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https://www.embibe.com/exams/madhya-pradesh-board-class-6-syllabus/ | 1,680,118,387,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00464.warc.gz | 815,981,954 | 52,752 | • Written By evangeline
Madhya Pradesh Board Class 6 Syllabus 2023: The Madhya Pradesh Board of Secondary Education (MPBSE) administers the class 6 exams and is also responsible for creating the curriculum. Topics in the Class 6 syllabus is primarily intended to prepare students for their senior classes, as primary education is important in laying a solid foundation for higher education.
Therefore, gaining knowledge of the exam syllabus is crucial. Knowing the syllabus can help in developing a productive study strategy and score better. For more information on the Madhya Pradesh Board Class 6 syllabus, keep reading the article.
At the conclusion of each semester or term, the school administers an exam to measure the progress of Class 6 students; passing these exams is essential. Therefore, in addition to comprehending the study material, students should adequately prepare for their Madhya Pradesh Board Class 6 exams. To aid them in their studies, we provide a detailed subject-wise syllabus.
### MPBSE Class 6 Syllabus: Subject-wise
Candidates must have a thorough understanding of the syllabus in order to develop a productive study plan. Check the class 6 Madhya Pradesh Board subject-wise syllabus in the following section:
Check the list below for the Maths syllabus for class 6 MPBSE:
• Knowing Our Numbers
• Whole Numbers
• Playing With Numbers
• Basic Geometrical Ideas
• Understanding Elementary Shapes
• Integers
• Fractions
• Decimals
• Data Handling
• Mensuration
• Algebra
• Ratio and Proportion
• Symmetry
• Practical Geometry
#### MPBSE Class 6 Science Syllabus 2023
Here is a the Science syllabus for the Class 6th Madhya Pradesh Board:
• Food: Where Does it Come from?
• Components of Food
• Fibre to Fabric
• Sorting Materials into Groups
• Separation of Substances
• Changes Around Us
• Getting to know Plants
• Body Movements
• The Living Organisms and their Surrounding
• Motion and Measurement of Distances
• Electricity and Circuits
• Fun with Magnets
• Water
• Air Around Us
• Garbage In, Garbage Out
The table below provides the general and special English syllabus for MPBSE class 6:
The special English Grammar Syllabus for class 6 Madhya Pradesh Board is as follows:
• Paragraph Writing
• Vocabulary
• Translation
• Grammar
• Letter and Application Writing
• Short Essay Writing
Also, check the General English Grammar Syllabus for class 6 Madhya Pradesh Board in the list below:
• Essay Writing
• Grammar
• Paragraph Writing
• Letters and Applications
#### MPBSE Class 6 Sanskrit Syllabus
The list below provides the Class 6 Sanskrit syllabus for the Madhya Pradesh Board:
• स्वराभ्यासः (संस्कृत शब्द परिचयः)
• कर्त्तृक्रिर्त्तृयासम्बन्धः
• सर्वनामशब्दाः
• सङ्ख्याबोधः
• विद्या-महिमा
• मम दिनचर्या
• संहतिः कार्यसाधिका
• परोपकारः
• उज्जयिनीदर्शनम्
• परिचयः
• अस्माकं प्रदेशः
• रामचरितम्
• चतुरः वानरः
• जन्तुशाला
• स्वतन्त्रतादिवसः
• भोजस्य शिक्षाप्रियता
• चरामेति चरामेति
• दीपावलिः
• विज्ञानस्य आविष्काराः
• श्रमस्य महत्वम्
• सुभाषितानि
A syllabus informs the students of the course objectives, learning goals, and assessment strategy. It lays out for the students what should be learned and what is anticipated to happen during the course. So, here are a few benefits of knowing the MPBSE class 6 syllabus.
• A syllabus informs the students of the course’s goals.
• Students and teachers attempt to accomplish the objectives by the end of the session by establishing the objectives in the syllabus.
• The steps needed to create a study plan are outlined in the syllabus.
• It also lists the responsibilities of the teachers to accomplish the objectives within the stipulated time.
Here are a few questions on the MPBSE Class 6 syllabus for 2023:
Q: What are the core subjects in Class 6 MPBSE?
Ans: The core subjects in MPBSE class 6 are: Science, Mathematics, and English Language.
Q: Who creates the curriculum for the Madhya Pradesh Board class 6?
Ans: The MPBSE prepares the curriculum for class 6 Madhya Pradesh Board.
Q: What are some of the chapters in the Science syllabus for class 6 MPBSE?
Ans: Some of the chapters in the Science syllabus for class 6 MPBSE are: Food: Where Does it Come from?, Components of Food, Fibre to Fabric, Sorting Materials into Groups, Separation of Substances, Changes Around Us, etc.,
Q: What are the benefits of following the Madhya Pradesh Board class 6 syllabus?
Ans: Some of the benefits of referring to the Madhya Pradesh Board class 6 syllabus are that it informs the students of the course objectives, learning goals, and assessment strategy. It lays out for the students what should be learned and what is anticipated to happen during the course.
Q: Where can I find the entire syllabus for class 6 Madhya Pradesh Board? | 1,210 | 4,742 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | latest | en | 0.842596 |
http://groundpotential.org/forum/viewtopic.php?f=8&t=14&view=print | 1,591,036,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419593.76/warc/CC-MAIN-20200601180335-20200601210335-00182.warc.gz | 56,742,178 | 2,819 | Page 1 of 1
### There are no infinities in the physical world.
Posted: Sun Aug 03, 2014 3:25 pm
Theory needs to confirm physical reality and if it doesn't we must concider it to be wrong, therefore any theory that suggest an infinite value for a physical mass must be wrong. Experience from our every day world tells us there are no infinities, or at least we can say if something has a beginning it also must also have an end.
In this forum we are concerned about potential energy, so let us take Coulombs force law and look at the integral from r to r = (infinity) and one can clearly see that something is wrong.
kqqr.png (9.9 KiB) Viewed 5355 times
Eq1. Electrical potential energy as a function of radius
As we can see this equation predicts all matter should have infinite energy, because protons with a charge of +1 and electrons with a charge of -1 are clearly a finite distance apart, and somehow had to get from zero to some distance r, without using an infinite amount of energy. Therefore we are looking for a minimum radius at which individual charges first come into play, beyond which this equation holds true.
So how do we determine this radius, and is this minimum radius a konstant, or is it a function of something else?
Well let us consider pair production, experiment shows that pairs are created when the energy of a photon exceeds 1022 keV, or 2 * 511 KeV per particle. It is surely no coincidence that this also happens to be the energy of an electron.
I think we can safely draw the conclusion that the minimum radius for the above equation is the radius where the potential energy equals 511 keV. Now if we are to believe Ground Potential theory, then this radius is a variable, and will be a function of the observers potential.
kqqø.png (8.74 KiB) Viewed 5354 times
Eq.2 Electron radius based on electron energy
What this says, is that the minimum radius between two charged particles is when the potential energy is equal to the energy of the electron, which just happens to 511 keV at this moment in time.
We can write a more general equation as follows;
general3.png (12.44 KiB) Viewed 5354 times
Eq3. General extpression for electron radius (∂ø = difference between proton potential and ground potential)
The interpretation of this result is that potential energies lower than the potential of the electron lies outside the observers domain, and are therefore not observable. This is consistent with Ground Potential theory, which postulates a minimum and maximum potential for any observer.
We can test this equation in Wolfram Alpha as follows;
(Coulombs Constant * (elementary charge)^2)/(8.18712225*10^-14 joules)
Which gives the result;
resultradius.png (8.71 KiB) Viewed 5354 times
Which agrees with the classical electron radius. | 628 | 2,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-24 | latest | en | 0.9372 |
https://numbersworksheet.com/how-to-multiply-two-digit-numbers-worksheet/ | 1,721,556,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00441.warc.gz | 381,726,911 | 13,899 | # How To Multiply Two Digit Numbers Worksheet
This multiplication worksheet concentrates on educating students how to psychologically multiply entire amounts. Pupils may use personalized grids to put specifically one particular issue. The worksheets also protectfractions and decimals, and exponents. You can even find multiplication worksheets using a handed out home. These worksheets are a must-have for the math concepts type. They may be used in course to learn how to mentally multiply total line and numbers them up. How To Multiply Two Digit Numbers Worksheet.
## Multiplication of whole figures
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets can help you learn this standard strategy. You may go for a single digit multipliers or two-digit and a few-digit multipliers. Capabilities of 10 will also be an incredible solution. These worksheets will allow you to training long practice and multiplication reading the numbers. Also, they are the best way to assist your son or daughter understand the value of understanding the different kinds of whole numbers.
## Multiplication of fractions
Getting multiplication of fractions over a worksheet may help educators program and get ready classes proficiently. Making use of fractions worksheets enables instructors to quickly examine students’ knowledge of fractions. College students might be pushed to complete the worksheet in just a particular time as well as then mark their strategies to see where by they want further more instruction. College students can usually benefit from term things that associate maths to true-lifestyle scenarios. Some fractions worksheets involve instances of comparing and contrasting phone numbers.
## Multiplication of decimals
If you multiply two decimal amounts, be sure to team them vertically. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. By way of example, 01 x (11.2) x 2 can be equivalent to 01 by 2.33 x 11.2 except when the product has decimal locations of below two. Then, this product is rounded towards the closest total number.
## Multiplication of exponents
A math concepts worksheet for Multiplication of exponents will allow you to training multiplying and dividing figures with exponents. This worksheet may also offer things that will need individuals to flourish two diverse exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. Apart from, you can also enter particular directions in the worksheet itself. When you’re finished, you can just click “Create” as well as the worksheet will be acquired.
## Department of exponents
The fundamental rule for department of exponents when multiplying figures is always to deduct the exponent within the denominator through the exponent inside the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. By way of example, \$23 separated by 4 will identical 27. This method is not always accurate, however. This method can bring about uncertainty when multiplying numbers which can be too big or not big enough.
## Linear capabilities
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So the total rent would be \$470. A linear function of this kind offers the kind f(by), exactly where ‘x’ is the volume of time the car was leased. Furthermore, they have the shape f(x) = ax b, where ‘b’ and ‘a’ are real phone numbers. | 707 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-30 | latest | en | 0.883088 |
https://www.jiakaobo.com/leetcode/438.%20Find%20All%20Anagrams%20in%20a%20String.html | 1,723,328,640,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00172.warc.gz | 671,785,632 | 6,774 | • ㊗️
• 大家
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• 多多！
## Problem
Given two strings s and p, return an array of all the start indices of p’s anagrams in s. You may return the answer in any order.
Example 1:
Input: s = "cbaebabacd", p = "abc"
Output: [0,6]
Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input: s = "abab", p = "ab"
Output: [0,1,2]
Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
## Code
567. Permutation in String
class Solution {
public List<Integer> findAnagrams(String s, String p) {
int[] dict = new int[26];
int[] curr = new int[26];
List<Integer> res = new ArrayList<>();
for(int i = 0; i < s.length(); i++){
curr[s.charAt(i) - 'a']++;
if(i < p.length()) {
dict[p.charAt(i) - 'a']++;
if(i == p.length() - 1 && helper(dict, curr)) {
}
} else {
curr[s.charAt(i - p.length()) - 'a']--;
if(helper(dict, curr)) {
}
}
}
return res;
}
private boolean helper(int[] dict, int[] curr) {
for(int i = 0; i < 26; i++) {
if(dict[i] != curr[i]) return false;
}
return true;
}
} | 402 | 1,282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.572799 |
http://oeis.org/A183757 | 1,601,045,310,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400226381.66/warc/CC-MAIN-20200925115553-20200925145553-00646.warc.gz | 97,051,947 | 3,950 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A183757 Unlabeled super-Catalan numbers: patterns of nonintersecting chords joining unlabeled points on a circle 2
1, 1, 2, 6, 17, 74, 324, 1558, 7640, 38245 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Interpret the super-Catalan sequence (A001003) as follows: Ways to insert parentheses in a string of n+1 symbols. The parentheses must be balanced but there is no restriction on the number of pairs of parentheses. The number of letters inside a pair of parentheses must be at least 2. Parentheses enclosing the whole string are ignored. Now picture the x's and parentheses as equally spaced unlabeled points on a circle with chords joining paired parentheses and x's having no chord. Circles thus produced may have n+1, n+3, ..., 3*n-1 points and up to n-1 chords. The circle may be rotated. a(n) is the count of unique patterns. LINKS EXAMPLE super-Catalan(3) = 11: (xx)xx, x(xx)x, xx(xx), (xx)(xx), (xxx)x, x(xxx),((xx)x)x, (x(xx))x, x((xx)x), x(x(xx)), xxxx. This sequence counts unique patterns up to rotation so a(3) = 6: (xx)xx, (xx)(xx), (xxx)x, ((xx)x)x, x(x(xx)), xxxx. CROSSREFS Cf. A001003 super-Catalan numbers. A183758 similar but with reflections discounted. A183759 this series decomposed by number of chords in the circles. Sequence in context: A264761 A253882 A327698 * A181490 A165325 A316765 Adjacent sequences: A183754 A183755 A183756 * A183758 A183759 A183760 KEYWORD nonn AUTHOR David Scambler, Jan 06 2011 STATUS approved
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Last modified September 25 10:38 EDT 2020. Contains 337337 sequences. (Running on oeis4.) | 568 | 2,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-40 | latest | en | 0.75909 |
http://mathhelpforum.com/trigonometry/118643-2-questions-proving-identity-reducing.html | 1,481,282,878,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542693.41/warc/CC-MAIN-20161202170902-00091-ip-10-31-129-80.ec2.internal.warc.gz | 172,232,576 | 10,556 | # Thread: 2 questions: proving identity and reducing
1. ## 2 questions: proving identity and reducing
The first question I think I have the correct answer to but wanted to double check.
The problem says prove the following identity...
sec x + csc x = csc x + csc x tan x
(1/cosx + 1/sin)=
sinx/(sinxcosx) + cosx/(sinxcosx)
(1/sinx)*(sinxcosx)= cscxtanx=secx
The second question I have I need to reduce this to a single term.
tan (x+y) - tan y
-----------------------
1+ tan (x+y) tan y
I know it reduces to tanx, but how would I show the work for a problem like this?
Thanks so much for any help!
2. Originally Posted by KarlosK
The first question I think I have the correct answer to but wanted to double check.
The problem says prove the following identity...
sec x + csc x = csc x + csc x tan x
(1/cosx + 1/sin)=
sinx/(sinxcosx) + cosx/(sinxcosx)
(1/sinx)*(sinxcosx)= cscxtanx=secx
Does this answer make sense? no
working from the right side ...
$\textcolor{red}{\csc{x} + \csc{x}\tan{x}}$
$\textcolor{red}{\csc{x} + \frac{1}{\sin{x}} \cdot \frac{\sin{x}}{\cos{x}}}$
$\textcolor{red}{\csc{x} + \frac{1}{\cos{x}}}$
$\textcolor{red}{\csc{x} + \sec{x}}$
The second question I have I need to reduce this to a single term.
tan (x+y) - tan y
-----------------------
1+ tan (x+y) tan y
use this identity ...
$\textcolor{red}{\tan(a-b) = \frac{\tan{a}-\tan{b}}{1+\tan{a}\tan{b}}}$
let $\textcolor{red}{a = (x+y)}$ and $\textcolor{red}{b = y}$
...
3. Skeeter thanks for the help on the first problem, I now understand how you did that. On the second problem I am still a little unclear as to how I should show the work to the professor.
So Tan A-B = x+y--y = x
tan x.
Is that enough?
4. $\frac{\tan(x+y) - \tan{y}}{1 + \tan(x+y)\tan{y}} = \tan[(x+y) - y] = \tan{x}$ | 596 | 1,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2016-50 | longest | en | 0.796733 |
https://www.scirra.com/forum/how-does-c2-calculate-number-precision-wise_t86531 | 1,524,141,066,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936914.5/warc/CC-MAIN-20180419110948-20180419130948-00242.warc.gz | 693,745,704 | 9,495 | # How does C2 calculate number, precision-wise?
Discussion and feedback on Construct 2
### » Sun Jul 21, 2013 5:44 am
Just curious here, are the number variables we are using double, float, or something else?
How many digits of number can we go for, until we start to have a little round off error, precision error, etc?
I got a game that you multiply, breath fire with two heads and brawl foes to oblivion with your clones: http://www.newgrounds.com/portal/view/660664 (use Chrome on Windows for best performance)
My sites:
http://twinblazar.deviantart.com
http://twinblazar.newgrounds.com
http://www.pixiv.net/member.php?id=15072448
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### » Sun Jul 21, 2013 5:52 am
Now Ashley can correct me if I'm wrong, but I believe C2 just uses JavaScript's logic and datatypes. Now JS is an untyped, or dynamically typed language. That means that you generally don't declare what type a variable is going to be, it's taken care of behind the scenes. This makes predicting round of errors and precision errors a nightmare because it's all guesswork on what type JS is choosing at that point in time. As a general rule, JS will pick the most suitable type, so:
"asdasd" = string
65161 = int
65651651684651654165168416517165141651 = long
1/3 = double
1/4 = float
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### » Sun Jul 21, 2013 4:15 pm
Construct 2 leaves number calculations to the Javascript engine, and Javascript is standardising as only supporting double precision floats for numbers.
Scirra Founder
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Reputation: 200,273 | 448 | 1,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-17 | latest | en | 0.821192 |
https://jonathanfrech.wordpress.com/tag/fractal/page/2/ | 1,539,911,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512161.28/warc/CC-MAIN-20181018235424-20181019020924-00558.warc.gz | 705,390,721 | 22,595 | ## Mandelbrot Set ASCII Viewer
The Mandelbrot Set is the set of all complex points which, when one iteratively and infinitely applies the function $f_c(z)=z^2+c$, converge to a value. This simple rule results in stunning complexity and beauty.
Many Mandelbrot Set animations use regularly colored pixels to represent the number of iterations needed at the fractal’s edges to escape converging. Yet this mathematical object can also be represented as ASCII characters — similar to what I did in my Curses Cam post. The characters are chosen according to their opaqueness. A full stop (‘.’) looks lighter than a dollar sign (‘$’), so they represent a smaller or larger number of iterations needed. The order of characters used is taken from this post by Paul Borke. As there are only 70 characters used, each frame is being rendered twice to determine the minimum number of iterations needed by every point in that frame. Thereby the full visual character range is used. The characters shown below represent a Mandelbrot Set still. To see the zoom in action, either run the program (listed below) or take a look at this Mandelbrot Set ASCII journey. ..................''''''''"">>II''''...... ..................''''''''^^,,ii::^^''''...... ..................''''''''^^::ww$$++,,''''...... ................''''''''^^^^""::$$::""^^''''...... ..............''''''""{{;;XXuuUU,,,,""''...... ............''''^^,,rr<<$$--........ ........''^^""LL$$$$__""''...... ..''''''^^!!"""",,""""::__$$ll""''........ ''''^^::__IIYYii::llpp^^''........ ''"";;[[$$++__$$^^''''...... ^^^^,,;;>>ww''''...... "",,,,II$$nn$$$$""''''...... "",,,,II$$nn""''''...... ^^^^,,;;>>ww''''...... ''"";;[[$$++__$$^^''''...... ''''^^::__IIYYii::llpp^^''........ ..''''''^^!!"""",,""""::__$$ll""''........ ........''^^""LL$$$$__""''...... ............''''^^,,rr$$$$<<$$--........ ..............''''''""{{;;XXuuUU,,,,""''...... ................''''''''^^^^""::$$::""^^''''...... ..................''''''''^^::ww$$++,,''''...... ..................''''''''^^,,ii::^^''''...... The fractal viewer is written in Python 2.7 and works by determining the terminal’s size and then printing a string of according size. This creates the illusion of a moving image, as the terminal will hopefully always perfectly scroll so that only one frame is visible at a time. In the code’s first non-comment line one can change the complex point at the image’s center, (really, its conjugate, which is partially irrelevant as the set is symmetric along the real axis) the initial zoom value (complex distance above the image’s center), the zoom factor (the factor by which the zoom value gets multiplied after a frame), the total number of frames (-1 means there is no upper limit), the delay between frames (in seconds, can be floating-point) and the color characters used. The program’s source code may not be particularly easy to read, yet it does its job and only requires seven non-comment lines! The code is shown below, though the .py file can also be downloaded. To achieve the JavaScript animation linked to above, I wrote a simple Python converter which takes in the fractal renderer’s output and it spits out an HTML page. This converter’s code is not listed, though the .py file can be downloaded. Instructions on how to use the converter can be seen in its source code. # Python 2.7 Code; Jonathan Frech, 15th and 16th of June 2017 P,Z,F,N,D,K=-.707+.353j,3,.9,-1,.1," .'^\",:;Il!i><~+_-?][}{1)(|\\/tfjrxnuvczXYUJCLQ0OZmwqpdbkhao*#MW&8%B@$"
def C(c):
global m;z,i=0j,-1
while abs(z)<=2 and i<len(K)-1+M:z,i=z*z+c,i+1
m=min(m,i);return K[i-M]*2
while n<N or N==-1:h=Z*2.;w=h*W/H;R=lambda:"\n\n"*(n!=0)+"\n".join("".join(C(P-complex(w/2-w*x/W,h/2-h*y/H))for x in range(W))for y in range(H));M,m=0,len(K);R();M=max(M,m);S.write(R());S.flush();Z,n=Z*F,n+1;time.sleep(D)
## Multibrot Set
The Mandelbrot Set is typically defined as the set of all numbers $c \in \mathbb{C}$ for which — with $z_0 = 0$, $z_{n+1} = f_c(z_n)$ and $f_c(z) = z^2 + c$ — the limit $\lim\limits_{n \to \infty} z_n$ converges. Visualizations of this standard Mandelbrot Set can be seen in three of my posts (Mandelbrot Set, Mandelbrot Set Miscalculations and Mandelbrot Set II).
However, one can extend the fractal’s definition beyond only having the exponent $2$ in the function to be $f_c(z)=z^\text{exp}+c$ with $\text{exp} \in \mathbb{R}$. The third post I mentioned actually has some generalization as it allows for $\text{exp} \in \{2,3,4,5\}$, although the approach used cannot be extended to real or even rational numbers.
The method I used in the aforementioned post consists of manually expanding $(a+b\cdot i)^n$ for each $n$. The polynomial $(a+b\cdot i)^3$, for example, would be expanded to $(a^3 - 3 \cdot a \cdot b^2) + (3 \cdot a^2 \cdot b - b^3) \cdot i$.
This method is not only tedious, error-prone and has to be done for every exponent (of which there are many), it also only works for whole-number exponents. To visualize real Multibrots, I had to come up with an algorithm for complex number exponentiation.
Luckily enough, there are two main ways to represent a complex number, Cartesian form $z = a+b\cdot i$ and polar form $z = k\cdot e^{\alpha\cdot i}$. Converting from Cartesian to polar form is simply done by finding the number’s vector’s magnitude $k = \sqrt{a^2+b^2}$ and its angle to the x-axis $\alpha = \mbox{atan2}(\frac{a}{b})$. (The function $\mbox{atan2}$ is used in favor of $\arctan$ to avoid having to divide by zero. View this Wikipedia article for more on the function and its definition.)
Once having converted the number to polar form, exponentiation becomes easy as $z^\text{exp} = (k \cdot e^{\alpha\cdot i})^\text{exp} = k^\text{exp} \cdot e^{\alpha \cdot \text{exp} \cdot i}$. With the exponentiated $z^\text{exp}$ in polar form, it can be converted back in Cartesian form with $z^\text{exp} = k^\text{exp} \cdot (\cos{(\alpha \cdot \text{exp})} + \sin{(\alpha \cdot \text{exp})} \cdot i \big)$.
Using this method, converting the complex number to perform exponentiation, I wrote a Java program which visualizes the Multibrot for a given range of exponents and a number of frames.
Additionally, I added a new strategy for coloring the Multibrot Set, which consists of choosing a few anchor colors and then linearly interpolating the red, green and blue values. The resulting images have a reproducible (in contrast to randomly choosing colors) and more interesting (in contrast to only varying brightness) look.
The family of Multibrot Sets can also be visualized as an animation, showing the fractal with an increasing exponent. The animated gif shown below was created using ImageMagick’s convert -delay <ms> *.png multibrot.gif command to stitch together the various .png files the Java application creates. To speed up the rendering, a separate thread is created for each frame, often resulting in 100% CPU-usage. (Be aware of this should you render your own Multibrot Sets!)
To use the program on your own, either copy the source code listed below or download the .java file. The sections to change parameters or the color palette are clearly highlighted using block comments (simply search for ‘/*’).
To compile and execute the Java application, run (on Linux or MacOS) the command javac multibrot.java; java -Xmx4096m multibrot in the source code’s directory (-Xmx4096m tag optional, though for many frames at high quality it may be necessary as it allows Java to use more memory).
If you are a sole Windows user, I recommend installing the Windows 10 Bash Shell.
// Java 1.8 Code
// Jonathan Frech, 11th of September 2016
// edited 17th of April 2017
// edited 18th of April 2017
// edited 20th of April 2017
// edited 21st of April 2017
// edited 22nd of April 2017
## Bifurcation Diagram
Generating the famous fractal, which can be used to model populations with various cycles, generate pseudo-random numbers and determine one of nature’s fundamental constants, the Feigenbaum constant $\delta$.
The fractal nature comes from iteratively applying a simple function, $f(x) = \lambda \cdot x \cdot (1-x)$ with $0 \leq \lambda \leq 4$, and looking at its poles.
The resulting image looks mundane at first, when looking at $0 \leq \lambda \leq 3$, though the last quarter section is where the interesting things are happening (hence the image below only shows the diagram for $2 \leq \lambda \leq 4$).
From $\lambda = 3$ on, the diagram bifurcates, always doubling its number of poles, until it enters the beautiful realm of chaos and fractals.
For more on bifurcation, fractals and $\delta$, I refer to this Wikipedia entry and WolframMathworld.
# Python 2.7.7 Code
# Jonathan Frech, 24th of March 2017` | 2,340 | 8,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 30, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-43 | latest | en | 0.626396 |
https://socratic.org/questions/what-is-the-slope-and-intercept-of-2y-5 | 1,723,503,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00261.warc.gz | 420,865,275 | 6,079 | # What is the slope and intercept of 2y = -5?
May 4, 2016
The slope is 0.
#### Explanation:
Consider the general equation of a straight line:
$y = m x + c$
Where $m$ refers to the value of the function's slope.
Let's now rearrange your equation to this general form:
$\frac{2 y}{2} = - \frac{5}{2}$
$y = - \frac{5}{2}$
Therefore, we can see that no $x$ coefficient exists. This is to say that:
$y = - \frac{5}{2} = 0 x - \frac{5}{2}$
This implies that no slope exists, and the graph instead exists parallel to the $x$ axis. | 169 | 535 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-33 | latest | en | 0.874936 |
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# How do I convert from a money datatype in SQL server?
I have a money data type in SQL Server. How do I reformat 0.00 to 0 in my query?
-
Provide more detail and you might get some answers. – Philip Morton Feb 10 '09 at 13:41
Normal money conversions will preserve individual pennies:
``````SELECT convert(varchar(30), moneyfield, 1)
``````
The last parameter decides what the output format looks like:
0 (default) No commas every three digits to the left of the decimal point, and two digits to the right of the decimal point; for example, 4235.98.
1 Commas every three digits to the left of the decimal point, and two digits to the right of the decimal point; for example, 3,510.92.
2 No commas every three digits to the left of the decimal point, and four digits to the right of the decimal point; for example, 4235.9819.
If you want to truncate the pennies, and count in pounds, you can use rounding to the nearest pound, floor to the lowest whole pound, or ceiling to round up the pounds:
``````SELECT convert(int, round(moneyfield, 0))
SELECT convert(int, floor(moneyfield))
SELECT convert(int, ceiling(moneyfield))
``````
-
This will work if you want to discard all penny values, and not JUST when it's 0.00. See my answer for that. – Dan Field Mar 20 '15 at 14:46
Would casting it to int help you? Money is meant to have the decimal places...
``````DECLARE @test AS money
SET @test = 3
SELECT CAST(@test AS int), @test
``````
-
Hi Tina, Now i got my expected output. thanks great..:-) – senthil Feb 10 '09 at 17:20
First of all, you should never use the money datatype. If you do any calculations you will get truncated results. Run the following to see what I mean
``````DECLARE
@mon1 MONEY,
@mon2 MONEY,
@mon3 MONEY,
@mon4 MONEY,
@num1 DECIMAL(19,4),
@num2 DECIMAL(19,4),
@num3 DECIMAL(19,4),
@num4 DECIMAL(19,4)
SELECT
@mon1 = 100, @mon2 = 339, @mon3 = 10000,
@num1 = 100, @num2 = 339, @num3 = 10000
SET @mon4 = @mon1/@mon2*@mon3
SET @num4 = @num1/@num2*@num3
SELECT @mon4 AS moneyresult,
@num4 AS numericresult
``````
Output: 2949.0000 2949.8525
Now to answer your question (it was a little vague), the money datatype always has two places after the decimal point. Use the integer datatype if you don't want the fractional part or convert to int.
Perhaps you want to use the decimal or numeric datatype?
-
Microsoft Dynamics uses the money data type. I saw that today. – Pierre-Alain Vigeant Jan 29 '10 at 0:33
How valid is it really to want to multiple \$450.50 * \$32.76, I sat here for a while thinking of any time I would actually want to multiple money values together, or divide them. I used your example sql and changed the */ to use +- and even wrapped those in a * and / and the values for each matched. – Salizar Marxx Mar 20 '15 at 18:04
I found this approach direct and useful.
CONVERT(VARCHAR(10), CONVERT(MONEY, fieldname)) AS PRICE
-
This looks like a formating issue to me.
As far as SQL Server's money type is concerned 0 == 0.00
If you're trying to display 0 in say c# rather then 0.00 you should convert it to a string, and format it as you want. (or truncate it.)
-
It seems despite the intrinsic limitations of the money datatype, if you're already using it (or have inherited it as I have) the answer to your question is, use DECIMAL.
-
You can try like this:
``````SELECT PARSENAME('\$'+ Convert(varchar,Convert(money,@MoneyValue),1),2)
``````
-
I had this issue as well, and was tripped up for a while on it. I wanted to display 0.00 as 0 and otherwise keep the decimal point. The following didn't work:
``````CASE WHEN Amount= 0 THEN CONVERT(VARCHAR(30), Amount, 1) ELSE Amount END
``````
Because the resulting column was forced to be a MONEY column. To resolve it, the following worked
``````CASE WHEN Amount= 0 THEN CONVERT(VARCHAR(30), '0', 1) ELSE CONVERT (VARCHAR(30), Amount, 1) END
``````
This mattered because my final destination column was a VARCHAR(30), and the consumers of that column would error out if an amount was '0.00' instead of '0'.
-
you could either use
``````SELECT PARSENAME('\$'+ Convert(varchar,Convert(money,@MoneyValue),1),2)
``````
or
``````SELECT CurrencyNoDecimals = '\$'+ LEFT( CONVERT(varchar, @MoneyValue,1),
LEN (CONVERT(varchar, @MoneyValue,1)) - 2)
``````
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# Darsana Thulasi
### MathWorks
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Technical Discussions
Nandesh Kumar A
Nandesh Kumar A
14:21 Feb-28-2010
Can anybody explain me for ISO 13847 ?
One of our code (ISO 13847 ) says the following.
"The total area, when projected radially through the weld, shall not exceed 2% of the projected weld area in the radiograph. The area shall be the length the weld affected by the porosity (with a minimum length of 150 mm ) times the maximum width of the weld."
The above criterion is in order to evaluate porosity on a radiograph. Now I've a uniformly distributed porosity of upto 45 mm length. The total length of the weld is 192 cm (24 inch OD, pipeline )Can anybody explain me (if possible with a sketch) how to proceed with it?
Replies are highly appreciated.
Nandesh Kumar
Nigel Armstrong
Engineering, - Specialist services
United Kingdom, Joined Oct 2000, 1096
Nigel Armstrong
Engineering, - Specialist services
United Kingdom,
Joined Oct 2000
1096
15:09 Mar-01-2010
Re: Can anybody explain me for ISO 13847 ?
In Reply to Nandesh Kumar A at 14:21 Feb-28-2010 (Opening).
Nandesh
If the weld length affected by porosity is 45mm, this is less than 150mm and so you must use the default 150mm figure. Multiply this by the maximum width of the weld in the affected region and then multiply by 0,02 (2%). Say the weld width is 25mm: 150mm x 25mm x 0,02 = 75 sq mm. Measure and add together all pore diameters in the affected region, multiply the total by pi. If the calcualted figure is less than your previous calculation (75 sq mm in my example) then the area is accpetable. Greater pore area sum total = rejectable.
Hope this helps
Nandesh
Nandesh
05:33 Mar-02-2010
Re: Can anybody explain me for ISO 13847 ?
In Reply to Nigel Armstrong at 15:09 Mar-01-2010 .
Thanks Nigel...but can ypu please explain me whats the terms " projected radially" and " projected weld area " in my post?
Regards,
Nigel Armstrong
Engineering, - Specialist services
United Kingdom, Joined Oct 2000, 1096
Nigel Armstrong
Engineering, - Specialist services
United Kingdom,
Joined Oct 2000
1096
06:34 Mar-02-2010
Re: Can anybody explain me for ISO 13847 ?
In Reply to Nandesh at 05:33 Mar-02-2010 .
Nandesh
The use of "projected" radially and weld area is to maintain proportionality with regard to image magnification. If the set-up used has a low ratio between source-to-object distance and material thickness, differential magnification will enlarge a source-side pore more than a film-side pore.
Assuming you used a panoramic exposure on a 24" sch 40 pipe (wt = 0,375") then differential magnification is not a problem you need worry about.
Hope that helps
Nigel
13:08 Feb-08-2014
Re: Can anybody explain me for ISO 13847 ?
In Reply to Nigel Armstrong at 15:09 Mar-01-2010 .
"The distribution of scattered porosity is such that when projected radially through the weld the total area of porosity exceeds 1% of any 600mm Sq of projected weld metal."
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this is debug window | 1,243 | 5,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-34 | latest | en | 0.862143 |
https://netlib.org/lapack/explore-html-3.6.1/dc/dd5/clapmr_8f_abb2a59d3f85587bafafd9d17c6374c0a.html | 1,669,820,205,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710764.12/warc/CC-MAIN-20221130124353-20221130154353-00021.warc.gz | 471,790,149 | 4,761 | LAPACK 3.6.1 LAPACK: Linear Algebra PACKage
subroutine clapmr ( logical FORWRD, integer M, integer N, complex, dimension( ldx, * ) X, integer LDX, integer, dimension( * ) K )
CLAPMR rearranges rows of a matrix as specified by a permutation vector.
Purpose:
``` CLAPMR rearranges the rows of the M by N matrix X as specified
by the permutation K(1),K(2),...,K(M) of the integers 1,...,M.
If FORWRD = .TRUE., forward permutation:
X(K(I),*) is moved X(I,*) for I = 1,2,...,M.
If FORWRD = .FALSE., backward permutation:
X(I,*) is moved to X(K(I),*) for I = 1,2,...,M.```
Parameters
[in] FORWRD ``` FORWRD is LOGICAL = .TRUE., forward permutation = .FALSE., backward permutation``` [in] M ``` M is INTEGER The number of rows of the matrix X. M >= 0.``` [in] N ``` N is INTEGER The number of columns of the matrix X. N >= 0.``` [in,out] X ``` X is COMPLEX array, dimension (LDX,N) On entry, the M by N matrix X. On exit, X contains the permuted matrix X.``` [in] LDX ``` LDX is INTEGER The leading dimension of the array X, LDX >= MAX(1,M).``` [in,out] K ``` K is INTEGER array, dimension (M) On entry, K contains the permutation vector. K is used as internal workspace, but reset to its original value on output.```
Date
September 2012
Definition at line 106 of file clapmr.f.
106 *
107 * -- LAPACK auxiliary routine (version 3.4.2) --
108 * -- LAPACK is a software package provided by Univ. of Tennessee, --
109 * -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
110 * September 2012
111 *
112 * .. Scalar Arguments ..
113 LOGICAL forwrd
114 INTEGER ldx, m, n
115 * ..
116 * .. Array Arguments ..
117 INTEGER k( * )
118 COMPLEX x( ldx, * )
119 * ..
120 *
121 * =====================================================================
122 *
123 * .. Local Scalars ..
124 INTEGER i, in, j, jj
125 COMPLEX temp
126 * ..
127 * .. Executable Statements ..
128 *
129 IF( m.LE.1 )
130 \$ RETURN
131 *
132 DO 10 i = 1, m
133 k( i ) = -k( i )
134 10 CONTINUE
135 *
136 IF( forwrd ) THEN
137 *
138 * Forward permutation
139 *
140 DO 50 i = 1, m
141 *
142 IF( k( i ).GT.0 )
143 \$ GO TO 40
144 *
145 j = i
146 k( j ) = -k( j )
147 in = k( j )
148 *
149 20 CONTINUE
150 IF( k( in ).GT.0 )
151 \$ GO TO 40
152 *
153 DO 30 jj = 1, n
154 temp = x( j, jj )
155 x( j, jj ) = x( in, jj )
156 x( in, jj ) = temp
157 30 CONTINUE
158 *
159 k( in ) = -k( in )
160 j = in
161 in = k( in )
162 GO TO 20
163 *
164 40 CONTINUE
165 *
166 50 CONTINUE
167 *
168 ELSE
169 *
170 * Backward permutation
171 *
172 DO 90 i = 1, m
173 *
174 IF( k( i ).GT.0 )
175 \$ GO TO 80
176 *
177 k( i ) = -k( i )
178 j = k( i )
179 60 CONTINUE
180 IF( j.EQ.i )
181 \$ GO TO 80
182 *
183 DO 70 jj = 1, n
184 temp = x( i, jj )
185 x( i, jj ) = x( j, jj )
186 x( j, jj ) = temp
187 70 CONTINUE
188 *
189 k( j ) = -k( j )
190 j = k( j )
191 GO TO 60
192 *
193 80 CONTINUE
194 *
195 90 CONTINUE
196 *
197 END IF
198 *
199 RETURN
200 *
201 * End of ZLAPMT
202 *
Here is the caller graph for this function: | 1,098 | 3,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-49 | latest | en | 0.574353 |
https://www.folkstalk.com/tech/pandas-create-new-column-conditional-on-other-columns-with-code-examples/ | 1,713,792,814,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00857.warc.gz | 686,466,789 | 6,433 | # Pandas Create New Column Conditional On Other Columns With Code Examples
Pandas Create New Column Conditional On Other Columns With Code Examples
With this article, we'll look at some examples of Pandas Create New Column Conditional On Other Columns problems in programming.
```df['color'] = ['red' if x == 'Z' else 'green' for x in df['Set']]
```
There are a variety of approaches that can be taken to solve the same problem Pandas Create New Column Conditional On Other Columns. The remaining solutions are discussed further down.
```# For creating new column with multiple conditions
conditions = [
(df['Base Column 1'] == 'A') & (df['Base Column 2'] == 'B'),
(df['Base Column 3'] == 'C')]
choices = ['Conditional Value 1', 'Conditional Value 2']
df['New Column'] = np.select(conditions, choices, default='Conditional Value 1')
```
```# np.where(condition, value if condition is true, value if condition is false)
df['hasimage'] = np.where(df['photos']!= '[]', True, False)
```
```conditions = [
df['gender'].eq('male') & df['pet1'].eq(df['pet2']),
df['gender'].eq('female') & df['pet1'].isin(['cat', 'dog'])
]
choices = [5,5]
df['points'] = np.select(conditions, choices, default=0)
print(df)
gender pet1 pet2 points
0 male dog dog 5
1 male cat cat 5
2 male dog cat 0
3 female cat squirrel 5
4 female dog dog 5
5 female squirrel cat 0
6 squirrel dog cat 0
```
As we've seen, a lot of examples were used to address the Pandas Create New Column Conditional On Other Columns problem.
## How do I create a new column based on another column condition in pandas?
• Step 1 - Import the library. import pandas as pd import numpy as np.
• Step 2 - Creating a sample Dataset. Here we have created a Dataframe with columns 'bond_name' and 'risk_score'.
• Step 3 - Creating a function to assign values in column.
• Step 5 - Converting list into column of dataset and viewing the final dataset.
## How will you create a new column whose value is calculated from two other columns?
Create a new column by assigning the output to the DataFrame with a new column name in between the [] . Operations are element-wise, no need to loop over rows. Use rename with a dictionary or function to rename row labels or column names.
## How do you create a new column by dividing two columns in pandas?
Method 2: Pandas divide two columns using div() function It divides the columns elementwise. It accepts a scalar value, series, or dataframe as an argument for dividing with the axis. If the axis is 0 the division is done row-wise and if the axis is 1 then division is done column-wise.
## How do you create a conditional column in Python?
You can create a conditional DataFrame column by checking multiple columns using numpy. select() function. The select() function is more capable than the previous methods. We can use it to give a set of conditions and a set of values.09-Dec-2021
## How do I get the value of a column based on another column value?
You can extract a column of pandas DataFrame based on another value by using the DataFrame. query() method. The query() is used to query the columns of a DataFrame with a boolean expression. The blow example returns a Courses column where the Fee column value matches with 25000.24-May-2022
## What is the difference between LOC () and ILOC ()?
The main distinction between loc and iloc is: loc is label-based, which means that you have to specify rows and columns based on their row and column labels. iloc is integer position-based, so you have to specify rows and columns by their integer position values (0-based integer position).17-Mar-2021
## Can you use a calculated field in another calculated field?
Unfortunately you cannot refer to calculated fields when creating a new calculated field.21-Jul-2016
## How do you create a new column in pandas and assign a value?
You can use the assign() function to add a new column to the end of a pandas DataFrame: df = df. assign(col_name=[value1, value2, value3, ])01-Jun-2021
## How do you add a conditional column?
Add a conditional column (Power Query)
• To open a query, locate one previously loaded from the Power Query Editor, select a cell in the data, and then select Query > Edit.
• Select Add Column > Conditional Column.
• In the New column name box, enter a unique name for your new conditional column.
## How do you divide a DataFrame based on a condition?
In the above example, the data frame 'df' is split into 2 parts 'df1' and 'df2' on the basis of values of column 'Weight'. Method 2: Using Dataframe. groupby(). This method is used to split the data into groups based on some criteria.20-Apr-2022 | 1,124 | 4,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.77484 |
https://www.wyzant.com/resources/answers/394055/is_final_velocity_a_vector_or_scalar | 1,537,909,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162385.84/warc/CC-MAIN-20180925202648-20180925223048-00305.warc.gz | 950,548,761 | 16,039 | 0
# Is final velocity a vector or scalar?
A ball starts from rest at point A and rolls .9 meters east to point B with an average velocity of 2 m/s. What was the velocity of the ball at B?
Given:
Vi=0 m/s
d (displacement)= .9m east
average velocity= 2 m/s
Vf= ???
average velocity = Vf + Vi divided by 2
is the answer 4 m/s or 4 m/s East?
Does the answer for final velocity include direction or no?
### 1 Answer by Expert Tutors
Arturo O. | Experienced Physics Teacher for Physics TutoringExperienced Physics Teacher for Physics ...
5.0 5.0 (66 lesson ratings) (66)
0
vi = 0 (started from rest)
<v> = average velocity = 2 m/s (given)
<v> = (vi + vf)/2 ⇒
vf = 2<v> - vi = 2(2 m/s) - 0 = 4 m/s
Velocity is a vector, so the complete answer must have both a magnitude and direction. Therefore
vf = 4 m/s east
or
vf = (4 m/s)i (if you are familiar with the i, jk notation). | 279 | 884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-39 | longest | en | 0.830178 |
http://math.stackexchange.com/questions/257904/holder-continuity-of-fractional-brownian-motion | 1,397,770,582,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00207-ip-10-147-4-33.ec2.internal.warc.gz | 147,758,172 | 14,934 | # Hölder Continuity of Fractional Brownian Motion
I would like to prove the following theorem:
Let $H\in (0,1)$. The fractional Brownian motion $B_H$ admits a version whose sample paths are $a.s.$ Hölder continuous of order strict less than $H$.
For any $\alpha >0$ we have by self similarity $$\mathbb{E}[|B_H(t)-B_H(s)|^{\alpha}]=\mathbb{E}[|B_H(1)|^{\alpha}]|t-s|^{\alpha H}.$$
The proof is done after applying the criterion of Kolmogorov which says:
A process $(X_t)_{t\in\mathbb{R}}$ admits a continuous modification if there exist constants $a,b,k>0$ such that $$\mathbb{E}[|X(t)-X(s)|^a]\leq k|t-s|^{1+b}$$ for all $s,t\in\mathbb{R}$.
But I don't know how to apply this criterion. Any help please.
Edit:
Well, maybe I should try to state my problem more precisely. I just would like to understand the proof. If I use the criterion of Kolmogorov it should hold $$\mathbb{E}[|B_H(t)-B_H(s)|^{\alpha}]=\mathbb{E}[|B_H(1)|^{\alpha}]|t-s|^{\alpha H}\leq k|t-s|^{1+\beta}$$ for $\alpha,\beta,k>0$, right? I don't see any relationship to the Hölder continuity.
Is there nobody who can demonstrate the proof for me to understand?
Maybe I should set $k=\mathbb{E}[|B_H(1)|^{\alpha}]$ and $\beta=\alpha H$ and say that $B_H$ is $\gamma$-Hölder continuous for every $\gamma\in\big[0,{\alpha H\over \alpha}\big)$?
-
In fact, there Kolmogorov criterion in your formulation can not help apriori. Use this one galton.uchicago.edu/~lalley/Courses/385/GaussianProcesses.pdf on the page six – Tarasenya Dec 17 '12 at 22:17
or for instance, here statslab.cam.ac.uk/~beresty/teach/StoCal/sc3.pdf page 10. Good lecture notes, by the way. – Tarasenya Dec 17 '12 at 22:22
add comment | 548 | 1,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2014-15 | latest | en | 0.778183 |
https://jsur.in/posts/2022-09-25-ductf-2022-kyber-writeup/ | 1,709,182,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00504.warc.gz | 324,977,033 | 135,294 | DownUnderCTF 2022 - kyber±
kyber± is a cryptography challenge written for DUCTF 2022. The challenge revolves around Kyber and its reference implementation. A (crypto) bug is artificially introduced through a patch given in my.patch. Before we even look any further into the challenge, let's review Kyber.
The challenge source and solution files can be found here.
Kyber
Kyber is a lattice-based key encapsulation mechanism (KEM) which has been selected for standardisation in the NIST PQC competition. Its security is based on the LWE problem in module lattices. Although we give a brief overview in this section, the written specification is an excellent resource to learn more about Kyber.
Overview
It is important to note that Kyber itself is a KEM. It is built on top of a IND-CPA-secure public-key encryption scheme which is introduced in the specification as $\small{\text{KYBER}}.\normalsize{\textsf{CPAPKE}}$. As is common with lattice-based KEMs, the IND-CCA2-secure KEM is obtained by applying the Fujisaki-Okamoto transform (or some variant of it) to the IND-CPA-secure PKE.
At its core, all Kyber really does is just a bunch of polynomial multiplications. We work in the ring $R_q = \mathbb{Z}_q[X]/(X^n + 1)$ where $n$ and $q$ are fixed to $n = 256, q = 3329$ for all parameter sets. These parameters were chosen specifically because they lend themselves well to implementing very efficient multiplications in $R_q$. The design also makes implementation easier as the main security parameter $k$ (which is the dimension of the lattice or size of the module elements) can be adjusted without any changes in the code. For the parameter set used in the challenge ($\small{\text{KYBER}}512$), $k$ is set to $2$.
We work with three main mathematical objects: polynomials (elements of $R_q$), vectors of polynomials, and (square) matrices of polynomials. We will denote polynomials by lower-case letters, vectors by bold lower-case letters, and matrices by bold upper-case letters.
How to randomly sample these objects is also an important part of the algorithm. We will avoid the specifics and just note that "small" polynomials or vectors (such as noise) have their coefficients sampled from a centered binomial distribution with parameter of either $\eta_1$ or $\eta_2$. In $\small{\text{KYBER}}512$, we have $\eta_1 = 3$ and $\eta_2 = 2$. In all other parameter sets, $\eta_1 = \eta_2 = 2$. These parameters were selected to balance between security, ciphertext size, and decryption failure probability and are not really important for this challenge. The main takeaway is that polynomials whose coefficients are sampled from a $\mathsf{CBD}_\eta$ have (integer) coefficients in $[-\eta, \eta]$. Since coefficients can have magnitude up to $q/2$, the coefficients of polynomials sampled from these distributions can be considered "small".
We'll first give a very high level overview of the Kyber algorithms to understand the math behind it and then introduce some other important details which mostly pertain to optimistaion and implementation later.
$\small{\text{KYBER}}.\normalsize{\textsf{CPAPKE}}$
We only give brief outlines here and refer to the specification for the full details of the algorithms.
$\textsf{KeyGen}()$
In the Kyber PKE, the secret key $\mathbf{s} \in R_q^k$ is a vector consisting of $k$ polynomials from $R_q$ with small coefficients sampled from $\mathsf{CBD}_{\eta_1}$. The public key is $(\mathbf{t}, \mathbf{A})$ where $\mathbf{A} \in R_q^{k \times k}$ is a randomly sampled matrix and $\mathbf{t} \in R_q^k$ is computed as $\mathbf t = \mathbf{A} \mathbf{s} + \mathbf{e}$ where $\mathbf{e} \in R_q^k$ is a small error term with coefficients sampled from $\mathsf{CBD}_{\eta_1}$ (just like $\mathbf{s}$).
$\textsf{Enc}(pk, m, r)$
To encrypt a 32 byte message $m$, we use the public key $pk = (\mathbf{t}, \mathbf{A})$ and a randomness seed $r$. Using $r$ as a seed, a vector $\mathbf{r} \in R_q^k$ is sampled from $\mathsf{CBD}_{\eta_1}$, a noise vector $\mathbf{e}_1 \in R_q^k$ is sampled from $\mathsf{CBD}_{\eta_2}$ and a noise polynomial $e_2 \in R_q$ is sampled from $\mathsf{CBD}_{\eta_2}$. The ciphertext $(\mathbf{u}, v)$ is then computed as:
\begin{aligned} \mathbf{u} &= \mathbf{A}^T \mathbf{r} + \mathbf{e_1} \\ v &= \mathbf{t} \cdot \mathbf{r} + e_2 + \mu \end{aligned}
Here, $\mu$ is the polynomial obtained by using the 256 bits of the message $m$ as coefficients and multiplying by $(q+1)/2$.
$\textsf{Dec}(sk, c)$
Decryption is simple. Given the secret key $\mathbf{s} = sk$ and ciphertext $(\mathbf{u}, v) = c$, the message polynomial is recovered by computing
$\mu \approx v - \mathbf{s} \cdot \mathbf{u}$
To recover the actual message $m$, we look at the coefficients of $\mu$; if the coefficient is close to $0$, then the bit is a $0$, and if the coefficient is closer to $(q+1)/2$ then the bit is a $1$.
Correctness
By looking at the encryption and decryption equations, we can gain an intuition for the correctness of the scheme. Note that we also write $\mathbf{x} \cdot \mathbf{y} = \mathbf{x}^T \mathbf{y}$.
\begin{aligned} v - \mathbf{s} \cdot \mathbf{u} &= \mathbf{t} \cdot \mathbf{r} + e_2 + \mu - \mathbf{s} \cdot (\mathbf{A}^T \mathbf{r} + \mathbf{e}_1) \\ &= \mathbf{t} \cdot \mathbf{r} + e_2 + \mu - \mathbf{s}^T \mathbf{A}^T \mathbf{r} - \mathbf{s} \cdot \mathbf{e}_1 \\ &= \mathbf{t} \cdot \mathbf{r} + e_2 + \mu - (\mathbf{A} \mathbf{s}) \cdot \mathbf{r} - \mathbf{s} \cdot \mathbf{e}_1 \\ &= (\mathbf{t} - \mathbf{A} \mathbf{s}) \cdot \mathbf{r} + e_2 + \mu - \mathbf{s} \cdot \mathbf{e}_1 \\ &= \mathbf{e} \cdot \mathbf{r} + e_2 + \mu - \mathbf{s} \cdot \mathbf{e}_1 \\ &= \underbrace{\mathbf{e} \cdot \mathbf{r} + e_2 - \mathbf{s} \cdot \mathbf{e}_1}_{\text{small}} + \mu \\ \end{aligned}
The combined error term $\mathbf{e} \cdot \mathbf{r} + e_2 - \mathbf{s} \cdot \mathbf{e}_1$ contains only "small" polynomials, so the result is quite small too. It follows that the approximation holds and $m$ can be decoded correctly with quite high probability. The specification goes into more detail and includes calculations for the decryption failure probability in each parameter set.
$\small{\text{KYBER}}.\normalsize{\textsf{CCAKEM}}$
The KEM involves three algorithms: KeyGen, Encapsulate, Decapsulate. All of these algorithms make use of symmetric primitives. Of interest are two hash functions $H$ (256 bits) and $G$ (512 bits), and a key-derivation function $\mathsf{KDF}$. In Kyber (not the "90s" variant), these are instantiated as follows:
Primitive Instantiation
$H$ $\text{SHA3-256}$
$G$ $\text{SHA3-512}$
$\mathsf{KDF}$ $\text{SHAKE-256}$
Since we will be hashing abstract math objects, there is a need to define a way to encode (and decode) such objects as bytes. For now, we'll ignore this detail and focus on the main ideas behind the KEM algorithms.
$\textsf{KeyGen}()$
The KEM key generation involves generating a PKE key pair $(pk, sk')$. The public key is just $pk$, and the secret key is $sk = (sk'||pk||H(pk)||z)$ where $z$ is a pseudorandom 32 byte value (which is used during decapsulation). The hash of the public key is included in the secret key to speed up decapsulation.
$\textsf{Enc}(pk)$
The goal of encapsulation is to generate a symmetric key material as well as a ciphertext which can be decapsulated using the secret key. A 32 byte value $m$ is randomly generated (the specification actually takes $m$ to be the hash of a 32 byte random value to avoid using the output of the system RNG), then a pre-key $\bar{K}$ and coins $r$ are computed as $(\bar{K}, r) = G(m||H(pk))$. The ciphertext $c$ is obtained by encrypting $m$ with $pk$ and randomness $r$. Finally, the shared secret $K$ is obtained by computing $K = \mathsf{KDF}(\bar{K}||H(c))$. The output is $(c, K)$.
