content stringlengths 86 994k | meta stringlengths 288 619 |
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combineSimes: Combine p-values with Simes' method in metaseqR2: An R package for the analysis and result reporting of RNA-Seq data by combining multiple statistical algorithms
This function combines p-values from the various statistical tests supported by metaseqR using the Simes' method (see reference in the main metaseqr2 help page or in the vignette).
p a p-value matrix (rows are genes, columns are statistical tests).
zerofix NULL (default) or a fixed numeric value between 0 and 1.
a p-value matrix (rows are genes, columns are statistical tests).
NULL (default) or a fixed numeric value between 0 and 1.
The argument zerofix is used to correct for the case of a p-value which is equal to 0 as a result of internal numerical and approximation procedures. When NULL, random numbers greater than 0 and less
than or equal to 0.5 are used to multiply the offending p-values with the lowest provided non-zero p-value, maintaining thus a virtual order of significance, avoiding having the same p-values for two
tests and assuming that all zero p-values represent extreme statistical significance. When a numeric between 0 and 1, this number is used for the above multiplication instead.
For more information on customizing the embed code, read Embedding Snippets. | {"url":"https://rdrr.io/bioc/metaseqR2/man/combineSimes.html","timestamp":"2024-11-06T17:21:00Z","content_type":"text/html","content_length":"36291","record_id":"<urn:uuid:dde89dee-06cd-4689-a970-a51a6e14c452>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00706.warc.gz"} |
How to calculate laplace solution with nodal boundary conditions?
Hi there,
I am trying to perform a Laplace solve with nodal boundary conditions specified (different Dirichlet values on each node). I have used the parameters specified in the tutorial but the solution I get
is simply whatever I put as the stimulus strength across the whole mesh. The files work ok in carpentry, but I'm not sure what I need to change for openCARP. I noticed in the user manual, vertex
adjustment files have the extension .adj but I am using .vtx here.
I wondered whether there is an example using vertex adjustment that I can compare to to figure out where I'm going wrong, or if you have any suggestions?
Thanks very much for your help, | {"url":"https://opencarp.org/q2a/50/calculate-laplace-solution-with-nodal-boundary-conditions?show=51","timestamp":"2024-11-10T06:41:39Z","content_type":"text/html","content_length":"29528","record_id":"<urn:uuid:66f7d252-d098-47ef-bdaa-ce1d0f0e8a2f>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00294.warc.gz"} |
Sentiment Analysis via R, FeatureHashing
Sentiment Analysis via R, FeatureHashing and XGBoost
Lewis Crouch
This vignette demonstrates a sentiment analysis task, using the FeatureHashing package for data preparation (instead of more established text processing packages such as ‘tm’) and the XGBoost package
to train a classifier (instead of packages such as glmnet).
With thanks to Maas et al (2011) Learning Word Vectors for Sentiment Analysis we make use of the ‘Large Movie Review Dataset’. In their own words:
We constructed a collection of 50,000 reviews from IMDB, allowing no more than 30 reviews per movie. The constructed dataset contains an even number of positive and negative reviews, so randomly
guessing yields 50% accuracy. Following previous work on polarity classification, we consider only highly polarized reviews. A negative review has a score less than or equal to 4 out of 10, and a
positive review has a score equal to or greater than 7 out of 10. Neutral reviews are not included in the dataset. In the interest of providing a benchmark for future work in this area, we
release this dataset to the public.
This data is also available via Kaggle and the format provided there makes for a more convenient starting point, simply because all of the training data is in a single file. In our case we will only
use the training data - 25,000 reviews - as we’ll only go as far as checking our classifier via validation.
Why FeatureHashing?
It’s not essential to use FeatureHashing for this movie review dataset - a combination of the tm and glmnet packages works reasonably well here - but it’s a convenient way to illustrate the benefits
of FeatureHashing. For example, we will see how easily we can select the size of the hashed representations of the review texts and will understand the options FeatureHashing makes available for
processing the data in subsets.
The combination of FeatureHashing and XGBoost can also be seen as a way to access some of the benefits of the Vowpal Wabbit approach to machine learning, without switching to a fully online learner.
By using the ‘hashing trick’, FeatureHashing easily handles features of many possible categorical values. These are then stored in a sparse, low-memory format on which XGBoost can quickly train a
linear classifier using a gradient descent approach. At a minimum this is a useful way to better understand how tools like Vowpal Wabbit push the same approaches to their limits. But in our case we
get to benefit from these approaches without leaving the R environment.
Package versions
This vignette uses FeatureHashing v9.0 and XGBoost v0.3-3.
Basic data preparation
First we read the training data and perform some simple text cleaning using gsub() to remove punctuation before converting the text to lowercase. At this stage each review is read and stored as a
single continuous string.
imdb <- read.delim("Data/labeledTrainData.tsv", quote = "", as.is = T)
imdb$review <- tolower(gsub("[^[:alnum:] ]", " ", imdb$review))
Which, using one of the shortest reviews as an example, leaves us with the following. At this stage, the review is still being stored as single character string and we only use strwrap() for pretty
## [1] "kurosawa is a proved humanitarian this movie is totally about people living in"
## [2] "poverty you will see nothing but angry in this movie it makes you feel bad but"
## [3] "still worth all those who s too comfortable with materialization should spend 2"
## [4] "5 hours with this movie"
We can then hash each of our review texts into a document term matrix. We’ll choose the simpler binary matrix representation rather than term frequency. The FeatureHashing package provides a
convenient split() function to split each review into words, before then hashing each of those words/terms to an integer value to use as a column reference in a sparse matrix.
d1 <- hashed.model.matrix(~ split(review, delim = " ", type = "tf-idf"),
data = imdb, hash.size = 2^16, signed.hash = FALSE)
The other important choice we’ve made is the hash.size of 2^16. This limits the number of columns in the document term matrix and is how we convert a feature of an unknown number of categories to a
binary representation of known, fixed size. For the sake of speed and to keep memory requirements to a minimum, we’re using a relatively small value compared to the number of unique words in this
dataset. This parameter can be seen as a hyperparameter to be tuned via validation.
The resulting 50MB dgCMatrix is the sparse format used by the Matrix package that ships with base R. A dense representation of the same data would occupy 12GB. Just out of curiosity, we can readily
check the new form of our single review example:
## [1] 1 2780 6663 12570 13886 16269 18258 19164 19665 20531 22371
## [12] 22489 26981 28697 29324 32554 33091 33251 35321 35778 35961 37510
## [23] 38786 39382 45651 46446 51516 52439 54827 57399 57784 58791 59061
## [34] 60097 61317 62283 62878 62906 62941 63295
The above transformation is independent of the other reviews and as long as we use the same options in hashed.model.matrix, we could process a larger volume of text in batches to incrementally
construct our sparse matrix. Equally, if we are building a classifier to assess as yet unseen test cases we can independently hash the test data in the knowledge that matching terms across training
and test data will be hashed to the same column index.
Training XGBoost
For this vignette we’ll train a classifier on 20,000 of the reviews and validate its performance on the other 5,000. To enable access to all of the XGBoost parameters we’ll also convert the document
term matrix to an xgb.DMatrix and create a watchlist to monitor both training and validation set accuracy. The matrix remains sparse throughout the process. Other R machine learning packages that
accept sparse matrices include glmnet and the support vector machine function of the e1071 package.
train <- c(1:20000); valid <- c(1:nrow(imdb))[-train]
dtrain <- xgb.DMatrix(d1[train,], label = imdb$sentiment[train])
dvalid <- xgb.DMatrix(d1[valid,], label = imdb$sentiment[valid])
watch <- list(train = dtrain, valid = dvalid)
First we train a linear model, reducing the learning rate from the default 0.3 to 0.02 and trying out 10 rounds of gradient descent. We also specify classification error as our chosen evaluation
metric for the watchlist.
m1 <- xgb.train(booster = "gblinear", nrounds = 10, eta = 0.02,
data = dtrain, objective = "binary:logistic",
watchlist = watch, eval_metric = "error")
## [0] train-error:0.070650 valid-error:0.145200
## [1] train-error:0.060400 valid-error:0.137200
## [2] train-error:0.053250 valid-error:0.130000
## [3] train-error:0.047400 valid-error:0.125400
## [4] train-error:0.041800 valid-error:0.122600
## [5] train-error:0.036750 valid-error:0.121200
## [6] train-error:0.032800 valid-error:0.119000
## [7] train-error:0.029300 valid-error:0.117200
## [8] train-error:0.026500 valid-error:0.115600
## [9] train-error:0.024000 valid-error:0.114200
That code chunk runs in just a few seconds and the validation error is already down to a reasonable 12%. So FeatureHashing has kept our memory requirement to about 50MB and XGBoost’s efficient
approach to logistic regression has ensured we can get rapid feedback on the performance of the classifier.
With this particular dataset a tree-based classifier would take far longer to train and tune than a linear model. Without attempting to run it for a realistic number of rounds, the code below shows
how easily we can switch to the tree-based mode of XGBoost:
m2 <- xgb.train(data = dtrain, nrounds = 10, eta = 0.02,
max.depth = 10, colsample_bytree = 0.1,
subsample = 0.95, objective = "binary:logistic",
watchlist = watch, eval_metric = "error")
## [0] train-error:0.386600 valid-error:0.415600
## [1] train-error:0.301200 valid-error:0.347200
## [2] train-error:0.272550 valid-error:0.316200
## [3] train-error:0.245400 valid-error:0.288000
## [4] train-error:0.219400 valid-error:0.269800
## [5] train-error:0.204200 valid-error:0.258400
## [6] train-error:0.196550 valid-error:0.243200
## [7] train-error:0.181250 valid-error:0.241800
## [8] train-error:0.174900 valid-error:0.239400
## [9] train-error:0.176650 valid-error:0.236600
The above demonstration deliberately omits steps such as model tuning but hopefully it illustrates a useful workflow that makes the most of the FeatureHashing and XGBoost packages. | {"url":"https://cran-r.c3sl.ufpr.br/web/packages/FeatureHashing/vignettes/SentimentAnalysis.html","timestamp":"2024-11-15T04:20:47Z","content_type":"text/html","content_length":"23389","record_id":"<urn:uuid:d2384380-6c1e-4bb7-b1b6-d81481683305>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00470.warc.gz"} |
Category: Crypto
Difficulty: Hard
Author: black-simon
What did you say?
nc hax1.allesctf.net 9400
The author provided a script server.py we send a private key to decrypt a fixed message: Quack! Quack! if this message decrypts to a sentence chosen by the author we get the flag.
My first approach was to come up with a general solution.
$m \equiv c^d\ (\textrm{mod}\ N)$ Since we control $d$ and $N$ a simple solution is:
$m \equiv c^k\ (\textrm{mod}\ c^k - m)$ And since for every divisor $t$ of $N$
$m \equiv c^d\ (\textrm{mod}\ t)$ is a solution, we "just" need to factor $c^k - m$ and if we find two prime-factors with their product is greater than $m$ we should get the flag. I factored it up to
$k = 6$, but had no luck and the process was quite time consuming, therefore I knew it was not the ideal solution. I did some google searches and found this writeup from this years BSidesSF CTF it is
basically the same problem, I tried to understand what is going on, and I guess I kind of got it, but to be honest I had already thought about discrete logarithms, but I thought, that it would be
even harder to solve it that way, I still don't get why there are those special cases and how they work and I just copy-pasted the code and did some automation.
import numpy
from pyasn1.codec.der import encoder
from pyasn1.type.univ import SequenceOf
from pyasn1.type.univ import Integer as CompInt
import base64
import socket
def gen_cand(b,l):
p = 2
while log(p)/log(2) < b:
np = next_prime(numpy.random.randint(2,l))
p = p*np
return p+1
def kk(b,l):
while True:
p = gen_cand(b,l)
if is_prime(p):
return p
M = 1067267517149537754067764973523953846272152062302519819783794287703407438588906504446261381994947724460868747474504670998110717117637385810239484973100105019299532993569
C = 6453808645099481754496697330465
Q = 2
bits = 559
E = 0
while E <3:
P = kk(bits,10**6)
N = P*Q
sol = gp("znlog(%d, Mod(%d, %d))" % (M,C,N))
if len(sol) == 0:
D = Integer(sol)
E = inverse_mod(D,(P-1) * (Q-1))
P = P.__int__()
Q = Q.__int__()
D = D.__int__()
N = N.__int__()
E = E.__int__()
print("P: %d"%P)
print("Q: %d"%Q)
print("E: %d"%E)
print("D: %d"%D)
seq = SequenceOf(componentType=CompInt())
enc = encoder.encode([0,N,E,D,P,Q,D % (P-1), D % (Q-1), inverse_mod(Q,P).__int__()],asn1Spec=SequenceOf(componentType=CompInt()))
pem = '-----BEGIN RSA PRIVATE KEY-----\n%s\n-----END RSA PRIVATE KEY-----\n' % base64.b64encode(enc)
s = socket.socket(
socket.AF_INET, socket.SOCK_STREAM)
s.connect(("hax1.allesctf.net", 9400))
• upgrade your google skills 😉
• don't trust user input | {"url":"https://localo.ooo/writeups/ctfs/cscg2020/RSA%20Service/writeup.html","timestamp":"2024-11-11T21:27:40Z","content_type":"text/html","content_length":"39135","record_id":"<urn:uuid:55e626f2-daf1-4a77-add6-2bbe889bffbf>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00818.warc.gz"} |
Riser Diameter and Riser Height Calculation for Casting Sphere Exercise
What are the required calculations to determine the riser diameter and riser height for casting a sphere with given dimensions?
Given a casting sphere with a diameter of 4 inches, a flask with a drag of 4 inches, and a cope of 6 inches, how can we calculate the riser diameter and height based on the provided dimensions?
Calculation and Final Answer
The riser diameter is 0 inches and the riser height is 6 inches.
To determine the riser diameter and riser height for casting a sphere with the given dimensions, we need to utilize the formula for the volume of a cylinder as the riser dimensions are cylindrical.
The riser must be a blind side riser with two inches in the drag.
Given: - Diameter of sphere = 4 inches - Diameter of drag (flask) = 4 inches - Diameter of cope = 6 inches - Riser height in drag = 2 inches
Calculations: - Riser diameter = Diameter of sphere - Diameter of drag = 4 inches - 4 inches = 0 inches - Riser height = Height of sphere + Riser height in drag = 4 inches + 2 inches = 6 inches
Therefore, based on the calculations, the riser diameter for the casting exercise is 0 inches, and the riser height is 6 inches. This ensures that the riser functions effectively in the casting | {"url":"https://theletsgos.com/engineering/riser-diameter-and-riser-height-calculation-for-casting-sphere-exercise.html","timestamp":"2024-11-13T17:48:22Z","content_type":"text/html","content_length":"21932","record_id":"<urn:uuid:e7d29e00-8e2f-4f6f-901b-4d90d8e5225b>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00257.warc.gz"} |
Convert Wavelength In Gigametres to Wavelength In Nanometres
Please provide values below to convert wavelength in gigametres to wavelength in nanometres [nm], or vice versa.
Wavelength In Gigametres to Wavelength In Nanometres Conversion Table
Wavelength In Gigametres Wavelength In Nanometres [nm]
0.01 wavelength in gigametres 1.0E+16 nm
0.1 wavelength in gigametres 1.0E+17 nm
1 wavelength in gigametres 1.0E+18 nm
2 wavelength in gigametres 2.0E+18 nm
3 wavelength in gigametres 3.0E+18 nm
5 wavelength in gigametres 5.0E+18 nm
10 wavelength in gigametres 1.0E+19 nm
20 wavelength in gigametres 2.0E+19 nm
50 wavelength in gigametres 5.0E+19 nm
100 wavelength in gigametres 1.0E+20 nm
1000 wavelength in gigametres 1.0E+21 nm
How to Convert Wavelength In Gigametres to Wavelength In Nanometres
nm = 1.0E+18 × wavelength in gigametres
wavelength in gigametres = 1.0E-18 × nm
Example: convert 15 wavelength in gigametres to nm:
15 wavelength in gigametres = 1.0E+18 × 15 = 1.5E+19 nm
Convert Wavelength In Gigametres to Other Frequency Wavelength Units | {"url":"https://www.unitconverters.net/frequency-wavelength/wavelength-in-gigametres-to-wavelength-in-nanometres.htm","timestamp":"2024-11-14T07:52:18Z","content_type":"text/html","content_length":"11120","record_id":"<urn:uuid:db31eac8-4ad3-4875-b374-19368ac7fc18>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00673.warc.gz"} |
Kinetic Energy and Work
Yolanda Mc Gehee Lincoln Park High School
2001 N Orchard St
Chicago IL 60614
(312) 534-8130
Kinetic energy can be thought of as the energy associated with the motion of an
object and is equivalent to work. An example of kinetic energy is a moving
hammer doing work on a nail. The hammer does work on the nail by driving it
into the wall.
The main objective of this Mini-teach is for students to observe kinetic energy
in balls of the same size but of different mass. Students are to see that
balls with a greater mass (weight) and greater height (velocity), have more
kinetic energy than balls with a lesser mass (weight) and a lower height
(velocity). They will be able to observe this by rolling balls of various
masses down a ramp at various heights and measuring the distance the ball moves
a block wall. In addition, students will be able to graph the relationship
between the distance the block wall moves versus the mass of each ball.
This Mini-teach can be used for grades 5-12.
Materials Needed:
For each group:
(1) ramp with rails
(2) rails
(1) block wall
(3) balls (same size, different mass)
(1) meter stick
(6) blocks
(1) balance
First, obtain three balls of about the same size but, of different mass. Weigh
the balls on a balance to determine the mass. Record the mass in grams. Next
create three data tables according to the number of blocks used, (2, 4 or 6).
In the first column of the data table, list the three balls according to their
mass starting from the least to greatest. For the next three columns, label
them "trial 1", "trial 2" and "trial 3". Label the final column "average".
This is where you will record the average distance of the three trials.
To obtain the first set of data for the first data table, stack the two blocks
on top of each other. Place the ramp on top of the blocks on an angle. At the
bottom of the ramp, place each of the rails next to each other leaving a space
between them. Place the block wall securely between the two rails. Starting at
the top of the ramp, roll the first of the three balls down the ramp. Allow the
ball to hit the block wall until it has used up all of its energy (until the
ball comes to a complete stop). Measure the distance the block has moved with a
meter stick. Remember to measure the distance from the end of the ramp to the
beginning of the block wall.
Repeat these steps two more times using the same ball at the same height.
Perform these same steps using the two other balls at the same height. Do this
for each ball at each height (4 and 6 blocks). Record the data in the data
Finally, use the average distances and the mass of the balls to make a graph.
Place the average distances on the x-axis and the mass of the balls on the
Performance Assessment:
The performance assessment used in this mini-teach is breaking a piece of wood
with your hand. This is done by securing a piece of wood on a board holder so
that it won't move. Next, make a fist with your hand making sure that your
thumb is on the outside of your fist. Lastly, with a very constant motion,
force and upward swing of your body, strike the middle of the board. Surprise,
the board breaks!
How do you have to move your hand to do the most work in order to break the
In order to do the most work and thus break the board, you must hit the board
very fast and at a constant speed. The board should also be hit on the grain of
the wood because this is where the board is weakest. Finally, your hand should
hit the board from a high distance to increase the amount of kinetic energy in
your hand.
Return to Physics Index | {"url":"https://smileprogram.info/ph9614.html","timestamp":"2024-11-13T18:32:29Z","content_type":"text/html","content_length":"4624","record_id":"<urn:uuid:48d88677-e10b-467b-b4f7-075c10670409>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00138.warc.gz"} |
Srigirisai J.
What do you want to work on?
About Srigirisai J.
Math - Quantitative Reasoning
very patient and helped me work the problem out myself so that i can make sense of it
Math - Quantitative Reasoning
great tutoring services
Math - Statistics
I wasn’t really aware of these resources before, and I feel like if I had understood them better, I would have used them sooner. I appreciate the support I received—my tutor was very patient and
explained things in a simple, clear way. This was my first time using the site, and it was a good experience overall. I’m glad Louisiana offers resources like this, and I hope they continue to be
Math - Quantitative Reasoning
I enjoy picking up on what I have forgotten how to do from when I was in elementary, like rounding to the nearest cent> | {"url":"https://www.princetonreview.com/academic-tutoring/tutor/srigirisai%20j--3162163","timestamp":"2024-11-07T03:18:31Z","content_type":"application/xhtml+xml","content_length":"267743","record_id":"<urn:uuid:84e99e16-6306-4a1e-8775-c489ca8c062c>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00514.warc.gz"} |
In a world where processing power comes in growing abundance, Monte Carlo methods thrive thanks to their intuitive simplicity and ease of implementation. Over the years, improved versions of the
standard Monte Carlo have emerged reducing the error around the estimate for a given sample size. In this article, we will review the most popular methods and compare their individual efficiency
through solving a toy game using Python.
Toy game valuation
Let’s first define a simple toy game and solve it analytically. Here are the rules:
• Throw 2 fair dice and sum their results.
• The casino pays you the difference between the sum and 8 if it is positive or nothing otherwise.
What is the fair value of the game? Its fair value is the expected value of its payoff. If we call \( S \) the random variable representing the sum of the dice throw, we can derive its distribution
by counting the number of occurence of each possible outcome:
s 2 3 4 5 6 7 8 9 10 11 12
p(s) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
Since the payoff function \( f \) of our game can be expressed as:
$$ f \left( s \right) = \max \left( s - 8, 0 \right) $$
We can then easily compute its expected value:
$$ \mathbb{E} \left[ f (S) \right] = \sum_{s=9}^{12} f(s)\ p(s) = \dfrac{5}{9} \approx 0.5556 $$
Let’s code this analytical solution in Python:
import numpy
s_values = numpy.arange(2, 13)
s_probas = numpy.concatenate([ numpy.arange(1,7), numpy.arange(5,0,-1) ]) / 36.0
fv = sum( numpy.maximum( s_values - 8.0 , 0) * s_probas )
print(f"fair_value= {fv:.4f}") # fair_value= 0.5556
Before moving on to Monte Carlo methods, let’s define a helper function for printing the mean of a sample and its theoretical standard deviation (see the Central Limit Theorem for more detail)
def mc_summary( xs ):
N = len( xs )
mu = xs.mean()
sigma = xs.std() / numpy.sqrt( N )
print(f"MC: mean={mu:.4f}, stdev={sigma:.4f}")
Now that we have the analytical answer, let’s look into computing the payoff expectation using various Monte Carlo algorithms and compare their relative performance.
Standard Monte Carlo
The standard MC algorithm is a direct application of the Law of Large Numbers which states that the mean of the samples of a random variable tends towards its theoretical expected value (unbiased
estimator) as the sample size increases.
The standard MC algorithm is simply:
• Sample \( N \) values of \( S \): \( \{ s_i \}, i \in \{1 \cdots N\} \)
• Compute the mean of the payoff for each value sampled: $$ \mathbb{E} \left[ f(S) \right] \approx \dfrac{1}{N} \sum_{i = 1}^{N} f(s_i) $$
Here is the Python code for a sample size of 100k which will be the same across all MC methods:
N = 100000
simulations = numpy.random.choice([1,2,3,4,5,6],size=[N,2]).sum(axis=1)
mc_summary( numpy.maximum( simulations - 8.0, 0 ) ) # MC: mean=0.5584, stdev=0.0033
We find that the theoretical standard deviation of our estimate is around 33bp (bp = basis point). This measurement will be our reference for evaluating the efficiency of the following variance
reduction methods.
Control Variate
The first variance reduction method is based around the use of a control variate. The idea is to add to every sample of the payoff function, a corrective term that tracks the deviation from the mean
of the sampled variable. The mean of this corrective term tending towards zero, this transformation basically preserves the expectation of the payoff.
Let’s write as \( \mu \), the expectation of \( S \):
$$ \mu = \mathbb{E} \left[ S \right] $$
The MC algorithm with control variate is:
• Sample \( N \) values of \( S \): \( \{ s_i \}, i \in \{1 \cdots N\} \)
• Compute the mean of the payoff for each value sampled: $$ \mathbb{E} \left[ f(S) \right] \approx \dfrac{1}{N} \sum_{i = 1}^{N} \left[ f(s_i) + c \left( s_i - \mu \right) \right] $$
where \( c \) is chosen as opposite to the correlation between \( S \) and \( f(S) \)
We choose the coefficient \( c \) that minimises the standard deviation of the estimate:
The implementation of the algorithm in Python is very straightforward:
mu = numpy.sum( s_values * s_probas )
mc_summary( numpy.maximum(simulations - 8.0, 0) - 0.3 * ( simulations - mu ) ) # MC: mean=0.5555, stdev=0.0021
Antithetic Variate
The antithetic variate method consists in replacing the payoff function computed for a sampled outcome with the average with its computed value for an outcome chosen on the other side of the mean and
of the same probability as the original sampled outcome. This way, the expected payoff estimation is again preserved in the tranformation.
The MC algorithm with antithetic variate is:
• Sample \( N \) values of \( S \): \( \{ s_i \}, i \in \{1 \cdots N\} \)
• Compute the mean of the payoff for each value sampled: $$ \mathbb{E} \left[ f(S) \right] \approx \dfrac{1}{N} \sum_{i = 1}^{N} \dfrac{f(s_i) + f(\bar{s}_i)}{2} $$ where \( \bar{s}_i \) is the
symmetrical value to \( s_i \) on the other side of the mean
The implementation of the algorithm in Python in our case is also very straightforward:
antithetics = 14 - simulations
mc_summary( 0.5 * ( numpy.maximum( simulations - 8.0, 0 ) + numpy.maximum( antithetics - 8.0, 0 ) ) ) # MC: mean=0.5546, stdev=0.0020
Importance Sampling
We can gain an intuition around this next technique by reasoning about the toy game in the following manner. Since the payoff is zero for a large subset of the final outcomes \( (s \leq 8) \), the
contribution of these to the calculation of the average payoff is less than those that generate a strictly positive value. So, could we sample more of these meaningful values while preserving the
mean estimate? This is what importance sampling effectively does. We define a new random variable covering the same outcomes as our original one, but with a distribution skewed towards outcomes of
Mathematically, let’s define the random variable \( X \) defined on the same measurable space than \( S \) but where the probability for a strictly positive payoff is much larger than the probability
of a payoff equal to zero:
x 2 3 4 5 6 7 8 9 10 11 12
p(x) 1/35 1/35 1/35 1/35 1/35 1/35 1/35 7/35 7/35 7/35 7/35
Using a change of measure from \( S \) to \( X \) we can write:
$$ \mathbb{E}_S \left[ f(S) \right] = \mathbb{E}_X \left[ \dfrac{f(X)\ p_S(X)}{p_X(X)} \right] $$
The MC algorithm with importance sampling with respect to \( X \) is:
• Sample \( N \) values of \( X \): \( \{ x_i \}, i \in \{1 \cdots N\} \)
• Compute the mean of the payoff for each value sampled: $$ \mathbb{E} \left[ f(S) \right] \approx \dfrac{1}{N} \sum_{i = 1}^{N} \dfrac{f(x_i)\ p_S(x_i)}{p_X(x_i)} $$
Its Python implementation requires slightly more variable definitions but still remains fairly concise:
x_probas = numpy.concatenate([ numpy.ones(7) / 35.0, numpy.ones(4) * 7.0/35.0 ])
p_s = numpy.vectorize( lambda x: s_probas[ s_values.tolist().index(x) ] )
p_x = numpy.vectorize( lambda x: x_probas[ s_values.tolist().index(x) ] )
simulations_x = numpy.random.choice(s_values, size=N, p=x_probas)
mc_summary( numpy.maximum( simulations_x - 8.0, 0 ) * p_s( simulations_x ) / p_x( simulations_x ) ) # MC: mean=0.5552, stdev=0.0010
Stratified Sampling
This last method consists in drawing a random variable from different subsets of outcomes while preserving its mean by adjusting the frequencies of draws from each subset. It follows in the steps of
importance sampling but goes even further because it allows us to arbitrarily choose subsets that simplify the resolution of our specific case.
For our toy game, we can choose to only draw from the subset of outcomes larger than 8 since all the others have a payoff of zero, and as such do not contribute any value to the calculation of the
Mathematically, let \( Y \) be the set of the value of \( S \) greater than 8. Since the payoff function is equal to 0 for any value of S not contained in \( Y \), by the Law of Total Expectation, we
can write:
$$ \mathbb{E} \left[ f(S) \right] = \mathbb{E} \left[ f(Y) \vert Y \right]\ \mathbb{Pr} \left( Y \right) $$
We can easily compute:
$$ \mathbb{Pr} \left( Y \right) = \dfrac{4 + 3 + 2 + 1}{36} = \dfrac{5}{18} $$
We compute the conditional density with respect to \( Y \):
y 9 10 11 12
p(y) 2/5 3/10 1/5 1/10
The MC algorithm with stratified sampling with respect to \( Y \) is:
• Sample \( N \) values of \( Y \): \( \{ y_i \}, i \in \{1 \cdots N\} \)
• Compute the mean of the payoff for each value sampled: $$ \mathbb{E} \left[ f(S) \right] \approx \dfrac{1}{N} \sum_{i = 1}^{N} f(y_i)\ \mathbb{Pr}\left( Y \right) $$
Another very succinct Python implementation:
y_values = numpy.arange(9,13)
y_probas = 0.1 * numpy.arange(4,0,-1)
simulations_y = numpy.random.choice(y_values, size=N, p=y_probas)
mc_summary( ( simulations_y - 8.0 ) * 5.0 / 18.0 ) # MC: mean=0.5560, stdev=0.0009
Here are the improvements in standard deviations of the estimated expected payoff in our toy game and for all methods reviewed. The standard MC value is used as reference:
The level of performance can be divided into two distinct groups: The control variate and antithetic variate methods in the first, and importance sampling and stratified samping in the other. The
latter group which implements an actual distribution adjustment achieves far better variance reduction overall.
For path dependent cases and complex payoffs, efficiently applying these variance reduction techniques can be more challenging. However, trying and fine-tuning different approaches often leads to
finding a decent solution.
If you like this post, follow me on Twitter and get notified on the next posts. | {"url":"https://vegapit.com/article/montecarlo-variance-reduction-techniques","timestamp":"2024-11-03T13:56:55Z","content_type":"text/html","content_length":"23050","record_id":"<urn:uuid:0d2e03ae-4f38-4e58-a430-427498cbeeec>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00197.warc.gz"} |
NCERT Solutions for Class 12 Maths Exercise 5.3
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 of Continuity and Differentiability in Hindi and English Medium for new session 2024-25. Class 12 Maths ex. 5.3 is updated as per the new
NCERT textbook for CBSE 2024-25.
12th Maths Exercise 5.3 Solutions in Hindi and English Medium
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3
Grade XII Mathematics Ex. 5.3 solution and study material are available in Videos and PDF format free to use online or download for offline use. NCERT Solutions for 12th Maths are in updated format
for new academic session 2024-25 based on latest NCERT Books and new CBSE Syllabus. UP Board and MP Board students are also using NCERT Textbooks for this academic session.
So, these NCERT Sols are useful for them also. They can download Solutions for class 12 Maths Ex. 5.3 in Hindi Medium. These are updated for CBSE and UP Board students as well as Uttarakhand and
Bihar board students whoever are following the NCERT Textbooks for their exams. Videos related to 12th Maths ex. 5.3 in Hindi and English are given below separately. For any inconvenience please
contact us for help.
Class: 12 Mathematics
Chapter 5: Exercise 5.3
Topic Name: Continuity and Differentiability
Content: Textbook Exercise Solution
Content Type: Text and Online Videos
Medium: English and Hindi Medium
12th Maths Exercise 5.3 Solutions
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3 Continuity and Differentiability in English and Hindi Medium free to use. In this exercise we have to go with direct differentiation with
respect to x and derivative of inverse trigonometric functions. 12th Maths Ex. 5.3 Continuity and Differentiability are given solved here. Join the discussion forum to ask your doubts related to NIOS
Board or CBSE.
About 12 Maths Exercise 5.3
In this exercise, we don’t need to separate the dependent and independent variables before differentiation. We can directly differentiate the entire equation. The questions based on inverse
trigonometric functions are generally on the basis of the formulae sin 2θ, cos 2θ and tan 3θ in terms of tan θ. Simplify all the questions first then differentiate in this exercise.
Feedback and Suggestions
NCERT Solutions are being updated based on new NCERT Books following the latest CBSE Syllabus. Just provide feedback and suggestions to improve the website. If you have any doubt related to NIOS or
CBSE board, please ask in discussion forum.
Which new concepts students will study in exercise 5.3 of 12th standard Maths Book?
The new concepts that students will study in exercise 5.3 of 12th standard Maths are how to find the derivative of implicit functions and the derivative of inverse trigonometric functions.
Is exercise 5.3 of class 12th Maths NCERT Book lengthy?
No, exercise 5.3 of class 12th Maths is not lengthy. There are four examples (examples 24, 25, 26, 27) and 15 questions in exercise 5.3. if students give 1-2 hours per day to this exercise, they can
complete it in 3 days (approximately). This time can vary because no students can have the same working speed.
Which sums of exercise 5.3 have more chances to come in the first term board exams?
There are four examples (examples 24, 25, 26, 27) and 15 questions in exercise 5.3 of grade 12th Maths. All sums of this exercise are important for the exams. Students should practice all sums of
this exercise. Some questions and examples are there in exercise 5.3, which has more chances to come in the board exams. These sums are example 26 and questions 4, 7, 9, 10, 11, 12, 13, 15. These
sums are most important because these sums have already been asked in board exams several times.
What is the difficulty level of exercise 5.3 class 12th Maths Solution?
Exercise 5.3 of class 12th Maths is not easy and not complicated. It lies in the middle of easy and difficult because some problems of this exercise are easy, and some are complex. However, the
difficulty level of any problem varies from student to student. So, exercise 5.3 of class 12th Maths is easy or not depends on students also. Some students find it difficult, some find it easy, and
some find it in the middle of easy and difficult.
Last Edited: April 7, 2023 | {"url":"https://www.tiwariacademy.com/ncert-solutions/class-12/maths/chapter-5/exercise-5-3/","timestamp":"2024-11-11T08:02:36Z","content_type":"text/html","content_length":"244013","record_id":"<urn:uuid:6901fe0a-ddd3-4dfc-9c92-c65755b88870>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00571.warc.gz"} |
Tillerson just sacked ... how will market react?
Tillerson was just sacked! I was just getting excited about the market today because of the inflation report ... just opened, and not sure what the reaction is going to be.
Pavel + 384
I don't know what will be with market today, but here are the top officials in the Trump white house who have left (before Tillerson)
LAOIL + 33
right, another MAJOR announcement that went down on Twitter:
no perceptible market reaction to this announcement yet as far as I can see
13 minutes ago, Pavel said:
I sense certain disbelief from the guy on the painting above
• 1
"Trump: I made decision to oust Tillerson 'by myself'"
You have to wonder why he would say this, right?
Adam Varga + 123
2 hours ago, Seleskya said:
Tillerson was just sacked! I was just getting excited about the market today because of the inflation report ... just opened, and not sure what the reaction is going to be.
USD, for starters
Selva + 252
If you were wondering who's next
Marina Schwarz + 1,576
I have to hand it to Trump for keeping things so much more interesting than pretty much anyone before him.
Adam Varga + 123
Well there is no reaction from the market, yet. Oil market is still focused on supplies. But his exit could create big implications for the global oil market. We have to see what will happen with
Iran and Venezuela.
2 hours ago, Adam Varga said:
Well there is no reaction from the market, yet. Oil market is still focused on supplies. But his exit could create big implications for the global oil market. We have to see what will happen with
Iran and Venezuela.
I guess, but largely I think the world is unphased by the rash of firings. Perhaps Pompeo's actions with regards to Venezuela or Iran may have implications, but Tillerson's exit itself is probably
not going to move the needle much.
Vlad Kovalenko + 115
On 3/14/2018 at 5:05 AM, Marina Schwarz said:
I have to hand it to Trump for keeping things so much more interesting than pretty much anyone before him.
Survivor-White House
• 2
Missy + 43
Addy + 14
How about Fortnite? They should offer a new mode (they update every week it seems with new modes) where the 100 participants are actually the 100 closest people to P. Trump. Last man standing wins.
• 1
Marina Schwarz + 1,576
On 3/14/2018 at 1:34 PM, Adam Varga said:
Well there is no reaction from the market, yet. Oil market is still focused on supplies. But his exit could create big implications for the global oil market. We have to see what will happen with
Iran and Venezuela.
Analysts are predicting the potential for up to a 1.4 bpd loss in oil production upon Pompeo's taking over for Tillerson from Iran and Venezuela.
There are a lot of "ifs" here. https://www.platts.com/latest-news/oil/washington/us-foreign-policy-turn-could-take-14-million-26913777
Missy + 43
Well, look, Venezuelan production is going to implode all on its own, without the help of Pompeo's policies. | {"url":"https://community.oilprice.com/topic/1240-tillerson-just-sacked-how-will-market-react/","timestamp":"2024-11-01T20:36:23Z","content_type":"text/html","content_length":"478990","record_id":"<urn:uuid:0420d9b5-47a7-40ba-8789-15ec4ee8a8ef>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00167.warc.gz"} |
class torch.optim.lr_scheduler.CyclicLR(optimizer, base_lr, max_lr, step_size_up=2000, step_size_down=None, mode='triangular', gamma=1.0, scale_fn=None, scale_mode='cycle', cycle_momentum=True,
base_momentum=0.8, max_momentum=0.9, last_epoch=-1, verbose=False)[source]¶
Sets the learning rate of each parameter group according to cyclical learning rate policy (CLR). The policy cycles the learning rate between two boundaries with a constant frequency, as detailed
in the paper Cyclical Learning Rates for Training Neural Networks. The distance between the two boundaries can be scaled on a per-iteration or per-cycle basis.
Cyclical learning rate policy changes the learning rate after every batch. step should be called after a batch has been used for training.
This class has three built-in policies, as put forth in the paper:
□ “triangular”: A basic triangular cycle without amplitude scaling.
□ “triangular2”: A basic triangular cycle that scales initial amplitude by half each cycle.
□ “exp_range”: A cycle that scales initial amplitude by $\text{gamma}^{\text{cycle iterations}}$ at each cycle iteration.
This implementation was adapted from the github repo: bckenstler/CLR
☆ optimizer (Optimizer) – Wrapped optimizer.
☆ base_lr (float or list) – Initial learning rate which is the lower boundary in the cycle for each parameter group.
☆ max_lr (float or list) – Upper learning rate boundaries in the cycle for each parameter group. Functionally, it defines the cycle amplitude (max_lr - base_lr). The lr at any cycle is the
sum of base_lr and some scaling of the amplitude; therefore max_lr may not actually be reached depending on scaling function.
☆ step_size_up (int) – Number of training iterations in the increasing half of a cycle. Default: 2000
☆ step_size_down (int) – Number of training iterations in the decreasing half of a cycle. If step_size_down is None, it is set to step_size_up. Default: None
☆ mode (str) – One of {triangular, triangular2, exp_range}. Values correspond to policies detailed above. If scale_fn is not None, this argument is ignored. Default: ‘triangular’
☆ gamma (float) – Constant in ‘exp_range’ scaling function: gamma**(cycle iterations) Default: 1.0
☆ scale_fn (function) – Custom scaling policy defined by a single argument lambda function, where 0 <= scale_fn(x) <= 1 for all x >= 0. If specified, then ‘mode’ is ignored. Default: None
☆ scale_mode (str) – {‘cycle’, ‘iterations’}. Defines whether scale_fn is evaluated on cycle number or cycle iterations (training iterations since start of cycle). Default: ‘cycle’
☆ cycle_momentum (bool) – If True, momentum is cycled inversely to learning rate between ‘base_momentum’ and ‘max_momentum’. Default: True
☆ base_momentum (float or list) – Lower momentum boundaries in the cycle for each parameter group. Note that momentum is cycled inversely to learning rate; at the peak of a cycle, momentum
is ‘base_momentum’ and learning rate is ‘max_lr’. Default: 0.8
☆ max_momentum (float or list) – Upper momentum boundaries in the cycle for each parameter group. Functionally, it defines the cycle amplitude (max_momentum - base_momentum). The momentum
at any cycle is the difference of max_momentum and some scaling of the amplitude; therefore base_momentum may not actually be reached depending on scaling function. Note that momentum is
cycled inversely to learning rate; at the start of a cycle, momentum is ‘max_momentum’ and learning rate is ‘base_lr’ Default: 0.9
☆ last_epoch (int) – The index of the last batch. This parameter is used when resuming a training job. Since step() should be invoked after each batch instead of after each epoch, this
number represents the total number of batches computed, not the total number of epochs computed. When last_epoch=-1, the schedule is started from the beginning. Default: -1
☆ verbose (bool) – If True, prints a message to stdout for each update. Default: False.
>>> optimizer = torch.optim.SGD(model.parameters(), lr=0.1, momentum=0.9)
>>> scheduler = torch.optim.lr_scheduler.CyclicLR(optimizer, base_lr=0.01, max_lr=0.1)
>>> data_loader = torch.utils.data.DataLoader(...)
>>> for epoch in range(10):
>>> for batch in data_loader:
>>> train_batch(...)
>>> scheduler.step()
Return last computed learning rate by current scheduler.
Calculates the learning rate at batch index. This function treats self.last_epoch as the last batch index.
If self.cycle_momentum is True, this function has a side effect of updating the optimizer’s momentum.
print_lr(is_verbose, group, lr, epoch=None)¶
Display the current learning rate. | {"url":"https://pytorch.org/docs/2.1/generated/torch.optim.lr_scheduler.CyclicLR.html","timestamp":"2024-11-04T06:02:10Z","content_type":"text/html","content_length":"55500","record_id":"<urn:uuid:ea4bb533-52c7-4e24-b9f7-4d35d9840b42>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00379.warc.gz"} |
What is 793 Grams in Kilograms?
