text:
# AutoCAD Activation Code With Keygen [32|64bit]

Last updated February 6th 2019

AutoCAD 2019.1.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2020.1.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2020.2.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2020.3.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2020.4.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2020.5.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2019.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2019.1.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2019.2.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2019.3.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

AutoCAD 2019.4.x can be downloaded from Autodesk.com as a single Mac or Windows installer file, which then installs AutoCAD as a stand-alone app.

SBS Essentials is an operating system-independent, AutoCAD add-on that allows users to perform and automate tasks within the Autodesk DesignCenter environment. There are four editions of the product, each providing a specific set of capabilities.

References

Category:Computer-aided design software
Category:Autodesk
Category:Proprietary commercial software for Linux
Category:Proprietary cross-platform software
Category:1998 softwareQ:

Induction doesn’t seem to make sense here (can someone explain)

Assume the induction hypothesis to be that for all natural numbers, n, $\{(x^n+1): x \in \mathbb{R}\}$ has at least $n$ elements (and we’re in $\mathbb{R}$)
Let $P(k)$ be the statement that $\{(x^k+1): x \in \mathbb{R}\}$ has at least $k$ elements
For $n=1$, $\{(x^1+1): x \in \mathbb{R}\}$ has exactly one element, namely $1$. So $P(1)$ holds.
Assume $P(n)$ to be true. We have $\{(x^n+1): x \in \mathbb{R}\}$ has at least $n$ elements. I don’t really know where to go from here. I can’t seem to “migrate” the “at least” in $P(n)$ to the statement $P(k)$. I’ve looked at this post: Find the number of elements in each term of a certain sequence
It seems like the answer says that since $\{(x^n+1): x \in \mathbb{R}\}$ has at least $n$ elements, and since $\mathbb{R}$ has at least $n$ elements, we know that $\{(x^k+1): x \in \mathbb{R}\}$ has at least $n$ elements. But I don’t see why this is true.

A:

The assumption that $\mathbb R$ has $n$ elements is used when you write that the power set $\wp(P)$ has $n$ elements. But you don’t need to use this assumption when you
5b5f913d15

Open a CAD software and make a drawing, then do not save it.
Select the drop-down menu View and open the database file.
Search for “db_version” and it will be displayed, which means it’s a 2007 version.

Double-click the Excel file to open it.
Go to data -> xl and search for “db_version” and it will be displayed, which means it’s a 2007 version.

Converter
The converter is stored in the installer folder.

Go to the folder where the executable file is installed.
Go to File > Convert.
Make sure the “Convert to.cad” option is selected.
Select the application you want to convert.
Click the Convert button.

Fully Automatic
Autodesk releases three types of automatic conversion:
Free: three methods
Full conversion: three methods, including the Sub-categories in the drop-down menu View
OEM: With a very basic interface

The software will automatically convert files.

There are three steps to converting files:
The Source file is selected from the list of files.
Click on the “Convert” button.
Go to Data > Export and choose which file format you want to convert your files to, such as the ABBYY Font Manager (.pfm) and Text (.txt).

The result will appear on the right.

Supported formats

Autodesk

References

In addition to the free converter and converter key, Autodesk offers a small fee to use their support program.
Autocad 2007 Converter, Activation, and Keygen

Category:Microsoft Office
Category:Autodesk
Category:Desktop engineering software
Category:Free software programmed in Microsoft Visual Basic
Category:2007 software
Category:Windows-only software
Category:Microsoft Office-related software// Copyright (C) 2017-2020 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library. This library is free
// software; you can

## What’s New in the?

Document Creation

Combine text and object-related information for better context and readability. Find text and edit it together with other objects, similar to combining an image and its text. (video: 2:07 min.)

Release Product

Proprietary file format now supports encoding multiple engineering file formats into one design. Preview and download engineering file formats in their original or AutoCAD-embedded versions. (video: 1:36 min.)

Product Viewer

Easily view pre-defined views of drawings without having to create them manually. Create predefined views, and quickly navigate to them from the Design tab. (video: 2:00 min.)

Storyboarding

Experience a second life for your slide decks by incorporating annotations and other user-drawn content.

2D

Create user-drawn annotations for 2D objects. Annotations can be freehand, snap-to, or linked to sections and views of a 2D drawing. (video: 2:22 min.)

3D

Create user-drawn annotations for 3D objects. Annotations can be freehand, snap-to, or linked to sections and views of a 3D drawing. (video: 2:55 min.)

Use the Simulation panel to simulate physics interactions between objects. Choose simulation types from the drop-down menu. The application determines a simulation type for an object by the type of collision it experiences, and by the speed of its movement. (video: 1:16 min.)

Fitting and Render

Revisit the Fitting and Rendering panels to improve your design workflow. Select fitting models, or create new ones, to determine the 3D shape of model parts. Select the type of solid to render, and the rendering type. Rendering options include UV Mapping and Light Field Rendering. (video: 2:38

## System Requirements:

Minimum:
OS:
Intel Pentium 4 2.0 GHz CPU or AMD Athlon 1.6 GHz CPU
Hard Drive:
500MB free space for installation
500MB free space for saved games
Memory:
128MB of RAM
Graphics:
128MB of graphics memory (NVIDIA/ATI)
DirectX:
9.0
Audio:
Windows Media Audio 9 (or later) compatible sound card


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## Elementary Geometry for College Students (6th Edition)

The proof is in the following form: Column 1 of proof; Column 2 of proof. Thus, we have: 1. $XY \parallel RT$ and $YZ \parallel RS$; Given 2. $\frac{YT}{SY} = \frac{RX}{XS}$ and $\frac{ZT}{RZ}= \frac{YT}{SY}$; lines parallel to one side of a triangle always proportionally divides the other side of the triangle. 3. $\frac{RX}{XS} = \frac{ZT}{RZ}$; transitive property (two things equal to the same thing are equal to each other.)

format: plain


text:
Select Page

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.

## Probability Biased and Unbiased – AIME 2010

Ramesh and Suresh have two fair coins and a third coin that comes up heads with probability $\frac{4}{7}$,Ramesh flips the three coins, and then Suresh flips the three coins, let $\frac{m}{n}$ be the probability that Ramesh gets the same number of heads as Suresh, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

But try the problem first…

Source

AIME, 2010, Question 4

Combinatorics by Brualdi

## Try with Hints

First hint

No heads TTT is $\frac{1.1.1}{2.2.7}=\frac{3}{28}$and $(\frac{3}{28})^{2}=\frac{9}{784}$

One Head HTT THT TTH with $\frac{3}{28}$ $\frac {3}{28}$ and $\frac{4}{28}$ then probability is $\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}$=$\frac{100}{784}$

Second Hint

Two heads HHT $\frac{4}{28}$ HTH $\frac{4}{28}$ THH $\frac{3}{28}$ then probability is $\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}$=$\frac{121}{784}$.

Final Step

Three heads HHH is $\frac{4}{28}$ then probability $\frac{16}{784}$

Then sum is $\frac{9+100+121+16}{784}=\frac{123}{392}$ then 123+392=515.


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This page describes the steps required to create a simple synchronous interaction between a sender and receiver.

Page contents

### Overview

The scenario shows a simple synchronous interaction between sender and receiver, as previously described in the Spine mini-service provider (SMSP) examples. This example is drawn out to show how the ITK Reference Implementation manages the transmission of messages in non-xml format. In this case an HL7v2 ADT message in traditional “pipe and hat” format is packaged up and sent over the wire with no implications for the sending application and very little configuration. Although not shown here, the implications for the receiver are similarly simple.

### Sender view

The Sender builds a SimpleMessage using a String containing the p/h ADT message. Configuration of the service tells the R.I. that this messages requires base64 encoding so the RI performs the encoding and builds the DistributionEnvelope with the appropriate tagging

The Receiver consumes the Simple Object Access Protocol (SOAP) message. This validates the SOAP headers and then extracts the payload. The SOAP payload is then split further into Distribution Envelope and business payload. The Distribution Envelope is validated and extracted into MessageProperties. The R.I. then looks up the service properties from the configuration and – in this case – as the payload is in base64 form it is decoded before being packaged up as a SimpleMessage and passed onto the Application Message Handler.

### Request message

[code lang=”xml”]

<soap:Envelope xmlns:itk="urn:nhs-itk:ns:201005" … >

<soap:Header> … </soap:Header>

<soap:Body>

<itk:DistributionEnvelope … >

<itk:header service="urn:nhs-itk:services:201005:admitPatientPH-v1-0" …>

…

</itk:header>

<itk:payloads count="1">

<itk:payload id="…">TVNIfF5+X … 4MTIzNHwNCg== </itk:payload>

</itk:payloads>

</itk:DistributionEnvelope>

</soap:Body>

</soap:Envelope>

[/code]

### Response message

[code lang=”xml”]

<soap:Envelope xmlns:itk="urn:nhs-itk:ns:201005" …

<soap:Header>

<wsa:MessageID>21BE8367-ADE7-4CD3-BB6A-0378AC9C2A91</wsa:MessageID>

<wsa:Action>urn:nhs-itk:services:201005:admitPatientPH-v1-0Response</wsa:Action>

<wsa:To/>

</soap:Header>

<soap:Body>

<itk:SimpleMessageResponse>OK</itk:SimpleMessageResponse>

</soap:Body>

</soap:Envelope>

[/code]

## How it all happens - Step 1

The scenario shows a simple synchronous interaction between sender and receiver, as previously described in the SMSP examples. This example is drawn out to show how the ITK Reference Implementation manages the transmission of messages in non-xml format. In this case an HL7v2 ADT message in traditional “pipe and hat” format is packaged up and sent over the wire with no implications for the sending application and very little configuration. Although not shown here, the implications for the receiver are similarly simple.

[code lang=”xml”]

// Create the message

ITKMessage msg = new SimpleMessage();

msg.setBusinessPayload(phMessage);

…

mp.setServiceId("urn:nhs-itk:services:201005:admitPatientPH-v1-0");

…

// Send this message synchronously.