$\textsf{Dec}(c, sk)$
Decapsulation allows one with the secret key to recover the shared secret from an encapsulation ciphertext. The ciphertext is decrypted using the PKE to obtain $m'$. Then, a candidate pre-key and coins are computed as $(\bar{K}', r') = G(m'||H(pk))$. The decrypted message $m'$ is then re-encrypted with randomness $r'$ to obtain $c'$. If the two ciphertexts match, the shared secret is computed as $K = \mathsf{KDF}(\bar{K}'||H(c))$. Otherwise, it is computed as $K = \mathsf{KDF}(z||H(c))$. In the latter case, the decapsulation fails and the shared secret is meaningless.
Implementation Details
While mostly accurate and hopefully easy to understand, our previous descriptions miss a few important details. This section will outline the main implementation details relevant in the challenge.
Serialisation
Serialisation (and deserialisation) is the process of converting objects to and from bytes so that they can be stored, hashed, transmitted, etc. In Kyber, there is really only one object which needs this: elements of $R_q$. Since each coefficient is smaller than $q = 3329$, they can each be represented with $12$ bits. Therefore, to encode a polynomial $f \in R_q$, we simply write the $12$ bits of each coefficients (starting from the constant coefficient) and convert the bit string to bytes. It takes $\frac{12 \cdot 256}{8}$ bytes to encode an element of $R_q$.
Compression
The Kyber specification defines compression and decompression functions for $x \in \mathbb{Z}_q$ and $d < \lceil \log_2(q) \rceil$ as follows:
\begin{aligned} \mathsf{Compress}_q(x, d) &= \lceil (2^d/q) \cdot x \rfloor \mod 2^d \\ \mathsf{Decompress}_q(x,d) &= \lceil (q/2^d) \cdot x \rfloor \end{aligned}
These functions do as they sound, and more specifically, if
$x' = \mathsf{Decompress}_q(\mathsf{Compress}(x, d), d)$
then
$|x' - x \mod q| \leq \left \lceil \frac{q}{2^{d+1}} \right \rfloor$
Note that the compression and decompression functions are actually used with $d = 1$ to implement encoding and decoding the message polynomial.
Their main use, however, is to reduce the ciphertext size. Compression parameters $d_u$ and $d_v$ are specified for each parameter set, and for $\small{\text{KYBER}}512$, they are $d_u = 10, d_v = 4$. This means that each coefficient of the polynomials in $\mathbf{u}$ are compressed to $10$ bits, and each coefficient in $v$ is compressed to $4$ bits. So the size of the compressed $\mathbf{u}$ vector is $\frac{2 \cdot 10 \cdot 256}{8} = 640$ bytes and the size of the compressed $v$ polynomial is $\frac{4 \cdot 256}{8} = 128$ bytes.
In fact, compressing the ciphertext seems to add some security as well. In the case of LWE, a random small error is added and in this case we effectively have extra (deterministic) noise by dropping some bits. This also has a name, and it's called "Learning with Rounding" (LWR).
NTT
One of Kyber's (and other lattice-based schemes) main attractions is its fast speeds. Aside from the use of symmetric primitives, polynomial multiplication is the most time-consuming operation in the scheme. Fortunately, it can be implemented very efficiently using the so-called number-theoretic transform (NTT). Although it isn't necessary to fully understand NTT and the associated algorithms for this challenge, we need to at least be aware of its existence.
To grossly simplify, we can multiply two polynomials $f, g \in R_q$ by transforming them into elements of the NTT domain where multiplication is cheaper, and then transforming them back into their normal form after the cheap multiplication. That is, we have two functions $\mathsf{NTT} : R_q \rightarrow R_q$ and $\mathsf{invNTT} : R_q \rightarrow R_q$ and we perform polynomial multiplication by computing $f \cdot g = \mathsf{invNTT}(\mathsf{NTT}(f) \circ \mathsf{NTT}(g))$ where $\circ$ denotes multiplication in the NTT domain.
If we simply reuse the library code for the challenge, we don't need to worry about implementing this. But the main idea is that $q - 1 = 2^8 \cdot 13$ and $X^{256} + 1$ factors into $128$ degree $2$ polynomials, so we can consider $f$ and $g$ each as vectors of $128$ degree $1$ polynomials (i.e. modulo each of the degree $2$ factors of $X^{256} + 1$) and multiply them componentwise. The actual algorithm itself is essentially the Cooley-Tukey FFT algorithm with minor modifications.
Challenge Overview
With preliminaries out of the way, we are finally ready to take a look at the challenge itself. We are given a kyber.py file which implements the server, a libpqcrystals_kyber512_ref.so library which is built from the Kyber reference implementation, my.patch which outlines the changes made to the reference implementation for the challenge, and a build-kyber.sh script which automates pulling and building the libpqcrystals_kyber512_ref.so file for the player's convenience (this also indicates that the Kyber reference implementation is being used as well as the specific commit).
kyber.py
#!/usr/bin/env python3
import ctypes
MAX_QUERIES = 7681
kyber_lib = ctypes.CDLL('./libpqcrystals_kyber512_ref.so')
class Kyber:
def __init__(self):
self.pk_buf = ctypes.c_buffer(800)
self.sk_buf = ctypes.c_buffer(1632)
kyber_lib.pqcrystals_kyber512_ref_keypair(self.pk_buf, self.sk_buf)
def kem_enc(self):
ct_buf = ctypes.c_buffer(1024)
ss_buf = ctypes.c_buffer(32)
kyber_lib.pqcrystals_kyber512_ref_enc(ct_buf, ss_buf, self.pk_buf)
return bytes(ct_buf), bytes(ss_buf)
def kem_dec(self, c):
assert len(c) == 1024
ct_buf = ctypes.c_buffer(c)
ss_buf = ctypes.c_buffer(32)
kyber_lib.pqcrystals_kyber512_ref_dec(ss_buf, ct_buf, self.sk_buf)
return bytes(ss_buf)
def main():
kyber = Kyber()
print('pk:', bytes(kyber.pk_buf).hex())
print('H(pk):', bytes(kyber.sk_buf)[-64:].hex())
for _ in range(MAX_QUERIES):
try:
inp = input('> ')
if inp.startswith('enc'):
ct, ss = kyber.kem_enc()
print('ct:', ct.hex())
print('ss:', ss.hex())
elif inp.startswith('dec '):
ct = bytes.fromhex(inp[4:])
ss = kyber.kem_dec(ct)
print('ss:', ss.hex())
else:
break
except:
print('>:(')
exit(1)
enc = bytes([a ^ b for a, b in zip(FLAG, bytes(kyber.sk_buf))])
print('flag_enc:', enc.hex())
if __name__ == '__main__':
main()
The server code itself is quite short and seems to not do much other than interface with the C Kyber library. It generates a Kyber key pair and gives us the public key as well as the hash of the public key. We then have access to 7681 queries where we can call either the KEM's encapsulation function or the decapsulation function with any ciphertext. After the queries, we are given the flag XORed with the secret key (its bytes serialisation).
my.patch
diff --git a/ref/indcpa.c b/ref/indcpa.c
index 60f4059..f822b0d 100644
--- a/ref/indcpa.c
+++ b/ref/indcpa.c
@@ -89,7 +89,7 @@ static void unpack_sk(polyvec *sk, const uint8_t packedsk[KYBER_INDCPA_SECRETKEY
static void pack_ciphertext(uint8_t r[KYBER_INDCPA_BYTES], polyvec *b, poly *v)
{
polyvec_compress(r, b);
- poly_compress(r+KYBER_POLYVECCOMPRESSEDBYTES, v);
+ poly_tobytes(r+KYBER_POLYVECCOMPRESSEDBYTES, v);
}
/*************************************************
@@ -105,7 +105,7 @@ static void pack_ciphertext(uint8_t r[KYBER_INDCPA_BYTES], polyvec *b, poly *v)
static void unpack_ciphertext(polyvec *b, poly *v, const uint8_t c[KYBER_INDCPA_BYTES])
{
polyvec_decompress(b, c);
- poly_decompress(v, c+KYBER_POLYVECCOMPRESSEDBYTES);
+ poly_frombytes(v, c+KYBER_POLYVECCOMPRESSEDBYTES);
}
/*************************************************
diff --git a/ref/params.h b/ref/params.h
index 3d02a0f..b0d929c 100644
--- a/ref/params.h
+++ b/ref/params.h
@@ -58,7 +58,7 @@
#define KYBER_INDCPA_MSGBYTES (KYBER_SYMBYTES)
#define KYBER_INDCPA_PUBLICKEYBYTES (KYBER_POLYVECBYTES + KYBER_SYMBYTES)
#define KYBER_INDCPA_SECRETKEYBYTES (KYBER_POLYVECBYTES)
-#define KYBER_INDCPA_BYTES (KYBER_POLYVECCOMPRESSEDBYTES + KYBER_POLYCOMPRESSEDBYTES)
+#define KYBER_INDCPA_BYTES (KYBER_POLYVECCOMPRESSEDBYTES + KYBER_POLYBYTES)
#define KYBER_PUBLICKEYBYTES (KYBER_INDCPA_PUBLICKEYBYTES)
/* 32 bytes of additional space to save H(pk) */
diff --git a/ref/verify.c b/ref/verify.c
index ed4a654..1e88e16 100644
--- a/ref/verify.c
+++ b/ref/verify.c
@@ -19,9 +19,9 @@ int verify(const uint8_t *a, const uint8_t *b, size_t len)
uint8_t r = 0;
for(i=0;i<len;i++)
- r |= a[i] ^ b[i];
+ r = r == 0xff ? r : r + (a[i] != b[i]);
- return (-(uint64_t)r) >> 63;
+ return r;
}
/*************************************************
@@ -41,7 +41,6 @@ void cmov(uint8_t *r, const uint8_t *x, size_t len, uint8_t b)
{
size_t i;
- b = -b;
for(i=0;i<len;i++)
r[i] ^= b & (r[i] ^ x[i]);
}
The patch consists of two main changes:
1. The ciphertext polynomial $v$ no longer uses compression. Before, it would be compressed to 4 bits and the ciphertext would have size 768 bytes, now the ciphertext size is 1024 bytes.
2. The verify function is slightly changed to return values other than 0 or 1. This function is used for the equality check during decapsulation. The cmov function is changed to not negate the b parameter. This function is also used during decapsulation. verify and cmov are both intended to be constant time as not to leak any information about the equality of the two ciphertexts.
Solution
With basics done and an understanding of the challenge, we can finally think about solving it. We start by looking for bugs, and the patch stands out the most so let's analyse it further.
Bugs
We saw that the patch introduces two major changes to the KEM. The first one simply removes some compression and while that does lower security a bit (since we lose the deterministic noise which acts as an error term), its main purpose was really just for making the ciphertext size smaller (at least the specification argues for Kyber's security without relying on this rounding too much). So we turn to the second change, which seems a lot more dangerous.
We want to understand what weaknesses the changes to verify and cmov introduce, so it's helpful to see how they are used for context. Their only usage is in the KEM decapsulation function (in kem.c):
int crypto_kem_dec(uint8_t *ss,
const uint8_t *ct,
const uint8_t *sk)
{
size_t i;
int fail;
uint8_t buf[2*KYBER_SYMBYTES];
/* Will contain key, coins */
uint8_t kr[2*KYBER_SYMBYTES];
uint8_t cmp[KYBER_CIPHERTEXTBYTES];
const uint8_t *pk = sk+KYBER_INDCPA_SECRETKEYBYTES;
indcpa_dec(buf, ct, sk);
/* Multitarget countermeasure for coins + contributory KEM */
for(i=0;i<KYBER_SYMBYTES;i++)
buf[KYBER_SYMBYTES+i] = sk[KYBER_SECRETKEYBYTES-2*KYBER_SYMBYTES+i];
hash_g(kr, buf, 2*KYBER_SYMBYTES);
/* coins are in kr+KYBER_SYMBYTES */
indcpa_enc(cmp, buf, pk, kr+KYBER_SYMBYTES);
fail = verify(ct, cmp, KYBER_CIPHERTEXTBYTES);
/* overwrite coins in kr with H(c) */
hash_h(kr+KYBER_SYMBYTES, ct, KYBER_CIPHERTEXTBYTES);
/* Overwrite pre-k with z on re-encryption failure */
cmov(kr, sk+KYBER_SECRETKEYBYTES-KYBER_SYMBYTES, KYBER_SYMBYTES, fail);
/* hash concatenation of pre-k and H(c) to k */
kdf(ss, kr, 2*KYBER_SYMBYTES);
return 0;
}
Just as in the algorithm description, we see that verify is used to check the equality of the given ciphertext ct and the re-encrypted ciphertext cmp. Before the patch, if these two values were equal then verify would return 0, otherwise it would return 1. The result is put into the fail variable which is passed as the last argument (the b parameter) to the cmov function. Without the patch applied, we see that if this argument is 0, then the cmov function essentially does nothing. Otherwise, if the argument is 1, then because of the b = -b line, and the fact that b is an unsigned 8 bit integer, we would have b = 0xff and the kr buffer would be overwritten by the $z$ value from the secret key.
The patch changes this behaviour in that it essentially makes verify return the number of byte inequalities there are between ct and cmp, with a max value of 0xff. The change in cmov makes it so that the overwriting just uses the value of fail directly as the & operand. At a higher level, we see that if the two ciphertexts are exactly equal, then fail will still be 0 and cmov will still do nothing. However, if fail any value greater than 0, then some bits of the pre-key value will be overwritten with $z$ according to the bit mask given by fail. Ultimately, this means that the behaviour is only the same in the case that the ciphertexts are exactly the same, or differ in at least 0xff positions. This differing behaviour is clearly a bug as it may allow us to distinguish between cases when the two ciphertexts are the same, only slightly different, or vastly different.
There is another bug, and it occurs in kyber.py. Because the server code is quite minimal and we are given a patch for the C library, it might seem innocent. But there is a very suspicious line:
print('H(pk):', bytes(kyber.sk_buf)[-64:].hex())
This gives us the last 64 bytes of the secret key serialisation, which if we look at the KeyGen function, is $H(pk)||z$. This means we have some extra information about the secret key (although not the secret vector itself). We'll see how to use this next.
Leaking Information
The encapsulation functionality of the server seems useless as we can simply perform that operation ourself using the public key. The goal is secret key recovery, so we need to somehow leak information from decapsulation queries. We are given the result of each decapsulation, but this is just a hash, so we don't necessarily have a decryption oracle. Thinking back to the patch and the fact that we have $z$ however, maybe we do have a way of learning some information about the decryption result...
The Oracle
The main idea behind our information leaking oracle is that we can distinguish between the case when a ciphertext and its re-encryption during decapsulation is slightly different, or completely different.
Suppose we run encapsulation (locally) to get a ciphertext $c = (\mathbf{u}, v)$ for a generated message $m$. If we pass $c$ to the decapsulation oracle, we expect it to return the correct shared secret; that is, we expect $c$ and the re-encryption to be exactly the same. What happens if we query the decapsulation oracle with $(\mathbf{u}, v+1)$? Because the decryption process can tolerate a bit of error, it is very likely that the decryption result will still be the original message $m$. Because the coins (the randomness used for encryption) depends on $m$, this means that the re-encryption result would be precisely $c$. Therefore, the verification will fail as $c \neq (\mathbf{u}, v+1)$. In fact, in this case we would only expect there to be an inequality in one byte of the two ciphertexts' serialisations.
Now suppose we instead query the decapsulation oracle with $c' = (\mathbf{u}, v+o)$ for some larger integer value $o$. It's entirely possible that decrypting this ciphertext yields a different message (i.e. one that differs by one bit, in this case the bit corresponding to the constant coefficient). If the decrypted message is even slightly different to the original message $m$, then the coins will be different, and so the re-encryption is also very likely to be completely different to $c$. In this case, fail is almost certainly going to be 0xff, and the resulting shared secret will be overwritten with $\mathsf{KDF}(z||H(c'))$. Since we know $z$ and $c'$, we can compute this value and hence determine when this occurs!
So in summary, our oracle allows us to check whether a given perturbed ciphertext decrypts to the same message as the original unperturbed ciphertext. We'll see how powerful this is a bit later.
Recovering $\mathbf{s}$
Before seeing some more details about how to use the oracle, we try and motivate it a bit. Suppose we have two message/ciphertext pairs $(m_1, (\mathbf{u}_1, v_1))$ and $(m_2, (\mathbf{u}_2, v_2))$. From the decryption equation, we have
\begin{aligned} v_1 - \mathbf{s} \cdot \mathbf{u}_1 &= \mathbf{e} \cdot \mathbf{r}_1 + e_{1,2} - \mathbf{s} \cdot \mathbf{e}_{1,1} + \mu_1 \\ v_2 - \mathbf{s} \cdot \mathbf{u}_2 &= \mathbf{e} \cdot \mathbf{r}_2 + e_{2,2} - \mathbf{s} \cdot \mathbf{e}_{2,1} + \mu_2 \\ \end{aligned}
Writing the combined error terms as $E_1$ and $E_2$:
\begin{aligned} v_1 - \mathbf{s} \cdot \mathbf{u}_1 &= E_1 + \mu_1 \\ v_2 - \mathbf{s} \cdot \mathbf{u}_2 &= E_2 + \mu_2 \\ \end{aligned}
Writing $y_i = v_i - E_i - \mu_i$:
\begin{aligned} \mathbf{s} \cdot \mathbf{u}_1 &= y_1 \\ \mathbf{s} \cdot \mathbf{u}_2 &= y_2 \\ \end{aligned}
Writing $\mathbf{s} = (s_0, s_1)$, $\mathbf{u}_1 = (u_{1,0}, u_{1,1})$, $\mathbf{u}_2 = (u_{2,0}, u_{2,1})$:
\begin{aligned} s_0 u_{1, 0} + s_1 u_{1, 1} &= y_1 \\ s_1 u_{2, 0} + s_1 u_{2, 1} &= y_2 \\ \end{aligned}
And so, solving for $s_0$ and $s_1$, we get
\begin{aligned} s_1 &= (y_2 - y_1 u_{1,0}^{-1} u_{2,0})(u_{2,1} - u_{1,0}^{-1} u_{1,1} u_{2,0})^{-1} \\ s_0 &= (y_1 - s_1 u_{1,1}) u_{1,0}^{-1} \end{aligned}
But of course, to actually compute this, we need to know $E_1$ and $E_2$, which depend on $\mathbf{s}$! At least now we know that if we can recover these small error terms, we can recover the secret key. So let's see how to do that using the oracle.
Recovering the Error
Suppose we run the encapsulation function locally to obtain a ciphertext $c = (\mathbf{u}, v)$ for a random message $m$. From the decryption equation, we have
\begin{aligned} v - \mathbf{s} \cdot \mathbf{u} &= \mathbf{e} \cdot \mathbf{r} + e_2 - \mathbf{s} \cdot \mathbf{e}_1 + \mu \\ &= E + \mu \end{aligned}
Our goal, as established in the previous section, is to recover $E$. The ciphertext polynomial $v$ sits by itself as a term, so any changes to coefficients of $v$ is reflected in the right hand side of this equation. Note that this is possible because the compression on $v$ was removed; otherwise we would lose fine-grained control and the attack may not work.
The coefficients of the message polynomial $\mu$ are either $0$ or $1665$ and the message itself is obtained by running the $\mathsf{Compress}_q$ function on the noisy message polynomial with $d = 1$. Note that
$\mathsf{Compress}_q(x, 1) = \begin{cases} \ 1 \qquad \text{for}\ 833 \leq x \leq 2496, \\ \ 0 \qquad \text{otherwise} \end{cases}$
The idea is to try different integer values of $o$ such that the decryption of $(\mathbf{u}, v + X^i o_i)$ results in a different message to $m$. When this is the case, we know that the combined error from the corresponding $X^i$ coefficient in $E$ (call it $E_i$) and $o_i$ is enough to cause that message bit to flip.
More concretely, consider the case where the $i$th message bit $b_i$ is $0$. If we find an $o_i$ such that $(\mathbf{u}, v + X^i o_i)$ decrypts to a different message to $m$, but $(\mathbf{u}, v + X^i (o_i - 1))$ does decrypt to $m$, then we know that $E_i + o_i = 833$. Therefore, we recover $E_i = 833 - o_i$.
On the other hand, if $b_i$ is $1$ and we find $o_i$ which satisfies this property, then we know that $E_i + o_i + 1665 = 2497$. Therefore, we recover $E_i = 2496 - 1665 - o_i = 832 - o_i$.
We can neatly summarise both cases in the single equation:
$E_i = 833 - b_i - o_i$
To recover the error polynomial, we simply run this search for each of the $256$ coefficients!
We must also consider the query complexity since we have a limited number of queries. Experiments show that the coefficients in the combined error polynomial can range from as small as $0$ to as large as $150$ or greater in magnitude. Even with a liberal estimate of $50$ queries per coefficient, a naive linear search still requires around $256 \cdot 50 = 12800$ queries. It is necessary to speed this up by implementing the search as a binary search instead. With a binary search approach, we can comfortably recover the coefficients using only around $3500$ queries most of the time.
Recovering the Flag
With the error polynomials recovered, we can easily recover the secret polynomial vector $\mathbf{s}$ by solving a system of two linear equations as we saw earlier. The hard part of the challenge is more or less done, and all that's left is to XOR the secret with the encrypted flag!
There is one final detail however. Serialising the recovered secret polynomials and XORing the result with the encrypted flag won't work. If we look at the specification, and more specifically the PKE KeyGen algorithm, we see that the secret key is encoded as an element in NTT domain. We've recovered $\mathbf{s}$ in the normal domain, so it is necessary to run the transformation algorithm on it before serialising it. Fortunately, we can just use the ntt function provided by the Kyber reference implementation to do this for us.
At last, we get the flag!
Solve Script
Note that the library is patched to also return the generated message in the encapsulation function.
diff --git a/ref/kem.c b/ref/kem.c
index f376bd2..2a41974 100644
--- a/ref/kem.c
+++ b/ref/kem.c
@@ -1,4 +1,5 @@
#include <stddef.h>
+#include <string.h>
#include <stdint.h>
#include "params.h"
#include "kem.h"
@@ -50,7 +51,7 @@ int crypto_kem_keypair(uint8_t *pk,
**************************************************/
int crypto_kem_enc(uint8_t *ct,
uint8_t *ss,
- const uint8_t *pk)
+ const uint8_t *pk, uint8_t *hm)
{
uint8_t buf[2*KYBER_SYMBYTES];
/* Will contain key, coins */
@@ -59,6 +60,7 @@ int crypto_kem_enc(uint8_t *ct,
randombytes(buf, KYBER_SYMBYTES);
/* Don't release system RNG output */
hash_h(buf, buf, KYBER_SYMBYTES);
+ memcpy(hm, buf, KYBER_SYMBYTES);
/* Multitarget countermeasure for coins + contributory KEM */
hash_h(buf+KYBER_SYMBYTES, pk, KYBER_PUBLICKEYBYTES);
diff --git a/ref/kem.h b/ref/kem.h
index 3f3eff6..50bb7e5 100644
--- a/ref/kem.h
+++ b/ref/kem.h
@@ -33,7 +33,7 @@
int crypto_kem_keypair(uint8_t *pk, uint8_t *sk);
#define crypto_kem_enc KYBER_NAMESPACE(enc)
-int crypto_kem_enc(uint8_t *ct, uint8_t *ss, const uint8_t *pk);
+int crypto_kem_enc(uint8_t *ct, uint8_t *ss, const uint8_t *pk, uint8_t *hm);
#define crypto_kem_dec KYBER_NAMESPACE(dec)
int crypto_kem_dec(uint8_t *ss, const uint8_t *ct, const uint8_t *sk);
kyber_util.py
from sage.all import *
import ctypes
import hashlib
kyber_lib = ctypes.CDLL('./libpqcrystals_kyber512_ref_patched.so')
q = 3329
F = GF(q)
P = PolynomialRing(F, 'X')
P.inject_variables()
R = P.quotient_ring(X**256 + 1, 'Xbar')
def hash_h(m):
return hashlib.sha3_256(m).digest()
def kdf(m):
return hashlib.shake_256(m).digest(32)
def poly_to_bytes(p):
buf = ctypes.c_buffer(int(384))
poly = (ctypes.c_int16 * int(256))(*list(p))
kyber_lib.pqcrystals_kyber512_ref_poly_tobytes(buf, poly)
return bytes(buf)
def bytes_to_poly(b):
poly = (ctypes.c_int16 * int(256))()
kyber_lib.pqcrystals_kyber512_ref_poly_frombytes(poly, ctypes.c_buffer(b))
return R(list(poly))
def polyvec_to_bytes(pv):
buf = ctypes.c_buffer(int(2 * 384))
polyvec = (ctypes.c_int16 * int(2 * 256))(*(list(pv[0]) + list(pv[1])))
kyber_lib.pqcrystals_kyber512_ref_polyvec_tobytes(buf, polyvec)
return bytes(buf)
def compressed_bytes_to_polyvec(b):
polyvec = (ctypes.c_int16 * int(2 * 256))()
kyber_lib.pqcrystals_kyber512_ref_polyvec_decompress(polyvec, ctypes.c_buffer(b))
return vector(R, [R(list(polyvec)[:256]), R(list(polyvec)[256:])])
def poly_frommsg(m):
poly = (ctypes.c_int16 * int(256))()
kyber_lib.pqcrystals_kyber512_ref_poly_frommsg(poly, ctypes.c_buffer(m))
return R(list(poly))
def kem_enc(pk):
ct_buf = ctypes.c_buffer(int(1024))
ss_buf = ctypes.c_buffer(int(32))
hm_buf = ctypes.c_buffer(int(32))
kyber_lib.pqcrystals_kyber512_ref_enc(ct_buf, ss_buf, ctypes.c_buffer(pk), hm_buf)
return bytes(ct_buf), bytes(ss_buf), bytes(hm_buf)
solv.sage
from pwn import *
from kyber_util import *
import time
import ctypes
import hashlib
q = 3329
F = GF(q)
P.<X> = PolynomialRing(F)
R.<Xbar> = P.quotient_ring(X^256 + 1)
ORACLE_QUERIES = 0
def oracle(z, ct):
global ORACLE_QUERIES
ORACLE_QUERIES += 1
conn.sendlineafter(b'> ', b'dec ' + ct.hex().encode())
ss = conn.recvline().decode().strip().split('ss: ')[1]
expected_ss_if_different = kdf(z + hash_h(ct)).hex()
return ss == expected_ss_if_different
def recover_error_coefficient(z, ct, hm, idx):
lower = 0
upper = 1200
v = bytes_to_poly(ct[-384:])
while lower <= upper:
middle = (upper + lower) // 2
v_ = v + middle * Xbar^idx
v_ = poly_to_bytes(v_)
r = oracle(z, ct[:-384] + v_)
if r:
v_ = v + (middle - 1) * Xbar^idx
v_ = poly_to_bytes(v_)
r = oracle(z, ct[:-384] + v_)
if not r:
b = (hm[idx // 8] >> (idx % 8)) & 1
return 833 - b - middle
else:
upper = middle - 1
else:
lower = middle + 1
def recover_error_polynomial(z, ct, hm):
coeffs = []
for i in range(256):
c = recover_error_coefficient(z, ct, hm, i)
coeffs.append(c)
return R(coeffs)
conn = remote('0.0.0.0', 1337)
pk = bytes.fromhex(conn.recvline().decode().strip().split('pk: ')[1])
hpk = bytes.fromhex(conn.recvline().decode().strip().split('H(pk): ')[1])
z = hpk[-32:]
ct1, _, hm1 = kem_enc(pk)
ct2, _, hm2 = kem_enc(pk)
timer_start = time.time()
E1 = recover_error_polynomial(z, ct1, hm1)
print(f'[!] used {ORACLE_QUERIES} in {time.time() - timer_start:.2f}s to recover E1: {list(E1)}')
timer_start = time.time()
E2 = recover_error_polynomial(z, ct2, hm2)
print(f'[!] used total {ORACLE_QUERIES} in {time.time() - timer_start:.2f}s to recover E2: {list(E2)}')
u1 = compressed_bytes_to_polyvec(ct1[:640])
v1 = bytes_to_poly(ct1[-384:])
m1 = poly_frommsg(hm1)
y1 = v1 - E1 - m1
u2 = compressed_bytes_to_polyvec(ct2[:640])
v2 = bytes_to_poly(ct2[-384:])
m2 = poly_frommsg(hm2)
y2 = v2 - E2 - m2
s1 = (y2 - y1 * u1[0]^-1 * u2[0]) * (u2[1] - u1[0]^-1 * u1[1] * u2[0])^-1
s0 = (y1 - s1 * u1[1]) * u1[0]^-1
s = vector(R, [s0, s1])
print(f'[+] recovered secret key:', list(s[0]), list(s[1]))
s = (ctypes.c_int16 * int(2 * 256))(*(list(s[0]) + list(s[1])))
kyber_lib.pqcrystals_kyber512_ref_polyvec_ntt(s)
s = vector(R, [R(s[:256]), R(s[256:])])
s_bytes = polyvec_to_bytes(s)
conn.sendlineafter(b'> ', b'hax')
flag_enc = bytes.fromhex(conn.recvline().decode().strip().split('flag_enc: ')[1])
flag = xor(flag_enc, s_bytes[:len(flag_enc)])
print(flag.decode()) | 9,866 | 33,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 232, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.906642 |
https://beta.geogebra.org/m/cqQS4Q6F | 1,638,458,702,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00168.warc.gz | 190,704,362 | 9,691 | # Endless cubes
Author:
Judith12
Topic:
Cube
Have a look at the construction. It shows cubes beeing put into a square-based pyramid with lenght of the sides=4 cm and heigth=6 cm. Note: It's always the biggest possible cube, that you put in the pyramid.
Calculate the volume of all the cubes, if you would contiune that pattern endlessly. Use the hints, if you get stuck! | 93 | 371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.90068 |
https://discuss.codechef.com/questions/35343/fgfs-editorial?sort=oldest | 1,553,435,424,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203448.17/warc/CC-MAIN-20190324124545-20190324150545-00131.warc.gz | 460,206,477 | 19,972 | You are not logged in. Please login at www.codechef.com to post your questions!
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# PROBLEM LINK:
Author: Vivek Hamirwasia
Tester: Mahbub
Editorialist: Jingbo Shang
Easy
Sort, Greedy.
# PROBLEM:
Given some sets of open intervals (exclusive at two ends), for each set, find the maximum number of disjoint intervals.
# EXPLANATION:
First of all, we need to observe that the intervals in different sets are independent (in the original statement, the customers preferred in different compartments are independent).
For a single set, it is a classical greedy problem, called Activity Selection problem. The greedy method is as following:
1. Sort the intervals by their right end ascending.
2. Initialized the select intervals as an empty set
3. Consider the sorted intervals one by one, add it if it is possible (only need to check the last select interval and the current one).
The intuition of this greedy is that, we need to make sure the space remained for the later intervals is as large as possible. The proof is also easy. For the first interval, we definite choose the interval which ends earliest. Otherwise, we can replace the first interval in the best solution with that earliest ended interval (with at least same best solution). Therefore, we can achieve the maximum interval selections with choosing the earliest ended interval. Recursively, we can see that the greedy is correct.
In summary, we can solve this problem in O(N log N), where N is the total number of intervals in all sets.
# AUTHOR'S AND TESTER'S SOLUTIONS:
Author's solution can be found here.
Tester's solution can be found here.
This question is marked "community wiki".
asked 13 Jan '14, 15:11
161446376
accept rate: 0%
0★admin ♦♦
19.8k350498541
1
"exclusive at two ends", it was left closed and right opened [from, to ) but probably it changes nothing...
(13 Jan '14, 15:33)
I have implemented the same method and solution is accepted with 2.02s but I can see solution with similar approach accepted in less than 1s. Is it only because of the fast IO or some other tweak in the code ?
My solution : http://www.codechef.com/viewsolution/3228255
(14 Jan '14, 11:22) 2★
I think it should be caused by different implementation. STL is always a little slower.
(15 Jan '14, 12:10)
15 Answers:
1 Very Strict Time Limit! Even with the sort functions removed, it gets a TLE. Probably because O(k^2) solutions were forced NOT to pass the time limits. http://www.codechef.com/viewplaintext/3241391 answered 13 Jan '14, 15:18 8.7k●19●48●98 accept rate: 9% Hm? I was able to do that in Java and had no problem with TLE (but I used DP)... (13 Jan '14, 15:30) Probably due to the doubled time limit for Java? (13 Jan '14, 15:35) My sol. was getting TLE due to sort function. http://www.codechef.com/viewsolution/3250997 But I have seen some sol. using the same approach and getting accepted. Can you explain the flaw in my sol.? (13 Jan '14, 15:56) sumitb2★ Even O(K) will be slow, so O(K log(K)) and O(K^2) are slow too... But similar to my approach, there are customers for at most N compartments, so you have to take care only those only... Just try to replace your for(int i=0; i> Ofcourse, that would never go 10^9 iterations. Damn! :( :D (13 Jan '14, 16:06) You didn't even need the k, i used a greedy approach as well and passed within time O(N). http://www.codechef.com/viewplaintext/3170241 (13 Jan '14, 17:39) f03nix2★ Basically, my algo was as follows: Sort the customers segregated by first the compartments, then by their arrival time. Hold one customer with us,start by the first customer : If the next customer needs a different slot(all requests for current slot is done) accept the one with us and hold on to the new one. Else, if the next customer arrives after the customer with us would leave - we can cater to the customer with us. So increment the count and hold the new customer now. Else, if the next customer can leave earlier than our current - shoo away the current and keep the new one. (13 Jan '14, 17:42) f03nix2★ @f03nix >> Nice implementation :) (13 Jan '14, 17:56) showing 5 of 8 show all
1 I have sometimes problem with greedy algorithms, because sometimes its not easy to see if it works or not... From my point of view the DP is more "deterministic". The key idea is that compartments are independent as written in editorial. You can use this DP: for max to time MT, we know that max guests starting at MT + 1 is 0 for all times T = MT downto 0 we can do: if there comes guest at time T and leaves at time L, the max for time T is max( T + 1, 1 + dp[L] ) at the end the result is dp[0] for each compartment. Implementation was tricky - times were up to 109 (so not possible to use array for DP), but N is up to 105, so at most 2 * 105 different times. My first idea was to do "normalization" -> map those at most 2 * 105 different times to numbers 0..200000 and use array for DP (a lot of work needed here), but than I made it using tree maps - see my Java solution for details and ask if something is unclear... answered 13 Jan '14, 15:45 16.9k●49●115●225 accept rate: 11% "times were up to 10^9 (so not possible to use array for DP), but N is up to 10^5" Exactly! This was the area that required to be hit in the problem. (13 Jan '14, 15:48) Yes, DP is feasible for this problem. Congratulations for you Accept! I think greedy is much easier to implement, since we don't need to remap the times. (13 Jan '14, 19:05) I have sometimes problem with greedy algorithms, because sometimes its not easy if it works or not... Just a comment, in this specific case, the correctness of the greedy approach can be proven. (cf. the linked wiki page about Activity Selection Problem) (13 Jan '14, 22:55) Shouldn't it be dp[T]=max(dp[T+1],1+dp[L]); ?? (16 Jan '14, 19:04) h1h33★
1 I have used map data structure, coupled with fast I/O and good code organization and got AC with relative ease. If you know the algorithm and some good data structures, using scanf()/printf() is almost always enough! As a bonus, if you know your way around how the I/O system works, you can even get AC with cin/cout as my solution submitted few seconds ago in practice section proves: http://www.codechef.com/viewsolution/3251027 Think algorithmically above all, Bruno :) answered 13 Jan '14, 16:06 3★kuruma 17.7k●72●143●209 accept rate: 8% 2 Good to see you playing with the STL. Cheers! (13 Jan '14, 16:10) Thanks @bugkiller, actually it was the first time I used an iterator to iterate over a map, so I was really happy with it :D (13 Jan '14, 16:17) kuruma3★ I was using similar approach, but was new with STL's, yet got to learn very new things from this question. My approaches: finally saw your solution just after completion, liked the way of your coding. (13 Jan '14, 16:30) hrculiz1★ Nice job though, playing with the multimap :) Usually, whenever I see a multimap question I always tend to "degenerate it" into a map of map > as it is more intuitive to me. Regarding your approach, I guess you can separate the multimap from the data structure and thus speed up your code quite considerably (13 Jan '14, 16:34) kuruma3★ Will you please help me checking why am I getting WA for the problem. http://www.codechef.com/viewsolution/3256692 (15 Jan '14, 09:40) hrculiz1★ hello :) the main differences I can see between both of our solutions is the type used to store the answer... I've used long long :) EDIT: After changing the data types I now get TLE veredict with your code... Try to encapsulate the main procedure to get the answer on a separate function.. It's easier to debug and spot bottlenecks :) (17 Jan '14, 20:13) kuruma3★ showing 5 of 6 show all
0 I've used the same approach given in the editorial but I thought of it as Earliest Deadline First(EDF) scheduling, allot a compartment to the customer who leaves earliest :) And in order to keep track of free/occupied compartments I used map, using compartment number as key and the leaving time as data. STL sure makes programming in C++ a lot simpler. Here's my solution. answered 13 Jan '14, 17:42 91●2●8 accept rate: 0%
0 @shangjingbo i just wanted to know what's wrong in the solution here. In this i sort the array two times. First i sort it with respect to "room no" then in the chunks of same room no's i sort them wrt deadlines. At least can you tell me which test case does it fail. Please help answered 13 Jan '14, 17:48 3★paramvi 16●5 accept rate: 0%
0 Could someone post an AC solution in Python ? I don't even know if it's possible since no pythonista managed to get AC for this problem. I would love to learn how to solve it answered 13 Jan '14, 18:58 3★anhtuann 1 accept rate: 0%
0 I tried the same approach after finding an article on Interval scheduling at wiki. To maintain all stuff sorted, I completely ignored the value of K and managed a map mapping compartment value to a set containing all input values of entry and exit time. Although first solution could have passed, I kept on optimizing it after TLEs only to realize the input data is so large that I need to read input fast. answered 13 Jan '14, 19:10 5★mjbpl 419●2●7 accept rate: 6%
0 Two similar codes(almost), one getting RUNTIME error & the other got AC first i submitted using vector (i have done this earlier in other problems successfully :) ) but got runtime error & same code i submitted using structure(everything was same except i stored the value in struct variable instead of vector ) & .... got AC . lol !! http://www.codechef.com/viewsolution/3160692 (using vector) http://www.codechef.com/viewsolution/3160968 (using structure) i'd appreciate if anyone willing to answer .... :) answered 13 Jan '14, 19:15 4★sagar55 1●3 accept rate: 0% Vectors of vector is extremely time consuming for such large data. POD works better here. Your struct is supposed to take 3 members at compile time, so it is faster. On the other hand, a vector is generic and ready to accommodate more elements (members) although you don't supply it. Plus vector keep track of currently allocated memory and size behind the scene that is more expensive than PODs(more@http://en.wikipedia.org/wiki/Plain_old_data_structure) without any constructor/destructor cost. (14 Jan '14, 16:36) mjbpl5★
0 I am trying to solve Bon appetit problem from Jan challenge 2014 from 6 days.Still couldn't solve it out.. So it would be really nice if someone could help me to solve out this. I am trying to solve it in the following way. 1.Sort customers according to departure time.(quicksort) 2.Then go through each customer and directly retrieve a recent customer in that compartment. if it results in an overlap(discard the customer). else increment count and update the recent customer in the specified compartment. I came to know that a hashmap would serve the purpose.Each compartment would be mappped to certain index by the hashing function and then we would directly go there. Questions. 1.How can i build a hash function so that given a compartment it would map it to certain index and when again a customer of this compartment appears i would directly go to this compartment in 1 operation. 2.Would this approach give me WA or TLE? Please help... answered 14 Jan '14, 10:46 296●8●21●22 accept rate: 11% I used radix sort for sorting finishing time and preferred components, and then used the activity selection problem, coupled with fast I/O and got AC. Link: http://www.codechef.com/viewsolution/3235970 (14 Jan '14, 11:29) but is my approach correct and if it is how to implement it? (14 Jan '14, 14:40) @anonymousins read my answer above, I've used the same idea. When searching for hash functions I came across STL maps, read about them, implemented them in my code and voila! AC! :D STL map takes care of hasing, I read that its insert, delete and lookup operations are of O(logn) complexity. (14 Jan '14, 22:30) how can we do it in c? (15 Jan '14, 11:35) If you want to do it in C you'll have to write your own code, but I suggest you start using C++ with STL, C++11 library is huge and its also the standard. (15 Jan '14, 18:01)
0 My solution did not get accepted because TLE. (I used JAVA) Here is the code for that solution. But now i canged the data type values. (I used long data type and i changed them to int`) Then my code got accepted. Here is that accepted code. (This is not a problem in the algorithm i suppose, The code is almost same, I think to overcome the TLE problems we should aware of these kind of problems too :D Changing data types will get you accepted :D I would really like to know why this is happening ) Can anybody explain why this is happening. thanks in advance :) answered 14 Jan '14, 12:00 3★vidudaya 1●2 accept rate: 0%
0 Can someone explain the time complexity given in the editorial? Does it take the time taken to sort the intervals into account? answered 17 Jan '14, 19:37 223●6 accept rate: 0%
0 I got runtime error please help me .solution id: https://www.codechef.com/viewsolution/9963237. answered 23 Apr '16, 17:53 2★vickycod 21●1●1●2 accept rate: 6%
0 https://www.codechef.com/viewsolution/11861897 https://www.codechef.com/viewsolution/11861658 Can someone please point the errors answered 17 Oct '16, 02:39 4★adi_1996 21●1●5 accept rate: 0%
0 https://www.codechef.com/viewsolution/14299209 What's wrong with the approach? Please help answered 20 Jun '17, 23:16 1 accept rate: 0%
0 I am getting TLE. Can someone please find the reason. Thanks a lot for giving your time. Here is my code. answered 16 Jul '18, 13:56 10●1 accept rate: 0%
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question asked: 13 Jan '14, 15:11
question was seen: 14,273 times
last updated: 16 Jul '18, 13:56 | 3,713 | 14,214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-13 | latest | en | 0.935834 |
https://science.thewire.in/the-sciences/covid-19-pandemic-case-fatality-rate-calculation/ | 1,723,447,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00170.warc.gz | 406,329,911 | 17,890 | # Why It’s Not so Simple to Calculate COVID-19’s Case Fatality Rate
Image: geralt/pixabay.
It’s essential but utterly difficult to estimate the case fatality rate (CFR) of an ongoing pandemic. The data is continuously updated and we constantly receive new knowledge of the disease, especially if it’s a new disease caused by a new virus.
On March 3, the WHO director-general Dr Tedros Adhanom Ghebreyesus said that about 3.4% of people reported to have COVID-19 around the world have died. However, the estimated fatality rates in different countries – or the world as a whole – could change dramatically once the pandemic ends and the dust settles.
We usually have data on the number of cases, called C(t); the number of deaths, D(t); and the number of people who have recovered, R(t) – where ‘t‘ is the time over which we’re measuring these numbers. The CFR is the proportion of people who eventually die of the disease. Once the pandemic ends, the CFR will be equal to D divided by C. But while the pandemic is still underway, the formula CFR = D(t)/C(t) could be misleading because the final outcome of many patients is still unknown, and we’re assuming that none of them will die eventually.
This in turn could have serious consequences. A research article published in the American Journal of Epidemiology in September 2005 discussed this problem. For example, when the 2003 SARS epidemic was still ongoing, the WHO reported in April 2003 that “data … indicates that 96% of persons developing SARS recover spontaneously. The focus now is on the roughly 4% who are dying.”
The estimated CFR of SARS, obtained by dividing D(t) by C(t), was 3-5% in the first few weeks of the global outbreak. But when researchers used the appropriate statistical techniques and accounted for the discrepancy between the cohorts of D(t) and C(t), the estimated CFR jumped to 6-8%.
Today, long after the outbreak ended, we know that SARS’s final death rate was 9.55% (774 of the 8,098 people who had the infection died). Such a shift in the estimates has nothing to do with the severity of the disease.
The September 2005 article suggested estimating the CFR by dividing D(t) by D(t) + R(t) – that is, the number of people who died by the number of people who died or recovered. However, with the novel coronavirus, many individuals who eventually die are not excluded from the cohort of those still under treatment until later.
So an alternative way to estimate the CFR is to calculate the number of people who have died as a fraction of the number of people who had the infection T days earlier. Here, T is the average number of days from case confirmation to death.
As on May 4, there had been 44,237 cases in India, 1,513 had died and 12,235 had recovered. If we assume that T = 7 days, we note that there had been 29,458 cases on April 27. Thus, the three estimates of CFR for India at the moment are:
i. ‘First’ way to measure: D(May 4)/C(May 4) = 3.42%
ii. The research article’s way: D(May 4)/[D(May 4) + R(May 4)] = 11.01%
iii. The alternative way: D(May 4)/C(April 27) = 5.14%
These estimates vary considerably, and we can know which one is closer to reality only once the pandemic is over.
In my opinion, the actual CFR of COVID-19 could be lower. To calculate the CFR, we need to know how many people were infected – and this figure hasn’t been easy to pin down with the novel coronavirus. We already know that there are many people with mild symptoms as well as no symptoms at all. These two ‘truths’ have become important considerations for governments to design effective distancing and quarantining strategies. We should also account for them when calculating the CFR.
Mild or asymptomatic patients could considerably enhance the C(t) but leave D(t) unchanged, so the CFR for COVID-19 will decrease considerably.
This said, it’s not easy to determine the extent to which the actual C(t) may be greater than the observed C(t). We will need to conduct a survey for this purpose.
Neil Ferguson, a public health expert at Imperial College London, told The Guardian on January 26 that his “best guess” was that 100,000 people could have been infected by the virus at the time even though there were only 2,000 confirmed cases. If this sounds weird, consider the recent example of Sweden.
It’s well-known that Sweden didn’t impose a lockdown like many other countries, and instead rolled out voluntary measures. The Swedish government advised older people and others particularly vulnerable to the virus to avoid social contact; recommended people work from home, wash their hands regularly and avoid nonessential travel; and kept the country’s borders, some schools and most businesses open. On April 22, Anders Tegnell, the chief epidemiologist of Sweden’s public health agency, said that about 20% of the people of Stockholm had developed immunity (through infection) to the virus by then and that the city would achieve herd immunity in a few weeks.
Five days later, Sweden’s ambassador to the US said that about 30% of people in Stockholm had achieved immunity, and that there would be herd immunity in May.
The population of Sweden is around 10 million. About a fourth – 2.5 million – live in Stockholm. So by April 22, about 500,000 Stockholmers had COVID-19. However, according to the daily cumulative data, only 5,071 Stockholmers had positive by April 21. So an astonishing 100-times the number of people who tested positive actually had COVID-19. These people must have been mostly asymptomatic.
Of course, this does not mean the actual number of cases is 100-times the number of those who have tested positive in other countries as well. Sweden deliberately instigated controlled social-mixing to achieve herd immunity. This ‘factor of amplification’ also depends on the rate of testing in a country. Iceland, for example, tested about 6% of its population, so it can’t expect the total number of cases within its borders to be 100-times the number of those who tested positive.
However, for countries that have tested a smaller fraction of their populations, the number of people with COVID-19 (most of them asymptomatic) could be higher than the number that tested positive. Even other factors like high population density, social discipline and adherence to healthcare protocols could affect the actual C(t) figure.
Thus, although it’s nearly impossible to determine the exact number of people in India already affected by COVID-19, it shouldn’t be surprising if that number turns out to be 10-100-times the number of cases detected thus far – most of them asymptomatic many of whom would have recovered. And if that is so, the CFR’s denominator should use that estimated C(t) instead of the observed C(t). This should be true for other countries as well. Ultimately, the CFR of COVID-19 may not be 3.4/100 but more like 3.4/1,000 or even 3.4/10,000.
Atanu Biswas is a professor of statistics at the Indian Statistical Institute, Kolkata.
Scroll To Top | 1,582 | 6,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.961827 |
https://homeworkandessays.com/week-2-practice-quiz-qnt-561-my-mathlab-university-of-phoenix/ | 1,624,064,821,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643354.47/warc/CC-MAIN-20210618230338-20210619020338-00482.warc.gz | 281,360,524 | 9,970 | # Week 2 Practice Quiz QNT/561 My Mathlab (University Of Phoenix)
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Week 2 Practice Quiz QNT/561 My Mathlab (University Of Phoenix)
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QNT/561 My Mathlab (University Of Phoenix) (I am only willing to pay 25 so please do not message me if you want more).
Week 2 Practice Quiz
Questions 1-16
It is a new quiz so I do not need an old one because it will not be the same answers.
Some sample questions that are on the Quiz:
1. A random sample of 88 observations produced a mean x=25.9 and a standard deviation s=2.8.
a. Find a 95% confidence interval for μ.
b. Find a 90% confidence interval for μ.
c. Find a 99% confidence interval for μ.
2. Health care workers who use latex gloves with glove powder on a daily basis are particularly susceptible to developing a latex allergy. Each in a sample of 50 hospital employees who were diagnosed with a latex allergy based on a skin-prick test reported on their exposure to latex gloves. Summary statistics for the number of latex gloves used per week are x=19.7 and s=11.8. Complete parts (a)−(d).
3. Each child in a sample of 63 low-income children was administered a language and communication exam. The sentence complexity scores had a mean of 7.64 and a standard deviation of 8.94. Complete parts a through d.
4. The random sample shown below was selected from a normal distribution. 3, 8, 8, 9, 7, 1 Complete parts a and b.
5. Periodically, a town water department tests the the drinking water of homeowners for contaminants such as lead. The lead levels in water specimens collected for a sample of 10 residents of the town had a mean of 2.7 mg/L and a standard deviation of 1.8 mg/L. Complete parts a through c.
6. An accounting firm annually monitors a certain mailing service’s performance. One parameter of interest is the percentage of mail delivered on time. In a sample of 311,000 items mailed between Dec. 10 and Mar. 3—the most difficult delivery season due to bad weather and holidays—the accounting firm determined that 280,600 items were delivered on time. Use this information to make a statement about the likelihood of an item being delivered on time by that mailing service.
7. Suppose you’re given a data set that classifies each sample unit into one of four categories: A, B, C, or D. You plan to create a computer database consisting of these data, and you decide to code the data as A=1, B=2, C=3, and D=4. Are the data consisting of the classifications A, B, C, and D qualitiative or quantitative? After the data are input as 1, 2, 3, or 4, are they qualitative or quantitative?
8. In one university, language professors incorporated a 10-week extensive reading program to improve students’ Japanese reading comprehension. The professors collected 269 books originally written for Japanese children and required their students to read at least 40 of them as part of the grade in the course. The books were categorized into reading levels (color-coded for easy selection) according to length and complexity. Complete parts a through c.
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Order your paper today and save upto 15% with the discount code 15BEST | 760 | 3,227 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2021-25 | latest | en | 0.911411 |
https://rpg.stackexchange.com/questions/22361/is-there-a-plainer-english-explanation-of-bravery-and-poltroonery | 1,721,054,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00345.warc.gz | 426,668,300 | 35,406 | # Is there a plainer english explanation of bravery and poltroonery?
Despite having reread the rules section multiple times, I don't understand how death numbers, death rolls and bravery / poltroonery settings other than flat numerical modifiers work. Is there an idiots guide? How do they work?
Limited Bravery and how you get to modify your death rolls.
You place in your orders Bravery: 8 (we're going to be silly brave)
Scenario 1: The death number set by the GM is equal to or lower than 8.
What happens: Nothing, he rolls 2d6 and if that number is equal to or higher than the set death number, you die. If it is less, your character lives and then rolls are made for loot, MiD, etc.
Scenario 2: The death number is Higher than 8.
What happens: The GM takes the death number minus your bravery number, then adds that to his death roll. So if the death number was 10. He would take 10-8= 2 and add 2 to his death roll. He would then roll 2d6+2 and if that number was equal to or higher than 10, your character would die. If it was lower, then your character would live. AND get a +2 to all his loot, MiD, etc.
For Completion
Poltroonery goes to 11 (yes, I'll make that pun)
Scenario 1: The Death number set by GM is 11 or higher
What happens: Nothing, rolls 2d6 if number is equal to or greater than, you die. Same as scenario 1 on the bravery side.
Scenario 2: Death number is 10 or lower.
What happens: GM takes your Poltroonery number minus the death number, then subtracts that number from the death roll. So if the death number was 2. you take 11-2=9. Then roll 2d6-9 and if that number was equal to/greater than 2 you then die. If it was lower your character would still be alive, but you would probably be called a coward for running away.Disgraced, kicked out of your unit, etc etc.
• This helps incredibly! I was unable to parse the rules as written, whereas this makes sense! Commented Feb 23, 2013 at 1:21
• Just to be sure: this means that the lower the death number is, the more likely you are to die, right? Commented Feb 23, 2013 at 2:52
• @Scrollmaster The more reckless bravery there is than necessary, the more likely the fool is going to go and make the situation dangerous enough to get killed, is how I understand it. Of course, fortune favours the bold, so if you win the gamble and don't get yourself killed you get the rewards of going so far above and beyond. Commented Feb 23, 2013 at 3:16
• Correct, To be more D&D non-thac0 the Death number is your AC and the death roll is his attack against your AC. Commented Feb 23, 2013 at 3:39 | 672 | 2,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-30 | latest | en | 0.954095 |
https://www.analystforum.com/t/economic-pension-expense/35464 | 1,660,843,625,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573242.55/warc/CC-MAIN-20220818154820-20220818184820-00755.warc.gz | 576,130,149 | 6,517 | # economic pension expense
ppl, i looked at a cfai problem and they calc economic pension expense as: Service Cost + Interest Cost + Actual Return i used the formula I know: Ending PBO - Beginning PBO + Benefits Paid - Actual Return The \$ value I get by my way of doing it isn’t even an answer choice. ideas?
can u post the problem #. I can look at it after I get home. Do not have the CFAI book with me to look.
niraj_a Wrote: ------------------------------------------------------- > ppl, > > i looked at a cfai problem and they calc economic > pension expense as: Service Cost + Interest Cost + > Actual Return > > i used the formula I know: Ending PBO - Beginning > PBO + Benefits Paid - Actual Return > > The \$ value I get by my way of doing it isn’t even > an answer choice. ideas? I don’t know your formula. The one they use is the one in the Stalla book, too. Usually you don’t start with the PBO to calculate PE. Is this because you are mixing the IS PBO and the true balance sheet PE?
Quick note: its less actual return (i.e. we subtract actual returns from the service and interests costs in order to get economic pension expense).
I think it’s adjusted PE = SC+IC-actual returns
to help you guys out - i think a littly birdy told me that this problem is in last year’s sample 1.
Econ pension exp = Change in Funded Status - Firm’s contribution. (It’s not the change in PBO.)
eltia, yes, your method and my formula both can be used to calc economic pension expense - as per schweser and an april cpk post.
Two formulae are applicable for Economic Pension expense: 1. End PBO - Beg PBO + Benefits Paid - Actual Return on Plan Assets 2. End FS - Beg FS - Employer Contributions. Both these methods should give you the same number - maybe because of the way we are calculating - signs of the amount (positive vs. negative) could be different.