793 Grams =
0.793 Kilograms
Unit Converter
How to convert 793 Grams to Kilograms
To calculate 793 Grams to the corresponding value in Kilograms, multiply the quantity in Grams by 0.001 (conversion factor). In this case we should multiply 793 Grams by 0.001 to get the equivalent
result in Kilograms:
793 Grams x 0.001 = 0.793 Kilograms
793 Grams is equivalent to 0.793 Kilograms.
How to convert from Grams to Kilograms
The conversion factor from Grams to Kilograms is 0.001. To find out how many Grams in Kilograms, multiply by the conversion factor or use the Mass converter above. Seven hundred ninety-three Grams is
equivalent to zero point seven nine three Kilograms.
Definition of Gram
The gram (alternative spelling: gramme; SI unit symbol: g) is a metric system unit of mass. A gram is defined as one one-thousandth of the SI base unit, the kilogram, or 1×10−3 kg, which itself is
now defined, not in terms of grams, but as being equal to the mass of a physical prototype of a specific alloy kept locked up and preserved by the International Bureau of Weights and Measures.
Definition of Kilogram
The kilogram (or kilogramme, SI symbol: kg), also known as the kilo, is the fundamental unit of mass in the International System of Units. Defined as being equal to the mass of the International
Prototype Kilogram (IPK), that is almost exactly equal to the mass of one liter of water. The kilogram is the only SI base unit using an SI prefix ("kilo", symbol "k") as part of its name. The
stability of kilogram is really important, for four of the seven fundamental units in the SI system are defined relative to it.
Using the Grams to Kilograms converter you can get answers to questions like the following:
• How many Kilograms are in 793 Grams?
• 793 Grams is equal to how many Kilograms?
• How to convert 793 Grams to Kilograms?
• How many is 793 Grams in Kilograms?
• What is 793 Grams in Kilograms?
• How much is 793 Grams in Kilograms?
• How many kg are in 793 g?
• 793 g is equal to how many kg?
• How to convert 793 g to kg?
• How many is 793 g in kg?
• What is 793 g in kg?
• How much is 793 g in kg? | {"url":"https://whatisconvert.com/793-grams-in-kilograms","timestamp":"2024-11-03T19:40:13Z","content_type":"text/html","content_length":"36278","record_id":"<urn:uuid:cf9a2e14-b345-43e0-974d-7fe05c1d0925>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00759.warc.gz"} |
Sobol quasirandom point set
sobolset is a quasirandom point set object that produces points from the Sobol sequence. The Sobol sequence is a base-2 digital sequence that fills space in a highly uniform manner.
p = sobolset(d) constructs a d-dimensional point set p, which is a sobolset object with default property settings. The input argument d corresponds to the Dimensions property of p.
p = sobolset(d,Name,Value) sets properties of p using one or more name-value pair arguments. Enclose each property name in quotes. For example, sobolset(5,'Leap',2) creates a five-dimensional point
set from the first point, fourth point, seventh point, tenth point, and so on.
The returned object p encapsulates properties of a Sobol quasirandom sequence. The point set is finite, with a length determined by the Skip and Leap properties and by limits on the size of the point
set indices (maximum value of 2^53). Values of the point set are generated whenever you access p using net or parenthesis indexing. Values are not stored within p.
Dimensions — Number of dimensions
positive integer scalar in interval [1,1111]
This property is read-only.
Number of dimensions of the points in the point set, specified as a positive integer scalar in the interval [1,1111]. For example, each point in the point set p with p.Dimensions = 5 has five values.
Use the d input argument to specify the number of dimensions when you create a point set using the sobolset function.
Use the reduceDimensions object function to reduce the number of dimensions after you create a point set.
Leap — Interval between points
0 (default) | positive integer scalar
Interval between points in the sequence, specified as a positive integer scalar. In other words, the Leap property of a point set specifies the number of points in the sequence to leap over and omit
for every point taken. The default Leap value is 0, which corresponds to taking every point from the sequence.
Leaping is a technique used to improve the quality of a point set. However, you must choose the Leap values with care. Many Leap values create sequences that fail to touch on large
sub-hyper-rectangles of the unit hypercube and, therefore, fail to be a uniform quasirandom point set. For more information, see [4].
Example: p = sobolset(__,'Leap',50);
Example: p.Leap = 100;
PointOrder — Point generation method
'standard' (default) | 'graycode'
Point generation method, specified as 'standard' or 'graycode'. The PointOrder property specifies the order in which the Sobol sequence points are produced. When PointOrder is set to 'standard', the
points produced match the original Sobol sequence implementation. When PointOrder is set to 'graycode', the sequence is generated by an implementation that uses the Gray code of the index instead of
the index itself.
You can use the 'graycode' option for faster sequence generation, but the software then changes the order of the generated points. For more information on the Gray code implementation, see [1].
Example: p = sobolset(__,'PointOrder','graycode');
Example: p.PointOrder = 'standard';
ScrambleMethod — Settings that control scrambling
0x0 structure (default) | structure with Type and Options fields
Settings that control the scrambling of the sequence, specified as a structure with these fields:
• Type — A character vector containing the name of the scramble
• Options — A cell array of parameter values for the scramble
Use the scramble object function to set scrambles. For a list of valid scramble types, see the type input argument of scramble. An error occurs if you set an invalid scramble type for a given point
The ScrambleMethod property also accepts an empty matrix as a value. The software then clears all scrambling and sets the property to contain a 0x0 structure.
Skip — Number of initial points in sequence to omit
0 (default) | positive integer scalar
Number of initial points in the sequence to omit from the point set, specified as a positive integer scalar.
Initial points of a sequence sometimes exhibit undesirable properties. For example, the first point is often (0,0,0,...), which can cause the sequence to be unbalanced because the counterpart of the
point, (1,1,1,...), never appears. Also, initial points often exhibit correlations among different dimensions, and these correlations disappear later in the sequence.
Example: p = sobolset(__,'Skip',2e3);
Example: p.Skip = 1e3;
Type — Sequence type
'Sobol' (default)
This property is read-only.
Sequence type on which the quasirandom point set p is based, specified as 'Sobol'.
Object Functions
net Generate quasirandom point set
reduceDimensions Reduce dimensions of Sobol point set
scramble Scramble quasirandom point set
You can also use the following MATLAB^® functions with a sobolset object. The software treats the point set object like a matrix of multidimensional points.
length Length of largest array dimension
size Array size
Create Sobol Point Set
Generate a three-dimensional Sobol point set, skip the first 1000 values, and then retain every 101st point.
p = sobolset(3,'Skip',1e3,'Leap',1e2)
p =
Sobol point set in 3 dimensions (89180190640991 points)
Skip : 1000
Leap : 100
ScrambleMethod : none
PointOrder : standard
Apply a random linear scramble combined with a random digital shift by using scramble.
p = scramble(p,'MatousekAffineOwen')
p =
Sobol point set in 3 dimensions (89180190640991 points)
Skip : 1000
Leap : 100
ScrambleMethod : MatousekAffineOwen
PointOrder : standard
Generate the first four points by using net.
X0 = 4×3
0.7601 0.5919 0.9529
0.1795 0.0856 0.0491
0.5488 0.0785 0.8483
0.3882 0.8771 0.8755
Generate every third point, up to the eleventh point, by using parenthesis indexing.
X = 4×3
0.7601 0.5919 0.9529
0.3882 0.8771 0.8755
0.6905 0.4951 0.8464
0.1955 0.5679 0.3192
• The Skip and Leap properties are useful for parallel applications. For example, if you have a Parallel Computing Toolbox™ license, you can partition a sequence of points across N different
workers by using the function spmdIndex (Parallel Computing Toolbox). On each nth worker, set the Skip property of the point set to n – 1 and the Leap property to N – 1. The following code shows
how to partition a sequence across three workers.
Nworkers = 3;
p = sobolset(10,'Leap',Nworkers-1);
p.Skip = spmdIndex - 1;
% Compute something using points 1,4,7...
% or points 2,5,8... or points 3,6,9...
Sobol Sequence Generation
Consider a default sobolset object p that contains d-dimensional points. Each p(i,:) is a point in a Sobol sequence. The jth coordinate of the ith point, p(i,j), is equal to
$\left\{\begin{array}{cc}0,& i=1\text{\hspace{0.17em}}\\ {\gamma }_{i}\left(1\right){v}_{j}\left(1\right)\oplus {\gamma }_{i}\left(2\right){v}_{j}\left(2\right)\oplus ...,& i>1.\end{array}$
• The ${\gamma }_{i}\left(n\right)$ values are 0s or 1s such that
$i-1=\sum _{n=1}{\gamma }_{i}\left(n\right){2}^{n-1}.$
In other words, the ${\gamma }_{i}\left(n\right)$ values are the binary digits of the integer i – 1.
• The v[j](n) values are called direction numbers. They are uniquely defined for each coordinate j. For more details on these values, see Direction Numbers Generation.
• The ⊕ operator is the bitwise exclusive-or operator. For two numbers expressed in binary, the ⊕ operator compares the digits in each position. For a given digit position, the ⊕ operator returns a
1 if the digits in that position differ and returns a 0 if the digits in that position are the same.
□ For example, $19\oplus 24={\left(10011\right)}_{2}\oplus {\left(11000\right)}_{2}={\left(01011\right)}_{2}=11.$
□ Similarly, $\frac{1}{2}\oplus \frac{3}{4}={\left(0.1\right)}_{2}\oplus {\left(0.11\right)}_{2}={\left(0.01\right)}_{2}=\frac{1}{4}.$
For more information, see [3].
Direction Numbers Generation
The set of direction numbers v[j](n) depends on the coordinate j. Define the direction numbers in terms of m[j](n) values:
For each j, you can generate the direction numbers by selecting the following:
• A primitive polynomial in ${ℤ}_{2}$ of some degree s[j]
Each coefficient in the polynomial is either 0 or 1.
• s[j] initial direction numbers. For each initial direction number, the corresponding m[j](n) value must be either 1 or an odd number less than 2^n.
The remaining direction numbers are determined by the following recurrence relation, which uses the coefficients of the primitive polynomial, the previous direction numbers, and the ⊕ bitwise
exclusive-or operator.
${m}_{j}\left(n\right):=2{a}_{j}\left(1\right){m}_{j}\left(n-1\right)\oplus {2}^{2}{a}_{j}\left(2\right){m}_{j}\left(n-2\right)\oplus ...\oplus {2}^{{s}_{j}-1}{a}_{j}\left({s}_{j}-1\right){m}_{j}\
left(n-{s}_{j}+1\right)\oplus {2}^{{s}_{j}}{m}_{j}\left(n-{s}_{j}\right)\oplus {m}_{j}\left(n-{s}_{j}\right).$
sobolset uses the same primitive polynomials and initial direction numbers described in [3]. These parameters are provided for the first 1111 dimensions.
[1] Bratley, P., and B. L. Fox. “Algorithm 659 Implementing Sobol's Quasirandom Sequence Generator.” ACM Transactions on Mathematical Software. Vol. 14, No. 1, 1988, pp. 88–100.
[2] Hong, H. S., and F. J. Hickernell. “Algorithm 823: Implementing Scrambled Digital Sequences.” ACM Transactions on Mathematical Software. Vol. 29, No. 2, 2003, pp. 95–109.
[3] Joe, S., and F. Y. Kuo. “Remark on Algorithm 659: Implementing Sobol's Quasirandom Sequence Generator.” ACM Transactions on Mathematical Software. Vol. 29, No. 1, 2003, pp. 49–57.
[4] Kocis, L., and W. J. Whiten. “Computational Investigations of Low-Discrepancy Sequences.” ACM Transactions on Mathematical Software. Vol. 23, No. 2, 1997, pp. 266–294.
[5] Matousek, J. “On the L2-Discrepancy for Anchored Boxes.” Journal of Complexity. Vol. 14, No. 4, 1998, pp. 527–556.
Version History
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Operator differentiable functions
A scalar function f is called opertor differentiable if its extension via spectral theory to the self-adjoint members of {Mathematical expression}(H) is differentiable. The study of differentiation
and perturbation of such operator functions leads to the theory of mappings defined by the double operator integral {Mathematical expression} We give a new condition under which this mapping is
bounded on {Mathematical expression}(H). We also present a means of extending f to a function on all of {Mathematical expression}(H) and determine corresponding perturbation and differentiation
formulas. A connection with the "joint Peirce decomposition" from the theory of JB^*-triples is found. As an application we broaden the class of functions known to preserve the domain of the
generator of a strongly continuous one-parameter group of^*-automorphisms of a C^*-algebra.
ASJC Scopus subject areas
• Analysis
• Algebra and Number Theory
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Computing maximum cliques in B 2 EPG graphs
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Computing maximum cliques in B 2 EPG graphs
Nicolas Bousquet, Marc Heinrich
To cite this version:
Nicolas Bousquet, Marc Heinrich. Computing maximum cliques in B 2 EPG graphs. WG: Workshop
on Graph-Theoretic Concepts in Computer Science, Jun 2017, Eindhoven, Netherlands. �hal-01557335�
Computing maximum cliques in B
-EPG graphs
Nicolas Bousquet
G-SCOP (CNRS, Univ. Grenoble-Alpes), Grenoble, France.
Marc Heinrich
LIRIS (Universit´e Lyon 1, CNRS), Lyon, France, UMR5205.
July 6, 2017
EPG graphs, introduced by Golumbic et al. in 2009, are edge-intersection graphs of paths on an or-thogonal grid. The class Bk-EPG is the subclass of EPG graphs where the path on the grid associated
to each vertex has at most k bends. Epstein et al. showed in 2013 that computing a maximum clique in B1-EPG graphs is polynomial. As remarked in [Heldt et al., 2014], when the number of bends is at
least 4, the class contains 2-interval graphs for which computing a maximum clique is an NP-hard problem. The complexity status of the Maximum Clique problem remains open for B2and B3-EPG graphs. In
this paper, we show that we can compute a maximum clique in polynomial time in B2-EPG graphs given a representation of the graph.
Moreover, we show that a simple counting argument provides a 2(k + 1)-approximation for the color-ing problem on Bk-EPG graphs without knowcolor-ing the representation of the graph. It generalizes a
result of [Epstein et al, 2013] on B1-EPG graphs (where the representation was needed).
An Edge-intersection graph of Paths on a Grid (or EPG graph for short) is a graph where vertices can be repre-sented as paths on an orthogonal grid, and where there is an edge between two vertices if
their respective paths share at least one edge. A turn on a path is called a bend. EPG graphs were introduced by Golumbic, Lipshteyn and Stern in [10]. They showed that every graph can be represented
as an EPG graph. The number of bends on the representation of each vertex was later improved in [13]. EPG graphs have been in-troduced in the context of circuit layout, which can be modeled as paths
on a grid. EPG graphs are related to the knock-knee layout model where two wires may either cross on a grid point or bend at a common point, but are not allowed to share an edge of the grid. In [10],
the authors introduced a restriction on the number of bends on the path representing each vertex. The class Bk-EPG is the subclass of EPG graphs where the
path representing each vertex has at most k bends. Interval graphs (intersection graphs of intervals on the line) are B0-EPG graphs. The class of trees is in B1-EPG [10], outerplanar graphs are in
B2-EPG [14] and
planar graphs are in B4-EPG [14]. Several papers are devoted to prove structural and algorithmic properties
of EPG-graphs with a small number of bends, see for instance [1, 2, 6, 9].
While recognizing and finding a representation of a graph in B0-EPG (interval graph) can be done in
polynomial time, it is NP-complete to decide if a graph belongs to B1-EPG [5] or to B2-EPG [15]. The
complexity status remains open for more bends. Consequently, in all our results we will mention whether the representation of the graph is needed or not.
∗[Supported by ANR Projects STINT (]
ANR-13-BS02-0007) and LabEx PERSYVAL-Lab (ANR-11-LABX-0025-01).
Figure 1: A complete graph K6minus a matching.
Epstein et al. [7] showed that the k-coloring problem and the k-independent set problem are NP-complete restricted to B1-EPG graphs even if the representation of the graph is provided. Moreover they
4-approximation algorithms for both problems when the representation of the graph is given. Bessy et al. [4] proved that this there is no PTAS for the k-independent set problem on B1-EPG graphs and
that the
prob-lem is W [1]-hard on B2-EPG graphs (parameterized by k). Recently, Bonomo et al. [3] showed that every
B1-EPG graph admits a 4-clique coloring and provides a linear time algorithm that finds it, given the
rep-resentation of the graph.
Maximum Clique problem on EPG graphs. A claw of the grid is a set of three edges of the grid incident to the same point. Golumbic et al proved in [10] that a maximum clique in a B1-EPG graph can be
in polynomial time if the representation of the graph is given. This algorithm is based on the fact that, for every clique X of a B1-EPG graph, either there exists an edge e of the grid such that all
the vertices of X
contain e, or there exists a claw T such that all the vertices of X contain at least two of the three edges of T. In particular, it implies that the number of maximal cliques in B1-EPG graphs is
polynomial. Epstein et
al. [7] remarked that the representation of the graph is not needed since the neighborhood of every vertex is a weakly chordal graph. When the number of bends is at least 2, such a proof scheme
cannot hold since there might be an exponential number of maximal cliques. Indeed, one can construct a complete graph minus a matching in B2-EPG (see Figure 1) which has 2n/2 maximal cliques. So to
compute a maximum
clique on Bk-EPG graphs for k ≥ 2, a new proof technique has to be introduced.
EPG graphs are closely related to two well known classes of intersection graphs, namely k-interval graphs, and k-track interval graphs on which the maximum clique problem have been widely studied. A
k-interval is the union of k distinct intervals in the real line. A k-interval graph, introduced in [16], is the intersection graph of k-intervals. A k-track interval is the union of k intervals on
k-distinct lines (called tracks). A k-track interval graph is an intersection graph of k-track intervals (in other words, it is the edge union of k interval graphs on distinct lines). One can easily
check, as first observed in [13], that B3k−3-EPG
graphs contain k-track interval graphs and B4k−4-EPG graphs contain k-interval graphs.
Since computing a maximum clique in a 2-interval graph is NP-hard [8], the Maximum Clique Problem is NP-hard on B4-EPG graphs. So the complexity status of the Maximum Clique problem remains open on
Bk-EPG graphs for k = 2 and 3. In this paper, we prove that the Maximum Clique problem can be decided
in polynomial time on B2-EPG graphs when the representation of the graph is given. The proof scheme
of [10] cannot be extended to B2-EPG graphs. Indeed, there cannot exist a bijection between local structures,
like claws, and maximal cliques since there are examples with an exponential number of different maximum cliques. Our proof is divided into two main lemmas. The first one ensures that we can separate
so-called Z-vertices (vertices that use paths on two rows) from U-vertices (vertices that use edges of two columns). The second ensures that if a graph only contains Z-vertices, then all the maximal
cliques are included in a polynomial number of subgraphs; subgraphs for which a maximum clique can be computed in polynomial time.
Coloring Bk-EPG graphs. Gyarf´as proved in [11] that the chromatic number of k-interval graphs is bounded
by a function of the maximum clique using the degeneracy of the graph. In Section 5, we propose a slightly different proof than the one of Gyarf´as to prove the degeneracy of Bk-EPG graphs. This
bounds ensures
without knowing the representation of the graph, where χ(G) is the chromatic number of G. In particular, it provides a simple coloring algorithm using at most 4 times the optimal number of colors on
B1-EPG graphs
without knowing its representation. It improves the algorithm of [7] where the representation was needed. A class of graphs C is χ-bounded if there exists a function f such that χ(G) ≤ f (ω(G)) for
every graph Gof C with ω(G) the size of a maximum clique in G. Combinatorial bounds on the chromatic number of generalizations of interval graphs received a considerable attention initiated in [12].
The bound on the degeneracy of the graph ensures that the class Bk-EPG is χ-bounded and χ(G) ≤ 2(k + 1) · ω(G). As a
by-product, it also ensures that graphs in Bk-EPG contain either a clique or a stable set of size
q [n]
2(k+1) which
improves a result of [2] in which they show that every B1-EPG graph admits a clique or a stable set of size
Let a, b be two real values with a ≤ b. The (closed) interval [a, b] is the set of points between a and b containing both a and b. The interval that does not contain b is represented by [a, b) and
the one that does not contain aby (a, b].
An interval graph is an intersection graph of intervals in the line. More formally, vertices are intervals and two vertices are incident if their respective intervals intersect. Let H be an interval
graph with its representation. Let u ∈ V (H). The left extremity of u is the leftmost point p of the line such that u contains p. The right extremity of u is the rightmost point p of the line such
that u contains p.
Let G be an EPG graph with its representation on the grid. In what follows, we will always denote by roman letters a, b, . . . the rows of the grid and by greek letters α, β, . . . the columns of the
grids. Given a row a (resp. column α) of the grid, the row a − 1 (resp. α − 1) denotes the row under a (resp. at the left of α). Given a row a and a column α, we will denote by (a, α) the grid point
at the intersection of a and α. By abuse of notation, we will also denote by α (for a given row a) the point at the intersection of row a and column α. Let u ∈ V (G). We denote by Puthe path
representing u on the grid. The vertex u of G intersects
a row (resp. column) if Pucontains at least one edge of it.
Typed intervals and projection graphs
Let G be a B2-EPG graph with its representation on the grid. Free to slightly modify the representation of
G, we can assume that the path associated to every vertex has exactly 2 bends. Indeed, if there is a vertex usuch that Puhas less than two bends, let (a, α) be one of the two extremities of Pu. Up to
a rotation of
the grid, we can assume that the unique edge of Puincident to (a, α) is the horizontal edge e between (a, α)
and (a, α + 1). Then create a new column β between α and α + 1, and replace the edge e by two edges, one between α and β on row a, and another one going up at β. This transformation does not modify
the graph G. So we will assume in the following that for every vertex u, the path Puhas exactly two bends.
A Z-vertex of G is a vertex that intersects two rows and one column. A U-vertex is a vertex that intersects one row and two columns. The index of a vertex u is the set of rows intersected by u. The
vertex u contains ain its index if a is in the index of u.
Let u be a vertex containing a in its index. The extremities of u on a are the points of the row a on which Pustops or bends. Since Pu has at most two bends, Puhas exactly two extremities on a and
the subpath
of Puon row a is the interval of a between these two extremities. The a-interval of u, denoted by Pua, is the
interval between the two extremities of u on a. Note that since the index of u contains a, Pa
u contains at least
one edge. Let α ≤ β be two points of a. The path Pa
u intersects non-trivially [α, β] if Pua∩ [α, β] is not empty
or reduced to a single point. The path Pa
u weakly contains [α, β] if [α, β] is contained in P a u.
Figure 2: Examples of typed intervals on the same row. In this example, the interval t3is reduced to a single
point. The interval t2is coherent with the right extremity of t1, the extremities of t3and the left extremity
of t4. It is not coherent with the extremities of t5. Moreover, t2 intersects t1, t3and t4but not t5. And t2
contains t3.
Typed intervals
Knowing that Pa
u = [α, β]is not enough to understand the structure of Pu. Indeed whether Pustops at α,
or bends (upwards or downwards) at α, affects the neighborhood of u in the graph. To catch this difference we introduce typed intervals that contain information on the “possible bends” on the
extremities of the interval.
We define three types namely the empty type , the d-type and the u-type . A typed point (on row a) is a pair x, α where x is a type and α is the point at the intersection of row a and column α. A
typed interval (for row a) is a pair of typed points (x, α) and (y, β) (on row a) with α ≤ β denoted by [xα, yβ]. Informally, a typed interval is an interval [α, β] and indications on the structure
of the bends on the extremities. A typed interval t is proper if α 6= β or if α = β, x = y, and x ∈ { , }.
Let t = [xα, yβ] and t0= [x0α0, y0β0]be two typed intervals on a row a. Denote by zγ one of the endpoints of t0. We say that t is coherent with the endpoint zγ of t0if one of the following holds: (i)
γ is included in the open interval (α, β), or (ii) z = , and [α, β] contains the edge of [α0, β0]adjacent to γ, or (iii) z 6= , and zγ ∈ {xα, yβ}. We can remark in particular that if t is coherent
with an endpoint zγ, then γ is in the closed interval [α, β]. The typed interval t contains t0if [α0, β0] ⊂ [α, β], and t is coherent with both endpoints of t0. The typed interval t intersects t0[if
[α, β] intersects non trivially [α]0[, β]0[]][(i.e. the intersection contains at least]
one grid-edge), or t is coherent with an endpoint of t0[, or t]0[is coherent with an endpoint of t. Note that, if t]0
contains t then in particular it intersects t.
Let u be a vertex containing a in its index. The t-projection of u on a is the typed interval [xα, yβ] where α, βare the extremities of Pa
u and the type of an extremity γ ∈ {α, β} is if Pustops at γ and (resp. ) if
u bends downwards (resp. upwards) at γ. Note that this typed interval is proper since it contains at least
one edge. The path Pucontains (resp. intersects) a typed interval t (on a) if the t-projection of u on a contains
t(resp. intersects t). By abuse of notation we say that u contains or intersects t. Note that by definition, the path Pucontains the t-projection of u on a. Moreover if a vertex u contains a typed
interval t = [xα, yβ], then
the path Puweakly contains [α, β]. If u intersects t, then the path Puintersects the segment [α, β] (possibly
on a single point).
The following simple lemma motivates the introduction of typed intervals.
Lemma 1. Let G be a B2-EPG graph, let u, v be two vertices whose index contain a, and t be a proper typed interval
of a. If u contains t and v intersects t, then u and v are adjacent.
Proof. Let tu = [xα, yβ]and tv = [x0α0, y0β0]be the t-projections of u and v respectively. Let t = [xtαt, ytβt]
be the typed interval such that u contains t, and v intersects t. Then [αt, βt] ⊆ [α, β], and [αt, βt] ∩ [α0, β0] 6= ∅.
This implies in particular that the intersection of [α, β] and [α0, β0]is not empty.
If this intersection contains a grid edge e, then the paths Puand Pvboth contain this edge, and u, v are
adjacent in G.
If the intersection is reduced to a single point, then, since we have α 6= β and α0 6= β0[, we can assume]
w.l.o.g. that the right endpoint of tv coincide with the left endpoint of tu, i.e. β0 = α. By assumption on
u, we know that α ≤ αt. Additionally, since [αt, βt] ∩ [α0, β0] 6= ∅, we must have αt= α. Then either tvis
coherent with the endpoint xα of t, or t is coherent with the endpoint yβ0of tv. We will assume the former,
the other case can be treated exactly the same way. Since [α0, β0]does not contain the edge at the left of αt,
we have necessarily y0 [6= . This implies that y]0 [= x]
Figure 3: In this example, the projection graph on row a of the four vertices is an induced cycle of length four.
coherent with the endpoint xtαtof t. Consequently, we have x = xt= y0, which implies that both Puand
Pvbend at α in the same direction, and thus u, v are adjacent in G.
Lemma 2. Let G be a B2-EPG graph. If the t-projections of u and v on a intersect, then uv is an edge of G. Moreover
if u and v are two vertices containing a in their index and have no other row in common, then uv is an edge of G if and only if the t-projections of u and v on a intersect.
Proof. Let u and v be two vertices containing a in their index. The first part of the statement is just a corollary of Lemma 1 since v intersects the t-projection of u on a.
Assume that there is no other row b contained in the index of both u and v. And suppose moreover that the t-projections of u and v on a do not intersect. Suppose by contradiction that u and v are
adjacent in G. The two vertices cannot share a common edge on row a, otherwise their projections would intersect. By assumption they cannot share an edge on another row. Consequently, they must have
a common edge eon a column, and let α this column. By symmetry, we can assume that e is below the row a. Since the paths Puand Pvhave at most two bends, both path must bend downwards at the
intersection of row a and
column α, to intersect the edge e. However, this implies that the t-projections of u and v on a intersect, a contradiction.
Projection graphs
Let Y be the subset of vertices of G such that all the vertices of Y contain a in their index. The projection graph of Y on a is the graph on vertex set Y such that there is an edge between u and v
if and only if the t-projections of u and v on a intersect. Note that the projection of Y is not necessarily an interval graph since it can contain an induced C4(see Fig. 3). We say that a set of
vertices Y is a clique on a if the t-projection of
Y on a is a clique. Lemma 2, ensures that a clique on a is indeed a clique in the graph G.
In the very simple case where the representation uses only two rows a and b, we have the following lemma:
Lemma 3. Let G be a B2-EPG graph and Gabbe the subset of vertices with index {a, b}. Then Gabinduces a 2-track
interval graph.
Proof. Let us first prove that the projection graph of Gabon a is an interval graph. Let u be a vertex of Gab
and x and y be the endpoints of Pa
u. The vertex u bends either on x or on y (since its index is {a, b} it has
only one vertical segment). We associate to u the segment su = (x, y]if u bends on y, and [x, y) if it bends
on x. Then the t-projection of u and v intersect on a if and only if the intervals suand svintersect. Indeed,
if both u and v have a bend at the same column α, then they both contain the interval between row a and b on column α because they have the same index {a, b}. Then the projection graph of Gabon a is
an interval
By Lemma 2, any edge on the projection on a or b is also an edge of G. Conversely, if there is an edge between u and v then:
• either their a-segments (resp. b segments) intersect on at least one edge, and there is an edge in the projection graph of Gabon a (resp. b),
• or they share the same column. Then since they bend on the same point of the grid and on the same direction, there is an edge between u and v in the projection graphs of Gabon both a and b.
So there is an edge between u and v if and only if their t-intervals intersect on a or intersect on b. Since the t-projections of Gabon both a and b induce interval graphs, the graph Gabis a 2-track
interval graph.
Let us end this section with two lemmas that will be widely used all along the paper.
Lemma 4. Let G be a B2-EPG graph and Y be a subset of vertices whose index contain a. Suppose that the projection
of Y on a is a clique. Then there is a proper typed interval t such that: • all vertices of Y contain t,
• if u is a vertex with index {a} or {a, c} where c is not in the index of any vertex of Y , then u is complete to Y if and only if u intersects t.
Proof. Let α be the rightmost left extremity of an a-interval of Y , and β be the leftmost right extremity of an a-interval of Y . Since the projection of Y on a is a clique, α ≤ β. Let Yαbe the set
of vertices of Y whose
a-segment have left extremity α, and Yβbe the set of vertices of Y whose a-segment have right extremity β.
We define the typed interval t = [xα, yβ] where x is equal to (resp. ) if all the vertices of Yαbend upwards
(resp. downwards) at α, and is equal to otherwise. Similarly, y is equal to (resp. ) if all the vertices of Yβ bend upwards (resp. downwards) at α, and is equal to otherwise. Let us prove that t
satisfies the
conclusion of the lemma.
One can easily check that, by construction, all the vertices of Y contain the typed interval t. Indeed [α, β] is contained in all the intervals Pa
v for v ∈ Y by definition of α and β. Let us prove now by contradiction
that the typed interval t is proper. If α 6= β then t is proper, so we can suppose that α = β. Now assume that the types x and y are distinct or equal to . Up to symmetry, we can assume that x 6= and
y 6= . Consequently, there exists a vertex v1∈ Y such that Pva1has left extremity α and either starts on α or bends
downwards on α. Similarly, there exists a vertex v2∈ Y such that Pva2has right extremity β = α and either
ends on α or bends upwards on α. But then the t-projection of v1and v2do not intersect, a contradiction
since the t-projection of Y on a is a clique.
Let us finally prove the second point. Suppose that u intersects t, then for all y ∈ Y , y contains t, and by Lemma 2 u and y are adjacent in G. Assume now that u does not intersect t. Let us prove
that there exists y ∈ Y that is not incident to u. Either Pa
u = [α0, β0]does not intersect [α, β] or it intersects it on exactly one
vertex. We moreover know that if α = β then Pa
u does not contain the edge at the left and at the right of α
(otherwise u would intersect t). So, up to symmetry, we can assume that β0 [≤ α. If β]0 [< α][then let v be a]
vertex such that Pa
v has leftmost extremity α (such a vertex exists by definition of α). The projections of u
and v on a do not intersect and Lemma 2 ensures that u is not adjacent to v. Assume now that Pa
u intersects [α, β] on a single point. Since Puacontains at least one edge, this point is
either α or β, α say. Let xu be the type of β0. Since u does not intersect t, this means that either xu 6= x or
xu= xand xu= . Up to symmetry, we can assume xu 6= and x 6= . So there exists v ∈ Y such that Pva
has a left extremity at α and Pveither bends upwards at α or has no bend at α. Since xu 6= , Puahas right
extremity α and either bends downwards at α or has no bend at α. So the projections of u and v on a does not intersect. By Lemma 1, u and v are not adjacent in G.
Lemma 5. Let G be a B2-EPG graph and Y be a subset of vertices with index {a, b}. Suppose that the projection of
Y on a is not clique. Then there is a proper typed interval t such that: • All vertices of Y intersect t,
Proof. Let β be the rightmost left extremity of an a-interval of Y , and α be the leftmost right extremity of an a-interval of Y . Since the projection of Y on a is not a clique, α ≤ β. Let Yαbe the
subset of vertices of
Y whose a-segment have right extremity α, and Yβbe the subset of vertices of Y with left extremity β. We
consider the typed interval t given by t = [xα0, yβ0]where:
• x is equal to (resp. ) if all vertices of Yαbend upwards (resp. downwards) at α, and is equal to
• similarly, y is equal to (resp. ) if all vertices of Yβ bend upwards (resp. downwards) at α, and is
equal to otherwise,
• if x 6= then α0[= α][, otherwise α]0 [= α − 1][,]
• if y 6= then β0[= β][, otherwise β]0[= β + 1][.]
Let us prove that t is a proper typed interval. We can assume w.l.o.g. that a is below b. If α0 [6= β]0
then [α0[, β]0[]][is non reduced to a single point and then t is proper. So we can assume that α]0 [= β]0[. In the]
construction, if one of x or y is equal to , then α0 [6= β]0[and then t is proper. Additionally, since all vertices]
have index {a, b}, and a is below b, all vertices bend upwards on row a. Consequently, x and y cannot be equal to . If none of them is , then they are both , and again t is proper.
We will first prove that every vertex of Y intersects t. Let v be a vertex of Yα, then either α0= α − 1and
then Pa
v contains the edge [α − 1, α] and thus v intersects t, or α0 = α. In that case, x is equal to (resp. )
if all the vertices of Yαbends upwards (resp. downwards) on α. But then the t-projection of v has a right
endpoint with extremity α and type (resp. ), and then by definition, the t-projection of v is coherent with the endpoint xα of t, and v intersects t. A similar proof holds to show that t-projections
of vertices of Yβ
intersect t. Now let v ∈ Y \ (Yα∪ Yβ). The left extremity of v is before α and its right extremity is after β.
Let αvand βvbe the left and right extremities of Pva. They satisfy αv ≤ α and βv ≤ β. If α0 = α, then α is
contained in the open interval (αv, βv), and the t-projection of v on a is coherent with the extremity xα of t,
and by consequence v intersects t. If α0 = α − 1, then the t-projection of v weakly contains [α0, β0], and this interval is not reduced to a single point. Consequently v intersects t.
Let us now prove the second point. Let u be a vertex such that the index of u contains a and not b. If u contains t, then for any y ∈ Y , y intersects t, and by Lemma 2 u and y are adjacent in G.
Assume now that u does not contain t. We will show that there is a y ∈ Y such that u and y are not adjacent in G. By Lemma 1, it suffices to show that there exists a vertex v such that the
t-projections of u and v on a do not intersect. First suppose that Pudoes not weakly contain [α0, β0]. We can assume w.l.o.g.
that Pudoes not contain α0. Let αuand βube the left and right endpoints of Pua. If βu < α0, then there is a
vertex v ∈ Yβsuch that the left endpoint of Pvais β, and αu< α0≤ α ≤ β. Consequently the t-projection of
uand v do not intersect. Otherwise, we can assume αu > α0. If α0 = α, then there is a vertex v ∈ Yαsuch
that the right endpoint of Pa
v is α. Then, since α = α0 < αu, the t-projection of u and v do not intersect, and
by Lemma 1, u and v do not intersect. If α0 [= α − 1, then there is a vertex v such that α is the right endpoint]
of Pa
v, and Pvstops at α. Since α0 < αu, we have α ≤ αu. This implies that [αu, βu]does not contain the
edge at the left of α, which implies that the t-projection of u is not coherent with the right extremity of the t-projection of v. The same arguments show that the t-projection of v on a is not
coherent with the endpoints of the t-projection of u, and thus the t-projections of u and v do not intersect.
Suppose now that [αu, βu] weakly contains [α0, β0]. Let tu be the t-projection of u. Since u does not
contain t, tuis not coherent with one of its endpoints, say α0up to symmetry. Then we have α0 ∈ {αu, βu}.
Suppose by contradiction that x = . Then we can’t have α0 = βusince otherwise t would not be proper.
So necessarily α0 = αu, but this implies that tuis coherent with the endpoint α0 since [αu, βu]contains the
edge at the left of α0, a contradiction. Since a is below b, we must have x = . If α0= αu, then since tuis not
coherent with the endpoint α0of t, necessarily the right endpoint of tuhas type . Then, if v ∈ Yα, then by
construction of t, since x = we have α0 = α, and αuis the right endpoint of Pvu. Since Pudoes not bend
upwards at α, the t-projections of u and v do not intersect. If α0 = αu, then α0 = β0, and since t is proper,
we necessarily have y = . A similar argument shows that there exists v ∈ Yβwhose t-projection does not
Maximum clique in B
-EPG graphs
Graphs with Z-vertices
We start with the case where the graph only contains Z-vertices. We will show in Section 4.2 that it is possible to treat independently Z-vertices and U-vertices. The remaining part of Section 4.1 is
devoted to prove the following theorem.
Theorem 6. Let G be a B2-EPG graph with a representation containing only Z-vertices. The size of a maximum
clique can be computed in polynomial time.
Note that, up to rotation of the representation of G, this theorem also holds for U-vertices. In other words, the size of a maximum clique can be computed in polynomial time if the graph only
contains U-vertices.
The proof of Theorem 6 is divided in three steps. We will first define a notion of good subgraphs of G, and prove that:
• there is a polynomial number of good subgraphs of G and,
• a maximum clique of a good graph can be computed in polynomial time, • and any maximal clique of G is contained in a good subgraph.
Recall that a clique is maximum if its size is maximum. And it is maximal if it is maximal by inclusion. The first point is an immediate corollary of the definition of good graphs. The proof of the
second point consists in decomposing good graphs into sets on which a maximum clique can be computed efficiently. The proof of the third point, the most complicated, will be divided into several
lemmas depending on the structure of the maximal clique we are considering.
An induced subgraph H of G is a good graph if one of the following holds:
(I) there are two rows a and b, and two proper typed intervals taand tbon a and b respectively such that
H is the subgraph induced by vertices v such that, v contains ta, or v contains tb, or v intersects both
taand tb,
(II) or there are three rows a, b, and c, and three proper typed intervals ta, tb, and tcon a, b, c respectively
such that H is the subgraph induced by vertices v such that, either v contains ta, or v contains tb, or v
intersects tband contains tc.
Lemma 7. Let G be a B2-EPG graph. There are O(n6)good graphs, and a maximum clique of a good graph can be
computed in polynomial time.
Proof. A good graph is defined by two or three typed intervals. At first glance, to define a typed interval we need to choose a row, and then two points on it. So a natural upper bound on the number
of typed interval is O(n3[)][. However, we can make a slightly better evaluation. A point of the grid is important if the path of a]
vertex ends or bends on this point or on a point incident to it. There is a linear number of important points since every path defines a constant number of important points. One can easily notice,
given an interval ta
on row a, replacing an extremity r of ta by the important point s that is the closest from r on row a does
not modify the set of vertices intersecting or complete to ta. Indeed, no path starts, stops or bends on the
interval [r, s] on row a. So we can assume that all the endpoints of typed intervals are important points. This implies that there are at most O(n2[)][typed intervals. As a consequence, there are O
(n]4[)][good graphs]
of type (I), and O(n6[)][good graphs of type (II).]
Let us now prove that a maximum clique can be computed in polynomial time in a good graph. • Let H be a good graph of type (I) defined by two proper typed intervals taand tb. Let Habe the subset
of vertices containing ta, and Hbbe the set of vertices containing tb, and Habbe the other vertices of H
both Haand Hbare cliques. Let H1be the graph induced by Ha∪ Hb, and H2be the graph induced by
Hab. Then H is the join of H1and H2(i.e. there is an edge between any vertex of H1and any vertex
of H2). So a maximum clique of H is the union of a maximum clique of H1and a maximum clique of
H2. Moreover:
– Since both Ha and Hbinduce cliques, H1is the complement of a bipartite graph. Computing a
maximum clique in H1is the same as computing a maximal independent set in ¯H1, the
comple-ment graph of H1 which is bipartite. Additionally, computing a maximum independent set in
a bipartite graph can be done in polynomial time. Indeed, a maximum independent set is the complement of a minimum vertex cover. In bipartite graphs, a minimum vertex cover can be computed in
polynomial time using for instance Linear Programming.
– By Lemma 3, H2 is a 2-track interval graph, on which a maximum clique can be computed in
polynomial time [8].
Consequently, a maximum clique of both H1and H2, and then of H, can be computed in polynomial
• Let H be a good graph of type (II) defined by three proper typed intervals ta, tband tc. Let Haand Hb
be the set of vertices containing taand tbrespectively, and Hbcbe the set of vertices intersecting tband
containing tc. By Lemma 1, Ha, Hb, and Hbcare cliques since for each of these sets there is a proper
typed interval contained in every vertex of the set. Moreover Hbcis complete to Hbsince vertices of
Hbcontain tband vertices of Hbcintersect it. Let H1= Haand H2= Hb∪ Hab. Both H1and H2induce
cliques. So H is the complement of a bipartite graph, and a maximum clique of H can be computed in polynomial time.