ITKMessage responseMsg = sender.sendSync(msg);[/code]

## Step 2

A raw HL7v2 message in p/h format would not be valid data if just wrapped up in XML, so the ITK R.I. needs to know that this message is p/h so that it can construct the on-the-wire message accordingly. The R.I. uses a service.properties file to contain service configurations such as this.

[code lang=”xml”]

urn\:nhs-itk\:services\:201005\:admitPatientPH-v1-0.isBase64=Y

urn\:nhs-itk\:services\:201005\:admitPatientPH-v1-0.mimeType=text/plain

[/code]

## Step 3

The Reference Implementation uses the service configuration to encode the message using core java language conversions.

[code lang=”java”]

if(service.isBase64()){

String b64DocText = javax.xml.bind.DatatypeConverter.

printBase64Binary(request.getBusinessPayload().getBytes());

request.setBusinessPayload(b64DocText);

}

message = new DEWrappedMessage(service, request, true);

[/code]

## Step 4

Now the R.I. sets the mime type, base64 encoding to the appropriate values and embeds the encoded message as the payload.

[code lang=”xml”]

<itk:manifest count="1">

<itk:manifestitem base64="true"

id="uuid_6e9306ba-9a09-4479-a719-fcc23234b9ec"

mimetype="text/plain" profileid="ITKv1.0"/>

</itk:manifest>

<itk:senderAddress uri="urn:nhs-uk:addressing:ods:TESTORGS:ORGA"/>

</itk:header>

<itk:payloads count="1">

<itk:payload id="uuid_6e9306ba-9a09-4479-a719-fcc23234b9ec">TVNIfF5+XCZ8…

….HwNCg==</itk:payload>

</itk:payloads>

[/code]

Last edited: 12 August 2021 9:41 am


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# GMAT Critical Reasoning Practice Question & Explanation

March 6, 2014 by

You may not love doing GMAT Critical Reasoning practice, but it’s good for your score. We initially posted this stimulus and question on Facebook – give it a try:

While the average American reads only two books per year, researchers have recently concluded that by reading  two books per month, people can expect their memorizing capacity to double. The most effective way for Americans to begin to read two books per month – thus increasing their memory capacity – is to support Proposition 75, which will require students to read at least two books per month beginning in 2nd grade and through their senior year of high school.

Which of the following can be most properly drawn, if the statements above are true, about future reading habits and memorization capacity?

• A)  If Proposition 75 passes, all teen-agers will see a significant increase in their ability to memorize for tests.

# Get Your Fresh GMAT Practice Question Here!

February 27, 2014 by

Yesterday, we posted a GMAT practice question on Facebook. It got a lot of attention and many responses. Here it is again:

What is the value of b if a = $\frac{b}{c^{2}}$ ?

(1) $ac^{2}&space;=&space;50$

(2) $c^{2}=&space;10$ and $a^{2}=&space;25$

First, you can multiply both sides of the equation in the question stem by $c^{2}$ to make it clear that $ac^{2}&space;=&space;b$. Then, look for values a, c and $ac^{2}$.

• Statement (1) is exactly what is needed – it gives you a precise value for $ac^{2}$. Statement (1) is sufficient, so eliminate answer choices C and E.
• Statement (2) alone, however, leads to two possible values for b, because you’d have to substitute the square roots of 25 for a, and those square roots are BOTH positive AND negative 5 (remember this! The GMAT likes

# GMAT Problem Solving Practice: The Solution

February 3, 2014 by

Did you try out Friday’s GMAT problem solving practice question? If not, give it a try before you look at the solution. Here’s a reminder:

• (A) 90
• (B) 110
• (C) 120
• (D) 130

# GMAT Study Tip: Try Coffivity

January 17, 2014 by

I’ve got a quick GMAT study tip for you today. We tell our students to think critically and strategically when approaching problems on the GMAT. So, for those of you who need a cognitive boost or who like a little ambient noise while you’re studying, I found a helpful little website with associated apps. It’s called Coffivity.

According to the site,

Research shows it’s pretty hard to be creative in a quiet space. And a loud workplace can be frustrating and distracting. But, the mix of calm and commotion in an environment like a coffee house is proven to be just what you need to get those creative juices flowing. Our team has delivered the vibe of a coffee shop right to your desktop, which means when your workspace just isn’t quite cutting it, we’ve got you covered. Coffitivity. Enough noise to work.” (Read the research here: http://www.jstor.org/stable/10.1086/665048Read full post

# GMAT Critical Reasoning: How to Make Predictions

January 13, 2014 by

If you still need to be convinced about the value of making predictions on the GMAT, then read this: Beating GMAT Verbal by Making Predictions. Now that we are all on board, let’s learn how to do it…

When a Critical Reasoning (CR) question pops up on the screen, adept test takers know to read the actual question first. The Question is always found in the middle between the Stimulus and the Answer Choices. By reading the question first and, thus, depending solely on the type of CR question posed, the test taker knows how to most efficiently and effectively untangle the stimulus above.

There are many different types of CR questions, but most of them will fall under the category we at Kaplan like to call the Argument Family. The members of the Argument Family are Assumption, Strengthen, Weaken, Flaw, and Evaluation. The correct answer to every question … Read full post

# GMAT Data Sufficiency Practice Question – The Solution

January 9, 2014 by

Did you try out our GMAT Data Sufficiency practice question? If not, take a couple minutes now to give it a try before reviewing the explanation.

Now, let’s get down to brass tacks on the Geometry and DS skills you need to solve this one…

One way to find the area of a quadrilateral is to divide it into triangles and add the areas of the triangles, which can be found using the formula for the area of a triangle: (1/2)(Base)(Height).
If you add dashed lines to the diagram connecting points A and C and points B and D, you will see that the quadrilateral is composed of 4 right triangles:

You can see that one side of each triangle is a radius of one of the circles; for example, AB is a radius of circle A and is the hypotenuse of triangle ABE. Also, you’ll … Read full post

# Using Common Knowledge on GMAT Critical Reasoning

December 27, 2013 by

Here is a very simple argument:

A – B = C. Over time, A has increased. Therefore, C has increased.

If the argument seems simplistic, that’s because it is—it’s not very test-like or very challenging. It gives evidence about changes in A and concludes about a change in C; the assumption (the missing piece necessary for the conclusion to be logical) is that B has not also increased. After all, if A went up but B went up as well, who knows what C will turn out to be!

But this simple example underlines a pattern that can turn into big points on Critical Reasoning questions. The GMAT testmakers expect students to understand certain common knowledge. And some of that common knowledge takes the form of simple three-part equations just like the one above. For instance:

Revenue – Cost = Profit

Price * Sales = Revenue

Imports – Exports = Read full post

# Free GMAT Question of the Week: Discover Data Sufficiency!

December 9, 2013 by

GMAT Question:

If b ≠ 0 and a > b, is a > c?

(1) a/b> c/b

(2) 5ab > 6bc

From a cursory glance, you can see that GMAT math takes you back to concepts that you learned in high school. Look a bit closer and you see that it actually takes you back much further than that, to math you learned in elementary school – integers, positives/negatives, etc. One of the interesting things about the GMAT is that sometimes these throwbacks to simple math are used to create challenging critical thinking problems. The problem above is one of those.

Post your answer and your method in the comments below. We’ll share the full Kaplan explanation, with secrets for how to master GMAT Data Sufficiency, in tomorrow’s blog entry.

# GMAT Quantitative Section: Stacking Percents

September 15, 2012 by

The Wrentham Village Premium Outlets are a great place to stop for cheap brand-name clothes, and they’re a popular tourist destination for visitors to Massachusetts.  Like all tourist/retail locations, they need to get people in the door. They’ve tried lot of things, but their latest gimmick has interesting implications for GMAT students. They’ve started stacking discounts.

Nearly every store in the mall has signs that say something like, “65% off, PLUS take an additional 20% off!” Moreover, a coupon book gives additional discounts—the particular store with that sign also offered 15% off purchases over a certain value.

To the unenlightened, this seems too good to be true. After all, 65% + 20% + 15% = 100%. Are we seriously to believe that the outlet store is giving away things for free?

Well, that might be a trap answer on the GMAT—and it’s a trap answer for the unwary … Read full post

Drag/Scroll

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# How do you find critical points of a logarithmic function y = e^x - 2e^(-x) - 3x?

Oct 29, 2015

Correction: That is not a logarithmic function, but here is how to find the critical numbers for the function.

#### Explanation:

$f \left(x\right) = {e}^{x} - 2 {e}^{- x} - 3 x$

Note that "Dom:(f) = (-oo,oo)

$f ' \left(x\right) = {e}^{x} - 2 {e}^{- x} \left(- 1\right) - 3$

$= {e}^{x} + 2 {e}^{- x} - 3$

$f ' \left(x\right)$ is never undefined, so we need only find its zeros.

We need to solve

${e}^{x} + 2 {e}^{- x} - 3 = 0$

As a first step, try getting rid of negative exponents:

${e}^{x} + \frac{2}{e} ^ x - 3 = 0$

Get a common denomiator, noting that ${e}^{x} \cdot {e}^{x} = {e}^{2 x}$, so we have

$\frac{{e}^{2 x} + 2 - 3 {e}^{x}}{e} ^ x = 0$.

So now we need to solve:

${e}^{2 x} - 3 {e}^{x} + 2 = 0$

Again note that ${e}^{2 x} = {\left({e}^{x}\right)}^{2}$, so we can factor:

$\left({e}^{x} - 1\right) \left({e}^{x} - 2\right) = 0$

${e}^{x} = 1$ $\text{ }$ OR $\text{ }$ ${e}^{x} = 2$

$x = 0$ $\text{ }$ OR $\text{ }$ $x = \ln 2$

Both are in $\text{Dom} \left(f\right)$, so both are critical numbers.


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# Tag Archives: i ching

## Letting the I Ching Write Your Story for You with Henry Lien

Philip K. Dick did not write one of his greatest novels, The Man in the High Castle. Dick discovered the I Ching when he began the novel and he let the I Ching write his novel for him. The I Ching is … Continue reading

## Letting the I Ching Write Your Story for You with Henry Lien

Philip K. Dick did not write one of his greatest novels, The Man in the High Castle. Dick discovered the I Ching when he began the novel and he let the I Ching write his novel for him. The I Ching is … Continue reading


text:
Revision history [back]

Starting from the NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)

Test whether any array element along a given axis evaluates to True.

