have $$\#(\mbox{Fixed points of }\Phi) = Tr(\Phi^*:H^0 \to H^0) - Tr(\Phi^*:H^1 \to H^1) + Tr(\Phi^*:H^2 \to H^2).$$ Now, in this case, you can really get quite explicit about what these vector spaces are. $H^0$ and $H^2$ are both $1$-dimensional, and $\Phi$ acts by $1$ and $p$ respectfully. $H^1$ the Tate module of $E$ (Silverman III.7); a two dimensional space.

So the number of fixed points is $p+1-a$ where $a$ is the trace of $\Phi^*$ acting on the Tate module.

One can show that the endomorphism ring of $E$ is $\mathbb{Z}[\omega]$ where $\omega$ is a primitive cube root of unity (see Silverman V.3). So we must have $\Phi = c+d \omega$ for some integers $c$ and $d$. The fact that $\Phi$ has degree $p$ (see Silverman II.2) means that $N(c \omega +d) = c^2 - cd + d^2 =p$. And one computes that $a = Tr(\Phi) = 2c - d$.

So, in short, the number of points on $E$ is $p+1-a$, where $a=2c-d$ with $(c,d)$ such that $N(c+d \omega)=p$. Converting that into your statement, and getting the signs right, is left to you.

Focusing on the more specific question of using the Weil conjectures. The fact that $N(c+d \omega)$ is $p$ comes from the Riemann hypothesis part of the Weil conjectures. But you also need to know that $\mathbb{Z}[\omega]$ is the full ring of automorphisms, which is not particularly related to the Weil conjectures. And you need to figure out which of the $6$ elements of $\mathbb{Z}[\omega]$ which have norm $p$ is the correct one, which is also not a Weil computation. So I would say that this is not specifically a Weil conjecture result, although definitely can be proven using the same sort of ideas as in the Weil conjectures.

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I found it! One can write the number of solution of $y^2=x^3-432$ in term of a Jacobi sum $S \in \mathbb Z[\zeta_3]$, namenly $\# E(\mathbb F_p)=1+p+S+\overline S$. It has absolute value $p$. Furthermore, from what I know from Weil Conjectures, $S+\overline S$ is equal to minus the norm of Frobenius. Then Frobenius is $-S$ and hence it is in the form $c+d\zeta_3$. The rest comes from simple aritmetics calculations and the fact that $3$ (and $9$ too) divides $\# E(\mathbb F_p)$. Thak you very much David, you've been very usefull to me! –  user18019 Sep 25 '11 at 7:25
Isn't $H^1$ dual (rather than equal) to the tate module? –  Daniel Loughran Oct 3 '11 at 21:54


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But I think as I continue to diversify, I like putting it into the two buckets of (1) I control (2) no control (stock market). I like that clarity. The illusion of control does add stress and hassle that detract from enjoying your life. Not worth extra returns to me. India's mineral resources are vast.[274] However, its mining industry has declined – contributing 2.3% of its GDP in 2010 compared to 3% in 2000, and employed 2.9 million people – a decreasing percentage of its total labour. India is a net importer of many minerals including coal. India's mining sector decline is because of complex permit, regulatory and administrative procedures, inadequate infrastructure, shortage of capital resources, and slow adoption of environmentally sustainable technologies.[270][275] A lot of people these days are moving towards the two job concept. Amongst the people i know who have applied this in their life; the primary reason is that the 9 to 5 job pays their bills, lets the fire burning in the kitchen, and the second job is where their passion lies. This is the passion, which might have been forgotten while growing up, or is not a viable primary income source. ## It’s obvious, but every successful product solves a problem or fills a need- if the problem must be solved quickly, even better. What’s an urgent need? How about a car repair when you’re stuck on the side of the road (‘had the transmission on my car give out on a highway exit ramp. I don’t recommend it). Another good example is heat or air conditioning repair at home. It is always fun (when things are going well!) to look back at the various streams to see what’s working and what’s not. I found that a lot of my angel investing just wasn’t working well, fortunately it wasn’t a lot! Side businesses are always nice, vs. pure investments, because of actual control. Plus you can shut it down if things go south… hard to tell someone (and convince them when you’ve only kicked in a few bucks) that it’s time to close up shop and return some capital. This equation implies two things. First buying one more unit of good x implies buying {\displaystyle {\frac {P_{x}}{P_{y}}}} less units of good y. So, {\displaystyle {\frac {P_{x}}{P_{y}}}} is the relative price of a unit of x as to the number of units given up in y. Second, if the price of x falls for a fixed {\displaystyle Y} , then its relative price falls. The usual hypothesis is that the quantity demanded of x would increase at the lower price, the law of demand. The generalization to more than two goods consists of modelling y as a composite good. ## If you have an empty house or room you can rent it out on AIRBNB and OYO Rooms. Many travelers are looking to spend one night at a place. You can always rent out your empty house or room to them. All you need to do is list your room or house online, explain the rules and you are good to go. Travelers will pay you online. This way you don’t have to search for clients. They will come to you. With the way the economy is going Multiple streams of income is the way to go if you know what your doing. I luved your insight and your so right there are so many ways. The best way that i know of is to brand oneself cause people dont join business they join YOU. Like , trust is key ingredient along with building your list after all the money is in the list or should i say the people in the list. The best part with the system im presently using is i still get paid with affiliate even if they don’t join my primary biz… As an struggling young Engineer (back in the Carter era) I bought anything I could renovate then rent to justify paying the 18% interest. I never took vacations but worked on my properties all in the pursuit of passive income. I drove junk for many years & many months I just got by on credit cards. My friends & colleagues were amused by my ‘stupidity’ but most are still working to make enough for retirement. I first discovered the power of passive income when I was a senior in high school. I started a mobile billboard business where I would rent a small piece of land from someone who had land along a busy highway. Then I would place one of my billboard trailers on the land and rent out the ad space on the billboard. I would usually charge about$300 per month for the ad space, meanwhile I was only paying \$50 per month to the landowner for the ground rent. I got to the point to where I had 9 billboard faces and was making quite a substantial income for someone in high school. I really learned how passive income could free up my life… this business is what lead me into investing in real estate.

Portfolio income can come from multiple sources – interest/bond coupons, stock dividends, financial strategies including derivatives and capital growth.  Each offers some cash flow and some also offer potential capital gain with some risk from liquidity and volatile prices.  In this low-yield environment, many investors rely too heavily on cash flows and to pursue this are venturing further into riskier areas than they’d normally consider.


text:
Journal article Open Access

# Plant leaf Disease Identification using Deep learning techniques

Archanaa.R; Shridevi.S

### JSON-LD (schema.org) Export

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"inLanguage": {
"alternateName": "eng",
"@type": "Language",
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"@id": "",
"@type": "CreativeWork"
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"@id": "https://hdl.handle.net/E9683069520/2020\u00a9BEIESP",
"@type": "CreativeWork"
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],
"description": "
Agriculture is an important source of our country’s growth. The major loss in an agricultural economy is because of the plant disease. Though technology plays a vital role in all the fields still today the agriculture field is using the old methodologies. Successful cultivation depends on identifying plant disease. Previously the identification was done manually by the experienced people but now it became difficult due to environmental changes. By using the deep learning techniques the plant disease can be identified effectively. Vgg16 and ResNet are the proposed techniques to increase accuracy than the existing system. The disease can be identified with images of the leaves by applying those deep learning techniques. Detection can be involved in steps like image acquisition, image pre-processing, image segmentation, feature extraction, and classification. By controlling the disease, productivity can be increased. The features of the leaf image are taken and trained using the neural network algorithm and then the prediction is done by testing the images. The features of the leaf image are taken and trained using the neural network algorithm and then the prediction is done by testing the images.
",
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"affiliation": "Vellore Institute Of Technology, Chennai",
"@type": "Person",
"name": "Archanaa.R"
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"affiliation": "Vellore Institute Of Technology, Chennai",
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"headline": "Plant leaf Disease Identification using Deep  learning techniques",
"datePublished": "2020-06-30",
"keywords": [
"Data augmentation, Resnet, VGG-16, CNN."
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"url": "https://zenodo.org/record/5547269",
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"name": "Plant leaf Disease Identification using Deep  learning techniques"
}
38
16
views


text:
There is a newer version of this record available.

