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RASESMA 2023 Nairobi
The Venue and How to reach it
The event will be hosted by the Technical and Applied Physics Department, school of Physics and Earth Science in lab N211 on the second floor of the N block, the first building on your left on
entering through the public entrance on Workshop Road. Please follow the following navigation map link to arrive a the university. Access to the university requires registration with security at the
gate, your destination will be N301, 3rd Floor, N-Block in the Physics Department, there is no access restrictions to enter the laboratory.
Setting up Yambo, QE and the tutorials
To be able to follow the school you need a running version of the yambo/QE codes, and the files and databases needed to run the tutorials.
Detailed instructions about the code(s) installation can be found in this dedicated page.
After installing the code you can setup the tutorial files following the instructions provided in the Tutorial files page.
Sunday 19 Feb
For those who will be already in Nairobi we will have a get together & Yambo installation session at the YMCA South C starting from 18:00.
If you are interested drop an E-mail at rasesma2023@tukenya.ac.ke
Monday 20 Feb
Time Lecture Speaker
09:00 - 09:20 Welcome & Introduction The Organizers
09:30 - 10:20 General Introduction to Density Functional Theory Omololu Akin-Ojo
10:20 - 11:10 Kohn-Sham, Exchange-Correlation functionals, approximations Korir Kipronoh
11:10 - 11:30 Break
11:30 - 12:20 Density Functional Theory in practice: Plane-Waves, pseudopotentials Omololu Akin-Ojo or Korir Kipronoh
12:20 - 13:30 Lunch
Tuesday 21 Feb
Wednesday 22 Feb
Thursday 23 Feb
Friday 24 Feb
Time Tutorial Tutor(s)
13:00 - 15:00 Students projects discussion The Organizers
15:00 - 15:30 Break
15:30 - 17:00 Students projects discussion The Organizers
After the School
Basic (and slightly advanced) studying material
Quantum Mechanics
• Harmonic oscillator
• Second quantization
• Electron-Phonon
• Photons
• (...)
1. Many-Particle Physics, G.D. Mahan, Chapter 1
2. Modern Quantum Mechanics, J.J. Sakurai, Chapter 1-2
• Ordinary differential equations (1st and 2nd order)
• Complex analysis
• Fourier and Laplace transformations
• Basic functional analysis
1. Mathematical methods in the physical sciences M.L. Boas
2. Time-Dependent Density Functional Theory: An Advanced Course, Appendix A, Engel, Eberhard / Dreizler, Reiner M.
Theoretical Approaches
• Many-Body Perturbation Theory
• Density Functional Theory
• Density Functional Perturbation Theory
• Time-Dependent Density Functional Theory
1. Many-Body Approach to Electronic Excitations F. Bechstedt
2. Electronic excitations: density-functional versus many-body Green's-function approaches G. Onida et al, Rev. Mod. Phys. 74 (2002)
3. Phonons and related crystal properties from density-functional perturbation theory, S. Baroni et al, Rev. Mod. Phys. 73 (2001)
4. Application of the Green's Functions Method to the Study of the Optical Properties of Semiconductors G. Strinati, La Rivista del Nuovo Cimento (1978-1999) volume 11, pages1–86 (1988)
5. Density Functional Theory: An Advanced Course, Engel, E. / Dreizler, R. M.
• Hartree-Fock
• Random-Phase Approximation (Linear Response)
• GW/BSE
1. Many-Body Approach to Electronic Excitations F. Bechstedt
2. Application of the Green's Functions Method to the Study of the Optical Properties of Semiconductors G. Strinati, La Rivista del Nuovo Cimento (1978-1999) volume 11, pages1–86 (1988)
The following exercises should be possible after having studied the basic concepts described described in the Studying Material section.
(Some) Exchange Programs
Here some Exchange programs where students and lecturers can apply. Please note that all these programs can be accessed via a selection. | {"url":"https://wiki.yambo-code.eu/wiki/index.php?title=RASESMA_2023_Nairobi","timestamp":"2024-11-02T02:55:31Z","content_type":"text/html","content_length":"38999","record_id":"<urn:uuid:f7570d62-a80f-4df5-8dde-ea35f8d58d83>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00403.warc.gz"} |
Interactive Math Activity - Multiplication with Arrays
Students will solve basic multiplication facts using their understanding of multiplication arrays in this interactive math game. Geared toward a third-grade math ability level, this online math
activity will challenge students to use arrays to solve multiplication facts. The math problems in this activity are presented in fill-in-the-blank format and multiple-choice format.
Here are a few examples of the math problems children may be asked to solve in this third grade multiplication activity: "Write a multiplication fact for the array below," for which students will
type "4 x 4 = 16" in the answer box; "Which set of objects shows an array?" for which students will choose the correct answer from the options shown; or a word problem, "Luke and Alison collected
rocks on the beach. When they got home, they set them up on the table in 8 rows. Each row had 8 rocks. How many rocks did they collect?" for which students will type in the answer, "64."
If students need a little extra help solving one of the math problems, they can click on the "Hint" button. They will be shown a helpful hint in the form of a multiplication array or written clue
that will point them in the right direction without giving away the answer. When students answer a question incorrectly, a detailed explanation page will show them the correct answer accompanied by
an easy-to-understand explanation.
This third grade multiplication game, as well as all the math activities on iKnowIt.com, includes built in features to help children make the most of their math practice session. For example, the
progress-tracker in the upper-right corner of the practice screen shows children how many questions they have answered out of the total number of questions in the lesson. The score-tracker beneath
that allows students to see how many points they have achieved for answering questions correctly. Opposite, the speaker icon indicates the read-aloud feature. Students can click on this button to
hear the question read out loud to them in a clear voice. It's an excellent resource for ESL/ELL students and children who are auditory processors. All of our lesson features were designed with your
students' success in mind.
I Know It Is Top Pick for Teachers and Students
Whether you are an elementary teacher, homeschool educator, or school administrator, we're hoping you will love using the I Know It math practice program in your elementary education journey!
Teachers enjoy integrating our program into their math lessons alongside a comprehensive curriculum for additional practice and concept reinforcement. The I Know It website features hundreds of math
practice activities for students from kindergarten through fifth grade. Each of our math lessons has been written by accredited elementary math teachers just like you to meet at least one (often
more) Common Core Standard. Math activities are arranged on our website by grade level and topic, making it easy for you to find and assign lessons in just a few clicks.
Children, too, love using the I Know It online math program to sharpen their math skills. When kids practice multiplication with arrays using this fun, engaging activity, we're sure they will love
the adorable, animated characters waiting to cheer them on in their math practice! The bright, kid-friendly design of our math lessons engages and delights, and plenty of positive feedback messages
encourage students to "Keep going!" even when they make mistakes. As a bonus, kids can earn math awards to add to their virtual trophy case for each new skill mastered in their practice sessions.
Learning math skills is fun and exciting with I Know It!
We hope you and your third-grade students will love practicing multiplication with arrays in this interactive math game! Be sure to check out the hundreds of other 3rd grade math lessons we have
available on our website as well.
Test out the I Know It Program for Free
Looking for a way to try out the I Know It math program with your class? We've got you covered! Sign up for a free thirty-day trial of I Know It and explore this multiplication arrays activity, as
well as any math lesson on our website, at no cost for a full thirty days! We're confident you and your students will love the difference interactive math practice can make! When your free trial runs
out, we hope you will consider joining the I Know It community as a member, so you can continue to enjoy the benefits of interactive math practice for a full calendar year. We have membership options
for families, individual teachers, schools, and school districts. Visit our membership information page for details: https://www.iknowit.com/order.html.
Your I Know It membership unlocks our website's handy administrative tools. You can access these features in your administrator account. They help you create a class roster for all your students,
assign unique login credentials to each of your students, give different math practice assignments to individual students, view your students' math practice progress with detailed reports, change
basic lesson settings, print, download, and email student progress reports, and much more.
When your students log into the I Know It website using their unique username and password, they will be shown a kid-friendly version of the homepage. From here they will be able to quickly find and
access the math activities you have assigned to them for practice. If you choose to give them the option through your administrator account, your students can also explore other math lessons at their
grade level and beyond for an added challenge or extra practice. Grade levels in the student mode of I Know It are labeled with letters instead of numbers, making it easy for you to assign math
lessons based on each child's needs and skill level.
This online math lesson is classified as Level C. It may be ideal for a third-grade class.
Common Core Standard
3.OA.1, MA.3.NSO.2.2, 3.4E
Operations And Algebraic Thinking
Represent And Solve Problems Involving Multiplication And Division.
Interpret products of whole numbers, e.g., interpret 5 x 7 as the total number of objects in 5 groups of 7 objects each. For example, describe a context in which a total number of objects can be
expressed as 5 x 7.
You might also be interested in...
Basic Multiplication (0-10) (Level C)
In this third grade-level math lesson, students will solve basic multiplication facts with factors from 0 to 10. Questions are presented in fill-in-the-blank format and multiple-choice format.
Basic Multiplication (0-5) (Level C)
In this math lesson geared toward third-grade, students will practice basic multiplication with factors from 0 to 5. Questions are presented in fill-in-the-blank format. | {"url":"https://www.iknowit.com/lessons/c-multiplication-with-arrays.html","timestamp":"2024-11-04T23:38:27Z","content_type":"text/html","content_length":"428163","record_id":"<urn:uuid:1739a799-b65b-48d1-918c-74736ca24184>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00541.warc.gz"} |
[SOLVED] Factors of 11 STEP by STEP Easy Method
Factors of 11 are 1, 11. Now understand the basic concept of factor and how to find the factors in easy way.
Factors are integers you multiply together to get another integer. Also remember that the factors always include 1 and itself.
NOTE: When finding the factors of a number, ask yourself “what numbers can be multiplied together to give me this number?.
How to find the factors of 11?
THINK: What pairs of numbers can be multiplied together to give me 11?
Step 1: 1 x 11 = 11, so put these numbers in the factor list.
Step 2: Take 2 and divide with 11. The remainder will be 1. But factors always give 0 remainder. This means 2 is NOT a factor of 11.
Step 3:Take 3 and divide with 11. The remainder will be 2. But factors always give 0 remainder. This means 3 is NOT a factor of 11.
Step 4:Take 4 and divide with 11. The remainder will be 3. But factors always give 0 remainder. This means 4 is NOT a factor of 11.
Step 5: Take 5 and divide with 11. The remainder will be 1. But factors always give 0 remainder. This means 5 is NOT a factor of 11.
Same way you can check all the numbers upto 11. So, factors of 11 is 1, 11.
TIPS: Try to keep tables (at least till 20) on your finger tips. Revise tables every day
Is 11 a prime or composite number?
Yes, 11 is a prime number because it has only two distinct divisors: 1 and 11 (itself).
What is prime number – An integer number that is divisible by two distinct integers, 1 and itself.
What is the prime factorization of 11
The prime factorization of 11 is 1 × 11. The number 11 is a prime number, because its only factors are 1 and itself.
Factors are often given as pairs of numbers, which multiply together to give the original number. These are called factor pairs
What is factor pairs: Factor pairs are combinations of two factors that multiply together to give the original number.
So for 11, there are 4 factor pairs
1 × 11 = 11
11 × 1 = 11
-1 × -11 = 11
-11 × -1 = 11
No, 11 is not a square number.
The square root of 11 is 3.317.
The square of 11 is 121.
Hope you learned to solve the factors of 11.
Factors of another number | {"url":"https://smartclass4kids.com/factors-of-11/","timestamp":"2024-11-07T22:19:57Z","content_type":"text/html","content_length":"208503","record_id":"<urn:uuid:28fb437d-74a9-40a7-b44c-7a6d556d04d9>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00748.warc.gz"} |
The unreasonable effectiveness of mathematics in large scale deep learning
# 152
Recently, the theory of infinite-width neural networks led to the first technology, muTransfer, for tuning enormous neural networks that are too expensive to train more than once. For example, this
allowed us to tune the 6.7 billion parameter version of GPT-3 using only 7% of its pretraining compute budget, and with some asterisks, we get a performance comparable to the original GPT-3 model
with twice the parameter count. In this talk, I will explain the core insight behind this theory. In fact, this is an instance of what I call the *Optimal Scaling Thesis*, which connects
infinite-size limits for general notions of 'size' to the optimal design of large models in practice. I'll end with several concrete key mathematical research questions whose resolutions will have
incredible impact on the future of AI.
Greg Yang, Microsoft Research
Greg Yang is a researcher at Microsoft Research in Redmond, Washington. He joined MSR after he obtained Bachelor's in Mathematics and Master's degrees in Computer Science from Harvard University,
respectively advised by ST Yau and Alexander Rush. He won the Hoopes prize at Harvard for best undergraduate thesis as well as Honorable Mention for the AMS-MAA-SIAM Morgan Prize, the highest honor
in the world for an undergraduate in mathematics. He gave an invited talk at the International Congress of Chinese Mathematicians 2019. | {"url":"https://cni.iisc.ac.in/seminars/2023-07-10/","timestamp":"2024-11-08T14:29:16Z","content_type":"text/html","content_length":"13844","record_id":"<urn:uuid:914bfd4e-6b01-4732-80b0-b22aed566325>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00799.warc.gz"} |
10-10-08 : On the Art of Good Arithmetic Coder Use
On the Art of Good Arithmetic Coder Use
I think it's somewhat misleading to have these arithmetic coder libraries. People think they can take them and just "do arithmetic coding". The reality is that the coder is only a small part of the
arithmetic coding process, and it's the easiest part. The other parts are subtle and somewhat of an art.
The other crucial parts are how you model your symbol, how you present symbols to the coder, how you break up the alphabet, and different type of adaptation.
If you have a large alphabet, like 256 symbols or more, you don't want to just code with that alphabet. It has few problems. For one, decoding is slow, because you need to do two divides, and then
you have to binary search to find your symbol from a cumulative probability. Aside from the speed issue, you're not compressing super well. The problem is you have to assign probabilities to a whole
mess of symbols. To get good information on all those symbols, you need to see a lot of characters to code. It's sort of like the DSP sampling problem - to get information on a big range of
frequencies in audio you need a very large sampling window, which makes your temporal coherence really poor. In compression, many statistics change very rapidly, so you want quick adaptation, but if
you try to do that on a large alphabet you won't be gathering good information on all the symbols. Peter Fenwick has some good old papers on multi-speed adaptation by decomposition for blocksorted
In compression most of our alphabets are highly skewed, or at least you can decompose them into a highly skewed portion and a flat portion. Highly skewed alphabets can be coded very fast using the
right approaches. First of all you generally want to sort them so that the most probable symbol is 0, the next most probable is 1, etc. (you might not actually run a sort, you may just do this
conceptually so that you can address them in order from MPS to LPS). For example, blocksort output or wavelet coefficients already are sorted in this order. Now you can do cumulative probability
searches and updates much faster by always starting at 0 at summing towards 0, so that you rarely walk very far into the tree.
You can also decompose your alphabet into symbols that more accurately represent its "nature" which will give you better adaptation. The natural alphabet is the one that mimics the actual source that
generated your code stream. Of course that's impossible to know, but you can make good guesses. Consider an example :
The source generates symbols thusly :
It chooses Source1 or Source2 with probability P
Source1 codes {a,b,c,d} with fixed probabilities P1(a),P1(b),...
Source2 codes {e,f,g,h} with fixed probabilities P2(e),P2(f),...
The probability P of sources 1 and 2 changes over time with a semi-random walk
After each coding event, it either does P += 0.1 * (1 - P) or P -= 0.1 * P
If we tried to just code this with an alphabet {a,b,c,d,e,f,g,h} and track the adaptation
we would have a hell of a time and not do very well.
Instead if we decompose the alphabet and code a binary symbol {abcd,efgh} and then code
each half, we can easily do very well. The coding of the sub-alphabets {a,b,c,d} and
{e,f,g,h} can adapt very slowly and gather a lot of statistics to learn the probabilities
P1 and P2 very well. The coding of the binary symbol can adapt quickly and learn the
current state of the random decision probability P.
This may seem rather contrived, but in fact lots of real world sources look exactly like this. For example if you're trying to compress HTML, it switches between basically looking like English text
and looking like markup code. Each of those is a seperate set of symbols and probabilities. The probabilites of the characters within each set are roughly constantly (not really, but they're
relatively constant compared to the abruptness of the switch), but where a switch is made is random and hard to predict so the probability of being in one section or another needs to be learned very
quickly and adapt very quickly.
We can see how different rates of adaptation can greatly improve compression.
Good decomposition also improves coding speed. The main way we get this is by judicious use of binary coding. Binary arithmetic coders are much faster - especially in the decoder. A binary arithmetic
decoder can do the decode, modeling, and symbol find all in about 30 clocks and without any division. Compare that to a multisymbol decoder which is around 70 clocks just for the decode (two
divides), and that doesn't even include the modeling and symbol finding, which is like 200+ clocks.
Now, on general alphabets you could decompose your multisymbol alphabet into a series of binary arithmetic codes. The best possible way to do this is with a Huffman tree! The Huffman tree tries to
make each binary decision as close to 50/50 as possible. It gives you the minimum total code length in binary symbols if you just wrote the Huffman codes, which means it gives you the minimum number
of coding operations if you use it to do binary arithmetic coding. That is, you're making a binary tree of coding choices for your alphabet but you're skewing your tree so that you get to the more
probable symbols with fewer branches down the tree.
(BTW using the Huffman tree like this is good for other applications. Say for example you're trying to make the best possible binary search tree. Many people just use balanced trees, but that's only
optimal if all the entries have equal probability, which is rarely the case. With non-equal probabilities, the best possible binary search tree is the Huffman tree! Many people also use
self-balancing binary trees with some sort of cheesy heuristic like moving recent nodes near the head. In fact the best way to do self-balancing binary trees with non equal probabilities is just an
adaptive huffman tree, which has logN updates just like all the balanced trees and has the added bonus of actually being the right thing to do; BTW to really get that right you need some information
about the statistics of queries; eg. are they from a constant-probability source, or is it a local source with very fast adaptation?).
Anyhoo, in practice you don't really ever want to do this Huffman thing. You sorted your alphabet and you usually know a lot about it so you can choose a good way to code it. You're trying to
decompose your alphabet into roughly equal probability coding operations, not because of compression, but because that gives you the minimum number of coding operations.
A very common case is a log2 alphabet. You have symbols from 0 to N. 0 is most probable. The probabilities are close to geometric like {P,P^2,P^3,...} A good way to code this is to write the log2 and
then the remainder. The log2 symbol goes from 0 to log2(N) and contains most of the good information about your symbol. The nice thing is the log2 is a very small alphabet, like if N is 256 the log2
only goes up to 9. That means coding it is fast and you can adapt very quickly. The remainder for small log2's is also small and tracks quickly. The remainder at the end is a big alphabet, but that's
super rare so we don't care about it.
Most people now code LZ77 offsets and lengths using some kind of semi-log2 code. It's also a decent way to code wavelet or FFT amplitudes. As an example, for LZ77 match lengths you might do a
semi-log2 code with a binary arithmetic coder. The {3,4,5,6} is super important and has most of the probability. So first code a binary symbol that's {3456} vs. {7+}. Now if it's 3456 send two more
binary codes. If it's {7+} do a log2 code.
Another common case is the case that the 0 and 1 are super super probable and everything else is sort of irrelevant. This is common for example in wavelets or DCT images at high compression levels
where 90% of the values have been quantized down to 0 or 1. You can do custom things like code run lengths of 0's, or code binary decisions first for {01},{2+} , but actually a decent way to
generally handle any highly skewed alphabet is a unary code. A unary code is the huffman code for a geometric distribution in the case of P = 50% , that is {1/2,1/4,1/8,1/16,...} ; we code our
symbols with a series of binary arithmetic codings of the unary representation. Note that this does not imply that we are assuming anything about the actual probability distribution matching the
unary distribution - the arithmetic coder will adapt and match whatever distribution - it's just that we are optimal in terms of the minimum number of coding operations only when the probability
distribution is equal to the unary distribution.
In practice I use four arithmetic coder models :
1. A binary model, I usually use my rung/ladder but you can use the fixed-at-pow2 fractional modeller too.
2. A small alphabet model for 0-20 symbols with skewed distribution. This sorts symbols from MPS to LPS and does linear searches and probability accumulates. It's good for order-N adaptive context
modeling, N > 0
3. A Fenwick Tree for large alphabets with adaptive statistics. The Fenwick Tree is a binary tree for cumulative probabilities with logN updates. This is what I use for adaptive order-0 modeling, but
really I try to avoid it as much as possible, because as I've said here, large alphabet adaptive modeling just sucks.
4. A Deferred Summation semi-adaptive order-0 model. This is good for the semi-static parts of a decomposed alphabet, such as the remainder portion of a log2 decomposition.
Something I haven't mentioned that's also very valuable is direct modeling of probability distributions. eg. if you know your probabilites are Laplacian, you should just model the laplacian
distribution directly, don't try to model each symbol's probability directly. The easiest way to do this usually is to track the average value, and then use a formula to turn the average into
probabilities. In some cases this can also make for very good decoding, because you can make a formula to go from a decoded cumulative probabilty directly to a symbol.
ADDENDUM BTW : Ascii characters actually decompose really nicely. The top 3 bits is a "selector" and the bottom 5 bits is a "selection". The probabilities of the bottom 5 bits need a lot of accuracy
and change slowly, the probabilities of the top 3 bits change very quickly based on what part of the file you're in. You can beat 8-bit order0 by doing a separated 3-bit then 5-bit. Of course this is
how ascii was intentionally designed :
0 1 2 3 4 5 6 7 8 9 A B C D E F 0 1 2 3 4 5 6 7 8 9 A B C D E F
0 NUL SOH STX ETX EOT ENQ ACK BEL BS HT LF VT FF CR SO SI DLE DC1 DC2 DC3 DC4 NAK SYN ETB CAN EM SUB ESC FS GS RS US
1 SP ! " # $ % & ' ( ) * + , - . / 0 1 2 3 4 5 6 7 8 9 : ; < = > ?
2 @ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z [ \ ] ^ _
3 ` a b c d e f g h i j k l m n o p q r s t u v w x y z { | } ~ DEL
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Calculating confidence intervals for Cohen’s d and eta-squared using SPSS, R, and Stata
[Now with update for STATA by my colleague
+Chris Snijders
[Now with update about using the MBESS package for within-subject designs]
[Now with an update on using ESCI]
Confidence intervals are confusing intervals. I have nightmares where my students will ask me what they are, and then I try to
them, and I mumble something nonsensical, and they all point at me and laugh.
Luckily, I have had extensive training in reporting statistics I don’t understand completely when I studied psychology, so when it comes to simply reporting confidence intervals, I’m fine. Although
these calculations are really easy to do, for some reason I end up getting a lot of e-mails about them, and it seems people don’t know what to do to calculate confidence intervals for effect sizes.
So I thought I’d write one clear explanation, and save myself some time in the future. I’ll explain how to do it in 2 ways, the first using SPSS, the second using R, the third by my colleague Chris
Snijders on using Stata, and some brief comments about using ESCI.
CI for eta-squared in SPSS
First, download
from the website of
Karl L Wuensch
. His website is extremely useful (the man deserves an award for it) especially for the step-by-step guides he has created. The explanations he has written to accompany the files are truly excellent
and if this blog post is useful, credit goes to Karl Wuensch.
This example focusses on designs where all factors in your ANOVA are fixed (e.g., manipulated), not random (e.g., measured), in which case you need to go
. All you need to do is open NoncF.sav (which refers to the non-central
-distribution, for an introduction, see the
OSC blog
), fill in some numbers in SPSS, and run a script. You’ll see an empty row of numbers, except .95 in the conf column (which happens to be a value you probably don’t want to use, see the end of this
Let’s say you have the following in your results section: F(1,198) = 5.72. You want to report partial η² and a confidence interval around it. You type in the F-value 5.72 in the first column, and
then the degrees of freedom (1 in the second column, 198 in the third), and you change .95 into .90 (see below for the reason). Then, you just open NoncF3.sps, run the script, and you get the output
in the remaining columns of your SPSS file:
We are only interested in the last three columns. From the r2 column, we get r² or η², = .028, with the lower (lr2) and upper (ur2) values of the confidence interval to the right that give us 90% CI
[.003; .076]. Easy peasy.
CI for eta-squared in R (or R Studio)
I’m still not very good with R. I use it as a free superpowered calculator, which means I rarely use it to analyze my data (for which I use SPSS) but I use it for stuff SPSS cannot do (that easily).
To calculate confidence intervals, you need to install the MBESS package (installing
and MBESS might take less time than starting up SPSS, at least on my computer).
To get the confidence interval for the proportion of variance (r², or η², or partial η²) in a fixed factor analysis of variance we need the ci.pvaf function. MBESS has loads more options, but you
just need to copy paste and run:
ci.pvaf(F.value=5.72, df.1=1, df.2=198, N=200, conf.level=.90)
This specifies the F-value, degrees of freedom, and the sample size (which is not needed in SPSS), and the confidence level (again .90, and not .95, see below). You’ll get the following output:
Here we see the by now familiar lower limit and upper limit (.003 and .076). Regrettably, MBESS doesn’t give partial η², so you need to request it in SPSS (or you can use my
effect size spreadsheet
I found out that for within designs, the MBESS package returns an error. For example:
Error in ci.pvaf(F.value = 25.73, df.1 = 2, df.2 = 28, N = 18, conf.level = 0.9) : N must be larger than df.1+df.2
This error is correct in between-subjects designs (where the sample size is larger than the degrees of freedom) but this is not true in within-designs (where the sample size is smaller than the
degrees of freedom for many of the tests). Thankfully, Ken Kelley (who made the MBESS package) helped me out in an e-mail by pointing out you could just use the R Code within the ci.pvaf function and
adapt it. The code below will give you the same (at least to 4 digits after the decimal) values as the Smithson script in SPSS. Just change the F-value, confidence level, and the df.1 and df.2.
CI for Cohen’s d in SPSS
Karl Wuensch adapted the files by Smithson (2001) and created
a zip file to compute effect sizes around Cohen’s d
which works in almost the same way as the calculation for confidence intervals around eta-squared (except for a dependent
). Open the file NoncT.sav. You’ll again see an almost empty row where you only need to fill in the
-value and the degrees of freedom. Note that (as explained in Wuensch’s help file) there’s a problem with the SPSS files if you fill in a negative
-value, so fill in a positive
-value, and reverse the signs of the upper and lower CI if needed.
If you have a t-test that yielded t(198) = 2.39, then you fill in 2.39 in the first column, and 198 in the second column. For a one-sample t-test this would be enough, in a two-sample t-test you need
to fill in the sample sizes n1 (100 participants) and n2 (100 participants). Open T-D-2samples.sps and run it. In the last three columns, we get Cohen’s d (0.33) and the upper and lower limits, 95%
CI [0.06, 0.62].
CI for Cohen’s d in R
In MBESS, you can calculate the 95% confidence interval using:
ci.smd(ncp=2.39, n.1=100, n.2=100, conf.level=0.95)
The ncp (or non-centrality parameter) sounds really scary, but it’s just the t-value (in our example, 2.39). n.1 and n.2 are the samples sizes in both groups. You’ll get the following output:
Yes, that’s really all there is to it. The step-by step guides by Wuensch, and the help files in the MBESS package should be able to help you if you run into some problems.
Why should you report 90% CI for eta-squared?
Again, Karl Wuensch has gone out of his way to explain this in
a very clear document, including examples, which you can find here
. If you don’t want to read it, you should know that while Cohen’s
can be both positive and negative, r² or η² are squared, and can therefore only be positive. This is related to the fact that
-tests are always one-sided (so no, don’t even think about dividing that
= .08 you get from an
-test by two and reporting
= .04,
). If you calculate 95% CI, you can get situations where the confidence interval includes 0, but the test reveals a statistical difference with a
< .05. For a paper by Steiger (2004) that addresses this topic,
. This means that a 95% CI around Cohen's
equals a 90% CI around
for exactly the same test. Furthermore, because eta-squared cannot be smaller than zero, a confidence interval for an effect that is not statistically different from 0 (and thus that would normally
'exclude zero') necessarily has to start at 0. You report such a CI as 90% CI [.00; .XX] where the XX is the upper limit of the CI.
Confidence intervals are confusing intervals.
@Stata afterburner - a confidence interval around eta-squared
by Chris Snijders
In the fall semester I will be teaching a statistics class with Daniel. I have sleepless nights already about what that will do to my evaluations – who wants to appear old and retarded next to such
youthful brilliance? [Note that Chris has little to worry about - he was chosen as the best teacher at Eindhoven University of Technology a few years ago - DL]. As Daniel is SPSS and R, and I am just
Stata, this also implies some extra work: trying to translate Daniel’s useful stuff to Stata - our students have had their basic course in Stata and I want them to be able to keep up. Luckily, in the
case of (confidence intervals around) eta-squared I have an easy way out: Stata 13 includes several effect size measures by default.
The below example is heavily “inspired” by the Stata manual. It shows how to get eta-squared and its confidence interval in Stata. First, read in your data. I use the standard apple data set from
use http://www.stata-press.com/data/r13/apple
The details of the data do not matter here. Running an Anova goes like this:
anova weight treatment (output omitted)
To get eta-squared and its confidence interval, type
estat esize
to get
Effect sizes for linear models
Source | Eta-Squared df [95% Conf. Interval]
Model | .9147383 3 .4496586 .9419436
treatment | .9147383 3 .4496586 .9419436
If you insist on 90% confidence interval, as does Daniel, type:
estat esize, level(90)
Stata does several additional things too, such as calculating bootstrapped confidence intervals if you prefer, or calculate effect sizes directly.
Useful additional info can be found at
ESCI update
For people who prefer to use ESCI software by Cumming, please note that ESCI also has an option to provide 95% CI around Cohen's d, both for independent as for dependent t-tests. However, the option
is slightly hidden - you need to scroll to the right, where you can check a box which is placed out of view. I don't know why such an important option is hidden away like that, but I've been getting
a lot of e-mails about this, so I've added a screenshot below to point out where you can find it. After clicking the check box, a new section appears on the left that allows you to calculate 95% CI
around Cohen's d (see second screenshot below)
I think you should report confidence intervals because calculating them is pretty easy and because they tell you something about the accuracy with which you have measured what you were interested in.
I don’t feel them in the way I feel p-values
, but that might change. Psychological Science is pushing confidence intervals as ‘new’ statistics, and I’ve seen some bright young minds being brainwashed to stop reporting
-values altogether and
report confidence intervals. I’m pretty skeptical about the likelihood this will end up being an improvement. Also, I don’t like brainwashing. More seriously, I think we need to stop kidding
ourselves that statistical inferences based on a single statistic will ever be enough. Confidence intervals, effect sizes, and
-values (all of which can be calculated from the test statistics and degrees of freedom) present answers to different but related questions. Personally, I think it is desirable to report as many
statistics that are relevant to your research question as possible.
Smithson, M. (2001). Correct confidence intervals for various regression effect sizes and parameters: The importance of noncentral distributions in computing intervals. Educational And Psychological
Measurement, 61(4), 605-632. doi:10.1177/00131640121971392^
32 comments:
1. Hi Daniel,
as always I'm not satisfied. Consider this as a question from a grad student in the first row :)
The computation you outline is all nice and good but how do I interpret the effect size? how do I interpret the eta^2 in particular? What am I supposed to write in my paper.
Consider a study which varies the room temperature and measures the reaction time in a visual detection task. We have four groups with temperature 20,22,24,26 degrees. We perform an Anova and
discover that eta^2 is 0.28 with [.03; .76] CI. How am I supposed to interpret this quantity? You tell me about intuition building. But is not quantitative analysis supposed to be objective? What
if other researcher builds different intuition than I did? Do we need standards/rules? What will they look like?
Actually, some insight can be gained by considering the computation of eta^2. This is SS of the temperature divided by total SS. But wait, this means that the effect size can be artificially
bloated. We just need to increase the variance due to temperature. We can do this by adding levels to our manipulation (e.g. another group with 18 deg). Or by spacing the groups further apart
(e.g. 15,20,25,30 deg). So it seems we need different intuition/rules for different designs, different number of levels and in fact different choice of group spacing. This is an overkill. If
don't think this can work.
My solution is to avoid any standardized (and especially squared) quantities and provide estimates of causal effects. In the above case we are interested in the regression coefficient that
predicts reaction time from temperature. Then I can say that the reaction time increases by 40 [10,70] CI ms per degree Celsius. Why can't we report this? Any layman understands this quantity and
the CI. We don't need any special intuitions/rules. This quantity is independent of the number of groups and the temperature spacing.
Once we set the estimation of causal effects as the goal of our analysis, p-values and hypothesis testing becomes redundant and can be disposed of without loss. I encourage everyone to do this.
1. Hi Matus, keep your comments coming, they're great! You are right that standardized effect sizes are less interesting when you can interpret scales in a meaningful manner. In psychology, this
is sometimes difficult. What does the difference between 5 and 6 on a 7-point scale measuring self-confidence mean?
Moreover, if you want to compare theoretically similar effects, measured using different paradigms, standardized effect sizes help. We don't have fixed scales we are measuring on (like
temperature). That's why meta-analyses practically always use standardized effect sizes.
That's not to say this situation is desirable. it's just difficult to change. Someone once pointed me to Rasch models as a preferable alternative : https://www.researchgate.net/publication/
45185202_Effect_sizes_can_be_misleading_is_it_time_to_change_the_way_we_measure_change?ev=prf_pub. However, I'm not too good at Rasch models.
Also, people have remarked we should only use estimation, but that argument has (I think) successfully been countered (e.g., http://www.ejwagenmakers.com/inpress/
2. "What does the difference between 5 and 6 on a 7-point scale measuring self-confidence mean?" Self-confidence is a psychological construct and on it's own it doesn't mean anything. Rather it
provides a (causal) link between the actual quantities. If study A shows an increase of self-confidence by 0.2 points due to treatment and study B shows that increase in self-confidence by 1
point decreases the BDI score (depression) by 50 %, then we can say that treatment decreases the BDI score by 10%. This of course means that study A and B use the same tool to measure
self-confidence. That's why clinicians developed standardized tools such as BDI and DSM - if every researcher uses different tool to measure the success of treatment how should we compare the
different treatments? Researchers need to invest time to develop reliable and valid tools if they wish to introduce some psychological construct. Otherwise we can't really say what they are
I've saw the comment by Morey et al. (There is also a response by Cumming somewhere already.) I remain unconvinced. Actually I'm currently summing my view in a blog post that will target their
criticism. Though, I think Morey et al. are right when they say that new statistics implies abandonment of hypothesis testing. This is a bullet that Cumming et al. will have to bite at some later
point if they want to remain true to their principles.
In general, a criticism of estimation in press is rare (although I can imagine it is rampant behind the scenes), so if you know more such cases just throw it at me. Thanks!
1. Sorry, that was meant as response to Daniel, June 7, 2014 at 6:03 AM
3. I think you linked the CI for Cohen’s d in SPSS zip to your local folder instead of where it's hosted online!
1. Thanks! Must be due to the fact I prepared it in Word - was wrong in 3 other places, fixed them all now I hope. Thanks for letting me know.
4. This comment has been removed by the author.
5. Hi Daniel, nice post. Just a suggestion: there is nothing wrong in dividing by 2 the p.value of the F test and report it one-side, if one knowns what s/he is doing and explain why.
6. Hi Daniel,
I'm just wondering if any these programs can calculate effect sizes/95% CIs for planned contrasts? My research question is not concerned with the overall one-way ANOVA.
Thank you,
1. Hi Lisa, I'm pretty sure you have either an F-test or t-test. So see above.
2. Ok thanks. Appreciate the re-assurance that I am in the right place!
7. Hi Daniel,
First of all, congratulation (and thanks) for the interesting and very useful work on methods and stats that you have been producing. Your effect sizes spreadsheets are lifesaving.
So, I have question regarding the ESCI software and the calculation of the 95% CI around Cohen's d (one sample): I can't even find the ESCI module where I can preform this. I searched in all the
modules (at least I believe I did) and can't find nothing that looks like the screenshot you have above. Can you please tell me where I can ask for this? Basically I would like to calculate the
95% CI around Cohen's dz, for my results vs. published results.
Thanks for taking the time to reply.
Tomás Palma
1. Hi, ESCI does not do a one-sample t-test. I'm not sure if it is formally correct to do a dependent t-test, but set all the values in the second condition to 0 - I think that should work, but
I think you should be able to see if it works. Let me know if it does.
8. Hi Daniel,
Thanks for the fast reply. Let me give you a bit more of details because I now realize that maybe my question was not very clear. What I have is a 2x3 repeated measures design and I'm testing for
differences between specific pairs of cells via contrasts. I read in your paper (Lakens, 2013, p. 8, Frontiers in Psych) that the Cohen's dz 95% CI can be calculated with the ESCI (Cumming and
Finch, 2005). I downloaded the ESCI module that accompanies the paper Cumming's paper but I can't find the options to calculate this CI. I calculated the Cohen's dz (and other effect sizes) in
your spreadsheet but I would like to have the CI too.
Thanks again!
9. Hi Daniel,
Thank you for the great blog! Do you know if there is a way to calculate CI around Cramer's V. I looked at the MBESS package and there is a function conf.limits.nc.chisq but it doesn't work for
me (says effect size too small). Chisq = 2.39, N=66, 2x2. Any suggestions what I should do?
10. Hi
This is really interesting. I am a little in the dark about the CIs around the effect in the repeated-measures example. Given that eta square and partial eta square are different what is it
exactly that CI is respect to.
Thanks in anticipation,
1. the CI is around partial eta squared.
11. The ci.smd() can only really be used for between-subject design. It is simple to use however my data is within-subjects. For the life of me I cannot find how to calculated the CI around Cohen's D
for within-subject (paired) data. Is there any adjustment to the ci.smd that can be made?
1. Hi, check out my code here for how I calculate CI for a within design using MBESS in R: https://github.com/Lakens/perfect-t-test/blob/master/Perfect_dependent_t-test.Rmd
2. This comment has been removed by the author.
3. Dear Daniel,
Thank you for this informative page! it really helps make things clearer.
Regarding the question mentioned here about within-subject designs; I found that when I calculate Cohen's d using the MBESS package, as you suggested, I get the value of the effect size you
termed "Cohen's d average" which is not influenced by the correlation between my paired observations (I used the "effect size calculation" spreadsheet attached to your great Frontiers in
Psychology paper from 2013). So, I assume this is how they calculate Cohen d in the package (right?).
Do you think its ok to report this value? or should i try to convert it in someway to a within-design?
This is how I performa the calculation in R:
cohend <- smd (Mean.1=mean_group1, Mean.2=mean_group2, s.1=sd_group1, s.2=sd_group2, n.1=23, n.2=23)
ci.smd (smd=cohend, n.1=23, n.2=23, conf.level=.95)
Thank you!!
12. Hi Daniel, thanks for this page. I'm trying to learn about the CI's. Just to make sure I'm doing the right things: 1) don't look at the LL and UL's SPSS provides, because these refer to means and
differences between the means, while the LL and UL's for the effects sizes are something completely different. 2) Not all parameters, e.g., the observed power, will perfectly overlap between the
F-test results in SPSS and the results from the Weunsch syntax, because the former is based on the .05 alpha level, and the latter is based on the 90%CI.
Regards, and thanks for the help,
13. This comment has been removed by the author.
14. hi Daniel, Thank you for this wonderful page. I had a problem while running the syntax, it gives me this:
>Error # 4070. Command name: END IF
>The command does not follow an unclosed DO IF command. Maybe the DO IF
>command was not recognized because of an error. Use the level-of-control
>shown to the left of the SPSS Statistics commands to determine the range of
>LOOPs and DO IFs.
>Execution of this command stops.
I used the syntax before and it run well, I don't know whats wrong this time, can you please help? Thanks a lot!
15. Hi, Daniel, Thank you for keep updating this post.
You mentioned that for with-in subject design, the code of MBESS give the confidence interval of ANOVA was same as Smithson script in SPSS. But Smithson's script calculated the CI for partial eta
squared, instead of generalized eta squared (I have check it by using you excel sheet).
So this means that for the generalized eta squared of repeated-measure ANOVA, we still have no idea how to calculated it ?
1. Hi, indeed, I don't have the formula's for generalized eta-squared (although you should use omega squared! http://daniellakens.blogspot.nl/2015/06/why-you-should-use-omega-squared.html). It's
on my to do list, but CI are not a priority for me at the moment.
2. Hi, Daniel, Thank you for your reply and link!
16. Hello-
Pardon my ignorance. I have a partial eta2 of .014, and CI ranging from .00 to .016. This asymmetry seems weird. Did I mess up? The reviewers thought so.
17. Hello Daniel,
First of all, thanks a lot for your efforts to help people with their statistical problems. I think that both this blog and your publications (especially your 2013 paper on effects sizes) are
extremely helpful.
I have the following question: I calculated a 2 x 4 x 3 x 3 MANOVA with two within-subject factors and two between-subjects factors. Importantly, I only have one dependent variable. So I could
have calculated a mixed-design ANOVA as well, but I decided to use the multivariate tests (Pillai's trace) provided by SPSS to circumvent the problem of violated sphericity.
Can the packages you describe above (CI-R2-SPSS and MBESS) also calculate confidence intervals around partial eta squared in my MANOVA design or are these algorithms limited to ANOVAs?
Best regards,
18. Great post.
19. Hi, thank you very much for this page, this is very helpful!
I used the SPSS script to calculate the CIs for eta squared in a MANOVA.
However, in some cases, mostly for the main effects in the MANOVA, I obtained an eta squared that was not covered by the CI: For instance I had F (34, 508) = 1.72, partial η2 =.103, 90% CI =
[.012; .086].
Is it possible that the multivariate design causes the problem here? And would you have any suggestions on how to fix this?
Thanks a lot and best regards,
20. >(so no, don’t even think about dividing that p = .08 you get from an
>F-test by two and reporting p = .04, one-sided) | {"url":"https://daniellakens.blogspot.com/2014/06/calculating-confidence-intervals-for.html?showComment=1410207773182","timestamp":"2024-11-12T02:46:10Z","content_type":"application/xhtml+xml","content_length":"148950","record_id":"<urn:uuid:31727657-d618-4081-af98-16a7309e224e>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00669.warc.gz"} |
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Project Example: Network-based kernel for the analysis of genome-wide association studies
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inverter is filtered, most of the high frequency harmonics are filtered out, which has little impact on the power system.
Electromagnetic Transient Simulation Research on Operation Characteristics of Power Grid with Large-Scale New Energy
Accelerating the connection of wind power, photovoltaic and other new energy units to the grid, new power system presents more and more technical characteristics of high proportion of renewable
energy and high proportion of power electronic equipment. The internal mechanism of its stability changes, which brings new challenges to security and stability
Electromagnetic Transient Equivalent Modeling Method of MMC with Supercapacitor-based Energy Storage
MMC-ESS(modular multilevel converter with energy storage system) has broad prospects on engineering application in the field of renewable energy consumption. However, MMC with higher levels has the
problem of low efficiency in EMT(electromagnetic transient) simulation on offline simulation platforms such as PSCAD/EMTDC and Simulink, which
Introduction to Electromagnetic Transient Analysis of Power Systems | part of Transient Analysis of Power
<P>The analysis and simulation of electromagnetic transients has become a fundamental methodology for understanding the performance of power systems, determining power component ratings, explaining
equipment failures or testing protection devices. Power system transients can be electromagnetic, when it is necessary to
Research on modeling technology of PV/storage integrated
The proposed modeling method for electromagnetic transient simulation of PV/storage power stations can iteratively adjust the model parameters related to short-circuit
Active Reactive Power Control Strategy Based on Electrochemical Energy
Abstract: In order to resolve the key problem of continuous rectification fault, this paper proposes a joint control strategy based on electrochemical energy storage power station. Firstly, the
influence of commutation failure on the AC system was analyzed, and a mathematical model with the minimum power grid fluctuation as the objective function
GB/T 42716-2023 English PDF (GBT 42716-2023).
Standards related to: GB/T 42716-2023 GB/T 42716-2023 GB NATIONAL STANDARD OF THE PEOPLE''S REPUBLIC OF CHINA ICS 27.180 CCS F 19 Guide for Modeling of Electrochemical Energy Storage Power Station
ISSUED ON: MAY 23, 2023 IMPLEMENTED ON: DECEMBER 1, 2023 Issued by: State Administration for Market
Research on Calculation Method of Energy Storage Capacity Configuration for Primary Frequency Control of Photovoltaic Power Station
An energy storage capacity allocation method is proposed to support primary frequency control of photovoltaic power station, which is difficult to achieve safe and stable operation after a high
proportion of photovoltaic connected to public power grid. In this paper, by
Simulation of Transient Overvoltage Under Different Operation Modes of Disconnectors in 500 kV Pumped Storage Power Station
In order to calculate and analyze the VFTO (ultra-fast transient overvoltage) and influencing factors of 500kV GIS Substation of a pumped storage power station when the motor is operated by
disconnector at different positions under the mode of generator and motor operation, the circuit simulation model is established by using EMTP-ATP
The energy storage mathematical models for simulation and comprehensive analysis of power
With increasing power of the energy storage systems and the share of their use in electric power systems, their influence on operation modes and transient processes becomes significant. In this case,
there is a need to take into account their properties in mathematical models of real dimension power systems in the study of
Control and simulation of STATCOM integrated energy storage
STATCOM integrated energy storage system can realize the coordinated control of active power and reactive power, that is, the system can be compensated in four quadrants, which can quickly compensate
the active power and reactive power required by the system, flexibly solve some power quality problems in the power system, and
Frontiers | Electromagnetic transient characteristics
New energy systems, such as wind power and photovoltaic and distributed power supply, in the power generation and grid-connected transmission system basically will use stronger insulation design,
Is electromagnetic transient modelling and simulation of large power
Situations prompting the need for electromagnetic transient models The Australian Energy Market Operator (AEMO) has been using EMT simulation models for several years, including for black start
studies, sub-synchronous control interactions between series compensated lines and IBRs, and stability analysis of one to two remote
Very fast over voltage simulation and analysis in 500kV GIS substation of Xiangshuijian pumped storage power station
three kinds of digital programs used in power system simulation are summarized in this paper. The electromagnetic and electromechanical transient simulation software ADPSS is introduced
Large-Scale Renewable Energy Transmission by HVDC: Challenges
The analysis methods and design principles of traditional power systems are no longer suitable for HPPESs. In this paper, the mechanisms of broadband oscillation and transient over-voltage are
revealed, and analytical methods are proposed for HPPESs, including small-signal impedance analysis and electromagnetic transient simulation.
Parallel electromagnetic transient simulation of power
Inspired by the latency insertion method used in the design of large-scale integrated circuits, this paper proposes a fine-grained parallel modelling and simulation method for the main components in
the power
Modelling and simulation analysis of MMC‐HVDC system
In Section 3, transient modelling method of MMC based on energy conservation is given, which can accurately cal-culate the bridge arm current and DC side current. In Section 4, the transient energy
analysis method based on electromagnetic transient simulation and related energy suppression strategies are introduced.
GB/T 42716-2023 English PDF (GBT 42716-2023).
It can be used for the analysis of the dynamic response characteristics between the energy storage power station and the power system in the full electromagnetic transient simulation of large power
grids, as well as the simulation calculation of fast action characteristics simulation of power electronic equipment in the
Research on Electromagnetic Transient Modeling and Simulation of Power
At present, there is a lack of refined modeling and simulation of electromagnetic transient in large-scale power grid with large-scale energy storage devices. EMT simulation has
Research on Energy Storage System Modeling Method Based on
In power system simulation research, it is necessary to develop an electromagnetic transient model for the energy storage system and conduct an accurate simulation.
FPGA-Based Real-Time Simulation for Multiple Energy Storage
Combining the renewable energy system, the Energy Storage (ES) station can maintain stable power transfer between renewable energy systems and
Design of EV Charging Station with Integrated Renewable Energy
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Hydraulic disturbance characteristics and power control of
1. Introduction. In power system, the pumped storage power plant (PSPP) undertakes the task of peak load and frequency regulation [[1], [2], [3]].Based on the features of quick start-up, quick
shutdown and flexible regulation, the PSPP plays an important role in enhancing and improving the stability of power system [4, 5].For the
GB/T 42716-2023 related PDF English
It can be used for the analysis of the dynamic response characteristics between the energy storage power station and the power system in the full electromagnetic transient simulation of large power
grids, as well as the simulation calculation of fast action characteristics simulation of power electronic equipment in the
Frontiers | Electromagnetic transient characteristics and
New energy systems, such as wind power and photovoltaic and distributed power supply, in the power generation and grid-connected transmission system basically will use stronger insulation design,
higher operational safety, and smaller footprint GIL equipment, while the new energy transmission system of the smart grid development
A review of the energy storage system as a part of power system
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Research on Self-adaptive Network Partitioning Algorithm for Electromagnetic Transient Parallel Simulation of Power
The electromagnetic transient simulation of large-scale power system is often accelerated by parallel simulation with sub-networks. However, the general network partitioning method based on
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Electromagnetic Transient (EMT) Simulation Algorithms for
The integration of hybrid photovoltaic (PV) and energy storage system (ESS) based plants has become a promising way of solving the intermittency of PV plants and providing frequency support to the
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The program modeling of the electromagnetic transient processes with the output parameters of solar power plants in parallel operation with the power system has been carried in the MatLab/Simulink
and Power System Blockset software environment, which allows simulating the parallel operation of stations and power systems. 7.
Review of Methods to Accelerate Electromagnetic Transient Simulation of
The modern power system is evolving with increasing penetration of power electronics introducing complicated electromagnetic phenomenon. Electromagnetic transient (EMT) simulation is essential to
understand power system behavior under disturbance which however is one of the most sophisticated and time-consuming applications in power
GB/T 42716-2023 related PDF English
Preview PDF: GB/T 42716-2023 GB/T 42716-2023: PDF in English (GBT 42716-2023) GB/T 42716-2023 GB NATIONAL STANDARD OF THE PEOPLE''S REPUBLIC OF CHINA ICS 27.180 CCS F 19 Guide for Modeling of
Electrochemical Energy Storage Power Station ISSUED ON: MAY 23, 2023 IMPLEMENTED ON: DECEMBER
Electromagnetic transient simulation models for
Figure 1 compares the response of a large-scale power system with RMS and full electromagnetic transient (EMT) models. An important difference demonstrated in this figure is the presence of
Transient Stability Simulation Analysis of Multi Node Power Network with Variable Speed Pumped Storage
The output characteristics of variable speed pumped storage are different from conventional hydropower and constant speed pumped storage units. The continuous increase of installed capacity of
variable speed pumped storage, poses a severe challenge to the safe and stable operation of the local power grid. Proposed in this paper is a kind
Is electromagnetic transient modelling and
Situations prompting the need for electromagnetic transient models. The Australian Energy Market Operator (AEMO) has been using EMT simulation models for several years, including for black
Modelling and simulation analysis of MMC‐HVDC
First, the transient model based on transient energy conservation can accurately solve the bridge arm current and DC current
Portal Analysis Approach Used for the Efficient Electromagnetic
Efficient electromagnetic transient (EMT) simulation is crucial for addressing the challenges associated with the modularity, cascading, and complex topologies of power electronics (PE) systems.
Hardware-based Advanced Electromagnetic Transient Simulation
This paper presents the implementation of hardware-based high-fidelity EMT dynamic model of a large-scale PV plant, accomplished through custom model
Parallel electromagnetic transient simulation of power systems with a high proportion of renewable energy
Traditional EMT simulation is limited by the computing power of the simulation platform and the parallel structure of the simulation algorithm. However, in recent years, with the emergence of
high-performance computing platforms, EMT simulation platforms have developed from single computer to computer cluster [ 4 ],
Full electromagnetic transient simulation for large power
Key routes for implementing full electromagnetic transient simulation of large-power systems are described in this paper, and a top framework is designed. A combination of the new large time step
algorithm and the traditional small-time step algorithm is proposed where both parts A and B are calculated independently. | {"url":"https://project-sarah.eu/electromagnetic/transient/simulation/of/energy/storage/power/station/24387.html","timestamp":"2024-11-06T21:24:47Z","content_type":"text/html","content_length":"40789","record_id":"<urn:uuid:1a4bb241-ebb6-4edc-8eb0-e8c8bfae5e94>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00749.warc.gz"} |
Capability of Visible-Near Infrared Spectroscopy in Estimating Soils Carbon, Potassium and Phosphorus
Capability of Visible-Near Infrared Spectroscopy in Estimating Soils Carbon, Potassium and Phosphorus ()
1. Introduction
For the large majority of countries, agriculture is an area of particular interest, in social, economic and environmental points of view [1] . It plays a key role in Malian economy and the country
food security. Quantification of soils nutrient levels available to plants contributes to the success of productive cropping systems and healthy environment [2] . In Mali, chemical methods are widely
used to make measure soil physical and chemical properties. However, these methods require the use of chemical extractants which are expensive and harmful to human and his environment. In this
context, the search for an alternative solution to replace or reduce conventional methods for soil property analyses becomes necessary.
Advantages of spectroscopy methods are justified for several reasons. For example, sample preparation involves only drying and grinding, thus the sample chemical properties are not affected by the
analysis. Also, the measurement is fast and several soil properties can be estimated from a single scan. Moreover, the technique can be performed in the laboratory or directly in situ.
First publications on the potential of VIS-NIR spectroscopy for soil analysis appeared in the early 1990 [3] [4] . Since this period, many works have been carried out on the use of this technique,
especially during the last decades [5] . Particular emphasis has been placed on the classical soil properties such as soil organic matter (SOM), clay content, mineralogy, chemical nutrients,
structure and microbial activity [5] . According to [6] , calibrations for total organic carbon as well as total soil nitrogen are the most likely to be successful. Previous findings indicated a
coefficient of determination ranging from 0.23 to 0.92 for P and 0.11 to 0.55 for exchangeable K [7] [8] [9] . However, predictions for P and K by spectroscopy remained unacceptable [10] [11] .
Regarding the development of spectroscopic methods, few studies have been done in some African countries like Mali. Other studies [12] used soil samples from across the Lake Victoria basin of Kenya
to investigate the potential of near infrared diffuse reflectance spectroscopy to estimate some properties. However, they suggested further calibrations using more diverse soil types and testing
alternative infrared diffuse reflectance based methods.
The interest of the VIS-PIR spectroscopy method is justified for several reasons. For instance, sample preparation involves only drying and grinding, the sample is not affected by the analysis in any
way, also no chemicals (environmental hazard) are required. In addition, the measurement is fast because it takes only a few seconds and several soil properties can be estimated simultaneously from a
single scan. Moreover, the technique can be done in the laboratory or directly in situ.
A network of infrared spectroscopy laboratories is supported by the World Agroforestry Center (ICRAF) in national institutions in Africa, currently in Cote d’Ivoire, Kenya, Malawi, Mali, Mozambique,
Nigeria, and Tanzania. Despite all this effort the spectroscopy method is poorly used for soil analysis in different research institutions. The present study contributes to document and to
demonstrate the potential of the diffuse reflectance spectroscopy to estimate some soil properties in Mali.
The general objective of this study is to evaluate the performance of VIS-NIR spectroscopy in comparing the estimates of two regression models, namely Principal Component Regression (PCR) and Partial
Least Squared Regression (PLSR) for the determination of total carbon (C), available phosphorus (P) and exchangeable potassium (K).
2. Material and Methods
2.1. Sample Collection
Soil sampling and preparation were carried out by the soil laboratory “Laboratoire Sol-Eau-Plante” in Mali. Areas covered by the study are the administrative regions of Koulikoro and Sikasso where
755 and 122 samples, were collected respectively. The soil samples were collected from 0 - 10 cm depth. Figure 1 shows the geographical characteristics of each sampling site.
2.2. Measurements of Soil Reference Data
Soil reference data measurements were carried out in the soil laboratory “Laboratoire Sol-Eau-Plante” (LSEP) using standard laboratory methods. Soil samples were priory air dried, crushed and sieved
to 2 mm. The total carbon was measured by an automatic titrator using the modified Anne method which is the oxidation of soil carbon by potassium dichromate. The available phosphorus of the soil was
extracted with a combined solution of 0.1 M HCl and 0.03 M NH4F and the measurements were made using an ultraviolet (UV) spectrophotometer or a colorimeter. For the exchangeable K, the soil was
leached with a 1 M ammonium acetate solution at pH 7. It was determined directly in the ammonium acetate percolate using a flame photometer.
2.3. Sample Selection Method
The 877 soils samples were partitioned into two sub-samples constituting sets of calibration and validation. The selection was made in the way that all sampling sites are represented in both
calibration and validation sets. Approximately 2/3 of samples from each sampling site are selected to form the calibration sample set (587 samples) and the remaining 1/3 is used to form the
validation sample set (290 samples).
2.4. Spectral Measurements
Spectral measurements and their processing were carried out at the Laboratory of Optics, Spectroscopy and Atmospheric Sciences (LOSSA). These measurements consist of recording the soils reflectance
over the wavelength range of 342 - 1060 nm. Soils samples were priory air dried and crushed to pass a 2-mm sieve. A Miniature Fiber Optic Spectrometer working on UV-VIS-NIR spectral range (BLUE-Wave
Miniature Fiber Optic Spectrometers for UV-VIS-NIR & OEM, StellarNet Inc.) was used to perform the spectral measurements. The spectrometer is connected to a PC on which the Spectra Wiz software is
installed for controlling the data acquisition. The samples were scanned using a halogen tungsten
Figure 1. Maps representing the distribution of soil sampling sites in Mali.
SL1 lamp source manufactured by StellarNet Inc. This light source has a wide spectral Range for 350 - 2500 nm-effective for color, reflectance, transmittance, and absorbance measurements. The
Y-shaped optical fiber was used to transport light from the source to the sample and from sample to the spectrometer (Figure 2(a)) with the same incident angle of illumination.
2.5. Pre-Treatment of Spectra
Before being used, the raw spectral data are undergoing various pretreatments. The most common strategy for pre-processing spectra is to submit the raw data to one or more mathematical
transformations intended to make them suitable for modeling.
Figure 2. (a) Scheme describing the spectroscopic equipment used for scanning soil samples and spectrum acquisition; (b) Raw spectrum; (c) filtered spectrum.
Sample spectra were filtered using the RunMean function under “caTools” package of the R statistical software. The filtering consists in eliminating the interference related to the experimental
conditions and the electronic noise of the measuring instrument.
Spectral reflectance was measured in the wavelength range of 342 - 1060 nm. Since the UV band is of a little interest for the soil spectral study, we have restrained the spectral band to 400 - 1000
nm with a spectral resolution of 0.5 nm. Figure 2(b) and Figure 2(c) shows the effect of filtering on the spectrum of a sample chosen randomly. The filtered spectra were then converted to spectral
absorbance (A) using the following relationship:
where R is the actual spectral Reflectance.
2.6. Chemometric Analysis of the Soil Chemistry and Spectral Data
The first part of the analysis is the calibration which consists of developing a mathematical model to determine the chemical properties Y (unknown concentration) from the available spectral
measurements X (spectral absorbance). The model is setup using variable X and Y information of the calibration sample set. Once the model is established, it can be used to estimate the chemical
properties of unknown samples. Two regressions models have been involved in this analysis: the principal component regression (PCR) and the partial least square regression (PLSR).
Validation is the second step of the analysis which consists in evaluating the performance of the calibrated model by comparing its estimates with the reference values. During this phase, the
accuracy of the model is evaluated on a set of independent samples meaning samples that did not participate in the calibration process. Thus, a good prediction implies a good quality of the
calibration model. If the model appears satisfactory, it can be applied in routine analysis to analyze unknown samples.
2.6.1. PCR Model
Principal component regression (PCR) is a two-step multivariate analysis method. The first step consists of performing a principal component analysis (PCA) of the explanatory data matrix X to convert
them into new data matrix: the matrix T (X-scores) and a matrix P' (X-loadings). PCA creates new orthogonal variables T (latent variables) that are linear combinations of the original X variables
with the coefficients a[i].
$X=T{P}^{\prime }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}T=X{a}_{i}$(2)
In the second step, a multiple linear regression (MLR) is established between the scores obtained and the measure (known) variable Y.
The Principal Component Analysis is a way of dealing with the problem of poorly conditioned matrices. The objective is to obtain a certain number of components capturing the maximum variation
relative to the variables of the matrix X while assuring the model a certain quality of prediction. PCR can be considered as a linear regression method in which the response variable is regressed on
new components.
2.6.2. PLS Model
The partial least squares regression (PLSR) is based on principal components on both the independent variable X, and the dependent variable Y. The PLS model links an unknown variable Y to a block of
explanatory variables X through latent variables that are linear combinations of the initial explanatory variables [13] . The latent variables explain as much as possible the covariance between both
variables X and Y. The approach is to calculate the principal scores of X and Y and to set up a regression model between the scores.
The equation system bellow (Equation (3)) highlights how the independent data matrix X can be decomposed into a matrix T (X-score), and a matrix, P' (X-loading), plus an error matrix, E. Similarly,
the dependent data matrix Y is decomposed into the matrix U (Y-scores) and a matrix Q' (Y-loadings) plus the error term F. These decompositions are made so as to maximize the covariance between the
scores matrix T and U.