Also, in that case, you gave to this function just the array, so in Python the function is called with some default parameters, so axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true, you have just to check each element of the array/Mat

Starting from the NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)keepdims=novalue)
Test whether any array element along a given axis evaluates to True.
True.

Also, in that case, you gave to this function just the array, so in Python the function is called with some default parameters, so axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true, you have just to check each element of the array/Mat

Starting from the From NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)

Test whether any array element along a given axis evaluates to True.

Also, in that case, you gave to this function just the array, so in Python the function is called with some default parameters, so axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true, you have just to check each element of the array/Mat

From NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)

Test whether any array element along a given axis evaluates to True.

Also, in that case, you gave give to this thie function just the input array, so in Python the function is called with some default parameters, so parameters: axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true, true: you have just to check each element of the array/Mat

From NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)

Test whether any array element along a given axis evaluates to True.

Also, in that case, you give to thie function just the input array, so in Python the function is called with some default parameters: axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true: you have just to check each element of the array/Matarray/Mat to find if at least one element is positive, so it return true, otherwise false.

From NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)

Test whether any array element along a given axis evaluates to True.

Also, in that case, you give to thie function just the input array, so in Python the function is called with some default parameters: axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true: you have just to check each element of the array/Mat to find if at least one element is positive, so it return true, otherwise false.

From NumPy official documentation, the behavior of the function is the following:

numpy.any(a, axis=None, out=None, keepdims=novalue)

Test whether any array element along a given axis evaluates to True.

Also, in that case, you give to thie function just the input array, so in Python the function is called with some default parameters: axis= None, out=None and keepdims=novalue. Just the first one is interesting for you, so:

The default (axis = None) is to perform a logical OR over all the dimensions of the input array

So I suppose that if just one element of the array is positive, this function return true: you have just to check each element of the array/Mat to find if at least one element is positive, positive (greter than 0, because 0 is false), so it return true, otherwise false.


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9
Q:

# Adam and Smith are working on a project. Adam takes 6 hrs to type 36 pages on a computer, while Smith takes 5 hrs to type 40 pages. How much time will they take, working together on two different computers to type a project of 120 pages?

 A) 8 hrs 45 min B) 8 hrs 42 min C) 8 hrs D) 8 hrs 34 min

Answer:   D) 8 hrs 34 min

Explanation:

Number of pages typed by Adam in 1 hour = 36/6 = 6
Number of pages typed by Smith in 1 hour = 40/5 = 8
Number of pages typed by both in 1 hour = (6 + 8) = 14
Time taken by both to type 110 pages = (120 * 1/14) = $8\frac{4}{7}$ = 8 hrs 34 min.

Q:

P is 30% more efficient than Q. How much time will they, working together, take to complete a job which P alone could have done in 23 days?

 A) 16 days B) 13 days C) 15 days D) 12 days

Explanation:

Ratio of times taken by P & Q = 100 : 130 = 10:13

Let Q takes x days to do the work

Then, 10:13 :: 23:x

=> x = 23x13/10

=> x = 299/10

P's 1 day's work = 1/23

Q's 1 day's work = 10/299

(P+Q)'s 1 day's work = (1/23 + 10/299) = 23/299 = 1/13

Hence, P & Q together can complete the work in 13 days.

13 857
Q:

If 2 men and 3 women can do a piece of work in 8 days and 3 men and 2 women in 7 days. In how many days can the work be done by 5 men and 4 women working together?

 A) 3 days B) 6 days C) 4 days D) 2 days

Explanation:

From the given data,

=> (2 M + 3W) 8 = (3M + 2W)7

=> 16M + 24W = 21M + 14 W

=> 10W = 5M

=> 2W = M

=> 14W × ? = 7W × 8

? = 4 days

11 852
Q:

28 Men and 52 women working together completes a work in 22.5 days. 35 men and 65 women working together on same work will complete it in how many days?

 A) 16 B) 18 C) 19 D) 21

Explanation:

Clearly total persons are increased in  28/35 :: 52/65 = 4:5.

As time is inversely proportional to men, so total time will decrease in the ratio 5:4.

Hence, 22.5 x 4/5 = 18 days.

8 542
Q:

X can complete the work in 10 days, Y can do the same in 15 days. If they are hired for 5 days to do the work together, what is the work that left unfinished?

 A) 1/3 B) 2/3 C) 1/6 D) 5/6

Explanation:

Given X can do in 10 days

=> 1 day work of X = 1/10

Y can do in 15 days

=> 1 day work of Y = 1/15

1day work of (X + Y) = 1/10 + 1/15 = 1/6

Given they are hired for 5 days

=> 5 days work of (X + Y) = 5 x 1/6 = 5/6

Therefore, Unfinished work = 1 - 5/6 = 1/6

9 693
Q:

If 10 men take 15 days to complete a work. In how many days will 37 men complete the work?

 A) 3 days B) 4 days C) 5 days D) 6 days

Explanation:

Given 10 men take 15 days to complete a work

=> Total mandays = 15 x 10 = 150

Let the work be 150 mandays.

=> Now 37 men can do 150 mandays in 150/37 =~ 4 days

13 566
Q:

Three taps P, Q and R can fill a tank in 12 hrs, 15 hrs and 20 hrs respectively. If P is open all the time and Q and R are open for one hour each alternately, starting with Q, then the tank will be full in how many hours ?

 A) 9 hrs B) 7 hrs C) 13 hrs D) 11 hrs

Explanation:

Given,

P can fill in 12 hrs

Q can fill in 15 hrs

R can fill in 20 hrs

=> Volume of tank = LCM of 12, 15, 20 = 60 lit

=> P alone can fill the tank in 60/12 = 5 hrs

=> Q alone can fill the tank in 60/15 = 4 hrs

=> R alone can fill the tank in 60/20 = 3 hrs

Tank can be filled in the way that

(P+Q) + (P+R) + (P+Q) + (P+R) + ....

=> Tank filled in 2 hrs = (5+4) + (5+3) = 9 + 8 = 17 lit

=> In 6 hrs = 17 x 6/2 = 51 lit

=> In 7th hr = 51 + (5+4) = 51 + 9 = 60 lit

=> So, total tank will be filled in 7 hrs.

9 954
Q:

P and Q can complete a job in 24 days working together. P alone can complete it in 32 days. Both of them worked together for 8 days and then P left. The number of days Q will take to complete the remaining work is ?

 A) 56 days B) 54 days C) 60 days D) 64 days

Explanation:

(P+Q)'s 1 day work = 1/24

P's 1 day work = 1/32

=> Q's 1 day work = 1/24 - 1/32 = 1/96

Work done by (P+Q) in 8 days = 8/24 = 1/3

Remainining work = 1 - 1/3 = 2/3

Time taken by Q to complete the remaining work = 2/3 x 96 = 64 days.

8 949
Q:

A can complete 9/10 part of a work in 31 days. After that, with the help of B he can complete the remaining work in 4 days. In how many days will  A and B together can complete that work ?

 A) 36 B) 40 C) 42 D) 38

Explanation:

Remaining work = 1 - 9/10 = 1/10

=> A & B together completes 1/10 of work in 4 days

=> 1 work can completed in ------ ? days

Let it be x days

=>  $\frac{4}{\frac{1}{10}}=\frac{x}{1}$

=> x = 40 days.

Hence, A & B together can complete the work in 40 days.


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Properties of Rational Numbers

## Rational numbers are real numbers that can be written in the form of a fraction. Use these exercises to determine properties of rational numbers.

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Properties of Rational Numbers

Eva just learned about rational numbers in math class. She learned that rational numbers are numbers that be written as a ratio of two integers. She understands that numbers like  and  are rational numbers. Eva's friend Mike says that all integers are rational numbers too, but Eva isn't sure if that is true. After all, integers don't look like fractions. How could Mike help to convince Eva that integers are rational numbers too?

In this concept, you will learn how to identify rational numbers.

### Identifying Rational Numbers

A ratio is a comparison between two numbers. Ratios can be written in words, in fraction form, or using a colon.

Here is an example.

Suppose your class has 12 boys and 13 girls. You could say that the ratio of boys to girls is:

• 12 to 13
• 12:13

All three forms are equivalent ways of expressing the same ratio.

The rational numbers are the set of numbers that can be written as a ratio of two integers. In other words, any number that can be written as a fraction where both the numerator and the denominator are integers is a rational number. Any number that cannot be written as a fraction in this way is not a rational number and is considered an irrational number.

Let's look at an example.

This number is already written as a fraction so it is definitely a rational number.

The answer is that  is a rational number because it is the ratio of -2 to 3.

Let's look at another example.

10

This number is not already written as a fraction; however, you could rewrite it as a fraction. Remember that any number divided by 1 is just itself. So you have

The answer is that because 10 can be written as the ratio of 10 to 1, it is a rational number. In fact, all integers are rational numbers because they can all be written as a ratio of themselves to 1.

Numbers written in decimal form can also be rational numbers as long as they are repeating or terminating decimals.

Here is an example.

0.687

This number goes out to the thousandths place. You can read this number as 687 thousandths. Reading it in this way can help you to rewrite this number in fraction form:

Because 0.687 can be written as the ratio of 687 to 1000, it is a rational number.

### Examples

#### Example 1

Earlier, you were given a problem about Eva, who just learned about rational numbers.

Mike is trying to convince Eva that integers are rational numbers too. Eva understands that a rational number is a number that can be written in fraction form as a ratio of two integers.

If Mike wanted to convince Eva that integers are rational numbers too, he could give a specific example. For example, the integer -20.

-20 does not look like the ratio of two integers. However, it can be rewritten in fraction form.

Because any integer could be similarly written as a ratio of itself to 1, all integers are rational numbers. In fact, integers make up a subset of the set of rational numbers.

#### Example 2

Show that the following number is rational by writing it as a ratio in fraction form.

.85

You read this number as eighty-five hundredths. You can convert it to a fraction:

The answer is that .85 is a rational number because it can be written as the ratio .

#### Example 3

Determine whether or not the following number is rational. If it is rational, show how it can be written as a ratio in fraction form.