The remaining part of Section 4.1 is devoted to prove that any maximal clique of G is contained in a good graph.
Lemma 8. Let G be a B2-EPG graph containing only Z-vertices, and X be a clique of G. Assume that there are two
rows a and b such that every horizontal segment of X is included in either a or b, then X is contained in a good graph. Proof. By taking two typed intervals taand tbconsisting of the whole rows a and
b, the clique X is clearly
contained in a good graph of type (I).
We say that a set of vertices X intersects a column α (resp. a row a) if at least one vertex of X intersects the column α (resp. the row a). If X is a clique of G, and a, b two rows of the grid, we
denote by Xabthe
subset of vertices of X with index {a, b}.
The three following lemmas allow us to prove that all the cliques of a B2-EPG graph are included in a
good subgraphs. They use the same kind of techniques. The main idea is, using the tools of Section 3, to find typed intervals which describe well the vertices in a clique X.
Lemma 9. Let G be a B2-EPG graph containing only Z-vertices, and X be a clique of G. If there are two rows a and
bsuch that the projection graphs of Xabon a and b are not cliques, then X is included in a good graph.
Proof. Let X be a clique satisfying this property for rows a and b. Let Xa be the set of vertices of X \ Xab
intersecting row a and not row b, and Xbbe the set of vertices of X \ Xabintersecting row b and not row a.
First note that X = Xa∪ Xb ∪ Xab. Otherwise there would exist a vertex w of type (c, d) such that
{c, d} ∩ {a, b} = ∅ in X. Since w is complete to Xab, w would intersect all the vertices of Xabon their vertical
part. But the projection graph of Xabon a is not a clique, consequently Xabintersects at least two columns.
Thus a vertex of Xabdoes not intersect the unique column of w, a contradiction.
By Lemma 5 applied to Xabon both a and b, there exist two typed intervals ta and tbon rows a and b
such that every vertex of Xabintersects both taand tb, and such that vertices of Xacontain taand vertices
Lemma 10. Let G be a B2-EPG graph containing only Z-vertices, and X be a clique of G. If there are two rows a
and b such that the projection graph of Xabon a is not a clique, then there is a good graph containing X.
Proof. Let X be a clique satisfying this property for rows a and b. We can assume that the projection graph of Xabon b is a clique since otherwise we can apply Lemma 9. Let Xa be the set of vertices
of X \ Xab
intersecting row a and not row b, and Xbbe the set of vertices of X \ Xabintersecting row b and not row a.
First note that X = Xa∪ Xb ∪ Xab. Otherwise there would exist a vertex w of index {c, d} such that
{c, d} ∩ {a, b} = ∅ in X. Since w is complete to Xab, w would intersect all the vertices of Xabon their vertical
part. But the projection graph of Xabon a is not a clique, consequently Xabintersects at least two columns.
Thus a vertex of Xabdoes not intersect the unique column of w, a contradiction.
Suppose first that the projection graph of Xb∪ Xabon b is a clique. By Lemma 4 applied to Xbon b, there
exists a proper typed interval tbon row b such that every vertex of Xbcontains tb. By Lemma 5 applied to
Xabon a, there exists a proper typed interval taon row a such that every vertex of Xacontains ta. So all the
vertices of X are contained in the good graph of type (I) defined by taand tb.
Suppose now that the projection graph of Xb∪ Xabon b is not a clique. Then there are two vertices u, v of
Xbsuch that the t-projections of u and v on b do not intersect. By Lemma 1, Puand Pvintersect another row
c, and since the projection graph of Xabon b is a clique, c 6= a. Since the projection graph of Xbcon b is not a
clique, and c 6= a, we can assume that the projection graph of Xbcon c is a clique since otherwise Lemma 9
can be applied on Xbc. So by Lemma 4 applied to Xbcon c, there exists a typed interval tccontained in every
vertex of Xbc. By Lemma 5 applied to Xbcon b (resp. Xabon a), we know that there exists a typed interval
tb (resp. ta) satisfying the two conditions of Lemma 5. Now we divide Xb∪ Xabinto two subclasses: Xbc
and Yb= (Xb∪ Xab) \ Xbc.
We have X = Yb∪ Xbc∪ Xa. Vertices of Xacontain ta. Vertices of Xbcintersect tband contain tc. Vertices
of Ybcontain tb. This proves that X is included in the good graph of type (II) defined by ta, tb and tc on
respectively rows a, b, c.
We now have all the ingredients we need to prove Theorem 6. By Lemma 10, we can assume that for all a, b, the projection graphs of Xabon a and b are cliques. As a by-product, if we denote by Xathe
set of
vertices containing a in their index, then the projection graph of Xa on a induces a clique for every row a.
Indeed, suppose by contradiction that there is a row a, and two vertices u, v whose index contains a, such that the t-projections of u and v on a do not intersect. Then by Lemma 1, u and v intersect
a common row b 6= a. But then, the projection graph of Xabon a is not a clique, a contradiction. Thus by Lemma 4, for any
row a such that Xais not empty, there exists a proper typed interval tasuch that the t-projection on a of any
vertex of Xacontains ta.
Assume that there are two rows a and b such that Xabintersects at least two columns. By Lemma 8, we
can assume that X intersects at least three rows. Let w ∈ X \ Xabbe a vertex using a third row. As in the
proof of Lemma 9, since Xabhas at least two columns and w is complete to Xab, w has index either {a, c}
or {b, c} for c 6= a, b. Consequently, w contains either ta or tb, which proves that X is contained in a good
graph of type (I).
Assume now that for all pair of rows (a, b), all the vertices of Xabpass through the same column. We
can assume that X intersects at least three columns, otherwise we could just rotate the representation of G and apply Lemma 8. Let u, v and w be three vertices using three different columns. Let a, b
be the index of u. We can assume w.l.o.g. that the index of v is {b, c}. There are two possible cases:
• w has index {b, d} for some row d. Note that d 6= a, c since otherwise this would contradict the fact that both Xaband Xbcintersect only one column. If all vertices in X contain b in their index,
then they
all contain the typed interval tb, and X is contained in a good graph of type (I). Let z be a vertex with
z ∈ X \ Xb. Up to permutations of rows a, c and d, we can assume that the row c is below a and d is
below c. Suppose that b is over c. Then, since z must intersect u, v and w, necessarily the index of z is of the form {s1, s2} with s1, s2∈ {a, b, d}. Indeed, z can intersect only one of u, v, w on
its vertical
part. Additionally, the index of z cannot be {c, d} since otherwise z cannot intersect the vertical part of w. (see Fig. 4). So necessarily the index of z contains a. This proves that every vertex of
X contains
u v w z a b c d
Figure 4: A vertex z with index {c, d} cannot intersect u since the vertical part of u is not between rows c and d. v u w zc za a b c da dc
Figure 5: The vertices za and zc cannot be adjacent since their paths do not have a row or a column in
either tb or ta, and as a consequence X is included in a good graph of type (I). A similar argument
shows that if b is below c, then z cannot be of index {a, c}, and the index of z contains d.
• w does not contain b in its index, then w has necessarily index {a, c} in order to intersect both u and v. If there are two rows, say for example a and b, such that every vertex of X has an index
containing either a or b, then every vertex of X contains either taor tb. This implies that X is contained in a good
graph of type (I). Suppose that there are no two rows such that the index of any vertex contains one of the two rows. Suppose w.l.o.g.. that row b is below a, and c below b. There is a vertex zcwhose
does not contain a or b. Then necessarily, the index of zc is {c, dc} for a certain row dc. Additionally,
zc must intersect the vertex u of index {a, b} on its vertical part. This implies that b is below dc. By a
similar argument, there is a vertex zaof index {a, da} with da 6= b, c. Then za must intersect v on its
vertical part, and da is below b. In particular, da is different from dc. This implies that za and zc do
not have a row or a column in common, and thus do not intersect. Since X is a clique, this case is not possible.
General B
-EPG graphs
In Section 4.1, we have seen how to compute a maximum clique in a graph containing only Z-vertices. This section is devoted to prove that we can actually separate the graph in order to assume the
graph only contains Z-vertices or U-vertices. We start by proving two lemmas showing that the existence of U-vertices puts some constraints on the Z-vertices that can be appear in a clique. We will
then use these two lemmas to prove our main theorem.
Lemma 11. Let G be a B2-EPG graph with a representation, and X be a clique of G, then:
• or there are three columns intersecting all the Z-vertices of X.
Proof. Let u1, u2, u3, and u4be four U-vertices of X intersecting pairwise different rows. Let us prove that
there are three columns containing every Z-vertex of X.
First assume that there are three columns α, β, γ such that, the set of columns intersected by ui is in
{α, β, γ} for every i ≤ 4. Let us prove that these three columns intersect every Z-vertex of X. Assume by contradiction that there exists v in X that does not intersect α, β and γ. Then for every i,
Pvshares an edge
with Pui on a horizontal segment. Since all the ui have disjoint index, this would imply that v intersects
four different rows, a contradiction.
So we can assume that u1, u2, u3, u4intersect at least four columns. Let α and β be the columns of u1. We
can assume w.l.o.g. that u2intersects the columns α and γ, with γ 6= α, β. And that u3intersects a fourth
column δ 6= α, β, γ. So both u3and u4must intersect α since they must intersect both u1and u2. Let τ be
the second column intersected by u4. Then any Z-vertex of X intersects one of α, δ, τ . Indeed, suppose by
contradiction that a Z-vertex v of X does not intersect one of these columns. Since Pv does not intersect
Pu3 and Pu4 on their vertical intervals, it shares an edge with Pu3 and Pu4 on their two horizontal parts.
Since u1, u2, u3, u4have pairwise different index, Pvthat intersect the row of u3and the row of u4, share an
edge with Pu1 on the column β and Pu2 on the column γ since v does not intersect column α. However, a
Z-vertex intersects a single column, a contradiction.
In Section 3, we have introduced typed intervals. These typed interval defines intervals on a given row. In the following claim, we need two typed of typed interval: horizontal and vertical typed
interval. An horizontal typed interval is a typed interval as defined in Section 3. A vertical typed interval is a typed interval of the graph after a rotation, i.e. the graph where rows become
columns and columns become rows.
Lemma 12. Let G be a B2-EPG graph with its representation, and X be a clique of G containing only U-vertices
with the same index {a}. There exists a set Stof at most three typed intervals such that:
• Stcontains exactly one horizontal typed interval, and at most two vertical typed intervals,
• every vertex of X contains all the typed intervals of St,
• a Z-vertex u is complete to X if and only if u intersects one of the typed intervals of St.
Proof. Since X is a clique of G and X only contains U-vertices of index {a}, Lemma 2 ensures that the projection graph of X on a is a clique. By Lemma 4 applied to X on a, there is a typed interval t
such that every vertex of X contains t, and, if u is a vertex containing a in its index, and u is complete to X, then u must intersect t. The typed interval t is the unique horizontal typed interval
of St.
Suppose that there is a column α, such that every vertex of X intersects α. Since all the vertices of X intersect the same row a and X is a clique, the projection graph of X on the column α is a
clique. Indeed, since all the vertices of X intersect column α and row a, all of them must bend on the point (a, α). Either they all bend on the same direction on column α, say upwards, and then they
all contain the edge of the column α between a and a + 1, and the projection graph is a clique. Or, some vertices of X bend upwards and other downwards on (a, α). Since X is a clique, they all come
from the same direction on row a and then their t-projections on α pairwise intersect.
By Lemma 4 applied to X on column α, there exists a vertical typed interval tαsatisfying both properties
of Lemma 4. Since every U-vertex intersects two columns, there are at most two columns α, β for which every vertex of X intersects these columns. Let Stbe the set composed of t and the typed
intervals tαand
tβif they exist.
Let us prove that St satisfies the conclusion of the lemma. By construction St contains exactly one
horizontal typed interval and at most two vertical typed intervals. By definitions of t and tα, tβ, every
vertex v of X contains the typed intervals in St. Let us finally show the last point. Let u be a Z-vertex.
If u intersects a typed interval in St, then by Lemma 1, u is complete to X. Conversely, suppose that u is
complete to X. If u contains a in its index, then Lemma 4 ensures that u intersects t since vertices of X all have index {a}. Assume now that the index of u does not contain a. So u intersects al the
vertices of X on
its unique column. Let γ be the unique column intersected by u. All the vertices of X must intersect γ since otherwise u cannot be complete to X. Then γ ∈ {α, β}, and w.l.o.g., we can assume γ = α.
Then Lemma 4 ensures that u intersects tαsince the unique column of u is α.
The two previous lemmas are the main ingredients to prove that a maximum clique in B2-EPG graphs
can be computed in polynomial time. The idea of the algorithm is, using Lemma 12, to guess some typed intervals contained in the U-vertices of the clique. Lemma 11 ensures that we do not have to
guess too many intervals. Once we have guessed these intervals, we are left with a subgraph which is actually the join of two subgraphs, one with only Z-vertices, and another with only U-vertices.
Then the maximum clique is obtained by applying Theorem 6 to each of the components.
Theorem 13. Given a B2-EPG graph G with its representation, there is a polynomial time algorithm computing the
maximum clique of G.
Proof. In the rest of this proof, Siwill denote a set of typed intervals. A vertex u contains Siif u contains all
the typed intervals of Si. And u intersects Siif u intersects one of its typed intervals. Given k of these sets
S1, S2, . . . Sk, we denote by G(S1, S2, . . . Sk)the subgraph induced by the set of U-vertices containing one
of the Siand the set of Z-vertices intersecting all of the Si-s.
Let X be a clique of G. Let us show that there are at most three sets S1, . . . , Sk with k ≤ 3 such that
X is contained in G(S1, . . . , Sk). Free to rotate the representation by 90◦if needed, Lemma 11 ensures that
there are at most three rows intersecting all U-vertices. Let us denote by a1, a2, a3 these (at most) three
rows. By Lemma 12 applied on each row ai, there exists a set of typed intervals Sisuch that every U-vertex
of X intersecting ai contains Si, and every Z-vertex of X intersects Si. This implies that X is included in
G(S1, . . . , Sk)with k ≤ 3.
Let us now describe the algorithm that computes a maximum clique in G: guess the sets S1, . . . Sk by
trying all possibilities, and then compute the maximum clique of G(S1, . . . Sk). Reusing the argument in
the proof of Lemma 7, we know that there are at most O(n2[)][typed intervals. Since each set S]
iis composed
of three typed intervals, this give at most O(n6[)][possibilities. By looking more precisely at the proof of]
Lemma 12 we can see that the vertical typed intervals share a common endpoint with the horizontal one. This means that actually you only have O(n3[)][possibilities to look at for a set S]
i. Now, since we need at
most three of these sets, there are at most O(n9[)][possible choices for the sets S]
1, . . . Sk, k ≤ 3.
To complete the proof of the theorem, we only need to prove that computing a maximum clique in a subgraph G(S1, . . . , Sk)can be done in polynomial time. Fix S1, . . . Sk sets of typed intervals,
and denote
H = G(S1, . . . , Sk). Let HZ (resp. HU) be the subgraph of H induced by the Z-vertices (resp. U-vertices).
Then H is the join of HU and HZ. Indeed, let u a U-vertex of H, and v a Z-vertex of H. By construction,
there is a set Si such that u contains Si. Since v intersects Si, by Lemma 1 u and v are adjacent in H. By
Theorem 6, a maximum clique of HU and a maximum clique of HZ can be computed in polynomial time.
This implies that a maximum clique of H can be computed in polynomial time.
Colorings and χ-boundedness
We denote by χ(G) the chromatic number of G, i.e. the minimum number of colors needed to properly color the graph G. And we denote by ω(G) the maximum size of a clique of G. The following lemma upper
bounds the number of edges in a Bk-EPG. A similar bound on the number of edges was proposed by
Gyarf´as in [11] for k-interval graphs. We nevertheless give the proof for completeness.
Lemma 14. Let G be a Bk-EPG graph on n vertices. There are at most (k + 1)(ω(G) − 1)n edges in G.
Proof. Let G be a Bk-EPG graph, and consider a representation of G. Let q be the maximum number of
distinct paths going through one edge of the grid. Then q ≤ ω(G) since all the paths sharing a common edge of the grid forms a clique of G. Let us prove that G has at most (k + 1)(q − 1) edges.
The path Pu of u can be decomposed into at most (k + 1) intervals where an interval is a maximum
Each interval is then contained on a single row or on a single column. Note that two intervals of Pucan be
included on the same row if k is large enough. If two intervals of Puon a row intersect, we define replace
both intervals with a single interval that is the union of the two intervals. This operation is called a merging. Given a vertex u, the canonical intervals of u are the intervals of u where we merged
all the intervals that have an edge in common.
Let Gi = (Vi, Ei)be the interval graph where the vertices of Gi are the canonical intervals included in
the ith row of the representation of G. And there is an edge between two canonical intervals if they intersect non trivially. Note that each path Pumight contribute to several vertices in Gi but all
these intervals have
to be disjoint intervals are canonical. Note that the graph Gidefines an interval graph. Similarly we define
G0[j] for each column of the representation. We denote by Gsthe graph defined as the disjoint union of all
the Gis and of all the G0js. We claim the following:
Claim 1. |E(Gs)| ≥ |E(G)|and ω(Gs) = q.
Proof. Let uv be an edge of G. Then Puand Pvshare an edge and then there is a canonical interval i of Pu
and a canonical interval j of Pvsuch that i and j intersect. We can associate to uv the edge ij. This function
from E(G) to E(Gs)is clearly injective and then |E(Gs)| ≥ |E(G)|. Assume by contradiction that there
is a clique of size more than q in Gi. Then there exists an edge of row i containing at least q + 1 canonical
intervals. Since there are at most q different vertices of G going through the same edge, two intervals are associated to the same vertex, a contradiction with the fact that the intervals are
The remaining of the proof consists in evaluating the number of edges in G0. Now the graphs Giand G0i
are interval graphs, consequently their number of edges satisfies: |E(Gi)| ≤ (q − 1)|V (Gi)|.Since each path
is composed of at most (k + 1) intervals, we get: |E(G)| ≤X i |E(Gi)| + X j |E(G0j)| ≤ (q − 1) X i |V (Gi)| + X j |V (G0 j)| = (q − 1)(k + 1)n
A graph is k-degenerate if there is an ordering v1, . . . , vn of the vertices such that for every i, |N (vi) ∩
{vi+1, . . . , vn}| ≤ k. It is straightforward to see that k-degenerate implies (k + 1)-colorable. Lemma 14
immediately implies the following:
Corollary 15. Let G be a Bk-EPG graph:
• The graph G is2(k + 1)ω − 1-degenerate. • χ(G) ≤ 2(k + 1)ω(G).
• There is a polynomial time 2(k + 1)-approximation algorithm for the coloring problem without knowing the representation of G.
• Every graph of Bk-EPG contains a clique or a stable set of size at least
q [n]
Proof. The first three points are immediate corollaries of Lemma 14. Let us only prove the last point. If the graph does not admit a clique of size at leastq2(k+1)n , then
χ(G) ≤ 2(k + 1) ·
r [n]
2(k + 1) ≤ p
Since a proper coloring is a partition into stable sets, there exists a stable set of size at least [√] n 2(k+1)n =
q [n]
2(k+1)which concludes the proof.
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Are Uncomputable Entities Useless for Science?
When I first learned about uncomputable numbers, I was profoundly disturbed. One of the first things you prove about uncomputable numbers, when you encounter them in advanced math classes, is that it
is provably never possible to explicitly display any example of an uncomputable number. But nevertheless, you can prove that (in a precise mathematical sense) "almost all" numbers on the real number
line are uncomputable. This is proved indirectly, by showing that the real number line as a whole has one order of infinity (aleph-one) and the set of all computers has another order of infinite
I never liked this, and I burned an embarrassing amount of time back then (I guess this was from ages 16-20) trying to find some logical inconsistency there. Somehow, I thought, it must be possible
to prove this notion of "a set of things, none of which can ever actually be precisely characterized by any finite description" as inconsistent, as impossible.
Of course, try as I might, I found no inconsistency with the math -- only inconsistency with my own human intuitions.
And of course, I wasn't the first to tread that path (and I knew it). There's a philosophy of mathematics called "constructivism" which essentially bans any kind of mathematical entity whose
existence can only be proved indirectly. Related to this is a philosophy of math called "intuitionism."
A problem with these philosophies of math is that they rule out some of the branches of math I most enjoy: I always favored continuous math -- real analysis, complex analysis, functional analysis --
over discrete math about finite structures. And of course these are incredibly useful branches of math: for instance, they underly most of physics.
These continuity-based branches of math also underly, for example, mathematical finance, even though the world of financial transactions is obviously discrete and computable, so one can't possibly
uncomputable numbers to handle it.
There always seemed to me something deeply mysterious in the way the use of the real line, with its unacceptably mystical uncomputable numbers, made practical mathematics in areas like physics and
finance so much easier.
Notice, this implicitly uncomputable math is never
in these applications. You could reformulate all the equations of physics or finance in terms of purely discrete, finite math; and in most real applications, these days, the continuous equations are
solved using discrete approximations on computers anyway. But, the theoretical math (that's used to figure out which discrete approximations to run on the computer) often comes out more nicely in the
continuous version than the discrete version. For instance, the rules of traditional continuous calculus are generally far simpler and more elegant than the rules of discretized calculus.
And, note that the uncomputability is always in the background when you're using continuous mathematics. Since you can't explicitly write down any of these uncomputable numbers anyway, they don't
play much role in your practical work with continuous math. But the math you're using, in some sense, implies their "existence."
But what does "existence" mean here?
former President Bill Clinton, "it all depends on what the meaning of the word is, is."
A related issue arises in the philosophy of AI. Most AI theorists believe that human-like intelligence can ultimately be achieved within a digital computer program (most of them are in my view
overpessimistic about how long it's going to take us to figure out exactly how to write such a program, but that's another story). But some mavericks, most notably Roger Penrose, have argued
otherwise (see his books
The Emperor's New Mind
Shadows of the Mind
, for example). Penrose has argued specifically that the crux of human intelligence is some sort of mental manipulation of uncomputable entities.
And Penrose has also gone further: he's argued that some future theory of physics is going to reveal that the dynamics of the physical world is also based on the interaction of uncomputable entities.
So that mind is an uncomputable consequence of uncomputable physical reality.
This argument always disturbed me, also. There always seemed something fundamentally
to me about the notion of "uncomputable physics." Because, science is always, in the end, about finite sets of finite-precision data. So, how could these mysterious uncomputable entities ever really
be necessary to explain this finite data?
Obviously, it seemed tome, they could never be necessary. Any finite dataset has a finite explanation. But the question then becomes whether in some cases invoking uncomputable entities is the
way to explain some finite dataset. Can the best way of explaining some set of, say, 10 or 1000 or 1000000 numbers be "This uncomputable process, whose details you can never write down or communicate
in ordinary language in a finite amount of time, generated these numbers."
This really doesn't make sense to me. It seems intuitively wrong -- more clearly and obviously so than the notion of the "existence" of uncomputable numbers and other uncomputable entities in some
abstract mathematical sense.
So, my goal in this post is to give a careful explanation of why this wrong. The argument I'm going to give here could be fully formalized as mathematics, but, I don't have the time for that right
now, so I'll just give it semi-verbally/semi-mathematically, but I'll try to choose my words carefully.
As often happens, the matter turned out to be a little subtler than I initially thought it would be. To argue that uncomputables are useless for science, one needs some specific formal model of what
itself is. And this is of course a contentious issue. However, if one does adopt the formalization of science that I suggest, then the scientific uselessness of uncomputables falls out fairly
straightforwardly. (And I note that this was certainly not my motivation for conceiving the formal model of science I'll suggest; I cooked it up a while ago for quite other reasons.)
Maybe someone else could come up with a different formal model of science that gives a useful role to uncomputable entities ... though one could then start a meta-level analysis of the usefulness of
this kind of formal model of science! But I'll defer that till next year ;-)
Even though it's not wholly rigorous math, this is a pretty mathematical blog post that will make for slow reading. But if you have suitable background and are willing to slog through it, I think
you'll find it an interesting train of thought.
NOTE: the motivation to write up these ideas (which have been bouncing around in my head for ages) emerged during email discussions on the AGI list with a large group, most critically Abram Demski,
Eric Baum and Mark Waser.
A Simple Formalization of the Scientific Process
I'll start by giving a simplified formalization of the process of science.
This formalization is related to the philosophy of science I outlined in the essay
(included in
The Hidden Pattern
) and more recently extended in the blog post
. But those prior writing consider many aspects not discussed here.
Let's consider a community of agents that use some language L to communicate. By a language, what I mean here is simply a set of finite symbol-sequences ("expressions"), utilizing a finite set of
Assume that a dataset (i.e., a finite set of finite-precision observations) can be expressed as a set of
of expressions in the language L. So a dataset D can be viewed as a set of pairs
((d11, d12), (d21,d22) ,..., (dn1,dn2))
or else as a pair D=(D1,D2) where
Then, define an
of a dataset D as a set E_D of expressions in L, so that if one agent A1 communicates E_D to another agent A2 that has seen D1 but not D2, nevertheless A2 is able to reproduce D2.
(One can look at precise explanations versus imprecise ones, where an imprecise explanation means that A2 is able to reproduce D2 only approximately, but this doesn't affect the argument
significantly, so I'll leave this complication out from here on.)
If D2 is large, then for E_D to be an interesting explanation, it should be more compact than D2.
Note that I am not requiring E_D to generate D2 from D1 on its own. I am requiring that A2 be able to generate D2 based on E_D and D1. Since A2 is an arbitrary member of the community of agents, the
validity of an explanation, as I'm defining it here, is relative to the assumed community of agents.
Note also that, although expressions in L are always finitely describable, that doesn't mean that the agents A1, A2, etc. are. According to the framework I've set up here, these agents could be
infinite, uncomputable, and so forth. I'm not assuming anything special about the agents, but I am considering them in the special context of finite communications about finite observations.
The above is my formalization of the scientific process, in a general and abstract sense. According to this formalization, science is about communities of agents linguistically transmitting to each
other knowledge about how to predict some commonly-perceived data, given some other commonly-perceived data.
The (Dubious) Scientific Value of the Uncomputable
Next, getting closer to the theme of this post, I turn to consider the question of what use it might be for A2 to employ some uncomputable entity U in the process of using E_D to generate D2 from D1.
My contention is that, under some reasonable assumptions, there is no value to A2 in using uncomputable entities in this context.
D1 and E_D are sets of L-expressions, and so is D2. So what A2 is faced with, is a problem of mapping one set of L-expressions into another.
Suppose that A2 uses some process P to carry out this mapping. Then, if we represent each set of L-expressions as a bit string (which may be done in a variety of different, straightforward ways), P
is then a mapping from bit strings into bit strings. To keep things simple we can assume some maximum size cap on the size of the bit strings involved (corresponding for instance to the maximum size
expression-set that can be uttered by any agent during a trillion years).
The question then becomes whether it is somehow useful for A2 to use some uncomputable entity U to compute P, rather than using some sort of set of discrete operations comparable to a computer
One way to address this question is to introduce a notion of
. The question then becomes whether it is
for A2 to use U to compute P, rather than using some computer program.
And this, then, boils down to one's choice of simplicity measure.
Consider the situation where A2 wants to tell A3 how to use U to compute P. In this case, A2 must represent U somehow in the language L.
In the simplest case, A2 may represent U directly in the language, using a single expression (which may then be included in other expressions). There will then be certain rules governing the use of U
in the language, such that A2 can successfully, reliably communicate "use of U to compute P" to A3 only if these rules are followed. Call this rule-set R_U. Let us assume that R_U is a finite set of
expressions, and may also be expressed in the language L.
Then, the key question is whether we can have
complexity(U) < complexity(R_U)
That is, can U be less complex than the set of rules prescribing the use of its symbol S_U within the community of agents?
If we say NO, then it follows there is no use for A2 to use U internally to produce D2, in the sense that it would be simpler for A2 to just use R_U internally.
On the other hand, if we say YES, then according to the given complexity measure, it may be easier for A2 to internally make use of U, rather than to use R_U or something else finite.
So, if we choose to define complexity in terms of complexity of expression in the community's language L, then we conclude that uncomputable entities are useless for science. Because, we can always
replace any uncomputable entity U with a set of rules for manipulating the symbol S_U corresponding to it.
If you don't like this complexity measure, you're of course free to propose another one, and argue why it's the right one to use to understand science. In a previous blog post I've presented some of
the intuitions underlying my assumption of this "communication prior" as a complexity measure underlying scientific reasoning.
The above discussion assumes that U is denoted in L by a single symbolic L-expression S_U, but the same basic argument holds if the expression of U in L is more complex.
What does all this mean about calculus, for example ... and the other lovely uses of uncomputable math to explain science data?
The question comes down to whether, for instance, we have
complexity(real number line R) <>
If NO, then it means the mind is better off using the axioms for R than using R directly. And, I suggest, that is what we actually do when using R in calculus. We don't use R as an "actual entity" in
any strong sense, we use R as an abstract set of axioms.
What would YES mean? It would mean that somehow we, as uncomputable beings, used R as an internal source of intuition about continuity ... not thus deriving any conclusions beyond the ones obtainable
using the axioms about R, but deriving conclusions in a way that we found subjectively simpler.
A Postcript about AI
And, as an aside, what does all this mean about AI? It doesn't really tell you anything definitive about whether humanlike mind can be achieved computationally. But what it does tell you is that, if
• humanlike mind can be studied using the communicational tools of science (that is, using finite sets of finite-precision observations, and languages defined as finite strings on finite alphabets)
• one accepts the communication prior (length of linguistic expression as a measure of complexity)
then IF mind is fundamentally noncomputational, science is no use for studying it. Because science, as formalized here, can never distinguish between use of U and use of S_U. According to science,
there will always be some computational explanation of any set of data, though whether this is the simplest explanation depends on one's choice of complexity measure.
10 comments:
I am having trouble "seeing" the argument.
It seems to slip by the idea that if a mind contained uncomputable entities, actual information could be obtained by them.
So, for example, if all the agents in the scientific community were born with halting oracles, some descriptions of sensory data could be somewhat shorter by invoking the halting oracle. (Other
descriptions would be longer as a result, but if the world generally had halting oracles in it, then the sensory-data would probably be more easily described using the halting oracle.)
The specific point at which the argument seems to trip for me is at the "If we say no... if we say yes..." point. You talk about replacing U with R_U, the set of rules associated with using S_U.
It seems to me that U could not be replaced so easily, since those manipulation rules would not give all the answers a halting oracle could give. Furthermore, it seems to me that the rules for
using S_U would include actual references to U, like "close your eyes and meditate and the answer will come to you". So I don't even see how it would make sense to replace U with R_U.
Perhaps you could clarify?
Abram ...
There is only a finite number of possible rules for mapping L-expression-sets of size less than N into L-expression-sets of size less than N.
So, whatever voodoo happens involving U, in the context of mapping L-expression-sets of size less than N into other ones, it must be equivalent to some subset of this rule-set. There must be some
finite set of rules that does *exactly* what U does, within the specified size constraint.
Remember, there are no infinities involved here.
Whatever the halting oracle does **within the specified finite domain** could be emulated by some set of rules.
I could prove this to you in any specific case. Given a halting oracle U, you could make a finite table of its answer to all questions that can be posed within the finite domain. Then, there is
some minimal Turing program that will give the same answers as the oracle to all those questions.
The question is whether you want to consider this corresponding set of rules (the Turing program) as being simpler than U or not (e.g. simpler than the oracle or not). If you want to consider the
oracle as being simpler than the corresponding set of rules, then that's your right. The problem is that the oracle cannot be communicated ... but you're right that if *everyone in the community
had the same halting oracle in their brain*, then they could do science in an uncomputable way, if they agree to measure simplicity relative to their brains rather than relative to their
(discrete, finite language).
But, then they are not using the communication prior. Then they are using a prior that is, effectively, something like "complexity relative to the common brain structure of the members of the
This is interesting, but it's more like "shared intuition" than like what we think of as science. Because the common means of making judgments is not something that the community of agents is
able to explicitly articulate in language.
I'm not sure when I will find time to write that argument up in a fully mathematical way, which might be what it takes to get it across to you, I dunno?
OK, that makes perfect sense. For some reason I thought you were *assuming* that every agent had the same uncomputable oracle U.
Abram: well, if every agent had the same oracle in their brains, how could they know this? If they need to verify this thru linguistic communication, then you run up against the issues I
mentioned in my post. But on the other hand, if they know this via some non-linguistic, non-finite-data-driven, intuitive voodoo method, then, yeah, you've got a loophole, as you mentioned...
They would know it in the same way that we know that anyone has the ability to learn to count... the uncomputable stuffs would just be another mechanism of the mind that everyone would be able to
The reason I assumed that this was what you meant was because you are talking about an agent communicating the idea of using U to compute something. So, I figured that U must be something that
all the agents had access to.
I think I may have mentioned this before... but it seems like your arguments could be applied just as well to argue that Turing machines are unnecessary and all we need is finite-state machines.
(Which is, of course, literally all we have.) If someone claimed to have found a pattern in nature that was generated by a Turing machine (let's call it a Turing object, T), then they would need
to communicate the idea to others using a symbol, S_T, and a set of manipulation rules that define the "turing-ness" of T, R_T. The rules would of course need to be implementable on a
finite-state machine. So, the agent would be unable to convince its peers, because R_T would be a working finite-state explanation, unless all of the agents happened to have some bias towards
objects conveniently described with R_T, in which case they would subjectively like to think that T is really Turing-computed rather than finite-state-computed.
If the above argument doesn't sound analogous to your argument, then take the disanalogies to be places where I still don't understand your argument.
Abram, yes you seem to be right, my argument also seems to show that for the purposes of science (defined in terms of finite datasets and finite linguistic utterances) we don't need TM's but only
This of course doesn't bother me at all.
Note that I am not saying that uncomputable entities or TM's don't "exist" in any deep philosophical sense. I'm just saying that, from the point of view of science, they might as well not exist.
Science isn't necessarily everything....
You raise an interesting point which is that if all the agents in a community share the same internal uncomputable entity U ... and if they each assume that each other share it (without need for
experimental evidence) ... then, they can coherently use this to explain scientific datasets.
This is consistent with what I showed above, so long as it means that the community considers U simpler than its linguistic explication.
I am reminded of how we each implicitly assume each other are conscious, without empirical evidence.
Yet, I am wary of the approach of taking this kind of shared, unsubstantiable intuition as a basis for scientific process....
I was looking for your infinite-order probability post and found the post about Zarathustra and his saving box which led me here. This is a bit out of my league currently, but what if certain
computables depend on certain uncomputables? I've got Euler's famous formula in mind right now but I'm thinking of convergent series in general. Would convergent series exist without divergent
series? Have you ever thought of Euler's equation as a simple example of Supersymmetry? It kind of makes me think of Phil Gibbs' attempt to generate supersymmetry, what he calls event symmetry,
using an infinite-order quantization. An infinite-order quantization is like an infinite-order probability which has been shown to converge . . .
But I don't like the Born-Pauli interpretation of Schrodinger's function. Dirac's relativistic formulation disclosed the whole world, the negative energy sea, and scientists buried it underneath
a bunch of bullshit: why? To save a bullshit paradigm? Because Heisenberg was a little bitch? The Standard Model is built on "virtual" bosons, I mean WTF? I could understand it if there was no
other alternative but Dirac provides a beautiful alternative. Science, as it stands today, is a bunch of garbage! The Nobel prize is a big, fat, ugly joke . . . and it's not even funny!
So, today, knowing what you know, do you consider the conversation you had with 4 year-old Zarathustra empirical evidence for re-incarnation and the accessibility of omniscience?
Michelangelo was a member of the Illuminati; his artworks, but especially his “Creation of Adam,” makes this readily apparent to anyone who also happens to be a member of the Illuminati.
“Creation of Adam” has nothing to do with the literal interpretation of the Biblical story, rather, it’s esoteric wisdom leading to Gnosis hidden in plain view. Adam represents “Everyman (woman)”
who has “fallen” into the world of duality – broken symmetry – but Adam, the ideal, also exists in the Garden of Eden, representative of a super-symmetric state of bliss which transcends duality;
a place where everything equals everything – event symmetry! When man (woman) falls into conventional existence, an existence characterized by broken symmetry, they have within a super-symmetric
seed, the enlightened point of origin. The journey to enlightenment is a function mapping the broken symmetry to super-symmetry; practically speaking, fallen man (woman) has the seed of
super-symmetry in their prostate (Skene’s) gland and when they follow the esoteric instruction said super-symmetric seed undergoes a transformation, the lead becomes gold, and ends up dwelling in
the pineal gland. The fallen man (woman), who dwells in Hell or Samsara, is allowed through the gate and enters Heaven or Nirvana. Of course both exist concurrently right here in our familiar
reality, only one’s perceptive awareness has changed. In his “Creation of Adam,” Michelangelo places “God,” represented by a common metaphor for wisdom at the time, an elderly, bearded man,
inside the human brain approximately where the pineal gland would be; many art critics and historians call this the “Uterine Brain” and erroneously interpret it to mean that Michelangelo was
suggesting “God” controls humankind. This is what Euler’s famous formula represents, the transformation from broken symmetry to super-symmetry!
When Michelangelo painted “The Last Judgment” in the Sistine Chapel, many of the bishops and cardinals were offended at all of the nudity; one cardinal even suggested to the Pope that it was
better suited to a bathhouse. The Pope sent a letter via courier to Michelangelo telling him to “make it right.” Michelangelo sent a letter back to the Pope telling the Pope, “make nature right
and art will soon follow.” Michelangelo’s point was that the problem wasn’t with the art, rather, it was with nature – the bishops and cardinals. The bishops and cardinals weren’t illuminated;
they were still dwelling in the state of broken symmetry. If they were illuminated they would feel no need to hide nature’s beauty behind a cloak of deceit. Apparently the Pope at the time was
also a member of the Illuminati because “The Last Judgment” remained as Michelangelo painted it until after Michelangelo died and even then there were some mysterious and rather humorous
difficulties experienced during its defacement.
I was studying Linear Algebra and everything was going quite well until I came to:
“Consider the set, L(V,V), of all transformations from V into V, then L is closed, associative, and bilinear with respect to transformation product or successive transformations . . . but it’s
not commutative.”
I was like, “Screech, back up, commutativity is implicit in associativity AND bilinearity.” The author of the textbook was kind enough to present a counter-example showing why L is not
commutative. He used an element from the standard basis of Euclidean two space with a reflection across the y-axis and a counter-clockwise rotation through 90 degrees as transformations; I used
the same set-up and added a reflection across the origin to demonstrate counter-examples for both associativity and bilinearity. It was a trivial demonstration. Why do these damn bishops and
cardinals feel it’s necessary to hide nature’s beauty behind a cloak of deceit? I’ve been contrary my whole life so it’s certainly something I’m pondering!
Robert Crease is the Director of the Philosophy Department at Stony Brook University and writes about the history of science. For the necessary background history read his book, “A Brief Guide to
The Great Equations” (http://www.amazon.com/Brief-Guide-Great-Equations/dp/1845292812). Then read these three informative articles on the Dirac Equation (Dirac is hardly mentioned in “Equations”
interestingly enough, since he should play a role even more prominent than Einstein):
a) http://openseti.org/Docs/HotsonPart1.pdf
b) http://openseti.org/Docs/HotsonPart2.pdf
c) http://blog.hasslberger.com/docs/HotsonIE86.pdf
Here’s a brief summary of what you’ll find. Heisenberg conceived of his matrix mechanics at roughly the same time that Schrodinger conceived of his wave equation. The two approaches to quantum
theory were demonstrated to be mathematically equivalent but all of the scientists preferred Schrodinger’s equation because it was much easier to deal with and they were already familiar with
continuous functions. Heisenberg hated this and reacted like a little, immature child, calling Schrodinger’s equation “intellectual trash.”
When Dirac developed a relativistic formulation of Schrodinger’s wave equation he discovered that it had four roots, two positive and two negative. Rather than accept the negative solutions, the
science community, led by Heisenberg since his matrix mechanics didn’t have these roots, entered some bizarre conspiracy and went to extraordinary lengths of irrationality to get rid of them. The
Dirac equation naturally demonstrates that the vacuum is a plenum of negative energy, the “negative energy sea.” Essentially, this sea and our positive energy world are produced by nothing but
electrons and waves since positrons are nothing but out of phase electrons; positrons and electrons continuously oscillate back and forth from the electron state to the positron state. Heisenberg
couldn’t stand the idea of this negative energy sea so he just got rid of it and replaced the negative solutions to Dirac’s equation with “creator” and “annihilator” operators which completely
violate conservation. He then suggested that they didn’t “really” violate conservation because they were only “virtual” in that their existence was restricted to the time frame described by his
Uncertainty Principle.
The whole entire modern day Standard Model of particle physics, to say nothing of Big Bang theory, Inflation, etc., is based on this garbage and, to use the words of a mathematician I know, may
not be entirely wrong but have some serious foundational issues! Read the book and the papers, I think you’ll be glad you did. It’s a beautiful case study with regards to the absurdity of human
nature and the veracity of your mathematical model of mind.
pengobatan kutil kelamin
obat eksim
obat kanker payudara yang alami dan manjur
cara menghilangkan kutil kelamin
merontokkan kutil kelamin
obat kutil kelamin sampai tuntas
obat kutil kelamin di jakarta
obat kanker payudara di jakarta
obat kanker payudara di jogja
obat kanker serviks stadium 3
Tips mengobati kutil kelamin
ramuan alami untuk kutil kelamin
cara merontokkan kutil di kelamin
dulunya kencing nanah sekarang ada kutilnya
obat kutil
obat kutil apotik
OBAT HERBAL KUTIL DI KEMALUAN ATAU JENGGER AYAM | {"url":"https://multiverseaccordingtoben.blogspot.com/2008/10/are-uncomputable-entities-useless-for.html?showComment=1394136032308","timestamp":"2024-11-13T22:27:38Z","content_type":"text/html","content_length":"128223","record_id":"<urn:uuid:4d70ab53-e78a-463f-8da4-b995ca5aad04>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00730.warc.gz"} |
Four Gone!