^ this is Economic Pension Expense. I was talking about (last year’s concept) called Adjusted Pension expense, which was SC + IC - ARPA. If I remember correctly now. EPE is indeed the 2 formulas stated above by CP.
I thought Adjusted Pension Expense = Economic Pension Expense?
yes it has been changed. Last year it was SC + IC - actual pension expense.
niraj_a Wrote: ------------------------------------------------------- > ppl, > > i looked at a cfai problem and they calc economic > pension expense as: Service Cost + Interest Cost + > Actual Return > > i used the formula I know: Ending PBO - Beginning > PBO + Benefits Paid - Actual Return > > The \$ value I get by my way of doing it isn’t even > an answer choice. ideas? Change in PBO + Benefits Paid - (Change in plan assets + Benefits Paid - Employer contributions)
ive got it down in my notes as (this is from schweser): current service cost + interest cost + plan ammendments - actual return on assets OR Beginning funded status - ended funded status - contributions
nvm, im gonna chuck this older question. our formulas seem to be right. sorry for the unnecessary brain f*ck. i had to go through it too.
the show NY Wrote: ------------------------------------------------------- > ive got it down in my notes as (this is from > schweser): > > > current service cost + interest cost + plan > ammendments - actual return on assets > > OR > > Beginning funded status - ended funded status - > contributions What includes in plan ammendment?
paid benefits or things like change in return / age etc. estimates
yo niraj, if possible can you please write down the formula or elaborate this
cpk123 Wrote: ------------------------------------------------------- > Two formulae are applicable for Economic Pension > expense: > > 1. End PBO - Beg PBO + Benefits Paid - Actual > Return on Plan Assets > > 2. End FS - Beg FS - Employer Contributions. > > Both these methods should give you the same number > - maybe because of the way we are calculating - > signs of the amount (positive vs. negative) could > be different. the first formula cpk mentions is what i have. from what i know, they will provide us with the \$\$ amount of benefits paid or plan amendments in the PBO / Plan Asset reconciliation amendment. HTH/
pensions are one of the most difficult topics in the whole course for me to grasp. its hard for me to actually udnerstand whats going on and what all the components of the formulas mean and i think i am going to have to spend some time actually understanding the formulas. does anyone have any good notes they can share that can help with this topic? | 1,014 | 4,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-33 | latest | en | 0.884497 |
http://www.justintools.com/unit-conversion/frequency.php?k1=radian-per-day | 1,568,733,437,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573080.8/warc/CC-MAIN-20190917141045-20190917163045-00100.warc.gz | 264,496,081 | 14,836 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
Unit of: FREQUENCY
FREQUENCY's base unit: hertz (Non-SI/Derived Unit)
In relation to the base unit (hertz), 1 Radian Per Day = 1.84207E-6 hertz.
1 rad/d= 1.84207E-6 1 per second (1/s)
1 rad/d= 1.84207E-6 cycle per second (cps)
1 rad/d= 57.2957452111 degree per day (°/day)
1 rad/d= 2.38610103627 degree per hour (°/hr)
1 rad/d= 0.0397889666494 degree per minute (°/min)
1 rad/d= 0.000663145146948 degree per second (°/sec)
1 rad/d= 1.84207E-6 frames per second (FPS) | 269 | 821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-39 | latest | en | 0.768121 |
https://financeband.com/how-many-401k-millionaires-are-there | 1,726,444,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00233.warc.gz | 227,677,138 | 14,300 | # How many 401k millionaires are there?
Asked by: Mrs. Nadia Torphy III | Last update: February 9, 2022
Fidelity Investments reported that the number of 401(k) millionaires—investors with 401(k) account balances of \$1 million or more—reached 233,000 at the end of the fourth quarter of 2019, a 16% increase from the third quarter's count of 200,000 and up over 1000% from 2009's count of 21,000.
## At what age should you be a 401k millionaire?
Recommended 401k Amounts By Age
Middle age savers (35-50) should be able to become 401k millionaires around age 50 if they've been maxing out their 401k and properly investing since the age of 23.
## How many 401k millionaires are there in 2021?
According to Fidelity Investments, one of the largest 401(k) providers in America today, the number of 401k millionaires reached roughly 180,000 in 2021 thanks to a prolonged bull market.
## What is a good 1 year rate of return on 401k?
The average 401(k) rate of return ranges from 5% to 8% per year for a portfolio that's 60% invested in stocks and 40% invested in bonds. Of course, this is just an average that financial planners suggest using to estimate returns.
## How many Americans have \$1000000 in their 401k?
Fidelity Investments reported that the number of 401(k) millionaires—investors with 401(k) account balances of \$1 million or more—reached 233,000 at the end of the fourth quarter of 2019, a 16% increase from the third quarter's count of 200,000 and up over 1000% from 2009's count of 21,000.
## How many 401k millionaires are there?
25 related questions found
### Can I retire at 60 with 500k?
Can I retire on \$500k plus Social Security? Yes, you can! The average monthly Social Security Income check-in 2021 is \$1,543 per person.
### How long will 300k last in retirement?
The amount of time it will take for \$300,000 to dwindle down to zero is based on the amount a retiree withdraws and the average growth rate. For example, if a retiree withdrew \$30,000 a year with no growth to their account, the \$300k would be totally spent in 9 to 10 years if including fees spent in the account.
### How much will a 401k grow in 20 years?
You would build a 401(k) balance of \$263,697 by the end of the 20-year time frame. Modifying some of the inputs even a little bit can demonstrate the big impact that comes with small changes. If you start with just a \$5,000 balance instead of \$0, the account balance grows to \$283,891.
### How much should a 30 year old have in 401k?
If you are earning \$50,000 by age 30, you should have \$50,000 banked for retirement. By age 40, you should have three times your annual salary. By age 50, six times your salary; by age 60, eight times; and by age 67, 10 times. 8 If you reach 67 years old and are earning \$75,000 per year, you should have \$750,000 saved.
### How much does the average American retire with?
According to this survey by the Transamerica Center for Retirement Studies, the median retirement savings by age in the U.S. is: Americans in their 20s: \$16,000. Americans in their 30s: \$45,000. Americans in their 40s: \$63,000.
### Does 401k double every 7 years?
The most basic example of the Rule of 72 is one we can do without a calculator: Given a 10% annual rate of return, how long will it take for your money to double? Take 72 and divide it by 10 and you get 7.2. This means, at a 10% fixed annual rate of return, your money doubles every 7 years.
### What net worth is considered rich?
The vast majority of Americans do not meet commonly held definitions of what it means to be rich in the U.S. Respondents to Schwab's 2021 Modern Wealth Survey said a net worth of \$1.9 million qualifies a person as wealthy.
### What percentage of Americans have a net worth of over \$1000000?
A new survey has found that there are 13.61 million households that have a net worth of \$1 million or more, not including the value of their primary residence. That's more than 10% of households in the US.
### What is the 4% rule?
The 4% rule assumes your investment portfolio contains about 60% stocks and 40% bonds. It also assumes you'll keep your spending level throughout retirement. If both of these things are true for you and you want to follow the simplest possible retirement withdrawal strategy, the 4% rule may be right for you.
### How long will \$500000 last retirement?
It may be possible to retire at 45 years of age, but it will depend on a variety of factors. If you have \$500,000 in savings, according to the 4% rule, you will have access to roughly \$20,000 for 30 years.
### Can I retire at 62 with 400k?
Yes, you can retire at 62 with four hundred thousand dollars. At age 62, an annuity will provide a guaranteed level income of \$21,000 annually starting immediately, for the rest of the insured's lifetime. ... The longer you wait before starting the lifetime income payout, the higher the income amount to you will be.
### Can a couple retire on 1 million dollars?
Is a million dollars enough money to ensure a financially secure retirement today? A recent study determined that a \$1 million retirement nest egg will last about 19 years on average. Based on this, if you retire at age 65 and live until you turn 84, \$1 million will be enough retirement savings for you.
### Can you retire off 3 million dollars?
Most folks would agree retiring early brings a lot of perks. ... Retire fully at age 60, and you could be sitting on a \$2 million nest egg. Keep working—and investing—for another five years, and you could retire with more than \$3 million at age 65!
### Can I retire at 64 with \$600000?
No. You can retire comfortably on a sum like \$600,000 if you take the right steps (and don't confuse “comfortable” with “luxurious”). With the right financial choices, a \$600,000 nest egg might be enough for an adequately funded retirement without depleting your savings at a dangerous rate.
### What is the largest 401k balance?
You're in very good company: A seven-figure 401(k) balance is the exception, not the rule. In fact, the average 401(k) balance at Fidelity — which holds 16.2 million 401(k) accounts and is consistently ranked as the largest defined contribution record-keeper — was \$103,700 as of March 2019.
### How much does a person need in a 401k to retire at 55?
How Much Money Do I Need To Retire At 55? If your goal is to retire at age 55, Fidelity recommends that you save at least seven times your annual income. That means if your annual income is \$70,000 a year, you need to save \$490,000. | 1,657 | 6,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.923231 |
https://www.texasgateway.org/resource-index/?f%5B0%5D=sm_field_resource_type%3Astudent_activity&f%5B1%5D=sm_field_resource_grade_range%3A4&f%5B2%5D=sm_field_resource_grade_range%3A7&%3Bf%5B1%5D=sm_field_resource_grade_range%3A-1&%3Bamp%3Bf%5B1%5D=sm_field_resource_grade_range%3A0 | 1,563,845,287,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528635.94/warc/CC-MAIN-20190723002417-20190723024417-00158.warc.gz | 852,037,318 | 15,153 | 1. ### Gravity
• Resource ID: S8M3L2
• Subject: Science
Using interactives, students will demonstrate that gravity is the force that governs the motion of our solar system.
2. ### Using Theoretical and Experimental Probability to Make Predictions
• Resource ID: M7M2L17
• Subject: Math
Given an event to simulate, the student will use theoretical probabilities and experimental results to make predictions and decisions.
3. ### Potential and Kinetic Energy
• Resource ID: R4SCI0001
• Subject: Science
This resource provides Tier I instruction ideas for Grade 6+ science teachers in the area of potential and kinetic energy.
4. ### What's Happening with the Weather?
• Resource ID: R4SCI0010
• Subject: Science
This resource shows how to observe and describe weather changes from day to day and over seasons.
5. ### Introduction to Plate Tectonics
• Resource ID: R4SCI0011
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This resource is intended to use for Tier I classroom instruction.
6. ### Human Impact
• Resource ID: R4SCI0012
• Subject: Science
This resource can be used, in conjunction with best practices, for Tier I classroom instruction.
7. ### Creating Nets for Three-Dimensional Figures
• Resource ID: GM3L2A
• Subject: Math
Given nets for three-dimensional figures, the student will apply the formulas for the total and lateral surface area of three-dimensional figures to solve problems using appropriate units of measure.
8. ### Interactive Math Glossary
• Resource ID: MATH_IMG_001
• Subject: Math
The Interactive Math Glossary is provided by the Texas Education Agency to help teachers explore and understand mathematics vocabulary.
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• Resource ID: M7M2L16
• Subject: Math
Given problem situations, the student will find the probability of the dependent and independent events.
10. ### Recognizing Misuses of Graphical or Numerical Information
• Resource ID: M7M2L20
• Subject: Math
Given a problem situation, the student will analyze data presented in graphical or tabular form by evaluating the predictions and conclusions based on the information given.
11. ### Dichotomous Keys Murder Mystery
• Resource ID: R4SCI0038
• Subject: Science
A resource for the instruction of (7)(11)(A) involving the use of dichotomous keys for identification.
12. ### Using Multiplication by a Constant Factor
• Resource ID: M7M2L6
• Subject: Math
Given problems involving proportional relationships, the student will use multiplication by a constant factor to solve the problems.
13. ### Predicting, Finding, and Justifying Data from a Table
• Resource ID: M7M2L2
• Subject: Math
Given data in table form, the student will use the data table to interpret solutions to problems.
14. ### Predicting, Finding, and Justifying Data from Verbal Descriptions
• Resource ID: M7M2L4b
• Subject: Math
Given data in a verbal description, the student will use equations and tables to solve and interpret solutions to problems.
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Given descriptions or illustrations, students will investigate how evidence of chemical reactions indicates that new substances with different properties are formed.
16. ### Understand New Vocabulary Using Roots and Affixes (English 7 Reading)
• Resource ID: E7RdM1L1
You will learn how to determine the meaning of grade-level academic English words derived from Latin, Greek, or other linguistic roots and affixes.
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You will learn how to analyze different points of view, including first person, third-person omniscient, and third-person limited.
19. ### Understanding Drama (English 7 Reading)
• Resource ID: E7RdM2L6 | 925 | 4,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2019-30 | latest | en | 0.772792 |
http://www.jiskha.com/display.cgi?id=1298697728 | 1,498,369,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00353.warc.gz | 585,031,661 | 4,082 | Math
posted by .
Consider a polygon ABCDE with coordinates A(0,4), B(3,0), C(7,2) , D(11,6) and E(8,15)
Choose all possible pairs from below at which a maximum over the region can be attained by a non constant linear function.
[(0,4),(7,2)] [(0,4),(8,15)]
[(7,2),(11,6)] [(11,6),(8,15)]
[(0,4),(3,0)] [(3,0),(7,2)]
[(3,0),(11,6)] [(0,4),(11,6)]
[(7,2),(8,15)] [(3,0),(8,15)]
Can someone at least help get me started?
Thanks so much!
• Math -
i don't no how to do it
• Math -
16 | 187 | 485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2017-26 | latest | en | 0.795178 |
https://docs.lfortran.org/pt/ast_and_asr/ | 1,716,794,620,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00005.warc.gz | 166,409,293 | 9,029 | # Diferenças entre uma AST e uma ASR¶
Tomemos um simples código Fortran:
integer function f(a, b) result(r)
integer, intent(in) :: a, b
integer :: c, d
c = a + b - d
r = c * a
end function
and look at what the AST and ASR look like.
## AST¶
[1]:
%%showast
integer function f(a, b) result(r)
integer, intent(in) :: a, b
integer :: c, d
c = a + b - d
r = c * a
end function
(TranslationUnit
[(Function
f
[(a)
(b)]
[(AttrType
TypeInteger
[]
()
()
None
)]
r
()
()
[]
[]
[]
[(Declaration
(AttrType
TypeInteger
[]
()
()
None
)
[(AttrIntent
In
)]
[(a
[]
[]
()
()
None
())
(b
[]
[]
()
()
None
())]
()
)
(Declaration
(AttrType
TypeInteger
[]
()
()
None
)
[]
[(c
[]
[]
()
()
None
())
(d
[]
[]
()
()
None
())]
()
)]
[(Assignment
0
c
(- (+ a b) d)
()
)
(Assignment
0
r
(* c a)
()
)]
[]
[]
)]
)
A AST não possui nenhuma informação semântica, mas sim «nós» que representam declarações do tipo integer, intent(in) :: a. Variáveis como a são representadas pelo nó Name e ainda não estão conectadas às suas declarações.
The AST can also be exported in JSON, including source file name, line and column information: lfortran example.f90 --show-ast --json
## ASR¶
[2]:
%%showasr
integer function f(a, b) result(r)
integer, intent(in) :: a, b
integer :: c, d
c = a + b - d
r = c * a
end function
(TranslationUnit
(SymbolTable
1
{
f:
(Function
(SymbolTable
2
{
a:
(Variable
2
a
[]
In
()
()
Default
(Integer 4)
()
Source
Public
Required
.false.
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.false.
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c:
(Variable
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.false.
),
d:
(Variable
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(Integer 4)
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.false.
),
r:
(Variable
2
r
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ReturnVar
()
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Default
(Integer 4)
()
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.false.
)
})
f
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[(Integer 4)
(Integer 4)]
(Integer 4)
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Implementation
()
.false.
.false.
.false.
.false.
.false.
[]
.false.
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[]
[(Var 2 a)
(Var 2 b)]
[(Assignment
(Var 2 c)
(IntegerBinOp
(IntegerBinOp
(Var 2 a)
(Var 2 b)
(Integer 4)
()
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Sub
(Var 2 d)
(Integer 4)
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(Assignment
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(Integer 4)
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(Var 2 r)
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Uma ASR possui toda a informação semântica (tipos, etc.), nós do tipo Function possuem uma tabela de símbolos e não tem nenhuma nó de declaração. Variáveis são simplesmente ponteiros para a tabela de símbolos.
The ASR can also be exported in JSON, including source file name, line and column information: lfortran example.f90 --show-asr --json
## Discussão¶
O exemplo acima foi simples. As coisas ficam mais evidentes para exemplos mais complicados, como:
integer function f2b(a) result(r)
use gfort_interop, only: c_desc1_int32
integer, intent(in) :: a(:)
interface
integer function f2b_c_wrapper(a) bind(c, name="__mod1_MOD_f2b")
use gfort_interop, only: c_desc1_t
type(c_desc1_t), intent(in) :: a
end function
end interface
r = f2b_c_wrapper(c_desc1_int32(a))
end function
As AST devem representar todos os comandos use e blocos de interface, bem como manter as coisas semanticamente consistentes.
As ASR, por outro lado, devem manter os registros de c_desc1_int32, c_desc1_t e f2b_c_wrapper na tabela de símbolos e sabe que eles foram definidos no módulo gfort_interop e portanto a ASR não possui nenhuma desses nós de declaração.
Ao converter uma ASR em uma AST, o LFortran criará todos os nós de declaração da AST apropriados de forma automática e correta. | 1,120 | 3,482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.344759 |
https://www.assignmentexpert.com/homework-answers/economics/microeconomics/question-76016 | 1,643,463,947,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00602.warc.gz | 680,852,707 | 67,257 | 104 414
Assignments Done
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# Answer to Question #76016 in Microeconomics for Claire
Question #76016
Given the demand function :
P = 500 - 4Q^2
Calculate arc elasticity averaged along the arc joining Q = 8 and Q = 10
1
2018-04-13T15:50:14-0400
E = (Q_2 - Q_1)*(P_1+P_2) / ((Q_2 + Q_1)*(P_2-P_1))
Q_1 = 8
Q_2 = 10
P_1 = 500 - 4*8^2 = 244
P_2 = 500 - 4*10^2 = 100
Then
E = (10 - 8)*(244+100) / ((10 + 8)*(100-244)) = - 2*344/(18*144) = -0.27
Demand is relatively inelastic
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for any assignment or question with DETAILED EXPLANATIONS! | 258 | 657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2022-05 | latest | en | 0.792994 |
https://www.homeworkmarket.com/questions/dependent-samples-t-test-or-anova | 1,579,856,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250616186.38/warc/CC-MAIN-20200124070934-20200124095934-00280.warc.gz | 896,173,003 | 51,550 | Dependent-Samples t Test or ANOVA”
MAROSA5913
submit a basic research proposal that calls for the use of a dependent-samples t test or repeated-measures ANOVA.
The paper should be APA formatted as a research proposal, and contain approximately 990-1320 words of content. Include a title page, and a reference page that includes any resources utilized.
Please include the following in the research proposal:
1. Introduction (1-2 paragraphs)
• Present the research question of interest.
• Explain how the chosen statistical test applies to this research question.
• Provide the statistical notation and written explanations for the null and alternative hypotheses.
2. Methods (1 paragraph)
• Participants
• List how many participants will be selected.
• Identify who will be the participants and their major demographic characteristics (e.g., sex, age, etc.).
• Explain how participants will be selected for the study.
3. Procedures (1-2 paragraphs)
• Identify the variables in the study.
• Describe each variable’s scale of measurement (nominal, ordinal, interval, or ratio) and characteristics (i.e., discrete vs. continuous, qualitative vs. categorical, etc.).
• Provide an operational definition for each variable, explaining how the variables will be measured.
4. Results (2-3 paragraphs)
• Describe the statistical test that will be conducted. Be sure to include why the test was chosen and why it is appropriate for this study. Include in the discussion the necessary assumptions that should be met for the chosen test and how these will be addressed.
• Identify the information that will be obtained from the results of this test and what will be needed to draw conclusions regarding the hypotheses. Be sure to include a discussion of applicable critical and calculated values, p levels, confidence intervals, effect sizes, post-hoc tests, and/or tables.
5. Discussion (1 paragraph)
• Identify any expected biases, assumptions, or faults with the proposed study and the use of the identified statistical test.
• Explain what conclusions can and cannot be made for this study, and using this statistical test.
• Describe the practical significance or importance of the results.
• Posted: 3 months ago
• Due:
• Budget: \$15
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Assignment 2
This written assignment is based on the work conducted in the “Independent-Samples t Test or ANOVA” discussion forum. Based on this initial work, feedback received, and additional …
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Mini Research Proposal
This written assignment is based on the work conducted in the “Z, T, or Chi-Square Test Study” discussion forum from last week. Based on this initial work, feedback received, …
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Correlation and Regression”
This written assignment is based on the work conducted in the “Correlation and Regression” discussion forum. Based on this initial work, feedback received, and additional research, students should …
• Not rated
Simple Linear, Multiple, Or Logistic Correlation/Regression Proposal
This written assignment is based on the work conducted in the “Correlation and Regression” discussion forum. Based on this initial work, feedback received, and additional research, students should …
• Not rated
“Complex ANOVA, ANCOVA, or MANOVA
his written assignment is based on the work conducted in the “Complex ANOVA, ANCOVA, or MANOVA” discussion forum. Based on this initial work, feedback received, and additional research, …
• Not rated
Mini Research Proposal ANOVA
This written assignment is based on the work conducted in the “Basic ANOVA Study” discussion forum. Based on this initial work, feedback received, and additional … | 782 | 3,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-05 | latest | en | 0.85784 |
http://mathandreadinghelp.org/middle_school_math_problems.html | 1,571,327,021,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00105.warc.gz | 131,248,797 | 8,178 | # Middle School Math Problems with Solutions
Math is a cumulative subject; as students advance from grade to grade, the concepts they learn in math class advance, too. If your child is struggling with middle school math, you can introduce sample problems and model how to solve them as a way to boost his confidence in his abilities.
## How Are Middle School Math Problems Different from Elementary Math Problems?
As your child enters middle school, he likely will be introduced to math concepts that are quite a big tougher than he's used to. While first through fifth grade generally are spent mastering basic skills in addition, subtraction, multiplication and division, middle school math typically requires more complex application of these skills in the form of ratios, proportions and statistical thinking.
If your child is struggling with a particular math concept, you might work through a sample question with him. Then, stand by while he attempts to solve a similar problem on his own, which can help him gain confidence in his math skills.
## Middle School Math Problems and Solutions by Grade Level
### Sixth
1. If a car is traveling 45 miles per hour, how many miles will it travel in seven hours?
To solve, your child should create a proportional relationship: 45/1 = x/7. Then, he or she should cross multiply to reach the answer: x = 315 miles.
2. 4/5 ÷ 2/1
To divide fractions, your child first must turn the second fraction into a reciprocal fraction, so 2/1 becomes 1/2. Then, he or she should multiply across: 4/5 x 1/2 = (4 x 1)/(5 x 2) = 4/10, which can be reduced to 2/5.
3. -41 + 8
In middle school, students begin to work with negative numbers, which can be confusing at first. You might have your child use a number line to visualize the negative number system. The answer to this problem is -33.
### Seventh
1. On his quiz, John answered 12 out of 15 questions correctly. What percentage of questions did he get right?
First, your child should establish the fraction 12/15, which equals 0.80. The, he should multiply the decimal by 100 to find the percentage: 0.80 x 100 = 80%.
2. A building is 510 feet high. On a scale drawing, one inch equals three feet. How large would the building be on the drawing?
Your child should use proportional relationships to solve this problem. The first ratio is 1/3 because three feet are represented by one inch. The second ratio is x/510, with the variable representing the height of the building in the scale drawing. Next, your child can establish a proportional relationship like this: 1/3 = x/510. Then, cross multiply so that 3x = 510, and divide both sides by three. The building would be 170 inches on the drawing.
### Eighth
1. Solve for x: 23 = 4x - 1.
Your eighth-grader should add one to both sides, so that the problem looks like this: 24 = 4x. Then, he can isolate the variable by dividing both sides by four. The answer is x = 6.
2. In a right triangle, the base is three inches, and the height is four inches. Find the length of the hypotenuse.
For this problem, your child will have to use the Pythagorean Theorem, which is A^2 + B^2 = C^2. He should first insert the known values: (3)^2 + (4)^2 = C^2. Next, he should square the numbers so that 9 + 16 = 25 = C^2. Finally, your child must find the square root of 25, which is 5. The hypotenuse is five inches long.
Did you find this useful? If so, please let others know!
## Other Articles You May Be Interested In
• Middle School Math Help: Sixth Grade Square Roots
Learning square roots can be tricky at first, because it's different from all the math you've learned so far. However, with some practice and review, you can master this concept, too. Keep reading to learn more.
• Middle School Math Help
Middle school students often need help when completing their mathematics homework. Read on to learn how you can help your child with his or her middle school math assignments.
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Online Only | 1,254 | 5,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2019-43 | longest | en | 0.962579 |
http://www.jiskha.com/display.cgi?id=1390443598 | 1,498,357,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320386.71/warc/CC-MAIN-20170625013851-20170625033851-00098.warc.gz | 593,991,717 | 3,691 | # calculus
posted by .
limit as n approaches infinity for the function ((-1)^n)((5n+7)/(6n+5))
• calculus -
multiply both numerator and denominator by 1/n
then
Lim (-1)^n (5+7/n)/(6+5/n)
so as n gets large, the second part becomes (5/6)
the first part is +-1, depending if n is odd or even.
there is not a limit. however, the funcion is +- 5/6 and is bounded by those two numbers. | 116 | 387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-26 | latest | en | 0.926666 |
https://plainmath.net/72008/sailing-ships-used-to-send-messages-w | 1,653,454,044,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00759.warc.gz | 483,682,579 | 12,056 | # 1. Sailing ships used to send messages with signal flags flown from their masts. How many different
1. Sailing ships used to send messages with signal flags flown from their masts. How many different signals are possible with a set of four distinct flags if a minimum of two flags is used for each signals?
2. A Gr. 9 students may build a timetable by selecting one course for each period, with no duplication of courses. Period 1 must be science, geography, or physical education. Period 2 must be art, music, French, os business. Period 3 and 4 must be math or English. How many different timetables could a student choose?
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Frida Wilkinson
For question 1, we must understand that with a minimum of 2 flags and a maximum of 4 flags being used for each signal configuration, the following configurations are possible:
1. 2 Flags Used
2. 3 Flags Used
3. 4 Flags Used
In the case of only 2 flags being used, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag. Thus, the number of different signals that can be made here is:
$4\cdot \left(4-1\right)=4\cdot 3=12$
In the case of using 3 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag. Thus, the number of different signals that can be made here is:
$4\cdot \left(4-1\right)\cdot \left(4-2\right)=4\cdot 3\cdot 2=24$
In the final case of using all 4 flags, any 4 of the flags can be picked initially as the possible first flag, leaving 3 flags remaining as the possible second flag, leaving 2 flags remaining as the possible third flag, leaving only 1 flag as the final flag to be chosen. Thus the number of different signals that can be made here is:
$4\cdot \left(4-1\right)\cdot \left(4-2\right)\cdot \left(4-3\right)=4\cdot 3\cdot 2\cdot 1=24$
$12+24+24=60$
For question 2, we can think of a tree-diagram to represent this. Starting with 3 separate trees: Science, Geography, and Phys. Ed at the top. From here, there are 4 possible classes to branch off of the first class: Art, Music, French, or business. We then branch off of these classes to Math or English, and the final period will be Math if the third period was English, or English if the third period was Math, in order to ensure no duplicates.
From here, $3\cdot 4\cdot 2=24$ possible timetables can be chosen. | 685 | 2,650 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 38, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-21 | latest | en | 0.914024 |
https://docs.juliahub.com/General/ClausenFunctions/stable/autodocs/ | 1,721,254,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00364.warc.gz | 188,983,540 | 4,046 | ClausenFunctions.clFunction
cl(n, z)
Returns the value of the Clausen function $\operatorname{Cl}_n(z)$ for all integers $n\in\mathbb{Z}$ and arguments $z$ of type Real or Complex. For complex $z\in\mathbb{C}$ this function is defined as
\begin{aligned} \operatorname{Cl}_n(z) &= \frac{i}{2}\left[\operatorname{Li}_n(e^{-iz}) - \operatorname{Li}_n(e^{iz})\right], \qquad \text{for}~n~\text{even}, \\ \operatorname{Cl}_n(z) &= \frac{1}{2}\left[\operatorname{Li}_n(e^{-iz}) + \operatorname{Li}_n(e^{iz})\right], \qquad \text{for}~n~\text{odd}. \end{aligned}
For real $z\in\mathbb{R}$ the function simplifies to
\begin{aligned} \operatorname{Cl}_n(z) &= \Im[\operatorname{Li}_n(e^{iz})] = \sum_{k=1}^\infty \frac{\sin(kz)}{k^n}, \qquad \text{for even}~n>0, \\ \operatorname{Cl}_n(z) &= \Re[\operatorname{Li}_n(e^{iz})] = \sum_{k=1}^\infty \frac{\cos(kz)}{k^n}, \qquad \text{for odd}~n>0. \end{aligned}
Note: $\operatorname{Cl}_1(z)$ is not defined for $z=2k\pi$ with $k\in\mathbb{Z}$.
For $z$ of type Float16, Float32 or Float64 the implementation follows the approach presented in [Jiming Wu, Xiaoping Zhang, Dongjie Liu, "An efficient calculation of the Clausen functions Cl_n(θ)(n >= 2)", Bit Numer Math 50, 193-206 (2010) https://doi.org/10.1007/s10543-009-0246-8].
Author: Alexander Voigt
Example
julia> cl(10, 1.0)
0.8423605391686301
julia> cl(10, big"1")
0.8423605391686302305168624869816554653186479602725762955247227248477087749849797
julia> cl(10, 1.0 + 1.0im)
1.301796548970136 + 0.6333106255561783im
julia> cl(10, big"1" + 1im)
1.301796548970136401486390838171927382622047488046695826621336318796926878116022 + 0.6333106255561785282041229828047700199791464296534659278461150656661132141323808im
ClausenFunctions.cl1Function
cl1(z)
Returns the value of the Clausen function $\operatorname{Cl}_1(z)$ for an argument $z$ of type Real or Complex. This function is defined as
$$$\operatorname{Cl}_1(z) = -\frac{1}{2}\left[\log(1 - e^{iz}) + \log(1 - e^{-iz})\right]$$$
Note: $\operatorname{Cl}_1(z)$ is not defined for $z=2k\pi$ with $k\in\mathbb{Z}$.
Author: Alexander Voigt
Example
julia> cl1(1.0)
0.04201950582536895
julia> cl1(big"1")
0.04201950582536896172579838403790203712453892055703441769956888996856898991572203
julia> cl1(1.0 + 1.0im)
-0.3479608285425303 - 0.7021088550913619im
julia> cl1(big"1" + 1im)
-0.3479608285425304681676697311626225254052481291939705626677618712960309352197459 - 0.7021088550913618982002640571209884840333848751442483836897991093694470903764252im
ClausenFunctions.cl2Method
cl2(x::Real)::Real
Returns the value of the Clausen function $\operatorname{Cl}_2(x)$ for a real angle $x$ of type Real. The Clausen function is defined as
$$$\operatorname{Cl}_2(x) = -\int_0^x \log|2\sin(t/2)| dt$$$
Author: Alexander Voigt
Example
julia> cl2(1.0)
1.0139591323607684
ClausenFunctions.cl3Method
cl3(x::Real)::Real
Returns the value of the Clausen function $\operatorname{Cl}_3(x)$ for a real angle $x$ of type Real. This function is defined as
$$$\operatorname{Cl}_3(x) = \Re[\operatorname{Li}_3(e^{ix})] = \sum_{k=1}^\infty \frac{\cos(kx)}{k^3}$$$
Author: Alexander Voigt
Example
julia> cl3(1.0)
0.44857300728001737
ClausenFunctions.cl4Method
cl4(x::Real)::Real
Returns the value of the Clausen function $\operatorname{Cl}_4(x)$ for a real angle $x$ of type Real. This function is defined as
$$$\operatorname{Cl}_4(x) = \Im[\operatorname{Li}_4(e^{ix})] = \sum_{k=1}^\infty \frac{\sin(kx)}{k^4}$$$
Author: Alexander Voigt
Example
julia> cl4(1.0)
0.8958052386793799
ClausenFunctions.cl5Method
cl5(x::Real)::Real
Returns the value of the Clausen function $\operatorname{Cl}_5(x)$ for a real angle $x$ of type Real. This function is defined as
$$$\operatorname{Cl}_5(x) = \Re[\operatorname{Li}_5(e^{ix})] = \sum_{k=1}^\infty \frac{\cos(kx)}{k^5}$$$
Author: Alexander Voigt
Example
julia> cl5(1.0)
0.5228208076420943
ClausenFunctions.cl6Method
cl6(x::Real)::Real
Returns the value of the Clausen function $\operatorname{Cl}_6(x)$ for a real angle $x$ of type Real. This function is defined as
$$$\operatorname{Cl}_6(x) = \Im[\operatorname{Li}_6(e^{ix})] = \sum_{k=1}^\infty \frac{\sin(kx)}{k^6}$$$
Author: Alexander Voigt
Example
julia> cl6(1.0)
0.855629273183937
ClausenFunctions.slMethod
sl(n, z)
Returns the value of the Glaisher-Clausen function $\operatorname{Sl}_n(z)$ for all integers $n\in\mathbb{Z}$ and arguments $z$ of type Real or Complex. For complex $z\in\mathbb{C}$ this function is defined as
\begin{aligned} \operatorname{Sl}_n(z) &= \frac{1}{2}\left[\operatorname{Li}_n(e^{-iz}) + \operatorname{Li}_n(e^{iz})\right], \qquad \text{for}~n~\text{even}, \\ \operatorname{Sl}_n(z) &= \frac{i}{2}\left[\operatorname{Li}_n(e^{-iz}) - \operatorname{Li}_n(e^{iz})\right], \qquad \text{for}~n~\text{odd}. \end{aligned}
For real $z\in\mathbb{R}$ the function simplifies to
\begin{aligned} \operatorname{Sl}_n(z) &= \Re[\operatorname{Li}_n(e^{iz})] = \sum_{k=1}^\infty \frac{\cos(kz)}{k^n}, \qquad \text{for even}~n>0, \\ \operatorname{Sl}_n(z) &= \Im[\operatorname{Li}_n(e^{iz})] = \sum_{k=1}^\infty \frac{\sin(kz)}{k^n}, \qquad \text{for odd}~n>0. \end{aligned}
Note: We set $\operatorname{Sl}_1(0) = 0$ for consistency with the series expansion.
Author: Alexander Voigt
Example
julia> sl(10, 1.0)
0.5398785706335891
julia> sl(10, big"1")
0.5398785706335893473201701431352733846690266746377581980285682505050227015352517
julia> sl(10, 1.0 + 1.0im)
0.832020890646937 - 0.9921163924162678im
julia> sl(10, big"1" + 1im)
0.832020890646937194384242195240640660735825609698824394209500978844042810286907 - 0.9921163924162683284596904585334878021171116435958891901780686447162617670776108im | 2,142 | 5,674 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.366134 |
https://ask.learncbse.in/t/calculate-the-total-energy-consumed/12106 | 1,726,036,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00705.warc.gz | 90,042,213 | 3,559 | Calculate the total energy consumed
Calculate the total energy consumed in the month of November in a household in which four devices of power 500 W each are used daily for 10 h.
Given, power, P = 500 W, time, t = (4 x 10 X 30) h
Energy consumed in the month of November = 500 Wx 4x 10 hx 30 = 600000 Wh = 600 kWh = 600 units | 96 | 327 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-38 | latest | en | 0.911706 |
https://www.physicsforums.com/threads/coexistance-curve.393662/ | 1,544,830,260,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826530.72/warc/CC-MAIN-20181214232243-20181215014243-00264.warc.gz | 997,883,816 | 12,347 | # Homework Help: Coexistance Curve
1. Apr 9, 2010
### subtletuna
1. The problem statement, all variables and given/known data
Consider the liquid-gas coexistence curve. Show that C$$_{(g)}$$ − C$$_{(l)}$$ = T($$\stackrel{d}{dT}$$ $$\Delta$$$$_{(vap)}$$S) for T, P along the coexistence curve. Now use the fact that
$$\Delta$$(vap)S = $$\Delta$$(vap)H/T along the coexistence curve, and the fact that $$\Delta$$(vap)H is positive and decreases with increasing T, to show that C(g) < C(l) . Note that this inequality applies only at Tvap.
2. Relevant equations
3. The attempt at a solution
2. Apr 10, 2010
### Gokul43201
Staff Emeritus
Please show us what you have tried. We can not help you unless you first put in some effort. | 224 | 734 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.811621 |
https://www.sqlshack.com/mathematics-sql-server-fast-introduction-set-theory/ | 1,718,353,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00232.warc.gz | 902,888,486 | 58,125 | From mathematics to SQL Server, a fast introduction to set theory
April 25, 2017
Introduction
In the previous article of this series “An introduction to set-based vs procedural programming approaches in T-SQL”, we’ve seen from a simple example that we could find actual benefit from learning set-based approach when writing T-SQL code.
In this article, we will carry on in this way by having a look at what a set is and what we can do with it in a mathematical point of view and how it’s implemented and provided to us in SQL Server. We will also have a look at more “realistic” examples using Microsoft’s AdventureWorks database.
Set Theory and fundamentals
Set definition
In mathematics, we define set theory is a branch of mathematics and more particularly mathematical logic that studies collections of objects we refer to as sets.
Don’t worry, we won’t do a lot of maths here as we will focus on practical aspects that we will use when writing T-SQL queries. Let’s just review some fundamentals of this theory:
• The elementary set is the empty set. It’s a collection of zero objects and you will find in some references, it’s also called the null set. Its notation is ∅ or { }.
• A non-empty set contains the empty set plus one or more objects. This also means that a set can contain a set.
• There is a fundamental binary relation between an object and a set: object to set membership. This is equivalent to the IN operation in a T-SQL query.
• As a set can contain another set, last binary relation can be extended to set to set membership also known as subset relation or set inclusion.
• Just like any arithmetic theory, set theory defines its own binary operations on sets. As an example, in number theory, we can find operations like addition or division.
Graphical representation of a set
In order to graphically represent an operation on sets, it’s common to use Venn diagrams, which show all possible logical relations between a finite number of different sets.
You will find an example of representation for a single set and its objects in following figure. This set contains the following objects: A, B, D, L, o, Q and z.
In following, we will use this representation in order to provide a good understanding of the product of each operation we will define.
Set operations and their equivalent in SQL Server
Now, let’s talk about operations on two sets called A and B. We will also translate them to T-SQL statements where A and B will be either tables or the results of a query.
As almost all readers should know, the way to get the content of either set A or set B is performed as followed using T-SQL (example with set A):
Note
In following, except where otherwise stipulated, T-SQL « sets » A and B have the same number of columns of same types. This is the reason why we used SELECT * notation above.
The union operation
The union operation will produce set made up with all the objects from A and all objects from B. It’s denoted as A∪B.
For instance, if A is composed of {1,3,6} and B of {3,9,10} then A∪B is a set composed of {1,3,6,9,10}.
Graphical representation of A∪B is as follows. Set A is represented by red circle while green circle represents Set B. A∪B is the grayscaled parts of those circles. Further representations will use this color usage convention.
In SQL Server, we will find an implementation of the UNION operator. You will find below an equivalent T-SQL statement to the A∪B set:
There is a little difference between T-SQL and set theory: Microsoft provides the possibility for their user to keep duplicate records that should normally be erased by UNION operator. To do so, we will append the word ALL after UNION.
Back to the previous example using sets of numbers, the UNION ALL of set A with set B will generate following set:
{1,3,3,6,9,10}
Now, let’s take an example using SQL Server and check that there is an actual difference between UNION and UNION ALL operators.
Union Example
Union all Example
As we can see in the above example, UNION operation is translated to MERGE JOIN operator in SQL Server while UNION ALL operator simply takes up all rows from each set and concatenates it.
So far, we haven’t seen what a « join » is. Basically, a join is a way to get a set based on two or more tables. This results set has either the same columns as base tables or column from both tables implied in join operation.
The intersection operation
The intersection of set A and set B is denoted A∩B and is the set of elements that can be found in both sets.
Back to the example with numeric sets,
• A = {1,3,6}
• B = {3,9,10}
• A∩B = {3}
Graphically, it looks like:
In SQL Server, there is also an INTERSECT T-SQL operator that implements this set operation. You will find below an equivalent T-SQL statement to the A∪B set:
Now, let’s look at a concrete T-SQL example where A is the set of persons with firstname starting with a « J » and B is the set of persons with lastname starting with a « E ». A∩B is the set of persons with « J. E. » as initials.
The execution plan for this particular query does not represent the equivalent operator in SQL Server database engine. As we can see, the INTERSECT operation is translated into a chain of Nested Loop operators as we have a WHERE clause in each sub-query.
If we comment those WHERE clauses, we will get following execution plan, that uses a MERGE JOIN operator, like for the UNION operation but in an « inner join » mode:
Set difference operation
The difference between a set A and a set B is denoted A \ B and will take all the elements composing set A that are not in set B.
Back to the example with numeric sets,
• A = {1,3,6}
• B = {3,9,10}
• A \ B = {1,6}
Graphically, it looks like:
In SQL Server, this operation is also implemented and available to users via EXCEPT operator. You will find below an equivalent T-SQL statement to the A \ B set:
So, if we want to get a concrete example, let’s say we want to get the identifier of all persons that do not have a contact phone number. To do this, we will take table Person.Person as set A and Person.PersonPhone table as set B. This gives us following statement:
If we take a look at its execution plan, we see that the operator used by SQL Server is called « Hash Match (Left Anti Semi Join) ».
The Cartesian product operation
Operation explanation
The Cartesian product is denoted A × B and is the set made up with all possible ordered pairs (a,b) where a is member of set A and b is member of set B.
Back to our example using numbers where:
• A = {1,3,6}
• B = {3,9,10}
In order to get the elements of A × B, we can make a table with an element of A by row and an element of B by column. Each combination of row and column values will be an element of A × B.
Elements of set B Elements of set A 3 9 10 1 (1,3) (1,9) (1,10) 3 (3,3) (3,9) (3,10) 6 (6,3) (6,9) (6,10)
So, A × B = {(1,3),(1,9),(1,10),(3,3),(3,9),(3,10),(6,3),(6,9),(6,10)}
Well, we might be very confused when seeing this operation and be asking ourselves « what the hell can I do with that? ». Actually, this operation is very useful in a wealth of situations and we will use it extensively in last article of this series.
We will first have a look at the way to run a cross join using T-SQL.
Corresponding implementation in SQL Server
In SQL Server, we can write a Cartesian product using CROSS JOIN command as follows.
Note:
• Here, star will return all columns from A and from B
• If we would like to cross join A with itself, we would get following error message except if we provide an alias for at least one of table A occurrences.
Alternatively, we can simply use a comma to replace CROSS JOIN notation:
Cross join real life application example usages
Now we know how to write a query using a cross join or Cartesian product, well, we should know in which cases we could use it.
Example 1: Compute/Generate all possible cases for a particular situation
Let’s assume we are in a clothing factory and we want to know how many different kinds of pieces we can create and at which cost, based on clothing size and clothing color.
If we have a ClothingSizes and a ClothingColors tables, then we can take advantage of CROSS JOIN operation as follows.
The results:
Example 2: Generate test data
With the example above, you can imagine a solution with a list of first names and a list of last names. When performing a cross join on both, we would get candidates for a Contact or a Person table. This can also be extended to addresses and any kinds of data. Your imagination is the limit.
Example 3: Generate charts data (X axis)
This example is part of an advanced example for set-based approach so it won’t be developed in details here. We will just present the situation. Let’s say we have a tool that logs any abnormal behavior with timestamp inside a SQL Server table but does not log anything when everything works like expected. In such a case, if we want, for instance, to plot number of abnormalities occurrences by day, we will face a problem as there are « holes » in data.
So, we cannot plot a chart directly and we have to first generate a timeline with the appropriate step (here hours). To generate this timeline, we would need to use CROSS JOIN.
In summary, we would consider a set containing short dates of interest, then cross join them with a collection of 24 numbers from 0 to 23, representing hours in a day.
If we would need to report by minutes in a day, we would add another CROSS JOIN operation with a collection of 60 numbers from 0 to 59 representing minutes in an hour.
Mathematical operations with no equivalent in SQL Server
These set operations are not that easy to implement so that it will work for every single case in an efficient manner. I think that’s the reason why Microsoft did not implement them.
In following, we will present the operation itself, an example use case where they could be useful and an implementation specific to this use case.
The symmetric difference operation
Symmetric difference operation is equivalent to a logical XOR. It’s denoted as A⊕B and contains all the elements that are in set A but not in set B and those that are in set B but not in set A.
Graphically, this operation can be represented as:
We can implement it in different ways:
Implementation 1 – simply like its definition
Implementation 2 – Using IN operator for key columns
The power set operation
Power set of a set A is the set composed of all possible subsets of set A.
In our former example using sets of numbers, A = {1,3,6}. This means that the power set of A is composed of following elements:
• The empty set
• Sets of one element {1}{3}{6}
• Sets of two elements {1,3}{1,6}{3,6}
• Sets of three elements (which is actually set A).
I haven’t found any particular reason to use this set operation in real life, but feel free to contact me if you find one!
Summary
In this article, we’ve seen that SQL Server implements most of mathematical operations on sets. We can add the set operators defined here on the right of the table create in previous article of this series –“An introduction to set-based vs procedural programming approaches in T-SQL”. As a reminder, this table summarizes instructions and objects we can use in both procedural and set-based approaches.
Procedural Approach Set-Based Approach SELECT and other DML operations, WHILE, BREAK, CONTINUE, IF…ELSE, TRY…CATCH Cursors (OPEN, FETCH, CLOSE) DECLARE SELECT and other DML operations, Aggregate functions (MIN, MAX, AVG, SUM…) UNION and UNION ALL EXCEPT and INTERSECT CROSS JOIN | 2,694 | 11,597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-26 | longest | en | 0.926016 |
https://exceed.hr/blog/dax-handbook-filtering-directions/ | 1,717,104,333,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971684053.99/warc/CC-MAIN-20240530210614-20240531000614-00279.warc.gz | 204,835,842 | 58,705 | # DAX HANDBOOK5.3 Filtering directions
If you wish to follow along, you can find PBIX/Excel files at the bottom of the article.
### Guide Topics (red is currently selected one)
#### Krešimir Ledinski
Microsoft certified expert in the field of Business Intelligence. His biggest passions are DAX, M, and data modeling.
Experienced data analyst. Advanced in SQL, PowerApps and M language.
## Explanation
When we talk about filtering directions, we are referring to the arrows in the relationships.
The way they are pointing is the way the filters can move through the data model. Most of the time you will have a single way of filtering data, and that is from 1 to many side of the relationship (the left orange arrow in the picture above). You can also allow bidirectional filtering, meaning filters can move from many to one side of the relationship too (the right orange arrow).
## Single Filter Direction
Let’s say we want to calculate the distinct number of Cities our products are sold in.
The calculation we will use is
```DistinctNumberOfCities = DISTINCTCOUNT(DimCustomer[City])
```
We will first try with a single-direction filter.
We want to show [DistinctNumberOfCities] filtered by the Colors column coming from DimProduct Table.
The result is always the same number, which is the total number of Cities in the DimCustomer table. Even though there is a relationship between the two tables through the FactSales table, the filters are unable to access the DimCustomer table through the relationship FactSales[CustomerKey] -> DimCustomer[CustomerKey] because the arrow is pointing only in the direction of the FactSales table. This means that filter can travel from DimCustomer to the FactSales table, but the opposite direction is not possible, so the filter reaches the FactSales, but when it has to “jump” onto DimCustomer one, it is stopped. Since no filter reaches the DimCustomer table, the values for each color are DistinctCount of DimCustomer[City] with no filters applied.
## Setting Cross-filter Direction to Both
Now let’s change the direction between the tables and observe the results.
This time the calculation shows the expected figures. The filter from the DimProduct table can reach the DimCustomer table and is able to filter its row so that only the ones corresponding to a specific color survive. We will now explain in detail the internals of the filter flow for the [DistinctNumberOfCities] calculation. We will focus only on the row in the visual containing filter “Black”.
The Black filter on the Color column filters the DimProduct so that only black color rows survive.
In the Fact table, Product keys that survived condition “= Black” color cross filter CustomerKeys column. Only CustomerKeys of products with Black color will survive and will form a list of CustomerKeys which will jump to the DimCustomer table and filter it.
Crossfilter is a special DAX capability to form a list of active values for column based on filters applied to different columns of the same table. In the example above. We can see that black color will filter out certain ProductKeys from FactSales table. Those product keys will have only corresponding CustomerKeys survive the filtering. CustomerKeys that survive based on ProductKey filter are the one who are being crossfiltered.
After the DimCustomer is filtered so it contains only CustomerKey-s of sales of Black products, does the calculation kick in and count distinct values of the column City of the rows that survived the condition.
The distinct count of the column City for the Color “Black” equals 371 and that number is returned to the corresponding cell in the visual.
## Use Cases
Using bidirectional filters is not a good idea, especially on high cardinality columns, since it tends to slow down calculations and can also introduce ambiguity in a data model. It’s often a better idea to always keep the filtering direction from 1 to many sides, and apply bidirectional filters only when strictly needed for specific calculations with the use of CALCULATE function. The same calculation could be written the following way, without the need for applying a bidirectional filter on the relationships.
```DistinctNumberOfCitiesCalculate =
CALCULATE(
DISTINCTCOUNT( DimCustomer[City] ),
CROSSFILTER ( FactSales[CustomerKey], DimCustomer[CustomerKey], BOTH )
)
```
CROSSFILTER is a CALCULATE modifier function that can activate different types of relationship filtering directions. Those directions can be:
• Both
• None
• One Way
• One Way left to right
• One Way right to left
The advantage of using CROSSFILTER function instead of activating bidirectional relationship is flexibility to use bidirectional filters only in specific calculations and without potential negative impact on the whole data model.
## Active vs Non-active
Data Model takes good care of relationships, especially with regard to Ambiguity. Ambiguity means that filters can travel from one table to the other through 2 or more different paths. In the picture below we can see that there are two possible paths between DimDate and DimCustomer tables. One is through the Fact table, and the other is a direct connection through many 2 many relationship based on Year columns. In this case, the model automatically makes one of the relationships non-active. But we can activate it on demand with the use of CALCULATE + USERELATIONSHIP function modifier.
We will add 2 calculations and observe their results.
```NumberOfStoresSelling = COUNTROWS(DimCustomer)
NumberOfStoresOpened = CALCULATE(COUNTROWS(DimCustomer),USERELATIONSHIP(DimDate[Year],DimCustomer[YearOpened]))
```
[NumberOfStoresSelling] is using active relationships in the model, going through the FactSales table to reach DimCustomer and perform the calculation. [NumberOfStoresOpened] measure uses CALCULATE modifier USERELATIONSHIP to activate the direct M2M relationship between DimDate and DimCustomer, therefore, ignoring the active relationship path.
## Materials
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### DAX HANDBOOK7. VARIABLES
Variables are used in almost every measure you will create. The reason we introduce them last is that they use all other parts of DAX code to produce faster, more powerful and maintainable code. Variables are like containers of a part of the DAX code which can be used throughout...
### DAX HANDBOOK6.8 ALLSELECTED
Explanation ALLSELECTED is one of the most complex functions in DAX. When used improperly it can lead to unexpected results. As a rule of thumb, you should not use it in iterative functions. It is preferable to use it only as a CALCULATE filter remover, not as a table function....
### DAX HANDBOOK6.7 Lineage
What is Lineage? Lineage is a part of DAX mechanics that enables us to use tables as filter arguments for the CALCULATE function. It’s also used in row2filter context transition and other aspects of the data model that involve filter propagation throughout the model. We can state that lineage is...
### DAX HANDBOOK6.6 Crossfilter
Crossfilter is a feature of DAX when it filters the underlying dataset even though there aren’t any visual filters present. Introduction In this example, we will explain a very important feature of CALCULATE filter arguments. We will also explain why you should always prefer the combination of ALL/REMOVEFILTER + VALUES...
0 | 1,599 | 7,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.90441 |
https://hoven-in.appspot.com/Home/Aptitude/Problems-on-Numbers/maths-aptitude-quiz-questions-and-mock-test-on-problems-on-numbers-013.html | 1,721,421,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00512.warc.gz | 256,224,869 | 6,246 | # Problems on Numbers Quiz Set 013
### Question 1
The sum of three consecutive integer numbers is -273. The smallest of the three is?
A
-92.
B
-91.
C
-89.
D
-90.
Soln.
Ans: a
Let the numbers be n - 1, n and n + 1. The sum is 3n. We are given 3n = -273, ⇒ n = \$-273/3\$, i.e., n = -91. So the smallest is -92.
### Question 2
The sum of three consecutive even integer numbers is 2556. The middle among the three is?
A
852.
B
853.
C
851.
D
854.
Soln.
Ans: a
Let the numbers be 2n - 2, 2n and 2n + 2. The sum is 6n = 3 x 2n = 3 x middle. We are given 3 x middle = 2556, ⇒ middle = \$2556/3\$, i.e., middle = 852.
### Question 3
How many prime factors does 34650 have?
A
7.
B
8.
C
6.
D
9.
Soln.
Ans: a
We can see that \$34650 = 2^1 × 3^2 × 5^2 × 7^1 × 11^1\$. The number of prime factors is sum of the powers = 7.
### Question 4
The sum of three consecutive integer numbers is 1014. The smallest of the three is?
A
337.
B
338.
C
336.
D
339.
Soln.
Ans: a
Let the numbers be n - 1, n and n + 1. The sum is 3n. We are given 3n = 1014, ⇒ n = \$1014/3\$, i.e., n = 338. So the smallest is 337.
### Question 5
What is the digit at units place: 4540 × 7403 × 8829?
A
0.
B
1.
C
3.
D
2.
Soln.
Ans: a
The digit at units place = (last digit of 4540) × (last digit of 7403) × (last digit of 8829) = 0 × 3 × 9 = 0, which = 0. | 527 | 1,368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.750577 |
http://research.stlouisfed.org/fred2/series/RGDPCHGAA625NUPN?rid=258 | 1,433,240,541,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195035572.9/warc/CC-MAIN-20150601214355-00038-ip-10-180-206-219.ec2.internal.warc.gz | 157,129,280 | 19,857 | # Purchasing Power Parity Converted GDP Per Capita (Chain Series) for Gabon
2010: 9,895.85838 2005 International Dollars per Person (+ see more)
Annual, Not Seasonally Adjusted, RGDPCHGAA625NUPN, Updated: 2012-09-17 10:39 AM CDT
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(a) Purchasing Power Parity Converted GDP Per Capita (Chain Series) for Gabon, 2005 International Dollars per Person, Not Seasonally Adjusted (RGDPCHGAA625NUPN)
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``` University of Pennsylvania, Purchasing Power Parity Converted GDP Per Capita (Chain Series) for Gabon [RGDPCHGAA625NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/RGDPCHGAA625NUPN/, June 2, 2015. ```
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# 28 - Kapoor(mk9499 oldhomework 28 Turner(60230 This...