-4

This number is not already written as a fraction; however, it can be rewritten as a fraction.

The answer is that because -4 can be written as the ratio of -4 to 1, it is a rational number.

#### Example 4

Determine whether or not the following number is rational. If it is rational, show how it can be written as a ratio in fraction form.

This is a repeating decimal and all repeating decimals are rational. This means you can rewrite this number in fraction form as the ratio of two integers.

The answer is that because  can be written as the ratio of 1 to 3, it is a rational number.

#### Example 5

Determine whether or not the following number is rational. If it is rational, show how it can be written as a ratio in fraction form.

, pronounced “pi” is an example of a number that is not rational. It is an irrational number. Its decimals continue on forever, but never repeat. It cannot be rewritten in fraction form.

### Review

Rewrite each number as the ratio of two integers to prove that each number is rational.

1. −11
2. 9
3. 0.08
4. -0.34
5. 0.678
6. -19
7. 25
8. 0.17
9. 0.2347
10. -17
11. 347
12. 87
13. -97

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

common denominator

The common denominator is the least common multiple of the denominators of multiple fractions. Each fraction can be rewritten as an equivalent fraction using the common denominator.

Denominator

The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.

Equivalent Fractions

Equivalent fractions are fractions that can each be simplified to the same fraction. An equivalent fraction is created by multiplying both the numerator and denominator of the original fraction by the same number.

Irrational Number

An irrational number is a number that can not be expressed exactly as the quotient of two integers.

Least Common Denominator

The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.

Least Common Multiple

The least common multiple of two numbers is the smallest number that is a multiple of both of the original numbers.

Lowest Common Denominator

The lowest common denominator of multiple fractions is the least common multiple of all of the related denominators.

Mixed Number

A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.

Numerator

The numerator is the number above the fraction bar in a fraction.

proper fraction

A proper fraction has a numerator that is a lesser absolute value than the denominator. Proper fractions always represent values between -1 and 1.

rational number

A rational number is a number that can be expressed as the quotient of two integers, with the denominator not equal to zero.

reduce

To reduce a fraction means to rewrite the fraction so that it has no common factors between numerator and denominator.

Repeating Decimal

A repeating decimal is a decimal number that ends with a group of digits that repeat indefinitely. 1.666... and 0.9898... are examples of repeating decimals.

Terminating Decimal

A terminating decimal is a decimal number that ends. The decimal number 0.25 is an example of a terminating decimal.


text:
Convert a gene into amino sequences
2
1
Entering edit mode
12 months ago
vpsev3 ▴ 20

Hello,

I'm trying to convert a gene (DNA) into a protein (amino acids)

I tried to convert them but I have a lot more amino acids than here for example https://alphafold.ebi.ac.uk/entry/Q8W3K0

This is what I tried https://web.expasy.org/translate/

I try to find a software where I just have to enter the sequence and I will have the result

alphafold • 515 views
3
Entering edit mode
12 months ago
GenoMax 120k

If this is a known gene then you don't need to do this manually. e.g. https://www.ncbi.nlm.nih.gov/nuccore/NM_000207.3/ is human insulin. If you look at the record you will find a link for protein sequence NP_000198.1 in that record. Or you could simply search for protein sequence directly using the name of the gene.

0
Entering edit mode

Thank you it works

0
Entering edit mode
6 days ago

For known genes you can use gget seq: https://github.com/pachterlab/gget


text:
# Add a new folder to each subfolder

I have main a folder say E:\donuts and there are hundreds of folders inside it. e.g.

E:\donuts\yellow\
E:\donuts\green\
...
E:\donuts\blue\

I want to create a new folder in each subfolder simply using some DOS command. Something like this.

E:\donuts\yellow\big
E:\donuts\green\big

and so on. How can I achieve this?

Also, would to be possible move the content of each subfolder into the corresponding big folder? For instance all the files and folders in E:\donuts\yellow\ should move to E:\donuts\yellow\big and so on.

-

I haven't tested this, so you'll want to try it out first

FOR /d %A IN (e:\donuts\*) DO mkdir "%A\big"

This should work to do the move as well:

FOR /d %A IN (e:\donuts\*) DO mkdir "%A\big" & mv "%A\*.*" "%A"

-
Wow @paul, the first one worked like a charm. I hope the second one will work too. Thanks a lot. I have been try various combos the whole evening but now I see my syntax was wrong. – nomidesigns Nov 20 '11 at 4:22
Slight modification on the second command worked for me. It moved all the files sans folders. How can I change it to move the folders too? FOR /d %A IN (e:\donuts*) DO mkdir %A\big & move %A*.* %A\big\ – nomidesigns Nov 20 '11 at 4:48
Oops yeah, well done. – Paul Nov 20 '11 at 13:10
Yeah @Paul thanks to you I'm almost there but the above command only moves files not folders. How can i achieve that? – nomidesigns Nov 20 '11 at 19:30
You can't easily unfortunately, the move command doesn't accept wildcard directory names – Paul Nov 20 '11 at 21:21


text:
# Math Help - algebraic fractions

1. true....... im looking for money and around couple hundred thousands a year for my future life...
i wonder if there is a job that has math in it that would also have an average of couple hundred thousand
i know i havent seen real math yet but i might want to get ready right?

2. To learn LaTeX thoroughly could take a long time, but for basic things, you could learn everything you need to know very quickly.

For example, double click on skeeter's expression to see what was typed to get it:

\frac{3a}{2a-1} - \frac{1}{2a}

There is not too much going on here.

In order to get it to render on this forum, you use  tags. There is a LaTeX Help subforum on MHF here, along with a tutorial. You can start making pretty expressions pretty much instantly, just look at the tutorial and play around with it..

Edit: For some reason, the PDF in the most recent tutorial thread is a broken link. You can get the PDF from this thread. I will notify the mods about the broken link.

3. If you're going to be doing a job that makes money, you'll pretty much never be doing real math. Some jobs like quantitative finance make a lot of money (if you're good, maybe 200,000 a year--I think, I don't really know, this is a guess based on what friends have suggested to me) and use some hard math. But it's still not real math. By real math, I mean like mathematical research. Maybe that's a definition of "real math" based on a prejudice.

4. mathematical research :O
like invent equations?

5. Originally Posted by buck
mathematical research :O
like invent equations?
Yeah. To me that's real math. I'm probably being too much of a purist and too philosophical about the point, but that's what I think of as real math. Nobody does that and makes huge money except some of the absolute smartest people in the world. Everyone else is just a teacher making $\approx 120,000$ which is not bad unless you have huge student loans and then it's still not bad but not great.

If you're looking at mech engineering, probably doing a much easier job and making a bit more money, you just need calc, differential equations, and linear algebra. If you're smart, you could knock that out by your first year of college.

6. my asian parents are like uni uni! college dont get the mo pay job! :@
i dont even know whats up
ill be on the forums later, im going to go upstairs.

7. What grade are you in NOW ? Attending math classes?

By the way, Latex is NOT math: it's simply a way of typing differently.

8. Originally Posted by ragnar
It makes your writing look much nicer. If you take advanced college courses in mathematics, professors will encourage you to use LaTeX. If you get university degree in mathematics you will need to write a piece of original research which must be produced in LaTeX. If you publish anything in the sciences or mathematics, it must be produced in LaTeX.

If you're not going to go very far beyond calculus, though, you'll never really need it.
What!?!? you use Latex in Uni? Wow now I'm so happy I learnt it!

Page 2 of 2 First 12


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10.2. Functionals, extremums and variations (continued)

## 10.2. Functionals, extremums and variations (continued)

### Boundary conditions

Let us consider functional (10.1.4) but now instead of constrain (10.1.7) we put a less restrictive $$\delta q (t_0)=0. \label{eq-10.2.1}$$ Then Euler-Lagrange equation (10.1.9) must still be fulfilled but it does not guarantee that $\delta S=0$ for all admissible variations as according to (10.1.6) $$\delta S= \frac{\partial L}{\partial \dot{q}}\delta q\Bigr|_{t=t_1}, \label{eq-10.2.2}$$ where we used Euler-Lagrange equation.

Now we need to have it be $0$ as well an since $\delta q(t_1)$ is arbitrary we need $$\frac{\partial L}{\partial \dot{q}}\Bigr|_{t=t_1}=0. \label{eq-10.2.3}$$ However under assumption (\ref{eq-10.2.1}) it makes sense to consider more general functional than (10.1.4): $$S[q]= \int_I L(q,\dot{q},t)\,dt + M_1(q(t_1)), \label{eq-10.2.4}$$ which includes a boundary term. One can see easily that variation of the boundary term is $\frac{\partial M_1}{\partial q}\delta q$ which should be added to (\ref{eq-10.2.2}) which becomes $$\delta S= \Bigl(\frac{\partial L}{\partial \dot{q}} + \frac{\partial M}{\partial q}\Bigr)\delta q(t_1); \label{eq-10.2.5}$$ then (\ref{eq-10.2.3}) becomes $$\Bigl( \frac{\partial L}{\partial \dot{q}}+ \frac{\partial M_1}{\partial q} \Bigr)\Bigr|_{t=t_1}=0. \label{eq-10.2.6}$$ Then we arrive to the following generalization of Theorem 10.1.1:

Theorem 1. Let us consider a functional (\ref{eq-10.2.4}) and consider as admissible all $\delta u$ satisfying (\ref{eq-10.2.1}). Then $u$ is a stationary point of $S$ if and only if it satisfies Euler-Lagrange equation (10.1.9) and a boundary condition (\ref{eq-10.2.6}).

Remark 1.

1. If instead of (\ref{eq-10.2.1}) we consider restriction $\delta q(t_1)=0$ then (\ref{eq-10.2.6}) is replaced by $$\Bigl( \frac{\partial L}{\partial \dot{q}}+ \frac{\partial M_0}{\partial q} \Bigr)\Bigr|_{t=t_10}=0. \label{eq-10.mathjax/2.7.5}$$
2. If we have no restrictions to $\delta q(t_0)$, $\delta q(t_1)$, then we have both (\ref{eq-10.2.6}) and (\ref{eq-10.mathjax/2.7.5}).
3. Thus, no matter what, we have on each endpoint one condition: either $q(t_0)=q^0$ or (\ref{eq-10.mathjax/2.7.5}) at $t=t_0$, and either $q(t_1)=q^0$ or (\ref{eq-10.2.6}).

