text:
Since 18 of December 2019 conferences.iaea.org uses Nucleus credentials. Visit our help pages for information on how to Register and Sign-in using Nucleus.

# 28th IAEA Fusion Energy Conference (FEC 2020)

10-15 May 2021
Nice, France
Europe/Vienna timezone
The Conference will be held virtually from 10-15 May 2021

## Characteristics of the extracted negative-ion beam in a cesium-free negative-ion source using TPDsheet-U

14 May 2021, 08:30
4h
Nice, France

#### Nice, France

Regular Poster Fusion Energy Technology

### Speaker

Prof. Akira TONEGAWA (Department of Physics, School of Science, Tokai University)

### Description

This study reports a progress toward realizing a high performance Cs-free negative ion source, that produces negative ions by volume production in a magnetized-sheet plasma device (TPDsheet-U). The experimental results show that this performance of the current density of the negative hydrogen ion beam J$_{c}$ without Cs on TPDsheet-U is a pproximately comparable to that of the negative ion source on ITER-NBI with Cs. Experimental observations show that (i) negative hydrogen ions are successfully extracted from the sheet plasma by single/multi-aperture grids, and (ii) the ratio of the extracted electron current I$_{EG}$(e) and hydrogen negative-ion current I$_{C}$(H-), I$_{EG}$(e)/I$_{C}$(H-) decreases from 5.0 to 1.0 with an increase in the height of the electron fence (H$_{EF}$), which is a physical filter that prevents electron diffusion from the extraction region. The J$_{c}$ is approximately 10-20 mA/cm$^{2}$ at the extraction voltage V$_{EX}$ of 9.0 kV, the discharge current I$_{d}$ of 80 A, the gas pressure of 0.3 Pa, and the magnetic field strength of 40 mT.

Most NBI systems in fusion devices use surface production with Cs. Without Cs, the current density that is based on volume production is < 2 mA/cm$^{2}$. If it is operated with Cs, the source can deliver a negative ion current density of over 20 mA/cm$^{2}$, which is 10–15 times higher. However, careful Cs control is required for the continuous operation of the NBI, which must be also regularly maintained. Thus, it is important to find alternative materials or to develop Cs-free methods.
To produce and control negative hydrogen ions in plasma, we proposed to use magnetized-sheet plasma, which is produced by TPDsheet-U [1-3]. Magnetized-sheet plasma is suitable for generating negative hydrogen ions during dissociative attachment processes because high-energy (10–15 eV) and low-energy (1 eV) electron regions are narrowly spaced (e.g., 10–25 mm). This study is a work in progress toward realizing a Cs-free negative-ion source. This paper presents the characteristics of the extracted ion beam current using TPDsheet-U.
The experimental setup is a Cs-free negative-ion source using TPDsheet-U, which is schematically illustrated in Fig. 1. The hydrogen sheet plasma is introduced to the extraction region under the stationary magnetic field generated by rectangular coils and terminated by an electrically floating target. The discharge current is varied between 40 A and 80 A. The neutral pressure in the extraction region is controlled by feeding the secondary hydrogen gas at the pressure of approximately 0.3 Pa. To extract negative-ion beams, two different sets of grids are located in the extraction region (Fig. 1): single-aperture and multi-aperture extraction grids. The negative hydrogen ion beam is extracted by the two-grid extraction system [i.e., plasma grid (PG) and extraction grid (EG)] located at the periphery of the sheet plasma, as shown in Fig. 2. The diameter of PG is Φ 4 mm and that of EG is Φ 6 mm, and the gap between grids, d, is 4 mm. In the multi-aperture grid, each grid has 39 apertures, and the beam group is made of 13 × 3 apertures. With a field strength of 40 mT and V$_{EX}$ of 3–10 kV, the extracted current densities J$_{c}$ and J$_{EG}$ are measured at the collector (C) and EG, respectively.
The extracted currents (J$_{EG}$ and J$_{c}$), as a function of discharge current I$_{d}$, with a single-aperture grid at V$_{EX}$ of 3 kV, 6 kV, and 9 kV are shown in Fig. 3. The current of J$_{EG}$ is the electron current, which is deflected onto EG by the external magnetic field. The current of J$_{c}$ is the extracted negative-ion current. At V$_{EX}$ of 9.0 kV and I$_{d}$ of 80 A, J$_{c}$ is 20 mA/cm$^{2}$. However, J$_{EG}$ is five times larger than that of J$_{c}$. An increase in the co-extracted electrons with a possibility of damaging the EG is an important issue to be solved in the negative-ion source.
The electron fence (EF) bar is located on PG to suppress co-extracted electrons (Fig. 2). The dependence of the ratio of the extracted electron current I$_{EG}$(e) and hydrogen negative-ion current I$_{C}$(H-), I$_{EG}$(e)/I$_{C}$(H-), on the height of the EF bar, H$_{EF}$, is shown in Fig. 4. I$_{d}$ changes from 40 A to 70 A at V$_{EX}$ of 9.0 kV. The ratio of I$_{EG}$(e)/I$_{C}$(H−) gradually decreases with an increase in H$_{EF}$ and is approximately 0.5–1.3. Near PG, electrons are trapped with the EF bar owing to the smaller gyro-radius of the electrons. The gyro-radius of hydrogen negative ions is larger than that of electrons, and hydrogen negative ions can pass through the electron fence bar. This result shows that electrons are effectively suppressed by the EF bar.
Multi-aperture grids have 39 holes to extract negative ions, which results in an increase in the extracted current depending on the number of apertures. In the multi-aperture setup with four EF bars (H$_{EF}$ = 6 mm, 4 mm, 4 mm, 6 mm), the extracted negative ion beam current I$_{C}$ increases with an increase in I$_{d}$. At an extraction voltage of 9.5 kV and a discharge current of 80 A, the current density of the negative hydrogen ion beam is approximately 23 mA. These shows that hydrogen negative ions are successfully extracted from sheet plasma.

This study indicates the importance of investigating the characteristics of extracted beam current in a magnetized-sheet plasma device (TPDsheet-U) without Cs and the possibility of achieving the continuous operation of the NBI without Cs.

References
[1] R. Endo, et al., AIP Conference Proceedings (2018) 020009.
[2] S. Ishihara, et al., AIP Conference Proceedings (2018) 050015.
[3] K. Hanai, et al., Fusion Eng. Des. 146 (2019) 2721.

Country or International Organization Japan Department of Physics, School of Science, Tokai University, 4-1-1

### Primary author

Prof. Akira TONEGAWA (Department of Physics, School of Science, Tokai University)

### Co-authors

Mr Hiroki KAMINAGA (Department of Physics, School of Science, Tokai University) Mr Toshikio TAKIMOTO (Graduate School of Science and Technology, Tokai University) Mr Keito HANAI (Graduate School of Science and Technology, Tokai University) Prof. Kohnosuke SATO (Tokyo University of Science) Prof. Kazutaka KAWAMURA (Tokai University)

### Presentation Materials

There are no materials yet.


text:
# CAT Quantitative Aptitude Questions | CAT Number Systems - Factorial questions

###### CAT Questions | Number Theory | Trailing zeroes
CAT Questions

The question is also about Trailing zeroes. Find the least number n such that no factorial has n trailing zeroes, or n + 1 trailing zeroes or n + 2 trailing zeroes. Dealing with factorials of a number is a vital component in CAT Number Theory. A range of CAT questions can be asked based on this simple concept of CAT Number Systems in the CAT exam. Make use of 2IIMs Free CAT Questions, provided with detailed solutions and Video explanations to obtain a wonderful CAT score. If you would like to take these questions as a Quiz,head on here to take these questions in a test format, absolutely free.

1. 153
2. 126
3. 624
4. 18

## Best CAT Coaching in Chennai

#### CAT Coaching in Chennai - CAT 2020Online Batches Available Now!

##### Method of solving this CAT Question from Number Theory - Factorial: We can find the number of trailing zeroes a factorial have. Can you find a number that has a specific number of trailing zeroes? Can you give a certain number of trailing zeroes that are not present in any factorial?