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Kapoor (mk9499) – oldhomework 28 – Turner – (60230) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A resistance R and a 1 . 8 H inductance are in series across a 55 Hz AC voltage. The voltage across the resistor is 40 V and the voltage across the inductor is 60 V. What is the resistance R ? Correct answer: 414 . 69 Ω. Explanation: Let : L = 1 . 8 H , f = 55 Hz , V R = 40 V , and V L = 60 V . The potential difference across an inductor is V L = I X L = I 2 π f L and the potential difference across a resistor is V R = I R . Thus we have V R V L = I R I 2 π f L = R 2 π f L , so the resistance is R = V R V L 2 π f L = parenleftbigg 40 V 60 V parenrightbigg 2 π (55 Hz) (1 . 8 H) = 414 . 69 Ω . 002 (part 2 of 2) 10.0 points What is the AC input voltage? Correct answer: 72 . 111 V. Explanation: Because V R leads V L by 90 in an LR circuit, the ac input voltage is V = radicalBig V 2 R + V 2 L = radicalBig (40 V) 2 + (60 V) 2 = 72 . 111 V .
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https://vzn1.wordpress.com/2015/01/04/ga-tuning-a-very-good-looking-solution-verging-on-breakthrough/ | 1,529,857,258,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00602.warc.gz | 735,558,745 | 40,266 | GA tuning & a very good-looking solution verging on breakthrough!?
❗ 💡 ❓ now the big fun begins, tuning the algorithm & looking for a real solution. did some further experimenting and came up with some new metrics. the prior metrics succeed in improving $x_a$ and $x_b$ but on further runs & thoughts (as is commonly the case with GAs), maybe these metrics that dont take into account slope of the sampled function are not sufficient to find the desired convergence behavior.
in fact to have below-constant max nonmonotonic run length $x_b$, the “sequential decrease/ nonreversal density” $x_a$ needs to increase gradually as the size of the starting seed increases (because average trajectory lengths increase as graphed previously) so merely maximizing $x_a$ is not enough. another basic metric is “average distance between local minima” and that needs to be constant, but again minimizing it over a small local region does not necessarily ensure the solutions will have a constant trend as is the case in some experiments. yet another simple (new) metric/ constraint worth looking at is “total count of local minima” and that also needs to increase at least as fast as the trajectory lengths to get the desired constant-bounded distance between local minima.
so then a natural idea is to look at trends of lines and a corresponding fitness function. how can this be measured/ quantified? the standard/ straightforward method is least squares linear regression, and that method was next employed.
experiments also showed that even with average distance between local minima nearly constant over a small optimizing region, sometimes there are massive irregular distances/ spikes in that metric. so next it seems like the variance in average distance between local minima is a key metric to minimize. of course small average distance between local minima is preferred but tuning GAs to optimize more than one thing at a time can sometimes be a can of worms.
the final evaluation function evaluates the linear regression of the variance in average distance between local minima fitted to the weighted walk starting at random starting seeds in the range $n \in \{40...60\}$ at offsets of 2 (10 points). in other words the algorithm is searching for a constant average distance but, while it is preferrably smaller, there is no fitness directly based on/ connected to its magnitude.
some other bug fixes: after some troubleshooting, discovered the prior code sometimes gets “NaN” “not-a-number” values for averages and standard deviations in rare cases, in one case where there are all ‘1’s in the binary expansion ie numbers in the form $2^n-1$.
so, after some not-difficult adjustments, ran the algorithm and found some striking results possibly verging on yet another breakthrough ❗ outlined in the following graphs. the algorithm found 9 improved solutions over ~7 hours. it is easy to throw out some for which $x_a$ is visually constant-bounded and not increasing (or declining) (even though $x_b$ may often be quite noisy in those cases and difficult to interpret unambiguously individually). but whats left?
a very strange-yet-dramatic (or vice versa?) solution pattern emerged over at least two different GA runs as documented in the following graphs, seems to be a strong vindication of the techniques, and apparently fits the desired criteria over a large test region $n \in \{1...750\}$ although in a somewhat unpredictable way. the weights are as below. note that this code has not been tested for strict resilience/ repeatability in floating point ops and controlled random number generator starting seeds, that will require some further analysis and code adjustments, although the current inherent seed randomness in the trend analysis seems to indicate some significant resilience/ repeatability of the solution.
the 1st graph is the whole “bottom line” of the max nonmonotonic distance ($x_b$). for many GA solutions these are very gradually increasing graphs as in the prior graphs. however, this solution is anomalous in many ways but also shows that the overall idea (linear weighting of nonlinear iterate analysis statistics) can apparently lead to quite structured behavior.
there are three very distinct regions. the leftmost region $n \in \{0...{\sim}150\}$ has the training region $n \in \{40...60\}$ and manages to stay below a constant under about 100, very desirable. but then it “blows up” around $n \approx 150$ and climbs rapidly in the 2nd region. then in the 3rd region starting around $n \approx 250$ its quite noisy but still apparently “sideways”, less than a ceiling of about ~1800 although with several large spikes.
the 2nd graph plots average (green) and standard deviation (red) in the nonmonotonic run lengths and has the same 3 corresponding regions. the 1st region is low. the 2nd region has rapidly climbing average/ deviation. the 3rd region shows both falling rapidly and then gradually! this is completely unexpected behavior in the test regions 2/3 outside of the training region 1 yet in the “final” region 3, desired, even “better than expected” or almost even “too good to be true” because only constant bounding was sought after. 😀 leading to a wild wonder/ conjecture, is an asymptotically declining average really theoretically possible? ❓ ❗
the 3rd graph plots sequential decrease/ nonreversal density $x_a$. in region 1 it is noisy and slowly climbing. in region 2 it falls rapidly. in region 3 it climbs gradually, the desired behavior (although does seem to plateau near the end right, which is maybe problematic).
the 4th graph plots total trajectory length in red (divided by 5 to scale it in the graph but which does not affect slope) and counts of local minima in green. the former climbs quite predictably but the latter shows the same 3-region behavior differences. in region 1 it climbs, in region 2 it falls, both in a single curved, concave-downward manner. then in region 3 it increases linearly. note in the 3rd region the local minima count climbs faster than the trajectory lengths, in line with the other statistics. wow!
the GA found “a” desired solution in the long run analysis although not in the form expected. it is nearly ideal except for some large nonmonotone distances and region 2; regions 1/3 are acceptable in their dynamics, with region 3 as the presumed asymptotic behavior. the big question is if there are any more region changes possible after the large analysis region. the limitations of empirical analysis are emphasized with these results. no hint of the unexpected test region 2/3 behavior is hinted in the training region 1 statistics. but the test-divergence-from-training is not at all typical of machine learning apps either. its not just a simplistic increase in error. its almost as if its the output of a program/ algorithm that has 3 different conditional cases…! 😮
actually the wondrous nature of this could be emphasized even further. a bunch of random walks are weighted and added together to get a distinctly less random walk. its almost unbelievable or magical. a dramatic, maybe vista-opening reaffirmation of the ancient/ deep principle, “the whole is greater than the sum of the parts.” or like a rabbit being pulled out of a hat. 😎 ❗
4
w0 -1.2965207298328
w1 0.00881255269287658
w2 0.834874460020838
w3 0.488768570929333
w4 -0.841996911397055
w5 0.461962697182684
w6 0.624682934548628
w7 0.182038750890474
w8 1.11402628548688
w9 0.853460506803983
w10 0.0276695282228376
w11 -0.428453944156957
w12 0.179899938636543
w13 -0.0481205801620707
#!/usr/bin/ruby1.8
def stat(l)
return 0, 0 if (l.empty?)
t = t2 = 0
l.each \
{
|x|
t += x
t2 += x ** 2
}
c = l.size
a = t.to_f / c
z = t2.to_f / c - a ** 2
sd = Math.sqrt(z < 0 ? 0 : z)
raise if (sd.nan?)
return a, sd
end
def avg(l)
t = 0
l.each { |x| t += x }
return t.to_f / l.size
end
def d(s)
c = s.split('').select { |x| x == '1' }.size
return c.to_f / s.length, s.length, c, s.size - c, c % 2, c % 3
end
def rank(n, l)
s = n.to_s(2)
m1 = Math.log(n)
m2 = m1 ** 2
m3 = m1 ** 0.5
d1, w, c1, c0, c2, c3 = d(s)
d2 = (0.5 - d1).abs
a1, sd1 = stat(s.split('0').map { |x| x.length })
a0, sd0 = stat(s.split('1').map { |x| x.length })
l2 = [m1, m2, m3, d1, w, c1, c0, c2, c3, d2, a1, sd1, a0, sd0]
t = 0.0
l.each_with_index { |w, i| t += w * l2[i] }
raise(l2.join(' ')) if (t.nan?)
return t
end
def f3(n)
return sprintf("%.3f", n).to_f
end
def f3m(l)
l.map { |x| f3(x.to_f) }
end
def coef(x, y)
x1 = x2 = y1 = y2 = xy = 0.0
n = x.size
n.times \
{
|i|
x1 += x[i]
y1 += y[i]
xy += x[i] * y[i]
x2 += x[i] ** 2
y2 += y[i] ** 2
}
d = n * x2 - x1 ** 2
a0 = (y1 * x2 - x1 * xy) / d
a1 = (n * xy - x1 * y1) / d
return a0, a1
end
def f(n, w)
l = []
t = 0
while (n != 1)
l << rank(n, w).abs
puts(l[-1]) if ($mode == 'x') n = (n % 2 == 0) ? n / 2 : (n * 3 + 1) / 2 t += 1 end c = 0 m = l[0] x = j = 0 l2 = [] (1...l.size).each \ { |i| c += 1 if (l[i] < l[i - 1]) if (l[i] < m) then l2 << i - j m, j = [l[i], i] end x = [x, i - j].max } r = c.to_f / (l.size - 1) return x, stat(l2), r, l2.size, t end def eval1(p, z) n = 2 ** p + rand(2 ** (p - 1)) return f(n, z) end def eval(z) l = [] x = [] p = 40 10.times { |i| x << p; l << eval1(x[-1], z).flatten; p += 2 } l1 = l.transpose l2 = stat(l1[0]) + l1[1..-1].map { |l0| avg(l0) } y = l1[2] return -(coef(x, y)[1].abs), l2 end def trend() l = File.open($fn).readlines
n = ARGV[0].nil? ? l.size - 1 : ARGV[0].to_i
d = l[n].split("\t")
s = f3m(d.slice!(0, 8))
w = f3m(d.slice!(0, $v))$stderr.puts(n, s.join("\t"), w.join("\t"), d.join("\t"))
f = File.open('out2.txt', 'w')
(2..750).each \
{
|p|
n = 2 ** p + rand(2 ** (p - 1))
l = f(n, w)
f.puts(l.join("\t"))
f.flush
}
end
def trend2()
exit
end
def rsign()
return rand(2) ? -1 : 1
end
$mode = ARGV[0]$v = 14
$fn = 'var.txt' if (!['a'].member?($mode))
trend2() if ($mode == 'x') trend() exit end l = [] 50.times \ { |i| w = (1..$v).map { rand() * 1.5 * ((i % 2 == 0) ? 1 : rsign()) }
l <<= [eval(w), w, "[#{i}]"]
}
f = File.open('out.txt', 'w')
f2 = File.open($fn, 'w') mhi = nil mlo = nil j = 0 loop \ { a = l[rand(l.size)] b = l[rand(l.size)] x = a[1] y = b[1] r = rand($v)
c = rand(4)
h = nil
case c
when 0
z = x.dup
z[r] = 0.0
when 1
z = x.dup
z[r] = rand() * 1.5 * rsign()
when 2
z = x.dup
z[r] *= rand() * 1.5 * rsign()
when 3
z = []
x.size.times { |i| z[i] = (rand(2) % 2 == 0) ? x[i] : y[i] }
h = "(#{a[2]}+#{b[2]})"
end
op = 'a'
c.times { op.succ! }
h = "#{a[2]}#{op}#{r}" if (h.nil?)
t = eval(z)
l.pop if (l.size > 200)
l << [t, z, h]
l = l.sort_by { |x| x[0][0] }.reverse
hi = l[0]
lo = l[-1]
fhi = flo = false
mhi, fhi = [hi, true] if (mhi.nil? || hi[0][0] >= mhi[0][0] && hi[2] != mhi[2])
mlo, flo = [lo, true] if (mlo.nil? || lo[0][0] > mlo[0][0])
ts = Time.now.to_s
f.puts([f3(t[0]).abs, f3(mhi[0][0]).abs, f3(mlo[0][0]).abs, h.length, j, ts].join("\t")) if (flo)
f.flush
f2.puts((mhi + [j, ts]).join("\t")) if (fhi)
f2.flush
j += 1
}
(1/10) this is some new code with some enhancements. oh, and oops the prior analysis code was truncating weights to 3 decimal digits! this was unintentional but on the flip side, the good news is that there is very good stability in the results, ie not much observed butterfly effect and even truncating a lot of decimal digits in the weights seems to have no noticeable effect on analysis statistics.
the code has some changes in the logging, a bunch of new genetic randomizing gene alterations too numerous to mention for now but easily understood by inspection of the code. it has new analysis code and other similar fitness functions, 3 types as labelled by letter modes. in each case it attempts to maximize upward slope after fitting by linear regression around the same training region:
1. the sequential decreasing/nonreversal density
2. the local minima count
3. the local minima density (prior count divided by trajectory length)
got some interesting/ similar yet somewhat different/ improved results in these 3 runs labelled a, b, c. the ‘b’ metric performed best and found a new solution with max nonmonotonic run length of less than ~650. here are the weights and max nonmonotonic run length plots for the best solutions found after over 24hr of GA iterations, ‘a’ red, ‘c’ green, ‘b’ blue. in some ways these seem to be a newly scaled version of the prior solution, ie variations on a theme. the 3-region phenomenon continues at roughly the same numeric boundaries.
however the improvement in the max nonmonotonic run length ~650 of best solution ‘b’ over prior ~1400 is about 53% which is quite encouraging and even dramatic. am hunting for a low value here but havent figured out how to do this (yet) combined with a constant trend in a fitness function, so am relying on presumed indirect effects here through the other 3 fitness functions. it is not known yet if the improvement was due to the change in fitness functions or if its due to starting in better random starting conditions or even just randomness in the search etc; its currently too computationally expensive to do multiple runs. (wheres that open science collaboration team and supercomputing cluster when they’re needed?)
12
0 -0.0621151129675874
1 0.027749950267922
2 0.0285131621680541
3 -0.84576774323482
4 -0.901049776447273
5 -1.61758279610741
6 -0.178417524589377
7 0.0
8 -0.0
9 -1.66351603027066
10 0.0
11 -0.832684135084786
12 0.0
13 0.19429829301678
21
0 -0.937405391284104
1 0.0670177950097334
2 0.0
3 -1.73005189948666
4 -1.38257016553931
5 -0.558885208274334
6 -0.175932470473479
7 0.00144369304594031
8 0.0
9 0.463002776570181
10 0.746205672068276
11 0.790480164773667
12 0.0415956136680239
13 -0.918165827279676
11
0 4.15867051538285
1 -0.13286683498869
2 -0.49242392058947
3 0.0986854597916064
4 2.8536377301632
5 0.239334156410213
6 -0.341876102500191
7 0.14631302770963
8 -0.361365204553417
9 0.472727150190803
10 0.411456277848107
11 0.0
12 0.306902447155646
13 1.43546799909965
#!/usr/bin/ruby1.8
def stat(l)
return 0, 0 if (l.empty?)
t = t2 = 0
l.each \
{
|x|
t += x
t2 += x ** 2
}
c = l.size
a = t.to_f / c
z = t2.to_f / c - a ** 2
sd = Math.sqrt(z < 0 ? 0 : z)
raise if (sd.nan?)
return a, sd
end
def avg(l)
t = 0
l.each { |x| t += x }
return t.to_f / l.size
end
def d(s)
c = s.split('').select { |x| x == '1' }.size
return c.to_f / s.length, s.length, c, s.size - c, c % 2, c % 3
end
def rank(n, l)
s = n.to_s(2)
m1 = Math.log(n)
m2 = m1 ** 2
m3 = m1 ** 0.5
d1, w, c1, c0, c2, c3 = d(s)
d2 = (0.5 - d1).abs
a1, sd1 = stat(s.split('0').map { |x| x.length })
a0, sd0 = stat(s.split('1').map { |x| x.length })
l2 = [m1, m2, m3, d1, w, c1, c0, c2, c3, d2, a1, sd1, a0, sd0]
t = 0.0
l.each_with_index { |w, i| t += w * l2[i] }
raise(l2.join(' ')) if (t.nan?)
return t
end
def f3(n)
return sprintf("%.3f", n).to_f
end
def f3m(l)
l.map { |x| f3(x.to_f) }
end
def coef(x, y)
x1 = x2 = y1 = y2 = xy = 0.0
n = x.size
n.times \
{
|i|
x1 += x[i]
y1 += y[i]
xy += x[i] * y[i]
x2 += x[i] ** 2
y2 += y[i] ** 2
}
d = n * x2 - x1 ** 2
a0 = (y1 * x2 - x1 * xy) / d
a1 = (n * xy - x1 * y1) / d
return a0, a1
end
def f(n, w)
l = []
l1 = []
t = 0
while (n != 1)
x = rank(n, w).abs
l << x
l1 << [Math.log(n), x]
n = (n % 2 == 0) ? n / 2 : (n * 3 + 1) / 2
t += 1
end
return l1 if ($mode == 'x') c = 0 m = l[0] x = j = 0 l2 = [] (1...l.size).each \ { |i| c += 1 if (l[i] < l[i - 1]) if (l[i] < m) then l2 << i - j m, j = [l[i], i] end x = [x, i - j].max } r = c.to_f / (l.size - 1) return x, stat(l2), r, l2.size, l2.size.to_f / l.size, t end def eval1(p, z) n = 2 ** p + rand(2 ** (p - 1)) return f(n, z) end def eval(z) l = [] x = [] p = 40 10.times { |i| x << p; l << eval1(x[-1], z).flatten; p += 2 } l1 = l.transpose l2 = stat(l1[0]) + l1[1..-1].map { |l0| avg(l0) } y = l1[3] if ($mode == 'a')
y = l1[4] if ($mode == 'b') y = l1[5] if ($mode == 'c')
return coef(x, y)[0], l2
end
def out(n, w)
puts(n)
w.each_with_index { |w, i| puts("#{i}\t#{w}") }
end
def info()
File.open($fn).readlines.each_with_index \ { |ln, i| d = ln.split("\t") puts([i, f3(d[0]), d[-2], d[-3].length, d[-1]].join("\t")) if ($mode == 'y')
puts([i, d[-3]].join("\t")) if ($mode == 'z') if ($mode == 'w') then
s = d.slice!(0, $s) w = d.slice!(0,$v).map { |x| x.to_f }
out(i, w)
end
}
exit
end
def trend()
l = File.open($fn).readlines n = ARGV[0].nil? ? l.size - 1 : ARGV[0].to_i d = l[n].split("\t") s = d.slice!(0,$s)
w = d.slice!(0, $v).map { |x| x.to_f } out(n, w) f = File.open("out_#{n}.txt", 'w') (2..750).each \ { |p| n = 2 ** p + rand(2 ** (p - 1)) l = f(n, w) f.puts(l.join("\t")) f.flush } end def trend2() l = File.open($fn).readlines
d = l[-1].split("\t")
s = d.slice!(0, $s) w = d.slice!(0,$v).map { |x| x.to_f }
f = File.open('out2.txt', 'w')
p = 10
19.times \
{
l = eval1(p, w)
l.each_with_index { |x, i| f.puts([l.size - i, x].join("\t")) }
p += 10
f.puts
}
exit
end
def rsign()
return (rand(2) % 2 == 0) ? -1 : 1
end
def rw()
return rand() * 1.5 * rsign()
end
def re()
return rand(2) % 2 == 0
end
$mode = ARGV[0]$s = 9
$v = 14$fn = 'var.txt'
if (!['a', 'b', 'c'].member?($mode)) trend2() if ($mode == 'x')
info() if (['w', 'y', 'z'].member?($mode)) trend() exit end l = [] 50.times \ { |i| w = (1..$v).map { rand() * 1.5 * rsign() }
l <<= [eval(w), w, "[#{i}]"]
}
f = File.open('out.txt', 'w')
f2 = File.open($fn, 'w') mhi = nil mlo = nil j = 0 loop \ { a = l[rand(l.size)] b = l[rand(l.size)] x = a[1] y = b[1] r = rand($v)
c = rand(12)
op = 'a'
c.times { op.succ! }
h = "#{a[2]}#{op}#{r}"
case op
when 'a'
z = []
x.size.times { |i| z[i] = re() ? x[i] : y[i] }
h = "(#{a[2]}+#{b[2]})"
when 'b'
z = x.dup
z[r] = 0.0
when 'c'
z = x.dup
z[r] += rw()
when 'd'
z = x.dup
z[r] *= rw()
when 'e'
z = x.dup
z[r] = rw()
when 'f'
z = x.map { |w| re() ? w : w + rw() }
when 'g'
z = x.map { |w| re() ? w : w * rw() }
when 'h'
z = x.map { |w| re() ? w : w * rsign() }
when 'i'
z = x.map { |w| re() ? w : rw() }
when 'j'
z = x.map { |w| w + rw() }
when 'k'
z = x.map { |w| w * rw() }
when 'l'
z = x.map { |w| w * rsign() }
end
t = eval(z)
l.pop if (l.size > 200)
l << [t, z, h]
l = l.sort_by { |x| x[0][0] }.reverse
hi = l[0]
lo = l[-1]
fhi = flo = false
mhi, fhi = [hi, true] if (mhi.nil? || hi[0][0] >= mhi[0][0] && hi[2] != mhi[2])
mlo, flo = [lo, true] if (mlo.nil? || lo[0][0] > mlo[0][0])
ts = Time.now.to_s
f.puts([$mode, f3(t[0]).abs, f3(mhi[0][0]).abs, f3(mlo[0][0]).abs, h.length, j, ts].join("\t")) if (flo) f.flush f2.puts((mhi + [j, ts]).join("\t")) if (fhi) f2.flush j += 1 } there are so many interesting ways to improve the algorithm and analyze this data right now its a bit overwhelming. the blog writeup is also very time consuming at this pt. the GAs throw off a lot of auxiliary data related to convergence etc, but also there are so many different/ similar metrics and fitness functions now. it could take awhile to sort through all this. also need to do weight analysis to figure out whether to cut any metrics, and also which metrics are most effective by weight and thereby suggest new metrics. right now the weights seem to be generally nonzero for optimal solutions. but some different weights are zero and not in consistent positions. this is somewhat surprising that totally different weight configurations seem to be leading to similar structured solutions. a big possibly key, even critical mystery right now. 😮 ❓ ⭐ 😎 oh some other way cool news. am formulating some rough ideas about computing density over FSMs as hinted in the last post. not sure exactly how to do it (yet) but it seems computable and maybe even in P time. was able to formulate it into a cstheory.se question that got a few votes but more importantly, hit the jackpot with a quick answer by Shallit. its always a very good day when someone with a wikipedia biography replies in cyberspace. maybe even the highlight of the week. thanks much JS! 😀 (maybe if this all works out mine will be forthcoming soon) 😉 … JS has answered one of my other questions before also… and is also notable in that it takes something different for him to respond; he doesnt have too much activity on the site with currently only 42 answers and zero questions. a real life math guru at the top of the mountain! (1/11) ❗ just realized on editing my Collatz page that Shallit introduced the concept of use of regular languages in understanding the Collatz problem in 1991, ref [1] on that page. also added two newly googled interesting refs [27][28] apparently written by advanced undergraduates as senior theses, LaTourette and Chamberland. alas it seems a pity so few undergraduates seem to be exposed to the problem… (1/14) ❗ 😮 some deeper eyepopping/ jawdropping analysis of what turn out to be remarkable solutions. a basic analysis is to run the multiple evaluation statistics (in general measuring random walk monoticity) on the “unconditioned/ raw” collatz trajectories and compare those statistics directly with the “conditioned” walks. those reveal the strong/ definitive differences, but that is to be expected. it turns out, even more dramatic is to plot the original 1d walk vs conditioned 1d random walk as a curve (ie correlation between the two) to figure out how the algorithm is optimizing and “smoothing” the latter. these following plots are for “weight set 1”. samples are taken at starting trajectory bit lengths at increasing lengths of 10 between 10 and 300. starting positions are plotted in red. then up to 20, 50, or 100 of the initial trajectory values are plotted on top in green. the 3 corresponding regions are now apparent. region 1, 2 are the rising and falling sections of the concave-downward curve, roughly parabolic. then there is a climbing region 3. the curve is clearly smoothed but is surprising and unexpected and seems to represent some attractor in the optimization strategy. note the trajectories are plotted right-to-left. def stat(l) return 0, 0 if (l.empty?) t = t2 = 0 l.each \ { |x| t += x t2 += x ** 2 } c = l.size a = t.to_f / c z = t2.to_f / c - a ** 2 sd = Math.sqrt(z < 0 ? 0 : z) raise if (sd.nan?) return a, sd end def avg(l) t = 0 l.each { |x| t += x } return t.to_f / l.size end def d(s) c = s.split('').select { |x| x == '1' }.size return c.to_f / s.length, s.length, c, s.size - c, c % 2, c % 3 end def rank(n, l) s = n.to_s(2) m1 = Math.log(n) m2 = m1 ** 2 m3 = m1 ** 0.5 d1, w, c1, c0, c2, c3 = d(s) d2 = (0.5 - d1).abs a1, sd1 = stat(s.split('0').map { |x| x.length }) a0, sd0 = stat(s.split('1').map { |x| x.length }) l2 = [m1, m2, m3, d1, w, c1, c0, c2, c3, d2, a1, sd1, a0, sd0] t = 0.0 l.each_with_index { |w, i| t += w * l2[i] } raise(l2.join(' ')) if (t.nan?) return t end def maxrun(l) m = l[0][1] l2 = [] l1 = [] l.each \ { |x| if (x[1] < m) then l2 = [] m = x[1] else l2 << x end l1 = l2 if (l2.size > l1.size) } return l1 end def f(n, w) l = [] while (n != 1) x = rank(n, w).abs l << [Math.log(n) / Math.log(2), x] n = (n % 2 == 0) ? n / 2 : (n * 3 + 1) / 2 end return l end def g1(w) [1,20,50,100].each_with_index \ { |t, i| f = File.open("out#{t}.txt", 'w') p = 10 d = 10 while (p <= 300) n = 2 ** p + rand(2 ** (p - 1)) l = f(n, w) l[0...t].each { |xy| f.puts(xy.join("\t")) } f.puts if (t != 1) f.flush p += d end f.close } end def g2(w) c = 5 f = File.open("out.txt", 'w') f2 = File.open("outx.txt", 'w') p = 10 d = 10 h = 300 v = 500 (c * c).times \ { |i| x = (i % c) * h y = (i / c) * v n = 2 ** p + rand(2 ** (p - 1)) l = f(n, w) l.each { |xy| f.puts([-xy[0] + x, xy[1] - y].join("\t")) } f.puts f2.puts([x, -y].join("\t")) f2.puts([x, -y + v].join("\t")) f2.puts f2.puts([x, -y].join("\t")) f2.puts([x - h, -y].join("\t")) f2.puts f.flush f2.flush p += d } f.close f2.close end def g3(w) f = File.open("outy.txt", 'w') f2 = File.open("outz.txt", 'w') p = 150 n = 2 ** p + rand(2 ** (p - 1)) l = f(n, w) l.each_with_index { |x| f.puts(x.join("\t")) } maxrun(l).each { |x| f2.puts(x.join("\t")) } end def trend() w = File.open("w#{ARGV[0]}.txt").readlines.map { |x| x.split[1].to_f } g1(w) g2(w) g3(w) end trend() the next 3 plots contain 5×5 subplots of full trajectories starting at increasing bit lengths by 10, so from 10 to the 250 range, and arrange from top left to bottom right. the x axis is mirrored so that trajectories move left-to-right instead. they are for weight sets 1, 2, 3. here there is a bigger story, a lot to take in and it would definitely make sense as an animation! the weights 2, 3 are strikingly much smoother, the regions are slightly different, getting compressed, and also tend to climb much quicker, and the regions seem to correspond to the changing “bump” size/ shape at the end. ⭐ ⭐ ⭐ the sheer smoothness of these final 2 plots is stunning! more strong, exciting evidence of a real, unmitigated breakthrough. and it is not at all hinted in the random walk monotonicity statistics which generally still look quite noisy. showing how differences in quantitative metrics might not reveal fundamental shifts in qualitative worth and how critical different types of analysis/ pictures are for understanding the big(ger) picture. 💡 conjecture: interpreting the diagrams it is natural to wonder where the longest nonmonotone run is occuring. maybe the longest nonmonotone run for different trajectories correlates/ corresponds to the left upward side of the bump in the final left-to-right plots (right before the steep falling section). this should be easy to determine but havent gotten to it yet. (1/15) the conjecture was too easy! it is simple to demonstrate empirically (but to say nothing of proof). it explains the prior graphs in a very elegant way. the code has been modified to add the 3rd analysis routine that tracks the longest nonmonotone run and allows plotting it superimposed on the full run, and it always coincides as the “ending bump”. some more interpretation. imagine computing running averages of the original collatz random walks. that would definitely also smooth out the function but from computation, the longest nonmonotone run lengths seem to be unbounded. therefore no finite averaging would lead to finite max nonmonotone run lengths. the weighting/ conditioning function seems to be “folding up” the random walk in some currently highly mysterious/ inexplicable way to achieve bounded nonmonotone run lengths. concepts of topology theory seem to be relevant here. it seems to be some kind of topological folding. the plots also seem to immediately suggest a new natural fitness metric that might behave much more smoothly than the other metrics and lead to new solutions: the euclidean distance from the origin of (x, y) where x is the collatz trajectory value and y is the weighted/ conditioned trajectory value! cant wait to try it ❗ (1/22) ❗ 😮 😦 😳 😥 bad news! heres some analysis code that does a bit-wise greedy search for long nonmonotonic runs, successfully finding large ones. then it looks at the unconditioned vs conditioned random walks and the longest nonmonotonic run for each (red, green respectively). both tend to trend upward. if the conditioned results levelled off while unconditioned walks climbed as in earlier plots it would indicate (potential) success, but they are roughly correlated so the formula is essentially failing on “hard” seeds. (wince) another indication of failure is that the longest nonmonotone runs are as high as ~900, which were not seen in the earlier analyses, which didnt find runs longer than ~650. this is for weight set #3 (1st arg). there does seem to be a minor improvement effect by the algorithm after x=~250 bits. overall, it looks like the GA is training well on random inputs but not generalizing to “hard” seeds which seem to have some different distribution. aka the needles in the haystack throw a curve ball at the conditioning formula & it strikes out. oh and the code gets a numeric overflow in the Math.log function when the # of bits hits 1024. 😳 never pushed ruby that far before, bummer. there is maybe some idea to salvage here. if the hard seeds can be differentiated from easier seeds somehow in the initial bit structure then those would be better to train the GA with. def stat(l) return 0, 0 if (l.empty?) t = t2 = 0 l.each \ { |x| t += x t2 += x ** 2 } c = l.size a = t.to_f / c z = t2.to_f / c - a ** 2 sd = Math.sqrt(z < 0 ? 0 : z) raise if (sd.nan?) return a, sd end def avg(l) t = 0 l.each { |x| t += x } return t.to_f / l.size end def d(s) c = s.split('').select { |x| x == '1' }.size return c.to_f / s.length, s.length, c, s.size - c, c % 2, c % 3 end def rank(n, l) s = n.to_s(2) m1 = Math.log(n) m2 = m1 ** 2 m3 = m1 ** 0.5 d1, w, c1, c0, c2, c3 = d(s) d2 = (0.5 - d1).abs a1, sd1 = stat(s.split('0').map { |x| x.length }) a0, sd0 = stat(s.split('1').map { |x| x.length }) l2 = [m1, m2, m3, d1, w, c1, c0, c2, c3, d2, a1, sd1, a0, sd0] t = 0.0 l.each_with_index { |w, i| t += w * l2[i] } raise(l2.join(' ')) if (t.nan?) return t end def f(n, w) l = [] while (n != 1) l << [Math.log(n), rank(n, w).abs] n = (n % 2 == 0) ? n / 2 : (n * 3 + 1) / 2 end return l end def maxrun(l) return 0 if (l.empty?) m = l[0] l2 = [] l1 = [] l.each \ { |x| if (x < m) then l2 = [] m = x else l2 << x end l1 = l2 if (l2.size > l1.size) } return l1.size end def fmt(h, c) h = h.dup h['n#'] = h.delete('n').to_s(2).length h['p#'] = h.delete('p').to_s(2).length h['l#'] = c return h.to_s.gsub('=>', ' => ') end def sample(t, w) l = [{'x' => 0, 'n' => 1, 'p' => 1, 'x2' => 0}] loop \ { t = (0...l.size).max_by { |j| l[j]['x'] } z = l.delete_at(t) x, n, p, x2 = [z['x'], z['n'], z['p'], z['x2']] puts(fmt(z, l.size))$stdout.flush
p <<= 1
l.push({'x' => x, 'n' => n, 'p' => p, 'x2' => x2})
n2 = n + p
l2 = f(n2, w)
x1 = maxrun(l2.map { |z| z[0] })
x2 = maxrun(l2.map { |z| z[1] })
l.push({'x' => x1, 'n' => n2, 'p' => p, 'x2' => x2})
}
end
w = File.open("w#{ARGV[0]}.txt").readlines.map { |x| x.split[1].to_f }
m, n, l = sample(nil, w)
(later) on immediate inspection of the binary seed values, the “pattern” is almost trivial! the greedy code is generating a long bitstring of mostly 0’s and some scattered 1’s as seeds. the 0 runs follow some irregular pattern. this simplified code (removing the formula weighting logic) outputs a list of the lengths of the 0 runs separated by 1s for bitwise greedy search of long nonmonotone runs. this suggests quite simply that (deja vu) a better random seed generator will range over all the different 0/1 densities! the current scheme is centered around 50% density with low deviation.
def f(n)
l = []
while (n != 1)
l << Math.log(n)
n = (n % 2 == 0) ? n / 2 : (n * 3 + 1) / 2
end
return l
end
def maxrun(l)
return 0 if (l.empty?)
m = l[0]
l2 = []
l1 = []
l.each \
{
|x|
if (x < m) then
l2 = []
m = x
else
l2 << x
end
l1 = l2 if (l2.size > l1.size)
}
return l1.size
end
def fmt(h, c)
h = h.dup
h['n#'] = h.delete('n').to_s(2).length
h['p#'] = h.delete('p').to_s(2).length
h['l#'] = c
return h.to_s.gsub('=>', ' => ')
end
def sample()
l = [{'x' => 0, 'n' => 1, 'p' => 1}]
h = {}
loop \
{
t = (0...l.size).max_by { |j| l[j]['x'] }
z = l.delete_at(t)
x, n, p = [z['x'], z['n'], z['p']]
if (!h.member?(n)) then
puts("\t" + fmt(z, l.size))
h[n] = nil
ns = n.to_s(2)
p(ns.split('1').map { |s| s.length })
\$stdout.flush
break if (ns.length > 1000)
end
p <<= 1
l.push({'x' => x, 'n' => n, 'p' => p})
n2 = n + p
x1 = maxrun(f(n2))
l.push({'x' => x1, 'n' => n2, 'p' => p})
}
end
sample()
{"x" => 0, "n#" => 1, "p#" => 1, "l#" => 0}
[]
{"x" => 4, "n#" => 2, "p#" => 2, "l#" => 1}
[]
{"x" => 7, "n#" => 3, "p#" => 3, "l#" => 2}
[]
{"x" => 7, "n#" => 4, "p#" => 4, "l#" => 4}
[]
{"x" => 56, "n#" => 5, "p#" => 5, "l#" => 5}
[]
{"x" => 80, "n#" => 11, "p#" => 11, "l#" => 11}
[0, 5]
{"x" => 80, "n#" => 25, "p#" => 25, "l#" => 26}
[0, 13, 5]
{"x" => 126, "n#" => 27, "p#" => 27, "l#" => 28}
[0, 15, 5]
{"x" => 130, "n#" => 80, "p#" => 80, "l#" => 81}
[0, 52, 15, 5]
{"x" => 181, "n#" => 82, "p#" => 82, "l#" => 83}
[0, 1, 52, 15, 5]
{"x" => 208, "n#" => 122, "p#" => 122, "l#" => 123}
[0, 39, 1, 52, 15, 5]
{"x" => 214, "n#" => 131, "p#" => 131, "l#" => 132}
[0, 8, 39, 1, 52, 15, 5]
{"x" => 286, "n#" => 162, "p#" => 162, "l#" => 163}
[0, 30, 8, 39, 1, 52, 15, 5]
{"x" => 298, "n#" => 213, "p#" => 213, "l#" => 214}
[0, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 308, "n#" => 238, "p#" => 238, "l#" => 239}
[0, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 329, "n#" => 243, "p#" => 243, "l#" => 244}
[0, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 384, "n#" => 251, "p#" => 251, "l#" => 252}
[0, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 401, "n#" => 260, "p#" => 260, "l#" => 261}
[0, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 406, "n#" => 277, "p#" => 277, "l#" => 278}
[0, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 433, "n#" => 329, "p#" => 329, "l#" => 330}
[0, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 444, "n#" => 333, "p#" => 333, "l#" => 334}
[0, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 481, "n#" => 374, "p#" => 374, "l#" => 375}
[0, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 497, "n#" => 376, "p#" => 376, "l#" => 377}
[0, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 512, "n#" => 393, "p#" => 393, "l#" => 394}
[0, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 524, "n#" => 409, "p#" => 409, "l#" => 410}
[0, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 533, "n#" => 449, "p#" => 449, "l#" => 450}
[0, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 657, "n#" => 465, "p#" => 465, "l#" => 466}
[0, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 706, "n#" => 495, "p#" => 495, "l#" => 496}
[0, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 722, "n#" => 578, "p#" => 578, "l#" => 579}
[0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 742, "n#" => 579, "p#" => 579, "l#" => 580}
[0, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 749, "n#" => 597, "p#" => 597, "l#" => 598}
[0, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 752, "n#" => 618, "p#" => 618, "l#" => 619}
[0, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 764, "n#" => 671, "p#" => 671, "l#" => 672}
[0, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 793, "n#" => 680, "p#" => 680, "l#" => 681}
[0, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 809, "n#" => 695, "p#" => 695, "l#" => 696}
[0, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 972, "n#" => 706, "p#" => 706, "l#" => 707}
[0, 10, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 996, "n#" => 785, "p#" => 785, "l#" => 786}
[0, 78, 10, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 1024, "n#" => 797, "p#" => 797, "l#" => 798}
[0, 11, 78, 10, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 1050, "n#" => 967, "p#" => 967, "l#" => 968}
[0, 169, 11, 78, 10, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 1051, "n#" => 981, "p#" => 981, "l#" => 982}
[0, 13, 169, 11, 78, 10, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5]
{"x" => 1062, "n#" => 1014, "p#" => 1014, "l#" => 1015}
[0, 32, 13, 169, 11, 78, 10, 14, 8, 52, 20, 17, 0, 82, 29, 15, 39, 15, 16, 1, 40, 3, 51, 16, 8, 7, 4, 24, 50, 30, 8, 39, 1, 52, 15, 5] | 13,478 | 36,134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-26 | latest | en | 0.94617 |
https://music.stackexchange.com/questions/52790/roman-numeral-analysis | 1,716,300,337,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058484.54/warc/CC-MAIN-20240521122022-20240521152022-00733.warc.gz | 336,160,497 | 42,694 | # Roman Numeral Analysis
So I've been learning about the scales (major, minor etc) and how all the scale degrees are related to the major scale.
So in a minor scale, if you saw m3 it would mean a flattened 3rd degree in relation to the major scale defined semitones.
For example...
That all seems fine, but I noticed that there's a degree notation using roman numerals and even with the major scale it suggests that some of the degrees are actually minor?
So in this ^^ example (in the key of A), instead of playing all major chords, I would play a mixture of major and minor chords:
• A major
• B minor
• C# minor
• D major
• E major
• F# minor
• G# diminished
Why is this?
## Update
I left the following comment on the top response so far:
so to recap: a chord is typically built from thirds of the scale. The quality of the chord associated with a melodic degree is determined by these thirds. So I counted the semitones between the thirds and found the quality (minor/major) matched the harmonic degrees indicated by the roman numerals. But what I don't understand is why the 7th is 'diminished' and not minor? As the semitones between the 7th degree and the degree a 3rd way from it (which would be a 2nd/9th) is 3 semitones; suggesting a minor third.
By looking back at the full list of interval qualities (e.g. P1, m2, M2...all the way up to...P8) I can see that 6 semitones would be an augmented/diminished quality `A4/d5`, but I can't see how that distance in semitones get's applied by counting semitone distances based on 3rds
Ps, what's the correct terminology to use above where I've labelled it 'QUALITY' and 'ROMAN QUALITY'?
That's because those are the chords that are built from the scale. To find a chord built from the scale degree in a scale we need a formula - for example 1-3-5 (which doesn't exclusively mean the major third and perfect fifth, just the third and fifth of the note, regardless of their quality), 1-3-5-7, 1-2-5, 1-4-5-7...
When talking about the basic triads, aka I ii iii IV V vi viio, we use the formula (1-3-5), and see that from the 1st, 4th and 5th scale degrees we get a major triad, from the 2nd, 3rd and 6th we get a minor triad and from the 7th scale degree we get a diminished triad.
It is important to note that the chords come from scale degrees, not the other way round - so the scale degrees aren't "minor", just the chords built from them.
A few more minor details (Pun not intended) - altered (not in the major scale) scale degrees are written with b and #, and the triads built from the scale degrees are upper case if major, lower case if minor, lower case with an o to the top right if diminished and upper case with a large + to the right if augmented.
UPDATE: NEW QUESTION
The basic chords (Major, minor, diminished, augmented...) are built by stacking thirds (this is called tetrian harmony) on top of eachother, of course being in the scale. By third we mean not the interval of either a M/m third, but x degree +2 (1->3, 2->4), which could result in diminished thirds (#2-4), augmented thirds (b2-#4) and double (and even triple) aug/dim intervals. So the major chord is built by a stack of M3-m3, the minor is mM, the diminished being mm and augmented MM. So when stacking the thirds from the 7th scale degree, we get 7-2-4, which is mm, thus a diminished triad.
The correct terminology is: Quality->Interval from 1 (in a single octave), which is useless and shouldn't be in such a table (also, the B above the A which you called P8 is called M9, not M2, as it is a M2 an octave up), and Roman Quality->Chord, which could be better described as the full 7 note chord (7th chord with extensions) - Imaj7(9,11,13) iim7(9,11,13) iiim7(b9,11,b13) IVmaj7(9,#11,13) V7(9,11,13) vim7(9,11,b13) and viim7b5(b9,11,b13).
• 'Altered scale degrees'. 'Written with # and b'. Alter the maj 3 in A, and we lose the #, and it turns natural. In fact, altered notes from maj to min will only become flat or natural, as they can only go downwards. Or have I missed the point?
– Tim
Jan 30, 2017 at 16:12
• When talking about scale degrees, we are talking about something that is relative to the tonic, and has nothing to do with the actual names of the notes being played, nor with their absolute pitch. Scale degrees are infact a matter of only relative pitch. We call the scale degrees by their number in a major scale, not by their interval from the tonic, so instead of calling the 3rd a Maj 3, we call it a 3. When we want a min 3, we call it a b3. An example of sharpening a note is the tritone in the lydian mode - #4. Jan 30, 2017 at 20:13
• It was probably the opening gambit - 'a few more minor details'...
– Tim
Jan 30, 2017 at 20:38
• Oh wow, didn't even notice it :P Jan 30, 2017 at 21:26
• @dudwhuknowstheory hello again, I've left an update on this question - you were amazing before with helping me identify the missing pieces in my understanding. Hoping you could possibly help again if you have time. Many thanks Feb 2, 2017 at 10:18
Watch that you don't start confusing melodic degrees with harmonic degrees. The chords built on those degrees are major or minor or diminished, so the Roman numeral reflects that - it is not major or minor relative to the tonic, but to the root of the chord on that degree as created by using notes from that scale.
I'll explain:
In major, your melodic degrees are set relative to the tonic, and they are 1 through 7 (8 if you include the octave, but that is usually just considered a transposition of 1). So to ascertain the chord with 3 as the root, you build the chord by choosing 3, 5 and 7 (piling up thirds, yes?). The distance between 3 and 5 is a semitone plus a whole tone = a minor third, hence the chord is minor, so the harmonic degree on 3 is iii, i.e., a minor chord.
In minor, your melodic degrees are 1, 2, ♭3, 4, 5, ♭6 and ♭7 (with ♭6 and ♭7 often being raised to 6 and 7 in practice). This is the more usual notation in the classical world, and it says that 3, 6 & 7 are flatted by comparison to the major scale. So, the chord on ♭3 would be created from ♭3, 5 & ♭7. ♭3 to 5 is two whole tones, i.e., a major third, hence the chord on ♭3 is ♭III, which signifies that it's a major chord.
• Thanks @Patrx2 so to recap: a chord is typically built from thirds of the scale. The quality of the chord associated with a melodic degree is determined by these thirds. So I counted the semitones between the thirds and found the quality (minor/major) matched the harmonic degrees indicated by the roman numerals. But what I don't understand is why the 7th is 'diminished' and not minor? As the semitones between the 7th degree and the degree a 3rd way (which would be a 2nd/9th) is 3 semitones, so a minor third. Feb 1, 2017 at 9:04
• You didn't talk about the fifth - The fifth of a basic 1-3-5 chord is very important for defining its quality: Both major and augmented triads have a major third, and both minor and diminished triads have a minor third. Feb 2, 2017 at 16:53
When you play in a key (A major in your example) the basic chords of the key are built using the notes of the scale. From each starting point you take the note that 2 steps higher as the third and the note that's two steps above that as the fifth. The result is that some chords are major (Tonic (A), Subdominant (D) and Dominant (E)), some are minor (supertonic (B), mediant (C#) and submediant (F#)) and the chord on the leading note (G#) is diminished.
Of course there are no rules about having to use these chords. That just how the theory works in describing the scale and the chords obtained from it.
Hope that helps | 2,073 | 7,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-22 | latest | en | 0.955811 |
https://www.bartleby.com/essay/Bus-640-Economics-PKQVDKS8J3DW | 1,600,773,727,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00725.warc.gz | 732,500,918 | 13,528 | # Bus. 640 Economics
942 Words4 Pages
Chapter 3: Problem 2: Appalachian Coal Mining believes that it can increase labor productivity and, there- fore, net revenue by reducing air pollution in its mines. It estimates that the marginalcost function for reducing pollution by installing additional capital equipment is MC 40P where P represents a reduction of one unit of pollution in the mines. It also feels that for every unit of pollution reduction the marginal increase in revenue (MR) isMR 1,000 10P How much pollution reduction should Appalachian Coal Mining undertake? MC = MR and solve for P. That is 40P = 1000-10P 40P+10P=1000 50P=1000 P=20 units. Problem 4: Twenty first Century Electronics has discovered a theft problem at its warehouse and has decided to hire…show more content…
Yes, at the 5% level, there is statistical significance Explain, using the appropriate p-value. With a 0.0128 p-value this means the exact level of significance for a T-Ratio of 2.63 is 1 % and the level of confidence is 99%. Stating b is statistically significant. d. Does advertising by its three largest rivals affect sales of Bright Side detergent in a statistically significant way? P-Value and T-Ratio show that the competitor’s advertising has a negative effect. Explain, using the appropriate p-value. The high P-value indicates that the negative T-ratio has a high probability of competitor’s advertising effecting sales of Bright Side negatively. e. What fraction of the total variation in sales of Bright Side remains unexplained? 22% What can the marketing director do to increase the explanatory power of the sales equation? He could look competitor’s prices and then Vanguard can add this to the equation as well as log variables on advertising expenses. What other explanatory variables might be added to this equation? Other variables might include family size, loads of laundry done , is there more laundry done in the summer versus the winter f. What is the expected level of sales each week when Vanguard spends \$40,000 per week and the combined advertising expenditures for the three rivals are \$100,000 per week? S = a + b(\$40,000) + c(\$100,000) S = 175086.0 + 0.85550(\$40,000) + - 0.284(\$100,000) S = 175086.0 + \$34,000 + | 515 | 2,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-40 | latest | en | 0.917688 |
https://questions.kunduz.com/math/linear-programming/3-consider-the-stochastic-differential-equation-sde-dx-tx-dt-x-dbt-xo-1-a-let-yt-ln-xt-compute-the-stochastic-d-20423666 | 1,656,464,985,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00396.warc.gz | 516,206,073 | 19,980 | Question:
# 3. Consider the stochastic differential equation (SDE) dX= tX+ dt + X{ dBť, Xo = 1. (a) Let Yt = ln Xt, compute the stochastic d
3. Consider the stochastic differential equation (SDE) dX= tX+ dt + X{ dBť, Xo = 1. (a) Let Yt = ln Xt, compute the stochastic differential dYt. (b) Find a formula for Yt, hence solve the SDE for Xt. | 116 | 342 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-27 | latest | en | 0.743224 |
https://www.biglearners.com/worksheets/grade-3/math/geometry-patterns | 1,632,497,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057558.23/warc/CC-MAIN-20210924140738-20210924170738-00466.warc.gz | 706,853,230 | 13,987 | # Geometry and Patterns : Third Grade Math Worksheets
This page contains all our printable worksheets in section Geometry and Patterns of Third Grade Math. As you scroll down, you will see many worksheets for plane figures, solid figures, congruence and symmetry, patterns, and more.
A brief description of the worksheets is on each of the worksheet widgets. Click on the images to view, download, or print them. All worksheets are free for individual and non-commercial use.
View the full list of topics for this grade and subject categorized by common core standards or in a traditional way.
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Plane Figures
Plane Figures
Identify, classify, and describe plane figures.
Category: Geometry and Patterns Plane Figures Identify and Combine Plane Figures
Plane Figures
Students will identify, classify, and describe the plane figures.
Category: Geometry and Patterns Plane Figures Identify and Combine Plane Figures
Combining and Dividing Plane Figures
Objective : Divide and combine plane figures.
Category: Geometry and Patterns Plane Figures Identify and Combine Plane Figures
Combining and Dividing Plane Figures
Practice dividing and combining plane figures.
Category: Geometry and Patterns Plane Figures Identify and Combine Plane Figures
Angles
Practice identifying angles (acute, obtuse, right, or straight).
Category: Geometry and Patterns Plane Figures Angles and Line Segments
Types of Angles
Practice identifying angles.
Category: Geometry and Patterns Plane Figures Angles and Line Segments
Identify points, lines, line segments, and rays.
Category: Geometry and Patterns Plane Figures Angles and Line Segments
Identify points, lines, line segments, and rays.
Category: Geometry and Patterns Plane Figures Angles and Line Segments
Types of Lines
Identify and classify pairs of lines as intersecting, parallel, or perpendicular.
Category: Geometry and Patterns Plane Figures Types of Lines
Types of Lines
Identifying and classify pairs of lines as intersecting, parallel, or perpendicular.
Category: Geometry and Patterns Plane Figures Types of Lines
Types of Lines
Practice identifying, classifying, and drawing pairs of lines as intersecting, parallel, or perpendicular.
Category: Geometry and Patterns Plane Figures Types of Lines
What's My Name?
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Types of Triangles
Practice identifying triangles.
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Types of Triangles
Identify, describe, and classify triangles.
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Parts of a Circle
Practice naming the parts of a circle.
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Parts of a Circle
Identify the parts of a circle.
Category: Geometry and Patterns Plane Figures Triangles, Quadrilaterals, and Circles
Solid Figures
Combine Solid Figures
Students will identify common solid figures in complex solid figures.
Category: Geometry and Patterns Solid Figures Identify and Combine Solid Figures
Combine Solid Shapes
Practice identifying common solid figures in complex solid figures.
Category: Geometry and Patterns Solid Figures Identify and Combine Solid Figures
Identify Solid Figures
Write the name of each 3-D figure.
Category: Geometry and Patterns Solid Figures Identify and Combine Solid Figures
Identify Solid Shapes
Name the solid shape that each object looks like.
Category: Geometry and Patterns Solid Figures Identify and Combine Solid Figures
Drawing Shapes
Draw and name the plane and solid figures on the dot paper.
Category: Geometry and Patterns Solid Figures Drawing and Modeling Figures
Drawing Shapes
Copy the shapes (plane and solid) on the dot paper. Then write the name of each shape.
Category: Geometry and Patterns Solid Figures Drawing and Modeling Figures
Model Solid Shapes
Identify the solid shape that can be made from each net. Write the name of each shape in the space provided.
Category: Geometry and Patterns Solid Figures Drawing and Modeling Figures
Model Solid Shapes
Which solid figure could be made from each net? Match each net with the correct solid shape and write the name in the space provided.
Category: Geometry and Patterns Solid Figures Drawing and Modeling Figures
Faces, Vertices, and Edges
Name the common solid figures. Then write the number of faces, vertices, and edges.
Category: Geometry and Patterns Solid Figures Faces, Vertices, and Edges
Congruence and Symmetry
Congruent Figures
Determine if the pair of figures appear to be congruent.
Category: Geometry and Patterns Congruence and Symmetry Congruent Figures and Similar Figures
Congruent and Similar Figures
Look at each pair of shapes in the box. Tell how the figures are related. Choose the correct answer.
Category: Geometry and Patterns Congruence and Symmetry Congruent Figures and Similar Figures
Similar Figures
Tell if each pair of figures appear to be similar.
Category: Geometry and Patterns Congruence and Symmetry Congruent Figures and Similar Figures
Symmetry
Tell if the line appears to be a line of symmetry. Write yes or no.
Category: Geometry and Patterns Congruence and Symmetry Symmetry and Line of Symmetry
Symmetry
Practice identifying figures that have a line of symmetry.
Category: Geometry and Patterns Congruence and Symmetry Symmetry and Line of Symmetry
Lines of Symmetry
Draw the line or lines of symmetry .