### Vector and complex valued functions

If our function is vector–valued: $\mathbf{q}=(q_1,\ldots,q_m)$, then we can consider variations with respect to different components and derive corresponding Euler-Lagrange equations $$\frac{\partial L}{\partial q_k} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_k} =0 \qquad k=1,\ldots,m. \label{eq-10.2.8}$$ We also get $m$ boundary conditions on each endpoint of the interval.

Remark 2. (Compare with Remark 10.1.4)

1. If $L_{q_k}=0$ then the corresponding equation integrates to $$\frac{\partial L}{\partial \dot{q}}=C_k. \label{eq-10.2.9}$$
2. The following equality holds (Beltrami identity): $$\frac{d}{dt} \left(\sum_{1\le k\le m} \frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L\right)=-\frac{\partial L}{\partial t}. \label{eq-10.2.10}$$ Indeed, the left hand expression equals to \begin{equation*} \sum_{1\le k\le m} \Bigl[ \Bigl(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}\Bigr)\dot{q}_k + \frac{\partial L}{\partial \dot{q}_k}\ddot{q}_k \Bigr]- \sum_{1\le k\le m}\Bigl[\frac{\partial L}{\partial q_k} \dot{q}_k + \frac{\partial L}{\partial \dot{q}_k} \ddot{q}_k\Bigr]- \frac{\partial L}{\partial t}, \end{equation*} where to differentiate $L$ we used a chain rule. Some terms cancel immediately, some due to Euler-Lagrange equations, and we are left with $\frac{\partial L}{\partial t}$.
3. In particular, if $\frac{\partial L}{\partial t}=0$ ($L$ does not depend explicitely on $t$), then $$\frac{d}{dt} \left(\sum_{1\le k\le m}\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L\right)=0\implies \sum_{1\le k\le m}\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L=C. \label{eq-10.2.11}$$

Remark 2. Complex-valued functions $q$ could be considered as vector-valued functions $\mathbf{q}=\left(\begin{smallmatrix}\Re q\\ \Im q\end{smallmatrix}\right)$. Similarly we can treat functions which are vector–valued with complex components: we just double $m$.

### Extremals under constrains. I

We can consider extremals of functionals under constrains $$\Psi_1[q]=\Psi_2[q]=\ldots=\Psi_s[q]=0 \label{eq-10.2.12}$$ where $\Psi_j$ are other functionals. This is done in the same way as for extremums of functions of several variables: instead of $S[q]$ we consider Lagrange functional $$S^*[q]:= S[q]-\lambda_1\Psi_1[q]-\lambda_2\Psi_2[q]-\ldots-\lambda_s \Psi_s[q] \label{eq-10.2.13}$$ and look for it extremals without constrains; factors $\lambda_1,\ldots,\lambda_s$ are Lagrange multipliers.

Remark 3. This works provided $\delta \Psi_1,\ldots,\delta\Psi_s$ are linearly independent which means that if $\alpha_1\delta\Psi_1[q]+\alpha_2\delta\Psi_2[q]+\ldots +\alpha_s \delta\Psi_s[q]=0$ for all admissible $\delta q$, then $\alpha_=\ldots=\alpha_s=0$.

### Extremals under constrains. II

However constrains could be different from those described in the previous subsection. They can be not in the form of functionals but in the form of functions. In this case we have continuum conditions and Lagrange multipliers became functions as well. Let us consider this on a example.

Example 1. Consider our standard functional under constrains $$F_j(\mathbf{q}(t),t)=0\qquad j=1,\ldots, s. \label{eq-10.2.14}$$ Then we consider functional \begin{equation*} S^* [\mathbf{q}]=\int_I \bigl( L(\mathbf{q},\dot{\mathbf{q}},t)- \sum_{1\le j\le s} \lambda_j (t)F_j(\mathbf{q}(t),t) \bigr) \,dt. \end{equation*} Then Euler-Lagrange equations are $$\frac{\partial L}{\partial q_k}- \frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k} - \sum_{1\le j \le s} \lambda_j(t) \frac{\partial F_j}{\partial q_k}=0 \label{eq-10.2.15}$$ with unknown functions $\lambda_j(t)$.

Remark 4. Obviously we can combine different types of constrains and different types of Lagrange multipliers.

### Higher order functionals

We could include into functional higher-order derivatives. Let us consider functional (for simplicity we consider scalar functions and second-order ) $$S[q]= \int_I L(q,\dot{q}, \ddot{q},t)\,dt. \label{eq-10.2.16}$$ Then the same arguments as before (but term $\frac{\partial L}{\partial\ddot{q}}\delta \ddot{q}$ should be integrated by parts twice) lead us to Euler-Lagrange equation $$\frac{\delta S}{\delta q}:= \frac{\partial L}{\partial q} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) + \frac{d^2}{dt^2} \left(\frac{\partial L}{\partial \ddot{q}}\right)=0 \label{eq-10.2.17}$$ which is the fourth order ODE.

But what about boundary conditions? We need to have now two of them at each endpoint. Consider, f.e. $t=t_1$. Then we have there in $\delta S$ terms $$\Bigl[ \Bigl(-\frac{d}{dt}\frac{\partial L}{\partial \ddot{q}}+ \frac{\partial L}{\partial \dot{q}} + \frac{\partial M_1}{\partial q}\Bigr) \delta q+ \Bigl(\frac{\partial L}{\partial \ddot{q}} + \frac{\partial M_1}{\partial \dot{q}}\Bigr) \delta \dot{q} \Bigr] \Bigr|_{t=t_1}, \label{eq-10.2.18}$$ if functional includes a term $M_1 (q(t_1), \dot{q}(t_1))$.

Obviously we have them if we are looking for solution satisfying $$q(t_1)=a, \qquad \dot{q}(t_1)=b. \label{eq-10.2.19}$$ Otherwise we need to consider $\delta S$. If we have only the first restriction, then $\delta q(t_1)=0$, but $\delta \dot{q}(t_1)$ is arbitrary and we get the second boundary condition $$\Bigl(\frac{\partial L}{\partial \ddot{q}} + \frac{\partial M_1}{\partial \dot{q}}\Bigr)\Bigr|_{t=t_1}=0. \label{eq-10.2.20}$$ If there are no restriction, $\delta q(t_1)$ is also arbitrary and in addition to (\ref{eq-10.2.20}) we get $$\Bigl(-\frac{d}{dt}\frac{\partial L}{\partial \ddot{q}}+ \frac{\partial L}{\partial \dot{q}} + \frac{\partial M_1}{\partial q}\Bigr) \Bigr|_{t=t_1}=0. \label{eq-10.2.21}$$


text:
Why is it a bad idea to store methods in Entities and Components? (Along with some other Entity System questions.)

This is a followup to this question, which I answered, but this one tackles with a much more specific subject.

This answer helped me understand Entity Systems even better than the article.

I've read the (yes, the) article about Entity Systems, and it told me the following:

Entities are just an id and an array of components (the articles says that storing entities in components isn't a good way of doing things, but doesn't provide an alternative).
Components are pieces of data, that indicate what can be done with a certain entity.
Systems are the "methods", they perform manipulation of data on entities.

This seems really practical in many situations, but the part about components being just data classes is bothering me. For example, how could I implement my Vector2D class (Position) in an Entity System?

The Vector2D class holds data: x and y coordinates, but it also has methods, which are crucial to its usefulness and distinguish the class from just a two element array. Example methods are: add(), rotate(point, r, angle), substract(), normalize(), and all other standard, useful, and absolutely needed methods that positions (which are instances of the Vector2D class) should have.

If the component was just a data holder, it wouldn't be able to have these methods!

One solution that could probably pop up would be to implement them inside systems, but that seems very counter-intuitive. These methods are things that I want to perform now, have them be complete and ready to use. I don't want to wait for the MovementSystem to read some expensive set of messages that instruct it to perform a calculation on the position of an entity!

And, the article very clearly states that only systems should have any functionality, and the only explanation for that, which I could find, was "to avoid OOP". First of all, I don't understand why should I refrain from using methods in entities and components. The memory overhead is practically the same, and when coupled with systems these should be very easy to implement and combine in interesting ways. The systems, for example, could only provide basic logic to entities/components, which know the implementation themselves. If you ask me - this is basically taking the goodies from both ES and OOP, something that can't be done according to the author of the article, but to me seems like a good practice.

Think about it this way; there are many different types of drawable objects in a game. Plain old images, animations (update(), getCurrentFrame(), etc), combinations of these primitive types, and all of them could simply provide a draw() method to the render system, which then doesn't need to care about how the sprite of an entity is implemented, only about the interface (draw) and the position. And then, I would only need an animation system which would call animation-specific methods that have nothing to do with the rendering.

And just one other thing... Is there really an alternative to arrays when it comes to storing components? I see no other place for components to be stored other than arrays inside an Entity class...

Maybe, this is a better approach: store components as simple properties of entities. For example, a position component would be glued to entity.position.

The only other way would be to have some kind of a strange lookup table inside systems, that references different entities. But that seems very inefficient and more complicated to develop than simply storing components in the entity.

• Alexandre are you just making a lot of edits to get another badge? Because that's naughty naughty, it keeps bumping a ton of ancient threads. – jhocking Jun 20 '17 at 20:30

I think it's totally fine to have simple methods for accessing, updating or manipulating the data in components. I think the functionality that should stay out of components is logical functionality. Utility functions are just fine. Remember, the entity-component system is just a guideline, not strict rules you need to follow. Don't go out of your way to follow them. If you think it makes more sense to do it one way, then do it that way :)

EDIT

To clarify, you're goal is not to avoid OOP. That would be pretty difficult in most of the common languages used these days. You're trying to minimize inheritance, which is a large aspect of OOP, but not required. You want to get rid of the Object->MobileObject->Creature->Bipedal->Human type inheritance.