$X=T{P}^{\prime }+E\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Y=U{Q}^{\prime }+F$(3)
The X-scores (T[i]) are orthogonal and they are estimated as linear combinations of the original variables X[i]. Thus, the matrix of latent variables T is a linear transformation of X.
2.7. Statistics Assessing Model Performance
Both multivariate models were validated with an independent data set representing about 1/3 of the total sample. The models performances were assessed by using some standard statistics: coefficient
of determination (R^2), the Bias and the root mean square error (RMSE). These statistical measures were computed using the following formulations:
${R}^{2}=\frac{{\sum }_{i=1}^{n}{\left({\stackrel{^}{y}}_{i}-\stackrel{¯}{y}\right)}^{2}}{{\sum }_{i=1}^{n}{\left({y}_{i}-\stackrel{¯}{y}\right)}^{2}},BIAS=\frac{1}{n}{\sum }_{i=1}^{n}\left({y}_{i}-
{\stackrel{^}{y}}_{i}\right),RMSE=\sqrt{\frac{1}{n}{\sum }_{i=1}^{n}{\left({y}_{i}-{\stackrel{^}{y}}_{i}\right)}^{2}}$(4)
With ${\stackrel{^}{y}}_{i}$ the values of the measurements, ${y}_{i}$ the predicted values and $\stackrel{¯}{y}$ the average of the measurements.
3. Results and Discussion
Descriptive statistics of three soil properties C, P and K are summarized in Table 1. The difference between the minimum and maximum values of the concentrations for these elements demonstrates their
strong variability in across the sampling sites. Also statistics for the validation sample set were within the range of the statistics for calibration sample set for all the three soil properties (
Table 1).
3.1. Performance of Models Cross-Validation
By checking at all the statistical criteria R^2, Bias and RMSE, both models show
Table 1. Descriptive statistics of the soil reference data constituting of 587 soil samples of calibration and 290 soil samples of validation.
*STD: Standard deviation.
good calibration qualities. They have strong coefficients of determination and weak bias for all elements. The PCR gives the best calibration quality with R^2 stronger than 0.80 for all elements (
Table 2). This performance is better than that found by [14] for the carbon, with R^2 = 0.68 in the NIR region (700 - 2498). The PLSR model also has good calibration quality with coefficients of
0.801 for potassium, 0.872 for carbon and 0.881 for phosphorus (Table 2).
However, the coefficient of determination is not the only parameter to be considered for assessing the performance of a model. The root mean square error and the Bias between the measured and
predicted values are also used as statistics to evaluate the robustness of a model.
The RMSE obtained from the PCR method is 0.22, 1.00 and 0.17 respectively for carbon, phosphorus and potassium. And for the PLSR method, the RMSE is 0.21 for the C, 1 for the P and 0.15 for the K.
The bias obtained from both models (0.0004 for C; 0.0017 for P and 0.0003 for K) are relatively weaker compared to their respective average values (0.51, 1.33 and 0.25).
These results can be further improved by proceeding to homogeneous distribution of the samples from different sampling sites, as some sampling sites are represented by nearly 60 samples while others
sites had only 4 samples. It has been argued that the calibration of predictive models on a limited number of samples or on fairly homogeneous samples may limit the scope of the calibration model
[15] .
3.2. Independent Validation
It can be seen that the performance of the prediction models varies from one chemical property to another and also from one model to another. An independent validation of both multivariate models
calibrated for the three chemical properties (C, K and P) reveals lower performance of prediction with regards to the cross-validation performance. The PCR model has coefficients of determination of
0.17 for carbon, 0.34 for potassium, and 0.50 for phosphorus (Figures 3(a)-(c)). These values are comparatively lesser than 0.87 (C) and 0.64 (K) obtained respectively by [16] with the band 1100 -
2500 nm and [17] with the band 400 - 2498 nm.
The independent validation of the PLSR method yielded R^2 = 0.29 for C, R^2 = 0.42 for K and R^2 = 0.57 for P (Figure 3(d), Figure 3(e), Figure 3(f)). Comparing
Table 2. Statistics showing the performances of the PCR and PLSR models for cross-validation (calibration) and for independent validation. All the coefficients of determination exceed the 95% of
level of significant.
Figure 3. Scatter plot with regression line showing the statistics comparing the reference values of C, P and K with their respective estimations from PCR model ((a)-(c)) and PLSR model ((d)-(f)).
to some previous finding, these performances are lower compared to R^2 = 0.66 (C) and R^2 = 0.61 (K) obtained respectively by Sorensen and [18] with the band 408 - 2492 nm and [19] with the band 400
- 2498 nm. This low performance can be attributed to the wavelength band used and also to the large extension of the study area. Indeed, the predictions of the soil components by spectroscopy may
fail if the samples are collected in a very large geographical area or from different morpho-pedological contexts. Some previous findings [20] explain the failure by the difference between the
original soil parent materials. However, for the potassium, our result is better than R^2 = 0.40 obtained by [19] in the same conditions.
Although, for some chemical, the estimates gave low performances for both models, but it has to be noted that the PLSR assures qualities of prediction of the phenomenon better than the PCR. This is
due to the fact that the PLSR components capture the information carried by the explanatory variables while paying attention to the link between the two variables.
The RMSE and the bias found are very low compared to the average of the reference data. For PCR, the RMSE obtained was 0.23 for C; 1 for P and 0.17 for K and the bias was 0.0125 for total carbon,
0.0709 for phosphorus and 0.002 for potassium. The PLSR model shows RMSE values of 0.22, 0.95 and 0.16, respectively for C, P and K; the respective biases found were −0.0108; 0.2023 and 0.009.
5. Conclusion
This study documents the potentiality of the VIS-NIR diffuse reflectance spectroscopy in soil study. This method of analysis is a very promising tool for soils study: rapidity, ease of measurement,
without the use of chemicals and even in situ measurements in the field can be envisaged. Results show that the PLSR estimation over-performs the prediction of the PCR model. The independent
validation reveals that VIS-NIR spectroscopy over 400 - 1000 nm has limited performance for estimating some soil properties. The prospect of using the entire spectral band of the VIS-NIR (400 - 2500
nm) for the analysis of soil properties may be considered. The creation of a spectral database by selected zone to limit the study area can be a promising solution to achieve good results. For
instance, spectral soils database can be realized for specific key areas for Malian agriculture, such as “Office du Niger” and “zone CMDT” dedicated respectively for rice and cotton cultivation.
Faster and easier prediction of the soil properties of these areas can be very promising. This will contribute to the development of agriculture in these areas which constitutes the major
agricultural basins of Mali and have a considerable impact on the country economy.
We acknowledge the International Science Program (ISP/IPPS) for supporting the Laboratory of Optics, spectroscopy and Atmospheric Science (LOSSA) of the “Faculté des Sciences et Techniques de
Bamako”. Our gratitude goes to the “Laboratoire Sol-Eau-Plante de l’IER” for providing soils samples and reference chemical properties. | {"url":"https://www.scirp.org/journal/paperinformation?paperid=85042","timestamp":"2024-11-10T02:02:36Z","content_type":"application/xhtml+xml","content_length":"127246","record_id":"<urn:uuid:5830130d-94eb-4e8a-89aa-69fcd2d3a797>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00169.warc.gz"} |
Two-wheel self-balancing robot
After my first fast made self balancing robot, I want to better understand it and especially to try different control methods. For me it is necessary to simulate the whole stuff on computer (robot +
control law).
The first step is therefore to model the robot with mathematical equations. Because of the complexity of the model I do make some assumption and simplification.
The model is separated into three part: the DC motor, the wheel and the inverted pendulum
The DC motor
electrical model:
equation of motion for the motor:
If the motor inductance and friction torque are neglected, then
in state space form:
The inputs to the motor is the applied voltage and applied torque on the rotor.
The wheel
in horizontal (x direction):
rotation equation around the center of the wheel:
by using motor dynamic equations, one can get
Above equation is valid for both wheels.
With following equations, one can translate the angular rotate to linear motion:
After rearrangement, one has:
The inverted pendulum
The sum of forces perpendicular to the pendulum:
The sum of motions around the center of mass of pendulum:
one has:
after arrangement, one can get the following state space expression:
On my first two-wheel self-balancing robot, I didn’t use the encoder for speed measurement.
So it can basically stand still and balance its weight with small movements of its two
wheels simultaneously. I preferred that the robot would work very soon without much effort
and then I could improve it by adding speed control, RC, etc.
It’s not difficult to gather all the parts. Here is my list:
1.glass fiber board
2.control board (arduino uno)
3.sensor board (MPU6050 gyroscope + accelerameter)
4.two-channel motor driver (L298N)
5.two geared DC motors (25GA370)
6.two toy wheels and fixation kit
Following is my assembly diagram:
Following is the wiring diagram:
In L298N, each motor is controlled by a full bridge circuit (or H-bridge) composed of two legs. Following figure is a simplified diagram coming from wikipedia. S1 and S2 are always controlled in
complementary manner. When S1 closes, S2 opens. S1 and S2 close simultaneously is not allowed since that creates a short circuit of the power supply. Idem for S3 and S4.
When S1 and S4 close simultaneously, the motor rotates in one direction. When S3 and S2 close simultaneously, the motor rotates in the other direction. I use PWM control to set the motors’ speed. the
switching frequency is about 1kHz which is the default value of Arduino module.
After the wiring, the robot looks like:
Then comes the control part. The following figure shows the control diagram. Only the inner
loop in blue is implemented this time. I use it to regulate the tilt angle of the robot. As
I want the robot stands straight, I simply set the reference angle (REF_ANGLE) to 0. The
gyro+acc sensor gives the information needed to calculate the actual tilt angle. The error
between the reference and the measurement is used by the PID controller to give the command
to the robot (the two wheel motors).
2.1 PID controller
The PID controller is a complexe topic. With google you can find tons of material about it.
Fortunately it is relatively simple to implement it in a microcontroller. With some basic
understanding and tuning techniques, one can make it work very easily.
The general mathematical form of the PID controller is (extracted from wikipedia):
Here is a simple software loop that implements a PID algorithm (extracted from wikipedia):
previous_error = 0
integral = 0
error = setpoint - measured_value
integral = integral + error*dt
derivative = (error - previous_error)/dt
output = Kp*error + Ki*integral + Kd*derivative
previous_error = error
goto start
The purpose of the PID controller is to generate a suitable command to the system (the self
balancing robot in this case) so that the system is stable and reactive. There are some
creteria of both the static and dynamic performance. But they are out of the scope. The PID
controller uses three parameters to do the job, the proportional (Kp), integral (Ki),
derivative(Kd) parameter. The big Kp makes the tilt angle error small and the reaction fast. The
Ki makes the static angle error dispear. This is important since I want the robot stands
straight (tilt angle=0) all the time. I choose to put Kd=0 because Ki and Kp achieve
perfectly my desired performance. So in fact I use a PI controller. For the PI parameter
tuning, I follow Ziegler-Nichols method that you can find on wikipedia. Following table is
copied from wikipedia. It works well for me.
2.2 Tilt angle measurement
A sensor combining a gyroscope and an accelerometer is used to measure the tilt angle. The sensor don’t give the angle directly. Some calculation is needed. I use MP6050 as sensor. You can check its
datasheet for more information.
The accelerometer gives the acceleration (unit g) in three dimension. If the robot is perpendicular to the earth (z axis for example), you should have only Z axis which is not zero. If you have
another axis not equaling zero besides Z axis in static state, it means the robot has a none zero tilt angle. And you can get easily the tilt angle based on these two axis acceleration. However using
accelerometer to get the tilt angle in this way is valid only for static state since we suppose that only gravity acts on the robot. When the two wheels act a force on the robot, the accelerometer
can measure the resulting acceleration as well! In consequence, the measures are very noised. The good point is it is accurate in long period of time (or in DC after filtering).
The gyroscope measures the angle speed (unit deg/s) in three dimension. In order to get the angle, we need to do the integration of it. It is accurate dynamically (in short period). But the
integration action accumulates the error of measurement, which makes the result drift in long period of time.
I use a complementary filter to combine the two sensor measurements. It is simple and needs little microcontroller calculation. It works well for me (so far). One can implement kalman filter for
optimal control. But the knowledge of the system model and more calculation capability are required. It needs definitely more effort. I use following complementary filter to get the tilt angle
The sensor communicates to the microcontroller through I2C bus. By using the WIRE library of arduino, it is easy to read the data. I use analogWrite function to set PWM duty cycle for the motor
driver L298N. The command (PID controller’s output) is a value between -255 and 255. Positive sign means the robot moves forward. 255 represents the max speed (100% duty cycle for the first half
bridge). I use anti-windup code to prevent the command to be higher than 255 (or lower than -255) which is the limit of PWM register (8 bits) max value.
The calibration of the sensor should be done generally. I calibrated the gyroscope because I see big static offset. My accelerometer’s offset is not obvious. I use 10ms as sampling time.
The total code has only about 50 lines which I believe possible to be optimized. It’s simple, right?
The robot resists relatively well the disturbance from both sides. The stability and dynamic response are satisfactory. The next step is to install an encoder on each motor, to make the robot
possible to move and to add a remote control. Following is a video of it standing still.
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Study of the transient phase of the forgetting factor RLS
In this paper, we consider the convergence properties of the forgetting factor RLS algorithm in a stationary data environment. More precisely, we study the dependence of the speed of convergence of
RLS with respect to the initialization of the input sample covariance matrix and with respect to observation noise level. By obtaining estimates of the settling time, that is, the time required by
RLS to converge, we conclude that the algorithm can exhibit a variable performance. Specifically, when the observation noise level is low (high SNR environment), RLS, when initialized with a matrix
of small norm, has a very fast convergence. Convergence speed decreases as we increase the norm of the initialization matrix. In a medium SNR environment, the optimum convergence speed of the
algorithm is reduced as compared with the previous case, but on the other hand, RLS becomes more insensitive to initialization. Finally, in a low SNR environment, we show that it is preferable to
start the algorithm with a matrix of large norm.
All Science Journal Classification (ASJC) codes
• Signal Processing
• Electrical and Electronic Engineering
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What Is The Value Of The Expression Shown?24 62A.-6B.-12C.12
None of the choices are correct
Step-by-step explanation:
f cannot have a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex] and [tex]$$ \lim _{x \uparrow x_0} f(x) < \lim _{x \mid x_0} f(x) .$$[/tex]
f cannot have a removable discontinuity at [tex]$$x_0 \in(a, b) $$[/tex] and [tex]\alpha=\lim _{x \rightarrow x_0} f(x) < f\left(x_0\right)[/tex]
Let f be a function defined on (a, b) satisfies intermediate value property.
Claim: f ca not have removable on jump discontinuity.
Suppose f has a jump discontinuity at [tex]$x_0 \in(a, b)$[/tex]
We take [tex]$\theta$[/tex] such that
[tex]$$\lim _{x \rightarrow x_0} f(x) < \theta < \lim _{x \downarrow x_0} f(x) \text { and } \theta \neq f\left(x_0\right)$$[/tex]
Now there exist [tex]$\delta > 0$[/tex] such that [tex]$f(x) < \theta$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right)$[/tex] and [tex]$f(x) > \theta$[/tex] for all [tex]$x \in\left(x_0, x_0+\
Now [tex]$f\left(x_0-\delta\right)[/tex][tex]< \theta < f\left(x_0+\delta\right)$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0+\delta\right] \backslash\left\{x_2\right\}$[/tex] and [tex]$f\left(x_0
\right) \neq \theta$[/tex].
Therefore the point [tex]$\theta$[/tex] has no preimage under f
that is, there does not exists [tex]$y \in\left[x_0-\delta, x_0+\delta\right][/tex] for which
[tex]$$f(y)=\theta[/tex] because [tex]\left\{\begin{array}{l}y=x_0 \Rightarrow f(y) \neq \theta \\y > x_0 \Rightarrow f(y) > \theta \\y < x_0 \Rightarrow f(y) < \theta\end{array}\right.$$[/tex]
Therefore f does not satisfies intermediate value property on [tex]$\left[x_0-\delta, x_0+\delta\right]$[/tex],
Hence f does not satisfies IVP on (a, b) which is not possible because we assume f satisfies IVP on (a, b),
Therefore f can not have a jump discontinuity.
Suppose f has a removable point of discontinuity at [tex]$x_0 \in(a, b)$[/tex],
Let [tex]$\alpha=\lim _{\alpha \rightarrow x_0} f(x)$[/tex],
Let [tex]\alpha < f\left(x_0\right)$[/tex] so [tex]$f\left(x_0\right)-\alpha > 0$[/tex].
Now [tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex] then [tex]\exists$ \delta > 0$[/tex] such that
[tex]$$\begin{aligned}& |f(x)-\alpha| < \frac{f\left(x_0\right)-\alpha}{2} \text { for all } x \in\left\{x_0-\delta, x_0-\alpha\right]-\left\{x_0\right\} \\& \Rightarrow \quad f(x) < \alpha+\frac{f\
left(x_0\right)-\alpha}{2} \text { for all } x \in\left[x_0-\delta, x_0+\delta\right]-\left\{x_0\right\}\end{aligned}$$[/tex]
So [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2}$[/tex] for all [tex]$x \in\left[x_0-\delta, x_0\right]-\left\{x_0\right\}$[/tex]
Now [tex]$f\left(x_0\right) > \alpha$[/tex].
And [tex]$f(x) < \frac{f\left(x_0\right)+\alpha}{2} < f\left(x_0\right)$[/tex] for all [tex]$x \in\left[\left(x_0 \delta, x_0\right)\right.$[/tex]
Let [tex]$\mu=\frac{f\left(x_0\right)+\alpha}{2}$[/tex].
Then there does not exist [tex]$e \in\left[x_0-\delta, c\right]$[/tex] such that [tex]$f(c)=\mu$[/tex]
Because for [tex]$e=x_0 \quad f(e) > \mu$[/tex]
for [tex]$c < x_0 \quad f(c) < \mu$[/tex].
Therefore f does not satisfy IVP on [tex]$\left[x_0-\delta_1 x_0\right]$[/tex] which contradict our hypothesis,
therefore [tex]$\alpha \geqslant f\left(x_0\right)$[/tex]
Let [tex]$\alpha > f\left(x_0\right)$[/tex]. so [tex]$\alpha-f\left(x_0\right) > 0$[/tex]
[tex]$\lim _{x \rightarrow x_0} f(x)=\alpha$[/tex]
Then [tex]\exists $ \varepsilon > 0$[/tex] such that
[tex]$|f(x)-\alpha| < \frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$\left.x \in\left[x_0-\varepsilon_0 x_0+\varepsilon\right]\right\}\left\{x_i\right\}$[/tex]
[tex]$\Rightarrow f(x) > \alpha-\frac{\alpha-f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0+\varepsilon\right] \backslash\left\{x_0\right\}$[/tex]
[tex]$\Rightarrow f(x) > \frac{\alpha+f\left(x_0\right)}{2}$[/tex] for all [tex]$x \in\left[x_0-\varepsilon_1, x_0\right)$[/tex]
Now [tex]$f\left(x_0\right) < \alpha$[/tex]
The [tex]$f(x) > \frac{f\left(x_0\right)+\alpha}{2} > f\left(x_0\right)$[/tex].
So [tex]$f\left(x_0\right) < \frac{f\left(x_0\right)+\alpha}{2} < f(x)$[/tex] for all [tex]$x \in\left[x_0 \varepsilon, \varepsilon_0\right)$[/tex]
Let [tex]$\eta=\frac{f\left(x_e\right)+\alpha}{2}$[/tex]
Then there does not exist [tex]$d \in\left[x_0-\varepsilon, x_0\right]$[/tex] such that [tex]$f(d)=\xi$[/tex].
Because if [tex]$d=x_0, f(d)=f\left(x_0\right) < \eta$[/tex] if [tex]$d E\left[x_0-\varepsilon, x_0\right)$[/tex]
Then [tex]$f(d) > \eta$[/tex]
Therefore f does not satisfies IVP on [tex]$\left[x_0-\varepsilon, x_0\right]$[/tex] which contradict olio hypothesis.
Therefore [tex]$\alpha \leq f\left(x_0\right)$[/tex] (b) From (a) and (b) it follows [tex]$\alpha=f\left(x_0\right)=\lim _{x \rightarrow x_0} f(x)$[/tex]. Therefore f can not have a removable
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A ball with a mass of 4 kg moving at 7 ms^(-1) hits a still ball with a mass of 7 kg. If the first ball stops moving, how fast is the second ball moving? | HIX Tutor
A ball with a mass of #4# #kg# moving at #7# #ms^(-1)# hits a still ball with a mass of #7# #kg#. If the first ball stops moving, how fast is the second ball moving?
Answer 1
Before the collision, the first ball has all the momentum (since the second is stationary). $p = m v = 4 \times 7 = 28$$k g m {s}^{- 1}$.
Momentum is conserved. After the collision the second ball has all the momentum. $v = \frac{p}{m} = \frac{28}{7} = 4$$m {s}^{- 1}$.
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Answer 2
To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the
total momentum after the collision. The momentum ( p ) of an object is given by the product of its mass ( m ) and velocity ( v ), i.e., ( p = m \times v ).
Before the collision: ( p_{\text{initial}} = m_1 \times v_1 + m_2 \times v_2 )
After the collision: ( p_{\text{final}} = 0 \times v_1' + (m_1 + m_2) \times v_2' )
Since momentum is conserved, ( p_{\text{initial}} = p_{\text{final}} ), thus: ( m_1 \times v_1 + m_2 \times v_2 = (m_1 + m_2) \times v_2' )
Substituting the given values: ( (4 , \text{kg} \times 7 , \text{m/s}) + (7 , \text{kg} \times 0 , \text{m/s}) = (4 , \text{kg} + 7 , \text{kg}) \times v_2' ) ( 28 , \text{kg m/s} = 11 , \text{kg} \
times v_2' ) ( v_2' = \frac{28 , \text{kg m/s}}{11 , \text{kg}} ) ( v_2' \approx 2.55 , \text{m/s} )
Therefore, the second ball is moving at approximately ( 2.55 , \text{m/s} ) after the collision.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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How to know if equation is one-to-one
• Thread starter intenzxboi
• Start date
In summary, the problem the student is having is being able to explain in an equation what is happening when they take the derivative of the function. They are not able to see that the function is
decreasing on the interval (-\infty, 8) and increasing on the interval (8, \infty).
i know for a equation to be 1-1 they have to pass the horizontal line test.
so say for example
(-8 + x) ^4
i know this is not a one to one because it gives me a parabola.
The problem that I am having is being able to explain in a equation.
First derive the equation so that i know where it is increasing/decreasing
so i take -8+x>0 is increasing
x>8 <----- I am sure that's correctnow to find it decreasing you flip the sign
and i take -8+x<0 is decreasing?
but then i also get x<8
What I'm i doing wrong??So if I am doing this right then that means that for a function to be one to one.. It either has to be increasing OR decreasing?? but cannot be both?
Last edited:
First off, it makes no sense to say that an equation is one-to-one. A function can be one-to-one.
Second, an equation has an = sign in it, and makes a statement about two expressions being equal.
Your first example, (x - 8)^4 is not an equation. What you probably mean is y = (x - 8)^4, which is an equation, and is the equation of a function. This can also be represented as f(x) = (x - 8)^4.
The graph of this function is not a parabola, but it has a similar U shape.
Let's take the derivative: f'(x) = 4(x - 8)^3
The function f is increasing when f'(x) > 0, or when (x - 8)^3 > 0, which is when x > 8.
The function f is decreasing when f'(x) < 0, or when (x - 8)^3 < 0, which is when x - 8 < 0. IOW, when x < 8.
So the function is decreasing on the interval ([itex]-\infty[/itex], 8) and is increasing on the interval (8, [itex]\infty[/itex]). This means that for some y values, there are two x values, so this
function is not one-to-one.
You could also show this by picking an appropriate y-value and showing that there are two x values.
For example, if (x - 8)^4 = 1, then x - 8 = 1 or x - 8 = -1, so x = 9 or x = 7.
You are perhaps confusing yourself by using unnecessary machinery that you will need for other cases that are not so obvious.
f is one-to-one if f(a) = f(b) implies a = b.
From what you say about the function f(x) = (x - 8)^4 you seem to know well enough its shape. I think I glimpsed somewhere there is something called the horizontal line test. If necessary draw
pictures, or calculate some numbers. I think it is pretty obvious where this function has its minimum. If you can't see it, calculate f(a) for some number a, then think whether any other number,
which would be your b, gives you the same output number.
(A quibble that does not affect this issue is that this function is not a parabola - that term is reserved for a second degree function like (x - 8)^2. If you plot it you will notice a difference to
parabolas you have seen. However for the essential properties you are concerned with here it is working just like a parabola.)
Last edited:
FAQ: How to know if equation is one-to-one
1. How do I determine if an equation is one-to-one?
To determine if an equation is one-to-one, you can use the horizontal line test. This involves drawing a horizontal line anywhere on the graph of the equation. If the line intersects the graph at
more than one point, the equation is not one-to-one. If the line only intersects at one point, then the equation is one-to-one.
2. Can an equation be both one-to-one and not one-to-one?
No, an equation can only be classified as either one-to-one or not one-to-one. If the equation passes the horizontal line test, it is one-to-one. If it fails the horizontal line test, it is not
3. Are all linear equations one-to-one?
Yes, all linear equations are one-to-one. This is because they have a constant rate of change and will never intersect with a horizontal line more than once.
4. What is the significance of an equation being one-to-one?
An equation being one-to-one means that each input has exactly one unique output. This can be useful in determining inverse functions and finding solutions to certain problems.
5. Can an equation be one-to-one without being a function?
No, for an equation to be one-to-one, it must also be a function. This means that each input has only one corresponding output. If an equation is not a function, it cannot be one-to-one. | {"url":"https://www.physicsforums.com/threads/how-to-know-if-equation-is-one-to-one.286971/","timestamp":"2024-11-07T10:03:18Z","content_type":"text/html","content_length":"82488","record_id":"<urn:uuid:fec82529-f4ac-40d6-aac0-ba1e5d0a126b>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00842.warc.gz"} |
How to find order of operations - SAT Math
All SAT Math Resources
Example Questions
Example Question #21 : Order Of Operations
Correct answer:
When doing operations of subtraction and addition, there is no priority but we must work from left to right. The difference on the left is
Example Question #21 : How To Find Order Of Operations
Correct answer:
There is multiplication and addition present. Remember PEMDAS. Multiplication comes first followed by addition.
Example Question #22 : How To Find Order Of Operations
Correct answer:
There is multiplication and addition present. Remember PEMDAS. Multiplication comes first followed by addition. Since there are two multiplication operations, we work from left to right then finally
add the products.
Example Question #23 : How To Find Order Of Operations
Correct answer:
When there's multiplication and division, they have the same priority. We will work from left to right.
Example Question #24 : How To Find Order Of Operations
Correct answer:
When there's multiplication and division, they have the same priority. We will work from left to right.
Example Question #25 : How To Find Order Of Operations
Correct answer:
Although there is just addition and subtraction, a paranthesis is present. According to PEMDAS, paranthesis has priority over all operations.
Example Question #26 : How To Find Order Of Operations
Correct answer:
Even there is multiplication, addition and subtraction, a paranthesis is present. According to PEMDAS, paranthesis has priority over all operations.
Example Question #27 : How To Find Order Of Operations
Correct answer:
In PEMDAS, the paranthesis comes first followed by the exponent. We have subtraction so the difference is
Example Question #28 : How To Find Order Of Operations
Correct answer:
In PEMDAS, the paranthesis comes first.
Example Question #29 : How To Find Order Of Operations
Correct answer:
In PEMDAS, we have parantheses so those come first.
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Mitio Takano
Search results for: Mitio Takano
Items from 1 to 7 out of 7 results
A modified subformula property for the modal logic KD with the additionalaxiom □ ◊(A ∨ B) ⊃ □ ◊ A ∨ □ ◊B is shown. A new modification of the notion of subformula is proposed for this purpose. This
modification forms a natural extension of our former one on which modified subformula property for the modal logics K5, K5D and S4.2 has been shown ([2] and [4]). The finite model property as well as
The modal logic S4.2 is S4 with the additional axiom ◊□A ⊃ □◊A. In this article, the sequent calculus GS4.2 for this logic is presented, and by imposing an appropriate restriction on the application
of the cut-rule, it is shown that, every GS4.2-provable sequent S has a GS4.2-proof such that every formula occurring in it is either a subformula of some formula in S, or the formula □¬□B or ¬□B,
A sequential axiomatization is given for the 16-valued logic that has been proposed by Shramko-Wansing (J Philos Logic 34:121–153, 2005) as a candidate for the basic logic of logical bilattices.
Sequent calculi for trilattice logics, including those that are determined by the truth entailment, the falsity entailment and their intersection, are given. This partly answers the problems in
Shramko-Wansing (J Philos Logic 34:121–153, 2005).
Abstract.Strong completeness of S. Titanis system for lattice valued logic is shown by means of Dedekind cuts.
The modal logic S4.2 is S4 with the additional axiom ◊□A ⊃ □◊A. In this article, the sequent calculus GS4.2 for this logic is presented, and by imposing an appropriate restriction on the application
of the cut-rule, it is shown that, every GS4.2-provable sequent S has a GS4.2-proof such that every formula occurring in it is either a subformula of some formula in S, or the formula □¬□B or ¬□B,
A sequential axiomatization is given for the 16-valued logic that has been proposed by Shramko-Wansing (J Philos Logic 34:121–153, 2005) as a candidate for the basic logic of logical bilattices.
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Becker, T, Burt, D, Corcoran, TC, Greaves-Tunnell, A, Iafrate, JR, Jing, J, Miller, SJ, Porfilio, JD, Ronan, R, Samranvedhya, J, Strauch, FW and Talbut, B (2018). Benford's Law and Continuous
Dependent Random Variables. Annals of Physics 388, pp. 350–381. DOI:10.1016/j.aop.2017.11.013.
Becker, T, Corcoran, TC, Greaves-Tunnell, A, Iafrate, JR, Jing, J, Miller, SJ, Porfilio, JD, Ronan, R, Samranvedhya, J and Strauch, FW (2013). Benford's Law and Continuous Dependent Random
Variables. Preprint arXiv:1309.5603 [math.PR]; last accessed October 23, 2018. DOI:10.1016/j.aop.2017.11.013.
Cuff, V , Lewis, A and Miller, SJ (2015). The Weibull distribution and Benford’s law. Involve Vol. 8 No. 5, pp. 859–874. DOI:10.2140/involve.2015.8.859.
Ileanu, B-V, Ausloos, M, Herteliu, C and Cristescu, MP (2019). Intriguing behavior when testing the impact of quotation marks usage in Google search results. Quality & Quantity 53(5), pp.
2507-2519. DOI:10.1007/s11135-018-0771-0.
Jing, J (2013). Benford’s Law and Stick Decomposition. Undergraduate thesis, Williams College, Williamstown, Massachusetts .
Johnson, GG (2005). Financial Sleuthing Using Benford's Law to Analyze Quarterly Data with Various Industry Profiles. Journal of Forensic Accounting 6(2), pp. 293-316.
McCarville, D (2021). A data transformation process for using Benford’s Law with bounded data. Preprint [version 1; peer review: awaiting peer review], Emerald Open Research 3(29).
Miller, SJ (2008). Benford’s Law and Fraud Detection, or: Why the IRS Should Care About Number Theory!. Presentation for Bronfman Science Lunch Williams College, October 21.
Miller, SJ (2016). Can math detect fraud? CSI: Math: The natural behavior of numbers. Presentation at Science Cafe, Northampton, September 26; last accessed July 4, 2019.
Miller, SJ (ed.) (2015). Benford's Law: Theory and Applications. Princeton University Press: Princeton and Oxford. ISSN/ISBN:978-0-691-14761-1.
Suh, I, Headrick, TC and Minaburo, S (2011). An Effective and Efficient Analytic Technique: A Bootstrap Regression Procedure and Benford's Law. Journal of Forensic & Investigative Accounting,
Vol.3, No. 3. | {"url":"https://benfordonline.net/references/up/1126","timestamp":"2024-11-05T22:47:07Z","content_type":"application/xhtml+xml","content_length":"17350","record_id":"<urn:uuid:92f341b1-2f06-4ed7-b290-d2f714a596f6>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00320.warc.gz"} |
Does the lot coverage include overhangs?
by admin
Does the lot coverage include overhangs?
Plot coverage is the percentage of plot area covered by the first floor of all buildings (including outbuildings) on the plot, including all projections, but excluding canopies, overhang Eaves, decks
and unenclosed porches.
What does batch coverage include?
Lot coverage is Percentage of total plots covered by buildings and impervious surfacesFor example, houses, garages, sheds, gazebos, swimming pools, driveways, parking lots and covered patios all
count as part of the lot coverage.
Does Lot Coverage Include Roof Overhang?
Is this overhang included in the building coverage calculation? A: Yes. If any portion of the overhang exceeds 3′, the entire overhang counts toward the building coverage (not just the portion that
exceeds 3′.)
What is the definition of batch coverage?
Batch coverage is The footprint size of buildings and/or structures on the parcel divided by the parcel size, expressed as a decimal number. Lot Coverage is used to calculate the intensity of use of
the development’s parcels. For example, 1000 square feet. 5000 square feet.
How to calculate total lot coverage?
Lot Coverage: Determined Percentage Divide (a) lot area by total area (square feet) (1) The floor area of the main building; (2) The floor area of the ancillary buildings (only for buildings with
a floor area greater than one hundred and fifty (150) square feet, or two or more storeys…
Floor area ratio explained by architect Jorge Fontan
39 related questions found
What is the maximum batch coverage?
Plot coverage refers to the percentage of plot area that is allow Covered by all buildings above ground, excluding swimming pools, and excluding parts of the lot occupied by buildings or parts
entirely below ground, and for the purposes of this…
Is the garage included in the floor area ratio?
floor area ratio Accounts for the entire floor area of the building, not just the floor area of the building. Unoccupied areas that are not included in the area calculation, such as basements,
parking lots, stairs, and elevator shafts.
What is the difference between FAR and batch coverage?
Answer: one Volume rate (FAR) is the sum of all floor area in a dwelling divided by the lot size; lot coverage is simply the sum of the floor area covered or the « bird’s eye view » of the lot (see
graph below).
How to calculate site coverage?
Site Coverage: Percentage figure of total projected area of all buildings and structures divided by site area.
Is the garage included in the FAR?
25.08. 265 Floor Area Ratio (FAR). … (1) When calculating the FAR of a batch, measurements shall include Gross floor area of primary residenceattached garage and all attached structures on the
foundation and shall include all basements with a ceiling height of six (6) feet or higher.
Does site coverage include eaves?
Floor area measurements are designed to capture the usable area within a building and therefore exclude: • Stairs, eaves, exterior awnings (including shutters and canopies), elevator shafts and void
What does it mean to cover impervious areas?
Impervious cover is Any surface in the landscape that cannot effectively absorb or penetrate rainfall. This includes driveways, roads, parking lots, roofs and sidewalks. When the natural landscape is
intact, rainwater is absorbed by soil and vegetation.
What does the plot ratio not include?
The floor area ratio considers not only the footprint of the building, but also the entire floor area of the building. Uninhabited areas such as parking lots, elevator shafts and basements Not
included in the calculation of the floor area ratio.
Are decks included in site coverage?
deck included in site coverage calculation Because it is more than 800mm higher than the existing ground level.
What is ground cover in construction?
ground cover means Zoning area occupied by all buildings Percentage of the total area of the partition. … ground coverage refers to the area of the ground covered by the building directly above
the level of the base.
How do you calculate the gross floor area?
Multiply the floor area by the number of floors in the building. Deduct the square footage of any elevator shafts, lobbies (except on the first floor), or rooms that only house equipment used for
building operations. The result is the total floor area.
Does the floor area include the garage?
Attached garages should not be included in the gross floor area A house – a garage is not a living space. …when talking about gross floor area, we can only count houses and additions – nothing else
is counted. If there is a basement, it is not included because it is not above ground.
Does the floor area include decks?
The following areas will be included in measurements provided for residential properties, but will be measured and identified separately from dedicated occupied areas: Balconies and Decksbalcony and
pergola, garage, carport, attic, cellar and separate storage area.
Are balconies included in the FAR?
Some important exceptions to FAR are public spaces, parking lots, any interior open spaces (such as balconies), basements dedicated to parking, attics, exterior spaces, sports fields, etc.These
Region not included in FAR.
What is an allowable FAR?
FAR is calculated by a simple formula – the total area covered by all floors divided by the lot area. Assuming the builder has a 1,000 sqm lot, the allowable FAR is 1.5 according to the development
plan. He was allowed to build a building on 1,500 square meters of the plot.
Are stairs included in the gross floor area?
The gross floor area is not used in the lease agreement. … floor area within the inner perimeter of the building’s exterior walls under consideration, excluding ventilation shafts and courtyards,
without deductions for corridors, stairs, ramps, closets, interior wall thickness, columns or other features.
Does the gross floor area include balconies?
Total surface area. The sum of the floor area of all spaces within a building, without exception. …the following spaces are considered outside the building and are not part of the floor area:
What is the maximum volume ratio?
The Floor Area Ratio (FAR) is a mathematical formula used to determine how many square feet a property can develop in proportion to the lot size.This Property size multiplied by FAR factor; The
result is the maximum floor area allowed for buildings on the plot.
How to calculate allowable distance?
Typically, FAR is calculated as Divide the gross floor area of the building by the gross buildable area of the land on which it is built.
What counts as building coverage?
Building coverage can be considered Footprints of all buildings on the site. Building coverage includes all roofed structures including semi-solid roofs such as trellis and arbors. In many zones,
building coverage was measured in addition to FAL/GFA.
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Blog - Investigations3
Our staff has been thinking hard about how teachers are using Investigations 3 to teach math in all of the different scenarios they are faced with this year. We’ve been visiting the remote classrooms
of teachers we’ve collaborated with previously, to learn from teachers and students who are teaching and learning math online, and to see how the rigor and coherence of the curriculum is supporting
them in that work. This series of blogs will share some of what we are learning. (Read an...
On a recent site visit, I was observing in a fourth grade classroom. The teacher started the lesson (Unit 6, Session 2.1) by writing “3/2″ on the board and asking students to name the fraction. Most
said “three halves” although one or two said “two thirds.” The teacher then displayed two blank 4 x 6 rectangles. She established that one rectangle was the whole, and asked students to use their
copy of the rectangles to draw a representation that showed 3/2. The math coach called me...
Over the last decade, much of my work has been focused on mathematical argument in the elementary classroom. Observing in our collaborating classrooms, I was struck again and again by how teachers
supported students to build on each other’s incomplete ideas. Constructing a mathematical argument is difficult and challenging for elementary students and, therefore, necessarily collaborative. When
students are learning what it means to make an argument, not just about the solution to a single... | {"url":"https://investigations.terc.edu/blog/?cat=42","timestamp":"2024-11-13T06:26:00Z","content_type":"text/html","content_length":"71978","record_id":"<urn:uuid:67319c30-e669-4250-8bf7-dd995cba651c>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00447.warc.gz"} |
Blokus Puzzles: 2 Colors 12x14
In 2016, Evan O'Dorney proposed filling a 12x14 rectangle with two colors, omitting the I5's. Presented below are different solutions with two colors, with only corner-touching of the same color, and
all pieces of each color in one continuous chain, on a 12x14 board, with no empty squares.
All of these were found by computer and are based on symmetric halves. In most cases, the two halves can be exchanged across a central axis to produce another solution.
Because of the symmetry, the two halves can be rotated 90 degrees each and rejoined along the top or bottom, creating 6x28 or 7x24 shaped solutions, as shown here:
This example, like some others, can be played in Bernard Tavitian's preferred order: which is descending by size (pentominoes first, then tetrominoes, and so on). Here is one possibility for this
U, N, T5, Y, W, Z5, F, P, X, V, L5,
L4, Z4, O4, I4, T4, V3, I3, I2, I1
Two sets of solutions are presented, one based on a 7x12 half, the other on 6x14. It is also possible to start with a 4x21 half, resulting in an 8x21 or 4x42 rectangle.
The solutions are presented based on the location of the X pentomino in (x,y) coordinates, with the upper left being (0,0).
It may be possible to find an asymmetric solution, but that would be a very difficult task.
Related pages:
10x18 with all the pieces, created out of two 10x9 halves.
12x15 with all the pieces, created out of two 6x15 halves.
13x14 with all the pieces, a slightly larger rectangle. | {"url":"http://puzzlesland.com/blokus/12x14all.html","timestamp":"2024-11-02T02:54:17Z","content_type":"text/html","content_length":"3043","record_id":"<urn:uuid:118e1326-bcf0-42e9-9763-9785721f31d0>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00132.warc.gz"} |
Stochastic Control
Stochastic Control Theory 2016
PhD course in Stochastic Control Theory based on Karl Johan Åström (2006): Introduction to Stochastic Control Theory, Dover Publications. (The first edition of the book was published by Academic
Press in 1970.) There will be nine lectures/seminars covering different aspects of stochastic control both continuous-time and discrete-time. The main emphasis is, however, on the discrete-time case.
The course is open for all PhD students and gives 7.5 ECTS credits.
Preliminary Schedule
• Monday January 18, 13:15-15:00
• Wednesday January 20, 10:15-12:00
• Monday January 25, 13:15-15:00
• Wednesday January 27, 10:15-12:00
• Wednesday February 17, 10:15-12:00
• Wednesday February 24, 10:15-12:00
• Wednesday March 2, 10:15-12:00
• Wednesday March 9, 10:15-12:00
• Monday March 14, 13:15-15:00
• Wednesday March 16, 10:15-12:00
Course Material (subject to change)
Good examples
There is a list of good examples from Åström: Introduction to Stochastic Control to practice your skills on.
Solutions to examples are available by courtesy from Tore Hägglund.
Some corrections to Åström: Introduction to Stochastic Control.
Ahlén, A. and M. Sternad (1991): "Wiener filter design using polynomial equations", IEEE Trans. on Signal Processing, SP-39, 2387-2399
Åström, K. J. (1970): Introdiction to Stochastic Control Theory, Academic Press
Åström, K. J. and B. Wittenmark (1971): "Problems of identification and control", Journal of Mathematical Analysis and Applications, 34, 90-113
Åström, K. J. and B. Wittenmark (1997): Computer-Controlled Systems, 3rd ed., Prentice Hall
Kalman, R. E. (1960): "A new approach to linear filtering and prediction problems", ASME J. Basic Eng., 82, 35-45
Levinson, N. (1947): "The Wiener RMS (Root Mean Square) error criterion in filter design and prediction", in Wiener, N: Extrapolation, Interpolation, and Smoothing of Stationary Time Series, MIT
Press, 129-148
Söderström, T.(2002): Discrete-time Stochastic Systems, 2nd ed., Springer Verlag | {"url":"http://archive.control.lth.se/Education/DoctorateProgram/stochasticcontrol2016.html","timestamp":"2024-11-13T03:18:18Z","content_type":"application/xhtml+xml","content_length":"12536","record_id":"<urn:uuid:b9fc7ba3-0fed-4464-ab30-eb64a1409212>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00796.warc.gz"} |
The Chronicles of math websites
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Providing Secrecy with Lattice Codes - IEEE Xplore - P.PDFKUL.COM
Forty-Sixth Annual Allerton Conference Allerton House, UIUC, Illinois, USA September 23-26, 2008
Providing Secrecy with Lattice Codes Xiang He
Aylin Yener
Wireless Communications and Networking Laboratory Electrical Engineering Department The Pennsylvania State University, University Park, PA 16802
[email protected] [email protected]
Abstract—Recent results have shown that lattice codes can be used to construct good channel codes, source codes and physical layer network codes for Gaussian channels. On the other hand, for Gaussian
channels with secrecy constraints, efforts to date rely on random codes. In this work, we provide a tool to bridge these two areas so that the secrecy rate can be computed when lattice codes are
used. In particular, we address the problem of bounding equivocation rates under nonlinear modulus operation that is present in lattice encoders/decoders. The technique is then demonstrated in two
Gaussian channel examples: (1) a Gaussian wiretap channel with a cooperative jammer, and (2) a multi-hop line network from a source to a destination with untrusted intermediate relay nodes from whom
the information needs to be kept secret. In both cases, lattice codes are used to facilitate cooperative jamming. In the second case, interestingly, we demonstrate that a non-vanishing positive
secrecy rate is achievable regardless of the number of hops.
I. I NTRODUCTION Information theoretic secrecy was first proposed by Shannon in [1]. In this classical model, Bob wants to send a message to Alice, which needs to be kept secret from Eve. Shannon’s
notion of secrecy requires the average rate of information leaked to Eve to be zero, with no assumption made on the computational power of Eve. Wyner, in [2], pointed out that, more often than not,
the eavesdropper (Eve) has a noisy copy of the signal transmitted from the source, and building a useful secure communication system per Shannon’s notion is possible [2]. Csiszar and Korner [3]
extended this to a more general channel model. Numerous channel models have since been studied under Shannon’s framework. The maximum reliable transmission rate with secrecy is identified for several
cases including the Gaussian wiretap channel [4] and the MIMO wiretap channel [5], [6], [7]. Sum secrecy capacity for a degraded Gaussian multiple access wiretap channel is given in [8]. For other
channels, upper bounds, lower bounds and some asymptotic results on the secrecy capacity exist. For the achievability part, Shannon’s random coding argument proves to be effective in majority of
these works. On the other hand, it is known that the random coding argument may be insufficient to prove capacity theorems for certain channels [9]. Instead, structured codes like lattice codes are
used. Using structured codes has two benefits. First, it is relatively easy to analyze large networks under these codes. For example, in [10], [11], the lattice code allows the relaying scheme to be
equivalent to a modulus sum operation, making it easy to trace the signal over a multi-hop relay network. Secondly, the structured nature of these codes makes
978-1-4244-2926-4/08/$25.00 ©2008 IEEE
it possible to align unwanted interference, for example, for the interference channel with more than two users [12], [13], and the two way relay channel [10], [11]. A natural question is therefore
whether structured codes are useful for secure communication as well. In particular, in this work, we are interested in answering two questions: 1) How do we bound the secrecy capacity when
structured codes are used? 2) Are there models where structured codes prove to be useful in providing secrecy? Relevant references in this line of thinking includes [14] and [15]. Reference [14]
considers a binary additive twoway wiretap channel where one terminal uses binary jamming signals. Reference [15] examines a wiretap channel where the eavesdropping channel is a modulus-Λ channel.
Under the proposed signaling scheme therein, the source uses a lattice code to convey the secret message, and, the destination jams the eavesdropper with a lattice code. The eavesdropper sees the sum
of these two codes, both taking value in a finite group, where the sum is carried under the addition defined over the group. It is known that if the jamming signal is sampled from a uniform
distribution over the group, then the sum is independent from the message. While these are encouraging steps in showing the impact of structured jamming signals, as commented in [15], using this
technique in Gaussian channels is a non-trivial step. In the Gaussian channel, also, the eavesdropper receives the sum of the signal from the source and the jamming signal. However, the addition is
over real numbers rather than over a finite group. The property of modulus sum is therefore lost and it is difficult to measure how much information is leaked to the eavesdropper. Most lattice codes
for power constrained transmission have a similar structure to the one used in [15]. First, a lattice is constructed, which should be a good channel code under the noise/interference. Then, to meet
the power constraint, the lattice, or its shifted version, is intersected with a bounded set, called the shaping set, to create a set of lattice points with finite average power. The lattice is
shifted to make sure sufficiently many lattice points fall into the shaping set to maintain the codebook size and hence the coding rate [16]. The decoder at the destination is called a lattice decoder
if it is only asked to find the most likely lattice point under the received signals, and is not aware of shaping set. Because of the structured nature of the lattice, a lattice decoder has lower
complexity compared to the maximum likelihood decoder
ThC5.1 where the knowledge of shaping set is used. Also, under the lattice decoder, the introduction of shaping set does not pose any additional difficulty to the analysis of decoding performance.
Commonly used shaping sets include the sphere [12] and the fundamental region of a lattice [17]. A key observation is that, from the viewpoint of an eavesdropper, the shaping set actually provides
useful information, since it reduces the set of lattice points the eavesdropper needs to consider. The main aim of this work, therefore, is to find a shaping set and lattice code construction under
which the information leaked to the eavesdropper can be bounded. This shaping set, as we shall see, turns out to be the fundamental region of a “coarse” lattice in a nested lattice structure. Under
this construction, we show that at most 1 bit is leaked to the eavesdropper per channel use. This enables us to lower bound the secrecy rate using a technique similar to the genie bound from [18]. To
demonstrate the utility of our approach, we then apply our technique to two channel models: a Gaussian wiretap channel with a cooperative jammer, and a multi-hop line network, where a source can
communicate a destination only through a chain of untrusted relays. In the second case, we demonstrate that a non-vanishing positive secrecy rate is achievable regardless of the number of hops. The
following notation is used throughout this work: We use H to denote the entropy. εk is used to denote any variable that goes to 0 when n goes to ∞. We define C(x) = 21 log2 (1 + x). a denotes the
largest integer less than or equal to a. II. T HE R EPRESENTATION T HEOREM In this section, we present a result about lattice codes which will be useful in the sequel. Let Λ denote a lattice in RN
[17], i.e., a set of points which is a group closed under real vector addition. The modulus operation x mod Λ is defined as x mod Λ = x − arg miny∈Λ d(x, y), where d(x, y) is the Euclidean distance
between x and y. The fundamental region of a lattice V is defined as the set {x : x mod Λ = 0}. It is possible that there are more than one lattice points that have the same minimal distance to x.
Breaking a tie like this is done by properly assign the boundary of V [17]. Let tA and tB be two numbers taken from V. For any set A, define 2A as 2A = {2x : x ∈ A}. Then we have: {tA + tB : tA , tB ∈
V} = 2V
Define Ax as Ax = {tA + tB + x, tA , tB ∈ V}. Then from (1), we have Ax = x + 2V. With this preparation, we are ready to prove the following representation theorem: Theorem 1: There exists a random
integer T , such that 1 ≤ T ≤ 2N , and tA +tB is uniquely determined by {T, tA + tB mod Λ}. Proof: By definition of the modulus Λ operation, we have tA + tB mod Λ = tA + tB + x, x ∈ Λ
The theorem is equivalent to finding the number of possible x meeting equation (2) for a given tA + tB mod Λ.
To do that, we need to know a little more about the structure of lattice Λ. Every point in a lattice, by definition, can be N represented in the following form [19]: x = ai vi , vi ∈ i=1
RN , ai ∈ Z. {ai } is said to be the coordinates of the lattice point x under the basis {vi }. Based on this representation, we can define the following relationship: Consider two points x, y ∈ Λ,
with coordinates {ai } and {bi } respectively. Then we say x ∼ y if ai = bi mod 2, i = 1...N . It is easy to see the relationship ∼ is an equivalence relationship. Therefore, it defines a partition
over Λ. 1) Depending on the values of ai − bi mod 2, there are 2N sets in this partition. 2) The sub-lattice 2Λ is one set in the partition, whose members have even coordinates. The remaining 2N − 1
sets are its cosets. Let Ci denote any one of these cosets or 2Λ. Then Ci can expressed as Ci = 2Λ + yi , yi ∈ Λ. It is easy to verify that Ax = x + 2V, x ∈ Ci is a partition of 2RN + yi , which
equals RN . We proceed to use the two partitions derived above: Since Ci , i = 1...2N is a partition of Λ, (2) can be solved by considering the following 2N equations: tA + tB mod Λ = tA + tB + x, x
∈ Ci
From (1), this means tA + tB mod Λ ∈ x + 2V for some x ∈ Ci . Since x + 2V, x ∈ Ci is a partition of RN , there is at most one x ∈ Ci that meets this requirement. This implies for a given tA + tB mod
Λ, and a given coset Ci , (3) only has one solution for x. Since there are 2N such equations, (2) has at most 2N solutions. Hence each tA + tB mod Λ corresponds to at most 2N points of tA + tB .
Remark 1: Theorem 1 implies that modulus operation looses at most one bit per dimension of information if tA , tB ∈ V. The following crypto lemma is useful and is provided here for completeness.
Lemma 1: [15] Let tA , tB be two independent random variables distributed over the a compact abelian group, tB has a uniform distribution, then tA + tB is independent from tA . Here + is the addition
over the group. In the remainder of the paper, (Λ, Λ1 ) denotes a nested lattice structure where Λ1 is the coarse lattice. Let V and V1 be their respective fundamental regions. We shall use a ⊕ b,
short for a + b mod Λ1 . Then from Lemma 1, we have the following corollary: Corollary 1: Let tA ∈ Λ ∩ V1 . tB ∈ Λ ∩ V1 and tB is uniformly distributed over Λ ∩ V1 . Let tS = tA ⊕ tB . Then tS is
independent from tA . III. W IRETAP C HANNEL WITH A C OOPERATIVE JAMMER In this section, we demonstrate the use of lattice codes for secrecy in the simple model depicted in Figure 1. Nodes S, D, E
form a wiretap channel where S is the source node, D is the destination node, E is the eavesdropper. Let the average power constraint of node S be P . Now suppose
ThC5.1 Z1 1
Fig. 1.
N N N N N N =H(tN A |tA ⊕ dA + tB ⊕ dB , dA , dB ) N N N N N N =H(tN A |tA ⊕ dA ⊕ tB ⊕ dB , dA , dB , T ) N N N N =H(tN A |tA ⊕ tB , dA , dB , T ) N N N =H(tA |tA ⊕ tB , T ) N N N N N =H T |tN A ⊕ tB
, tA + H tA |tA ⊕ tB −
D 1
N N N N N N N N ≥H(tN A |tA ⊕ dA + tB ⊕ dB + Z2 , dA , dB , Z2 )
Z2 E
Wiretap Channel with a Cooperative Jammer, CJ
that there is another transmitter CJ in the system, also with power constraint P , as shown in Figure 1. We assume that the interference caused by CJ to node D is either too weak or too strong that
it can be ignored or removed, and consequently there is no link between CJ and D. In this model, node CJ may choose to help S by transmitting a jamming signal to confuse the eavesdropper E. Below, we
derive the secrecy rate for this case when the jamming signal is chosen from a lattice codebook. A. Gaussian Noise We first consider the case when Z1 and Z2 are independent Gaussian random variables
with zero mean and unit variance. In this case, we have the following theorem: Theorem 2: A secrecy rate of [C(P ) − 1]+ is achievable. Proof: The codebook is constructed as follows: Let (Λ, Λ1 ) be
a properly designed nested lattice structure in RN as described in [17]. The codebook is all the lattice points within the set Λ ∩ V1 . Let tN A be the lattice point transmitted by node S. Let N dA
be the dithering noise uniformly distributed over V1 . The N transmitted signal is given by tN A ⊕ dA . The receiver receives the above signal corrupted by Gaussian noise and tries to ˆN decode tN A
. Let the decoding result be tA . Then as shown in [17, Theorem 5], there exists a sequence of properly designed (Λ, Λ1 ) with increasing dimension, such that 1 log2 |Λ ∩ V1 | < C(P ) N 1 C(P ) =
log2 (1 + P ) 2 lim
N →∞
(8) (9) (10) (11) N H T |tA ⊕ tN B (12) (13) (14) (15)
In (9), we introduce the N bit information T that will help N N N N N N N to recover tN A ⊕ dA + tB ⊕ dB from tA ⊕ dA ⊕ tB ⊕ dB . In N N (14), we use the fact that tA is independent from tA ⊕ tN B
based on Corollary 1. N N N N N N N Let c = N1 I tN A ; tA ⊕ dA + tB ⊕ dB + Z2 , dA , dB . Then from (15), since H(T ) ≤ N , we have c ≤ 1. Therefore, if the message is mapped one-to-one to tN A ,
then an equivocation rate of at least C(P ) − 1 is achievable under a transmission rate of C(P ) bits per channel use. We note that to obtain perfect secrecy, some additional effort is required.
First, we define a block of channel uses as the N channel uses required to transmit a N dimensional lattice point. A perfect secrecy rate of C(P ) − 1 can then be achieved by coding across multiple
blocks: A codeword in this case is composed of Q components, each component is an N dimensional lattice point sampled from a uniform distribution over V1 ∩ Λ in an i.i.d. fashion. The resulting
codebook C contains 2N QR codewords with R < C(P ). Like wiretap codes, the codebook is then randomly binned into several bins, where each bin contains 2N Qc codewords. The secret message W is mapped
to the bins. The actual transmitted codeword is chosen from that bin according to a uniform distribution. Let YeN Q denote the signals available to the eavesdropper: Q Q Q Q Q NQ YeN Q = {tN ⊕ dN +
tN ⊕ dN + Z N Q , dN A A B B A , dB }. Then we have
ˆN and limN →∞ Pr(tN A = tA ) = 0. The cooperative jammer CJ uses the same codebook as node S. Let the lattice point transmitted by CJ be tN B and the dithering noise be dN . The transmitted signal
is given B N N by tN ⊕ d . As in [17], we assume that d is known by B B A node S, the legitimate receiver node D and the eavesdropper node E. dN B is known by node S, and the eavesdropper node E.
Hence, there is no common randomness between the legitimate communicating pairs that is not known by the eavesdropper. Then the signal received by the eavesdropper can be N N N N N represented as tN
A ⊕ dA + tB ⊕ dB + Z2 , where Z2 is the Gaussian channel noise over N channel uses. Then we have N N N N N N N H(tN A |tA ⊕ dA + tB ⊕ dB + Z2 , dA , dB )
N N N N ≥H tN A |tA ⊕ tB − H T |tA ⊕ tB N N =H tN A − H T |tA ⊕ tB ≥H tN A − H (T )
H(W |YeN Q , C)
Q Q NQ NQ =H(W |tN , C) + H(tN , C) A , Ye A |Ye Q NQ , C) − H(tN A |W, Ye
Q NQ ≥H(tN , C) A |Ye NQ NQ =H(tA |Ye , C)
− N Qε −
Q H(tN A |C)
(17) +
Q H(tN A |C)
Q NQ NQ =H(tN |C) − N Qε A |C) − I(tA ; Ye Q Q N ≥H tN I tN A ; Ye |C − N Qε A |C −
− N Qε (18) (19) (20)
Q |C − QN c − N Qε = QN (R − c) − N Qε =H tN A (21) In (17), we use Fano’s inequality to bound the last term in (16). This is because the size of each bin is kept small enough Q such that given W ,
the eavesdropper can determine tN A from
ThC5.1 its received signal YeN Q . Using the standard random coding argument and (21), it can then be shown a secrecy rate of C(P ) − c is achievable. Since c < 1, this means a secrecy rate of at
least C(P ) − 1 bits per channel use is achievable. Remark 2: It is interesting to compare the secrecy rate obtained here with that obtained by cooperative jamming with Gaussian noise [20]. The
latter is given by C(P ) − C( PP+1 ). limP →∞ C( PP+1 ) = 0.5. Therefore there is at most 0.5 bit per channel use of loss in secrecy rate at high SNR by using a structured code book as the jamming
signal. B. Non-Gaussian Noise The performance analysis in [17] requires Gaussian noise. This is not always the case, for example, in the presence of interference, which is not necessarily Gaussian.
For nonGaussian noise, in principle, the analysis in [16] can be used instead. On the other hand, in [16], a sphere is used as the shaping set, making it difficult to computing the equivocation rate
via Theorem 1. We show below, if the code rate R has the form log2 t, t ∈ Z+ , then a scaled lattice tΛ of the fine lattice Λ can be used for shaping instead. Theorem 3: If Z1 , Z2 are i.i.d.
continuous random variables with differential entropy√h(E), such that 22h(E) = 2πe, then a secrecy rate of [log2 P − 1]+ is achievable. Proof: We need to show that there exists a fine lattice Λ that
has a good decoding performance [16, Theorem 6], and Λ is close to a sphere in the sense that 1 lim h(S) = log2 (2πeP ) (22) N →∞ 2 where h(S) = N1 log2 |V|, |V| isthe volume of the fundamen2 tal
region of Λ, and P = N 1|V| x∈V x dx. It is shown in [21] that when a lattice is sampled from the lattice ensemble defined therein, it is close to a sphere in the sense of (22). The lattice ensemble
is generally called construction A [16], whose generation matrices are all matrix of size K × N over finite group GF(q), with q being a prime. The lattice sampled from the ensemble is “good” in
probability when q, N → ∞ and K grows faster than log2 N [21, (25)-(28)]. Note that this property of “goodness” is invariant under scaling. Therefore, we can scale the lattice so that the volume of
its fundamental region remains fixed when its dimension N → ∞. This gives us a sequence of lattice ensembles that meet the condition of [13, Lemma 1]: (1) N → ∞ (2) q → ∞. (3) Each lattice ensemble of
a given dimension is balanced [16]. This means when N → ∞, at least 3/4 of the lattice ensemble is good for channel coding [13, Lemma 1]. The lattice decoder will have a positive decoding error
exponent as long as |V| > 2N h(E) . Combined, this means there must exist a lattice Λ∗ that is close to a sphere and is a good channel code at the same time. Hence we have N1 log2 |V| → 12 log2 (2πeP
) as N → ∞. Since we assume h(E) = 12 log2 (2πe) and require |V| > 2N h(E) , this means as long as P > 1, the decoding error will decrease exponentially when N → ∞. Now pick the shaping set to be the
fundamental region of tΛ∗ , t ∈ Z+ . Then the code rate R = log2 (t) [17]. With the dithering and modulus operation from [17], the average
power of the transmitted signal per dimension is t2 P . Note that the modulus operation at the destination, required in order to remove the dithering noise, may distort the additive channel noise.