Category: Geometry and Patterns Congruence and Symmetry Symmetry and Line of Symmetry
Slides, Flips, and Turns
Identify the position of a shape/figure after a slide, flip, or turn.
Category: Geometry and Patterns Congruence and Symmetry Slide, Flips, and Turns
Slides, Flips, and Turns
Tell how each figure was moved.
Category: Geometry and Patterns Congruence and Symmetry Slide, Flips, and Turns
Reflection, Rotation, and Translation
Practice identifying how each figure has been transformed.
Category: Geometry and Patterns Congruence and Symmetry Slide, Flips, and Turns
Patterns
Number patterns
Students will identify and extend whole-number patterns to find rules and solve problems.
Category: Geometry and Patterns Patterns Number Patterns
Number Patterns
Students will find the rule and follow the rule to complete each number pattern.
Category: Geometry and Patterns Patterns Number Patterns
Number Patterns
Identify and extend number patterns to find rules and solve problems.
Category: Geometry and Patterns Patterns Number Patterns
Repeating Geometric Patterns
Identify and extend the geometric patterns.
Category: Geometry and Patterns Patterns Geometric Pattern
Growing Geometric Patterns
Draw the missing figure in each growing pattern.
Category: Geometry and Patterns Patterns Geometric Pattern | 1,595 | 7,440 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-39 | longest | en | 0.742286 |
https://www.dprg.org/history/mail_list_archives/dprglist/2001-June/015457.html | 1,568,615,471,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572491.38/warc/CC-MAIN-20190916060046-20190916082046-00353.warc.gz | 803,150,270 | 3,288 | DPRG List
[DPRG] Researching Robotics Design Tools Message index sorted by: [ date ] [ thread ] [ subject ] [ author ] Previous message: [DPRG] Researching Robotics Design Tools Next message: [DPRG] Researching Robotics Design Tools Subject: [DPRG] Researching Robotics Design Tools From: R. Bickle rbickle at swbell.net Date: Tue Jun 12 10:08:44 CDT 2001 ```Kip, I have read several good articles on the internet on how to tune PID controllers. In fact, a friend of mine from UTD gave me a spreadsheet on tuning calculations for various plants. (I can send it to you if you're interested). The thing I was trying to work out is: How to measure the actual motor speed as a function of the set speed. It would be great if you could see a simple curve on a scope so that you could see the effect of changes to each PID tuning parameter. The only idea I have so far is to stream PWM numbers out on the serial port and have a PC program to graph them. My plan is to make a motor control module that has programmable tuning paramters to that anyone could use it with any motor. Rick Bickle -----Original Message----- >From: Kipton Moravec [mailto:kip at kdream.com] Sent: Tuesday, June 12, 2001 8:10 AM To: R. Bickle Cc: DPRG List Subject: Re: [DPRG] Researching Robotics Design Tools I just got a book "Step-by-Step Design of Motion Control Systems" by Jacob Tal from Amazon (It took 6 weeks!) That goes through the math step by step and lets you calculate what the PID parameters should be. They do it for voltage and current amplifiers, but not for PWM, so I am currently at a loss as how to model the PWM for their formulas. I am thinking it is like the voltage model, but I will probably have to try it. > I am currently > working on a PID motor control module for my robot. I need to find a good > way to tune the parameters of the PID algorithm for different plants. Once > this is done, I plan to actally make my robot move... :-) > > Rick Bickle ``` Previous message: [DPRG] Researching Robotics Design Tools Next message: [DPRG] Researching Robotics Design Tools Message index sorted by: [ date ] [ thread ] [ subject ] [ author ] More information about the DPRG mailing list | 533 | 2,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2019-39 | latest | en | 0.912151 |
https://classroomsecrets.co.uk/mixed-age-year-3-and-4-fractions-step-8-resource-pack/ | 1,723,342,245,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640843545.63/warc/CC-MAIN-20240811013030-20240811043030-00885.warc.gz | 134,948,595 | 80,021 | Mixed Age Year 3 and 4 Fractions Step 8 Resource Pack – Classroom Secrets | Classroom Secrets
All › Mixed Age Year 3 and 4 Fractions Step 8 Resource Pack
# Mixed Age Year 3 and 4 Fractions Step 8 Resource Pack
## Step 8: Mixed Age Year 3 and 4 Fractions Resource Pack
Mixed Age Year 3 and 4 Fractions Step 8 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 3 Subtract Fractions & Year 4 Subtract 2 Fractions for Spring Block 3.
### What's included in the Pack?
This Mixed Age Year 3 and 4 Fractions Step 6 pack includes:
• Mixed Age Year 3 and 4 Fractions Step 6 Teaching PowerPoint with examples for both year groups.
• Year 3 Subtract Fractions Varied Fluency with answers.
• Year 3 Subtract Fractions Reasoning and Problem Solving with answers.
• Year 4 Subtract 2 Fractions Varied Fluency with answers.
• Year 4 Subtract 2 Fractions Reasoning and Problem Solving with answers.
#### National Curriculum Objectives
Mathematics Year 3: (3F10) Solve problems that involve the above objectives
Mathematics Year 4: (4F4) Add and subtract fractions with the same denominator
Differentiation for Year 3 Subtract Fractions:
Varied Fluency
Developing Questions to support subtracting fractions. Subtracting one fraction and the denominator less than 10.
Expected Questions to support subtracting fractions. Subtracting one/two fractions and the denominator less than 20.
Greater Depth Questions to support subtracting fractions. Subtracting two fractions and using improper fractions.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Problem Solving)
Developing Subtract 1 fraction from a fraction using a visual image. Denominator less than 10.
Expected Subtract 2 fractions from a fraction with no visual image, Denominator less than 20.
Greater Depth Subtract 2 or 3 fractions from a fraction with no visual image. Using improper fractions.
Questions 2, 5 and 8 (Reasoning)
Developing Identify and explain if the statement is correct when subtracting 1 fraction from a fraction and comparing whether it is more or less than another fraction. Denominator less than 10.
Expected Identify and explain if the statement is correct when subtracting 2 fractions from a fraction and comparing whether it is more or less than another fraction. Denominator less than 20.
Greater Depth Identify and explain if the statement is correct when subtracting 2 fractions from a fraction and comparing whether it is more or less than another fraction. Using improper fractions.
Questions 3, 6 and 9 (Problem Solving)
Developing Choose the fraction to make the part whole model correct. 1 fraction from a possible 3.
Expected Choose the fractions to make the part whole model correct. 2 fractions from a possible 4.
Greater Depth Choose the fractions to make the part whole model correct. 3 fractions from a possible 5.
Differentiation for Year 4 Subtract 2 Fractions:
Varied Fluency
Developing Questions to support subtracting 2 fractions with the same denominator; no mixed numbers or improper fractions. Images provided for support.
Expected Questions to support subtracting 2 fractions with the same denominator, with mixed numbers and improper fractions. Images provided for support.
Greater Depth Questions to support subtracting 2 fractions where some of the denominators are double or half the starting fraction. Using mixed numbers and improper fractions. No images provided.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Reasoning)
Developing Explain a mistake in a word problem involving subtracting 2 fractions with the same denominator, no mixed numbers or improper fractions.
Expected Explain a mistake in a word problem involving subtracting 2 fractions with the same denominator, with mixed numbers or improper fractions.
Greater Depth Explain a mistake in a word problem involving subtracting 2 fractions where some of the denominators are double or half the starting fraction. Using mixed numbers and improper factions.
Questions 2, 5 and 8 (Problem Solving)
Developing Choose from 3 digit cards to solve a subtraction problem with 2 fractions with the same denominator, no mixed numbers or improper fractions.
Expected Choose from 4 digit cards to solve a subtraction problem with 2 fractions with the same denominator, with mixed numbers or improper fractions.
Greater Depth Choose from 5 digit cards to solve a subtraction problem with 2 fractions where some of the denominators are double or half the starting fraction. Using mixed numbers and improper fractions.
Questions 3, 6 and 9 (Reasoning)
Developing Explain a mistake in a calculation involving subtracting 2 fractions with the same denominator, no mixed numbers or improper fractions. Images provided for support.
Expected Explain a mistake in a calculation involving subtracting 2 fractions with the same denominator, with mixed numbers and improper fractions. Images provided for support.
Greater Depth Explain a mistake in a calculation involving subtracting 2 fractions where some of the denominators are double or half the starting fraction. Using mixed numbers and improper fractions. No images provided. | 1,088 | 5,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-33 | latest | en | 0.811531 |
https://www.simtrade.fr/blog_simtrade/black-litterman-model/ | 1,638,187,978,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358705.61/warc/CC-MAIN-20211129104236-20211129134236-00543.warc.gz | 1,123,433,322 | 10,937 | # Black-Litterman Model
In this article, Youssef Louraoui (Bayes Business School, MSc. Energy, Trade & Finance, 2021-2022) presents the Black-Litterman model, used to determine optimal asset allocation in a portfolio. Black-Litterman Model takes the Markowitz Model one step further: It incorporates an investor’s own views in determining asset allocations.
This article is structured as follows: we introduce the Black-Litterman model. We then present the mathematical foundations of the model to understand how the method is derived. We finish with an example to illustrate how we can implement a Black-Litterman asset allocation in practice.
## Introduction
The Black-Litterman asset allocation model, developed by Fischer Black and Robert Litterman for the first time in the early 90’s, is a complex method for dealing with unintuitive, highly concentrated, input-sensitive portfolios. The most likely reason why more portfolio managers do not employ the Markowitz paradigm, in which return is maximized for a given level of risk, is input sensitivity, which is a well-documented problem with mean-variance optimization.
The Black-Litterman Model employs a Bayesian technique to integrate an investor’s subjective views on expected returns for one or more assets with the market equilibrium vector (prior distribution) of expected returns to get a new, mixed estimate of expected returns. The new vector of returns (the posterior distribution) is a complex, weighted average of the investor’s views and the market equilibrium.
The purpose of the Black-Litterman model is to develop stable, mean-variance efficient portfolios based on an investor’s unique insights that overcome the problem of input sensitivity. The Black-Litterman Model, according to Lee (2000), “essentially mitigates” the problem of estimating error maximization (Michaud, 1989) by dispersing errors throughout the vector of expected returns.
The vector of expected returns is the most crucial input in mean-variance optimization; yet, Best and Grauer (1991) demonstrate that this input can be very sensitive in the final result. Black and Litterman (1992) and He and Litterman (1999) investigate various potential projections of expected returns in their search for a fair starting point: historical returns, equal “mean” returns for all assets, and risk-adjusted equal mean returns. They demonstrate that these alternate forecasts result in extreme portfolios, which have significant long and short positions concentrated in a small number of assets.
## Mathematical foundation of Black-Litterman model
It is important to introduce the Black-Litterman formula and provide a brief description of each of its elements. K is used to represent the number of views and n is used to express the number of assets in the formula (NB: ’ indicates the transpose and -1 indicates the inverse).
Where:
• E[R] =New (posterior) Combined Return Vector (n x 1 column vector)
• τ = Scalar
• Σ = Covariance Matrix of Returns (n x n matrix)
• P = Identifies the assets involved in the views (k x n matrix or 1 x n row vector in the special case of 1 view)
• Ω = Diagonal covariance matrix of error terms in expressed views representing the level of confidence in each view (k x k matrix)
• П = Implied Equilibrium Return Vector (n x 1 column vector)
• Q = View Vector (k x 1 column vector)
Traditionally, personal views are used for prior distribution. Then observed data is used to generate a posterior distribution. The Black-Litterman Model assumes implied returns as the prior distribution and personal views alter it. The basic procedure to find the Black-Litterman model is: 1) Find implied returns 2) Formulate investor views 3) Determine what the expected returns are 4) Find the asset allocation for the optimal portfolio.
## Black-Litterman asset allocation in practice
An investment manager’s views for the expected return of some of the assets in a portfolio are frequently different from the the Implied Equilibrium Return Vector (Π), which represents the market-neutral starting point for the Black-Litterman model. representing the uncertainty in each view. Such views can be represented in absolute or relative terms using the Black-Litterman Model. Below are three examples of views stated in the Black and Litterman model (1990).
• View 1: Merck (MRK) will generate an absolute return of 10% (Confidence of View = 50%).
• View 2: Johnson & Johnson (JNJ) will outperform Procter & Gamble (PG) by 3% (Confidence of View = 65%).
• View 3: GE (GE) will beat GM (gm), Wal-Mart (WMT), and Exxon (XOM) by 1.5 percent (Confidence of View = 30%).
An absolute view is exemplified by View 1. It instructs the Black-Litterman model to set Merck’s return at 10%.
Views 2 and 3 are relative views. Relative views are more accurate representations of how investment managers feel about certain assets. According to View 2, Johnson & Johnson’s return will be 3 percentage points higher than Procter & Gamble’s. To determine if this will have a good or negative impact on Johnson & Johnson in comparison to Procter & Gamble, their respective Implied Equilibrium returns must be evaluated. In general (and in the absence of constraints and other views), the model will tilt the portfolio towards the outperforming asset if the view exceeds the difference between the two Implied Equilibrium returns, as shown in View 2.
View 3 shows that the number of outperforming assets does not have to equal the number of failing assets, and that the labels “outperforming” and “underperforming” are relative terms. Views that include several assets with a variety of Implied Equilibrium returns are less intuitive, generalizing more challenges. In the absence of constraints and other views, the view’s assets are divided into two mini-portfolios: a long and a short portfolio. The relative weighting of each nominally outperforming asset is proportional to that asset’s market capitalization divided by the sum of the market capitalization of the other nominally outperforming assets of that particular view. Similarly, the relative weighting of each nominally underperforming asset is proportional to that asset’s market capitalization divided by the sum of the market capitalizations of the other nominally underperforming assets. The difference between the net long and net short positions is zero. The real outperforming asset(s) from the expressed view may not be the mini-portfolio that receives the good view. In general, the model will overweight the “outperforming” assets if the view is greater than the weighted average Implied Equilibrium return differential.
## Why should I be interested in this post?
Modern Portfolio Theory is at the heart of modern finance and its core foundations are structuring the modern investing panorama. MPT has established itself as the foundation for modern financial theory and practice. MPT’s premise is that beating the market is difficult, and those that do it by diversifying their portfolios appropriately and accepting higher-than-average investment risks.
MPT has been around for almost sixty years, and its popularity is unlikely to wane anytime soon. His theoretical contributions have laid the groundwork for more theoretical research in the field of portfolio theory. Markowitz’s portfolio theory, however, is vulnerable to and dependent on continuing ‘probabilistic’ development and expansion. This article shed light on an enhancement of the initial Markowitz work by going a step further: to incorporate the views of the investors in the asset allocation process.
## Related posts on the SimTrade blog
▶ Louraoui Y. Portfolio
▶ Louraoui Y. Alpha
▶ Louraoui Y. Factor Investing
▶ Louraoui Y. Origin of factor investing
▶ Louraoui Y. Markowitz Modern Portfolio Theory
▶ Walia J.Capital Asset Pricing Model (CAPM)
## Useful resources
Best, M.J., and Grauer, R.R. 1991. On the Sensitivity of Mean-Variance-Efficient Portfolios to Changes in Asset Means: Some Analytical and Computational Results.The Review of Financial Studies, 315-342.
Black, F. and Litterman, R. 1990. Asset Allocation: Combining Investors Views with Market Equilibrium. Goldman Sachs Fixed Income Research working paper
Black, F. and Litterman, R. 1991. Global Asset Allocation with Equities, Bonds, and Currencies. Goldman Sachs Fixed Income Research working paper
Black, F. and Litterman, R. 1992. Global Portfolio Optimization.Financial Analysts Journal, 28-43.
He, G. and Litterman, R. 1999. The Intuition Behind Black-Litterman Model Portfolios. Goldman Sachs Investment Management Research, working paper.
Idzorek, T.M. 2002. A step-by-step guide to Black-Litterman model. Incorporating user-specified confidence levels. Working Paper, 2-11.
Lee, W., 2000, Advanced theory and methodology of tactical asset allocation. Fabozzi and Associates Publications.
Markowitz, H., 1952. Portfolio Selection. The Journal of Finance, 7(1): 77-91.
Michaud, R.O. 1989. The Markowitz Optimization Enigma: Is Optimized Optimal?. Financial Analysts Journal, 31-42.
Mossin, J. 1966. Equilibrium in a Capital Asset Market. Econometrica, 34(4): 768-783.
Sharpe, W.F. 1963. A Simplified Model for Portfolio Analysis. Management Science, 9(2): 277-293.
Sharpe, W.F. 1964. Capital Asset Prices: A Theory of Market Equilibrium under Conditions of Risk. The Journal of Finance, 19(3): 425-442. | 2,087 | 9,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.859208 |
https://www.archyde.com/viral-challenge-whats-wrong-in-the-picture-only-2-out-of-7-solved-the-visual-puzzle-trends-viral-challenge-social-networks-trends-mexico-usa-nnda-nnrt-mexico/ | 1,660,789,329,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00692.warc.gz | 561,657,401 | 26,378 | # viral challenge | What’s wrong in the picture: only 2 out of 7 solved the visual puzzle | trends | viral challenge | social networks | trends | Mexico | USA | nnda nnrt | MEXICO
Los viral challenges continue to capture the attention of Internet users Facebook and other social networks. These challenges have become the favorites of millions of users around the world. Then we will leave you a visual puzzle that only managed could be solved by 2 out of 7 who tried it.
We can only point you in this visual challenge, that you look carefully at all the details. Even those that you consider to be useless and that may help you to solve the viral challenge. Do you need clues? Look at the kettle, did you notice anything strange?
Well, if despite the help you cannot find the error in this viral challenge, then surely you gave up. If that happened, we will show you below the solution. But if you still think you have a chance, then try again.
## Image of the viral challenge 2022
If you got to this part of the note, you probably already have the answer. Did you find the bug? Not yet? Well, as we pointed out at the beginning of the note, to solve this viral challenge you must have a privileged view. The mistake in the illustration is that the spout of the kettle is too low.
## What is a viral challenge?
A viral challenge is a perfect entertainment alternative for users who have free time and want to make the most of it. It consists of finding a person, animal, object or number in an image. Some have a time limit and some do not. They are also known as challenges, visual tests, visual or logic puzzles.
## Visual riddles what are they for
They serve to relax, work our minds and test our power of observation. Thus, in social networks, the tests have become viral and more and more people are trying to solve them.
## How do you do a riddle?
Use simple and forceful words. Riddles were originally a form of oral rather than written literature, so think about how the riddle sounds as you recite it. Try not to get entangled with elaborate words or concepts that are too abstract, recommends the portal WikiHow.
## How do you solve a puzzle?
To solve the most common riddles you have to use your imagination and the ability to deduce. The resolution has to be given with the mere statement of the statement, so questions are not allowed, he points out Wikipedia.
### practicing sports intensively would allow you to eat less fat
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 548 | 2,536 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-33 | latest | en | 0.958187 |
https://www.geeksforgeeks.org/maximum-number-of-diamonds-that-can-be-gained-in-k-minutes/ | 1,670,448,718,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711218.21/warc/CC-MAIN-20221207185519-20221207215519-00150.warc.gz | 812,334,130 | 28,395 | # Maximum number of diamonds that can be gained in K minutes
• Difficulty Level : Easy
• Last Updated : 28 Oct, 2021
Given an array arr[] consisting of N positive integers such that arr[i] represents that the ith bag contains arr[i] diamonds and a positive integer K, the task is to find the maximum number of diamonds that can be gained in exactly K minutes if dropping a bag takes 1 minute such that if a bag with P diamonds is dropped, then it changes to [P/2] diamonds, and P diamonds are gained.
Examples:
Input: arr[] = {2, 1, 7, 4, 2}, K = 3
Output: 14
Explanation:
The initial state of bags is {2, 1, 7, 4, 2}.
Operation 1: Take all diamonds from third bag i.e., arr[2](= 7), the state of bags becomes: {2, 1, 3, 4, 2}.
Operation 2: Take all diamonds from fourth bag i.e., arr[3](= 4), the state of bags becomes: {2, 1, 3, 2, 2}.
Operation 3: Take all diamonds from Third bag i.e., arr[2](= 3), the state of bags becomes{2, 1, 1, 2, 2}.
Therefore, the total diamonds gains is 7 + 4 + 3 = 14.
Input: arr[] = {7, 1, 2}, K = 2
Output: 10
Approach: The given problem can be solved by using the Greedy Approach with the help of max-heap. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
## C++
// C++ program for the above approach#include using namespace std; // Function to find the maximum number// of diamonds that can be gained in// exactly K minutesvoid maxDiamonds(int A[], int N, int K){ // Stores all the array elements priority_queue pq; // Push all the elements to the // priority queue for (int i = 0; i < N; i++) { pq.push(A[i]); } // Stores the required result int ans = 0; // Loop while the queue is not // empty and K is positive while (!pq.empty() && K--) { // Store the top element // from the pq int top = pq.top(); // Pop it from the pq pq.pop(); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = top / 2; pq.push(top); } // Print the answer cout << ans;} // Driver Codeint main(){ int A[] = { 2, 1, 7, 4, 2 }; int K = 3; int N = sizeof(A) / sizeof(A[0]); maxDiamonds(A, N, K); return 0;}
## Java
// Java program for the above approachimport java.util.*; class GFG{ // Function to find the maximum number// of diamonds that can be gained in// exactly K minutesstatic void maxDiamonds(int A[], int N, int K){ // Stores all the array elements PriorityQueue pq = new PriorityQueue<>( (a, b) -> b - a); // Push all the elements to the // priority queue for(int i = 0; i < N; i++) { pq.add(A[i]); } // Stores the required result int ans = 0; // Loop while the queue is not // empty and K is positive while (!pq.isEmpty() && K-- > 0) { // Store the top element // from the pq int top = pq.peek(); // Pop it from the pq pq.remove(); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = top / 2; pq.add(top); } // Print the answer System.out.print(ans);} // Driver Codepublic static void main(String[] args){ int A[] = { 2, 1, 7, 4, 2 }; int K = 3; int N = A.length; maxDiamonds(A, N, K);}} // This code is contributed by 29AjayKumar
## Python3
# Python3 program for the above approach # Function to find the maximum number# of diamonds that can be gained in# exactly K minutesdef maxDiamonds(A, N, K): # Stores all the array elements pq = [] # Push all the elements to the # priority queue for i in range(N): pq.append(A[i]) pq.sort() # Stores the required result ans = 0 # Loop while the queue is not # empty and K is positive while (len(pq) > 0 and K > 0): pq.sort() # Store the top element # from the pq top = pq[len(pq) - 1] # Pop it from the pq pq = pq[0:len(pq) - 1] # Add it to the answer ans += top # Divide it by 2 and push it # back to the pq top = top // 2; pq.append(top) K -= 1 # Print the answer print(ans) # Driver Codeif __name__ == '__main__': A = [ 2, 1, 7, 4, 2 ] K = 3 N = len(A) maxDiamonds(A, N, K) # This code is contributed by SURENDRA_GANGWAR
## C#
// C# program for the above approachusing System;using System.Collections;using System.Collections.Generic; class GFG{ // Function to find the maximum number// of diamonds that can be gained in// exactly K minutesstatic void maxDiamonds(int []A, int N, int K){ // Stores all the array elements var pq = new List(); // Push all the elements to the // priority queue for(int i = 0; i < N; i++) { pq.Add(A[i]); } // Stores the required result int ans = 0; // Loop while the queue is not // empty and K is positive while (pq.Count!=0 && K-- > 0) { pq.Sort(); // Store the top element // from the pq int top = pq[pq.Count-1]; // Pop it from the pq pq.RemoveAt(pq.Count-1); // Add it to the answer ans += top; // Divide it by 2 and push it // back to the pq top = top / 2; pq.Add(top); } // Print the answer Console.WriteLine(ans);} // Driver Codepublic static void Main(string[] args){ int []A= { 2, 1, 7, 4, 2 }; int K = 3; int N = A.Length; maxDiamonds(A, N, K);}} // This code is contributed by rrrtnx.
## Javascript
Output:
14
Time Complexity: O((N + K)*log N)
Auxiliary Space: O(N)
My Personal Notes arrow_drop_up | 1,747 | 5,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-49 | latest | en | 0.770156 |
https://in.mathworks.com/matlabcentral/cody/problems/8-add-two-numbers/solutions/1986808 | 1,590,822,246,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00463.warc.gz | 402,290,463 | 15,547 | Cody
# Problem 8. Add two numbers
Solution 1986808
Submitted on 23 Oct 2019 by Chai Yat Fung
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
a = 1; b = 2; c_correct = 3; assert(isequal(add_two_numbers(a,b),c_correct))
2 Pass
a = 17; b = 2; c_correct = 19; assert(isequal(add_two_numbers(a,b),c_correct))
3 Pass
a = -5; b = 2; c_correct = -3; assert(isequal(add_two_numbers(a,b),c_correct)) | 168 | 518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-24 | latest | en | 0.676081 |
https://www.jiskha.com/questions/1020062/On-an-18-hole-golf-course-there-are-par-3-holes-par-4-holes-and-par-5-holes-A | 1,537,926,832,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162809.73/warc/CC-MAIN-20180926002255-20180926022655-00181.warc.gz | 767,222,606 | 5,290 | # Algebra
On an 18-hole golf course, there are par-3 holes, par-4 holes, and par-5 holes. A golfer who shoots par on every hole has a total of 66. There are twice as many par-4 holes as there are par-5 holes. How many of each type of hole are there on the golf course?
How many par-3 holes are there__?
How many par-4 holes are there__?
How many par-5 holes are there__?
1. let there b x par-3, y par-4, z par-5 holes. Then
x+y+z = 18
3x+4y+5z = 66
y = 2z
Now just solve for x,y,z
posted by Steve
2. I'm not sure if I did this right, but I followed your formula, and got 9,6, and 3 as the answer. Is this correct?
posted by Curious
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More Similar Questions | 868 | 3,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-39 | latest | en | 0.914489 |
http://www.onlinemathlearning.com/applications-pythagorean-theorem.html | 1,512,964,756,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00109.warc.gz | 421,301,883 | 9,529 | # Applications of the Pythagorean Theorem
Videos to help Grade 8 students learn how to apply the Pythagorean Theorem to real world and mathematical problems in two dimensions.
New York State Common Core Math Grade 8, Module 7, Lesson 18
Related Topics:
Lesson Plans and Worksheets for Grade 8
Lesson Plans and Worksheets for all Grades
Lesson 18 Student Outcomes
• Students apply the Pythagorean Theorem to real world and mathematical problems in two dimensions.
Lesson 18 Summary
• We know some basic applications of the Pythagorean Theorem in terms of measures of a television, length of a ladder, area and perimeter of right triangles, etc.
• We know that there will be some three-dimensional applications of the theorem beyond what we have already seen.
Lesson 18 Classwork
Exercises 1–5
1. The area of the right triangle shown below is 36.46 in2. What is the perimeter of the right triangle? Round your answer to the tenths place.
2. The diagram below is a representation of a soccer goal.
a. Determine the length of the bar, c, that would be needed to provide structure to the goal. Round your answer to the tenths place.
b. How much netting (in square feet) is needed to cover the entire goal?
3. The typical ratio of length to width that is used to produce televisions is 4:3.
a. A TV with those exact measurements would be quite small, so generally the size of the television is enlarged by multiplying each number in the ratio by some factor of x. For example a reasonably sized television might have dimensions of 4 × 5:3 × 5, where the original ratio 4:3 was enlarged by a scale factor of 5. The size of a television is described in inches, such as a 60” TV, for example. That measurement actually refers to the diagonal length of the TV (distance from an upper corner to the opposite lower corner). What measurement would be applied to a television that was produced using the ratio of 4 × 5:3 × 5?
b. A 42” TV was just given to your family. What are the length and width measurements of the TV?
c. Check that the dimensions you got in part (b) are correct using the Pythagorean Theorem.
d. The table that your TV currently rests on is 30” in length. Will the new TV fit on the table? Explain.
4. Determine the distance between the following pairs of points. Round your answer to the tenths place. Use graph paper if necessary.
a. (7, 4) and (-3, -2)
b. (-5, 2) and (3, 6)
5. What length of ladder will be needed to reach a height of 7 feet along the wall when the base of the ladder is 4 feet from the wall? Round your answer to the tenths place.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 693 | 2,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2017-51 | latest | en | 0.94328 |
http://www.usingenglish.com/forum/ask-teacher/50212-bring-forward.html | 1,386,551,872,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163839270/warc/CC-MAIN-20131204133039-00007-ip-10-33-133-15.ec2.internal.warc.gz | 605,250,614 | 11,859 | 1. ## bring forward
We are going to bring our first midterm forward for two days.
We'll bring our meeting forward for an hour.
I try to make two samples with "bring forward." Correct me if I am wrong. Thanks.
2. Junior Member
Join Date
Sep 2007
Posts
64
## Re: bring forward
We are going to bring our first midterm forward for two days.
I am not exactly sure what you mean by this????? If you mean that you will write the exam in two days, I guess you could say that with the one small change I made.
We are going to bring our first midterm forward in two days. or
Our first mid-term will be brought forward in two days.
We'll bring our meeting forward for an hour.
We'll move our meeting ahead one hour. Then we can bring forward your idea
3. ## Re: bring forward
I am not exactly sure what you mean by this?????
I mean "We are going to bring our first midterm ahead/forword two days earlier than the date on the schedule."
If you mean that you will write the exam in two days, I guess you could say that with the one small change I made.
We are going to bring our first midterm forward in two days. or
Our first mid-term will be brought forward in two days.
Does my new version mean the same as these two of yours?
We'll bring our meeting forward an hour.
We'll move our meeting ahead one hour.
Are the above two interchangeable?
4. Junior Member
Join Date
Sep 2007
Posts
64
## Re: bring forward
I am not exactly sure what you mean by this?????
I mean "We are going to bring our first midterm ahead/forword two days earlier than the date on the schedule."
If you mean that you will write the exam in two days, I guess you could say that with the one small change I made.
We are going to bring our first midterm forward in two days. or
Our first mid-term will be brought forward in two days.
Does my new version mean the same as these two of yours?
No. When I use "bring forward" , I usually intend it to mean "introduce for the first time". In the situation you describe, I would use move instead of bring.
We'll bring our meeting forward an hour.
We'll move our meeting ahead one hour.
Are the above two interchangeable?
No, I don't think so. I would use the latter... although , if someone said the former to me, I would probably "get it". As I said, I use "bring forward" to mean "introduce", as in a topic at a meeting of some sort.
We'll move our meeting ahead one hour. Then we can bring forward (introduce) your idea about......blah, blah.
5. ## Re: bring forward
Thanks, redcrows, for your time and kindness.
I get it now.
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In Wiring Diagram291 views
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Don't ask me why I have such of an obsession with wires, but I do. My mother always said that ever since I've been able to walk, I would find things with wires and play with them and tear them apart, figure out how they worked and would be totally fascinated.
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https://byjusexamprep.com/defence/find-the-total-number-of-protons-in-10g-of-calcium-carbonate-caco3 | 1,713,755,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00854.warc.gz | 129,215,759 | 28,068 | # Find the total number of protons in 10g of calcium carbonate (CaCO3).
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The total number of protons in 10g of calcium carbonate (CaCO3) is 3.0115 x 1024 protons. The positively charged particles called protons are found in the hydrogen atom’s nucleus. A proton has the same mass as an atom of hydrogen.
### Calculate the Number of Moles
Step-1 Now we have to Calculate the number of moles
• The molar mass of calcium carbonate is = 100 g/mol
• A given weight of calcium carbonate is = 10g
Formula to find number of moles = Given weight/ Molar mass
• Number of moles = 10/100
• Number of moles = 0.1
Step-2 Now calculate the number of molecules using Avogadro Number
• As one mole of molecules = 6.022 x 1023 molecules
• 0.1 mole of molecules = 6.022 x 1022 molecules
Step-3 Calculating the number of Proton
• One calcium atom, one carbon atom, and three oxygen atoms make up calcium carbonate.
• The calcium atom possesses 20 protons compared to the carbon atom’s 6 protons.
• Oxygen atoms have eight protons each.
Therefore, three oxygen atoms will have = 8 x 3 = 24 protons. We know the number of protons in 10g of Calcium Carbonate = the Number of protons in 6.022 x 1022 molecules. Then, we can use it to calculate
Total number of protons in 10 mg of Calcium Carbonate = Total number of protons x 6.022 x 1022
= 50 x 6.022 x 1022
= 3.0115 x 1024 protons
Hence, the Total Number Of Protons In 10g Of Calcium Carbonate (CaCO3) is 3.0115 x 1024 protons
Summary:
## Find the total number of protons in 10g of calcium carbonate (CaCO3).
Therefore the total number of protons in 10g of calcium carbonate (CaCO3) is 3.0115 x 1024 protons. The nucleus of a hydrogen atom contains protons and positively charged subatomic particles.
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I'm thinking you might not be a "pear".
[–] 2 points3 points (10 children)
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I tend to agree with /u/mockingjay Have you used to body type calculator on the side bar? Hw did you determine that you are pear, as opposed to hourglass? I think it may be easier to find suggestions on how to dress if you recalculate your measurements.
[–][S] 0 points1 point (8 children)
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I was always told if the bottom half is considerably wider than the top, that's a pear shape?
Sorry if I'm wrong.
And hour glasses are normally evenly proportioned, where I have just noticed how proportionally larger my hip/bum is to the rest of me.
[–] 0 points1 point (7 children)
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Well is usually based on measurements. If your best measurement is 44 inches (as someone below figured out, I dont know if thats correct), and your hips are 39, then the body calculator comes out with a top hourglass. Its not actual width that this is based on, but over all measurements and proportion.
[–][S] 2 points3 points (6 children)
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Right ok. So how do I dress for width then. I just stated in another comment that my bust goes forwards not sidewards and my hip/bum area go sidewards not backwards.
I'm so damn sorry for posting now, because it seems that numbers are what it's all about, and that me saying my hips are wider means nothing.
Sorry for wasting all our times.
[–]actual tiger 4 points5 points (5 children)
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That's actually true of almost all hourglasses, as an hourglass usually has a fuller bust (forward not sideways) rather than huge lats (would give torso width). It would be very rare to have boobs that go sideways, right?
This may be unusual but I think it would help to know your shoulder size. In my experience, shoulders+bust is a better indication of how much you need to balance out your hips than just bust because broader shoulders (sideways measurement) can give the illusion of smaller hips whereas smaller shoulders can make hips look larger.
In general, the stock advice given to folks with your body type (37-26-36 personally so similar) is to play up chest and hips and define waist. However, it sounds like you'd rather downplay your chest and hips. It's hard to do that while also playing up your waist, but some things you can try:
1. Emphasize legs if you like your legs. Even if you don't like your thighs, brightly colored shoes or fun tights paired with knee length skirts can help.
2. Read advice for short women on how to elongate their torso. Although you may not be short at all (no height info above) you may want to focus on creating a long, line line to draw the eye up and down rather than having it stop at the horizontal breaks in your body. Ways to do this: all one color, dark colors, not wearing pieces that break up your torso (dresses or longer tops), creating a v (low button blazer, long pendant necklace, etc).
[–][S] 2 points3 points (4 children)
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This. This is an amazing reply. Thank you.
Sorry, I forgot my height! I'm around 5'4/5'5. (Don't know if its just me, but some days I'm as short as 5'3, I attribute it to back problems but god only knows!) basically I'm not tiny, but not model material either.
I don't have a measure tape handy, but asked my SO and he said I do have wide shoulders, followed by 'you have a large frame' (what a keeper? Jk) I think it'd be fair to say my shoulders wouldn't be to narrow compared to my hips.
I'll definitely look into creating a longer looking frame, I think this will help a lot. Do you know of any good resources off hand? It's fine if you don't I'll google about when I'm at home again.
Another q for you, if you don't mind, for elongating my legs (which I don't mind doing) is there a good way of doing this while still wearing flat shoes? I normally wear bright red doc martins so I have the bright fun shoe part down! But any other advice would be greatly appreciated.
I feel completely out of depth even remotely trying to be fashion conscious. I've never been immersed in it or brought up in it or shown it. So thank you for helping, sincerely.
[–]actual tiger 2 points3 points (3 children)
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I am not sure of any specific resources though you might like this on hourglasses. It's mostly about emphasizing the hourglass rather than downplaying curves, though. If you google "how to look taller" or "how to look thinner" most of the tips that appear will be advice on creating a longer line.
Typically, to elongate the legs while wearing flat shoes in a skirt, you want to match the color of your tights to either your shoes or your skirt. Because your shoes are red, probably you want to match your tights and your skirt-- so, navy skirt, navy tights, red shoes, etc. Shoes with a lower vamp will also help elongate the leg.
[–][S] 1 point2 points (2 children)
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Thank you so much. I'll have a read of that article and search for those terms you've suggested.
Thanks for explaining the shoes/tights/skirt thing, I never knew that or heard of it before. It will help loads. Would it be good idea to stay away from red shoes/red skirt combo?
[–]actual tiger 2 points3 points (1 child)
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Nothing wrong with red shoes and a red skirt although they should either match or be very very intentionally different reds. However, if your top goal is leg elongation, you are creating two strong lines across your leg: one where your shoes end and one where your skirt ends. The more horizontal lines you have the shorter your legs or torso look.
[–][S] 0 points1 point (0 children)
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Ok got it. Thanks very much for all your help :)
[–] 0 points1 point (0 children)
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I used the body calculator and it told me I was pear shaped. I definitely have big boobs though. My hips/butt are probably just a couple inches bigger. Still said I was pear. I started to think I had been wrongly classifying myself until the husband said the website must be wrong.
[–] 2 points3 points (23 children)
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That puts you in the realm of a 44 chest (without the bra). I agree you're not a pear, if anything you're probably top heavy.
[–][S] 0 points1 point (22 children)
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44 chest? Where are you getting that? And I'm looking at myself, my hip/bum is considerably larger/wider than the upper half of my body.
[–] 0 points1 point (20 children)
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I think she got that from the 34H bra size. If that is incorrect, what is your bust measurement?
[–][S] 0 points1 point (19 children)
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Ok so I majorly herped and was counting my band size 34inch to be my bust size. Don't burn me at the steak!
Ok with that said. My upper body is still slimmer than my lower body, so what should I say I am to address that?
My breasts protrude from my body, they go forwards not sidewards, my hips/bum are going sidewards not backwards
Sorry I'm really frustrated now, and feel quite attacked. I didn't realise I wasn't allowed to misunderstand things.
[–] 1 point2 points (2 children)
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Don't burn me at the steak!
I know you're feeling frustrated now but this just made me giggle. Stake is what you're looking for; steaks are delicious.
Either way, I agree with everyone else. You're probably an hourglass shape. Since you want to try to downplay your hips, I would suggest getting some skirts or dresses that are fitted around your natural waist but flare around the hips. Circle skirts are great for that sort of thing.
[–][S] 1 point2 points (0 children)
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I have loads of circle skirts/ dresses. Love them. But they tend to run off my bum/hips and end up making my rear look humongous.
It's been established, that on paper I'm an hour glass. But in terms of width, my top is a lot skinner than my bottom thus making me appear like a pear shape, but yet I have a large chest(which causes the hourglass to be the on paper hourglass) but my breasts go forward, not sideways, where my bum/hips go sideways not backward. This gives the impression that I'm a pear. It's really frustrating, and so far no one has been able to give advice for it.
I don't want to have my chest on show to try bring attention away from my wide hip, it's already difficult enough to find suitable tops that don't give cleavage. I don't like being on show. It's just my preference.
[–][S] 0 points1 point (0 children)
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I have loads of circle skirts/ dresses. Love them. But they tend to run off my bum/hips and end up making my rear look humongous.
It's been established, that on paper I'm an hour glass. But in terms of width, my top is a lot skinner than my bottom thus making me appear like a pear shape, but yet I have a large chest(which causes the hourglass to be the on paper hourglass) but my breasts go forward, not sideways, where my bum/hips go sideways not backward. This gives the impression that I'm a pear. It's really frustrating, and so far no one has been able to give advice for it.
I don't want to have my chest on show to try bring attention away from my wide hip, it's already difficult enough to find suitable tops that don't give cleavage. I don't like being on show. It's just my preference.
Also steaks ARE delicious. Now I want steak.
[–] 0 points1 point (4 children)
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I dont think anyone here is trying to attack you. It will just be much easier to answer your question once your actual body type has been established.
[–][S] 1 point2 points (3 children)
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Well it's been established, hourglass. Which I have generally always adhered to. But my bum/hips are now much wider than my top half. So what do I do?
[–] 0 points1 point (2 children)
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Well are you trying to downplay your hips? Or do you just want to look proportioned over all? Items like pencil skirts or high waisted pants (though these can be touch and go) work well to accentuate your smaller waist and curves. But these items wont work well at hiding your hips (if thats your goal).
[–][S] 1 point2 points (0 children)
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Yeah that is my goal. I don't want to accentuate my bum or my breasts, but I do want to look well presented, in general. I just want tips for this figure, cause I could not find any.
[–][S] 0 points1 point (0 children)
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Also, personally, I don't think it's nice to see tight clinging clothes on larger bottoms. I thought that was a general rule
[–] 0 points1 point (1 child)
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Most breasts go forward, not sideways.
[–][S] 0 points1 point (0 children)
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Yes, I was making this point to emphasise why I said pear, and why I am asking for help with it. Having a much wider hip/bum area is noticeable even if I am top heavy also.
And it's frustrating to be met with "no your am hourglass, it's fine".
[–] 0 points1 point (0 children)
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An H cup is usually a 10 inch difference between your under bust measurement and the fullest part of the breast.
[–] 2 points3 points (2 children)
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If you want to show off the features you feel most confident in, I would go with a fit and flare style of dress. In order to 'rein in' the girls up top, so to speak, I would recommend going with a more structured and tailored look with a higher neckline. Here are some examples of what I mean: 1 2 3
And if it gets cold enough in your area, these are some coats that follow the same principle: 1 2
For the best possible advice though, I think a picture would be necessary. For example, I would totally rock some of these looks myself, but I have kind of a short neck so I feel that the high neckline makes me look stumpy on top. So your mileage may vary!
[–][S] 1 point2 points (1 child)
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Thank you for the reply! Ok, those dresses make sense. I'll have to eventually invest in some. Any I've tried similar to that so far, don't fit the ladies in too well. But I'm sure a good one will fit better.
Thanks very much :)
[–] 1 point2 points (0 children)
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No problem! If you can't find anything exactly in your size, you can also do separates. If you find a nice structured top, you can pair it with a high waisted A-line skirt, or a pencil skirt if you like the way those look on you. | 3,351 | 13,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2015-35 | latest | en | 0.968964 |
https://www.12000.org/my_notes/CAS_integration_tests/reports/rubi_4_16_1_graded/test_cases/1_Algebraic_functions/1.2_Trinomial_products/1.2.1_Quadratic/1.2.1.2-d+e_x-%5Em-a+b_x+c_x%5E2-%5Ep/rese2069.htm | 1,695,864,543,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510334.9/warc/CC-MAIN-20230927235044-20230928025044-00587.warc.gz | 680,490,665 | 5,719 | ### 3.2069 $$\int \frac{1}{\sqrt{d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx$$
Optimal. Leaf size=196 $-\frac{3 c d \sqrt{d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{\sqrt{d+e x} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{3 c d \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}}$
[Out]
1/((c*d^2 - a*e^2)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (3*c*d*Sqrt[d + e*x])/((c*d^2
- a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (3*c*d*Sqrt[e]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 +
a*e^2)*x + c*d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/(c*d^2 - a*e^2)^(5/2)
________________________________________________________________________________________
Rubi [A] time = 0.131511, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {672, 666, 660, 205} $-\frac{3 c d \sqrt{d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{1}{\sqrt{d+e x} \left (c d^2-a e^2\right ) \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac{3 c d \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}}$
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]
[Out]
1/((c*d^2 - a*e^2)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (3*c*d*Sqrt[d + e*x])/((c*d^2
- a*e^2)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (3*c*d*Sqrt[e]*ArcTan[(Sqrt[e]*Sqrt[a*d*e + (c*d^2 +
a*e^2)*x + c*d*e*x^2])/(Sqrt[c*d^2 - a*e^2]*Sqrt[d + e*x])])/(c*d^2 - a*e^2)^(5/2)
Rule 672
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*(m + 2*p + 2))/((m + p + 1)*(2*c*d - b*e)), I
nt[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ
[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]
Rule 666
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((2*c*d - b*e)*(d +
e*x)^m*(a + b*x + c*x^2)^(p + 1))/(e*(p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*c*d - b*e)*(m + 2*p + 2))/((p + 1)*
(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]
Rule 660
Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(
2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^
2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
Rule 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac{1}{\left (c d^2-a e^2\right ) \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac{(3 c d) \int \frac{\sqrt{d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{2 \left (c d^2-a e^2\right )}\\ &=\frac{1}{\left (c d^2-a e^2\right ) \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{3 c d \sqrt{d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{(3 c d e) \int \frac{1}{\sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 \left (c d^2-a e^2\right )^2}\\ &=\frac{1}{\left (c d^2-a e^2\right ) \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{3 c d \sqrt{d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{\left (3 c d e^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 x^2} \, dx,x,\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}}\right )}{\left (c d^2-a e^2\right )^2}\\ &=\frac{1}{\left (c d^2-a e^2\right ) \sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{3 c d \sqrt{d+e x}}{\left (c d^2-a e^2\right )^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac{3 c d \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{c d^2-a e^2} \sqrt{d+e x}}\right )}{\left (c d^2-a e^2\right )^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.023945, size = 77, normalized size = 0.39 $-\frac{2 c d \sqrt{d+e x} \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{e (a e+c d x)}{a e^2-c d^2}\right )}{\left (c d^2-a e^2\right )^2 \sqrt{(d+e x) (a e+c d x)}}$
Antiderivative was successfully verified.
[In]
Integrate[1/(Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]
[Out]
(-2*c*d*Sqrt[d + e*x]*Hypergeometric2F1[-1/2, 2, 1/2, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/((c*d^2 - a*e^2)^
2*Sqrt[(a*e + c*d*x)*(d + e*x)])
________________________________________________________________________________________
Maple [A] time = 0.214, size = 235, normalized size = 1.2 \begin{align*}{\frac{1}{ \left ( cdx+ae \right ) \left ( a{e}^{2}-c{d}^{2} \right ) ^{2}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}xcd{e}^{2}+3\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) \sqrt{cdx+ae}c{d}^{2}e-3\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}xcde-\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}a{e}^{2}-2\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}c{d}^{2} \right ) \left ( ex+d \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x)
[Out]
(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(1/2)*(3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1
/2)*x*c*d*e^2+3*arctanh(e*(c*d*x+a*e)^(1/2)/((a*e^2-c*d^2)*e)^(1/2))*(c*d*x+a*e)^(1/2)*c*d^2*e-3*((a*e^2-c*d^2
)*e)^(1/2)*x*c*d*e-((a*e^2-c*d^2)*e)^(1/2)*a*e^2-2*((a*e^2-c*d^2)*e)^(1/2)*c*d^2)/(e*x+d)^(3/2)/(c*d*x+a*e)/(a
*e^2-c*d^2)^2/((a*e^2-c*d^2)*e)^(1/2)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{3}{2}} \sqrt{e x + d}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)*sqrt(e*x + d)), x)
________________________________________________________________________________________
Fricas [B] time = 2.25803, size = 1555, normalized size = 7.93 \begin{align*} \left [\frac{3 \,{\left (c^{2} d^{2} e^{2} x^{3} + a c d^{3} e +{\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x^{2} +{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2}\right )} x\right )} \sqrt{-\frac{e}{c d^{2} - a e^{2}}} \log \left (-\frac{c d e^{2} x^{2} + 2 \, a e^{3} x - c d^{3} + 2 \, a d e^{2} - 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{-\frac{e}{c d^{2} - a e^{2}}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (3 \, c d e x + 2 \, c d^{2} + a e^{2}\right )} \sqrt{e x + d}}{2 \,{\left (a c^{2} d^{6} e - 2 \, a^{2} c d^{4} e^{3} + a^{3} d^{2} e^{5} +{\left (c^{3} d^{5} e^{2} - 2 \, a c^{2} d^{3} e^{4} + a^{2} c d e^{6}\right )} x^{3} +{\left (2 \, c^{3} d^{6} e - 3 \, a c^{2} d^{4} e^{3} + a^{3} e^{7}\right )} x^{2} +{\left (c^{3} d^{7} - 3 \, a^{2} c d^{3} e^{4} + 2 \, a^{3} d e^{6}\right )} x\right )}}, -\frac{3 \,{\left (c^{2} d^{2} e^{2} x^{3} + a c d^{3} e +{\left (2 \, c^{2} d^{3} e + a c d e^{3}\right )} x^{2} +{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2}\right )} x\right )} \sqrt{\frac{e}{c d^{2} - a e^{2}}} \arctan \left (-\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (c d^{2} - a e^{2}\right )} \sqrt{e x + d} \sqrt{\frac{e}{c d^{2} - a e^{2}}}}{c d e^{2} x^{2} + a d e^{2} +{\left (c d^{2} e + a e^{3}\right )} x}\right ) + \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (3 \, c d e x + 2 \, c d^{2} + a e^{2}\right )} \sqrt{e x + d}}{a c^{2} d^{6} e - 2 \, a^{2} c d^{4} e^{3} + a^{3} d^{2} e^{5} +{\left (c^{3} d^{5} e^{2} - 2 \, a c^{2} d^{3} e^{4} + a^{2} c d e^{6}\right )} x^{3} +{\left (2 \, c^{3} d^{6} e - 3 \, a c^{2} d^{4} e^{3} + a^{3} e^{7}\right )} x^{2} +{\left (c^{3} d^{7} - 3 \, a^{2} c d^{3} e^{4} + 2 \, a^{3} d e^{6}\right )} x}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")
[Out]
[1/2*(3*(c^2*d^2*e^2*x^3 + a*c*d^3*e + (2*c^2*d^3*e + a*c*d*e^3)*x^2 + (c^2*d^4 + 2*a*c*d^2*e^2)*x)*sqrt(-e/(c
*d^2 - a*e^2))*log(-(c*d*e^2*x^2 + 2*a*e^3*x - c*d^3 + 2*a*d*e^2 - 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*
x)*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-e/(c*d^2 - a*e^2)))/(e^2*x^2 + 2*d*e*x + d^2)) - 2*sqrt(c*d*e*x^2 + a*d
*e + (c*d^2 + a*e^2)*x)*(3*c*d*e*x + 2*c*d^2 + a*e^2)*sqrt(e*x + d))/(a*c^2*d^6*e - 2*a^2*c*d^4*e^3 + a^3*d^2*
e^5 + (c^3*d^5*e^2 - 2*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x^3 + (2*c^3*d^6*e - 3*a*c^2*d^4*e^3 + a^3*e^7)*x^2 + (c^3
*d^7 - 3*a^2*c*d^3*e^4 + 2*a^3*d*e^6)*x), -(3*(c^2*d^2*e^2*x^3 + a*c*d^3*e + (2*c^2*d^3*e + a*c*d*e^3)*x^2 + (
c^2*d^4 + 2*a*c*d^2*e^2)*x)*sqrt(e/(c*d^2 - a*e^2))*arctan(-sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(c*d^2
- a*e^2)*sqrt(e*x + d)*sqrt(e/(c*d^2 - a*e^2))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) + sqrt(c*d*e*x^
2 + a*d*e + (c*d^2 + a*e^2)*x)*(3*c*d*e*x + 2*c*d^2 + a*e^2)*sqrt(e*x + d))/(a*c^2*d^6*e - 2*a^2*c*d^4*e^3 + a
^3*d^2*e^5 + (c^3*d^5*e^2 - 2*a*c^2*d^3*e^4 + a^2*c*d*e^6)*x^3 + (2*c^3*d^6*e - 3*a*c^2*d^4*e^3 + a^3*e^7)*x^2
+ (c^3*d^7 - 3*a^2*c*d^3*e^4 + 2*a^3*d*e^6)*x)]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac{3}{2}} \sqrt{d + e x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)**(1/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)
[Out]
Integral(1/(((d + e*x)*(a*e + c*d*x))**(3/2)*sqrt(d + e*x)), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")
[Out]
sage0*x | 6,280 | 11,586 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-40 | latest | en | 0.134526 |
https://www.gamedev.net/forums/topic/623423-math-classes-relevant-for-graphics-programming/?forceDownload=1&_k=880ea6a14ea49e853634fbdc5015a024 | 1,503,182,388,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105927.27/warc/CC-MAIN-20170819220657-20170820000657-00227.warc.gz | 888,198,222 | 26,906 | # Math classes relevant for graphics programming?
## Recommended Posts
sanjaygir 102
hi Everyone,
What math courses do you think will be important if you want to specialize in computer graphics. I assume, for one, graph theory will be important. Thanks for suggestions!
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alvaro 21246
I challenge that assumption. What part of CG do you think would benefit from graph theory?
The most important math courses for computer graphics are linear algebra and geometry (affine, euclidean and projective, although I think many universities don't teach projective geometry these days).
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mrjones 612
Graph theory is important for almost any software development field. In computer graphics optimization structures are based on graph theory. Although whether you'll ever need to implement these depends on how deep you want to go. Also meshes are rooted in graph theory and describeable as such. Even fractals can be thought of in terms of graph theory. So I'd say there are benefits, but of course it depends on whether one wants to go so low-level with graphics.
For graphics specifically I completely agree about algebra and geometry. Maybe functional analyses as well. It is also not so much graphics specific, but it's just generally useful.
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Promit 13246
Short answer: pretty much all of them.
Long answer: Linear algebra and (computational) geometry are pretty much crucial, in that you basically cannot function without them. Calculus/differential equations is fairly important, especially as modern lighting equations are derived from multi-dimensional surface integrals. Digital signal processing is also important for a lot of the basics. Topology is a big deal for a lot of things involving geometry, especially if you want to be working with parametric surfaces (which have become hardware accelerated recently). Probability and statistics drive a lot of the stochastic graphics techniques. Information theory (eg entropy) comes in big time if you want to look at compression, numerical stability, etc. Mathematical optimization is very useful for spatial partitioning algorithms.
The list goes on. It's difficult to learn any math that [i]isn't[/i] relevant.
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alvaro 21246
Is someone going to mention category theory again?
Don't get overwhelmed by Promit's list. Although he does have a point, you can get by with just linear algebra and geometry, and perhaps a bit of calculus so you don't run scared when you see an integral in some paper about lighting. I didn't mention computational geometry because I don't think of it as math, although I can see how other people would. And yes, it is important for computer graphics as well.
I am curious as to what part of topology might be useful for CG. I took three topology classes in college (point topology, algebraic topology and differential topology) and I can't think of any of it that might help with CG (actually, I can't think of anything I learned in differential topology, but has to do with bad teachers and not with this thread ).