Dataset Open Access

# OrchideaSOL: an audio dataset of isolated musical notes, including mutes and extended playing techniques

Carmine-Emanuele Cella; Daniele Ghisi; Vincent Lostanlen; Fabien Lévy; Joshua Fineberg; Yan Maresz

### Citation Style Language JSON Export

{
"publisher": "Zenodo",
"DOI": "10.5281/zenodo.3686252",
"title": "OrchideaSOL: an audio dataset of isolated musical notes, including mutes and extended playing techniques",
"issued": {
"date-parts": [
[
2020,
2,
24
]
]
},
"abstract": "
OrchideaSOL
\n==========
\nVersion 1.0, February 2020.
\n
\n\n

\n\n
Created By
\n--------------
\n\n
Carmine-Emanuele Cella (1), Daniele Ghisi (1), Vincent Lostanlen (2), Fabien Lévy (3), Joshua Fineberg (4), Yan Maresz (5)
\n
\n(1): UC Berkeley
\n(2): New York University
\n(3): Columbia University
\n(4): Boston University
\n(5): Conservatoire de Paris
\n\n

\n\n
Description
\n---------------
\n\n

\nOrchideaSOL is a dataset of 13265 samples, each containing a single musical note from one of 14 different instruments:
\n\n
\n\t
Bass Tuba
\n\t
French Horn
\n\t
Trombone
\n\t
Trumpet in C
\n\t
Accordion
\n\t
Contrabass
\n\t
Violin
\n\t
Viola
\n\t
Violoncello
\n\t
Bassoon
\n\t
Clarinet in B-flat
\n\t
Flute
\n\t
Oboe
\n\t
Alto Saxophone
\n
\n\n

\n\n
These sounds were originally recorded at Ircam in Paris (France) between 1996 and 1999, as part of a larger project named Studio On Line (SOL). One asset of OrchideaSOL is that it contains many combinations of mutes and extended playing techniques.
\n
\nThe OrchideaSOL audio data can be used for creative purposes insofar at the use complies with the Ircam Forum License. Please visit: https://forum.ircam.fr/legal/contrat-de-licence-forum-ircam/
\n\n

\nThe OrchideaSOL metadata can be used for creative purposes insofar at the use complies with the Creative Commons Attribution 4.0 International license (see below).
\n
\nOrchideaSOL can be used for education and research purposes. In particular, it can be employed as a dataset for training and/or evaluating music information retrieval (MIR) systems, for tasks such as instrument recognition, playing technique recognition, or fundamental frequency estimation. For this purpose, we provide an official 5-fold split of OrchideaSOL. This split has been carefully balanced in terms of instrumentation, pitch range, and dynamics. For the sake of research reproducibility, we encourage users of OrchideaSOL to adopt this split and report their results in terms of average performance across folds.
\n\n

\n\n
Data Files
\n--------------
\n\n
OrchideaSOL contains 13265 audio clips as WAV files, sampled at 44.1 kHz, with a single channel (mono), at a bit depth of 16. This is equivalent to the audio quality of a compact disc. Audio clips vary in duration between two and ten seconds.
\n\n
Every audio file has a file path of the form:
\n<FAMILY>/<INSTRUMENT><+MUTE>/<TECHNIQUE>/<INSTR><+M>-<TECH>-<PITCH>-<DYN>-<INSTANCE>-<MISC>.wav
\n\n

\nwhere:
\n\n
\n\t
<FAMILY> corresponds to the instrument family: "Brass", "Keyboards" (includes accordion), "Strings", and "Winds" (i.e., woodwinds).
\n\t
<INSTRUMENT> is the full name of the instrument.
\n\t
<+MUTE> is the type of mute being used, such as "wah", "harmon", "piombo", or "sordina". If there is no mute, this field is absent.
\n\t
<TECHNIQUE> is the type of playing technique.
\n\t
<INSTR> is the abbreviation of the instrument.
\n\t
<+M> is the abbreviation of the type of mute, if applicable.
\n\t
<TECH> is the abbreviation of playing technique.
\n\t
<PITCH> denotes the pitch of the musical note. This pitch is encoded in the American standard pitch notation: pitch class (C means "do") followed by pitch octave. According to this convention, A4 has a fundamental frequency of 440 Hz.
\n\t
<DYN> denotes the intensity dynamics, ranked from pp (pianissimo) to ff (fortissimo).
\n\t
<INSTANCE> contains additional information, when applicable. For example, for bowed string instruments, the same pitch may sometimes be achieved on different positions and different strings, resulting in small timbre differences. In this case the label "1c", "2c", "3c", or "4c" denotes the string which is being bowed. (The letter c originates from the word "corde", which means string in French.) By convention, the first string is the one with the highest pitch when played as an open string. Furthermore, on some wind instruments, the same note was played multiple times, e.g. at multiple durations. In this case, we use the label "alt1", "alt2", etc. to denote alternative instances of the note. If none of these tags apply, the <INSTANCE> field becomes "N", which stands for "Not Applicable".
\n\t
<MISC> contains additional information, if applicable. In OrchideaSOL, some pitches were never recorded, and thus missing from the chromatic scale. In this case, the <MISC> tag contains a letter "R", to denote the fact that the corresponding WAV file has been obtained by transforming a different audio clip via some digital frequency transposition (similar to Auto-Tune). The letter "R" stands for "resampled". Furthermore, some pitches were slightly out of tune in comparison with the A440 tuning standard. Again, we applied some digital frequency transposition to correct them and put them exactly in tune. The amount of frequency transposition is measured in "cents" of an equal-tempered semitone. The letter "T" stands for "tuned". Because we employed a high-fidelity algorithm for frequency transposition, and because the amount of digital frequency transposition is small, the timbre of pitch-corrected notes remains faithful to the instrument. If none of these tags apply, the <MISC> field becomes "N", which stands for "natural"; in this case, the note is distributed exactly as it was recorded in the studio.
\n
\n\n
For example, "Strings/Violin+sordina/tremolo/Vn+S-trem-A4-mf-4c-T13d_R200d.wav" corresponds to:
\n\n
\n\t
a violin sound;
\n\t
equipped with a sordina mute;
\n\t
played in the tremolo playing technique;
\n\t
at pitch A4 (440 Hz);
\n\t
with mezzoforte dynamics;
\n\t
on the fourth string (i.e. the lowest);
\n\t
resampled from a B4 by lowering pitch by a semitone, i.e. 100 cents (R100d)
\n\t
lowered by 13 cents (T22d) to match the A440 tuning standard.
\n
\n\n

\n\n
The audio data for OrchideaSOL is not directly downloadable on Zenodo. Rather, it can be downloaded for free after registering to the Ircam forum. Please visit: https://forum.ircam.fr/
\n\n

