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# How to automatically run an elevated program shortcut at Windows startup, without login or UAC prompts

Posted by Paul Braren on Nov 10 2016 (updated on Nov 14 2016) in
• Windows
• Productivity
• ## Backstory

Why did I want to do this? Well, I was setting up to record this one-long-shot 4K test video, where I wanted my Window Server 2016 test system to automatically go right into maximum CPU and RAM abuse mode, without requiring a keyboard or monitor. A little strange, I know, and certainly a corner case. That said, the little tricks featured in this intermediate-skills article can be applied to many other use cases too.

So the first hurtle I had to clear was to grab the new 64 bit version of the simple, oldie-but-goodie Prime 95 that bangs away at all CPU cores and RAM. But alas, the dang thing also invokes that UAC prompt. Requires it actually. What to do? How about a mashup of three techniques:

1. get past the UAC (User Access Control) prompt
3. set that special shortcut to auto-start

This is one of those articles I'm writing both for you and for my future self, because I know I'll be using these techniques again.

## Warning

This technique is done at your own risk, see also the usual disclaimer found with every article.

Not that this article doesn't suggest disabling UAC is a good idea, even if it is admittedly simpler than this work-around which only creates a single exception to the usual UAC prompt. It may be more helpful for the Windows Server family of operating systems.

I realize that setting Windows to automatically log you in is not exactly a great idea from a security perspective, but remember, I said test system. This is a private, temporary, isolated home-lab. I'm using the free 180 trial of Windows Server 2016. This auto-login behavior requires an encrypted, saved password. It's also handy for test VMs as well, including reboot time on SSD tests.

## Prerequisites

• an executable that requires elevated rights, and you want to run it unattended, automatically, at system reboot, and be able to interact with it
• Windows 7
• Windows 8 / 8.1
• Windows 10, including latest Anniversary Edition
• Windows Server 2012
• Windows Server 2016
• if you don't mind external downloads or third party software, per James Senecal's suggestion below, you may find it simpler to use the portable RunAsTool by Sordum as a substitute for the Task Scheduler first step below

## Instructions

1. ### Create the special shortcut using Task Scheduler

by following along with this excellent set of instructions:

How to Create an Elevated Program Shortcut without a UAC Prompt in Windows
Jul 28 2016 by Shawn Brink at Windows TenForums
For my use, as demonstrated in the video below, I only had to make the following minor tweak to the essentially identical Windows 7 version of the same procedure:
a. paste the following line line of code into the Actions tab, with -t being a Prime95.exe unattended test startup option (your optional startup parameters will vary):

%windir%\System32\cmd.exe /c start "Prime95" "C:\util\prime95.exe" -t

When you're done creating and testing your special desktop shortcut, run it to test that it works, then continue with these instructions.

2. ### Open your Startup folder

b. type Win + R, brings up Run dialogue
c. press Del key, to delete whatever was there
d. paste the following one-liner into the Run dialogue:

shell:common startup

then press Enter
d. Windows Explorer opens right up, to the special Startup Folder

3. ### Move the special shortcut to your Startup folder

that you created in step 1 to this special folder, using drag-and-drop

4. ### Launch User Accounts

it's built-in to Windows, and it saves your credentials, so Windows no longer prompts you for a password after reboot, here's how to set it up:
b. type Win + R, brings up Run dialogue
c. press Del key, to delete whatever was there
d. paste the following one-liner into the Run dialogue:

netplwiz.exe

then press Enter

5. ### Use User Accounts to configure automatic login

on netplwiz's Users tab, Uncheck:
Users must enter a user name and password to use this computer.
then click OK, you'll now be prompted to enter admin credentials, then press Enter

6. ### Reboot to test

watch what happens after rebooting

Tada, you're done! Your special shortcut automatically runs your program shortcut as Administrator after that reboot, without the User access Control pop-up. Nice!

Have a problem following along with these written instructions? Worry not, my detailed, step-by-step video below should help.


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# C++ implementation of Hackerrank's “Maximum Element in a Stack”

I just attempted the Maximum Element Challenge on Hackerrank. This question has been posted before in swift Swift Hackerrank Maximum Element in a Stack but it has attracted low views/no answers. Moving forward, based on the requirements of finding the maximum element, I used a vector<int> as my underlying data structure as opposed to using Stack as I can't iterate through a stack. This exercise is meant to be a simple one but the time limit gets exceeded for larger input.

How do I improve the performance? Please feel free to criticise my approach as I'm new to C++ with 5 days of experience.

You have an empty sequence, and you will be given $N$ queries. Each query is one of these three types:

• 1 x -Push the element x into the stack
• 2 -Delete the element present at the top of the stack
• 3 -Print the maximum element in the stack

Input Format

The first line of input contains an integer, $N$ . The next $N$ lines each contain an above mentioned query. (It is guaranteed that each query is valid.)

Constraints

• $1 \leq N \leq 10^5$
• $1 \leq x \leq 10^9$
• $1 \leq \text{type} \leq 3$

Output Format

For each type $3$ query, print the maximum element in the stack on a new line.

Sample Input

10
1 97
2
1 20
2
1 26
1 20
2
3
1 91
3

Sample Output

26
91

#include <vector>
#include <iostream>
#include <algorithm>

//using namespace std;
bool compareValues(int i, int j) { return i<j; }

int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int N;
int queryType;
int element;
std::vector<int> stackOfNumbers;

std::cin >> N;
for(int i =0; i < N ; i++){
std::cin >> queryType;

switch(queryType) {
case 1 :
std::cin >> element;
stackOfNumbers.push_back(element);
break;
case 2 :
stackOfNumbers.erase(stackOfNumbers.end()-1);
break;
case 3 :
std::cout<< *std::max_element(std::begin(stackOfNumbers), std::end(stackOfNumbers),compareValues ) << std::endl;
break;
default :
std::cout << "Invalid queryType" << std::endl;
}

}
return 0;
}

This is one of the questions, that is directly related to the underlying data structure. So you can actually be quite sure, that this problem is best solved with a stack.

The trick here is, that a stack hahttp://codereview.stackexchange.com/questions/146882/c-implementation-of-hackerranks-maximum-element-in-a-stacks a well defined order in which the elements are retrieved.

In that particular problem, lets say you know the current maximum of the stack. If the new element is smaller than that, you can push the maximum value instead of the actual one, as you are only interested in the maximum value of the whole stack. This is only true, as a stack guaranties, that the new element is removed before the previous maximum element. Only if it is greater you push the new element.

This has the distinct disadvantage, that you do not preserve the original data. So there are two solutions, one for tracking the maximum element and one for also preserving the data.

#include <iostream>
#include <stack>

int main () {
size_t numQuerries;
std::cin >> numQuerries;
std::stack<int> maxStack;

int inputType;
for (size_t querry = 0; querry < numQuerries; ++querry) {
int input;
std::cin >> inputType;
switch(inputType) {
case 1:
std::cin >> input;
if (maxStack.empty()) {
maxStack.push(input);
} else {
maxStack.push(input > maxStack.top() ? input : maxStack.top());
}
break;
case 2:
if (!maxStack.empty()) {
maxStack.pop();
}
break;
case 3:
std::cout << maxStack.top() << "\n";
break;
default:
break;
}
}
}

So now if you want to keep the original data you simply need two stacks

#include <iostream>
#include <stack>

int main () {
size_t numQuerries;
std::cin >> numQuerries;
std::stack<int> maxStack;
std::stack<int> dataStack;

int inputType;
for (size_t querry = 0; querry < numQuerries; ++querry) {
int input;
std::cin >> inputType;
switch(inputType) {
case 1:
std::cin >> input;
if (maxStack.empty()) {
maxStack.push(input);
} else {
maxStack.push(input > maxStack.top() ? input : maxStack.top());
}
dataStack.push(input);
break;
case 2:
if (!maxStack.empty()) {
maxStack.pop();
dataStack.pop();
}
break;
case 3:
std::cout << maxStack.top() << "\n";
break;
default:
break;
}
}
}

## Better Solution

So these were the intuitive solutions. However, is there some room for further improvement?. Obviously yes. The point is, that you do not need to store the same element every time because you know what the maximum is. You can circumvent this by using two stacks. One with the current maximum and one with the number of its occurences.