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taby 1265
Surely meshes have something to do with computer graphics. As for category theory... ;)
Seriously... to get a good idea of what's going on with modern graphics, check out the SIGGRAPH proceedings, or the abstracts in journals like Computers & Graphics or IEEE Transactions on Visualization and Computer Graphics. Spot the buzzwords, do some googling, see what suits *your* fancy.
Like they're saying... the possibilities are huge. Do you like biology? Then work on software to procedurally generate plants. Do you like physics? Do you like chemistry? Do you like art?
You will most likely need linear algebra, no matter what type of graphics you get into.
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sanjaygir 102
Thanks everyone for the valuable input!
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Register a new account | 851 | 4,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-34 | latest | en | 0.954013 |
https://techqa.club/v/q/solver-tolerance-and-residual-error-when-using-sweep-function-in-fipy-c3RhY2tvdmVyZmxvd3w1NDYzNDI2OA== | 1,642,548,789,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00693.warc.gz | 547,185,208 | 5,754 | # Solver tolerance and residual error when using sweep function in FiPy
I was trying to use FiPy to solve a set of PDEs when I realized the command sweep was not working the way I thought it would. Here goes a sample with part of my code:
``````from pylab import *
import sys
from fipy import *
viscosity = 5.55555555556e-06
Pe =5.
pfi=100.
lfi=0.01
Ly=1.
Nx =200
Ny=100
Lx=Ly*Nx/Ny
dL=Ly/Ny
mesh = PeriodicGrid2DTopBottom(nx=Nx, ny=Ny, dx=dL, dy=dL)
x, y = mesh.cellCenters
xVelocity = CellVariable(mesh=mesh, hasOld=True, name='X velocity')
xVelocity.constrain(Pe, mesh.facesLeft)
xVelocity.constrain(Pe, mesh.facesRight)
var1 = DistanceVariable(name='distance to center', mesh=mesh, value=numerix.sqrt((x-Nx*dL/2.)**2+(y-Ny*dL/2.)**2))
pi_fi= CellVariable(mesh=mesh, value=0.,name='Fluid-interface energy map')
xVelocityEq = DiffusionTerm(coeff=viscosity) - ImplicitSourceTerm(pi_fi)
xres=10.
while (xres > 1.e-6) :
xVelocity.updateOld()
mySolver = LinearGMRESSolver(iterations=1000,tolerance=1.e-6)
xres = xVelocityEq.sweep(var=xVelocity,solver=mySolver)
print 'Result = ', xres
#Thats it
``````
In short, I am declaring a function called xVelocityEq and solving it using sweep. Here is my output:
``````Result = 0.0007856742013190237
Result = 6.414470433257661e-07
``````
As you can see, the while loop ends after two iterations. My first question is: why is my first residual error (=0.0007856742013190237) higher than the solver's tolerance? I thought that, since xVelocityEq corresponds to a linear system, solver tolerance and residual error would mean the same thing.
If I increase the no. of iterations in mySolver from 1000 to 10000, I get the following output:
``````Result = 0.0007856742013190237
Result = 2.4619110931978988e-09
``````
Why did the second residual change, given that the first remained the same?
If I increase the tolerance in mySolver from 1.e-6 to 7.e-4, I get the following output:
``````Result = 0.0007856742013190237
Result = 6.414470433257661e-07
``````
Note that these residuals are the same as in the first output. Now if I try to further increase the tolerance to 8.e-4, here's what I get as output:
``````Result = 0.0007856742013190237
Result = 0.0007856742013190237
Result = 0.0007856742013190237
Result = 0.0007856742013190237
Result = 0.0007856742013190237
...
``````
At this point I was completely lost. Why the residuals have the same values for all solver tolerances smaller than 7.e-4? And why these residuals are constant and equal to 0.0007856742013190237 for solver tolerances higher than 7.e-4?
If I change the mySolver to LinearLUSolver (iterations=1000, tolerance=1.e-6), here's what I get:
``````Result = 0.0007856742013190237
Result = 1.6772757200988522e-18
``````
Why in the world is my first residual the same as before, even though I have changed the solver? | 892 | 2,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-05 | latest | en | 0.750027 |
https://www.magischvierkant.com/two-dimensional-eng/4x4/sudoku-method/ | 1,714,009,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00418.warc.gz | 779,386,921 | 32,280 | ### Sudoku method
How can I use a 4x4 Sudoku to construct a 4x4 (pan)magic square?
Most times a Sudoku consists of 9 rows and 9 columns. In each row and in each column (and in each 3x3 sub-square) you find all the numbers from 1 up to 9. Use a 4x4 Sudoku to construct a 4x4 (pan)magic square in only 4 steps:
(1st) Fill in the 4x4 Sudoku with the numbers 0, 1, 2, 3. Take care that all the numbers from 0 up to 3 are in each row, column and diagonal.
(2nd) Construct the second 4x4 Sudoku by rotating the first Sudoku by a quarter to the right.
(3rd) Take 4x number from the first Sudoku and add 1x number from the same cell of the second Sudoku.
(4th) Finally add 1 to each cell.
4x number + 1x number = +1 = magic square
0 1 2 3 2 1 3 0 2 5 11 12 3 6 12 13 3 2 1 0 3 0 2 1 15 8 6 1 16 9 7 2 1 0 3 2 0 3 1 2 4 3 13 10 5 4 14 11 2 3 0 1 1 2 0 3 9 14 0 7 10 15 1 8
This magic square happens to be panmagic!
Use this method [Sudoku method (1)] to construct magic squares of order is 2^n (= 2x2, 2x2x2, 2x2x2x2, ...) from 4x4 to infinity. See 4x48x816x1632x32
4x4, Sudoku method.xls | 452 | 1,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-18 | longest | en | 0.779323 |
https://mattcoaty.com/2019/09/10/data-landmarks-and-context/ | 1,686,241,476,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00638.warc.gz | 421,407,491 | 35,516 | # Data Landmarks and Context
One of my classes is working on a unit related to data displays and number systems. Around a week ago the class was putting together sets of numbers to match data landmarks. This was a challenge as students had to think differently. The class was also asked which data landmark better represents a student’s performance. I was meaning to write a post then, but a number of things came up and it never happened. Fast forward a week and here we are.
Students were given two sets of scores from two different students.
Jack’s scores: 85, 81, 78, 100, 84, 89
Sonja’s scores: 55, 87, 91, 92, 68, 93
Students were asked to find the median and mean for each student. For the most part, students were able to identify both of these landmarks. Here comes the kicker … now students needed to determine which landmark better represents each student’s performance, mean or median? This was a challenging prompt for a couple reasons.
• Students weren’t accustomed to using the word represent in this context. Students were taking the view that the students should get the higher grade and that would be the mean or median. They explained that the student should receive the higher grade because they (the person) is a hard worker and deserves to be rewarded with the highest score.
• Students thought of the word represents as the typical score. When discussing the mean earlier in the year the word typical would often come up as a synonym.
• Students looked at the last score as the most recent and thought that should be the final representation. My school is heading in the direction of standards-based grading so that’s maybe why students took that approach. I don’t know.
• Students looked at the lowest and highest score of each set of data and reviewed the range to help them pick the median or mean
After struggling a bit, the class came together and we discussed a few possible solutions. The class agreed that the question allows a lot of room for interpretation and context certainly matters. The fruitful conversation brought about a change in perspective for some as students started to see this type of math differently than just numbers sprawled across a page. The numbers had meaning and the context drives the answer.
A little later in the week students were asked the following prompt:
If you were the teacher in Jack and Sonja’s class, would you use the median or the mean to calculate students’ grades? Explain.
This was a bit confusing at first, but students made progress in understanding the context and how it helped determine which landmark to use. Again, I had answers related to the teacher wanting to give the higher score to help students with confidence. Other students used the data landmarks to find the average. I felt like students were more comfortable using the average as they could say that they used every data point, therefore making sure all assignments counted for something.
I’m looking forward to next week as we dive into histograms.
## Author: Matt Coaty
I've taught elementary students for the past 14 years. I enjoy reading educational research and learning from my PLN. Words on this blog are my own. | 666 | 3,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2023-23 | latest | en | 0.968782 |
https://www.myclassbook.org/2017/03/performance-of-evaporator.html | 1,721,535,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00327.warc.gz | 769,998,890 | 38,005 | # Performance of Evaporator
This post provides some information about what is mean by performance of evaporator, how performance of evaporator can be increased.
Basically the performance of evaporator depends upon its capacity and economy. Let us discuss these terms one by one.
## 1) Capacity of evaporator:
Capacity of evaporator is defined as the number of kilograms of water vaporized/evaporated per hour.
The rate of heat transfer Q through the heating surface of evaporator is the product of heat transfer coefficient, heat transfer surface area and the overall temperature drop.
Therefore Q = U×A×?T
Where
• Q = Rate of heat transfer
• A = area of the heat transfer surface
• ?T = overall temperature drop
The capacity of an evaporator depends upon the temperature of the feed solution. If the feed solution is at the boiling temperature corresponding to the pressure in vapor space of an evaporator, all the heat supplied will be utilized for evaporation, thus increasing the capacity of evaporator.
Similarly if the cold feed solution is fed in the evaporator, initially some energy will be required to increase the temperature of the feed solution to the boiling point of the solution corresponding to the vapor pressure inside an evaporator. In some cases this amount energy may be very high. Thus the capacity of evaporator will be reduced.
## 2) Evaporator Economy:
Economy of the evaporator is another important parameter which decides the performance of evaporator. It may be defined as the amount of steam used and is expressed in terms of pounds of vapor produced per pound of steam supplied to the evaporator train.
Following are some of the methods to increase the economy of evaporator;
a) Use of multiple effect evaporation system ( Vapor recompression )
In multiple effect evaporation system, the vapor produced in the first stage is used as energy source (heat) for the second stage and so on. Thus increasing the economy of evaporator.
For example, in three effect evaporator if 1Kg. of steam is supplied to the first stage 2.5 Kg. steam is produced.
In vapor recompression method, vapor from the evaporator is compressed to increase its temperature so that it will condensate at temperature higher enough to permit its use as a heating media in the same evaporator.
You may also like: | 482 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.892825 |
https://mathoverflow.net/questions/40745/how-to-recognize-subgroups-through-dynkin-diagram | 1,656,990,282,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00439.warc.gz | 412,484,486 | 25,868 | # how to recognize subgroups through Dynkin diagram?
Fix $\mathbb{C}$ as the base field, and reductive groups are assumed to be connected.
Consider the example $SO_N\subset SL_N$. $SO_N$ is its own normalizer in $SL_N$, and the rank is much smaller than the rank of $SL_N$. How can this be reflected in the Dynkin diagram of $SL_N$?
In Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group one sees that Levi subgroups $L$ of a given semi-siimple group $G$ can be recovered from Dynkin subdiagrams of the diagram of $G$, and subgroups of maximal rank can be recovered similarly from extended Dynkin diagrams. As is commented below, having maximal rank means being of the same rank as $G$ is, namely containing a maximal torus of $G$.
Contrary to the discussions in loc.cit, one considers a reductive subgroup $H\subset G$, such that the normalizer $N(H,G)$ of $H$ in $G$ is of lower rank than $G$. Its connected component, denoted as $N^\circ$, is not covered in loc.cit. Say $SO_{2n}\subset SL_{2n}$, the extended Dynkin diagram of $SL_{2n}$ looks like a loop with dots, while the one for $SO_{2n}$ contains branching vertices. It is not clear that the latter is produced from the former by removing vertices.
And furthermore when one passes to a general base field, say perfect or of characteristic zero for simplicity, with separable closure $\bar{k}$, it becomes more complicated if one restricts to the notion of $k$-rank. Consider a reductive $k$-subgroup $H\subset G$ that is self-normalizing, in the sense that it equals the neutral connected component of $N(H,G)$. By comparing the $k$-ranks and $\bar{k}$-ranks of $H$ and $G$, one is led to several different cases. Does one still have arguments similar to the operations on Dynkin diagrams as is in Reductive subgroup corresponding to a subdiagram of the Dynkin diagram of a simple group ?
Sorry for any misunderstanding about the cited mathoverflow discussions above, and references on extended Dynkin diagrams are also welcome.
• It isn't really a question of relating Dynkin diagrams (whether over an algebraically closed or smaller field), since this subgroup of the special linear group isn't of maximal rank and since its intrinsic root system isn't so closely related to the bigger one. There may be problems such as branching rules of representations where it's useful to compare the group and subgroup here. But a vast number of smaller reductive groups can live in a given special linear group of large rank. In any case, your rank description is off a bit for even $N$. Oct 1, 2010 at 17:00
• P.S. Perhaps you are thinking about the technique of "folding" a Dynkin diagram which plays a role in the work of Tits and others on forms of algebraic groups over various fields. (A basic article by Tits is freely available from AMS online in the old Proc. Symp. Pure Math. 9 at e-math.org.) But this seems far away from the question you ask in your next-to-last paragraph, so clarification would help. In any case, a subgroup of maximal rank has the same rank as the big group (thus shares a maximal torus). Oct 1, 2010 at 17:26
• @Jim:thanks a lot for the comments and the reference! I modify a little bit the statements above, and is starting reading the article by Dynkin mentioned below. Oct 2, 2010 at 14:08
Not easy!. If you want to know classification of $A_1$-subalgebras then it is the same as classification of nilpotent elements and it can be reduced to classifying all distinguished parabolics in all your Levi subalgebras. | 857 | 3,545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | latest | en | 0.909355 |
https://www.mi.fu-berlin.de/inf/groups/ag-ti/projekte/application-oriented/index.html | 1,726,516,388,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651710.86/warc/CC-MAIN-20240916180320-20240916210320-00741.warc.gz | 798,172,897 | 14,088 | Springe direkt zu Inhalt
# Design and Analysis of Application Oriented Geometric Algorithms
Projektleitung:
Mitarbeiter/innen:
Förderung:
Deutsche Forschungsgemeinschaft (DFG)
Projektlaufzeit:
01.01.2008 — 31.08.2013
## In collaboration with
Christian Knauer
A project within the DFG Priority Program 1307 Algorithm Engineering.
# Summary
The goal of this project is to close a gap between theoretically designed and studied algorithms and the methods that are widely used in practice, especially heuristics. We consider primarily but not exclusively the problem of the geometric shape matching.
There are many algorithms and heuristic methods that work well and are widely used in practice but lack theoretical analysis. We analyze some of such heuristics with respect to the following questions:
1. What is computed by the method? The quality of the computed results is often not known or even not clearly defined.
2. What is the running time of the method? In many cases there is a certain trade-off between the running time and the quality of results.
It might be necessary to constraint the inputs of the methods and to characterize the cases that occur in practice in order to be able to perform the analysis.
Besides these theoretical studies we also plan to do prototype implementations of the designed algorithms to verify their practicability. The research focuses on the following problems:
## Patterns and shapes
### Probabilistic shape matching
We designed and analyzed a generic probabilistic approach for the shape matching problem. The studied approach is based on an intuitive definition of the shape matching task: Given two shapes A and B find that transformation within the class of allowable transformations which maps B to A in a best possible way. A mapping is considered to be good if large parts of the two shapes coincide within some tolerance distance δ.
The main idea of the probabilistic algorithm is to take random samples of points from both shapes and give a "vote" for that transformation matching one sample with the other. If that experiment is repeated frequently, we obtain by the votes a certain probability distribution in the space of transformations. Maxima of this distribution indicate which transformations give the best match between the two figures. The matching step of the algorithm is, therefore, a voting scheme.
Related methods in the image processing community are the generalized Hough transform, the generalized Radon transform, and the RANSAC algorithm. In contrast to those methods we do not consider a discrete set of features that describe shapes, but work with continuous objects, e.g., curves, regions, and surfaces in 3D. Our method is independent of the choice of the parameterization and the discretization grid in the transformation space.
Our important contribution is the theoretical analysis since it leads to a better understanding of this kind of heuristic techniques. We studied this probabilistic method for several types of shapes in combination with different classes of transformations. We developed and analyzed exactly probabilistic shape matching for two dimensional shapes modeled as sets of line segments and various classes of transformations, such as translations, rigid motions, similarity maps, and affine transformations in general. Prototype implementations experimentally confirmed our results. We showed that the technique can be extended to matching regions instead of curves and to matching higher dimensional volumes.
We started to create a software package for shape matching that has a graphic user interface and can be made publicly available.
### Symmetry detection
The ongoing Ph.D. thesis of Claudia Dieckmann investigates algorithms for the detection of approximate symmetries in point sets. Here, probabilistic matching can be applied successfully, as implementations show, and also under certain assumptions on the input other efficient algorithms can be found.
### Application oriented methods
In the Ph.D. thesis of Fabian Stehn, geometric hybrid registration problems are studied. Registration problems are closely related to geometric matching problems. The term geometric registration problem describes the task of mapping points from one space (pattern space) to their corresponding points in a deformed copy of that space called model space.
This research is motivated by a real world application: navigated surgery. Here, the goal is to register an operation theatre space (pattern space) to the internal coordinate system (model space) of a medical navigation system. The purpose of a medical navigation system is to support surgeons by visualizing the used surgical instruments at their correct position in a 3D-model of a patient. The models are generated beforehand based on CT or MRT scans.
Hybrid registration is a novel strategy to compute solutions for this alignment problem. Geometric hybrid registrations reduce the spatial synchronization problem to a series of (at least two) geometric matching problems that are solved interdependently. Usually, a computationally involved point-to-surface matching is combined with a comparably simpler but underdefined point-to-point matching. The point-to-surface matching is computed for a sufficiently large and suitably distributed set of points (called surface points) measured in the pattern space to a geometric surface in the model space. For the point-to-point matching, a small set of (one to three) characteristic points are measured in the pattern space and are defined in the model space. In the context of the intended application, these points are called anatomic landmarks - anatomically exposed spots within the field of interest.
In the Ph.D. thesis of Sven Scholz the ideas of probabilistic shape matching are applied to a real world question in collaboration with AKTOR KNOWLEDGE TECHNOLOGY NV in Antwerp, Belgium: automatically determining the similarity between trademarks of companies. In particular, after some preprocessing shapes are extracted from the images of trademarks and compared using probabilistic shape matching enhanced with further heuristics. In order to demonstrate the usability of this approach in practice, all these algorithms have been implemented and tested on various sets of figurative images and real-world trademark images. The results of the experiments carried out show a large conformity with human perception of similarity and the operating figures obtained are clearly better than the ones of other existing systems.
## Packing and stacking geometric objects
One topic connected with shape matching, which we plan to address, are efficient algorithms for packing and stacking geometric objects. In the plane geometric objects would be, e.g., simple polygons which can be transformed by translations or rigid motions. The objective is to find transformations placing the objects such that some cost function is minimized, for example the area or the perimeter of the convex hull of the union of all transformed figures. The objects may either be overlapping (stacking) or overlapping is not allowed (packing).
Optimal packing of an arbitrary number of objects, even axes-parallel rectangles under translations, can easily shown to be NP-hard. This is an important practical problem in textile or steel industry, namely, cutting out given parts from a large cloth or sheet metal and minimizing the waste. Therefore, numerous heuristic approaches exist from the optimization community. Rigorously analyzed methods with a guaranteed performance have been found so far in the computational geometry community only in special cases.
## Realistic input models
### Geometric shape matching
For many other distance measure (e.g., the Hausdorff-distance or the area of the symmetric difference) the situation is similar. For the Earth-Mover's-distance (a distance measure for weighted point sets which is very popular in multimedia-retrieval no exact algorithmic solutions are known for the shape matching problem (even for translations in the plane). Again one is forced to turn to approximation methods - unfortunately, even these turn out to be fairly complicated and are not useful for real applications.
The goal is to investigate if and how restrictions to the structure of the input data can be exploited to develop more efficient and conceptually simpler algorithms for this kind of problems.
### Geometric search structures
Many search structures for geometric objects are based on the following general (heuristic) approach: For each object a (high-dimensional) feature vector is determined and the similarity of two objects is simply determined by the distance (in some Lp-metric) of the corresponding feature vectors. The vantage-point methods suggest to use the distance (w.r.t. a suitable geometric distance measure δ, like, e.g., the Hausdorff distance) to a small set of fixed so-called vantage-objects from the data set as a feature vector. These methods seem to work very well in practice, but they have not been fully analyzed from a theoretical point of view.
Worst-case time bounds are known only in special cases (like when δ is "close" to the Euclidean metric) and in existing applications the vantage-objects are chosen according to some ad-hoc heuristic.
We aim to analyze this approach in a broader context by first identifying and formalizing the appropriate theoretical questions and concepts, like, e.g., "What is the relation between δ and the Lp-distance of the feature vectors?", "What is a good set of vantage-objects?", or "What is the algorithmic complexity of finding such a good set (approximately)?". In a second step we will try to answer these questions for selected scenarios (geometric objects and distance measures δ) again with the emphasis on appropriate assumptions about realistic input models.
## PCA
Principal component analysis (PCA)is one of the oldest and best known techniques of multivariate analysis. Numerous applications of PCA can be found in data compression, visualization, image processing, pattern and image recognition, time series prediction, detecting symmetry, or dimension detection. Most of the applications of PCA are non-geometric in their nature. However, there are also few purely geometric applications like estimation of the undirected normals of the point sets or computing PCA bounding boxes (bounding boxes determined by the principal components of the point set).
We presented closed-form solutions for efficiently updating the principal components of a set of n points, when m points are added or deleted from the point set. For both operations performed on a discrete point set in Rd, we can compute the new principal components in O(m) time for fixed d. This is a significant improvement over the commonly used approach of recomputing the principal components from scratch, which takes O(n + m)time. An important application of the above result is the dynamical computation of bounding boxes based on principal component analysis. We have implemented and evaluated several algorithms for computing dynamically PCA bounding boxes in R3. In addition, we present closed-form solutions for computing dynamically principal components of continuous point sets in R2 and R3. In both cases, discrete and continuous, to compute the new principal components, no additional data structures or storage are needed. | 2,132 | 11,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-38 | latest | en | 0.917933 |
https://www.jiskha.com/display.cgi?id=1291436095 | 1,498,715,819,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323870.46/warc/CC-MAIN-20170629051817-20170629071817-00613.warc.gz | 908,110,365 | 3,786 | # Physics
posted by .
A plane drops a hamper of medical supplies from a height of 3410 m during a practice run over the ocean. The plane's horizontal velocity was 121 m/s at the instant the hamper was dropped. What is the magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean? The acceleration of gravity is 9.8 meters per second squared
• Physics -
Are you supposed to estimate aerodynamic drag? When dropped from that that altitude, it will definitely not be negligible.
Both the horizontal and vertical drag will be appreciable.
I would complain to the teacher | 131 | 614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2017-26 | longest | en | 0.943333 |
https://81018.com/csh/ | 1,685,986,043,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00069.warc.gz | 99,073,422 | 27,336 | # Assumptions, first principles, presuppositions
There are three facets of pi that are not finite or quantitative so we assume (hypothesize and/or hypostatize) these facets define the infinite and the qualitative (unique classes and categories).
Continuity is our first facet of infinity. It is the very nature of order. Within the finite it looks like a string of numbers and feels like time. Pi qualifies; it’s an equation that has never-ending results that are always the same and always changing.
Symmetry is the second facet of infinity. It looks like geometries and is the very nature of a relation. Within the finite it feels like space. Pi qualifies; it’s a symmetry that generates symmetries. It’s an equation that generates equations.
Harmony is the third facet of infinity. It is the very nature of dynamics; and within the finite, it is always cyclical (periodicity) and experienced as space-time moments. Pi’s numbers, geometries, and equations (Fourier transform and others) are here within an eternal dance and there’s a domain of perfection which may be experienced as a moment of perfection.
All other definitions of the infinite are put on hold. Most are personal definitions that come from personal experiences and family history. That is one’s own business, not ours. If those beliefs help you through life, that is great. Our goal here is to engage those principles and functions that give rise to mathematics, physics, and eventually all the other sciences.
Review: In this model the infinite is profoundly within the finite. It is not finite, but actively imparts qualities to the finite. For those who follow David Hilbert, please stay open. Pi’s three facets of the infinite are really real. These are not just abstractions, but realities of every circle and sphere. These three qualities condition the finite. Everything-everywhere-for all time, is in accordance with numbers, geometries, and equations; and, it all has some manifestation of these infinite qualities.
A rather different start to grasp the finite-infinite relation, our understanding of the infinite starts with pi and her most infinitesimal circles and spheres.
_____
This document is the key part of the following homepages:
https://81018.com/almost/#1f
_____
Assumptions: https://81018.com/assumptions/
First principles: https://81018.com/principles/
Presuppositions: https://81018.com/presuppositions/ | 513 | 2,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-23 | latest | en | 0.941708 |
https://honeycreekpres.org/and-pdf/1081-sean-carroll-spacetime-and-geometry-pdf-306-483.php | 1,642,482,281,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300722.91/warc/CC-MAIN-20220118032342-20220118062342-00063.warc.gz | 358,202,650 | 7,230 | and pdfThursday, April 29, 2021 3:16:35 PM1
# Sean Carroll Spacetime And Geometry Pdf
File Name: sean carroll spacetime and geometry .zip
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Published: 29.04.2021
These lecture notes are a lightly edited version of the ones I handed out while teaching Physics 8.
Please note: In order to keep Hive up to date and provide users with the best features, we are no longer able to fully support Internet Explorer. The site is still available to you, however some sections of the site may appear broken. We would encourage you to move to a more modern browser like Firefox, Edge or Chrome in order to experience the site fully.
## Physics 411: General Relativity
Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs and how to get involved. General Relativity and Quantum Cosmology. Authors: Sean M. Comments: pages, numerous figures. Bibliographic Explorer What is the Explorer?
Geometry of a Spacetime Unit Cell. The smallest components of spacetime will never be seen with the human eye as it is orders of magnitudes smaller than an atom. If an atom was the size of the Milky Way galaxy, a granule of Planck length radius would be roughly the size of a grain of sand on Earth. Spacetime and Geometry: An Introduction to General Relativity provides a lucid and thoroughly modern introduction to general relativity. Sean Carroll Spacetime and Geometry: An Introduction to General Relativity provides a lucid and thoroughly modern introduction to general relativity for advanced undergraduates and graduate students.
## Spacetime and Geometry: An Introduction to General Relativity
General relativity is the most beautiful physical theory ever invented. It describes one of the most pervasive features of the world we experience-gravitation-in terms of an elegant mathematical structure-the differential geometry of curved spacetime-leading to unambiguous predictions that have received spectacular experimental confirmation. If you are contemplating reading this book, that point is here. Download your book. Post Pagination Next Post Next. Like it? Share with your friends!
Sean M. Carroll manifolds, Riemannian geometry, Einstein's equations, and three updated versions, can be found at honeycreekpres.org~carroll/notes/. vature — the Newtonian limit — physics in curved spacetime.
## SPACETIME AND GEOMETRY pdf
Albert Einstein - spacetime diagram for two black holes colliding to become one Einstein with Tagore General Introduction The purpose of this class: This class will provide an overview of the theory of general relativity, Einstein's theory of relativistic gravity, as well as some basic applications, including at least the solar-system tests of gravitational theories,some of the more interesting properties of black holes and gravitational waves, along with some surveys of cosmology. This class will not completely prepare you for research in this area: it will be an overview with insufficient depth for that purpose. However, that is more likely than not exactly what you wanted anyway. The first third to half of the course will focus primarily on the basic structure of the theory, with relevant physical motivation and insight thrown in along the way, and also provide a reasonable introduction to the needed mathematics. You do NOT need to already know more physics and mathematics than is described in the Prerequisite section just below.
#### General Introduction
Она с самого начала возражала против его кандидатуры, но АНБ посчитало, что другого выхода. Хейл появился в порядке возмещения ущерба. После фиаско Попрыгунчика. Четыре года назад конгресс, стремясь создать новый стандарт шифрования, поручил лучшим математикам страны, иными словами - сотрудникам АНБ, написать новый супералгоритм. Конгресс собирался принять закон, объявляющий этот новый алгоритм национальным стандартом, что должно было решить проблему несовместимости, с которой сталкивались корпорации, использующие разные алгоритмы.
Бринкерхофф посмотрел на мониторы, занимавшие едва ли не всю стену перед ее столом. На каждом из них красовалась печать АНБ. - Хочешь посмотреть, чем занимаются люди в шифровалке? - спросил он, заметно нервничая.
Беккер вошел в телефонную будку и начал набирать номер Стратмора. Не успел он набрать международный код, как в трубке раздался записанный на пленку голос: Todos los circuitos estan ocupados - Пожалуйста, положите трубку и перезвоните позднее.
Три… три… Беккера словно еще раз ударило пулей, выпущенной из пистолета. Мир опять замер. Три… три… три… 238 минус 235. | 1,104 | 4,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.87839 |
https://community.jmp.com/t5/Discussions/Multiple-comparisons-test-on-a-Generalized-Linear-Model/td-p/188363 | 1,600,615,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198213.25/warc/CC-MAIN-20200920125718-20200920155718-00243.warc.gz | 334,804,315 | 75,121 | Share your ideas for the JMP Scripting Unsession at Discovery Summit by September 17th. We hope to see you there!
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Level I
## Multiple comparisons test on a Generalized Linear Model
Hello,
I was wondering whether it is possible to adjust for multiple comparisons (for instance with Tukey's test) after creating a generalized linear model using the Fit Model option. The goal would be to compare a volume of interest (continuous variable) between three groups of patients (class variable), after adjusting for a bunch of other covariates.
When I create (fit) the model, the volumes between the three groups seem to be significantly different from one another (as can be seen from the p-values reported under the Parameter Estimates), but I wanted to apply correction for multiple comparisons and make sure that the significant associations persist. I used the "contrast" option under the red triangle, but I wasn't quite clear on what the reported results represented.
Thanks in advance for any help!
Best,
Panos
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Staff
## Re: Multiple comparisons test on a Generalized Linear Model
If your response is normally distributed, then you could use Bivariate platform (Analyze > Fit Y by X) with volume in the Y role and patient in the X role. The platform menu includes a Compare Means sub-menu for multiple comparison methods.
What is the reason for using a GLM? What distribution and link function did you use?
The contrast report is not difficult to understand and use. I illustrate it here using the sample data set Big Class and a GLM using the normal distribution and identity link function. I tested height against age. Then I specified a series of comparisons between age=12 and all the other ages separately. The result is:
The bottom of the report is the place to start. The likelihood ratio test (L-R ChiSquare) is for any significant contrast. This statistic is your control of type I errors with multiple comparisons. The result here is significant at alpha=0.05 so we conclude that at least one contrast is significant. The individual contrasts show that the mean height for age=12 is not significantly different from that of age=13 but it is significant for the other ages.
NOTE: if you select the Help tool (?) and click on the Contrast report, you will be taken to a full description of the method and explanation of the interpretation.
Learn it once, use it forever!
4 REPLIES 4
Highlighted
Staff
## Re: Multiple comparisons test on a Generalized Linear Model
If your response is normally distributed, then you could use Bivariate platform (Analyze > Fit Y by X) with volume in the Y role and patient in the X role. The platform menu includes a Compare Means sub-menu for multiple comparison methods.
What is the reason for using a GLM? What distribution and link function did you use?
The contrast report is not difficult to understand and use. I illustrate it here using the sample data set Big Class and a GLM using the normal distribution and identity link function. I tested height against age. Then I specified a series of comparisons between age=12 and all the other ages separately. The result is:
The bottom of the report is the place to start. The likelihood ratio test (L-R ChiSquare) is for any significant contrast. This statistic is your control of type I errors with multiple comparisons. The result here is significant at alpha=0.05 so we conclude that at least one contrast is significant. The individual contrasts show that the mean height for age=12 is not significantly different from that of age=13 but it is significant for the other ages.
NOTE: if you select the Help tool (?) and click on the Contrast report, you will be taken to a full description of the method and explanation of the interpretation.
Learn it once, use it forever!
Highlighted
Level I
## Re: Multiple comparisons test on a Generalized Linear Model
Hi Mark,
Thank you for your prompt response!
I wanted to use the Fit Model option rather than the Fit Y by X solely because there are some other covariates that I would want/need to adjust for. Moreover, the only reason why I am using the Generalized Linear Model is because a reviewer suggested it for this particular analysis. The distribution I'm using is "Normal" and the link function is "identity."
Also, thank you for the very clear explanation on the contrast! So just to be clear, the contrast option performs the multiple comparison analysis, right? And should I just report it as "control of type 1 errors with mulriple comparisons" or does this particular correction applied here have a specific name (e.g. Tukey's, Bonferroni, etc).
Thanks again for all the help!
Best,
Panos
Highlighted
Staff
## Re: Multiple comparisons test on a Generalized Linear Model
Glad to help. You had not mentioned the covariates in the first post so your choice makes perfect sense now.
By the way, using a GLM with the normal distribtution and the identity link function is the same as the ordinary least squares regression with the linear predictor. You will have more choices of methods, such as Tukey HSD, if you use Fit Least Squares instead.
Learn it once, use it forever!
Highlighted
Level I
## Re: Multiple comparisons test on a Generalized Linear Model
The Fit Least Squares method sounds great in that case!
Thanks again,
Panos
Article Labels
There are no labels assigned to this post. | 1,140 | 5,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-40 | latest | en | 0.910266 |
https://web2.0calc.com/questions/psl4-18 | 1,596,994,947,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00552.warc.gz | 571,195,848 | 7,491 | +0
PSL4#18
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305
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+935
I have no idea how to approach this.
Jul 28, 2019
#1
+25481
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PSL4#18
$$\begin{array}{|c|c|c|c|c|} \hline & \text{penny} & \text{nickel} & \text{dime}& \text{quater} \\ & 1\ \text{cent} & 5\ \text{cent} & 10\ \text{cent}& 25\ \text{cent} \\ \hline &0&0&0&3 \\ &0&0&3&0 \\ &0&3&0&0 \\ &3&0&0&0 \\ \hline &1&2&0&0 \\ &1&0&2&0 \\ &1&0&0&2 \\ &0&1&2&0 \\ &0&1&0&2 \\ &0&0&1&2 \\ \hline &2&1&0&0 \\ &2&0&1&0 \\ &2&0&0&1 \\ &0&2&1&0 \\ &0&2&0&1 \\ &0&0&2&1 \\ \hline &1&1&1&0 \\ &1&1&0&1 \\ &1&0&1&1 \\ &0&1&1&1 \\ \hline sum &15&15&15&15 \\ \hline \end{array}$$
$$\begin{array}{|rcll|} \hline \text{sum} &=& 15\times 1\ \text{cent}+ 15\times 5\ \text{cent} + 15\times 10\ \text{cent}+ 15\times 25\ \text{cent} \\ &=& 15 + 75 + 150 + 375 \\ &=& 615 \\ &=& \mathbf{6.15} \\ \hline \end{array}$$
Jul 28, 2019
edited by heureka Jul 29, 2019
#4
+935
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dgfgrafgdfge111 Jul 28, 2019
#5
+25481
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Thank you dgfgrafgdfge111 !
heureka Jul 28, 2019
#2
+2
$$\binom{3+4-1}{3}*\frac{3}{4}*(0.01+0.05+0.1+0.25)=6.15$$
.
Jul 28, 2019
#3
+935
+2
How did you get this?
dgfgrafgdfge111 Jul 28, 2019
#6
+6196
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$$\text{The first term comes from the stars and bars solution to the number of ways to allot coins into bins.~\\ In this case we allot 3 coins into 4 bins labeled pennies, nickels, dimes, quarters.}\\~\\ \text{The term \dfrac{0.01+0.05+0.10+0.25}{4} represents the average value of a coin}\\~\\ \text{There are then 3 coins so the value is 3 times the average coin value}\\~\\ \text{Put all that together and you get the very clever formula posted}$$
btw I'm not the original poster of the solution, but who knows if they'll be back and it warrented explanation.
Rom Jul 28, 2019
edited by Rom Jul 28, 2019
#7
+935
+1
Ah, nice one. I would never have thought of that.
dgfgrafgdfge111 Jul 29, 2019 | 867 | 1,838 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-34 | latest | en | 0.58344 |
https://www.spoj.com/problems/QN02/ | 1,620,510,462,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988927.95/warc/CC-MAIN-20210508211857-20210509001857-00114.warc.gz | 1,045,597,907 | 9,718 | ## QN02 - Seating Arrangement
### Problem Statement
In DA-IICT, the end sems are just around the corner and it seems like all the students are working very hard. But it's not just students, the professors are hard at work too. In particular, the Dean (Students) is very busy working out the exam seating arrangements for all the students.
Now normally, as we all know, the benches in each of the exam halls can seat exactly 2 students. Also, it is ensured that every bench contains exactly 2 students from different batches (so that there is no copying). But inspite of this, the Dean has noticed that even if the 2 students are from totally different batches, if they are friends, then they tend to help each other (or atleast try to, depending on how much they both know). The Dean wants to prevent this, so the seating arrangement is such that no two friends sit side by side during any exam, so that the students prepare well for the exams. But he is falling short of ideas to work this out.
Please help him out. You are given a list of pairs of IDs ( A, B ), such that the student with ID A is friends with the student with ID B. For every query ( C, D ) please print out whether or not these 2 students are friends ( meaning they cannot be seated with each other ).
Note: In this case, please assume friendship to be both symmetric and transitive. That is, if A is friends with B, B is also friends with A. Moreover, if A and B are friends and B and C are friends, this implies that A and C are also friends.
### Input
The input comprises of several lines. The first line contains 2 integers - n and m, where n is the number of students who will be writing the exam ( 2 <= n <= 100000 ) and m is the number of pairs of student IDs in the input. ( 0 <= m <= 100000. Also m <= n*( n+1 ) / 2 ).
This is followed by m lines of the form : A B
This indicates that the student with ID A is friends with the student with ID B. ( 0 <= A,B < n ).
This is followed by a line containing a single integer q ( 1 <= q <= 100000 ) indicating the number of queries you have to answer. q lines of queries follow. Each query consists of a single line containing 2 space separated integers C and D. ( 0 <= C,D < n ). These are the 2 student IDs for which you have to state whether or not they are friends.
Note
: All student IDs are unique, and range from 0 to n-1.
### Output
For each query, output a single line with "YES" if the 2 students are friends, and "NO" otherwise. Please note that the quotes are just for clarity, and that the output is case-sensitive.
### Example
```Input:
7 41 22 31 34 541 34 65 15 4Output:YESNONOYES``` | 647 | 2,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-21 | latest | en | 0.964011 |
https://datacadamia.com/data/type/number/computer/floating_point | 1,653,029,362,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00577.warc.gz | 248,831,943 | 60,663 | # Computer - Floating-point (Representation and usage)
Computer representations of floating point numbers typically use a form of rounding to significant figures, but with binary numbers. The number of correct significant figures is closely related to the notion of relative error (which has the advantage of being a more accurate measure of precision, and is independent of the radix of the number system used).
Floating-point is ubiquitous (everywhere) in computer systems
• Almost every language has a floating-point datatype (Javascript, Python, Java, Oracle (SQL), …)
• Computers from PCs to supercomputers have floating-point accelerators (???)
• Most compilers will be called upon to compile floating-point algorithms from time to time;
• Every operating system must respond to floating-point exceptions such as overflow
Generally, the numbers represented in float are to big to fit in their physical representation (typically 32 bit). Therefore the result of a floating-point calculation must often be rounded in order to fit back into its finite representation.
This rounding error is the characteristic feature of floating-point computation.
## Management
### Approximate numeric
Avoid float and double if exact answers are required
If you need precise numbers (e.g. money), see fixed-point number (exact numeric).
Float are great, for geometry (2D, 3D,…).
### Rounding Error
Floating-point arithmetic can only produce approximate results, rounding to the nearest representable real number.
Floating-point numbers offer a trade-off between accuracy and performance.
With a 52 bits of precision , if you're trying to represent numbers whose expansion repeats endlessly, the expansion is cut off after 52 bits.
Unfortunately, most software needs to produce output in base 10, and common fractions in base 10 are often repeating decimals in binary.
For example:
• 1.1 decimal is binary 1.0001100110011 …;
• .1 = 1/16 + 1/32 + 1/256 plus an infinite number of additional terms.
IEEE 754 has to chop off that infinitely repeated decimal after 52 digits, so the representation is slightly inaccurate.
Sometimes you can see this inaccuracy when the number is printed:
>>> 1.1
1.1000000000000001
#### Guard Digits
Guard Digits are a means of reducing the error when subtracting two nearby numbers.
### Performance
Floats (doubles) are fast because they are native type. Floats are usable with vector registers (xmm etc.) whereas decimals aren't.
In general, processors execute integer operations much faster than floating-point operations.
Example
• the first loop is easily twice as fast compared to the second loop.
// Integer
for (let i = 0; i < 1000; ++i) {
// fast 🚀
}
// Float
for (let i = 0.1; i < 1000.1; ++i) {
// slow 🐌
}
• The performance of the modulo operator code depends on whether you’re dealing with integers or not.
const remainder = value % divisor;
// Fast 🚀 if value and divisor are represented as integers,
// slow 🐌 otherwise.
### Specification
The IEEE standard gives an algorithm for addition, subtraction, multiplication, division and square root, and requires that implementations produce the same result as that algorithm.
### Equal
Due to the rounding error, equality function has always a delta parameter to define the permissible rounding error.
Delta or epsilon is defined as been: $$| expected - actual |< epsilon$$
Example: AssertEquals of double
### Associativity Error
real numbers are associative but this is not always true of floating-point numbers:
console.log( (0.1 + 0.2) + 0.3 ); // 0.6000000000000001
console.log( 0.1 + (0.2 + 0.3) ); // 0.6
console.log( ( (0.1 + 0.2) + 0.3 ) == ( 0.1 + (0.2 + 0.3) ) ); // false
### Inexact representations
Always remember that floating point representations using float and double are inexact. Floating-point numbers offer a trade-off between accuracy and performance.
For example, consider these Javascript number expressions (Javascript supports only float)
console.log(999199.1231231235 == 999199.1231231236) // true
console.log(1.03 - 0.41) // 0.6200000000000001
In Java, for exactness, you want to use BigDecimal.
### to Integer
Doubles (float) can represent integers perfectly with up to 53 bits of precision.
All of the integers from -9,007,199,254,740,992 (–2^53) to 9,007,199,254,740,992 (2^53) are then valid doubles. | 1,021 | 4,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-21 | longest | en | 0.915009 |
https://socratic.org/questions/omar-wants-to-buy-chairs-for-his-new-office-each-chair-costs-12-and-there-is-a-f | 1,638,419,111,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00017.warc.gz | 594,928,043 | 6,230 | # Omar wants to buy chairs for his new office. Each chair costs $12 and there is a flat delivery fee of$10. If he has $80, how many chairs can he buy? ##### 1 Answer Oct 27, 2016 This is of the form $a x + b = c$#### Explanation: $a =$price per chair ($12)
$x =$number of chairs
$b =$delivery fee ($10) Now the problem can be filled in: $12 \cdot x + 10 = 80 \to$(subtract 10 from both sides) $12 \cdot x = 70 \to$(divide by 12) $x = \frac{70}{12} = 5 \frac{10}{12}$So he can buy 5 chairs and have$10 left. He is \$2 short for the 6th chair. | 180 | 541 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-49 | latest | en | 0.947965 |
https://www.jiskha.com/questions/943530/A-rock-is-dropped-into-the-Grand-Canyon-It-takes-18-seconds-to-hit-the-bottom-Calculate | 1,558,519,005,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256778.29/warc/CC-MAIN-20190522083227-20190522105227-00190.warc.gz | 823,649,546 | 5,005 | # physics
A rock is dropped into the Grand Canyon. It takes 18 seconds to hit the bottom. Calculate how deep the canyon is.
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1. s = 1/2 at^2 = 4.9 * 18^2 = ...
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# Hardware-in-the-loop simulation
Embed a dynamic system for hardware-in-the loop simulation
# Introduction
Hardware-in-the-loop (HIL) simulation is a simulation that uses a mathematical model of the physical system in a real-time environment. This is especially useful in scenarios when the physical system is inaccessible for cost or safety reasons. Because of the real-time nature of the simulation, a controller or person will be essentially interacting with the physical system if the mathematical model is of adequate fidelity. Typical applications of HIL simulation are in testing controllers before deploying them to control the actual physical system, in analyzing the physical system, and in training.
In this project, we will deploy the nonlinear model of a two-tank system to an Arduino Micro microcontroller board. The input to the system is the flow rate into the upper tank. This controls the liquid levels which will be considered as the outputs. A potentiometer will be used to adjust the flow rate, and the liquid levels will be output as serial-over-usb signals. This can then be read using the device framework. Thus we can simulate and analyze the behavior of the two-tank system in real-time.
# Parts
Arduino Micro [link]
400-point breadboard [link]
Micro USB cable [link]
Misc. jumper wires [link]
# Electrical wirings
Connect the wiper terminal of the potentiometer to pin A0 and one terminal to GND and the other to 5V.
The Arduino Micro must also be connected to the computer using a micro USB cable.
# Mathematical model
In the two-tank system, the flow into the upper tank is used to control the liquid levels in both tanks.
The physical parameters of the system.
Use Bernoulli's law and mass balance to derive the resulting differential equations.
Use steady-state operating values .
Define the corresponding nonlinear system with input and outputs and .
# Embed the model
Discretize the model with a sampling period of 0.25 seconds:
The input flow is controlled by a potentiometer and the flow is scaled to vary from 1 to 2:
Set up the model to return the input flow along with the heights in the two tanks.
Load the package.
Embed the model:
# Read the simulated signals
Open a serial connection to the target:
The start, delimiter, and end bytes.
A function to parse the data coming though the serial connection:
Set up a task to read the values every second (and apply a step input using the potentiometer):
Remove the task and terminate the device connectionafter about 30 minutes:
The height of the liquid in the two tanks to a step input from the potentiometer:
Retrieve the step input that was applied to the system:
Compare the real-time and simulated results:
We see that the real-time simulation results are on par with that of the simulation in the Wolfram system. However, the real-time simulation was running for the approximately 30 minutes that it took for the liquid levels to rise from 2.7 meters to 3.3 meters when the flow rate was increased from 1 to 1.11 .
# More things to try
• Try other ways to visualize the liquid levels.
• Send the output signals using some other channel.
• Simulate disturbances to the liquid levels.
• Design a controller that regulates the liquid levels and do a HIL simulation.
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# In 1430, Duke Philip the Good of Burgundy founded the Order of the
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In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding from the 24 knights nominated to join it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East.
A) it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East
B) him uncompromising fealty, devotion to the glory of God, and defense of Christianity, if need be, by going on a crusade to the East
C) it to be uncompromisingly loyal, to devote themselves to the glory of God, and to defense the Christian faith, if need be, on an Eastern crusade
D)it uncompromising fealty, devotion to the glory of God, and a commitment to defend the Christian faith, if need be, by going on a crusade to the East
E) him uncompromising fealty, devoting themselves to the glory of God, and defending the Christian faith by going on a crusade to the East, if necessary
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In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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25 Jun 2019, 16:43
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Helium wrote:
chetan2u wrote:
saswata4s wrote:
In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding from the 24 knights nominated to join it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East.
A) it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East
B) him uncompromising fealty, devotion to the glory of God, and defense of Christianity, if need be, by going on a crusade to the East
C) it to be uncompromisingly loyal, to devote themselves to the glory of God, and to defense the Christian faith, if need be, on an Eastern crusade
D) it uncompromising fealty, devotion to the glory of God, and a commitment to defend the Christian faith, if need be, by going on a crusade to the East
E) him uncompromising fealty, devoting themselves to the glory of God, and defending the Christian faith by going on a crusade to the East, if necessary
Hi..
look at the 3-2 split first
1) IT vs HIM
the knights were to join " the order of golden fleece", so pronoun should be IT and not him
B and D eliminated
I have a doubt -
If I read the sentence as below :
Original sentence - In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding from the 24 knights nominated to join it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East.
To be read as - In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East, from the 24 knights nominated to join him/it.
Can you help me understand selection between him/it ?
souvik101990, pushpitkc, abhimahna, Skywalker18
generis
Helium , sorry for the delay, Catching up!
• Short answer - use 4 steps (I did not use this method)
1) what Philip demands is directly connected to founding the Order.
Demanding modifies the previous clause.
At minimum, we can say that "demanding X, Y, and Z" IS connected to joining the Order.
2) we cannot tell whether Philip was a member of the Order, so the knights might have been nominated to join him in a general or overall sense.
That interpretation, though, may not be as certain as the "at minimum" concept in (1). If you understand what fealty means, you probably should read about the approach that I used, which is below.
3) nominated to join usually connotes nominated to join a prestigious organization such as the Order. Choose D and confirm by checking parallelism.
4) Check parallelism. Fealty, devotion, and commitment are similar nouns. They describe character traits.
Fealty, devotion, and defense are not similar nouns. Defense does not describe a character trait.
Answer D
• My actual approach
I did not think that the IT/HIM decision was a straightforward choice.
I analyzed parallelism first and got rid of 3 choices; examined IT/HIM but was not convinced; returned to parallelism and chose D; and made sure that D was not nonsensical.
• Split #1 - demands are not parallel in A, C, and E
A) uncompromising fealty, devotion to the glory of God, and to defend the Christian faith
C) uncompromisingly loyal, to devote themselves to the glory of God, and to defense the Christian faith
In addition to lacking parallelism, there is no such thing as "demanding ... to defense." To defend is correct.
E) uncompromising fealty, devoting themselves to the glory of God, and defending the Christian faith
-- the ___ING words seem to control (there are two), but they are very awkward and have noun forms (devotion and defense). GMAC usually prefers noun forms.
Awkward: demanding ... devoting themselves to the glory of God...demanding . . . defending is also awkward
Eliminate A, C, and E
Eliminating B on the basis of parallelism was harder. I turned momentarily to HIM/IT, but found issues.
• Split #2: IT or HIM?T
Issue: FEALTY complicates the IT/HIM choice
I could not choose immediately between HIM in B and IT in D.
-- tension exists between fealty (to HIM) and nominated (probably for membership in an order, IT)
Fealty is a particular kind of loyalty.
Fealty refers to a feudal underling's sworn loyalty or allegiance TO a lord [TO an aristocrat].
Philip was demanding fealty, which suggests that he was demanding that the knights join HIM with an oath of allegiance.
So it's possible, maybe even likely, that the knights were nominated to join HIM as members in the order.
Just one problem.
We do not know whether Philip was a member of the order. If he was not a member, perhaps we must say that the knights were not really nominated to join Philip directly.
We do know that the knights were nominated to join the order, because the demands are connected to the setup of the order.
(demanding modifies the previous clause about setting up the order)
Maybe, I thought to myself, I am importing my own knowledge about fealty and defining it too narrowly. (Then again, it's a vocabulary word.)
I could not decide. I returned to parallelism.
• The parallelism in B is not as good as that in D.
( abhi2707 , the analysis of parallelism above and this analysis may address your sense that parallelism is off. Not sure. In a separate post I addressed, X, Y, and a Z.)
I really wish I had caught this issue in the first round of parallelism analysis.
I would have simply checked to make sure that D was grammatical and made sense.
Parallelism in B
(B) fealty, devotion to the glory of God, and defense of Christianity
defense is an action noun.
defense is the action of defending from or resisting attack. HERE
fealty and devotion are not action nouns. They are "abstract" nouns.
fealty and devotion are nouns that refer to human qualities.
They involve character, emotions, and will.
They do describe the character trait itself (a person is loyal and devoted.)
But fealty and devotion also suggest commitment, which in turn suggests an ongoing promise in word and deed to possess and display a character trait.
defense does not describe a character trait or suggest commitment.
Parallelism in D
(D) fealty, devotion to the glory of God, and a commitment to defend the Christian faith
a commitment to defend is an abstract noun.
More importantly, a commitment to defend describes a character trait.
a commitment to defend is similar to fealty and devotion
In a list whose first two items are fealty and devotion, a commitment to defend (D) is better than defense is (B).
That is, parallelism in Option D is better than option B.
Eliminate B, but check.
• CONFIRM? Check nominated [to join] in context
I decided that I could defend the IT in (D) based on nominated in overall context coupled with logic based on what I knew for sure.
Nominated to join the Order (thus IT) may be stronger than nominated to join Philip (HIM) in part because we don't know whether Philip was a member of the order and in part because of syntax. (Fealty, though, still complicates matters, and ultimately parallelism decided the issue.)
demanding tells us what happened along with the setup of the order.
-- [COMMA + present participle] = [COMMA + verbING] refers to the previous clause in which Philip set up the Order.
demanding is connected to the setup of the order.
We know that the knights were invited to join the Order.
We do not know whether PHILIP was a member of the order.
If he was not a member, then perhaps in a strict sense the knights were not nominated to join HIM.
That is, if Philip was not a member of the Order, perhaps the knights were not nominated to join Philip (him) as a co-member. Still, the order was Philip's. He was deeply committed to the whole sordid enterprise of the crusades.
NOMINATED?
Nominate typically connotes a formal proposal that someone be a candidate for an award such as joining an elite order.
I was not willing to depend on that definition alone at the outset.
Fealty complicated matters.
Plus, I could make the case for both IT and HIM.
Both of these sentences are defensible.
Duke: Knight, become my vassal. Because you are a knight of demonstrated character, I nominate you to join me in a general sense and to join this Order in a particular sense—and I demand X, Y, and Z.
Duke: Knight, I nominate you to join this Order, and I demand X, Y, and Z.
But parallelism analysis had led me to pick D.
In the end:
I did not know whether Philip was a member of the order.
I did know that the knights had been invited to join the order.
Nominated usually connotes becoming a candidate for an honor such as joining an elite order. At the least, "nominated" for the order is not outlandish.
Having just decided that the parallelism in D was better than that in B, I simply had to confirm that the choice of D was okay: Did D make sense?
Philip set up the Order, demanding fealty, devotion, and a commitment to defend . . . FROM the knights [who had been] nominated to join the Order.
That construction (mostly) makes sense. Fealty is a little weird, but parallelism trumps that weirdness. (D) is confirmed.
This logical analysis was not easy to describe.
Thinking about issues this subtle is a lot easier than describing those thoughts.
The answer is D.
I hope that analysis helps.
If it does not, please ask another question. I'll be happy to try to help.
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Re: In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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04 Nov 2017, 08:30
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saswata4s wrote:
In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding from the 24 knights nominated to join it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East.
A) it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East
B) him uncompromising fealty, devotion to the glory of God, and defense of Christianity, if need be, by going on a crusade to the East
C) it to be uncompromisingly loyal, to devote themselves to the glory of God, and to defense the Christian faith, if need be, on an Eastern crusade
D)it uncompromising fealty, devotion to the glory of God, and a commitment to defend the Christian faith, if need be, by going on a crusade to the East
E) him uncompromising fealty, devoting themselves to the glory of God, and defending the Christian faith by going on a crusade to the East, if necessary
Hi..
look at the 3-2 split first
1) IT vs HIM
the knights were to join " the order of golden fleece", so pronoun should be IT and not him
B and D eliminated
2) a crusade to east vs eastern crusade..
as per meaning a crusade to the east is correct
C eliminated
3) parallelism
the three items are noun..
uncompromising fealty, devotion and a commitment
Only D left
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In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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Updated on: 23 Jun 2019, 12:20
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chetan2u wrote:
saswata4s wrote:
In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding from the 24 knights nominated to join it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East.