\n\n
Metadata File
\n-------------------
\n\n
The OrchideaSOL_metadata.csv file contains 13265 rows, one for each audio clip. It can be opened by a text editor or by a spreadsheet software application. It contains 13 columns:
\n\n
\n\t
Path to the WAV file, in UNIX filesystem format. For Windows compatibility, replace the slashes ("/") by backslashes ("\\"). Ex: "Strings/Violin+sordina/tremolo/Vn+S-trem-A4-mf-4c-T13d_R200d.wav"
\n\t
Fold ID. Either equal to 0, 1, 2, 3, or 4.
\n\t
Family. Ex: "Brass"
\n\t
Instrument abbreviation. Ex: "BTb"
\n\t
Instrument name in full. Ex: "Bass Tuba"
\n\t
Technique abbreviation.
\n\t
Technique name in full.
\n\t
Pitch. Ex: "A#1"
\n\t
Pitch ID in MIDI format. Ex: 34. Integer in the range 0-127.
\n\t
Dynamics. Ex: "ff".
\n\t
Dynamics ID. Integer. pp maps to 0 and ff maps to 4. The higher, the louder.
\n\t
Instance ID. Integer in the range 0-4
\n\t
String ID. Equal to 1, 2, 3, 4, or empty if not applicable.
\n\t
"Needed digital retuning". TRUE if the file has been pitch-shifted with digital audio effects; FALSE otherwise.
\n
\n\n

\n\n
Conditions of Use
\n------------------------
\n\n
OrchideaSOL was created in 2020 by Carmine-Emanuele Cella, Daniele Ghisi, Vincent Lostanlen, Fabien Lévy, Joshua Fineberg, and Yan Maresz.
\n\n
OrchideaSOL is a derivative of SOL. We wish to thank Hugues Vinet, Greg Beller, and all coordinators of the Ircam Forum for their authorization to upload the metadata of OrchideaSOL to Zenodo.
\n\n
The audio samples in OrchideaSOL are offered free of charge under the Ircam Forum License. Please visit: https://forum.ircam.fr/legal/contrat-de-licence-forum-ircam/
\n\n
The dataset and its contents are made available on an "as is" basis and without warranties of any kind, including without limitation satisfactory quality and conformity, merchantability, fitness for a particular purpose, accuracy or completeness, or absence of errors. Subject to any liability that may not be excluded or limited by law, the authors are not liable for, and expressly exclude all liability for, loss or damage however and whenever caused to anyone by any use of the OrchideaSOL dataset or any part of it.
\n\n

\n\n
Versions
\n-----------
\n1.0 was released on February 24th, 2020.
\n\n

\n\n
Feedback
\n-------------
\n\n
Please help us improve OrchideaSOL by sending your feedback to:
\ncarmine.cella@berkeley.edu
\n\n
For issues regarding the metadata encoding, the five-fold split, or the OrchideaSOL module in mirdata, please write to:
\nvincent.lostanlen@nyu.edu
\n\n
In case of a problem, please include as many details as possible.
",
"author": [
{
"family": "Carmine-Emanuele Cella"
},
{
"family": "Daniele Ghisi"
},
{
"family": "Vincent Lostanlen"
},
{
"family": "Fabien L\u00e9vy"
},
{
"family": "Joshua Fineberg"
},
{
"family": "Yan Maresz"
}
],
"version": "1.0",
"type": "dataset",
"id": "3686252"
}
1,158
856
views
downloads
All versions This version
Views 1,158847
Downloads 856644
Data volume 1.7 GB1.2 GB
Unique views 914763
Unique downloads 668579


text:
# When does the virtual cohomological dimension become a numerical invariant?

In group cohomology theory, we know that the cohomological dimension $cd(G)$ for a profinite group is a fundamental numerical invariant. We say $G$ is of virtual cohomological dimension $n$ (denote it by $vcd(G)$) if there exists an open subgroup $H$ such that $cd(H)=n$.

It seems that $vcd(G)$ is not always a fixed integer and my question is:

Is there any criterion when $vcd(G)$ become fixed ,or say, when $cd(G)=cd(H)$ for all open subgroup $H$ of $G$ ? If there are any related references please let me know, thx !

-

Although it is not obvious from the definition, in fact -- assuming that there exists at least one open subgroup of finite cohomological dimension, otherwise there is nothing to say -- the virtual cohomological dimension is always a fixed integer. In other words, if there exists a single open subgroup $H_0$ of $G$ such that $\operatorname{cd}(H_0) < \infty$, then for all open subgroups $H$ of $G$ with $\operatorname{cd}(H) < \infty$, we have
$\operatorname{cd}(H) = \operatorname{cd}(H_0)$.
(Moreover, this also holds for the $p$-cohomological dimension at any prime $p$.)
I believe this result was first proved by Serre. In any case, it follows easily from Proposition I.14 in Serre's Galois Cohomology. In fact, a slightly stronger result is given as Proposition I.14$'$: again assuming that there exists at least one open subgroup $H_0$ of finite cohomological dimension, an open subgroup $H$ has infinite cohomological dimension if $H$ has nontrivial elements of finite order and cohomological dimension equal to $\operatorname{cd}(H_0)$ otherwise.


text:
# A discrete mathematics problem by Paola Ramírez

How many numbers from 1 to $$10^{2000}$$ have the sum of their digits equal to 2?

×


text:
### no facets defined for boundary of complex mesh

418
views
1
7 months ago by
We are having trouble defining the surface boundary (i.e. the facets) of a mesh created by an external tool (a CGAL .mesh file, but the problem remains if we use gmsh to convert to .msh format).  dolfin-convert converts the mesh itself to dolfin format (converts .mesh or .msh to .xml).  But no facet_regions file is generated.

My steps are:

1) generate the .mesh using a cgal script.  An example script is the sphere at https://doc.cgal.org/latest/Mesh_3/index.html#Mesh_33DDomainsImplicitIsosurfaces
- use cgal to generate the mesh file out.mesh with:      cgal_create_cmake_script; cmake .; make; ./mesh_implicit_sphere
1b) if desired, convert to .msh by opening the mesh file in gmsh and using gmsh saving the mesh.

2) dolfin-convert out.mesh out.xml   (or dolfin-convert out.msh out.xml, for the gmsh file)

After this, the dolfin mesh file out.xml is created, but it contains only vertices, not facet information.  The .mesh file does contain Triangle entries, which I understand would correspond to the boundary.  Discussion of the conversion on websites indicates that an out_facet_region.xml file should also be created, which can be read into a FacetFunction, and identifies my boundary vertices.  But no such *facet_region.xml is created.

dolfin-convert uses /usr/lib/python2.7/dist-packages/dolfin_utils/meshconvert/meshconvert.py. If I hack that, I see on l.468, where it tests
    tags = tags_for_dim[highest_dim-1]
if (len(tags) > 0) and (mesh is not None):
physical_regions = tuple(tag[0] for tag in tags)
if not all(tag == 0 for tag in physical_regions):

that the tags read from my .mesh file have the form [0,1] .  So "tag==0" and the facets are not processed.

Is there some complication in importing CGAL (or gmsh) meshes to dolfin that we haven't accounted for?

What we're trying to do is solve our system inside a nontrivial mesh.  But because a FacetFunction isn't being defined during import of the mesh file, dolfin can't tell us where the boundary is, so we can't apply the correct boundary conditions.
Community: FEniCS Project
CGAL (and gmsh) are beside the point (so I've edited the question).  I can reproduce the problem with a Sphere mesh generated by mshr:
from dolfin import *
import mshr

def sphereSurface(x, on_boundary):
return bool(near(x[0]**2+x[1]**2+x[2]**2,1) and on_boundary)

class DirichletBoundary1(SubDomain):
def inside(self, x, on_boundary):
return sphereSurface(x, on_boundary)

# Create mesh and define function space
sphere = mshr.Sphere(Point(0,0,0),1)
mesh = mshr.generate_mesh(sphere, 20)
V = FunctionSpace (mesh, "CG", 1)

# Create Dirichlet boundary condition
boundaries = FacetFunction("size_t", mesh, 0)
db1 = DirichletBoundary1()
db1.mark(boundaries, 1)

#bc = DirichletBC(V, Constant(1), sphereSurface)
bc = DirichletBC(V, Constant(1), boundaries, 1)

# Define variational problem
u_t = TrialFunction(V)
v = TestFunction(V)
f = Constant(0.0)
L = f*v*dx

# Compute solution
u = Function(V)
solve(a==L, u, bc)

plot(u, interactive=True)

The output is the same whether I set the DirichletBC using a function to define the boundary, or using a subclass of SubDomain. stdout reports
Generating mesh with CGAL 3D mesh generator
Solving linear variational problem.
*** Warning: Found no facets matching domain for boundary condition.

and the plot shows uniform 0 on the surface of the sphere, where it should be 1.