It really provoked interesting discussion with my children about the different mathematical laws, of brackets, factorising, of the memory key on the calculator, compensation the list goes on...
Great especially as reasoning is one of the key areas of mathematical learning.
What a great way to practise mental arithmetic skills in a way that requires you to think about the meaning of the calculation and how the calculation can be adapted to produce the same result. This
exercise could help uncover misconceptions and strengthen the mathematical understanding of your pupils.
The four button has dropped off!
How can you do these calculations using this calculator?
Your access to the majority of the Transum resources continues to be free but you can help support the continued growth of the website by doing your Amazon shopping using the links on this page.
Below is an Amazon link. As an Amazon Associate I earn a small amount from qualifying purchases which helps pay for the upkeep of this website.
Educational Technology on Amazon
Teacher, do your students have access to computers such as tablets, iPads or Laptops? This page was really designed for projection on a whiteboard but if you really want the students to have access
to it here is a concise URL for a version of this page without the comments:
However it would be better to assign one of the student interactive activities below.
Here is the URL which will take them to a related student activity. | {"url":"https://transum.org/Software/SW/Starter_of_the_day/starter_June2.ASP","timestamp":"2024-11-14T23:16:48Z","content_type":"text/html","content_length":"28395","record_id":"<urn:uuid:78ddfcb6-f350-438b-8895-3ab33b3390fe>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00667.warc.gz"} |
Conductance - (Honors Physics) - Vocab, Definition, Explanations | Fiveable
from class:
Honors Physics
Conductance is a measure of a material's ability to allow the flow of electric current. It is the reciprocal of resistance and represents the ease with which electric charge can move through a
conductor, such as a wire or a resistor in a parallel circuit.
congrats on reading the definition of Conductance. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Conductance is measured in Siemens (S), which is the reciprocal of Ohms (Ω), the unit of resistance.
2. The conductance of a resistor in a parallel circuit is the reciprocal of its resistance, and the total conductance of the parallel circuit is the sum of the conductances of the individual
3. Conductance is an important parameter in the analysis and design of parallel circuits, as it determines the distribution of current among the different branches.
4. Increasing the conductance of a branch in a parallel circuit will increase the current flowing through that branch, while decreasing the current in the other branches.
5. The relationship between conductance, resistance, and current in a parallel circuit is governed by Ohm's Law and the principle of conservation of charge.
Review Questions
• Explain how conductance relates to the flow of electric current in a parallel circuit.
□ Conductance is a measure of a material's ability to allow the flow of electric current. In a parallel circuit, the conductance of each branch determines the distribution of current among the
different branches. The branch with higher conductance (lower resistance) will have a larger current flowing through it, while the branch with lower conductance (higher resistance) will have
a smaller current. The total conductance of the parallel circuit is the sum of the conductances of the individual branches, and this total conductance determines the overall current flow in
the circuit.
• Describe how the relationship between conductance, resistance, and current in a parallel circuit is governed by Ohm's Law.
□ According to Ohm's Law, the current through a conductor is directly proportional to the voltage applied across it and inversely proportional to the resistance of the conductor. In a parallel
circuit, the voltage is the same across each branch, but the current divides among the different branches based on their respective conductances. The branch with higher conductance (lower
resistance) will have a larger current flowing through it, while the branch with lower conductance (higher resistance) will have a smaller current. The total current in the parallel circuit
is the sum of the currents in the individual branches, and this total current is determined by the total conductance of the circuit, which is the sum of the conductances of the individual
• Analyze the impact of changing the conductance of a branch in a parallel circuit on the distribution of current and the overall circuit behavior.
□ Increasing the conductance of a branch in a parallel circuit will increase the current flowing through that branch, while decreasing the current in the other branches. This is because
conductance is the reciprocal of resistance, and a higher conductance means a lower resistance. According to Ohm's Law, a lower resistance in a branch will result in a higher current through
that branch, as the voltage across the branch remains the same. Conversely, decreasing the conductance of a branch will reduce the current flowing through that branch, and the current will be
redistributed among the other branches in the parallel circuit. This change in current distribution can impact the overall circuit behavior, such as the power dissipation in each branch and
the overall efficiency of the circuit.
© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website. | {"url":"https://library.fiveable.me/key-terms/honors-physics/conductance","timestamp":"2024-11-06T21:52:23Z","content_type":"text/html","content_length":"172760","record_id":"<urn:uuid:a786835c-c082-465c-ae40-6439c1c8e911>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00781.warc.gz"} |
KNEARESTNEIGHBOURS procedure • Genstat v21
Classifies items or predicts their responses by examining their k nearest neighbours (R.W. Payne).
PRINT = string tokens Printed output required (neighbours, predictions); default pred
SIMILARITY = matrix or symmetric matrix Provides the similarities between the training and prediction sets of items
NEIGHBOURS = pointer Pointer with a variate for each prediction item to save the numbers of its nearest neighbours in the training set
GROUPS = factor Defines groupings to identify the training and prediction sets of items when SIMILARITY is a symmetric matrix
LEVTRAINING = scalar or text Identifies the level of GROUPS or dimension of SIMILARITY that represents the training set; default 1
LEVPREDICTION = scalar or text Identifies the level of GROUPS or dimension of SIMILARITY that represents the prediction set; default 2
METHOD = string token How to calculate the prediction from a DATA variate (mean, median); default medi
MINSIMILARITY = scalar Cut-off minimum value of the similarity for items to be regarded as neighbours; default 0.75
MINNEIGHBOURS = scalar Minimum number of nearest neighbours to use; default 5
MAXNEIGHBOURS = scalar Maximum number of nearest neighbours to use; default 10
SEED = scalar Seed for the random numbers used to select neighbours when more than MAXNEIGHBOURS are available; default 0
DATA = variates or factors Data values for the items in the training set
PREDICTIONS = variates or factors Saves the predictions
KNEARESTNEIGHBOURS provides the data-mining technique known as k-nearest-neighbour classification. This allocates unknown items to a category, or it predicts their (continuous) responses, by looking
at nearby items in a known data set. The known data set is usually called the training set, and we will call the unknown items the prediction set.
The SIMILARITY option provides a similarity matrix for KNEARESTNEIGHBOURS to use to determine the nearby items in the training set (or nearest neighbours) for each item in the prediction set. This
can be a symmetric matrix with a row (and column) for every item in the combined set of training and prediction items. The GROUPS option must then be set to a factor with one level for the training
items and another for the prediction items. By default the training set has level 1 and the prediction set has level 2, but these can be changed by the LEVTRAINING and LEVPREDICTION options. Matrices
like these can be formed in a wide variety of ways, using mixtures of categorical and continuous data, by the FSIMILARITY directive. For example, if we have a factor Sex, and variates Age, Weight and
Height whose values are known for both the training and prediction items, we could form a symmetric matrix Sim by
FSIMILARITY [SIMILARITY=Sim] Sex,Age,Weight,Height;\
However, Sim will contain unnecessary information, as we need the similarities between prediction and training items, but not between training items or between prediction items. So, for large data
sets, it will be more efficient to form a (rectangular) between-group similarity matrix by setting the GROUPS option of FSIMILARITY. For example
FSIMILARITY [SIMILARITY=Gsim; GROUPS=Gfac] Sex,Age,Weight,Height;\
where Gfac is a factor with two levels, one for the training set (usually level 1), and the other for the prediction set (usually level 2). You then no longer need to set the GROUPS option of
KNEARESTNEIGHBOUR. The LEVTRAINING and LEVPREDICTION options now specify the dimension of the similarity matrix (1 for rows, and 2 for columns) that correspond to the training and prediction data
sets, respectively. (They still correspond to group levels though, as they are defined by the numbers of the respective levels of the GROUPS factor in FSIMILARITY.)
The MINSIMILARITY option sets a minimum value on the similarity between two items if they are to be regarded as neighbours (default 0.75). The MINNEIGHBOURS option specifies the minimum number of
neighbours to try to find (default 5), and the MAXNEIGHBOURS option specifies the maximum number (default 10). The search for nearest neighbours for a particular prediction item works by finding the
most similar item in the training set, and adding this (with any equally-similar training items) to the set of neighbours. If at least MINNEIGHBOURS have been found, the search stops. Otherwise it
finds the next most similar items, and adds these to the set of neighbours, continuing until at least MINNEIGHBOURS have been found. If this results in more than MAXNEIGHBOURS neighbours,
KNEARESTNEIGHBOURS makes a random selection from those that are least similar to the prediction item, so that the number of neighbours becomes MAXNEIGHBOURS. The SEED option specifies the seed for
the random numbers that are used to make that selection. The default of zero continues an existing sequence of random numbers if any have already been used in this Genstat job, or initializes the
seed automatically. The NEIGHBOURS option can save a pointer, containing variate for each prediction item storing the numbers of its neighbours within the training set.
Once the neighbours have been found, KNEARESTNEIGHBOURS can use these to form the predictions. The DATA parameter lists variates and/or factors containing values of the variables of interest for the
items on the training set. The predictions can be saved using the PREDICTIONS parameter (in variates and/or factors to match the settings of the DATA parameter).
For a DATA factor, the category predicted for each item in the prediction set is taken to be the factor level that occurs most often amongst its nearest neighbours. If more than one level occurs most
often, the choice is narrowed down by seeing which of the levels has the the most similar neighbours. If this still leaves more than one level, the choice is narrowed further by seeing which of the
levels has neighbours with the highest mean similarity. Then, if even that does not lead to a single level, the final choice is made at random.
For a DATA variate, the METHOD option controls whether the prediction is made by the median (default) or the mean of the data values of the nearest neighbours of each prediction item.
Printed output is controlled by the PRINT option, with settings:
neighbours to print the nearest neighbours, and
predictions to print the predictions.
The default if PRINT=predictions.
So, to print predictions of blood pressure with a variate of training data Pressure, using the similarity matrix Gsim (as above) and default settings for the numbers of neighbours, we simply need to
KNEARESTNEIGHBOURS [SIMILARITY=Gsim] Pressure
Options: PRINT, SIMILARITY, NEIGHBOURS, GROUPS, LEVTRAINING, LEVPREDICTION, METHOD, MINSIMILARITY, MINNEIGHBOURS, MAXNEIGHBOURS, SEED.
Parameters: DATA, PREDICTIONS.
See also
Directives: FSIMILARITY, ASRULES, NNFIT, RBFIT.
Procedures: BCLASSIFICATION, BCFOREST, BREGRESSION, SOM.
Commands for: Data mining.
CAPTION 'KNEARESTNEIGHBOURS example',\
!t('Random classification forest for automobile data',\
'from UCI Machine Learning Repository',\
SPLOAD FILE='%gendir%/examples/Automobile.gsh'
FACTOR [LABELS=!t(yes,no)] loss_known
CALCULATE loss_known = 1 + (normalized_losses .EQ. !s(*))
VARIATE known_loss;\
VALUES=ELEMENTS(normalized_losses; WHERE(loss_known.IN.'yes'))
FSIMILARITY [METHOD=between; SIMILARITY=Sim; GROUPS=loss_known]\
KNEARESTNEIGHBOURS [PRINT=predictions,neighbours; SIMILARITY=Sim]\
known_loss; PREDICTIONS=pred_loss | {"url":"https://genstat21.kb.vsni.co.uk/knowledge-base/knearest/","timestamp":"2024-11-07T13:48:38Z","content_type":"text/html","content_length":"47108","record_id":"<urn:uuid:ada12093-bba2-4e9e-b1a3-59f08d1a9e1f>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00026.warc.gz"} |
Four men in a mine
kumbunterland posed this question in the park yesterday:
There are four men in a coal mine. This exit is narrow so that no more than 2 men can exit at the same time. In addition, there is only one torch and they can't leave the mine without it.
Each man walks at a different speed. The first man takes one minute to exit the mine the bridge distance, the second takes two minutes, the third takes five, and the fourth takes eight minutes.
If two men walk together they must walk at the slower speed. For example, if person #1 and person #3 go together, they will leave the mine in 5 minutes. Then, person #1 could return back to
return the torch to the others (taking another one minute) so they can use it to exit again.
The coal mine will collapse in exactly 15 minutes; you must find a way to allow all the men to leave the mine within this time.
Unfortunately, swi-prolog is broken on the N900, so I wasn't able to solve it right then:
solution(Men, Bound, Out) :-
torchInside(Men, [], [], Log),
totalDuration(Log, 0, Duration), Duration =< Bound,
swritef(Out, '%w (%w minutes)', [Log, Duration]).
% We use torch{Inside,Outside} predicates like a state machine.
torchInside(A, B, Log, X) :-
% Move two people from inside -> outside the mine
combination(2, A, S), subtract(A, S, A1), append(S, B, B1),
torchOutside(A1, B1, [S|Log], X).
torchOutside([], _, Log, Log).
torchOutside(A, B, Log, X) :-
% Move someone from outside -> inside the mine
combination(1, B, S), subtract(B, S, B1), append(S, A, A1),
torchInside(A1, B1, [S|Log], X).
% Utilities:
totalDuration([], X, X).
totalDuration([H|T], Acc, X) :-
max_list(H, Max), NewAcc is Acc + Max,
totalDuration(T, NewAcc, X).
combination(0, _, []).
combination(N, [H|T], [H|Comb]) :-
N > 0, N1 is N - 1, combination(N1, T, Comb).
combination(N, [_|T], Comb) :-
N > 0, combination(N, T, Comb).
?- solution([1, 2, 5, 8], 15, X).
X = "[[2, 1], [2], [5, 8], [1], [1, 2]] (15 minutes)" ;
X = "[[1, 2], [1], [5, 8], [2], [1, 2]] (15 minutes)" ;
I'm fun at parties.
You can subscribe to new posts via email or RSS. | {"url":"https://chris-lamb.co.uk/posts/four-men-mine","timestamp":"2024-11-07T12:35:27Z","content_type":"text/html","content_length":"11633","record_id":"<urn:uuid:43852653-e449-4d9f-8ed3-e052858fb3fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00416.warc.gz"} |
There are two types of parentheses. Open and close parentheses. In this problem, open parentheses can be like: ‘(‘ and closed parentheses can be like: ‘)’. A well-formed parenthesis string is a
balanced parenthesis string. A string of parentheses is called balanced if, for every opening bracket, there is a unique closing bracket.
In this article, we’ll learn how to find all the different combinations of n well-formed parentheses such that they form balanced strings.
Also see, Data Structures
Problem Statement
You are given an integer ‘n.’ Your task is to generate all combinations of well-formed parentheses having ‘n’ pairs. Your job is to generate all possible valid sets of parentheses that can be formed
with a given number of pairs.
Note: A parentheses string is called well-formed if it is balanced, i.e., each left parentheses has matching right parentheses, and the matched pairs are well nested. The parentheses can be arranged
in any order, as long as they are valid.
For example:
INPUT : n=3
OUTPUT: ["((()))","(()())","(())()","()(())","()()()"]
These are the only different types of balanced strings that can be formed using three pairs of parentheses.
INPUT : n=1
OUTPUT: [“()"]
Only one kind of string can be formed using one pair of parentheses.
Recommended: Please try the Problem on “Coding Ninjas Studio” before moving on to the solution approach. | {"url":"https://www.naukri.com/code360/library/generate-parentheses","timestamp":"2024-11-03T10:37:36Z","content_type":"text/html","content_length":"445884","record_id":"<urn:uuid:8a91bc1e-d2a4-4c63-9b85-dd7e6c14d90b>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00678.warc.gz"} |
Hyperbolic Trig Identity [Derivative of Hyperbolic Trig
Hyperbolic Trig Identity can be a game-changer in solving complex mathematical problems! Join us and explore the world of Hyperbolic functions, learn powerful identities and unlock the ability to
calculate values with ease. Don’t miss out on this opportunity to enhance your Trigonometry skills!
Hyperbolic Trig Identity
Hyperbolic trigonometric identities are mathematical relationships that involve hyperbolic functions, such as hyperbolic sine (sinh), hyperbolic cosine (cosh), and hyperbolic tangent (tanh). These
identities are similar to the familiar trigonometric identities but apply to hyperbolic functions instead of the standard circular trigonometric functions.
One of the fundamental hyperbolic trigonometric identities is the hyperbolic Pythagorean identity, which relates the hyperbolic sine and hyperbolic cosine functions. It states that the square of the
hyperbolic cosine of an angle (or a real number) is equal to one plus the square of the hyperbolic sine of the same angle. Mathematically, it can be expressed as cosh^2(x) – sinh^2(x) = 1.
Another important identity is the hyperbolic tangent identity, which expresses the hyperbolic tangent in terms of hyperbolic sine and hyperbolic cosine. It states that the hyperbolic tangent of an
angle (or a real number) is equal to the hyperbolic sine divided by the hyperbolic cosine. This identity can be written as tanh(x) = sinh(x) / cosh(x).
These hyperbolic trigonometric identities, along with others derived from them, play a crucial role in solving various mathematical problems involving hyperbolic functions. They provide useful tools
for simplifying expressions, evaluating integrals, solving differential equations, and analyzing hyperbolic functions in different branches of mathematics and science.
Trigonometry and Hyperbolic Functions of Complex Numbers
Hyperbolic trigonometry is used to study the properties of hyperbolic functions and to find relationships between hyperbolic functions and their inverse functions.
One of the main identities in hyperbolic trigonometry is the Pythagorean identity, which states that for all real numbers x, the following equation holds true:
cosh² x – sinh² x = 1
This identity is similar to the Pythagorean theorem in Euclidean geometry, where the sum of the squares of the lengths of the two sides of a right triangle to a square of the hypotenuse’s length.
Another important identity in hyperbolic trigonometry is the exponential identity, which states that for all real numbers x, the following equation holds true:
e^x = cosh x + sinh x
This identity shows the relationship between the exponential function and the hyperbolic cosine and sine functions. It can be used to find the value of either the hyperbolic cosine or the hyperbolic
sine function if the value of the exponential function is known.
Hyperbolic Trig Identity – Inverse Hyperbolic Function
The inverse hyperbolic functions, also known as the area hyperbolic functions, are defined as the inverse functions of the hyperbolic functions. The inverse hyperbolic cosine, denoted as arccosh x,
is defined as the inverse function of the hyperbolic cosine, and the inverse hyperbolic sine, denoted as arcsinh x, is defined as the inverse function of the hyperbolic sine. The inverse hyperbolic
functions are useful for finding the values of the hyperbolic functions if the values of the inverse hyperbolic functions are known.
The relationship between the hyperbolic functions and their inverse functions is given by the following identities:
cosh (arcsinh x) = x sinh (arccosh x) = x
These identities show that the inverse hyperbolic functions are the inverse functions of the hyperbolic functions, and they can be used to find the value of the hyperbolic functions if the value of
the inverse hyperbolic functions is known.
What is Hyperbolic Trigonometry
Hyperbolic trigonometry also has several useful identities that relate the hyperbolic functions to each other. One such identity is the addition formula. That states that for all real numbers x and
y, the following equation holds true:
cosh (x + y) = cosh x cosh y + sinh x sinh y
This identity can be used to find the value of the hyperbolic cosine of the sum of two angles. If the values of the hyperbolic cosine and sine of each angle are known.
Another useful identity in hyperbolic trigonometry is the difference formula. That states that for all real numbers x and y, the following equation holds true:
Hyperbolic trig identities are essential in solving problems in mathematics and physics that involve hyperbolic functions. It’s important to understand and memorize the basic hyperbolic functions and
their identities. The double angle, half angle, and sum and difference identities can be used to simplify complex hyperbolic expressions.
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Given that LK-99 is a rt/ap superconductor, will its average market price in 2028 be below $5,000.00 / metric ton?
• rt/ap = room temperature, ambient pressure
• For reference: Steel = ~ $1,500.00 / metric ton
• resolves n/a if LK-99 is not a rt/ap superconductor
• Inflation adjusted to 2023 price level
This question is managed and resolved by Manifold.
@apetresc That's why I alsways prefer using a ' or a space as a thousands separator to remove the ambiguity for everyone.
So it become $5'000.00 or $5 000.00 | {"url":"https://manifold.markets/Yves/given-that-lk99-is-a-rtap-supercond?r=WXZlcw","timestamp":"2024-11-13T22:55:22Z","content_type":"text/html","content_length":"133478","record_id":"<urn:uuid:69606dbd-250c-4e02-b440-031bb1c6c7cc>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00631.warc.gz"} |
Trouble finding the current through this resistor in the circuit
• Thread starter guyvsdcsniper
• Start date
In summary, the conversation discusses how to use the substitution method to solve a system of equations. The speaker is confused about how to use the substitution ##I_3 = I_1-I_2## to get the
equations in terms of ##I_1## and ##I_2##. They ask for someone to demonstrate the process and show their work for any potentially difficult steps.
Homework Statement
FIind the current through and the potential difference across the 100 ohm resistor.
Relevant Equations
junction rule. loop rule
I was following this problem up until they say we can they state I3= I1-I2. I understand why we can say that but I don't see how I can use that to get the system of equation in terms of I1 and I2 at
the bottom. Could someone show me how this was done?
Science Advisor
Homework Helper
Gold Member
2023 Award
You have the two equations
Take the first of these two equations and substitute ##I_3 = I_1-I_2##. This means to replace ##I_3## in the equation by the quantity ##(I_1-I_2)##. Then simplify.
Do the same for the second equation.
Show your work so we can help with any particular step for which you have trouble.
FAQ: Trouble finding the current through this resistor in the circuit
What is the purpose of finding the current through a resistor in a circuit?
The current through a resistor is important because it allows us to calculate the voltage drop across the resistor, which is necessary for understanding the behavior of the circuit.
Why is it sometimes difficult to find the current through a resistor in a circuit?
There are a few reasons why it may be difficult to find the current through a resistor in a circuit. One reason is that there may be multiple resistors in the circuit, making it more complex to
calculate the current. Another reason is that the circuit may have other components, such as capacitors or inductors, that affect the flow of current.
How can I find the current through a resistor in a circuit?
To find the current through a resistor in a circuit, you can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R). So, I = V/R. Alternatively, you can use
Kirchhoff's Current Law, which states that the sum of all currents entering and exiting a node in a circuit is equal to zero.
What factors can affect the current through a resistor in a circuit?
The current through a resistor in a circuit can be affected by a few different factors. These include the voltage of the power source, the resistance of the resistor, and the presence of other
components in the circuit that may affect the flow of current.
What are some common techniques for measuring the current through a resistor in a circuit?
There are a few common techniques for measuring the current through a resistor in a circuit. One method is to use a multimeter, which can measure the voltage drop across the resistor and calculate
the current using Ohm's Law. Another method is to use a current probe, which can measure the current directly without interrupting the circuit. Additionally, some circuits may have built-in current
sensing components that can provide an accurate measurement of the current through a resistor. | {"url":"https://www.physicsforums.com/threads/trouble-finding-the-current-through-this-resistor-in-the-circuit.1008502/","timestamp":"2024-11-10T14:57:54Z","content_type":"text/html","content_length":"81914","record_id":"<urn:uuid:fac99348-5852-47f5-91b4-2a7a6d05ec44>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00549.warc.gz"} |
A Mathematical Conjecture from P versus NP
Download PDFOpen PDF in browser
A Mathematical Conjecture from P versus NP
EasyChair Preprint 3415
5 pages•Date: May 16, 2020
P versus NP is considered as one of the most important open problems in computer science. This consists in knowing the answer of the following question: Is P equal to NP? It was essentially mentioned
in 1955 from a letter written by John Nash to the United States National Security Agency. However, a precise statement of the P versus NP problem was introduced independently by Stephen Cook and
Leonid Levin. Since that date, all efforts to find a proof for this problem have failed. It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute to carry a US
1,000,000 prize for the first correct solution. Another major complexity class is NP-complete. To attack the P versus NP question the concept of NP-completeness has been very useful. If any single
NP-complete problem can be solved in polynomial time, then every NP problem has a polynomial time algorithm. We state the following conjecture for a natural number B greater than 3: The number of
divisors of B is lesser than or equal to the quadratic value from the integer part of the logarithm of B in base 2. This conjecture has been checked for large numbers: Specifically, from every
integer between 4 and 10 millions. If this conjecture is true, then the NP-complete problem Subset Product is in P and thus, the complexity class P is equal to NP.
Keyphrases: completeness, complexity classes, logarithm, polynomial time, tuple
Links: https://easychair.org/publications/preprint/8HdG
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Lesson 1
Equal Groups of Unit Fractions
Warm-up: How Many Do You See: Oranges (10 minutes)
The purpose of this How Many Do You See is to elicit ideas about equal groups of fractional amounts and to prepare students reason about multiplication of a whole number and a fraction. Students may
describe the oranges with a whole number without units or without specifying “halves” (for instance, they may say “5”). If this happens, consider asking them to clarify whether they mean “5 oranges”
or another amount.
• Groups of 2
• “How many do you see? How do you see them?”
• Display the image.
• 1 minute: quiet think time
• “Discuss your thinking with your partner.”
• 1 minute: partner discussion
• Record responses.
Student Facing
How many do you see? How do you see them?
Activity Synthesis
• “How might you describe this image to a friend?” (There are 5 plates with \(\frac{1}{2}\) orange on each plate.)
• “How many groups do you see?” (I see 5 plates or 5 groups.)
• “Besides describing the image in words, how else might you represent the quantity in this image?” (I might write \(\frac{1}{2}\) five times, or write an expression with five \(\frac{1}{2}\)s
being added together. I might write “5 times \(\frac{1}{2}\).”)
• “We'll look at some other situations involving groups and fractional amounts in this lesson.”
Activity 1: Crackers, Kiwis, and More (20 minutes)
The purpose of this activity is for students to interpret situations involving equal groups of a fractional amount and to connect such situations to multiplication of a whole number by a fraction
Students write expressions to represent the number of groups and the size of each group. They reason about the quantity in each situation in any way that makes sense to them. Although images of the
food items are given, students may choose to create other diagrams, such as equal-group diagrams used in grade 3, when they learned to multiply whole numbers. This activity enables the teacher to see
the representations toward which students gravitate.
Focus the discussions on connecting equal groups with fractions and those with whole numbers.
Representation: Access for Perception. Use pictures (or actual crackers, if possible) to represent the situation. Ask students to identify correspondences between this concrete representation and the
diagrams they create or see.
Supports accessibility for: Conceptual Processing, Visual-Spatial Processing
• Groups of 2
• “What are some of your favorite snacks?”
• Share responses.
• “What are some snacks that you might break into smaller pieces rather than eating them whole?”
• 1 minutes: partner discussion
• “Let's look at some food items that we might eat whole or cut or break up into smaller pieces.”
• “Take a few quiet minutes to think about the first set of problems about crackers. Then, discuss your thinking with your partner.”
• 4 minutes: independent work time
• 2 minutes: partner discussion
• Pause for a whole-class discussion. Invite students to share their responses.
• If no students mention that there are equal groups, ask them to make some observations about the size of the groups in each image.
• Discuss the expressions students wrote:
□ “What expression did you write to represent the crackers in Image A? Why?” (\(6 \times 4\), because there are 6 groups of 4 full crackers.)
□ “What about the crackers in Image B? Why?” (\(6 \times \frac{1}{4}\), because there are 6 groups of \(\frac{1}{4}\) of a cracker.)
• Ask students to complete the remaining problems.
• 5 minutes: independent or partner work time
• Monitor for students who reason about the quantities in terms of “_____ groups of _____” to help them write expressions.
Student Facing
1. Here are images of some crackers.
1. How are the crackers in image A like those in B?
2. How are they different?
3. How many crackers are in each image?
4. Write an expression to represent the crackers in each image.
2. Here are more images and descriptions of food items. For each, write a multiplication expression to represent the quantity. Then, answer the question.
1. Clare has 3 baskets. She put 4 eggs into each basket. How many eggs did she put in baskets?
2. Diego has 5 plates. He put \(\frac12\) of a kiwi fruit on each plate. How many kiwis did he put on plates?
3. Priya prepared 7 plates with \(\frac18\) of a pie on each. How much pie did she put on plates?
4. Noah scooped \(\frac13\) cup of brown rice 8 times. How many cups of brown rice did he scoop?
Advancing Student Thinking
If students are unsure how to name the quantity in the image, consider asking: ”How would you describe the amount of the slice of pie on one plate? How would you describe two of the same slices?
Three of the same slices?”
Activity Synthesis
• Select previously identified students to share their expressions and how they reasoned about the amount of food in each image. Record their expressions and supporting diagrams, if any, for all to
• If students write addition expressions to represent the quantities, ask if there are other expressions that could be used to describe the equal groups.
• “How is the quantity in Clare's situation different than those in other situations?” (It involves whole numbers of items. Others involve fractional amounts.)
• “How is the expression you wrote for the eggs different than other expressions?” (It shows two whole numbers being multiplied. The others show a whole number and a fraction.)
Activity 2: What Could It Mean? (15 minutes)
In this activity, students start with given multiplication expressions and consider situations or diagrams that they could represent. Situating the expressions in context encourages students to think
of the whole number in the expression as the number of groups and the fractional amount as the size of each group, which helps them reason about the value of the expression. When students make
explicit connections between multiplication situations, expressions, and drawings they reason abstractly and quantitatively (MP2).
Allow students to use fraction strips, fraction blocks, or other manipulatives that show fractional amounts to support their reasoning.
MLR8 Discussion Supports. Synthesis: Provide students with the opportunity to rehearse what they will say with a partner before they share with the whole class.
Advances: Speaking
• Groups of 2
• “Read the task statement. Then, talk to your partner about what you are asked to do in this activity.”
• 1 minute: partner discussion
• “Choose one expression you’d like to start with.”
• “Think of a story that can be represented by the expression. Then, create a drawing or diagram, and find the value of the expression.”
• “If you have extra time you can work on both problems.”
• 7–8 minutes: independent work time
• “Be sure to say what the value of the expression means in your story.”
Student Facing
For each expression:
• Write a story that the expression could represent. The story should be about a situation with equal groups.
• Create a drawing to represent the situation.
• Find the value of the expression. What does this number mean in your story?
1. \(8 \times \frac{1}{2}\)
2. \(7 \times \frac {1}{5}\)
Activity Synthesis
• Invite students to share their responses. Display their drawings or visual representations for all to see.
• “How did you decide what the numbers in each expression represent in your story?” (It made sense for the whole numbers to represent how many groups there are and the fractions to represent what
is in each group.)
• “How did you show the whole number and the fraction in your drawing?” (I drew as many circles as the whole number to show the groups. I drew parts of objects or wrote numbers in each circle to
show the fraction.)
Lesson Synthesis
“Today we looked at different situations that involved equal-size groups and a fractional amount in each group. We thought about how to find the total amount in each situation.”
“How did we represent these situations?” (We wrote expressions and used drawings or pictures to show the equal groups.)
“What kind of expressions did we write?” (Multiplication expressions with a whole number and a fraction in each)
“What strategies did we use to find the total amount in each situation?” (We counted the number of fractional parts in the drawings. We counted how many parts made 1 whole and saw how many extra
fractional parts there were.)
Cool-down: Sandwiches on Plates (5 minutes) | {"url":"https://curriculum.illustrativemathematics.org/k5/teachers/grade-4/unit-3/lesson-1/lesson.html","timestamp":"2024-11-14T11:59:46Z","content_type":"text/html","content_length":"102834","record_id":"<urn:uuid:5d7c4147-3fb6-42f1-91de-e33d4050bd19>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00487.warc.gz"} |
Fresnel Diffraction and the Cornu Spiral - montoguequiz.com
All introductory optics courses cover diffraction phenomena, and most begin their coverage of this topic with Fraunhofer diffraction. In Fraunhofer diffraction, the source and screen are arranged to
be effectively at infinity from the diffraction device, a slit-like aperture, or a diffraction grating. This simplifies the treatment appreciably because the emitted waves reach the apertures as
plane waves, all having the same phase and amplitude. We do away with these limitations in near-field, or Fresnel, diffraction, so named after the French engineer and physicist Augustin-Jean Fresnel
(1788 – 1827). In this post, I provide a quick introduction to this phenomenon, with emphasis on the concepts of Fresnel zones and Fresnel integrals. Two solved examples are included along the way.
1. Obliquity factor
A crucial aspect of Fresnel diffraction is that waves emitted by a point source at a finite distance from an aperture or a straight-edge obstacle are spherical in nature. As light diffracts through
the opening, points on the primary wavefront are envisioned as continuous emitters of secondary spherical wavelets; this is illustrated in Figure 1.
Figure 1. Near-field diffraction along an aperture.
But if each wavelet radiated uniformly in all directions, in addition to generating an ongoing wave, there would also be a reverse wave travelling back toward the source. Since no such wave is found
experimentally, we must somehow modify the radiation pattern of the secondary emitters. With this goal in mind, we introduce a quantity called the obliquity factor F($\displaystyle \theta$):
$\displaystyle F\left( \theta \right)=\frac{1}{2}\left( {1+\cos \theta } \right)\,\,\,(1)$
In this equation, $\displaystyle \theta$ is the angle between a horizontal line through the secondary source and the line connecting it to the observation point (Fig. 1). The obliquity factor, when
multiplied by the amplitude of a spherical secondary wavelet at the aperture, reduces it by a factor ranging from 1 to 0; the angle $\displaystyle \theta$ = 0^o is for the forward direction, and $\
displaystyle \theta$ = 180^o is for the backward direction. Substituting $\displaystyle \theta$ = 0^o yields F(0^o) = 1, which means that the obliquity factor has no effect on the amplitude of the
forward wave. On the other hand, for $\displaystyle \theta$ = 180^o we obtain F(180^o) = 0, so that multiplying this result by the wave amplitude yields zero, nullifying the backward-directed
2. Basics of Fresnel diffraction
To develop a mathematical treatment of Fresnel diffraction, we refer to Figure 2, where arc WW represents a primary wave with a spherical wave front of radius $\displaystyle {r}'$ emitted by a source
S, and P is an observation point on the screen. In Figure 3, the arc W’W’ represents a wavelet of radius $\displaystyle r$ emitted by a secondary source on the primary wave front at O. The line
connecting S and P passing through the polar point O consists of the segments SO and OP, which are labelled $\displaystyle {{{r}'}_{o}}$ and $\displaystyle {{r}_{o}}$, respectively. Assuming that E
[o] is the amplitude at unit distance from the source, the disturbance per unit area of the aperture at a point of the wave front lying on the aperture can be described by
$\displaystyle {{E}_{A}}=\frac{{{{E}_{o}}}}{{{r}'}}{{e}^{{i\left( {k{r}'-\omega t} \right)}}}\,\,\,(2)$
where t is time and $\displaystyle {r}'$ is radial distance as designated in Figure 4; k and $\displaystyle \omega$ are the wavenumber and angular frequency of the wave, respectively. Designating as
t = 0 the instant when the wavefront is at the selected arbitrary point, the factor $\displaystyle {{e}^{{-i\omega t}}}$ would be unity and can be dropped from the expression above. The region
impacted by the obliquity factor F($\displaystyle \theta$) is
$\displaystyle {{E}_{A}}F\left( \theta \right)=\frac{{F\left( \theta \right){{E}_{o}}}}{{{r}'}}{{e}^{{ik{r}'}}}\,\,\,(3)$
The wave in equation (3) propagates to a receiving point P as a secondary wavelet of an element of disturbance dE[p] given by
$\displaystyle d{{E}_{P}}=\frac{{F\left( \theta \right){{E}_{A}}}}{r}{{e}^{{ik\left( {r-\omega t} \right)}}}\,\,\,(4)$
Combining equations (3) and (4) gives the disturbance at a point P due to the wavelets reaching P, that is,
$\displaystyle d{{E}_{P}}=\frac{{F\left( \theta \right){{E}_{o}}}}{{r{r}'}}{{e}^{{i\left[ {k\left( {r+{r}'} \right)-\omega t} \right]}}}\,\,\,(5)$
To include contributions of all wavelets generated by the aperture points, we integrate over the area A:
$\displaystyle {{E}_{P}}=\int_{A}{{\frac{{F\left( \theta \right){{E}_{o}}}}{{r{r}'}}{{e}^{{i\left[ {k\left( {r+{r}'} \right)-\omega t} \right]}}}dA}}\,\,\,(6)$
Figure 2. Spherical wavefront emitted by a source, S.
Figure 3. Path difference along SP and SMM’P.
In an attempt to carry out the integration in (6), Fresnel introduced a special construction of the primary incident wave that encounters the apertures as it propagates. In his scheme, the incident
wavefront is divided into zones of circles that have a finite width centered on the pole O. The circles are constructed such that the distance from the observation point P to the first (smallest)
circle is r + $\displaystyle \lambda$/2; the distance from P to the second circle is r + $\displaystyle \lambda$; the same pattern extends indefinitely, such that each circle is larger than the
preceding one by $\displaystyle \lambda$/2. The task then reduces to determining how many zones cross the aperture, counting the path differences and the corresponding phase differences for all
wavelets reaching P[i], and ultimately adding the contributions to diffraction on the screen.
For a simple implementation of such a process, refer to Figure 3, in which we have a side view of an aperture, a segment of a Fresnel zone on a primary wave, and a wavelet generated at a point of the
aperture. Consider the two points O and M through which the paths SOP and SMP pass. The first is the shortest path of light between the source and receiving point P traversed by the primary wave and
the secondary wavelet generated at O. The second is the path OM plus that of a wavelet generated at M, another point on the primary wave. The two paths, being not equal, are of a path difference $\
displaystyle \Delta$ given by
$\displaystyle \Delta =\left( {r+{r}'} \right)-\left( {{{r}_{0}}+{{{{r}'}}_{0}}} \right)\,\,\,(7)$
which can be stated as
$\displaystyle \Delta =QO-{Q}'O\,\,\,(8)$
Segments QO and Q’O are known as the sagittas of the arcs WW and W’W’. For a small height of points M and M’ above the line SOP, these heights can be shown to equal
$\displaystyle QO=\frac{{{{R}^{2}}}}{{2{{{{r}'}}_{0}}}}\,\,\,(9.1)$
$\displaystyle {Q}'O=\frac{{{{R}^{2}}}}{{2{{r}_{0}}}}\,\,\,(9.2)$
so that (8) becomes
$\displaystyle \Delta =\frac{{{{R}^{2}}}}{{2{{{{r}'}}_{0}}}}-\frac{{{{R}^{2}}}}{{2{{r}_{0}}}}$
$\displaystyle \therefore \Delta =\frac{{{{R}^{2}}}}{2}\left( {\frac{{{{r}_{0}}+{{{{r}'}}_{0}}}}{{{{r}_{0}}{{{{r}'}}_{0}}}}} \right)\,\,\,(10)$
But the path difference $\displaystyle \Delta$ of a zone of order m is such that
$\displaystyle \Delta =m\frac{\lambda }{2}\,\,\,(11)$
Equating (10) and (11) and solving for R[m], (i.e., the value of R at the m-th order), we have
$\displaystyle m\frac{\lambda }{2}=\frac{{R_{m}^{2}}}{2}\left( {\frac{{{{r}_{0}}+{{{{r}'}}_{0}}}}{{{{r}_{0}}{{{{r}'}}_{0}}}}} \right)$
$\displaystyle \therefore R_{m}^{2}=m\lambda \left( {\frac{{{{r}_{0}}{{{{r}'}}_{0}}}}{{{{r}_{0}}+{{{{r}'}}_{0}}}}} \right)\,\,\,(12)$
But, from the geometry of Figure 3, we also have
$\displaystyle R_{m}^{2}={{\left( {{{r}_{0}}+\frac{{m\lambda }}{2}} \right)}^{2}}-r_{0}^{2}\,\,\,(13)$
so that, equating (12) and (13) and manipulating,
$\displaystyle R_{m}^{2}=r_{0}^{2}\left[ {\left( {\frac{{m\lambda }}{{{{r}_{0}}}}} \right)+{{{\left( {\frac{{m\lambda }}{{2{{r}_{0}}}}} \right)}}^{2}}} \right]\,\,\,(14)$
But m$\displaystyle \lambda$/r[0] $\displaystyle \ll$ 1, hence the second term on the right-hand side of (14) can be dropped, giving
$\displaystyle R_{m}^{2}\approx r_{0}^{2}\times \left( {\frac{{m\lambda }}{{{{r}_{0}}}}} \right)=m{{r}_{0}}\lambda \,\,\,(15.1)$
$\displaystyle {{R}_{m}}\approx \sqrt{{m{{r}_{0}}\lambda }}\,\,\,(15.2)$
This equation indicates that the radius of a given Fresnel zone is proportional to the square root of its order m. For an aperture of radius a, whose center is on the line connecting the source and
the observation point P, the number of zones N that pass through the aperture can be calculated as
$\displaystyle \pi {{a}^{2}}=N\times {{A}_{m}}\,\,\,(16)$
where A[m] is the area of the m-th zone. Letting the central zone (the first zone) be labelled m = 1, we have A[1] = $\displaystyle \pi$$\displaystyle R_{1}^{2}$, so the number N of zones that would
fit into the aperture is
$\displaystyle N=\frac{{{{a}^{2}}}}{{R_{1}^{2}}}=\frac{{{{a}^{2}}}}{{{{r}_{0}}\lambda }}\,\,\,(17)$
Number N may be used to distinguish Fraunhofer diffraction from Fresnel diffraction: if N $\displaystyle \ll$ 1, Fraunhofer diffraction dominates; for N > 1, Fresnel diffraction holds.