The previous question includes a detailed discussion on how to find the number of trailing zeroes of n!, for any natural number n.
We see that 24! has $$frac{24}{5}\\$ = 4 zeroes 25! ends with [$\frac{25}{5}\\$] + [$\frac{25}{25}\\$] = 6 zeroes. There is no natural number m such that m! has exactly 5 zeroes. Similarly, we see that 49! ends with[$\frac{49}{5}\\$] + [$\frac{49}{25}\\$] = 10 zeroes, whereas 50! ends with [$\frac{50}{5}\\$] + [$\frac{50}{25}\\$] = 12 zeroes. No factorial ends with 11 zeroes. So, any time we have a multiple of 25, we 'skip' a zero. This is because a multiple of 25 adds two zeroes to the factorial. Extrapolating this, we can see that 125 might actually 'skip' two zeroes. 124! ends with [$\frac{124}{5}\\$] + [$\frac{124}{25}\\$] = 24 + 4 = 28 zeros, whereas 125! has [$\frac{125}{5}\\$] + [$\frac{125}{25}\\$] + [$\frac{125}{125}\\$] = 25 + 5 + 1 = 31 zeros. There is no factorial with 29 or 30 zeros. In order to jump three zeros, think about what we need to look at. Every multiple of 25 gives us one 'skipped' zero. Every multiple of 125 gives us two 'skipped' zeroes. In order to have three skipped zeroes, we need to look at 624! and 625! 624! has [$\frac{624}{5}\\$] + [$\frac{624}{25}\\$] + [$\frac{624}{125}\\$] = 124 + 24 + 4 = 152 zeros 625! has [$\frac{625}{5}\\$] + [$\frac{625}{25}\\$] + [$\frac{625}{125}\\$] + [$\frac{625}{625}\\$] = 125 + 25 + 5 + 1 = 156 zeros There is no factorial with 153, 154 or 155 zeros. Or the least value of n such that no factorial ends with n,$n + 1) or (n + 2) zeroes is 153.

The question is "Find the least number n such that no factorial has n trailing zeroes, or n + 1 trailing zeroes or n + 2 trailing zeroes."

##### Hence the answer is "153"

Choice A is the correct answer.

###### CAT Coaching in ChennaiCAT 2020Enroll at 35,000/-

Online Classroom Batches Starting Now!

###### Best CAT Coaching in ChennaiRegister Online, get Rs 7000/- off

Attend a Demo Class

## CAT Online Coaching | CAT Number Systems questions Videos On YouTube

#### Other useful sources for Number System Questions | Number Theory Factorials Sample Questions

##### Where is 2IIM located?

2IIM Online CAT Coaching
A Fermat Education Initiative,
58/16, Indira Gandhi Street,
Kaveri Rangan Nagar, Saligramam, Chennai 600 093

##### How to reach 2IIM?

Phone: (91) 44 4505 8484
Mobile: (91) 99626 48484
WhatsApp: WhatsApp Now
Email: prep@2iim.com

format: plain


text:
# How do you simplify (20b^10) /( 10b^20)?

May 27, 2018

$2 {b}^{-} 10$

#### Explanation:

You can split the fraction into two:
$\frac{20}{10}$ and ${b}^{10} / {b}^{20}$ And you can work out each separate.

$\frac{20}{10} = 2$

Using the formula
${a}^{b} / {a}^{c} = {a}^{b - c}$ you can work out the second fraction.

${b}^{10} / {b}^{20} = {b}^{10 - 20} = {b}^{-} 10$

Then you can multiply $2$ by ${b}^{-} 10$ to get $2 {b}^{-} 10$

May 27, 2018

$\frac{2}{b} ^ 10$

#### Explanation:

$\frac{20 {b}^{10}}{10 {b}^{20}}$

$\frac{2 \times 10 \times {b}^{10}}{1 \times 10 \times {b}^{20}}$

$\frac{2 \times \cancel{10} \times {b}^{10}}{1 \times \cancel{10} \times {b}^{20}}$

$\frac{2 \times 1 \times {b}^{10}}{1 \times 1 \times {b}^{20}}$

$\frac{2 {b}^{10}}{b} ^ 20$

$2 {b}^{10} \div {b}^{20}$

Recall;

${x}^{a} \div {x}^{b} = {x}^{a - b}$

Hence;

$2 {b}^{10} \div {b}^{20} = 2 {b}^{10 - 20}$

$2 {b}^{-} 10$

But: ${x}^{-} 1 = \frac{1}{x}$

Therefore;

$\frac{2}{b} ^ 10$

May 27, 2018

$2 {b}^{-} 10$

#### Explanation:

$\frac{20 {b}^{10}}{10 {b}^{20}}$

It might be helpful to rewrite this as two separate fractions.

$\frac{20}{10} \cdot {b}^{10} / {b}^{20}$

$2 \cdot {b}^{10} / {b}^{20}$

We can simplify the second fraction if we remember a little rule: ${n}^{a} / {n}^{b} = {n}^{a - b}$. In other words, ${b}^{10} / {b}^{20}$ will equal ${b}^{10 - 20}$ or ${b}^{-} 10$.

$2 \cdot {b}^{-} 10$

$2 {b}^{-} 10$


text:
12/4/2021: In the previous version of this post there were some incorrect details that didn’t allow me to sleep well at night. So, I rewrote it in the correct formalism, but please let me know if you see mistakes.

In this post, we talk about solving saddle-point problems with online convex optimization (OCO) algorithms. Next time, I’ll show the connection to two-person zero-sum games and boosting.

We want to solve the following saddle-point problem

$\displaystyle \label{eq:saddle_point} \inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x},{\boldsymbol y})~. \ \ \ \ \ (1)$

Let’s say from the beginning that we need inf and sup rather than min and max because the minimum or maximum might not exist. Everytime we know for sure the inf/sup are attained, I’ll substitute them with min/max.

While for the minimization of functions is clear what it means to solve it, it might not be immediate to see what is the meaning of “solving” the saddle-point problem in (1). It turns out that the proper notion we are looking for is the one of saddle-point.

Definition 1. Let ${X \subseteq {\mathbb R}^n}$, ${Y \subseteq {\mathbb R}^m}$, and ${f: X \times Y \rightarrow {\mathbb R}}$. A point ${({\boldsymbol x}^\star, {\boldsymbol y}^\star) \in X \times Y}$ is a saddle-point of ${f}$ in ${X\times Y}$ if

$\displaystyle f({\boldsymbol x}^\star, {\boldsymbol y}) \leq f({\boldsymbol x}^\star, {\boldsymbol y}^\star) \leq f({\boldsymbol x}, {\boldsymbol y}^\star), \quad \forall {\boldsymbol x} \in X, {\boldsymbol y} \in Y~.$

We will now state conditions under which there exists a saddle-point that solves (1). First, we need an easy lemma.

Lemma 2. Let ${f}$ is any function from a non-empty product set ${X \times Y}$ to ${{\mathbb R}}$. Then,

$\displaystyle \inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x},{\boldsymbol y}) \geq \sup_{{\boldsymbol y} \in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})~.$

Proof: For any ${{\boldsymbol x}' \in X}$ and ${{\boldsymbol y} \in Y}$ we have that ${f({\boldsymbol x}',{\boldsymbol y}) \geq \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})}$. This implies that ${\sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x}', {\boldsymbol y}) \geq \sup_{{\boldsymbol y} \in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})}$ for all ${{\boldsymbol x}' \in X}$ that gives the stated inequality. $\Box$

We can now state the following theorem.

Theorem 3. Let ${f}$ any function from a non-empty product set ${X \times Y}$ to ${{\mathbb R}}$. A point ${({\boldsymbol x}^\star, {\boldsymbol y}^\star)}$ is a saddle-point of ${f}$ if and only if the supremum in

$\displaystyle \sup_{{\boldsymbol y} \in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x}, {\boldsymbol y})$

is attained at ${{\boldsymbol y}^\star}$, the infimum in

$\displaystyle \inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x}, {\boldsymbol y})$

is attained at ${{\boldsymbol x}^\star}$, and these two expressions are equal.

Proof: If ${({\boldsymbol x}^\star,{\boldsymbol y}^\star)}$ is a saddle-point, then we have

\displaystyle \begin{aligned} f({\boldsymbol x}^\star,{\boldsymbol y}^\star) &= \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x}, {\boldsymbol y}^\star) \leq \sup_{{\boldsymbol y} \in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x}, {\boldsymbol y})\\ f({\boldsymbol x}^\star,{\boldsymbol y}^\star) &= \sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x}^\star, {\boldsymbol y}) \geq \inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x}, {\boldsymbol y})~. \end{aligned}

From Lemma 2, we have that these quantities must be equal, so that the three conditions in the theorem are satisfied.