#include <iostream>
#include <stack>

int main () {
size_t numQuerries;
std::cin >> numQuerries;
std::stack<int> maxStack;
std::stack<size_t> occurenceStack;

int inputType;
for (size_t querry = 0; querry < numQuerries; ++querry) {
int input;
std::cin >> inputType;
switch(inputType) {
case 1:
std::cin >> input;
if (maxStack.empty()) {
maxStack.push(input);
occurenceStack.push(1);
} else {
if (input > maxStack.top()) {
maxStack.push(input);
occurenceStack.push(1);
} else {
occurenceStack.top()++;
}
}
break;
case 2:
if (!maxStack.empty()) {
if (occurenceStack.top() > 1) {
occurenceStack.top()--;
} else {
maxStack.pop();
occurenceStack.pop();
}
}
break;
case 3:
std::cout << maxStack.top() << "\n";
break;
default:
break;
}
}
}

I believe to finish this within the time limit, you need to complete each of the three operations with $O(1)$ time complexity.

There are a number of variations in how you do that, but nearly all of them involve keeping track of the maximum value on the stack as you build the stack, rather than searching through all the existing stack items each time you receive a query. Given that this is an online challenge, I hesitate to go into a lot more detail on this point than that.

As far as the code style goes, the most obvious point would be that your compareValues isn't really necessary--std::less<T> will be used by default, and your comparison adds nothing it doesn't already do perfectly well. In fact, std::less<T> may be superior in one respect: it's defined as a function object--a class that overloads operator() to do the comparison. You'r defining an actual function, and passing a pointer to a function. Passing a function object tends to make it easier for a compiler to generate inline code, so letting std::less<T> do the job may well improve speed at least slightly.

One more (somewhat less obvious) point is that in the process of improving efficiency to the point that you can finish within the time limit, I'm (reasonably) confident that you'll eliminate the need to traverse the stack, so you should be able to use std::stack<int>.

It is possible to query for the maximum element, pushing and popping in constant time. The idea is that you maintain a partial linked list whithin the stack:

max_stack.h

#pragma once
#ifndef MAX_STACK_H
#define MAX_STACK_H

#include <stdexcept>

template<typename T>
class MaxStack {
public:

MaxStack()
:
top_node{nullptr},
maximum{nullptr}
{}

void push(T datum)
{
MaxStackNode* new_node = new MaxStackNode(datum, top_node);

if (top_node == nullptr)
{
maximum = new_node;
}
else if (maximum->datum < datum)
{
new_node->previousMaximum = maximum;
maximum = new_node;
}

top_node = new_node;
}

void pop()
{
check_not_empty();
MaxStackNode* ret = top_node;

if (maximum == ret)
{
maximum = ret->previousMaximum;
}

top_node = top_node->previous;
delete ret;
}

T top()
{
check_not_empty();
}

T max()
{
check_not_empty();
return maximum->datum;
}

private:

struct MaxStackNode {
T datum;
MaxStackNode* previous;
MaxStackNode* previousMaximum;

MaxStackNode(T datum, MaxStackNode* previous)
:
datum{datum},
previous{previous},
previousMaximum{nullptr}
{}
};

MaxStackNode* top_node;
MaxStackNode* maximum;

void check_not_empty()
{
if (top_node == nullptr)
{
throw std::runtime_error{"The stack is empty."};
}
}
};

#endif // MAX_STACK_H

What comes to your coding style, you are not quite consistent:

for(int i =0; i < N ; i++){

Please, put a single space character before (, after ) and after =. Also, remove the space character after N; like this:

for (int i = 0; i < N; i++) {

You might want to fix the indentation as well.

main.cpp

#include "max_stack.h"
#include <iostream>

using std::cout;
using std::endl;

int main() {
int N;
int queryType;
int element;
MaxStack<int> stack;

std::cin >> N;

for(int i = 0; i < N ; i++) {
std::cin >> queryType;

switch(queryType) {
case 1 :
std::cin >> element;
stack.push(element);
break;

case 2 :
stack.pop();
break;

case 3 :
cout << stack.max() << endl;
break;
}
}

return 0;
}

Hope that helps.

format: plain


text:
## Results (14 matches)

Label Base field Conductor Isogeny class Weierstrass coefficients
81.1-a1 $$\Q(\sqrt{17})$$ 81.1 81.1-a $$\bigl[0$$ , $$0$$ , $$1$$ , $$7152 a - 18327$$ , $$-472886 a + 1211329\bigr]$$
81.1-a2 $$\Q(\sqrt{17})$$ 81.1 81.1-a $$\bigl[0$$ , $$0$$ , $$1$$ , $$-48 a + 123$$ , $$-134 a + 343\bigr]$$
81.1-b1 $$\Q(\sqrt{17})$$ 81.1 81.1-b $$\bigl[0$$ , $$0$$ , $$1$$ , $$-6 a - 12$$ , $$14 a + 19\bigr]$$
81.1-b2 $$\Q(\sqrt{17})$$ 81.1 81.1-b $$\bigl[0$$ , $$0$$ , $$1$$ , $$-54 a - 108$$ , $$-378 a - 520\bigr]$$
81.1-b3 $$\Q(\sqrt{17})$$ 81.1 81.1-b $$\bigl[0$$ , $$0$$ , $$1$$ , $$6 a - 18$$ , $$-14 a + 33\bigr]$$
81.1-b4 $$\Q(\sqrt{17})$$ 81.1 81.1-b $$\bigl[0$$ , $$0$$ , $$1$$ , $$54 a - 162$$ , $$378 a - 898\bigr]$$
81.1-c1 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$435 a - 1115$$ , $$-7202 a + 18448\bigr]$$
81.1-c2 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$285 a - 725$$ , $$3676 a - 9366\bigr]$$
81.1-c3 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$525 a + 820$$ , $$-10064 a - 15716\bigr]$$
81.1-c4 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$195 a - 500$$ , $$1442 a - 3694\bigr]$$
81.1-c5 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$15 a - 50$$ , $$58 a - 132\bigr]$$
81.1-c6 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$-15 a - 35$$ , $$-58 a - 74\bigr]$$
81.1-c7 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$-435 a - 680$$ , $$7202 a + 11246\bigr]$$
81.1-c8 $$\Q(\sqrt{17})$$ 81.1 81.1-c $$\bigl[1$$ , $$-1$$ , $$1$$ , $$-285 a - 440$$ , $$-3676 a - 5690\bigr]$$


text:
# Gauss's Law Calculator

Created by Kenneth Alambra
Last updated: Sep 14, 2022

If you're curious about how to calculate the electric flux passing through a surface given the electric charge it has, this Gauss law calculator is for you. Performing Gauss's law calculations is surprisingly easy, and this tool will show you exactly how. Keep on reading to learn:

• What Gauss's Law is;
• How to calculate electric flux; and
• How to use this Gauss law calculator.

## What is Gauss's law?

Gauss's law states that the electric flux coming out of a closed surface of any size is proportional to the electric charge around it and varies inversely due to the permittivity of free space constant.

Electric flux is a representation of the electric field passing through a surface. The more electric field lines we can observe radiating a surface, the stronger the electric flux is. We use Gauss's law to quantify the strength of that electric field by determining its equivalent electric flux.

We usually find the total electric flux through an object by integrating the electric flux through infinitesimally small flat areas on a surface. In this tool, we consider a closed flat surface enclosed by a uniform electric field from a single electric charge.

In the next section of this text, let us learn about the Gauss law equation we can use in calculating electric flux.

## How to calculate electric flux using the Gauss law formula

When determining the electric flux through a closed flat surface, we use the Gauss law formula, as shown below:

$\phi = \frac{Q}{\varepsilon_0}$

Where:

• $\phi$ is the electric flux through the closed surface in volt-meters ($\small{\text{V}\cdot\text{m}}$);
• $Q$ is the total electric charge inside the surface in coulombs ($\small\text{C}$); and
• $\varepsilon_0$ is permittivity of free space electric constant with the value of $\small{8.854\times 10^{-12}\ \tfrac{\text{C}^2}{\text{N}\cdot\text{m}^2}}$.