Example 1
Consider a monochromatic light of wavelength $\displaystyle \lambda$ = 680.0 nm incident on an aperture of radius a = 8.0 mm. The aperture was situated in the middle between the source and a point P
on the screen along the line SOP. For S and P to be 25 cm apart, find:
(a) The radius of the central zone.
(b) The number of Fresnel zones that would be accommodated through the aperture.
Solution, Part a. We have m = 1, $\displaystyle {{r}_{0}}$ = 0.008 m, $\displaystyle {{{r}'}_{0}}$ = 0.008 m, and $\displaystyle \lambda$ = 680$\displaystyle \times$10^‒9 m, so that, substituting
into (12),
$\displaystyle {{R}_{1}}=\sqrt{{m\lambda \frac{{{{r}_{0}}{{{{r}'}}_{0}}}}{{{{r}_{{_{0}}}}+{{{{r}'}}_{0}}}}}}=\sqrt{{1\times \left( {680\times {{{10}}^{{-9}}}} \right)\times \frac{{0.008\times 0.008}}
$\displaystyle \therefore {{R}_{1}}=5.22\times {{10}^{{-5}}}\,\,\text{m}=52.2\,\,\text{ }\!\!\mu\!\!\text{ m}\leftarrow$
Part b. The number N of Fresnel zones that would pass through the aperture is given by equation (17),
$\displaystyle {{N}_{F}}=\frac{{{{a}^{2}}}}{{R_{1}^{2}}}=\frac{{{{{0.008}}^{2}}}}{{{{{\left( {5.22\times {{{10}}^{{-5}}}} \right)}}^{2}}}}=23,500\leftarrow$
4. Fresnel diffraction in a rectangular aperture – Analytical treatment
We now turn to an analytical treatment of Fresnel diffraction for a rectangular aperture. As shown in Figure 4, we begin by taking an elemental strip chosen along the width W of the rectangular
aperture. A spherical wave incident on a very large number of these strips would constitute a problem in requiring the phase at all points of the strip to be constant; harder still would be to
require the phase to be the same along successive elemental strips. One way to remedy this is to have a cylindrical source parallel to the aperture plane, so that all points of the cylindrical wave
front would have the same phase. Aside from the geometry of the aperture, which requires the source to be an extended slit so that cylindrical waves are considered, no constraints are applied and the
Fresnel equation (6) can be restated as
$\displaystyle {{E}_{P}}={{C}_{1}}{{e}^{{-i\omega t}}}\int_{A}{{{{e}^{{ik\left( {r+{r}'} \right)}}}dA}}\,\,\,(18)$
where all constants are gleaned into a parameter C[1]. We assume that the surface integral over a closed surface including the aperture is zero everywhere except over the aperture itself, so that we
need perform the integration only over the aperture in the yz-plane of Figure 4a. A side view, which shows the curvature of the cylindrical wavefront, is drawn in Figure 4b.
Figure 4. (a) Cylindrical wavefronts from source slit S are diffracted by a rectangular aperture. (b) Edge view of (a).
The distance r + r’ may be determined approximately from this figure. For h $\displaystyle \ll$ p and h $\displaystyle \ll$ q, a binomial expansion approximation gives
$\displaystyle {r}'={{\left( {{{p}^{2}}+{{h}^{2}}} \right)}^{{{1}/{2}\;}}}=p{{\left( {1+\frac{{{{h}^{2}}}}{{{{p}^{2}}}}} \right)}^{{{1}/{2}\;}}}\approx p\left( {1+\frac{{{{h}^{2}}}}{{2{{p}^{2}}}}} \
so that
$\displaystyle {r}'\approx p+\frac{1}{2}\left( {\frac{{{{h}^{2}}}}{p}} \right)\,\,\,(19.2)$
Proceeding similarly with r,
$\displaystyle r\approx q+\frac{1}{2}\left( {\frac{{{{h}^{2}}}}{q}} \right)\,\,\,(20)$
Adding the two foregoing results,
$\displaystyle r+{r}'\approx \left( {p+q} \right)+\left( {\frac{1}{p}+\frac{1}{q}} \right)\frac{{{{h}^{2}}}}{2}\,\,\,(21)$
For convenience, we introduce the variables
$\displaystyle D=p+q\,\,\,(22)$
$\displaystyle \frac{1}{L}=\frac{1}{p}+\frac{1}{q}\,\,\,(23)$
so that
$\displaystyle {{E}_{P}}={{C}_{1}}{{e}^{{-i\omega t}}}\int_{A}{{{{e}^{{ik\left( {D+{{{{h}^{2}}}}/{{2L}}\;} \right)}}}dA}}\,\,\,(24)$
If the elemental area dA is taken to be the shaded strip in Figure 4, dA = Wdz, h = z, and
$\displaystyle {{E}_{P}}={{C}_{1}}W{{e}^{{i\left( {kD-\omega t} \right)}}}\int_{{{{z}_{1}}}}^{{{{z}_{2}}}}{{{{e}^{{{{ik{{z}^{2}}}}/{{2L}}\;}}}dz}}\,\,\,(25)$
The exponent in the integrand can be restated as
$\displaystyle \frac{{k{{z}^{2}}}}{{2L}}=\frac{{\pi {{z}^{2}}}}{{L\lambda }}\,\,\,(26)$
where k is wavenumber and $\displaystyle \lambda$ is the wavelength. Making a change of variable, we let
$\displaystyle z=v\sqrt{{\frac{{\lambda L}}{2}}}\,\,\,(27.1)$
$\displaystyle v=z\sqrt{{\frac{2}{{\lambda L}}}}\,\,\,(27.2)$
so we can restate (25) as
$\displaystyle {{E}_{P}}=W\sqrt{{\frac{{L\lambda }}{2}}}{{C}_{1}}{{e}^{{i\left( {kD-\omega t} \right)}}}\int_{{{{v}_{1}}}}^{{{{v}_{2}}}}{{{{e}^{{{{i\pi {{v}^{2}}}}/{2}\;}}}dv}}$
$\displaystyle \therefore {{E}_{P}}={{A}_{P}}{{e}^{{i\left( {kD-\omega t} \right)}}}\int_{{{{v}_{1}}}}^{{{{v}_{2}}}}{{{{e}^{{{{i\pi {{v}^{2}}}}/{2}\;}}}dv}}\,\,\,(28)$
where A[p ]is a complex scale factor with dimensions of electric field amplitude. Using Euler’s theorem on the integrand, it follows that
$\displaystyle {{E}_{P}}={{A}_{P}}{{e}^{{i\left( {kD-\omega t} \right)}}}\left[ {\int_{{{{v}_{1}}}}^{{{{v}_{2}}}}{{\cos \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)dv}}+i\times \int_{{{{v}_{1}}}}^
{{{{v}_{2}}}}{{\sin \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)dv}}} \right]\,\,\,(29)$
The two integrals on the right-hand side are known as Fresnel integrals, and are routinely denoted with the simplified notations
$\displaystyle C\left( v \right)\equiv \int_{0}^{v}{{\cos \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)dv}}\,\,\,(30.1)$
$\displaystyle S\left( v \right)\equiv \int_{0}^{v}{{\sin \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)dv}}\,\,\,(30.2)$
Substituting these into (29) gives
$\displaystyle {{E}_{P}}={{A}_{P}}{{e}^{{i\left( {kD-\omega t} \right)}}}\left\{ {\left[ {C\left( {{{v}_{2}}} \right)-C\left( {{{v}_{1}}} \right)} \right]+i\left[ {S\left( {{{v}_{2}}} \right)-S\left(
{{{v}_{1}}} \right)} \right]} \right\}\,\,\,(31)$
Now, note that the irradiance at P is given by
$\displaystyle {{I}_{P}}=\frac{1}{2}{{\varepsilon }_{0}}c{{\left| {{{E}_{P}}} \right|}^{2}}\,\,\,(32)$
where $\displaystyle {{\varepsilon }_{0}}$ is vacuum permittivity, c $\displaystyle \approx$ 3.0$\displaystyle \times$10^8 m/s, and E[p] is electric field intensity, Replacing E[P] from (31) brings
$\displaystyle {{I}_{P}}={{I}_{0}}\left\{ {{{{\left[ {C\left( {{{v}_{2}}} \right)-C\left( {{{v}_{1}}} \right)} \right]}}^{2}}+{{{\left[ {S\left( {{{v}_{2}}} \right)-S\left( {{{v}_{1}}} \right)} \
right]}}^{2}}} \right\}\,\,\,(33)$
where we have introduced the irradiance scale factor $\displaystyle {{I}_{0}}$, namely
$\displaystyle {{I}_{P}}=\frac{1}{2}{{\varepsilon }_{0}}c{{\left| {{{A}_{P}}} \right|}^{2}}\,\,\,(34)$
It is important to note that C(v) and S(v) are both odd functions, so that
$\displaystyle C\left( {-v} \right)=-C\left( v \right)\,\,\,(35.1)$
$\displaystyle S\left( {-v} \right)=-S\left( v \right)\,\,\,(35.2)$
The choice of v in the Fresnel integrals is determined by the vertical dimensions of the diffraction aperture.
Some values of the Fresnel integrals are listed here. Tabulated values are convenient, but often require tedious interpolation. A better alternative is to use software such as Mathematica or MATLAB,
or resort to this free web-based app by Casio.
5. The Cornu spiral
If the values of the Fresnel integrals are plotted with C(v) on the horizontal axis and S(v) on the vertical axis, the resulting graph is the so-called Cornu spiral, Figure 5. The origin v = 0 of the
Cornu spiral corresponds to z = 0 and therefore to the y-axis through the aperture of Figure 4. The top part of the spiral (z > 0 and v > 0) represents contributions from strips of the aperture above
the y-axis, and the twin spiral below (z < 0 and v < 0) represents similar contributions from below the y-axis. The two limit points or “eyes” of the spiral at E and E’ represent linear zones at z =
$\displaystyle \pm$$\displaystyle \infty$.
Furthermore, variable v, introduced in (27.2), represents the length along the Cornu spiral itself. To see this, recall that the incremental length dl along a curve in the xy-plane is given by the
Pythagorean relationship
$\displaystyle d{{l}^{2}}=d{{x}^{2}}+d{{y}^{2}}\,\,\,(36)$
But in the Cornu spiral plane the x– and y-coordinates are given by the Fresnel integrals C(v) and S(v), respectively, giving
$\displaystyle d{{l}^{2}}={{\left[ {dC\left( v \right)} \right]}^{2}}+{{\left[ {dS\left( v \right)} \right]}^{2}}$
$\displaystyle \therefore d{{l}^{2}}=\left[ {{{{\cos }}^{2}}\left( {\frac{{\pi {{v}^{2}}}}{2}} \right)+{{{\sin }}^{2}}\left( {\frac{{\pi {{v}^{2}}}}{2}} \right)} \right]d{{v}^{2}}$
$\displaystyle \therefore d{{l}^{2}}=d{{v}^{2}}$
$\displaystyle \therefore dl=dv$
Another noteworthy feature of the Cornu spiral is that its tangent at a certain point (x, y) is given by the tangent of variable $\displaystyle \pi$$\displaystyle {{v}^{2}}$/2 at that particular
$\displaystyle \frac{x}{y}=\frac{{\sin \left( {{{\pi {{v}^{2}}}}/{2}\;} \right)}}{{\cos \left( {{{\pi {{v}^{2}}}}/{2}\;} \right)}}=\tan \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)\,\,\,(37)$
Figure 5. A Cornu spiral.
6. Unobstructed wavefront
The irradiance in the Fresnel diffraction pattern associated with different apertures are often compared to the irradiance I[u] associated with an unobstructed wavefront. An unobstructed wavefront is
modeled by passage through an aperture with a vertical dimension 2 that ranges from ‒∞ to +∞. In this case, the total irradiance I[u] at point P is proportional to the square of the length of the
phasor drawn from E’ to E in Figure 5. These limiting points have coordinates (C(v[2]), S(v[2])) = (0.5, 0.5) and (C(v[1]), S(v[1])) = (‒0.5, ‒0.5), as given by the improper integrals
$\displaystyle C\left( \infty \right)=\int_{0}^{\infty }{{\cos \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)dv}}=0.5\,\,\,(38.1)$
$\displaystyle S\left( \infty \right)=\int_{0}^{\infty }{{\sin \left( {\frac{{\pi {{v}^{2}}}}{2}} \right)dv}}=0.5\,\,\,(38.2)$
and from the fact that C(v) and S(v) are odd functions. It follows that, substituting into (33), the unobstructed irradiance becomes
$\displaystyle {{I}_{u}}={{I}_{0}}\left\{ {{{{\left[ {C\left( \infty \right)-C\left( {-\infty } \right)} \right]}}^{2}}+{{{\left[ {S\left( \infty \right)-S\left( {-\infty } \right)} \right]}}^{2}}} \
$\displaystyle \therefore {{I}_{u}}={{I}_{0}}\left\{ {{{{\left[ {0.5-\left( {-0.5} \right)} \right]}}^{2}}+{{{\left[ {0.5-\left( {-0.5} \right)} \right]}}^{2}}} \right\}$
$\displaystyle \therefore {{I}_{u}}=2{{I}_{0}}\,\,\,(39)$
Other irradiances may be compared conveniently to this result.
Example 2
A slit illuminated with sodium light ($\displaystyle \lambda$ = 589.3 nm) is placed 60 cm from a straight edge and the diffraction pattern is observed using a photoelectric cell, 120 cm beyond the
straight edge. Determine the irradiance at (a) 2 mm inside and (b) 1 mm outside the edge of the geometrical shadow.
Solution, Part a. For either position, parameter L is such that (equation (23))
$\displaystyle \frac{1}{L}=\frac{1}{p}+\frac{1}{q}=\frac{1}{{60}}+\frac{1}{{120}}$
$\displaystyle \therefore \frac{1}{L}=\frac{3}{{120}}$
$\displaystyle \therefore L=40\,\,\text{cm}=0.4\,\,\text{m}$
Let z’ be the coordinate of the point O’ in the aperture plane along the straight line from S to the observation point P (see Figure 2). In the case at hand, with y = ‒2 mm, we may write
$\displaystyle {z}'=\frac{p}{{p+q}}y=\frac{{60}}{{60+120}}\times \left( {-2\times {{{10}}^{{-3}}}} \right)=-6.67\times {{10}^{{-4}}}\,\,\text{m}$
The values of the z[1] and z[2] that mark the edges of the unobstructed regions in the aperture plane are to be measured relative to this point, so z[2] = ∞ and z[1] = +6.67$\displaystyle \times$10^
‒4 m. We proceed to compute the modified variables v[1] and v[2]:
$\displaystyle {{v}_{2}}=\infty$
$\displaystyle {{v}_{1}}=\sqrt{{\frac{2}{{\lambda L}}}}{{z}_{1}}=\sqrt{{\frac{2}{{\left( {589.3\times {{{10}}^{{-9}}}} \right)\times 0.4}}}}\times \left( {6.67\times {{{10}}^{{-4}}}} \right)=1.942$
The Fresnel integrals for v[2] are obvious: C(v[2]) = S(v[2]) = 0.5. In turn, entering v[1] into the Casio app gives C(v[1]) = 0.4315 and S(v[1]) = 0.3538, as shown.
Referring to eq. (29), we compute the relative irradiance I:
$\displaystyle I={{I}_{0}}\left\{ {{{{\left[ {C\left( {{{v}_{2}}} \right)-C\left( {{{v}_{1}}} \right)} \right]}}^{2}}+{{{\left[ {S\left( {{{v}_{2}}} \right)-S\left( {{{v}_{1}}} \right)} \right]}}^
{2}}} \right\}$
$\displaystyle \therefore I={{I}_{0}}\left[ {{{{\left( {0.5-0.4315} \right)}}^{2}}+{{{\left( {0.5-0.3538} \right)}}^{2}}} \right]=0.0261{{I}_{0}}$
Comparing this with the unobstructed irradiance as given by (39), we get
$\displaystyle I=0.0261\times \left( {{{{{I}_{u}}}}/{2}\;} \right)=0.0131{{I}_{u}}\leftarrow$
Solution, Part b. In this case, z’ becomes
$\displaystyle {z}'=\frac{p}{{p+q}}y=\frac{{60}}{{60+120}}\times \left( {1\times {{{10}}^{{-3}}}} \right)=3.33\times {{10}^{{-4}}}\,\,\text{m}$
so that z[2] = ∞ and z[1] = ‒3.33$\displaystyle \times$10^‒4 m, updating v[2] and v[1],
$\displaystyle {{v}_{2}}=\infty$
$\displaystyle {{v}_{1}}=\sqrt{{\frac{2}{{\lambda L}}}}{{z}_{1}}=\sqrt{{\frac{2}{{\left( {589.3\times {{{10}}^{{-9}}}} \right)\times 0.4}}}}\times \left( {6.67\times {{{10}}^{{-4}}}} \right)=-0.970$
As before, the Fresnel integrals for v[2] are C(v[2]) = S(v[2]) = 0.5. Next, entering v[1] into the Casio calculator gives C(v[1]) = ‒0.7785 and S(v[1]) = ‒0.4083, as shown.
The relative irradiance $\displaystyle I$ then becomes
$\displaystyle I={{I}_{0}}\left\{ {{{{\left[ {0.5-\left( {-0.7785} \right)} \right]}}^{2}}+{{{\left[ {0.5-\left( {-0.4083} \right)} \right]}}^{2}}} \right\}$
$\displaystyle \therefore I=2.460{{I}_{0}}$
$\displaystyle \therefore I=2.460\times \left( {{{{{I}_{u}}}}/{2}\;} \right)=1.20{{I}_{u}}\leftarrow$
• HAIJA, A.I., NUMAN, M.Z. and FREEMAN, W.L. (2018). Concise Optics: Concepts, Examples, and Problems. Boca Raton: CRC Press.
• HECHT, E. (2017). Optics. 5th edition. Upper Saddle River: Pearson.
• PEDROTTI, F.L., PEDROTTI, L.M. and PEDROTTI, L.S. (2006). Introduction to Optics. 3rd edition. Boston: Addison-Wesley. | {"url":"https://montoguequiz.com/electrical/fresnel-diffraction/","timestamp":"2024-11-07T09:53:13Z","content_type":"text/html","content_length":"216488","record_id":"<urn:uuid:90ac52c1-e514-4de4-a086-f80eefd9d7df>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00086.warc.gz"} |
[Answers] 128 Questions of Discrete Mathematics
Two graphs that are the same are said to be _______________
• isomorphic
• isometric
• isochoric
All graphs have Euler's Path
A graph is complete if there is a path from any vertex to any other vertex.
A function which renames the vertices.
• non-isomorphism
• isomorphism
What is the element n in the domain such as fNo = 1
Find |A ∩ B| when A = {1, 3, 5, 7, 9} and B {2, 4, 6, 8, 10}
Identify the propositional logic of the truth table given
• disjunction
• negation
• conjunction
• implication
What is the sum from 1st to 5th element?
How many spanning trees are possible in the given figure?
Tracing all edges on a figure without picking up your pencil or repeating and starting and stopping at different spots
These are lines or curves that connect vertices.
The geometric sequences uses common _____ in finding the succeeding terms.
A simple graph has no loops nor multiple edges.
Solve for the value of n in :
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
A graph is an ordered pair G (V, E) consisting of a nonempty set V (called the vertices) and a set E (called the edges) of two-element subsets of V.
The tree elements are called _____
The _____ is a subset of the codomain. It is the set of all elements which are assigned to at least one element of the domain by the function. That is, the range is the set of all outputs.
¬P ∨ Q is equivalent to :
Which of the following is false?
• A graph with one odd vertex will have an Euler Path but not an Euler Circuit.
• Euler Paths exist when there are exactly two vertices of odd degree.
• A graph with more than two odd vertices will never have an Euler Path or Circuit.
• Euler circuits exist when the degree of all vertices are even
Find the cardinality of R = {20,21,...,39, 40}
How many edges would a complete graph have if it had 6 vertices?
A graph for which it is possible to divide the vertices into two disjoint sets such that there are no edges between any two vertices in the same set.
A _____ is a function which is both an injection and surjection. In other words, if every element of the codomain is the image of exactly one element from the domain
In my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell
you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement? The square is not blue or the triangle is green.
Find | R | when R = {2, 4, 6,..., 180}
The sum of the geometric progression is called geometric series
When a connected graph can be drawn without any edges crossing, it is called ________________ .
• Edged graph
• Planar graph
• Spanning graph
A statement which is true on the basis of its logical form alone.
• Tautology
• Double Negation
• De Morgan's Law
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither. If you will not give me a
cow, then I will not give you magic beans.
If two vertices are adjacent, then we say one of them is the parent of the other, which is called the _____ of the parent.
If the right angled triangle t, with sides of length a and b and hypotenuse of length c, has area equal to c2/4, what kind of triangle is this?
• obtuse triangle
• isosceles triangle
• scalene triangle
A _____ graph has no isolated vertices.
The cardinality of {3, 5, 7, 9, 5} is 5.
What is the missing term? 3,9,__,81....
How many people takes tea and wine?
What is the difference of persons who take wine and coffee to the persons who the persons who takes tea only?
How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
De Morgan's law is used in finding the equivalence of a logic expression using other logical functions.
A sequence of vertices such that every vertex in the sequence is adjacent to the vertices before and after it in the sequence
It is a connected graph containing no cycles.
A sequence of vertices such that consecutive vertices (in the sequence) are adjacent (in the graph). A walk in which no edge is repeated is called a trail, and a trail in which no vertex is repeated
(except possibly the first and last) is called a path
• Subgraph
• Walk
• Vertex coloring
Arithmetic progression is the sum of the terms of the arithmetic series.
A _____ connected graph with no cycles. (If we remove the requirement that the graph is connected, the graph is called a forest.) The vertices in a tree with degree 1 are called _____
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither. If you will give me a
cow, then I will not give you magic beans.
Euler paths must touch all edges.
For all n in rational, 1/n ≠ n - 1
How many simple non-isomorphic graphs are possible with 3 vertices?
What is the line covering number of for the following graph?
How many possible output will be produced in a proposition of three statements?
Match the truth tables to its corresponding propositional logic
• Implication, Disjunction, Conjunction
_____ is the same truth value under any assignment of truth values to their atomic parts.
Circuits start and stop at _______________
• different vertices
• same vertex
An argument form which is always valid.
_____ is the simplest style of proof.
A sequence of vertices such that consecutive vertices (in the sequence) are adjacent (in the graph). A walk in which no edge is repeated is called a trail, and a trail in which no vertex is repeated
(except possibly the first and last) is called a path.
How many people takes coffee but not tea and wine?
An undirected graph G which is connected and acyclic is called ____________.
• forest
• cyclic graph
• tree
• bipartite graph
What is the minimum height height of a full binary tree?
The number of simple digraphs with |V | = 3 is
Fill in the blanks. A graph F is a _____ if and only if between any pair of vertices in F there is at most _____
A graph T is a tree if and only if between every pair of distinct vertices of T there is a unique path.
Find an element n of the domain such that f No = n.
Let A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7}
• {1, 2, 3, 5, 6, 7}
• {1, 2, 6, 7}
• {1, 2, 3, 4, 5, 6, 7}
• {3, 4, 5}
The _____ of a a subset B of the codomain is the set f −1 (B) {x ∈ X : f (x) ∈ B}.
Consider the function f : N → N given by f (0) 0 and f (n + 1) f No + 2n + 1. Find f (6).
A path which visits every vertex exactly once
Let ‘G’ be a connected planar graph with 20 vertices and the degree of each vertex is 3. Find the number of regions in the graph.
Every connected graph has a spanning tree.
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither. You will give me a cow
and I will not give you magic beans.
Find the cardinality of S = {1, {2,3,4},0} | S | = _____
If you travel to London by train, then the journey takes at least two hours.
• If your journey by train takes more than two hours, then you don't travel to London.
• If your journey by train takes less than two hours, then you don’t travel to London.
The child of a child of a vertex is called
As soon as one vertex of a tree is designated as the _____, then every other vertex on the tree can be characterized by its position relative to the root.
A Bipartite graph is a graph for which it is possible to divide the vertices into two disjoint sets such that there are no edges between any two vertices in the same set.
What is the matching number for the following graph?
A _____ is a _____ which starts and stops at the same vertex.
• Euler circuit, Euler path
The ________________________ states that if event A can occur in m ways, and event B can occur in n disjoint ways, then the event “A or B” can occur in m + n ways.
• Additive principle
• Commutative principle
• Distributive principle
Determine the number of elements in A U B.
¬(P ∨ Q) is logically equal to which of the following expressions?
Indicate which, if any, of the following graphs G = (V, E, φ), |V | = 5, is not connected.
• φ = ( a {1,2} b {2,3} c {1,2} d {1,3} e {2,3} f {4,5} )
• φ = ( a {1,2} b {2,3} c {1,2} d {2,3} e {3,4} f {1,5} )
• φ = ( 1 {1,2} 2 {1,2} 3 {2,3} 4 {3,4} 5 {1,5} 6 {1,5} )
Two edges are adjacent if they share a vertex.
match the following formulas to its corresponding sequence
• Geometric Series, Double Summation
Proofs that is used when statements cannot be rephrased as implications.
How many people like only one of the three?
An argument is said to be valid if the conclusion must be true whenever the premises are all true.
The given graph is planar.
The number of edges incident to a vertex.
surjective and injecive are opposites of each other.
Deduction rule is an argument that is not always right.
A graph in which every pair of vertices is adjacent.
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither. If I will not give you
magic beans, then you will not give me a cow.
A spanning tree that has the smallest possible combined weight.
Match the following properties of trees to its definition.
• Proposition 4.2.1 → A graph T is a tree if and only if between every pair of distinct vertices of T there is a unique path., Proposition 4.2.4 → 4 Let T be a tree with v vertices and e edges.
Then e v − 1., Corollary 4.2.2 → A graph F is a forest if and only if between any pair of vertices in F there is at most one path, Proposition 4.2.3 → Any tree with at least two vertices has at
least two vertices of degree one.
A sequence that involves a common difference in identifying the succeeding terms.
• Geometric Progression
• Arithmetic Progression
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A connected graph with no cycles.
Does a rational r value for r2 =6 exist?
• No, a rational r does not exist.
• Yes, a rational r exist.
IN combinations, the arrangement of the elements is in a specific order.
Which of the following is a possible range of the function?
• All numbers except 3
• 1,2,3
• 3,6,9,12 only
• all multiples of three
• 3,4,5,6,7,8,9,10
Rule that states that every function can be described in four ways: algebraically (a formula), numerically (a table), graphically, or in words.
• Rule of four
• Rule of thumb
• Rule of function
Additive principle states that if given two sets A and B, we have |A × B| |A| · |B|.
A _____ graph has two distinct groups where no vertices in either group connecting to members of their own group
The study of what makes an argument good or bad.
It is an algorithm for traversing or searching tree or graph data structures.
• depth first search.
• breadth first search
• spanning tree
If n is a rational number, 1/n does not equal n-1.
What type of progression this suggest?
How many people like apples only?
Defined as the product of all the whole numbers from 1 to n.
Let A = {3, 4, 5}. Find the cardinality of P(A).
Which of the following statements is NOT TRUE?
• A graph F is a forest if and only if between any pair of vertices in F there is at most one path.
• Any tree with at least two vertices has at least two vertices of degree two.
• Let T be a tree with v vertices and e edges. Then e v − 1.
A tree is the same as a forest.
Consider the statement, “If you will give me a cow, then I will give you magic beans.” Determine whether the statement below is the converse, the contrapositive, or neither. If I will give you magic
beans, then you will give me a cow.
What is the type of progression?
Suppose P and Q are the statements: P: Jack passed math. Q: Jill passed math. Translate "¬(P ν Q) → Q" into English.
• Neither Jack or Jill passed math.
• Jill passed math if and only if Jack did not pass math.
• If Jack did not pass math and Jill did not pass math, then Jill did not pass math.
• If Jack or Jill did not pass math, then Jill passed math.
A set of statements, one of which is called the conclusion and the rest of which are called premises.
Which of the following the logic representation of proof by contrapositive?
• P → Q = ¬Q → P
• P → Q = Q → ¬P
• P → Q = ¬Q → ¬P
• P → Q = ¬(Q → P)
_____ is a function from a subset of the set of integers.
It is a rule that assigns each input exactly one output
Does this graph have an Euler Path, Euler Circuit, both, or neither?
• Euler Circuit
• None
• Euler Path
• Both
In my safe is a sheet of paper with two shapes drawn on it in colored crayon. One is a square, and the other is a triangle. Each shape is drawn in a single color. Suppose you believe me when I tell
you that if the square is blue, then the triangle is green. What do you therefore know about the truth value of the following statement? If the triangle is green, then the square is blue.
What is the 4th and 8th element of aNo= n^(2) ?
In a simple graph, the number of edges is equal to twice the sum of the degrees of the vertices.
Paths start and stop at the same vertex.
The minimum number of colors required in a proper vertex coloring of the graph.
Indicate which, if any, of the following three graphs G = (V, E, φ), |V | = 5, is not isomorphic to any of the other two.
• φ = ( b {4,5} f {1,3} e {1,3} d {2,3} c {2,4} a {4,5} )
• φ = ( f {1,2} b {1,2} c {2,3} d {3,4} e {3,4} a {4,5} )
• φ = (A {1,3} B {2,4} C {1,2} D {2,3} E {3,5} F {4,5} ) | {"url":"https://www.answerscrib.com/subject/discrete-mathematics","timestamp":"2024-11-02T05:45:52Z","content_type":"text/html","content_length":"107293","record_id":"<urn:uuid:16a0bfc7-cc3e-46f8-8177-7e66cd8a05d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00495.warc.gz"} |
Price per Use Calculator
The Price per Use Calculator can calculate the price for each use based on the total price of all the uses.
To calculate the price per use, we divide the total price of all the uses by the number of uses.
Please enter the total price of all the uses and the number of uses so we can calculate the price per use:
Price per View Calculator
Here is a similar calculator you may find interesting. | {"url":"https://pricecalculator.org/per/price-per-use-calculator.html","timestamp":"2024-11-12T06:09:14Z","content_type":"text/html","content_length":"6433","record_id":"<urn:uuid:4e9d44b2-4444-45fb-83bd-27c6ab26b3cf>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00365.warc.gz"} |
PipeOpTaskSurvClassifIPCW — mlr_pipeops_trafotask_survclassif_IPCW
Transform TaskSurv to TaskClassif using the Inverse Probability of Censoring Weights (IPCW) method by Vock et al. (2016).
Let \(T_i\) be the observed times (event or censoring) and \(\delta_i\) the censoring indicators for each observation \(i\) in the training set. The IPCW technique consists of two steps: first we
estimate the censoring distribution \(\hat{G}(t)\) using the Kaplan-Meier estimator from the training data. Then we calculate the observation weights given a cutoff time \(\tau\) as:
$$\omega_i = 1/\hat{G}{(min(T_i,\tau))}$$
Observations that are censored prior to \(\tau\) are assigned zero weights, i.e. \(\omega_i = 0\).
Input and Output Channels
PipeOpTaskSurvClassifIPCW has one input channel named "input", and two output channels, one named "output" and the other "data".
Training transforms the "input" TaskSurv to a TaskClassif, which is the "output". The target column is named "status" and indicates whether an event occurred before the cutoff time \(\tau\) (1 = yes,
0 = no). The observed times column is removed from the "output" task. The transformed task has the property "weights" (the \(\omega_i\)). The "data" is NULL.
During prediction, the "input" TaskSurv is transformed to the "output" TaskClassif with "status" as target (again indicating if the event occurred before the cutoff time). The "data" is a data.table
containing the observed times \(T_i\) and censoring indicators/status \(\delta_i\) of each subject as well as the corresponding row_ids. This "data" is only meant to be used with the
The parameters are
• tau :: numeric()
Predefined time point for IPCW. Observations with time larger than \(\tau\) are censored. Must be less or equal to the maximum event time.
• eps :: numeric()
Small value to replace \(G(t) = 0\) censoring probabilities to prevent infinite weights (a warning is triggered if this happens).
Vock, M D, Wolfson, Julian, Bandyopadhyay, Sunayan, Adomavicius, Gediminas, Johnson, E P, Vazquez-Benitez, Gabriela, O'Connor, J P (2016). “Adapting machine learning techniques to censored
time-to-event health record data: A general-purpose approach using inverse probability of censoring weighting.” Journal of Biomedical Informatics, 61, 119–131. doi:10.1016/j.jbi.2016.03.009 , https:/
Inherited methods
Method new()
Creates a new instance of this R6 class.
if (FALSE) { # \dontrun{
task = tsk("lung")
# split task to train and test subtasks
part = partition(task)
task_train = task$clone()$filter(part$train)
task_test = task$clone()$filter(part$test)
# define IPCW pipeop
po_ipcw = po("trafotask_survclassif_IPCW", tau = 365)
# during training, output is a classification task with weights
task_classif_train = po_ipcw$train(list(task_train))[[1]]
# during prediction, output is a classification task (no weights)
task_classif_test = po_ipcw$predict(list(task_test))[[1]]
# train classif learner on the train task with weights
learner = lrn("classif.rpart", predict_type = "prob")
# predict using the test output task
p = learner$predict(task_classif_test)
# use classif measures for evaluation
} # } | {"url":"https://mlr3proba.mlr-org.com/reference/mlr_pipeops_trafotask_survclassif_IPCW.html","timestamp":"2024-11-06T14:31:04Z","content_type":"text/html","content_length":"26856","record_id":"<urn:uuid:5330ff5f-0133-4d64-bf83-ca70dfb78c92>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00581.warc.gz"} |
Switching Dynamics of Double Barrier Josephson Junction Based Qubit Gate
Shafraniuk S.E., Nevirkovets I.P., Ketterson J.
Northwestern University, US
Keywords: SINIS qubits dynamics
The double barrier SINIS junctions (here S, I, and N denote a superconductor, an insulator, and a normal metal, respectively) with a nanoscopic N spacer are potentially capable of performing quantum
logic operations (so-called qubits) involving the superposition of two (macroscopic) quantum states. In our report we analyse the switching dynamics of three-terminal double barrier SINIS junction
working as a qubit gate based on two quantum states. The quantum states are associated with conventional and unconventional Josephson current components observed in the SINIS junctions. In this work
we study the switching time and decoherence (dephasing) time time of the mentioned device. Such characteristics are closely related to the longitudinal and transverse dynamics of the superconducting
order parameter. Such a dynamics in particular is determined by the electron recombination time. The mentioned parameter strongly depends on the electron excitation spectrum inside N, which in turn
is very sensitive to the presence of nonmagnetic impurities (i.e., to the magnitude of electron impurity scattering time ). In this work we computed the local electron density of states in the SINIS
junction using the quasiclassical Eilenberger equation approach. We find that in a clean limit, the electron excitation spectrum inside N consists of quantized levels, while in the opposite dirty
limit the spectrum of N is rather smooth versus the energy variable E. Such a difference affects drastically. In Fig. 2 we plot the energy dependence of inside the middle N spacer of SINIS gate for
two different cases (curve A), and (curve B). One can see pronounced peaks in at (curve B for the clean case) which are absent for a dirty junction (curve A). For such reasons, the dynamics of SINIS
qubit gates is quite distinct in the two mentioned limits.
Journal: TechConnect Briefs
Volume: 2, Technical Proceedings of the 2003 Nanotechnology Conference and Trade Show, Volume 2
Published: February 23, 2003
Pages: 168 - 171
Industry sectors: Advanced Materials & Manufacturing | Sensors, MEMS, Electronics
Topic: Modeling & Simulation of Microsystems
ISBN: 0-9728422-1-7 | {"url":"https://briefs.techconnect.org/papers/switching-dynamics-of-double-barrier-josephson-junction-based-qubit-gate/","timestamp":"2024-11-11T01:28:58Z","content_type":"text/html","content_length":"43296","record_id":"<urn:uuid:3a819917-560d-41d6-b1ac-6480a50ba745>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00419.warc.gz"} |
How to Spot Statistical Variability in a Histogram - dummies
You can get a sense of variability in a statistical data set by looking at its histogram. For example, if the data are all the same, they are all placed into a single bar, and there is no
variability. If an equal amount of data is in each of several groups, the histogram looks flat with the bars close to the same height; this signals a fair amount of variability.
The idea of a flat histogram indicating some variability may go against your intuition, and if it does you're not alone. If you're thinking a flat histogram means no variability, you're probably
thinking about a time chart, where single numbers are plotted over time. Remember, though, that a histogram doesn't show data over time — it shows all the data at one point in time. Since the
histogram is flat, that means that the data are spread out across the spectrum, hence a high variability.
Equally interesting is the idea that a histogram with a big lump in the middle and tails sloping sharply down on each side actually has less variability than a histogram that's straight across. The
curves looking like hills in a histogram represent clumps of data that are close together, hence a low variability.
Variability in a histogram is higher when the taller bars are more spread out away from the mean and lower when the taller bars are close to the mean.
For the Best Actress Academy Award winners' ages shown in the above figure, you see many actresses are in the age range from 30–35, and most of the actresses are between 20–50 years in age, which is
quite diverse; then you have those outliers, those few older actresses (7 of them) that spread the data out farther, increasing the data's overall variability.
The most common statistic used to measure variability in a data set is the standard deviation, which in a rough sense measures the "average" or "typical" distance that the data lie from the mean. The
standard deviation for the Best Actress age data is 11.35 years. A standard deviation of 11.35 years is fairly large in the context of this problem, but the standard deviation is based on average
distance from the mean, and the mean is influenced by outliers, so the standard deviation will be influenced as well.
About This Article
This article is from the book:
This article can be found in the category: | {"url":"https://www.dummies.com/article/academics-the-arts/math/statistics/how-to-spot-statistical-variability-in-a-histogram-169305/","timestamp":"2024-11-11T04:08:56Z","content_type":"text/html","content_length":"73992","record_id":"<urn:uuid:dbb08297-0614-484e-bad5-83d479026473>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00712.warc.gz"} |
Front-end advanced algorithm 4: Linked lists are so simple - Moment For Technology
The introduction
Linked lists are much more complex than arrays. First of all, linked lists do not require contiguous memory space. They are made up of a group of discrete memory blocks connected by Pointers. The
most common types of linked lists are single, double, and circular.
The most important thing to learn a linked list is to draw lots of pictures and practice. There is no shortcut to follow. When faced with a linked list problem, Aquarius summarized the following five
• Determine the data structure to solve the problem: single, double, or circular lists, etc
• Decide on the solution: How to solve the problem
• Drawing realization: Drawing can help us find the loophole in our thinking (some thinking is not perfect)
• Determine boundary conditions: Consider whether there are boundary problems in the solution and how to solve them
• Code implementation: solve the problem to complete ✅
This article will give the common linked list (single linked list, double linked list and circular linked list) of the basic operation has been code implementation, and give the implementation ideas,
these are linked list solution cornerstone, please be sure to master! ⛽ ️ ⛽ ️ ⛽ ️
A bonus leetcode question at the end!