For the other direction, if the conditions are satisfied, then we have

$\displaystyle f({\boldsymbol x}^\star, {\boldsymbol y}^\star) \leq \sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x}^\star, {\boldsymbol y}) = \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x}, {\boldsymbol y}^\star) \leq f({\boldsymbol x}^\star, {\boldsymbol y}^\star)~.$

Hence, ${({\boldsymbol x}^\star,{\boldsymbol y}^\star)}$ is a saddle-point. $\Box$

Remark 1. The above theorem tells a surprising thing: If a saddle-point exists, then there might be multiple ones, and all of them must have the same minimax value. This might seem surprising, but it is due to the fact that the definition of saddle-point is a global and not a local property. Moreover, if the ${\inf\sup}$ and ${\sup\inf}$ problem have different values, no saddle-point exists.

Remark 2. Consider the case that the value of the ${\inf\sup}$ and ${\inf\sup}$ are different. If your favorite optimization procedure to find the solution of a saddle-point problem does not distinguish between a ${\inf\sup}$ and ${\sup\inf}$ formulation, then it cannot be correct!

Let’s show a couple of examples that show that the conditions above are indeed necessary.

Example 1. Let ${f(x,y)=(x-y)^2}$, ${X=[-1,1]}$, and ${Y=[-1,1]}$. Then, we have

$\displaystyle \inf_{x \in X} \sup_{y \in Y} \ (x-y)^2 = \inf_{x \in X} \ (1+|x|)^2 = 1,$

while

$\displaystyle \sup_{y \in Y} \inf_{x \in X} \ (x-y)^2 = \sup_{y \in Y} 0 = 0~.$

Indeed, from Figure 1 we can see that there is no saddle-point.

Example 2. Let ${f(x,y)=x y }$, ${X=(0,1]}$, and ${Y=(0,1]}$. Then, we have

$\displaystyle \inf_{x \in X} \sup_{y \in Y} \ x y = \inf_{x \in X} \ x = 0$

and

$\displaystyle \sup_{y \in Y} \inf_{x \in X} \ xy = 0~.$

Here, even if inf sup is equal to sup inf, the saddle-point does not exist because the inf in the first expression is not attained in a point of ${X}$.

This theorem also tells us that in order to find a saddle-point of a function ${f}$, we need to find the minimizer in ${{\boldsymbol x}}$ of ${\sup_{{\boldsymbol y} \in Y} \ f({\boldsymbol x},{\boldsymbol y})}$ and the maximizer in ${y}$ of ${\inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})}$. Let’s now use this knowledge to design a proper measure of progress towards the saddle-point.

We might be tempted to use ${f({\boldsymbol x}', {\boldsymbol y}')-f({\boldsymbol x}^\star,{\boldsymbol y}^\star)}$ as a measure of suboptimality of ${({\boldsymbol x}',{\boldsymbol y}')}$ with respect to the saddle-point ${({\boldsymbol x}^\star,{\boldsymbol y}^\star)}$. Unfortunately, this quantity can be negative or equal to zero for an infinite number of points ${({\boldsymbol x}',{\boldsymbol y}')}$ that are not saddle-points. We might then think to use some notion of distance to the saddle-point, like ${\|{\boldsymbol x}'-{\boldsymbol x}^\star\|^2_2+\|{\boldsymbol y}'-{\boldsymbol y}^\star\|^2_2}$, but this quantity in general can go to zero at an arbitrarily slow rate. To see why consider the case that ${f({\boldsymbol x},{\boldsymbol y})=h({\boldsymbol x})}$, so that the saddle-point problem reduces to minimize a convex function. So, assuming only convexity, the rate of convergence to a minimizer of ${f}$ can be arbitrarily slow. Hence, we need something different.

Observe that the Theorem 3 says one of the problems we should solve is

$\displaystyle \inf_{{\boldsymbol x} \in X} \ h({\boldsymbol x}),$

where ${h({\boldsymbol x}) = \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x},{\boldsymbol y})}$. In this view, the problem looks like a standard convex optimization problem, where the objective function has a particular structure. Moreover, in this view we only focus on the variables ${{\boldsymbol x}}$. The standard measure of convergence in this case for a point ${{\boldsymbol x}'}$, the suboptimality gap, can be written as

$\displaystyle h({\boldsymbol x}') - \inf_{{\boldsymbol x} \in X} h({\boldsymbol x}) = \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x}',{\boldsymbol y}) - \inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x},{\boldsymbol y})~.$

We also have to find the maximizer in ${y}$ of ${\inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})}$, hence we have

$\displaystyle \sup_{{\boldsymbol y} \in Y} \ g({\boldsymbol y}),$

where ${g({\boldsymbol y}) = \inf_{{\boldsymbol x}\in X} \ f({\boldsymbol x},{\boldsymbol y})}$. This also implies another measure of convergence in which we focus only on the variable ${{\boldsymbol y}}$:

$\displaystyle \sup_{{\boldsymbol y} \in Y} \ g({\boldsymbol y}) - g({\boldsymbol y}') = \sup_{{\boldsymbol y}\in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y}) - \inf_{{\boldsymbol x}\in X} \ f({\boldsymbol x},{\boldsymbol y}') ~.$

Finally, in case we are interested in studying the quality of a joint solution ${({\boldsymbol x}',{\boldsymbol y}')}$, a natural measure is a sum of the two measures above:

$\displaystyle \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x}',{\boldsymbol y}) - \inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x},{\boldsymbol y}) + \sup_{{\boldsymbol y}\in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y}) - \inf_{{\boldsymbol x}\in X} \ f({\boldsymbol x},{\boldsymbol y}') = \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x}',{\boldsymbol y}) - \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x}, {\boldsymbol y}'),$

where we assumed the existence of a saddle-point to say that ${\inf_{{\boldsymbol x} \in X} \sup_{{\boldsymbol y}\in Y} \ f({\boldsymbol x},{\boldsymbol y}) = \sup_{{\boldsymbol y}\in Y} \inf_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})}$ from Theorem 3. This measure is called duality gap. From the definition it should be clear that the duality gap is always non-negative. Let’s add even more intuition of the duality gap definition, using the one of ${\epsilon}$-saddle-point.

Definition 4. Let ${X \subseteq {\mathbb R}^n}$, ${Y \subseteq {\mathbb R}^m}$, and ${f: X \times Y \rightarrow {\mathbb R}}$. A point ${({\boldsymbol x}^\star, {\boldsymbol y}^\star) \in X \times Y}$ is a ${\epsilon}$-saddle-point of ${f}$ in ${X\times Y}$ if

$\displaystyle f({\boldsymbol x}^\star,{\boldsymbol y}) -\epsilon \leq f({\boldsymbol x}^\star,{\boldsymbol y}^\star) \leq f({\boldsymbol x},{\boldsymbol y}^\star) + \epsilon, \quad \forall {\boldsymbol x} \in X, {\boldsymbol y} \in Y~.$

This definition is useful because we cannot expect to numerically calculate a saddle-point with infinite precision, but we can be find something that satisfies the saddle-point definition up to a ${\epsilon}$. Obviously, any saddle-point is also an ${\epsilon}$-saddle-point.

Now, the notion of ${\epsilon}$-saddle-point is equivalent up to a multiplicative constant to the ${\epsilon}$ duality gap. In fact, it is easy to see that if ${({\boldsymbol x}^\star,{\boldsymbol y}^\star)}$ is an ${\epsilon}$-saddle-point then its duality gap of ${2\epsilon}$. In turn, a duality gap of ${2 \epsilon}$ and the existence of a saddle-point imply that the point is a ${2\epsilon}$-saddle.

The above reasoning told us that finding the saddle-point of the function ${f}$ is equivalent to solving a maximization problem and a minimization problem. However, is it always the case that the saddle-point exists? So, let’s now move to easily checkable sufficient conditions for the existence of saddle-point. For this, we can state the following theorem.

Theorem 5. Let ${X}$, ${Y}$ be compact convex subsets of ${{\mathbb R}^n}$ and ${{\mathbb R}^m}$ respectively. Let ${f : X \times Y \rightarrow {\mathbb R}}$ a continuous function, convex in its first argument, and concave in its second. Then, we have that

$\displaystyle \min_{{\boldsymbol x} \in X} \max_{{\boldsymbol y}\in Y} \ f({\boldsymbol x},{\boldsymbol y}) = \max_{{\boldsymbol y}\in Y} \min_{{\boldsymbol x} \in X} \ f({\boldsymbol x},{\boldsymbol y})~.$

This theorem gives us sufficient conditions to have the min-max problem equal to the max-min one. So, for example, thanks to the Weierstrass theorem, the assumptions in Theorem 5, in light of Theorem 3, are sufficient conditions for the existence of a saddle-point.

We defer the proof of this theorem for a bit and we now turn ways to solve the saddle-point problem in (1).

2. Solving Saddle-Point Problems with OCO

Let’s show how to use Online Convex Optimization (OCO) algorithms to solve saddle-point problems.