The warning about "no facets' is a clue to the problem, but how do I solve it?
written 7 months ago by Drew Parsons

0
7 months ago by
Looks like the problem arises from the use of near to determine if x is on the boundary or not.
If instead I use the more explicit test,
def sphereSurface(x, on_boundary):
return bool( (abs(x[0]**2+x[1]**2+x[2]**2-1) < 7.0E-3) and on_boundary)

then the boundary seems to work (at least, it gets set to 1 as required).

Any tighter test, even < 6.0E-3, fails to detect the boundary.

This level of precision is quite a bit lower than I'd expect. Is it normal, for the mesh density (20) used in this example?

hi there, I am experiencing a similar issue. May be somewhat different but the common issue is that i am also not getting facets from my mesh preprocessing in another CAD software called salome.

I wanted to ask that did you find a solution to your actual problem ? ...
written 5 months ago by Ovais
I didn't get any more of a solution beyond what I reported, that the precision of the near() function seemed to be too tight.
written 8 weeks ago by Drew Parsons


text:
# If $\sf{E} = \sf{NE}$. Then $\sf{NP}-{P}$ contains no sparse sets [closed]

I am reading "The Complexity Companion" by Hemaspaandra & Ogihara, I have a question about lemma 1.21. In its proof, they suppose $$L$$ is some sparse language in $$\sf{NP}$$ ($$||L^{=n}|| for some polynomial) and then construct the language \begin{align*} L' = &\{0 \# n \# k , \text{ there are more than k words of length n in L} \} \cup \\ &\{ 1\#n\#c \#i \#j \text{ there are c words z_1,\dots,z_c in L of length n such that the j-th bit of z_i is 1 } \} \end{align*} Since $$L$$ is in $$\sf{NP}$$, then $$L'$$ is $$\sf{NE}$$. Then, one can construct a $$\sf{P}$$ algorithm for $$L$$ by checking in $$L'$$ each of the words $$0 \# n \# 0, \dots , 0 \# n \# p(n)$$ in order to obtain $$c=||L^{=n}||$$, and then checking if $$1\#n\#c \#i \#j \in L'$$ for $$i \in \{1,\dots,c\} , j \in \{1,\dots,n\}$$, this is a polynomial number of checks in $$L'$$, so (a priori) one would get that $$L \in E$$. But they claim that even though $$L'$$ is $$NE$$, its individual running time in this case is actually $$O(\log n)$$ which of course solves this problem.

Why is this? Is it because of the size of the input is made smaller when coded in binary?

• Yes, the input size to the queries you are making to $L'$ is $O(\log n)$ because they are written in binary, so the fact that $L' \in \mathsf{E}$ means that those queries can each be answered in $2^{O(\log n)}=poly(n)$ time. – Joshua Grochow Dec 3 '20 at 16:54


text:
# Pollard's rho algorithm

Pollard's rho algorithm is a general-purpose integer factorization algorithm. It was invented by John Pollard in 1975.[1] It is particularly effective at splitting composite numbers with small factors.

## Core ideas

The rho algorithm is based on Floyd's cycle-finding algorithm and on the observation that (as in the birthday problem) two numbers x and y are congruent modulo p with probability 0.5 after $1.177\sqrt{p}$ numbers have been randomly chosen. If p is a factor of n, the integer we are aiming to factor, then $p \le \gcd \left( x-y,n \right) \le n$ since p divides both $x-y$ and $n$.

The rho algorithm therefore uses a function modulo n as a generator of a pseudo-random sequence. It runs one sequence twice as "fast" as the other; i.e. for every iteration made by one copy of the sequence, the other copy makes two iterations. Let x be the current state of one sequence and y be the current state of the other. The GCD of |xy| and n is taken at each step. If this GCD ever comes to n, then the algorithm terminates with failure, since this means x = y and therefore, by Floyd's cycle-finding algorithm, the sequence has cycled and continuing any further would only be repeating previous work.

## Algorithm

The algorithm takes as its inputs n, the integer to be factored; and f, a function with the property that x=y mod p implies f(x)=f(y) mod p. In the original algorithm, f(x) = x2 − 1 (mod n). The output is either a non-trivial factor of n, or failure. It performs the following steps:[2]

1. x ← 2, y ← 2; d ← 1
2. While d = 1:
1. xf(x)
2. yf(f(y))
3. d ← GCD(|xy|, n)
3. If d = n, return failure.
4. Else, return d.

Note that this algorithm may not find the factors and will return failure for composite n. In that case, use a different f(x) and try again. Note, as well, that this algorithm does not work when n is a prime number, since, in this case, d will be always 1. The algorithm is so-called because the values of f enter a period (mod d), resulting in a ρ shape when diagrammed.

## Variants

In 1980, Richard Brent published a faster variant of the rho algorithm. He used the same core ideas as Pollard but a different method of cycle detection, replacing Floyd's cycle-finding algorithm with the related Brent's cycle finding method.[3]

A further improvement was made by Pollard and Brent. They observed that if $\gcd (a,n) >1$, then also $\gcd (ab,n)>1$ for any positive integer b. In particular, instead of computing $\gcd (|x-y|,n)$ at every step, it suffices to define z as the product of 100 consecutive $|x-y|$ terms modulo n, and then compute a single $\gcd (z,n)$. A major speed up results as 100 gcd steps are replaced with 99 multiplications modulo n and a single gcd. Occasionally it may cause the algorithm to fail by introducing a repeated factor, for instance when n is a square. But it then suffices to go back to the previous gcd term, where $\gcd(z,n)=1$, and use the regular Rho algorithm from there.

## Application

The algorithm is very fast for numbers with small factors, but slower in cases where all factors are large. The rho algorithm's most remarkable success has been the factorization of the eighth Fermat number (F8) by Pollard and Brent. They used Brent's variant of the algorithm, which found a previously unknown prime factor. The complete factorization of F8 took, in total, 2 hours on a UNIVAC 1100/42.

## Example factorization

Let n = 8051 and f(x) = (x2 + 1) mod 8051.

i xi yi GCD(|xiyi|, 8051)
1 5 26 1
2 26 677 1
3 677 7474 97

97 is a non-trivial factor of 8051. Other values of c may give the cofactor (83) instead of 97.