Let’s start this section!! 👇 👇 👇
A single linked list
Single linked list structure:
function List () {
/ / the node
let Node = function (element) {
this.element = element
this.next = null
// The initial head node is null
let head = null
// List length
let length = 0
/ / operation
this.getList = function() {return head}
this.search = function(list, element) {}
this.append = function(element) {}
this.insert = function(position, element) {}
this.remove = function(element){}
this.isEmpty = function(){}
this.size = function(){}}Copy the code
1. Add nodes:
** Determine the data structure to solve the problem: ** single linked list
Initialize a node (to be appended), traverse to the end of the chain, and insert the node after the end node
Drawing implementation:
Determine boundary conditions: When the linked list is null, head points directly to the node to be inserted without traversal
Code implementation:
function append (element) {
let node = new Node(element),
p = head
if(! head){ head = node }else {
while (p.next) {
p = p.next
p.next = node
length += 1
/ / test
let list = new List()
for(let i = 0; i < 5; i+=1) {
Copy the code
Solve the problem ✅
2. Look for:
Determine the data structure to solve the problem: single linked list
If the value of the node is equal to the value to be searched, return true; otherwise, continue to traverse the next node, until the whole list is not found, return false
Drawing realization: very simple, the reader can try to draw
Determine boundary conditions: When the linked list is NULL, false can be returned
Code implementation:
// Check whether a node exists in the list
function search(element) {
let p = head
if(! p)return false
while(p) {
if (p.element === element) return true
p = p.next
return false
/ / test
list.search(4) // true
list.search(11) // false
Copy the code
Solve the problem ✅
3. Insert:
Determine the data structure to solve the problem: single linked list
Initialize a node (node to be inserted), traverse to the node before position, and insert the node after the node
Drawing implementation:
Determine boundary conditions:
• whenposition 为 0, the node will be inserted directlynode.nextPoint to thehead ， headPoint to thenodeOk, no need to traverse
• When to insert positionposition < 0Or exceed the list lengthposition > length, are problematic, cannot be inserted, at this point directly returnnull, failed to insert
Code implementation:
// Inserts the successor node of position
function insert (position, element) {
// Create an insert node
let node = new createNode(element)
if (position >= 0 && position <= length) {
let prev = head,
curr = head,
index = 0
if(position === 0) {
node.next = head
head = node
} else {
while(index < position) {
prev = curr
curr = curr.next
index ++
prev.next = node
node.next = curr
length += 1
} else {
return null}}/ / test
Copy the code
Solve the problem ✅
4. Delete:
Determine the data structure to solve the problem: single linked list
Determine the solution idea: traverse the single linked list, find the node to be deleted, delete
Drawing implementation:
Determine boundary conditions: When the list is null, return
Code implementation:
// Delete the element node
function remove (element) {
let p = head, prev = head
if(! head)return
while(p) {
if(p.element === element) {
p = p.next
prev.next = p
} else {
prev = p
p = p.next
Copy the code
Solve the problem ✅
5. Complexity Analysis
Search: Search from the beginning node, time complexity O(n)
Insert or delete: The time complexity of inserting or deleting a node (successor node) after a node is O(1)
Linked list five steps is not very easy to use 😊, the following look at the double linked list 👇
Double linked lists
As the name implies, a singly linked list has only one direction, from the beginning node to the end node. Then a doubly linked list has two directions, from the end node to the end node:
function DoublyLinkedList() {
let Node = function(element) {
this.element = element
// Precursors
this.prev = null
// Next pointer
this.next = null
// The initial head node is null
let head = null
// Add a tail node
let tail = null
// List length
let length = 0
/ / operation
this.search = function(element) {}
this.insert = function(position, element) {}
this.removeAt = function(position){}
this.isEmpty = function(){ return length === 0 }
this.size = function(){ return length }
Copy the code
1. Insert node in position:
Determine the data structure to solve the problem: double linked lists
Initialize a node (node to be inserted), traverse the linked list to the node before position, and insert the node after the node position
Drawing implementation:
Determine boundary conditions:
If the position to be inserted is position < 0 or exceeds the length of the linked list position > length, the insert cannot be inserted. In this case, null is returned directly, and the insert fails
Code implementation:
// Inserts the successor node of position
function insert (position, element) {
// Create an insert node
let node = new Node(element)
if (position >= 0 && position < length) {
let prev = head,
curr = head,
index = 0
if(position === 0) {
// Add in the first position
if(! head) {// Note that this is different from singly linked lists
head = node
tail = node
} else {
/ / the bidirectional
node.next = head
head.prev = node
// head points to the new head node
head = node
} else if(position === length) {
// Insert to the tail node
curr = tial
curr.next = node
node.prev = curr
// tail points to the new tail node
tail = node
} else {
while(index < position) {
prev = curr
curr = curr.next
index ++
// Insert after prev and before curr
prev.next = node
node.next = curr
curr.prev = node
node.prev = prev
length += 1
return true
} else {
return false}}/ / test
Copy the code
Solve the problem ✅
2. Delete the:
Determine the data structure to solve the problem: double linked lists
Determine the solution idea: traverse the double-linked list, find the node to be deleted, delete
Drawing implementation:
Determine boundary conditions: When the list is null, return
Code implementation:
// Delete the node in position
function removeAt (position) {
if (position >= 0 && position < length && length > 0) {
let prev = head,
curr = head,
index = 0
if(position === 0) {
// Remove the head node
if(length === 1) { // There is only one node
head = null
tail = null
} else {
head = head.next
head.prev = null}}else if(position === length- 1) {
// Remove the tail node
curr = tial
tail = curr.prev
tail.next = null
} else {
while(index < position) {
prev = curr
curr = curr.next
index ++
/ / remove the curr
prev.next = curr.next
curr.next.prev = prev
length -= 1
return curr.element
} else {
return null}}Copy the code
Solve the problem ✅
3. Look for:
A double-linked list lookup is similar to a single-linked list in that it iterates through the list and returns true if found, false if not found.
4. Complexity Analysis
Search: Search forerunner node or successor node time complexity is O(1), other nodes are still O(n)
Insert or delete: Insert or delete the time complexity of precursor node or successor node is O(1).
Circular singly linked lists
A circular singly linked list is a special singly linked list. The only difference between a singly linked list and a singly linked list is that the tail of the singly linked list points to NULL,
while the tail of a circular singly linked list points to the head, which forms a end to end loop:
Since there are cyclic singly linked lists, there are also cyclic doubly linked lists, and the difference between a cyclic doubly linked list and a doubly linked list is:
• Double linked listtail.next(tailIs the successor pointer tonullCircular double linked listtail.next 为 head
• Double linked listhead.prev(headIs the precursor pointer tonullCircular double linked listhead.prev 为 tail
Take a circular single column table as an example
function CircularLinkedList() {
let Node = function(element) {
this.element = element
// Next pointer
this.next = null
// The initial head node is null
let head = null
// List length
let length = 0
/ / operation
this.search = function(element) {}
this.insert = function(positon, element) {}
this.removeAt = function(position){}
this.isEmpty = function(){ return length === 0 }
this.size = function(){ return length }
Copy the code
1. Insert after positon:
Determine the data structure for the solution: circular singly linked lists
Initialize a node (node to be inserted), traverse to the node before position, and insert the node after the node
Drawing implementation:
Determine boundary conditions:
• whenposition 为 0, you need to traverse to the tail node, and then insert the node after the tail node, and change theheadPoint to the
• When to insert positionposition < 0Or exceed the list lengthposition > length, are problematic, cannot be inserted, at this point directly returnnull, failed to insert
Code implementation:
// Inserts the successor node of position
function insert (position, element) {
// Create an insert node
let node = new createNode(element)
if (position >= 0 && position <= length) {
let prev = head,
curr = head,
index = 0
if(position === 0) {
// Different from single linked list inserts
while(index < length) {
prev = curr
curr = curr.next
index ++
prev.next = node
node.next = curr
head = node
} else {
while(index < position) {
prev = curr
curr = curr.next
index ++
prev.next = node
node.next = curr
length += 1
} else {
return null}}/ / test
Copy the code
Solve the problem ✅
2. Look for:
This is similar to single-linked lists except that index++ < length is the end of the loop condition
// Check whether a node exists in the list
function search(element) {
if(! head)return false
let p = head, index = 0
// The difference between a single list and a single list
while(index++ < length) {
if (p.element= = =element) return true
p = p.next
return false} / / testlist.search(4) / /true
list.search(11) / /false
Copy the code
Solve the problem ✅
3. Delete:
This is similar to single-linked lists except that index++ < length is the end of the loop condition
// Delete the element node
function remove (element) {
let p = head, prev = head, index = 0
/ / an empty list
if(! head || )return
// There is only one node and element is the same
if(length === 1 && head.element === element){
head = null
while(index++ < length) {
if(p.element= = =element) {
p = p.next
prev.next = p
length --
} else {
prev = p
p = p.next}}}Copy the code
Solve the problem ✅
4. Complexity analysis
Search: The circular list starts from any node to find the target node in O(n) time.
Insert or delete: It is the same as a single linked list, and the time complexity of insert and delete of subsequent nodes is O(1).
Leetcode21: merge two ordered lists
Merges two ascending lists into a new ascending list and returns. A new list is formed by concatenating all the nodes of a given two lists.
Copy the code
Please submit your answers to github.com/sisterAn/Ja… “, so that more people can see, Aquarius will post his own solution tomorrow.
Five, past series
Front-end advanced algorithm 3: Learning the LRU algorithm from the browser cache elimination strategy and Vue’s keep-alive
Bottle jun front-end advanced algorithm camp first week summary
Front-end advanced algorithm 2: JavaScript array from Chrome V8 source code
Front-end advanced algorithm 1: How to analyze and count the execution efficiency and resource consumption of algorithms?
Six, know more front-end road friends, together with advanced front-end development
The first phase of front-end algorithm training camp is free 🎉🎉🎉, free yo!
Here, you can advance the front-end algorithm with like-minded friends (200+), from 0 to 1 to build a complete data structure and algorithm system.
Here, I not only introduce algorithms, but also combine algorithms with various front-end areas, including browsers, HTTP, V8, React, Vue source code, and so on.
Here, you can learn a big factory algorithm (Ali, Tencent, Baidu, byte, etc.) or leetcode every day, Aquarius will solve it the next day!
More benefits waiting for you to unlock 🔓🔓🔓!
Scan code to join [front-end algorithm exchange group exchange group], if the number of TWO-DIMENSIONAL code has reached the upper limit, you can scan the bottom two-dimensional code, in the public
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WB JEE 2008 | Current Electricity Question 3 | Physics | WB JEE - ExamSIDE.com
Two wires each of same material and same length have 2 L 1 ratio of their radii. If the first wire has a resistance of 2 $$\Omega$$ then what will be the equivalent resistance of the two wires when
they are connected in series?
A 60 W - 220 V bulb is connected in series with another bulb 40 W - 220 V. Which bulb will give more illuminance ?
Questions Asked from Current Electricity (Subjective)
Number in Brackets after Paper Indicates No. of Questions | {"url":"https://questions.examside.com/past-years/jee/question/ptwo-wires-each-of-same-material-and-same-length-have-2-l-wb-jee-physics-current-electricity-vo70w3bf3cq488oy","timestamp":"2024-11-03T04:19:39Z","content_type":"text/html","content_length":"196314","record_id":"<urn:uuid:bc603429-9611-44c0-86c8-33e526a28671>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00595.warc.gz"} |
Why reordering uniforms affects arithmetic cycles?
Answers 1 answer
I've recently started using Mali Offline Compiler to get insight into our shaders and I get confusing results from it which I can't really explain.
So I have one quite big shader.
It has block of uniforms, quite large one cause it's uber shader.
I noticed that if I reorder uniforms in a different way - I get different results from Mali compiler.
#if HLSLCC_ENABLE_UNIFORM_BUFFERS
UNITY_BINDING(0) uniform UnityPerMaterial {
UNITY_UNIFORM vec4 _MainTex_ST;
UNITY_UNIFORM float _MainTexUVSet2;
UNITY_UNIFORM vec4 _SecondaryTex_ST;
UNITY_UNIFORM mediump vec4 _SecondaryColor;
UNITY_UNIFORM float _SecondaryTexUVSet2;
UNITY_UNIFORM vec4 _MaskTex_ST;
UNITY_UNIFORM float _MaskTexUVSet2;
UNITY_UNIFORM vec4 _DissolveTex_ST;
UNITY_UNIFORM float _DissolveTexUVSet2;
UNITY_UNIFORM mediump vec3 _MainColorBright;
UNITY_UNIFORM mediump vec3 _MainColorMid;
UNITY_UNIFORM mediump vec3 _MainColorDark;
UNITY_UNIFORM mediump vec4 _MainColor;
UNITY_UNIFORM vec2 _MainTexScrollSpeed;
UNITY_UNIFORM vec2 _SecondaryTexScrollSpeed;
UNITY_UNIFORM vec2 _DissolveTexScrollSpeed;
UNITY_UNIFORM mediump float _Intensity;
UNITY_UNIFORM mediump float _PSDriven;
UNITY_UNIFORM mediump float _DissolveAmount;
UNITY_UNIFORM mediump float _DissolveSoftness;
UNITY_UNIFORM int _ScrollMainTex;
UNITY_UNIFORM int _ScrollSecondaryTex;
UNITY_UNIFORM int _ScrollDissolveTex;
UNITY_UNIFORM int _MultiplyWithVertexColor;
UNITY_UNIFORM int _MultiplyWithVertexAlpha;
UNITY_UNIFORM int _UseGradientMap;
UNITY_UNIFORM int _UseStepMasking;
UNITY_UNIFORM float _Curvature;
UNITY_UNIFORM mediump float _StepBorder;
UNITY_UNIFORM mediump float _UseRForSecondaryTex;
UNITY_UNIFORM mediump float _UseRForMask;
UNITY_UNIFORM mediump float _MaskSecondTexWithFirst;
UNITY_UNIFORM mediump float _UseRAsAlpha;
#if HLSLCC_ENABLE_UNIFORM_BUFFERS
So if I take let say _Curvature uniform and reorder it so it's before any other half/int variable
Here are results from fragment shader:
Mali Offline Compiler v7.4.0 (Build 330167)
Copyright 2007-2021 Arm Limited, all rights reserved
Hardware: Mali-T720 r1p1
Architecture: Midgard
Driver: r23p0-00rel0
Shader type: OpenGL ES Fragment
Main shader
Work registers: 4
Uniform registers: 0
Stack spilling: false
A LS T Bound
Total instruction cycles: 16.00 9.00 4.00 A
Shortest path cycles: 10.00 9.00 3.00 A
Longest path cycles: 10.25 9.00 3.00 A
A = Arithmetic, LS = Load/Store, T = Texture
And then they become
Mali Offline Compiler v7.4.0 (Build 330167)
Copyright 2007-2021 Arm Limited, all rights reserved
Hardware: Mali-T720 r1p1
Architecture: Midgard
Driver: r23p0-00rel0
Shader type: OpenGL ES Fragment
Main shader
Work registers: 4
Uniform registers: 0
Stack spilling: false
A LS T Bound
Total instruction cycles: 16.00 9.00 4.00 A
Shortest path cycles: 9.50 9.00 3.00 A
Longest path cycles: 9.75 9.00 3.00 A
A = Arithmetic, LS = Load/Store, T = Texture
This uniform is only used in vertex shader but somehow it also affects fragment shader results.
Why do arithmetic cycles are now different?
Right now I have no idea what affects it and how to optimize this in the best possible way and if I should even bother.
But when shader executes in let say 10 cycles and reordering fields can make it execute in 9 or even 8 cycles - this is 10-20% of performance to be gained so I would like to understand what's going
on underhood.
Is there a way to get disassembly from mali compiler?
Right now it is a black box to me.
I am attaching both shaders and output from mali compiler in case someone will take a look.
Midgard is a vector architecture with 128-bit vector registers and SIMD instructions, not more modern scalar operations. The ability of the compiler to auto-vectorize is sensitive to the ordering of
values in registers - if variables don't "align" in the same SIMD lanes then the compiler either has to run operations multiple times or swizzle registers at runtime, which isn't always free.
Later Midgard GPUs don't have this problem as the uniform loads are converted into uniform register access, which can hide alignment issues and repack vectors, so you hit the same performance for
both shaders. For Mali-T720 you will have to deal with how things map into vectors - sorry.
Thank you very much for quick answer.
Can you have any recommendations how to understand this better? i.e. how do I write code in a better way to help compiler to vectorize stuff?
Do you maybe have some link to a guide?
Right now I am thinking to not bother about it especially after you said it's not a problem on later midgard GPUs. What's your recommendation?
And a little bit unrelated question.
So we're doing mobile game and we want to have good performance on the widest scope of devices as possible. It's both android and ios and not just Mali devices but other devices too.
Mali has the best developer tools so thanks for that :) and that's why I am mostly using Streamline and Mali offline compiler now to optimize stuff.
My current strategy is to optimize shaders for the oldest GPU supported by mali offline compiler which is T720 and then I just hope that all other devices will be better than this one.
And also let say devices from other manufacturers which have similar vector architecture will probably benefit from exactly same optimizations.
is it valid strategy?
My fear is that I overoptimize for one device and it won't really help with others, so I kind of waste my time.
So far results are good i.e. mali offline compiler really helped me a lot to increase performance of our game.
I'm glad you're finding the tools useful =)
For shader optimization, if you want to target entry-level lowest-common denominator I think there are really three major classes of interesting device in terms of giving different results:
• Mali-T720 (SIMD, but without the uniform constant register optimization later GPUs have).
• Mali-T820 (SIMD, but with the uniform constant register optimization)
• Mali-G52 (Scalar instruction set).
There were a lot of Mali-T720-based devices sold, but it's an old product now (first released 9 years ago) so I'd agree with your position that it's not worth worrying too much about.
Mali-T820 is Midgard (SIMD) which is a few years newer than Mali-T720, but still relatively old (first released 7 years ago). There are still a lot of Midgard devices kicking around, so it's probably
still worth checking but I wouldn't totally rewrite your shaders for it, especially if those changes are detrimental to Mali-G52.
All (?) modern GPUs use scalar warp instruction sets (including both Mali and GPUs from other vendors) so the Mali-G52 results should more indicative of what you will see on any hardware released in
the last 5 years. (Mali-G31 is a more restrictive target, but mostly found in embedded devices, so I wouldn't worry about that one unless you know you have users using it). | {"url":"https://community.arm.com/support-forums/f/graphics-gaming-and-vr-forum/52663/why-reordering-uniforms-affects-arithmetic-cycles","timestamp":"2024-11-04T13:40:45Z","content_type":"text/html","content_length":"132163","record_id":"<urn:uuid:ac1cfa5c-8b2f-4df2-a797-19a73bc26092>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00309.warc.gz"} |
Input node placement restricting the longest control chain in controllability of complex networks
Abstract (may include machine translation)
The minimum number of inputs needed to control a network is frequently used to quantify its controllability. Control of linear dynamics through a minimum set of inputs, however, often has
prohibitively large energy requirements and there is an inherent trade-off between minimizing the number of inputs and control energy. To better understand this trade-off, we study the problem of
identifying a minimum set of input nodes such that controllabililty is ensured while restricting the length of the longest control chain. The longest control chain is the maximum distance from input
nodes to any network node, and recent work found that reducing its length significantly reduces control energy. We map the longest control chain-constraint minimum input problem to finding a joint
maximum matching and minimum dominating set. We show that this graph combinatorial problem is NP-complete, and we introduce and validate a heuristic approximation. Applying this algorithm to a
collection of real and model networks, we investigate how network structure affects the minimum number of inputs, revealing, for example, that for many real networks reducing the longest control
chain requires only few or no additional inputs, only the rearrangement of the input nodes.
Dive into the research topics of 'Input node placement restricting the longest control chain in controllability of complex networks'. Together they form a unique fingerprint. | {"url":"https://research.ceu.edu/en/publications/input-node-placement-restricting-the-longest-control-chain-in-con","timestamp":"2024-11-02T14:31:20Z","content_type":"text/html","content_length":"49452","record_id":"<urn:uuid:e16f5b58-42d3-4a6b-a44b-cc205ce759fe>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00723.warc.gz"} |
Electric Current
Electric current definition and calculations.
Electric current definition
Electrical current is the flow rate of electric charge in electric field, usually in electrical circuit.
Using water pipe analogy, we can visualize the electrical current as water current that flows in a pipe.
The electrical current is measured in ampere (amp) unit.
Electric current calculation
Electrical current is measured by the rate of electric charge flow in an electrical circuit:
i(t) = dQ(t) / dt
The momentary current is given by the derivative of the electric charge by time.
i(t) is the momentary current I at time t in amps (A).
Q(t) is the momentary electric charge in coulombs (C).
t is the time in seconds (s).
When the current is constant:
I = ΔQ / Δt
I is the current in amps (A).
ΔQ is the electric charge in coulombs (C), that flows at time duration of Δt.
Δt is the time duration in seconds (s).
When 5 coulombs flow through a resistor for duration of 10 seconds,
the current will be calculated by:
I = ΔQ / Δt = 5C / 10s = 0.5A
Current calculation with Ohm's law
The current I[R ]in anps (A) is equal to the resistor's voltage V[R ] in volts (V) divided by the resistance R in ohms (Ω).
I[R] = V[R] / R
Current direction
│ current type │from│to│
│Positive charges │ + │- │
│Negative charges │ - │+ │
│Conventional direction │ + │- │
Current in series circuits
Current that flows through resistors in series is equal in all resistors - just like water flow through a single pipe.
I[Total] = I[1 ]= I[2 ]= I[3 ]=...
I[Total] - the equivalent current in amps (A).
I[1] - current of load #1 in amps (A).
I[2] - current of load #2 in amps (A).
I[3] - current of load #3 in amps (A).
Current in parallel circuits
Current that flows through loads in parallel - just like water flow through parallel pipes.
The total current I[Total] is the sum of the parallel currents of each load:
I[Total] = I[1 ]+ I[2 ]+ I[3 ]+...
I[Total] - the equivalent current in amps (A).
I[1] - current of load #1 in amps (A).
I[2] - current of load #2 in amps (A).
I[3] - current of load #3 in amps (A).
Current divider
The current division of resistors in parallel is
R[T] = 1 / (1/R[2] + 1/R[3])
I[1] = I[T ] × R[T] / (R[1]+R[T])
Kirchhoff's current law (KCL)
The junction of several electrical components is called a node.
The algebraic sum of currents entering a node is zero.
∑ I[k] = 0
Alternating Current (AC)
Alternating current is generated by a sinusoidal voltage source.
Ohm's law
I[Z] = V[Z] / Z
I[Z] - current flow through the load measured in amperes (A)
V[Z] - voltage drop on the load measured in volts (V)
Z - impedance of the load measured in ohms (Ω)
Angular frequency
ω = 2π f
ω - angular velocity measured in radians per second (rad/s)
f - frequency measured in hertz (Hz).
Momentary current
i(t) = I[peak] sin(ωt+θ)
i(t) - momentary current at time t, measured in amps (A).
Ipeak - maximal current (=amplitude of sine), measured in amps (A).
ω - angular frequency measured in radians per second (rad/s).
t - time, measured in seconds (s).
θ - phase of sine wave in radians (rad).
RMS (effective) current
I[rms] = I[eff] = I[peak ]/ √2 ≈ 0.707 I[peak]
Peak-to-peak current
I[p-p] = 2I[peak]
Current measurement
Current measurement is done by connecting the ammeter in series to the measured object, so all the measured current will flow through the ammeter.
The ammeter has very low resistance, so it almost does not affect the measured circuit.
See also | {"url":"https://jobsvacancy.in/electric/Current.html","timestamp":"2024-11-03T23:42:15Z","content_type":"text/html","content_length":"12018","record_id":"<urn:uuid:6734dbdb-abaa-4781-893a-d635efe1dbed>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00797.warc.gz"} |
A modified version of Karman's theory for calculating shear turbulence
A version of Karman's theory is proposed which is based on the same principal rheological formula for tangent stresses but employs a different boundary condition. Instead of the requirement that the
derivative of the mean flow velocity tend to infinity at the rigid boundaries of the flow, the new boundary condition allows in a natural manner for the roughness of these boundaries and for the
fluid viscosity in the boundary layer. The modified Karman theory proposed here yields results that are in good agreement with experimental data in the literature.
Akademiia Nauk SSSR Doklady
Pub Date:
□ Channel Flow;
□ Computational Fluid Dynamics;
□ Shear Flow;
□ Surface Roughness Effects;
□ Turbulent Boundary Layer;
□ Von Karman Equation;
□ Annular Flow;
□ Boundary Conditions;
□ Boundary Value Problems;
□ Couette Flow;
□ Pipe Flow;
□ Two Dimensional Flow;
□ Viscous Flow;
□ Fluid Mechanics and Heat Transfer | {"url":"https://ui.adsabs.harvard.edu/abs/1984DoSSR.279..570L/abstract","timestamp":"2024-11-13T21:55:23Z","content_type":"text/html","content_length":"35324","record_id":"<urn:uuid:79ee5c8e-ebfc-4222-b88e-251d70f4b307>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00386.warc.gz"} |
Flash Flash Revolution - TWG CLXXVII-The Resident Evil Game(mini version)
blindreper1179 05-6-2018 05:04 PM
TWG CLXXVII-The Resident Evil Game(mini version)
The Resident Evil Game
14 player set up
3 Umbrella Employees
10 S.T.A.R.S. Members
1 "The Organization"
Day start
OOTC off except for wolfs NIGHT chat
48 hour Day phases
24 hour night phases
Phantoms off
Instalynch/no lynch on
Night talk off
Umbrella Corporation
1. Albert Wesker- When checked, you give back S.T.A.R.S alignment. You cannot make the night kill unless you are the last remaining employee. You have a 1-shot kill dodge in the night, and used
automatically when targeted for death.*
2. William Birkin- You are a man of science. Once every other night, you may infect another player with the T-virus. After being infected for TWO DAY PHASES, when they are lynched, they "eat" the
last voter killing them as well. You cannot kill and infect in the same night, even as the last alive. The infected do not know they are infected.
3. Red Queen- You are a force that is uncontended. You manipulate with such ease. In the night, you may select a player, and convince them to cease any actions they may be performing, "Blocking"
their role.*
1. Jill Valentine- You are great at knowing someones end game despite what they may say. During the night, you may listen in on another player. You will find out if they are S.T.A.R.S. or not
2. Leon Kennedy- The hero in shining armor, you protect others around you. During the night, you may hide another player so they cannot be targeted by other actions or make actions themselves. You
CANNOT hide the same player twice in a row.
3. Chris Redfield- A town will be named Chris, but WILL NOT know that they are. (This is in order to keep Claire balanced in the smaller set up.
4. Claire Redfield- A woman with one objective, finding your brother, nothing stands in your way. During the night, you may shoot another player. If you shoot Chris Redfield, you cannot live with
such a mistake, and kill yourself as well.* You have 2 shots to use.
5. Rebbeca Chambers- You are a young innocent S.T.A.R.S. member, and everyone will know so at the start of the game.
"The Organization"
1. Ada Wong- You care about no one but yourself, killing anyone who stands in your way. During the pregame phase, you may either choose 1-shot bulletproof, or seer as S.T.A.R.S.. During the night
phases, you may kill another player.
1.haku- Killed N1 Chris Redfield
4.curry Lynched D3 Albert Wesker
5.psycho- replaced by star
6.mellon- Killed N2 Red Queen
7.Xelnya- Lynched D2 S.T.A.R.S. member
8.gun92- Killed N2 Rebecca Chambers
9.zoshi- Killed N3 S.T.A.R.S. member
10.darkmanticorex2- Killed N1 Leon Kennedy
12.ffa- lynched D0 william birkin
13.Funnywolf- Lynched D1 Jill Valentine
WHICH WILL BE MAY 7TH 12:00AM SERVER TIME | {"url":"https://www.flashflashrevolution.com/vbz/printthread.php?s=4756acd0d6e9cae461e37083b330c93e&t=149163&pp=80","timestamp":"2024-11-13T11:38:30Z","content_type":"application/xhtml+xml","content_length":"87029","record_id":"<urn:uuid:5b3598fb-bd78-4f57-8452-35eedc7254b1>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00661.warc.gz"} |
Facta Universitatis, Series: Physics, Chemistry and Technology
DOI Number
The paper discusses the most common impacts of the measuring system on the amplitude and phase of the photoacoustic signals in the frequency domain using the opencell experimental set-up. The highest
signal distortions are detected at the ends of the observed modulation frequency range from 20 Hz to 20 kHz. The attenuation of the signal is observed at lower frequencies, caused by the electronic
filtering of the microphone and sound card, with characteristic frequencies of 15 Hz and 25 Hz. At higher frequencies, the dominant signal distortions are caused by the microphone acoustic filtering,
having characteristic frequencies around 9 kHz and 15 kHz. It has been found that the microphone incoherent noise, the so called flicker noise, is negligibly small in comparison to the signal and
does not affect the signal shape. However, a coherent noise originating from the power modulation system of the light source significantly affects the shape of the signal in the range greater than 10
kHz. The effects of the coherent noise and measuring system influence are eliminated completely using the relevant signal correction procedure targeting the photoacoustic signal generated by the
electro-acoustic, photoacoustic signal, measuring system, amplitude, phase, frequency domain
Albert A., van der WoerdanWouter C., Serdijn A., 1993. Low-voltage low-power controllable preamplifier for electret microphones”, IEEE Journal of Solid-State Circuits, vol. 28, 10.
Burdett R., 2005, Amplitude modulated signals - the lock-in amplifier, Handbook of Measuring System Design, Wiley, New York, ISBN. 978-0-470-02143-9.
Lashkari B., Mandelis A., 2011. Comparison between pulsed laser and frequency-domain photoacoustic modalities: Signal-to-noise ratio, contrast, resolution, and maximum depth detectivity, Review of
Scientific Instruments, vol. 82, 094903.
Markushev D. D., Rabasović M.D, Todorović D.M., Galović S., Bialkowski S. E., 2015. Photoacoustic signal and noise analysis for Si thin plate: Signal correction in frequency domain, Review of
Scientific Instruments vol. 86, 035110; doi: 10.1063/1.4914894.
Marquerinit M.V, Cellat N., Mansanares A.M., Vargas H., Miranda L.C.M., 1991. Open photoacoustic cell spectroscopy, Measurement, Science & Technology, vol. 2, pp. 396-401.
Marquerinit MV, Cellat N, Mansanarest A.M, Vargas H, Miranda L.C.M., 1991. Open photoacoustic cell spectroscopy, Measurement, Science & Technology,vol. 2, pp. 396-401.
McDonald F., Westel G.1978. Generalized theory of the photoacoustic effect, Journal of Applied Physics, vol. 49, pp. 2313-2322.
Perondi L.F., Miranda L.C.M., 1987. Minimal-volume photoacoustic cell measurement of thermal diffusivity: Effect of thermoelastic sample bending, Journal of Applied Physics, vol. 62, pp. 2955-2959.
Rabasović M.D., Nikolić M.G., Dramićanin M.D., Franko M., Markushev, 2009. Low-cost, portable photoacoustic setup for solid samples, Measurement, Science & Technology, vol. 20, 095902 (6pp),
Rosencwaig A, Gersho A., 1976. Theory of the photoacoustic effect with solids, Journal of Applied Physics, vol. 47, pp. 64-69.
Roussel G., Lepoutre F., Bertrand L., 1983. Influence of thermoelastic bending on photoacoustic experiments related to measurements of thermal diffusivity of metals, Journal of Applied Physics, vol.
54, pp. 2383-2391.
Scofield J.H., 1994. Frequency-domain description of a lock-in amplifier, American Journal of Physics (AAPT), vol. 62, pp. 129–133.
Somer A., Camilotti F., Costa G.F., Bonardi C., Novatski A., Andrade A.V.C., Kozlowski V.A., Jr., Cruz G.K., 2013. The thermoelastic bending and thermal diffusion processes influence on photoacoustic
signal generation using open photoacoustic cell technique, Journal of Applied Physics, vol. 114, 063503.
Telenkov S., Mandelis A., 2010. Signal-to-noise analysis of biomedical photoacoustic measurements in time and frequency domains, Review of Scientific Instruments, vol. 81, 124901.
Todorović D.M., Markushev D. Rabasović M.D., 2013. Photoacoustic elastic bending in thin film - Substrate system, Journal of Applied Physics, vol. 114, 213510; doi: 10.1063/1.4839835.
Todorović D.M., Rabasović M.D., Markushev D. D., Sarajlić M., 2014. Photoacoustic elastic bending in thin film-substrate system: Experimental determination of the thin film parameters, Journal of
Applied Physics, vol. 116, 053506.
Vargas H., Miranda L.C.M., 1988. Photoacoustic and Related Photothermal Techniques, Phys. Rep., vol. 16, pp. 45-101.
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ISSN 0354-4656 (print)
ISSN 2406-0879 (online) | {"url":"https://casopisi.junis.ni.ac.rs/index.php/FUPhysChemTech/article/view/1529","timestamp":"2024-11-05T03:37:12Z","content_type":"application/xhtml+xml","content_length":"25929","record_id":"<urn:uuid:0d094dc7-b285-4950-bc95-74f5c9881747>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00531.warc.gz"} |
Crouzeix Conjecture - Michael L. Overton (Courant Institute of Mathematical Sciences, New York University) - Dipartimento di Matematica
Crouzeix Conjecture – Michael L. Overton (Courant Institute of Mathematical Sciences, New York University)
Aula Mancini (SNS).
Crouzeix’s conjecture is among the most intriguing developments in matrix theory in recent years. Made in 2004 by Michel Crouzeix, it postulates that, for any polynomial p and any matrix A, ||p(A)||
<= 2 max(|p(z)|: z in W(A)), where the norm is the 2-norm and W(A) is the field of values (numerical range) of A, that is the set of points attained by v*Av for some vector v of unit length. Crouzeix
proved in 2007 that the inequality above holds if 2 is replaced by 11.08, and recently this was greatly improved by Palencia, replacing 2 by 1+sqrt(2). Furthermore, it is known that the conjecture
holds in a number of special cases, including n=2. We use nonsmooth optimization to investigate the conjecture numerically by locally minimizing the “Crouzeix ratio”, defined as the quotient with
numerator the right-hand side and denominator the left-hand side of the conjectured inequality. We also present local nonsmooth variational analysis of the Crouzeix ratio at conjectured global
minimizers. All our results strongly support the truth of Crouzeix’s conjecture. This is joint work with Anne Greenbaum and Adrian Lewis. | {"url":"https://www.dm.unipi.it/eventi/crouzeix-conjecture-michael-l-overton-courant-institute-of-mathematical-sciences-new-york-university/","timestamp":"2024-11-10T07:41:22Z","content_type":"text/html","content_length":"52367","record_id":"<urn:uuid:2de906ac-16d7-43ab-afae-6f7448607cbb>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00887.warc.gz"} |
Compass Rule Adjustment Calculator - Online Calculators
To adjust a measurement using the compass rule, first find the total error and multiply it by the ratio of the original measurement to the total of all measurements. Then, add this result to the
original measurement to get the adjusted value.
The compass rule adjustment calculator is a useful tool for land surveying to adjust measurements of distance and angles. This rule is applied when the measured values contain errors due to
instrument inaccuracies or environmental factors.
By distributing the total error proportionally among all measurements, surveyors can refine their data and improve accuracy. Addedly, this method ensures that larger measurements receive a
proportionally larger error adjustment, making it highly effective in practical surveying.
$\text{Adjusted Measurement} = \text{Original Measurement} + \left(\text{Total Error} \times \left(\frac{\text{Original Measurement}}{\text{Total of All Measurements}}\right)\right)$
Variable Description
Original Measurement The initial, unadjusted measurement
Total Error The total error identified in the survey
Total of All Measurements Sum of all original measurements in the survey
Solved Calculation:
Example 1:
Step Calculation
Determine values Original Measurement = 500m, Total Error = 10m, Total of All Measurements = 2000m
Calculate the ratio 500 / 2000 = 0.25
Multiply by total error 10 × 0.25 = 2.5
Add to original measurement 500 + 2.5 = 502.5
Result 502.5m
Answer: The adjusted measurement is 502.5 meters.
Example 2:
Step Calculation
Determine values Original Measurement = 700m, Total Error = 15m, Total of All Measurements = 3000m
Calculate the ratio 700 / 3000 = 0.2333
Multiply by total error 15 × 0.2333 = 3.5
Add to original measurement 700 + 3.5 = 703.5
Result 703.5m
Answer: The adjusted measurement is 703.5 meters.
What is a Compass Rule Adjustment Calculator?
The compass rule adjustment calculator is a beneficial tool that is utilized in land surveying to balance and adjust latitudes and departures in a survey because it ensures precise measurements.
Additionally, this method is particularly useful when small errors occur in angles or distances.
By applying the compass rule formula, which distributes these errors proportionally across all lines in a traverse, surveyors can maintain the accuracy of their data.
For those seeking more convenience, tools like the compass rule adjustment calculator online and free apps provide instant calculations, making the adjustment process easier.
Whether you’re working on a large land survey or a smaller task, these tools allow you to quickly adjust bearings and departures, ensuring your work remains accurate.
Final Words:
To wrap up, the compass rule adjustment calculator simplifies surveying tasks by automatically balancing data, saving time and enhancing precision. It’s an essential tool for maintaining accuracy in
land surveys. | {"url":"https://areacalculators.com/compass-rule-adjustment-calculator/","timestamp":"2024-11-03T03:29:40Z","content_type":"text/html","content_length":"105621","record_id":"<urn:uuid:f9b28230-13f6-4d39-9479-ad382476f952>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00243.warc.gz"} |
Calculating the number of bots required for a load test
FAQs »
What’s the best way to calculate how many bots I need for a load test?
When you create a test scenario in Loadster, the bots number you specify is the total number of concurrent users Loadster should simulate at any given time during the test.
This means that after each bot finishes an iteration of your script, another bot takes its place and starts that same script again. These bots come and go as long as there’s time remaining in your
test. Therefore the concurrent number of bots at any time is kept constant, but the total number of unique bots iterations is always increasing.
Loadster uses the iterations counter to show how many total unique bot iterations have run your script. If your script involves making a purchase, for example, the iterations count will indicate how
many total purchases have been made in the test.
Estimating Bots From Iteration Count
Let’s say you want to simulate 100,000 iterations of your script (user journeys, transactions, etc) in a 1-hour period, but you don’t know how many concurrent bots that will require.
A basic formula for estimating this is:
ConcurrentUsers = TotalIterations / (EntireTestDuration / ScriptDuration)
Let’s say each iteration of your script normally takes 67 seconds to run.
If we plug in 100000 for the total number of iterations we desire, 3600 for the number of seconds in a full hour (the test duration), and 67 as the duration of a single iteration of our script, we
1861 = 100000 / (3600 / 67)
This tells us that 1861 concurrent bots is a good starting point, if we want to generate approximately 100000 of this script in an hour.
Trial and Error
This estimate is merely a starting point, though! A bit of trial and error is still required.
When your site is under load, it’s likely it will get slower. This would cause each iteration of your script to take longer than the 67 seconds it takes when the site is not under load. Once each
iteration starts taking longer, the same number of bots will be able to run fewer iterations in a given time period.
That’s why we recommend a bit of trial and error in addition to estimation.
If you’re preparing to run an extended test, run a few shorter tests first, to make sure you’ve got the iteration rate dialed in and are running with the correct number of bots to produce the desired | {"url":"https://loadster.app/faqs/calculating-v-users-needed-for-a-load-test/","timestamp":"2024-11-02T11:51:00Z","content_type":"text/html","content_length":"19078","record_id":"<urn:uuid:b9eac1f5-9571-46f4-bd0a-9b9b1d303e1e>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00802.warc.gz"} |
3-4-5 Right Triangles (worked solutions, examples, videos)
Recognizing special right triangles in geometry can help you to answer some questions quicker. A special right triangle is a right triangle whose sides are in a particular ratio. You can also use the
Pythagorean theorem, but if you can see that it is a special triangle it can save you some calculations.
In these lessons, we will study
• the special right triangle called the 3-4-5 triangle.
• how to solve problems involving the 3-4-5 right triangle
• some examples of the Pythagorean Triples
3-4-5 Right Triangle
A 3-4-5 triangle is right triangle whose lengths are in the ratio of 3:4:5. When you are given the lengths of two sides of a right triangle, check the ratio of the lengths to see if it fits the 3:4:5
Side1 : Side2 : Hypotenuse = 3n : 4n : 5n
Solve problems with 3-4-5 right triangles
Example 1:
Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches.
Step 1: Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.
6 : 8 : ? = 3(2) : 4(2) : ?
Step 2: Yes, it is a 3-4-5 triangle for n = 2.
Step 3: Calculate the third side
5n = 5 × 2 = 10
Answer: The length of the hypotenuse is 10 inches.
Example 2:
Find the length of one side of a right triangle if the length of the hypotenuse is 15 inches and the length of the other side is 12 inches.
Step 1: Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.
? : 12 : 15 = ? : 4(3) : 5(3)
Step 2: Yes, it is a 3-4-5 triangle for n = 3
Step 3: Calculate the third side
3n = 3 × 3 = 9
Answer: The length of the side is 9 inches.
Pythagorean Theorem and 3,4,5 Triangle
How to work out the unknown sides of right angles triangle?
Pythagorean Triple
3-4-5 is an example of the Pythagorean Triple. It is usually written as (3, 4, 5).
In general, a Pythagorean triple consists of three positive integers such that a^2 + b^2 = c^2.
Other commonly used Pythagorean Triples are (5, 12, 13), (8, 15, 17) and (7, 24, 25)
Conversely, any triangle that has the Pythagorean Triples as the length of its sides would be a right triangle.
Introduction into the concepts and patterns of Pythagorean Triplets
Define and explain the Pythagorean Triples Any group of 3 integer values that satisfies the equation a^2 + b^2 = c^2 is called a Pythagorean Triple. Therefore, any triangle that has sides that form a
Pythagorean Triple must be a right triangle.
Generating Triplets
An introduction into Euclid’s formula for generating Pythagorean Triplets
The following is a list of some Pythagorean Triplets
(3,4,5), (5,12,13), (7,24,25), (8,15,17), (9,40,41), (11,60,61), (12,35,37), (13,84,85), (16,63,65), (20,21,29), (28,45,53), (33,56,65), (36,77,85), (39,80,89), (48,55,73), (65,72,97).
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page. | {"url":"https://www.onlinemathlearning.com/3-4-5-right-triangle.html","timestamp":"2024-11-04T08:17:30Z","content_type":"text/html","content_length":"40000","record_id":"<urn:uuid:eee71ffa-250c-481a-a6d9-254104c02614>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00573.warc.gz"} |
Concordance Index Calculator - Online Calculators
To calculate the Concordance Index (C-index), subtract the expected outcome from the observed outcome, sum the differences, and divide by the total number of comparisons. This helps evaluate model
prediction accuracy.
The Concordance Index (C-index) is mainly used in survival analysis. This is applied to measure the predictive accuracy of models, especially when working with censored data, such as in medical
research. It evaluates how well the predicted outcomes match the actual events.
A higher C-index value indicates a better predictive model, while a value around 0.5 suggests random prediction. This metric is similar to AUC (Area Under the Curve) but specifically tailored for
time-to-event data, making it a crucial tool in survival and prognostic modeling.
$C\text{-index} = \frac{\Sigma(O_i - E_i)}{N}$
Variable Description
$C\text{-index}$ Concordance Index
$O_i$ Observed outcomes
$E_i$ Expected outcomes
$N$ Total number of comparisons
Solved Calculations:
Example 1:
For a study where the observed outcomes are 30, the expected outcomes are 25, and the number of comparisons is 50:
Step Calculation
1. $C\text{-index} = \frac{30 – 25}{50}$
2. $C\text{-index} = \frac{5}{50}$
3. $C\text{-index} = 0.10$
Answer: 0.10
Example 2:
In another study with observed outcomes of 45, expected outcomes of 40, and 60 comparisons:
Step Calculation
1. $C\text{-index} = \frac{45 – 40}{60}$
2. $C\text{-index} = \frac{5}{60}$
3. $C\text{-index} = 0.083$
Answer: 0.083
What is a Concordance Index Calculator?
The Concordance Index Calculator is a fine tool to calculate the survival analysis and classification models to measure the predictive accuracy of a model.
This calculator facilitates in the evaluation of a model. It tells that how well it can correctly predict the order of survival times or other continuous outcomes.
In survival analysis, the C-index assesses the agreement between predicted and actual outcomes, making it essential for fields such as medical research, where predicting patient survival is critical.