Let's say we have a closed surface inside an electric charge of 20 nanocoulombs or $\small{2.0\times 10^{-8}\ \text{coulombs}}$, substituting that to the Gauss's law equation gives us:

\begin{align*} \phi &= \frac{Q}{\varepsilon_0}\\[1.0em] &=\frac{2.0\times 10^{-8}\ \text{C}}{8.854\times 10^{-12}\ \tfrac{\text{C}^2}{\text{N}\cdot\text{m}^2}}\\[1.5em] &= 2258.866049\ \tfrac{\text{N}\cdot\text{m}^2}{\text{C}}\\[0.8em] &\approx 2259\ \text{V}\cdot\text{m} \end{align*}

Note that we can use $\small{\text{V}\cdot\text{m}}$ to replace $\small{\tfrac{\text{N}\cdot\text{m}^2}{\text{C}}}$ since $\small{1\ \tfrac{\text{N}\cdot\text{m}}{\text{C}}}$ is equivalent to $\small{1\ \text{V}}$. You can learn more about the unit of volts (V) by checking out our electric potential calculator.

💡 Newtons per coulomb or $\tfrac{\text{N}}{\text{C}}$, as shown occurring in the units we use above, is the unit of electric field. You can expound your knowledge of the topic by exploring our electric field calculator.

As additional knowledge, we can also find the electric charge around a surface given the electric flux passing through it by rearranging the Gauss law equation, expressed in equation form as shown below:

$Q = \phi\times\varepsilon_0$

## How to use this Gauss law calculator

Our Gauss law calculator is very simple and works both ways. That means you can either enter a value for the electric charge to find the electric flux, or vice versa. That's how easy-to-use our Gauss law calculator is!

Suppose you need to experiment on other dielectric materials and, therefore, need to enter different values for permittivity. In that case, you can do that by clicking on the Advanced mode button below our Gauss's law calculator. Doing so will display the permittivity variable where you can enter other values.

If you're still craving more electromagnetic stuff, perhaps you'll find our Coulomb's law calculator interesting. There you'll learn how to determine the electrostatic force acting between two charges.

Kenneth Alambra
Electric charge (Q)
nC
Electric flux (ϕ)
Vm
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## Problem 1A. Archery Tournament ≡

 Author: NEERC 2017 Time limit: 3 sec Input file: Standard input Memory limit: 512 Mb Output file: Standard output

### Statement

You were invited to the annual archery tournament. You are going to compete against the best archers from all of the Northern Eurasia. This year, a new type of competition is introduced, where a shooting range is dynamic and new targets might appear at any second.

As the shooting range is far enough from you, it can be represented as a 2D plane, where y = 0 is the ground level. There are some targets in a shape of a circle, and all the targets are standing on the ground. That means, if a target's center is (x, y) (y > 0), then its radius is equal to y, so that it touches the line y = 0. No two targets simultaneously present at the range at any given time intersect (but they may touch).

Initially, the shooting range is empty. Your participation in this competition can be described as n events: either a new target appears at the range, or you shoot an arrow at some point at the range. To hit a target, you must shoot strictly inside the circle (hitting the border does not count). If you shoot and hit some target, then the target is removed from the range and you are awarded one point.

### Input format

The first line of the input contains integer n (1 ≤ n ≤ 2 ⋅ 105). Next n lines describe the events happening at the tournament. The i-th line contains three integers ti, xi, and yi (ti = 1, 2;  − 109 ≤ xi, yi ≤ 109; yi > 0).

• If ti = 1, then a new target with center (xi, yi) and radius yi appears at the range.
• If ti = 2, then you perform a shot, which hits the range at (xi, yi).

### Output format

For each of your shots, output a separate line with the single integer. If the shot did not hit any target, print -1. If the shot hit a target, print the number of event when that target was added to the range. Events are numbered starting from 1.

### Sample tests

No. Standard input Standard output
1
8
1 0 12
2 -11 22
1 24 10
1 12 3
2 12 12
2 16 14
1 28 15
2 3 6
-1
-1
3
1

0.188s 0.092s 13


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# PrimitiveOp¶

class PrimitiveOp(primitive, coeff=1.0)[ソース]

ベースクラス: qiskit.opflow.operator_base.OperatorBase

A class for representing basic Operators, backed by Operator primitives from Terra. This class (and inheritors) primarily serves to allow the underlying primitives to 「flow」 - i.e. interoperability and adherence to the Operator formalism - while the core computational logic mostly remains in the underlying primitives. For example, we would not produce an interface in Terra in which QuantumCircuit1 + QuantumCircuit2 equaled the Operator sum of the circuit unitaries, rather than simply appending the circuits. However, within the Operator flow summing the unitaries is the expected behavior.

Note that all mathematical methods are not in-place, meaning that they return a new object, but the underlying primitives are not copied.

パラメータ
• primitive (Union[QuantumCircuit, Operator, Pauli, SparsePauliOp, OperatorBase]) – The operator primitive being wrapped.

• coeff (Union[complex, ParameterExpression]) – A coefficient multiplying the primitive.

Methods Defined Here

 add Return Operator addition of self and other, overloaded by +. adjoint Return a new Operator equal to the Operator’s adjoint (conjugate transpose), overloaded by ~. assign_parameters Binds scalar values to any Terra Parameters in the coefficients or primitives of the Operator, or substitutes one Parameter for another. compose Return Operator Composition between self and other (linear algebra-style: A@B(x) = A(B(x))), overloaded by @. equals Evaluate Equality between Operators, overloaded by ==. eval Evaluate the Operator’s underlying function, either on a binary string or another Operator. exp_i Return Operator exponentiation, equaling e^(-i * op) log_i Return a MatrixOp equivalent to log(H)/-i for this operator H. mul Returns the scalar multiplication of the Operator, overloaded by *, including support for Terra’s Parameters, which can be bound to values later (via bind_parameters). permute Permutes the qubits of the operator. primitive_strings Return a set of strings describing the primitives contained in the Operator. reduce Try collapsing the Operator structure, usually after some type of conversion, e.g. tensor Return tensor product between self and other, overloaded by ^. tensorpower Return tensor product with self multiple times, overloaded by ^. to_circuit Returns a QuantumCircuit equivalent to this Operator. to_circuit_op Returns a CircuitOp equivalent to this Operator. to_instruction Returns an Instruction equivalent to this Operator. to_matrix Return NumPy representation of the Operator. to_matrix_op Returns a MatrixOp equivalent to this Operator. to_pauli_op Returns a sum of PauliOp s equivalent to this Operator.

Attributes

INDENTATION = ' '
coeff

The scalar coefficient multiplying the Operator.

Union[complex, ParameterExpression]

The coefficient.

instance_id

Return the unique instance id.

int

num_qubits

int

parameters
primitive

The primitive defining the underlying function of the Operator.

Union[QuantumCircuit, Operator, Pauli, SparsePauliOp, OperatorBase]

The primitive object.

settings

Return operator settings.

Dict


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# 2007A&A...468..529G

 other querymodes : Identifierquery Coordinatequery Criteriaquery Referencequery Basicquery Scriptsubmission TAP Outputoptions Help

 Query : 2007A&A...468..529G

2007A&A...468..529G - Astronomy and Astrophysics, volume 468, 529-540 (2007/6-3)

X-rays from T Tauri: a test case for accreting T Tauri stars.

GUEDEL M., SKINNER S.L., MELNIKOV S.Y., AUDARD M., TELLESCHI A. and BRIGGS K.R.

Abstract (from CDS):

The generation of X-rays in accreting T Tauri stars (TTS) is thought to be predominantly due to energy dissipation in magnetic fields, but alternative X-ray generation mechanisms have been proposed, such as heating in accretion shocks near the stellar surface, or in shocks forming in jets. We test models and trends discussed in the literature using X-ray data from the classical TTS T Tau. High-resolution spectroscopy from the Reflection Grating Spectrometers on XMM-Newton is used to infer electron densities, element abundances and the thermal structure of the X-ray source. We also discuss the ultraviolet light curve obtained by the Optical Monitor, and complementary ground-based photometry. A high-resolution image from Chandra constrains contributions from the two companions of T Tau N. The X-ray grating spectrum is rich in emission lines, but shows an unusual mixture of features from very hot (≃30MK) and very cool (1-3MK) plasma, both emitted by similar amounts of emission measure. The cool plasma confirms the picture of a soft excess in the form of an enhanced Ovii/Oviii Lyα flux ratio, similar to that previously reported for other accreting TTS. Diagnostics from lines formed by this plasma indicate low electron densities (≲1010cm–3). The Ne/Fe abundance ratio is consistent with a trend in pre-main sequence stars in which this ratio depends on spectral type, but not on accretion. On the basis of line density diagnostics, we conclude that the density of the cool soft-excess'' plasma is orders of magnitude below that predicted for an accretion shock, assuming previously determined accretion rates of (3-6)x10–8M/yr. We argue that loading of magnetic field lines with infalling material suppresses the heating process in a part of the corona. We thus suggest that the X-ray production of T Tau is influenced by the accretion process although the X-rays may not form in the bulk of the accretion footpoints.