Suppose to use an OCO algorithm fed with losses ${\ell_t({\boldsymbol x})= f({\boldsymbol x},{\boldsymbol y}_t)}$ that produces the iterates ${{\boldsymbol x}_t}$ and another OCO algorithm fed with losses ${h_t({\boldsymbol y})=-f({\boldsymbol x}_t,{\boldsymbol y})}$ that produces the iterates ${{\boldsymbol y}_t}$. Then, we can state the following Theorem.

Theorem 6. Let ${f:X\times Y\rightarrow {\mathbb R}}$ be convex in the first argument and concave in the second. Then, with the notation in Algorithm 2, for any ${{\boldsymbol x} \in X}$, we have

$\displaystyle \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x}_t, {\boldsymbol y}_t) - \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x},{\boldsymbol y}_t) = \frac{\text{Regret}^{X}_T({\boldsymbol x})}{T},$

where ${\text{Regret}^{X}_T({\boldsymbol x})=\sum_{t=1}^T \ell_t({\boldsymbol x}_t) - \sum_{t=1}^T \ell_t({\boldsymbol x})}$.

For any ${{\boldsymbol y} \in Y}$, we have

$\displaystyle \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x}_t,{\boldsymbol y}) - \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x}_t, {\boldsymbol y}_t) = \frac{\text{Regret}^{Y}_T({\boldsymbol y})}{T},$

where ${\text{Regret}^{Y}_T({\boldsymbol y})=\sum_{t=1}^T h_t({\boldsymbol y}_t) - \sum_{t=1}^{T} h_t({\boldsymbol y})}$.

Also, if ${f}$ is continuous and ${X}$, ${Y}$ compact, we have

$\displaystyle \max_{{\boldsymbol y}\in Y} f(\bar{{\boldsymbol x}}_T, {\boldsymbol y}) - \min_{{\boldsymbol x}\in X} f({\boldsymbol x},\bar{{\boldsymbol y}}_T) \leq \frac{\text{Regret}^{X}_T({\boldsymbol x}'_T)+\text{Regret}^{Y}_T({\boldsymbol y}'_T)}{T},$

for any ${{\boldsymbol x}_T'\in\arg\min_{{\boldsymbol x}\in X}f({\boldsymbol x},\bar{{\boldsymbol y}}_T)}$ and ${{\boldsymbol y}_T'\in\arg\max_{{\boldsymbol y}\in Y}f(\bar{{\boldsymbol x}}_T,{\boldsymbol y})}$.

Proof: The first two equalities are obtained simply observing that ${\ell_t({\boldsymbol x})=f({\boldsymbol x},{\boldsymbol y}_t)}$ and ${h_t({\boldsymbol y})=-f({\boldsymbol x}_t,{\boldsymbol y})}$.

For the inequality, using Jensen’s inequality, we obtain

\displaystyle \begin{aligned} f(\bar{{\boldsymbol x}}_T,{\boldsymbol y})- f({\boldsymbol x},\bar{{\boldsymbol y}}_T) &\leq \frac{1}{T}\sum_{t=1}^{T}f({\boldsymbol x}_t,{\boldsymbol y})- \frac{1}{T} \sum_{t=1}^{T}f({\boldsymbol x},{\boldsymbol y}_t)~. \end{aligned}

Summing the first two equalities, using the above inequality, and taking ${{\boldsymbol x}={\boldsymbol x}'_T\in\arg\min_{{\boldsymbol x}\in\mathcal{X}} f({\boldsymbol x},\bar{{\boldsymbol y}}_T)}$ and ${{\boldsymbol y}={\boldsymbol y}'_T\in\arg\max_{{\boldsymbol y}\in\mathcal{Y}} f(\bar{{\boldsymbol x}}_T,{\boldsymbol y})}$, we get the stated inequality. $\Box$

From this theorem, we can immediately prove the following corollary.

Corollary 7. Let ${f:X\times Y\rightarrow {\mathbb R}}$ be continuous, convex in the first argument, and concave in the second. Assume that ${X}$ and ${Y}$ are compact and that the ${X}$-player and ${Y}$-player use no-regret algorithms, possibly different, than

$\displaystyle \lim_{T\rightarrow\infty}\max_{{\boldsymbol y}\in\mathcal{Y}} f(\bar{{\boldsymbol x}}_T, {\boldsymbol y}) - \min_{{\boldsymbol x}\in\mathcal{X}} f({\boldsymbol x},\bar{{\boldsymbol y}}_T)=0~.$

Example 3. Consider the saddle-point problem

$\displaystyle \min_{|x|\leq 2} \max_{|y|\leq 2} \ (x-1) (y-1)~.$

The saddle-point of this problem is ${(x,y)=(1,1)}$. We can find it using, for example, Projected Online Gradient Descent. So, setting ${{\boldsymbol x}_1={\boldsymbol y}_1=0}$, we have the iterations

\displaystyle \begin{aligned} x_{t+1} &= \max\left(\min\left(x_t - \frac{1}{\sqrt{t}} (y_t-1),2\right),-2\right)\\ y_{t+1} &= \max\left(\min\left(y_t + \frac{1}{\sqrt{t}} (x_t-1),2\right),-2\right)~. \end{aligned}

According to Theorem 6, the duality gap in ${(\frac1T \sum_{t=1}^T x_t, \frac1T \sum_{t=1}^T y_t)}$ converges to 0.

Surprisingly, we can even prove a (simpler version of the) minimax theorem from the above result! In particular, we will use the additional assumption that there exist OCO algorithms that minimize ${\ell_t}$ and ${h_t}$ have sublinear regret.
Proof with OCO assumption: From Lemma 2, we have one inequality. Hence, we now have to prove the other inequality.

We will use a constructive proof. Let’s use Algorithm 2 and Theorem 6. For the first player, for any ${{\boldsymbol x} \in X}$ we have

$\displaystyle \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x}_t, {\boldsymbol y}_t) = \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x},{\boldsymbol y}_t) + \frac{\text{Regret}^{X}_T({\boldsymbol x})}{T} \leq f\left({\boldsymbol x},\frac{1}{T}\sum_{t=1}^T {\boldsymbol y}_t\right) + \frac{\text{Regret}^{X}_T({\boldsymbol x})}{T}~.$

Observe that

$\displaystyle \min_{{\boldsymbol x} \in X} f\left({\boldsymbol x},\frac{1}{T}\sum_{t=1}^T {\boldsymbol y}_t\right) \leq \max_{{\boldsymbol y} \in Y} \min_{{\boldsymbol x} \in X} f({\boldsymbol x},{\boldsymbol y})~.$

Hence, using an OCO algorithm that has ${o(T)}$ regret for each ${{\boldsymbol x} \in X}$, we have

$\displaystyle \frac{1}{T}\sum_{t=1}^T f({\boldsymbol x}_t, {\boldsymbol y}_t) \leq \max_{{\boldsymbol y} \in Y} \min_{{\boldsymbol x} \in X} f({\boldsymbol x},{\boldsymbol y})+o(1)~.$

In the same way, we have

$\displaystyle -\frac{1}{T}\sum_{t=1}^T f({\boldsymbol x}_t, {\boldsymbol y}_t) \leq -\min_{{\boldsymbol x} \in X} \max_{{\boldsymbol y} \in Y} f({\boldsymbol x},{\boldsymbol y})+o(1)~.$

Summing the two inequalities, taking ${T\rightarrow \infty}$, and using the sublinear regret assumption, we have

$\displaystyle \min_{{\boldsymbol x} \in X} \max_{{\boldsymbol y} \in Y} f({\boldsymbol x},{\boldsymbol y}) \leq \max_{{\boldsymbol y} \in Y} \min_{{\boldsymbol x} \in X} f({\boldsymbol x},{\boldsymbol y})~.$

$\Box$

2.1. Variations with Best Response and Alternation

In some cases, it is easy to compute the max with respect to ${{\boldsymbol y} \in Y}$ of ${f({\boldsymbol x}_t,{\boldsymbol y})}$ for a given ${{\boldsymbol x}_t}$. For example, this is trivial for bilinear games over the simplex. In these cases, we can remove the second learner and just use its best response in each round. Note that in this way we are making one of the two players “stronger” through the knowledge of the loss of the next round. However, this is perfectly fine: The proof in Theorem 6 is still perfectly valid.

In this case, the ${Y}$-player has an easy life: it knows the loss before making the prediction, hence it can just output the minimizer of the loss in ${Y}$. Hence, we also have that the regret of the ${Y}$-player will be non-positive and it will not show up in Theorem 6. Putting all together, we can state the following corollary.