A) it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East
B) him uncompromising fealty, devotion to the glory of God, and defense of Christianity, if need be, by going on a crusade to the East
C) it to be uncompromisingly loyal, to devote themselves to the glory of God, and to defense the Christian faith, if need be, on an Eastern crusade
D)it uncompromising fealty, devotion to the glory of God, and a commitment to defend the Christian faith, if need be, by going on a crusade to the East
E) him uncompromising fealty, devoting themselves to the glory of God, and defending the Christian faith by going on a crusade to the East, if necessary
Hi..
look at the 3-2 split first
1) IT vs HIM
the knights were to join " the order of golden fleece", so pronoun should be IT and not him
B and D eliminated
I have a doubt -
If I read the sentence as below :
Original sentence - In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding from the 24 knights nominated to join it uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East.
To be read as - In 1430, Duke Philip the Good of Burgundy founded the Order of the Golden Fleece, demanding uncompromising fealty, devotion to the glory of God, and to defend the Christian faith, if need be, by going on a crusade to the East, from the 24 knights nominated to join him/it.
Can you help me understand selection between him/it ?
souvik101990, pushpitkc, abhimahna, Skywalker18
generis
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Originally posted by Helium on 12 Aug 2018, 04:51.
Last edited by Helium on 23 Jun 2019, 12:20, edited 1 time in total.
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Re: In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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09 Jun 2019, 03:04
By POE the best possible option is D. But the parallelism seems odd because of X,Y, and a Z
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In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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25 Jun 2019, 17:00
abhi2707 wrote:
By POE the best possible option is D. But the parallelism seems odd because of X,Y, and a Z
abhi2707 , I can see how the "a Z" might appear odd.
Please see my analysis of parallelism above.
fealty, devotion to the glory of God, and A commitment to defend
are parallel nouns [the last one is a phrasal noun].
These three are called abstract nouns. The jargon does not matter.
Do understand the content, though: they are NOT action nouns. They are not like defense.
These three nouns describe character traits.
A commitment to defend is a very good way to mimic the other nouns, fealty and devotion.
The person has a character trait: a commitment to defend something.
He is protective.
The "a" before commitment is just a quirk; some nouns don't work too well without an article.
If we did not have the word "a," we would probably say,
. . . and commitment to defending Christianity.
The trait is now not as purposeful.
A commitment explicitly suggests A promise—a choice to possess and demonstrate a certain kind of character.
In any event, defense in B (action noun not describing a character trait) is not parallel with fealty and devotion.
Option B is not as good as D is on that front.
Hope that helps.
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Re: In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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25 Jun 2019, 18:13
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Thank you for the detailed reply generis.
I will summerise the takeaways from your analysis :
a) IT/HIM is not the decision point in above question.
Both could have been correct depending on context as you analyzed.
b) Parallelism helps to zero down on choice D.
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In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink]
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26 Jun 2019, 09:28
Helium wrote:
Thank you for the detailed reply generis.
I will summerise the takeaways from your analysis :
a) IT/HIM is not the decision point in above question. [As I see the matter, that statement is correct. IT/HIM did not work for me as a primary decision point.]
Both could have been correct depending on context as you analyzed.
b) Parallelism helps to zero down on choice D.
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Helium , yes, and almost. (I think you assumed that this next step is implied . . . just don't want to confuse others.)
After (b), then
(c) confirm that D is the right choice by ascertaining that (D) makes sense (or is not nonsensical).
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In 1430, Duke Philip the Good of Burgundy founded the Order of the [#permalink] 26 Jun 2019, 09:28
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# In 1430, Duke Philip the Good of Burgundy founded the Order of the
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https://indiafreenotes.com/network-topology/ | 1,713,619,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00567.warc.gz | 280,999,001 | 53,470 | Network Topology
28/03/2020
Network topology is the arrangement of the elements (links, nodes, etc.) of a communication network. Network topology can be used to define or describe the arrangement of various types of telecommunication networks, including command and control radio networks, industrial fieldbusses and computer networks.
Network topology is the topological structure of a network and may be depicted physically or logically. It is an application of graph theory wherein communicating devices are modeled as nodes and the connections between the devices are modeled as links or lines between the nodes. Physical topology is the placement of the various components of a network (e.g., device location and cable installation), while logical topology illustrates how data flows within a network. Distances between nodes, physical interconnections, transmission rates, or signal types may differ between two different networks, yet their topologies may be identical. A network’s physical topology is a particular concern of the physical layer of the OSI model.
Examples of network topologies are found in local area networks (LAN), a common computer network installation. Any given node in the LAN has one or more physical links to other devices in the network; graphically mapping these links results in a geometric shape that can be used to describe the physical topology of the network. A wide variety of physical topologies have been used in LANs, including ring, bus, mesh and star. Conversely, mapping the data flow between the components determines the logical topology of the network. In comparison, Controller Area Networks, common in vehicles, are primarily distributed control system networks of one or more controllers interconnected with sensors and actuators over, invariably, a physical bus topology.
Types of Network Topology
The arrangement of a network which comprises of nodes and connecting lines via sender and receiver is referred as network topology. The various network topologies are:-
1. Mesh Topology
In mesh topology, every device is connected to another device via particular channel.
Every device is connected with another via dedicated channels. These channels are known as links.
• If suppose, N number of devices are connected with each other in mesh topology, then total number of ports that is required by each device is N-1. In the Figure 1, there are 5 devices connected to each other, hence total number of ports required is 4.
• If suppose, N number of devices are connected with each other in mesh topology, then total number of dedicated links required to connect them is NC2 i.e. N(N-1)/2. In the Figure 1, there are 5 devices connected to each other, hence total number of links required is 5*4/2 = 10.
• It is robust.
• Fault is diagnosed easily. Data is reliable because data is transferred among the devices through dedicated channels or links.
• Provides security and privacy.
Problems with this topology
• Installation and configuration is difficult.
• Cost of cables are high as bulk wiring is required, hence suitable for less number of devices.
• Cost of maintenance is high.
1. Star Topology
In star topology, all the devices are connected to a single hub through a cable. This hub is the central node and all others nodes are connected to the central node. The hub can be passive in nature i.e. not intelligent hub such as broadcasting devices, at the same time the hub can be intelligent known as active hubs. Active hubs have repeaters in them.
A star topology having four systems connected to single point of connection i.e. hub.
• If N devices are connected to each other in star topology, then the number of cables required to connect them is N. So, it is easy to set up.
• Each device require only 1 port i.e. to connect to the hub.
Problems with this topology
• If the concentrator (hub) on which the whole topology relies fails, the whole system will crash down.
• Cost of installation is high.
• Performance is based on the single concentrator i.e. hub.
1. Bus Topology
Bus topology is a network type in which every computer and network device is connected to single cable. It transmits the data from one end to another in single direction. No bi-directional feature is in bus topology.
A bus topology with shared backbone cable. The nodes are connected to the channel via drop lines.
• If N devices are connected to each other in bus topology, then the number of cables required to connect them is 1 which is known as backbone cable and N drop lines are required.
• Cost of the cable is less as compared to other topology, but it is used to built small networks.
Problems with this topology
• If the common cable fails, then the whole system will crash down.
• If the network traffic is heavy, it increases collisions in the network. To avoid this, various protocols are used in MAC layer known as Pure Aloha, Slotted Aloha, CSMA/CD etc.
1. Ring Topology
In this topology, it forms a ring connecting a devices with its exactly two neighbouring devices.
A ring topology comprises of 4 stations connected with each forming a ring..
The following operations takes place in ring topology are:-
One station is known as monitor station which takes all the responsibility to perform the operations.
To transmit the data, station has to hold the token. After the transmission is done, the token is to be released for other stations to use.
When no station is transmitting the data, then the token will circulate in the ring.
There are two types of token release techniques : Early token release releases the token just after the transmitting the data and Delay token release releases the token after the acknowledgement is received from the receiver. | 1,163 | 5,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-18 | latest | en | 0.922606 |
http://swmm5.posthaven.com/tag/%23FEATURE | 1,540,227,570,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00377.warc.gz | 359,156,403 | 9,237 | ## How to Use Two Control Curves in SWMM 5 to Simulate a Head Difference Rule
Subject: How to Use Two Control Curves in SWMM 5 to Simulate a Head Difference Rule
The SWMM 5 control rules for Real Time Control (RTC) do not allow the rule to be governed by the head across the orifice but does allow rules based on the depth, head and inflow at any node. If you have an Orifice in which you want the Orifice to be open when the head difference across the Orifice is either less than or greater than zero then and closed when the head difference is close to zero then you can use two Orifices (Figure 1) and two rules (Figure 2) to control the orifice setting for Orifice1 and Orifice2. In attached file the two rules have the settings set to two control curves. Orifice1 will start open and close gradually as the depth at Node UPNode increases, Orifice2 will start closed and gradually open when depth in Node DNode increases. Possible variations are to control Orifice1 based on the DNode and node UPNode to control Orifice2.
RULE Orifice1
IF NODE UPNode Depth >= 0
THEN ORIFICE ORIFICE1 SETTING = Curve RuleOrf1
PRIORITY 10
RULE Orifice2
IF NODE DNode Depth >= 0
THEN ORIFICE ORIFICE2 SETTING = Curve RuleOrf2
PRIORITY 10
Figure 1. Two Orifice Solution
Figure 2. Two Orifice solution to have control over the Orifice(s) at both the upstream and downstream nodes.
// Posted
## How to Use the Map Display for the Maximum Adjusted d/D or Maximum q/Q in an EPS InfoSewer Simulation
How to Use the Map Display for the Maximum Adjusted d/D or Maximum q/Q in an EPS InfoSewer Simulation
You can do a Map Display of the adjusted d/D values (Figure 1) from the Gravity Main Range Report (Figure 2) to show those pipes that are full thematically. For example, the links in red in Figure 1 show the effect of the pump blockage and the links in green are those NOT full due to the pump blockage. You will need to copy the adjusted d/D or the maximum q/Q values from the Range report to the Link Information Table to have some values to Map (Figure 3 and 4). The maximum adjusted d/D or the Maximum q/Q can be mapped using the new link information (Figure 5).
Figure 1 Map Display of the Maximum Adjusted d/D from the Gravity Range Report.
Figure 2. Maximum Adjusted d/D or Maximum q/Q can be copied from the EPS Range Gravity Main Report.
Figure 3. Create a new variable In the Link Information Table.
Figure 4. New variables for the Map Display from the Range Report in the Pipe Information Tables for Each Link.
Figure 5. Link Information new Parameters of Variables can be used to Display the maximum d/D or q/Q during the EPS simulation.
Posted
## Import of Sections from SWMM 5 into InfoSWMM and H2oMAP SWMM
Subject: Import of Sections from SWMM 5 into InfoSWMM and H2oMAP SWMM
A very useful hidden feature of the import SWMM 5 to InfoSWMM and H2OMAP SWMM is the ability to import all of the data or just one section. For example, you can import the LID data, DWF patterns, control rules, pollutants, transects and other data that is transferable between different networks.
Posted
## How to Combine InfoSWMM 2D Output Report Manager and Arc GIS Tools
Note: How to Combine InfoSWMM 2D Output Report Manager and Arc GIS Tools
You can combine the 2D Map Display of InfoSWMM 2D, the Label Properties Expression Labels and the Point Element to show the depths on a mesh at a particular time AND the time series of the mesh depth in the Output Report Manager of InfoSWMM. You will need to use the Query Feature In Arc GIS to show only those mesh depths greater than 0 otherwise you will end up with a lot of 0 depth labels.
Posted | 916 | 3,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-43 | longest | en | 0.844693 |
http://forums.nesdev.com/viewtopic.php?f=2&t=16512&view=print | 1,542,430,065,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743282.69/warc/CC-MAIN-20181117040838-20181117062838-00065.warc.gz | 133,148,153 | 12,641 | nesdev.comhttp://forums.nesdev.com/ NES Blaster Master Boss Skiphttp://forums.nesdev.com/viewtopic.php?f=2&t=16512 Page 1 of 1
Author: CLChambers00 [ Tue Sep 19, 2017 4:11 pm ] Post subject: NES Blaster Master Boss Skip The Boss Skip that had been observed during a live run during AGDQ 2017 can be seen here: https://www.youtube.com/watch?v=g5Hq2eA ... e&t=28m55sI have observed that \$03FB will increment after you defeat a boss. From the beginning of the game this is set at value 0, then value 1 after the first boss, and then at value 3 after the second boss. In other words, if you overwrite the value of 3 into \$03FB just before entering the second boss fight then you will observe what we see in the video. The game thinks the boss was already defeated. I have observed that \$03FB increments the same frame that \$03FD reaches a value of 128. \$03FD starts at 0 from the beginning of the game and only does a 255 to 0 count down at the beginning of each boss fight, and then counts up from 0 to 128 upon the death of the boss. When this value reaches 128 then \$03FB, the address that tracks what bosses have been defeated, will increment. Observation #1 The first boss loads normally which means that \$03FB is at value 0 when entering the first boss fight. If this value was 1 or higher then the boss would not spawn. The first moment that \$03FB normally increments is upon the death of the first boss, in which case it increments to 1. Observation #2 Upon entering the second boss fight the boss did not load which means that \$03FB was already at a value of 3.Conclusion: At some point this address is incremented again. My first suspicion is that it occurs after the first boss fight, that some how it is incremented twice. The later portion of this message shows "conditions" upon an overflow may occur that could possibly corrupt this address but it is still unclear exactly how these conditions could be met.This is the most pertinent information that I have been able to find at this time.I played around a little with the Chou-Wakusei Senki - MetaFight (J).nes ROM (GoodNES 3.14) after seeing that happen and will summarize some info here. Hopefully this will serve as a starting point for figuring out the glitch. I used FCEUX 2.2.3 and only played as far as the second boss fight.After the scrolling ends when you enter a boss room, the following code is executed:Code:\$971A:A5 14 LDA \$0014\$971C:29 07 AND #\$07\$971E:AA TAX\$971F:BD 2B B6 LDA \$B62B,X\$9722:60 RTS\$9627:2D FB 03 AND \$03FB <-- This is the important check\$962A:D0 0F BNE \$963B <-- Branch taken if boss defeated\$962C:A9 C0 LDA #\$C0\$962E:85 51 STA \$0051\$9630:A9 08 LDA #\$08\$9632:20 8C C1 JSR \$C18C\$9635:20 CD 97 JSR \$97CD\$9638:E6 46 INC \$0046\$963A:60 RTS\$963B:20 1A 97 JSR \$971A\$0014 is the value (current area - 1) at this point. At \$B62B, you find a sequence of increasing powers of two, starting with 0x01. Per Data Crystal's RAM map, \$03FB indicates which bosses have been defeated. So this checks if the boss of the current area has been defeated. If it has but the item has not been obtained, then the item is displayed. That seems to be what happened during the race.If the boss has not been defeated, then the room flashes, the boss loads (including its health), and you proceed with the fight. It appears that the health of the first two bosses is hard-coded into the ROM. For area 1, the pertinent code is:Code:\$9C15:A9 80 LDA #\$80\$9C17:85 53 STA \$0053For area 2, the pertinent code is:Code:\$91FD:A9 40 LDA #\$40\$91FF:85 53 STA \$0053The value from \$0053 eventually ends up somewhere around \$0470, but the exact RAM address varies. If you change the health to 0x00 before starting the fight (e.g., set execution breakpoint on \$9C17, change A from 0x80 to 0x00, then continue emulation), the boss graphics load but immediately go into the death animation. This leads me to believe that whatever Skavenger did may have corrupted \$03FB. There are likely other possibilities, but I will let someone with more time and experience take over.Actually, one other thing. This is the code that is executed when a boss is defeated:Code:\$971A:A5 14 LDA \$0014\$971C:29 07 AND #\$07\$971E:AA TAX\$971F:BD 2B B6 LDA \$B62B,X\$9722:60 RTS\$97A7:0D FB 03 ORA \$03FB\$97AA:8D FB 03 STA \$03FBMost if it should be familiar from that first block, but the last two lines update the value at \$03FB rather than testing it.And another analysis:I haven't looked at Blaster Master before, but this glitch got me curious, so a few observations from looking at the code in case they give anyone any ideas. Note that these are all from the English ROM (the only version I have handy), so if the glitch turns out to be JP-only then none of this may be of much use...- The only stores to \$03FB are the one Dacicus quoted which is done on boss kill, and a few that store literal 0 or 255 (and the 255 clearly can't be the case because the rest of the bosses showed up properly). The store after the area 1 boss kill in that run can't have set the wrong bit, because then the boss would immediately respawn when the room was reloaded after the boss death.- The only indexed stores that could reach \$03FB are the stores to the PPU RAM write buffer at \$0300. I played up to area 2 to check, but the buffer high water mark was \$0340, and a cursory review of the code suggests it's pretty good at flushing the buffer before it overflows, so it seems unlikely to me that a buffer overflow is the cause. (Also, in order to have disabled only the second boss, the store value would have to have been \$02 or \$03, which seems unlikely for an arbitrary PPU data store; and if the buffer overflowed all the way up to \$03FB then it would stomp on lots of other stuff in the \$03xx range, which would probably cause more problems than just a missing boss.) On the other hand, Personman's example of a case when both a boss and its item were missing could indicate a buffer overflow that wrongly set bits in both \$03FB and \$03FC, so I dunno.- The boss-defeated store to \$03FB has this stack trace (again, addresses are from the English ROM - bank 4 is mapped to \$8000 here):Code:(0) A109: sta \$03FB(1) A58C: jsr \$A07B(2) C9D3: jmp (\$007A) = \$A58C(3) C987: jsr \$C9A4(4) C3D8: jsr \$C971For \$A07B to reach the store to \$03FB, the following have to be true:- \$0053 = 0- The low bit of \$0010 is clear- \$03FD >= 127It does look like this routine is called for other purposes as well, so conceivably there could be a case where all those preconditions are (incorrectly) met, which would trigger the glitch.Here's the relevant part of that routine:Code:A07B: A5 53 lda \$53A07D: F0 10 beq \$A08FA07F: AD FD 03 lda \$03FDA082: F0 7B beq \$A0FFA084: CE FD 03 dec \$03FDA087: D0 76 bne \$A0FFA089: 20 18 A1 jsr \$A118A08C: 4C FC A0 jmp \$A0FCA08F: A5 10 lda \$10A091: 4A lsr aA092: B0 19 bcs \$A0ADA094: AD FD 03 lda \$03FDA097: D0 0A bne \$A0A3A099: A9 0A lda #\$0AA09B: 20 16 C2 jsr \$C216A09E: A9 0B lda #\$0BA0A0: 20 16 C2 jsr \$C216A0A3: EE FD 03 inc \$03FDA0A6: AD FD 03 lda \$03FDA0A9: C9 80 cmp #\$80A0AB: B0 53 bcs \$A100(...)A100: 20 0E C1 jsr \$C10E // Only reachable from branch at \$A0ABA103: 20 72 A0 jsr \$A072 // Get bit for current area (encapsulates "lda \$14" etc.)A106: 0D FB 03 ora \$03FB // Set boss-killed flag for current areaA109: 8D FB 03 sta \$03FB // Store updated flag byteA10C: A9 5A lda #\$5AA10E: 85 46 sta \$46A110: A9 00 lda #\$00A112: 20 1A C1 jsr \$C11AA115: 68 plaA116: 68 plaA117: 60 rts
Author: CLChambers00 [ Wed Sep 20, 2017 6:51 pm ] Post subject: Re: NES Blaster Master Boss Skip This is my first crack at this issue today. Thoughts are always appreciated. At the beginning of the game when the memory is initialized, address \$03FB is set to #\$00. Perhpas a simple hack of the rom could have this value set to 2 and this would still allow the first boss to spawn and cause the second boss not to be present since \$03FB would be at value 3. Since the starting value is 2, then the amount it will increment is based upon the value 2 to the power of \$0014 value. Since area 1 boss is in \$0014 value zero, then 2 to the power of zero is 1, and therefore the resulting value of \$03FB after Boss one is 3. This of course is assuming that the rom was hacked.Code:STA \$0300,X @ \$03FB = #\$00 ;Initializing memory, setting address to zeroUpon entering the Boss 1 room the following executes. Notable is that it loads the value of address \$0014 to the A Register. When I used a Lua Script to change \$0014 to a value of 1, the value of Boss 2 location, I was able to get the Boss 2 to spawn in the Boss 1 area. The value of \$03FB is checked at this time. If the value is a 0, as it is supposed to be, then the Boss will spawn. If the value is 1, which it should be after the Boss fight, then the boss will not spawn. I also checked and even if this value was a 2 then the boss will still spawn. Basically here after the AND \$03FB instruction it will Branch on Not Equal as seen BNE \$963B.Code:\$961F:F0 03 BEQ \$9624 A:00 X:0A Y:00 S:FB P:nvUbdIZC \$9624:20 1A 97 JSR \$971A A:00 X:0A Y:00 S:FB P:nvUbdIZC \$971A:A5 14 LDA \$0014 = #\$00 A:00 X:0A Y:00 S:F9 P:nvUbdIZC \$971C:29 07 AND #\$07 A:00 X:0A Y:00 S:F9 P:nvUbdIZC \$971E:AA TAX A:00 X:0A Y:00 S:F9 P:nvUbdIZC \$971F:BD 2B B6 LDA \$B62B,X @ \$B62B = #\$01 A:00 X:00 Y:00 S:F9 P:nvUbdIZC \$9722:60 RTS (from \$971A) --------------------------- A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$9627:2D FB 03 AND \$03FB = #\$00 A:01 X:00 Y:00 S:FB P:nvUbdIzC \$962A:D0 0F BNE \$963B A:00 X:00 Y:00 S:FB P:nvUbdIZC \$962C:A9 C0 LDA #\$C0 A:00 X:00 Y:00 S:FB P:nvUbdIZC \$962E:85 51 STA \$0051 = #\$00 A:C0 X:00 Y:00 S:FB P:NvUbdIzC \$9630:A9 08 LDA #\$08 A:C0 X:00 Y:00 S:FB P:NvUbdIzC \$9632:20 8C C1 JSR \$C18C A:08 X:00 Y:00 S:FB P:nvUbdIzC At the beginning of the Boss 1 fight when the boss begins to load, address 03FD is set to the value 255 and then decrements to 0, at which time the boss begins to move. Upon the death of the boss address 03FD increments up to 128 from zero. Once 03FD reaches 128 (#\$80)then we see the following execute when the Comparision is made. It is at this time that the addition to \$03FB will be 2 to the power of the value of \$0014, the area locator. If at any time that \$0014 is changed via Lua Script then the game starts to glitch between where you are and where you are telling the game where you are. I have observed some funny things as a result of this. What this means is that I think that it is unlikely that the game was confused as to your location and therefore gave a different value in \$03FB, because it thought that you were in a different \$0014 location. If \$0014 was anything but 0 during the boss 1 fight then you would see graphical distortions. Since the addition to \$03FB only occurs after a boss is defeated it would be curious how it might occur at any other time, and if so, then how?Code: \$9C2D:20 23 97 JSR \$9723 A:2D X:2A Y:00 S:FB P:nvUbdIzc \$9723:A5 53 LDA \$0053 = #\$00 A:2D X:2A Y:00 S:F9 P:nvUbdIzc \$9725:F0 10 BEQ \$9737 A:00 X:2A Y:00 S:F9 P:nvUbdIZc \$9737:A5 10 LDA \$0010 = #\$14 A:00 X:2A Y:00 S:F9 P:nvUbdIZc \$9739:4A LSR A:14 X:2A Y:00 S:F9 P:nvUbdIzc \$973A:B0 19 BCS \$9755 A:0A X:2A Y:00 S:F9 P:nvUbdIzc \$973C:AD FD 03 LDA \$03FD = #\$7F A:0A X:2A Y:00 S:F9 P:nvUbdIzc \$973F:D0 0A BNE \$974B A:7F X:2A Y:00 S:F9 P:nvUbdIzc \$974B:EE FD 03 INC \$03FD = #\$7F A:7F X:2A Y:00 S:F9 P:nvUbdIzc \$974E:AD FD 03 LDA \$03FD = #\$80 A:7F X:2A Y:00 S:F9 P:NvUbdIzc \$9751:C9 80 CMP #\$80 A:80 X:2A Y:00 S:F9 P:NvUbdIzc \$9753:B0 4C BCS \$97A1 A:80 X:2A Y:00 S:F9 P:nvUbdIZC \$97A1:20 C9 C0 JSR \$C0C9 A:80 X:2A Y:00 S:F9 P:nvUbdIZC \$C0C9:4C 5B D0 JMP \$D05B A:80 X:2A Y:00 S:F7 P:nvUbdIZC (...) \$D066:60 RTS (from \$C0C9) --------------------------- A:0F X:FF Y:00 S:F7 P:NvUbdIzC \$97A4:20 1A 97 JSR \$971A A:0F X:FF Y:00 S:F9 P:NvUbdIzC \$971A:A5 14 LDA \$0014 = #\$00 A:0F X:FF Y:00 S:F7 P:NvUbdIzC \$971C:29 07 AND #\$07 A:00 X:FF Y:00 S:F7 P:nvUbdIZC \$971E:AA TAX A:00 X:FF Y:00 S:F7 P:nvUbdIZC \$971F:BD 2B B6 LDA \$B62B,X @ \$B62B = #\$01 A:00 X:00 Y:00 S:F7 P:nvUbdIZC \$9722:60 RTS (from \$971A) --------------------------- A:01 X:00 Y:00 S:F7 P:nvUbdIzC \$97A7:0D FB 03 ORA \$03FB = #\$00 A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$97AA:8D FB 03 STA \$03FB = #\$00 A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$97AD:A9 5A LDA #\$5A A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$97AF:85 46 STA \$0046 = #\$63 A:5A X:00 Y:00 S:F9 P:nvUbdIzC \$97B1:A9 00 LDA #\$00 A:5A X:00 Y:00 S:F9 P:nvUbdIzC \$97B3:20 D5 C0 JSR \$C0D5 A:00 X:00 Y:00 S:F9 P:nvUbdIZC Twenty nine frames after this addition to \$03FB occurs the \$971A sub routine runs again (twice). After the first sub routine it checks the \$03FB address, and then it will branch if not equal BNE \$963B, the same as when we entered the room. And the second routine is followed by a check to \$03FC, seeing if you have already collected the item, and then a branch if not equal BNE \$965C.Code: \$961F:F0 03 BEQ \$9624 A:00 X:0A Y:00 S:FB P:nvUbdIZC \$9624:20 1A 97 JSR \$971A A:00 X:0A Y:00 S:FB P:nvUbdIZC \$971A:A5 14 LDA \$0014 = #\$00 A:00 X:0A Y:00 S:F9 P:nvUbdIZC \$971C:29 07 AND #\$07 A:00 X:0A Y:00 S:F9 P:nvUbdIZC \$971E:AA TAX A:00 X:0A Y:00 S:F9 P:nvUbdIZC \$971F:BD 2B B6 LDA \$B62B,X @ \$B62B = #\$01 A:00 X:00 Y:00 S:F9 P:nvUbdIZC \$9722:60 RTS (from \$971A) --------------------------- A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$9627:2D FB 03 AND \$03FB = #\$01 A:01 X:00 Y:00 S:FB P:nvUbdIzC \$962A:D0 0F BNE \$963B A:01 X:00 Y:00 S:FB P:nvUbdIzC \$963B:20 1A 97 JSR \$971A A:01 X:00 Y:00 S:FB P:nvUbdIzC \$971A:A5 14 LDA \$0014 = #\$00 A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$971C:29 07 AND #\$07 A:00 X:00 Y:00 S:F9 P:nvUbdIZC \$971E:AA TAX A:00 X:00 Y:00 S:F9 P:nvUbdIZC \$971F:BD 2B B6 LDA \$B62B,X @ \$B62B = #\$01 A:00 X:00 Y:00 S:F9 P:nvUbdIZC \$9722:60 RTS (from \$971A) --------------------------- A:01 X:00 Y:00 S:F9 P:nvUbdIzC \$963E:2D FC 03 AND \$03FC = #\$00 A:01 X:00 Y:00 S:FB P:nvUbdIzC \$9641:D0 19 BNE \$965C A:00 X:00 Y:00 S:FB P:nvUbdIZC \$9643:A5 14 LDA \$0014 = #\$00 A:00 X:00 Y:00 S:FB P:nvUbdIZC \$9645:C9 07 CMP #\$07 A:00 X:00 Y:00 S:FB P:nvUbdIZC \$9647:D0 0E BNE \$9657 A:00 X:00 Y:00 S:FB P:NvUbdIzc \$9657:A9 6A LDA #\$6A A:00 X:00 Y:00 S:FB P:NvUbdIzc \$9659:85 46 STA \$0046 = #\$5A A:6A X:00 Y:00 S:FB P:nvUbdIzc \$965B:60 RTS (from \$C8EC) --------------------------- A:6A X:00 Y:00 S:FB P:nvUbdIzcWhile there may not be anything revelatoty here beyond the previous posts concerning the matter, but it is where I am at with it at the moment.
Author: B00daW [ Thu Sep 21, 2017 9:53 pm ] Post subject: Re: NES Blaster Master Boss Skip I want to provide you with key information that may help you with this... Values from \$8000+ cannot change in value unless they have been bankswapped through bankswitching. Find out which "mapper" MetaFight or Blaster Master is... Research the bankswitching portion for PRG ROM to figure out if you're seeing values change because of bankswitching.\$8000-\$FFFF is ROM. These values CANNOT be written to... They can be addressed for external functions, such as mappers that perform bankswitching.The address values for subroutines that you describe between \$8000-\$FFFF can change if they have been bankswapped.Bankswitching is a way for the NES/Famicom to address more than 32K of PRG ROM in a quick fashion.Please attach savestates and movie files for those of us who are not as good at videogames as you are.
Author: tepples [ Fri Sep 22, 2017 6:35 am ] Post subject: Re: NES Blaster Master Boss Skip Blaster Master is SLROM: 128 KiB PRG ROM, 128 KiB CHR ROM, no WRAM, MMC1.
Author: CLChambers00 [ Fri Sep 22, 2017 10:30 am ]
Post subject: Re: NES Blaster Master Boss Skip
Thank you B00daW for that explanation, that \$8000+ is used for Rom explains why I could not add \$B62C to the Ram watch, and why I was not able to write a value to it via Lua Script. As for bankswitching/bankswapping I am not sure I quite understand except that I did observe that addresses like \$B62C were temporarily used when it loads the screen after death and loading the screen when unpausing it. Is this an example of bankswitching/bankswapping?
Thank you for the link Tepples.
Attached you will find the disassembly for Chou-Wakusei Senki - MetaFight (J).nes. This is the Rom on which this issue occurred so I am sticking with it. I did notice that the J version has only 6 references to \$03FB while the US version has 7 references. I did not check to see what was added to the US version. The J version was released first so I wonder why the developers thought to modify the US version to involve one more reference to \$03FB. Were they trying to fix something and if so is it relevant. See Zeronoid's analysis of some key differences between these versions, if interested, either way I am focusing on the J version. http://zeronoin.com/bm/0002.html
Attached you will also find three text files compiled by Trace Logger taking a one frame snapshot of when \$03FB is referenced. First when it is checked to see if you have beaten the boss, second after you beat the boss and if so it writes an updated value to it, and third, immediately after to see if this is still true before checking to spawn the item or not.
Shortly I will post the FM2 file for FCEUX which can be played in the TAS Editor up to and defeating the second boss.
Attachments: Blaster Master J Disassemble.zip [824.81 KiB] Downloaded 45 times Boss 2 logger 3.log [827.28 KiB] Downloaded 41 times Boss 2 logger 2.log [826.88 KiB] Downloaded 43 times Boss 2 logger 1.log [858.54 KiB] Downloaded 49 times
Author: CLChambers00 [ Fri Sep 22, 2017 10:50 am ]
Post subject: Re: NES Blaster Master Boss Skip
Attached is the FM2 file of a poor attempt at good game play It plays up to and defeats Boss 2. This Boss 2 is the one that did not spawn but the item did on one live occasion. I have not been able to duplicate this without a Lua Script to \$03FB. I used controller input recording in the TAS Editor and I have to say that FCUEX lags horridly when doing this. It is like playing the game at 1/8 the normal speed. I think that BizHawk does not lag in this same way and I wonder why that is.
Attachments: MetaFight (J) TASed Example.fm2 [377.45 KiB] Downloaded 43 times
Author: CLChambers00 [ Fri Sep 22, 2017 11:13 am ]
Post subject: Re: NES Blaster Master Boss Skip
The following are 5 save states. I did not know how many would be appropriate so I created a save state just before entering a new \$0014 value location, and just before entering the Boss.
FC1 = Just before entering into \$0014 value 0 location from the starting 8 value at beginning of the game. This is when you are playing with the large walking character.
FC2 = Just before entering the Boss 1 fight in \$0014 value 0 location.
FC3 = Just before entering the \$0014 value 9 location. After the Boss one fight in \$0014 value 0 location we back track to value 8 and that is where we are before entering 9.
FC4 = Just before entering the \$0014 value 1 location. This is when you are playing with the large walking character again.
FC5 = Just before entering the Boss 2 fight in \$0014 value 1 locaiton. | 6,893 | 21,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-47 | latest | en | 0.95071 |
https://gmatclub.com/forum/a-store-owner-decided-to-raise-the-price-of-a-particular-128875.html?kudos=1 | 1,510,947,922,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803906.12/warc/CC-MAIN-20171117185611-20171117205611-00146.warc.gz | 622,483,529 | 60,506 | It is currently 17 Nov 2017, 12:45
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# A store owner decided to raise the price of a particular
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11 Mar 2012, 00:04
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I could figure this one out in about two seconds if I had a calculator, but I hear that's not allowed on the GMAT, so I need to know how to get this answer quick by hand. Thanks!
A store owner decided to raise the price of a particular item by exactly 10%. Of the following which is NOT the new price?
A. \$1.10
B. \$8.80
C. \$11.00
D. \$57.30
E. \$78.10
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Re: Price Percentage Increase, Find Original Price. [#permalink]
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11 Mar 2012, 00:17
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The new price should be divisible by 11 , because new price = 1.1 * old price. Answer is D, because it's not divisible by 11.
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11 Mar 2012, 00:39
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Stoneface wrote:
I could figure this one out in about two seconds if I had a calculator, but I hear that's not allowed on the GMAT, so I need to know how to get this answer quick by hand. Thanks!
A store owner decided to raise the price of a particular item by exactly 10%. Of the following which is NOT the new price?
A. \$1.10
B. \$8.80
C. \$11.00
D. \$57.30
E. \$78.10
Old Price * 1.1 = New Price -> Old Price * 11 = New Price *10 --> the new price in cents must be a multiple of 11 (assuming that the new price in cents is an integer);
To check which price is a multiple of 11 and which is not you should use divisibility rule for 11: if you sum every second digit of a number and then subtract the sum of all other digits and the answer is divisible by 11, then the number is divisible by 11. Only answer choice D, 5730 cents, is not divisible by 11: (7+0)-(5+3)=-1, which is not divisible by 11
Check for more divisibility rules here: math-number-theory-88376.html
Hope it helps.
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14 Apr 2016, 11:16
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stonecold wrote:
Abhishek009 wrote:
stonecold wrote:
Completely DISAGREE
For non integers every answer works..
Is it so ? I never bothered to check that please provide us with your explanation..
Lets check again.
see
Every answer is valid Except D (as you put it) ..
D will be valid if => x= 573/11
Hence D will be valid too
Remember No one told us that x needs be an integer ..
Let price be x { we do not know whether it is even / odd/ integer / whole number}
The store owner decided to raise the price of a particular item by exactly 10%, so the new price must be x *110 /100 =>1.1x
So, The new number must be 1.1 times the original price.
Now, check option (D)
1.1x = 57.30
So, x = 57.30/1.1 ~ 52.09 (Approximately)
But the Same is not with option (E)
1.1x = 78.10
x = 71
So IMHO (D) does not fall into the group....
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A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 11:44
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Hi stonecold,
Abhishek is right. Option D is correct.
When 573 is divided by 11 we get a remainder 1.
1/11 is non terminating and the question states that the increase is exactly 10%.
Since the value is non terminating, original price cannot be accurately determined. The original value can be 52.0909090909.........................
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Re: A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 12:07
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Hi stonecold,
This question is still 'restricted' to the reality of 'dollars and cents.' While Answer D could be divided by 1.1, you would end up with a repeating decimal in the original price of that item. Regardless of how hard you look, you will never find a store that charges you a price that includes a repeating decimal.
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Re: A store owner decided to raise the price of a particular [#permalink]
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11 Mar 2012, 00:50
But (8+0)-(7+1)=0, and zero isn't divisble by 11, either. Is zero exempt from the rule?
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Re: A store owner decided to raise the price of a particular [#permalink]
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11 Mar 2012, 00:53
Stoneface wrote:
But (8+0)-(7+1)=0, and zero isn't divisble by 11, either. Is zero exempt from the rule?
Zero is divisible by EVERY integer except zero itself, since 0/integer=integer.
Check for more number properties hints/tips/rules the link in my previous topic.
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Re: A store owner decided to raise the price of a particular [#permalink]
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25 Nov 2013, 11:33
Stoneface wrote:
I could figure this one out in about two seconds if I had a calculator, but I hear that's not allowed on the GMAT, so I need to know how to get this answer quick by hand. Thanks!
A store owner decided to raise the price of a particular item by exactly 10%. Of the following which is NOT the new price?
A. \$1.10
B. \$8.80
C. \$11.00
D. \$57.30
E. \$78.10
10% increase = 1.1x where x is the original amount.
1.1 = 11/10. So the answer must be a multiple of 11.
Only answer choice that is not a multiple of 11 is D.
Hence D
Hope it helps!
Cheers
J
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Re: A store owner decided to raise the price of a particular [#permalink]
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18 Dec 2013, 04:55
jlgdr wrote:
Stoneface wrote:
I could figure this one out in about two seconds if I had a calculator, but I hear that's not allowed on the GMAT, so I need to know how to get this answer quick by hand. Thanks!
A store owner decided to raise the price of a particular item by exactly 10%. Of the following which is NOT the new price?
A. \$1.10
B. \$8.80
C. \$11.00
D. \$57.30
E. \$78.10
10% increase = 1.1x where x is the original amount.
1.1 = 11/10. So the answer must be a multiple of 11.
Only answer choice that is not a multiple of 11 is D.
Hence D
Hope it helps!
Cheers
J
A 1 * 1.1 = 1.10 \$
B 8 * 1.1 = 8.80 \$
C 10* 1.1 = 11 \$
D - strange Number
E 71 * 1.1 = 78.1 \$
since D is the only answer, where I couldn't calculate the percents right away, I chose D.
BUT if the store owner had strange prices like 52.09 \$ the new price would be 57.30 \$. Is this a real GMAT question? Because it says nowhere that the original price has to be integer.
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18 Dec 2013, 05:11
No, it doesn't and thats why the question is somewhat flawed
Cheers
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Re: A store owner decided to raise the price of a particular [#permalink]
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12 Jan 2015, 14:39
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Hi All,
This question has a great "pattern-matching" shortcut built into it, but you'll have to pay attention to the answer choices to see it.
We're told that a store owner raises the price of an item by 10%. We're asked which of the 5 answers could NOT possibly be the new price.
In this scenario, it's real easy to come up with a simple example of what the store owner did.
IF....
price = \$1, then new price = \$1.10
Answer B is just 8 times \$1.10, which makes sense because 10% of 8 is 0.80...
IF...
price = \$8, then new price = \$8.80
So the shortcut is just to find the answers that are multiples of \$1.10
Answer C is another quick "find" - it's 10 times \$1.10
Between D and E, you have to do a little work, but if you "break down" E, you get....
78.10 = 77.00 + 1.10
77.00 = 70 + 7 (which is 10% of 70)
1.10 = 1 + 0.1 (which is 10% of 1)
So E is another multiple of \$1.10
[Reveal] Spoiler:
D
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Re: A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 04:45
Hey EMPOWERgmatRichC You are assuming here that the price is an INteger..
What if its NOT...
My Guess is => THIS IS A WRONG QUESTION AS ALL OPTIONS ARE POSSIBLE..
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Re: A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 10:52
stonecold wrote:
Hey EMPOWERgmatRichC You are assuming here that the price is an INteger..
What if its NOT...
My Guess is => THIS IS A WRONG QUESTION AS ALL OPTIONS ARE POSSIBLE..
Let price be \$ 10
The store owner decided to raise the price of a particular item by exactly 10%, so the new price must be 110 *10 /100 =>11
So, The new price must be a multiple of 11
From the given options check for all the multiples of 11
Test of divisibility of 11 is
" if the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11."
A. \$1.10
B. \$8.80
C. \$11.00
By a quick glance even a 2nd grader can eliminate options A,B and C , because they are all divisible by 11 , let with only 2 options check for the divisibility rule for options D and E
D. \$57.30
Sum of even digits is 5+3 = 8
Sum of odd digits is 7 +0 = 7
Difference of the even and odd digits is 8 - 7 = 1 ; not divisible by 11
E. \$78.10
Further inspection is not required since we have already found out the answer at (D)
stonecold No question of treating the number as integers / not take it as it is given in the question stem.
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Re: A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 10:56
Abhishek009 wrote:
stonecold wrote:
Hey EMPOWERgmatRichC You are assuming here that the price is an INteger..
What if its NOT...
My Guess is => THIS IS A WRONG QUESTION AS ALL OPTIONS ARE POSSIBLE..
Let price be \$ 10
The store owner decided to raise the price of a particular item by exactly 10%, so the new price must be 110 *10 /100 =>11
So, The new price must be a multiple of 11
From the given options check for all the multiples of 11
Test of divisibility of 11 is
" if the difference of the sum of digits at odd places and the sum of its digits at even places, is either 0 or divisible by 11, then clearly the number is divisible by 11."
A. \$1.10
B. \$8.80
C. \$11.00
By a quick glance even a 2nd grader can eliminate options A,B and C , because they are all divisible by 11 , let with only 2 options check for the divisibility rule for options D and E
D. \$57.30
Sum of even digits is 5+3 = 8
Sum of odd digits is 7 +0 = 7
Difference of the even and odd digits is 8 - 7 = 1 ; not divisible by 11
E. \$78.10
Further inspection is not required since we have already found out the answer at (D)
stonecold No question of treating the number as integers / not take it as it is given in the question stem.
Completely DISAGREE
For non integers every answer works..
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Re: A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 11:02
stonecold wrote:
Completely DISAGREE
For non integers every answer works..
Is it so ? I never bothered to check that please provide us with your explanation..
Lets check again.
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A store owner decided to raise the price of a particular [#permalink]
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14 Apr 2016, 11:07
Abhishek009 wrote:
stonecold wrote:
Completely DISAGREE
For non integers every answer works..
Is it so ? I never bothered to check that please provide us with your explanation..
Lets check again.
see
Every answer is valid Except D (as you put it) ..
D will be valid if => x= 573/11
Hence D will be valid too
Remember No one told us that x needs be an integer ..
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14 Apr 2016, 11:20
Yes..Abhishek009
But the Question is Which of the following Cannot be the value of x
hence we cannot say which one.
x needs to be an integer here
P.S - We can play along all day .. Every value is valid unless stated otherwise
CC :- Vyshak
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Display posts from previous: Sort by | 5,549 | 19,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-47 | latest | en | 0.892665 |
https://edslash.com/leetcode-problem-145-binary-tree-postorder-traversal/ | 1,725,928,599,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00594.warc.gz | 199,473,460 | 32,757 | ### LeetCode Challenge #145. Binary Tree Postorder Traversal
Given the `root` of a binary tree, return the postorder traversal of its nodes’ values.
Example 1:
```Input: root = [1,null,2,3]
Output: [3,2,1]
```
Example 2:
```Input: root = []
Output: []
```
Example 3:
```Input: root = [1]
Output: [1]
```
Constraints:
• The number of the nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`
##### Video Solution
###### Java Solution
``` ```
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
helper(root,list);
return list ;
}
public void helper(TreeNode root , List<Integer> list ){
if(root==null){
return ;
}
helper(root.left , list );
helper(root.right, list); | 211 | 759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.390835 |
https://www.tutorialkart.com/python/python-membership-operators/ | 1,709,282,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475203.41/warc/CC-MAIN-20240301062009-20240301092009-00042.warc.gz | 1,031,852,115 | 18,885 | ## Python – Membership Operators
Membership Operators are used to check if an element or item is present in the given collection or sequence.
## Different Membership Operators
The following table presents the different membership operators in Python.
## Examples
### 1. in Operator
In the following program, we check if an element e is present in the collection x.
main.py
e = 5
x = [0, 5, 10, 15, 20]
if e in x :
print(f"{e} is present in {x}")
else:
print(f"{e} is not present in {x}")
Try Online
Output
5 is present in [0, 5, 10, 15, 20]
Let us rerun the above program with e = 12. Since, this value is not present in x, the expression e in x returns False.
main.py
e = 12
x = [0, 5, 10, 15, 20]
if e in x :
print(f"{e} is present in {x}")
else:
print(f"{e} is not present in {x}")
Try Online
Output
5 is present in [0, 5, 10, 15, 20]
### 2. not in Operator
“not in” Operator is the inverse of the “in” operator.
In the following program, we check if an element e is not present in the collection x.
main.py
e = 12
x = [0, 5, 10, 15, 20]
if e not in x :
print(f"{e} is not present in {x}")
else:
print(f"{e} is present in {x}")
Try Online
Output
12 is not present in [0, 5, 10, 15, 20]
## Conclusion
In this Python Tutorial, we learned about Membership operators in Python, and how to use them with the help of example programs. | 414 | 1,358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-10 | latest | en | 0.870435 |
https://gmatclub.com/forum/in-the-1960s-surveys-of-florida-s-alligator-population-indicated-that-322539-20.html | 1,721,017,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514659.22/warc/CC-MAIN-20240715040934-20240715070934-00845.warc.gz | 267,570,006 | 153,149 | Last visit was: 14 Jul 2024, 21:32 It is currently 14 Jul 2024, 21:32
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# In the 1960s, surveys of Florida's alligator population indicated that
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Hi Xolmuhammad
Note that you need to weaken the conclusion, not strengthen it. E seems to do the latter.
If it’s tough to analyze whether E weakens or strengthens the conclusion, try to evaluate the opposite of E. Change “few” to “many”.
The opposite of E: Most of the sightings on golf courses and lawns occurred when MANY people were present on those golf courses and lawns.
You can see that the reversed choice E looks like choice A. If indeed there were MANY people on those courses and lawns, then the conclusion is weakened. Suppose, there were 100 people and they all saw just one alligator. This means 100 sightings of an alligator, but doesn’t mean 100 alligators. There was only one. So, the opposite of E weakens the conclusion, not E itself. Now take E as it is:
Original E: Most of the sightings on golf courses and lawns occurred when FEW people were present on those golf courses and lawns.
If there were few people, then probably sightings increased NOT because many people see the same alligator, BUT because their population indeed increased. So, this choice seems to strengthen the conclusion. But we need to weaken it.
A helpful tactic: if you can’t see whether a certain choice weakens or strengthens the conclusion, try to analyze the opposite of that choice. Analyzing the opposite is sometimes much easier. If the opposite weakens, then the original can NOT weaken. And if the opposite strengthens, then the original can NOT strengthen.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Hi , JonShukhrat, First of all ,thanks for the explanation.
there is one concern left. I exclude A because there is no mention of golf courses or lawns. So should I just assume that when population increases , the people visiting golf courses and lawns increase?
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Xolmuhammad wrote:
Hi , JonShukhrat, First of all ,thanks for the explanation.
there is one concern left. I exclude A because there is no mention of golf courses or lawns. So should I just assume that when population increases , the people visiting golf courses and lawns increase?
You are right. The increase in the human population doesn’t necessarily mean an increase in the number of golf club people. The number of such people on courses and lawns might have remained the same or even decreased. But these people are not the all people who saw the alligators. In other words, not only golf club people can see alligators on golf lawns. There can be many people who walk around or near those lawns. There can be many buildings near the golf clubs and people in those buildings can see alligators from afar. Most golf lawns border with rivers and alligators actually live in these rivers. So, many people sailing in these rivers can also be the witnesses. So forth and so on.
Do not make the assumption that ONLY people in golf courses and lawns can see the alligators.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
parkhydel wrote:
In the 1960s, surveys of Florida's alligator population indicated that the population was dwindling rapidly. Hunting alligators was banned. By the early 1990s, the alligator population had recovered, and restricted hunting was allowed. Over the course of the 1990s, reports of alligators appearing on golf courses and lawns increased dramatically. Therefore, in spite of whatever alligator hunting went on, the alligator population must have increased significantly over the decade of the 1990s.
Which of the following, if true, most seriously weakens the argument?
A. The human population of Florida increased significantly during the 1990s.
B. The hunting restrictions applied to commercial as well as private hunters.
C. The number of sightings of alligators in lakes and swamps increased greatly in Florida during the 1990s.
D. Throughout the 1990s, selling alligator products was more strictly regulated than hunting was.
E. Most of the sightings of alligators on golf courses and lawns in the 1990s occurred at times at which few people were present on those golf courses and lawns.
CR91630.02
We can easily narrow down to options A and E.
A. The human population of Florida increased significantly during the 1990s.
- This is correct. If the population of Florida increased "significantly", then it is possible that a lot more people are reporting the same sighting incident.
E. Most of the sightings of alligators on golf courses and lawns in the 1990s occurred at times at which few people were present on those golf courses and lawns.
- This option is tempting, but it is also incorrect. We cannot assume that people would be lying, as this is what is required to be assumed for this option to strengthen the conclusion.
Option A is correct.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
CrackVerbalGMAT wrote:
A tricky one!
A. The human population of Florida increased significantly during the 1990s.
This could probably be the reason for the increased sighting of alligators. Let’s look at the other options to see if we have a contender.
B. The hunting restrictions applied to commercial as well as private hunters.
Very much irrelevant. Was there any difference in the hunting restriction in the given period? Not sure. Eliminate.
C. The number of sightings of alligators in lakes and swamps increased greatly in Florida during the 1990s.
Only supports the conclusion that the alligator population in Florida during the 1990s increased. Eliminate.
D. Throughout the 1990s, selling alligator products was more strictly regulated than hunting was.
Irrelevant. Eliminate.
E. Most of the sightings of alligators on golf courses and lawns in the 1990s occurred at times at which few people were present on those golf courses and lawns.
Does that mean people were lying and the sightings were fake? Assumptions like these could often lead us to wrong answers. Eliminate.
Option A says that the human population of Florida increased significantly during the 1990s- More people- more sightings of alligators.
This tells us that the population of alligators has not increased but the sightings of alligators have increased because there has been an increase in the human population of Florida. Weakens the argument. Option A is correct.
Vishnupriya
GMAT Verbal SME
Hi, Request your clarification as to why this question is different from the another official question.
The answer to this question is in line with option E here.
https://gmatclub.com/forum/farmer-sever ... 94365.html
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Karthik740 wrote:
Hi, Request your clarification as to why this question is different from the another official question.
The answer to this question is in line with option E here.
https://gmatclub.com/forum/farmer-sever ... 94365.html
GMATNinja VeritasKarishma
Hello, Karthik740. I know your query was not addressed to me, and I myself would love to hear what either of the Experts you mentioned have to say on the matter, but I have addressed both questions in response to the same query way back on page one, in this post. I hope it helps to dispel your concerns.
Good luck with your studies.
- Andrew
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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AndrewN wrote:
Karthik740 wrote:
Hi, Request your clarification as to why this question is different from the another official question.
The answer to this question is in line with option E here.
https://gmatclub.com/forum/farmer-sever ... 94365.html
GMATNinja VeritasKarishma
Hello, Karthik740. I know your query was not addressed to me, and I myself would love to hear what either of the Experts you mentioned have to say on the matter, but I have addressed both questions in response to the same query way back on page one, in this post. I hope it helps to dispel your concerns.
Good luck with your studies.
- Andrew
Thank you Andrew. I went through your solution and understand better the solution.
To be honest, I feel that Option A is a very poorly worded weakener and we'd have to make multiple assumptions to link increase in population to increase in reports.
Thanks,
Karthik
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Karthik740 wrote:
Thank you Andrew. I went through your solution and understand better the solution.
To be honest, I feel that Option A is a very poorly worded weakener and we'd have to make multiple assumptions to link increase in population to increase in reports.
Thanks,
Karthik
I do not disagree, Karthik. However, we also have to keep in mind the question stem itself, which asks us to pick the answer that most seriously weakens the argument. In other words, this answer, as often occurs on harder CR questions, is allowed to fall into a grey area. It need not be unassailable; it just has to be the least debatable of the five options presented.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Karthik740 wrote:
CrackVerbalGMAT wrote:
A tricky one!
A. The human population of Florida increased significantly during the 1990s.
This could probably be the reason for the increased sighting of alligators. Let’s look at the other options to see if we have a contender.
B. The hunting restrictions applied to commercial as well as private hunters.
Very much irrelevant. Was there any difference in the hunting restriction in the given period? Not sure. Eliminate.
C. The number of sightings of alligators in lakes and swamps increased greatly in Florida during the 1990s.
Only supports the conclusion that the alligator population in Florida during the 1990s increased. Eliminate.
D. Throughout the 1990s, selling alligator products was more strictly regulated than hunting was.
Irrelevant. Eliminate.
E. Most of the sightings of alligators on golf courses and lawns in the 1990s occurred at times at which few people were present on those golf courses and lawns.
Does that mean people were lying and the sightings were fake? Assumptions like these could often lead us to wrong answers. Eliminate.
Option A says that the human population of Florida increased significantly during the 1990s- More people- more sightings of alligators.
This tells us that the population of alligators has not increased but the sightings of alligators have increased because there has been an increase in the human population of Florida. Weakens the argument. Option A is correct.
Vishnupriya
GMAT Verbal SME
Hi, Request your clarification as to why this question is different from the another official question.
The answer to this question is in line with option E here.
https://gmatclub.com/forum/farmer-sever ... 94365.html
GMATNinja VeritasKarishma
Karthik740
The different wordings of the two options will tell you why it works in one and not in the other.
(E) Most of the sightings of alligators on golf courses and lawns in the 1990s occurred at times at which few people were present on those golf courses and lawns.
The sightings occurred at times when few people were around. So all it makes me think is hopefully then, people did not get hurt. I am given that sightings occurred so I do not question whether alligators were actually seen or not.