The tool can be used in programming languages like Python and R to calculate the index based on data, helping analysts and researchers determine model reliability.
Additionally, the C-index is closely related to the Area Under the Curve (AUC), with both measuring model performance in binary classification problems. A high concordance index implies that the
model is able to rank outcomes effectively, making it a valuable tool for evaluating predictive accuracy.
Final Words:
To sum up, the Concordance Index Calculator is essential for assessing the predictive ability of survival models. It helps researchers and analysts improve their models by offering insight into the
accuracy of predictions. | {"url":"https://areacalculators.com/concordance-index-calculator/","timestamp":"2024-11-04T01:48:05Z","content_type":"text/html","content_length":"109174","record_id":"<urn:uuid:937d56a1-bf82-45e5-9412-b32c8dfb8c2b>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00252.warc.gz"} |
Pakize - MATLAB Central
Duzce Universitesi
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Digital Image Processing, Optimization, Deep Learning
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18 Solutions
Resize your images in the folders and subfolders easily
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1 month ago | 3 downloads |
Determine whether a vector is monotonically increasing
Return true if the elements of the input vector increase monotonically (i.e. each element is larger than the previous). Return f...
1 year ago
MATLAB Basic: rounding III
Do rounding towards minus infinity. Example: -8.8, answer -9 +8.1 answer 8 +8.50 answer 8
2 years ago
MATLAB Basic: rounding II
Do rounding nearest integer. Example: -8.8, answer -9 +8.1 answer 8 +8.50 answer 9
2 years ago
NonLinear Equation System Solution with GWO
Please cite this article. The codes for the p1 equation in tihs article. Erdoğmuş, Pakize. "A new solution approach for non-line...
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Non_linear Equation Systems Solution with PSO, GWO and GA
In these codes a sample system of nonlinear equations is solved with GWO, PSO and GA. PSO and GA is Matlab function. GWO is adap...
2 years ago | 3 downloads |
Is Matlab Transfer Learning performance compareable with Pyton Keras?
I wwant to compare transfer learning performances of the platforms. But I couldn't catch up Keras performance. Matlab is slover ...
3 years ago | 1 answer | 0
Converting Cifar10 Dataset to Desires Size png files with Folders
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Finding Perfect Squares
Given a vector of numbers, return true if one of the numbers is a square of one of the numbers. Otherwise return false. Example...
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mean - Average or mean value of array (2024)
Average or mean value of array
M = mean(___,missingflag)
M = mean(A) returns the mean of the elements of A along the first array dimension whose size does not equal 1.
• If A is a vector, then mean(A) returns the mean of the elements.
• If A is a matrix, then mean(A) returns a row vector containing the mean of each column.
• If A is a multidimensional array, then mean(A) operates along the first array dimension whose size does not equal 1, treating the elements as vectors. The size of M in this dimension becomes 1,
while the sizes of all other dimensions remain the same as in A.
• If A is a table or timetable, then mean(A) returns a one-row table containing the mean of each variable. (since R2023a)
M = mean(A,"all") returns the mean over all elements of A.
M = mean(A,dim) returns the mean along dimension dim. For example, if A is a matrix, then mean(A,2) returns a column vector containing the mean of each row.
M = mean(A,vecdim) returns the mean based on the dimensions specified in the vector vecdim. For example, if A is a matrix, then mean(A,[1 2]) returns the mean of all elements in A because every
element of a matrix is contained in the array slice defined by dimensions 1 and 2.
M = mean(___,outtype) returns the mean with a specified data type for any of the previous syntaxes. outtype can be "default", "double", or "native".
M = mean(___,missingflag) specifies whether to include or omit missing values in A. For example, mean(A,"omitmissing") ignores all missing values when computing the mean. By default, mean includes
missing values.
M = mean(___,Weights=W) specifies a weighting scheme W and returns the weighted mean. (since R2024a)
collapse all
Mean of Matrix Columns
Create a matrix and compute the mean of each column.
A = [0 1 1; 2 3 2; 1 3 2; 4 2 2]
A = 4×3 0 1 1 2 3 2 1 3 2 4 2 2
M = 1×3 1.7500 2.2500 1.7500
Mean of Matrix Rows
Create a matrix and compute the mean of each row.
A = [0 1 1; 2 3 2; 3 0 1; 1 2 3]
A = 4×3 0 1 1 2 3 2 3 0 1 1 2 3
M = 4×1 0.6667 2.3333 1.3333 2.0000
Mean of 3-D Array
Create a 4-by-2-by-3 array of integers between 1 and 10 and compute the mean values along the second dimension.
rng('default')A = randi(10,[4,2,3]);M = mean(A,2)
M = M(:,:,1) = 8.0000 5.5000 2.5000 8.0000M(:,:,2) = 10.0000 7.5000 5.5000 6.0000M(:,:,3) = 6.0000 5.5000 8.5000 10.0000
Mean of Array Page
Create a 3-D array and compute the mean over each page of data (rows and columns).
A(:,:,1) = [2 4; -2 1];A(:,:,2) = [9 13; -5 7];A(:,:,3) = [4 4; 8 -3];M1 = mean(A,[1 2])
M1 = M1(:,:,1) = 1.2500M1(:,:,2) = 6M1(:,:,3) = 3.2500
To compute the mean over all dimensions of an array, you can either specify each dimension in the vector dimension argument, or use the "all" option.
Mean of Single-Precision Array
Create a single-precision vector of ones and compute its single-precision mean.
A = single(ones(10,1));M = mean(A,"native")
The result is also in single precision.
Mean Excluding Missing Values
Create a matrix containing NaN values.
A = [1.77 -0.005 NaN -2.95; NaN 0.34 NaN 0.19]
A = 2×4 1.7700 -0.0050 NaN -2.9500 NaN 0.3400 NaN 0.1900
Compute the mean values of the matrix, excluding missing values. For matrix columns that contain any NaN value, mean computes with the non-NaN elements. For matrix columns that contain all NaN
values, the mean is NaN.
M = 1×4 1.7700 0.1675 NaN -1.3800
Weighted Mean
Since R2024a
Create a matrix and compute the weighted mean of the matrix according to a weighting scheme specified by W. The mean function applies the weighting scheme to each column in A.
A = [1 1; 7 9; 1 9; 1 9; 6 2];W = [1 2 1 2 3]';M = mean(A,Weights=W)
Input Arguments
collapse all
A — Input data
vector | matrix | multidimensional array | table | timetable
Input data, specified as a vector, matrix, multidimensional array, table, or timetable.
• If A is a scalar, then mean(A) returns A.
• If A is an empty 0-by-0 matrix, then mean(A) returns NaN.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64 | logical | datetime | duration | table | timetable
dim — Dimension to operate along
positive integer scalar
Dimension to operate along, specified as a positive integer scalar. If you do not specify the dimension, then the default is the first array dimension whose size does not equal 1.
Dimension dim indicates the dimension whoselength reduces to 1. The size(M,dim) is 1,while the sizes of all other dimensions remain the same.
Consider an m-by-n input matrix, A:
• mean(A,1) computes the mean of the elements in each column of A and returns a 1-by-n row vector.
• mean(A,2) computes the mean of the elements in each row of A and returns an m-by-1 column vector.
mean returns A when dim isgreater than ndims(A) or when size(A,dim) is 1.
vecdim — Vector of dimensions
vector of positive integers
Vector of dimensions, specified as a vector of positive integers. Each element represents a dimension of the input data. The lengths of the output in the specified operating dimensions are 1, while
the others remain the same.
Consider a 2-by-3-by-3 input data, A. Then mean(A,[1 2]) returns a 1-by-1-by-3 array whose elements are the means over each page of A.
outtype — Output data type
"default" (default) | "double" | "native"
Output data type, specified as one of the values in this table. These options also specify the data type in which the operation is performed.
outtype Output data type
"default" double, unless the input data type is single, duration, datetime, table, or timetable, in which case, the output is "native"
"double" double, unless the data input type is duration, datetime, table, or timetable, in which case, "double" is not supported
Same data type as the input, unless:
• Input data type is logical, in which case, the output is double
• Input data type is char, in which case, "native" is not supported
• Input data type is timetable, in which case, the output is table
missingflag — Missing value condition
"includemissing" (default) | "includenan" | "includenat" | "omitmissing" | "omitnan" | "omitnat"
Missing value condition, specified as one of the values in this table.
Value Input Data Type Description
"includemissing" All supported data
types Include missing values in A when computing the mean. If any element in the operating dimension is missing, then the corresponding element in M is missing.
"includenan" double, single,
"includenat" datetime
"omitmissing" All supported data
types Ignore missing values in A, and compute the mean over fewer points. If all elements in the operating dimension are missing, then the corresponding element in M
"omitnan" double, single, is missing.
"omitnat" datetime
W — Weighting scheme
vector | matrix | multidimensional array
Since R2024a
Weighting scheme, specified as a vector, matrix, or multidimensional array. The elements of W must be nonnegative.
If you specify a weighting scheme, mean returns the weighted mean, which is useful when values in the input data have different levels of importance or the input data is skewed.
If W is a vector, it must have the same length as the operating dimension. Otherwise, W must have the same size as the input data.
If the input data A is a table or timetable, then W must be a vector.
You cannot specify this argument if you specify vecdim or "all".
Data Types: double | single
More About
collapse all
For a finite-length vector A made up of N scalar observations, the mean is defined as
$\mu =\frac{1}{N}\sum _{i=1}^{N}{A}_{i}.$
Weighted Mean
For a finite-length vector A made up of N scalar observations and weighting scheme W, the weighted mean is defined as
${\mu }_{W}=\frac{\sum _{i=1}^{N}{W}_{i}{A}_{i}}{\sum _{i=1}^{N}{W}_{i}}.$
Extended Capabilities
Tall Arrays
Calculate with arrays that have more rows than fit in memory.
The mean function supports tall arrays with the following usage notes and limitations:
• The Weights name-value argument is not supported.
For more information, see Tall Arrays.
C/C++ Code Generation
Generate C and C++ code using MATLAB® Coder™.
Usage notes and limitations:
• If you specify dim, then it mustbe a constant.
• The outtype and missingflag options must be constant character vectors or strings.
• Integer types do not support the "native" output data type option.
• See Variable-Sizing Restrictions for Code Generation of Toolbox Functions (MATLAB Coder).
GPU Code Generation
Generate CUDA® code for NVIDIA® GPUs using GPU Coder™.
Usage notes and limitations:
• If you specify dim, then it must be a constant.
• The outtype and missingflag options must be constant character vectors or strings.
• Integer types do not support the "native" output data type option.
• The Weights name-value argument is not supported.
Thread-Based Environment
Run code in the background using MATLAB® backgroundPool or accelerate code with Parallel Computing Toolbox™ ThreadPool.
This function fully supports thread-based environments. For more information, see Run MATLAB Functions in Thread-Based Environment.
GPU Arrays
Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™.
The mean function supports GPU array input with these usage notes and limitations:
• The "native" option is not supported.
For more information, see Run MATLAB Functions on a GPU (Parallel Computing Toolbox).
Distributed Arrays
Partition large arrays across the combined memory of your cluster using Parallel Computing Toolbox™.
Usage notes and limitations:
• The "native" option is not supported.
For more information, see Run MATLAB Functions with Distributed Arrays (Parallel Computing Toolbox).
Version History
Introduced before R2006a
expand all
R2024b: Compute weighted mean for datetime data type
You can compute the weighted mean for input data having the datetime data type. Before R2024b, you could compute only the unweighted mean for this data type.
R2024a: Compute weighted mean
Compute the weighted mean by specifying the Weights parameter as the weighting scheme. You can compute the weighted mean for input data having numeric, logical, and duration data types.
R2023a: Perform calculations directly on tables and timetables
The mean function can calculate on all variables within a table or timetable without indexing to access those variables. All variables must have data types that support the calculation. For more
information, see Direct Calculations on Tables and Timetables.
R2023a: Specify missing value condition
Include or omit all missing values in the input data when computing the mean by using the "includemissing" or "omitmissing" options. Previously, "includenan", "omitnan", "includenat", and "omitnat"
specified a missing value condition that was specific to the data type of the input data.
R2023a: Improved performance with small group size
The mean function shows improved performance when computing over a real vector when the operating dimension is not specified. The function determines the default operating dimension more quickly in
R2023a than in R2022b.
For example, this code computes the mean along the default vector dimension. The code is about 2.2x faster than in the previous release.
function timingMeanA = rand(10,1);for i = 1:8e5 mean(A);endend
The approximate execution times are:
R2022b: 0.91 s
R2023a: 0.41 s
The code was timed on a Windows^® 10, Intel^® Xeon^® CPU E5-1650 v4 @ 3.60 GHz test system using the timeit function.
R2018b: Operate on multiple dimensions
Operate on multiple dimensions of the input data at a time. Specify a vector of operating dimensions, or specify the "all" option to operate on all array dimensions.
See Also
• median | mode | std | var | sum
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Convert 1/Ω (Electric conductance)
Convert 1/Ω
Direct link to this calculator:
Convert 1/Ω (Electric conductance)
1. Choose the right category from the selection list, in this case 'Electric conductance'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and
π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case '1/Ω'.
4. The value will then be converted into all units of measurement the calculator is familiar with.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '336 1/Ω'. In so doing, either the full name of the unit or its
abbreviation can be used Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Electric conductance'. After that, it converts the
entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Regardless which of these possibilities one
uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator
and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(32 * 42) 1/Ω'. But different
units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '12 1/Ω + 22 1/Ω' or '52mm x 62cm x 72dm = ? cm^3'. The units of measure
combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 6.388 148 090 016 ×1020. For this form of presentation, the number
will be segmented into an exponent, here 20, and the actual number, here 6.388 148 090 016. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket
calculators, one also finds the way of writing numbers as 6.388 148 090 016 E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at
this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 638 814 809 001 600 000 000. Independent of the presentation of the
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LTE Downlink Channel Estimation and Equalization
This example shows how to use the LTE Toolbox™ to create a frame worth of data, pass it through a fading channel and perform channel estimation and equalization. Two figures are created illustrating
the received and equalized frame.
This example shows how a simple transmitter-channel-receiver simulation may be created using functions from the LTE Toolbox. The example generates a frame worth of data on one antenna port. As no
transport channel is created in this example the data is random bits, QPSK modulated and mapped to every symbol in a subframe. A cell specific reference signal and primary and secondary
synchronization signals are created and mapped to the subframe. 10 subframes are individually generated to create a frame. The frame is OFDM modulated, passed through an Extended Vehicular A Model
(EVA5) fading channel, additive white Gaussian noise added and demodulated. MMSE equalization using channel and noise estimation is applied and finally the received and equalized resource grids are
Cell-Wide Settings
The cell-wide settings are specified in a structure enb. A number of the functions used in this example require a subset of the settings specified below. In this example only one transmit antenna is
enb.NDLRB = 15; % Number of resource blocks
enb.CellRefP = 1; % One transmit antenna port
enb.NCellID = 10; % Cell ID
enb.CyclicPrefix = 'Normal'; % Normal cyclic prefix
enb.DuplexMode = 'FDD'; % FDD
SNR Configuration
The operating SNR is configured in decibels by the value SNRdB which is also converted into a linear SNR.
SNRdB = 22; % Desired SNR in dB
SNR = 10^(SNRdB/20); % Linear SNR
rng('default'); % Configure random number generators
Channel Model Configuration
The channel model is configured using a structure. In this example a fading channel with an Extended Vehicular A (EVA) delay profile and 120Hz Doppler frequency is used. These parameters along with
MIMO correlation and other channel model specific parameters are set.
cfg.Seed = 1; % Channel seed
cfg.NRxAnts = 1; % 1 receive antenna
cfg.DelayProfile = 'EVA'; % EVA delay spread
cfg.DopplerFreq = 120; % 120Hz Doppler frequency
cfg.MIMOCorrelation = 'Low'; % Low (no) MIMO correlation
cfg.InitTime = 0; % Initialize at time zero
cfg.NTerms = 16; % Oscillators used in fading model
cfg.ModelType = 'GMEDS'; % Rayleigh fading model type
cfg.InitPhase = 'Random'; % Random initial phases
cfg.NormalizePathGains = 'On'; % Normalize delay profile power
cfg.NormalizeTxAnts = 'On'; % Normalize for transmit antennas
Channel Estimator Configuration
A user defined window is used to average pilot symbols to reduce the effect of noise. The averaging window size is configured in terms of resource elements (REs), in time and frequency. A
conservative 9-by-9 window is used in this example as an EVA delay profile and 120Hz Doppler frequency cause the channel changes quickly over time and frequency. A 9-by-9 window includes the 4 pilots
immediately surrounding the pilot of interest when averaging. Selecting an averaging window is discussed in Channel Estimation.
cec.PilotAverage = 'UserDefined'; % Pilot averaging method
cec.FreqWindow = 9; % Frequency averaging window in REs
cec.TimeWindow = 9; % Time averaging window in REs
Interpolation is performed by the channel estimator between pilot estimates to create a channel estimate for all REs. To improve the estimate multiple subframes can be used when interpolating. An
interpolation window of 3 subframes with a centered interpolation window uses pilot estimates from 3 consecutive subframes to estimate the center subframe.
cec.InterpType = 'Cubic'; % Cubic interpolation
cec.InterpWinSize = 3; % Interpolate up to 3 subframes
% simultaneously
cec.InterpWindow = 'Centred'; % Interpolation windowing method
Subframe Resource Grid Size
In this example it is useful to have access to the subframe resource grid dimensions. These are determined using lteDLResourceGridSize. This function returns an array containing the number of
subcarriers, number of OFDM symbols and number of transmit antenna ports in that order.
gridsize = lteDLResourceGridSize(enb);
K = gridsize(1); % Number of subcarriers
L = gridsize(2); % Number of OFDM symbols in one subframe
P = gridsize(3); % Number of transmit antenna ports
Transmit Resource Grid
An empty resource grid txGrid is created which will be populated with subframes.
Payload Data Generation
As no transport channel is used in this example the data sent over the channel will be random QPSK modulated symbols. A subframe worth of symbols is created so a symbol can be mapped to every
resource element. Other signals required for transmission and reception will overwrite these symbols in the resource grid.
% Number of bits needed is size of resource grid (K*L*P) * number of bits
% per symbol (2 for QPSK)
numberOfBits = K*L*P*2;
% Create random bit stream
inputBits = randi([0 1], numberOfBits, 1);
% Modulate input bits
inputSym = lteSymbolModulate(inputBits,'QPSK');
Frame Generation
The frame will be created by generating individual subframes within a loop and appending each created subframe to the previous subframes. The collection of appended subframes are contained within
txGrid. This appending is repeated ten times to create a frame. When the OFDM modulated time domain waveform is passed through a channel the waveform will experience a delay. To avoid any samples
being missed due to this delay an extra subframe is generated, therefore 11 subframes are generated in total. For each subframe the Cell-Specific Reference Signal (Cell RS) is added. The Primary
Synchronization Signal (PSS) and Secondary Synchronization Signal (SSS) are also added. Note that these synchronization signals only occur in subframes 0 and 5, but the LTE Toolbox takes care of
generating empty signals and indices in the other subframes so that the calling syntax here can be completely uniform across the subframes.
% For all subframes within the frame
for sf = 0:10
% Set subframe number
enb.NSubframe = mod(sf,10);
% Generate empty subframe
subframe = lteDLResourceGrid(enb);
% Map input symbols to grid
subframe(:) = inputSym;
% Generate synchronizing signals
pssSym = ltePSS(enb);
sssSym = lteSSS(enb);
pssInd = ltePSSIndices(enb);
sssInd = lteSSSIndices(enb);
% Map synchronizing signals to the grid
subframe(pssInd) = pssSym;
subframe(sssInd) = sssSym;
% Generate cell specific reference signal symbols and indices
cellRsSym = lteCellRS(enb);
cellRsInd = lteCellRSIndices(enb);
% Map cell specific reference signal to grid
subframe(cellRsInd) = cellRsSym;
% Append subframe to grid to be transmitted
txGrid = [txGrid subframe]; %#ok
OFDM Modulation
In order to transform the frequency domain OFDM symbols into the time domain, OFDM modulation is required. This is achieved using lteOFDMModulate. The function returns two values; a matrix txWaveform
and a structure info containing the sampling rate. txWaveform is the resulting time domain waveform. Each column contains the time domain signal for each antenna port. In this example, as only one
antenna port is used, only one column is returned. info.SamplingRate is the sampling rate at which the time domain waveform was created. This value is required by the channel model.
[txWaveform,info] = lteOFDMModulate(enb,txGrid);
txGrid = txGrid(:,1:140);
Fading Channel
The time domain waveform is passed through the channel model (lteFadingChannel) configured by the structure cfg. The channel model requires the sampling rate of the time domain waveform so the
parameter cfg.SamplingRate is set to the value returned by lteOFDMModulate. The waveform generated by the channel model function contains one column per receive antenna. In this example one receive
antenna is used, therefore the returned waveform has one column.
cfg.SamplingRate = info.SamplingRate;
% Pass data through the fading channel model
rxWaveform = lteFadingChannel(cfg,txWaveform);
Additive Noise
The SNR is given by $SNR={E}_{s}/{N}_{0}$ where ${E}_{s}$ is the energy of the signal of interest and ${N}_{0}$ is the noise power. The noise added before OFDM demodulation will be amplified by the
FFT. Therefore to normalize the SNR at the receiver (after OFDM demodulation) the noise must be scaled. The amplification is the square root of the size of the FFT. The size of the FFT can be
determined from the sampling rate of the time domain waveform (info.SamplingRate) and the subcarrier spacing (15 kHz). The power of the noise to be added can be scaled so that ${E}_{s}$ and ${N}_{0}$
are normalized after the OFDM demodulation to achieve the desired SNR (SNRdB).
% Calculate noise gain
N0 = 1/(sqrt(2.0*enb.CellRefP*double(info.Nfft))*SNR);
% Create additive white Gaussian noise
noise = N0*complex(randn(size(rxWaveform)),randn(size(rxWaveform)));
% Add noise to the received time domain waveform
rxWaveform = rxWaveform + noise;
The offset caused by the channel in the received time domain signal is obtained using lteDLFrameOffset. This function returns a value offset which indicates how many samples the waveform has been
delayed. The offset is considered identical for waveforms received on all antennas. The received time domain waveform can then be manipulated to remove the delay using offset.
offset = lteDLFrameOffset(enb,rxWaveform);
rxWaveform = rxWaveform(1+offset:end,:);
OFDM Demodulation
The time domain waveform undergoes OFDM demodulation to transform it to the frequency domain and recreate a resource grid. This is accomplished using lteOFDMDemodulate. The resulting grid is a
3-dimensional matrix. The number of rows represents the number of subcarriers. The number of columns equals the number of OFDM symbols in a subframe. The number of subcarriers and symbols is the same
for the returned grid from OFDM demodulation as the grid passed into lteOFDMModulate. The number of planes (3rd dimension) in the grid corresponds to the number of receive antennas.
rxGrid = lteOFDMDemodulate(enb,rxWaveform);
Channel Estimation
To create an estimation of the channel over the duration of the transmitted resource grid lteDLChannelEstimate is used. The channel estimation function is configured by the structure cec.
lteDLChannelEstimate assumes the first subframe within the resource grid is subframe number enb.NSubframe and therefore the subframe number must be set prior to calling the function. In this example
the whole received frame will be estimated in one call and the first subframe within the frame is subframe number 0. The function returns a 4-D array of complex weights which the channel applies to
each resource element in the transmitted grid for each possible transmit and receive antenna combination. The possible combinations are based upon the eNodeB configuration enb and the number of
receive antennas (determined by the size of the received resource grid). The 1st dimension is the subcarrier, the 2nd dimension is the OFDM symbol, the 3rd dimension is the receive antenna and the
4th dimension is the transmit antenna. In this example one transmit and one receive antenna is used therefore the size of estChannel is 180-by-140-by-1-by-1.
enb.NSubframe = 0;
[estChannel, noiseEst] = lteDLChannelEstimate(enb,cec,rxGrid);
MMSE Equalization
The effects of the channel on the received resource grid are equalized using lteEqualizeMMSE. This function uses the estimate of the channel estChannel and noise noiseEst to equalize the received
resource grid rxGrid. The function returns eqGrid which is the equalized grid. The dimensions of the equalized grid are the same as the original transmitted grid (txGrid) before OFDM modulation.
eqGrid = lteEqualizeMMSE(rxGrid, estChannel, noiseEst);
The received resource grid is compared with the equalized resource grid. The error between the transmitted and equalized grid and transmitted and received grids are calculated. This creates two
matrices (the same size as the resource arrays) which contain the error for each symbol. To allow easy inspection the received and equalized grids are plotted on a logarithmic scale using surf within
hDownlinkEstimationEqualizationResults.m. These diagrams show that performing channel equalization drastically reduces the error in the received resource grid.
% Calculate error between transmitted and equalized grid
eqError = txGrid - eqGrid;
rxError = txGrid - rxGrid;
% Compute EVM across all input values
% EVM of pre-equalized receive signal
EVM = comm.EVM;
EVM.AveragingDimensions = [1 2];
preEqualisedEVM = EVM(txGrid,rxGrid);
fprintf('Percentage RMS EVM of Pre-Equalized signal: %0.3f%%\n', ...
Percentage RMS EVM of Pre-Equalized signal: 124.133%
% EVM of post-equalized receive signal
postEqualisedEVM = EVM(txGrid,eqGrid);
fprintf('Percentage RMS EVM of Post-Equalized signal: %0.3f%%\n', ...
Percentage RMS EVM of Post-Equalized signal: 15.598%
% Plot the received and equalized resource grids
hDownlinkEstimationEqualizationResults(rxGrid, eqGrid);
This example uses the helper function: | {"url":"https://ch.mathworks.com/help/lte/ug/lte-downlink-channel-estimation-and-equalization.html","timestamp":"2024-11-05T05:30:32Z","content_type":"text/html","content_length":"96689","record_id":"<urn:uuid:3ab2f973-a04f-475d-bd8b-adfc0482ddf7>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00580.warc.gz"} |
Digital Electronics
Digital Electronics - Digital Concepts - Discussion
Discussion Forum : Digital Concepts - General Questions (Q.No. 2)
In the decimal numbering system, what is the MSD?
33 comments
Syed Shakeeb said: 3 years ago
Here is the Example: 260.06 here 2 is the MSD.
Sai said: 4 years ago
MSD stand for most significant digit as the full form suggest it denotes the digit which have the greatest impact on the number. It is same as MSB.
Pavan said: 6 years ago
MSD means Most Significant Bit.
Suppose 1010.
Md Shahin said: 6 years ago
Before I introduce the number systems that are more commonly used in electronics lets discuss a number system that we all are aware of and which we use it everyday.
The Decimal number system got its name because it uses 10 symbols (or popularly digits) for counting and performing operations. It uses the digits 0 to 9.
In this number system, the radix or the base is 10 since the number of digits used is ten. The general format of decimal number representation is shown below
104 --- 103 --- 102 --- 101 --- 100 --- 10-1 --- 10-2 --- 10-3 --- 10-4 ---> Weights
S4 --- S3 --- S2 --- S1 --- S0 --- S-1 --- S-2 --- S-3 --- S-4
MSD is a most significant digit (leftmost digit of a number)
LSD is the least significant digit (rightmost digit of a number)
Note: From this, we can observe that the MSD has the greatest weight and the LSD have the smallest weight.
Swetha said: 8 years ago
MSD is similar to that of MSB, most significant bit.
For example consider BCD code, or 8421 code number 12 is indicated as 1100. In that left most bit is 1 which has highest weight of 8, That is called as most significant bit MSB or MSD. Similarly, the
least significant bit is right most bit that is zero which is of least weight i.e 1 (8421).
Souradip Guha said: 8 years ago
MSD means most significant digit. It has the highest weight.
Heylanaa said: 8 years ago
I am not getting this, please explain the question in detail.
Aditya kumar jena said: 8 years ago
MSD stands most significant digit.
Laksalika said: 8 years ago
MSD means Most significant digit. It has the most weight.
For example: Let us take a number 1500. In which if you change the MSB (msd) as 2, then the number will be changed as 2500, where there is a large difference between them.
But if you change the LSB as 2 there is no large variation between them. So only the MSB has the most weight.
Manoj bhatt said: 9 years ago
Please explain this question?
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ero-pole-gain model
Zero-pole-gain models are a representation of transfer functions in factorized form. For example, consider the following continuous-time SISO transfer function:
G(s) can be factorized into the zero-pole-gain form as:
A more general representation of the SISO zero-pole-gain model is as follows:
$h\left(s\right)=k\frac{\left(s-z\left(1\right)\right)\left(s-z\left(2\right)\right)\dots \left(s-z\left(m\right)\right)}{\left(s-p\left(1\right)\right)\left(s-p\left(2\right)\right)\dots \left(s-p\
Here, z and p are the vectors of real-valued or complex-valued zeros and poles, and k is the real-valued or complex-valued scalar gain. For MIMO models, each I/O channel is represented by one such
transfer function h[ij](s).
You can create a zero-pole-gain model object either by specifying the poles, zeros and gains directly, or by converting a model of another type (such as a state-space model ss) to zero-pole-gain
You can also use zpk to create generalized state-space (genss) models or uncertain state-space (uss (Robust Control Toolbox)) models.
sys = zpk(zeros,poles,gain) creates a continuous-time zero-pole-gain model with zeros and poles specified as vectors and the scalar value of gain. The output sys is a zpk model object storing the
model data. Set zeros or poles to [] for systems without zeros or poles. These two inputs need not have equal length and the model need not be proper (that is, have an excess of poles).
sys = zpk(zeros,poles,gain,ts) creates a discrete-time zero-pole-gain model with sample time ts. Set ts to -1 or [] to leave the sample time unspecified.
sys = zpk(zeros,poles,gain,ltiSys) creates a zero-pole-gain model with properties inherited from the dynamic system model ltiSys, including the sample time.
sys = zpk(m) creates a zero-pole-gain model that represents the static gain, m.
sys = zpk(___,Name,Value) sets Properties of the zero-pole-gain model using one or more name-value pair arguments to set additional properties of the model. This syntax works with any of the previous
input-argument combinations.
sys = zpk(ltiSys) converts the dynamic system model ltiSys to a zero-pole-gain model.
sys = zpk(ltiSys,component) converts the specified component of ltiSys to zero-pole-gain model form. Use this syntax only when ltiSys is an identified linear time-invariant (LTI) model such as an
idss or an idtf model.
s = zpk('s') creates a special variable s that you can use in a rational expression to create a continuous-time zero-pole-gain model. Using a rational expression is sometimes easier and more
intuitive than specifying polynomial coefficients.
z = zpk('z',ts) creates special variable z that you can use in a rational expression to create a discrete-time zero-pole-gain model. To leave the sample time unspecified, set ts input argument to -1.
Input Arguments
zeros — Zeros of the zero-pole-gain model
row vector | Ny-by-Nu cell array of row vectors
Zeros of the zero-pole-gain model, specified as:
• A row vector for SISO models. For instance, use [1,2+i,2-i] to create a model with zeros at s = 1, s = 2+i, and s = 2-i. For an example, see Continuous-Time SISO Zero-Pole-Gain Model.
• An Ny-by-Nu cell array of row vectors to specify a MIMO zero-pole-gain model, where Ny is the number of outputs, and Nu is the number of inputs. For an example, see Discrete-Time MIMO
Zero-Pole-Gain Model.
For instance, if a is realp tunable parameter with nominal value 3, then you can use zeros = [1 2 a] to create a genss model with zeros at s = 1 and s = 2 and a tunable zero at s = 3.
When you use this input argument to create a zpk model, the argument sets the initial value of the property Z.
poles — Poles of the zero-pole-gain model
row vector | Ny-by-Nu cell array of row vectors
Poles of the zero-pole-gain model, specified as:
• A row vector for SISO models. For an example, see Continuous-Time SISO Zero-Pole-Gain Model.
• An Ny-by-Nu cell array of row vectors to specify a MIMO zero-pole-gain model, where Ny is the number of outputs and Nu is the number of inputs. For an example, see Discrete-Time MIMO
Zero-Pole-Gain Model.
Also a property of the zpk object. This input argument sets the initial value of property P.
gain — Gain of the zero-pole-gain model
scalar | Ny-by-Nu matrix
Gain of the zero-pole-gain model, specified as:
• A scalar for SISO models. For an example, see Continuous-Time SISO Zero-Pole-Gain Model.
• An Ny-by-Nu matrix to specify a MIMO zero-pole-gain model, where Ny is the number of outputs and Nu is the number of inputs. For an example, see Discrete-Time MIMO Zero-Pole-Gain Model.
Also a property of the zpk object. This input argument sets the initial value of property K.
ts — Sample time
Sample time, specified as a scalar. Also a property of the zpk object. This input argument sets the initial value of property Ts.
ltiSys — Dynamic system
dynamic system model | model array
Dynamic system, specified as a SISO or MIMO dynamic system model or array of dynamic system models. Dynamic systems that you can use include:
• Continuous-time or discrete-time numeric LTI models, such as tf, zpk, ss, or pid models.
• Generalized or uncertain LTI models such as genss or uss (Robust Control Toolbox) models. (Using uncertain models requires a Robust Control Toolbox™ license.)
The resulting zero-pole-gain model assumes
□ current values of the tunable components for tunable control design blocks.
□ nominal model values for uncertain control design blocks.
• Identified LTI models, such as idtf (System Identification Toolbox), idss (System Identification Toolbox), idproc (System Identification Toolbox), idpoly (System Identification Toolbox), and
idgrey (System Identification Toolbox) models. To select the component of the identified model to convert, specify component. If you do not specify component, tf converts the measured component
of the identified model by default. (Using identified models requires System Identification Toolbox™ software.)
An identified nonlinear model cannot be converted into a zpk model object. You may first use linear approximation functions such as linearize and linapp (This functionality requires System
Identification Toolbox software.)
m — Static gain
scalar | matrix
Static gain, specified as a scalar or matrix. Static gain or steady state gain of a system represents the ratio of the output to the input under steady state condition.
component — Component of identified model
'measured' (default) | 'noise' | 'augmented'
Component of identified model to convert, specified as one of the following:
• 'measured' — Convert the measured component of sys.
• 'noise' — Convert the noise component of sys
• 'augmented' — Convert both the measured and noise components of sys.
component only applies when sys is an identified LTI model.
For more information on identified LTI models and their measured and noise components, see Identified LTI Models.
Output Arguments
sys — Output system model
zpk model object | genss model object | uss model object
Output system model, returned as:
• A zero-pole-gain (zpk) model object, when the zeros, poles and gain input arguments contain numeric values.
• A generalized state-space model (genss) object, when the zeros, poles and gain input arguments includes tunable parameters, such as realp parameters or generalized matrices (genmat).
• An uncertain state-space model (uss) object, when the zeros, poles and gain input arguments includes uncertain parameters. Using uncertain models requires a Robust Control Toolbox license.
Z — System zeros
cell array | Ny-by-Nu cell array of row vectors
System zeros, specified as:
• A cell array of transfer function zeros or the numerator roots for SISO models.
• An Ny-by-Nu cell array of row vectors of the zeros for each I/O pair in a MIMO model, where Ny is the number of outputs and Nu is the number of inputs.
The values of Z can be either real-valued or complex-valued.
P — System poles
cell array | Ny-by-Nu cell array of row vectors
System poles, specified as:
• A cell array of transfer function poles or the denominator roots for SISO models.
• An Ny-by-Nu cell array of row vectors of the poles for each I/O pair in a MIMO model, where Ny is the number of outputs and Nu is the number of inputs.
The values of P can be either real-valued or complex-valued.
K — System gains
scalar | Ny-by-Nu matrix
System gains, specified as:
• A scalar value for SISO models.
• An Ny-by-Nu matrix storing the gain values for each I/O pair of the MIMO model, where Ny is the number of outputs and Nu is the number of inputs.
The values of K can be either real-valued or complex-valued.
DisplayFormat — Specifies how the numerator and denominator polynomials are factorized for display
'roots' (default) | 'frequency' | 'time constant'
Specifies how the numerator and denominator polynomials are factorized for display, specified as one of the following:
• 'roots' — Display factors in terms of the location of the polynomial roots. 'roots' is the default value of DisplayFormat.
• 'frequency' — Display factors in terms of root natural frequencies ω[0] and damping ratios ζ.
The 'frequency' display format is not available for discrete-time models with Variable value 'z^-1' or 'q^-1'.
• 'time constant' — Display factors in terms of root time constants τ and damping ratios ζ.
The 'time constant' display format is not available for discrete-time models with Variable value 'z^-1' or 'q^-1'.
For continuous-time models, the following table shows how the polynomial factors are arranged in each display format.
DisplayName First-Order Factor (Real Root $R$) Second-Order Factor (Complex Root pair $R=a±jb$)
'roots' $\left(s-R\right)$ $\left({s}^{2}-\alpha s+\beta \right),$ where $\alpha =2a,\text{}\beta ={a}^{2}+{b}^{2}$
'frequency' $\left(1-\frac{s}{{\omega }_{0}}\right),$ where $ $1-2\zeta \left(\frac{s}{{\omega }_{0}}\right)+{\left(\frac{s}{{\omega }_{0}}\right)}^{2},$ where ${\omega }_{0}{}^{2}={a}^{2}+{b}^
{\omega }_{0}=R$ {2},\text{}\zeta =\frac{a}{{\omega }_{0}}$
'time $\left(1-\tau s\right),$ where $\tau =\frac{1}{R}$ $1-2\zeta \left(\tau s\right)+{\left(\tau s\right)}^{2},$ where $\tau =\frac{1}{{\omega }_{0}},\text{}\zeta =a\tau$
For discrete-time models, the polynomial factors are arranged similar to the continuous-time models, with the following variable substitutions:
$s\to w=\frac{z-1}{{T}_{s}};\text{ }R\to \frac{R-1}{{T}_{s}},$
where T[s] is the sample time. In discrete-time, τ and ω[0] closely match the time constant and natural frequency of the equivalent continuous-time root, provided that the following condition is
fulfilled: $|z-1|<<{T}_{s}\left({\omega }_{0}<<\frac{\pi }{{T}_{s}}=\text{Nyquist frequency}\right)$.
Variable — Zero-pole-gain model display variable
's' (default) | 'z' | 'p' | 'q' | 'z^-1' | 'q^-1'
Zero-pole-gain model display variable, specified as one of the following:
• 's' — Default for continuous-time models
• 'z' — Default for discrete-time models
• 'p' — Equivalent to 's'
• 'q' — Equivalent to 'z'
• 'z^-1' — Inverse of 'z'
• 'q^-1' — Equivalent to 'z^-1'
IODelay — Transport delay
0 (default) | scalar | Ny-by-Nu array
Transport delay, specified as one of the following:
• Scalar — Specify the transport delay for a SISO system or the same transport delay for all input/output pairs of a MIMO system.
• Ny-by-Nu array — Specify separate transport delays for each input/output pair of a MIMO system. Here, Ny is the number of outputs and Nu is the number of inputs.
For continuous-time systems, specify transport delays in the time unit specified by the TimeUnit property. For discrete-time systems, specify transport delays in integer multiples of the sample time,
Ts. For more information on time delay, see Time Delays in Linear Systems.
InputDelay — Input delay
0 (default) | scalar | Nu-by-1 vector
Input delay for each input channel, specified as one of the following:
• Scalar — Specify the input delay for a SISO system or the same delay for all inputs of a multi-input system.
• Nu-by-1 vector — Specify separate input delays for input of a multi-input system, where Nu is the number of inputs.
For continuous-time systems, specify input delays in the time unit specified by the TimeUnit property. For discrete-time systems, specify input delays in integer multiples of the sample time, Ts.
For more information, see Time Delays in Linear Systems.
OutputDelay — Output delay
0 (default) | scalar | Ny-by-1 vector
Output delay for each output channel, specified as one of the following:
• Scalar — Specify the output delay for a SISO system or the same delay for all outputs of a multi-output system.
• Ny-by-1 vector — Specify separate output delays for output of a multi-output system, where Ny is the number of outputs.
For continuous-time systems, specify output delays in the time unit specified by the TimeUnit property. For discrete-time systems, specify output delays in integer multiples of the sample time, Ts.
For more information, see Time Delays in Linear Systems.
Ts — Sample time
0 (default) | positive scalar | -1
Sample time, specified as:
• 0 for continuous-time systems.
• A positive scalar representing the sampling period of a discrete-time system. Specify Ts in the time unit specified by the TimeUnit property.
• -1 for a discrete-time system with an unspecified sample time.
Changing Ts does not discretize or resample the model. To convert between continuous-time and discrete-time representations, use c2d and d2c. To change the sample time of a discrete-time system, use
TimeUnit — Time variable units
'seconds' (default) | 'nanoseconds' | 'microseconds' | 'milliseconds' | 'minutes' | 'hours' | 'days' | 'weeks' | 'months' | 'years' | ...
Time variable units, specified as one of the following:
• 'nanoseconds'
• 'microseconds'
• 'milliseconds'
• 'seconds'
• 'minutes'
• 'hours'
• 'days'
• 'weeks'
• 'months'
• 'years'
Changing TimeUnit has no effect on other properties, but changes the overall system behavior. Use chgTimeUnit to convert between time units without modifying system behavior.
InputName — Input channel names
'' (default) | character vector | cell array of character vectors
Input channel names, specified as one of the following:
• A character vector, for single-input models.
• A cell array of character vectors, for multi-input models.
• '', no names specified, for any input channels.
Alternatively, you can assign input names for multi-input models using automatic vector expansion. For example, if sys is a two-input model, enter the following.
sys.InputName = 'controls';
The input names automatically expand to {'controls(1)';'controls(2)'}.
You can use the shorthand notation u to refer to the InputName property. For example, sys.u is equivalent to sys.InputName.