However, it's OK to have some inheritance! You're dealing with a language that's heavily influenced by inheritance, it's very difficult to not use any of it. For example, you can have a Component class or interface that all your other components extend or implement. Same deal with your System class. This makes things a lot easier. I strongly recommend you take a look at the Artemis framework. It's open source and it has some example projects. Open those things up and see how it works.

For Artemis, entities are stored in an array, simple. However, their components are stored in an array or arrays (separate from the entities). The top level array groups the lower level array by component type. So each component type has its own array. The lower level array is indexed by entity ID. (Now I'm not sure if I'd do it that way, but that's the way it's done here). Artemis reuses entity IDs, so the max entity ID doesn't get larger than your current number of entities, but you can still have sparse arrays if the component is not a frequently used component. Anyway, I won't pick that apart too much. This method for storing entities and their components seems to work. I think it would make a great first pass at implementing your own system.

The entities and components are stored in an separate manager.

The strategy you mention, making entities store their own components (entity.position), is kind of against the entity component theme, but it's totally acceptable if you feel that makes the most sense.

• Hmm, that greatly simplifies the situation, thanks! I thought that there was some magic "you're gonna regret it later" thing going on, and I just couldn't see it! – jcora Jul 19 '12 at 22:54
• Na, I totally use them in my entity component system. I even have some components that inherit from a common parent, gasp. I think the only regretting you would do is if you tried to work around not using methods like that. It's all about doing what makes the most sense to you. If it makes sense to use inheritance or put some methods in a component, go for it. – MichaelHouse Jul 19 '12 at 22:58
• I've learned from my last answer on this subject. Disclaimer: I'm not saying this is the way to do it. :) – MichaelHouse Jul 19 '12 at 23:01
• Yeah, I know how daunting it can be to learn a new paradigm. Luckily, you get to use aspects of the old paradigm to make things easier! I've updated my answer with storage info. If you look at Artemis, check out the EntityManager as that's where things are stored. – MichaelHouse Jul 19 '12 at 23:44
• Nice! That's going to be a pretty sweet engine when it's done. Good luck with it! Thanks for asking interesting questions. – MichaelHouse Jul 20 '12 at 21:35

'That' article is not one I particularly agree with, so my answer will be somewhat critical I think.

This seems really practical in many situations, but the part about components being just data classes is bothering me. For example, how could I implement my Vector2D class (Position) in an Entity System?

The idea isn't to ensure that nothing in your program is anything other than an entity ID, component, or system - it's to ensure that entity data and behaviour is created through composing objects rather than either using a complex inheritance tree or worse trying to put all possible functionality into one object. To implement these components and systems you will certainly have normal data like vectors which are, in most languages, best represented as a class.

Ignore the bit in the article that suggests this is not OOP - it is just as OOP as any other approach. When most compilers or language runtimes implement object methods, it is basically just like any other function, except there is a hidden argument called this or self, which is a pointer to a place in memory where that object's data is stored. In a component based system, the entity ID can be used to find where the relevant components (and thus data) are for a given entity. Thus the entity ID is equivalent to a this/self pointer, and the concepts are basically the same thing, just rearranged a little.

And, the article very clearly states that only systems should have any functionality, and the only explanation for that, which I could find, was "to avoid OOP". First of all, I don't understand why should I refrain from using methods in entities and components.

Good. Methods are an effective way of organising your code. The important thing to take away from the "avoid OOP" idea is to avoid using inheritance everywhere to extend functionality. Instead, break the functionality up into components that can be combined to do the same thing.

Think about it this way; there are many different types of drawable objects in a game. Plain old images, animations (update(), getCurrentFrame(), etc), combinations of these primitive types, and all of them could simply provide a draw() method to the render system [...]

The idea of a component based system is that you wouldn't have separate classes for these, but would have a single Object/Entity class, and the image would be an Object/Entity that has an ImageRenderer, the Animations would be an Object/Entity that has an AnimationRenderer, etc. The relevant systems would know how to render these components and so there wouldn't need to be any base class with a Draw() method.

[...] which then doesn't need to care about how the sprite of an entity is implemented, only about the interface (draw) and the position. And then, I would only need an animation system which would call animation-specific methods that have nothing to do with the rendering.

Sure, but this doesn't work well with components. You have 3 choices:

• Every component implements this interface and has a Draw() method, even if nothing gets drawn. If you did this for every bit of functionality then components would look pretty ugly.
• Only components that have something to draw implement the interface - but who decides which components to call Draw() on? Does a system have to somehow query each component to see what interface is supported? That would be error-prone and potentially tricky to implement in some languages.
• Components are only ever handled by their owning system (which is the idea in the linked article). In which case, the interface is irrelevant because a system knows exactly what class or object type it's working with.

And just one other thing... Is there really an alternative to arrays when it comes to storing components? I see no other place for components to be stored other than arrays inside an Entity class...

You can store the components in the system. The array isn't the issue, but where you store the component is.

• +1 Thanks for another viewpoint. It's good to get a few when dealing with such an ambiguous subject! If you're storing components in the system, does that mean that components can only ever be modified by one system? For example, the drawing system and movement system would both access the position component. Where do you store it? – MichaelHouse Jul 20 '12 at 2:50
• Well they would only store a pointer to those components, which can, as long as I'm concerned be somewhere... Also, why would you store components in systems? Is there an advantage to that? – jcora Jul 20 '12 at 5:15
• Am I correct, @Kylotan? That's how I would do it, it seems logical... – jcora Jul 20 '12 at 11:25
• In Adam/T-Machine's example, the intention is that there is 1 system per component, but a system could certainly access and modify other components. (This hinders the multithreading benefits of components, but that's a different matter.) – Kylotan Jul 20 '12 at 15:02
• Storing components in the system allows better locality of reference for that system - that system only (generally) works with that data, so why go all over your computer's memory from entity to entity to get it? It also helps with concurrency because you could put a whole system and its data on one core or processor (or even a separate computer, in MMOs). Again, these benefits diminish when 1 system accesses more than 1 type of component, so that should be taken into account when deciding where to split component/system responsibilities. – Kylotan Jul 20 '12 at 15:06

A vector is data. The functions are more like utility functions - they're not specific to that instance of the data, they can be applied to all vectors independently. A nice way of thinking about it is: Can these functions be rewritten as static methods? If so, it's just utility.

• I know that, but the problem is that calling methods is faster and can be done on the spot by a system, or anything else that might need to manipulate an entity's position. I explained that, check it out, also, the question has a lot more to it than just this, I believe. – jcora Jul 19 '12 at 22:53


text:
Importance: Low ✭
 Author(s): Georgakopoulos, Agelos
 Subject: Graph Theory » Infinite Graphs
 Keywords: end ray
 Posted by: Agelos on: February 3rd, 2008
Problem   Let be a graph, a countable end of , and an infinite set of pairwise disjoint -rays in . Prove that there is a set of pairwise disjoint -rays that devours such that the set of starting vertices of rays in equals the set of starting vertices of rays in .

We say that a set of rays devours the end if every ray in meets some ray in . An end is countable if there is a countable set of rays devouring it.

If is a finite set of rays then it is not hard to prove (see [G]) that this problem has a positive answer:

Theorem   For every graph and every countable end of , if has a set of pairwise disjoint -rays, then it also has a set of pairwise disjoint -rays that devours . Moreover, can be chosen so that its rays have the same starting vertices as the rays in~.

## Bibliography

*[G] A. Georgakopoulos, Infinite Hamilton Cycles in Squares of Locally Finite Graphs, Preprint.

* indicates original appearance(s) of problem.


text:
Does a countable set generate a countable group?

Let $G$ be a group and let $A\subseteq G$ be infinite. Do $A$ and $\left< A\right>$ have the same cardinality?

-

Yes. Any element of $\langle A\rangle$ is a "word" in the alphabet $A\cup\{a^{-1}\mid a\in A\}$, which is again a set of size $|A|$. Any word is a finite string of symbols from this alphabet (but of course, different words may give the same element of $\langle A\rangle$), so $\langle A\rangle$ has size at most $|A|^{<\omega}$, which again is $|A|$. This gives $|A|$ as an upper bound. Of course, every element of $A$ is in this group, so it has size at least $|A|$, and equality follows.

This uses (some amount of) the axiom of choice, though. Without it, we cannot prove for an arbitrary infinite set $A$ that $|A|^{<\omega}=|A|$. Choice is not needed to prove this equality in concrete cases, such as if $|A|=|\mathbb N|$ or $|A|=|\mathbb R|$. There is however another, somewhat subtler, use of choice in the proof: Many different words may be equal, which is why a priori the number of words is just an upper bound, because this shows that $\langle A\rangle$ is a quotient of the set of words, under some equivalence relation. Without choice, a quotient of a set may even be larger than the set. I point this out here. (In that answer I link to a nice talk you may enjoy on this issue, by Mike Oliver.)

-
This is a very nice answer. –  Alexander Gruber Apr 9 '13 at 1:44

Let me give a proof of the statement in the title which has a different flavour from the other proofs. It certainly doesn't use choice, however it does use more group theory than the other answers (most notable, it uses the fact that subgroups of free groups are free).

Let $G$ be a countably-generated group, $G=\langle A\rangle$. We want to show that $G$ is countable. (I shall use "countable" to mean "infinite and countable", but it is difficult not to see how this result implies that finitely generated groups are countable also.)

To prove that $G$ is countable it suffices to prove that the free group on the generators $A$ is countable (can you see why this is?). This is a well-known result, and I shall give a relatively full proof. For more details, the book Combinatorial group theory by Magnus Karrass and Solitar will provide better detail/background. (There also exist topological proofs, but I do not know any good the references. Perhaps John Meier's book, or Pierre de la Harpe's?).

Write $F_{\infty}$ then for the free group on countably many generators.

Let $\langle a, b\rangle$ generate the free group on two generators.

Proposition: $H:=\langle a^{-i}ba^i, i\in\mathbb{Z}\rangle$ is free and infinitely generated, $H\cong F_{\infty}$.

This clearly proves the theorem.

Proof: As subgroups of free groups are free, $H$ is free. It therefore suffices to prove that the set $\{a^{-i}ba^i\}$ is a minimal generating set.