Journal keyword(s): stars: coronae - stars: formation - stars: pre-main sequence - X-rays: stars - stars: individual: T Tauri

Full paper

 Number of rows : 9
N Identifier Otype ICRS (J2000)
RA
ICRS (J2000)
DEC
Mag U Mag B Mag V Mag R Mag I Sp type #ref
1850 - 2022
#notes
1 V* BP Tau Or* 04 19 15.8339827632 +29 06 26.926959492   13.13 12.12 11.89   K5/7Ve 632 0
2 IRAS 04187+1927 TT* 04 21 43.2345350808 +19 34 13.311438132       15.41   M2.4 47 0
3 V* T Tau TT* 04 21 59.4319873992 +19 32 06.432393336   11.22 10.12 9.80   K0IV/Ve 1364 1
4 2MASS J04220496+1934483 TT* 04 22 04.9620264720 +19 34 48.392208264           M3.5 24 0
5 NAME Taurus Complex SFR 04 41.0 +25 52           ~ 3926 0
6 V* TW Hya TT* 11 01 51.9053285064 -34 42 17.033218380   11.94 10.50 10.626 9.18 K6Ve 1736 1
7 HD 98800 TT* 11 22 05.28975 -24 46 39.7571   10.20 9.28 9.26 7.386 K5V(e) 351 0
8 CD-33 7795 TT* 11 31 55.2608963280 -34 36 27.203892120   12.97 11.524 11.150 9.150 M2Ve 186 0
9 HD 319139 SB* 18 14 10.4818675368 -32 47 34.516836060   11.47 10.68   9.11 K5+K7 328 0

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2022.05.27-08:10:07


text:
# Category Archives: lakewood escort

## The main difference from the two kinds of relationship usually in long-lasting relationship

The main difference from the two kinds of relationship usually in long-lasting relationship However in short-term relationship, Islam gave the prospective spouses ideal of exercise its tasks and expectation designs. One example is, the partner isn’t obliged to keep the spouse unless it’s often hence specified during the union acquire. Also, the girlfriend can placed […]


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# Introduction

THIS POST CONSIDERS recent trends in house prices. Thinking a great deal about house prices, I like to look at many different DATA VISUALIZATIONS of house prices, at the national, state, and metro levels of aggregation. Also, in this post, I will include descriptions of how I built my visualizations, including code. These meditations as I call them, required some data wrangling and some thinking as to how to construct the charts.

As the post was getting long, I decided to split it up. In this post, I’ll handle data wrangling. We’ll prepare the data for analysis both in Excel and R. In following post(s), we’ll get into the details of creating the data visualizations. When we’re through we’ll build something like this:

Let us begin by gathering data.

## The data

We’re going to be using the Freddie Mac House Price Index (FMHPI), which is available to the public on Freddie Mac’s webpage. This index is available at monthly frequencies for the nation, all 50 states plus the District of Columbia, and over 300 metro areas (CBSA). The national index is available both seasonally-adjusted (SA), and non-seasonally adjusted (NSA), while the state and metro indices are only available NSA.

Disclosure my team at Freddie Mac helped assemble the Tableau data visualization on the Freddie Mac webpage. This blog posts represents my own view and does not necessarily represent the views of Freddie Mac. We’re using the FMHPI because it’s convenient to work with, has broad coverage, and is publicly available.

### FMHPI webpage:

We’ll need to navigate to the archive page to download the data we want. The data are in two excel spreadsheets : one for the national and state indices and : one for the metro indices. We’ll need them both for what follows.

There are R packages that can manipulate data in .xls or .xlsx file formats. I’m not planning on getting proficient at that, instead I’m going to open up my excel bag of tricks to handle some data preparation. First, let’s consider what we get when we download those house price spreadsheets.

### Comments on file layouts

Now these are not tidy data. Fortunately, the structure can be pretty quickly wrangled within excel to make our R life easier. My strategy is to employ some Excel tricks to tidy this data. I’ve used the gather tools in the tidyr pacakge and they do work well, but as I’m going to write out a .txt file anyway so I can avoid using a .xls file I might as well use my excel trick to tidy these data.

## An excel trick to tidy data

Our goals is to convert the files to a tidy dataset.

And we’re going to do it in about 10 seconds:

This is one of my favorite Excel tricks. You could of course, achieve this just as fast using the gather function in tidyr. But in case you are an Excel user, you can see the details of what I did in the this youtube video.

Because the metro data is laid out in two worksheets, you’ll have to do a little more leg work to get the metro data laid out (and you might have convert the files to .xlsx to handle the extra rows). Also, I’m going to convert the date field, which isn’t a date in the Excel file to a year, month, and date column. Ultimately, we’ll want a file that looks like this:

and this for the metros:

For the metros I included variables called “states” and “state1”, which were created by parsing the metro name. The variable “states” contains a list of all states that the metro area includes, while “state1” includes the primary state (first one listed by OMB). For example, the New York City metro area (New York-Newark-Jersey City, NY-NJ-PA) includes parts of New York, New Jersey, and Pennsylvania, but the primary state is New York. This field could be helpful if we want to group the metros.

Then I save these files as .txt files to read into R. You can download the .txt files here (note I only included the NSA national index)

#Load some packages
library(data.table, warn.conflicts = FALSE, quietly=TRUE)
library(ggplot2, warn.conflicts = FALSE, quietly=TRUE)
library(dplyr, warn.conflicts = FALSE, quietly=TRUE)
## Warning: package 'dplyr' was built under R version 3.4.2
library(zoo, warn.conflicts = FALSE, quietly=TRUE)
library(ggrepel, warn.conflicts = FALSE, quietly=TRUE)
## Warning: package 'ggrepel' was built under R version 3.4.2
library(ggthemes, warn.conflicts = FALSE, quietly=TRUE)
library(scales, warn.conflicts = FALSE, quietly=TRUE)
library(animation, warn.conflicts = FALSE, quietly=TRUE)
library(grid, warn.conflicts = FALSE, quietly=TRUE)
library(tidyr, warn.conflicts = FALSE, quietly=TRUE)
## Warning: package 'tidyr' was built under R version 3.4.2

Now we’ll need to load and prepare some data. We’ll do this using the data.table package to set up our data.

#load national & state data
statedata$date<-as.Date(statedata$date, format="%m/%d/%Y")

#Now uses some data table caclulations to compute percent changes in house prices by state/metro
statedata<-statedata[,hpa:=c(NA,((1+diff(hpi)/hpi))^12)-1,by=state]
statedata<-statedata[,hpa12:=c(rep(NA,12),((1+diff(hpi,12)/hpi))^1)-1,by=state]
statedata<-statedata[,hpa3:=c(rep(NA,12),((1+diff(hpi,3)/hpi))^4)-1,by=state]

#create lags of state
statedata<-statedata[, hpi12 :=  shift(hpi,12), by=state]

#compute rolling min/max
statedata<-statedata[, hpi12min:=rollapply(hpi, 12, min,fill=NA, na.rm=FALSE,align='right'), by=state]
statedata<-statedata[, hpi12max:=rollapply(hpi, 12, max,fill=NA, na.rm=FALSE,align='right'), by=state]

#Do the same for metro data:

metrodata$date<-as.Date(metrodata$date, format="%m/%d/%Y")

metrodata<-metrodata[,hpa:=c(NA,((1+diff(hpi)/hpi))^12)-1,by=metro]
metrodata<-metrodata[,hpa12:=c(rep(NA,12),((1+diff(hpi,12)/hpi))^1)-1,by=metro]
metrodata<-metrodata[,hpa3:=c(rep(NA,12),((1+diff(hpi,3)/hpi))^4)-1,by=metro]

#create lags of metros
metrodata<-metrodata[, hpi12 :=  shift(hpi,12), by=metro]

#compute 12-month rolling min/max
metrodata<-metrodata[, hpi12min:=rollapply(hpi, 12, min,fill=NA, na.rm=FALSE,align='right'), by=metro]
metrodata<-metrodata[, hpi12max:=rollapply(hpi, 12, max,fill=NA, na.rm=FALSE,align='right'), by=metro]

In the code above we take advantage of the data table structure and some functions to compute some time series calculations that will be helpful for the plots we’ll make later. For example, the diff function is used together with some exponentiation to calculate the monthly, quarterly, and annual house price appreciation (all annualized). I also used the rollapply function to caclulate a rolling 12-month min and max (backward looking by using align=‘right’).