Corollary 8. Let ${f:X\times Y\rightarrow {\mathbb R}}$ be continuous, convex in the first argument, and concave in the second. Assume ${X}$ compact and that the argmax of the ${Y}$-player is never empty. Then, with the notation in Algorithm 2, we have

$\displaystyle \sup_{{\boldsymbol y}\in\mathcal{Y}} f(\bar{{\boldsymbol x}}_T, {\boldsymbol y}) - \min_{{\boldsymbol x}\in\mathcal{X}} f({\boldsymbol x},\bar{{\boldsymbol y}}_T) \leq \frac{\text{Regret}^{X}_T({\boldsymbol x}'_T)}{T},$

for any ${{\boldsymbol x}_T'\in\arg\min_{{\boldsymbol x}\in X}f({\boldsymbol x},\bar{{\boldsymbol y}}_T)}$ and where ${\text{Regret}^{X}_T({\boldsymbol x})=\sum_{t=1}^T \ell_t({\boldsymbol x}_t) - \sum_{t=1}^{T} \ell_t({\boldsymbol x})}$.

This alternative seems interesting from a theoretical point of view because it allows to avoid the complexity of learning in the ${Y}$ space, for example removing the dependency on its dimension. Of course, an analogous result can be stated using best-response for the ${X}$-player and an OCO algorithm for the ${Y}$-player.

There is a third variant, very used in empirical implementations, especially of Counterfactual Regret Minimization (CFR) (Zinkevich et al., 2007). It is called alternation and it breaks the simultaneous reveal of the actions of the two players. Instead, we use the updated prediction of the first player to construct the loss of the second player. Empirically, this variant seems to greatly speed-up the convergence of the duality gap.

For this version, Theorem 6 does not hold anymore because the terms ${f({\boldsymbol x}_t,{\boldsymbol y}_t)}$ and ${f({\boldsymbol x}_{t+1},{\boldsymbol y}_t)}$ are now different, however we can prove a similar guarantee.

Theorem 9. Let ${f:X\times Y\rightarrow {\mathbb R}}$ continuous, convex in the first argument, and concave in the second. Assume that ${X}$ and ${Y}$ are compact. Then, with the notation in Algorithm 4, for any ${{\boldsymbol x}_T'\in\arg\min_{{\boldsymbol x}\in\mathcal{X}}f({\boldsymbol x},\bar{{\boldsymbol y}}_T)}$ and any ${{\boldsymbol y}_T'\in\arg\max_{{\boldsymbol y}\in\mathcal{Y}}f(\bar{{\boldsymbol x}}_T,{\boldsymbol y})}$ we have

$\displaystyle \max_{{\boldsymbol y}\in\mathcal{Y}} f(\bar{{\boldsymbol x}}_T, {\boldsymbol y}) - \min_{{\boldsymbol x}\in\mathcal{X}} f({\boldsymbol x},\bar{{\boldsymbol y}}_T) \leq \frac{\text{Regret}^{X}_T({\boldsymbol x}'_T)+\text{Regret}^{Y}_T({\boldsymbol y}'_T)+\sum_{t=1}^{T} (f({\boldsymbol x}_{t+1},{\boldsymbol y}_t)-f({\boldsymbol x}_{t},{\boldsymbol y}_t))}{T},$

where ${\text{Regret}^{Y}_T({\boldsymbol y})=\sum_{t=1}^T h_t({\boldsymbol y}_t) - \sum_{t=1}^{T} h_t({\boldsymbol y})}$ and ${\text{Regret}^{X}_T({\boldsymbol x})=\sum_{t=1}^T \ell_t({\boldsymbol x}_t) - \sum_{t=1}^T \ell_t({\boldsymbol x})}$.

Proof: Note that ${\ell_t({\boldsymbol x})=f({\boldsymbol x},{\boldsymbol y}_t)}$, and ${h_t({\boldsymbol y})=-f({\boldsymbol x}_{t+1},{\boldsymbol y})}$. By Jensen’s inequality, we have

\displaystyle \begin{aligned} T(f(\bar{{\boldsymbol x}}_T, {\boldsymbol y}) - f({\boldsymbol x},\bar{{\boldsymbol y}}_T)) &\leq \sum_{t=1}^{T}f({\boldsymbol x}_{t+1},{\boldsymbol y})- \sum_{t=1}^{T}f({\boldsymbol x},{\boldsymbol y}_t)\\ &=\sum_{t=1}^{T}f({\boldsymbol x}_{t+1},{\boldsymbol y})-\sum_{t=1}^{T}f({\boldsymbol x}_{t+1},{\boldsymbol y}_t) +\sum_{t=1}^{T}f({\boldsymbol x}_{t},{\boldsymbol y}_t)-\sum_{t=1}^{T}f({\boldsymbol x},{\boldsymbol y}_t)\\ &\quad +\sum_{t=1}^{T}f({\boldsymbol x}_{t+1},{\boldsymbol y}_t)-\sum_{t=1}^{T}f({\boldsymbol x}_{t},{\boldsymbol y}_t)\\ &= \sum_{t=1}^{T} (h_t({\boldsymbol y}_t)-h_t({\boldsymbol y}))+\sum_{t=1}^{T}(\ell_t({\boldsymbol x}_t)-\ell_t({\boldsymbol x})) + \sum_{t=1}^{T} (f({\boldsymbol x}_{t+1},{\boldsymbol y}_t)-f({\boldsymbol x}_{t},{\boldsymbol y}_t))~. \end{aligned}

Taking ${{\boldsymbol x}={\boldsymbol x}'_T\in\arg\min_{{\boldsymbol x}\in\mathcal{X}} f({\boldsymbol x},\bar{{\boldsymbol y}}_T)}$ and ${{\boldsymbol y}={\boldsymbol y}'_T\in\arg\max_{{\boldsymbol y}\in\mathcal{Y}} f(\bar{{\boldsymbol x}}_T,{\boldsymbol y})}$, we get the stated result. $\Box$

Remark 3. In the case that ${f(\cdot,\cdot)}$ is linear in the first argument, using OMD for the ${X}$-player we have that ${f({\boldsymbol x}_t,{\boldsymbol y}_t) = \ell_t({\boldsymbol x}_t) \geq \ell_t({\boldsymbol x}_{t+1})=f({\boldsymbol x}_{t+1},{\boldsymbol y}_t)}$. Hence, in this case the additional term in Theorem 9 is negative, showing a (marginal) improvement to the convergence rate.

Next post, we will show how to connect saddle-point problems with Game Theory.

3. History Bits

Theorem 3 is (Rockafellar, 1970, Lemma 36.2).

The proof of Theorem 6 is from (Liu and Orabona, 2021) and it is a minor variation on the one in (Abernethy and Wang, 2017): (Liu and Orabona, 2021) stressed the dependency of the regret on a competitor that can be useful for refined bounds, as we will show next time for Boosting. It is my understanding that different variant of this theorem are known in the game theory community as “Folk Theorems”, because such result was widely known among game theorists in the 1950s, even though no one had published it.

The celebrated minimax theorem for zero-sum two-person games was first discovered by John von Neumann in the 1920s (von Neumann, 1928)(von Neumann and Morgenstern, 1944). The version is state here is a simplification of the generalization due to (Sion, 1958). The proof here is from (Abernethy and Wang, 2017). A similar proof is in (Cesa-Bianchi and Lugosi, 2006) based on a discretization of the space that in turn is based on the one in (Freund, Y. and Schapire, 1996)(Freund, Y. and Schapire, 1999).

Algorithms 2 and 3 are a generalization of the algorithm for boosting in (Freund and Schapire, 1996)(Freund and Schapire, 1999). Algorithm 2 was also used in (Abernethy and Wang, 2017) to recover variants of the Frank-Wolfe algorithm (Frank and Wolfe, 1956).

It is not clear who invented alternation. Christian Kroer told me that it was a known trick used in implementations of CFR for the computer poker competition from 2010 or so. Note that in CFR the method of choice is Regret Matching (Hart and Mas-Colell, 2000). However, (Kroer, 2020) empirically shows that alternation improves a lot even OGD for solving bilinear games. (Tammelin et al., 2015) explicitly include this trick in their implementation of an improved version of CFR called CFR+, claiming that it would still guarantee convergence. However, (Farina et al., 2019) pointed out that averaging of the iterates in alternation might not produce a solution to the min-max problem, providing a counterexample. Theorem 9 is from (Burch et al., 2019) (see also (Kroer, 2020)).