It is an important fact about free groups that if $\langle S\rangle=F$ and $F$ is $n$-generated then there exists an $n$-element subset of $S$ which generates $F$. Thus, if we can show that every element of $\{a^{-i}ba^i\}$ is needed to generate $H$ then we are done.

We shall write $W(x, y)$ to mean a word in $x$ and $y$. For example, if $W(x, y)=x^2y^2$ then $W(a^{-1}ba, a^{-2}ba^2)=(a^{-1}ba)^2(a^{-2}ba^2)^2=(a^{-1}b^2a)(a^{-2}b^2a^2)=a^{-1}b^2a^{-1}b^2a^2$.

The trick is to count the number of $b$-terms. Define the $b$-length of a word over $\{a^{-i}ba^i\}$ to be the number of $b$-terms in the word. For example, $b^2a^{-1}b^{-1}a$ has $b$-length $3$. You can think of it as taking the sum of the absolute values of the $b$-exponents: $|2|+|-1|=2+1=3$.

To see that every element of $\{a^{-i}ba^i\}$ is needed, suppose $g:=a^{-j}ba^j$ is not needed. However, $g$ is in $H$, so $g\in\langle a^{-i}ba^i; i\in\mathbb{N}\setminus\{j\} \rangle$=H. Note that the $b$-length of $g$ is one, while in general the $b$-length of $W(\{a^{-i}ba^i\})$ is simply the length of the word $W$. Thus, as $g$ has $b$-length one it is not in $\langle a^{-i}ba^i; i\in\mathbb{N}\setminus\{j\} \rangle$, a contradiction.

-

I rarely can be sure that my proofs are correct. However I can't say this trial is not correct:

Let $$C=A\cup A^{-1}\cup\{1\}$$

then $$\left< A\right>=\bigcup_{n\in \Bbb N}C^n$$ $$\Rightarrow|\left<A\right>|\le|\Bbb N||C|=|\Bbb N||A|=|A|$$ $$\Rightarrow |\left<A\right>|=|A|$$

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I am not sure what you mean by $|\mathbb N||C^n|$ in the previous to last line, since this depends on $n$ now while nothing else does. Maybe you meant $|\mathbb N|\sup_n|C^n|$, from which the argument can proceed now as written. –  Andres Caicedo Apr 9 '13 at 0:28
thanks. $n$ is not needed. because $C$ is infinite, $C\times C\times C\times...\times C$ has the same size as $C$ (as $A$). So $C^n=CC...C$ has size $|C|$ too. now $sup_n |C^n|=sup_n |C|=|C|$. I think at most a countable version of Axiom of Choice is used. –  user59671 Apr 9 '13 at 1:52


text:
# How to bind two distinct actions to Ctrl-Shift-Key and Ctrl-Key in vi

keyboard shortcutsvimxterm

I am trying to map two different actions to e.g. CtrlShifta and Ctrla:

map <C-S-a>    :echo "ctrl-shift-a received"<CR>
map <C-a>      :echo "ctrl-a received"<CR>

But all I ever get when pressing either of the two, is the output "ctrl-a received" in the status line. Apparently, vi does not see a difference between a capitalized control character and a lowercase one.

Is there a way to change this?

I am running vi (v7.4) in an xterm (X.Org 6.8.99).

This is a terminal restriction, because vi sees the keycodes as being the same. One can verify this by going into insert mode i, then hitting Ctrlv to put vi in a mode to expect the literal input of a special character and then hit e.g. CtrlShifta. A ^A output will appear. Now do the sequence Ctrlv, Ctrla, and again a ^A will appear: Both input sequences result in the same keycode being seen by vi.

For terminals such as xterm you can change that. For gnome-terminal which ignores the X resources settings I don't know how to do the following:

This SO question covers the same topic, and this example shows this in use for mapping multiple keys in the .Xresources file. E.g.:

XTerm*vt100.translations: #override \
Ctrl ~Meta Shift <Key>a: string(0x1b) string("[65;5u") \n\
Ctrl ~Meta Shift <Key>b: string(0x1b) string("[66;5u") \n\
Ctrl ~Meta Shift <Key>f: string(0x1b) string("[70;5u")

Now we have different key sequences coming into vi for Ctrla vs. CtrlShifta, Ctrlb vs. CtrlShiftb, Ctrlf vs. CtrlShiftf.

We can now map them differently in ~/.vimrc:

map <ESC>[65;5u   :echo "ctrl-shift-a received"<CR>
map <C-a>         :echo "ctrl-a received"<CR>


text:
# Variance of periodogram estimate of the power spectrum

I have been reading chapter 13.4. ("Power Spectrum Estimation Using the FFT") of the Numerical Recipies Book.

Some things related to the expectation value of the "periodogram estimate of the power spectrum" became not clear to me though.

Background

Suppose we have an equally-spaced $N$-point sample $c_0 ... c_{N-1}$ of the discrete signal $c(t)$ and its Fourier transform $$C_k = \sum_{j=0}^{N−1} = c_j \exp{ 2πijk/N}$$ $$k = 0, . . . , N − 1$$ Then we can define the "periodogram estimate of the power spectrum" $P(f_k)$ as $$P(f_k) = \frac{1}{N^2} [{|C_k|^2 + |C_{N−k}|^2]}$$

with $k = 1, 2, . . . , \frac{N}{2}− 1$ and $f_k = \frac{k}{N\Delta}$ for $k = 0, 1, . . . ,\frac{N}{2}$ ($\Delta$ is the sampling interval).

So far everything is clear.

Now the book says

"... the variance of the periodogram estimate at a frequency $f_k$ is always equal to the square of its expectation value at that frequency. In other words, the standard deviation is always 100 percent of the value, independent of N!"

My questions

1. What is the expectation value of the periodogram estimate at a frequency $f_k$? Why (can you show me how to derive it)?
2. The expectation value doesn't change with $N$, so changing the length of the input doesn't improve the power spectrum estimation. But if I have a non-integral number of periods in my sample, more samples (longer sampling time) means less weight on the end of the signal (where there is a disconnect) so less smearing of frequency components measured due to the disconnect between the beginning and the end of the signal. So in this case a higher $N$ means a better estimation? Where am I wrong?
3. If the SD is always 100 percent of the expectation value, doesn't that make this simple periodogram estimation (using just a single DFT of $N$ samples) pretty much useless? Should one always opt for Welch's or Bartlett's method e.g.?

Thank you very much in advance!

• 3. yes-- this is the motivation to introduce Welch's method, which uses averaging to exchange frequency resolution for lower variance. – nibot Sep 19 '12 at 18:55

$\ var\{{I_{xx}}\} = S^2_{xx}e^{(j\Omega)}.[1 + (sinN\Omega / Nsin\Omega)^2]$
As $\ N \rightarrow \infty$ denominator of the second part in the parenthesis goes to infinity and the division goes to zero because of the N but summing with 1 prevents all equation to become zero. So variance converges to the square of the PSD, not the zero.


text:
# My Year Of Linux Inside The Desktop

Christmas has come and gone, and nothing much has happened other than an untimely viral flu, a fair amount of reading and the start of a well-deserved break, during which I noticed it’s been a year since I last posted about Elementary and built a rather nice Hackintosh, and that it’s probably worthwhile to do a little recap.

## Another Thing In The Closet

First off, about that Hackintosh, which I rather affectionately christened rogueone.

The long and short of it is that although I was able to get most of the niceties working (including Bluetooth input), I never had the time to fix dual display support under Sierra (which is critical for me) and ended up installing Elementary on it instead–which worked beautifully with everything, except that Linux power management and fan control lack finesse to a degree where the machine was substantially louder–and even with handpicked components, the fan noise was too much compared to my mini.

Since I needed to run Windows and have a native Linux server for testing, I set up KVM (Elementary is Ubuntu LTS under the hood, after all), stuffed the machine into a closet and turned my nearly eight-year-old Mac mini into a souped-up thin terminal, which has worked out quite well in many regards–there is now zero fan noise in the office (the mini also runs cooler), I get to use both my monitors and can still enjoy all my Bluetooth input devices.

## Thin and Big vs Fat and Locked Down

In fact, I’ve been pondering if I shouldn’t follow through–just get a cheap fanless PC to dual boot between Windows and Linux and call it a day. I’ve even drafted a shortlist (currently topped by the Mele PCG35 paired with an LG 43UD79), and the only two hard requirements are that it has to be utterly noise-free (so low-end fanless machines are mostly OK), and that it needs to be able to drive a 4K display.

Given these preferences, It’s pretty obvious to me that I’m not part of [Apple][a]’s elusive “Pro” segment–I’m a UNIX guy that codes and builds cloud solutions, I don’t do (much) audio or video editing, my photography needs aren’t really met by their software (although the OS does have great color management), and I find all-in-one machines to be too limiting, especially when they’re not expandable at all. The mini form factor suited me perfectly until they stopped being upgradable, but I’m getting round to accept that my hope for a decent compact Mac desktop is 90% likely to go unfulfilled.

On the other hand, it’s actually quite amazing to think that I spent a good portion of my time working inside the Linux subsystem in Windows and hardly missed the Mac during business hours–were it not for my reliance on Mail.app (something that Thunderbird never managed to solve) and the buttery smoothness of macOS input devices, and I’d probably have switched wholesale to Windows for office work and Linux for coding and photography.

## Less Entropy is More

And, let’s face it, that is an increasingly likely option given the mess macOS turned out to be during 2017. I kept running Elementary on my ancient Chromebook, and the thing is absolutely rock solid and consistently smooth (even with only 2GB of RAM, a crummy screen and a lousy touchpad), to the extent that it’s often the first laptop I reach for when inspiration strikes me.

Yeah, it’s Linux, and it’s impossible to file away all the jutting corners and kludges. But it’s been a year, and I can’t find anything (else) wrong with it that I can’t work around or live with. In fact, it’s aged better than macOS, and that alone should give you pause.


text:
qiskit.aqua.operators.expectations.AerPauliExpectation.compute_variance¶

AerPauliExpectation.compute_variance(exp_op)[ソース]

Compute the variance of the expectation estimator. Because Aer takes this expectation with matrix multiplication, the estimation is exact and the variance is always 0, but we need to return those values in a way which matches the Operator’s structure.