Before we close out this data preparation blog post, let’s at least make one plot from the data. The code below generates a time series plot for the national index from January 2000 to March 2016.

The chart shows the rise, fall, and recovery of national house prices. In this index house prices are still a little below their nominal peak, but trending higher. In the follow-up posts, we’ll get creative with the data visualizations.

Click here for Part 2: sparklines and dots (animated)


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• Positive Solutions and Eigenvalue Intervals for Nonlinear Systems

• # Fulltext

https://www.ias.ac.in/article/fulltext/pmsc/117/01/0085-0095

• # Keywords

Nonlinear system; 𝑝-Laplacian; positive solutions; eigenvalue intervals; fixed point theorem in cones.

• # Abstract

This paper deals with the existence of positive solutions for the nonlinear system

$$(q(t)\phi(p(t){u'}_i(t)))'+f^i(t,u)=0, \quad 0 < t < 1, \quad i=1,2,\ldots,n.$$

This system often arises in the study of positive radial solutions of nonlinear elliptic system. Here $u=(u_1,...,u_n)$ and $f^i,i=1,2,\ldots,n$ are continuous and nonnegative functions, $p(t), q(t):[0, 1]\to(0,\infty)$ are continuous functions. Moreover, we characterize the eigenvalue intervals for

$$(q(t)\phi(p(t){u'}_i(t)))'+\lambda h_i(t)g^i(u)=0,\quad 0 < t < 1, \quad i=1,2,\ldots,n.$$

The proof is based on a well-known fixed point theorem in cones.

• # Author Affiliations

1. Department of Mathematical Sciences, Tsinghua University, Beijing 100 084, China
2. Department of Applied Mathematics, Hohai University, Nanjing 210 098, China
3. Department of Mathematics, National University of Ireland, Galway, Ireland

• # Proceedings – Mathematical Sciences

Current Issue
Volume 129 | Issue 3
June 2019


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# Probabilistic halting problem

I’m a physics and math student working through Nielsen & Chuang’s text on quantum computation and information. I don’t have much experience in CS theory, so some of these exercises are confusing to me:

Suppose we number the probabilistic Turing machines and define the probabilistic halting function $$h_p(x)$$ to be 1 if machine $$x$$ halts on input of $$x$$ with probability at least 1/2 and 0 if machine $$x$$ halts on input of $$x$$ with probability less than 1/2. Show that there is no probabilistic Turing machine which can output $$h_p(x)$$ with probability of correctness strictly greater than 1/2 for all $$x$$.

I’m sure we’re supposed to assume that there is a probabilistic Turing machine that does this, and then show that we can solve the deterministic halting problem using this machine, thus obtaining a contradiction. I don’t have a clue how to go from this probabilistic Turing machine to a deterministic one, though.


text:
# Standard deviation and percentage

Standard deviation - wikipedia, the free strengths vs with a standard deviation of 20 percentage deviation - wikipedia, the free encyclopedia. It is expressed in percent and is obtained by relative standard deviation, calculate the average, standard devia tion, and relative standard deviation. Some analysts prefer to call this the percent relative standard deviation and to call the relative standard deviation the value that is.

Standard deviation and percentage essay 6614 words | 27 pages remington’s steakhouse is a single location restaurant in tampa florida. A simple shortcut for calculating the % under a normal curve the music is not mine the link below is where i got the music from . Summary of video in this video, we one standard deviation of the mean we know that the percent of women tall enough to join. Using the weight/height table to calculate the weight-for-height/length percentage or standard deviation (z-score) this reference table helps in the interpretation.

Two of the hardest cpcu 500 concepts to grasp are standard deviation and coefficient of variation read more to get a better explanation of both and see how they differ. Standard deviation is in the same units as the mean so standard deviation is the standard deviation of returns, which are expressed in percent. The standard normal distribution can also be useful for computing percentiles for example, the median is the 50 th percentile, the first quartile is the.

An r tutorial on computing the standard deviation of an observation variable in statistics. Practice sample and population standard deviation with khan academy's free online exercises. How to calculate mean, standard deviation, and standard error after collecting data, oftentimes the first thing you need to do is analyze it this usually. Sample size required for estimating the standard deviation as a percent of its true value sample size required for estimating the standard deviation as a percent. Stock b over the past 20 years had an average return of 12 percent but a higher standard deviation of 30 pp 9973 percent and four standard deviations.

Defining the standard score these z-scores have a distribution with a mean of 0 and a standard deviation of 1 if we look at this as a percentage,. Learn how to use excel's stdev function to estimate standard deviation for each number in a list from the overall average. When to use percent relative standard deviation—and how to do so correctly. Excel percent relative standard deviation %rsd - detailed tutorial with worked example and clear screen shots.

• Standard deviation (continued) lo: to find the mean and standard deviation from a frequency table the formula for the standard deviation of a set of data is [pic.
• Another is when the standard deviation is of the same order of magnitude as the mean, but the observations must be non-negative expressed as a percentage.

Standard deviation and variance deviation just means how far from the normal standard deviation the standard deviation is a measure of how spread out. Hello quick question: is this valid: (\frac{standard \left \left deviation}{average} 100) the standard deviation is just a measure of the spread of. Stock a over the past 20 years had an average return of 10 percent, with a standard deviation of 20 account for 9545 percent three standard.

Standard deviation and percentage
Rated 3/5 based on 10 review

2018.


text:
# First non repeated character

## Problem Statement :

Print the first non repeating character in a string.

## Example :

In the string somecharsjustdon'tliketorepeat, m is the first non-repeating charecter.

## My attempt :

New Code : Old one was not working as expected so I have updated the post. Old code can be found below new code

public class FirstNonRepeatedChar{
static boolean isRepeated(String str, char c){
int count = 0;
for(char ch: str.toCharArray()){
if(c == ch){
count++;
}
}

return count > 1;
}

static void printFirstNonRepeatedChar(String str){
boolean found = false;

for(char c : str.toCharArray()){

if(! isRepeated(str, c)){
found = true;
System.out.println(str + " : " +c);
break;
}
}

if(!found){
System.out.println(str + " : No non-repeated char");
}
}

public static void main(String [] args){
String []testCases = {"aaabbcc", "sss", "121", "test", "aabc"};

for(String str: testCases){
printFirstNonRepeatedChar(str);
}
}
}

Old code

public class FirstNonRepeatedChar{

public static void main(String [] args){
String str = "somecharsjustdon'tliketorepeat";

loopI:
for(int i = 0;i<str.length();i++){
for(int j = i+1; j<str.length();j++){
if(str.charAt(i)==str.charAt(j))
continue loopI;
}
System.out.println(str.charAt(i));
break;
}
}
}

• Is there better way to solve this problem? Especially, is there efficient way to do this using built-in classes?
• Can I reduce complexity?
• Is there any case where my code may fail to generate appropriate result? and how can I avoid these failures?