However, the decoding error event, defined as the noise pushing a lattice codeword into the set of typical noise sequence centered on a different lattice point [16], remains identical. Therefore, the
decoding error exponent 2 is the same. Hence we √ have P > 1 and t P < P√. The largest possible t is P , with the rate being log2 ( P ). With similar arguments√as in Theorem 2, we conclude that a
secrecy rate of [log2 ( P ) − 1]+ is achievable. IV. M ULTI - HOP L INE N ETWORK WITH U NTRUSTED R ELAYS A. System Model In this section, we examine a more complicated communication scenario, as
shown in Figure 2. The source has to communicate over K − 1 hops (K ≥ 3) to reach the destination. Yet the intermediate relaying nodes are untrusted and need to be prevented from decoding the source
information. Under this model, we will show that, using Theorem 1, with lattice codes for source transmission and jamming signals and an appropriate transmission schedule, an end-to-end secrecy rate
that is independent of the number of untrusted relay nodes is achievable. We assume nodes can not receive and transmit signals simultaneously. We assume that each node can only communicate to its two
neighbors, one on each side. Let Yi and Xi be the received and transmitted signal of the ith node respectively. Then they are related as Yi = Xi−1 +Xi+1 +Zi , where Zi are zero mean Gaussian random
variables with unit variance, and are independent from each other. Each node has S
Fig. 2.
A Line Network with 3 Un-trusted Relays
the same average power constraint: n1 nk=1 E Xi (k)2 ≤ P¯ where n is the total number of channel uses. The channel gains are normalized for simplicity. We consider the case where there is an
eavesdropper residing at each relay node and these eavesdroppers are not cooperating. This also addresses the scenario where there is one eavesdropper, but the eavesdropper may appear at any one
relay node that is unknown a priori. In either case, we need secrecy from all relays and the secrecy constraints for the K relay nodes are expressed as lim n1 H (W |Yin ) = 1 H n→∞ n
(W ) , i = 1...K.
B. Signaling Scheme Because all nodes are half duplex, a schedule is necessary to control when a node should talk. The node schedule is best represented by the acyclic directional graph as shown in
Figure 3. The columns in Figure 3 indicate the nodes and the rows in Figure 3 indicate the phases. The length of a phase is the number of channel uses required to transmit a lattice point, which
equals the dimension of the lattice. A
ThC5.1 node in a row has an outgoing edge if it transmits during a phase. The node in that row has an incoming edge if it can hear signals during the previous phase. It is understood, though not
shown in the figure, that the signal received by the node is a superposition of the signals over all incoming edges corrupted by the additive Gaussian noise. A number of consecutive phases is called
one block, as shown in Figure 3. The boundary of a block is shown by the dotted line in Figure 3. The data transmission is carried over M blocks.
One block of channel uses
J−1 J0
J0 t0 + J0 t0 + J1 t0 + t1 + J1
t0 + J1 t0 + J2
t0 + t1 + J2
J2 t0 + J2 t0 + J3
t0 + t1 + J3
J2 J3 t0 + J3 t0 + J4 t0 + t1 + J4
node index which transmit this signal. If this is the first time the relay transmits during this block, then tN k is drawn from a uniform distribution over Λ ∩ V1 , and all previous received signals
are ignored. Otherwise, tN k is computed from the signal it received during the previous phase. This will be clarified in the sequel. dN k again is the dithering noise uniformly distributed over V1 .
The signal received by the relay within a block can be categorized into the following three cases. Let z N denote the Gaussian channel noise. 1) If this is the first time the relay receives signals
during N N this block, then it has the form (tN A ⊕ dA )+ z . It only contains interference from its left neighbor. 2) If this is the last time the relay receives signals during N N this block, then
it has the form (tN B ⊕ dB )+ z . It only contains interference from its right neighbor. N N 3) Otherwise it has the form ykN = (tN A ⊕ dA ) + (tB ⊕ N N dB ) + z . N N N Here tN A , tB are lattice
points, and dA , dB are dithering noises. Following reference [10], if the lattice is properly designed and the cardinality of the set Λ ∩ V1 is properly chosen, then for case (3), the relay, with
the knowledge of N N N dN A , dB , will be able to decode tA ⊕ tB . For case (1) and N (2), the relay will be able to decode tA and tN B respectively. Otherwise, we say that a decoding error has
occurred at the relay node. The transmitted signal at the relay node is then computed as follows: N N N xN = tN A ⊕ tB ⊕ (−x ) ⊕ dC
Fig. 3.
One Block of Channel Uses
Again the nested lattice code (Λ, Λ1 ) from [10] is used within each block. The codebook is constructed in the same fashion as in Section III. 1) The Source Node: The input to the channel by the
source has the form tN ⊕ J N ⊕ dN . Here dN is the dithering noise which is uniformly distributed over V1 . tN and J N are determined as follows: If it is the first time the source node transmits
during this block, tN is the origin. J N is picked from the lattice points in Λ∩V1 under a uniform distribution. Otherwise, tN is picked by the encoder. J N is the lattice point decoded from the
jamming signal the source received during the previous phase. This design is not essential but it brings some uniformness in the form of received signals and simplifies explanation. 2) The Relay Node:
As this signal propagates toward the destination, each relay node, when it is its turn, sends a N jamming signal in the form of tN k +dk mod Λ, k = 2...K −1, where K is the number of nodes. Subscript
k denotes the
Here x is the lattice point contained in the jamming signal transmitted by this relay node during the previous phase. − is N the inverse operation defined over the group V1 ∩ Λ. tN A ⊕ tB are decoded
from the signal it received during the previous phase. In Figure 3, we labeled the lattice points transmitted over some edges. For clarity we omitted the superscript N . The + signs in the figure are
all modulus operations. The reason why we have (−xN ) in (23) is now apparent: it leads to a simple expression for the signal as it propagates from the relay to the destination. 3) The Destination:
As shown in Figure 3, the destination behaves identically to a relay node when it computes its jamming signal. It is also clear from Figure 3 that the destination will be able to decode the data from
the source. This is because the lattice point contained in the signal received by the destination has the form tN ⊕ J N , where tN is the lattice point determined by the transmitted data, and J N is
the lattice point in the jamming signal known by the destination. C. A Lower Bound to the Secrecy Rate Suppose the source transmits Q + 1 times within a block. Then each relay node receives Q+2
batches of signals within the block. An example with Q = 2 is shown in Figure 3. Given the inputs from the source of the current block, the signals received by the relay node are independent from
ThC5.1 the signals it received during any other block. Therefore, if a block of channel uses is viewed as one meta-channel use, with the source input as the channel input and the signal received by
the relay as the channel output, then the effective channel is memoryless. Each relay node has the
tN A1
xN A1 tN A2
xN A2 tN A3
xN A3
the subscript in product includes the indices of all the relay node and the indices of the phases in this block. For any given block length Q, we have limN →∞ P¯e = 0. Note that P¯e is just a
function of N and Q. Because there are only finite number of relay nodes, this convergence is uniform over all relay nodes. Let the equivocation under error free decoding be
The relay node under consideration
¯2 = H M (¯ xN Ai
xN A1
M NM dN αi , dβ(i−1) , i = 2...Q + 1 NM NM M NM NM NM (t¯N D(Q+1) ⊕ dβ(Q+1) ) + zQ+1 , dβ(Q+1) , tB1 , db1 ) (26)
tN B1
tN B1 xN A2
tN D1
M M where x ¯N equals the value xN takes with error free Ai Ai N M N M decoding. t¯D(i−1) and t¯D(Q+1) are defined in a similar fashion. Then we have the following lemma: ¯ 2 −ε1 where ¯ 2 +ε2 ≥ H2 ≥
H Lemma 2: For a given Q, H ε1,2 → 0 as N, M → ∞. Proof: Let cj , cˆj denote the part of signals received by the relay node within the jth block. More specifically, they have the following form:
tN B2
tN B2 xN A3
tN D2
tN B3
1 M NM NM M H(W |(xN , dN A1 ⊕ dα1 ) + z1 α1 NM M NM NM ¯N M , ⊕ dN αi ) + (tD(i−1) ⊕ dβ(i−1) ) + zi
tN B3 tN D3
N cˆj = {(xN Ai (j) ⊕ dαi (j))+
Fig. 4.
N N (tN D(i−1) (j) ⊕ dβ(i−1) (j)) + zi (j), i = 2...Q + 1}
Notations for Lattice Points contained in Signals, Q = 2
following side information regarding the source inputs within one block: 1) Q + 2 batches of received signals. 2) All the dithering noises {di }. 3) Signals transmitted from the relay node during
this block. Note that only the first batch of signals it transmitted may provide information because all subsequent transmitted signals are computed from received signals and dithering noises. Let W
be the secret message transmitted over M blocks. Following the notation in Figure 4, the equivocation with respect to the relay node is given by: 1 M NM NM M H(W |(xN , dN A1 ⊕ dα1 ) + z1 α1 NM M NM
NM NM NM (xN , Ai ⊕ dαi ) + (tD(i−1) ⊕ dβ(i−1) ) + zi H2 =
M NM dN αi , dβ(i−1) , i = 2...Q + 1 NM NM NM NM M NM (tN D(Q+1) ⊕ dβ(Q+1) ) + zQ+1 , dβ(Q+1) , tB1 , db1 ) (24)
Define the block error probability as P¯e = Pr(∃i ∈ {2...Q + 1}, s.t.xN Ai is in error,
N or tN D(i−1) is in error, or tD(Q+1) is in error.)
xN Ai
M xN that is within one Ai N for tN D(i−1) and tD(Q+1) .
where is the part of block. Similar notations are used Given the signaling scheme presented in section IV-B and [17, Theorem 2], the probability of decoding error at each relay node goes to zero as N
→ ∞. Let Pe (i, k) be the probability of decoding error at relay node i during phase k. Then P¯e ¯ is related to Pe (i, k) as Pe ≤ 1 − i,k (1 − Pe (i, k)), where
{(¯ xN Ai (j)
dN αi (j))+
c = ⊕ N N ¯ (tD(i−1) (j) ⊕ dN β(i−1) (j)) + zi (j), i = 2...Q + 1}
In this notation, we exclude the first and the last batch of received signals. The first batch of received signals does not undergo any decoding operation. For the last batch of received signals we
have the following notation: N N fˆj = (tN D(Q+1) (j) ⊕ dβ(Q+1) (j)) + zQ+1 (j) N N f j = (t¯N D(Q+1) (j) ⊕ dβ(Q+1) (j)) + zQ+1 (j)
(29) (30)
The block index (j) will be omitted in the following discussion for clarity. We first prove that cj − cˆj is a discrete random variable with a finite support. According to the notation of (28), cj −ˆ
cj has Q components. Each component can be expressed as N N N x ¯Ai ⊕ dN αi − xAi ⊕ dαi + N N N (t¯N (31) D(i−1) ⊕ dβ(i−1) ) − (tD(i−1) ⊕ dβ(i−1) ) For the first line of (31) we have N N N x ¯Ai ⊕ dN
αi − xAi ⊕ dαi N N N N N =¯ xN Ai + dαi + x1 − xAi + dαi + x2 =¯ xN Ai
xN Ai
xN 1
xN 2
(32) (33) (34)
N where xN 1 , x2 belong to the coarse lattice Λ1 . Applying N N Theorem 1, we note that xN 1 and x2 each has at most 2 N N possible solutions. x ¯Ai and xAi each take V1 ∩ Λ possible values. Let R =
N1 log2 V1 ∩ Λ . Then (32) takes at most 22N (R+1) possible values. Similarly, we can prove that the second line of (31) has at most 22N (R+1) possible values as well. Therefore cj − cˆj takes at
most 24N Q(R+1) possible values. Therefore H cj − cˆj ≤ 4N Q(R + 1). Similarly, it
ThC5.1 M NM NM t¯N ⊕ Jk+1 D2 = t0 M NM M NM ⊕ tN ⊕ Jk+2 t¯N D3 = t0 1
can be shown that f − fˆ has at most 2N (R + 1) solutions. This means that H(cj − cˆj , f j − fˆj ) ≤ (4Q + 2)N (R + 1)
Let c = {cj }, cˆ = {ˆ cj }, f = {f j } and fˆ = {fˆj } j = 1...M . Let b denote the remaining conditioning terms in H2 . Let E j denote the random variable cj = cˆj or f j = fˆj . Then with
probability P¯e that E j = 1. Otherwise E j = 0. Let W be the message transmitted over the M blocks. Then we have H(W |b, cˆ, fˆ) ≥H(W |b, c, cˆ, f, fˆ)
=H(W |b, c, f, c − cˆ, f − fˆ)
=H(W |b, c, f) + H(c − cˆ, f − fˆ|W, b, c, f) − H(c − cˆ, f − fˆ|b, c, f)
≥H(W |b, c, f) − H(c − cˆ, f − fˆ)
≥H(W |b, c, f) −
H(cj − cˆj , f j − fˆj )
H(cj − cˆj , f j − fˆj , E j )
=H(W |b, c, f ) −
≥H(W |b, c, f ) − −
... M NM M NM M ⊕ ...tN t¯N ⊕ tN 1 Q−1 ⊕ Jk+Q D(Q+1) = t0
M tN B1
NM Jk−1
the Given the lattice points transmitted by the source joint distribution of the side information for any relay node is the same. Hence we have the lemma. With these preparation, we are now ready to
present the following achievable rate. Theorem 4: For any ε > 0, a secrecy rate of at least 0.5(C(2P¯ − 0.5) − 1) − ε bits per channel use is achievable regardless of the number of hops. Proof:
According to Lemma 3, it suffices to design the coding scheme based on one relay node. We focus on one block of channel uses as shown in Figure 3. Let V (j) to denote all the side information
available to the relay node within the jth block. We start by lower bounding Q NQ H(tN 0 |V (j)) under ideal error free decoding, where t0 are the lattice points picked by the encoder at the source
node Q as described in Section IV-B within this block. H(tN 0 |V (j)) equals Q N N N ¯N H(tN xN Ai ⊕ dαi ) + (tD(i−1) ⊕ dβ(i−1) ) + zi , 0 |(¯ N N N dN αi , dβ(i−1) , i = 2...Q + 1, tB1 , db1 )
H(E j )
j=1 M
M , tN j
j=1 M
Pr(E j = 1)H(cj − cˆj , f j − fˆj )
≥H(W |b, c, f) − M − M P¯e (4Q + 2)N (R + 1)
By dividing N M on both sides and letting N, M → ∞, and ¯ 2 − ε1 . ε1 = 1/N + P¯e (4Q + 2)(R + 1) we get H2 ≥ H ¯ 2 ≥ H2 − ε 2 . Similarly we can prove H Remark 3: Lemma 2 says that if a particular
equivocation value is achievable with regard to one relay node, when all the other relay nodes do error free decoding, then the same equivocation value is achievable when other relay nodes do decode
and forward which is only error free in asymptotic sense. ¯ 2 is the same for all relay nodes. Lemma 3: H Proof: Lemma follows because relay nodes receive statistically equivalent signals if there
are no decoding errors. For the kth relay node, as shown by the edge labels in ¯ 2 in (26) is related to tN M Figure 3, the condition term of H j as follows: M NM xN A1 = Jk−2
M NM NM x ¯N ⊕ Jk−1 A2 = t0
M NM M x ¯N ⊕ tN ⊕ JkN M A3 = t0 1 ... M NM M M NM x ¯N ⊕ tN ⊕ ... ⊕ tN 1 Q−1 ⊕ JK+Q−2 A(Q+1) = t0 M NM t¯N D1 = Jk
Comparing (53) with the condition terms in (26), we see that we have removed the first batch and the last batch of received signals during a block from the condition terms because they are independent
from everything else. The last batch of received signals contains the lattice point of the most recent jamming signal observable by the relay node. Its independence follows from Lemma 1. We then
assume that the eavesdropper residing at the relay node knows the channel noise. This means (53) can be lower bounded by: Q N N ¯N xN H(tN Ai ⊕ dαi ) + (tD(i−1) ⊕ dβ(i−1) ), 0 |(¯ N N N dN αi , dβ
(i−1) , i = 2...Q + 1, tB1 , db1 )
Q N N ¯N xAi ⊕ dN H(tN αi ⊕ tD(i−1) ⊕ dβ(i−1) , Ti , 0 |¯ N N N dN αi , dβ(i−1) , i = 2...Q + 1, tB1 , db1 )
where Ti can be represented with N bits. Using the similar argument as in (9)-(13), (55) is lower bounded by: Q N N ¯N xAi ⊕ dN H(tN αi ⊕ tD(i−1) ⊕ dβ(i−1) , 0 |¯ N N N dN αi , dβ(i−1) ,i=2...Q+1 ,
tB1 , db1 ) − H(Ti ,i=2...Q+1 ) (56)
Q N N xAi ⊕ t¯N =H(tN 0 |¯ D(i−1) ,i=2...Q+1 , tB1 ) − H(Ti ,i=2...Q+1 ) (57)
Next, we invoke Theorem 1. Equation (54) can be lower bounded by:
It turns out that in the first term in (57), the conditional Q variables are all independent from tN 0 . This is because N t¯N D(i−1) contains Ji−2+k , which is a new lattice point not
ThC5.1 contained in previous t¯N ¯N Aj j < i. The new lattice D(j−1) or x point is uniformly distributed over V1 ∩ Λ. Therefore, from NQ ¯N Lemma 1, x¯N Ai ⊕ tD(i−1) is independent from t0 .
Therefore (57) equals Q H(tN 0 ) − H(Ti ,i=2...Q+1 )
Define c=
1 I(tN Q ; V (j)) NQ 0
Then from (58), we have c ∈ (0, 1). To achieve perfect secrecy, a similar argument of coding across different blocks as the one in Section III can be used. A codebook with rate R and size 2MN QR that
spans over M blocks is constructed as follows: Each codeword is a length M Q sequence. Each component of the sequence is an N -dimensional lattice point sampled in an i.i.d fashion from the uniform
distribution over V1 ∩ Λ. The codebook is then randomly binned into several bins. Each bin contains 2MN Qc codewords, with c given by (59). Denote the codebook with C. The transmitted codeword is
determined as follows: Consider a message set {W }, whose size equals the number of the bins. The message is mapped to the bins in a one-to-one fashion. The actual transmitted codeword is then
selected from the bin according to a uniform distribution. Let this codeword be uMN Q . Let V = {V (j), j = 1...M }. Then we have: H (W |V, C) =H W |uMN Q , V, C + H uMN Q |V, C − H uMN Q |W, V, C ≥H
uMN Q |V, C − M N Qε MN Q |C − I uMN Q ; V |C − M N Qε =H u
(60) (61) (62) (63)
M ≥H uMN Q |C − I uMN Q (j); V (j) − M N Qε (64) j=1
=H uMN Q |C − M N Qc − M N Qε
(62) follows from Fano’s inequality and the size of the bin is picked according to the rate of information leaked to the eavesdropper under the same input distribution used to sample the codebook.
(64) follows from C → uMN Q → V being a Markov chain. Divide (60) and (65) by M N Q and let 1 M → ∞, we have ε → 0 and limM→∞ MN Q H(W |V, C) = 1 limM→∞ MN Q H(W ). Therefore a secrecy rate of R − c
bits per channel use is achieved. According to [10], R can be arbitrarily close to C(P −0.5) by making N → ∞, where P is the average power per channel use spent to transmit a lattice point. For a
given node, during 2Q + 3 phases, it is active in Q + 1 phases. Since c ∈ [0, 1], a secrecy rate of Q+1 2Q+3 ¯ 2Q+3 (C( Q+1 P − 0.5) − 1) is then achievable by letting M → ∞. Taking the limit Q → ∞,
we have the theorem. V. C ONCLUSION Lattice codes were shown recently as a useful technique to prove information theoretic results. In this work, we showed
that lattice codes are also useful to prove secrecy results. This was done by showing that the equivocation rate could be bounded if the shaping set and the “fine” lattice forms a nested lattice
structure. With this new tool, we computed the secrecy rate for two models: (1) a wiretap channel with a cooperative jammer, (2) a multi-hop line network with untrusted relays. For the second model,
we have shown that a coding scheme can be designed to support a non-vanishing secrecy rate regardless of the number of hops. R EFERENCES [1] C. E.. Shannon. Communication Theory and Secrecy Systems.
Bell Telephone Laboratories, 1949. [2] A. D. Wyner. The Wire-tap Channel. Bell System Technical Journal, 54(8):1355–1387, 1975. [3] I. Csiszar and J. Korner. Broadcast Channels with Confidential
Messages. IEEE Transactions on Information Theory, 24(3):339–348, 1978. [4] S. Leung-Yan-Cheong and M. Hellman. The Gaussian Wire-tap Channel. IEEE Transactions on Information Theory, 24(4):451–456,
1978. [5] A. Khisti and G. Wornell. Secure Transmission with Multiple Antennas: The MISOME Wiretap Channel. Submitted to IEEE Transactions on Information Theory, 2007. [6] S. Shafiee, N. Liu, and S.
Ulukus. Towards the Secrecy Capacity of the Gaussian MIMO Wire-tap Channel: The 2-2-1 Channel. Submitted to IEEE Transactions on Information Theory, 2007. [7] F. Oggier and B. Hassibi. The Secrecy
Capacity of the MIMO Wiretap Channel. IEEE International Symposium on Information Theory, 2008. [8] E. Tekin and A. Yener. The Gaussian Multiple Access Wire-tap Channel. IEEE Transaction on
Information Theory, to appear, December 2008. [9] B. Nazer and M. Gastpar. The Case for Structured Random Codes in Network Capacity Theorems. European Transactions on Telecommunications, Special
Issue on New Directions in Information Theory, 2008. [10] K. Narayanan, M.P. Wilson, and A. Sprintson. Joint Physical Layer Coding and Network Coding for Bi-Directional Relaying. Allerton Conference
on Communication, Control, and Computing, 2007. [11] W. Nam, S-Y Chung, and Y.H. Lee. Capacity Bounds for Twoway Relay Channels. Internation Zurich Seminar on Communications, 2008. [12] G. Bresler,
A. Parekh, and D. Tse. the Approximate Capacity of the Many-to-one and One-to-many Gaussian Interference Channels. Allerton Conference on Communication, Control, and Computing, 2007. [13] S.
Sridharan, A. Jafarian, S. Vishwanath, and S.A. Jafar. Capacity of Symmetric K-User Gaussian Very Strong Interference Channels. 2008. http://www.citebase.org/abstract?id=oai:arXiv.org:0808.2314. [14]
E. Tekin and A. Yener. Achievable Rates for Two-Way Wire-Tap Channels. International Symposium on Information Theory, 2007. [15] L. Lai, H. El Gamal, and H.V. Poor. The Wiretap Channel with Feedback:
Encryption over the Channel. 2007. IEEE Transaction on Information Theory, to appear. [16] H.A. Loeliger. Averaging bounds for lattices and linear codes. IEEE Transaction on Information Theory, 43
(6):1767–1773, 1997. [17] U. Erez and R. Zamir. Achieving 1/2 log (1+ SNR) on the AWGN Channel with Lattice Encoding and Decoding. IEEE Transactions on Information Theory, 50(10):2293–2314, 2004.
[18] S.A. Jafar. Capacity with Causal and Non-Causal Side Information - A Unified View. IEEE Transactions on Information Theory, 52(12):5468–5475, 2006. [19] J.H. Conway and N.J.A. Sloane. Sphere
Packings, Lattices and Groups. Springer, 1999. [20] E. Tekin and A. Yener. The General Gaussian Multiple Access and Two-Way Wire-Tap Channels: Achievable Rates and Cooperative Jamming. IEEE
Transactions on Information Theory, 54(6):2735– 2751, June 2008. [21] U. Erez, S. Litsyn, and R. Zamir. Lattices Which Are Good for (Almost) Everything. IEEE Transactions on Information Theory, 51
(10):3401–3416, 2005. | {"url":"https://p.pdfkul.com/providing-secrecy-with-lattice-codes-ieee-xplore_59e269f21723dd500645a0d9.html","timestamp":"2024-11-07T16:05:46Z","content_type":"text/html","content_length":"100424","record_id":"<urn:uuid:37cb8d56-252e-4d6a-ab3a-52b1479a0d04>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00099.warc.gz"} |
Electronic Components Shipments Industrial Marketplace
From: John Conover <john@email.johncon.com>
Subject: Electronic Components Shipments Industrial Marketplace
Date: Wed, 14 Oct 1998 18:44:38 -0700
Although executive management is concerned with many issues, the most
significant issue in a corporation's durability is operational
strategy-which means controlling cash flow. It is a difficult
financial issue to address since cash inflow fluctuates month to
month, do to the market environment, etc. However, these fluctuations
do have general characteristics that can be used in strategic
financial planning.
To demonstrate a methodology for analyzing industrial market
fluctuations, I will derive the metrics for the electronic component
market in the United States. The data used readily available from the
US Department of Commerce, and contains the monthly shipments of
electronic components between January, 1979 and January, 1994,
inclusive. (The specific data used is labeled "MSEL367X, Electronic
Components and Accessories Shipments, by Millions of Dollars," and is
available on line from http://www.doc.gov/.)
The objective is to develop a conceptual model for the fluctuations in
industrial markets that:
1) Can be used in financial planning-specifically, we would like
to have an understanding of the duration of industrial market
expansions and contractions.
2) Should be conceptually useful-specifically, it must be
intuitive, and usable without elegant numerical
methods. Preferably, any calculations should be simple enough that
they could be done in one's head.
3) Should be capable of working with limited data-the reality is
that we do not have adequately concise data for industrial markets
in the US. Preferably, the model should be able to predict what
would be observed with such limited data.
What I will do is to state the model, and then offer a compelling and
lengthy argument as to why it is correct.
The model is that duration of the fluctuations in industrial markets
have a probability of 1 / sqrt (t). It is not complicated. What's the
probability of a "recession" lasting longer than four months? 1 / sqrt
(4) = 1 / 2 = 50%. And longer than two months? 1 / sqrt (2) = 0.707 =
71%. How about longer than seven months? 1 / sqrt (7) = 0.378 = 38%.
Note exactly what the model says-as an example, if you are two months
into a market's "recession", the chances of the "recession" continuing
two more months, or longer, is 1 in 2. The chances of it continuing 5
more months, or longer, is a little more than about 1 in 3. The same is
true for those times when the market is expanding, too. Simple as it
may be, it is astonishingly accurate.
The first and part of the second objectives have been satisfied. The
remaining question is why it works?
The reason is that industrial markets are fractal systems, and there
is a robust infrastructure in mathematics and economics that deals
with such things. However, I will approach the issue of intuitive
understanding through the construction of a single graph.
What we are discussing is the run length of industrial market
expansions and contractions. Let me propose the following prescription
for analyzing industrial market data. (You do not have to do this
analysis for other industrial markets-astonishingly, all markets have
the same contraction and expansion characteristics.) Suppose we have
market data that shows dollar shipments per month over a time period
of many years. For each month in our data, we simply count the number
of months until the market comes back to that value. That number would
be the run length of that market expansion or contraction, depending
on whether the market increased, or decreased respectively. We then
count how many of each run length we have, and make a graph of it.
Unfortunately, this is not exactly the information we want. Such a
graph would give us the likelihood that a run length's duration would
be EXACTLY 3 months, or 5 months, or whatever. For planning purposes,
we want to know the likelihood that a run length's duration would be
LONGER than 3 months, or 5 months, or whatever. We need the cumulative
sum of the run lengths, (which is a fancy statement that means we
simply want to add them up, and make a new graph.)
Although such data "munching" can be done on a spread sheet, there are
programs available that will do the same thing very expediently. (The
program I used for the generation of the attached graph was
tsrunlength, and available at
http://www.johncon.com/ndustrix/archive/fractal.tar.gz, and produced the data
for the attached graph in just a few seconds. The program is Open
Source, ie., free.)
The formula listed above, 1 / sqrt (t), is not precisely true. It
actually has an error function in it-the precise, theoretical form is
erf (1 / sqrt (t)). In the attached graph, I used the formal version
to demonstrate the accuracies. The error function has virtually no
significance for t much greater than 1.
I now have to address the issue of limited data set size-there were
181 "points" in the data for electronic component shipments in the
data from the Department of Commerce. If you think about it, finding a
run length of longer than 181 months in the data for 181 months is
impossible. But I know, that given a much larger data set size, I
would find some. What's the chances of that happening? About 1 / sqrt
(181) = 0.0743 = 7.4%. So, I can now compensate my theoretical form,
erf (1 / sqrt (t)), to make it look like my empirical data from the
Department of Commerce. I just subtract 0.0743 from it to compensate
for the limited data set size-and our third objective is complete.
We have one remaining issue. And that is how intuitive is it? The
graph is compelling:
All the graph means is that if you want to find the chances of the
duration of an industrial market expansion or contraction being longer
than, say four months, you find 4 on the x axis, move up to the graph,
and left to the y axis, and read about 0.5 = 50%. There are three
graphs displayed. The "real" graph is erf (1 / sqrt (t)), and is the
graph that should be used. The graph erf (1 / sqrt (t)) - 0.0743294146
is what we would expect to see if our data set was 181 months, and the
remaining graph is the empirical data from the Department of Commerce,
which consisted of 181 months. All in all, a respectable "fit" of
empirical data to our theoretical model.
There is one remaining question. What happens if we want to know the
characteristics of the duration of the run lengths of market
expansions and contractions in years, instead of months?
Astonishingly, it doesn't make any difference! The same rule holds
irregardless of the time scale-for which, I will offer compelling
John Conover, john@email.johncon.com, http://www.johncon.com/
Copyright © 1998 John Conover, john@email.johncon.com. All Rights Reserved. Last modified: Fri Mar 26 18:53:47 PST 1999 $Id: 981014184454.18095.html,v 1.0 2001/11/17 23:05:50 conover Exp $ | {"url":"http://www.johncon.com/john/correspondence/981014184454.18095.html","timestamp":"2024-11-05T10:06:37Z","content_type":"text/html","content_length":"8571","record_id":"<urn:uuid:513b52d2-64c4-4d94-9e07-3d3b8ab453ec>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00321.warc.gz"} |
Your explanations count for much more that simple correct
answers. Your wording must be your own;...
Your explanations count for much more that simple correct answers. Your wording must be your own;...
Your explanations count for much more that simple correct answers. Your wording must be your own; using my words will not earn any credit; your explanations must indicate that you understand the
material, not simply copy the explanations from somewhere else.
You must work on your own; collaboration will inevitably show up with similar wordings of the explana- tions and invalidate your answer.
Clear signs of cheating will be taken seriously. Blackboard is not forgiving of late submissions.
1. The following are samples of a normal distribution with mean θ and variance σ2, both are unknown: 54.1, 53.3, 55.9, 56.0, 55.7.
Find an unbiased estimate for θ.
2. With the data given in Problem 1, with θ unknown, find an unbiased estimate for σ2.
3. With the data given in Problem 1, with σ2 unknown, find a 95% confidende interval for θ. | {"url":"https://justaaa.com/statistics-and-probability/1308891-your-explanations-count-for-much-more-that-simple","timestamp":"2024-11-11T05:11:06Z","content_type":"text/html","content_length":"30656","record_id":"<urn:uuid:663134e7-17d2-49e4-ae56-96adb6307296>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00699.warc.gz"} |
Outer Method
Name Description
Outer(Byte, Byte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Decimal) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int16) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int64) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, SByte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Byte, Byte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Byte, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Byte, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Double, Byte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Int32, Byte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Single, Byte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Byte, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Decimal, Decimal) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Decimal, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Decimal, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Double, Decimal) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Int32, Decimal) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Single, Decimal) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Decimal, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Double, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Double, Int16) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int16, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int16, Int16) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int16, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int32, Int16) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Single, Int16) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int16, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int32, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Double, Int64) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int32, Int64) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int64, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int64, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Int64, Int64) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Single, Int64) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Int64, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Double, SByte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Int32, SByte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, SByte, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, SByte, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, SByte, SByte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Single, SByte) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(SByte, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Double, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Double, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Double, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Int32, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Int32, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Int32, Single) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Single, Double) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Single, Int32) Gets the outer product (matrix product) between two vectors (a*bT).
Outer(Single, Single, Single) Gets the outer product (matrix product) between two vectors (a*bT). | {"url":"https://accord-framework.net/docs/html/Overload_Accord_Math_Jagged_Outer.htm","timestamp":"2024-11-11T19:43:18Z","content_type":"text/html","content_length":"203608","record_id":"<urn:uuid:d080c9e4-2b70-4962-bd0c-420fcf06e083>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00607.warc.gz"} |
Fitting Integral Function with parametric limit using NAG Library
4.2.2.12 Fitting Integral Function with parametric limit using NAG Library
Before you start delving into this tutorial, you are recommended to read the relevant tutorial in Fitting with Integral using NAG Library. And as far as programming is concerned, the two tutorials
are basically the same, except that here you will learn to define Origin C fitting function with fitting parameters in the integral limit, while in the previous tutorial we in fact define a fitting
independent variable in the integral limit. Also note that a different NAG integrator is used here.
Minimum Origin Version Required: Origin 8.0 SR6
What you will learn
This tutorial will show you how to:
• Create a fitting function with Definite Integral using the NAG integration routine
• Create a fitting function with a parametric integral limit
• Use a log function to scale a large return value from the fitting function
Example and Steps
For example, we will fit the sample data at the bottom of this page with the following model:
$y=\int_{c}^{d} \frac { \cosh { ((x_i + b^2 \cdot x^2) /(b + x))}}{a+(x_i^2+x^2)}\, dx_i$
Note that we use $x_i \,$ to indicate the integral independent variable while $x \,$ indicates the fitting independent variable. The model parameters $a$, $b$, $c$, and $d$ are fitted parameters we
want to obtain from the sample data. To prepare the data, you just need to copy the sample data to an Origin Work sheet. The fitting procedure is similar to the previous tutorial:
Define Fitting Function in Fitting Function Organizer
Press F9 to open the Fitting Function Organizer and add the User-Defined integral fitting function nag_integration_fitting_cosh to the Category FittingWithIntegral, similar to the first tutorial.
Function Name: nag_integration_fitting_cosh
Function Type: User-Defined
Independent Variables: x
Dependent Variables: y
Parameter Names: a, b, c, d
Function Form: Origin C
Click the button beside the Function box to open the code builder and define and compile the fitting function as follows: (Note: Remember to save the Function after compiling it and returning to the
Function Organizer Dialog):
#include <origin.h>
// Add your special include files here.
// For example, if you want to fit with functions from the NAG library,
// add the header file for the NAG functions here.
#include <OC_nag.h>
// Add code here for other Origin C functions that you want to define in this file,
// and access in your fitting function.
struct user
double a, b, fitX; // fitX the independent variable of fitting function
static double NAG_CALL f_callback(double x, Nag_User *comm) // x is the independent variable of the integrand
struct user *sp = (struct user *)(comm->p);
double aa, bb, fitX; // temp variable to accept the parameters in the Nag_User communication struct
aa = sp->a;
bb = sp->b;
fitX = sp->fitX;
return cosh((x*x+bb*bb*fitX*fitX)/(bb+fitX))/(aa+(x*x+fitX*fitX));
// You can access C functions defined in other files, if those files are loaded and compiled
// in your workspace, and the functions have been prototyped in a header file that you have
// included above.
// You can access NLSF object methods and properties directly in your function code.
// You should follow C-language syntax in defining your function.
// For instance, if your parameter name is P1, you cannot use p1 in your function code.
// When using fractions, remember that integer division such as 1/2 is equal to 0, and not 0.5
// Use 0.5 or 1/2.0 to get the correct value.
// For more information and examples, please refer to the "User-Defined Fitting Function"
// section of the Origin Help file.
void _nlsfnag_integration_fitting_cosh(
// Fit Parameter(s):
double a, double b, double c, double d,
// Independent Variable(s):
double x,
// Dependent Variable(s):
double& y)
// Beginning of editable part
double epsabs = 0.00001, epsrel = 0.0000001, result, abserr;
Integer max_num_subint = 500;
// you may use epsabs and epsrel and this quantity to enhance your desired precision
// when not enough precision encountered
Nag_QuadProgress qp;
static NagError fail;
// the parameters parameterize the integrand can be input to the call_back function
// through the Nag_User communication struct
Nag_User comm;
struct user s;
s.a = a;
s.b = b;
s.fitX = x;
comm.p = (Pointer)&s;
d01sjc(f_callback, c, d, epsabs, epsrel, max_num_subint, &result, &abserr, &qp, &comm, &fail);
// you may want to exam the error by printing out error message, just uncomment the following lines
// if (fail.code != NE_NOERROR)
// printf("%s\n", fail.message);
// For the error other than the following three errors which are due to bad input parameters
// or allocation failure NE_INT_ARG_LT NE_BAD_PARAM NE_ALLOC_FAIL
// You will need to free the memory allocation before calling the integration routine again to
// avoid memory leakage
if (fail.code != NE_INT_ARG_LT && fail.code != NE_BAD_PARAM && fail.code != NE_ALLOC_FAIL)
y = log(result);
// note use log of the integral result as return as the integral result is large,
// you are not necessary to do so
// End of editable part
In the above code, we define the integrand as a callback function f_callback just outside the fitting function body _nlsfnag_integration_fitting_cosh. Note that we parametrize the integrand function
with the variables a, b and fitX, and pass them into the callback funtion through the Nag_User struct. After that we perform the integration using NAG integrator d01sjc. Besides, you can also use
other Quadrature Routines as you want. In the current example, we also use a log scale for the fitting function. (The sample data are already scaled by a log function)
Compile the code, return to the dialog and then Save the fitting function in the function Organizer and open the Nonlinear Curve Fit dialog in the Analysis-Fitting menu. You can then select this
user-defined fitting function in the Function Selection page under Setting Tab.
Set the Initial Values for the Parameters
Similarly, as it is a user-defined fitting function, you have to supply the initial guess values for the parameters. You may manually set them in the Parameter tab in Nonlinear Curve Fit dialog. For
current example, you can just set the initial values for the parameters $a=1$, $b=10$, $c=3$, $d=4$. After the parameters are initialized, you can perform the fitting to obtain the fitting result, as
shown in the following.
Sample Data | {"url":"http://cloud.originlab.com/doc/Tutorials/Fitting-Integral-ParaLimit-NAG","timestamp":"2024-11-08T20:32:11Z","content_type":"text/html","content_length":"173280","record_id":"<urn:uuid:200bc1fc-1194-4f1f-9d13-2fddb51f433b>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00015.warc.gz"} |
On the final programming project I was having a little trouble with the ratio.
The maximum dimension will always be set to MAX_SIZE. You can determine the maximum dimension by subtracting minX from maxX to get the width of the points in the x direction. The same with the ys the
get the height of the points in the y direction.
If the width is greater than the height, then we'll make the width of the GraphWin to be MAX_SIZE. The window's height will then be the height determined by the points divided by the width determined
by the points times MAX_SIZE. So if the points gave a width of 500 and a height of 200, the height of the window should be 200/500 times MAX_SIZE or 2/5th of the MAX_SIZE (So if MAX_SIZE = 1000, 1000
wide and (2/5) * 1000 = 400 tall.)
Otherwise, make the height of the window, MAX_SIZE and the width a fraction of it. | {"url":"http://lovelace.augustana.edu/q2a/index.php/3405/final-programming-project-having-little-trouble-with-ratio","timestamp":"2024-11-08T06:28:29Z","content_type":"text/html","content_length":"23045","record_id":"<urn:uuid:217a7b7c-4e0c-445c-b7d3-a74157514c78>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00465.warc.gz"} |
Dilip Lalwani, Author at PuzzlersWorld.com
There are some children on a school ground, a square has been drawn and all children are standing on a squares four lines, they are standing with same distance, four children are standing on a four
corners. No 16 Is exactly opposite of No 6. How many children are there?
Check your answer:-
Assuming there are x children in every side of the square, thus a total of 4x -4 children will be there(not counting corner child twice).
Now we know there are atleast 16 children, thus 4x-4 >= 16. x >= 5
Lets assume first child is sitting on bottom left corner, than assume three cases
Case 1: x = 5
then 6th child will be standing on second position in second line of square, thus the exactly opposite child will be sitting on x + x-2 + x + x-1 = 4x-3th position, i.e. 17th position.
Which is not the case as per the question, thus x can’t be 5
Case 2: x = 6
then 6th child is sitting on the corner(point B), thus exactly opposite child is on x + x-2 + x= 3x-2th position, i.e. 16th.
Thus it satisfies all conditions of the question.
Thus there are a total of 4x-4, i.e. 20 children in this case.
Case 3: x > 6
then 6th child is standing somewhere in the bottom line (point A), as per this exactly opposite child will be standing at position x + x-2 + x -6 = 3x – 8
for 3x-8 = 16 => x = 24/3 = 8
Thus there are a total of 4x-4 = 28 children in this case. | {"url":"https://puzzlersworld.com/author/dilip_lalwani/","timestamp":"2024-11-08T18:11:25Z","content_type":"text/html","content_length":"85018","record_id":"<urn:uuid:b5babab7-31c6-4174-b322-d046b97e91c6>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00715.warc.gz"} |
Samacheer Kalvi 5th Maths Guide Term 2 Chapter 3 Patterns Ex 3.1
Students can download 5th Maths Term 2 Chapter 3 Patterns Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 5th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus,
helps students complete homework assignments and to score high marks in board exams.
Tamilnadu Samacheer Kalvi 5th Maths Solutions Term 2 Chapter 3 Patterns Ex 3.1
Question 1.
Find the angle of the given shapes using the equilateral triangle.
Question 2.
Find the angles of a rectangle using a circle.
Angle of a circle is 360° ○
= 360 ÷ 4 = 90 | {"url":"https://tnboardsolutions.com/samacheer-kalvi-5th-maths-guide-term-2-chapter-3-ex-3-1/","timestamp":"2024-11-08T20:59:45Z","content_type":"text/html","content_length":"52944","record_id":"<urn:uuid:ce459ab7-d515-46d1-9320-768316184695>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00635.warc.gz"} |
Quadratic Equations
• To define quadratic equation
• To solve the roots of a quadratic equation using
□ factoring/factorising
□ completing the square
□ quadratic formula
• To use discriminant to determine the nature of the roots of the quadratic equation
• To describe the properties of the graph of the quadratic function
• QUADRATIC EQUATION – the general form of quadratic equation is $ax^{2}+bx+c$, where a, b, and c are constants and $aeq 0$.
• FACTORING/FACTORISING- is the mathematical process of finding the factors of a number or expression.
• COMPLETING THE SQUARE – is a process used to solve a quadratic equation by rewriting the form of the equation so that the left side is a perfect square trinomial.
• QUADRATIC FORMULA – is a method that is used to find the roots of a quadratic equation from its coefficients. The quadratic formula is:
$x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
• DISCRIMINANT – the number $d= b^{2}-4ac$ determined from the coefficients of the equation $ax^{2}+bx+c$. The discriminant reveals what type of roots the equation has. It can be real or imaginary/
complex roots.
• MINIMUM/MAXIMUM VALUE- The minimum value of a function is the place where the graph has a vertex at its lowest point while the maximum value of a function is the place where the graph has a
vertex at its highest point.
• Quadratic equation has many applications in everyday life. It can be used when calculating areas, determining a product’s profit or formulating the speed of an object.
• Quadratic equations are also useful in calculating speeds. Avid kayakers, for example, use quadratic equations to estimate their speed when going up and down a river.
• In athletic events that involve throwing objects like the shot put, balls or javelin, quadratic equations become highly useful.
• In the real world, you can use the minimum value of a quadratic function to determine minimum cost or area. It has practical uses in science, architecture and business.
Quadratic equation in general form is $ax^{2}+bx+c$, where a, b, and c are constants and $aeq 0$. It is very important that the value of a should not be zero because that will make the equation
linear and not quadratic anymore.
Quadratic equations come in different forms.
Note: Vertex of the parabola – it is the turning point of the graph of a quadratic equation. It contains the minimum/maximum value. It can be the peak of the graph or the lowest point.
One of the important concepts about quadratic equations is finding its roots. There are many ways to solve for the roots of the quadratic equation —- factoring/factorising, completing the square, and
using quadratic formula.
Using factorising, find the roots of the following quadratics.
1) $x^{2}+6x+8$ 2) $x^{2}-5x+6$
1) For this given, a = 1, b = 6, c = 8. We should think of two numbers that when multiplied together, give a product of 8 and when added, they will be equal to 6. Thus, the numbers are 2 and 4. So,
$x^{2}+6x+8=(x+2)(x+4)$Rewrite the equation in factored form
(x + 2)(x + 4) = 0 Equate the expression to 0.
$f_{1}: (x+2)=0$$f_{2}: (x+4)=0$Equate each factor to 0.
$r_{1}: x=-2$$r_{2}: x=-4$
The roots of $x^{2}+6x+8$ are -2 and -4.
2) For this given, a = 1, b = -5, c = 6. We should think of two numbers that when multiplied together, give a product of 6 and when added, they will be equal to -5. Take note that the product must be
positive but the sum is negative. So we need two negative numbers. Thus, the numbers are -3 and -2. So,
$x^{2}-5x+6=(x-2)(x-3)$Rewrite the equation in factored form
(x – 2) (x – 3) = 0 Equate the expression to 0.
$f_{1}: (x-2)=0$$f_{2}: (x-3)=0$Equate each factor to 0.
$r_{1}: x=2$$r_{2}: x=3$
The roots of $x^{2}-5x+6$ are 2 and 3.
Note: Not all quadratics can be factored. So completing the square method can be an alternative process.
Completing the Square
Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.
To solve $ax^{2}+bx+c$, using completing the square:
1. Rewrite the equation in the form $ax^{2}+bx=-c$. In this case, the constant will be on the right side of the equation.
2. If $aeq 1$, divide both sides by a.
3. Divide b by 2a then square quotient $(\frac{-b}{2a})^{2}$. Add this to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Extract the square root of the both sides of the equation (Note: $(x+p)^{2}=r$ is equivalent to $x+p = \pm \sqrt{r}$.
6. Solve for the value of x.
1. Find the roots using completing the square method.
1) $x^{2}-8x+1=0$ 2) $2x^{2}+5x+4=0$
1) $x^{2}-8x+1=0$
$b= -8$$-\frac{-8}{2(1)}=4$$4^{2}=16$
$x^{2}-8x+16 =-1+16$
$x^{2}-8x+16 =15$
$(x-4)^{2} =15$
$x=4\pm \sqrt{15}$
2) $2x^{2}+5x+4=0$
$(x+\frac{5}{4})= \frac{\sqrt{-7}}{4}$
$x= -\frac{5}{4}\pm \frac{\sqrt{-7}}{4}$ or $x= \frac{-5\pm \sqrt{-7}}{4}$
Quadratic Formula
Another method to solve for the roots of a quadratic equation is using a quadratic formula. Let’s derive the formula:
$2ax+b = \pm \sqrt{b^{2}-4ac}$
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
Find the value of the following series.
1) $x^{2}+6x+8$ 2) $3x^{2}-5x+2$
1) Identify first the values of a, b, and c.
a = 1 b = 6 c = 8
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$x=\frac{-6\pm \sqrt{6^{2}-4(1)(8)}}{2(1)}$
$x=\frac{-6\pm \sqrt{36-32}}{2} = \frac{-6\pm \sqrt{4}}{2} =\frac{-6\pm 2}{2} = -2 \, and \, -4$
2) Identify first the values of a, b, and c.
a = 3 b = -5 c = 2
$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$
$x=\frac{5\pm \sqrt{(-5)^{2}-4(3)(2)}}{2(3)}$
$x=\frac{5\pm \sqrt{25-24}}{6}=\frac{5\pm \sqrt{1}}{6}=\frac{5\pm 1}{6} = 1 \, and \, 2/3$
Discriminant tells the nature of the roots of a quadratic equation. By determining this, it can be helpful in finding the roots. Using discriminant, you will be told ahead of time if the roots will
be real or complex. The table below shows the classification of the roots of a quadratic equation.
Determine the nature of the roots of the following quadratic equations.
1) $x^{2}-8x+1=0$ 2) $2x^{2}+5x+4=0$
1) Remember that is determining the nature of the roots using discriminant, use the formula: $b^{2}-4ac$. Let us first write the value of the coefficients.
a = 1 b = -8 c = 1
60>0. Thus the roots of the equation are real and unequal.
2) Let us first write the value of the coefficients.
a = 2 b = 5 c = 4
-7<0. Thus the roots of the equation are imaginary and unequal.
• The graph of a quadratic is called parabola, a U-shaped curve.
• The parabola contains the vertex— the turning point of the graph.
• A parabola may open upward if the value of a is positive. Thus, it has a minimum value.
• A parabola may open downward if the value of a is negative. Thus, it has a maximum value.
• The graph is symmetric with the vertical line that passes through the vertex. It is called the axis of symmetry. | {"url":"https://alevelmaths.co.uk/pure-maths/algebra/quadratic-equations/","timestamp":"2024-11-11T17:50:36Z","content_type":"text/html","content_length":"73815","record_id":"<urn:uuid:6066a4fe-2fa8-4677-a1ed-fc042a4f76b9>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00012.warc.gz"} |
1 Bit Full Adder
Introduction: 1 Bit Full Adder
An adder is a digital electronic circuit that performs addition of numbers. Adders are used in every single computer's processors to add various numbers, and they are used in other operations in the
processor, such as calculating addresses of certain data.
In this instructable, we are going to construct and test the one bit binary full adder. The attached figure shows the block diagram of a one bit binary full adder. A block diagram represents the
desired application and its various components, such as inputs and outputs.
Inputs: A, B, Carry in (Cin)
Outputs: Carry out (Cout), Sum (S)
Parts Needed:
9V battery
Batter Connector
5V regulator
IC chips: 74LS136, 74LS08, 74LS32.
(Optional: 74LS00) *If you do not have an XOR, 74LS136, IC Chip*
2 330 Ohm resistors
2 LED's (Different colors preferred)
DIP switch
10 Kilo Ohm resistor bank
Wires as needed
Step 1: Truth Table, Derived Boolean Function, and Schematic
The truth table of a one bit full adder is shown in the first figure; using the truth table, we were able to derive the boolean functions for both the sum and the carry out, as shown in the second
attached figure. Furthermore, the derived boolean function lead us to the schematic design of the one bit full adder. Finally, I did not have any XOR IC chips, so I used the XOR mixed gates
equivalent, which is shown in the last figure.
Step 2: Implementation on a Breadboard
If the switch is up, then it is off.
If the switch is down, then it is on.
The White wire represents A.
The Blue wire represents B.
The Yellow wire represents Carry in (Cin).
The Green LED represents the Sum.
and the Red LED represents the Carry Out (Cout). | {"url":"https://www.instructables.com/1-Bit-Binary-Full-Adder/","timestamp":"2024-11-02T01:41:52Z","content_type":"text/html","content_length":"88528","record_id":"<urn:uuid:47947d37-d229-4fa4-937e-bcd0ef6438f0>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00675.warc.gz"} |
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The most fascinating brain game
Sudoku is a fascinating brain game often called as Rubik's cube of 21st century. From a 3 year old child to an 80 year old person can enjoy Sudoku in her own way.
Strengths of Sudoku
• A Sudoku game has some of the 81 cells in a 9 by 9 matrix filled with digits. Rest you have to fill using digits 1 to 9. No digit can appear twice in a row, a column or in any of the 3 by 3 major
• No need to know maths or any other subject to play Sudoku. You should only be able to find useful patterns in the digits for breakthrough.
• Sudoku improves brain function, specifically problem solving and pattern recognition abilities.
• A nearly endless supply of Sudoku games of varieties of difficulty levels are available in popular websites and mobile apps.
• Sudoku games appear in difficulty levels of easy, medium, hard and expert. One can attain expertise step by step.
Sudoku solutions hard to find
Very little is available in the area of "How to solve Sudoku of various difficulty levels". To address this vacuum, we have created Sudoku solutions of,
Use the resources to become a Sudoku expert from a Sudoku beginner. | {"url":"https://suresolv.com/sudoku?page=8","timestamp":"2024-11-06T05:02:41Z","content_type":"text/html","content_length":"50519","record_id":"<urn:uuid:714d9de8-a367-41f2-bc93-07f293c84963>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00161.warc.gz"} |
15,352 research outputs found
This introductory paper studies a class of real analytic functions on the upper half plane satisfying a certain modular transformation property. They are not eigenfunctions of the Laplacian and are
quite distinct from Maass forms. These functions are modular equivariant versions of real and imaginary parts of iterated integrals of holomorphic modular forms, and are modular analogues of
single-valued polylogarithms. The coefficients of these functions in a suitable power series expansion are periods. They are related both to mixed motives (iterated extensions of pure motives of
classical modular forms), as well as the modular graph functions arising in genus one string perturbation theory. In an appendix, we use weakly holomorphic modular forms to write down modular
primitives of cusp forms. Their coefficients involve the full period matrix (periods and quasi-periods) of cusp forms.Comment: Based on a talk given at Zagier's 65th birthday conference `modular
forms are everywhere'. What was formerly the appendix has now turned into arXiv:1710.0791
The values at 1 of single-valued multiple polylogarithms span a certain subalgebra of multiple zeta values. In this paper, the properties of this algebra are studied from the point of view of motivic
We introduce a new family of real analytic modular forms on the upper half plane. They are arguably the simplest class of `mixed' versions of modular forms of level one and are constructed out of
real and imaginary parts of iterated integrals of holomorphic Eisenstein series. They form an algebra of functions satisfying many properties analogous to classical holomorphic modular forms. In
particular, they admit expansions in $q, \overline{q}$ and $\log |q|$ involving only rational numbers and single-valued multiple zeta values. The first non-trivial functions in this class are real
analytic Eisenstein series.Comment: Introduction rewritten in version 2, and other minor edit
We define a generalisation of the completed Riemann zeta function in several complex variables. It satisfies a functional equation, shuffle product identities, and has simple poles along finitely
many hyperplanes, with a recursive structure on its residues. The special case of two variables can be written as a partial Mellin transform of a real analytic Eisenstein series, which enables us to
relate its values at pairs of positive even points to periods of (simple extensions of symmetric powers of the cohomology of) the CM elliptic curve corresponding to the Gaussian integers. In general,
the totally even values of these functions are related to new quantities which we call multiple quadratic sums. More generally, we cautiously define multiple-variable versions of motivic $L$
-functions and ask whether there is a relation between their special values and periods of general mixed motives. We show that all periods of mixed Tate motives over the integers, and all periods of
motivic fundamental groups (or relative completions) of modular groups, are indeed special values of the multiple motivic $L$-values defined here.Comment: This is the second half of a talk given in
honour of Ihara's 80th birthday, and will appear in the proceedings thereo
We study the depth filtration on multiple zeta values, the motivic Galois group of mixed Tate motives over $\mathbb{Z}$ and the Grothendieck-Teichm\"uller group, and its relation to modular forms.
Using period polynomials for cusp forms for $\mathrm{SL}_2(\mathbb{Z})$, we construct an explicit Lie algebra of solutions to the linearized double shuffle equations, which gives a conjectural
description of all identities between multiple zeta values modulo $\zeta(2)$ and modulo lower depth. We formulate a single conjecture about the homology of this Lie algebra which implies conjectures
due to Broadhurst-Kreimer, Racinet, Zagier and Drinfeld on the structure of multiple zeta values and on the Grothendieck-Teichm\"uller Lie algebra.Comment: Rewritten introduction, added brief section
explaining the depth-spectral sequence, and made a few proofs more user-friendly by adding some more detail
We prove that the category of mixed Tate motives over $\Z$ is spanned by the motivic fundamental group of \Pro^1 minus three points. We prove a conjecture by M. Hoffman which states that every
multiple zeta value is a \Q-linear combination of $\zeta(n_1,..., n_r)$ where $n_i\in \{2,3\}$ | {"url":"https://core.ac.uk/search/?q=author%3A(Brown%2C%20Francis)","timestamp":"2024-11-14T21:06:58Z","content_type":"text/html","content_length":"109104","record_id":"<urn:uuid:8b31b3d2-094d-4416-adaf-e7a090de3dcd>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00748.warc.gz"} |
Square Wave from Sine Waves
This example shows how the Fourier series expansion for a square wave is made up of a sum of odd harmonics.
Start by forming a time vector running from 0 to 10 in steps of 0.1, and take the sine of all the points. Plot this fundamental frequency.
t = 0:.1:10;
y = sin(t);
Next add the third harmonic to the fundamental, and plot it.
y = sin(t) + sin(3*t)/3;
Now use the first, third, fifth, seventh, and ninth harmonics.
y = sin(t) + sin(3*t)/3 + sin(5*t)/5 + sin(7*t)/7 + sin(9*t)/9;
For a finale, go from the fundamental all the way to the 19th harmonic, creating vectors of successively more harmonics, and saving all intermediate steps as the rows of a matrix.
Plot the vectors on the same figure to show the evolution of the square wave. Note that the Gibbs effect says it will never quite get there.
t = 0:.02:3.14;
y = zeros(10,length(t));
x = zeros(size(t));
for k = 1:2:19
x = x + sin(k*t)/k;
y((k+1)/2,:) = x;
title('The building of a square wave: Gibbs'' effect')
Here is a 3-D surface representing the gradual transformation of a sine wave into a square wave.
shading interp
axis off ij | {"url":"https://uk.mathworks.com/help/matlab/math/square-wave-from-sine-waves.html","timestamp":"2024-11-03T10:02:58Z","content_type":"text/html","content_length":"69629","record_id":"<urn:uuid:67dbb104-4788-49a1-943f-eb2e3e13467a>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00473.warc.gz"} |
Portfolio Optimization for 20 Securities Using Lagrange Multipliers, No Short-Selling, Weights Sum to 1
Portfolio Optimization for 20 Securities Using Lagrange Multipliers, No Short-Selling, Weights Sum to 1
Construct the Optimal Portfolio that:
delivers the target return (mu_Target)
with minimum risk
Minimize the risk of the portfolio (in this case, measured as half the variance)
While maintaining an expected return target of (mu_Target)
By adjusting the investment weights on each asset
Subject to the budget constraint that the weights sum to 1
Since constraints are equalities => We can use Method Lagrange
Supports up to 20 securities.
Able to do more if requested. Please contact us.
No short-selling (ie. No negative weights)
Solution 00:
Basic MPT with only budget constraint that weights sum to 1
Solution 01:
Tweaked solution where no negative weights are allowed,
but budget constraint fails, as sum of weights exceed 1.
Solution 02:
Maintain that no negative weights are allowed,
but normalize weights such that they sum to 1.
This yields a practical solution, but usually unable to meet target return.
There are no reviews yet.
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How would I interpolate between two tables?
How would I interpolate between two tables?
Say I have a table A, representing the output for an input variable X=1.
Now say I also have a table B, representing the output for an input variable of X=2.
How would I create a patch that can interpolate between A and B (and so on with tables C, D, etc...) for any X (i.e. X = 1.5 would be halfway between tables A and
Turns out list-inter and list-inter-many do what I need... | {"url":"https://forum.puredata.info/topic/8236/how-would-i-interpolate-between-two-tables/2","timestamp":"2024-11-05T11:51:17Z","content_type":"text/html","content_length":"45323","record_id":"<urn:uuid:85d01123-742b-4c0b-9b1f-a675e38d0f6b>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00039.warc.gz"} |
Area under the precision recall curve — average_precision
Area under the precision recall curve
average_precision() is an alternative to pr_auc() that avoids any ambiguity about what the value of precision should be when recall == 0 and there are not yet any false positive values (some say it
should be 0, others say 1, others say undefined).
It computes a weighted average of the precision values returned from pr_curve(), where the weights are the increase in recall from the previous threshold. See pr_curve() for the full curve.
average_precision(data, ...)
# S3 method for data.frame
estimator = NULL,
na_rm = TRUE,
event_level = yardstick_event_level(),
case_weights = NULL
estimator = NULL,
na_rm = TRUE,
event_level = yardstick_event_level(),
case_weights = NULL,
A data.frame containing the columns specified by truth and ....
A set of unquoted column names or one or more dplyr selector functions to choose which variables contain the class probabilities. If truth is binary, only 1 column should be selected, and it
should correspond to the value of event_level. Otherwise, there should be as many columns as factor levels of truth and the ordering of the columns should be the same as the factor levels of
The column identifier for the true class results (that is a factor). This should be an unquoted column name although this argument is passed by expression and supports quasiquotation (you can
unquote column names). For _vec() functions, a factor vector.
One of "binary", "macro", or "macro_weighted" to specify the type of averaging to be done. "binary" is only relevant for the two class case. The other two are general methods for calculating
multiclass metrics. The default will automatically choose "binary" or "macro" based on truth.