There is also a complementary view on alternation: (Zhang et al., 2021) link alternating updates to Gauss-Seidel methods in numerical linear algebra, in contrast to the simultaneous updates of the Jacobi method. Also, they provide a good review of the optimization literature on the advantages of alternation, but this paper and the papers they cite do not seem to be aware of the use of alternation in CFR.

Acknowledgements

Thanks to Christian Kroer for the history of alternation in CFR, and to Gergely Neu for the references on alternation in the optimization community. Thanks to Nicolò Campolongo for hunting down my typos. Also, thanks to Christian Kroer for his great lecture notes that helped me getting started on this topic.


text:
Euler's sum of powers conjecture, if a sum of $k$th powers is a $k$th power, then the sum has at least $k$ terms.
Proposed by Euler in 1769. Counterexample for $k=5$ found in 1966, for $k=4$ in 1986.
Euler's sum of powers conjecture, if a sum of $k$th powers is a $k$th power, then the sum has at least $k$ terms.
Proposed by Euler in 1769. Counterexample for $k=5$ found in 1966, for $k=4$ in 1986.


text:
# Problem with Biblatex, Latex does not compile the pdf

I have a big problem with Latex. I tried to install biblatex to create a bibliography. Since I had some trouble with the .bib file, I deleted the last command I inserted, such as `\usepackage{biblatex}` and `\addbibresource{sample.bib}, \printbibliography`. Now Latex does not compile the pdf document anymore and I get the following error. What should I do? Maybe trying to uninstall the package biblatek? Thank you.

``````19
Undefined control sequence.
l.19 \abx@aux@refcontext
{nty/global//global/global}
``````
• that is presumably line 19 of the aux file, first try deleting the aux file and see if that fixes the problem. – David Carlisle May 1 at 20:51
• If you are not loading `biblatex` any more the command `\abx@aux@refcontext` and many other `biblatex`-defines stuff will be undefined. Those commands can, however, still persist in the temporary files (mainly `.aux`). Delete all auxiliary files and recompile from scratch. The error should be gone. – moewe May 1 at 20:54
• Thank you, that fortunately worked! – Nenne May 1 at 20:54
• I am a novice and I didn't know that. Thank you, you helped me a lot. – Nenne May 1 at 20:56

LaTeX basically reads your document from beginning to end and can not skip forward when processing your file. In order to be able to act on stuff that happens later in your document, LaTeX remembers certain things from earlier LaTeX runs. That is the reason why for some things you need to run LaTeX multiple times to get things right. (For example for labels, see Understanding how references and labels work.) LaTeX remembers things from an earlier run by writing the relevant information into so-called auxiliary or temporary files. The most important auxiliary file is the `.aux` file. The `.aux` file is read quite early on when your document is processed and contains information from the previous LaTeX run.
The fact that the `.aux` file was produced by the previous run becomes crucial here. When you load `biblatex`, that package writes certain commands into the `.aux` file to help typesetting the bibliography and citations as expected. Those commands are defined by `biblatex` and will be executed on the next run to make sure everything looks nice.
If you now drop `biblatex` and run LaTeX with an `.aux` file produced by the previous run where `biblatex` was still loaded, the `.aux` file will contain references to `biblatex` macros that are now unknown to LaTeX (you haven't loaded `biblatex`, after all).
Errors like this can swiftly be resolved by deleting all temporary and auxiliary files (`.aux`, `.bbl`, `.bcf`, ...) and running a full compilation cycle from scratch. Deleting the auxiliary files is safe since they will be recreated in the next run of LaTeX (or the relevant helper program).


text:
# Enumerating Items within a Longtable w/ Section Numbers

I am using longtable and enumitem and was hoping to enumerate items inside a longtable. I am making a template for RFQ and need to enumerate all points with the section/subsection number. I had a MWE example, which is shown below, and an example that does not work. I would like to use enumitem's label function to identify items while keeping consistent formatting (i.e., such as used in the rest of the document) which is why the MWE is not robust enough to be sufficient. I tried making a counter and then defining a label but it does not compile. I am open to other solutions. Thanks, all!

Here is where I got my idea: Automatic table row numbers

\documentclass{article}

\usepackage{longtable}
\usepackage{enumitem}
\setenumerate[1]{label=\thesection.\arabic*.}
\setenumerate[2]{label*=\arabic*.}

%% Counters
\newcounter{magicrownumbers}
\newcommand\rownumber{\stepcounter{magicrownumbers}\arabic{magicrownumbers}}

\begin{document}
\begin{center}
\renewcommand{\arraystretch}{1.2}
\begin{longtable}{| l | l|}

\hline
\multicolumn{1}{|c|}{\textbf{Description}} & \multicolumn{1}{c|}{\textbf{File Types}} \\
\hline

\multicolumn{2}{c}%
{{\bfseries \tablename\ \thetable{} -- continued from previous page}} \\
\hline
\multicolumn{1}{|c|}{\textbf{Description}} & \multicolumn{1}{c|}{\textbf{File Types}}\\ \hline

\hline
\multicolumn{2}{|r|}{{Continued on next page}} \\ \hline
\endfoot

\hline
\endlastfoot

\thesection.\rownumber. Controls One-line & Visio or AutoCAD 2015 \\ %%<-- Works by has issues when text takes two lines
\begin{enumerate}[label = {\thesection.\rownumber.}] \item Mechanical Layout \end{enumerate} & AutoCAD 2015  \\ %%<----Doesn't work

\end{longtable}
\end{center}
\end{document}

• longtable environments are centered already, so no need of using \begin{center}...\end{center}, but actually, I can't compile your 'MWE' – user31729 Apr 11 '17 at 3:44

A version with an automatic row counting and a specialized enumerate environment.

In my point of view, the enumerate environment does nothing healthy there.

\documentclass{article}

\usepackage{longtable}
\usepackage{enumitem}
\usepackage{array}
\usepackage{chngcntr}

\newlist{rownumbers}{enumerate}{2}

\setlist[rownumbers,1]{label=\thesection.\arabic*,resume}
\setlist[rownumbers,2]{label*=\arabic*}

% Counter is named rownumbersi !!!!
\counterwithin{rownumbersi}{table}

%% Counters

%\setenumerate[1]{label=\thesection.\arabic*.}
%\setenumerate[2]{label*=\arabic*.}
%\newcounter{magicrownumbers}
%\newcommand\rownumber{\stepcounter{magicrownumbers}\arabic{magicrownumbers}}
\newcommand\rownumber{\stepcounter{rownumbersi}\therownumbersi}

\newcolumntype{N}[1]{>{\raggedleft\rownumber}p{#1}}
\newcolumntype{L}[1]{>{\raggedright}p{#1}}
\newcolumntype{R}[1]{>{\raggedleft}p{#1}}
\newcolumntype{C}[1]{>{\centering\arraybackslash}p{#1}}

\begin{document}
\renewcommand{\arraystretch}{1.2}
\begin{longtable}{|L{5cm} | L{2in}|}

\hline
\multicolumn{1}{|c|}{\textbf{Description}} & \multicolumn{1}{c|}{\textbf{File Types}} \\
\hline

\multicolumn{2}{c}%
{{\bfseries \tablename\ \thetable{} -- continued from previous page}} \\
\hline
\multicolumn{1}{|c|}{\textbf{Description}} & \multicolumn{1}{c|}{\textbf{File Types}}\\ \hline

\hline
\multicolumn{2}{|r|}{{Continued on next page}} \\ \hline
\endfoot

\hline
\endlastfoot

%\thesection.\rownumber. Controls One-line & Visio or AutoCAD 2015 \\ %%<-- Works by has issues when text takes two lines
\begin{rownumbers}
\item Mechanical Layout
\item Foo
\item Some stuff
\end{rownumbers} & AutoCAD 2015  \tabularnewline %%<----Doesn't work

\end{longtable}
\end{document}

• That was exactly as needed! Thank you soooo much! – Bryan Manning Apr 11 '17 at 17:27
• @BryanManning: Alright, happy TeXing then! ;-) – user31729 Apr 11 '17 at 17:28
• I noticed a small problem, I want to assign one element of a row with another, thus, I would need only one item inside the \begin{rownumbers} ... \end{rownumbers}. What ends up happening is I get 1.1 for each value and using [resume] does not seem to help since each bullet point is at the same level. Could I amend the environment to take the row number instead of the sub-section number? – Bryan Manning Apr 11 '17 at 17:48
• @BryanManning: That's changing the question ;-) – user31729 Apr 11 '17 at 17:50
• Would you like me to resubmit the question? I might've mistyped after a few hours of failure. :P – Bryan Manning Apr 11 '17 at 18:15


text:
# Are there any simply connected parallelizable 4-manifolds?