• Try abab or ababx. Your alrogithm outputs a for both of them, where no output and x is expected respectively. Use an ordered map<char, count> to count the occurences of each character, then lookup the first character whose occurence is 1. Sep 25 '20 at 7:06
• @slepic, Thanks for your reply. I think i have made very silly mistakes. I'll try to solve it. Thanks for suggesting orderedmap I'll check that. Sep 25 '20 at 7:16
• Please fix the code and write some tests to make sure your code actually works. Also include the complete requirements to the question such as restrictions on input and such. I.e. post the exact assignment text. Sep 25 '20 at 8:06
• @TorbenPutkonen I have updated post to provide correct code. Sorry but I was not given any input restrictions or any other rules. Sep 25 '20 at 8:23
• Now I have also added some test cases. Sep 25 '20 at 8:33

Welcome to Code Review. There are already excellent answers, I am adding an alternative solution running in $$\O(n)\$$ and relatively simple to understand.

private static void printFirstNonRepeatedChar(String str) {
// LinkedHashMap maintains the insertion order
Map<Character, Integer> freq = new LinkedHashMap<>();

// Create map of frequencies
for (Character c : str.toCharArray()) {
if (freq.containsKey(c)) {
freq.put(c, freq.get(c) + 1);
} else {
freq.put(c, 1);
}
}

// Find first character with frequency 1
for (Entry<Character, Integer> entry : freq.entrySet()) {
if (entry.getValue() == 1) {
System.out.println(entry.getKey());
break;
}
}
}

The map freq contains the frequencies of the characters. For example, given the string aabbc, the map will be:

• a -> 2
• b -> 2
• c -> 1

LinkedHashMap maintains the insertion order so after building the map it's enough to find the first entry with frequency 1.

This is the same but using Streams:

private static Character findFirstNonRepeatedChar(String str) {
Map<Character, Long> freq = str.codePoints()
.mapToObj(c -> (char) c)
.collect(Collectors.groupingBy(Function.identity(),
Collectors.counting()));
return freq.entrySet().stream()
.filter(e -> e.getValue()==1)
.map(Map.Entry::getKey)
.findFirst().orElse(null);
}

## Time complexity

Your solution runs in $$\O(n^2)\$$, where $$\n\$$ is the length of the input string. I highlighted in your code the relevant parts:

static void printFirstNonRepeatedChar(String str){
// O(n)
for(char c : str.toCharArray()){
// isRepeated runs in O(n)
if(! isRepeated(str, c)){
// O(1)
}
}
//..
}

As @Joop Eggen said, for small input strings you need to benchmark the solutions to really see the difference, but for large inputs the Big O notation must be considered.

## Testing and reusability

Generally, it's better to return a result instead of printing in the method. It's easier to test and reuse. In your case the method could return the character or null. For example:

static Character printFirstNonRepeatedChar(String str){
//...
if(! isRepeated(str, c)){
return c;
}
return null;
}

Now the method is easy to test:

@Test
public void findFirstNonRepeatedCharTest() {
assertEquals(Character.valueOf('c'),findFirstNonRepeatedChar("aabbc"));
assertNull(findFirstNonRepeatedChar("aa"));
}

## Minor changes

• In the method printFirstNonRepeatedChar you can just return; instead of using the flag found and break.
• First of all thanks for warm welcomes Marc :) Very nice solution and explaination. It's easy to understand. codereview seems to best community so far ;). I'm confused whose answer to should I mark as best. All are good. Sep 25 '20 at 11:43
• @Omkar76 I am glad I could help. My suggestion is to wait until the question is not "active" anymore, to be sure that you don't miss any other good review. Then pick the answer that is best for you ;)
– Marc
Sep 25 '20 at 12:08
• @Omkar76 FYI I added the solution using Java Streams and few other advices.
– Marc
Sep 26 '20 at 9:58
• Something to keep in mind is autoboxing, to a certain degree. Every time you assign a char to a Character, the compiler silently adds Character.valueOf(c), which might or might not give you a cached instance. Same goes for your unit test, instead of creating a new instance, you should call Character.valueOf. Sep 26 '20 at 14:24
• @Bobby good point. And I just noticed that new Character() has been deprecated since Java 9.
– Marc
Sep 26 '20 at 15:19

There are some weak spots, but an optimal algorith would need:

• To check all characters for unique ones.

So:

Optional<Character> firstNonRepeatingChar(String s) {
Set<Character> duplicates = new HashSet<>();
for (char ch : s.toCharArray()) {
if (!duplicates.contains(ch)) {
candidates.remove(ch);
}
}
}
return candidates.stream().findFirst();
}

if (!duplicates.contains(ch)) treats a new candidate char or existing one. When !candidates.add(ch) the char already existed in candidates, hence we encountered a new duplicate.

This uses for candidate chars a LinkedHashSet which keeps the order of insertion. The return of findFirst hence will return the first added unique char.

At some time duplicates need to be removed from the candidates to find the solution. Here it is done immediately.

Note that add returns whether indeed added; else there was already such an element.

One might have used a Map<Character, Boolean> for both char states (isCandidate/isDuplicate).

• isRepeated starts from 0, not utilising the position where you found c.
• Calling toCharArray would create a char array, which will happen for every c.
• When count reaches 2 one can return true.

Even when repairing this, the complexity is quadratic: O(N²).

My solution would be: O(N.log N) as one may assume log N for contains/adding/removing from a Set. Considerably better. Especially as HashSet is fast, almost O(1).

However

For small strings of say approx. 30 characters one would need benchmarking to determine the fastest algorithm. Data structures like Set are very valuable - also for code quality - but a brute force for-loop might still be faster.

• Thanks @Joop Eggen. Interesting approach. I got a bit confused looking at those if statements but overall nice solution. Thanks for pointing out my mistakes. I really appriciate that. Sep 25 '20 at 10:39
• Interesting. Compared to @Marc's answer using one LinkedHashMap to count occurrences (and then search it for the first entry with a count of 1), this has the advantage of being read-only when scanning through already-duplicate characters, so it's great for long strings with a small alphabet (like English or ASCII). Although if you know you have ASCII, bool seen[256] and int firstpos[256] would remove the cost of hashing. Sep 25 '20 at 21:20
• Over a long string with a large alphabet with a mix of repeats and unique characters (where the answer appears fairly early in the string), Marc's might win: this answer would be removing from one set and adding to another for a lot of inputs, vs. just finding and incrementing counters. But that's probably the worst case for this answer. Also for small strings, one linkedhashmap might be cheaper to init and tear down than two sets (one of them linked). As you say, you'd need to benchmark for any given use-case if it was a significant time sink or source of GC churn. Sep 25 '20 at 21:26
• Something to keep in mind is autoboxing here. Ever time you use a char as Character, the comiler silently adds a call to Character.valueOf, which might or might not give you a cached instance. Sep 26 '20 at 14:25

My take is entirely similar to his, isolating determination of the result from use:

 /** @return first non-repeated code point from str */
static Optional<Integer> firstNonRepeatedChar(CharSequence str) {
Set<Integer>
all = new java.util.LinkedHashSet<>(9 + Integer.highestOneBit(str.length())),
repeated = new java.util.HashSet<>();
str.codePoints().forEachOrdered(c -> {
all.removeAll(repeated);  // "all non-repeated" hereafter
return all.stream().findFirst();
}

• +1 Thanks for help but your code is not compiling as all.removeAll() is returning boolean value. Please edit your code. Sep 25 '20 at 10:50
• I find this answer compelling, especially as using int code points (UTF-32, pure Unicode numbers) is nice, IMHO state of the art, rather than char (UTF-16). You still might consider OptionalInt and then a better name firstUniqueCodePoint. Sep 25 '20 at 10:57
• (One of my problems is I didn't yet upgrade my local IDE from Java 8: infant streams, no Optional at all, … - lame excuse with all the web IDEs around.) (Hey, and Oracle ruined the use of their online-javadoc with firefox.) Sep 25 '20 at 11:17
• (@JoopEggen Optional<Integer> has the advantage of being the type of Collection<Integer>.stream().findFirst(). I want(ed) to help making the connection to printFirstNonRepeatedChar().) Sep 25 '20 at 12:12
• Why do you think that this method needs a documentation comment? Every word from that comment is already contained in the function signature, except for "Code Point", but that's due to the inappropriate method name. Sep 26 '20 at 11:23

I would do it without the additional set as follows. Basically same approach as here, but with a different return value.

public class FirstNonRepeatedChar {
public static void main(String[] args) {
System.out.println(findFirstNonRepeatedChar("somecharsjustdon'tliketorepeat"));
}