A logical value indicating whether NA values should be stripped before the computation proceeds.
A single string. Either "first" or "second" to specify which level of truth to consider as the "event". This argument is only applicable when estimator = "binary". The default uses an internal
helper that defaults to "first".
The optional column identifier for case weights. This should be an unquoted column name that evaluates to a numeric column in data. For _vec() functions, a numeric vector,
hardhat::importance_weights(), or hardhat::frequency_weights().
If truth is binary, a numeric vector of class probabilities corresponding to the "relevant" class. Otherwise, a matrix with as many columns as factor levels of truth. It is assumed that these are
in the same order as the levels of truth.
A tibble with columns .metric, .estimator, and .estimate and 1 row of values.
For grouped data frames, the number of rows returned will be the same as the number of groups.
For average_precision_vec(), a single numeric value (or NA).
The computation for average precision is a weighted average of the precision values. Assuming you have n rows returned from pr_curve(), it is a sum from 2 to n, multiplying the precision value p_i by
the increase in recall over the previous threshold, r_i - r_(i-1).
$$AP = \sum (r_{i} - r_{i-1}) * p_i$$
By summing from 2 to n, the precision value p_1 is never used. While pr_curve() returns a value for p_1, it is technically undefined as tp / (tp + fp) with tp = 0 and fp = 0. A common convention is
to use 1 for p_1, but this metric has the nice property of avoiding the ambiguity. On the other hand, r_1 is well defined as long as there are some events (p), and it is tp / p with tp = 0, so r_1 =
When p_1 is defined as 1, the average_precision() and roc_auc() values are often very close to one another.
Macro and macro-weighted averaging is available for this metric. The default is to select macro averaging if a truth factor with more than 2 levels is provided. Otherwise, a standard binary
calculation is done. See vignette("multiclass", "yardstick") for more information.
Relevant Level
There is no common convention on which factor level should automatically be considered the "event" or "positive" result when computing binary classification metrics. In yardstick, the default is to
use the first level. To alter this, change the argument event_level to "second" to consider the last level of the factor the level of interest. For multiclass extensions involving one-vs-all
comparisons (such as macro averaging), this option is ignored and the "one" level is always the relevant result.
# ---------------------------------------------------------------------------
# Two class example
# `truth` is a 2 level factor. The first level is `"Class1"`, which is the
# "event of interest" by default in yardstick. See the Relevant Level
# section above.
# Binary metrics using class probabilities take a factor `truth` column,
# and a single class probability column containing the probabilities of
# the event of interest. Here, since `"Class1"` is the first level of
# `"truth"`, it is the event of interest and we pass in probabilities for it.
average_precision(two_class_example, truth, Class1)
#> # A tibble: 1 × 3
#> .metric .estimator .estimate
#> <chr> <chr> <dbl>
#> 1 average_precision binary 0.947
# ---------------------------------------------------------------------------
# Multiclass example
# `obs` is a 4 level factor. The first level is `"VF"`, which is the
# "event of interest" by default in yardstick. See the Relevant Level
# section above.
# You can use the col1:colN tidyselect syntax
hpc_cv %>%
filter(Resample == "Fold01") %>%
average_precision(obs, VF:L)
#> # A tibble: 1 × 3
#> .metric .estimator .estimate
#> <chr> <chr> <dbl>
#> 1 average_precision macro 0.617
# Change the first level of `obs` from `"VF"` to `"M"` to alter the
# event of interest. The class probability columns should be supplied
# in the same order as the levels.
hpc_cv %>%
filter(Resample == "Fold01") %>%
mutate(obs = relevel(obs, "M")) %>%
average_precision(obs, M, VF:L)
#> # A tibble: 1 × 3
#> .metric .estimator .estimate
#> <chr> <chr> <dbl>
#> 1 average_precision macro 0.617
# Groups are respected
hpc_cv %>%
group_by(Resample) %>%
average_precision(obs, VF:L)
#> # A tibble: 10 × 4
#> Resample .metric .estimator .estimate
#> <chr> <chr> <chr> <dbl>
#> 1 Fold01 average_precision macro 0.617
#> 2 Fold02 average_precision macro 0.625
#> 3 Fold03 average_precision macro 0.699
#> 4 Fold04 average_precision macro 0.685
#> 5 Fold05 average_precision macro 0.625
#> 6 Fold06 average_precision macro 0.656
#> 7 Fold07 average_precision macro 0.617
#> 8 Fold08 average_precision macro 0.659
#> 9 Fold09 average_precision macro 0.632
#> 10 Fold10 average_precision macro 0.611
# Weighted macro averaging
hpc_cv %>%
group_by(Resample) %>%
average_precision(obs, VF:L, estimator = "macro_weighted")
#> # A tibble: 10 × 4
#> Resample .metric .estimator .estimate
#> <chr> <chr> <chr> <dbl>
#> 1 Fold01 average_precision macro_weighted 0.750
#> 2 Fold02 average_precision macro_weighted 0.745
#> 3 Fold03 average_precision macro_weighted 0.794
#> 4 Fold04 average_precision macro_weighted 0.757
#> 5 Fold05 average_precision macro_weighted 0.740
#> 6 Fold06 average_precision macro_weighted 0.747
#> 7 Fold07 average_precision macro_weighted 0.751
#> 8 Fold08 average_precision macro_weighted 0.759
#> 9 Fold09 average_precision macro_weighted 0.714
#> 10 Fold10 average_precision macro_weighted 0.742
# Vector version
# Supply a matrix of class probabilities
fold1 <- hpc_cv %>%
filter(Resample == "Fold01")
truth = fold1$obs,
c(fold1$VF, fold1$F, fold1$M, fold1$L),
ncol = 4
#> [1] 0.6173363 | {"url":"https://yardstick.tidymodels.org/reference/average_precision.html","timestamp":"2024-11-03T23:25:50Z","content_type":"text/html","content_length":"35127","record_id":"<urn:uuid:5abfa3c2-aab0-4bc9-80c7-0a2d2dc67825>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00082.warc.gz"} |
How to teach child to tell timeHow to teach child to tell time 🚩 how to teach a child to tell the time 🚩 Education and training of the child.
Explain to your child that the clock consists of a dial, numbers, and arrow – hour and minute. When he will remember, leave only the hour hand and numbers. Show how slowly the arrow moves. Explain
that if the pointer is on the number one then one hour. If a little farther then a few more hours. You can take a few months. Opposite numbers draw event. For example, next to 7 - the child wakes up.
9 – Breakfast in the garden. First draw a few pictures. Don't rush time. But always ask the child what time it is.
Now go on to dealing with the minute hand. The child must understand that it is a long time and moves faster. Then explain that the minute hand goes all round in one hour. That is, moves to the next
digit. Ask him to show you how to put the arrow to get half an hour, hour and half hour, two hours, and so on.
Around the face, draw small figures to determine the minutes from 1 to 60. Explain that the division between the two numbers includes 5 minutes and that the minute hand goes full circle in 60 minutes
or one hour. Give your child tasks to show you the movements of hands, 10 minutes, 15, 20.
Introduce such concepts as a quarter of an hour, half an hour.
Now explain how to determine when one minute passed. Show how the clock will look like 7 minutes and like 12. Draw with the child and his routine. Next to the event draw, the dial depicting the time.
If you think that the child is already quite prepared, go to a real clock. Do not expect that at this age the child will be able to understand all the details. But it will be able to remember the
main points. Ask as many questions as possible. Play with the child in determining time every day. Otherwise he will quickly forget all that appears, the skills and you have the whole process start
How to teach a child time. To understand time is a necessary skill for all people. Although most of us have constant access to electronic watch, mobile phone or computer monitor, most parents remain
convinced that the child needs to understand traditional analog clock.
Useful advice
How to teach a child time? The child at preschool age it is still very difficult to grasp and understand synchronous clock numerology from 1-12 and 0-60. He's difficult to understand what is a
spatial notion of time. Often I meet first graders who are poorly oriented in time. But if You've already decided and are ready to give your child knowledge of how to teach a child to watch, pay
attention to these recommendations | {"url":"https://eng.kakprosto.ru/how-46744-how-to-teach-child-to-tell-time","timestamp":"2024-11-11T14:31:57Z","content_type":"text/html","content_length":"30697","record_id":"<urn:uuid:cc99aadd-cb6b-4d3a-a2d5-a7de1b55b16d>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00456.warc.gz"} |
A statistical analysis of the theoretical yield of ethanol from corn starch
This paper analyzes the Illinois State Variety Test results for total and extractable starch content in 708 samples of 401 commercial varieties of corn. It is shown that the normally distributed
extractable starch content has the mean of 66.2% and the standard deviation of 1.13%. The corresponding maximum theoretical yield of ethanol is 0.364 kg EtOH/kg dry corn, and the standard deviation
is 0.007. In the ethanol industry units, this yield translates to 2.64 gal EtOH/nominal wet bushel, and the standard deviation is 0.05 gal/bu. The U.S. ethanol industry consistently has inflated its
ethanol yields by counting 5 volume percent of # 14 gasoline denaturant (8% of energy content) as ethanol. Also, imports from Brazil and higher alcohols seem to have been counted as U.S. ethanol. The
usually accepted USDA estimate of mean ethanol yield in the U.S., 2.682 gal EtOH/bu, is one standard deviation above the rigorous statistical estimate in this paper.
• Distribution
• Extractable
• Monte Carlo
• Starch
• Statistics
• Total
ASJC Scopus subject areas
• General Environmental Science
Dive into the research topics of 'A statistical analysis of the theoretical yield of ethanol from corn starch'. Together they form a unique fingerprint. | {"url":"https://faculty.kaust.edu.sa/en/publications/a-statistical-analysis-of-the-theoretical-yield-of-ethanol-from-c","timestamp":"2024-11-11T19:20:30Z","content_type":"text/html","content_length":"49917","record_id":"<urn:uuid:acbfb1f9-117d-4dbc-818e-55fd92b1a71f>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00085.warc.gz"} |
[Solved] Consider the following system differentia | SolutionInn
Answered step by step
Verified Expert Solution
Consider the following system differential equation: d+y(t) dy(t) dt3 dt4 +5 7 dy(t) dt +36 dy(t) dt - 100y(t) b) Find the transfer function
Consider the following system differential equation: d+y(t) dy(t) dt3 dt4 +5 7 dy(t) dt +36 dy(t) dt - 100y(t) b) Find the transfer function for the system, G(s) = = du(t) dt + 5u(t) a) Find the
Laplace transform of the ODE, assuming zero initial conditions. Y(s) U (s)* c) Find the Laplace transform for the output, Y(s), if the input is u(t) = sin(6t). Note: It is okay to express you answer
as the multiplication of several polynomials; no need to expand.
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Authors: Gerald Albaum , Alexander Josiassen , Edwin Duerr
8th Edition
1292016922, 978-1292016924
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Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities - LindasHelp
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities: 1 0.8+ 0.6 0.4+ 0.2 + 0.4 + 0
0.2 0.6 0.8 (a) P(0.1 X 0.8) = (b) P(X 0.8) Share this:TwitterFacebook
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities: 1 0.8+ 0.6 0.4+ 0.2 + 0.4 + 0
0.2 0.6 0.8 (a) P(0.1 X 0.8) = (b) P(X 0.8) Share this:TwitterFacebook
Let X be a random number between 0 and 1 produced by the idealized uniform random number generator. Use the density curve for X, shown below, to find the probabilities
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On Query-efficient Planning in MDPs under Linear Realizability of the Optimal State-value Function
We consider the problem of local planning in fixed-horizon Markov Decision Processes (MDPs) with a generative model under the assumption that the optimal value function lies close to the span of a
feature map. The generative model provides a restricted, “local” access to the MDP: The planner can ask for random transitions from previously returned states and arbitrary actions, and the features
are also only accessible for the states that are encountered in this process. As opposed to previous work (e.g. Lattimore et al. (2020)) where linear realizability of all policies was assumed, we
consider the significantly relaxed assumption of a single linearly realizable (deterministic) policy. A recent lower bound by Weisz et al. (2020) established that the related problem when the
action-value function of the optimal policy is linearly realizable requires an exponential number of queries, either in H (the horizon of the MDP) or d (the dimension of the feature mapping). Their
construction crucially relies on having an exponentially large action set. In contrast, in this work, we establish that poly(H, d) planning is possible with state value function realizability
whenever the action set has a constant size. In particular, we present the TENSORPLAN algorithm which uses poly((dH/δ)^A) simulator queries to find a δ-optimal policy relative to any deterministic
policy for which the value function is linearly realizable with some bounded parameter (with a known bound). This is the first algorithm to give a polynomial query complexity guarantee using only
linear-realizability of a single competing value function. Whether the computation cost is similarly bounded remains an interesting open question. We also extend the upper bound to the
near-realizable case and to the infinite-horizon discounted MDP setup. The upper bounds are complemented by a lower bound which states that in the infinite-horizon episodic setting, planners that
achieve constant suboptimality need exponentially many queries, either in the dimension or the number of actions.
ASJC Scopus subject areas
• Artificial Intelligence
• Software
• Control and Systems Engineering
• Statistics and Probability
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Cisco Paper-3
SECTION 1 — BASIC DIGITAL SECTION
1. In order to find out stack fault of a three input nand gate how many necessary input vectors are
needed ?
2. What is parity generation ?
3. A nand gate becomes ___ gate when used with negative logic ?
4. What is the advantage of cmos over nmos ?
5. What is the advantage of syncronous circuits over asynchronous circuits ?
6. What is the function of ALE in 8085 ?
7. A voice signal sample is stored as one byte. Frequency range is 16 Hz to 20 Hz. What is the
memorysize required to store 4 minutes voice
8. What will the controller do before interrupting CPU?
9. In a normalised floating point representation, mantissa is represented using 24 bits and
exponent with 8 bits using signed representation.
What is range ?
10. The stack uses which policy out of the following– LIFO, FIFO, Round Robin or none of these ?
11. Where will be the actual address of the subroutine is placed for vectored interrupts?
12. Give the equivalent Gray code reprasentation of AC2H.
13.What is the memory space required if two unsigned 8 bit numbers are multiplied ?
14. The vector address of RST 7.5 in 8085 processor is _______.
Ans. 003C (multiply 7.5 by 8 and convert to hex)
15. Subtract the following hexadecimal numbers— 8416 – 2A16
16. Add the following BCD numbers— 1001 and 0100
17. How much time does a serial link of 64 Kbps take to transmit a picture with 540 pixels.
18. Give the output when the input of a D-flip flop is tied to the output through the XOR gate.
19. Simplify the expression AB + A( B + C ) + B ( B + C )
20. Determine the logic gate to implement the foolowing terms–ABC, A+B+C
21. Implement the NOR gate as an inverter.
22. What is the effect of temperature on the Icb in a transistor
23. What is the bit storage capacity of a ROM with a 512*4 organisation?
24. What is the reason of the refresh operation in dynamic RAM’s ?
25. Suppose that the D input of a flip flop changes from low to high in the middle of a clock
pulse.Describe what happens if the flip flop is a
positive edge triggered type?
26. How many flip flops are required to produce a divide by 32 device ?
27. An active HIGH input S-R latch has a 1 on the S input and a 0 on the R input. What state is the
latch in?
28. Implement the logic equation Y = C^BA^ + CB^A + CBA with a multiplexer. (where C^ stands
for C complement)
29.Equivalent Gray code reprasentation of AC2H.
30. What does a PLL consist of ?
We advice you to know the design of PLL as questions pertaining to this may be asked
SECTION 2 – SOFTWARE SECTION
1. The starting location of an array is 1000. If the array[1..5/…4] is stored in row major order, what
is the location of element [4,3]. Each word
occupies 4 bytes.
2. In a tertiary tree, which has three childs for every node, if the number of internal nodes are N,
then the total number of leaf nodes are
3. Explain the term “locality of reference” ?
4. What is the language used for Artificial Intelligence
Ans: lisp
5. What is the character set used in JAVA 2.0 ?
Ans: Unicode
6. char a =0xAA ;
int b ;
b = (int) a ;
b = b >> 4 ;
What is the output of the above program segment ?
7. struct s1 { struct { struct { int x; } s2 } s3 }y;
How does one access x in the above given structure definition ?
8. Why there is no recursion in Fortran ?
Ans. There is no dynamic allocation.
9. What is the worst case complexity of Quick sort?
Ans. O(n2)
10. What will be sequence of operating system activities when an interrupt occurs ?
11. In a sequential search, what is the average number of comparisons it takes to search through n
elements ?
Ans: (n+1)/2.
12. What is the size of the array declared as double * X[5] ?
Ans. 5 * sizeof ( double * )
13. A binary search tree with node information as 1,2,3,4,5,6,7,8 is given. Write the result obtained
on preorder traversal of the binary search tree
Ans : 53124768
14. If size of the physical memory is 232-1, then what is the size of the virtual
memory ?
15. S -> A0B
A-> BB|0
B-> AA|1
How many strings of length 5 are possible with the above productions?
16. (3*4096+15*256+3*16+3). How many 1’s are there in the binary representation of the result ?
Ans. 10
17. In memory mapped I/O how is I/O is accessed ?
18. What is the use of ALE in 8085 ?
Ans To latch the lower byte of the address.
19. If the logical memory of 8 X 1024 is mapped into 32 frames, then the number of bits for the
logical address are____ ?
Ans. 13
20. Context free grammar is useful for which purpose ?
21. In ternary number representation, numbers are represented as 0,1,-1.(Here -1 is represented
as 1 bar.) How is 352/9 represented in ternary
number representation?
22. There are processes which take 4,1,8,1 machine cycles respectively. If these are executed in
round robin fashion with a time quantum of 1,
what is the time it take for process 4 to complete ?
Ans. 9
23. The minimum frequency of operation is specified for every processor because……
a)for interfacing slow peripherals
b)dynamic memory refreshing.
c)to make compatible with other processor.
24. For linked list implementation , which search is not applicable ?
Ans: Binary search | {"url":"https://placement.freshershome.com/cisco/cisco-paper-3_158.html","timestamp":"2024-11-14T21:08:53Z","content_type":"text/html","content_length":"84425","record_id":"<urn:uuid:3ce25237-99bc-412b-acd6-e10d95cb0e65>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00116.warc.gz"} |
Multivariate Analysis using SAS | Towards AI
Multivariate Analysis using SAS
Last Updated on February 23, 2022 by Editorial Team
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The difference between univariate, multivariate, and multivariable is often overlooked. As such, multivariate and multivariable are used interchangeably although they mean completely different
There is no such thing as a univariate, multivariate model, but you canΒ have:
1. Univariate, multivariable
2. Multivariate, multivariable
3. Univariate, univariable
4. Multivariate, univariable
Multivariate means multiple dependent variables (Yβ s), multivariable means multiple independent variables (Xβ s). The difference between multivariable and univariable is probably known to most
since the majority of the models that you run have more than one independent variable. This means you have a single outcome and multiple predictors.
The difference between univariate and multivariate will have a steeper learning curve since multivariate analysis often leads to a reduction or reframing of the original data to handle the multiple
outcomes you are trying toΒ model.
A multivariable model can be thought of as a model in which multiple independent variables are found on the right side of the model equation. This type of statistical model can be used to attempt to
assess the relationship between a number of variables; one can assess independent relationships while adjusting for potential confounders.
Multivariate Modeling refers to the modeling of data that are often derived from longitudinal studies, wherein an outcome is measured for the same individual at multiple time points (repeated
measures), or the modeling of nested/clustered data, wherein there are multiple individuals in eachΒ cluster
A multivariate linear regression model is a model where the relationships between multiple dependent variables (i.e., Ys)β β β measures of multiple outcomesβ β β and a single set of predictor
variables (i.e., Xs) are assessed.
Multivariate analysis refers to a broad category of statistical methods used when more than one dependable variable at a time is analyzed for aΒ subject.
Although many physical and virtual systems studied in scientific and business research are multivariate most analyses are univariate in practice. What often happens is that these relationships are
merged in a new variable (e.g., Feed Conversion Rate). Often, some dimension reduction is possible, enabling you to see patterns in complex data using graphical techniques.
In univariate statistics, performing separate analyses on each variable provides only a limited view of what is happening in the data as means and standard deviations are computed only one variable
at aΒ time.
If a model has more than one dependent variable, analyzing each dependent variable separately also increases the probability of type-I error in the set of analyses (which is normally set at 5%
orΒ 0.05).
And, if you did not realize it yet, longitudinal data can be analyzed both in an univariate and multivariate wayβ β β it depends on where you want to place the variance-covariance matrix.
Examples of multivariate analysisΒ are:
1. Factor Analysis can examine complex intercorrelations among many variables to identify a small number of latentΒ factors.
2. A Discriminant Function Analysis maximizes the separation among groups on a set of correlated predictor variables, and classifies observations based on their similarity to the overall
groupΒ means.
3. Canonical Correlation Analysis examines associations among two sets of variables and maximizes the between-set correlation in a small number of canonical variables (to be discussed laterΒ on).
So, start thinking about unseen or latent variables.
Univariate & univariable
Univariate & multivariable
Multivariate & multivariable
Multivariate analyses are amongst the most dangerous analyses you can conduct as they are capable of dealing with situations that areΒ rare:
1. Large column Nβ β β datasets that have >100Β columns
2. N<Pβ β β datasets that have less rows (N) than columnsΒ (P)
3. Multicollinearityβ β β datasets containing highly correlated data
However, before just applying multivariate analysis on any dataset you see, you must make sure that you know your data first. The model could not care less if it does not make sense biologically.
Just look at the datasets below and see if you can spot issues that will make analyzing the data difficult. I promise you there are definitely issues to beΒ found.
Selecting before analyzing, screening for abnormalities or looking for potential convergence errors is especially important when dealing with multivariate data. Once included, and the model runs, you
are often not able to find back what you put in the model because of dimension reduction.
In SAS there are many tools you can use to explore the data, summarize it, tabulate it, and create associations.
Never forget that to conduct meaningful analysis, you need to spend a considerable amount of time to wrangle yourΒ data.
Bubble plots are the equivalents of scatterplots but then for 4 variables.
Scatterplot and scatterplot matrix. The scatterplot matrix is a great way of looking at the relationship between a large variety of variables, and between groups. Each cell in the matrix shows the
relationship between two variables. The upper and lower side of the diagonal are mirrored. The diagonal shows the variableΒ names.
Plot shows that outliers influence scatter matrices.
Correlation matrixβ β β a combination of a heatmap and a scatterplot matrix.
A good first step is to create several correlational matrices to identify the largest correlations. Heat maps will help as well. Once you have identified places for zooming in, use scatter matrices
and bubble plots to lookΒ closer.
So, yes, you will create a lot of graphs and so it is best to think about which graphs you would like to make before actually making them so you do not get lost! Since graphing your data is the first
step to gaining insights, a multifold of variables means that you need to graph smart. Start by graphing biologically connected data. Then look at the more unknownΒ data.
Paradox: many of the analytical methods that will be introduced actually require interconnected data. So, if you find a lot of multicollinearities, do not worry. Actually, embrace it! This is where
multivariate models are at theirΒ best.
So, let's start with correlations first, since they are the de facto measurement of association. Remember, we WANT to find the correlation. However, in this post, I will discuss more than just good
old Pearson correlations. I will alsoΒ discuss:
1. good old Pearson correlationsβ β β Can variable 1 predict variableΒ 2?
2. canonical Correlationsβ β β Can set 1 predict setΒ 2?
3. discriminant Analysisβ β β Can a combination of variables be used to predict group membership?
Does variable 1 have a connection to variableΒ 2?
Does a set of variables have a connection to another set of variables?
Can a combination of variables be used to predict group membership?
Correlations are the easiest way of looking for potential relationships between two variables. Look for absolute values >Β 0.7.
And the scatterplot matrix. You are looking for clouds, preferably very diagonal clouds. Anything that is not a diagonal cloud is not really worth your time. The direction matters, but first find a
diagonalΒ one.
When dealing with correlations, you will also have to deal with outliers since outliers can make a correlationβ s life very difficult. Outliers add noise to theΒ signal.
There are two possible ways to deal with outliers:
1. winsorizingβ replacing a value with another value based on the range of values in the dataset. A value at the 4th percentile is replaced by a value on the 5th percentile; a value on the 97th
percentile is replaced by a value on the 95th percentile.
2. trimmingβ β β deleting the value outside of the boundary.
Both winsorizing & trimming are done by variable.
You can immediately see the difference, but is not as great as you might expect. The clouds to the right only have a little bit more patterns.
Raw data vs. winsorized Data. You must always be careful when transforming the data. Sometimes you introduce bias where you would like to free aΒ signal.
In summary, correlation analysis is probably not new to you and despite its drawbacks, it offers a good start to explore large quantities of variables. Especially to seeΒ if:
1. relationships exist
2. known relationships are confirmed
3. clusters exist
4. there are unknown relationships
Next-up is is the Canonical Correlation Analysis (CCA) which is used to identify and measure the associations among two sets of variables. These sets are defined by the analyst. No strings attached.
Canonical correlation is especially appropriate when there is a high level of multicollinearity. This is because CCA determines a set of canonical variates which are orthogonal linear combinations of
the variables within each set that best explains the variabilityβ β β both within and betweenΒ sets.
In short Canonical Correlations allow youΒ to:
1. interpret how the predictors are related to the responses.
2. interpret how the responses are related to the predictors.
3. examine how many dimensions the variable sets share inΒ common.
Canonical variates colored by a groupingΒ factor.
These plots show how each of the observations in the dataset load on the two sets of variables, and how these two sets areΒ related.
PROC CANCORR is the procedure to go towards forΒ CCA.
These statistics test the null hypothesis that all canonical correlations are zero.The small p-values for these tests (< 0.0001) are evidence for rejecting the null hypothesis that a CCA is not
warranted. There is enough shared variance!
Coefficient interpretation can beΒ tricky:
1. Standardized coefficients address the scaling issue, but they do not address the problem of dependencies among variables.
2. Coefficients are useful for scoring but not for interpretationβ β β the analysis method is aimed at prediction!
These are correlational tables. Look for relationships that exceed the absolute value ofΒ 0.7
The CCA created 11 canonical dimensions because 11 variables are included. This is because canonical variates are similar to latent variables that are found in factor analysis, except that the
canonical variates also maximize the correlation between the two sets of variables. They are linear functions of the variables included. And thus, automatically, an equal amount of canonicals are
made as there are variables included
In this graph you can see if all 11 are worth theΒ effort.
However, a more useful way to interpret the canonical correlation in terms of the input variables is to look at simple correlation statistics. For each pair of variates, look at the canonical
structure tables.
Below, is the correlation between each variable and its canonical variate.
Below, is the correlation between each variable and the canonical variate for the other set of variables.
We can even go further and apply the canonical redundancy statistics which indicate the amount of shared variance explained by each canonical variate. It provides youΒ with:
1. the proportion of variance in each variable is explained by the variableβ s own variates.
2. the proportion of variance in each variable explained by the other variablesβ variates.
3. RΒ² for predicting each variable from the first M variates in the otherΒ set.
Each of the variables are better explained by their own canonicals instead of the others, but is not a land-slide.
The output for redundancy analysis enables you to investigate the variance in each variable explained by the canonical variates. In this way, you can determine not only whether highly correlated
linear combinations of the variables exist, but whether those linear combinations are actually explaining a sizable portion of the variance in the original variables. This is not the caseΒ here!
You can also perform Canonical Regression Analysis by which one set regresses on a second set. Together with the Redundancy Statistics, Regression Analysis will provide you with more insight into the
predictive ability of the sets specified.
The regression results (average RΒ²) do not hint at a strong relationship between the Ileum and the Jejenum. Do not look too much at p-values, rather look at how each variables is contributing to
RΒ². You do not have to be a rocket scientist to figure out that the Jejenum will also not be very predictive for theΒ Ileum.
In summary, Canonical Correlation Analysis is a descriptive method trying to relate one set of variables to another set of variables byΒ using:
1. correlation
2. regression
3. redundancy Analysis
As a first method, it gives you a good idea about the level of multicollinearity involved and how much the two specified sets relate to themselves and each other. Do not forgetβ β β CCA is mainly
used for prediction, not interpretation.
And the last of the trio is the Discriminant Function Analysis (DFA) which is used to answer the question: Can a combination of variables be used to predict group membership? Because, if a set of
variables predicts group membership, it is also connected to thatΒ group.
DFA is a dimension-reduction technique related to Principal Component Analysis (PCA) and Canonical Correlation AnalysisΒ (CCA).
DFA in actionβ β β the big rings are are the center points, predicted center points, for each of the groups we are trying to predict for based on the two canonical variates.
In SAS, there are three procedures for conducting Discriminant Function Analysis:
1. PROC CANDISCβ β β canonical discriminant analysis.
2. PROC DISCRIMβ β β develops a discriminant criterion to classify each observation into one of theΒ groups.
3. PROC STEPDISCβ β β performs a stepwise discriminant analysis to select a subset of the quantitative variables for use in discriminating among theΒ classes.
PROC STEPDISC and PROC DISCRIM can be used together to enable selection methods on top of Discriminant Analysis.
An example of PROC CANDISC. The class variable is important hereβ β β you are trying to predict for thatΒ group.
The results clearly show that the Canonical Variables created from the dataset will not provide a powerful set to predict group attribution.
Two ways to plot the same data. Not really impressive results as the ellipsesΒ overlap.
A one standard deviation increase on the IFA_Y variable will result in a -0.089 standard deviation decrease in the predicted values on discriminant function 1. As you can see, the relationships are
not impressive, which was already clear from the previous graphs. This means that the Canonical Variates do not really represent a set of variables!
A much much better split due to canonical variates.
The way to get into PROCΒ DISCRIM.
The difference between the left and right plot is because of mathematical differences AND because of some preliminary steps I took in DISCRIM. Lets lookΒ closer!
PROC DISCRIM in combination with crossvalidation.
Easy to include cross-validation inΒ SAS.
The option β Validationβ seems to be a small request, but it is not. It is actually an introduction to the topic of overfitting. Overfitting occurs when the model is β overfittedβ on the data
that itβ s using to come to a solution. It mistakes noise for signal. An over-fitted model will predict nicely on the trained dataset but horribly on new test data. Hence, as a prediction model, it
is limited. REMEMBER: DFA is a prediction method. Hence, safeguarding from overfitting makes a lot ofΒ sense!
And the inclusion of selection methods.
You can augment PROC DISCRIM by first using PROC STEPDISC which includes algorithms for variable selection. These are mostly traditional methods:
1. Backward Elimination: Begins with the full model and at each step deletes the effect that shows the smallest contribution to theΒ model.
2. Forward Selection: Begins with just the intercept and at each step adds the effect that shows the largest contribution to theΒ model.
3. Stepwise Selection: Modification of the forward selection technique that differs in that effects already in the model do not necessarily stayΒ there.
I am asking SAS to include variables that meet a certain threshold for adding or retaining a variable.
Save your output forΒ graphs.
And the complete code to include selection methods and crossvalidation for a discriminant analysis.
Canonical 1 is defined by IL10_1. Canonical 2 is defined byΒ IFG_Y.
Classification matrices showing model performance. NOT THATΒ GOOD!
The big circles indicate the groups and their mean loadings on both canonical variables. The smaller circles show the individual animals and the group they were assigned to IN THE DATABASE (no
prediction). If you see a plot like this, then you know that the canonical variates have no real discriminant ability.
Hence, what this graph shows is how much these canonical variables are able to predict group assignment based on the models included. Good predictive power would show that animals in a certain group
would cluster at the group mean canonical loading. This is not theΒ case.
A better way to show the discriminant power of the model is to create a new dataset containing all the variables included and add ranges to them so you can do a grid search. You can then ask each DFA
model, using different algorithms, to show you how theyΒ perform.
Discriminant analysis on testΒ data.
Different classification and discriminant functions can be combined leading to six different algorithms. Lets see if it also leads to six different models.
And the results. As you can see, the functions provide different models with different separation lines. What sticks out is that the model is not able to separate en that the separation line actually
runs right through the clusters of dataΒ points.
As you can see above, there is no universal best method. It all depends on the data since the non-parametric methods estimate normality and the parametric methods assume it. The most important
distinction is the use of a linear or quadratic discriminant function. This clearly changes the prediction model and thus the classification matrices. As in many things in life, try different
flavors, but never forget to check your assumptions, the model, and its performance.
In summary, Discriminant function analysis is usually used to predict membership in naturally occurring groups. It answers the question: β Can a combination of variables be used to predict group
membership?β In SAS, there are two procedures for conducting Discriminant Function Analysis:
1. PROC STEPDISCβ β β select a subset of variables for discriminating among theΒ classes.
2. PROC CANDISCβ β β perform canonical discriminant analysis.
3. PROC DISCRIMβ β β develop a discriminant criterion to classify each observation into one of theΒ groups.
Lets venture further into the world of dimension reduction and ask ourselves the following questions:
1. Can I reduce what I have seen to variables that are invisible?
2. Can I establish an underlying taxonomy?
Examples of output that you can obtain from SAS when running dimension reduction techniques.
There are carious PROCS available in SAS to conduct dimension reduction. Of course, the examples shown before in this post are also examples of dimension reduction.
1. PROC PRINCOMP performs principal component analysis on continuous data and outputs standardized or unstandardized principal component scores.
2. PROC FACTOR performs principal component analysis and various forms of exploratory factor analyses with rotation and outputs estimates of common factor scores (or principal component scores).
3. PROC PRINQUAL performs principal component analysis of qualitative data and multidimensional preference analysis. This procedure performs one of three transformation methods for a nominal,
ordinal, interval, or ratio scale data. It can also be used for missing data estimation with and without constraints.
4. PROC CORRESP performs simple and multiple correspondence analyses, using a contingency table, Burt table, binary table, or raw categorical data as input. Correspondence analysis is a weighted
form of principal component analysis that is appropriate for frequency data.
5. PROC PLS fits models using any one of a number of linear predictive methods, including partial least squares(PLS). Although it is used for a much broader variety of analyses, PLS can perform
principal components regression analysis, although the regression output is intended for prediction and does not include inferential hypothesis testing information.
Probably the most widely known clustering technique is Principal Components Analysis (PCA) and as you saw before, there is a heavy relationship with Canonical Correlation Analysis. A PCA tries to
answer a practical question: β How can I reduce a set of many correlated variables to a more manageable number of uncorrelated variables?β
PCA is a dimension reduction technique that creates new variables that are weighted linear combinations of a set of correlated variables β principal components. It does not assume an underlying
latent factor structure.
PCA displayed.
PCAβ s work with components which are orthogonal regression lines, created to minimize theΒ errors.
The third component is constructed in the same manner, and each subsequent component accounts for less and less of the total variability in the data. Typically, a relatively small number of created
variables, or components, can account for most of the total variability in theΒ data.
PCA creates as many components as there are input variables by performing an eigenvalue decomposition of a correlation or covariance matrix. It creates components that consolidate more of the
explained variance into the first few PCs than in any variable in the original data. They are mutually orthogonal and therefore mutually independent. They are generated so that the first component
accounts for the most variation in the variables, followed by the second component, and soΒ on.
As with many multivariate techniques, PCA is typically a preliminary step in a larger data analytics plan. For example, PCA could be usedΒ to:
1. explore data and detect patterns among observations.
2. find multivariate outliers.
3. determine the overall extent of collinearity in a dataΒ set.
Partial Least Squares also uses PCA as an underlying engine.
SAS Studio code to run PROCΒ PRINCOMP
The first plot you are going to look atβ β β how many components will give you a good split between variance explained and remaining to be explainable.
The Scree plot shows how many principal components you need to reach a decent level of the variance explained. The trick is to look at the Scree plot and see when the drop levels off. Here, this is
after eight components. Let's plot those eight components.
This Component Pattern Profiles plot shows how each component is loading on the variables included. Hence, it shows what each component represents. What you can immediately see is that it is quite a
mess. Many variables load on many components to some degree. So you must, for interpretationβ s sake, limit the number of components toΒ use.
Three components. You can clearly see that component 1 and 2 are represented by several distinct variables.
Hence, observations that load highly on those components are probably also distinctly different on those variables. Component 3 looks like a bit of a garbage component.
These plots show how the variables load on each component. Some clusters stick out, but it is also clear that a lot of variables do not really load on a component. This is reflected on the low
percentage of variance explained.
The difference between PCA and DFA analysis. As you can see, the DFA does a much better job, but was already made to separate the data based on theΒ groups.
This plot shows how observations load on all three components using both the x and y axis, and color. As you can see, component 1 (mostly blue) and 2 (everything around 0) are quite informative.
Component 3 adds some dimension, but not as clear as 1 andΒ 2.
In summary, PCA is a dimension reduction technique that creates new variables that are weighted linear combinations of a set of correlated variables Γ principal components. PCA tries to answer a
practical question: β How can I reduce a set of many correlated variables to a more manageable number of uncorrelated variables?β PCA is typically a preliminary step in a larger data analytics
plan, and a part of many regression techniques to ease analysis.
From PCA, it is quite straightforward to move further towards Principal Factor Analysis (PFA). The difference is that, in PCA, the unique factors are uncorrelated with each other. They are linear
combinations of the data. In PFA, the unique factors are uncorrelated with the common (latentβ β β Yx) factors. They are estimates of latent variables that are partially measured by theΒ data.
Component analysis is actually restructuring the same data. Exploratory factor analysis is modeling.
PROC FACTOR is the de factor procedure for factor analysis.
Factor Analysis is used when you suspect that the variables that you observe (manifest variables) are functions of variables that you cannot observe directly (latent variables). Hence, factor
analysis is usedΒ to:
1. Identify the latent variables to learn something interesting about the behavior of your population.
2. Identify relationships between different latent variables.
3. Show that a small number of latent variables underlies the process or behavior that you have measured to simplify yourΒ theory.
4. Explain inter-correlations among observed variables.
The difference between Factor Analysis and Principal Component Analysis.
Look for absolute factor loadings >0.5. PBMC clearly loads very high on Factor 1. CO and Y seem to load high on FactorΒ 2.
The initial factor loadings are just the first step, and rotation methods need to be used to interpret the results ad they will help you a lot in understanding the results coming from factor
There are two general classifications of rotationΒ methods
1. Assume orthogonal factors.
2. Relax orthogonality assumption.
As you can see there are a lot of options available to do rotation of factorΒ models.
Orthogonal rotation maintains mutually uncorrelated factors that fall on perpendicular axes. For this, the Varimax-Orthogonal method is often used which maximizes the variance of columns of the
factor pattern matrix. The axes are rotated, but the distance between the axes remains orthogonal.
Then we also have Oblique rotation which allows factors to be correlated with each other. Because factors might be theoretically correlated, using an oblique rotation method can make it much easier
to interpret the factors. For this, the Promax-Oblique is often used, which performs:
1. varimax rotation
2. relaxes orthogonality constraints and rotatesΒ further
3. rotate axes that can converge/diverge
But as you can see below, there are a lot of combinations possible. The major part is in relaxing or not relaxing the orthogonal assumption, meaning factors can have a covariance matrix orΒ not.
Original Results EFA. Rotation of the axes. Rotated resultsΒ EFA.
The difference between the initial and rotated factor pattern is clear to see. There are now three clusters instead of two. The variance loadings on the factors changed a bit, because the distance
between the grid and the observations also differ due toΒ rotation
Having 18 factors will make this PFA quite the challenge. It also tells you that these variables will not be so easy to load on an underlying latent variable. To counteract this a bit, we could also
limit the number ofΒ factors.
The results will be the same, I just cut it off atΒ 4.
To check for Orthogonality, the factors should not correlate highly with the others. The loading table to the right clearly shows what the factors stand for, eat least factor 1 (PBMC) and factor 2
(CO). Then, it becomesΒ blurry.
You would like to see a high number on the diagonal and not so much in the otherΒ cells.
Rotations has led to three nice clusters.
In the previous example, I just decided out of the blue to downsize the number of factors from 18 to 4. Selecting the numbers of factors can be done more elegantly using parallel analysis which is a
form of simulation.
Parallel analysis requesting 10000 simulations to see how many factors I need to establish a decent factor analysis.
Parallel analysis shows that 6 factors should be retained.
Exploratory Factor Analysis can also provide you with path diagrams which are a visual representation of the model. If you find yourself getting a model like the graph on the left, something is wrong
with the model being able to really identify the latent variables.
In summary, exploratory factor analysis (EFA) is a variable identification technique. Factor analytic methods are used when an underlying factor structure is presumed to exist but cannot be
represented easily with a single (observed) variable. In EFA, the unique factors are uncorrelated with the latentβ β β factors. They are estimates of latent variables that are partially measured
by the data. Hence, not 100% of the variance is explained inΒ EFA.
EFA is an exploratory step towards full-fledged causal modeling.
Clustering is all about measuring the distance between variables and observations, between themselves, and the clusters that are made. A lot of methods for clustering data are available in SAS. The
various clustering methods differ in how the distance between two clusters is computed.
In general, clustering works likeΒ this:
1. Each observation begins in a cluster byΒ itself.
2. The two closest clusters are merged to form a new cluster that replaces the two old clusters.
3. The merging of the two closest clusters is repeated until only one cluster isΒ left.
Examples of using clustering to find patterns in theΒ data.
SAS offers a variety of procedures to help you clusterΒ data:
1. PROC CLUSTER performs hierarchical clustering of observations.
2. PROC VARCLUS performs clustering of variables and divides a set of variables by hierarchical clustering.
3. PROC TREE draws tree diagrams using output from the CLUSTER or VARCLUS procedures.
4. PROC FASTCLUS performs k-means clustering on the basis of distances computed from one or more variables.
5. PROC DISTANCE computes various measures of distance, dissimilarity, or similarity between the rows (observations).
6. PROC ACECLUS is useful for processing data prior to the actual cluster analysis by estimating the pooled within-cluster covariance matrix.
7. PROC MODECLUS performs clustering by implementing several clustering methods instead ofΒ one.
Although I have labeled a lot of possibilities for clustering methods, the most powerful procedure is by far the Partial Least Squares (PLS) methodβ β β a multivariate multivariable algorithm. The
PLS balances between principal components regression (explain total variance) and canonical correlation analysis (explain shared variance). It extracts components from both the dependent and
independent variables, and searches for explained variance within sets, and shared variance betweenΒ sets.
PLS is a great regression technique when N < P as it extracts factors / components / latent vectorsΒ to:
1. explain response variation.
2. explain predictor variation.
Hence, partial least squares balance two objectives:
1. seeking factors that explain response variation.
2. seeking factors that explain predictor variation.
The PLS procedure is used to fit models and to account for any variation in the dependent variables. The techniques used by the Partial Least Squares Procedure are:
1. Principal component regression (PCR) technique, in which factors are extracted to explain the variation of predictor sample.
2. Reduced rank regression (RRR) technique, in which factors are extracted to explain response variation.
3. Partial least squares (PLS) regression technique, where both response variation and predictor variation are accounted for.
PCR, RRR, and PLS regression are all examples of biased regression techniques. This means using information from k variables but reducing them to <k dimensions in the regression model, making the
error DF larger than would be the case if Ordinary least Squares (OLS) regression were used on all the variables. PLS is commonly confused with PCR and RRR, although there are the following key
1. PCR and RRR only consider the proportion of variance within a set explained by PCs. The linear combinations are formed without regard to association among the predictors and responses.
2. PLS seeks to maximize association between the sets while considering the explained variance in each set of variables.
The PLS procedure is used to fit models and to account for any variation in the dependent variables.
However, PLS does not fit the sample data better than OLSβ it can only fit worse or as well. As the number of extracted factors increases, PLS approaches OLS. However, OLS over fits sample data,
where PLS with fewer factors often performs better than OLS in predicting future data. PLS uses cross-validation as a technique for determining how many factors should be retained to prevent
SAS Studio task to begin using the PLS regression.
You are able to create some wildΒ models.
Many of the options I leave as default to see if a model can be run. The PLS can become very complex very fast and to safeguard from pouring variables in them without consideration it is good that
you first understand the defaultΒ results.
And the code, straightforward. I have many outcomes and many predictors. The hallmark of a multvariate multivariable model.
This model went absolutely nowhere!
So let's start a bitΒ simpler:
1. One dependent variable
2. More independent variables
3. No cross-validation to ensure I have all the data to train aΒ model.
Univariate multivariable model.
You can see the number of extract factors and how RΒ² behaves. Note that two RΒ² β s are provided. This is because the PLS is building two models at the same time. One that can explain the
predictors and one that can explain the outcome(s) included.
At least we have a result. However, it is not a result you would like to haveβ β β 15 factors explain 87% of the variance of the independent variables and 16% of the dependent variables.
Preferably, you would like a small number of extracted factors to be able to predict at a solid level. RΒ² is not the best method for assessing thisΒ though.
Below are some slides on how to interpret the most interesting plot ever providedβ β β the correlation loading plot. It is quite a handful to look at but wonderfully simple once you getΒ it.
Through cross-validation, the model selects 2 factors. Always check if the model makes biological sense. Remember, statistical models could not care less what they include. They donβ tΒ know.
PLS provides you with a long list of results and plots if you want them. Some of the most important ones you will find here. First and foremost, look at the fit diagnostics of the model. Then look at
the Variable importance plots. Do they make biological sense? Additionally, you can use the results to create diagnostic plots of yourΒ own.
Here, we have another correlation loading plot. They are quite heavy to digest First, look at the factors and the percentage variance explained in the model (X RΒ²) and the prediction (Y RΒ²). The
outcome variable variance is 75% explained by the two factors. As you can see, many variables load very highly (range of 75%β β β 100%). All in all, the observations show 3 clusters which indicate
that a classification variable might help explain even more variance.
This plot shows the ability of each factor to distinguish the granularity of the response variable. If you look at Factor 1 you see that the complete range of the response variables is included. The
more straight the diagonal line is, from left bottom to upper right, the better. This is clearly show in Factor 2β β β much more diverging. Numbers indicate observations.
In summary, the PLS procedure is a statistically advanced, output-heavy procedure. Know what you are doing before you start! PLS strikes a balance between principal components regression and
canonical correlation analysis. It extracts components from the predictors/responses that account for explained variance within the sets and shared variation between sets. PLS is a great regression
technique when N < P, unlike multivariate multiple regression or predictive regression methods.
I hope you enjoyed this post. Let me know if something isΒ amiss!
Multivariate Analysis using SAS was originally published in Towards AI on Medium, where people are continuing the conversation by highlighting and responding to this story.
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Published via Towards AI | {"url":"https://towardsai.net/p/l/multivariate-analysis-using-sas","timestamp":"2024-11-11T06:34:03Z","content_type":"text/html","content_length":"570968","record_id":"<urn:uuid:adb767e6-b445-473d-a972-a3f4a47ddf91>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00555.warc.gz"} |
RGLUEANN package available on GitHub | R-bloggersRGLUEANN package available on GitHub
RGLUEANN package available on GitHub
[This article was first published on
Bart Rogiers' blog
, and kindly contributed to
]. (You can report issue about the content on this page
) Want to share your content on R-bloggers?
if you have a blog, or
if you don't.
The RGLUEANN package is now available on GitHub at
. The package provides an R implementation of the coupling between general likelihood uncertainty estimation (GLUE) and artificial neural networks (ANNs), as presented in our
2012 Mathematical Geosciences paper
. It is basically a probabilistic non-linear data-driven modelling tool. The package can be installed using the following R code:
• demo(“RGLUEANN_training_and_prediction”) shows how to train a GLUE-ANN ensemble, and make predictions with it;
• demo(“RGLUEANN_cross-validation”) provides an example of cross-validation.
Any feedback is welcome! | {"url":"https://www.r-bloggers.com/2014/10/rglueann-package-available-on-github/","timestamp":"2024-11-06T05:09:20Z","content_type":"text/html","content_length":"85297","record_id":"<urn:uuid:4ca7c882-6d88-4fca-9f4d-f72aa636fb2c>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00661.warc.gz"} |
Arxiv trawling, May 19
Tensor recovery and completion — a novel approach to recovering low-rank tensors that does not involve matrix unfolding, and has reasonable sample complexities (better than square deal)
Partial sharing of latent factors in multiview data — explicitly take advantage of the fact that not only should information be shared across modes, but also each mode should have its own unique
Analysis of a nonconvex low-rank matrix estimator — not sure what the regularizer is, but looks readable, analysis perhaps uses the restricted isometry assumption
Exponential families on Manifolds — fun light reading. natural adaption of exponential families to (Lie?) manifolds; not much experimental results
Exponential families in high dimensions — learning convergence rate under sparsity assumptions. useful for anything I do? looks like fun/informative reading, though
Graph partitioning for parallel machine learning — uses submodularity. how compares to the sampling scheme from Cyclades?
Column selection from matrices with missing data — looks like a long read, but worthwhile, and well written. READ
Minkowski symmetrization of star-shaped sets — looks readable; applicable to using Householder rotations to do range space estimation?
Distributed preconditioning for sparse linear systems — A Saad paper. Is this useful for us anywhere? Nice to read to soak in some knowledge on preconditioning
Another preconditioner for sparse systems — Another Saad paper. Nice to read to get some preconditioner knowledge
Using bootstrap to test equality of covariance matrices — relevant to the science problems we’re working on (e.g. diagnostics)? might be worth reading just for general statistical sophistication;
they apply the method to a biology problem
A second-order active set algorithm for l1-regularized optimization — with short, sweet, convergence analysis. looks like a fun, informative optimization read
Boosting in linear regression is related to subgradient optimization — how is this related to AdaBoost and Forward Stagewise Regression are First-Order Convex Optimization Methods by the same
And an oldie:
Duality between subgradient and conditional gradient methods — exactly as stated. Looks like a nice piece of work in terms of convergence analyses and ideas to purpose, as well as general pedagogical
value. A Bach paper. | {"url":"http://www.thousandfold.net/readingblog/?p=50","timestamp":"2024-11-05T21:26:11Z","content_type":"text/html","content_length":"30160","record_id":"<urn:uuid:c2386411-399b-4adb-8cab-5fab27c7bd4c>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00145.warc.gz"} |
Direction of Electric Field in context of Electric Fields
30 Aug 2024
Direction of Electric Field: Understanding the Fundamentals
Electric fields are a fundamental concept in physics, and understanding their direction is crucial for grasping many electrical phenomena. In this article, we will delve into the direction of
electric fields, exploring the underlying principles and providing formulas to help you better comprehend this important topic.
What is an Electric Field?
An electric field is a region around a charged particle or object where the force on another charge can be detected. It is a vector quantity that has both magnitude (strength) and direction. The
direction of an electric field is defined as the direction in which a positive test charge would move if it were placed in the field.
Coulomb’s Law
The direction of an electric field is governed by Coulomb’s law, which states that the force between two charges is proportional to the product of their magnitudes and inversely proportional to the
square of the distance between them. The formula for Coulomb’s law is:
F = k * (q1 * q2) / r^2
where F is the force between the two charges, k is Coulomb’s constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them.
Electric Field Lines
To visualize the direction of an electric field, we can use electric field lines. These lines emerge from positive charges and enter negative charges, indicating the direction of the force on a test
charge. The density of the lines around a charge is proportional to the magnitude of the electric field.
Direction of Electric Field Around a Point Charge
The direction of the electric field around a point charge can be determined using Coulomb’s law. For a positive point charge, the electric field lines emerge from the charge and spread out radially
in all directions. The direction of the electric field is away from the charge, as shown in Figure 1.
Direction of Electric Field Around a Dipole
A dipole is a pair of charges with equal magnitude but opposite sign. The direction of the electric field around a dipole can be determined by considering the forces on positive and negative test
charges. As shown in Figure 2, the electric field lines emerge from the positive charge and enter the negative charge, indicating that the force on a positive test charge is away from the dipole and
towards the negative charge.
Direction of Electric Field Around a Conductor
When a conductor is placed in an electric field, it can become charged. The direction of the electric field around a conductor depends on the type of charging that occurs. If the conductor becomes
positively charged, the electric field lines emerge from the conductor and spread out radially. If the conductor becomes negatively charged, the electric field lines enter the conductor and spread
out radially.
In this article, we have explored the direction of electric fields in the context of electric fields. We have seen how Coulomb’s law governs the force between two charges and how electric field lines
can be used to visualize the direction of the electric field. By understanding the direction of electric fields around point charges, dipoles, and conductors, you will gain a deeper appreciation for
the fundamental principles of electricity.
• Coulomb’s law: F = k * (q1 * q2) / r^2
• Electric field strength: E = k * q / r^2
Key Takeaways:
• The direction of an electric field is defined as the direction in which a positive test charge would move if it were placed in the field.
• Coulomb’s law governs the force between two charges and determines the direction of the electric field.
• Electric field lines emerge from positive charges and enter negative charges, indicating the direction of the force on a test charge.
• The direction of the electric field around a dipole is away from the positive charge and towards the negative charge.
Related articles for ‘Electric Fields’ :
• Reading: Direction of Electric Field in context of Electric Fields
Calculators for ‘Electric Fields’ | {"url":"https://blog.truegeometry.com/tutorials/education/b3adaf10612d509acaa489ee2960d154/JSON_TO_ARTCL_Direction_of_Electric_Field_in_context_of_Electric_Fields.html","timestamp":"2024-11-06T08:00:49Z","content_type":"text/html","content_length":"17434","record_id":"<urn:uuid:fffdc88b-6603-474a-8f81-c14535383cb5>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00150.warc.gz"} |
CSL-101 Computer Programming (Assignment – 5) solved
Q.1 Write a Python program for printing Fibonacci series for ‘n’ terms.
Q.2 Write a Python program to Reverse a Given Number.
Q.3 Write a Python program to Check if a Number is a Palindrome. A number is
Palindrome if it equals to its reversed number.
Q.4 Write a Python program to find the LCM of two numbers.
Q.5 Write a Python program to check whether a number is a strong number or not. A
number is called a strong number if the sum of the factorials of its digits is equal to
the number itself. For example: 145 is strong because 1! + 4! + 5! = 1 + 24 + 120 =
Q.6 Write a Python program to check if two numbers are amicable numbers. Amicable
numbers are two different numbers so related that the sum of the proper divisors of
each is equal to the other number. For example, 220 and 284 are amicable because
the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110, of which the
sum is 284; and the proper divisors of 284 are 1, 2, 4, 71 and 142, of which the sum
is 220.
Q.7 Write a Python program to check if a given number is a Fibonacci number or not.
Make use of following hint. Hint: A number is a Fibonacci number if either 5n2
+ 4
or 5n2
– 4 is a perfect square.
Q.8 Write a Python program to print the Collatz sequence. The Collatz sequence is
defined as: start with a number n. The next number in the sequence is n/2 if n is even
and 3n + 1 if n is odd. Repeat above steps, until it becomes 1.
For example, The Collatz sequence for n = 6 should print 6, 3, 10, 5, 16, 8, 4, 2, 1. | {"url":"https://codeshive.com/questions-and-answers/csl-101-computer-programming-assignment-5-solved/","timestamp":"2024-11-04T02:45:08Z","content_type":"text/html","content_length":"98738","record_id":"<urn:uuid:58dc46b3-39f3-4b2c-a4b4-6b170ff46c95>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00834.warc.gz"} |
Mortgage Payment Calculator
A fast and simple mortgage payment calculator online web app that gives you data fast & easy. Perfect if you are in search of a reliable, fast and intuitive free mortgage calculator with taxes &
mortgage calculator with PMI.
Can you afford the mortgage?
This specific mortgage loan calculator, or also known as a home loan calculator, is the tool you want to use prior to getting a mortgage loan. The reason is simple: it'll tell you if you can afford
the mortgage or not. Also, if you can afford it, it will precisely calculate your loan with taxes and PMI so you can know what to expect each month.
So, this neat payment calculator will help you determine a realistic budget that suits your lifestyle and expected earnings.
How to use our mortgage payment calculator?
Enter the amount of the monthly payment you want to pay or you think you can afford. Fill out the other important data (taxes, start date, PMI etc.) only if they are different then the default data
in the mortgage payment calculator and hit enter.
Then our free mortgage calculator will give precise data about monthly principal & interest, a number of total payments, the total interest that you need to pay and payout date. Not only that, there
is a complete amortization schedule up to the final year of payment. Additionally, the amortization schedule can be set to monthly or yearly.
No matter your needs and the type of mortgage loan, the precise and thorough calculations done by our advanced mortgage calculator can save you from a lot of frustration and uncertainties.
Try the online mortgage calculator now for free!
Figuring Out What You Can Afford
Buying a home is a huge investment, and the decisions you make now could haunt you for a long time, 30 years to be exact. Before you enter into any mortgage agreement, you should know what type of
home you can afford and be familiar with loan terms and how they affect the repayment of the loan. At the very least, you should have a good idea of what kind of payment you can realistically afford
each month. Be sure to calculate insurance and land taxes into the payment as well.
A great tool
A mortgage calculator is a great tool that you can use to see how much you can realistically afford. Before you start punching numbers into a calculator, however, you need to have a budget. To create
a realistic budget, keep a notebook with you and jot down everything that you spend. Include bills, restaurant tabs, transportation expenses, entertainment, etc. Track everything for an entire month.
This will give you a realistic budget. You may be wondering why you can’t simply write down your bills and formulate a budget that way. You can, but you will probably leave out daily expenses that
will affect your ability to make your mortgage payment.
Remember: be sure to allow for expenses that will be tacked to your mortgage payment. Your payment could end up being hundreds of dollars more than what you figured with the calculator after you add
on land taxes and insurance payments.
After you formulate a budget, use a mortgage calculator to see what you can afford. If you think you can afford a $700 monthly payment, enter this amount into the payment field of the calculator and
it will then automatically fill in the other fields so that you can see how much you can borrow.
You should always use a mortgage calculator when shopping for a home. It can help you compare the cost of buying different homes which will help you immensely during the selection process. A
calculator can also give you all of the information that you need regarding a loan and may prompt you to seek more favorable terms.
Whenever you shop for a new home, you should shop for a new home loan as well. Gather as many loan offers as you can and compare each using a loan calculator. Doing your homework can save you a lot
of money and heartache in the long run. Think about this: a difference of only 1.5% interest on a 30 year, $100,000 will cost you $39,980 in interest over the course of the loan. It’s your money. Use
a mortgage calculator to learn how you can hold onto more of it.
How do we calculate?
If you would like to know how to calculate mortgage payment on your own, the equation is:
• MP = monthly payment;
• P = principal;
• r = monthly interest rate**
• n = number of months you will have to repay your loan for.
**To calculate your monthly interest rate simply divide the annual interest rate by 12.
Read more
Example calculation
Let's do an example calculation. To do that, we need to know: the principal amount, monthly interest rate, loan period/number of payments. You can find this information in your mortgage loan
agreement. For our purposes, we will assume the following numbers:
• our principal (P) equals 100 000 EUR;
• our loan period is 20 years - that is 240 months, therefore "n" = 240;
• the annual interest rate amounts to 5%, this divided;
• by 12 equals 0,004 (0,05/12) and this is our "r".
Now, we can get on with the calculation:
MP=100 000[0,004(1+0,004) ^ 240/(1+0,004)^240-1]
To make it easier, we will add 1 to the "r"
MP=100 000(0,004*1,004 ^ 240/(1,004^240)-1)
In the next step we have to raise the "(1+r)" (in our example 1,004) to the power of "n" (in our example 240). It is best to use a calculator (put in the value to be raised, than press the xy button
and enter the "n" value, then press "=") or an excel sheet (use the POWER function: =power(number to be raised,power). The number in our case is: 2,607. Now our equation would look like this:
MP=100 000(0,004*2,607 / 2,607-1)
Let's simplify again and multiply the "r" times the result of raising to power (the top value) and subtract "1" from the result of raising to power on the bottom:
MP=100 000(0,01043)/1,607
All that is left to do now is to divide the numerator by the denominator...
MP=100 000*0,006490
...and there you go: your monthly payment is 649,03. If you want to know what the total sum of all your payments will amount to, just multiply your monthly payment (MP) by the number of months you
will pay your loan (n). In our example it would be:
When you know what your total payments will be, you can also calculate how much you will pay the bank for loaning you money. Just subtract your principal from your total payments. In our case the
costs of our loan would amount to 55 767,2 EUR.
You can also forget about all this long counting and use our mortgage calculator. | {"url":"https://www.mortgagecalculatorplus.com/","timestamp":"2024-11-11T20:50:27Z","content_type":"text/html","content_length":"237343","record_id":"<urn:uuid:47cdcdbc-d217-40ce-a7b1-2f69f79dccbb>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00393.warc.gz"} |
The Whetstone of Witte
The first instance of the concept of a symbolic mathematical equation is usually attributed to Robert Recorde, a Welsh mathematician, in a 1557 book with the rather verbose title ‘The Whetstone of
Witte, whiche is the seconde parte of Arithmeteke: containing the extraction of rootes; the cossike practise, with the rule of equation; and the workes of Surde Nombers’.
That is to say, Recorde invented the equals sign, which is still in use today. However, over the centuries it has undergone significant length contraction. As Piers Bursill-Hall remarked (you may
enjoy reading his History of Maths lecture notes or, even better, attending the aforementioned lectures), the symbol for equality was a pair of ridiculously elongated parallel lines, like so:
Even though he invented one of the most useful mathematical notations in existence, Recorde had very bizarre reasons for doing so. In The Whetsone of Witte, he explained his reasoning for choosing
the symbol to represent equality, namely that ‘no two things can be more equal than a pair of parallel lines’ (!).