On pp. 166 of Scorpan's "The Wild World of 4--manifolds", he gives an example of a parallelizable 4-manifold ($S^1\times S^3$) and then asserts: "there are no simply connected examples". It's confusing to me whether he means that there are no such examples in general, or examples that are 4-manifolds. In any case, here is my question:

Do there exist (compact, smooth, oriented) simply connected manifolds that are parallelizable? 4-manifolds?

(Recall that a manifold is called parallelizable if it has a trivial tangent bundle.)

• At the very least "compact" should be added, otherwise just take $\mathbb{R}^n$. In fact, any contractible manifold will be both simply connected and parallelizable. – Matt Jun 13 '13 at 18:34
• @Matt you're absolutely right. I've edited the question. – Aru Ray Jun 13 '13 at 18:59

At least $S^3$ is a lie group so parallelizable.


text:
Please use this identifier to cite or link to this item: http://hdl.handle.net/2289/6346
DC FieldValueLanguage
dc.contributor.authorSathyanarayana Rao, Mayuri-
dc.contributor.authorSubrahmanyan, Ravi-
dc.contributor.authorUdaya Shankar, N.-
dc.contributor.authorChluba, Jens-
dc.date.accessioned2015-11-17T11:24:25Z-
dc.date.available2015-11-17T11:24:25Z-
dc.date.issued2015-09-
dc.identifier.citationAstrophysical Journal, 2015, Vol. 810, p3en_US
dc.identifier.issn1538-4357-(Online)-
dc.identifier.issn0004-637X-
dc.identifier.urihttp://hdl.handle.net/2289/6346-
dc.descriptionRestricted Access. An open-access version is available at arXiv.org (one of the alternative locations)en_US
dc.description.abstractPhotons emitted during cosmological hydrogen ($500\lesssim z\lesssim 1600$) and helium recombination ($1600\lesssim z\lesssim 3500$ for He ii $\to$ He i, $5000\lesssim z\lesssim 8000$ for He iii $\to$ He ii) are predicted to appear as broad, weak spectral distortions of the cosmic microwave background. We present a feasibility study for a ground-based detection of these recombination lines, which would uniquely probe astrophysical cosmology beyond the last scattering surface and provide observational constraints on the thermal history of the universe. We find that including sufficient signal spectral structure and maximizing signal-to-noise ratio, an octave band in the 2–6 GHz window is optimal; in this band the predicted signal appears as an additive quasi-sinusoidal component with amplitude about 8 nK embedded in a sky spectrum some nine orders of magnitude brighter. We discuss algorithms to detect these tiny spectral fluctuations in the sky spectrum by foreground modeling and introduce a maximally smooth function capable of describing the foreground spectrum and distinguishing the signal of interest. We conclude that detection is in principle feasible in realistic observing times provided that radio frequency interference and instrument bandpass calibration are controlled in this band at the required level; using Bayesian tests and mock data, we show that 90% confidence detection is possible with an array of 128 radiometers observing for 255 days of effective integration time. We propose APSERa—Array of Precision Spectrometers for the Epoch of Recombination—a dedicated radio telescope to detect these recombination lines.en_US
dc.language.isoenen_US
dc.publisherIOP Sciences for The American Astronomical Societyen_US
dc.relation.urihttp://arxiv.org/abs/1501.07191en_US
dc.relation.urihttp://dx.doi.org/10.1088/0004-637X/810/1/3en_US
dc.rights2015, The American Astronomical Society.en_US


text:
## Tuesday, 18 August 2015

### Time Conversion Hacker Rank Problem Solution Using C++.

/*
Time Conversion

Problem Statement
You are given time in AM/PM format. Convert this into a 24 hour format.
Note Midnight is 12:00:00AM or 00:00:00 and 12 Noon is 12:00:00PM.
Input Format
Input consists of time in the AM/PM format i.e. hh:mm:ssAM or hh:mm:ssPM
where
01hh12
00mm59
00ss59
Output Format
You need to print the time in 24 hour format i.e. hh:mm:ss
where
00hh23
00mm59
00ss59
Sample Input
07:05:45PM

Sample Output
19:05:45

TAG: C++,Implementation , Simple Program , Hacker Rank , Algorithm Easy.

*/

Solution:

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<cstring>
using namespace std;
/*
*
* Prosen Ghosh
* American International University - Bangladesh (AIUB)
*
*/
int main() {
int h,m,s;
char c[2];
scanf("%d:%d:%d%s",&h,&m,&s,c);
if(!strcmp(c,"PM") && h != 12)h+=12;
if(!strcmp(c,"AM") && h == 12)h = 0;
printf("%02d:%02d:%02d\n",h,m,s);
return 0;
}

1. for this line,printf("%02d:%02d:%02d\n",h,m,s); what is the 02 for?

1. The 0 means to pad the field using zeros and the 2 means that the field is two characters wide, so for any numbers that take less than 2 characters to display, it will be padded with a 0.

2. I didn't get the strcmp line plzz explain it to me

3. so am I what is this strcmp

4. https://www.tutorialspoint.com/c_standard_library/c_function_strcmp.htm

2. How do I do it in C++

1. You can use:
cout << setfill('0') << setw(2) << your_value;

Don't forget to include iomanip :)

2. #include dosent work

3. I hope
char c[2];
is too short
char c[3] = {}; is better.

Try this with clang address sanitizer:

4. The first strcmp statement working isn't clearing explained you must explain it brief in comments.

5. This comment has been removed by the author.

6. #include
#include
#include
#include
#include
#include
#include
using namespace std;

int main(){
string time,temp;
cin >> time;
int pos =0;
pos = time.find(":");
int h = stoi(time.substr(0, pos));

temp = time.substr(pos+1);
pos = temp.find(":");
int m = stoi(temp.substr(0,pos));

int s = stoi(temp.substr(pos+1, 2));

if((string::npos != time.find("PM")) && (h != 12))h+=12;
if((string::npos != time.find("AM")) && (h == 12))h = 0;
cout << setfill('0') << setw(2) << h << ":" << setfill('0') << setw(2)<< m << ":" << setfill('0') << setw(2) << s <<endl;
return 0;
}

7. Can you explain the logic of
If(!strcmp(c,"AM") && h!=12)H=H+12;

1. The code doesn't have that statement.

2. we are comparing c with pm if its pm then strcmp will return 0 but for and case we need 1 so we are using (!)not here.Further if h is suppose 7 then in case of pm we have to make it 19 so we are doing +12 to it.

8. I'm don't understand the code


text:
# StringCount in Mathematica

StringCount["one letters three letters letters one letters letters three letters",
"one" ~~ __ ~~ "three"]

gives

1

because there is the one occurence of "one [letters three letters letters one letters letters] three [letters]"

It took me some time to understand that Mathematica "thinks" in this way because I expected the value

2

due to "one letters three letters letters one letters letters three letters"

Now I am facing the problem of finding the code that solves this problem in my desired way.

-
Use Shortest[__] – R. M. May 14 '14 at 21:40
– Jacob Akkerboom May 15 '14 at 15:35


text:
# Age of The Universe!

1. Mar 30, 2010

### NeedPhysHelp8

Hi all,
Hoping you can help with this problem!

1. The problem statement, all variables and given/known data
For a flat universe, show that $$t_{0}=2/3 (H_{0})^{-1}$$ ? Seems like a simple problem I just don't really know where to start.

2. Relevant equations
Hubble's law $$v= H_{0}r$$
and then $$t= 1/H_{0}$$

3. The attempt at a solution
I just need guidance where to start. Flat universe means that expansion is constant in any direction right? What do I do?

2. Mar 30, 2010

### mgb_phys

flat universe means the hubble constant is constant.

Then 1/hubble constant has units time.
Think of the equations for something that doubles/halves with a constant time interval

3. Mar 30, 2010

### NeedPhysHelp8

How about the kinematic equation: x(t)= vt + 1/2at^2
Not sure if this is right and what to do with it? maybe take a derivatives of both sides???