パラメータ

exp_op (OperatorBase) – The full expectation value Operator after sampling.

Union[list, float]

The variances or lists thereof (if exp_op contains ListOps) of the expectation value estimation, equal to 0.


text:
Small harbor detection in polarimetric SAR images based on coastline feature point merging
LIU Chun, YIN Junjun, YANG Jian
Department of Electronic Engineering, Tsinghua University, Beijing 100084, China
Abstract: A method was developed to automatic detect small harbors in polarimetric synthetic aperture radar (polSAR) images using coastline feature point merging based on analyses of structural characteristics of harbors. The coastline is accurately extracted by level set segmentation algorithm of polSAR with the coastline feature points then detected with a split and merge algorithm for digital curves. Then, the algorithm takes advantage of the characteristic that feature points along small harbor contour are denser than those along other coastline contours using a merging algorithm to detect the small harbors. The detection scheme was tested using polarimetric SAR images acquired by RADARSAT-2 over Singapore and the Zhanjiang area of China. The results show that almost all the harbors along the coastline are correctly detected by this method.
Key words: synthetic aperture radar    harbor detection    polarization    level set segmentation    point merge

1 基于水平集的海陆分割算法

 $\begin{array}{l} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\left( {\Gamma ,\left\{ {{\Omega _1},{\Omega _2}} \right\}} \right) = \\ \nu \left| \Gamma \right| - \int\limits_{{\Omega _1}} {\ln f\left( {x|{\Omega _1}} \right){\rm{d}}x - } \int\limits_{{\Omega _2}} {\ln f\left( {x|{\Omega _2}} \right){\rm{d}}x.} \end{array}$ (1)

 $\hat \Gamma = \mathop {\min }\limits_\Gamma E\left( {\Gamma ,\left\{ {{\Omega _1},{\Omega _2}} \right\}} \right).$ (2)

 $\begin{array}{l} \;\;\;E\left( \Phi \right) = \nu \int\limits_\Omega {\left| {\nabla H\left( \Phi \right)} \right|} {\rm{d}}x - \\ \;\;\;\int\limits_\Omega {\left( {H\left( \Phi \right)\ln \left( {f\left( {x|{\Omega _1}} \right)} \right) + } \right.} \\ \left. {\left( {1 - H\left( \Phi \right)} \right)\ln f\left( {x|{\Omega _2}} \right)} \right){\rm{d}}x. \end{array}$ (3)

 $\frac{{\partial \Phi }}{{\partial t}} = - \delta \left( \Phi \right)\left( {\nu \kappa + \ln \frac{{f\left( {x|{\Omega _2}} \right)}}{{f\left( {x|{\Omega _1}} \right)}}} \right).$ (4)

 $\begin{array}{*{20}{c}} {f\left( {C|\Sigma ,L,p} \right) = }\\ {\frac{{{L^{pL}}{{\left( {\left| C \right|} \right)}^{L - p}}\exp \left( { - L{\rm{tr}}\left( {{\Sigma ^{ - 1}}C} \right)} \right)}}{{K\left( {L,p} \right){{\left( {\left| \Sigma \right|} \right)}^L}}}.} \end{array}$ (5)

 $\begin{array}{*{20}{c}} {\frac{{\partial \Phi }}{{\partial t}} = - \delta \left( \Phi \right) \cdot }\\ {\left( {\nu \kappa + L\left( {\ln \left| {{\Sigma _1}} \right| + {\rm{tr}}\left( {{\Sigma _1}^{ - 1}C} \right)} \right) - } \right.}\\ {\left. {L\left( {\ln \left| {{\Sigma _2}} \right| + {\rm{tr}}\left( {{\Sigma _2}^{ - 1}C} \right)} \right)} \right).} \end{array}$ (6)

2 基于岸线特征点合并的港口检测算法

 图 1 点对点的特征点合并算法

 图 2 闭合海岸轮廓线起始与终止特征点距离过近情况

3 实验结果

 图 3 新加坡部分海岸区域港口检测结果

 图 4 湛江部分海岸区域港口检测结果

4 结 论

 [1] 陈琪. SAR图像港口目标提取方法研究 [D].长沙:国防科学技术大学, 2011.CHEN Qi. Harbor extraction from SAR imagery [D]. Changsha: NUDT, 2011. (in Chinese) [2] 李艳,彭嘉雄.港口目标特征提取与识别 [J].华中科技大学学报:自然科学版, 2001, 29(6): 9-11.LI Yan, PENG Jiaxiong. Feature extraction of the harbor target and its recognition [J]. Journal of Huazhong University of Science and Technology, 2001, 29(6): 9-11. (in Chinese) [3] 陈琪,陆军,赵凌君,等.基于特征的 SAR遥感图像港口检测方法 [J].电子与信息学报, 2010, 32(6): 2873-2878.CHEN Qi, LU Jun, ZHAO Lingjun, et al. Harbor detection method of SAR remote sensing images based on feature [J]. Journal of Electronics & information technology, 2010, 32(6): 2873-2878. (in Chinese) [4] 张志龙,张焱,沈振康.基于特征谱的高分辨率遥感图像港口识别方法 [J].电子学报, 2010, 38(9): 2184-2188.ZHANG Zhilong, ZHANG Yan, SHEN Zhenkang. Port recognition in high resolution remote sensing images based on feature spectrum [J]. Chinese Journal of Electronics, 2010, 38(9): 2184-2188. (in Chinese) [5] 邢坤,付宜利.基于内港区域的港口目标识别 [J].电子与信息学报, 2009, 31(6): 1275-1278.XING Kun, FU Yili. Harbor target recognition based on inside region [J]. Journal of Electronics & information technology, 2009, 31(6): 1275-1278. (in Chinese) [6] Douglas D H, Peucker T K. Algorithms for the reduction of the number of points required to represent a digitized line or its caricature [J]. Cartographica: The International Journal for Geographic Information and Geovisualization, 1973, 10(2): 112-122. [7] Mason D C, Davenport I J. Accurate and efficient determination of the shoreline in ERS-1 SAR images [J]. IEEE Transactions on Geoscience and Remote Sensing, 1996, 34(5): 1243-1253. [8] Niedermeier A, Romaneessen E, Lehner S. Detection of coastlines in SAR images using wavelet methods [J]. IEEE Transactions on Geoscience and Remote Sensing, 2000, 38(5): 2270-2281. [9] Ben Ayed I, Mitiche A, Belhadj Z. Polarimetric image segmentation via maximum-likelihood approximation and efficient multiphase level-sets [J]. IEEE Transactions on Pattern Analysis and Machine Intelligence, 2006, 28(9): 1493-1500. [10] Sonka M, Hlavac V, Boyle R. Image processing, analysis, and machine vision [M]. Toronto, Canada: Cengage Learning, 2014. [11] Rosenfeld A, Johnston E. Angle detection on digital curves [J]. IEEE Transactions on Computers, 1973, 100(9): 875-878. [12] Rosenfeld A, Weszka J S. An improved method of angle detection on digital curves [J]. IEEE Transactions on Computers, 1975, 24(9): 940-941.


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# cpython-withatomic / Doc / libwww.tex

The branch 'legacy-trunk' does not exist.
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 \chapter{Internet and WWW Services} \nodename{Internet and WWW} \index{WWW} \index{Internet} \index{World-Wide Web} The modules described in this chapter provide various services to World-Wide Web (WWW) clients and/or services, and a few modules related to news and email. They are all implemented in Python. Some of these modules require the presence of the system-dependent module \code{sockets}, which is currently only fully supported on Unix and Windows NT. Here is an overview: \begin{description} \item[cgi] --- Common Gateway Interface, used to interpret forms in server-side scripts. \item[urllib] --- Open an arbitrary object given by URL (requires sockets). \item[httplib] --- HTTP protocol client (requires sockets). \item[ftplib] --- FTP protocol client (requires sockets). \item[gopherlib] --- Gopher protocol client (requires sockets). \item[nntplib] --- NNTP protocol client (requires sockets). \item[urlparse] --- Parse a URL string into a tuple (addressing scheme identifier, network location, path, parameters, query string, fragment identifier). \item[sgmllib] --- Only as much of an SGML parser as needed to parse HTML. \item[htmllib] --- A parser for HTML documents. \item[xmllib] --- A parser for XML documents. \item[formatter] --- Generic output formatter and device interface. \item[rfc822] --- Parse RFC-822 style mail headers. \item[mimetools] --- Tools for parsing MIME style message bodies. \item[binhex] --- Encode and decode files in binhex4 format. \item[uu] --- Encode and decode files in uuencode format. \item[binascii] --- Tools for converting between binary and various ascii-encoded binary representation \item[xdrlib] --- The External Data Representation Standard as described in RFC 1014, written by Sun Microsystems, Inc. June 1987. \item[mailcap] --- Mailcap file handling. See RFC 1524. \item[base64] --- Encode/decode binary files using the MIME base64 encoding. \item[quopri] --- Encode/decode binary files using the MIME quoted-printable encoding. \item[SocketServer] --- A framework for network servers. \item[mailbox] --- Read various mailbox formats. \item[mimify] --- Mimification and unmimification of mail messages. \item[BaseHTTPServer] --- Basic HTTP server (base class for SimpleHTTPServer and CGIHTTPServer). \end{description}


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# A pressure-temperature graph of an ideal gas is shown below. Choose the correct alternative (assume equal number of moles of different volumes).

$(A)\; V_1 = V_2, V_3 = V_4, V_2 \gt V_3$ $(B)\; V_1 = V_2, V_3 = V_4, V_2 \lt V_1$ $(C)\; V_1 = V_2 = V_3 = V_4$ $(D)\; V_4 \gt V_3 \gt V_2 \gt V_1$
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## 1 Answer

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Since $PV = nRT$, the Pressure-Temperature graph at constant volume has a greater slope at lower volume.
Since pt. 2 lies on the line with lesser slope, $V_2 > V_3$.
Since pt. 2 and pt. 1 are on the same line and pts. 3 and 4 are on the same line, $V_1 = V_2$ and $V_3 = V_4$ must hold true, since the P-T graph is a linear one.
answered Jan 22, 2014

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