Actually, the hyperbolic Fermat equation shown above is not entirely in Recorde’s notation, since the Cartesian notation for powers had not been adopted. Recorde had a brilliantly clumsy method of
expressing powers of quantities. We still had the Hindu-Arabic numerals at this point thanks to Fibonacci’s text, Liber Abaci (which Bursill-Hall accidentally spoonerised, resulting in him saying ‘
[…] the great mathematician, Liberace’). Hence, Recorde would have expressed the quantities in this equation as:
30042907 square
So far, so good. And you can probably guess the next one:
96222 cube
Again, nothing out of the ordinary. But once we get to higher powers, the notation appears slightly comical:
43 zenzizenzizenzic
What? Zenzizenzizenzic???
I have to admit that this is definitely my favourite obsolete mathematical notation. It is based on a Germanised spelling of the Italian word censo (meaning ‘squared’), and exists in the Oxford
English Dictionary, noted for possessing more copies of the letter ‘z’ than any other word. Some books feature 16th powers, described as zenzizenzizenzizenzic (starting to get ridiculous now), but
this term unfortunately did not make it into the OED.
The fourth power, of course, was merely denoted zenzizenzic. The fifth power was called the first sursolid, and subsequent prime exponents were described similarly. There’s almost a
sub-exponential-time algorithm for converting the Nth power into Recorde’s notation:
• Factorise the number N into a product of primes (with possible multiplicity). This can be done in sub-exponential time using the General Number Field Sieve.
• Sort the prime factors into ascending order. This can be accomplished in O(log N log log N) time with von Neumann’s merge sort.
• If there is only one instance of ‘2’ or ‘3’, replace it with ‘square’ or ‘cube’, respectively. If there is more than one instance, replace each ‘2’ with ‘zenzi’, each ‘3’ with ‘cubi’, and
concatenate them with a final ‘c’ appended.
• Replace any remaining primes p with ‘(π(p) − 2)th sursolid’, where π is the prime-counting function. This can be accomplished in time O(p^(½ + ε)) with the Lagarias-Odlyzyko algorithm.
Unfortunately, unless N is a smooth number, this final step is exponential in the input length, destroying what would otherwise be a sub-exponential-time algorithm.
Contrast this with linear time for expressing a power in Descartes’ notation, and it is clear that moving away from Recorde’s notation was probably a good idea.
0 Responses to The Whetstone of Witte
1. How does one acquire a copy of the History of Maths lecture notes? I feel these could be appropriate entertainment for an early morning ferry trip tomorrow.
□ Nevermind, I found the link. 😛
☆ Then again, I can’t access the pdf…
○ Oh, sorry, I think you have to be inside the .cam.ac.uk network. You’ll have to remind me on or after the 6th October, when I regain absolute power. 😛
■ If you can’t wait that long, you could always pester me on the 5th instead.
■ Alternatively, I could pester Ben Green… Oh, wait, he’s not at Cambridge. ;P
2. Being picky here, but these aren’t even lines. These are line segments 😛
3. I am enjoying thinking of Robert Recorde as reading his equations out loud in the style of legendary boxing announcer Michael Buffer.
“One pluus one equaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaals TWO!”
This entry was posted in Uncategorized. Bookmark the permalink. | {"url":"https://cp4space.hatsya.com/2013/09/27/the-whetstone-of-witte/","timestamp":"2024-11-04T09:20:56Z","content_type":"text/html","content_length":"74540","record_id":"<urn:uuid:363da538-2118-4df4-a989-45d19717755d>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00188.warc.gz"} |
HyperLogLog sketches
This topic describes how to use HyperLogLog sketches in Amazon Redshift. HyperLogLog is an algorithm for the count-distinct problem, approximating the number of distinct elements in a data set.
HyperLogLog sketches are arrays of uniqueness data about a data set.
HyperLogLog is an algorithm used for estimating the cardinality of a multiset. Cardinality refers to the number of distinct values in a multiset. For example, in the set of {4,3,6,2,2,6,4,3,6,2,2,3},
the cardinality is 4 with distinct values of 4, 3, 6, and 2.
The precision of the HyperLogLog algorithm (also known as m value) can affect the accuracy of the estimated cardinality. During the cardinality estimation, Amazon Redshift uses a default precision
value of 15. This value can be up to 26 for smaller datasets. Thus, the average relative error ranges between 0.01–0.6%.
When calculating the cardinality of a multiset, the HyperLogLog algorithm generates a construct called an HLL sketch. An HLL sketch encapsulates information about the distinct values in a multiset.
The Amazon Redshift data type HLLSKETCH represents such sketch values. This data type can be used to store sketches in an Amazon Redshift table. Additionally, Amazon Redshift supports operations that
can be applied to HLLSKETCH values as aggregate and scalar functions. You can use these functions to extract the cardinality of an HLLSKETCH and combine multiple HLLSKETCH values.
The HLLSKETCH data type offers significant query performance benefits when extracting the cardinality from large datasets. You can preaggregate these datasets using HLLSKETCH values and store them in
tables. Amazon Redshift can extract the cardinality directly from the stored HLLSKETCH values without accessing the underlying datasets.
When processing HLL sketches, Amazon Redshift performs optimizations that minimize the memory footprint of the sketch and maximize the precision of the extracted cardinality. Amazon Redshift uses two
representations for HLL sketches, sparse and dense. An HLLSKETCH starts in sparse format. As new values are inserted into it, its size increases. After its size reaches the size of the dense
representation, Amazon Redshift automatically converts the sketch from sparse to dense.
Amazon Redshift imports, exports, and prints an HLLSKETCH as JSON when the sketch is in a sparse format. Amazon Redshift imports, exports, and prints an HLLSKETCH as a Base64 string when the sketch
is in a dense format. For more information about UNLOAD, see Unloading the HLLSKETCH data type. To import text or comma-separated value (CSV) data into Amazon Redshift, use the COPY command. For more
information, see Loading the HLLSKETCH data type.
For information about functions used with HyperLogLog, see HyperLogLog functions. | {"url":"https://docs.aws.amazon.com/redshift/latest/dg/hyperloglog-overview.html","timestamp":"2024-11-04T00:05:14Z","content_type":"application/xhtml+xml","content_length":"15677","record_id":"<urn:uuid:8481bc23-072b-492a-a24c-ef1df7634169>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00793.warc.gz"} |
# Understanding Light Pressure and Its Role in Spacecraft Propulsion
Written on
Chapter 1: Introduction to Light Pressure
Have you ever wondered how light, despite having no mass, can exert pressure? This intriguing concept is the foundation of solar sails, a unique method of spacecraft propulsion. By deploying a sail,
a spacecraft harnesses sunlight's pressure to generate thrust, much like traditional sailboats rely on wind. This article delves into the complexities of how photons manage to exert pressure on solar
sails, despite their lack of mass.
Chapter 2: Historical Perspectives on Light Pressure
The idea that light can exert pressure dates back to the 17th century, when Johannes Kepler first proposed it after observing comet tails influenced by the Sun's gravity. Later, in the 19th century,
James Clerk Maxwell provided a theoretical framework for light pressure. Russian physicist Pyotr Lebedev conducted experiments that further explored this phenomenon.
Conventional sailboats experience wind pressure, which imparts momentum to their sails. Momentum, defined as the product of mass and velocity, raises the question: how can light, which is massless,
transfer momentum?
Chapter 3: The Physics of Light and Momentum
In classical Newtonian physics, the notion of light pressure and its application to sails seems implausible. Fortunately, Newton's laws have limitations, and this is where solar sails find their
place in modern physics. For instance, in 2010, the IKAROS spacecraft successfully reached Venus using a solar sail.
The resolution to the momentum transfer mystery lies within the framework of special relativity. In this theory, physical properties such as momentum and velocity are represented as four-vectors
within four-dimensional Minkowski space, known as four-momentum and four-velocity. This approach is particularly beneficial for relativistic calculations, as these vectors remain consistent across
various reference frames (known as Lorentz covariance).
Section 3.1: The Link Between Energy and Momentum
The transformation of four-momentum during Lorentz transformations illustrates the deep connection between momentum and energy. Just as space and time merge into a single concept called "spacetime,"
energy and momentum are also interlinked. Consequently, observers in different reference frames may perceive energy and momentum differently. However, the relationship between these two entities is
defined by a fundamental equation:
By substituting the mass of a photon (which is zero) into this equation, we find that photons can still possess momentum, which can be calculated as follows:
Chapter 4: Practical Applications of Light Pressure
When a photon collides with a solar sail, it transfers its momentum to the sail, thus generating pressure in accordance with the law of conservation of momentum. But how much light is required to
accelerate a spacecraft effectively?
Each photon carries a minuscule amount of momentum, and the energy of a photon is tied to the light's frequency. For instance, a single photon in a beam of monochromatic light with a frequency of (5
times 10^{14}) hertz imparts a momentum of about (1.11 times 10^{-22}) kgm/s. Fortunately, the Sun emits an immense number of photons, enabling a thrust of 10 newtons on the sail's surface,
necessitating approximately (9 times 10^{22}) photons per second.
This first video, titled "Why light has energy, but no mass? (Understanding E = mc2)," explores the relationship between light and mass in detail, providing further insights into this fascinating
In the second video, "Why Doesn't Light Have Mass?" the concept of massless photons and their implications are discussed, enhancing our understanding of light's properties.
If you're interested in more content about space and physics, please subscribe to our channel and stay tuned for more articles. Your support helps us create better content, so consider becoming a
member for just $5 a month! | {"url":"https://garyprinting.com/understanding-light-pressure-spacecraft-propulsion.html","timestamp":"2024-11-11T11:48:54Z","content_type":"text/html","content_length":"14057","record_id":"<urn:uuid:6b1bed29-e142-44cb-94db-452ab531d905>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00269.warc.gz"} |
The miracle of the appropriateness of the language of mathematics for the formulation of the laws of physics is a wonderful gift which we neither understand nor deserve.
Perhaps I didn’t explain myself clearly. A differential equation describing piano physics, programmed into a computer, sounds like a real piano.
If that’s not beauty and wonderment, I don’t know what is. | {"url":"https://davepeck.org/2011/12/28/the-unreasonable-effectiveness-of-mathematics/","timestamp":"2024-11-13T22:57:00Z","content_type":"text/html","content_length":"8649","record_id":"<urn:uuid:a62de41a-2595-49a5-8ec7-7a3a9643cec6>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00258.warc.gz"} |
Saha Ionization Equation - Examples, Definition, Formula, FAQ\'S
Saha Ionization Equation
The Saha Ionization Equation, a fundamental concept in the laws of atomic and molecular physics, quantifies the degree of ionization of gases under equilibrium conditions based on temperature and
pressure. It integrates physics principles, including thermodynamics and quantum mechanics, to predict how atoms ionize in stellar atmospheres and other plasmas, illuminating crucial aspects of the
laws of physics as they apply to astrophysical phenomena.
What Is Saha Ionization Equation?
The Saha Ionization Equation is a pivotal tool in physics that describes how the ionization state of gases depends on the temperature and pressure. This equation explains the equilibrium between
electrons, ions, and neutral atoms in a plasma, particularly highlighting how conditions like temperature influence ionization rates. Essentially, it enables astronomers to interpret the physical
conditions of stars and nebulae by relating the degree of ionization to the environmental factors.
Saha Ionization Equation Formula
The Saha Ionization Equation formula is given by:
𝑁ᵢ𝑁ₑ/𝑁ᵢ ₋ ₁=(2𝜋𝑚ₑ𝑘𝑇)3^2/ℎ³2𝑍ᵢ/𝑍ᵢ ₋ ₁𝑒⁻ᴱᵢ/ᴷᵀ
• 𝑁ᵢ is the number of atoms in the ionization state 𝑖,
• 𝑁ₑ is the number of free electrons,
• 𝑁ᵢ ₋ ₁ is the number of neutral atoms,
• 𝑚ₑ is the electron mass,
• 𝑘 is the Boltzmann constant,
• 𝑇 is the temperature,
• ℎ is the Planck constant,
• 𝑍ᵢ is the partition function for the ionized state,
• 𝑍ᵢ ₋ ₁ is the partition function for the neutral state,
• 𝐸ᵢ is the ionization energy from state 𝑖−1 to state i,
• 𝑒 is the base of natural logarithms.
This equation models the ionization balance in astrophysical plasmas and helps in determining the physical conditions of stellar atmospheres.
Saha Ionization Equation Derivation
Understanding the Basic Concept
The Saha Ionization Equation is derived by considering how atoms in a gas can exist in different states—either as neutral atoms or as ionized atoms (atoms missing one or more electrons). The
transition between these states involves either the absorption or release of energy, specifically the energy required to remove an electron from an atom.
Applying Thermodynamics
To derive the equation, we start by applying the principles of thermodynamics and statistical mechanics. These principles tell us that the behavior of atoms and electrons in a gas at a given
temperature and pressure can be predicted by understanding the distribution of energy among them.
Considering Equilibrium
In a state of thermal equilibrium, the arrangement of electrons and ions in a gas reaches a balance. This balance isn’t static; electrons are continually being knocked out of atoms (ionizing them)
and re-captured (re-forming neutral atoms), but the rate of ionization and the rate of recombination are equal, so the overall system remains in equilibrium.
Relating to Energy Levels
The likelihood of an atom being ionized or remaining neutral depends on the energy available to it (from the environment, such as thermal energy at a given temperature) and the energy required to
ionize the atom. The higher the temperature, the more thermal energy is available, increasing the likelihood of ionization.
Statistical Considerations
Statistical mechanics provides a framework to calculate the relative populations of neutral and ionized atoms based on their energy states and the temperature of the surrounding environment. It
considers all possible energy states of atoms and the statistical likelihood of each state being occupied.
Equating Conditions
The final step in deriving the equation involves setting up an equality condition where the rate at which atoms are ionized equals the rate at which ions recombine into neutral atoms. This involves
considering the density of neutral atoms, the density of ions, and the density of free electrons, along with how these quantities change with temperature.
Uses of Saha Ionization Equation
• Astrophysics: It helps determine the ionization states of elements in stellar atmospheres, crucial for understanding star temperatures and luminosities.
• Plasma Physics: The equation is used to analyze the ionization equilibrium in laboratory plasmas, important for fusion research and industrial applications.
• Spectroscopy: It assists in interpreting spectral lines from hot gases, aiding in the identification of elements and their states in various environments.
• Material Science: The equation is employed to study ionized gases used in the manufacturing of semiconductors and other materials.
• Cosmology: It plays a role in understanding the ionization history of the early universe, particularly during the reionization epoch.
• Astrochemistry: The equation aids in exploring the chemical composition and reactions within interstellar clouds where stars form, impacting the study of molecular clouds and star formation.
Examples for Saha Ionization Equation
1. Stellar Atmospheres: Astronomers use the Saha Ionization Equation to determine the temperature of stars. By analyzing the ionization levels of different elements in a star’s atmosphere, they can
infer the star’s surface temperature and luminosity.
2. Spectroscopic Analysis: In laboratories, spectroscopists apply the Saha Ionization Equation to interpret the spectral lines of elements heated to high temperatures. This helps in identifying
elements and their ionization states, crucial for materials science and chemical analysis.
3. Plasma Diagnostics: Plasma physicists use the equation to understand the properties of plasma in both natural settings (like the solar corona) and in controlled environments (such as fusion
reactors). The ionization states calculated help in modeling plasma behavior and optimizing reactor conditions.
4. Understanding Early Universe Chemistry: Cosmologists apply the Saha Ionization Equation to model the ionization processes occurring in the early universe, particularly during the recombination
era. This is key to understanding the formation of the cosmic microwave background radiation.
5. Planetary Atmospheres: The equation is used in planetary science to model the ionization of gases in the atmospheres of planets and moons. This helps in predicting atmospheric composition,
weather patterns, and potential for electrical phenomena like auroras.
What is the importance of the Saha equation?
The Saha equation is crucial for determining ionization states in plasmas, essential in astrophysics for analyzing stellar atmospheres and understanding cosmic phenomena.
What is the Saha Boltzmann method?
The Saha-Boltzmann method combines the Saha Ionization Equation and Boltzmann distribution to analyze the thermal and ionization equilibrium in stellar atmospheres, thereby decoding star compositions
and temperatures.
At what temperature is hydrogen ionized?
Hydrogen begins to ionize at about 10,000 Kelvin, a temperature where enough energy is available to start stripping electrons from hydrogen atoms, significantly seen in stars. | {"url":"https://www.examples.com/physics/saha-ionization-equation.html","timestamp":"2024-11-07T05:35:46Z","content_type":"text/html","content_length":"108889","record_id":"<urn:uuid:48c52cfc-dd74-4d23-a376-13cb0d876755>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00842.warc.gz"} |
1060 - Change
Sue is waiting in line at the grocery store. Being in a hurry, she wants to pay with exact change when she
gets to the front of the line. However, she does not know how much her items are going to cost; instead, she
only knows an upper bound C on their total cost. Given a list of the various coins Sue has in her pocket,
your goal is to determine the minimum number of coins she must take out in order to ensure that she can
make exact change for every amount from 1 to C.
The input test file will contain multiple cases. Each test case begins with a single line containing two integers
C (where 1 ≤ C ≤ 1000000000) and m (where 1 ≤ m ≤ 1000), where C is the maximum amount for which
Sue must be able to make change, and m is the number of unique coin denominations Sue has in her pocket.
The next m lines each contain two numbers, vi (where 1 ≤ vi ≤ 1000) and ni (where 1 ≤ ni ≤ 1000), where
vi is the value of the ith coin denomination, and ni is the number of coins of that denomination that Sue
has in her pocket. Input is terminated by a single line containing the number 0; do not process this line.
For each test case, either print a single line containing the number of coins Sue must use in order to make
exact change for all amounts up to C, or print “Not possible” if exact change cannot always be made with
any combination of coins in Sue’s pocket.
sample input
sample output
Not possible
Stanford Local 2007 | {"url":"http://hustoj.org/problem/1060","timestamp":"2024-11-13T16:05:12Z","content_type":"text/html","content_length":"8694","record_id":"<urn:uuid:829e660a-780a-4f03-a52c-879dfee9c5ba>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00508.warc.gz"} |
Incompressible Poiseuille flows of Newtonian liquids with a pressure-dependent viscosity
The pressure-dependence of the viscosity becomes important in flows where high pressures are encountered. Applications include many polymer processing applications, microfluidics, fluid film
lubrication, as well as simulations of geophysical flows. Under the assumption of unidirectional flow, we derive analytical solutions for plane, round, and annular Poiseuille flow of a Newtonian
liquid, the viscosity of which increases linearly with pressure. These flows may serve as prototypes in applications involving tubes with small radius-to-length ratios. It is demonstrated that, the
velocity tends from a parabolic to a triangular profile as the viscosity coefficient is increased. The pressure gradient near the exit is the same as that of the classical fully developed flow. This
increases exponentially upstream and thus the pressure required to drive the flow increases dramatically. (C) 2011 Elsevier B.V. All rights reserved.
• Newtonian flow
• Poiseuille flow
• Pressure-dependent viscosity
• Annular Poiseuille flow
• NAVIER-STOKES EQUATIONS
• POLYMER MELTS
• FLUIDS
• SHEAR
• MANTLE | {"url":"https://research-portal.uea.ac.uk/en/publications/incompressible-poiseuille-flows-of-newtonian-liquids-with-a-press","timestamp":"2024-11-05T09:40:27Z","content_type":"text/html","content_length":"44861","record_id":"<urn:uuid:839bdbb2-9afd-4cdd-b513-541d318b1ca3>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00266.warc.gz"} |
kindergarten math question Archives : JoJo Teacher - Korean mathematics
[태그:] kindergarten math question
Kindergarten math question. Can you solve this math question? Hello, I’m jojo, a math teacher in Korea. I have brought a math question that I solved with a kindergarten student today. This is a
math question I used during a lesson with a 6-year-old child. What is the price of the cola? Can you solve… | {"url":"https://www.jojoteacher.com/tag/kindergarten-math-question/","timestamp":"2024-11-15T03:41:26Z","content_type":"text/html","content_length":"50204","record_id":"<urn:uuid:eda61d4c-9444-4409-b936-6d4ac2ce3290>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00615.warc.gz"} |
algebra over an operad
Categorical algebra
An operad is a structure whose elements are formal operations, closed under the operation of plugging some formal operations into others. An algebra over an operad is a structure in which the formal
operations are interpreted as actual operations on an object, via a suitable action.
Accordingly, there is a notion of module over an algebra over an operad.
Let $M$ be a closed symmetric monoidal category with monoidal unit $I$, and let $X$ be any object. There is a canonical or tautological operad $Op(X)$ whose $n^{th}$ component is the internal hom $M
(X^{\otimes n}, X)$; the operad identity is the map
$1_X: I \to M(X, X)$
and the operad multiplication is given by the composite
$\array{ M(X^{\otimes k}, X) \otimes M(X^{\otimes n_1}, X) \otimes \ldots \otimes M(X^{\otimes n_k}, X) & \stackrel{1 \otimes func_\otimes}{\to} & M(X^{\otimes k}, X) \otimes M(X^{\otimes n_1 + \
ldots + n_k}, X^{\otimes k}) \\ & \stackrel{comp}{\to} & M(X^{\otimes n_1 + \ldots + n_k}, X) }$
Let $O$ be any operad in $M$. An algebra over $O$ is an object $X$ equipped with an operad map $\xi: O \to Op(X)$. Alternatively, the data of an $O$-algebra is given by a sequence of maps
$O(k) \otimes X^{\otimes k} \to X$
which specifies an action of $O$ via finitary operations on $X$, with compatibility conditions between the operad multiplication and the structure of plugging in $k$ finitary operations on $X$ into a
$k$-ary operation (and compatibility with actions by permutations).
An algebra over an operad can equivalently be defined as a category over an operad which has a single object.
If $M$ is cocomplete, then an operad in $M$ may be defined as a monoid in the symmetric monoidal category $(M^{\mathbb{P}^{op}}, \circ)$ of permutation representations in $M$, aka species in $M$,
with respect to the substitution product $\circ$. There is an actegory structure $M^{\mathbb{P}^{op}} \times M \to M$ which arises by restriction of the monoidal product $\circ$ if we consider $M$ as
fully embedded in $M^{\mathbb{P}^{op}}$:
$i: M \to M^{\mathbb{P}^{op}}: X \mapsto (n \mapsto \delta_{n 0} \cdot X)$
(interpret $X$ as concentrated in the 0-ary or “constants” component), so that an operad $O$ induces a monad $\hat{O}$ on $M$ via the actegory structure. As a functor, the monad may be defined by a
coend formula
$\hat{O}(X) = \int^{k \in \mathbb{P}} O(k) \otimes X^{\otimes k}$
An $O$-algebra is the same thing as an algebra over the monad $\hat{O}$.
Remark If $C$ is the symmetric monoidal enriching category, $O$ the $C$-enriched operad in question, and $A \in Obj(C)$ is the single hom-object of the O-category with single object, it makes sense
to write $\mathbf{B}A$ for that $O$-category. Compare the discussion at monoid and group, which are special cases of this.
Over single-coloured operads
Over coloured operads
• There is a coloured operad $Mod_P$ whose algebras are pairs consisting of a $P$-algebra $A$ and a module over $A$;
• For a single-coloured operad $P$ there is a coloured operad $P^1$ whose algebras are triples consisting of two $P$ algebras and a morphism $A_1 \to A_2$ between them.
• Let $C$ be a set. There is a $C$-coloured operad whose algebras are $V$-enriched categories with $C$ as their set of objects.
• S. N. Tronin, Algebras over multicategories, Russ Math. (2016) 60: 52. doi; Rus. original: С. Н. Тронин, Об алгебрах над мультикатегориями, Изв. вузов. Матем., 2016, № 2, 62–74 | {"url":"https://ncatlab.org/nlab/show/algebra+over+an+operad","timestamp":"2024-11-05T13:55:37Z","content_type":"application/xhtml+xml","content_length":"39754","record_id":"<urn:uuid:f91cefd6-9f66-4305-b326-1e05b93ccf22>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00317.warc.gz"} |
John von Neumann: Early life, Discoveries, and Accomplishments - Malevus
John von Neumann was born in Budapest. He was the eldest of the three sons of a wealthy and cultured Jewish banking family. He took lessons from a private teacher until he was 10 years old, and then
he started studying at the Lutheran High School in the capital of Hungary. His remarkable ability was evident from an early age; he had an almost photographic memory and the ability to quickly
perform arithmetic calculations in his mind.
Who Was John von Neumann?
At the age of 18, he enrolled in the mathematics department of the University of Budapest but spent most of his time in Berlin getting to know the European scientific elite. He then started his
doctorate at the University of Budapest, but also studied chemical engineering at the Eidgenössische Technische Hochschule (ETH) in Zurich, due to the insistence of his father, who wanted his son to
have a professional education. In 1925, ETH gave him a bachelor’s degree in chemical engineering. In 1926, the University of Budapest gave him a Ph.D. in mathematics.
John von Neumann received the Rockefeller Scholarship from the University of Göttingen in Germany in 1926. The following year, he was appointed as a Privatdozent (faculty member) at the University of
Berlin, making him the university’s youngest Privatdozent in its history. He conducted extensive research in the 1920s on mathematical logic, set theory, operator theory, and quantum mechanics. In
the 1930s, he became a guest professor at Princeton University, dividing his time between Berlin and Princeton for several years. However, he wanted a permanent position in the United States due to
the deteriorating political situation in Europe.
This opportunity came from the newly established Institute for Advanced Study in Princeton in 1933, which appointed me as a founding professor (the other professor was Albert Einstein). He became an
American citizen in 1937. Von Neumann achieved fundamental results in institutional and applied mathematics at the Institute and also developed his game theory. Together with Oskar Morganstern, he
wrote “The Theory of Games and Economic Behavior” in 1944. This book was a major step forward in the field of mathematical economics.
Wartime calculations and the first electronic computer
John von Neumann and J. Robert Oppenheimer, former director of the Manhattan Project, in front of the IAS machine in 1952.
John von Neumann had a pleasant personality, great social skills, and brilliant political intelligence. When the United States entered World War II after the Pearl Harbor attack on December 7, 1941,
there was a huge increase in demand for consulting services. John von Neumann served in this field thanks to his adaptability, legendary mental abilities, and talent to solve complex math problems
easily. In 1943, he turned his attention to war-related work, especially numerical computational problems.
Most importantly, he was a consultant to the Manhattan Project in Las Alamos. There, he consulted on implosion techniques to detonate the nuclear material in the center of the atomic bomb. Complex
math system equations had to be solved numerically as part of this process, so he looked for the most advanced calculators he could find.
John von Neumann was also a consultant to the US Army’s Ballistic Research Laboratory at the Aberdeen Proving Ground in Maryland. One of the lab’s main tasks was the production of ballistic charts
and the founding of the first electronic computer, ENIAC (Electronic Numerical Integrator and Computer), which was developed by the Moore School of Electrical Engineering at the University of
Pennsylvania. Due to technical and design limitations, Neumann’s calculations for the atomic bomb couldn’t be done on the ENIAC. Neumann and the group at Moore worked together to design EDVAC
(Electronic Discrete Variable Automatic Computer), which replaced ENIAC.
In June 1945, he summarized the group’s findings in his report, First Draft of a Report on the EDVAC. The report provided the logical definition of what is known as a “stored-program computer” that
all subsequent computer developments would be based on. The computer was called by this name because both the program and the numbers were using the same electronic memory. This made the computer
much more powerful and flexible because now a program could run its instructions without having to use cumbersome ways to program like plugboards, punched cards, or paper.
John von Neumann and the hydrogen bomb
In 1946, von Neumann returned to the Institute for Advanced Study, leading the construction of one of the first practical computers. With the advent of general-purpose computers, he began to be
concerned less with numerical weather forecasting and more with philosophy, cybernetics, and automata. At the same time, he kept giving advice in Los Alamos about how to make the hydrogen bomb.
In 1954, President Eisenhower took over the Atomic Energy Commission. Here, he made an impact on science and military policy with an inclination toward war. Neumann was diagnosed with bone cancer in
1955, and he eventually died of this disease.
His last major competence was the preparation of the Silliman Conferences for Yale University, which were published in 1958 after his death under the title The Computer and the Brain. He died in 1957
at the age of 53.
• Ayoub, Raymond George (2004). Musings Of The Masters: An Anthology Of Mathematical Reflections. Washington, D.C.: MAA. ISBN 978-0-88385-549-2. OCLC 56537093.
• Blair, Clay, Jr. (February 25, 1957). “Passing of a Great Mind”. Life. pp. 89–104.
• Blume, Lawrence E. (2008). “Convexity”. In Durlauf, Steven N.; Blume, Lawrence E. (eds.). The New Palgrave Dictionary of Economics (Second ed.). New York: Palgrave Macmillan. pp. 225–226. doi:
10.1057/9780230226203.0315. ISBN 978-0-333-78676-5. | {"url":"https://malevus.com/john-von-neumann/","timestamp":"2024-11-14T20:32:00Z","content_type":"text/html","content_length":"95169","record_id":"<urn:uuid:02f6c72f-28d8-42e7-9af7-e3cb64a30c55>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00038.warc.gz"} |
String theory and classical absorption by three-branes
Low-energy absorption cross sections for various particles falling into extreme non-dilatonic branes are calculated using string theory and world-volume field theory methods. The results are compared
with classical absorption by the corresponding gravitational backgrounds. For the self-dual three-brane, earlier work by one of us demonstrated precise agreement of the absorption cross sections for
the dilaton, and here we extend the result to Ramond-Ramond scalars and to gravitons polarized parallel to the brane. In string theory, the only absorption channel available to dilatons and
Ramond-Ramond scalars at leading order is conversion into a pair of gauge bosons on the three-brane. For gravitons polarized parallel to the brane, scalars, fermions and gauge bosons all make
leading-order contributions to the cross section, which remarkably add up to the value predicted by classical gravity. For the two-brane and five-brane of M-theory, numerical coefficients fail to
agree, signaling our lack of a precise understanding of the world-volume theory for large numbers of coincident branes. In many cases, we note a remarkable isotropy in the final state particle flux
within the brane. We also consider the generalization to higher partial waves of minimally coupled scalars. We demonstrate agreement for the three-brane at ℓ = 1 and indicate that further work is
necessary to understand ℓ > 1.
All Science Journal Classification (ASJC) codes
• Nuclear and High Energy Physics
• Coincident membranes
• Extremal black holes
Dive into the research topics of 'String theory and classical absorption by three-branes'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/string-theory-and-classical-absorption-by-three-branes","timestamp":"2024-11-04T19:03:45Z","content_type":"text/html","content_length":"53247","record_id":"<urn:uuid:e398e090-6c0a-4d79-a9e0-0242a0938074>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00639.warc.gz"} |
Seminars and Endowed Lectures | Department of Applied Mathematics and Statistics
The Department hosts weekly colloquia at 1:30 p.m. every Thursday.
Endowed Lectures
Three endowed lecture series bring leading researchers and scientists from academia, government, and more, to campus every year.
Past Seminars
2023 – Mateo Diaz, Johns Hopkins University – Clustering a mixture of Gaussians with unknown covariance
2023 – Luana Ruiz, Johns Hopkins University – Manifold neural networks for large-scale geometric information processing
2023 – Ewan Davies, Colorado State University – Counting independent sets in dense bipartite graphs
2023 – Nicolas Trillos, University of Wisconsin Madison – Adversarial machine learning and clustered federated learning: a collective dynamics perspective
2023 – Jincheng Yang, University of Chicago – Recent developments in the Navier-Stokes equation
2023 – Javier Pena, Carnegie Mellon University – On the affine invariance of the conditional gradient algorithm
2023 – Ilias Zadik, Yale University – Revisiting the Metropolis Process for the Planted Clique Model Revisiting Process
2023 – Chunmei Wang, University of Florida – Finite expression methods for discovering physical laws from data
2023 – Richard M. Low, San Jose State University – ZP-magic graph labelings and the combinatorial nullstellensatz
2023 – Lorin Crawford, Microsoft Research New England – Probabilistic methods to identify multi-scale enrichment in genomic sequencing studies
2023 – Amin Karbasi, Yale University – When we talk about reproducibility, what are we talking about?
2024 – Aïda Ouangraoua, Université de Bordeaux – Transcript Homology Relationships and Phylogeny Reconstruction
2024 – Wei Zhu, University of Massachusetts Amherst – Symmetry-Preserving Machine Learning: Theory and Applications
2024 – Eric Vanden-Eijnden, New York University – Scientific Computing in the Age of Generative AI
2024 – Jeff Calder, University of Minnesota – PDEs and graph-based semi-supervised learning
2024 – Govind Menon, Brown University – A Bayesian View of Geometry
2024 – Christopher Musco, NYU Tandon School of Engineering – The Lanczos method, matrix functions, and the quest for optimality
2024 – Zhengling Qi, The George Washington University – Offline Data-Driven Decision-Making: Some Challenges and Solutions
2024 – Cristopher Moore, Sante Fe Institute – The physics of inference, phase transitions, and community detection
2024 – Jared Tanner, University of Oxford Mathematics Institute – Deep neural network stability at initialization: Nonlinear activations impact on the Gaussian process
2024 – Agostino Capponi, Columbia University – Virtual Trading in Multi-Settlement Electricity Markets
2024 – Marc A. Suchard, University of California, Los Angeles – Stochastic Compartmental Models for Infectious Disease Dynamics without Tears
2024 – Jeremias Sulam, Johns Hopkins University – Yay, my deep network works! But.. what did it learn?
2024 – Krishnakumar Balasubramanian, University of California, Davis – Geometry-aware algorithms for statistical data science
2024 – Francesco Sanna Passino, Imperial College London – Low-rank models for dynamic multiplex graphs and vector autoregressive processes
Past Seminars
2022 – Kevin Leder, University of Minnesota – Computational techniques for understanding heterogeneous tumors
2022 – Haihao Lu, University of Chicago – First Order Methods for Linear Programming: Theory, Computation, and Applications
2022 – Evelyn Sander, George Mason University – Rigorous bifurcation methods for diblock and triblock copolymer models
2022 – Edinah Gnang, Johns Hopkins University – The composition lemma and its application in graph theory
2022 – Anthony Yezzi, Georgia Institute of Technology – Accelerated Gradient Descent in the PDE Framework
2022 – Victor Bailey, Georgia Institute of Technology – Frames via Unilateral Iterations of Bounded Operators
2022 – Maria Cameron, University of Maryland – Quantifying rare events with the aid of diffusion maps
2022 – Miranda Holmes-Cerfon, New York University – Numerically simulating particles with short-ranged interactions
2022 – Fei Liu, Johns Hopkins University – Statistical learning and inverse problems for interacting particle systems
2022 – Philip Thompson, Purdue University – A spectral least-squares-type method for heavy-tailed corrupted regression with unknown covariance & heterogeneous noise
2023 – Santanu Dey, Georgia Technology – Theoretical and computational analysis of sizes of branch-and-bound trees
2023 – James Schmidt, Applied Physics Laboratory Johns Hopkins University – Relations of Dynamical Systems through their Maps
2023 – Natalia Trayanova, Johns Hopkins University – AI-Powered Personalized Computational Cardiology
2023 – Carina Curto, Pennsylvania State University – An introduction to threshold-linear networks for neuroscience
2023 – Vince Lyzinski, University of Maryland – Lost in the Shuffle: Testing Power in the Presence of Errorful Network Vertex Labels
2023 – Ben Grimmer, Johns Hopkins University – Scalable, Projection-Free Optimization Methods
2023 – Joshua Agterberg, Johns Hopkins University – Estimating Higher-Order Mixed Memberships via the Two to Infinity Tensor Perturbation Bound
2023 – Yannis Kevrekidis, Johns Hopkins University – Data and the modeling of complex dynamical systems
2023 – Luhao Zhang, University of Texas at Austin – Model uncertainty, robust control, and costly information acquisition
2023 – Qiuqi Wang, University of Waterloo – E-backtesting risk measures
2023 – Haoyang Cao, Centre de Mathématiques Appliquées (CMAP), École Polytechnique –Bridging GANs and Stochastic Analysis
2023 – Yu Wang, University of Florida – Statistical verification algorithms for logical specifications on autonomous systems
2023 – Dennice Gayme, Johns Hopkins University – Toward wind farm control for power tracking
Past Seminars
2021 – James Fill, Johns Hopkins University – Breaking Multivariate Records
2021 – Aki Nishimura, Johns Hopkins University – Bayesian sparse regression for large-scale observational health data
2021 – Brittany Hamfeldt, New Jersey Institute of Technology – Numerical Optimal Transport on the Sphere
2021 – Nadia Drenska, Johns Hopkins University – A PDE Interpretation of Prediction with Expert Advice
2021 – Alberto Bietti, New York University – On the Sample Complexity of Learning under Invariance and Geometric Stability
2021 – Dimitrii Ostrovskii, University of Southern California – Nonconvex-Nonconcave Min-Max Optimization with a Small Maximization Domain
2021 – Tom Fletcher, University of Virginia – The Riemannian Geometry of Deep Neural Networks
2021 – Eric Tchetgen, University of Pennsylvania – Proximal Causal Inference
2021 – Stephanie Hicks, Johns Hopkins University – Scalable statistical methods and software for single-cell data science
2021 – Dustin Mixon, Ohio State – Neural collapse with unconstrained features
2021 – Bruno Jedynak, Portland State University – A Convergent RKHS Algorithm for Estimating Nonparametric ODEs
2021 – Clare Lau, Johns Hopkins University Applied Physics Laboratory – Wasserstein Gradient Flows for Potentials in Frame Theory
2022 – Alex Wein, Georgia Institute of Technology – Understanding Statistical-vs-Computational Tradeoffs via Low-Degree Polynomials
2022 – Mike Dinitz, Johns Hopkins University – Faster Matchings via Learned Duals
2022 – Maxim Bichuch, Johns Hopkins University – Deep PDE Solution of BSDE
2022 – Yufei Zhao, Massachusetts Institute of Technology – Equiangular lines and eigenvalue multiplicities
2022 – Ian Tobasco, University of Illinois Chicago – The many, elaborate wrinkle patterns of confined elastic shells
2022 – Jasmine Foo, University of Minnesota Twin Cities – Spatial evolution and phenotypic switching in cancer
2022 – Mike Dinitz, Johns Hopkins University – Faster Matchings via Learned Duals
2022 – Christian Kuemmerle, Johns Hopkins University – Iteratively Reweighted Least Squares: New Formulations and Guarantees
2022 – Monika Nitsche, University of New Mexico – Evaluating near-singular integrals with application to vortex sheet and multi-nested Stokes flow
2022 – Jose Perea, Northeastern University – DREiMac: Dimensionality Reduction with Eilenberg-MacLane Coordinates
2022 – Michael Perlmutter, University of California, Los Angeles – Deep Learning on Graphs and Manifolds
2022 – Baba Vemuri, University of Florida – Nested Homogeneous Spaces: Construction, Learning and Applications
2022 – Julien Guyon, Bloomberg L.P., New York – Dispersion-Constrained Martingale Schrödinger Problems and the Joint S&P 500/VIX Smile Calibration Puzzle
2022 – Tyrus Berry, George Mason University – Beyond Regression: Operators and Extrapolation in Machine Learning
2022 – Maxim Bichuch, Johns Hopkins University – Introduction to Decentralized Finance
2022 – Cencheng Shen, University of Delaware – Graph Encoder Embedding
The Alan Goldman Lecture Series in Operations Research was established in 1999 to honor the highly respected professor when he was named Professor Emeritus at JHU.
About Alan J. Goldman
Alan J. Goldman was an expert in operations research – the use of mathematics to improve decisions on the design and operation of complex systems – whose favorite application areas included facility
siting, transportation systems, and mathematical game theory. Goldman received his BA from Brooklyn College in mathematics and physics in 1952. He earned his MA and PhD in mathematics from Princeton
University in 1954 and 1956, respectively. His dissertation area was topology, and the title of his dissertation is A Cech Theory of Fundamental Groups and Covering Spaces. From 1956 to 1961 he was
an evening lecturer at American University and Catholic University of America, but his principal pre-JHU affiliation was with the National Bureau of Standards (now the National Institute of Standards
and Technology), where he was founder and chief of operations research and also deputy chief of applied mathematics. Goldman joined Hopkins in 1979; earned the status of Professor Emeritus in 1999
and continued to teach until his death in 2010.
Past Lectures
2024 – David Blei, Columbia University – Scaling and Generalizing Approximate Bayesian Inference
2022 – Anna Gilbert, Yale University – Metric Representations: Algorithms and Geometry
2021 -Maria Chudnovsky, Princeton University – Induced Subgraphs and Tree Decompositions
2020 – Rekha Thomas, University of Washington, Seattle – “Lifting for Simplicity: Concise Descriptions of Convex Sets” Goldman Lecture 11-19-2020(pdf)
2019 – Jack Edmonds, “Matroids and Optimum Branching Systems” edmonds goldman – slides
2019 – Gerard Cornuejols, Carnegie Mellon University – “Min-Max Relations for Packing and Covering”
2018 – Bill Cook, Johns Hopkins University
2016 – Jorge Nocedal, Northwestern University – “Stochastic Newton Methods for Machine Learning” goldman-lecture-slides
2015 – Daniel Bienstock, Columbia University – “Recent Results on Polynomial Optimization Problems”
2014 – Stephen Wright, University of Wisconsin-Madison – “The Revival of Coordinate Descent Methods” SLIDES – Video of Lecture
2013 – Michael Todd, Cornell University – “Exponential Gaps in Optimization Algorithms”
2013 – David Shmoys, Cornell University – “Improving Christofides’ Algorithm for the s-t Path Traveling Salesman Problem”
2011 – Dimitris Bertsimas, Massachusetts Institute of Technology – “A Computationally Tractable Theory of Performance Analysis in Stochastic Systems”
2010 – Arthur Benjamin, Harvey Mudd College – “Combinatorial Trigonometry”
2009 – Richard Francis, University of Florida – “Aggregation Error for Location Models: Survey and Analysis”
2009 – Lisa Fleischer, Dartmouth University – “Submodular Approximation: Sampling-based Algorithms and lower Bounds”
2007 – Eva Tardos, Cornell University – “Games in Networks”
2006 – Christine Shoemaker, Cornell University – “Optimization, Calibration, and Uncertainty Analysis of Multimodal, Computationally Expensive Models with Environmental Applications”
2005 – George Nemhauser, Georgia Institute of Technology – “Scheduling an Air Taxi Service”
2004 – Karla Hoffman, George Mason University
2000 – Tom Magnanti, MIT
1999 – Alan J. Goldman, Johns Hopkins University – “Reflections and Translations”
The Acheson J. Duncan Lecture Series
In 1986, an anonymous donor established the Acheson J. Duncan Distinguished Visitor Fund to honor the internationally recognized leader in quality control and industrial statistics. The endowment
supports an annual visit and lecture by a distinguished mathematical scholar.
About Acheson J. Duncan
Acheson J. Duncan spent 25 years as a faculty member at Johns Hopkins. His extensive writings in the field include the text, Quality Control and Industrial Statistics, published in 1952 and now in
its fifth edition with several international translations. The late dean of the Whiting School of Engineering, Robert H. Roy noted that Duncan and his work were revered in Japan, where Duncan
frequently lectured in the years following World War II. A native of New Jersey, Duncan received his PhD in economics from Princeton in 1936, and was a faculty member there for 13 years before coming
to Hopkins. Duncan died in 1995 at age 90.
Past Lectures
2023 – Leslie Greengard, New York University – Adaptive methods for the simulation of diffusion in complex geometries
2023 – Karen Willcox, The University of Texas at Austin – Learning physics-based models from data: Perspectives from projection-based model reduction
2020 – Kavita Ramanan, Brown University- “Beyond Mean-field Limits for Large-scale Stochastic Systems” Duncan Flyer Lecture 12-3-2020
2019 – Susan Murphy, Harvard University- “Online Experimentation and Learning Algorithms in a Clinical Trial”
2019 – Rina Foygel Barber, University of Chicago- “Robust inference with the knockoff filter.”
2018 – Stuart Geman, Brown University- “Real and Artificial Neural Networks”
2017 – René Carmona, Princeton University- “Mean Field Games with Major and Minor Players: Theory and Numerics”
2016 – Bin Yu, University of California – “Movie Reconstruction from Brain Signals : Mind-Reading”
2014 – Laurent Saloff-Coste, Cornell University – “Groups and Random Walks” and “Random Walk Invariants of Groups”
2013 – David Siegmund, Stanford University – “The Intersection of Operations Research, Kinetic Theory, and Genetics” and “Detection of Local Signals in Genomics”
2012 – Gerard Ben Arous, New York University – “Counting Critical Points of Random Functions of Many Variables” and “RMT^2: Random Morse Theory Meets Random Matrix Theory”
2011 – Joel Zinn, Texas A&M University – “A Meandering ‘Trip’ through High Dimensions” and “Limit Theorems in High Dimensions”
2010 – Andreas Buja, University of Pennsylvania – “Seeing is Believing: Statistical Visualization for Teaching and Data Analysis” and “Statistical Inference for Exploratory Data Analysis and Model
2009 – Jonathan Taylor, Stanford University – “Deformation Based Morphometry, Random Fields and Multivariate Linear Models” and “Integral Geometry of Random Level Sets”
2008 – Yali Amit, University of Chicago – “Statistical Models in Computer Vision” and “Estimation of Deformable Object Models”
2007 – Robert Azencourt, University of Houston – “Automatic Learning and Multi-Sensors Diagnosis” and “Ultrasound Image Analysis: Speckle Tracking for Recovery of Cardiac Motion”
2006 – Lawrence A. Shepp, Rutgers University – “Applications of Convexity” and “Problems in Convexity”
2005 – Gregory F. Lawler, Cornell University – “Random Walks: Simple and Self-Avoiding” and “Conformal Invariance, Brownian Loops, and Measures on Random Paths”
2004 – Leo Breiman, University of California, Berkeley – “Random Forests: A Statistical Tool for the Sciences” and “Statistics, Machine Learning, and Data Mining”
2003 – Oded Schramm, Microsoft Research – “Emergence of Symmetry: Conformal Invariance of Scaling Limits of Random Systems” and “Random Triangulations”
2002 – Steven E. Shreve, Carnegie-Mellon University – “Probability Models for Derivative Securities” and “A Unified Model for Credit Derivatives”
2001 – David Donoho, Stanford University – “Interactions Between Data Analysis of Natural Images, Biological Vision, and Mathematical Analysis” and “Beyond Wavelets: Ridgelets, Curvelets, Beamlets”
2000 – Roger J-B Wets, University of California, Davis – “Limit Theorems for Random Lower Semicontinuous Functions with Applications to Statistics, Stochastic Optimization, Probability, and
Stochastic Homogenization” and “Stability Issues for Equilibrium Points”
1999 – Ken Alexander, University of Southern California – “Power-Law Corrections to Exponential Decay of Correlations and Connectivities in Lattice Models” and “Droplets and Bubbles: The Mathematical
Description of Phase Separation”
1998 – David Pollard, Yale University – “Some Statistical Issues in the Construction of Jury Arrays” and “What is Randomization?”
1997 – Rick Durrett, Cornell University
1996 – Michael Saks, Rutgers University – “Randomness as a Scarce Resource” and “Extractors, Dispersers, and Pseudorandom Generators”
1995 – Michael J. Todd, Cornell University
1994 – David J. Aldous, University of California, Berkeley
1993 – Rudolph Beran, University of California, Davis
1992 – Peter Ney, University of Wisconsin
1991 – Paul D. Seymour, University of Waterloo
1990 – Persi Diaconis, Harvard University
1989 – Ralph L. Disney, Texas A&M University
The John C. and Susan S. G. Wierman Lecture Series in Air Quality Data Analysis features talks on developments in air quality data analysis that are relevant for policy development. It seeks to bring
together faculty and researchers in engineering and natural sciences with state and local air quality officials, to enhance understanding and stimulate collaboration on important air quality issues.
The lectures are intended to showcase new developments, to encourage the quantitative analysis of scientific issues related to air quality, and to elucidate the policy implications of recent
research. The lecture series was established with a permanent endowment by Prof. John C. Wierman and Susan S. G. Wierman.
About the Sponsors
John C. Wierman, a professor of Applied Mathematics and Statistics at Johns Hopkins University since 1981, served as department chair from 1988 to 2000. The founder of the W. P. Carey Program in
Entrepreneurship & Management, he was director of the program and its successor, the Center for Leadership Education, from 1996 until 2009. His mathematical research is published in probability,
discrete mathematics, and statistics journals, with applied articles in physics, computer science, molecular biology, education, and business journals. He received his BS. and PhD from the University
of Washington and is a Fellow of the Institute of Mathematical Statistics and the Institute of Combinatorics and its Applications.
Susan S.G. Wierman was executive director of Mid-Atlantic Regional Air Management Association from 1996 to 2017, where she worked to improve regional air quality. She earned urban planning degrees
from the University of Washington, and a certificate in Continuing Engineering Studies from Johns Hopkins University. She is a Fellow of the international Air and Waste Management Association, and
was the 2012 recipient of its S. Smith Griswold Outstanding Air Pollution Official award.
Past Lectures
2023 – Cory Zigler, University of Texas at Austin – “Causal Inference in Air Quality Regulation: An Overview and Two Topics in Statistics and Machine Learning”
2022 – Susan Anenberg, George Washington University – “Climate change, air pollution, and public health impacts: From science to policy”
2021 – Roger Peng, Johns Hopkins Bloomberg School of Public Health – “Statistical Approaches to Studying Air Pollution Mixtures and Health”
2020 – Dr. Doug Dockery, Harvard University – “Air Pollution Accountability Studies: Lessons Learned and Future Opportunities”
2019 – Dr. Lianne Sheppard, University of Washington – “Modeling Particulate Air Pollution for Inference About Neurodegenerative Effects”
2017 – Brian Duncan, NASA Goddard Space Flight Center – “The Growing Importance of Satellite Data for Air Quality Applications”
2017 – Amy Herring, University of North Carolina Chapel Hill – “Spatial-temporal Modeling of the Association between Air Pollution Exposures and Birth Outcomes: Identifying Critical Exposure Windows”
2015 – Francesca Dominici, Harvard University – “Comparative Effectiveness Research of Environmental Exposures: Connecting the Dots with Big Data”
2014 – Michelle Bell, Yale University – “Exposure to Air Pollution during Pregnancy and Risk of Adverse Birth Outcomes”
2013 – Montse Fuentes, North Carolina State University – “Calibration of deterministic numerical models using nonparametric spatial density functions”
2012 – Richard L. Smith, University of North Carolina, Chapel Hill/SAMSI – “Attribution of Extreme Climatic Events”
2011 – C. Arden Pope III, Brigham Young University – “Human Health Effects of Air Pollution” Statistics and Public Policy”
2009 – Katherine Bennett Ensor, Rice University – “Houston Air Quality: A Simultaneous Examination of Multiple Pollutants”
2008 – William Christensen, Brigham Young University – “Identifying Pollution Source Locations for Air Quality Monitoring”
2008 – Barry D. Nussbaum, US Environmental Protection Agency – “Greenhouse, White House, and Environmental Statistics: The Use of Statistics in Environmental Decision Making”
2006 – William F. Hunt, Jr., North Carolina State University – “Environmental Statistics: A New Source of Discovery for Tomorrow’s Problem-Solvers”
2004 – Philip K. Hopke, Clarkson University – “Advanced Factor Analysis Methods for Receptor Modeling”
Past Lectures
2023 – Gitta Kutyniok, Ludwig-Maximilians Universität München – Reliable AI: Successes, Challenges, and Limitations
2023 – Deanna Needell, The University of California, Los Angeles – Towards Transparency, Fairness, and Efficiency in Machine Learning
2023 – Edo Airoldi, Temple University – Designing experiments on social, healthcare, and information networks | {"url":"https://engineering.jhu.edu/ams/seminars-and-endowed-lectures/","timestamp":"2024-11-12T19:26:53Z","content_type":"text/html","content_length":"156052","record_id":"<urn:uuid:be34a648-7644-48d6-8ce4-9740902c1591>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00173.warc.gz"} |
JuNoLo - Jülich nonlocal code for parallel post-processing evaluation of vdW-DF correlation energy
Lazić, Predrag; Atodiresei, Nicolae; Alaei, Mojtaba; Caciuc, Vasile; Blügel, Stefan; Brako, Radovan (2010) JuNoLo - Jülich nonlocal code for parallel post-processing evaluation of vdW-DF correlation
energy. Computer Physics Communications, 181 (2). pp. 371-379. ISSN 0010-4655
PDF - Accepted Version
Download (288kB)
Nowadays the state of the art Density Functional Theory (DFT) codes are based on local (LDA) or semilocal (GGA) energy functionals. Recently the theory of a truly nonlocal energy functional has been
developed. It has been used mostly as a post-DFT calculation approach. i.e. by applying the functional to the charge density calculated using any standard DFT code, thus obtaining a new improved
value for the total energy of the system. Nonlocal calculation is computationally quite expensive and scales as N-2 where N is the number of points in which the density is defined, and a massively
parallel calculation is welcome for a wider applicability of the new approach. In this article we present a code which accomplishes this goal.
Item Type: Article
Uncontrolled Keywords: Electronic structure; Density functional theory; Van der Waals interaction; nonlocal correlation
Subjects: NATURAL SCIENCES > Physics
Divisions: Theoretical Physics Division
│ Project title │ Project leader │ Project code │ Project type │
Projects: ├────────────────────────────────────────────────────────────────────┼──────────────────────┼──────────────────┼──────────────┤
│ Površine i nanostrukture: Teorijski pristupi i numerički proračuni │ [5180] Radovan Brako │ 098-0352828-2863 │ MZOS │
Depositing User: Predrag Lazić
Date Deposited: 15 Jan 2014 16:02
URI: http://fulir.irb.hr/id/eprint/1282
DOI: 10.1016/j.cpc.2009.09.016
Actions (login required) | {"url":"http://fulir.irb.hr/1282/","timestamp":"2024-11-13T09:41:39Z","content_type":"application/xhtml+xml","content_length":"27871","record_id":"<urn:uuid:e1741153-3dd5-4f4a-a92d-d1cd739d13c5>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00593.warc.gz"} |
Conclusion - Tracking, Correcting and Absorbing Water Surface Waves
We have presented a novel approach that takes a sequence of arbitrary closed surfaces and produces as output a temporally coherent sequence of meshes aug-mented with vertex correspondences. The
output of our algorithm is useful for a variety of applications such as (dynamic) displacement maps, texture propagation, template-free tracking and morphs. We have also demonstrated the robustness
of the method to parameters as well as input. In the future we would like to ex-tend the method to handle non-closed surfaces, as well as explore problem-specific applications of our general-purpose
Liquid Surface Tracking with Error Compensation
Our work concerns the combination of an Eulerian liquid simulation with a high-resolution surface tracker (e.g. the level set method or a Lagrangian triangle mesh).
The naive application of a high-resolution surface tracker to a low-resolution ve-locity field can produce many visually disturbing physical and topological artifacts that limit their use in
practice. We address these problems by defining an error function which compares the current state of the surface tracker to the set of physi-cally valid surface states. By reducing this error with a
gradient descent technique, we introduce a novel physics-based surface fairing method. Similarly, by treating this error function as a potential energy, we derive a new surface correction force that
mimics the vortex sheet equations. We demonstrate our results with both level set and mesh-based surface trackers.
4.1 Detailed surface tracking
This paper addresses the problem of tracking a liquid surface in an Eulerian fluid simulation. Within the field of computer graphics, Eulerian fluid simulation has become commonplace, with standard
methods relying on a rectilinear grid or tetrahedral mesh for solving the Navier-Stokes equations [Bridson 2008]. The problem becomes significantly more complicated when we wish to simulate a free
surface, such as when animating liquid. Correct treatment of this free surface requires special boundary conditions as well as some additional computational machinery called asurface tracker, such as
the level set method [Osher and Fedkiw 2003] or a moving triangle mesh [Wojtan et al. 2011].
When animating a free surface, almost all of the visual detail is directly depen-dent on this surface tracker, because the surface is often the only visible part of the resulting fluid simulation. In
order to make a simulation as detailed and visually
(a) Original simulation (b) Smoothed
(c) Our smoothing (d) Our dynamics
Figure 4.1: Our method permits high-resolution tracking of a low-resolution fluid simulation, without any visual or topological artifacts. The original simulation (a) exhibits sharp details and
low-resolution banding artifacts. Smoothing the surface tracker (b) hides the artifacts but corrodes important surface features. We propose a smoothing technique (c) that preserves sharp details
while selectively removing surface tracking artifacts, and a force generation method (d) that removes visual artifacts with strategically placed surface waves. Our algorithms are general and apply to
both level sets as well as mesh-based surface tracking techniques.
Figure 4.2: If the surface tracker (orange) is much more detailed than the simula-tion grid (black squares), then the simulasimula-tion can only work with a rough approxi-mation of the surface
(blue). The mismatch between the orange and blue surfaces can create visual artifacts like permanent surface kinks and floating droplets.
rich as possible, we must add detail to the surface tracker. The computational cost of solving the Navier-Stokes equations scales with the volume of the simulation.
Therefore, adding details to the surface by simply increasing the number of com-putational elements quickly becomes intractable. The problem can be somewhat alleviated by speeding up computational
bottlenecks like the pressure projection step [Lentine et al. 2010; McAdams et al. 2010], but ultimately the volumetric complexity remains an obstacle. On the other hand, the costs of surface
tracking only scales with the surface area, so the immediate temptation here is to increase the resolution of the surface tracker while keeping the fluid simulation resolution fixed. This strategy of
only increasing the surface resolution has produced some beautiful results in the past [Goktekin et al. 2004; Bargteil, Goktekin, et al. 2006;
Heo and Ko 2010; Kim et al. 2009; Wojtan et al. 2009], but it introduces visual and topological errors that limit its usefulness with extremely detailed surfaces (Figure 4.2).
To see where these errors come from, we consider the relationship between the surface tracker and the fluid simulation. While the surface tracker certainly acts as the source of visual detail, it is
also responsible for communicating the location of the free surface to the fluid simulation. The fluid simulation then converts the shape of this free surface into Dirichlet boundary conditions for a
Poisson equation.
After solving this Poisson equation, the fluid simulation then adds pressure forces to ensure that any subtle variations near the free surface are accounted for in a manner consistent with the
Navier-Stokes equations. However, a problem occurs if we lose information when conveying the free surface shape from the surface tracker to the fluid simulation; if the surface tracker is
significantly more detailed than the fluid simulation, then there is no way to adequately encode all of the subtleties of the free surface into the boundary conditions. As a result of these
mismatched levels of detail, the fluid simulation cannot recognize highly detailed
surface features, and it cannot supply the necessary high-resolution pressure forces.
Consequently, high resolution surface structures will clearly violate natural fluid motion, because they ignore the pressure term of the Navier-Stokes equations — the fluid simulation simply does not
have enough degrees of freedom to prevent unphysical states in the surface tracker.
Previous methods have either ignored these errors, applied surface smooth-ing, or added additional detail to the fluid simulation in order to address these problems. While surface smoothing
eventually removes unphysical high-resolution details, it also removes important physically valid motions, and it has no physical basis (or is based on unphysically strong and over-damped surface
tension). Re-fining the fluid simulation detail near the surface is certainly a valid strategy, but perfectly matching the resolution of a detailed surface tracker often comes with significant extra
implementation effort and computational overhead.
Our paper presents a fundamentally different approach for reconciling the difference between a high resolution surface tracker and a low resolution velocity field. We first propose a novel error
metric that identifies and quantifies any unphysical surface behaviors by contrasting the current state of the surface tracker with the set of physically valid surface states. Once we have this
information, we can use the gradient of this error function in a couple of different ways. We first introduce a novel physics-based surface fairing method that quickly removes surface artifacts while
preserving physically-valid surface details. Next, we derive a novel surface correction force that removes artifacts with strategically placed gravity and surface tension waves. We show that this
approach conveniently mimics the vortex sheet form of the Navier-Stokes equations while seamlessly integrating into an Eulerian simulation.
The contributions of our paper are as follows:
• Theoretical insight into the problem of coupling a high resolution surface tracker to a low resolution fluid simulation.
• A novel error metric for quantifying the physical validity of a fluid surface tracker.
• A surface fairing algorithm for fluid-like surfaces that clearly out-performs standard smoothing techniques.
• A surface correction force that removes high-resolution artifacts while pre-serving physically-valid details.
• Applications to both level set and mesh-based surface trackers. | {"url":"https://9pdf.net/article/conclusion-tracking-correcting-absorbing-water-surface-waves.q2n989e6","timestamp":"2024-11-11T15:07:44Z","content_type":"text/html","content_length":"68311","record_id":"<urn:uuid:107bb5c6-077b-4747-9b0c-43a5c4377b8a>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00593.warc.gz"} |
How to Apply the stopifnot() Function in R (Example Code)How to Apply the stopifnot() Function in R (Example Code)
How to Apply the stopifnot() Function in R (Example Code)
In this article, I’ll show how to apply the stopifnot function to test for the truth of certain expressions in the R programming language.