4. Mar 31, 2010

### NeedPhysHelp8

I still can't figure out the answer, can someone give me a hint please


text:
## average current density

Also contains definition of: local current density
https://doi.org/10.1351/goldbook.A00539
The average current density, j, is defined by $j = A^{-1}\ \int _{A}j_{x}\ \text{d}A$ where A is the electrode area, jx is the local current density and dA is an infinitesimal surface element.
Source:
PAC, 1981, 53, 1827. 'Nomenclature for transport phenomena in electrolytic systems' on page 1836 (https://doi.org/10.1351/pac198153101827)


text:
Open accessJournal Article

# Hurwitz Theory of Elliptic Orbifolds, I

02 Mar 2021-Geometry & Topology (Mathematical Sciences Publishers)-Vol. 25, Iss: 1, pp 229-274
Abstract: An elliptic orbifold is the quotient of an elliptic curve by a finite group. In 2001, Eskin and Okounkov proved that generating functions for the number of branched covers of an elliptic curve with specified ramification are quasimodular forms for SL2(ℤ). In 2006, they generalized this theorem to branched covers of the quotient of an elliptic curve by ±1, proving quasimodularity for Γ0(2). We generalize their work to the quotient of an elliptic curve by ⟨ζN⟩ for N=3, 4, 6, proving quasimodularity for Γ(N), and extend their work in the case N=2. It follows that certain generating functions of hexagon, square and triangle tilings of compact surfaces are quasimodular forms. These tilings enumerate lattice points in moduli spaces of flat surfaces. We analyze the asymptotics as the number of tiles goes to infinity, providing an algorithm to compute the Masur–Veech volumes of strata of cubic, quartic, and sextic differentials. We conclude a generalization of the Kontsevich–Zorich conjecture: these volumes are polynomial in π.

Topics: Elliptic curve (63%), Quotient (56%), Moduli space (54%) ... read more
##### Citations
More

6 results found

Open accessPosted Content
Abstract: We express the Masur-Veech volume and the area Siegel-Veech constant of the moduli space of meromorphic quadratic differential with simple poles as polynomials in the intersection numbers of psi-classes supported on the boundary cycles of the Deligne-Mumford compactification of the moduli space of curves. Our formulae are derived from lattice point count involving the Kontsevich volume polynomials that also appear in Mirzakhani's recursion for the Weil-Petersson volumes of the moduli space of bordered hyperbolic Riemann surfaces. A similar formula for the Masur-Veech volume (though without explicit evaluation) was obtained earlier by Mirzakhani through completely different approach. We prove further result: up to an explicit normalization factor depending only on the genus and on the number of cusps, the density of the orbit of any simple closed multicurve computed by Mirzakhani coincides with the density of square-tiled surfaces having horizontal cylinder decomposition associated to the simple closed multicurve. We study the resulting densities in more detail in the special case when there are no cusps. In particular, we compute explicitly the asymptotic frequencies of separating and non-separating simple closed geodesics on a closed hyperbolic surface of genus g for all small genera g and we show that in large genera the separating closed geodesics are exponentially less frequent. We conclude with detailed conjectural description of combinatorial geometry of a random simple closed multicurve on a surface of large genus and of a random square-tiled surface of large genus. This description is conditional to the conjectural asymptotic formula for the Masur-Veech volume in large genera and to the conjectural uniform asymptotic formula for certain sums of intersection numbers of psi-classes in large genera.

Topics: Asymptotic formula (55%), Moduli space (55%),  ... read more

28 Citations

Open accessJournal ArticleDOI: 10.5802/JEP.159
30 Mar 2021-
Abstract: Let $S$ be a connected closed oriented surface of genus $g$. Given a triangulation (resp. quadrangulation) of $S$, define the index of each of its vertices to be the number of edges originating from this vertex minus $6$ (resp. minus $4$). Call the set of integers recording the non-zero indices the profile of the triangulation (resp. quadrangulation). If $\kappa$ is a profile for triangulations (resp. quadrangulations) of $S$, for any $m\in \mathbb{Z}_{>0}$, denote by $\mathscr{T}(\kappa,m)$ (resp. $\mathscr{Q}(\kappa,m)$) the set of (equivalence classes of) triangulations (resp. quadrangulations) with profile $\kappa$ which contain at most $m$ triangles (resp. squares). In this paper, we will show that if $\kappa$ is a profile for triangulations (resp. for quadrangulations) of $S$ such that none of the indices in $\kappa$ is divisible by $6$ (resp. by $4$), then $\mathscr{T}(\kappa,m)\sim c_3(\kappa)m^{2g+|\kappa|-2}$ (resp. $\mathscr{Q}(\kappa,m) \sim c_4(\kappa)m^{2g+|\kappa|-2}$), where $c_3(\kappa) \in \mathbb{Q}\cdot(\sqrt{3}\pi)^{2g+|\kappa|-2}$ and $c_4(\kappa)\in \mathbb{Q}\cdot\pi^{2g+|\kappa|-2}$. The key ingredient of the proof is a result of J. Kollar on the link between the curvature of the Hogde metric on vector subbundles of a variation of Hodge structure over algebraic varieties, and Chern classes of their extensions. By the same method, we also obtain the rationality (up to some power of $\pi$) of the Masur-Veech volume of arithmetic affine submanifolds of translation surfaces that are transverse to the kernel foliation.

5 Citations

Open accessPosted Content
Abstract: We study the moduli spaces of flat surfaces with prescribed conical singularities. Veech showed that these spaces are diffeomorphic to the moduli spaces of marked Riemann surfaces, and endowed with a natural volume form depending on the orders of the singularities. We show that the volumes of these spaces are finite. Moreover we show that they are explicitely computable by induction on the Euler characteristics of the punctured surface for almost all orders of the singularities. The proof relies on the computation of the large $k$ asymptotics of intersection numbers on moduli spaces of $k$-canonical divisors. This analysis was made possible by recent progress in the study of the intersection theory of the universal Jacobian by Bae, Holmes, Pandharipande, Schmitt, and Schwarz.

Topics: Moduli space (62%), Riemann surface (54%),  ... read more

5 Citations

Open accessPosted Content
Abstract: The volume $\mathscr{B}_{\Sigma}^{{\rm comb}}(\mathbb{G})$ of the unit ball -- with respect to the combinatorial length function $\ell_{\mathbb{G}}$-- of the space of measured foliations on a stable bordered surface $\Sigma$ appears as the prefactor of the polynomial growth of the number of multicurves on $\Sigma$. We find the range of $s \in \mathbb{R}$ for which $(\mathscr{B}_{\Sigma}^{{\rm comb}})^{s}$, as a function over the combinatorial moduli spaces, is integrable with respect to the Kontsevich measure. The results depends on the topology of $\Sigma$, in contrast with the situation for hyperbolic surfaces where Arana-Herrera and Athreya (arXiv:1907.06287) recently proved an optimal square-integrability.

Topics: Unit sphere (56%),

Open accessJournal Article
Abstract: We express the Masur–Veech volume and the area Siegel–Veech constant of the moduli space Qg,n of genus g meromorphic quadratic differentials with at most n simple poles and no other poles as polynomials in the intersection numbers ∫M‾ g′,n′ψ1d1⋯ψ n′dn′ with explicit rational coefficients, where g′

Topics: Teichmüller space (55%), Moduli space (53%),  ... read more

##### References
More

22 results found

Open accessJournal Article
Abstract: Die grundlegende Bedeutung des vorliegenden Themas fur die Riemann’sche Theorie der algebraischen Funktionen brauche ich wohl kaum hervorzuheben. Geht doch diese Theorie von der graphisch uber der komplexen Zahlenebene konstruierten Riemann’schen Flache aus, um erst sodann die Funktionen, welche durch diese Flache bestimmt sind, zu untersuchen.

555 Citations

Open accessJournal Article
Andrei Okounkov1, Rahul Pandharipande1Institutions (1)
Abstract: We establish an explicit equivalence between the stationary sector of the Gromov-Witten theory of a target curve X and the enumeration of Hurwitz coverings of X in the basis of completed cycles. The stationary sector is formed, by definition, by the descendents of the point class. Completed cycles arise naturally in the theory of shifted symmetric functions. Using this equivalence, we give a complete description of the stationary Gromov-Witten theory of the projective line and elliptic curve. Toda equations for the relative stationary theory of the projective line are derived.

Topics: Projective line (59%), Elliptic curve (54%), Algebraic geometry (53%) ... read more

341 Citations

Book Chapter
Robbert Dijkgraaf1Institutions (1)
01 Jan 1995-
Abstract: I review how recent results in quantum field theory confirm two general predictions of the mirror symmetry program in the special case of elliptic curves: (1) counting functions of holomorphic curves on a Calabi-Yau space (Gromov-Witten invariants) are ‘quasimodular forms’ for the mirror family; (2) they can be computed by a summation over trivalent Feynman graphs.

Topics: , Mirror symmetry (60%), Schoof's algorithm (58%) ... read more

232 Citations

##### Performance
###### Metrics
No. of citations received by the Paper in previous years
YearCitations
20214
20201
20191

Copyright @ 2022 | PubGenius Inc. | 3260 Hillview Avenue, Palo Alto, California 94304