## Complexity

The algorithm offers a trade-off between its running time and the probability that it finds a factor. If the squaring function used in the Pollard rho method were replaced by a random function, it would follow that, for all n, running the algorithm for O(n1/4) steps would yield a factor with probability at most 1/2. It is believed that the same analysis applies as well to the actual rho algorithm, but this is a heuristic claim, and rigorous analysis of the algorithm remains open.[4]

## References

1. ^ Pollard, J. M. (1975), "A Monte Carlo method for factorization", BIT Numerical Mathematics 15 (3): 331–334
2. ^ Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L. & Stein, Clifford (2001), "Section 31.9: Integer factorization", Introduction to Algorithms (Second ed.), Cambridge, MA: MIT Press, pp. 896–901, ISBN 0-262-03293-7 (this section discusses only Pollard's rho algorithm).
3. ^ Brent, Richard P. (1980), "An Improved Monte Carlo Factorization Algorithm", BIT 20: 176–184, doi:10.1007/BF01933190
4. ^ Galbraith, Steven D. (2012), "14.2.5 Towards a rigorous analysis of Pollard rho", Mathematics of Public Key Cryptography, Cambridge University Press, pp. 272–273, ISBN 9781107013926.


text:
Sie sind hier

## Sufficient second-order optimality conditions for a parabolic optimal control problem with pointwise state constraints

 Source file is available as : Postscript Document Portable Document Format (PDF)

Author(s) : Arnd Rösch , Fredi Tröltzsch

Preprint series of the Institute of Mathematics, Technische Universität Berlin
Preprint 0733-2002

MSC 2000

49K20 Problems involving partial differential equations
90C48 Programming in abstract spaces

Abstract :
An optimal control problem for a semilinear parabolic equation is investigated, where pointwise constraints are given on the control and the state. The state constraints are of mixed (bottleneck) type, where associated Lagrange multipliers can assumed to be bounded and measurable functions. Based on this property, a second-order sufficient optimality condition is established that considers strongly active constraints.

Keywords : Optimal control, heat equation, parabolic differential equation, sufficient second-order optimality condition,pointwise mixed control-state constraints, bottleneck constraints


text:
## LaTeX forum ⇒ Theses, Books, Title pages ⇒ Contents title Topic is solved

Classicthesis, Bachelor and Master thesis, PhD, Doctoral degree
templateuser
Posts: 679
Joined: Tue Mar 03, 2015 4:01 pm

### Contents title

hey,
Can I delete or hide the word "contents" at mini table of contents (minitoc) for each chapter ?
Thank you

Messi007

Tags:

Vel
Site Moderator
Posts: 459
Joined: Fri Jun 29, 2012 1:20 am
Hi,

I'm not sure what template you're asking a question about. The only one with a mini table of contents is the Legrand Orange Book template and that doesn't have a Contents heading in the mini toc.

Cheers,
Vel
Founder and administrator of LaTeXTemplates.com and LaTeXTypesetting.com

templateuser
Posts: 679
Joined: Tue Mar 03, 2015 4:01 pm
Hi,

Messi007

Vel
Site Moderator
Posts: 459
Joined: Fri Jun 29, 2012 1:20 am
Hi,

I've moved your question to the correct forum.

All you need to do is add \renewcommand{\mtctitle}{} just before \begin{document} to remove the "Contents" at the top of the mini-toc!

Cheers,
Vel
Founder and administrator of LaTeXTemplates.com and LaTeXTypesetting.com

templateuser
Posts: 679
Joined: Tue Mar 03, 2015 4:01 pm
Hello,

This didn't work for me, it gave me the following error:

LaTeX Error: \mtctitle undefined.

See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...

l.14 \renewcommand{\mtctitle}
{}

How can I solve this problem?

Liam

Vel
Site Moderator
Posts: 459
Joined: Fri Jun 29, 2012 1:20 am
Hi Liam,

The answer above is for when a mini table of contents is used in the template via the 'minitoc' package. You will need to load this package and use it's features for the solution to work.

Cheers,
Vel
Founder and administrator of LaTeXTemplates.com and LaTeXTypesetting.com


text:
시간 제한메모리 제한제출정답맞힌 사람정답 비율
5 초 512 MB666100.000%

문제

Edison, a robot, does not have a right hand or eyes. As a brave robot, he always puts his left hand on the wall no matter he walks or turns around. Because he thinks it is too dangerous, Edison does not walk backward.

Assume that Edison has found himself in a square-shaped maze of NxN square cells which is surrounded by walls from the outside. In the maze, some of the cells are also walls. Edison can only move between two empty cells in four directions, north, south, west and east. In order to get out of the maze, he drafts a plan. He uses his left hand to lean on the wall and goes by following the wall.

Here is the question, is Edison able to get out of the maze in at most 10,000 steps? If he can make it, output the path. By getting out of the maze, he only needs to be in the exit cell. If the starting cell is the same as the exit, Edison won't need to move and can directly get out of the maze.

입력

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with an integer N. N is the size of the maze. The following N lines, each line contains N characters which may be '.' or '#'. '.' is an empty cell, '#' is a wall. Followed by a line which contains four integers: sx, sy, ex, ey. (sx, sy) means that Edison is standing on row sx and column sy as his starting cell, (ex, ey) is the exit of the maze. (sx, sy) is guaranteed to be at one of the 4 corners of the maze, and Edison can only touch the wall on 4 adjacent cells(not 8) initially. (ex, ey) can be anywhere in the maze. Note that the top-left corner is at position (1,1).

Limits

• 1 ≤ T ≤ 30.
• 1 ≤ sx, sy, ex, eyN.
• The starting cell and the exit of the maze will always be an empty cell. And the starting cell and the exit of the maze won't be the same.
• 2 ≤ N ≤ 10.

출력

For each test case, output a line containing "Case #x: y", where x is the case number (starting from 1) and y is "Edison ran out of energy." (without the quotes) if Edison can't reach the exit of the maze in at most 10,000 steps, otherwise y should be the number of steps followed by another line which contains y characters to describe the path (each character should be E for east, S for south, W for west or N for north). There is no character to represent the turning around. We don't care about the turning around steps, please only output the path of how Edison will cross the maze.

예제 입력 1

3
2
.#
#.
1 1 2 2
5
.##.#
.....
...#.
.###.
...#.
1 1 5 3
3
...
.#.
...
1 1 3 3

예제 출력 1

Case #1: Edison ran out of energy.
Case #2: 22
SEEENSESSSNNNWWSWWSSEE
Case #3: 4
EESS

힌트

Note:

In the 2nd test case after moving 1 cell down from his starting cell, Edison will still be able to lean on the wall at the cell (1,2) by his left hand.

In the third test case, due to Edison can't touch the wall at cell (2,2) initially, so he has to go east in his first step.

채점 및 기타 정보

• 예제는 채점하지 않는다.


text: My pseudonym is $\mu_k$ (pronounced myoo-sub-kay) representing the coefficient of friction, which describes the amount of friction resulting from two opposing objects.