How to Apply the stopifnot() Function to Test the Truth of Expressions
stopifnot("x" == "x", # Test if all expressions are TRUE
length(1:3) > 2,
5 == 4)
# Error: 5 == 4 is not TRUE
stopifnot("x" == "x", # Test if all expressions are TRUE length(1:3) > 2, 5 == 4) # Error: 5 == 4 is not TRUE | {"url":"https://data-hacks.com/stopifnot-r-function","timestamp":"2024-11-10T22:08:41Z","content_type":"text/html","content_length":"209923","record_id":"<urn:uuid:f0060f5f-2578-4ac2-86a3-b968c85f537a>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00257.warc.gz"} |
Approximate the Value of Π in Your Kitchen
Introduction: Approximate the Value of Π in Your Kitchen
A few introductory facts about π.
The number π is probably the most well known mathematical constant. It is defined as the ratio of a circle's circumference to its diameter and is commonly approximated as 3.14159. Hippocrates Of
Chios was the first to prove that this ratio is constant for any given circle.
Being an irrational number, π cannot be expressed exactly as a fraction. Or equivalently, its decimal representation never ends and never settles into a permanent repeating pattern (as we understand
so far).
It has been represented by the Greek letter π since the mid-18th century, though it is also sometimes spelled out as p-ai.
The first recorded algorithm for rigorously calculating the value of π was a geometrical approach using polygons, devised around 250 BC by the Greek mathematician Archimedes. This polygonal algorithm
dominated for over 1000 years. That's the reason why π is sometimes referred to as Archimedes' constant.
Archimedes computed upper and lower bounds of π by drawing a regular hexagon inside and outside a circle. He successively doubled the number of sides until he reached a 96-sided regular polygon. By
calculating the perimeters of these polygons, he proved that 223 / 71 < π < 22 / 7 (that is approximately 3.1408 < π < 3.1429). It is said that Archimedes used 22 / 7 as an approximation of π.
Step 1: Materials
In this instructable you will learn how to approximate Archimedes' Constant π in the comfort of your kitchen.
This is a fun, educational project to carry out with your kids, grandchildren, nieces or nephews.
We are going to use a small model and then we are going to scale it up.
You will need :
• several Oreo type biscuits (or any other circular shaped ones)
• an A4 sheet of paper
• a pencil
• a ruler
• a calculator
• cling foil (optional)
• printer to print the circular template (download my template or draw your own)
• a beam compass (for the bigger model)
For the construction of the beam compass look at my other instructable :
Step 2: Download the Template for the Small Model
The biscuits I used measured approximately 4 cm in diameter (and were delicious!).
I created a simple template to put the cookies on.
The template was created using Xfig 3.2 . It is a GNU/Linux schematics program.
Download my template or create your own using paper and compass.
Step 3: Small Model - Measure the Circumference
Print the template and put it on the kitchen table.
Begin by putting the biscuits in circular order, following the guides.
Just for now, resist the temptation to eat them. Leave it for the end of the activity.
Look at the above pictures. In my case the last biscuit leaves a small space. Measure that distance with your ruler and write it down. My measurement was 0.7 cm.
Also measure how many whole biscuits you used. I used 12.
So the circumference of the dashed circle will be :
C = 4 x 12 + 0.7 = 48.7 cm.
Step 4: Small Model - Measure the Diameter
Now put some biscuits along the diameter line of the template.
The edge of the first biscuit must barely touch the dashed line.
In my case, I used 3 whole biscuits and a portion of the fourth.
The fourth biscuit was cut at 3.6 cm in order to fit.
So the diameter of the dashed circle will be :
d = 4 x 3 + 3.6 = 15.6 cm.
Step 5: Small Model - Data Crunching
To calculate π, divide C with d. In my case :
π = C / d =
= 48.7 / 15.6 =
= 3.12179...
Since the real value for π is 3.14159... we get :
(3.12179... - 3.14159...) x (100 %) / 3.14159... =
0.63025 %... or roughly 0.63 % deviation from the real value. That's a decent approximation!
Step 6: Scale Up the Model
Let's repeat the measurement, only this time we will use a bigger template and a lot more biscuits.
The bigger circle has diameter 60 cm and the smaller 56 cm. They were drawn using my big beam compass on a piece of cardboard. You can see how I made this device, viewing my other instructable :
Step 7: Big Model - Measure the Circumference
Before laying the biscuits I applied several pieces of cling film along the path because the cardboard was dirty.
As before, lay the biscuits along the circular path. I used 42 whole and 3.6 cm of the 43th.
So now the circumference will be :
C = 4 x 42 + 3.6 = 171.6 cm.
Step 8: Big Model - Measure the Diameter
Now put some biscuits along the diameter.
In my case, I used 13 whole biscuits and 2.7 cm of the 14th.
So the diameter will be :
d = 4 x 13 + 2.7 = 54.7 cm.
Step 9: Big Model - Data Crunching
To calculate π, again we divide C with d. In my case :
π = C / d =
= 171.6 / 54.7 =
= 3.13711...
Since the real value for π is 3.14159... we get :
(3.13711...- 3.14159...) x (100 %) / 3.14159... =
or roughly 0.14 % deviation from the real value.
The more biscuits, the merrier!
Step 10: Conclusions
In this activity we used the polygon approximation method in order to approximate the value of π. This method converges slowly and requires polygons with a large number of sides (more biscuits and/or
smaller biscuits in our example). There are other algorithms that converge quicker towards the answer such as infinite series, continued fractions, computer iterative algorithms, etc.
The deviation from the real value of π is justified, if we consider the fact that the cookies were not perfect cylinders and our crude measurements have finite precision.
But this can't lessen the fact that we measured the constant πin our kitchen with decent accuracy. Bravo!
Step 11: Enjoy Your Biscuits
Now, that the mathematical session is over, you can enjoy your biscuits.
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nForum - Discussion Feed (locally cartesian closed (infinity,1)-category)Urs comments on "locally cartesian closed (infinity,1)-category" (93318)Urs comments on "locally cartesian closed (infinity,1)-category" (93091)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (59445)Urs comments on "locally cartesian closed (infinity,1)-category" (59442)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (59441)Urs comments on "locally cartesian closed (infinity,1)-category" (59440)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (59439)Urs comments on "locally cartesian closed (infinity,1)-category" (59438)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (53898)Urs comments on "locally cartesian closed (infinity,1)-category" (53897)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (53895)Urs comments on "locally cartesian closed (infinity,1)-category" (53885)Urs comments on "locally cartesian closed (infinity,1)-category" (53464)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (42028)Urs comments on "locally cartesian closed (infinity,1)-category" (42023)Zhen Lin comments on "locally cartesian closed (infinity,1)-category" (42022)Urs comments on "locally cartesian closed (infinity,1)-category" (42021)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (42016)Urs comments on "locally cartesian closed (infinity,1)-category" (42015)Urs comments on "locally cartesian closed (infinity,1)-category" (42011)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (42006)Urs comments on "locally cartesian closed (infinity,1)-category" (42003)Urs comments on "locally cartesian closed (infinity,1)-category" (41802)Mike Shulman comments on "locally cartesian closed (infinity,1)-category" (41794)
Most aspects of type theory aside from univalence (e.g. Σ-types, Π-types, and identity types) are now known to admit models in all (∞, 1)-toposes, and indeed in all locally presentable, locally
cartesian closed (∞, 1)-categories. By the coherence theorem of [LW13], it suffices to present such an (∞, 1)-category by a “type-theoretic model category” in the sense of [Shu12], such as a
right proper Cininski model category [Cis02,Cis06]. That this is always possible has been proven by Cisinski [Cis12] and by Gepner–Kock [GK12].
[Cis12] Denis-Charles Cisinski. Blog comment on post “The mysterious nature of right properness”. http://golem.ph.utexas.edu/category/2012/05/the_mysterious_nature_of_right.html#c041306, May
[GK12] David Gepner and Joachim Kock. Univalence in locally cartesian closed ∞-categories. arXiv:1208.1749, 2012. | {"url":"https://nforum.ncatlab.org/search/?PostBackAction=Search&Type=Comments&Page=1&Feed=ATOM&DiscussionID=3326&FeedTitle=Discussion+Feed+%28locally+cartesian+closed+%28infinity%2C1%29-category%29","timestamp":"2024-11-03T22:29:37Z","content_type":"application/atom+xml","content_length":"31222","record_id":"<urn:uuid:5fc4bfc6-a790-46f6-a3a8-c9a79e2aa0ad>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00235.warc.gz"} |
per hour to Gallon
Conversion Liter per hour to Gallon per hour(US)
To convert liters per hour to gallons per hour, you must multiply liters per hour by approximately 0.2642.
Tool converts liters per hour to gallons per hour
This tool converts liters per hour to gallons per hour (l/h to gal/h) and vice versa. 1 liter per hour ≈ 0.2642 gallons per hour. The user must fill in one of the two fields, and the conversion will
occur automatically.
1 liters per hour = 0.2642 gallons per hour
Formula liters per hour in gallons per hour (lt/h in gal/h). Gal/h ≈ lt/h*0.2642
Formula to convert liters per hour to gallons per hour
Here is the formula to convert liters per hour to gallons per hour:
$\text{Gallons per hour (gal/h)}=\text{Liters per hour (l/h)}×0.2642$
For example, if you have 100 liters per hour:
100 l/h × 0.2642 = 26.42 gal/h
So, 100 liters per hour is approximately 26.42 gallons per hour.
How to convert 1 liter per hour to gallons per hour?
To convert 1 liter per hour to gallons per hour, you can use the following steps:
1. Understand the Conversion Factor: 1 liter per hour is approximately equal to 0.2642 gallons per hour.
2. Use the Formula: Multiply the number of liters per hour by 0.2642 to get the equivalent gallons per hour.
Example Calculation: If you have 1 liter per hour: 1 l/h × 0.2642 gal/h per l/h = 0.2642 gal/h
Therefore, 1 liter per hour is approximately 0.2642 gallons per hour.
How many gallons per hour are 255 liters per hour (255 lph to gph)?
To convert 255 liters per hour to gallons per hour, you use the conversion factor:
1 liter per hour (l/h) ≈ 0.2642 gallons per hour (gal/h).
So, for 255 liters per hour: 255 l/h × 0.2642 gal/h per l/h = 67.161 gal/h
Therefore, 255 liters per hour is approximately 67.161 gallons per hour.
Why multiply by 0.2642 to convert liters per hour to gallons per hour?
Multiplying by 0.2642 to convert liters per hour to gallons per hour is necessary because this factor represents the relationship between liters and gallons. Here’s the detailed explanation:
1. Volume Conversion:
□ 1 US gallon is equal to approximately 3.78541 liters.
□ To find out how many gallons are in 1 liter, you take the reciprocal of 3.78541 liters per gallon.
2. Calculate the Conversion Factor:
□ Since 1 gallon = 3.78541 liters, the conversion factor from liters to gallons is: $1\text{\hspace{0.17em}}\text{liter}=\frac{1\text{\hspace{0.17em}}\text{gallon}}{3.78541\text{\hspace
{0.17em}}\text{liters}}\approx 0.2642\text{\hspace{0.17em}}\text{gallons}$
3. Applying the Conversion Factor:
□ When converting a flow rate from liters per hour to gallons per hour, you use the volume conversion factor of 0.2642 gallons per liter.
□ This means you multiply the number of liters per hour by 0.2642 to get the equivalent number of gallons per hour.
Thus, multiplying by 0.2642 converts the volume from liters to gallons, which allows you to maintain the flow rate measurement in terms of gallons per hour.
Convert 340 liters per hour to gallons per hour
To convert 340 liters per hour to gallons per hour, you need to use the conversion factor:
1 liter per hour (l/h) is approximately equal to 0.2642 gallons per hour (gal/h).
Here's the calculation: 340 l/h × 0.2642 gal/h per l/h = 89.828gal/h
Thus, 340 liters per hour is roughly equivalent to 89.83 gallons per hour.
Common conversions from liters per hour to gallons per hour
• Convert 1 liters per hour to gallons per hour (1 lph to gph):
1 liters per hour is approximately 0.26 gallons per hour.
• Convert 2 liters per hour to gallons per hour (2 lph to gph):
2 liters per hour is approximately 0.53 gallons per hour.
• Convert 3 liters per hour to gallons per hour (3 lph to gph):
3 liters per hour is approximately 0.79 gallons per hour.
• Convert 4 liters per hour to gallons per hour (4 lph to gph):
4 liters per hour is approximately 1.06 gallons per hour.
• Convert 5 liters per hour to gallons per hour (5 lph to gph):
5 liters per hour is approximately 1.32 gallons per hour.
• Convert 6 liters per hour to gallons per hour (6 lph to gph):
6 liters per hour is approximately 1.59 gallons per hour.
• Convert 7 liters per hour to gallons per hour (7 lph to gph):
7 liters per hour is approximately 1.85 gallons per hour.
• Convert 8 liters per hour to gallons per hour (8 lph to gph):
8 liters per hour is approximately 2.11 gallons per hour.
• Convert 9 liters per hour to gallons per hour (9 lph to gph):
9 liters per hour is approximately 2.38 gallons per hour.
• Convert 10 liters per hour to gallons per hour (10 lph to gph):
10 liters per hour is approximately 2.64 gallons per hour.
• Convert 11 liters per hour to gallons per hour (11 lph to gph):
11 liters per hour is approximately 2.91 gallons per hour.
• Convert 12 liters per hour to gallons per hour (12 lph to gph):
12 liters per hour is approximately 3.17 gallons per hour.
• Convert 13 liters per hour to gallons per hour (13 lph to gph):
13 liters per hour is approximately 3.43 gallons per hour.
• Convert 14 liters per hour to gallons per hour (14 lph to gph):
14 liters per hour is approximately 3.7 gallons per hour.
• Convert 15 liters per hour to gallons per hour (15 lph to gph):
15 liters per hour is approximately 3.96 gallons per hour.
• Convert 16 liters per hour to gallons per hour (16 lph to gph):
16 liters per hour is approximately 4.23 gallons per hour.
• Convert 17 liters per hour to gallons per hour (17 lph to gph):
17 liters per hour is approximately 4.49 gallons per hour.
• Convert 18 liters per hour to gallons per hour (18 lph to gph):
18 liters per hour is approximately 4.76 gallons per hour.
• Convert 19 liters per hour to gallons per hour (19 lph to gph):
19 liters per hour is approximately 5.02 gallons per hour.
• Convert 20 liters per hour to gallons per hour (20 lph to gph):
20 liters per hour is approximately 5.28 gallons per hour.
• Convert 21 liters per hour to gallons per hour (21 lph to gph):
21 liters per hour is approximately 5.55 gallons per hour.
• Convert 22 liters per hour to gallons per hour (22 lph to gph):
22 liters per hour is approximately 5.81 gallons per hour.
• Convert 23 liters per hour to gallons per hour (23 lph to gph):
23 liters per hour is approximately 6.08 gallons per hour.
• Convert 24 liters per hour to gallons per hour (24 lph to gph):
24 liters per hour is approximately 6.34 gallons per hour.
• Convert 25 liters per hour to gallons per hour (25 lph to gph):
25 liters per hour is approximately 6.6 gallons per hour.
• Convert 26 liters per hour to gallons per hour (26 lph to gph):
26 liters per hour is approximately 6.87 gallons per hour.
• Convert 27 liters per hour to gallons per hour (27 lph to gph):
27 liters per hour is approximately 7.13 gallons per hour.
• Convert 28 liters per hour to gallons per hour (28 lph to gph):
28 liters per hour is approximately 7.4 gallons per hour.
• Convert 29 liters per hour to gallons per hour (29 lph to gph):
29 liters per hour is approximately 7.66 gallons per hour.
• Convert 30 liters per hour to gallons per hour (30 lph to gph):
30 liters per hour is approximately 7.93 gallons per hour.
• Convert 31 liters per hour to gallons per hour (31 lph to gph):
31 liters per hour is approximately 8.19 gallons per hour.
• Convert 32 liters per hour to gallons per hour (32 lph to gph):
32 liters per hour is approximately 8.45 gallons per hour.
• Convert 33 liters per hour to gallons per hour (33 lph to gph):
33 liters per hour is approximately 8.72 gallons per hour.
• Convert 34 liters per hour to gallons per hour (34 lph to gph):
34 liters per hour is approximately 8.98 gallons per hour.
• Convert 35 liters per hour to gallons per hour (35 lph to gph):
35 liters per hour is approximately 9.25 gallons per hour.
• Convert 36 liters per hour to gallons per hour (36 lph to gph):
36 liters per hour is approximately 9.51 gallons per hour.
• Convert 37 liters per hour to gallons per hour (37 lph to gph):
37 liters per hour is approximately 9.77 gallons per hour.
• Convert 38 liters per hour to gallons per hour (38 lph to gph):
38 liters per hour is approximately 10.04 gallons per hour.
• Convert 39 liters per hour to gallons per hour (39 lph to gph):
39 liters per hour is approximately 10.3 gallons per hour.
• Convert 40 liters per hour to gallons per hour (40 lph to gph):
40 liters per hour is approximately 10.57 gallons per hour.
• Convert 41 liters per hour to gallons per hour (41 lph to gph):
41 liters per hour is approximately 10.83 gallons per hour.
• Convert 42 liters per hour to gallons per hour (42 lph to gph):
42 liters per hour is approximately 11.1 gallons per hour.
• Convert 43 liters per hour to gallons per hour (43 lph to gph):
43 liters per hour is approximately 11.36 gallons per hour.
• Convert 44 liters per hour to gallons per hour (44 lph to gph):
44 liters per hour is approximately 11.62 gallons per hour.
• Convert 45 liters per hour to gallons per hour (45 lph to gph):
45 liters per hour is approximately 11.89 gallons per hour.
• Convert 46 liters per hour to gallons per hour (46 lph to gph):
46 liters per hour is approximately 12.15 gallons per hour.
• Convert 47 liters per hour to gallons per hour (47 lph to gph):
47 liters per hour is approximately 12.42 gallons per hour.
• Convert 48 liters per hour to gallons per hour (48 lph to gph):
48 liters per hour is approximately 12.68 gallons per hour.
• Convert 49 liters per hour to gallons per hour (49 lph to gph):
49 liters per hour is approximately 12.94 gallons per hour.
• Convert 50 liters per hour to gallons per hour (50 lph to gph):
50 liters per hour is approximately 13.21 gallons per hour.
• Convert 60 liters per hour to gallons per hour (60 lph to gph):
60 liters per hour is approximately 15.9 gallons per hour.
• Convert 70 liters per hour to gallons per hour (70 lph to gph):
70 liters per hour is approximately 18.5 gallons per hour.
• Convert 80 liters per hour to gallons per hour (80 lph to gph):
80 liters per hour is approximately 21.1 gallons per hour.
• Convert 90 liters per hour to gallons per hour (90 lph to gph):
90 liters per hour is approximately 23.8 gallons per hour.
• Convert 100 liters per hour to gallons per hour (100 lph to gph):
100 liters per hour is approximately 26.4 gallons per hour.
• Convert 110 liters per hour to gallons per hour (110 lph to gph):
110 liters per hour is approximately 29.1 gallons per hour.
• Convert 120 liters per hour to gallons per hour (120 lph to gph):
120 liters per hour is approximately 31.7 gallons per hour.
• Convert 130 liters per hour to gallons per hour (130 lph to gph):
130 liters per hour is approximately 34.3 gallons per hour.
• Convert 140 liters per hour to gallons per hour (140 lph to gph):
140 liters per hour is approximately 37 gallons per hour.
• Convert 150 liters per hour to gallons per hour (150 lph to gph):
150 liters per hour is approximately 39.6 gallons per hour.
• Convert 160 liters per hour to gallons per hour (160 lph to gph):
160 liters per hour is approximately 42.3 gallons per hour.
• Convert 170 liters per hour to gallons per hour (170 lph to gph):
170 liters per hour is approximately 44.9 gallons per hour.
• Convert 180 liters per hour to gallons per hour (180 lph to gph):
180 liters per hour is approximately 47.6 gallons per hour.
• Convert 190 liters per hour to gallons per hour (190 lph to gph):
190 liters per hour is approximately 50.2 gallons per hour.
• Convert 200 liters per hour to gallons per hour (200 lph to gph):
200 liters per hour is approximately 52.8 gallons per hour.
• Convert 210 liters per hour to gallons per hour (210 lph to gph):
210 liters per hour is approximately 55.5 gallons per hour.
• Convert 220 liters per hour to gallons per hour (220 lph to gph):
220 liters per hour is approximately 58.1 gallons per hour.
• Convert 230 liters per hour to gallons per hour (230 lph to gph):
230 liters per hour is approximately 60.8 gallons per hour.
• Convert 240 liters per hour to gallons per hour (240 lph to gph):
240 liters per hour is approximately 63.4 gallons per hour.
• Convert 250 liters per hour to gallons per hour (250 lph to gph):
250 liters per hour is approximately 66 gallons per hour.
• Convert 260 liters per hour to gallons per hour (260 lph to gph):
260 liters per hour is approximately 68.7 gallons per hour.
• Convert 270 liters per hour to gallons per hour (270 lph to gph):
270 liters per hour is approximately 71.3 gallons per hour.
• Convert 280 liters per hour to gallons per hour (280 lph to gph):
280 liters per hour is approximately 74 gallons per hour.
• Convert 290 liters per hour to gallons per hour (290 lph to gph):
290 liters per hour is approximately 76.6 gallons per hour.
• Convert 300 liters per hour to gallons per hour (300 lph to gph):
300 liters per hour is approximately 79.3 gallons per hour.
How many gallons per hour are X liters per hour?
• How many gallons per hour are 350 liters per hour (350 lph to gph)?
350 liters per hour is approximately 92.5 gallons per hour.
• How many gallons per hour are 400 liters per hour (400 lph to gph)?
400 liters per hour is approximately 105.7 gallons per hour.
• How many gallons per hour are 450 liters per hour (450 lph to gph)?
450 liters per hour is approximately 118.9 gallons per hour.
• How many gallons per hour are 500 liters per hour (500 lph to gph)?
500 liters per hour is approximately 132.1 gallons per hour.
• How many gallons per hour are 550 liters per hour (550 lph to gph)?
550 liters per hour is approximately 145.3 gallons per hour.
• How many gallons per hour are 600 liters per hour (600 lph to gph)?
600 liters per hour is approximately 158.5 gallons per hour.
• How many gallons per hour are 650 liters per hour (650 lph to gph)?
650 liters per hour is approximately 171.7 gallons per hour.
• How many gallons per hour are 700 liters per hour (700 lph to gph)?
700 liters per hour is approximately 184.9 gallons per hour.
• How many gallons per hour are 750 liters per hour (750 lph to gph)?
750 liters per hour is approximately 198.1 gallons per hour.
• How many gallons per hour are 800 liters per hour (800 lph to gph)?
800 liters per hour is approximately 211.3 gallons per hour.
• How many gallons per hour are 850 liters per hour (850 lph to gph)?
850 liters per hour is approximately 224.5 gallons per hour.
• How many gallons per hour are 900 liters per hour (900 lph to gph)?
900 liters per hour is approximately 237.8 gallons per hour.
• How many gallons per hour are 950 liters per hour (950 lph to gph)?
950 liters per hour is approximately 251 gallons per hour.
• How many gallons per hour are 1000 liters per hour (1000 lph to gph)?
1000 liters per hour is approximately 264.2 gallons per hour.
• How many gallons per hour are 1050 liters per hour (1050 lph to gph)?
1050 liters per hour is approximately 277.4 gallons per hour.
• How many gallons per hour are 1100 liters per hour (1100 lph to gph)?
1100 liters per hour is approximately 290.6 gallons per hour.
• How many gallons per hour are 1150 liters per hour (1150 lph to gph)?
1150 liters per hour is approximately 303.8 gallons per hour.
• How many gallons per hour are 1200 liters per hour (1200 lph to gph)?
1200 liters per hour is approximately 317 gallons per hour.
• How many gallons per hour are 1250 liters per hour (1250 lph to gph)?
1250 liters per hour is approximately 330.2 gallons per hour.
• How many gallons per hour are 1300 liters per hour (1300 lph to gph)?
1300 liters per hour is approximately 343.4 gallons per hour.
• How many gallons per hour are 1350 liters per hour (1350 lph to gph)?
1350 liters per hour is approximately 356.6 gallons per hour.
• How many gallons per hour are 1400 liters per hour (1400 lph to gph)?
1400 liters per hour is approximately 369.8 gallons per hour.
• How many gallons per hour are 1450 liters per hour (1450 lph to gph)?
1450 liters per hour is approximately 383 gallons per hour.
• How many gallons per hour are 1500 liters per hour (1500 lph to gph)?
1500 liters per hour is approximately 396.3 gallons per hour.
• How many gallons per hour are 1550 liters per hour (1550 lph to gph)?
1550 liters per hour is approximately 409.5 gallons per hour.
• How many gallons per hour are 1600 liters per hour (1600 lph to gph)?
1600 liters per hour is approximately 422.7 gallons per hour.
• How many gallons per hour are 1650 liters per hour (1650 lph to gph)?
1650 liters per hour is approximately 435.9 gallons per hour.
• How many gallons per hour are 1700 liters per hour (1700 lph to gph)?
1700 liters per hour is approximately 449.1 gallons per hour.
• How many gallons per hour are 1750 liters per hour (1750 lph to gph)?
1750 liters per hour is approximately 462.3 gallons per hour.
• How many gallons per hour are 1800 liters per hour (1800 lph to gph)?
1800 liters per hour is approximately 475.5 gallons per hour.
• How many gallons per hour are 1850 liters per hour (1850 lph to gph)?
1850 liters per hour is approximately 488.7 gallons per hour.
• How many gallons per hour are 1900 liters per hour (1900 lph to gph)?
1900 liters per hour is approximately 501.9 gallons per hour.
• How many gallons per hour are 1950 liters per hour (1950 lph to gph)?
1950 liters per hour is approximately 515.1 gallons per hour.
• How many gallons per hour are 2000 liters per hour (2000 lph to gph)?
2000 liters per hour is approximately 528.3 gallons per hour.
• How many gallons per hour are 2050 liters per hour (2050 lph to gph)?
2050 liters per hour is approximately 541.6 gallons per hour.
• How many gallons per hour are 2100 liters per hour (2100 lph to gph)?
2100 liters per hour is approximately 554.8 gallons per hour.
• How many gallons per hour are 2150 liters per hour (2150 lph to gph)?
2150 liters per hour is approximately 568 gallons per hour.
• How many gallons per hour are 2200 liters per hour (2200 lph to gph)?
2200 liters per hour is approximately 581.2 gallons per hour.
• How many gallons per hour are 2250 liters per hour (2250 lph to gph)?
2250 liters per hour is approximately 594.4 gallons per hour.
• How many gallons per hour are 2300 liters per hour (2300 lph to gph)?
2300 liters per hour is approximately 607.6 gallons per hour.
• How many gallons per hour are 2350 liters per hour (2350 lph to gph)?
2350 liters per hour is approximately 620.8 gallons per hour.
• How many gallons per hour are 2400 liters per hour (2400 lph to gph)?
2400 liters per hour is approximately 634 gallons per hour.
• How many gallons per hour are 2450 liters per hour (2450 lph to gph)?
2450 liters per hour is approximately 647.2 gallons per hour.
• How many gallons per hour are 2500 liters per hour (2500 lph to gph)?
2500 liters per hour is approximately 660.4 gallons per hour.
• How many gallons per hour are 2550 liters per hour (2550 lph to gph)?
2550 liters per hour is approximately 673.6 gallons per hour.
• How many gallons per hour are 2600 liters per hour (2600 lph to gph)?
2600 liters per hour is approximately 686.8 gallons per hour.
• How many gallons per hour are 2650 liters per hour (2650 lph to gph)?
2650 liters per hour is approximately 700.1 gallons per hour.
• How many gallons per hour are 2700 liters per hour (2700 lph to gph)?
2700 liters per hour is approximately 713.3 gallons per hour.
• How many gallons per hour are 2750 liters per hour (2750 lph to gph)?
2750 liters per hour is approximately 726.5 gallons per hour.
• How many gallons per hour are 2800 liters per hour (2800 lph to gph)?
2800 liters per hour is approximately 739.7 gallons per hour.
• How many gallons per hour are 2850 liters per hour (2850 lph to gph)?
2850 liters per hour is approximately 752.9 gallons per hour.
• How many gallons per hour are 2900 liters per hour (2900 lph to gph)?
2900 liters per hour is approximately 766.1 gallons per hour.
• How many gallons per hour are 2950 liters per hour (2950 lph to gph)?
2950 liters per hour is approximately 779.3 gallons per hour.
• How many gallons per hour are 3000 liters per hour (3000 lph to gph)?
3000 liters per hour is approximately 792.5 gallons per hour.
• How many gallons per hour are 3050 liters per hour (3050 lph to gph)?
3050 liters per hour is approximately 805.7 gallons per hour.
• How many gallons per hour are 3100 liters per hour (3100 lph to gph)?
3100 liters per hour is approximately 818.9 gallons per hour.
• How many gallons per hour are 3150 liters per hour (3150 lph to gph)?
3150 liters per hour is approximately 832.1 gallons per hour.
• How many gallons per hour are 3200 liters per hour (3200 lph to gph)?
3200 liters per hour is approximately 845.4 gallons per hour.
• How many gallons per hour are 3250 liters per hour (3250 lph to gph)?
3250 liters per hour is approximately 858.6 gallons per hour.
• How many gallons per hour are 3300 liters per hour (3300 lph to gph)?
3300 liters per hour is approximately 871.8 gallons per hour.
• How many gallons per hour are 3350 liters per hour (3350 lph to gph)?
3350 liters per hour is approximately 885 gallons per hour.
• How many gallons per hour are 3400 liters per hour (3400 lph to gph)?
3400 liters per hour is approximately 898.2 gallons per hour.
• How many gallons per hour are 3450 liters per hour (3450 lph to gph)?
3450 liters per hour is approximately 911.4 gallons per hour.
• How many gallons per hour are 3500 liters per hour (3500 lph to gph)?
3500 liters per hour is approximately 924.6 gallons per hour.
• How many gallons per hour are 3550 liters per hour (3550 lph to gph)?
3550 liters per hour is approximately 937.8 gallons per hour.
• How many gallons per hour are 3600 liters per hour (3600 lph to gph)?
3600 liters per hour is approximately 951 gallons per hour.
• How many gallons per hour are 3650 liters per hour (3650 lph to gph)?
3650 liters per hour is approximately 964.2 gallons per hour.
• How many gallons per hour are 3700 liters per hour (3700 lph to gph)?
3700 liters per hour is approximately 977.4 gallons per hour.
• How many gallons per hour are 3750 liters per hour (3750 lph to gph)?
3750 liters per hour is approximately 990.6 gallons per hour.
• How many gallons per hour are 3800 liters per hour (3800 lph to gph)?
3800 liters per hour is approximately 1003.9 gallons per hour.
• How many gallons per hour are 3850 liters per hour (3850 lph to gph)?
3850 liters per hour is approximately 1017.1 gallons per hour.
• How many gallons per hour are 3900 liters per hour (3900 lph to gph)?
3900 liters per hour is approximately 1030.3 gallons per hour.
• How many gallons per hour are 3950 liters per hour (3950 lph to gph)?
3950 liters per hour is approximately 1043.5 gallons per hour.
• How many gallons per hour are 4000 liters per hour (4000 lph to gph)?
4000 liters per hour is approximately 1056.7 gallons per hour.
In what categories do we use the unit liters per hour, and in which categories do we use gallons per hour?
The units liters per hour (L/h) and gallons per hour (gal/h) are primarily used to measure flow rates, each typically preferred based on regional measurement standards or specific industry practices.
Liters per hour are commonly used in sectors like water treatment, laboratory experiments, and healthcare, especially in regions that adhere to the metric system. This unit is ideal for applications
requiring precise measurement of fluid flow, such as in medical infusions or controlled scientific settings.
Gallons per hour, on the other hand, are often used in the United States for applications in agriculture, plumbing, and firefighting equipment. This unit is integral to industries where the imperial
system is predominant, measuring water flow in irrigation systems or the output of water heaters.
How many gallons (US) per hour are 1 liter per hour?
1 liter per hour = 0.2642 gallons (US) per minute. 1 lt/h is 0.2642 gph. Gallons (US) per hour and liters per hour are units that people measures the flow. To convert liters per hour to gallons per
hour, multiply liters per hour by 0.2642.
Conversions liters per hour to other units
Liter/h to Liter/s Liter/h to Liter/m Liter/h to Liter/d
Liter/h to Cubic meter/s Liter/h to Cubic meter/m Liter/h to Cubic meter/h
Liter/h to Cubic meter/d Liter/h to Gallon/s Liter/h to Gallon/m
Liter/h to Gallon/h Liter/h to Gallon/d Liter/h to Barrel/s
Liter/h to Barrel/m Liter/h to Barrel/h Liter/h to Barrel/d
Table lt/h to gal/h
1 lt/h = 0.2642 gal/h 11 lt/h = 2.9059 gal/h 21 lt/h = 5.5476 gal/h
2 lt/h = 0.5283 gal/h 12 lt/h = 3.1701 gal/h 22 lt/h = 5.8118 gal/h
3 lt/h = 0.7925 gal/h 13 lt/h = 3.4342 gal/h 23 lt/h = 6.076 gal/h
4 lt/h = 1.0567 gal/h 14 lt/h = 3.6984 gal/h 24 lt/h = 6.3401 gal/h
5 lt/h = 1.3209 gal/h 15 lt/h = 3.9626 gal/h 25 lt/h = 6.6043 gal/h
6 lt/h = 1.585 gal/h 16 lt/h = 4.2268 gal/h 26 lt/h = 6.8685 gal/h
7 lt/h = 1.8492 gal/h 17 lt/h = 4.4909 gal/h 27 lt/h = 7.1326 gal/h
8 lt/h = 2.1134 gal/h 18 lt/h = 4.7551 gal/h 28 lt/h = 7.3968 gal/h
9 lt/h = 2.3775 gal/h 19 lt/h = 5.0193 gal/h 29 lt/h = 7.661 gal/h
10 lt/h = 2.6417 gal/h 20 lt/h = 5.2834 gal/h 30 lt/h = 7.9252 gal/h
40 lt/h = 10.5669 gal/h 70 lt/h = 18.492 gal/h 100 lt/h = 26.4172 gal/h
50 lt/h = 13.2086 gal/h 80 lt/h = 21.1338 gal/h 110 lt/h = 29.0589 gal/h
60 lt/h = 15.8503 gal/h 90 lt/h = 23.7755 gal/h 120 lt/h = 31.7006 gal/h
200 lt/h = 52.8344 gal/h 500 lt/h = 132.086 gal/h 800 lt/h = 211.338 gal/h
300 lt/h = 79.2516 gal/h 600 lt/h = 158.503 gal/h 900 lt/h = 237.755 gal/h
400 lt/h = 105.669 gal/h 700 lt/h = 184.92 gal/h 1000 lt/h = 264.172 gal/h
Flow Conversions
Liter/s to Liter/m Liter/s to Liter/h Liter/s to Liter/d
Liter/s to Cubic meter/s Liter/s to Cubic meter/m Liter/s to Cubic meter/h
Liter/s to Cubic meter/d Liter/s to Gals/s Liter/s to Gallons/m
Liter/s to Gallons/h Liter/s to Gallons/d Liter/s to Barrels/s
Liter/s to Barrels/m Liter/s to Barrels/h Liter/s to Barrels/d
Liter/m to Liter/s Liter/m to Liter/h Liter/m to Liter/d
Liter/m to Cubic meter/s Liter/m to Cubic meter/m Liter/m to Cubic meter/h
Liter/m to Cubic meter/d Liter/m to Gals/s Liter/m to Gallons/m
Liter/m to Gallons/h Liter/m to Gallons/d Liter/m to Barrels/s
Liter/m to Barrels/m Liter/m to Barrels/h Liter/m to Barrels/d
Liter/d to Liter/s Liter/d to Liter/m Liter/d to Liter/h
Liter/d to Cubic meter/s Liter/d to Cubic meter/m Liter/d to Cubic meter/h
Liter/d to Cubic meter/d Liter/d to Gals/s Liter/d to Gallons/m
Liter/d to Gallons/h Liter/d to Gallons/d Liter/d to Barrels/s
Liter/d to Barrels/m Liter/d to Barrels/h Liter/d to Barrels/d
Cubic meter/s to Liter/s Cubic meter/s to Liter/m Cubic meter/s to Liter/h
Cubic meter/s to Liter/d Cubic meter/s to Cubic meter/m Cubic meter/s to Cubic meter/h
Cubic meter/s to Cubic meter/d Cubic meter/s to Gals/s Cubic meter/s to Gallons/m
Cubic meter/s to Gallons/h Cubic meter/s to Gallons/d Cubic meter/s to Barrels/s
Cubic meter/s to Barrels/m Cubic meter/s to Barrels/h Cubic meter/s to Barrels/d
Cubic meter/m to Liter/s Cubic meter/m to Liter/m Cubic meter/m to Liter/h
Cubic meter/m to Liter/d Cubic meter/m to Cubic meter/s Cubic meter/m to Cubic meter/h
Cubic meter/m to Cubic meter/d Cubic meter/m to Gals/s Cubic meter/m to Gallons/m
Cubic meter/m to Gallons/h Cubic meter/m to Gallons/d Cubic meter/m to Barrels/s
Cubic meter/m to Barrels/m Cubic meter/m to Barrels/h Cubic meter/m to Barrels/d
Cubic meter/h to Liter/s Cubic meter/h to Liter/m Cubic meter/h to Liter/h
Cubic meter/h to Liter/d Cubic meter/h to Cubic meter/s Cubic meter/h to Cubic meter/m
Cubic meter/h to Cubic meter/d Cubic meter/h to Gals/s Cubic meter/h to Gallons/m
Cubic meter/h to Gallons/h Cubic meter/h to Gallons/d Cubic meter/h to Barrels/s
Cubic meter/h to Barrels/m Cubic meter/h to Barrels/h Cubic meter/h to Barrels/d
Cubic meter/d to Liter/s Cubic meter/d to Liter/m Cubic meter/d to Liter/h
Cubic meter/d to Liter/d Cubic meter/d to Cubic meter/s Cubic meter/d to Cubic meter/m
Cubic meter/d to Cubic meter/h Cubic meter/d to Gals/s Cubic meter/d to Gallons/m
Cubic meter/d to Gallons/h Cubic meter/d to Gallons/d Cubic meter/d to Barrels/s
Cubic meter/d to Barrels/m Cubic meter/d to Barrels/h Cubic meter/d to Barrels/d
Gals/s to Liter/s Gals/s to Liter/m Gals/s to Liter/h
Gals/s to Liter/d Gals/s to Cubic meter/s Gals/s to Cubic meter/m
Gals/s to Cubic meter/h Gals/s to Cubic meter/d Gals/s to Gallons/m
Gals/s to Gallons/h Gals/s to Gallons/d Gals/s to Barrels/s
Gals/s to Barrels/m Gals/s to Barrels/h Gals/s to Barrels/d
Gallons/m to Liter/s Gallons/m to Liter/m Gallons/m to Liter/h
Gallons/m to Liter/d Gallons/m to Cubic meter/s Gallons/m to Cubic meter/m
Gallons/m to Cubic meter/h Gallons/m to Cubic meter/d Gallons/m to Gals/s
Gallons/m to Gallons/h Gallons/m to Gallons/d Gallons/m to Barrels/s
Gallons/m to Barrels/m Gallons/m to Barrels/h Gallons/m to Barrels/d
Gallons/h to Liter/s Gallons/h to Liter/m Gallons/h to Liter/h
Gallons/h to Liter/d Gallons/h to Cubic meter/s Gallons/h to Cubic meter/m
Gallons/h to Cubic meter/h Gallons/h to Cubic meter/d Gallons/h to Gals/s
Gallons/h to Gallons/m Gallons/h to Gallons/d Gallons/h to Barrels/s
Gallons/h to Barrels/m Gallons/h to Barrels/h Gallons/h to Barrels/d
Gallons/d to Liter/s Gallons/d to Liter/m Gallons/d to Liter/h
Gallons/d to Liter/d Gallons/d to Cubic meter/s Gallons/d to Cubic meter/m
Gallons/d to Cubic meter/h Gallons/d to Cubic meter/d Gallons/d to Gals/s
Gallons/d to Gallons/m Gallons/d to Gallons/h Gallons/d to Barrels/s
Gallons/d to Barrels/m Gallons/d to Barrels/h Gallons/d to Barrels/d
Barrels/s to Liter/s Barrels/s to Liter/m Barrels/s to Liter/h
Barrels/s to Liter/d Barrels/s to Cubic meter/s Barrels/s to Cubic meter/m
Barrels/s to Cubic meter/h Barrels/s to Cubic meter/d Barrels/s to Gals/s
Barrels/s to Gallons/m Barrels/s to Gallons/h Barrels/s to Gallons/d
Barrels/s to Barrels/m Barrels/s to Barrels/h Barrels/s to Barrels/d
Barrels/m to Liter/s Barrels/m to Liter/m Barrels/m to Liter/h
Barrels/m to Liter/d Barrels/m to Cubic meter/s Barrels/m to Cubic meter/m
Barrels/m to Cubic meter/h Barrels/m to Cubic meter/d Barrels/m to Gals/s
Barrels/m to Gallons/m Barrels/m to Gallons/h Barrels/m to Gallons/d
Barrels/m to Barrels/s Barrels/m to Barrels/h Barrels/m to Barrels/d
Barrels/h to Liter/s Barrels/h to Liter/m Barrels/h to Liter/h
Barrels/h to Liter/d Barrels/h to Cubic meter/s Barrels/h to Cubic meter/m
Barrels/h to Cubic meter/h Barrels/h to Cubic meter/d Barrels/h to Gals/s
Barrels/h to Gallons/m Barrels/h to Gallons/h Barrels/h to Gallons/d
Barrels/h to Barrels/s Barrels/h to Barrels/m Barrels/h to Barrels/d
Barrels/d to Liter/s Barrels/d to Liter/m Barrels/d to Liter/h
Barrels/d to Liter/d Barrels/d to Cubic meter/s Barrels/d to Cubic meter/m
Barrels/d to Cubic meter/h Barrels/d to Cubic meter/d Barrels/d to Gals/s
Barrels/d to Gallons/m Barrels/d to Gallons/h Barrels/d to Gallons/d
Barrels/d to Barrels/s Barrels/d to Barrels/m Barrels/d to Barrels/h | {"url":"https://www.advancedconverter.com/unit-conversions/flow-conversion/liters-per-hour-to-gallons-per-hour","timestamp":"2024-11-10T05:12:53Z","content_type":"application/xhtml+xml","content_length":"115951","record_id":"<urn:uuid:52359516-666b-4ecc-b7cc-1db1ff083c66>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00734.warc.gz"} |
Riddle of the Day
If there are fifteen crows on a fence and the farmer shoots a third of them, how many crows are left?
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Riddle of the Day
A farmer in California owns a beautiful pear tree. He supplies the fruit to a nearby grocery store. The store owner has called the farmer to see how much fruit is available for him to purchase. The
farmer knows that the main trunk has 24 branches. Each branch has exactly 12 boughs and each bough has exactly 6 twigs. Since each twig bears one piece of fruit, how many plums will the farmer be
able to deliver?
Riddle of the Day
While on my way to St. Ives, I saw a man with 7 wives. Each wife had 7 sacks. Each sack had 7 cats. Each cat had 7 kittens. Kitten, cats, sacks, wives, How many were going to St. Ives?
Problem of the Week
While on my way to St. Ives, I saw a man with 7 wives. Each wife had 7 sacks. Each sack had 7 cats. Each cat had 7 kittens. Kitten, cats, sacks, wives, How many were going to St. Ives?
Weekly Enigma
If a wheel has 64 spokes, how many spaces are there between the spokes?
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The Importance of Math and Solutions
Across an array of areas, math and systems are crucial for that variety of applications. In fact , they will influence many techniques from public basic safety to welfare policies for the definition
of “full time” career. Even the intelligence levels are affected by mathematics and systems. Read on to find out more. Listed below are a few examples of applications. Let’s commence with an overview
of math. This article will teach you how mathematics and technology work together.
The earliest known use of mathematics is at weaving. Materials found in north Peru had been dated to around 12, 000 BC, and other sites possess found elderly imprints of woven material. While
technology progressed continuously over the centuries, the usage of mathematics in technology did not keep tempo. Until the second half of the nineteenth century, spinning control calculating devices
had been very rare. Today, computers execute most careers and are widespread.
Technology transfer is another crucial use of math and technology. Although it may not get much visibility, technology transfer is important to the scientific operation and economical competitiveness
of this U. S. economy. Good technology transfer requires both parties to talk about common desired goals and build working relationships. Numerical researchers will need to develop in depth
technology programs and pay attention to quantitative and quality control methodologies. Creating these tools is a sophisticated process, and requires decades of effort. However , the benefits will
be immense.
Even though the use of math creates miles within public groups, the perception of mathematical objectivity is one of the key reasons for the lack of contestation against mathematically prescribed
decisions. Objectivity is a have difficulty against subjectivity and accounts for the guru of scientific pronouncements in contemporary political affairs. http://ultiaction.com/
the-basics-of-mathematics In addition to defining the rights of controlled pronouncements, objectivity as well names the strategies for dealing with distance and distrust. In addition, it clarifies
the role of math in decision making, which might make it more difficult to engage in discussions.
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Reducing experiment numbers
DoE is a very powerful tool to efficiently look at the cause and effect relationship between factors such as temperature and concentration, and responses such as conversion and impurity formation.
DoE looks at each factor at 2 or more levels.
At a recent conference we saw some misconceptions about DoE which may add to the confusion and reluctance around the use of experimental design and would like to clarify some points. The relationship
between n factors and the number of experiments is 2^n for a full factorial design. Therefore, investigating 4 factors equates to 16 experiments, while 12 factors requires 4096 experiments if all
experimental combinations are carried out. It is important to also include a centre point which is repeated to help check for non-linear responses and to determine the experimental error for the
design. Typically, three repeated experiments are carried out to determine the experimental error.
Full factorial designs investigating 3-4 factors are commonly carried out in industry, yet there are many different experimental designs that serve different purposes and you do not need to run a
full factorial design where experiment numbers can still be high.
Fractional factorial designs look at the relationship between factors and responses, but higher resolution designs can also investigate and interpret interactions between factors (see below). In fact
you can investigate the effects of 4 to 7 factors in as few as 8 (+3) experiments while up to 16 factors can be investigated in 32 (+3) experiments. The additional 3 experiments are identical centre
│No. of factors│Resolution III│Resolution IV│Resolution V│Response surface │Full factorial│
│4 │ │8 │16 │24 │16 │
│6 │8 │12 │32 │44 │64 │
│12 │16 │32 │256 │280 │4096 │
│16 │ │32 │256 │288 │65536 │
A resolution III fractional factorial design investigates the effect of changing each factor on the responses but is unable to explain any interactions.
A resolution IV fractional factorial design investigates the effect of changing each factor on a response and identifies if interactions are present, but will not be able to fully identify each
interaction as there is some confounding.
A resolution V fractional factorial design investigates the effect of changing each factor on the responses and explains any 2-way interactions.
Response surface designs investigate the effect of changing each factor on the responses, identifies interactions and investigates any quadratic terms to explain non-linear effects by investigating
each factor at 3 or more levels.
There are other design types for different purposes and each design will allow you to get the information you require with fewer experiments than a full factorial design, especially for larger
numbers of factors.
When used correctly, DoE has the power to reveal the cause and effect relationship between factors and responses and deliver enhanced process understanding, while minimising experimental time and
resource. The careful selection of the correct design type for your purpose will enable the efficient investigation and optimisation of your process. We are happy to support you to make the right
choice for your needs. | {"url":"https://paulmurraycatalysisconsulting.com/reducing-experiment-numbers.html","timestamp":"2024-11-05T01:06:52Z","content_type":"text/html","content_length":"13306","record_id":"<urn:uuid:b76eb393-5df5-495b-a399-f2e1b64c9a8a>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00234.warc.gz"} |
Using the Calculus Quotient Rule: A Complete Guide - Calculus Help
Sometimes, one of the hardest things about calculus is remembering how to apply the different rules. For that reason, anything that helps you remember how to work out equations and makes calculus
easier is worth learning, right?
The calculus quotient rule is a great example of a concept in calculus that can be easily memorized to help you perform calculus quicker and more efficiently.
So let’s take a look at how to do it.
What is the quotient rule?
To put it simply, the quotient rule is a method for differentiating problems where one function is divided by another. The quotient rule allows you to find the derivative of a quotient of two
functions – hence the name.
The theory behind the calculus quotient rule goes like this:
Anytime you have two differentiable functions – let’s use f(x) and g(x) as an example – the quotient must also be differentiable.
If it makes it easier, you can think of it like this: the derivative of the quotient of these two functions must exist.
And there’s a formula to express this basic fact:
This formula can help you find the derivative of a function that is written in the form of a quotient, so any fraction where the numerator and the denominator are both functions of the variable x.
Looking at that formula, it might seem unlikely that you can commit that to memory. But actually, there is a way to remember this formula so you always have it at the tip of your fingers when you
need it.
Using the Quotient Rule
The trick to using the calculus quotient rule is to always start with the bottom function, and to end with the bottom function squared. Some people remember this with the phrase “Hi Dee Ho”!
The top function is ‘Hi’ in this scheme. That makes the bottom function ‘Ho,’ since it’s low.
So a way to remember this formula is with the phrase, “Ho Dee Hi minus Hi Dee Ho over Ho Ho.”
The ‘dee’ in the phrase stands for the derivative.
In words, it means: “The derivative of the quotient of two functions is equal to the bottom function times the derivative of the top function, minus the top function times the derivative of the
bottom function, all divided by the bottom function squared.”
Still confused? Let’s take a look at some examples to make things clearer.
Example 1
Example 2
Next, let’s look at an equation that demonstrates the common mistake people make when using the calculus quotient rule.
It’s important to remember that the derivative of a quotient is not equal to the quotient of the derivatives. That’s a mistake people often make because it seems like it should be, but it’s important
not to fall into this trap when using the quotient rule.
Hopefully, this equation will make it clear what I mean:
Example 3
Now let’s look at how the quotient rule can help you figure out the instantaneous rate of change at a given point:
Uses of the Quotient Rule
So often, calculus can seem abstracted from the real world. But as we all know, mathematics is the secret language the universe speaks, and calculus is no exception. The quotient rule does indeed
have some very useful applications in the real world for those who know how to use it correctly.
Here are some potential uses of the quotient rule:
1. Physics:
□ Velocity and Acceleration: In kinematics, the quotient rule can be used to find the rate of change of velocity with respect to time when velocity is given as a quotient of two functions.
□ Optics: The rule can be used in studying light refraction and the behavior of lenses where functions might be divided.
2. Economics:
□ Marginal Cost and Revenue: When cost or revenue functions are expressed as ratios, the quotient rule helps in determining the marginal values.
□ Elasticity of Demand: Elasticity is often expressed as a ratio of percentage changes in quantity and price, and its derivative can be found using the quotient rule.
3. Biology:
□ Population Dynamics: In models where population growth rate is expressed as a ratio of two functions (e.g., predator-prey models), the quotient rule can be used to analyze changes over time.
□ Enzyme Kinetics: In biochemical reactions, the rate of reaction might be given as a ratio of concentrations, and the quotient rule helps in studying these rates.
4. Engineering:
□ Control Systems: Transfer functions in control systems are often expressed as quotients of polynomials. The quotient rule is used to analyze system stability and response.
□ Signal Processing: In analyzing signals and systems, the quotient rule helps in differentiating transfer functions and other related expressions.
5. Mathematics:
□ Differential Equations: In solving differential equations where solutions might be in the form of a ratio of functions, the quotient rule aids in finding derivatives.
□ Optimization Problems: When dealing with optimization of ratios, such as maximizing efficiency or minimizing cost per unit, the quotient rule helps in finding critical points.
6. Chemistry:
□ Reaction Rates: In chemical kinetics, reaction rates are often expressed as ratios, and the quotient rule can be used to differentiate these expressions.
□ Equilibrium Constants: In thermodynamics, equilibrium constants are expressed as ratios, and their derivatives with respect to temperature or pressure can be analyzed using the quotient rule.
Mastering the Quotient Rule
Hopefully, this article has helped demystify the quotient rule. Once you’ve mastered the ‘Hi-Dee-Ho’ mnemonic device, you may find that this rule comes to you quicker than you might have expected
from looking at the formula alone.
And whatever you need to master this rule for, you’ll find it’s a useful way to find derivatives in calculus. | {"url":"https://calculus-help.com/calculus-quotient-rule/","timestamp":"2024-11-04T14:19:45Z","content_type":"text/html","content_length":"47162","record_id":"<urn:uuid:7596ab28-ccc3-4add-a42c-c9d96e70a119>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00744.warc.gz"} |
Tamás Róbert Mezei
About me
I am a mathematician specializing in combinatorics and algorithms. The research topics I have studied are diverse, their nature hovering around the boundary of theoretical and applied mathematics.
Extremal and graph theoretical themes are regularly featured in my papers.
To take a break from work, I enjoy riding around the hilly regions of the country. I also love cooking dishes of different cuisines. I find reading about antrohopology, philoshophy, physics, and
economics inspiring. Sometimes I make enamel art on copper.
Most Proud of
Academic Youth Award
Awarded by the Hungarian Academy of Sciences to recognise outstanding scientific achievements. Received in Feb 2022.
Co-authored 20 scientific papers
Contributed to a broad range of topics, including (extremal) graph theory, computational geometry, network theory, and bioinformatics.
Multiple finalist in Challenge 24
24 hour programming contest between teams of 3. Organised by the Budapest University of Technology (BME). Finalist in 2010 and in 2014.
Algorithm development
Markov chains
Monte Carlo MC method
Graph theory
Theory building
Scientific writing
Mental stamina
Team player
Interdisciplinary experience
Computer skills
Competitive programming
TOEFL iBT, Score 109/120
Sep 2012
ÖSD B2 (Mittelstufe)
Jun 2007
Road cycling
Home improvement
Enamel art
Research fellow
Young researcher grant
Worked on the following topics & grants
• Traditional & non-traditional methods in extremal combinatorics
• Synthetic networks
• Modern extremal combinatorial problems
• Graph theory & combinatorial scientific computing
• Clustering in wireless networks (industry collaboration)
Visiting researcher
Studied phylogenetic networks with Prof. Andrew Francis.
Visiting researcher
Studied network growth & dynamics with Prof. Zoltán Toroczkai.
Assistant research fellow
Pre-doc & post-doc
Ph.D. in Mathematics & its applications
with a certificate in Network Science
Central European University
Sep 2013 - Nov 2017
Thesis: Extremal solutions to some art gallery and terminal-pairability problems
Grade: Summa cum laude
(Received the Academic Achievement Award for first-year doctoral students in 2014)
Masters in Mathematics
Eötvös Loránd University
Sep 2011 - Jul 2013
Thesis: Seating couples and Tic-Tac-Toe
Grade: with honours
Bachelors in Mathematics
Eötvös Loránd University
Sep 2008 - Jul 2011
Thesis: Combinatorial Nullstellensätze
Grade: with distinction
P. L. Erdős, S. R. Kharel, T. R. Mezei, and Z. Toroczkai, “New Results on Graph Matching from Degree-Preserving Growth”, Mathematics, vol. 12 no. 22, Nov. 2024
DOI arXiv
A. Hubai, T. R. Mezei, F. Béres, A. Benczúr, and I. Miklós, “Constructing and sampling partite, 3-uniform hypergraphs with given degree sequence”, PLOS ONE, vol. 19, no. 5, p. e0303155, May 2024
DOI arXiv
P. L. Erdős, T. R. Mezei, and I. Miklós, “Approximate Sampling of Graphs with Near-P-Stable Degree Intervals”, Annals of Combinatorics, Dec. 2023
DOI arXiv
P. L. Erdős, S. R. Kharel, T. R. Mezei, and Z. Toroczkai, “Degree-preserving graph dynamics: a versatile process to construct random networks”, Journal of Complex Networks, vol. 11, no. 6, p.
cnad046, Dec. 2023
DOI arXiv
P. L. Erdős and T. R. Mezei, “Minimizing Interference-to-Signal Ratios in Multi-Cell Telecommunication Networks”, Algorithms, vol. 16 no. 7, Jul. 2023
DOI arXiv
T. R. Mezei, “Covering simple orthogonal polygons with 𝑟-stars”, arXiv, Apr. 26, 2023
P. L. Erdős, G. Harcos, S. R. Kharel, P. Maga, T. R. Mezei, and Z. Toroczkai, “The sequence of prime gaps is graphic”, Mathematische Annalen, Feb. 2023
DOI arXiv
S. R. Kharel, T. R. Mezei, S. Chung, P. L. Erdős, and Z. Toroczkai, “Degree-preserving network growth”, Nature Physics, vol. 18, no. 1, Jan. 2022
P. L. Erdős, C. Greenhill, T. R. Mezei, I. Miklós, D. Soltész, and L. Soukup, “The mixing time of switch Markov chains: A unified approach”, European Journal of Combinatorics, vol. 99, pp. 99–146,
Jan. 2022
DOI arXiv
P. L. Erdős, E. Győri, T. R. Mezei, I. Miklós, and D. Soltész, “Half-Graphs, Other Non-stable Degree Sequences, and the Switch Markov Chain”, The Electronic Journal of Combinatorics, vol. 28, no. 3,
p. P3.7, Jul. 2021
DOI arXiv
P. L. Erdős, A. Francis, and T. R. Mezei, “Rooted NNI moves and distance-1 tail moves on tree-based phylogenetic networks”, Discrete Applied Mathematics, vol. 294, pp. 205–213, May 2021
DOI arXiv
I. Hartarsky and T. R. Mezei, “Complexity of Two-dimensional Bootstrap Percolation Difficulty: Algorithm and NP-Hardness”, SIAM J. Discrete Math., vol. 34, no. 2, pp. 1444–1459, Jan. 2020
DOI arXiv
E. Győri and T. R. Mezei, “Mobile versus Point Guards”, Discrete & Computational Geometry, vol. 61, no. 2, pp. 421–451, Mar. 2019
DOI arXiv
E. Győri, T. R. Mezei, and G. Mészáros, “Terminal-Pairability in Complete Graphs”, Journal of Combinatorial Mathematics and Combinatorial Computing, vol. 107, pp. 221–231, Nov. 2018
P. L. Erdős, T. R. Mezei, I. Miklós, and D. Soltész, “Efficiently sampling the realizations of bounded, irregular degree sequences of bipartite and directed graphs”, PLOS ONE, vol. 13, no. 8, p.
e0201995, Aug. 2018
DOI arXiv
L. Colucci, P. L. Erdős, E. Győri, and T. R. Mezei, “Terminal-pairability in complete bipartite graphs with non-bipartite demands: Edge-disjoint paths in complete bipartite graphs”, Theoretical
Computer Science, vol. 775, pp. 16–25, Jul. 2019
DOI arXiv
L. Colucci, P. L. Erdős, E. Győri, and T. R. Mezei, “Terminal-pairability in complete bipartite graphs”, Discrete Applied Mathematics, vol. 236, pp. 459–463, Feb. 2018
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D. Dedinszki et al., “Oral administration of pyrophosphate inhibits connective tissue calcification”, EMBO Molecular Medicine, p. e201707532, Jul. 2017
E. Győri, T. R. Mezei, and G. Mészáros, “Note on terminal-pairability in complete grid graphs”, Discrete Mathematics, vol. 340, no. 5, pp. 988–990, May 2017
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E. Győri and T. R. Mezei, “Partitioning orthogonal polygons into ≤ 8-vertex pieces, with application to an art gallery theorem”, Computational Geometry, vol. 59, pp. 13–25, Dec. 2016
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Providing scientifically accurate yet easy to understand up-to date information about COVID-19 from the early stages of the pandemic on. Responsible for website design and structure. Proofread most
of the content.
It is cited in the recommendations for decision-makers by the Hungarian Academy of Sciences as a widely acclaimed website targeting laypeople with science-backed information.
The website was also featured by ELTE (my MSc and BSc alma mater).
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RRB JE 27th May 2019 Shift-2
A cuboid with dimensions $$l \times b \times h$$ is cut to get planks measuring $$l \times 0.5 b \times 0.4 h$$. How many planks will you get?
Find the missing group of alphabets in the following series.
LY, IP, GI, (…), TA
A clock started at noon. By 10 minutes past 7, the hour-hand will move through ............
Complete the series.
540, 270, 90, 45, 15, (…)
In a certain code, 'ABC DEF'is written as 'ZYX WVU". How would 'ADULT' be written in that code?
Rs.800 becomes Rs.956 in 3 years at a certain rate of Simple Interest. If the rate of interest is increased by 4%, what amount will Rs.800 become in 3 years?