/**
* Returns the first non-repeated char.
*
* @param input
* @return
*/
private static char findFirstNonRepeatedChar(final String input) {
final var len = input.length();
for (var index = 0; index < len; index++) {
final var ch = input.charAt(index);
// find next index of that char
final var firstIndex = input.indexOf(ch);
final var lastIndex = input.lastIndexOf(ch);
if (firstIndex == lastIndex) {
// this means there is no next char
return ch;
}
}
// No such char found.
return 0;
}
}

• Fine solution. Though complexity O(N²) it stops when the unique char is found, say in the beginning. So its actual speed can compare very good. And indexOf is ideal. Sep 25 '20 at 11:55
• This is one of the approaches I thought of but there is a trap! Your code is going to check for duplicates ahead of currecnt char so in case where there is no similar char ahead but there are some behind current char it'll fail. One such case will be "aasdds* - in second iteraion when loop checks for duplicates of second 'a' it won't find any ahead but there is one 'a' behind it. I hope it's clear what I'm trying to say. Thanks anyway :) Sep 25 '20 at 11:58
• (one way to fix "the first/last occurrence problem" would be to compare firstIndexOf() to lastIndexOf() - with or without the current index as a parameter.) Sep 25 '20 at 12:17
• I see the point. I admit it wasn't tested thoroughly. Sep 25 '20 at 14:10
• @greybeard Thanks, I updated my code. Sep 26 '20 at 9:36


text:
# Math Help - Analytic function, complex Taylor series and polynomials.

1. ## Analytic function, complex Taylor series and polynomials.

Hi all;
I need the solution of the following question

Suppose f is analytic function defined everywhere in C and such that for each $z_{0}$ in C at least one coefficient in the expansion
$f(z)=\sum_{n=0}^{\infty}c_{n}(z-z_{0})^n$
is equal to 0. prove that f is polynomail

Kind regards

2. Originally Posted by raed
Hi all;
I need the solution of the following question

Suppose f is analytic function defined everywhere in C and such that for each $z_{0}$ in C at least one coefficient in the expansion
$f(z)=\sum_{n=0}^{\infty}c_{n}(z-z_{0})^n$
is equal to 0. prove that f is polynomail

Kind regards
What do you think? You need to show that all but finitely many of those $z_0$ are zero. What if they aren't?


text:
# Amplitude modulation is used for broadcasting because

This question was previously asked in
ESE Electronics 2014 Paper 2: Official Paper
View all UPSC IES Papers >
1. It is more noise immune than other modulation systems
2. Compared with other systems it requires less transmitting power
3. Its use avoids receiver complexity
4. No other modulation system can provide the necessary BW for high fidelity

Option 3 : Its use avoids receiver complexity
Free
CT 3: Building Materials
3014
10 Questions 20 Marks 12 Mins

## Detailed Solution

• The advantages of AM radio are that it is relatively easy to detect with simple equipment, even if the signal is not very strong.
• It has a narrower bandwidth than FM, and wider coverage compared with FM radio.
• Its use avoids receiver complexity, thus reducing the cost at the receiver end. This makes them suitable for broadcasting purposes.

• In AM most of the transmitted power is in a carrier that contains no information.
• It is affected by noise.
• In AM if there are two more signals received at the same frequency, then both will be demodulated which can lead to interference.

Important differences between AM and FM are elaborated in the table below:

 Features: AM FM Noise immunity In AM, the message is stored in the form of variation in amplitude. Noise affects the amplitude of signal most so AM is less noise immune. In FM, the message is stored in the form of variation in frequency so it has better noise immunity. Bandwidth B.W. required in AM is = 2fm. Hence, less bandwidth is required in case of AM. B.W. required in FM is = 2(β+1)fm. Hence, more bandwidth is required in case of FM. Transmitted power Power transmitted in AM is given by: $${P_T} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)$$ As the modulation index ‘μ’ increases power in AM increases. In FM, power transmitted is always equal to the total power of carrier before modulation. Hence, FM requires less power than AM.


text:
# Why does Hydrogen molar heat capacity reach 7/2 R?

If a diatomic gas like Hydrogen has 6 maximum degrees of freedom why its molar heat capacity reaches at high temperatures $C_V = \frac{7}{2} R$ and not $C_V = \frac{6}{2} R= 3R$?

molar heat capacity of hydrogen

-

The degrees of freedom of a diatomic gas are as follows:

• 3 translational: The molecule can move in x, y and z-direction.
• 2 rotational: The molecule can in principle rotate around each axis. But consider rotations around the molecular axis (connecting the H atoms): in this case, the physics doesn't change. Another way of thinking about it is that the axial rotation mode only can store a vanishing amount of energy, compared to the others.

The rotational modes are only available at higher temeratures, since the molecule has a moment of inertia that has to be overcome to start rotating.

• 2 vibrational: The atoms can wiggle together and apart, which is one degree of freedom. But there is also another one, which is harder to see: Think of the molecule as an harmonic oszillator with kinetic and potential energy.

If both atoms are displaced towards the middle, the molecule has a higher potential energy. The equations of the harmonic oscillator would normally fix the kinetic energy of the atoms in this case. But in a gas with its random kinematics it is totally possible that they have the "wrong" kinetic energies for their relative displacement. So, for the purposes of statistical mechanics, these are 2 further DOFs, making seven.

-
A lot of words for a simple fact: Vibration is two degrees, textbooks explain that with kinetic and potential energy. I found that rather strange as a student too, but textbooks are textbooks. I think mistnim is asking on the same problem. Your explanation isn't a explanation to me. –  Georg Apr 23 '13 at 14:26
@Georg: I found it in need of an explanation that the 7th degree of freedom is really a separate DOF. Some simple arguments suggest otherwise: Say #6 is the wiggling, or $\dot p$ of the HO, and #7 is the motion of the center of mass, or $\dot x$. Then a) one variable is determined by the other through the HO DGL, thus they are not independent and b) isn't the movement of the center just a translation? –  jdm Apr 23 '13 at 14:37


text:
7 times table – Explanation & Examples

The 7 times table is a very important multiplication table. Seven is a prime number, and it has a great significance in our lives; for example, the number of days in a week and the total number of continents in the world are 7. Seven is also considered a lucky number in some cultures.

The 7 times table is a table that contains multiples of the number 7.

Learning and understanding the 7 times table is essential for solving multiplication, division, and factorization-related mathematical problems. It also helps in problems related to calculating the total number of weeks in a month or a year, etc.

Unlike previous tables, the 7 times table is somewhat difficult as it does not follow any simple patterns. This topic will present some tips that will help students learn and memorize the 7 times table.

You should refresh the following concepts to understand this topic better.

1. Basics of addition and multiplication.
2. Five times table
3. Six times table

7 multiplication table

We can write the table of 7 as:

• $7\times1 = 7$
• $7 \times 2 = 14$
• $7 \times 3 = 21$
• $7 \times 4 =28$
• $7 \times 5 =35$
• $7 \times 6 =42$
• $7 \times 7 = 49$
• $7 \times 8 = 56$
• $7 \times 9 = 63$
• $7 \times 10 = 70$

Different tips for 7 times table:

As we said earlier, the 7 times table is considered a difficult table to learn and memorize. So, whats makes it tricky and difficult? The answer is that 7 is a prime number, and it is only divisible by itself and 1.

Furthermore, the addition method (adding 7 to 0, then 7 to 7, 7 to 14, and so on.) generates numbers that follow no simple and identifiable pattern. Still, some tips and tricks are useful for learning the 7 times table, as discussed below.

Using the 6 Times table: If the students have memorized the six-time table, this is one of the easiest methods to learn the 7 times table. In this method, if we add natural numbers in increasing order to the multiples of 6, we get the 7 times table.

The first multiple of 6 is added with the first natural number, i.e., 1. The second multiple of 6 is added with the second natural number, i.e., 7 and so on. This method is presented in the table below.