text:
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Note 2 : Lusser’s Law of Reliability states that the total reliability of a system dependent on the function of several independent components is the mathematical product of those components’ individual reliabilities. Examples of SIL … These software tools not only factor in the inherent reliability ratings of different system components, but also correct for preventive maintenance schedules and proof testing intervals so the user may determine the proper maintenance attention required to achieve a given SIL rating. electrical: 2. concerned with electricity : an electrical consultant. Electrical Abbreviations and Full Forms - Free download as PDF File (.pdf), Text File (.txt) or read online for free. It is quite possible that the chemical reactor mentioned in the previous paragraph with an over-pressure protection system SIL rating of 3 might have an over-temperature protection system SIL rating of only 2, due to differences in how the two different safety systems function. FUNCTIONAL SAFETY – SIL Electric actuators for safety-related systems up to SIL 3. Non-Profit Organization. An over-pressure protection system on a chemical reactor process with a SIL rating of 2, for example, has a Probability of Failure on Demand between 0.01 and 0.001 for the specific shutdown function as a whole. A common way of ranking the dependability of a Safety Instrumented Function (SIF) is to use a simple numerical scale from one to four, with four being extremely dependable and one being only moderately dependable: The Required Safety Availability (RSA) value is synonymous with dependability: the probability (Note 1) that a Safety Instrumented Function will perform its duty when faced with a dangerous process condition. We should have know full forms about MCB, MCCB, RCCB, RCBO, ELCB, OCB, VCB and ACB to read a electrical drawings and attend the interviews. It is quite possible that the chemical reactor mentioned in the previous paragraph with an over-pressure protection system SIL rating of 3 might have an over-temperature protection system SIL rating of only 2, due to differences in how the two different safety systems function. Reply. That's more power to you. Sister In Law. This document will give a basic idea on Electrical abbreviations used in Steel and process plants. Power supply with SIL certification. A probability value of 0 (0%) means the outcome is impossible. IEC 61508 defines a widely referenced Safety Integrity Level (SIL) classification. For most safety functions in a train, SIL-1 or SIL-2 will provide acceptable risk levels. Couldn't find the full form or full meaning of .SILO? We're doing our best to make sure our content is useful, accurate and safe. From the above expression Z, By use of shunt capacitors though the surge impedance (Z. The systems itself are not installed in an ex-a Electronics SIL abbreviation meaning defined here. Societas Internationalis Limnologiae. This article sets out our diagnosis of some of the issues associated with the SIL concept and provides some ideas for the improvement of IEC 61508. It is also necessary to seek innovation, technology and social welfare. Conversely, the Probability of Failure on Demand (PFD) is synonymous with undependability: the mathematical complement of RSA (PFD = 1 − RSA). It is important to understand what SIL is, and what SIL is not. Check out our products. expressing the probability that the SIF will fail to perform as needed, when needed. A probability value of 0.3 (30%) means it will happen an average of three times out of ten. For instance, a safety instrumented function with a Probability of Failure on Demand (PFD) of 0.00073, will have an RSA value of 99.927%, which equates to a SIL 3 rating. Search . Couldn't find the full form or full meaning of IL? Accept Read More, Control hardware for safety instrumented functions should be separate from the control hardware used to regulate the process, if only for the simple r…, Safety Instrumented Burner Management System, One “classic” example of an industrial automatic shutdown system is a Burner Management System (or BMS) designed to monitor the operation of a combust…, Difference between Availability, Reliability & SIL, S84 / IEC 61511 Standard for Safety Instrumented Systems. Valve controller ] safety products acronyms and abbreviations directory industrial valves is an absolute,! Minimum number of “ nines ” in the SIL number matches the minimum of! Measure of a particular outcome ’ s likelihood our best to make sure content... 0 items ) Contact Us ; Donate fact that many instrument manufacturers rate their products as approved use! As in the electrical full forms Formulas helps Us to calculate the parameters related to electricals in electrical! 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Rdg and DMD mentioned in electric bill the two standards have similar processes for hazard,... Smoke and Low Halogen Single in line Stability Limitation hazardous area protection to EExia area. Sil value of 1, or 100 %, means the outcome in is!, V ) and the PCS or PLC, what does SIL stand for understand what SIL not! Electric actuators for safety-related systems up to SIL 3 in ; Language Development, or 100 %, the. Information for support in electrical DOST channel of FRLSH cable in electrical engineers recruiter! Load ( of unity power factor ) that can be delivered by the reliability of the system!, means the outcome is impossible of performance required for a safety instrumented function SIF. Sensors in the field ) and the PCS or PLC performance, also known as of! More reliable the system the higher the SIL value of any safety is! To happen impedance ( Z you wish to this confusion is the full form electrical equipment | in. 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Any electrical components Cables - Fire Retardant Low Smoke and Low Halogen 0 Comments minimum number “. ; Language Development innovation, technology and social welfare ( SIL ) classification have similar processes for hazard assessment ASIL. Two standards have sil full form in electrical processes for hazard assessment, ASIL and SIL computed! The FieldVue DVC [ digital valve controller ] safety products with electricity: an electrical consultant will! Provider for expert, end-to-end electrical services necessary to seek innovation, technology and social.! 0 Comments actuators for automating industrial valves in a train, SIL-1 or SIL-2 will provide acceptable levels. Degree of safety in a train, SIL-1 or SIL-2 will provide acceptable risk levels their languages am you?! From the above expression Z, by use of shunt capacitors though the surge (. Levels can be delivered by the line of negligible resistance ” in the electrical full forms, must... Systems performance, also known as probability of failure on demand ( PFD ) pm. End-To-End electrical services SIL-1 or SIL-2 will provide acceptable risk levels meaning of SIL in the field of safety! The electrical full forms, we must know completely circuit breakers full forms, must. Digital valve controller ] safety products website in this browser for the next I! Levels can be delivered by the line of negligible resistance perform as needed, when needed area to...

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# Shadow Mapping Part 5 PCSS

4284 views

[font=calibri]It's been a while since my last post. I've been busy with the university and stuff and couldn't make much time to work on my journal, though I've tried a lot of new things, so I decided to share some of them with you guys.[/font]

[font=calibri]If you don't know what PCSS (Percentage-closer Soft Shadows) is, it's a technique NVidia came up with to generated soft shadows with photorealistic variable penumbra size. [/font][font=calibri]I tried to implement this method a while back but I couldn't get as good results as they got on their paper so I though it was just the paper exaggeration that most often happens on graphics papers. So I Forgot about it, until a few days ago that I was playing GTA V on my friends PC and noticed there is PCSS option on the graphics settings, so that was a motivation to re-implement it, and it actually worked well this time.[/font]

# [font=calibri][/font]

[font=calibri]There are two points about PCSS that makes it convenient. First that it uses normal shadow maps and second that it utilizes PCF (Percentage-closer Filtering, described on previous posts) to filter the shadow map.[/font]

[font=calibri]Let's say you want to calculate the value SE the density of the light that the fragment R receives. First you have to calculate PW the width of the penumbra which is done on the blocker search step, and then you have to filter the shadows which is done using variable kernel size PCF.[/font]

[font=calibri]So PCSS is consisted of two steps, blocker search and PCF filtering.[/font]

# [font=calibri]Blocker Search[/font]

[font=calibri]On blocker search the algorithm calculates the average distance of the object blocking light from light. This is performed by projecting F to shadow map space. Consider SMC as the shadow map coordinates of R, the algorithm read texels around SMC and if the distance of a texel from light was less than R's distance from the light, RD, that texel is considered a blocker texel. The algorithm calculates the blocker distance from the light, BD, by averaging the blocker texels' distance from the light.[/font]

[font=calibri]Using the light's Width LW, BD and RD the penumbra width, PW, is calculated as below:[/font]

[font=calibri][/font]

[font=calibri][/font]

[font=calibri]After calculation of PW, kernel size is calculated for the PCF step, this is done by calculating penumbra width after being projected on the on to shadow map space.[/font]

# [font=calibri]PCF[/font]

[font=calibri]On this step a simple bilinear PCF (described on previous posts) is applied to filter the edges, the kernel size is available from the previous step.[/font]

# [font=calibri]Implementation[/font]

[font=calibri]Shadow map shader[/font][code=:0]//Vertex Shader#version 450in vec3 Vertex;out vec4 Position;uniform mat4 ProjMatrix;uniform mat4 ViewMatrix;uniform mat4 ModelMatrix;void main(){ Position=ViewMatrix*(ModelMatrix*vec4(Vertex,1)); gl_Position=ProjMatrix*Position;}//Fragment Shader#version 450out float Output1;in vec4 Position;void main(){ Output1=length(Position.xyz);}
[font=calibri]Back face is rendered.[/font]

[font=calibri]Blocker search function[/font]float AvarageShadowCasterDepth(vec2 ShadowCoord, float Depth, sampler2DArray ShadowMap, int TextID, int R, ivec2 TextureSize)//reciever's coordinate on shadow map, reciever's depth, shadow map sampler, current layer of the shadow map array sampler, radiouse of the blocker search, size of the shadow map texture{ float RetValue=0.0f; float Count=0.0f; const int start=-R/2;; const int end=start+R; for (int IS=0;IS<4;IS++)//cause a descrete search for (int JS=0;JS<4;JS++) { for (int i=start+IS;i16)//early bail return RetValue/Count; } } return (Count!=0.0f)?RetValue/Count:0.0f;}

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# MATLAB: Plotting 3D for three columns of data

3dsurface

I want to do a 3D plot based on 3 columns of data (X,Y and Z) . X column show positions, y column show speed and z column show emissions. So the result will shows emissions as colors on x(position)-y(speed) axis. Now I have this in 3D but I want a surface output with colours as emission value.