Two pipes P, Q can fill a cistern in 12 and 15 minutes separately. P is open for 4 minutes and turned off. In how much time will the remaining part be filled by Q?
The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/h completes one round in 8 minutes, find the area of
the park in square metres.
Read the following information carefully and answer the question given below.
Seven persons— P, Q, R, S, T, U and V like different colours—orange, blue, white, red, green, black and yellow, but not necessarily in the same order. They all have different hobbies (or they
like different activities) driving, swimming, sailing, shopping, running, dancing and camping, but not necessarily in the same order.
The hobby of S is driving. The one whose hobby is shopping likes red colour. P likes yellow colour. The one whose hobby is camping likes green colour. The hobby of U is sailing. Neither S nor U likes
black colour. The one whose hobby is dancing, and swimming does not like black colour. The hobby of P is not swimming. The one whose hobby is swimming does not like orange and white colour; S does
not like orange colour. The hobby of T is neither running nor shopping. T does not like green colour. V does not like green and black colour. The hobby of Q is not camping.
Who among the following has the hobby of camping?
Which of the following affects the stomach? | {"url":"https://cracku.in/rrb-je-27th-may-2019-shift-2-question-paper-solved?page=7","timestamp":"2024-11-09T00:28:39Z","content_type":"text/html","content_length":"156382","record_id":"<urn:uuid:fe981dff-42d4-4ebb-be24-59d99d5acfa1>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00680.warc.gz"} |
Ace LeetCode: Middle Node Linked List Guide
Mastering LeetCode's 'Middle of the Linked List': A Comprehensive Guide for Software Engineers
Unlock the secrets to efficiently finding the middle node in a singly linked list with walkthrough and solutions in Python, TypeScript, and Java.
Introduction to the Problem
In software engineering interviews, demonstrating proficiency in data structures is pivotal. Today, I'll unpack a common problem that often perplexes many: finding the middle node of a singly linked
list, as featured on LeetCode (LeetCode 876. Middle of Linked List).
This challenge tests your understanding of linked list traversal and two-pointer techniques. Consider a linked list where you're asked to identify the central element. For instance, given a list
[1,2,3,4,5], the output should be [3,4,5], pinpointing node 3 as the midpoint. In scenarios with an even number of nodes, say [1,2,3,4,5,6], the expectation is to return the second middle node,
resulting in [4,5,6].
Given the head of a singly linked list, return the middle node of the linked list.
If there are two middle nodes, return the second middle node.
Example 1:
Input: head = [1,2,3,4,5]
Output: [3,4,5]
Explanation: The middle node of the list is node 3.
Example 2:
Input: head = [1,2,3,4,5,6]
Output: [4,5,6]
Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one.
Solving the Problem
The crux of solving this problem lies in the two-pointer strategy: employing a slow pointer that moves one step at a time and a fast pointer advancing two steps per turn. This approach ensures that
when the fast pointer reaches the end of the list, the slow pointer will be at the middle, elegantly sidestepping the need to count elements beforehand.
The beauty of this solution is its efficiency, boasting a time complexity of O(n) where n is the number of nodes in the list, and a space complexity of O(1), as it only utilizes two pointers
regardless of the list's size.
The Solution in Python
Let's dive into the Python solution, which embraces simplicity and efficiency.
def middleNode(head: Optional[ListNode]) -> Optional[ListNode]:
slow = head # Starts at the beginning
quick = head # Also starts at the beginning
while quick and quick.next: # Continues until the fast pointer reaches the end
slow = slow.next # Moves one step
quick = quick.next.next # Moves two steps
return slow # When fast pointer is at the end, slow is at the middle
The Solution in TypeScript
TypeScript, often used for its strong typing features, offers a structured way to tackle this problem.
function middleNode(head: ListNode | null): ListNode | null {
let slow: ListNode | null = head; // Initialize slow pointer
let quick: ListNode | null = head; // Initialize fast pointer
while (quick !== null && quick.next !== null) {
slow = slow.next; // Move slow pointer one step
quick = quick.next.next; // Move fast pointer two steps
return slow; // Return the slow pointer as the middle node
The Solution in Java
Java offers a class-based approach to solving the problem, emphasizing readability and robustness.
public class Solution {
public ListNode middleNode(ListNode head) {
ListNode slow = head; // Starts at the beginning
ListNode quick = head; // Also starts at the beginning
while (quick != null && quick.next != null) {
slow = slow.next; // Moves one step
quick = quick.next.next; // Moves two steps
return slow; // Slow is at the middle when fast reaches the end
Mastering this problem not only boosts your confidence in handling linked lists but also sharpens your problem-solving strategy with the two-pointer technique.
Whether you prefer Python, TypeScript, or Java, understanding the underlying concept remains key. As you practice, remember that the elegance of a solution often lies in its simplicity and
Happy coding, and best of luck in your software engineering interviews!
Did you find this article valuable?
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Find the limit as x goes to 1 of (1 / log x - 1 / (x-1)) - Stumbling Robot
Find the limit as x goes to 1 of (1 / log x – 1 / (x-1))
Evaluate the limit.
We use this exercise (Section 7.11, Exercise #4) to evaluate,
5 comments
1. I have worked through exercises nr. 6 – 29 and It actually isn’t that complicated. I have one thing however that confuses me and caused some problems (especially at ex. nr. 20 -22). How do I know
to what degree I have to take the polynomial? The choices in the solutions sometimes seemed arbitrary to me. I couldn’t see a pattern there. Because of that I sometimes chose to many degrees
which resulted in an algebraic mess and sometimes to few degrees which when I evaluated the expression led me back to 0/0 or similiar. Is that something that comes with experience or is there a
fixed set or rules that tells me to what degree I have to take the polynomial in each case?
□ This clearly is trial and error until you get the right expansion. If you do 100 of these, maybe you can easily see the required expansion then.
☆ Second step I am not getting. I can’t understand that step… can u plz give more explanation for that 2nd step?
☆ I can’t understand 2nd step from that solution…. can u plz give more explanation about that 2nd step?
☆ It is probably simpler if you just substitythe variable x=t+1 and take the limit approaching 0.
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Variations of checking stack automata: Obtaining unexpected decidability properties - PDF Free Download
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Doctopic: Algorithms, automata, complexity and games
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Theoretical Computer Science ••• (••••) •••–•••
Contents lists available at ScienceDirect
Theoretical Computer Science www.elsevier.com/locate/tcs
Variations of checking stack automata: Obtaining unexpected decidability properties ✩ Oscar H. Ibarra a,1 , Ian McQuillan b,∗,2 a b
Department of Computer Science, University of California, Santa Barbara, CA 93106, USA Department of Computer Science, University of Saskatchewan, Saskatoon, SK S7N 5A9, Canada
a r t i c l e
i n f o
Article history: Received 23 February 2018 Accepted 13 April 2018 Available online xxxx Communicated by J. Karhumaki Keywords: Checking stack automata Pushdown automata Decidability Reversal-bounded
a b s t r a c t We introduce a model of one-way language acceptors (a variant of a checking stack automaton) and show the following decidability properties: 1. The deterministic version has a
decidable membership problem but has an undecidable emptiness problem. 2. The nondeterministic version has an undecidable membership problem and emptiness problem. There are many models of accepting
devices for which there is no difference with these problems between deterministic and nondeterministic versions, i.e., the membership problem for both versions are either decidable or undecidable,
and the same holds for the emptiness problem. As far as we know, the model we introduce above is the first one-way model to exhibit properties (1) and (2). We define another family of one-way acceptors
where the nondeterministic version has an undecidable emptiness problem, but the deterministic version has a decidable emptiness problem. We also know of no other model with this property in the
literature. We also investigate decidability properties of other variations of checking stack automata (e.g., allowing multiple stacks, two-way input, etc.). Surprisingly, two-way deterministic
machines with multiple checking stacks and multiple reversal-bounded counters are shown to have a decidable membership problem, a very general model with this property. © 2018 Elsevier B.V. All
rights reserved.
1. Introduction The deterministic and nondeterministic versions of most known models of language acceptors exhibit the same decidability properties for each of the membership and emptiness problems.
In fact, it is possible to define machine models in a general fashion by varying the allowed store types, such as with Abstract Families of Acceptors (AFAs) from [1], or a similar type of machine
model with abstract store types used in [2] and in this paper. Studying machine models defined in such
✩ A preliminary version of this paper has appeared in the Springer LNCS Proceedings of the 21st International Conference on Developments in Language Theory (DLT 2017), pp. 235–246. Corresponding
author. E-mail addresses:
[email protected]
(O.H. Ibarra),
[email protected]
(I. McQuillan). 1 Supported, in part, by NSF Grant CCF-1117708 (Oscar H. Ibarra). 2 Supported, in part, by Natural Sciences and Engineering Research Council of Canada Grant 2016-06172 (Ian
https://doi.org/10.1016/j.tcs.2018.04.024 0304-3975/© 2018 Elsevier B.V. All rights reserved.
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a general fashion is advantageous as certain decidability problems are equivalently decidable for arbitrary machine models defined using such a framework, and therefore it is possible to see which
problems could conceivably differ in terms of decidability. For arbitrary one-way machine models defined in the way used here, the emptiness problem for the nondeterministic machines of this class,
the membership problem for nondeterministic machines of this class, and the emptiness problem for the deterministic machines in this class, must all be either decidable or undecidable. Membership for
deterministic machines could conceivably differ from the other three decidability problems. However, as far as we know, no one-way model has been shown to exhibit different decidability properties
for deterministic and nondeterministic versions. The question arises of whether there is a model where membership for deterministic machines is decidable while it is undecidable for nondeterministic
machines? A second topic of interest here is that of studying decidability properties of different classes of machines when adding additional data stores. In [3], it was shown that for any one-way
machine model (defined using another method used there), if the languages accepted by these machines are all semilinear,3 then augmenting these machines with additional reversalbounded counters4
produces only semilinear languages. And, if semilinearity is effective with the original model, then it is also effective after adding counters, and therefore the emptiness problem is decidable.
However, it is unknown what can occur when augmenting a class of machines that accepts non-semilinear languages with reversal-bounded counters. Can adding such counters change decidability
properties? These two topics are both simultaneously studied in this paper. Of primary importance is the one-way checking stack automaton [5], which is similar to a pushdown automaton that cannot
erase its stack, but can enter and read the stack in two-way read-only mode, but once this mode is entered, the stack cannot change. This model accepts non-semilinear languages, but has decidable
emptiness and membership problems. Here, we introduce a new model of one-way language acceptors by augmenting a checking stack automaton with reversal-bounded counters, and the deterministic and
nondeterministic versions are denoted by DCSACM and NCSACM, respectively. The models with two-way input (with end-markers) are called 2DCSACM and 2NCSACM. These are generalized further to models with
k checking stacks: k-stack 2DCSACM and k-stack 2NCSACM. These models can be defined within the general machine model framework mentioned above. We show the following results concerning membership and
emptiness: 1. The membership and emptiness problems for NCSACMs are undecidable, even when there are only two reversalbounded counters. 2. The emptiness problem for DCSACM is decidable when there is
only one reversal-bounded counter but undecidable when there are two reversal-bounded counters. 3. The membership problem for k-stack 2DCSACMs is decidable for any k. Therefore, this machine model
provides the first known model where membership is decidable for deterministic machines, while the other decidability properties are undecidable, which is the only property that can conceivably
differ. It also shows one possible scenario that can occur when augmenting a machine model accepting non-semilinear languages with reversal-bounded counters: it can change the emptiness problem for
both nondeterministic and deterministic machines to be undecidable, as with the membership problem for nondeterministic machines, but membership for deterministic machines can remain decidable (and
therefore, all such languages accepted by deterministic machines are recursive). In addition, we define another family of one-way acceptors where the deterministic version has a decidable emptiness
problem, but the nondeterministic version has an undecidable emptiness problem. This model must necessarily not be defined using the general machine model framework, as emptiness for deterministic and
nondeterministic machine models are always equivalently decidable. But the model is still natural and demonstrates what must occur to obtain unusual decidable properties. Further, we introduce a new
family with decidable emptiness, containment, and equivalence problems, which is one of the most powerful families to have these properties (one-way deterministic machines with one reversal-bounded
counter and a checking stack that can only read from the stack at the end of the input). We also investigate the decidability properties of other variations of checking stack automata (e.g., allowing
multiple stacks, two-way input, etc.). 2. Preliminaries This paper requires basic knowledge of automata and formal languages, including finite automata, pushdown automata, and Turing machines [6]. An
alphabet is a (usually finite unless stated otherwise) set of symbols. The set ∗ is the set of all words over , which contains the empty word λ. A language is any set L ⊆ ∗ . Given a word w ∈ ∗ , | w
| is the length of w. A language L is bounded if there exists words w 1 , . . . , w k such that L ⊆ w ∗1 · · · w k∗ , and L is letter-bounded if w 1 , . . . , w k are letters. We use a variety of
machine models here, mostly built on top of the checking stack. It is possible to define each machine model directly. As discussed in Section 1, an alternate approach is to define “store types” first,
which describes
See [4] for the formal definition. Equivalently, a language is semilinear if it has the same commutative closure as some regular language. A counter stores a non-negative integer that can be tested
for zero, and it is reversal-bounded if there is a bound on the number of changes between non-decreasing and non-increasing. 4
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just the behavior of the store, including instructions that can change the store, and the manner in which the store can be read. This can capture standard types of stores studied in the literature,
such as a pushdown, or a counter. Defined generally enough, it can also define a checking stack, or a reversal-bounded counter. Then, machines using one or more store types can be defined, in a standard
fashion. A (1 , . . . , k )-machine is a machine with k stores, where i describes each store. This is the approach taken here, in a similar fashion to the one taken in [7] or [1] to define these same
types of automata. This generality will also help in illustrating what is required to obtain certain decidability properties; see e.g. Lemma 1 and Proposition 2 which are proven generally for
arbitrary store types. Furthermore, these two results are used many times within other proofs rather than having many ad hoc proofs. Hence, this generality in defining machines serves several purposes
for this work. First, store types, and machines using store types are defined formally using the same framework used by the authors in [2]. A store type is a tuple = (, I , f , g , c 0 , L I ), where
is the store alphabet (potentially infinite available to all machines using this type of store), I is the set of allowable instructions, c 0 is the initial configuration which is a word in ∗ , and L I
⊆ I ∗ is the instruction language (over possibly an infinite alphabet) of allowable sequences of instructions, f is the read function, a partial function from ∗ to , and g is the write function, a
partial function from ∗ × I to ∗ . We will study a few examples of store types. First, a pushdown store type is a tuple = (, I , f , g , c 0 , L I ), where is an infinite set of store symbols
available to pushdowns, where special symbol Z b ∈ is the bottom-of-stack marker, and 0 = − { Z b }, I = {push( y ) | y ∈ 0 } ∪ {pop, stay} is the set of instructions of the pushdown, where the first
set are called the push instructions, and the second set contains the pop and stay instruction, L I = I ∗ , c 0 = Z b , f (xa) = a, a ∈ , x ∈ ∗ with xa ∈ Z b ∗0 , and g is defined as:
• g ( Z b x, push( y )) = Z b xy for x ∈ ∗0 , y ∈ 0 , • g ( Z b xa, pop) = Z b x, for x ∈ ∗0 , a ∈ 0 , • g ( Z b x, stay) = Z b x, for x ∈ ∗0 . A counter store tape can be obtained by restricting the
pushdown store type to only having a single symbol c ∈ 0 (plus the bottom-of-stack marker). The instruction language L I in the definition of restricts the allowable sequences of instructions
available to the store type (that is, a computation can only accept if its sequence of instructions is in the instruction language). This restriction does not exist in the definition of AFAs, but can
be used to define many classically studied machine models, while still preserving many useful properties. For example, an l-reversal-bounded counter store type is a counter store type with L I equal
to the alternating concatenation of {push(c ), stay}∗ and {pop, stay}∗ with l applications of concatenation (this is more classically stated as, there are at most l alternations between
non-decreasing and non-increasing). Next, the more complicated stack store type is a tuple = (, I , f , g , c 0 , L I ), where
• is an infinite set of store symbols available to stacks, where special symbols ↓∈ are the position of the read/write head in the stack, Z b ∈ is the bottom-of-stack marker, and Z t ∈ is the
top-of-stack marker, with 0 = − {↓, Z b , Z t }, • I = {push( y ) | y ∈ 0 } ∪ {pop, stay} ∪ {D, S, U} is the set of instructions of the stack, where the first set are called the push instructions, the
second set is the pop and stay instruction, and the third set are the move instructions (down, stay, or up), • L I = I ∗ , c 0 = Z b ↓ Z t , f (xa ↓ x ) = a, a ∈ 0 ∪ { Z t , Z b }, x, x ∈ ∗ with xax
∈ Z b ∗0 Z t , • and g is defined as: – g ( Z b x ↓ Z t , push( y )) = Z b xy ↓ Z t for x ∈ ∗0 , y ∈ 0 , – g ( Z b xa ↓ Z t , pop) = Z b x ↓ Z t , for x ∈ ∗0 , a ∈ 0 , – g ( Z b x ↓ Z t , stay) = Z b
x ↓ Z t , for x ∈ ∗0 , – g ( Z b xa ↓ x , D) = Z b x ↓ ax , for x, x ∈ ∗ , a ∈ 0 ∪ { Z t }, with xax ∈ ∗0 Z t , – g ( Z b x ↓ x , S) = Z b x ↓ x , for x, x ∈ ∗ , xx ∈ ∗0 Z t , – g ( Z b x ↓ ax , U) =
Z b xa ↓ x , for x, x ∈ ∗ , a ∈ 0 ∪ { Z t }, xax ∈ ∗0 Z t . That is, a stack is just like a pushdown with the additional ability to read from the “inside” of the stack (but not change the inside) in
two-way read-only mode. Also, the checking stack store type is a restriction of stack store type above where L I is restricted to be in {push( y ), stay | y ∈ 0 }∗ {D, S, U}∗ . That is, a checking
stack has two phases, a “writing phase”, where it can push or stay (no pop), and then a “reading phase”, where it enters the stack in read-only mode. But once it starts reading, it cannot change the
stack again. Given store types (1 , . . . , k ), with i = (i , I i , f i , g i , c 0,i , L I i ), a two-way r-head k-tape (1 , . . . , k )-machine is a tuple M = ( Q , , , δ, , , q0 , F ) where Q is
the finite set of states, q0 ∈ Q is the initial state, F ⊆ Q is the set of final states, is the finite input alphabet, is a finite subset of the store alphabets of 1 ∪ · · · ∪ k , δ is the finite
transition relation from Q × []r × 1 × · · · × k to Q × I 1 × · · · × I k × [{−1, 0, +1}]r . An instantaneous description (ID) is a tuple (q, w , α1 , . . . , αr , x1 , . . . , xk ), where q ∈ Q is
the current state, w is the current input word (surrounded by left input end-marker and right input end-marker), 0 ≤ α j ≤ | w | + 1 is the current position of tape head j (this can be thought of as
0 scanning , and | w | + 1 scanning ), for 1 ≤ j ≤ r, and xi ∈ ∗i is the
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Doctopic: Algorithms, automata, complexity and games
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current word in the i store, for 1 ≤ i ≤ k. Then M is deterministic if δ is a partial function (i.e. it only maps each element to at most one element). Then (q, w , α1 , . . . , αr , x1 , . . . , xk
) M (q , w , α1 , . . . , αr , x1 , . . . , xk ), (two IDs) if there exists a transition (q , ι1 , . . . , ιk , γ1 , . . . , γr ) ∈ δ(q, a1 , . . . , ar , b1 , . . . , bk ), where a j is character α
j + 1 of w (1 is added since is the letter at position (ι ,...,ι )
k 0), and α j = α j + γ j , for 1 ≤ j ≤ r, b i = f i (xi ), and g i (xi , ιi ) = xi for 1 ≤ i ≤ k. Instead of M , we can also write M1 ∗ to show the instructions applied to each store on the
transition. We let M be the reflexive and transitive closure of M ,
(γ ,...,γ )
k and let M 1 , where γi ∈ I i∗ is the sequence of instructions applied to store i, 1 ≤ i ≤ k, in the sequence of transitions applied, and |γ1 | = · · · = |γk |. The language accepted by M, L ( M ),
is equal to
(γ ,...,γk )
{ w | (q0 , w , 1, . . . , 1, c 0,1 , . . . , c 0,k ) M 1
(q f , w , α1 , . . . , αr , x1 , . . . , xk ), q f ∈ F , γi ∈ L I i , 1 ≤ i ≤ k}.
Thus, the sequence of instructions applied to each store must satisfy its instruction language, and they must each be of the same length. The different machine modes are combinations of either
one-way or two-way, deterministic or nondeterministic, and r-head for some r ≥ 1. For example, one-way, 1-head, deterministic, is a machine mode. Given a sequence of store types 1 , . . . , k and a
machine mode, one can study the set of all (1 , . . . , k )-machines with this mode. The set of all such machines with a mode is said to be complete. Any strict subset is said to be incomplete. Given
a set of (complete or incomplete) machines M of this type, the family of languages accepted by these machines is denoted L(M). For example, the set of all one-way deterministic pushdown automata is
complete as it contains all one-way deterministic machines that use the pushdown store. But consider the set of all one-way deterministic pushdown automata that can only decrease the size of the
stack when scanning the right end-marker. This is a strict subset of all one-way deterministic machines that use the pushdown store, since the instructions available to such machines depend on the
location of the input (whether it has reached the end of the input or not). Therefore, this is an incomplete set of machines. The instruction language of a store does allow a complete class of
machines to restrict the allowable sequences of instructions, but it has to apply to all machines using the store. Later in the paper, we will consider variations of checking stack automata such as
one called no-read, which means that they do not read from the inside of the checking stack before hitting the right input end-marker. This is similarly an incomplete set of automata since the
instructions allowed differs depending on the input position. The class of one-way deterministic (resp. nondeterministic) checking stack automata is denoted by DCSA (resp., NCSA) [5]. The class of
deterministic (resp. nondeterministic), finite automata is denoted by DFA (resp., NFA) [6]. For k, l ≥ 1, the class of one-way deterministic (resp. nondeterministic) l-reversal-bounded k-counter
machines is denoted by DCM(k, l) (resp. NCM(k, l)). If only one integer is used, e.g. NCM(k), this class contains all l-reversal-bounded k counter machines, for some l, and if the integers is
omitted, e.g., NCM and DCM, they contain all l-reversal-bounded k counter machines, for some k, l. Note that a counter that makes l reversals can be simulated by l+21 1-reversal-bounded counters [8].
Closure and decidable properties of various machines augmented with reversal-bounded counters have been studied in the literature (see, e.g., [3,8]). For example, it is known that the membership and
emptiness problems are decidable for NCM [8]. Also, here we will study the following new classes of machines that have not been studied in the literature: one-way deterministic (resp.
nondeterministic) machines defined by stores consisting of one checking stack and k l-reversal-bounded counters, denoted by DCSACM(k, l) (resp. NCSACM(k, l)), those with k-reversal-bounded counters,
denoted by DCSACM(k) (resp. NCSACM(k)), and those with some number of reversal-bounded counters, denoted by DCSACM (resp. NCSACM). All models above also have two-way versions of the machines defined,
denoted by preceding them with 2, e.g. 2DCSA, 2NCSA, 2NCM(1), 2DFA, 2NFA, etc. We will also define models with k checking stacks for some k, which we will precede with the phrase “k-stack”, e.g.
k-stack 2DCSA, k-stack 2NCSA, k-stack 2DCSACM, k-stack 2NCSACM, etc. When k = 1, then this corresponds with omitting the phrase “k-stack”. 3. A checking stack with reversal-bounded counters Before
studying the new types of stores and machine models, we determine several properties that are equivalent for any complete set of machines. This helps to demonstrate what is required to potentially
have a machine model where the deterministic version has a decidable membership problem with an undecidable emptiness problem, while both problems are undecidable for the nondeterministic version.
First, we examine a machine’s behavior on one word. Lemma 1. Let M be a one- or two-way, r-head, for some r ≥ 1, (1 , . . . , k )-machine, and let w ∈ ∗ . We can effectively construct another (1 , .
. . , k )-machine M w that is one-way and 1-head which accepts λ if and only if M accepts w. Furthermore, M w is deterministic if M is deterministic. Proof. The input w is encoded in the state of M w
, and M w on input λ, simulates the computation of M and accepts λ if and only if M accepts w. This uses a subset of the sequence of transitions used by M (and thereby would satisfy the instruction
language of each store). Since M w is only on λ input, two-way input is not needed in M w , and the r-heads are simulated completely in the finite control. 2
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Then, for all machines with the same store types, the following decidability problems are equivalent: Proposition 2. Consider store types (1 , . . . , k ). The following problems are equivalently
decidable, for the stated complete sets of automata: 1. 2. 3. 4. 5.
the emptiness problem for one-way deterministic (1 , . . . , k )-machines, the emptiness problem for one-way nondeterministic (1 , . . . , k )-machines, the membership problem for one-way
nondeterministic (1 , . . . , k )-machines, acceptance of λ, for one-way nondeterministic (1 , . . . , k )-machines, the membership problem for two-way r-head (for r ≥ 1) nondeterministic (1 , . . .
, k )-machines.
Proof. The equivalence of 1) and 2) can be seen by taking a nondeterministic machine M. Let T = {t 1 , . . . , tm } be labels in bijective correspondence with the transitions of M. Then construct M
which operates over alphabet T . Then M reads each input symbol t and simulates t of M on the store, while always moving right on the input. However, if it is a stay transition on the input of M,
then M also checks that the next input symbol read (if any), t , is defined on the same letter of in M. Then M is deterministic, and changes its stores identically in sequence to M (thereby still
satisfying the same instruction language), and L ( M ) is therefore empty if and only if L ( M ) is empty. It is immediate that 5) implies 4), and it follows that 4) implies 5) from Lemma 1.
Similarly, 3) implies 4), and 4) implies 3) from Lemma 1. To show that 4) implies 2), notice that any complete set of nondeterministic one-way automata are closed under homomorphisms h where h(a) ≤
1, for all letters a. Considering the homomorphism that erases all letters, the resulting language is empty if and only if λ is accepted by the original machine. To see that 2) implies 4), take a
one-way nondeterministic machine and make a new one that cannot accept if there is an input letter. This new machine is non-empty if and only if λ is accepted in the original machine. 2 It is
important to note that this proposition is not necessarily true for incomplete sets of automata, as the machines constructed in the proof need to be present in the set. We will see some natural
restrictions later where this is not the case, such as sets of machines where there is a restriction on what instructions can be performed on the store based on the position of the input. And indeed,
to prove the equivalence of 1) and 2) above, the deterministic machine created reads a letter for every transition of the nondeterministic machine applied. Hence, consider a set of machines that is
only allowed to apply a strict subset of store instructions before the end-marker. Let M be a nondeterministic machine of this type, and say that M applies some instruction on the end-marker that is
not available to the machine before the end-marker. But the deterministic machine M created from M in Proposition 2 reads an input letter when every instruction is applied, even including those
applied on the end-marker of M. But since M is reading an input letter during this operation, it would violate the instructions allowed by M before the end-marker. The above proposition indicates
that for every complete set of one-way machines, membership for nondeterminism, emptiness for nondeterminism, and emptiness for determinism are equivalent. Thus, the only problem that can potentially
differ is membership for deterministic machines. Yet we know of no existing model where it differs from the other three properties. We examine one next. We will study NCSACMs and DCSACMs, which are
NCSAs and DCSAs (nondeterministic and deterministic checking stack automata) respectively, augmented by reversal-bounded counters. First, two examples will be shown, demonstrating a language that can
be accepted by a DCSACM. Example 1. Consider the language L = {(an #)n | n ≥ 1}. A DCSACM M with one 1-reversal-bounded counter can accept L as follows: M when given an input w (we may assume that
the input is of the form w = an1 # · · · ank # for some k ≥ 1 and ni ≥ 1 for 1 ≤ i ≤ k, since the finite control can check this), copies the first segment an1 to the stack while also storing number n1
in the counter. Then M goes up and down the stack comparing n1 to the rest of the input to check that n1 = · · · = nk while decrementing the counter by 1 for each segment it processes. Clearly, L ( M
) = L and M makes only 1 reversal on the counter. We will show in Proposition 3 that L cannot be accepted by an NCSA (or an NCM). Example 2. Let L = {ai b j ck | i , j ≥ 1, k = i · j }. We can
construct a DCSACM(1) M to accept L as follows. M reads ai and stores ai in the stack. Then it reads b j and increments the counter by j. Finally, M reads ck while moving up and down the stack
containing ai and decrementing the counter by 1 every time the stack has moved i cells, to verify that k is divisible by i and k/i = j. Then M accepts L, and M needs only one 1-reversal counter. We
will see in Proposition 27 that L cannot be accepted by a 2DCM(1). The following shows that, in general, NCSACMs and DCSACMs are computationally more powerful than NCSAs and DCSAs, respectively.
Proposition 3. There are languages in L(DCSACM(1, 1)) − (L(NCSA) ∪ L(NCM)). Hence, L(DCSA) L(DCSACM(1, 1)), and L(NCSA) L(NCSACM(1, 1)).
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Proof. Consider the language L = {(an #)n | n ≥ 1} from Example 1. L cannot be accepted by an NCSA; otherwise, L = 2 {an | n ≥ 1} can also be accepted by an NCSA (since NCSA languages are closed
under homomorphism), but it was shown in [5] that L cannot be accepted by any NCSA. However, Example 1 showed that L can be accepted by a DCSACM(1, 1). Furthermore, L is not semilinear, but NCM only
accepts semilinear languages [8]. 2 We now proceed to show that the membership problem for DCSACM s is decidable. In view of Lemma 1, our problem reduces to deciding, given a DCSACM M, whether it
accepts λ. For acceptance of λ, the next lemma provides a normal form. Lemma 4. Let M be a DCSACM. We can effectively construct a DCSACM M such that:
• all counters of M are 1-reversal-bounded and each must return to zero before accepting, • M always writes on the stack at each step during the writing phase, • the stack head returns to the left
end of the stack before accepting, whereby M accepts λ if and only if M accepts λ. Proof. It is evident that all counters can be assumed to be 1-reversal-bounded as with DCM [8], and that each
counter can be forced to return to zero before accepting. Similarly, the checking stack can be forced to return to the left end before accepting. We introduce a dummy symbol $ to the stack alphabet
so that if in a step, M does not write on the stack, then M writes $. When M enters the reading phase, M simulates M but ignores (i.e., skips over) the $’s. Then M accepts λ if and only if M accepts
λ. 2 In view of Lemma 4, we may assume that a DCSACM writes a symbol at the end of the stack at each step during the writing phase. This is important for deciding the following problem. Lemma 5. Let
M be a DCSACM satisfying the assumptions of Lemma 4. We can effectively decide whether or not M, on λ input, has an infinite writing phase (i.e., will keep on writing). Proof. Let s be the number of
states of M. We construct an NCM M which, when given an input w over the stack alphabet of M, does the following: simulates the computation of M on ‘stay’ transitions while checking that w could be
written by M on the stack at some point during the computation of the writing phase of w, while also verifying that there is a subword x of w of length s + 1 such that x was written by M without: 1.
incrementing a counter that has so far been at zero, and 2. decrementing a non-zero counter. If so, M accepts w. Next, it will be argued that L ( M ) is not empty if and only if M has an infinite
writing phase on λ, and indeed this is decidable since emptiness for NCM is decidable [8]. If L ( M ) is not empty, then there is a sequence of s + 1 transitions during the writing phase where no
counter during this sequence is increased from zero, and no counter is decreased. Thus, there must be some state q hit twice by the pigeonhole principle, and the sequence of transitions between q and
itself must repeat indefinitely in M. Thus, M has an infinite writing phase on λ input. Conversely, assume M has an infinite writing phase. Then there must be a sequence of s + 1 transitions where no
counter is decreased, and no counter is increased from zero. Thus, L ( M ) must be non-empty. 2 From this, decidability of acceptance of λ is straightforward. Lemma 6. It is decidable, given a DCSACM
M satisfying the assumptions of Lemma 4, whether or not M accepts λ. Proof. From Lemma 5, we can decide if M has an infinite writing phase. If so, M will not accept λ (as the stack must return to the
bottom before accepting). If M does not have an infinite writing phase, the (final) word w written in the stack is unique and hence has a unique length d. In this case, we can simulate faithfully the
computation of M (on λ input) and determine d. We then construct a DCM M d , which on λ input, encodes the stack in the state and simulates M. Thus, M d needs a buffer of size d to simulate the
operation of the stack, and M d accepts if and only if M accepts. The result follows, since the membership problem for DCM is decidable [8]. 2
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O.H. Ibarra, I. McQuillan / Theoretical Computer Science ••• (••••) •••–•••
From Lemmas 1, 4, and 6: Proposition 7. For r ≥ 1, the membership problem for r-head 2DCSACM is decidable. We now give some undecidability results. The proofs will use the following result in [8]:
Proposition 8. [8] It is undecidable, given a 2DCM(2) M over a letter-bounded language, whether L ( M ) is empty. Proposition 9. The membership problem for NCSACM(2) is undecidable. Proof. Let M be a
2DCM(2) machine over a letter-bounded language. Construct from M an NCSACM M which, on λ input (i.e. the input is fixed), guesses an input w to M and writes it on its stack. Then M simulates the
computation of M by using the stack and two reversal-bounded counters and accepts if and only if M accepts. Clearly, M accepts λ if and only if L ( M ) is not empty which is undecidable by
Proposition 8. 2 By Propositions 2 and 9, the following is true: Corollary 10. The emptiness problem for DCSACM(2) is undecidable. Combining together the results thus far demonstrates that NCSACM is
a model where,
• • • •
the the the the
deterministic version has a decidable membership problem, deterministic version has an undecidable emptiness problem, nondeterministic version has an undecidable membership problem, nondeterministic
version has an undecidable emptiness problem.
Moreover, this is the first (to our knowledge) model where these properties hold. The next restriction serves to contrast this undecidability result. Consider an NCSACM where during the reading phase,
the stack head crosses the boundary of any two adjacent cells on the stack at most d times for some given d ≥ 1. Call this machine a d-crossing NCSACM. Then we have: Proposition 11. It is decidable,
given a d-crossing NCSACM M, whether or not L ( M ) = ∅. Proof. Define a d-crossing NTMCM to be an nondeterministic Turing machine with a one-way read-only input tape and a d-crossing read/write
worktape (i.e., the worktape head crosses the boundary between any two adjacent worktape cells at most d times) augmented with reversal-bounded counters. Note that a d-crossing NCSACM can be
simulated by a d-crossing NTMCM. It was shown in [3] that it is decidable, given a d-crossing NTMCM M, whether L ( M ) = ∅. The proposition follows. 2 Although we have been unable to resolve the open
problem as to whether the emptiness problem is decidable for both NCSACM and DCSACM with one reversal-bounded counter, as with membership for the nondeterministic version, we show they are all
equivalent to an open problem in the literature. Proposition 12. The following are equivalent: 1. 2. 3. 4. 5.
the emptiness problem is decidable for 2NCM(1), the emptiness problem is decidable for NCSACM(1), the emptiness problem is decidable for DCSACM(1), the membership problem is decidable for r-head
2NCSACM(1), it is decidable if λ is accepted by a NCSACM(1).
Proof. The last four properties are equivalent by Proposition 2. It can be seen that 2) implies 1) because a NCSACM(1) machine can simulate a 2NCM(1) machine by taking the input, copying it to the
stack, then simulating the 2NCM(1) machine with the two-way stack instead of the two-way input. Furthermore, it can be seen that 1) implies 5) as follows: given a NCSACM(1) machine M, assume without
loss of generality, that M immediately and nondeterministically sets the stack and returns to the bottom of the stack in read-only mode in some special state q before changing any counter (as it can
verify that M would have pushed the stack contents). Then, build a 2NCM(1) machine M that on some input over the stack alphabet, simulates the stack using the input, and the counter using the counter
starting at state q. Then L ( M ) is non-empty if and only if λ is accepted by M. 2
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O.H. Ibarra, I. McQuillan / Theoretical Computer Science ••• (••••) •••–•••
It is indeed a longstanding open problem as to whether the emptiness problem for 2NCM(1) is decidable [8]. Now consider the following three restricted models, with k counters: For k ≥ 1, a DCSACM(k)
(or a NCSACM(k)) machine is said to be:
• no-read/no-counter if it does not read the checking stack nor use any counter before hitting the right input end-marker, • no-read/no-decrease if it does not read the checking stack nor decrease
any counter before hitting the right input endmarker,
• no-read if it does not read the checking stack before hitting the right input end-marker. We will consider the families of DCSACM(k) (NCSACM(k)) machines satisfying each of these three conditions.
Proposition 13. For any k ≥ 1, every 2DCM(k) machine can be effectively converted to an equivalent no-read/no-decrease DCSACM(k) machine, and vice-versa. Proof. First, a 2DCM(k) machine M can be
simulated by a no-read/no-decrease DCSACM(k) machine M that first copies the input to the stack, and simulates the input of M using the checking stack, while simulating the counters faithfully.
Indeed, the checking stack is not read and counters are not decreased until M reads the entire input. Next we will prove the converse. Let M be a no-read/no-decrease DCSACM(k) machine with input
alphabet and stack alphabet . A two-way deterministic gsm, 2DGSM, is a deterministic generalized sequential machine with a two-way input (surrounded by end-markers), accepting states, and output. It
is known that if L is a language accepted by a two-way k-head deterministic machine augmented with some storage/memory structure (such as a pushdown, checking stack, k checking stacks, etc.), then T
−1 ( L ) = {x | T (x) = y , y ∈ L } is also accepted by the same type of machine [7]. Let T be 2DGSM which, on input x ∈ ∗ , first outputs x#. Then it moves to the left end-marker and on the second
sweep of input x, simulates M and outputs the string z written on the stack during the writing phase of M. Note that T can successfully do this as M generates the checking stack contents from
left-to-right, and does not read the contents during the writing phase; and because the counters of M are not decreased during the writing phase of M, the counters can never empty during the writing
phase, thereby affecting the checking stack contents created. When T reaches its right end-marker, it outputs the state s of M at that time, and then T enters an accepting state. Thus, T (x) = x#zs.
Now construct a 2DCM(k) M which when given a string x#zs, reads x, and while doing so, M simulates the writing phase of M on x by only changing the counters as M would do. Then, M moves to the right
and stores the state s in the finite control. Then M simulates the reading phase of M on string z (which only happens after the end of the input has been reached), starting in state s and the current
counter contents, and accepts if and only if M accepts. It is straightforward to see that T −1 ( L ( M )) = L, which can therefore be accepted by a 2DCM(k) machine. 2 From this, the following is
immediate, since emptiness for 2DCM(1) is known to be decidable [9]. Corollary 14. The emptiness problem for no-read/no-decrease DCSACM(1) is decidable. In the first part of the proof of Proposition
13, the DCSACM(k) machine created from a 2DCM(k) machine was also no-read/no-counter. Therefore, the following is immediate: Corollary 15. For k ≥ 1, the family of languages accepted by the following
three sets of machines coincide:
• all no-read/no-decrease DCSACM(k) machines, • all no-read/no-counter DCSACM(k) machines, • 2DCM(k). One particularly interesting corollary of this result is the following: Corollary 16. 1. The
family of languages accepted by no-read/no-decrease (respectively no-read/no-counter) DCSACM(1) is effectively closed under union, intersection, and complementation. 2. Containment and equivalence
are decidable for languages accepted by no-read/no-decrease DCSACM(1) machines. This follows since this family is equal to 2DCM(1), and these results hold for 2DCM(1) [9]. Something particularly
noteworthy about closure of languages accepted by no-read/no-decrease 2DCSACM(1) under intersection, is that, the proof does not follow the usual approach for one-way machines. Indeed, it would be
usual to simulate two machines in parallel,
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each requiring its own counter (and checking stack). But here, only one counter is needed to establish intersection, by using a result on two-way machines. Later, we will show that Corollary 16, part
2 also holds for no-read DCSACM(1)s. Also, since emptiness is undecidable for 2DCM(2), even over letter-bounded languages [8], the following is true: Corollary 17. The emptiness problem for languages
accepted by no-read/no-counter DCSACM(2) is undecidable, even over letterbounded languages. Turning now to the nondeterministic versions, from the first part of Proposition 13, it is immediate that
for any k ≥ 1, every 2NCM(k) can be effectively converted to an equivalent no-read/no-decrease NCSACM(k). But, the converse is not true combining together the following two facts: Proposition 18. 1.
For every k ≥ 1, the emptiness problem for languages accepted by 2NCM(k) over a unary alphabet is decidable. 2. The emptiness problem for languages accepted by no-read/no-counter (also for no-read/
no-decrease) NCSACM(2) over a unary alphabet is undecidable. Proof. The first part was shown in [9]. For the second part, it is known that the emptiness problem for 2DCM(2) M (even over a
letter-bounded language) is undecidable by Proposition 8. We construct a no-read/no-counter NCSACM(2) M which, on a unary input, nondeterministically writes some string w on the stack. Then M
simulates M using w. The result follows since L ( M ) = ∅ if and only if L ( M ) = ∅. 2 In contrast to part 2 of Proposition 18: Proposition 19. For any k ≥ 1, the emptiness problem for languages
accepted by no-read/no-decrease DCSACM(k) machines over a unary alphabet, is decidable. Proof. If M is a no-read/no-decrease DCSACM(k) over a unary alphabet, we can effectively construct an
equivalent 2DCM(k) M (over a unary language) from Proposition 13. The result follows since the emptiness problem for 2NCM(k) over unary languages is decidable [9]. 2 Combining these two results
yields the following somewhat strange contrast: Corollary 20. Over a unary input alphabet and for all k ≥ 2, the emptiness problem for no-read/no-counter NCSACM(k)s is undecidable, but decidable for
no-read/no-counter DCSACM(k)s. As far as we know, this demonstrates the first known example of a family of one-way acceptors where the nondeterministic version has an undecidable emptiness problem,
but the deterministic version has a decidable emptiness problem. This presents an interesting contrast to Proposition 2, where it was shown that for complete sets of automata for any store types, the
emptiness problem of the deterministic version is decidable if and only if it is decidable for the nondeterministic version. However, the set of unary no-read/no-counter NCSACM(k) machines can be
seen to not be a complete set of machines, as a complete set of machines contains every possible machine involving a store type. This includes those machines that read input letters while performing
read instructions on the checking stack. And indeed, to prove the equivalence of 1) and 2) in Proposition 2, the deterministic machine created reads a letter for every transition applied, which can
produce machines that are not of the restriction no-read/no-counter. With only one counter, decidability of the emptiness problem for no-read/no-decrease NCSACM(1), and for no-read/nocounter NCSACM
(1) can be shown to be equivalent to all problems listed in Proposition 12. This is because 2) of Proposition 12 implies each immediately, and each implies 1) of Proposition 12, as a 2NCM(1) machine
M can be converted to a no-read/no-decrease, or no-read/no-counter NCSACM(1) machine where the input is copied to the stack, and then the 2NCM(1) machine simulated. Therefore, it is open as to
whether the emptiness problem for no-read/no-decrease (or no-read/no-counter) NCSACM(1) is decidable, as this is equivalent to the emptiness problem for 2NCM(1). One might again suspect that
decidability of emptiness for no-read/no-decrease DCSACM(1) implies decidability of emptiness for no-read/no-decrease NCSACM(1) by Proposition 2. However, it is again important to note that
Proposition 2 only applies to complete sets of machines, including those machines that read input letters while performing read instructions on the checking stack, again violating the ‘no-read/
no-decrease’ condition. Even though it is open as to whether the emptiness problem is decidable for no-read/no-decrease NCSACM(1)s, we have the following result, which contrasts Corollary 16, part 2:
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Proposition 21. The universe problem is undecidable for no-read/no-counter NCSACM(1)s. (Thus, containment and equivalence are undecidable.) Proof. It is known that the universe problem for a one-way
nondeterministic 1-reversal-bounded one-counter automaton M is undecidable [10]. Clearly, we can construct a no-read/no-counter NCSACM(1) M to simulate M. 2 In the definition of a no-read/no-decrease
DCSACM, we imposed the condition that the counters can only decrement when the input head reaches the end-marker. Consider the weaker condition no-read, i.e., the only requirement is that the machine
can only enter the stack when the input head reaches the end-marker, but there is no constraint on the reversalbounded counters. It is an interesting open question about whether no-read DCSACM(k)
languages are also equivalent to a 2DCM(k) (we conjecture that they are equivalent). However, the following stronger version of Corollary 14 can be proven. Proposition 22. The emptiness problem is
decidable for no-read DCSACM(1)s. Proof. Let M be a no-read DCSACM(1). Let T = {t 1 , . . . , tm } be symbols in bijective correspondence with transitions of M that can occur in the writing phase.
Then, build a 2DCM(1) machine M that, on input w over T , reads w while changing states as w does, and changing the counter as the transitions do. Let q be the state where the last transition symbol
ends. Then, at the end of the input, M simulates the reading phase of M starting in q by scanning w, and interpreting a letter t of w as being the stack letter written by t in M (while always
skipping over a letter t if t does not write to the stack in M). Then L ( M ) is empty if and only if L ( M ) is empty. 2 We can further strengthen Proposition 22 somewhat. Define a restricted no-read
NCSACM(1) to be a no-read NCSACM(1) which is only nondeterministic during the writing phase. Then the proof of Proposition 22 applies to the following, as the sequence of transition symbols used in
the proof can be simulated deterministically: Corollary 23. The emptiness problem is decidable for languages accepted by restricted no-read NCSACM(1) machines. While we are unable to show that the
intersection of two no-read DCSACM(1) languages is a no-read DCSACM(1) language, we can prove: Proposition 24. It is decidable, given two no-read DCSACM(1)s M 1 and M 2 , whether L ( M 1 ) ∩ L ( M 2
) = ∅. Proof. Let M 1 and M 2 be no-read DCSACM(1) over input alphabet . Let T i be symbols in bijective correspondence with transitions of M i that can occur in the writing phase, for each i ∈ {1,
2}. Let T be the set of all pairs of symbols (r , s), where r is a transition of M 1 , s is a transition of M 2 , and where both r and s read the same input letter of . Let T be all those symbols (r
, $) where r is a transition of M 1 that stays on the input, and let T be all those symbols ($, s) where s is a transition of M 2 that stays on the input. Build a 2DCM(1) machine M operating over
alphabet T ∪ T ∪ T . On input w, M verifies that the first component changes states as M 1 does (skipping over any $ symbol) and that if a stay transition is read, the next letter has a first component
on the same input letter, and changing the counter as M 1 does. Let q be the state where the last transition symbol ends. Then, at the end of the input, M simulates the reading phase of M 1 starting
in q by scanning w, and interpreting a letter t = $ in the first component of w as being the stack letter written by t in M, and skipping over $ or any t that does not write to the stack. After
completion, then M does the same thing with M 2 using the second component. Notice that the alphabet is structured such that a transition of M 1 on a letter a ∈ is used exactly when a transition of M
2 using a ∈ is used, since M 1 and M 2 are both no-read (so their entire input is used before the reading phases starts). For example, a word w = (s1 , r1 )(s2 , $)(s3 , $)($, r2 )(s4 , r3 ) implies
s1 reads the same input letter in M 1 as does r1 in M 2 , similarly with s4 and r3 , s2 and s3 are stay transitions in M 1 , and r2 is a stay transition in M 2 . Hence, L ( M ) is empty if and only
if L ( M 1 ) ∩ L ( M 2 ) is empty. 2 One can show that no-read DCSACM(1) languages are effectively closed under complementation. Thus, from Proposition 24: Corollary 25. The containment and
equivalence problems are decidable for no-read DCSACM(1)s. No-read DCSACM(1) is indeed quite a large family for which emptiness, equality, and containment are decidable. The proof of Proposition 24
also applies to the following: Proposition 26. It is decidable, given two restricted no-read NCSACM(1)s M 1 and M 2 , whether L ( M 1 ) ∩ L ( M 2 ) = ∅.
JID:TCS AID:11554 /FLA
Doctopic: Algorithms, automata, complexity and games
[m3G; v1.235; Prn:25/04/2018; 15:18] P.11 (1-12)
O.H. Ibarra, I. McQuillan / Theoretical Computer Science ••• (••••) •••–•••
Finally, consider the general model DCSACM(1) (i.e., unrestricted). While it is open whether no-read DCSACM(1) is equivalent to 2DCM(1), we can prove: Proposition 27. L(2DCM(1)) L(DCSACM(1)). Proof.
It is obvious that any 2DCM(1) can be simulated by a DCSACM(1) (in fact by a no-read/no-counter DCSACM(1)). Now let L = {ai b j ck | i , j ≥ 1, k = i · j }. We can construct a DCSACM(1) M to accept L
by Example 2. However, it was shown in [11] that L cannot be accepted by a 2DCM(1) by a proof that shows that if L can be accepted by a 2DCM(1), then one can use the decidability of the emptiness
problem for 2DCM(1)s to show that Hilbert’s Tenth Problem is decidable. 2 4. Multiple checking-stacks with reversal-bounded counters In this section, we will study deterministic and nondeterministic
k-checking-stack machines. These are defined by using multiple checking stack stores. Implied from this definition is that each stack has a “writing phase” followed by a “reading phase”, but these
phases are independent of each letter for each stack. A k-stack DCSA (NCSA respectively) is the deterministic (nondeterministic) version of this type of machine. The two-way versions (with input
end-markers) are called k-stack 2DCSA and k-stack 2NCSA, respectively. These k-stack models can also be augmented with reversal-bounded counters and are called k-stack DCSACM, k-stack NCSACM, k-stack
2DCSACM, and k-stack 2NCSACM. Consider a k-stack DCSACM M. By Lemma 1, for the membership problem, we need only investigate whether λ is accepted. Also, as in Lemma 4, we may assume that each stack
pushes a symbol at each move during its writing phase, and that all counters are 1-reversal-bounded. We say that M has an infinite writing phase (on λ input) if no stack enters a reading phase. Thus,
all stacks will keep on writing a symbol at each step. If M has a finite writing phase, then directly before a first such stack enters its reading phase, all the stacks would have written strings of
the same length. Lemma 28. Let k ≥ 1 and M be a (k + 1)-stack DCSACM M satisfying the assumption of Lemma 4. 1. We can determine if M has an infinite writing phase. If so, M does not accept λ. 2. If M
has a finite writing phase, we can construct a k-stack DCSACM M satisfying the assumption of Lemma 4 such that M accepts λ if and only if M accepts λ. Proof. Let M have s states and stack alphabets 1
, . . . , k+1 for the k + 1 stacks. Let = {[a1 , . . . , ak+1 ] | ai ∈ i , 1 ≤ i ≤ k + 1}. By assumption, each stack of M writes a symbol during its writing phase. We can determine if M has a finite
writing phase as follows: As in Lemma 5, we construct an NCM M which, when given an input w ∈ ∗ , does the following: simulates the computation of M on ‘stay’ transitions such that the input w was
written by M (in a component-wise fashion on each checking stack) and there is a subword x of w of length s + 1 such that the subword was written by M without: 1. incrementing a counter that has so
far been at zero, and 2. decrementing a non-zero counter. If so, M accepts w. So we need only check if L ( M ) is not empty, which is decidable since emptiness is decidable for NCM [8]. Then, M does
not accept λ if and only if M has an infinite writing phase, and if and only if L ( M ) is not empty, which is decidable. If L ( M ) is empty, we then simulate M faithfully to determine the unique
word w ∈ ∗ and its length d just before the reading phase of at least one of the stacks, say S i , is entered. Note that by construction, no stack entered its stack earlier. We then construct a
k-stack DCSACM M which, on λ input, encodes the operation of stack S i in the state and simulates M (also converted into satisfying the assumptions of Lemma 4). Thus, M needs a buffer of size d to
simulate the operation of stack S i . M accepts if and only if M accepts, and has one less stack than M. 2 Notice that M has fewer stacks than M. Then, from Proposition 7 (the result for a single
stack) and using Lemma 28 recursively: Proposition 29. The membership problem for k-stack DCSACMs is decidable. Then, by Lemma 1: Corollary 30. The membership problem for r-head k-stack 2DCSACM is
JID:TCS AID:11554 /FLA
Doctopic: Algorithms, automata, complexity and games
[m3G; v1.235; Prn:25/04/2018; 15:18] P.12 (1-12)
O.H. Ibarra, I. McQuillan / Theoretical Computer Science ••• (••••) •••–•••
Table 1 Summary of decision problem results, with the left half being nondeterministic machines, and the right half being the corresponding deterministic machines. The decision problems are listed on
the columns. A is placed when the problem is decidable for the class of machines, a × is placed when the property is undecidable, and a ? is placed when the problem is still open. In all cases where
it is open, the decidability is equivalent to the open problem of whether emptiness is decidable for 2NCM(1). The theorem proving each is listed. Nondeterministic class
Deterministic class
? Proposition 12
? Proposition 12
Proposition 7
? Proposition 12
DCSACM(k), k ≥ 2
Proposition 7
Corollary 10
NCSACM(k), k ≥ 2
Proposition 9
Proposition 9
r-head k-stack 2NCSACM
Proposition 9
r-head k-stack 2DCSACM
Proposition 9
Corollary 30
Corollary 10
no-read/no-decrease NCSACM(1)
? page 9
? page 9
no-read/no-decrease DCSACM(1)
Proposition 7
Corollary 14
no-read/no-counter NCSACM(2)
Proposition 18
no-read/no-counter DCSACM(2)
Proposition 18
Proposition 7
Corollary 17
This is one of the most general machine models known with a decidable membership problem. Although space complexity classes of Turing machines are also very general, the membership problem for both
deterministic and nondeterministic Turing machines satisfying some space complexity function are both decidable. However, for NCSACMs, membership is undecidable but is decidable for deterministic
machines. Moreover, unlike space-bounded Turing machines, r-head k-stack 2DCSACMs do not have a space restriction on their stacks. 5. Conclusions We introduced several variants of checking stack
automata and showed the difference between the deterministic and nondeterministic models with respect to the decidability of the membership and emptiness problems. The main decision problems are
summarized in Table 1. We believe the contrasting results obtained are the first of its kind. An interesting open question is the status of the emptiness problem for nondeterministic checking stack
automata augmented with one reversal-bounded counter which can only read the stack and decrease the counter at the end of the input. As shown in the paper, this problem is equivalent to a
long-standing open problem of whether emptiness for two-way nondeterministic finite automata augmented with one reversal-bounded counter is decidable. Furthermore, we investigated possible scenarios
that can occur when augmenting a machine model accepting non-semilinear languages with reversal-bounded counters. This contrasts known results on models accepting only semilinear languages.
Acknowledgements We thank the Editor and the referees for their expeditious handling of our paper and, in particular, the referees for their comments that improved the presentation of our results.
References [1] S. Ginsburg, Algebraic and Automata-Theoretic Properties of Formal Languages, North-Holland Publishing Company, Amsterdam, 1975. [2] O. Ibarra, I. McQuillan, On store languages of
language acceptors, submitted for publication, a preprint appears in https://arxiv.org/abs/1702.07388, 2017. [3] T. Harju, O. Ibarra, J. Karhumäki, A. Salomaa, Some decision problems concerning
semilinearity and commutation, J. Comput. System Sci. 65 (2) (2002) 278–294. [4] M. Harrison, Introduction to Formal Language Theory, Addison–Wesley Series in Computer Science, Addison–Wesley Pub.
Co., 1978. [5] S. Greibach, Checking automata and one-way stack languages, J. Comput. System Sci. 3 (2) (1969) 196–217. [6] J.E. Hopcroft, J.D. Ullman, Introduction to Automata Theory, Languages, and
Computation, Addison–Wesley, Reading, MA, 1979. [7] J. Engelfriet, The power of two-way deterministic checking stack automata, Inform. and Comput. 80 (2) (1989) 114–120. [8] O.H. Ibarra,
Reversal-bounded multicounter machines and their decision problems, J. ACM 25 (1) (1978) 116–133. [9] O.H. Ibarra, T. Jiang, N. Tran, H. Wang, New decidability results concerning two-way counter
machines, SIAM J. Comput. 23 (1) (1995) 123–137. [10] B.S. Baker, R.V. Book, Reversal-bounded multipushdown machines, J. Comput. System Sci. 8 (3) (1974) 315–332. [11] E.M. Gurari, O.H. Ibarra,
Two-way counter machines and diophantine equations, J. ACM 29 (3) (1982) 863–873. | {"url":"https://c.coek.info/pdf-variations-of-checking-stack-automata-obtaining-unexpected-decidability-properti841be16ebd88e74915cfd6b80730b4b030908.html","timestamp":"2024-11-11T04:39:54Z","content_type":"text/html","content_length":"96297","record_id":"<urn:uuid:55ecdea2-8789-447b-bb72-cc9af9de4599>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00007.warc.gz"} |
Loop for adding value to last n values in list
Asked by Canaan Daugherty on
5 Answers
Answer by Kylie Kerr
Deleting list members by rule ,I have a list where I would like to +1 to all the values after n in a loop, where n is a row index. I want to repeat this multiple times to achieve the following:,If
you are counting from 1 (instead of 0), then change i+1 to i-1 and your original_list becomes [1, 2, 4, 5, 6, 8, 9].,You don't need to do any fancy list slicing actually. You can accomplish this with
an additional index, j, that goes from i + 1 (after n) to the len of the original_list.
original_list = np.array([1,2,3,4,5,6,7])
multiples = np.array([3,6])
cs = np.zeros_like(original_list)
multiples = np.array(multiples) - 1 # since you are not using 0-index
cs[multiples] = 1
original_list + cs.cumsum()
array([1, 2, 4, 5, 6, 8, 9])
Source: https://stackoverflow.com/questions/67451785/loop-for-adding-value-to-last-n-values-in-list
Answer by Aliyah Medina
Method #2 : Using islice() + reversed()The inbuilt functions can also be used to perform this particular task. The islice function can be used to get the sliced list and reversed function is used to
get the elements from rear end. ,Accessing elements in a list has many types and variations. These are an essential part of Python programming and one must have the knowledge to perform the same.
This article discusses ways to fetch the last N elements of list. Let’s discuss certain solution to perform this task.Method #1 : Using list slicingThis problem can be performed in 1 line rather than
using a loop using the list slicing functionality provided by Python. Minus operator specifies slicing to be done from rear end. ,Python | Check if all elements in a list are identical,Python –
Common list elements and dictionary values
Source: https://www.geeksforgeeks.org/python-get-last-n-elements-from-given-list/
Answer by Etta Bradley
l = [1, 2, 3]
Source: https://www.codegrepper.com/code-examples/python/how%2Bto%2Bget%2Blast%2Bn%2Belements%2Bof%2Ba%2Blist%2Bin%2Bpython
Answer by Zhuri Lugo
numbers[-1] is the last element of the list, numbers[-2] is the second to last, and numbers[-3] doesn’t exist.,It is common to use a loop variable as a list index.,We can use the not in combination
with in to test whether an element is not a member of a list:,The easiest way to clone a list is to use the slice operator:
[10, 20, 30, 40]
["spam", "bungee", "swallow"]
Source: https://www.openbookproject.net/thinkcs/python/english2e/ch09.html | {"url":"https://www.devasking.com/issue/loop-for-adding-value-to-last-n-values-in-list","timestamp":"2024-11-13T06:32:29Z","content_type":"text/html","content_length":"125173","record_id":"<urn:uuid:9aa02168-f290-41d2-8789-4a771340e1c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00525.warc.gz"} |
GeolOil - How to calculate fracture porosity from well logs
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GeolOil - How to calculate fracture porosity from well logs
Fracture porosity is the ratio between the volume of open fractures and the rock's bulk volume. The fracture porosity, if present, is one of the contributions to the rock's porosity. The other
porosity contributions are other sources of secondary porosity like vugs or vugular porosity, effective matrix porosity, and non effective matrix porosity like clay cement porosity.
Secondary porosity is classically estimated from well logs as the difference of total density porosity minus total sonic porosity φ[f] ≅ φ[d] - φ[s]. It is applicable to either carbonate rocks or
clastic sandstones that may or may not contain clays. The porosity difference should be similar to that of effective density porosity minus the effective sonic porosity. If the only source of
secondary porosity are fractures, then the porosity difference could be a good estimator of the fracture porosity.
Since the sonic speed is higher for the solid rock's fabric or matrix (it's denser than fluid filled pores and fracture gaps), the compressive sonic pulse or wave is mostly transmitted through the
matrix. However, the bulk density log and its derived density porosity curve, measures all the porosity spaces. That's why the difference φ[d] - φ[s] is an estimator to the secondary porosity and
fracture porosity. If the rock is shaly and the effective density porosity is not available, it can be replaced by the effective neutron porosity log, but only after it is converted to the correct
matrix mineral blend.
Density - sonic porosity estimates fracture porosity: (brown + blue pore volume) - (brown pore volume) = blue pore volume
Fracture porosity estimation from the density porosity - sonic porosity difference φ[d] - φ[s] requires good quality logs, knowledge of the rock's minerals, and some properties. The figure below
shows the GeolOil Porosity Functions panel equations for:
• Density porosity
• Wyllie sonic porosity
• Raymer sonic porosity
• Modified Raymer sonic porosity
• Effective Porosity
• Secondary Porosity (Fracture porosity if there are no vugs)
The figure below shows the GeolOil porosity functions panel.