 Six Times Table Addition (Answer) Six Times Table $6\times {\color{#cccc00}1} = {\color{green}6}$ ${\color{green}6} +{\color{#cccc00} 1}$ ${\color{red}7}$ $7 \times 1 = 7$ $6\times {\color{#cccc00}2} = {\color{green}12}$ ${\color{green}12} +{\color{#cccc00} 2}$ ${\color{red}14}$ $7 \times 2 = 14$ $6\times {\color{#cccc00}3} = {\color{green}18}$ ${\color{green}18} +{\color{#cccc00} 3}$ ${\color{red}21}$ $7 \times 3 = 21$ $6\times {\color{#cccc00}4} = {\color{green}24}$ ${\color{green}24} +{\color{#cccc00} 4}$ ${\color{red}28}$ $7 \times 4 =28$ $6\times {\color{#cccc00}5} = {\color{green}30}$ ${\color{green}30} +{\color{#cccc00} 5}$ ${\color{red}35}$ $7 \times 5 =35$ $6\times {\color{#cccc00}6} = {\color{green}36}$ ${\color{green}36} +{\color{#cccc00} 6}$ ${\color{red}42}$ $7 \times 6 =42$ $6\times {\color{#cccc00}7} = {\color{green}42}$ ${\color{green}42} +{\color{#cccc00} 7}$ ${\color{red}49}$ $7 \times 7 = 49$ $6\times {\color{#cccc00}8} = {\color{green}48}$ ${\color{green}48} +{\color{#cccc00} 8}$ ${\color{red}56}$ $7 \times 8 = 56$ $6\times {\color{#cccc00}9} = {\color{green}54}$ ${\color{green}54} +{\color{#cccc00} 9}$ ${\color{red}63}$ $7 \times 9 = 63$ $6\times {\color{#cccc00}10} = {\color{green}60}$ ${\color{green}60} +{\color{#cccc00} 10}$ ${\color{red}70}$ $7 \times 10 = 70$
• Using the 5 times table: This method is quite similar to the above method. 5 times table is easier as compared to 6 times table, so those students who haven’t learned and memorized the 6 times table can use this method. In this method, the multiples of 5 are added with even numbers in ascending order from top to bottom.

For example, the first multiple of 5 will be added with the first even number, i.e., 2. The second multiple of 5, i.e.,10, will be added with the second even number, i.e., 4, and so on. The resulting table is the 7 times table.

 Five Times Table Addition of Even Numbers (Answer) Six Times Table $5\times 1 = {\color{green}5}$ ${\color{green}5} +{ \color{#cccc00}2}$ ${\color{red}7}$ $7 \times 1 = 7$ $5\times 2 = {\color{green}10}$ ${\color{green}10} +{ \color{#cccc00}4}$ ${\color{red}14}$ $7 \times 2 = 14$ $5\times 3 = {\color{green}15}$ ${\color{green}15} +{ \color{#cccc00}6}$ ${\color{red}21}$ $7 \times 3 = 21$ $5\times 4 = {\color{green}20}$ ${\color{green}20} +{ \color{#cccc00}8}$ ${\color{red}28}$ $7 \times 4 =28$ $5\times 5 = {\color{green}25}$ ${\color{green}25} +{ \color{#cccc00}10}$ ${\color{red}35}$ $7 \times 5 =35$ $5\times 6 = {\color{green}30}$ ${\color{green}30} +{ \color{#cccc00}12}$ ${\color{red}42}$ $7 \times 6 =42$ $5\times 7 = {\color{green}35}$ ${\color{green}35} +{ \color{#cccc00}14}$ ${\color{red}49}$ $7 \times 7 = 49$ $5\times 8 = {\color{green}40}$ ${\color{green}40} +{ \color{#cccc00}16}$ ${\color{red}56}$ $7 \times 8 = 56$ $5\times 9 = {\color{green}45}$ ${\color{green}45} +{ \color{#cccc00}18}$ ${\color{red}63}$ $7 \times 9 = 63$ $5\times 10 = {\color{green}50}$ ${\color{green}50} +{ \color{#cccc00}20}$ ${\color{red}70}$ $7 \times 10 = 70$

Paired Tables:
This method involves using paired tables and is only effective if the students have learned and memorized all the previous tables by heart. This method is the quickest method to learn the 7 times table, but it works only for the first six multiples of 7.

We know that $7\times 1 =7$ and $1\times 7$ is also equal to 7. Similarly, $7\times 2 =14$ and $2\times 7 = 14$. So, if we have memorized the previous tables, then we can draw a paired table of 7 up to $7 \times 6$, as shown below.

Using this method, students will quickly learn the 7 times table up to the sixth multiple of 7 and memorize the rest of the multiples separately.

• Addition: This is an easy and effective method, and this method can be used to learn any table, but the downside is that this method is lengthy, tedious, and the calculations take time. This method is good if students want to learn and memorize the first 10 multiples of 7.

Students can use this method along with the recitation of the 7 times table to help them memorize the table quickly.  As the name suggests, this method involves simple addition.

For example, we start with digit 0, and if we add the number 7 to it, we get the first multiple of 7. Now, we can work out the next multiple of 7 by adding 7 to the previous multiple and so on.

• Recitation: This is the most popular adopted method in schools to help students memorize the 7 times table. If students have difficulty grasping previous tips and tricks to learn the 7 times table, they can use this method. Students can recite the 7 times table loudly and repeatedly. Once they have memorized it, it will be easier for them to develop their basic understanding.

Recitation can be done like:

• Seven times one is 7
• Seven times two is 14
• Seven times three is 21
• Seven Times four is 28
• Seven times five is 35
• Seven times six is 42
• Seven times seven is 49
• Seven times eight is 56
• Seven times nine is 63
• Seven times ten is 70

Table of 7 from 1 to 20:

We can write a complete table of 7 from 1 to 20 as:

 Numerical Representation Descriptive Representation Product (Answer) $7 \times 1$ Seven times one $7$ $7 \times 2$ Seven times two $14$ $7 \times 3$ Seven times three $21$ $7 \times 4$ Seven times four $28$ $7 \times 5$ Seven times five $35$ $7 \times 6$ Seven times six $42$ $7 \times 7$ Seven times seven $49$ $7 \times 8$ Seven times eight $56$ $7 \times 9$ Seven times nine $63$ $7\times 10$ Seven times ten $70$ $7\times 11$ Seven times eleven $77$ $7\times 12$ Seven times twelve $84$ $7\times 13$ Seven times thirteen $91$ $7 \times 14$ Seven times fourteen $98$ $7 \times 15$ Seven times fifteen $105$ $7 \times 16$ Seven times sixteen $112$ $7 \times 17$ Seven times seventeen $119$ $7 \times 18$ Seven times eighteen $126$ $7 \times 19$ Seven times nineteen $133$ $7 \times 20$ Seven times twenty $140$

Example 1:  Calculate 7 times 1 times 7 minus 20

Solution:

7 times 1 times 7 minus 20 can be written as:

$7\times1 \times 7 – 20$

$= 7\times 7 – 20$

$= 49 – 20$

$= 29$

Example 2: What is the 6th multiple of 7?

Solution:

We know the first 10 multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, and 60.

So, the 6th multiple is 42.

Example 3: Using the 7 times table, calculate the number of weeks in a month.

Solution:

We know that a month can contain a minimum of 28 and a maximum of 31 days.

We know the first 5 multiples of the number 7 are 7, 14, 21, 28, and 35. So, the 4th multiple is 28, so a month contains 4 weeks.

Practice Questions:

1. Eve works 4 days a week. Using the 7 times table, determine how many days does she work in 7 weeks?
2. Calculate 7 times 2 minus 10 times 7?
3. Find the value of “Y” if “Y x 7 = 63”.
4. From the given table, select the numbers which are multiples of 7
 7 21 36 35 45 15 19 14 12 10 16 11 28 17 15 30 37 09 16 29 31 63 70 25 84 32 44 42 49 80 77 73 71 74 65 37 57 54 55 51 105 82 72 51 65 44 48 56 89 60

1) Eve works 4 days a week. A week contains 7 days. So total number of working days in 7 weeks

$4\times 7 =28$ days.

2) 10 times 7 minus 2 times 7 can be written as:

$10\times 7 – 2\times 7$

$= 70 – 14$

$= 56$

3) $Y \times 7 = 63$

We know $7\times 9 =63$, so

$Y = 9$.

4)

 7 21 36 35 45 15 19 14 12 10 16 11 28 17 15 30 37 09 16 29 31 63 70 25 84 32 44 42 49 80 77 73 71 74 65 37 57 54 55 51 105 82 72 51 65 44 48 56 89 60


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# Tag Info

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