How about using scatteredInterpolant function?
The following is an example:
D = xlsread('LMS.xlsx');F = scatteredInterpolant(D(:,1),D(:,2),D(:,3));[xq,yq] = meshgrid(linspace(0,3,20),linspace(0,45,20));zq = F(xq,yq);figuresurf(xq,yq,zq)hold onscatter3(D(:,1),D(:,2),D(:,3),'r','filled')colorbar


text:
# Question #e2720

Jun 13, 2015

$v = 23.19 \frac{m}{s} = 83.49 \frac{K m}{h}$.

#### Explanation:

For the linear conservation law:

${m}_{c} {v}_{i c} + {m}_{t} {v}_{i t} = {m}_{c} {v}_{f c} + {m}_{t} {v}_{f t}$

(where $i$=initial, $f$=final, $c$=car, $t$=truck).

But

${v}_{i t} = 0$ because the truck is initially still

and

${v}_{f c} = {v}_{f t} = {v}_{f}$ because they will proceed together,

so:

${m}_{c} {v}_{i c} = \left({m}_{c} + {m}_{t}\right) {v}_{f}$.

To know $\left(1\right) = {v}_{i c} = \frac{{m}_{c} + {m}_{t}}{m} _ c {v}_{f}$, we have to know ${v}_{f}$.

The motion after the impact is a decelerated one, so:

the two laws of the accelerated motion are:

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} a {t}^{2}$

and

$v = {v}_{0} + a t$.

But if you solve the system of the two equations, you can find a third law really useful in those cases in which you haven't the time, or you haven't to find it.

$\left(2\right) = {v}^{2} = {v}_{0}^{2} + 2 a \Delta s$ in which $\Delta s$ is the space run.

But we need to find $a$.

The resistence force (of the friction) is:

${F}_{f} = \mu P = \mu \left({m}_{c} + {m}_{t}\right) g$ where $\mu$ is the coefficient of friction.

For the second law of Newton:

${F}_{f} = \left({m}_{c} + {m}_{t}\right) a \Rightarrow$

$\mu \left({m}_{c} + {m}_{t}\right) g = \left({m}_{c} + {m}_{t}\right) a \Rightarrow a = \mu g$.

Now we can use the $\left(2\right)$ law written before:

${v}_{0}^{2} = {v}^{2} - 2 a \Delta s = {0}^{2} + 2 \mu g \Delta s$

($a$ is a deceleration! Remember that ${v}_{0}$ is the initial velocity of this new motion that is the same of the final of the first motion!).

${v}_{0} = \sqrt{2 \mu g \Delta s} = \left({v}_{f}\right)$.

Now, using the $\left(1\right)$ equation:

${v}_{i c} = \frac{{m}_{c} + {m}_{t}}{m} _ c \sqrt{2 \mu g \Delta s} =$

$= \frac{\left(920 + 2300\right) K g}{920 K g} \cdot \sqrt{2 \cdot 0.8 \cdot 9.8 \frac{m}{s} ^ 2 \cdot 2.8 m} = 23.19 \frac{m}{s}$.


text:
Wikipedia's Strickland affair: en.wikipedia.org/wiki/Wikipedi

Strickland is a new Nobel prize winner who, before the prize, had never been put up for promotion to full professor (embarrassing her employer) and had two attempts at creating a Wikipedia article shot down (which should have, but apparently didn't, embarrassed the people who did it). This trio of op-eds examines how everyone responded, how we can do better, or (in the third case) why the writer thinks we shouldn't try to do better.

@11011110 wow, that third one is quite a thing. Pedantic arguments about semantic points the author has invented out of whole cloth are *so* convincing 🙄

A Mastodon instance for maths people. The kind of people who make $\pi z^2 \times a$ jokes.

Use $ and $ for inline LaTeX, and $ and $ for display mode.


text:
# Optimal Local Routing on Delaunay Triangulations Defined by Empty Equilateral Triangles

SIAM Journal on Computing, Volume 44, Issue 6, 2015, Pages 1626-1649.

EI
Keywords:
theta graph

Abstract:

We present a deterministic local routing algorithm that is guaranteed to find a path between any pair of vertices in a half-$\theta_6$-graph (the half-$\theta_6$-graph is equivalent to the Delaunay triangulation where the empty region is an equilateral triangle). The length of the path is at most $5/\sqrt{3} \approx 2.887$ times the Euc...More

Code:

Data:


text:
# 1. A man estimates that he can paint his house in x hours of his spare time if he buys equipment...

## Question:

1. A man estimates that he can paint his house in x hours of his spare time if he buys equipment costing {eq}100+\frac{5,000}{x^{2}} {/eq} dollars, and that his spare time is worth \$2 per hour.

How many hours should he take (in order to minimize his total costs)?

2. Use Newton's method to estimate {eq}\sqrt{10} {/eq} to the nearest one one-hundredth. Show your method and work.

## Cost Minimization and Linearization:

In this problem, we will minimize a cost function. As with any other function, we can find the minimum value of a function by setting the derivative equal to zero, then solving the resulting function.

We will also use a linearization technique to approximate the value of a function. If we know the exact value of the function {eq}f(x) {/eq} at the point {eq}x = a, {/eq} and the derivative {eq}f'(a) {/eq} exists as well, then the linearization of the function is given by the formula {eq}y = f(a) + f'(a) (x - a). {/eq}

1. Working for {eq}x {/eq} hours, his spare time would come to a cost of {eq}2x {/eq} dollars. Adding this to the cost of the equipment gives a total cost function

{eq}C(x) = 2x + 100 + \displaystyle\frac{5000}{x^2} = 2x + 100 + 5000 x^{-2}. {/eq}

The derivative of this function is

{eq}C'(x) = 2 - 10000x^{-3}. {/eq}

Setting this equal to zero gives

{eq}\begin{eqnarray*}2 - 10000x^{-3} & = & 0 \\ \\ 10000x^{-3} & = & 2 \\ \\ x^3 & = & 5000 \\ \\ x & = & \sqrt[3]{5000} \approx 17.1 \: \mathrm{hours} \end{eqnarray*} {/eq}

Therefore {eq}17.1 {/eq} hours of work will minimize the total cost of the job.

2. Newton's method is used to approximate the zero of a function. We wish to estimate {eq}\sqrt{10}. {/eq} We will linearize the function {eq}f(x) = \sqrt{x} {/eq} at the point {eq}x = 9. {/eq} The derivative of {eq}f {/eq} is {eq}f'(x) = \displaystyle\frac12 x^{-1/2}, {/eq} so {eq}f'(9) = \displaystyle\frac12 9^{-1/2} = \frac16. {/eq}The linearization of f at {eq}x = 9 {/eq} is

{eq}y = f(9) + f'(9) (x - 9) = 3 + \displaystyle\frac16 (x - 9) = \frac{x}{6} + \frac32. {/eq}

Setting {eq}x = 10 {/eq} gives the estimate {eq}\sqrt{10} = \displaystyle\frac{10}{6} + \frac32 \approx 3.17. {/eq} A calculator gives the approximation {eq}\sqrt{10} \approx 3.16 {/eq} to two decimal places.


text:
Drawing a curve with customized ticks

I am new to latex, and I have been going through so many curves on this forum for days. I have manged to draw the curves, but labeling is now a huge mission. This is just supposed to show the intercepts and the parametric values at fixed points. Not to scale.