The Wyllie sonic porosity equation is one of most well known models to estimate sonic porosity. However, it depends upon the sonic travel time of pore fluid, which may be unknown if the fluid column
is variable, or if there is a transition zone with a blend of fluids. The modified Raymer sonic porosity removes the fluid dependency and could behave as a robust estimator.
The most critical parameter to calculate sonic porosity is perhaps the matrix compressional travel time Δt[matrix]. It is not only usually unknown, but in many reservoirs is not a constant parameter,
but a curve that depends on the variable column lithology, its matrix fabric, and rock's stiffness. Since typical values of fracture porosity barely range 0%-4%, its computation may be unstable,
requiring good quality curves, and precise parameters.
The following methodology provides some guidance to achieve a reasonable estimation of fracture porosity from logs:
1. Calculate a first draft of fracture porosity as the algebraic difference φ[d] - φ[s] without trimming or discarding any negative estimates.
2. Name this difference curve itself as Quality indicator curve Q(md)
3. If Q(md) yields too big negative values, or too big positive values, the fracture porosity estimation is completely unreliable and should be discarded at least for the problematic depths.
4. If Q(md) absolute value magnitude is not so big, try to shift it to the right to remove most of the negative estimates by slightly tunning the matrix compressional travel time Δt[matrix]
5. Finally, replace all negative values with 0 (no fractures at such depths), and trim values exceeding 4% to 4%. Make sure only few depths exceeds 4%, otherwise discard the fracture porosity
estimation as it looks unreliable large.
Estimation of fracture porosity with a quality index curve for a gas reservoir
Another estimation of fracture porosity: the more plastic the rock, less fractures
⚠ NOTE: The techniques presented in this article can only estimate fractures at the log resolution scale —that is, fractures that may be enclosed in the log scale of around few feet—. Any fractures
out of this volume would be overlooked, given the false impression that the reservoir has no fractures. ∎ | {"url":"https://geoloil.com/fracturePorosity.php","timestamp":"2024-11-09T13:31:07Z","content_type":"text/html","content_length":"52571","record_id":"<urn:uuid:73199457-377c-40a1-8a68-b61b398547c3>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00068.warc.gz"} |
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Theory and Publications
Hello all.
The last few months I had to stop my practical work for writing some reports and documentation.
The first part in my documentation was, to introduce IMUs in general, then the ones I use, the mathematic of quaternions and so on and so on.
the next thing i would like to do, is to show how multibody dynamics work, how quaternions are calculated out of imu data, and to calculate the kneeangle as example.
everything refers to opensense.
now i found some papers in theors and publications, but these are just for multibody dynamics.
I hoped to find some papers about the background of opensense. e.g. sensor position, syncing the differnet data types.
If you have anything publishes so far, I would be very thankfull if you let me know.
Hams-Jochen Edenhart
Re: Theory and Publications
There is no OpenSense paper yet.
There are plenty of papers on Pubmed that describe how IMUs compute their own quaternion orientation-- I would suggest doing a literature review there on data fusion of IMUs.
OpenSense doesn't compute quaternions-- it uses the available quaternion information from the IMU and transforms the quaternion to a Rotation object using Simbody. | {"url":"https://simtk.org/plugins/phpBB/viewtopic.php?p=32380&sid=ab4d61f7e0db8c30522259aa6e1062c1","timestamp":"2024-11-14T01:16:01Z","content_type":"text/html","content_length":"16958","record_id":"<urn:uuid:8afc9be4-924d-4611-b62f-f82c5e3beb30>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00757.warc.gz"} |
Researcher Directory
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Condensed Matter Theory
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Hatakenaka Noriyuki Interdisciplinary science and engineering; Applied physics; Optical engineering, Photon science Bell inequalities
Higashitani Seiji Mathematical and physical sciences; Physics; Condensed matter physics II
Mathematical and physical sciences; Physics; Condensed matter physics I
Higuchi Katsuhiko Mathematical and physical sciences; Physics; Condensed matter physics II Solid state physics / band theory / Density-functional theory
Mathematical and physical sciences; Physics; Mathematical physics / Fundamental condensed matter
ISHIZAKA SATOSHI Mathematical and physical sciences; Physics; Mathematical physics / Fundamental condensed matter quantum cryptography / quantum entanglement / quantum information
Mathematical and physical sciences; Physics; Mathematical physics / Fundamental condensed matter
MUNEJIRI Shuji physics Physics Education Research / Physics education
Complex systems; Science education / Educational technology; Science education
Nagato Yasushi Mathematical and physical sciences; Physics; Condensed matter physics II Superfluid / Superconductor / Rough Interface
Shimahara Hiroshi Mathematical and physical sciences; Physics; Condensed matter physics II Superconductivity / Solid State Physics / Theoretical Physics
TADA YASUHIRO Mathematical and physical sciences; Physics; Condensed matter physics II quantum many-body systems, superconductivity, magnetism, topological
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To Prove Atriangle Is Isosceles Without The Measures Of Each Side , You Need To Know The?
You can use one of the following properties to prove that a triangle is isosceles without knowing the measure of each side.
How to prove isosceles traiangle
Angle-Angle (AA): A triangle is equal if two angles of one triangle meet (equal in measure) two angles of another triangle. For an isosceles triangle, if you can prove that two angles are equal, then
the triangle must be an isosceles triangle. This is because two right angles are the defining feature of an isosceles angle, as opposed to two congruent sides.
Base Angles: If the base angles of a triangle are equal (equal in measure), then the triangle is isosceles. This is because in an isosceles triangle, the base angles are always congruent.
Vertex angle bisection: If the vertex angle bisector of a triangle is also a perpendicular bisector of the opposite side (base), then the triangle is isosceles. This is because in an isosceles
triangle, the two sides of the vertex angle always bisect the base and are perpendicular.
Read more on triangles here:https://brainly.com/question/1058720
1. The area of square C is 625 ft². 2. The perimeter of square C is 46 ft. 3. The area of square B is 1456 ft². 4. The length of square A is 4√15 ft.
What is Pythagoras theorem?
A key idea in geometry that connects the lengths of a right triangle's sides is known as the Pythagorean Theorem. It says that the hypotenuse's square length, which is the side that faces the right
angle, is equal to the sum of the squares of the lengths of the other two sides in a right triangle. It has the following mathematical expression:
a² + b² = c²
1. The area of square C can be found using the Pythagorean Theorem:
a² + b² = c²
Using the area the side is calculates as follows:
Side length of A = √225 = 15 ft
Side length of B = √400 = 20 ft
15² + 20² = c²
225 + 400 = c²
625 = c²
c = √625 = 25 ft
So the area of square C is:
Area of C = 25² = 625 ft²
2. The perimeter of square C can be found by adding up the side lengths of all three squares:
Perimeter of A = 36 ft, so each side length of A is 9 ft.
Perimeter of B = 48 ft, so each side length of B is 12 ft.
Perimeter of C = Side length of A + Side length of B + Side length of C
Perimeter of C = 9 + 12 + c
We found earlier that c = 25 ft, so we can substitute that in:
Perimeter of C = 9 + 12 + 25 = 46 ft
3. The area of square B can be found using the fact that the areas of squares A, B, and C are related by the equation:
Area of A + Area of B = Area of C
We know the area of A and the area of C, so we can solve for the area of B:
Area of A = 15² = 225 ft²
Area of C = 1681 ft²
Area of A + Area of B = Area of C
225 + Area of B = 1681
Area of B = 1456 ft²
4. Perimeter of B = 64 ft
Side length of B = 64 / 4 = 16 ft
Now we can use the Pythagorean Theorem to find the side length of C:
a² + b² = c²
a = 15 ft (side length of A)
b = 16 ft (side length of B)
c = √(a² + b²) = √(15² + 16²) = √481 = 22 ft
Finally, we can use the Pythagorean Theorem again to find the side length of A:
a² + b² = c²
b = 16 ft (side length of B)
c = 22 ft (side length of C)
a = √(c² - b²) = √(22² - 16²) = √240 = 4√15 ft
Learn more about Pythagoras Theorem here: | {"url":"https://brightideas.houstontx.gov/ideas/to-prove-abr-br-triangle-is-isosceles-without-the-measures-o-nnfp","timestamp":"2024-11-14T17:52:40Z","content_type":"text/html","content_length":"98519","record_id":"<urn:uuid:428abfe6-fd2a-40a6-a338-04877d31363c>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00576.warc.gz"} |
Rachael Morgan
• Sun October 7 - Lunchtime: Buy a new beanie hat
• Mon October 8 - 6pm: Go rollerblading
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Linear equations
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Linear grapghs are used to represent a group of data, they can be created through finding the coordinates/positions of each point by folowing the rule provided such as y=3x+8. Then each point is
joined up to form a perfectly straight line. In order to plot the graph, you must work out the point through the rule, the rule is broken down in different steps: 1. Y can equl any number but is
usually used as -3, -2, -1 0 1, 2 & 3 2. Then to find x, you multiply it by the number next to it in the rule, in this case it would be 3 3. you then add the end number to the number you have, and
that it x 4. when all of the coordinates are found, you plot them on the graph and join them. The equation y=3x+8 has the results...
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Triganommetry is used to find either missing sides or angles in triangles In order to answer triganometry questions, you must follow a formular that shows which buttons you must press on the
calcuator, either sin, cos or tan, this formular is: o a o s h c h t a soh, shows that you would have a letter (usually x) or number on either the hypotonuse or the opposite line to the angle, then
in order to find the opposite you enter (sin(ang)) x the hypotonuse lenght = opposite length. To find the hypotenuse, you do the same formular , but you multiply the sin by the opposite lenght which
equals the hypotenuse. In order to find the angle, you must divide the opposite by the hypotenuse, then enter (cos-1) and press = . In order to solve other problems using triganometry, you use the
same formular but using different basic formuars (soh cah toa) o = opposite length to the angle h = hypotenuse a = adjacent length to the angle
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in order to factorise an expression, you must identlfy the numerical similarities in the numbers given. An example of this would be the similarietes of 15 and 10, they are both multipuls of 5. There
for in order to to solve: 15x+10 we identify the similarieties and show them by placing the 5 outside some bracets to show that it is a multipul and the most common factor of the to numbers of the
nummbers inside, 5 x 3 equals 15 and 5 x 3 equals 10, therefor 15x+10 becomes 15x+10=5(3x)+5(2) then in order to finish the equation, we put the two bracets together into one so the answer reads:
Can you answer this question?
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Title : Classifying Connected f-Factor Problems based on f
Speaker : Rahul C S (IITM)
Details : Tue, 12 May, 2015 3:00 PM @ BSB 361
Abstract: : A graph is a mathematical structure widely used for modeling networks and in many other applications.
Finding subgraphs satisfying degree and connectivity constraints is a well studied problem over many
years. Given an undirected graph G=(V,E) having n vertices and a function f:V->mathbb{N}$, an $f$-factor
H is a spanning subgraph such that degree of each vertex v in H is $f(v)$, for every v in V. We consider
the problem of computing an $f$-factor which is connected. We observe that the computational hardness of
the problem vary depending on the smallest value in the range of $f$. When $f(v)$>=c for every v in V
and some constant c, the problem is shown to be NP-Complete. When $f(v)$>=|V|/2 for every v in V, the
problem is polynomial time solvable. Thus, when the lower bound on the range of $f$ is too small the
problem is hard and when it is too large, the problem is easy. Motivated by this observation, we attempt
to explore the codomain of $f$ and understand how the hardness of the problem vary along with change in
$f$. Further we come up with an algorithm and a hardness result.
We had shown that the problem of computing a connected $f$-factor is hard even when $f(v)$>= n^
{1-epsilon} for some constant 0= n/3 for every v in V.
In this talk we present a polynomial time algorithm for computing connected $f$-factor where $f(v)$>= n/
c for every v in V and some constant c. The same algorithm can be used to solve for the case where $f(v)
$>= n/polylog(n) for every v in V in time n^{O(polylog(n))}$. This means the problem cannot be
NP-Complete unless Exponential Time Hypothesis(ETH) is false. Similarly, we give a subexponential time
reduction from Hamiltonian cycle problem where $f(v)$>= n/{log^{1+epsilon}} n(epsilon>0) for each vertex
v. This implies there does not exist a polynomial time algorithm for the problem unless ETH is false.
Thus, this infinite class of problems parameterized by $epsilon$ is in NP-Intermediate. | {"url":"https://cse.iitm.ac.in/seminar_details.php?arg=NTM=","timestamp":"2024-11-07T03:53:59Z","content_type":"application/xhtml+xml","content_length":"15002","record_id":"<urn:uuid:bc2e6efe-d260-488a-9dca-0623a040673d>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00380.warc.gz"} |
CRMC - Practice Assessment Q 1.7 (b)
broom123 Registered Posts: 11 Regular contributor ⭐
How is the simple annual interest rate of the discount calculated?
Thanks for your help.
Why is there only one practice assessment??
• Dear broom123
How is the simple annual interest rate of the discount calculated?
In your question your company wants the clients to pay within 14 days rather than within the normal credit period of 28 days.
This is reduced credit period of 14 days.
As a reward for paying early your company discounts the total amount payable by clients by 5% and will receive 95% of the total invoice value. This constitutes a discount on the amount received
of 5/95 = just over 5.26%
To covert 5.26% for 14 days into an annual amount, find the number of 14 day periods in a year 365/14 = 26.07
And multiply 5.26 by 26.07 to find 137.13%. On a spreadsheet I got 137.218% because I'd not rounded the numbers, but your examiner will not set mulitple choice questions where this difference
exists. So, don't worry about it
As a formula simple annual interest rate of a early payment discount
Number of times the reduced credit period exists in a year x the discount as a % of the money actually received = rate | {"url":"https://forums.aat.org.uk/Forum/discussion/38584/crmc-practice-assessment-q-1-7-b","timestamp":"2024-11-12T07:11:01Z","content_type":"text/html","content_length":"286393","record_id":"<urn:uuid:fa5b9be0-9d2a-4c07-b671-f3c830e78ca4>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00066.warc.gz"} |
Math Colloquia - What happens inside a black hole?
Black holes are perhaps the most celebrated predictions of general relativity. Miraculously, these complicated spacetimes arise as explicit (i.e., exact expression can be written down!) solutions to
the vacuum Einstein equation. Looking these explicit black hole solutions leads to an intriguing observation: While the black hole exterior look qualitatively similar for every realistic black hole,
the structure of the interior, in particular the nature of the `singularity' inside the black hole, changes drastically depending on whether the black hole is spinning (Kerr) or not (Schwarzschild).
A proposed picture for what happens in general is the so-called strong cosmic censorship conjecture of R. Penrose, which is a central conjecture in general relativity. In this colloquium, I will give
a short introduction to general relativity and explain what this conjecture is. Then I will describe my recent joint work with J. Luk (Stanford Univ.), where we rigorously establish a version of
strong cosmic censorship for the Einstein-Maxwell-(real)-scalar-field system in spherical symmetry, which has long been studied by physicists and mathematicians as a model problem for the conjecture. | {"url":"http://my.math.snu.ac.kr/board/index.php?mid=colloquia&sort_index=speaker&order_type=asc&page=7&document_srl=772036&l=en","timestamp":"2024-11-13T01:02:45Z","content_type":"text/html","content_length":"46118","record_id":"<urn:uuid:f0d6c39f-c6b6-4043-b81c-1dd9d83f4fd2>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00019.warc.gz"} |
ball mill grinding simulation
Other subsignals cannot be provided with reasonable physical explanations considering the insufficient analysis on the ball mill grinding mechanism. Thus, numerical simulation and domain
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A method for simulating the motion of balls in tumbling ball mill under wet condition is investigated. The simulation method is based on the threedimensional discrete element method (DEM) and
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His approach has two key concepts: the breakage rate and the mill residence time. The model considers the mill equivalent to several grinding stages with internal classification in series,
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The grinding circuit, represented i Figure 1, comprises a semiautogenous (SAG) mill and a ball mill (BM). Both mills operate in close circuit with separate ydrocyclone clusters. Water is added
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1. Fill the container with small metal balls. Most people prefer to use steel balls, but lead balls and even marbles can be used for your grinding. Use balls with a diameter between ½" (13 mm)
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Planetary ball mills and stirred media mills are some of the popular ultrafine grinding equipment [ 1 ]. A planetary mill typically consists of four rotating pots installed on a revolving disk.
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Abstract. Talc powder samples were ground by three types of ball mill with different sample loadings, W, to investigate rate constants of the size reduction and structural change into the
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The simulation results demonstrate that the proposed method has a better disturbance rejection property than those of the MPC and PI methods in controlling ball mill grinding circuits.
Introduction. Ball mill grinding circuit is one of the most important mineral processing units in concentrator plants. The product particle size of the grinding ...
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Grinding tests were conducted using a pearshaped ball mill on an oxidized coppercobalt ore to determine the milling parameters. Twelve monosized fractions of the ore sample were prepared and ...
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In the ball mill, the main grinding method was the impact of the steel ball medium with the finer ore particles, ... Hulthen, E.; Asbjornsson, G. Dynamic modeling and simulation of a SAG
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The simulator implements the dynamic ball mill grinding model which formulates the dynamic responses of the process variables and the product particle size distribution to disturbances and
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Ball mill is a type of rotating drum, mainly used for grinding and the particle motion in a ball mill is a major factor affecting its final product. However, understanding of particle motion in
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Vertical stirred mills have been widely applied in the minerals industry, due to its greater efficiency in comparison with conventional tumbling mills. In this context, the agitator liner wear
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Nevertheless, as stated by Chen et al. (2007), a ball mill grinding circuit is essentially a multipleinputmultipleout (MIMO) system with strong coupling among process variables. ... The grind
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Open Circuit Grinding. The object of this test was to determine the crushing efficiency of the ballmill when operating in open circuit. The conditions were as follows: Feed rate, variable from 3
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This confirms that the grinding process of the ball mill follows the firstorder kinetics, and the particle size decays exponentially with time. ... Table 3 shows DEM simulation time for different
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Section snippets Laboratoryscale batch grinding tests. The tests were made in a steel cylindrical ball mill of 250 mm internal diameter (D T) and 250 mm length (L T) fitted with eight
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The article discusses the study of the possibility of applying nonstandard mill MSL14K for the Bond ball mill test. As a result of ore grinding experiments, the Bond ball mill work index was ...
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Grinding experiments were designed and executed by a laboratory ball mill, considering ball size, ball charge and solid content as variables. Grinding tests were performed changing these three
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standard Bond ball mill in time intervals of min, 1 min, 2 min, and 4 min. After each grinding run, the mass of the sample is measured, and the PSD is determined.
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Introduction where W is the specific energy for grinding in kW h/ton, Wi is the work index in kW h/ton, P80 is the mill product d80 Grinding process mathematical modelling and com size in lm and
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The objective of this paper is to investigate the scaleup method of the planetary ball mill by the computational simulation based on Discrete Element Method. Firstly, the dry grinding of a ...
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A laboratory scale planetary ball mill ( Retsch PM400) was equipped with a test rig which enables the observation and recording of the grinding ball motion inside the grinding chamber. A high
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The grinding process in the ball mill is due to the centrifugal force induced by the mill on the balls. This force depends on the weight of the balls and the ball mill rotational speed. ... J.
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Excessive mill speeds caused more power consumption but resulted in reduced grinding rate. Based on the simulation data, two scaleup models were proposed to predict power draw and grinding rate.
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Using a 254mmdiameter instrumented ball mill, a series of batchgrinding experiments was carried out dry by Herbst and Fuerstenau, 1968, Herbst and Fuerstenau, 1973 with 7×9 mesh (× mm) dolomite
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Math 340 Home
Textbook Contents
Online Homework Home
Numerical Methods For Higher Order Equations
Higher-Order Equations and First-Order Systems
In the last chapter we learned how to use the geometric interpretation of the solution of a first order equation as the integral curve following a slope field to compute numerical approximations to
initial value problems, even when we couldn't find the exact solution. The situation is more complicated for higher order equations because now instead of just dealing with slope, $dy/dx$, the
presence of terms like $d^2y/dx^2$ means a geometric analysis would include more involved concepts like curvature. Fortunately we can avoid this by transforming higher-order equations into
first-order systems, and then sticking to the methods we already know. Consider the equation $y''+3y'+2y=0$. We can convert this to a pair of first-order equations by introducing $v=dy/dx$. Then $dv/
dx=d^2y/dx^2$ and the equation becomes $$ y'' + 3y' + 2y = 0& \\ \frac{dv}{dx} + 3v + 2y = 0& \\ \frac{dv}{dx} = -3v-2y& $$ With this we see the second-order equation $y''+3y'+2y=0$ is the same as
first-order system
$$ \frac{dy}{dx}&=v \\ \frac{dv}{dx} &= -3v-2y $$ If you have initial values in the original equation, you just convert them to initial values for the system by making the same substitution. This
technique of converting a second-order equation to a first-order system works for general equations, not just linear equations. If we have the initial value problem $\displaystyle\frac{d^2y}{dx^2}=f
(x,y,y')$, $y(0)=y_0$, $y'(0)=y_1$, then this can be converted to the first-order system $$ \frac{dy}{dx}&=v,\qquad &y(0)=y_0 \\ \frac{dv}{dx}&=f(x,y,v),\qquad &v(0)=y_1 $$ We can also convert
higher-order equations to first order systems by making multiple substitutions. Consider the third-order equation $y'''+y''-2y'+xy=e^x$. Letting $v=dy/dx$ and $w=dv/dx=d^2y/dx^2$ we get $$ \frac{dy}
{dx}&=v \\ \frac{dv}{dx}&=w \\ \frac{dw}{dx}&=e^x-xy+2v-w $$ The technique of using Picard iteration to show that a "nice" first-order initial value problem has a solution applies to first order
systems as well. You just have to make the integral in the Picard iteration vector-valued. You can see the proof of the following theorem in Math 540.
If $f(x,t_0,t_1,\ldots,t_n)$ and all its first partial derivatives are continuous in $(-h,h)\times(y_0-h,y_0+h)\times(y_1-h,y_1+h)\times\cdots\times(y_{n-1}-h,y_{n-1}+h)$ for some $h>0$, then there
is an $\epsilon>0$ such that the initial value problem $$ \frac{d^ny}{dx^n}=f(x,y,y',\ldots,y^{(n-1)}),\qquad y(0)=y_0,\quad y'(0)=y_1,\quad\ldots\quad, y^{(n-1)}=y_{n-1} $$ has a unique solution for
$x\in(-\epsilon,\epsilon)$ (note this statement asserts both the existence and uniqueness of the solution to the initial value problem).
Euler's Method
Just as for first-order equations, we often won't be able to find explicit solutions to higher-order equations and will want to find numerical approximations. We do this by converting the
higher-order equations into first-order systems and then using the same techniques as we had in the last chapter. The difference is now we will apply these to all the different first-order equations
in the system at once. We begin with Euler's method. As before, this method is well suited to implementation on a spreadsheet. We will use the example $$ y'' + 3y'+ 2y = 0,\qquad y(0)=1,\quad y'(0)=
0. $$ Open Excel (or your favorite spreadsheet if you are working on your own computer). We'll start by labeling the columns to keep track of what goes where. Enter
into cell A1,
into cell B1,
into cell C1,
into cell D1,
into cell E1, and
into cell H1. Next put the initial values and step-size into the appropriate starting spots with
in cell A2,
in cell D2,
in cell E2, and
in cell H2. Then put in the formulas with
in cell A3,
in cell B3,
in cell C3,
in cell D3, and
in cell E3. You now have the
column set to add the step size
as you move down each row, the
columns to compute the slopes based on the starting
, and
values, and the
columns to add
to the previous values for y and v respectively. Now click and drag to select cells A3..E3 and then click on the small square at the lower right of the box around those cells and drag it down for
about 20 rows or so to copy the formulas down and automate the calculations.
Note that the approximate values of the solution function $y(x)$ are contained in the
column which is column D,
the last column in the tableau. So here we see $y(0.5)\approx 0.8533$ for this initial value problem. Since this is a nice constant coefficient linear homogeneous equation, we can find the exact
value is $y=2e^{-x}-e^{-2x}$ so we will be able to check our work. In cell F1 of the spreadsheet enter
y exact
and in cell G1 enter
rel error
(which stands for relative error). Then in cell F2 enter
and in cell G2 enter
. It will be convenient to right-click on cell G2 and Format Cells as Percentage since relative error is a percent. Now click and drag to select cells F2 and G2 and drag those formulas down. This
will show you how accurate the approximation produced by Euler's method is.
Euler's method isn't particularly accurate, though it works well enough in this problem with a relative error of about 1% (problems with exponential decay tend to work better than exponential
growth). We can improve the approximation by taking smaller steps. If you change cell H2 to
to cut the step size in half, you find that the error in the value of $y(0.5)$ has been cut to about 0.43%. Of course you had to take 10 steps of size 0.05 to get to 0.5 while it only took 5 steps of
size 0.1. So you do twice as much work but you cut the error roughly in half. This is the standard pattern for Euler's method, just as it was in for first-order equations. Naturally, you can use this
technique to handle other equations. The only thing you need to edit is the formula for
in column C. Do be sure to copy the edited formula down the whole column.
The Improved Euler Method
Just like for first-order equations, we can get more accuracy with the improved Euler's method. As before, you can use whatever tool you find most appropriate, but the simplest is probably the
spreadsheet. We set things up similarly to the Euler's method, but with a couple of extra columns now for the extra computations. Enter
in cell A1,
left dy/dx
in cell B1,
left dv/dx
in cell C1,
y tilde
in cell D1,
v tilde
in cell E1,
right dy/dx
in cell F1,
right dv/dx
in cell G1,
in cell H1,
in cell I1, and
in cell L1. Next enter the initial values
in cell A2,
in cell H2,
in cell I2, and
in cell L2. Then enter the formulas
in cell A3,
in cell B3,
in cell C3,
in cell D3,
in cell E3,
in cell F3,
in cell G3,
in cell H3, and
in cell I3. Now click and drag to select A3..I3 and then copy the formula down the sheet to repeat the calculations as far as you want.
As before we can set up columns for the exact values and the relative error to see how accurate this method is for this problem. Enter
y exact
in cell J1 and
rel error
in cell K1. Enter the formulas
in cell J2 and
in cell K2. As before it looks nicer if you set the format of the cells in column K to Percentage.
Note that the improved Euler's method is twice as much work as the original Euler's method, since you do an extra improvement step each row. On the other hand, the error for $y(0.5)$ is now just
0.21%, which is better than the -0.43% error we got by doing twice as much work with a smaller step size with the original Euler's method. We can improve the accuracy of our approximation further by
using a smaller step size of $h=.05$. The result of this calculation is off by only 0.05% of the true value. While halving the step size (hence doubling the work) halved the error for Euler's method,
it reduces the error in the improved Euler's method by about a factor of 4. Again, this is the same pattern as you saw for the first-order equations. And of course you can apply this method for other
equations just by editing the formulas for the dv/dx in columns C and G.
Concluding Discussion
As noted above, to adjust to a different second-order equation you just need to change the column(s) for
. It is up to you to convert the second-order equation to a first order system and determine what the formula needs to be for $dv/dx$. It should be emphasized that the spreadsheets we've designed
here are for
equations. If we had a third-order equation then, as discussed in the first section on this page, we would need a system of three equations. That would require columns for
in addition to those for
, and
. This isn't hard, but you do need to be sure you don't just blindly plug into the spreadsheets we've created if the problem doesn't fit. We can extend the Runge-Kutta and Runge-Kutta-Feldman
techniques to higher order equation by converting them to first-order systems in the same was as we have extended the Euler and Improved Euler method here. The same tradeoffs will apply. The RK and
RKF methods are more accurate and preferred for careful computations. But for situations in which a fast runtime is more important that precise accuracy you may choose to use Euler's method or the
improved Euler's method.
1. Why do you only have to adjust the column(s) for dv/dx and not for dy/dx or anything else to change your spreadsheet to approximate a different equation?
2. Approximate $y(2)$ for the linear homogeneous initial value problem $$ \frac{d^2y}{dx^2}-\frac{dy}{dx}-6y=0,\qquad y(0)=1,\quad y'(0)=0 $$ using Euler's method with $h=0.1$ and $h=0.05$. How does
cutting the step-size in half in Euler's method affect the relative error for this problem?
3. Approximate $y(2)$ for the linear homogeneous initial value problem $$ \frac{d^2y}{dx^2}+\frac{dy}{dx}-6y=0,\qquad y(0)=1,\quad y'(0)=0 $$ using the improved Euler's method with $h=0.1$ and $h=
0.05$. How does cutting the step-size in half in the improved Euler's method affect the relative error for this problem?
4. Approximate $y(2)$ for the linear inhomogeneous initial value problem $$ \frac{d^2y}{dx^2}+4\frac{dy}{dx}+3y=e^x,\qquad y(0)=1,\quad y'(0)=0 $$ using Euler's method with $h=0.1$ and $h=0.05$. How
does cutting the step-size in half in Euler's method affect the relative error for this problem?
5. Approximate $y(2)$ for the linear inhomogeneous initial value problem $$ \frac{d^2y}{dx^2}+4\frac{dy}{dx}+3y=e^x,\qquad y(0)=1,\quad y'(0)=0 $$ using the improved Euler's method with $h=0.1$ and
$h=0.05$. How does cutting the step-size in half in the improved Euler's method affect the relative error for this problem?
6. Approximate $y(2)$ for the third-order linear initial value problem $$ \frac{d^3y}{dx^3}+2\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=0,\qquad y(0)=1,\quad y'(0)=0,\quad y''(0)=-1 $$ using Euler's method
with $h=0.1$ and $h=0.05$. How does cutting the step-size in half in Euler's method affect the relative error for this problem?
7. Approximate $y(2)$ for the third-order linear initial value problem $$ \frac{d^3y}{dx^3}+2\frac{d^2y}{dx^2}-\frac{dy}{dx}-2y=0,\qquad y(0)=1,\quad y'(0)=0,\quad y''(0)=-1 $$ using the improved
Euler's method with $h=0.1$ and $h=0.05$. How does cutting the step-size in half in the improved Euler's method affect the relative error for this problem?
8. Approximate $y(0.5)$ for the non-linear initial value problem $$ \frac{d^2y}{dx^2}-2y\frac{dy}{dx}=0,\qquad y(0)=1,\quad y'(0)=1 $$ using Euler's method with $h=0.1$ and $h=0.05$. The exact
solution is $y(x)=\displaystyle\frac{1}{1-x}$ so $y(0.5)=2$. How does cutting the step-size in half in Euler's method affect the relative error for this non-linear problem.
9. Approximate $y(0.5)$ for the non-linear initial value problem $$ \frac{d^2y}{dx^2}-2y\frac{dy}{dx}=0,\qquad y(0)=1,\quad y'(0)=1 $$ using the improved Euler's method with $h=0.1$ and $h=0.05$.
The exact solution is $y(x)=\displaystyle\frac{1}{1-x}$ so $y(0.5)=2$. How does cutting the step-size in half in the improved Euler's method affect the relative error for this non-linear problem.
10. (TRICKY) Approximate $y(1.1)$ for the non-linear initial value problem $$ \frac{d^2y}{dx^2}-2y\frac{dy}{dx}=0,\qquad y(0)=1,\quad y'(0)=1 $$ using the improved Euler's method with $h=0.1$. The
exact solution is $y(x)=\displaystyle\frac{1}{1-x}$ so $y(1.1)=-10$. In this case the approximation is badly off. How small a step-size do you need to pick to get the relative error under 10%.
If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett | {"url":"https://onlinehw.math.ksu.edu/math340book/chap2/numerical2.php","timestamp":"2024-11-09T06:21:44Z","content_type":"text/html","content_length":"22598","record_id":"<urn:uuid:10c3e715-6f3b-4ca8-a61f-85cb2dbbb4b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00736.warc.gz"} |
best time to buy and sell a stock (clojure)
best time to buy and sell a stock (clojure)
we are given a list prices, which contains the prices of a stock on a given day, the goal is to find the maximum profit possible by choosing 1 day to buy and a subsequent day to sell the stock.
input output description
'[7,1,5,3,6,4] 5 if we buy on day 2 (price = 1) and sell on day 5 (price = 6) then we make a profit of 5.
'[7,6,4,3,1] 0 since the prices decrease day by day there is no way to buy the stock and sell it at a higher price.
the best time to buy the stock is at the lowest price and the best time to sell the stock is the highest price seen AFTER the lowest price. so we can simply process the prices sequentially, keeping
track of the lowest price we’ve seen and using that to calculate if the current price gives us the highest profit so far.
functional programs are naturally suited to a recursive approach which works well for this problem. our function will fit the simple formula below.
name description initial value
prices list of prices by day provided
minPrice the minimum price seen so far Infinity
maxProfit the max profit seen so far 0
base case
1. if prices is empty, return maxProfit.
this covers edge cases where the profit is zero, including an empty list, list of 1, and the second example above.
recursive case
1. remove the first element in prices and call it currentPrice.
2. find the minimum between currentPrice and minPrice.
3. calculate currentProfit as currentPrice - minPrice
4. find the maximum between the currentProfit and maxProfit.
5. call recursively with the updated values.
(defn max-profit
([prices] (max-profit prices ##Inf 0)) ; driver function to set proper initial values
([prices minPrice maxProfit] ; recursive function
(let [currentPrice (peek prices)] ; look at first element in prices
(if (nil? currentPrice) ; return maxProfit if empty/first is nil
(let [minPrice (min currentPrice minPrice)
currentProfit (- currentPrice minPrice)]
(pop prices) ; removes first from prices, returns new list
(max currentProfit maxProfit))))))) | {"url":"https://ckuhn.github.io/buy-and-sell-stock.html","timestamp":"2024-11-07T17:00:06Z","content_type":"text/html","content_length":"10646","record_id":"<urn:uuid:0072d2eb-a01e-40f8-83bc-50e67ea988dc>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00704.warc.gz"} |
Lesson 2
How Do We Measure Area?
Lesson Purpose
The purpose of this lesson is for students to use square tiles to build shapes and measure area.
Lesson Narrative
Previously, students compared the area of shapes informally—by cutting out and overlaying the shapes, by observing whether one shape would fit into another, and by covering the shapes with pattern
blocks and comparing the number of blocks used.
In this lesson, students learn that squares can be used to measure area: by tiling all of the shape. Each square represents one unit of area, or one square unit. Inch tiles are used, but are referred
to as “square tiles” with students to emphasize how the tiles are used to measure square units. Students learn that shapes that don’t have specific names can be referred to as “figures.”
In the next lesson, students will take a closer look at square tiles that overlap.
Provide inch tiles for students to use during the cool-down.
Learning Goals
Teacher Facing
• Explore area by building shapes with unit squares.
• Use unit squares to measure area.
Student Facing
• Let’s use square tiles to measure area.
Required Materials
Materials to Gather
Materials to Copy
• Use Square Tiles to Measure Area
Required Preparation
Activity 1:
• Each group of 4 needs 80 square tiles.
Activity 2:
• Each group of 2 needs 80 square tiles.
CCSS Standards
Building Towards
Lesson Timeline
Warm-up 10 min
Activity 1 20 min
Activity 2 15 min
Lesson Synthesis 10 min
Cool-down 5 min
Teacher Reflection Questions
What ideas and experiences do students have about area? How did they influence students' work?
Suggested Centers
• Can You Build It? (3–5), Stage 1: Rectangles (Addressing)
• Five in a Row: Multiplication (3–5), Stage 1: Factors 1–5 and 10 (Supporting)
Print Formatted Materials
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Blackline Masters zip
Additional Resources
Google Slides Log In
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Kinematics in physics: definition, movements and examples
Kinematics is the branch of physics that studies the movement of objects and their trajectory as a function of time without taking into account the causes that produce it.
In the study of kinematics, the chemical and physical properties of moving bodies are not taken into account, with the exception of dimensions, as well as the forces acting on them. For this reason,
it is also known as rigid solid kinematics.
This branch of physics is included within classical physics, so the formulas that we will present in the articles in this section do not take into account Einstein's theory of relativity. That is,
the equations that we will present are only valid for small speeds compared to the speed of light.
Basic concepts
To understand this branch of physics, it is essential to familiarize yourself with some basic concepts:
1. Position
The position of an object refers to its location in space relative to a reference system. It can be described using coordinates, such as a Cartesian coordinate system (x, y, z), where x, y, and z
represent the spatial dimensions.
Position is typically represented by a vector indicating both the magnitude (distance) and direction from the origin of the reference frame.
2. Displacement
Displacement is a change in the position of an object.
It is expressed as a vector that connects the initial position with the final position of the object. Displacement is independent of the path followed by the object and is measured in terms of length
and direction.
3. Velocity
The velocity of an object refers to the rate of change of its position with respect to time.
It is expressed in terms of displacement per unit of time and is represented as a vector. The speed can be constant (uniform motion) or change with time (accelerated motion).
4. Acceleration
Acceleration is the rate of change of an object's speed with respect to time.
Like velocity, it is represented as a vector. Acceleration can be positive (increase in speed), negative (decrease in speed), or zero (uniform motion).
5. Time
Time is a fundamental variable in kinematics that is used to measure the duration of an event. It allows us to describe how the other kinematic variables change over time.
Types of movements
types of movement that can be summarized as follows:
Examples of kinematics
Kinematics manifests itself in various situations in everyday life and in scientific fields.
Movement of a car
When this car accelerates or brakes, such as when starting from a traffic light or slowing down to stop, this is an example of uniformly accelerated rectilinear motion (MRUA).
At the moment when the car takes a curve, the formulas of uniform circular motion (MCU) or uniformly accelerated circular motion (MCUA) can be applied if inside the curve it is braking or
In the sports field, kinematics is applied when analyzing the throwing of a baseball, where the trajectory of the ball follows a parabolic movement .
In astronomy, it is used to study the movement of planets around the Sun and calculate their orbits.
Nuclear power plants and nuclear physics
In nuclear physics, kinematics is used to analyze the collision of subatomic particles in particle accelerators.
In addition, nuclear power plants have steam turbines that, when in operation, experience circular motion.
Difference between kinematics and dynamics
dynamics are two branches of physics that focus on different aspects of the motion of objects.
On the one hand, kinematics is concerned with describing movement in terms of position, speed and acceleration, without considering the causes that generate it. On the other hand, dynamics focuses on
the study of the forces and interactions that cause the movement of objects.
That is, while kinematics is concerned with "how" objects move, dynamics focuses on "why" they move and how they respond to applied forces. | {"url":"https://nuclear-energy.net/physics/kinematics","timestamp":"2024-11-15T03:33:06Z","content_type":"text/html","content_length":"70828","record_id":"<urn:uuid:f1a9e9b6-2a1d-4fdd-bfb7-ff57923463d0>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00383.warc.gz"} |
Geometry with three inclined planes
I’m trying to model the front part of a vehicle in threejs which has three inclined planes, one at the front and two oblique ones at each side:
It’s easy to model the front-middle part extruding a shape like this:
But I am having trouble on how to make the side parts, the only solution i could think of is making an extruded squared/rectangular mesh and trying to slice it with a plane (i’m guessing there is a
way to do this somehow on threejs) but maybe there’s a better way to do this.
I’m not quite sure I understand the shape from these snapshots, but some possible approaches are:
• it looks like a convex shape, so you might try ConvexGeometry
• use an object with the same number of vertices and similar shape ( CylinderGeometry looks like a good candidate); then modify the vertices
• construct the object “by hand”
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An Introduction to Tree Diagrams
What is a Tree Diagram?
A tree diagram is simply a way of representing a sequence of events. Tree diagrams are particularly useful in probability since they record all possible outcomes in a clear and uncomplicated manner.
First principles
Let's take a couple of examples back to first principles and see if we can gain a deeper insight into tree diagrams and their use for calculating probabilities.
Let's take a look at a simple example, flipping a coin and then rolling a die. We might want to know the probability of getting a Head and a 4.
If we wanted, we could list all the possible outcomes:
(H,1) (H,2) (H,3) (H,4) (H,5) (H,6)
(T,1) (T,2) (T,3) (T,4) (T,5) (T,6)
Probability of getting a Head and a 4:
P(H,4) = $\frac{1}{12}$
Here is one way of representing the situation using a tree diagram. To save time, I have chosen not to list every possible die throw (1, 2, 3, 4, 5, 6) separately, so I have just listed the outcomes
"4" and "not 4":
Each path represents a possible outcome, and the fractions indicate the probability of travelling along that branch. For each pair of branches the sum of the probabilities adds to 1.
"And" Means Multiply
So how might we work out P(H,4) from the tree diagram? We could word this as the probability of getting a Head and then a 4. This is the green path.
Half the time, I expect to travel along the first green branch.
Then, on one sixth of those occasions, I will also travel along the second green branch. We can think of this as $\frac{1}{6} \text{ of } \frac{1}{2}$.
$\frac{1}{6} \text{ of } \frac{1}{2} = \frac{1}{6}$ x $\frac{1}{2} = \frac{1}{12}$
So this is why Joe said that you multiply across the branches of the tree diagram.
"And" only means multiply if events are independent, that is, the outcome of one event does not affect the outcome of another. This is certainly true for our example, since flipping the coin has no
impact on the outcome of the die throw.
"Or" Means Add
Now let's consider the probability of getting a Head
a 4.
We are using the word "or" in its mathematical sense to mean "Head or 4 or both", as opposed to the common usage which often means "either a Head or a 4":
(H,1) (H,2) (H,3) (H,4) (H,5) (H,6) (T,4)
So P(H or 4) is $\frac{7}{12}$
Again, we can work this out from the tree diagram, by selecting every branch which includes a Head or a 4:
Each of the ticked branches shows a way of achieving the desired outcome. So P(H or 4) is the sum of these probabilities:
$P(H\text{ or }4) = P(H,4) + P(H, \text{ not }4 ) + P(T, 4) = \frac{1}{12} + \frac{5}{12} + \frac{1}{12} = \frac{7}{12}$.
So this is why Joe said that you add down the ends of the branches.
Picturing the Probabilities
Imagine I roll an ordinary die three times, and I'm interested in the probability of getting one, two or three sixes.
I might draw a tree diagram like this:
Check that you agree with the probabilities at the end of each branch before reading on.
We can now work out:
P(three sixes) = $\frac{1}{216}$
P(exactly two sixes) = $\frac{15}{216}$
P(exactly one six) = $\frac{75}{216}$
P(no sixes) = $\frac{125}{216}$
Again, check that you understand where these probabilities have come from before reading on.
To really check your understanding, think about the outcomes that contribute to each of the probabilities on the tree diagram. For example, P(6, not 6, 6) is $\frac{5}{216}$, because out of the 216
total outcomes there are five outcomes which satisfy (6, not 6, 6):
6, 1, 6
6, 2, 6
6, 3, 6
6, 4, 6
6, 5, 6
Can you explain why there are 25 outcomes that satisfy (not 6, not 6, 6)?
What about the other probabilities on the tree diagram?
I hope this article helps you to understand what's happening next time you come across a tree diagram, and that it helps you to construct your own tree diagrams to solve problems.
Click here for a selection of NRICH problems where tree diagrams can be used. | {"url":"https://nrich.maths.org/articles/introduction-tree-diagrams","timestamp":"2024-11-11T22:41:23Z","content_type":"text/html","content_length":"43781","record_id":"<urn:uuid:a6ef0361-ae25-45c5-9ab7-f0d5f2aa2946>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00160.warc.gz"} |
How To Buy Down Interest Rate On A Mortgage
With this offering, your borrowers can permanently reduce their interest rate by financing up to three discount points into the loan amount for fixed-rate. An interest rate buydown is when the home
seller, in this case, the home builder, pays the lender to decrease your mortgage rate for a certain period. With this offering, your borrowers can permanently reduce their interest rate by financing
up to three discount points into the loan amount for fixed-rate. A temporary buydown could be the answer. Ease into the homeownership journey with a lower starting monthly payment. Free up cash for
all the things new. This is also called “buying down the rate.” Essentially, you pay some interest up front in exchange for a lower interest rate over the life of your loan. Each.
Permanent buydowns offer borrowers an opportunity to get a lower interest rate over the life of their loan. This typically requires buying more mortgage points. To figure out the cost associated with
buying down the rate, multiply the loan amount by 1% (or whatever the percent of the buydown is), and that will give you. Enter the number of years of your loan term, the total loan amount, and the
interest rate percentage into the remaining calculator fields and click Calculate. The mortgage servicer still receives payment of the amount due for the actual interest rate. Who Qualifies for a
Buydown? Anyone looking for a home loan can. In a buydown structure, the rate for the first year is 2% lower than the note rate; and in the second year of the loan, the rate is 1% lower than the
note. The initial rate is lower for a set time. Borrowers can choose buydown plans with rates up to 3% lower than current mortgage rates. For example, if market rates. Each point buy down the rate
about% but the lender will usually cut you off on how many points you can get at somewhere around The formula for calculating buydown points is: buydown points = (loan amount x percentage) / For
example, you're buying a home for $,, and the total. Each mortgage discount point usually costs one percent of your total loan amount, and lowers the interest rate on your monthly payments by
percent. For. lenders like it because buying down the rate does more for your monthly payment than putting more money down on the house, makes you more likely. These points are optional fees you pay
to your lender to can reduce the interest rate on your a loan. Learn MoreAbout Us. The Buydown Method and Mortgage Points.
A buydown is a mortgage-financing technique that allows a homebuyer to obtain a lower interest rate for at least the first few years of the loan, or possibly. The easiest way to buy down your
mortgage rate is to buy discount points. Each point is percent of your mortgage amount, and reduces your mortgage rate by. Mortgage points shave off fractions of a percent from your rate, which can
save you thousands of dollars on a year mortgage. You'll typically reduce your. Paying for Discount Points · Temporary Mortgage Buy-Downs · Assumable Mortgages · Buy Now, Refinance Later · Other
State Resources. For each of the first three years of the mortgage, the buyer's interest rate would increase incrementally by 1% annually. The full interest rate would apply. Buydown: A Way To Reduce
Interest Rates. A Buydown is a method used by buyers and sellers to lower interest rates in the early years of a new mortgage. This means, using the example above, you would need to be approved for a
loan with a 6% interest rate. How much can buy-downs lower mortgage payments? Buy-downs. 1. Shop for mortgage rates · 2. Improve your credit score · 3. Choose your loan term carefully · 4. Make a
larger down payment · 5. Buy mortgage points · 6. Lock in. With a permanent mortgage rate buydown, you pay a fee known as discount points to lower your interest rate for the life of your loan. You
can purchase as little.
A temporary mortgage buydown is a financing option that allows you to obtain a lower interest rate for the first few years of your mortgage. Enter the number of years of your loan term, the total
loan amount, and the interest rate percentage into the remaining calculator fields and click Calculate. The buydown agreement may include an option for the buydown funds to be returned to the
borrower or to the lender, if it funded the buydown, if the mortgage is. If a builder bought down, took about 6% of the mortgage amount and paid that upfront in fees, they can buy down the interest
rate by a point and a half. The Temporary Buydown reduces the buyer's interest rate by 3% for the first year of their loan, 2% for the second year, and 1% for the third year. EXAMPLE.
The total buydown cost is the difference between the total payments made at the original monthly payment, and the total payments made at the rate-adjusted. Loan Amount. $ ; Term (Yrs) ; Interest Rate
(%). % ; Third-party Contribution toward. Buydown Fee (% of Loan Amount). %.
Can I Get Braces At 26 | Average 30 Yr Mortgage | {"url":"https://apbaskakov.ru/learn/how-to-buy-down-interest-rate-on-a-mortgage.php","timestamp":"2024-11-06T01:11:52Z","content_type":"text/html","content_length":"13173","record_id":"<urn:uuid:a78ac2b3-985d-4da7-b86f-f503c460f958>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00407.warc.gz"} |
interior of real numbers
I'll try to provide a very verbose mathematical explanation, though a couple of proofs for some statements that probably should be provided will be left out. We will now state the important geometric
representation of the absolute value with respect to the real number line. N. Bourbaki, a group of French … The Density of the Rational/Irrational Numbers. 1.1. Rational Expressions; Rational Numbers
for Class 8; Irrational Numbers; Rational And Irrational Numbers; Standard Form of Rational Numbers. The interior of an interval I is the largest open interval that is contained in I; it is also the
set of points in I which are not endpoints of I. The supremum or infimum of a set may or may not belong to the set. 1 Some simple results. Integers involve natural numbers(N). Free PDF download of
Chapter 1 - Real Numbers Formula for Class 10 Maths. Prove your answer. Real numbers include the integers (Z). This leads to a method of expressing the ratio of two complex numbers in the form x+iy,
where x and y are real complex numbers. They went up again in fiscal 2018 but decreased in fiscal 2019 and remain far lower than during President Barack Obama’s first term in … A useful identity
satisfied by complex numbers is r2 +s2 = (r +is)(r −is). 94 5. Show transcribed image text. (That is, the boundary of A is the closure of A with the interior points removed.) In the de nition of a A=
˙: Every … Open and Closed Sets; 5.2. Lectures by Walter Lewin. Find the best interior decorators in Sector 62 Noida on RealEstateIndia.com. Topology; 5.1. Sequences of Functions; 9. Enjoy the videos
and music you love, upload original content, and share it all with friends, family, and the world on YouTube. De nition. We don’t give proofs for most of the results stated here. • The closure of A
is the set c(A) := A∪d(A).This set is sometimes denoted by A. Let S be an … Every whole number is a rational number because every whole number can be expressed as a fraction. Lecture 15 : Topology of
Real Numbers: Limit Points, Interior Points, Open Sets and Compact Sets - Part III: Download: 16: Lecture 16 : Topology of Real Numbers: Compact Sets and Connected Sets - Part I: Download: 17:
Lecture 17 : Topology of Real Numbers: Compact Sets and Connected Sets - Part II: Download: 18: Lecture 18 : Topology of Real Numbers: Compact Sets and Connected Sets - Part III: … $\endgroup$ –
Catalin Zara Apr 3 '16 at 2:10. (b) Is 0 a boundary point of A? See the answer. If supA∈ Adoes belong to A, then we also denote it by maxAand refer to it as the maximum of A; if inf A∈ Athen we also
denote it by minAand refer to it as the … . 2.1.1 Proof; 2.2 Existence of Greatest Lower Bounds. 4 … Real numbers (R) include all the rational numbers (Q). We can as well consider a an algebraically
closed field $\mathbb C$ of characteristic $0$ given and … > Why is the closure of the interior of the rational numbers empty? First, here is the definition of a limit/interior point (not word to
word from Rudin) but these definitions are worded from me (an undergrad student) so please correct me if they are not rigorous. Countable. The interior of a set, [math]S[/math], in a topological
space is the set of points that are contained in an open set wholly contained in [math]S[/math]. Let m = sup N. … The distance between real numbers xand yis jx yj. Prove your answer. x1 +iy1 x2 +iy2
= (x1 +iy1)(x2 −iy2) (x2 +iy2)(x2 −iy2) = (x1x2 +y1y2)+i(−x1y2 +y1x2) x2 2 +y2 2. The set of real numbers R is a complete, ordered, field. The Real Numbers If m∈ R is a lower bound of Asuch that m≥ m′
for every lower bound m′ of A, then mis called the infimum or greatest lower bound of A, denoted m= inf A. Series of Numbers; 5. Symbols of Real Numbers and Integers. Get complete detail of interior
designers, phone numbers, address, service and service area. . De nition. real-analysis general-topology. Historical Tidbits; Java Tools; 5.1. In arithmetical terms, the Cantor set consists of all
real numbers of the unit interval [,] that do not require the ... of the Cantor set, but none is an interior point. A set of real numbers is open if and only if it is a countable union of disjoint
open intervals. An open subset of R is a subset E of R such that for every xin Ethere exists >0 such that B (x) is contained in E. For example, the open … As a set, real numbers are uncountable while
integers are countable. Are They Open, Closed Or Compact (or Several Or None)? 1. Proof: Suppose N is bounded above. Let A be a subset of the real numbers. Chapter 1 The Real Numbers 1 1.1 The Real
Number System 1 1.2 Mathematical Induction 10 1.3 The Real Line 19 Chapter 2 Differential Calculus of Functions of One Variable 30 2.1 Functions and Limits 30 2.2 Continuity 53 2.3 Differentiable
Functions of One Variable 73 2.4 L’Hospital’s Rule 88 2.5 Taylor’s Theorem 98 Chapter 3 Integral Calculus of Functions of One Variable 113 3.1 Definition of the Integral … Given a topological space X,
a subset A of X that can be expressed as the union of countably many nowhere dense subsets of X is called meagre. share | cite | improve this question | follow | asked Apr 3 '16 at 2:06. Basic proofs
. The number of interior arrests made by ICE (known as “administrative arrests”) rose 30% in fiscal 2017 after Trump signed an executive order giving the agency broader authority to detain
unauthorized immigrants, including those without criminal records. . Prove you answer. (d) Is 0 an isolated point of A? It can be constructed by taking the union of all the open sets contained in A.
(c) Is 0 a limit point of A? Denote by Aº the set of interior points of A, by bd(A) the set of boundary points of A and cl(A) the set of closed points of A. I am reading Rudin's book on real analysis
and am stuck on a few definitions. $\begingroup$ You have $\not\subset$ if you construct them one after another. on any two numbers in a set, the result of the computation is another number in the
same set. • The complement of A is the set C(A) := R \ A.The complement of A is sometimes … This problem has been solved! The Real Number Line One way to represent the real numbers $\mathbb{R}$ is on
the real number line as depicted below. A closed set in which every point is an accumulation point is also called a perfect set in topology, while a closed subset of the interval with no interior
points is nowhere dense in the interval. of complex numbers is performed just as for real numbers, replacing i2 by −1, whenever it occurs. The set of rational numbers Q, although an ordered field, is
not complete. On the contrary, integers are not considered as a field. Real numbers are symbolized as “R” while a set of integers is symbolized as “Z”. ... (possibly empty) open set; the maximum
(ordered under inclusion) such open set is called the interior of A. (a) S = Q N (0,1). Theorem 3-5. The rational numbers, while dense in the real numbers, are meagre as a subset of the reals. 2 1.
The … 1,516 3 3 gold badges 17 17 silver badges 35 35 bronze badges $\endgroup$ 4 $\begingroup$ You are right: the complement of $\mathbb{N}$ in $\mathbb{R}$ is open, hence, by definition, $\mathbb
{N}$ is a closed set. The interior of the complement of a nowhere dense set is always dense. We will now look at a theorem regarding the density of rational numbers in the real numbers, namely that
between any two real numbers there exists a rational number. Every point of the Cantor set is also an accumulation point of the … Noida Search from Over 2500 Cities - All India A point p is an
interior point of E if there is a neighborhood N of p such that N ⊂ E. E is open if every point of E is an interior point of E. E is perfect if E is closed and if every point of E is a limit point of
E. E is bounded if there is a real number M and a point q ∈ X such that d(p,q) < M for all p ∈ E. E is dense in X every point of X is a limit point of E or a point of E (or both). For a real number
xand >0, B (x) = fy2R : dist(x;y) < g: Of course, B (x) is another way of describing the open interval (x ;x+ ). Our understanding of the real numbers derives from durations of time and lengths in
space. Prove your answer. (b) {x € Ql2 = ' Where N,k E NU{0} And 0 Sk 5 2"}. For any set X of real numbers, the interval enclosure or interval span of X is the unique interval that contains X, and
does not properly contain any other … To Register Online Maths Tuitions on Vedantu.com to clear your doubts from our expert teachers and download the Real Numbers Formula to solve the problems easily
to score more marks in your CBSE Class 10 Board Exam. The standard form of a rational … 2.1 Uniqueness of Least Upper Bounds. Therefore, given a real number x, one can speak of the set of all points
close to that real number; that is, within ε of x. For example, dist( 4;3) = j( 4) (3)j= 7. Definition: A real number r is said to be rational if there are integers n and m (m≠0) such that r = with
greatest common divisor betwee n [n, m] = 1. Compact and Perfect Sets; 5.3. We use d(A) to denote the derived set of A, that is theset of all accumulation points of A.This set is sometimes denoted by
A′. They will make you ♥ Physics. A topological space … Given topological spaces X and Y, a function f from X to Y is continuous if the preimage of every open set in … Sequences of Numbers; 4.
Topology of the Real Numbers When the set Ais understood from the context, we refer, for example, to an \interior point." Note. For example, the set T = {r ∈Q: r< √ 2} is bounded above, but T does
not have a rational least upper bound. Open and Closed Sets Definition 5.1.5: Boundary, Accumulation, Interior, and Isolated Points. Recommended for you Limits, Continuity, and Differentiation; 7.
Jabernet Jabernet. Connected and Disconnected Sets ; 6. Real numbers are a kind of field which is an essential algebraic structure where arithmetic processes are defined. But already the fact that
there are several constructions possible (e.g. Derived Set, Closure, Interior, and Boundary We have the following definitions: • Let A be a set of real numbers. The Archimedean Property THEOREM 4. The
closure of I is the smallest closed interval that contains I; which is also the set I augmented with its finite endpoints. Completeness of R Intuitively, unlike the rational numbers Q, the real
numbers R form a continuum with no ‘gaps.’ There are two main ways to state this completeness, one in terms of the existence of suprema and the other in terms of the convergence of … Prove that bd(A)
= cl(A)\A°. The Real Numbers In this chapter, we review some properties of the real numbers R and its subsets. Properties of The Real Numbers: Exercises → Contents. With proofs please! 1.1.1 Theorem
(Square roots) 1.1.2 Proof; 1.1.3 Theorem (Archimedes axiom) 1.1.4 Proof; 1.1.5 Corollary (Density of rationals and irrationals) 1.1.6 Proof; 2 Properties of Least Upper Bounds. Question: For The
Following Sets Of Real Numbers, Calculate All Interior Points, Boundary Points, Accumulation Points And Isolated Points. Here, our concern is only with the closure property as it applies to real
numbers . Expert Answer . 2.2.1 Proof; … For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. Interior and isolated points of a set belong to the set, whereas boundary and
accumulation points may or may not belong to the set. Previous question Next … Theorem 3-5 allows us to completely describe an open set of real numbers in terms of open intervals. The Closure
Property states that when you perform an operation (such as addition, multiplication, etc.) We think of the real line, or continuum, as being composed of an (uncountably) in nite number of points,
each of which corresponds to a real number, and denote the set of real numbers by R. There are philosophical questions, going back at least to Zeno’s paradoxes, about whether the continuum can be
represented … (The Archimedean Property) The set N of natural numbers is unbounded above. The Integral; 8. We also call this an epsilon neighborhood of x. Dedekind cuts or Cauchy sequences for $\
mathbb R$) these ZFC models of $\mathbb R$ and the otger number sets are often not what we intuitively mean. Consider the set of real numbers A defined by A = 1 in EN n N} (a) Is O an interior point
of A? The complement of a closed nowhere dense set is a dense open set. 1.1 Applications. Derives from durations of time and lengths in space are defined the fact that there are several constructions
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Coding Interviews Searching and Sorting Problems and Solutions
Coding Interviews: Searching and Sorting
AllVideoEasyMediumHardPythonCode SolutionJavaJavascript
There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums
[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:
Every adjacent pair of words differs by a single letter.
Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
sk == endWord
Given two words, beginWord and endWord, and a dictionary wordList, return all the shortest transformation sequences from beginWord to endWord, or an empty list if no such sequence exists. Each
sequence should be returned as a list of the words [beginWord, s1, s2, ..., sk].
Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the
combinations in any order.
The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.
There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course
bi first if you want to take course ai.
For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.
Return true if you can finish all courses. Otherwise, return false.
Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.
Given a string s and an integer k, return the length of the longest substring of s that contains at most k distinct characters.
Given the head of a linked list, return the list after sorting it in ascending order.
The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle. You may return the answer in any order.
Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space, respectively.
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
The overall run time complexity should be O(log (m+n)).
Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.
Note that the same word in the dictionary may be reused multiple times in the segmentation.
Given an array nums with n objects colored red, white, or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white, and blue.
We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.
Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.
If target is not found in the array, return [-1, -1].
You must write an algorithm with O(log n) runtime complexity.
Given an m x n integers matrix, return the length of the longest increasing path in matrix.
From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
Given an m x n grid of characters board and a string word, return true if word exists in the grid.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
There is a new alien language that uses the English alphabet. However, the order among the letters is unknown to you.
You are given a list of strings words from the alien language's dictionary, where the strings in words are sorted lexicographically by the rules of this new language.
Return a string of the unique letters in the new alien language sorted in lexicographically increasing order by the new language's rules. If there is no solution, return "". If there are multiple
solutions, return any of them.
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You must write an algorithm with O(log n) runtime complexity.
Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:
Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom. | {"url":"https://www.practiceproblems.org/course/Coding_Interviews/Searching_and_Sorting/1/cls0suj4o0003oopi9utimemf","timestamp":"2024-11-11T12:00:15Z","content_type":"text/html","content_length":"228409","record_id":"<urn:uuid:61ed70ec-1124-4ffe-afa9-798ca5d1cb27>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00020.warc.gz"} |