A word on option (A) here - I was looking for something that told me that alligators are losing their natural habitat. That would explain why they are venturing beyond in search of food etc even though their numbers have not jumped. So my logic for (A) was two pronged - more people means more people around to see when an alligator slides in, and more people also could mean more construction and hence destruction of habitat. Yes, there is a leap involved here but NO other option is viable.
Now let's look at the other question given here: https://gmatclub.com/forum/farmer-sever ... 94365.html
C. No person who claimed to have seen a mountain lion had anyone else with them at the purported sighting.
The option hints that the authenticity may be questionable by saying "no person who claimed to have seen..."
It is not saying no person who saw a mountain lion had anyone else along. If it had said that, the option would have become irrelevant.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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It almost seems like everyone is trying to find reasons to prove the official answer correct rather than thinking about it logically.
Option B,C,D are rejected ofcourse.
Just like in case of:
Option A: the inference is that (population increase) implies (alligators leaving their habitats) implies (more sightings on golf courses and lawns),
one may argue that,
Option E: the inference is that (less people saw the alligators) implies (the reports may not be fully correct because people may be wrong/lying)
I think the options are pretty close. There have been similar questions in the GMAT (remember the mountain lion question???) with E as the logic.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
This got to be the one of the easiest CR question for the fact of the matter it had all boil down to thhe a and E now there are already a truck load of definition as of why all the other answers are wrong let me concentrate on A , what has increased no of population got to do with no of sighthing of animals yes if more no of peop;e are sighting a single crocodile then it means they are not reporting a multiple crocodiles but single crocodile is reported by multiple people , it might be a reason IMO A
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maheswariviresh wrote:
It almost seems like everyone is trying to find reasons to prove the official answer correct rather than thinking about it logically.
Option B,C,D are rejected ofcourse.
Just like in case of:
Option A: the inference is that (population increase) implies (alligators leaving their habitats) implies (more sightings on golf courses and lawns),
one may argue that,
Option E: the inference is that (less people saw the alligators) implies (the reports may not be fully correct because people may be wrong/lying)
I think the options are pretty close. There have been similar questions in the GMAT (remember the mountain lion question???) with E as the logic.
Hello, maheswariviresh. I have to say that the part I have highlighted above seems harsh. Have you read through the thread? In addition to several community responses that touch on (E), I also see those by all of the following Experts:
GMATNinjaThis post discusses that tricky duo, (E) and (A).
Myself—Here I talk about the mountain lion question and how its correct answer differs from what we see in this question.
VeritasPrepHailey—It looks to me as if each answer is treated thoroughly in this post.
CrackVerbalGMAT—The analysis of (E) in this post makes a good point about assumptions.
VeritasKarishma—In the post right above yours, she discusses both answer choice (E) and the mountain lion question.
In short, I am not sure what you mean by your claim. It seems, at best, hastily made. Moreover, I would be curious to hear your own take on either (A) or (E) beyond a four- or five-word commentary. If you do not like (A), fine, but it is the official answer, so the best we can do is to try to understand what makes it work for this particular question.
Good luck with your studies.
- Andrew
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In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
I swear, I truly despise this question....not because it’s bad, but because it is so damn good.
The subtle “GAP” and frameshift written in the passage makes A the obvious choice (once you “see” this frameshift from the facts to the conclusion).
The goal in a lot of the weaken-type questions is to obviously clearly identify the conclusion. This is what we are trying to weaken. But also we want to look for a new fact that makes it seem like the supporting premises......do not really support the conclusion as effectively as they did before we knew this new fact in the answer.
The “jump” from the facts to the conclusion that the argument makes is the following:
The facts mention an increase in the “reports” of alligators roaming lawns and golf courses in the 1990s. The author also introduces information about hunting bans and restrictions to make you look one way while this fact about “reports” is slid right under your nose.
The argument’s fatal logical flaw is that it takes a fact about “reports” having increased in the 1990s to mean that the alligator population itself must have increased.
“Reports” of alligators being seen on lawns does NOT necessarily indicate that there has been an actual increase in the alligator population.
Thus, if we have a lot more people in Florida (a “significant” increase in the population) during the 1990s, then the supporting premise about “reports” of sightings begins to look more suspect.
Perhaps the reason for more reported sightings isn’t because the alligator population increased in the 1990s. Instead, it could be the case that the alligator population remainder the same. The only thing that might have changed was that there were significantly more people filing reports of alligator sightings on lawns and golf courses.
A is the winner
And yes it got me good....
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In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Hey AndrewN,
Thank you for the reply.
First things first. Let us please leave out the "manner" in which you might have interpreted my reply. This is a written forum, and I am sure you judged the tone of what I said in your mind, rather than how i meant it to be. Apologize if I hurt sentiments here, but I guess we are all here to learn.
Further to the way you have made your case, that experts have tried to explain it and hence it is correct, is difficult to work with. If more number of experts have said the same thing, does that make the official answer correct? I dont agree with that approach.
I believe that this forum is for discussion, and I have no come across a single question/logic which is refuted by the experts saying that "No. The official anwer is incorrect, because it should be this." It is very hard to believe that out of the hundreds of thousands of questions that have official answers, none of them are incorrect. Beg your pardon if that sounds harsh, but its really hard to take that on face value that everything that the official answer suggests is correct, and has been so over the last 10-15 years, EVERY SINGLE TIME. By no means do I want to suggest that I am right or you are wrong. I just want to contest with the question "What if the official answer was E? Would we not have come up with reasons for the same, or would you have said, "No, the official answer is incorrect" ?
Coming back to the question (sorry I got deviated because the "manner" of questioning was itself in question here) -
The logic: "In choice (A), we are to understand that if the human population of Florida increased significantly during the 1990s (my emphases), then it would make sense that the number of reported sightings might also logically increase" is almost like "If population around a house deemed haunted increases, the number of reported sightings of a ghost will also increase" thereby implying that "people will always be sound and not go by hearsay." Sounds crazy, no?
Reports of the number of alligator sightings increased in no way implies that the reports are correct, dont you think? If there are lesser people when these sightings are made, even a patch of grass that in the shape of an alligator some distance off might induce someone to run and report an alligator sighting (since they are just thinking about the fact that this is an alligator prone area). I am just asking. Why can't it be true?
Another logic that I've read on the threads is this: "Human population increased>>> more people>>> more sightings of alligators (even if the population of them is same or decreased)"
Again, I see the same flaw. I disagree with the fact that human population increase leads to more people seeing alligators. Is that true in case of blue whales, or maybe other rarely found animal species? We are bringing outside knowledge to answer this question, which we have, as a community decided not to do when answering CR questions.
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In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Sorry, I hope I don’t step on toes by butting into a conversation that doesn’t involve me.
In some ways, I prefer the LSAT’s approach to formal logic. To me, the answers seem to be more clear cut and the logical fallacies committed by the authors more clearly identifiable.
The GMAT’s approach, though frustrating at times, is really about stepping back and thinking from a “common sense” perspective: making common sense connections between the answer choice to the passage in such a way that we answer the question.
Now where does the line between what is a “common sense” type of connection and what is an “unwarranted assumption” end and begin? On the upper-level questions, this line seems to get blurry.
I still struggle with the urge to want to rip through a lot of the official answers because, as you’ve shown, there are certain times in which it seems as if the answers cut a little close to the “line” I described above.
However, the test is the test. Arguing with the creators’ official explanation isn’t going to result in beating this demon (though it does feel good to vent). In the end, there are identifiable patterns from answer choice to answer choice. I believe finding and spotting these patterns is what makes the upper-level CR questions so difficult. In the end, it really is about learning how to “think like the test makers.”
The way we have to approach this test, if we want to conquer it, is to use the official answers as a kind of roadmap of sorts to navigate through all the little intricacies of each problem. We can learn which types of answers GMAC believes effectively weaken an argument and which do not. If GMAC tells us this is the answer, then this is the answer. If GMAC tells us that dinosaurs still exist, then dinosaurs still exist.
It is frustrating at times because I (and I’m sure everyone else) wish we could obtain a clearer picture or what constitutes an “unwarranted assumption.”
The best hope, I’ve found, is to delve into each official answer meticulously. There are patterns to how the writers of the test seem to “think”. Perhaps I shouldn’t write that in print for everyone to read. Someone from GMAC might read the post and then switch things up entirely to avoid any sort of “pattern” in the correct answers. LOL
The patterns are there. I’m sure there are knowledgeable experts who can pull up official questions in which the premise of “reports of X event changing” is used to support the opinion that X events actually changed. This type of logical flaw has appeared before.
In the end, the best we can do is figure out why the correct answers are what they are. Arguing with what GMAC says is the correct answer is ultimately futile: a fact that has lead to many fits of frustration throughout this personal journey.
By all means we should all question everything. The more questions that appear on this board, the more insight and knowledge we can hope to pull from each question.
However, I had to find out the hard way that arguing with the official answers just doesn’t help. Instead, looking for why each answer is correct will at least help us put together our own roadmap of sorts that will help us find correct answers in the future.
Best of luck to all!
Edit: Throughout my long message I forgot to tie things up with a Main Idea Statement.
Arguing with the official answers puts one on the road to nowhere. It’s the equivalent of repeatedly banging one’s head against the wall.
maheswariviresh wrote:
Hey AndrewN,
Thank you for the reply.
First things first. Let us please leave out the "manner" in which you might have interpreted my reply. This is a written forum, and I am sure you judged the tone of what I said in your mind, rather than how i meant it to be. Apologize if I hurt sentiments here, but I guess we are all here to learn.
Further to the way you have made your case, that experts have tried to explain it and hence it is correct, is difficult to work with. If more number of experts have said the same thing, does that make the official answer correct? I dont agree with that approach.
I believe that this forum is for discussion, and I have no come across a single question/logic which is refuted by the experts saying that "No. The official anwer is incorrect, because it should be this." It is very hard to believe that out of the hundreds of thousands of questions that have official answers, none of them are incorrect. Beg your pardon if that sounds harsh, but its really hard to take that on face value that everything that the official answer suggests is correct, and has been so over the last 10-15 years, EVERY SINGLE TIME. By no means do I want to suggest that I am right or you are wrong. I just want to contest with the question "What if the official answer was E? Would we not have come up with reasons for the same, or would you have said, "No, the official answer is incorrect" ?
Coming back to the question (sorry I got deviated because the "manner" of questioning was itself in question here) -
The logic: "In choice (A), we are to understand that if the human population of Florida increased significantly during the 1990s (my emphases), then it would make sense that the number of reported sightings might also logically increase" is almost like "If population around a house deemed haunted increases, the number of reported sightings of a ghost will also increase" thereby implying that "people will always be sound and not go by hearsay." Sounds crazy, no?
Reports of the number of alligator sightings increased in no way implies that the reports are correct, dont you think? If there are lesser people when these sightings are made, even a patch of grass that in the shape of an alligator some distance off might induce someone to run and report an alligator sighting (since they are just thinking about the fact that this is an alligator prone area). I am just asking. Why can't it be true?
Another logic that I've read on the threads is this: "Human population increased>>> more people>>> more sightings of alligators (even if the population of them is same or decreased)"
Again, I see the same flaw. I disagree with the fact that human population increase leads to more people seeing alligators. Is that true in case of blue whales, or maybe other rarely found animal species? We are bringing outside knowledge to answer this question, which we have, as a community decided not to do when answering CR questions.
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Originally posted by Fdambro294 on 13 Jun 2021, 10:03.
Last edited by Fdambro294 on 13 Jun 2021, 18:05, edited 1 time in total.
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Re: In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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Expert Reply
Hello again, maheswariviresh. Dissenting opinions are fine and should be a part of such a forum. But to make a blanket statement about what every other post in the thread (via "everyone") does or does not achieve, to suggest that not just one, but several Experts may be doing little more than peddling or aping the OA, is bold and most likely misguided. Maybe some of these people got the answer correct the first time because they had insights into the matter that you may not have had—i.e. maybe they were "thinking about it logically" and made an educated decision about separating one trap answer from another, correct answer to this particular question.
Quote:
I have no come across a single question/logic which is refuted by the experts saying that "No. The official anwer is incorrect, because it should be this."
Perhaps you have not spent enough time on the forum. I wrote a post just yesterday in which I debated an OA to a third-party question. I have even seen one post in which Bunuel, whose Quant expertise I do not question, has pointed out that an official GMAT Prep question has an incorrect OA... and has gone on to explain why. (I wish I could find the link.)
I appreciate your outlining your thought process about this question in your response. Nevertheless, there are a few points I would like to touch on so that other readers might not make similar conclusions.
Quote:
The logic: "In choice (A), we are to understand that if the human population of Florida increased significantly during the 1990s (my emphases), then it would make sense that the number of reported sightings might also logically increase" is almost like "If population around a house deemed haunted increases, the number of reported sightings of a ghost will also increase" thereby implying that "people will always be sound and not go by hearsay." Sounds crazy, no?
You have spun my words into something entirely different here. I write my responses very carefully to (generally) avoid using absolute language such as will or always. I said "it would make sense" and "might also." Cautious language is more difficult to challenge than definitive or absolute language, the type that is often found in incorrect answer choices in both CR and RC. In fact, many Experts often point out that a single word such as "never" or even "not" (without a cushioning "may" in front of it) can make all the difference between a correct line of thought and an incorrect one.
Quote:
Reports of the number of alligator sightings increased in no way implies that the reports are correct, dont you think? If there are lesser people when these sightings are made, even a patch of grass that in the shape of an alligator some distance off might induce someone to run and report an alligator sighting (since they are just thinking about the fact that this is an alligator prone area). I am just asking. Why can't it be true?
...
I disagree with the fact that human population increase leads to more people seeing alligators.
Of course, people could have seen something other than an alligator and reported having seen one. But we have to come to terms, either way, with why the number of these reports increased dramatically in the 1990s. (E) actually looks worse when you consider that part about few people having been present during sightings. Fewer people around to see alligators, whether real or imaginary, should lead to fewer reports, not more. Meanwhile, more people around to see alligators might reasonably lead to more reports. (This is not a fact.) So, while (E) could be true, against (A), it does not make a more compelling case. One final point on the matter: Do not lose sight of the question itself, which (also) does not deal in absolutes. We are not seeking an airtight, open-and-shut case (as we would in a "Must Be True" question). We are only looking for the answer choice that most seriously weakens the argument.
Again, you are free to disagree with my view. As long as you explain why, it will only benefit the community.
I wish you the best in your preparation. I may not use emoticons (even with friends or family), but my sentiments are just as genuine.
- Andrew
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In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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When we decide to take a standardised test, we have to conform to that standards that it sets. We can chose to question it, using all the intelligence we have, and sometimes rightly so, but, when you have decided to surrender yourself to standardised test, you’ll have to conform to the patterns of its use of logic and structure.
GMAT is to get to a business school. It’s not life, neither, it makes you less intelligent or more IMO, only makes you more attuned to what it demands. Most successful businessmen I know never went to a business school, or took a standardised test, and if we are to believe we are above the test and its maker’s logic, we shouldn’t take the test, I think. The very idea of testing yourself to a third party creation is to conform to its structure and you can only possibly succeed in it is by learning its patterns. There’s no point debating it’s correctness or feeling offended by being told to stick with its structure. This is the fight you have chosen, so you must conform to it. Because if you don’t, it’s ruthless in reminding you that you must.
So yeah, sweating over why official questions are right is better than perspiring over how flawed it is. Because gmat, verbal especially, tests your conformance to its logic pattern, not to a general pattern. A lesson I learned early on.
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In the 1960s, surveys of Florida's alligator population indicated that [#permalink]
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821 posts | 8,402 | 36,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.905402 |
http://www.conversion-website.com/speed/speed-of-light-to-kilometer-per-second.html | 1,695,651,804,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233508977.50/warc/CC-MAIN-20230925115505-20230925145505-00881.warc.gz | 57,346,972 | 4,401 | # Speed of light to kilometers per second (c to km/s)
## Convert speed of light to kilometers per second
Speed of light to kilometers per second converter on this page calculates how many kilometers per second are in 'X' speed of light (where 'X' is the number of speed of light to convert to kilometers per second). In order to convert a value from speed of light to kilometers per second (from c to km/s) type the number of c to be converted to km/s and then click on the 'convert' button.
## Speed of light to kilometers per second conversion factor
1 speed of light is equal to 299792.458 kilometers per second
## Speed of light to kilometers per second conversion formula
Speed(km/s) = Speed (c) × 299792.458
Example: Find the number of kilometers per second in 247 speed of light.
Speed(km/s) = 247 ( c ) × 299792.458 ( km/s / c )
Speed(km/s) = 74048737.126 km/s or
247 c = 74048737.126 km/s
247 speed of light equals 74048737.126 kilometers per second
## Speed of light to kilometers per second conversion table
speed of light (c)kilometers per second (km/s)
72098547.206
123597509.496
175096471.786
226595434.076
278094396.366
329593358.656
3711092320.946
4212591283.236
4714090245.526
5215589207.816
speed of light (c)kilometers per second (km/s)
25074948114.5
30089937737.4
350104927360.3
400119916983.2
450134906606.1
500149896229
550164885851.9
600179875474.8
650194865097.7
700209854720.6
Versions of the speed of light to kilometers per second conversion table. To create a speed of light to kilometers per second conversion table for different values, click on the "Create a customized speed conversion table" button.
## Related speed conversions
Back to speed of light to kilometers per second conversion
TableFormulaFactorConverterTop | 482 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-40 | latest | en | 0.649166 |
https://www.printablemultiplicationflashcards.com/grade-3-multiplication-chart/ | 1,632,422,798,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00421.warc.gz | 971,565,992 | 9,039 | Discovering multiplication soon after counting, addition, as well as subtraction is ideal. Youngsters learn arithmetic using a all-natural progression. This progress of understanding arithmetic is usually the pursuing: counting, addition, subtraction, multiplication, and finally department. This document contributes to the concern why find out arithmetic within this series? Moreover, why learn multiplication soon after counting, addition, and subtraction just before division?
## The next facts respond to these inquiries:
1. Children find out counting initially by associating visual things with their hands and fingers. A tangible case in point: How many apples exist in the basket? Much more abstract instance is just how outdated have you been?
2. From counting phone numbers, the subsequent rational stage is addition combined with subtraction. Addition and subtraction tables can be very helpful teaching helps for youngsters because they are visible resources producing the changeover from counting less difficult.
3. Which ought to be discovered after that, multiplication or division? Multiplication is shorthand for addition. At this time, kids have got a company understanding of addition. As a result, multiplication will be the up coming reasonable method of arithmetic to learn.
## Assess essentials of multiplication. Also, review the essentials how to use a multiplication table.
Let us assessment a multiplication illustration. Utilizing a Multiplication Table, increase several times about three and acquire a solution a dozen: 4 by 3 = 12. The intersection of row 3 and column four of your Multiplication Table is 12; a dozen is definitely the respond to. For the kids beginning to find out multiplication, this is simple. They may use addition to resolve the situation thus affirming that multiplication is shorthand for addition. Example: 4 x 3 = 4 4 4 = 12. It is an outstanding introduction to the Multiplication Table. The added advantage, the Multiplication Table is aesthetic and mirrors straight back to discovering addition.
## Where can we start learning multiplication while using Multiplication Table?
1. First, get knowledgeable about the table.
2. Start with multiplying by 1. Begin at row number 1. Proceed to line primary. The intersection of row one particular and column the initial one is the perfect solution: a single.
3. Replicate these steps for multiplying by one. Flourish row a single by posts one particular via twelve. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Repeat these techniques for multiplying by two. Flourish row two by columns 1 by means of several. The replies are 2, 4, 6, 8, and 10 respectively.
5. Let us hop forward. Perform repeatedly these steps for multiplying by 5 various. Flourish row five by posts one through 12. The replies are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now let us raise the level of issues. Replicate these steps for multiplying by a few. Increase row 3 by posts one via a dozen. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. If you are comfortable with multiplication so far, try out a analyze. Resolve the subsequent multiplication troubles in your mind then examine your answers on the Multiplication Table: increase six and 2, grow nine and 3, flourish a single and 11, flourish 4 and 4, and flourish several and 2. The trouble solutions are 12, 27, 11, 16, and 14 respectively.
When you obtained 4 out of several difficulties proper, design your personal multiplication assessments. Estimate the replies in your head, and view them making use of the Multiplication Table. | 824 | 3,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-39 | latest | en | 0.92907 |
https://www.mathworks.com/matlabcentral/cody/problems/46-which-doors-are-open/solutions/1908163 | 1,571,070,975,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653876.31/warc/CC-MAIN-20191014150930-20191014174430-00323.warc.gz | 990,229,779 | 15,355 | Cody
# Problem 46. Which doors are open?
Solution 1908163
Submitted on 21 Aug 2019 by Joshua Agarth
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = 1; assert(isequal(which_doors_open(x),y_correct))
2 Pass
x = 3; y_correct = 1; assert(isequal(which_doors_open(x),y_correct))
3 Pass
x = 100; y_correct = [1 4 9 16 25 36 49 64 81 100]; assert(isequal(which_doors_open(x),y_correct)) | 169 | 527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-43 | latest | en | 0.73767 |
https://sites.math.rutgers.edu/~zeilberg/EM14/hw9/hw9FranksCole.txt | 1,653,374,458,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00383.warc.gz | 448,274,808 | 2,617 | #hw9.txt Cole Franks 22/02/2014 read(`/Users/Cole/C9.txt`): #### problem 1. Help:=proc(): print(`ApplyUmbra(P,x,U,m),PnG(n, x, U, m),GuessCoeff(n,i,U,m), GuessOP(n,r,U,m)`): end: # ApplyUmbra(P,x,U,m) replaces x^m by the expression U(m) # and extends to U(P) by linearity. ApplyUmbra:=proc(P,x,U,m) local i, Q, d: Q:=expand(P): d:=degree(P,x): for i from 0 to d-1 do Q:=applyrule(x^(d-i)=subs(m=(d-i),U),Q): #print(Q): od: Q: end: #### problem 2. PnG:=proc(n, x, U, m) local eq,var,i,P,a: option remember: P:=x^n+add(a[i]*x^i,i=0..n-1): var:={seq(a[i],i=0..n-1)}: eq:={ seq( ApplyUmbra(expand(P*x^i), x, U,m)=0, i=0..n-1)}: #eq: sort(subs(solve(eq,var),P)): end: ###### Problem 3. GuessCoeff:=proc(n,i,U,m) local L, j, limit: option remember: #limit is just a reasonable large number of #inputs to guessRF that we would possibly want to make. limit:=100: L:=[]: for j from i to i+limit do L:=[op(L),coeff(PnG(j,x,U,m), x, (j-i))]: #print(L): if GuessRF(L,n)<>FAIL then #we have to substitute n=n-i+1 because #the 1st entry of the list gives the #value for n=i. return(factor(subs(n=n-i+1, GuessRF(L, n)))): fi: od: return(FAIL): end: #i, rational function: #U=1/m+1 # 0, 1 # 1 # 1, - - n # 2 # 2 # (n - 1) n # 2, ------------ # 4 (-1 + 2 n) # 2 # (n - 2) n (n - 1) # 3, - ------------------ # 24 (-1 + 2 n) # 2 2 # (n - 3) n (n - 1) (n - 2) # 4, --------------------------- # 96 (-1 + 2 n) (-3 + 2 n) # # 2 2 # (n - 4) n (n - 1) (n - 2) (n - 3) # 5, - ----------------------------------- # 960 (-1 + 2 n) (-3 + 2 n) # 2 2 2 # (n - 5) n (n - 1) (n - 2) (n - 3) (n - 4) # 6, -------------------------------------------- # 5760 (2 n - 5) (-1 + 2 n) (-3 + 2 n) # # 2 2 2 # (n - 6) n (n - 1) (n - 2) (n - 3) (n - 4) (n - 5) # 7, - ---------------------------------------------------- # 80640 (2 n - 5) (-1 + 2 n) (-3 + 2 n) #U=1/m+3 # 0, 1 # 1 # 1, - - n # 2 # 2 # (n - 1) n # 2, ------------ # 4 (-1 + 2 n) # 2 # (n - 2) n (n - 1) # 3, - ------------------ # 24 (-1 + 2 n) # 2 2 # (n - 3) n (n - 1) (n - 2) # 4, --------------------------- # 96 (-1 + 2 n) (-3 + 2 n) # 2 2 # (n - 4) n (n - 1) (n - 2) (n - 3) # 5, - ----------------------------------- # 960 (-1 + 2 n) (-3 + 2 n) # 2 2 2 # (n - 5) n (n - 1) (n - 2) (n - 3) (n - 4) # 6, -------------------------------------------- # 5760 (2 n - 5) (-1 + 2 n) (-3 + 2 n) # 2 2 2 # (n - 6) n (n - 1) (n - 2) (n - 3) (n - 4) (n - 5) # 7, - ---------------------------------------------------- # 80640 (2 n - 5) (-1 + 2 n) (-3 + 2 n) #U = eval(m!) # 0, 1 # 2 # 1, -n # 1 2 2 # 2, - n (n - 1) # 2 # 1 2 2 2 # 3, - - n (n - 1) (n - 2) # 6 # 1 2 2 2 2 # 4, -- n (n - 1) (n - 2) (n - 3) # 24 # 1 2 2 2 2 2 # 5, - --- n (n - 1) (n - 2) (n - 3) (n - 4) # 120 # 1 2 2 2 2 2 2 # 6, --- n (n - 5) (n - 1) (n - 2) (n - 3) (n - 4) # 720 # 1 2 2 2 2 2 2 2 #7, - ---- n (n - 1) (n - 2) (n - 3) (n - 4) (n - 5) (n - 6) # 5040 # note that the constant in the denominator is n!. #### problem 4 #it does not appear that these will be rational functions. # it does, however, appear that their ratios will be. #GuessOP(n,r,U,m) takes variables n,r and an umbra U in the variable m #and guesses a formula for the ratio #GuessCoeff(n,r,U,m)/GuessCoeff(n,r-1,U,m). GuessOP:=proc(n,r,U,m) local L, i, limit: #we will do this by making #a list of ratios of GuessCoeff(n, i, U,m) #for i ranging from 0 to some fairly large number (limit). #then we will use guess(RF) to guess this ratio. #I am considering a ratio (and an initial value, which here will always be 1) #to be just as good as a #closed form, because you can just multiply #the ratio lots of times to get the formula. limit:=100: L:=[]: for i from 1 to limit do L:=[op(L), GuessCoeff(n,i,U,m)/GuessCoeff(n,i-1,U,m)]: #ColeGuess is a slightly altered version of #GuessRF that only asks for 2d + 2 elements in the #list so that we don’t have to compute so many #expressions. if ColeGuessRF(L,r)<>FAIL then return(factor(ColeGuessRF(L,r))): fi: od: return(FAIL): end: ### old stuff ColeGuessRF1:=proc(L,d,n) local R,a,b,i,j,eq,var: if nops(L)<=2*d+2 then #print(`Make list bigger`): RETURN(FAIL): fi: R:=(n^d+add(a[i]*n^i,i=0..d-1))/add(b[j]*n^j,j=0..d): var:={seq(a[i],i=0..d-1), seq(b[j],j=0..d)}: eq:={seq( numer(L[i]-subs(n=i,R)) =0,i=1..2*d+2)}: var:=solve(eq,var): if var=NULL then RETURN(FAIL): fi: R:=subs(var , R): if {seq( numer(L[i]-subs(n=i,R)),i=1..nops(L))}<>{0} then print(`It worked up to`, 2*d+6, `terms, but that's life `): FAIL: else R: fi: end: #GuessRF(L,n): inputs a list of numbers (or expressions) #L, and a (discrete) variable name #n and tries to guess a rational function of n of degree #<=(nops(L)-6)/2 such that L[i]-R(i)=0 for all i from 1 to nops(L) ColeGuessRF:=proc(L,n) local d, katie: for d from 0 to (nops(L)-2)/2 do katie:=GuessRF1(L,d,n): if katie<>FAIL then RETURN(katie): fi: od: FAIL: end: | 2,045 | 4,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-21 | latest | en | 0.382247 |
https://grazeconomics.wordpress.com/tag/credit-markets/ | 1,582,151,946,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00272.warc.gz | 408,575,243 | 16,511 | # Intro to Econ: Ninth Lecture Aside – The Winner’s Curse
For one last time, I want to come back to the problem of whether you get a loan for your project under the assumption that the risk inherent in your project is stochastically independent of other investment risks. So this was our problem (see also here and here):
$\begin{tabular}{c|ccccc} Scenario & Income & Probability & you get & investor gets \\ \hline good & 200.000 & 80\% & 200.000-x & x \\ bad & -50.000 & 20\% & 0 & -50.000 \\ \end{tabular},$
where $x$ is the repayment amount that you pay back to the investor in case of the project being successful. We argued (in a previous post) that the range of feasible interest rates is 12,5% to 200%. Anything outside that will certainly not be accepted by either the investor or by you.
Suppose that you and the investor are close to agreeing to an interest rate of just over 12,5%. Put yourself in the shoes of the investor for a moment. What might worry you in this case?
# Intro to Econ: Ninth Lecture Aside – Moral Hazard
I want to briefly come back to the problem of whether you get a loan for your project under the assumption that the risk inherent in your project is stochastically independent of other investment risks. So this was our problem (see also here and here):
$\begin{tabular}{c|ccccc} Scenario & Income & Probability & you get & investor gets \\ \hline good & 200.000 & 80\% & 200.000-x & x \\ bad & -50.000 & 20\% & 0 & -50.000 \\ \end{tabular},$
where $x$ is the repayment amount that you pay back to the investor in case of the project being successful. We argued (in a previous post) that the range of feasible interest rates is 12,5% to 200%. Anything outside that will certainly not be accepted by either the investor or by you.
Suppose that you and the investor are close to agreeing to an interest rate of almost 200%. Put yourself in the shoes of the investor for a moment. What might worry you in this case?
# Intro to Econ: Ninth Lecture – Risk Premia under Independent Risks
In the previous post we had the following problem. We were wondering about which interest rate we could expect to see for a loan for a particular risky project. You would like to get a loan, and an investor might like to give it to you. The question was, under what conditions you would get this loan, if you get it at all. Recall, that your project can turn out to be good or bad and that investors generally agree about the chances and consequences of either outcome. The problem can be summarized by the following table, where $x$ is the repayment amount that you pay back to the investor in case of the project being successful. If it is unsuccessful you pay nothing, because you have nothing. You “default” on your loan in that case. This is the risk the investor takes on when she or he gives you this loan.
$\begin{tabular}{c|ccccc} Scenario & Income & Probability & you get & investor gets \\ \hline good & 200.000 & 80\% & 200.000-x & x \\ bad & -50.000 & 20\% & 0 & -50.000 \\ \end{tabular}$
We figured out that you will not accept the loan if the repayment amount $x$ is more than € 200.000 (that would be an interest rate of 200%). Because then you have nothing to gain from this project. In reality, you might not even accept anything close to 200%, but we will come back to this problem later.
We also figured out that the investor will (almost) certainly not accept an interest rate below 12.5%, as otherwise the investor expects a negative return on their investment and would then be better off just putting her or his money under a mattress or, I guess, in a safe or vault. By the way, for a very long time the Catholic Church (and other religions) considered positive interest rates morally wrong. In such a world, you probably wouldn’t get a loan for your great project, unless you find a way around this problem. And that would probably be a shame (see previous post).
In this post I want to think about whether an investor will really accept an interest of 12.5% (or slightly above) given that the investor now takes all the risk and at an interest rate of 12.5% only expects a zero return. The answer to this question, it turns out, all depends on whether the risk in this project is essentially stochastically independent of all other risks inherent in all other projects or not.
# Intro to Econ: Ninth Lecture – Credit Markets – Financial Markets
So far we talked a bit abstractly about markets. Yes, we used some specific products for examples, such as white wine, rental apartments, and perhaps airline pricing, but we have not yet developed a particular market model specifically for a particular product. In this post, I want to do this for a particularly important market: the market for money. This post gives a first account of the basic insights and ingredients that underlie our understanding of credit markets and financial markets. You will see, I hope, that what we have learned so far, especially about supply and demand, while not enough to understand these markets fully, was also not in vain. It will come in handy. | 1,186 | 5,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-10 | latest | en | 0.931954 |
http://betterlesson.com/lesson/resource/3240056/aa-similarity-criterion | 1,488,400,519,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174276.22/warc/CC-MAIN-20170219104614-00642-ip-10-171-10-108.ec2.internal.warc.gz | 28,655,620 | 21,347 | ## AA Similarity Criterion - Section 2: Establishing the AA Similarity Criterion
AA Similarity Criterion
AA Similarity Criterion
# Triangle Similarity Criteria
Unit 6: Similarity
Lesson 4 of 8
## Big Idea: Enough is enough...but what is enough? In this lesson, students will determine the criteria that are sufficient for determining triangle similarity.
Print Lesson
2 teachers like this lesson
Standards:
Subject(s):
Math, Similarity and Congruence, reasoning and proof
75 minutes
### Anthony Carruthers
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Environment: Urban | 293 | 1,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-09 | latest | en | 0.806102 |
https://www.education.com/resources/subtraction-facts/?page=5 | 1,623,963,654,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487633444.37/warc/CC-MAIN-20210617192319-20210617222319-00416.warc.gz | 679,234,453 | 29,022 | # Subtraction Facts
620 filtered results
620 filtered results
Subtraction Facts
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Two-Digit Subtraction
Worksheet
Two-Digit Subtraction
Give your child's math skills a boost with this 2nd grade subtraction worksheet that includes two-digit problems and a word puzzle at the end!
Math
Worksheet
Match: Subtraction Facts Within 5
Game
Match: Subtraction Facts Within 5
Kids hone their early math skills with this math matching game that focuses on subtraction within 5.
Kindergarten
Math
Game
Inverse Subtraction Practice
Worksheet
Inverse Subtraction Practice
Your child will practice subtraction, by using inverse operations. Find the missing numbers to solve the subtraction problems.
Math
Worksheet
Spring Kindergarten Math Assessment Subtraction within 20
Worksheet
Spring Kindergarten Math Assessment Subtraction within 20
Use these fun fruit and vegetable themed math problems to assess your students’ ability to solve subtraction problems within 20.
Kindergarten
Math
Worksheet
Solve the Story Problem
Worksheet
Solve the Story Problem
Practice solving different types of subtraction word problems with this worksheet. Plenty of space is included for students to show their thinking!
Math
Worksheet
Ski Racer: Subtraction and Finding 10 Less
Game
Ski Racer: Subtraction and Finding 10 Less
Kids boost their math skills while shredding the slope in this subtraction race to the finish line!
Math
Game
Beginning Word Problems
Worksheet
Beginning Word Problems
Bug out on word problems! These simple story problems are perfect for helping beginners master subtraction.
Kindergarten
Math
Worksheet
Fact Family Practice: Icy Fact Families
Worksheet
Fact Family Practice: Icy Fact Families
In this fact family practice worksheet, kids show that they understand the inverse relationship between addition and subtraction.
Math
Worksheet
Preschool Math: Take Away the Sea Creatures
Worksheet
Preschool Math: Take Away the Sea Creatures
Preschool Math: Take Away the Sea Creatures offers preschoolers an easy introduction to subtraction that will have them wanting to dive right in!
Preschool
Math
Worksheet
Worksheet
Math
Worksheet
Mid-Year Kindergarten Math Assessment: Subtraction within 10
Worksheet
Mid-Year Kindergarten Math Assessment: Subtraction within 10
Use these animal-themed math problems to assess your students’ ability to solve subtraction problems within 10.
Kindergarten
Math
Worksheet
Math With Bears
Lesson Plan
Math With Bears
Math
Lesson Plan
Match: Subtraction Within 10 and Missing Factors
Game
Match: Subtraction Within 10 and Missing Factors
In this matching game, first graders look at subtraction facts that are missing factors.
Math
Game
Subtraction Across the Number Line
Worksheet
Subtraction Across the Number Line
Bounce across the number line to solve these math problems! Your students will practice their subtraction skills by using the number line as a guide.
Math
Worksheet
At the Grocery Store: Addition and Subtraction
Worksheet
At the Grocery Store: Addition and Subtraction
Math
Worksheet
Worksheet
Math
Worksheet
Frosty Fact Families: Addition and Subtraction
Worksheet
Frosty Fact Families: Addition and Subtraction
Practice addition and subtraction with this fact families worksheet. Your 1st grader will be more prepared for school after the break!
Math
Worksheet
Subtraction Number Stories
Worksheet
Subtraction Number Stories
Can you find the equations hidden in these number stories? Have your little math sleuth write out each subtraction problem to go with the word problem.
Math
Worksheet
Double Digit Subtraction: Giraffe Jam!
Worksheet
Double Digit Subtraction: Giraffe Jam!
Give your second grader valuable at-home math practice with this worksheet that offers practice in double digit subtraction with borrowing and word problems.
Math
Worksheet
Addition and Subtraction Word Problems: Sea Life
Worksheet
Addition and Subtraction Word Problems: Sea Life
Help your child with his math skills with this printable worksheet that asks him to develop a logical sequence for baking a cake.
Math
Worksheet
Subtraction Practice in Outer Space
Worksheet
Subtraction Practice in Outer Space
Help your kiddo reach for the stars with this subtraction practice worksheet! For each problem, color in the star with the correct answer.
Kindergarten
Math
Worksheet
Treasure Diving: Subtraction Fact Families Within 20 (Game 1)
Game
Treasure Diving: Subtraction Fact Families Within 20 (Game 1)
In this underwater game, children choose the correct subtraction expression for a given fact family.
Math
Game
Subtraction Facts: Cupcake Subtraction
Worksheet
Subtraction Facts: Cupcake Subtraction
Want to soothe those subtraction woes? Cupcakes to the rescue! This neat, sweet worksheet will guide your kid through subtraction facts.
Math
Worksheet
Water Rafting: Subtraction Fact Families Within 10
Game
Water Rafting: Subtraction Fact Families Within 10
Build up subtraction skills in this online game about math fact families and quick thinking. | 1,038 | 4,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-25 | latest | en | 0.816782 |
https://www.airmilescalculator.com/distance/amh-to-dem/ | 1,656,266,382,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103271763.15/warc/CC-MAIN-20220626161834-20220626191834-00585.warc.gz | 676,380,026 | 33,427 | # Distance between Arba Minch (AMH) and Dembidollo (DEM)
Flight distance from Arba Minch to Dembidollo (Arba Minch Airport – Dembidolo Airport) is 255 miles / 410 kilometers / 222 nautical miles. Estimated flight time is 58 minutes.
Driving distance from Arba Minch (AMH) to Dembidollo (DEM) is 537 miles / 864 kilometers and travel time by car is about 15 hours 22 minutes.
255
Miles
410
Kilometers
222
Nautical miles
## How far is Dembidollo from Arba Minch?
There are several ways to calculate distances between Los Angeles and Chicago. Here are two common methods:
Vincenty's formula (applied above)
• 254.963 miles
• 410.323 kilometers
• 221.557 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 255.441 miles
• 411.093 kilometers
• 221.972 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Arba Minch to Dembidollo?
Estimated flight time from Arba Minch Airport to Dembidolo Airport is 58 minutes.
## What is the time difference between Arba Minch and Dembidollo?
There is no time difference between Arba Minch and Dembidollo.
## Flight carbon footprint between Arba Minch Airport (AMH) and Dembidolo Airport (DEM)
On average flying from Arba Minch to Dembidollo generates about 63 kg of CO2 per passenger, 63 kilograms is equal to 138 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Arba Minch to Dembidollo
Shortest flight path between Arba Minch Airport (AMH) and Dembidolo Airport (DEM).
## Airport information
Origin Arba Minch Airport
City: Arba Minch
Country: Ethiopia
IATA Code: AMH
ICAO Code: HAAM
Coordinates: 6°2′21″N, 37°35′25″E
Destination Dembidolo Airport
City: Dembidollo
Country: Ethiopia
IATA Code: DEM | 529 | 2,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.845045 |
http://fxcodebase.com/code/viewtopic.php?f=31&t=70760&p=139927 | 1,611,100,460,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00138.warc.gz | 41,078,359 | 7,668 | ## Stinger66 Strategy
Strategies are published here.
### Stinger66 Strategy
Based on request.
viewtopic.php?f=27&p=139826#p139826
Stinger66 Strategy.lua
Apprentice
FXCodeBase: Confirmed User
Posts: 28491
Joined: Thu Dec 31, 2009 11:59 am
Location: Zagreb, Croatia
### Re: Stinger66 Strategy
Hi Apprentice,
Thanks for quick reply and looking ok as it is closing the positions in profit at close of bar.
I am using D1 time period and have another strategy that opens trades.
What I am not seeing is the partial closing of positions in loss when the profit trade is closed.
Is this possible and not sure if you understood my original request?
What I am trying to achieve is, if there is more that one position open in the same direction, that the profitable positions are always closed but the losing positions in the same direction are partially closed by the profit amount achieved.
Logic is only for the same direction of positions.
Example.
1. Position 1 is long 100 with a loss of -\$1000
2. Position 2 is long 100 with a profit of \$500
Position 2 is closed for a profit for \$500
Position 1 is partially closed for 50 Units and loss of -\$500 and 50 units remains open at at a loss of -\$500.
Let me know if this is possible?
Stinger66
Posts: 3
Joined: Thu Dec 10, 2020 1:43 am
### Re: Stinger66 Strategy
Development reference 17.
Apprentice
FXCodeBase: Confirmed User
Posts: 28491
Joined: Thu Dec 31, 2009 11:59 am
Location: Zagreb, Croatia
### Re: Stinger66 Strategy
Stinger66 Strategy.lua
Try this version.
Apprentice
FXCodeBase: Confirmed User
Posts: 28491
Joined: Thu Dec 31, 2009 11:59 am
Location: Zagreb, Croatia
### Re: Stinger66 Strategy
Hello,
Strategy is not closing trades as per my request.
I am looking for immediate partial or full close of trades in loss for the opposite amount of the trade in profit that is closed.
The trades in loss that are closed should always be the oldest trades first.
If this can be achieved then I would like to see if this strategy could be converted to an indicator so that this indicator can be accessed from any strategy,.
What I am trying to achieve is, if there is more than one position open in the same direction, that the profitable positions are always closed but the losing positions in the same direction are always partially closed by the profit amount achieved.
Logic is only for the same direction of positions.
Example.
1. Position 1 is long 100 with a loss of -\$1000
2. Position 2 is long 100 with a profit of \$500
Position 2 is closed for a profit for \$500
Position 1 is partially closed for 50 Units and loss of -\$500 and 50 units remains open at at a loss of -\$500.
Frank,
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### Re: Stinger66 Strategy
Development reference 102.
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I’ve had a hard time wrapping my head around the way to code rotations using Euler angles. In this blog post, I’m trying to apply them to the task of rotating a bunny. | 368 | 1,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.851049 |
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Frequent Visitor
## Return Name of User With Highest Number of Clients
Hi,
Looking for some help please, im trying to create a measure to retrieve the name of the user with the highest number of clients, my columns are "Keyworker" (this is the user) and "carerid" (this is the client).
I have tried and failed to get this to work, it would be great if someone could point me in the right direction.
thanks
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Super Contributor
## Re: Return Name of User With Highest Number of Clients
Hi @j3sting
You can use a pattern like this (pattern taken from SQLBI - Alternative use of FIRSTNONBLANK and LASTNONBLANK):
```TopUser =
FIRSTNONBLANK (
TOPN (
1,
VALUES ( YourTable[Keyworker] ),
CALCULATE ( DISTINCTCOUNT ( YourTable[carerid] ) )
),
1
)```
The FIRSTNONBLANK is just there to break ties.
Owen
Proud to be a Datanaut!
4 REPLIES 4
Highlighted
Super Contributor
## Re: Return Name of User With Highest Number of Clients
Hi @j3sting
You can use a pattern like this (pattern taken from SQLBI - Alternative use of FIRSTNONBLANK and LASTNONBLANK):
```TopUser =
FIRSTNONBLANK (
TOPN (
1,
VALUES ( YourTable[Keyworker] ),
CALCULATE ( DISTINCTCOUNT ( YourTable[carerid] ) )
),
1
)```
The FIRSTNONBLANK is just there to break ties.
Owen
Proud to be a Datanaut!
Super Contributor
## Re: Return Name of User With Highest Number of Clients
@j3sting
You can also use the RANKX function to rank the client numbers and get the user name whose rank number is 1.
Assuming we have a table like below.
We can create two measures with following formulas.
```Rank_Client =
RANKX ( ALLSELECTED ( Table1 ), CALCULATE ( SUM ( Table1[carerid] ) ) )```
```Highest =
CALCULATE (
ALLSELECTED ( Table1[Keyworker] ),
FILTER (
ADDCOLUMNS ( VALUES ( Table1[Keyworker] ), "RankNum", [Rank_Client] ),
[RankNum] = 1
)
)```
Best Regards,
Herbert
Frequent Visitor
## Re: Return Name of User With Highest Number of Clients
Perfect thanks! and thank you for replying
Frequent Visitor
## Re: Return Name of User With Highest Number of Clients
thanks Herbert_Liu
another great sollution, thanks for the help much appreciated!
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https://www.analystforum.com/t/which-probability-formula-to-use/154121 | 1,716,569,616,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00766.warc.gz | 558,367,114 | 6,047 | # Which probability formula to use?
Two are provided:
odds / ( 1 + odds) and
1 / (1 + odds)
The second one uses this formula essentially:
If i get a standard logistic regression, I’m not sure how to find out probability. The first one would be to work out the model using the provided coefficients, then exp the outcome, and then do odds / (1 + odds).
The second one would be to work out the model, multiply it by minus, then exp the outcome, and then do 1 / (1 + odds)
Pr(rolling 6 on a dice with 6 faces) = 1/6
Odds rolling 6 = 1/5 = 0.2
Prob = odds / (1 + odds) = 0.2 / (1 + 0.2) = 0.2 / 1.2 = 1/6 prob for rolling a six
Prob = 1 / (1+ odds) = 1 / (1 +0.2) = 1 / 1.2 = 5/6 prob AGAINST rolling a six
2 Likes
Thanks, that makes sense, but as an example I had a question that asked me the probability of a winning fund, and it used the second example instead of the first, even though I didn’t want the ‘against’ probability.
There is a separate table with coefficients and independent variable means, but I don’t understand why I have to use this formula which does 1 / 1 + odds, instead of doing a regression that looks like this (and using this formula):
ln(p/1-p) = b0 + b1X1 + …
Sorry I don’t really understand what you are asking.
I can’ t make any comments on questions without seeing all the questions and answers
Note the second formula has e^ (negative odds) | 392 | 1,385 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-22 | latest | en | 0.927602 |
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0
84
1
+86
Suppose the function f is defined by the following list of ordered pairs:
{(2,3);(4,−2);(5,−3);(6,2)}{(2,3);(4,−2);(5,−3);(6,2)}
Which of the following represents the domain of f?
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# Conceptual Questions
## 16.1Hooke’s Law: Stress and Strain Revisited
1.
Describe a system in which elastic potential energy is stored.
## 16.3Simple Harmonic Motion: A Special Periodic Motion
2.
What conditions must be met to produce simple harmonic motion?
3.
(a) If frequency is not constant for some oscillation, can the oscillation be simple harmonic motion?
(b) Can you think of any examples of harmonic motion where the frequency may depend on the amplitude?
4.
Give an example of a simple harmonic oscillator, specifically noting how its frequency is independent of amplitude.
5.
Explain why you expect an object made of a stiff material to vibrate at a higher frequency than a similar object made of a spongy material.
6.
As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty? Explain your answer.
7.
Some people modify cars to be much closer to the ground than when manufactured. Should they install stiffer springs? Explain your answer.
## 16.4The Simple Pendulum
8.
Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.
## 16.5Energy and the Simple Harmonic Oscillator
9.
Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.)
## 16.7Damped Harmonic Motion
10.
Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)
11.
How would a car bounce after a bump under each of these conditions?
• overdamping
• underdamping
• critical damping
12.
Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics?
## 16.8Forced Oscillations and Resonance
13.
Why are soldiers in general ordered to “route step” (walk out of step) across a bridge?
## 16.9Waves
14.
Give one example of a transverse wave and another of a longitudinal wave, being careful to note the relative directions of the disturbance and wave propagation in each.
15.
What is the difference between propagation speed and the frequency of a wave? Does one or both affect wavelength? If so, how?
## 16.10Superposition and Interference
16.
Speakers in stereo systems have two color-coded terminals to indicate how to hook up the wires. If the wires are reversed, the speaker moves in a direction opposite that of a properly connected speaker. Explain why it is important to have both speakers connected the same way.
## 16.11Energy in Waves: Intensity
17.
Two identical waves undergo pure constructive interference. Is the resultant intensity twice that of the individual waves? Explain your answer.
18.
Circular water waves decrease in amplitude as they move away from where a rock is dropped. Explain why.
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http://cn.metamath.org/mpeuni/frcond2.html | 1,657,017,556,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00412.warc.gz | 11,010,848 | 4,452 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > frcond2 Structured version Visualization version GIF version
Theorem frcond2 27442
Description: The friendship condition: any two (different) vertices in a friendship graph have a unique common neighbor. (Contributed by Alexander van der Vekens, 19-Dec-2017.) (Revised by AV, 29-Mar-2021.)
Hypotheses
Ref Expression
frcond1.v 𝑉 = (Vtx‘𝐺)
frcond1.e 𝐸 = (Edg‘𝐺)
Assertion
Ref Expression
frcond2 (𝐺 ∈ FriendGraph → ((𝐴𝑉𝐶𝑉𝐴𝐶) → ∃!𝑏𝑉 ({𝐴, 𝑏} ∈ 𝐸 ∧ {𝑏, 𝐶} ∈ 𝐸)))
Distinct variable groups: 𝐴,𝑏 𝐶,𝑏 𝐺,𝑏 𝑉,𝑏
Allowed substitution hint: 𝐸(𝑏)
Proof of Theorem frcond2
StepHypRef Expression
1 frcond1.v . . 3 𝑉 = (Vtx‘𝐺)
2 frcond1.e . . 3 𝐸 = (Edg‘𝐺)
31, 2frcond1 27441 . 2 (𝐺 ∈ FriendGraph → ((𝐴𝑉𝐶𝑉𝐴𝐶) → ∃!𝑏𝑉 {{𝐴, 𝑏}, {𝑏, 𝐶}} ⊆ 𝐸))
4 prex 5058 . . . . 5 {𝐴, 𝑏} ∈ V
5 prex 5058 . . . . 5 {𝑏, 𝐶} ∈ V
64, 5prss 4496 . . . 4 (({𝐴, 𝑏} ∈ 𝐸 ∧ {𝑏, 𝐶} ∈ 𝐸) ↔ {{𝐴, 𝑏}, {𝑏, 𝐶}} ⊆ 𝐸)
76bicomi 214 . . 3 ({{𝐴, 𝑏}, {𝑏, 𝐶}} ⊆ 𝐸 ↔ ({𝐴, 𝑏} ∈ 𝐸 ∧ {𝑏, 𝐶} ∈ 𝐸))
87reubii 3267 . 2 (∃!𝑏𝑉 {{𝐴, 𝑏}, {𝑏, 𝐶}} ⊆ 𝐸 ↔ ∃!𝑏𝑉 ({𝐴, 𝑏} ∈ 𝐸 ∧ {𝑏, 𝐶} ∈ 𝐸))
93, 8syl6ib 241 1 (𝐺 ∈ FriendGraph → ((𝐴𝑉𝐶𝑉𝐴𝐶) → ∃!𝑏𝑉 ({𝐴, 𝑏} ∈ 𝐸 ∧ {𝑏, 𝐶} ∈ 𝐸)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1072 = wceq 1632 ∈ wcel 2139 ≠ wne 2932 ∃!wreu 3052 ⊆ wss 3715 {cpr 4323 ‘cfv 6049 Vtxcvtx 26094 Edgcedg 26159 FriendGraph cfrgr 27431 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-6 2054 ax-7 2090 ax-9 2148 ax-10 2168 ax-11 2183 ax-12 2196 ax-13 2391 ax-ext 2740 ax-sep 4933 ax-nul 4941 ax-pr 5055 This theorem depends on definitions: df-bi 197 df-or 384 df-an 385 df-3an 1074 df-tru 1635 df-ex 1854 df-nf 1859 df-sb 2047 df-eu 2611 df-clab 2747 df-cleq 2753 df-clel 2756 df-nfc 2891 df-ne 2933 df-ral 3055 df-rex 3056 df-reu 3057 df-rab 3059 df-v 3342 df-sbc 3577 df-csb 3675 df-dif 3718 df-un 3720 df-in 3722 df-ss 3729 df-nul 4059 df-if 4231 df-sn 4322 df-pr 4324 df-op 4328 df-uni 4589 df-br 4805 df-iota 6012 df-fv 6057 df-frgr 27432 This theorem is referenced by: frgreu 27443 frgrncvvdeqlem9 27482 frgr2wwlkeu 27502 numclwwlk2lem1 27558 numclwwlk2lem1OLD 27565
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• Subject: Re: LPEG > 0.10 regression: 'B' (pattern may not have fixed length)
• From: Roberto Ierusalimschy <roberto@...>
• Date: Fri, 15 Apr 2016 13:35:56 -0300
```> Consider this piece of code:
>
> ```
> local lpeg = require 'lpeg'
>
> local D = lpeg.R'09'
> local I = lpeg.R('AZ', 'az', '\127\255') + '_'
> local SOS = lpeg.P(function(s, i) return i == 1 end) -- start of string
> local EOS = -lpeg.P(1) -- end of string
> local B = -(I + D) -- word boundary
>
> -- Transform a string with keywords into an LPeg pattern that matches a keyword
> -- followed by a word boundary.
>
> local patt = lpeg.P("foo")+lpeg.P("bar")+lpeg.P("qux")
> local BB = lpeg.B(B) + SOS -- starting boundary
> local AB = B + EOS -- ending boundary
>
> patt = BB * patt * AB
>
> patt = lpeg.P{ patt + 1 * lpeg.V(1) }
>
> print(patt:match(" foo"))
> ```
>
> It works in lpeg 0.10. but not in lpeg 1.0
> To get it to work in lpeg 1.0 I applied this change:
>
> -local B = -(I + D) -- word boundary
> +local B = #lpeg.P(1) + -(I + D) -- word boundary
The "look behind" (lpeg.B) function changed from 0.10 to 0.11.
In 0.10, it received an explicit number to "go back" before the match,
and the default was one. In 0.11 and later versions, it uses the
"pattern length" as the length to go back. For instance,
lpeg.B("123") will try to match "123" three positions
behind the current one,
and lpeg.B("1") will go back one position.
In your code, BB goes back the size of B, which is zero.
Probably what you want is this:
- BB = lpeg.B(-(I + D)) + SOS -- starting boundary
+ BB = -peg.B(I + D) + SOS -- starting boundary
-- Roberto
```
• References: | 561 | 1,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-33 | latest | en | 0.716294 |
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