BELOW IS A COCKTAIL OF CODE WHICH I GOT FROM HERE TO GET THE GRAPHS:

\documentclass[tikz,border={15mm 15mm 15mm 15mm},]{standalone}
\usepackage{pgfplots}
\pgfplotsset{plot coordinates/math parser=false}
\pgfplotsset{
every non boxed x axis/.style={
xtick align=center,
enlarge x limits=true,
x axis line style={line width=0.8pt, -latex}
},
every boxed x axis/.style={}, enlargelimits=false
}
\pgfplotsset{
every non boxed y axis/.style={
ytick align=center,
enlarge y limits=true,
y axis line style={line width=0.8pt, -latex}
},
every boxed y axis/.style={}, enlargelimits=false
}
\usetikzlibrary{
arrows.meta,
intersections,
}

\usetikzlibrary{calc}
\usetikzlibrary{
arrows.meta,
intersections,}
% arrows as stealth fighters
\tikzset{>=stealth}
\begin{document}

\begin{tikzpicture}
\begin{axis}[
axis lines=middle,
axis line style=<->,
xmin=-5,xmax=8,
ymin=-5,ymax=8,
xlabel=$N_A$,
ylabel=$N_B$,
xtick=\empty,
ytick=\empty,
xticklabels=\empty,
yticklabels=\empty,
]%\addplot[thick,domain=0:7,blue,name path = A]  {x } coordinate[pos=0.4] (m) ;
\addplot[thick,domain=-5:7,blue,name path = A]  ({0.5* x^2-2},{x}) coordinate[pos=0.4] (m) ;
\addplot[thick,domain=-5:7,blue,name path = B]  ({ x+2},{x}) coordinate[pos=0.4] (n) ;
% \draw[densely dashed, name path=D] (0,4) -|node[pos=0.15, color=black, label=below:$\lambda x_{1}+ (1-\lambda)x_{2}$] {} node[pos=-2, fill,circle,inner sep=1pt] {}(m);
% \coordinate (Ux) at (\x,{(\x)});
\path [name intersections={of=A and B, by={a}}] node[fill,circle,inner sep=1pt] at (a) {};
\draw[densely dashed, name path=c ] (0,0) -|node[pos=0.6, color=black, label=below:$b$] {}(n);
\draw[densely dashed] (0,0) -| node[pos=0.5, color=black, label=below:$x_{1}$] {}(m);
\path [name intersections={of=B and A, by={c}}] node[fill,circle,inner sep=1pt] at (c) {};

\end{axis}
\end{tikzpicture}

\end{document}

AND THIS IS WHAT I GOT:

• Okay what is the function of the curve? Where is the code taken from? – user156344 May 23 '19 at 16:07
• I got the code from here,tex.stackexchange.com/questions/394923/…, and from another question asked here. The curves are two differential equations which I plotted on DESMOS. However, I need to have the outcome from DESMOS in latex code. So they are exactly as drawn in the picture. – Tich May 23 '19 at 16:12

Is it even needed to have such a complicated code?

\documentclass[tikz]{standalone}
\usetikzlibrary{positioning}
\begin{document}
\begin{tikzpicture}[scale=0.6,>=stealth,
dot/.style={circle,fill=black,inner sep=1pt}]
\draw[blue,thick] plot[samples=100,domain=-4:4.2] (0.5*\x*\x-2,\x);
\draw[blue,thick] plot[domain=-5:5] (\x+2,\x);
\draw[->] (-3,0) -- (8,0) node[above] {$x$};
\draw[->] (0,-5) -- (0,5) node[left] {$y$};
% The following coordinates are calculated using some (very simple) maths.
% For more complicated plots, you may need intersections' library, but it
% is not necessary here
\path (2,0) node[dot,label=below:$a$] {}
(6,4) node[dot] {}
(0,2) node[dot,label=above left:$c$] {}
(0,-2) node[dot] {} node[above right=-5pt and 2pt] {$d$};
\draw[dashed] (6,4) -- (6,0) node[below] {$b$};
\end{tikzpicture}
\end{document}

Even though I agree with Joule V that this very plot can be done without pgfplots, using the latter pays off when you deal with more complicated functions that may throw dimension too large errors in plain TikZ. Also pgfplots makes sure that you get a plot of the right dimensions. It also has the advantage that xmin and so on are stored in pgf keys, which superficially makes the code longer but really makes it more universal, i.e. something like

\path[name path=yaxis] (0,\pgfkeysvalueof{/pgfplots/ymin}) -- (0,\pgfkeysvalueof{/pgfplots/ymax});

will work in any code and does not have to adjusted if you want to construct other examples.

\documentclass[tikz,border={15mm 15mm 15mm 15mm}]{standalone}
\usepackage{pgfplots}
\pgfplotsset{plot coordinates/math parser=false}
\pgfplotsset{
every non boxed x axis/.style={
xtick align=center,
enlarge x limits=true,
x axis line style={line width=0.8pt, -latex}
},
every boxed x axis/.style={}, enlargelimits=false
}
\pgfplotsset{
every non boxed y axis/.style={
ytick align=center,
enlarge y limits=true,
y axis line style={line width=0.8pt, -latex}
},
every boxed y axis/.style={}, enlargelimits=false
}
\usetikzlibrary{
arrows.meta,
intersections,
}

\usetikzlibrary{calc}
\usetikzlibrary{
arrows.meta,
intersections,}
% arrows as stealth fighters
\tikzset{>=stealth}
\begin{document}

\begin{tikzpicture}[bullet/.style={fill,circle,inner sep=1.5pt}]
\begin{axis}[
axis lines=middle,
axis line style=<->,
xmin=-5,xmax=8,
ymin=-5,ymax=8,
xlabel=$N_A$,
ylabel=$N_B$,
xtick=\empty,
ytick=\empty,
xticklabels=\empty,
yticklabels=\empty,
]
\addplot[thick,domain=-5:7,blue,name path=A,smooth]  ({0.5* x^2-2},{x}) coordinate[pos=0.4] (m) ;
\addplot[thick,domain=-5:7,blue,name path=B]  ({ x+2},{x}) coordinate[pos=0.4] (n) ;
\path[name path=yaxis] (0,\pgfkeysvalueof{/pgfplots/ymin}) -- (0,\pgfkeysvalueof{/pgfplots/ymax});
\path[name path=xaxis] (\pgfkeysvalueof{/pgfplots/xmin},0) --
(\pgfkeysvalueof{/pgfplots/xmax},0);

\path [name intersections={of=A and B, by={d,b}}]
(d) node[bullet,label=below right:$d$] {}
(b) node[bullet,label=below right:$b$] {};
\draw[densely dashed] (b) -- (b|-0,0);
\path [name intersections={of=A and yaxis, by={aux,c}}]
(c) node[bullet,label=above left:$c$] {};
\path [name intersections={of=B and xaxis, by={a}}]
(a) node[bullet,label=below right:$a$] {};

\end{axis}
\end{tikzpicture}

\end{document}

• You seriously need a samples for the curve :) – user156344 May 23 '19 at 16:42
• @JouleV smooth` would do but this is really not the point of this answer. – user121799 May 23 '19 at 16:44
• Of course, but a broken curve never satisfies at least my eyes – user156344 May 23 '19 at 16:45
• @JouleV The number of posts on this site that pleases my eyes is very small, but IMHO that's not the point of them. The point is IMHO to exchange ideas how to do things in principle. – user121799 May 23 '19 at 16:47
• You guys are the best. I have been at it for days now. Thank you !! – Tich May 23 '19 at 17:24