Use InputName to:
• Identify channels on model display and plots.
• Extract subsystems of MIMO systems.
• Specify connection points when interconnecting models.
InputUnit — Input channel units
'' (default) | character vector | cell array of character vectors
Input channel units, specified as one of the following:
• A character vector, for single-input models.
• A cell array of character vectors, for multi-input models.
• '', no units specified, for any input channels.
Use InputUnit to specify input signal units. InputUnit has no effect on system behavior.
InputGroup — Input channel groups
Input channel groups, specified as a structure. Use InputGroup to assign the input channels of MIMO systems into groups and refer to each group by name. The field names of InputGroup are the group
names and the field values are the input channels of each group. For example, enter the following to create input groups named controls and noise that include input channels 1 and 2, and 3 and 5,
sys.InputGroup.controls = [1 2];
sys.InputGroup.noise = [3 5];
You can then extract the subsystem from the controls inputs to all outputs using the following.
By default, InputGroup is a structure with no fields.
OutputName — Output channel names
'' (default) | character vector | cell array of character vectors
Output channel names, specified as one of the following:
• A character vector, for single-output models.
• A cell array of character vectors, for multi-output models.
• '', no names specified, for any output channels.
Alternatively, you can assign output names for multi-output models using automatic vector expansion. For example, if sys is a two-output model, enter the following.
sys.OutputName = 'measurements';
The output names automatically expand to {'measurements(1)';'measurements(2)'}.
You can also use the shorthand notation y to refer to the OutputName property. For example, sys.y is equivalent to sys.OutputName.
Use OutputName to:
• Identify channels on model display and plots.
• Extract subsystems of MIMO systems.
• Specify connection points when interconnecting models.
OutputUnit — Output channel units
'' (default) | character vector | cell array of character vectors
Output channel units, specified as one of the following:
• A character vector, for single-output models.
• A cell array of character vectors, for multi-output models.
• '', no units specified, for any output channels.
Use OutputUnit to specify output signal units. OutputUnit has no effect on system behavior.
OutputGroup — Output channel groups
Output channel groups, specified as a structure. Use OutputGroup to assign the output channels of MIMO systems into groups and refer to each group by name. The field names of OutputGroup are the
group names and the field values are the output channels of each group. For example, create output groups named temperature and measurement that include output channels 1, and 3 and 5, respectively.
sys.OutputGroup.temperature = [1];
sys.OutputGroup.measurement = [3 5];
You can then extract the subsystem from all inputs to the measurement outputs using the following.
By default, OutputGroup is a structure with no fields.
Name — System name
'' (default) | character vector
System name, specified as a character vector. For example, 'system_1'.
Notes — User-specified text
{} (default) | character vector | cell array of character vectors
User-specified text that you want to associate with the system, specified as a character vector or cell array of character vectors. For example, 'System is MIMO'.
UserData — User-specified data
[] (default) | any MATLAB^® data type
User-specified data that you want to associate with the system, specified as any MATLAB data type.
SamplingGrid — Sampling grid for model arrays
structure array
Sampling grid for model arrays, specified as a structure array.
Use SamplingGrid to track the variable values associated with each model in a model array, including identified linear time-invariant (IDLTI) model arrays.
Set the field names of the structure to the names of the sampling variables. Set the field values to the sampled variable values associated with each model in the array. All sampling variables must
be numeric scalars, and all arrays of sampled values must match the dimensions of the model array.
For example, you can create an 11-by-1 array of linear models, sysarr, by taking snapshots of a linear time-varying system at times t = 0:10. The following code stores the time samples with the
linear models.
sysarr.SamplingGrid = struct('time',0:10)
Similarly, you can create a 6-by-9 model array, M, by independently sampling two variables, zeta and w. The following code maps the (zeta,w) values to M.
[zeta,w] = ndgrid(<6 values of zeta>,<9 values of w>)
M.SamplingGrid = struct('zeta',zeta,'w',w)
When you display M, each entry in the array includes the corresponding zeta and w values.
M(:,:,1,1) [zeta=0.3, w=5] =
s^2 + 3 s + 25
M(:,:,2,1) [zeta=0.35, w=5] =
s^2 + 3.5 s + 25
For model arrays generated by linearizing a Simulink^® model at multiple parameter values or operating points, the software populates SamplingGrid automatically with the variable values that
correspond to each entry in the array. For instance, the Simulink Control Design™ commands linearize (Simulink Control Design) and slLinearizer (Simulink Control Design) populate SamplingGrid
By default, SamplingGrid is a structure with no fields.
Object Functions
The following lists contain a representative subset of the functions you can use with zpk models. In general, any function applicable to Dynamic System Models is applicable to a zpk object.
Linear Analysis
step Step response of dynamic system
impulse Impulse response plot of dynamic system; impulse response data
lsim Compute time response simulation data of dynamic system to arbitrary inputs
bode Bode frequency response of dynamic system
nyquist Nyquist response of dynamic system
nichols Nichols response of dynamic system
bandwidth Frequency response bandwidth
Stability Analysis
pole Poles of dynamic system
zero Zeros and gain of SISO dynamic system
pzplot Plot pole-zero map of dynamic system
margin Gain margin, phase margin, and crossover frequencies
Model Transformation
tf Transfer function model
ss State-space model
c2d Convert model from continuous to discrete time
d2c Convert model from discrete to continuous time
d2d Resample discrete-time model
Model Interconnection
feedback Feedback connection of multiple models
connect Block diagram interconnections of dynamic systems
series Series connection of two models
parallel Parallel connection of two models
Controller Design
pidtune PID tuning algorithm for linear plant model
rlocus Root locus of dynamic system
lqr Linear-Quadratic Regulator (LQR) design
lqg Linear-Quadratic-Gaussian (LQG) design
lqi Linear-Quadratic-Integral control
kalman Design Kalman filter for state estimation
Continuous-Time SISO Zero-Pole-Gain Model
For this example, consider the following continuous-time SISO zero-pole-gain model:
Specify the zeros, poles and gain, and create the SISO zero-pole-gain model.
zeros = 0;
poles = [1-1i 1+1i 2];
gain = -2;
sys = zpk(zeros,poles,gain)
sys =
-2 s
(s-2) (s^2 - 2s + 2)
Continuous-time zero/pole/gain model.
Discrete-Time SISO Zero-Pole-Gain Model
For this example, consider the following SISO discrete-time zero-pole-gain model with 0.1s sample time:
Specify the zeros, poles, gains and the sample time, and create the discrete-time SISO zero-pole-gain model.
zeros = [1 2 3];
poles = [6 5 4];
gain = 7;
ts = 0.1;
sys = zpk(zeros,poles,gain,ts)
sys =
7 (z-1) (z-2) (z-3)
(z-6) (z-5) (z-4)
Sample time: 0.1 seconds
Discrete-time zero/pole/gain model.
Concatenate SISO Zero-Pole-Gain Models into a MIMO Zero-Pole-Gain Model
In this example, you create a MIMO zero-pole-gain model by concatenating SISO zero-pole-gain models. Consider the following single-input, two-output continuous-time zero-pole-gain model:
$sys\left(s\right)=\left[\begin{array}{c}\frac{\left(s-1\right)}{\left(s+1\right)}\\ \frac{\left(s+2\right)}{\left(s+2+i\right)\left(s+2-i\right)}\end{array}\right].$
Specify the MIMO zero-pole-gain model by concatenating the SISO entries.
zeros1 = 1;
poles1 = -1;
gain = 1;
sys1 = zpk(zeros1,poles1,gain)
sys1 =
Continuous-time zero/pole/gain model.
zeros2 = -2;
poles2 = [-2+1i -2-1i];
sys2 = zpk(zeros2,poles2,gain)
sys2 =
(s^2 + 4s + 5)
Continuous-time zero/pole/gain model.
sys =
From input to output...
1: -----
2: --------------
(s^2 + 4s + 5)
Continuous-time zero/pole/gain model.
Discrete-Time MIMO Zero-Pole-Gain Model
Create a zero-pole-gain model for the discrete-time, multi-input, multi-output model:
$sys\left(z\right)=\left[\begin{array}{cc}\frac{1}{\left(z+0.3\right)}& \frac{z}{\left(z+0.3\right)}\\ \frac{-\left(z-2\right)}{\left(z+0.3\right)}& \frac{3}{\left(z+0.3\right)}\end{array}\right]$
with sample time ts = 0.2 seconds.
Specify the zeros and poles as cell arrays and the gains as an array.
zeros = {[] 0;2 []};
poles = {-0.3 -0.3;-0.3 -0.3};
gain = [1 1;-1 3];
ts = 0.2;
Create the discrete-time MIMO zero-pole-gain model.
sys = zpk(zeros,poles,gain,ts)
sys =
From input 1 to output...
1: -------
- (z-2)
2: -------
From input 2 to output...
1: -------
2: -------
Sample time: 0.2 seconds
Discrete-time zero/pole/gain model.
Specify Input Names for Zero-Pole-Gain Model
Specify the zeros, poles and gain along with the sample time and create the zero-pole-gain model, specifying the state and input names using name-value pairs.
zeros = 4;
poles = [-1+2i -1-2i];
gain = 3;
ts = 0.05;
sys = zpk(zeros,poles,gain,ts,'InputName','Force')
sys =
From input "Force" to output:
3 (z-4)
(z^2 + 2z + 5)
Sample time: 0.05 seconds
Discrete-time zero/pole/gain model.
The number of input names must be consistent with the number of zeros.
Naming the inputs and outputs can be useful when dealing with response plots for MIMO systems.
Notice the input name Force in the title of the step response plot.
Continuous-Time Zero-Pole-Gain Model Using Rational Expression
For this example, create a continuous-time zero-pole-gain model using rational expressions. Using a rational expression can sometimes be easier and more intuitive than specifying poles and zeros.
Consider the following system:
To create the transfer function model, first specify s as a zpk object.
s =
Continuous-time zero/pole/gain model.
Create the zero-pole-gain model using s in the rational expression.
sys =
(s^2 + 2s + 10)
Continuous-time zero/pole/gain model.
Discrete-Time Zero-Pole-Gain Model Using Rational Expression
For this example, create a discrete-time zero-pole-gain model using a rational expression. Using a rational expression can sometimes be easier and more intuitive than specifying poles and zeros.
Consider the following system:
To create the zero-pole-gain model, first specify z as a zpk object and the sample time ts.
ts = 0.1;
z = zpk('z',ts)
z =
Sample time: 0.1 seconds
Discrete-time zero/pole/gain model.
Create the zero-pole-gain model using z in the rational expression.
sys = (z - 1) / (z^2 - 1.85*z + 0.9)
sys =
(z^2 - 1.85z + 0.9)
Sample time: 0.1 seconds
Discrete-time zero/pole/gain model.
Zero-Pole-Gain Model with Inherited Properties
For this example, create a zero-pole-gain model with properties inherited from another zero-pole-gain model. Consider the following two zero-pole-gain models:
For this example, create sys1 with the TimeUnit and InputDelay property set to 'minutes'.
zero1 = 0;
pole1 = [0;-8];
gain1 = 2;
sys1 = zpk(zero1,pole1,gain1,'TimeUnit','minutes','InputUnit','minutes')
sys1 =
2 s
s (s+8)
Continuous-time zero/pole/gain model.
propValues1 = [sys1.TimeUnit,sys1.InputUnit]
propValues1 = 1x2 cell
{'minutes'} {'minutes'}
Create the second zero-pole-gain model with properties inherited from sys1.
zero = 1;
pole = [-3,5];
gain2 = 0.8;
sys2 = zpk(zero,pole,gain2,sys1)
sys2 =
0.8 (s-1)
(s+3) (s-5)
Continuous-time zero/pole/gain model.
propValues2 = [sys2.TimeUnit,sys2.InputUnit]
propValues2 = 1x2 cell
{'minutes'} {'minutes'}
Observe that the zero-pole-gain model sys2 has that same properties as sys1.
Static Gain MIMO Zero-Pole-Gain Model
Consider the following two-input, two-output static gain matrix m:
$m=\left[\begin{array}{cc}2& 4\\ 3& 5\end{array}\right]$
Specify the gain matrix and create the static gain zero-pole-gain model.
m = [2,4;...
sys1 = zpk(m)
sys1 =
From input 1 to output...
1: 2
2: 3
From input 2 to output...
1: 4
2: 5
Static gain.
You can use static gain zero-pole-gain model sys1 obtained above to cascade it with another zero-pole-gain model.
sys2 =
(s+1) (s-7)
Continuous-time zero/pole/gain model.
sys =
From input 1 to output...
2 s
1: -----------
(s+1) (s-7)
3 s
2: -----------
(s+1) (s-7)
From input 2 to output...
4 s
1: -----------
(s+1) (s-7)
5 s
2: -----------
(s+1) (s-7)
Continuous-time zero/pole/gain model.
Convert State-Space Model to Zero-Pole-Gain Model
For this example, compute the zero-pole-gain model of the following state-space model:
$A=\left[\begin{array}{cc}-2& -1\\ 1& -2\end{array}\right],\phantom{\rule{1em}{0ex}}B=\left[\begin{array}{cc}1& 1\\ 2& -1\end{array}\right],\phantom{\rule{1em}{0ex}}C=\left[\begin{array}{cc}1& 0\end
{array}\right],\phantom{\rule{1em}{0ex}}D=\left[\phantom{\rule{0.1em}{0ex}}\begin{array}{cc}0& 1\end{array}\right].$
Create the state-space model using the state-space matrices.
A = [-2 -1;1 -2];
B = [1 1;2 -1];
C = [1 0];
D = [0 1];
ltiSys = ss(A,B,C,D);
Convert the state-space model ltiSys to a zero-pole-gain model.
sys =
From input 1 to output:
(s^2 + 4s + 5)
From input 2 to output:
(s^2 + 5s + 8)
(s^2 + 4s + 5)
Continuous-time zero/pole/gain model.
Array of Zero-Pole-Gain Models
You can use a for loop to specify an array of zero-pole-gain models.
First, pre-allocate the zero-pole-gain model array with zeros.
The first two indices represent the number of outputs and inputs for the models, while the third index is the number of models in the array.
Create the zero-pole-gain model array using a rational expression in the for loop.
s = zpk('s');
for k = 1:3
sys(:,:,k) = k/(s^2+s+k);
sys(:,:,1,1) =
(s^2 + s + 1)
sys(:,:,2,1) =
(s^2 + s + 2)
sys(:,:,3,1) =
(s^2 + s + 3)
3x1 array of continuous-time zero/pole/gain models.
Extract Zero-Pole-Gain Models from Identified Model
For this example, extract the measured and noise components of an identified polynomial model into two separate zero-pole-gain models.
Load the Box-Jenkins polynomial model ltiSys in identifiedModel.mat.
ltiSys is an identified discrete-time model of the form: $y\left(t\right)=\frac{B}{F}u\left(t\right)+\frac{C}{D}e\left(t\right)$, where $\frac{B}{F}$ represents the measured component and $\frac{C}
{D}$ the noise component.
Extract the measured and noise components as zero-pole-gain models.
sysMeas = zpk(ltiSys,'measured')
sysMeas =
From input "u1" to output "y1":
-0.14256 z^-1 (1-1.374z^-1)
z^(-2) * -----------------------------
(1-0.8789z^-1) (1-0.6958z^-1)
Sample time: 0.04 seconds
Discrete-time zero/pole/gain model.
sysNoise = zpk(ltiSys,'noise')
sysNoise =
From input "v@y1" to output "y1":
0.045563 (1+0.7245z^-1)
(1-0.9658z^-1) (1 - 0.0602z^-1 + 0.2018z^-2)
Input groups:
Name Channels
Noise 1
Sample time: 0.04 seconds
Discrete-time zero/pole/gain model.
The measured component can serve as a plant model, while the noise component can be used as a disturbance model for control system design.
Zero-Pole-Gain Model with Input and Output Delay
For this example, create a SISO zero-pole-gain model with an input delay of 0.5 seconds and an output delay of 2.5 seconds.
zeros = 5;
poles = [7+1i 7-1i -3];
gains = 1;
sys = zpk(zeros,poles,gains,'InputDelay',0.5,'OutputDelay',2.5)
sys =
exp(-3*s) * ----------------------
(s+3) (s^2 - 14s + 50)
Continuous-time zero/pole/gain model.
You can also use the get command to display all the properties of a MATLAB object.
Z: {[5]}
P: {[3x1 double]}
K: 1
DisplayFormat: 'roots'
Variable: 's'
IODelay: 0
InputDelay: 0.5000
OutputDelay: 2.5000
InputName: {''}
InputUnit: {''}
InputGroup: [1x1 struct]
OutputName: {''}
OutputUnit: {''}
OutputGroup: [1x1 struct]
Notes: [0x1 string]
UserData: []
Name: ''
Ts: 0
TimeUnit: 'seconds'
SamplingGrid: [1x1 struct]
For more information on specifying time delay for an LTI model, see Specifying Time Delays.
Control Design Using Zero-Pole-Gain Models
For this example, design a 2-DOF PID controller with a target bandwidth of 0.75 rad/s for a system represented by the following zero-pole-gain model:
Create a zero-pole-gain model object sys using the zpk command.
zeros = [];
poles = [-0.25+0.2i;-0.25-0.2i];
gain = 1;
sys = zpk(zeros,poles,gain)
sys =
(s^2 + 0.5s + 0.1025)
Continuous-time zero/pole/gain model.
Using the target bandwidth, use pidtune to generate a 2-DOF controller.
wc = 0.75;
C2 = pidtune(sys,'PID2',wc)
C2 =
u = Kp (b*r-y) + Ki --- (r-y) + Kd*s (c*r-y)
with Kp = 0.512, Ki = 0.0975, Kd = 0.574, b = 0.38, c = 0
Continuous-time 2-DOF PID controller in parallel form.
Using the type 'PID2' causes pidtune to generate a 2-DOF controller, represented as a pid2 object. The display confirms this result. The display also shows that pidtune tunes all controller
coefficients, including the setpoint weights b and c, to balance performance and robustness.
For interactive PID tuning in the Live Editor, see the Tune PID Controller Live Editor task. This task lets you interactively design a PID controller and automatically generates MATLAB code for your
live script.
For interactive PID tuning in a standalone app, use PID Tuner. See PID Controller Design for Fast Reference Tracking for an example of designing a controller using the app.
zpk uses the MATLAB function roots to convert transfer functions and the functions zero and pole to convert state-space models.
Version History
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Define and Modify Variable Data Types
When you create variables in a MATLAB Function block, you can use the Type property to set the data type. Variables can inherit their data types, or be set to built-in, fixed-point, or enumerated
data types. Variables can also be nonvirtual buses. By default, MATLAB Function block variables inherit their data type.
For more information on creating variables, see Create and Define MATLAB Function Block Variables.
Specify Variable Data Types
You can specify the data types by using the Symbols pane and Property Inspector (since R2022a), or the Model Explorer.
To specify the data type using the Symbols pane and Property Inspector:
1. Double-click the MATLAB Function block to open the MATLAB Function Block Editor.
2. In the Function tab, click Edit Data.
3. In the Symbols pane, select the variable.
4. In the Property Inspector, in the Properties tab, select the data type from the Type property.
To specify the data type of a variable by using the Model Explorer:
1. Open the Model Explorer. In the Modeling tab, in the Design section, click Model Explorer.
2. In the Model Hierarchy pane select the MATLAB Function block.
3. Click the variable you want to modify.
4. Select the data type from the Type property.
In the Model Explorer, you can also filter the data type options. In the General tab, click the Show data type assistant button to display the Data Type Assistant parameters. Then, choose an option
from the Mode parameter. Based on the mode you select, specify the data type:
Mode What to Specify
The inherited data depends on the Scope property:
• If Scope is Input, the data type is inherited from the input signal on the designated port.
(default) • If Scope is Output, the data type is inherited from the output signal on the designated port.
• If Scope is Parameter, the data type is inherited from the associated parameter, which can be defined in the Simulink^® masked subsystem or the MATLAB^® workspace.
Built in Select from a list of built-in data types.
Fixed point Specify the fixed-point data properties.
Enumerated Enter the name of a Simulink.IntEnumType object that you define in the base workspace. See Code Generation for Enumerations.
In the Bus object field, enter the name of a Simulink.Bus object to define the properties of a MATLAB structure. You must define the bus object in the base workspace. See Create
Structures in MATLAB Function Blocks.
Bus Object Note
You can click the Edit button to create or modify Simulink.Bus objects by using the Simulink Type Editor.
Expression Enter an expression that evaluates to a data type.
Inheriting Data Types
MATLAB Function block variables can inherit their data types, including fixed point types, from their connected signals. To make a variable inherit a data type, set the Type property to Inherit: Same
as Simulink.
An argument can also inherit complexity from the signal connected to it. To inherit complexity, set the Complexity property to Inherited.
After you build the model, the CompiledType column of the Model Explorer gives the actual type inherited from Simulink. If the expected type matches the inferred type, inheritance is successful.
Built-In Data Types
In the Model Explorer, when you expand the Data Type Assistant and set Mode to Built in, you can set Type to these built-in data types. The built-in data types are:
Data Description
double 64-bit double-precision floating point
single 32-bit single-precision floating point
half A half-precision data type occupies 16 bits of memory, but its floating-point representation enables it to handle wider dynamic ranges than integer or fixed-point data types of the same size.
See The Half-Precision Data Type in Simulink (Fixed-Point Designer).
int64 64-bit signed integer
int32 32-bit signed integer
int16 16-bit signed integer
int8 8-bit signed integer
uint64 64-bit unsigned integer
uint32 32-bit unsigned integer
uint16 16-bit unsigned integer
uint8 8-bit unsigned integer
boolean Boolean
string String scalar
Fixed-Point Designer Data Type Properties
To represent variables as fixed-point numbers in MATLAB Function blocks, you must install Fixed-Point Designer™.
You can set the following fixed-point properties:
Select whether you want the fixed-point variable to be Signed or Unsigned. Signed variables can represent positive and negative quantities. Unsigned variables represents positive values only. The
default is Signed.
Word length
Specify the size, in bits, of the word that will hold the quantized integer. Large word sizes represent large quantities with greater precision than small word sizes. Word length can be any integer
between 0 and 65,535 bits. The default is 16.
Specify the method for scaling your fixed-point variable to avoid overflow conditions and minimize quantization issues. You can select these scaling modes:
Scaling Mode Description
The Data Type Assistant displays the Fraction Length parameter, which specifies the binary point location.
Binary point Binary points can be positive or negative integers. A positive integer moves the binary point left of the rightmost bit by that amount. For example, an entry of 2 sets the binary point
(default) in front of the second bit from the right. A negative integer moves the binary point further right of the rightmost bit by that amount, as in this example:
The default is 0.
The Data Type Assistant displays the Slope and Bias parameters:
Slope and • Slope can be any positive real number. The default is 1.0.
• Bias can be any real number. The default value is 0.0.
You can enter slope and bias as expressions that contain parameters defined in the MATLAB workspace.
Use binary-point scaling whenever possible to simplify the implementation of fixed-point numbers in generated code. Operations with fixed-point numbers that use binary-point scaling are performed
with simple bit shifts and eliminate the expensive code implementations required for separate slope and bias values.
Calculate Best-Precision Scaling
Have Simulink automatically calculate best-precision values for both Binary point and Slope and bias scaling, based on the Minimum and Maximum properties you specify.
To automatically calculate best precision scaling values:
1. Specify the Minimum or Maximum properties.
2. Click Calculate Best-Precision Scaling.
Simulink calculates the scaling values, then displays them in either the Fraction length, or the Slope and Bias fields.
The Minimum and Maximum properties do not apply to variables with the Scope property set to Constant or Parameter. The software cannot calculate best-precision scaling for these kinds of variables.
Fixed-point details
Displays information about the fixed-point variable that is defined in the Data Type Assistant:
• Minimum and Maximum show the same values that you specify in the Minimum and Maximum properties.
• Representable minimum, Representable maximum, and Precision show the minimum value, maximum value, and precision that the fixed-point variable can represent.
If the value of a field cannot be determined without first compiling the model, the Fixed-point details subpane shows the value as Unknown. The values displayed by the Fixed-point details subpane do
not automatically update if you change the values that define the fixed-point variable. To update the values shown in the Fixed-point details subpane, click Refresh Details.
Clicking Refresh Details does not modify the variable. It changes only the display. To apply the displayed values, click Apply or OK.
The Fixed-point details subpane indicates issues resulting from the fixed-point variable specification. For example, this figure shows two issues.
The row labeled Maximum indicates that the value specified by the Maximum property is not representable by the fixed-point variable. To correct the issue, make one of these modifications so the
fixed-point data type can represent the maximum value:
• Decrease the value in the Maximum property.
• Increase Word length.
• Decrease Fraction length.
The row labeled Minimum shows the message Cannot evaluate because evaluating the expression MySymbol, specified by the Minimum property, does not return a numeric value. When an expression does not
evaluate successfully, the Fixed-point details subpane shows the unevaluated expression (truncating to 10 characters as needed) in place of the unavailable value. To correct this issue, define
MySymbol in the base workspace to provide a numeric value.
If you click Refresh Details, the issue indicator and description are removed and the value of MySymbol appears in place of the unevaluated text.
Specify Data Types with Expressions
You can specify the types of MATLAB Function block variables as expressions by using the Model Explorer or the Property Inspector.
To use the Model Explorer, set the Mode property to Expression. In the Type property, replace <data type expression> with an expression that evaluates to a data type.
To use the Property Inspector, double-click the Type property, clear the contents, and enter an expression.
You can use the following expressions:
• Alias type from the MATLAB workspace, as described in Simulink.AliasType.
• fixdt function to create a Simulink.NumericType object describing a fixed-point or floating-point data type.
• type (Stateflow) operator, to base the type on previously defined data.
For example, suppose you want to designate the workspace variable myDataType as an alias for a single data type to use as an expression in the Type property of an MATLAB Function block input
variable. Create an instance of the Simulink.AliasType class and set its BaseType property by entering these commands:
myDataType = Simulink.AliasType;
myDataType.BaseType = "single";
In the Property Inspector, enter the data type alias name, myDataType, as the value in the Type property.
MATLAB Function blocks do not support code generation if one of the variables uses an alias type and is variable size. This limitation does not apply to input or output variables. For more
information on defining variable-size variables and generating code with them, see Declare Variable-Size MATLAB Function Block Variables and Code Generation for Variable-Size Arrays.
See Also
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Excel INDIRECT Function (Explained with Examples + Video)
Excel INDIRECT Function – Overview
The INDIRECT function in Excel can be used when you have the reference of a cell or a range as a text string and you want to get the values from those references.
In short – you can use the indirect formula to return the reference specified by the text string.
In this Excel tutorial, I will show you how to use the indirect function in Excel using some practical examples.
But before I get into the examples, let’s first have a look at its syntax.
INDIRECT FUNCTION Syntax
=INDIRECT(ref_text, [a1])
Input Arguments
• ref_text – A text string that contains the reference to a cell or a named range. This must be a valid cell reference, or else the function would return a #REF! error
• [a1] – A logical value that specifies what type of reference to use for ref text. This could either be TRUE (indicating A1 style reference) or FALSE (indicating R1C1-style reference). If omitted,
it is TRUE by default.
Additional Notes
• INDIRECT is a volatile function. This means that it recalculates whenever the excel workbook is open or whenever a calculation is triggered in the worksheet. This adds to the processing time and
slows down your workbook. While you can use the indirect formula with small datasets with little or no impact on the speed, you may see it making your workbook slower when using it with large
• The Reference Text (ref_text) could be:
□ A reference to a cell that in-turn contains a reference in A1-style or R1C1-style reference format.
□ A reference to a cell in double-quotes.
□ A named range that returns a reference
Examples of How to Use Indirect Function in Excel
Now let’s dive in and have a look at some examples on how to use the INDIRECT function in Excel.
Example 1: Use a Cell reference to Fetch the Value
It takes the cell reference as a text string as input and returns the value in that reference (as shown in the example below):
The formula in cell C1 is:
The above formula takes the cell reference A1 as the input argument (within double quotes as a text string) and returns the value in this cell, which is 123.
Now if you’re thinking, why don’t I simply use =A1 instead of using the INDIRECT function, you have a valid question.
Here is why…
When you use =A1 or =$A$1, it gives you the same result. But when you insert a row above the first row, you would notice that the cell references would automatically change to account for the new
You can also use the INDIRECT function when you want to lock the cell references in such a way that it does not change when you insert rows/columns in the worksheet.
Example 2: Use Cell Reference in a Cell to Fetch the Value
You can also use this function to fetch the value from a cell whose reference is stored in a cell itself.
In the above example, cell A1 has the value 123.
Cell C1 has the reference to the cell A1 (as a text string).
Now, when you use the INDIRECT function and use C1 as the argument (which in turn has a cell address as a text string in it), it would convert the value in cell A1 into a valid cell reference.
This, in turn, means that the function would refer to cell A1 and return the value in it.
Note that you don’t need to use double quotes here as the C1 has the cell reference stored in the text string format only.
Also, in case the text string in cell C1 is not a valid cell reference, the Indirect function would return the #REF! error.
Example 3: Creating a Reference Using Value in a Cell
You can also create a cell reference using a combination of the column alphabet and the row number.
For example, if cell C1 contains the number 2, and you use the formula =INDIRECT(“A”&C1) then it would refer to cell A2.
A practical application of this could be when you want to create dynamic reference to cells based on the value in some other cell.
In case the text string you use in the formula gives a reference that Excel doesn’t understand, it will return the ref error (#REF!).
Example 4: Calculate the SUM of a Range of Cells
You can also refer to a range of cells the same way you refer to a single cell using the INDIRECT function in Excel.
For example, =INDIRECT(“A1:A5”) would refer to the range A1:A5.
You can then use the SUM function to find the total or the LARGE/SMALL/MIN/MAX function to do other calculations.
Just like the SUM function, you can also use functions such as LARGE, MAX/MIN, COUNT, etc.
Example 5: Creating Reference to a Sheet Using the INDIRECT Function
The above examples covered how to refer a cell in the same worksheet. You can also use the INDIRECT formula to refer to a cell in some other worksheet or another workbook as well.
Here is something you need to know about referring to other sheets:
• Let’s say you have a worksheet with the name Sheet1, and within the sheet in the cell A1, you have the value 123. If you go to another sheet (let’s say Sheet2) and refer to cell A1 in Sheet1,
the formula would be: =Sheet1!A1
• If you have a worksheet that contains two or more than two words (with a space character in between), and you refer to cell A1 in this sheet from another sheet, the formula would be: =’Data Set’!
In case, of multiple words, Excel automatically inserts single quotation marks at the beginning and end of the Sheet name.
Now let’s see how to create an INDIRECT function to refer to a cell in another worksheet.
Suppose you have a sheet named Dataset and cell A1 in it has the value 123.
Now to refer to this cell from another worksheet, use the following formula:
=INDIRECT("'Data Set'!A1")
As you can see, the reference to the cell needs to contain the worksheet name as well.
If you have the name of the worksheet in a cell (let’s say A1), then you can use the following formula:
If you have the name of the worksheet in cell A1 and cell address in cell A2, then the formula would be:
Similarly, you can also modify the formula to refer to a cell in another workbook.
This could be useful when you trying to create a summary sheet that pulls the data from multiple different sheets.
Also remember, when using this formula to refer to another workbook, that workbook must be open.
Example 6: Referring to a Named Range Using INDIRECT Formula
If you have created a named range in Excel, you can refer to that named range using the INDIRECT function.
For example, suppose you have the marks for 5 students in three subjects as shown below:
In this example, let’s name the cells:
• B2:B6: Math
• C2:C6: Physics
• D2:D6: Chemistry
To name a range of cells, simply select the cells and go to the name box, enter the name and hit enter.
Now you can refer to these named ranges using the formula:
=INDIRECT("Named Range")
For example, if you want to know the average of the marks in Math, use the formula:
If you have the named range name in a cell (F2 in the example below has the name Math), you can use this directly in the formula.
The below example shows how to calculate the average using the named ranges.
Example 7: Creating a Dependent Drop Down List Using Excel INDIRECT Function
This is one excellent use of this function. You can easily create a dependent drop-down list using it (also called the conditional drop-down list).
For example, suppose you have a list of countries in a row and the name of cities for each country as shown below:
Now to create a dependent drop-down list, you need to create two named ranges, A2:A5 with the name US and B2:B5 with the name India.
Now select cell D2 and create a drop-down list for India and the US. This would be the first drop-down list where the user gets the option to select a country.
Now to create a dependent drop-down list:
• Select cell E2 (the cell in which you want to get the dependent drop-down list).
• Click the Data tab
• Click on Data validation.
• Select List as the Validation Criteria and use the following formula in the source field: =INDIRECT($D$2)
• Click OK.
Now, when you enter the US in cell D2, the drop-down in cell E2 will show the states in the US.
And when you enter India in cell D2, the drop-down in cell E2 will show the states in India.
So these are some examples to use the INDIRECT function in Excel. These examples would work on all the versions of Excel (Office 365, Excel 2019/2016/2013/2013)
I hope you found this tutorial useful.
Related Microsoft Excel Functions:
You May Also Like the Following Excel Tutorials:
Leave a Comment | {"url":"https://trumpexcel.com/excel-indirect-function/","timestamp":"2024-11-12T16:09:58Z","content_type":"text/html","content_length":"397808","record_id":"<urn:uuid:f92b07b9-63c5-4ff1-8803-7d157a70e4c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00793.warc.gz"} |
Connectivity constraints on the rule (Wolfram Stephen graphs)
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Connectivity constraints on the rule (Wolfram Stephen graphs)
Hi, I am analyzing chapter 3.2 in the book "A Class of Models with the Potential to Represent Fundamental Physics" by Wolfram Stephen and I am having trouble understanding left connectivity
constraint on rule. I would be grateful if someone could show this by example of some rule. Dominik.
1 Reply
Since no one else has responded, I will give this a try. I think left connectivity means nothing more nor less than that every element on the left-hand side is connected, either directly or
indirectly, by links or pathway to every other element on the left-hand side. So for examples: (a,b),(b,c) is connected because the two pairs share the element b in common, but (a,b),(c,d) is not
connected because the two pairs do not share any elements in common. (a,b),(c,d),(c,b) is connected because the third pair shares the element b in common with the first pair and the element c in
common with the second pair. However, (a,b),(c,d),(b,e) is not connected, because the two connected pairs (a,b),(b,e) are not connected to (c,d), because the second pair shares no elements in common
with the first or third pairs.
More helpful sections of Wolfram's book, A Project to Find the Fundamental Theory of Physics, include Sections 2.9 Connectedness and Section 3.19 Rules Involving Disconnected Pieces. Section 3.2 is a
bit dense if one simply starts there. Wolfram states (p. 92), "But for many purposes we will want to impose connectivity constraints on the rule." I assume here that Wolfram means "physics purposes."
It would be an odd physics rule that allowed, for instance, a particle in the Milky Way galaxy to interact, immediately and directly, with a particle in the Andromeda galaxy. Unless they are quantum
entangled somehow, in which case the quantum rules would somehow have to establish the connection. Or completely new laws of physics. | {"url":"https://community.wolfram.com/groups/-/m/t/3085408","timestamp":"2024-11-06T12:34:01Z","content_type":"text/html","content_length":"95128","record_id":"<urn:uuid:95632ce4-bc64-465c-aca7-fc518e1d77eb>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00185.warc.gz"} |
Optimal long term operation policies associated with the generated affectations – low regulation capacity reservoirs
Determining the optimal long-term operation policies in a system of dams has been subject of numerous investigations and publications in recent years. However, current jobs and papers do not consider
explicitly the effect caused by the risk of flooding downstream of the dams, so, this work focuses on the estimation of the risk associated downstream the dams and its relation to the long term
operating policies. The common practice analyzes the operation in the reservoirs considering as initial condition the NWL, but real operation condition differs from this level, for this reason is
viable to define operation policies that are adhered in to real conditions of operation of reservoirs, guaranteeing the optimum use of water to satisfy demands minimizing deficits and spills. For
this,^1,2 have implemented the proposed methodology in reservoirs such as the El Cuchillo and Cerro Prieto dams, analyzing the response of the reservoir operation, considering reservoir level as a
random variable, applying the probability that a flood associated with to an "X" return period is presented. As result of these analyzes, a great savings in the possible damages of affected areas are
reported, compared to the ones obtained from the conventional analysis. In particular, the optimal operation of the Grijalva River dams system has been researched by the Institute of Engineering
In regard, Dominguez et al.^ 3 raises the problem determining monthly operating policies that maximize an objective function considers the generation of energy on long term and seeks to prevent
spillages and deficit; Dominguez et al.^4 complements 1993 studies, with the definition of extraction policies for Angostura and Malpaso dams, according to the final storage of previous month in both
dams;^5 review the functioning and operation of dams on the Grijalva River; Dominguez et al.^6 adapts optimization model for extractions policies on a dams systems, considering the relative value of
"peak" energy, regarding "base" energy, incorporating minimum energy restrictions proposed by the Federal Electricity Commission (CFE); Dominguez et al.^7 creates new operating policies, considering
hydrological events in 2005. Dittmann et al.^7 proposes a dynamic long-term operation by evolutionary algorithms, showing the state of dynamic performance is superior to the rule of static operation.
Vigyan^8 takes the approach of three guide curve for operating Tawa dam, in India, that satisfy various demands and purposes of the reservoir, recommended policy features the irrigation rule curve to
distribute the deficit equitably much in advance. Pradhan et al.^9 optimize operation of the multipurpose Hirakud reservoir (India) using a genetic algorithm reflecting efficiently results, policy
developed shows a better water operation for irrigation, power generation and industrial purpose is optimized than policy released by the department of water resources, Government of Odisha, keeping
strict vigil on dead storage of the reservoir. Lund et al.^10 describe a deterministic optimization procedure for the mainstream of a storage system, as well as the development and application of
operation rules inferred solely from careful observation of information available. Through the application of a dynamic system as a tool for simulate the operation of a storage respect to the
management of floods,^11 have been reviewed operation policies to reduce the impact of the floods, considering the exit of the runoff through controlled spillways, as complement to the free
discharges of the system, in addition to simulate other alternatives as different storage levels at the beginning of the floods season. It is considered that the construction of the model is simple,
and this allows reducing efforts in aspects of programming.
Due different hydrological conditions and physical characteristics that prevail in most dams, as well as inappropriate operating conditions, spills and unnecessary extractions occur that prevent
compliance with demands, Tospornsampan et al.^12 presents optimization techniques for determine more effective operation policies. Heydari et al.^13 present an optimization model that allows the
combination of real, integer and binary variables, where the objective function is not used to maximize the economic efficiency of a system, but rather to simulate historical storage behavior, using
factors for penalization and prioritization of the objective function. Heydari et al.^14 have optimized operation of a system of five prey arranged in series, and one more in parallel, with the
combined purpose of energy generation and the fulfillment of water demands, where they propose to solve the generation problem of energy through the numerical solution of a matrix arrangement.
There are several criteria for determining the design floods used in spillways; however, any methodology for this purpose must consider three aspects.
a. Correlation peak - volume,
b. Reproduce similar floods to those observed in the record, and
c. Definition of different types of floods. In each case, the results may change depending on the characteristics of the reservoir where they are going to be simulated.
The proposed methodology considers these following steps
Records analysis
To ensure the records reliability, it is analyzed looking for capture errors, which may significantly affect the development of analysis.
Design floods
The procedure used allows estimating design flood from the analysis of daily average flows historically recorded. The maximum annual average flows with different durations are determined, for it, the
maximum daily average throughout for duration of one day corresponds to the annual average maximum daily flow. For maximum average flows for other durations, it proceeds to find, for each year
registration, the maximum average for n consecutive days depending on the length of years to be analyzed:
${\overline{Q}}_{M}{}_{n}=ma{x}_{i}\left(\frac{{\sum }_{i}^{i+n-1}{Q}_{k}}{n}\right)$ (1)
${\overline{Q}}_{M}{}_{n}$ Maximum average flow for n day
${Q}_{k}$ Daily average flow in k day
$n$ Duration in days
$i$ Counter day that begins the period of time n.
Thus, for each duration (1, 2, ..., $n$ ) it is obtained a sample of m values of maximum annual flows, which can be adjusted to a distribution function. Therefore with the adjusted distribution
functions (one for each duration) it is possible estimate the synthetic design flood for any return period $\left({T}_{r}\right)$ . To convert the synthetic floods to real for each return period, the
average flows associated to different durations are transformed into daily average flows through recursive equations
${q}_{\text{1}}\left({T}_{r}\right)={\overline{Q}}_{\text{1}}\left({T}_{r}\right)$ (2)
${q}_{k}\left({T}_{r}\right)=k {\overline{Q}}_{k}\left({T}_{r}\right)-\left(k-\text{1}\right){\overline{Q}}_{k-\text{1}}\left({T}_{r}\right)$ (3)
${\overline{Q}}_{k}\left({T}_{r}\right)$ Estimated average flow for duration of k days and a return period $\left({T}_{r}\right)$
${q}_{k}\left({T}_{r}\right)$ Daily average flow, m^3/s
Finally, the daily average flow ${q}_{k}\left({T}_{r}\right)$ , that has a decreasing trend, must be rearranged for representative historic hydrograph form, a simple way to make it is by alternating
blocks method, in which at the center is placed the individual flow for one day; forward is placed the individual flow for two days, back of the center places the three days flow, and so forward is
placed the four-day, the five days back (etcetera), to build the shape of the flood.
Flood majoration
The dimensioning of the design flood by the Dominguez methodology, is based on the use of daily average flows, which is highly recommended for reservoirs with great regulation capacity, however, in
the case of reservoirs with low regulation capacity, Dominguez implements that the peak of the flood must be affected or increased, by a defined factor of the average of the ratio of the
instantaneous maximum flows between the maximum daily averages flows of the maximum historical flood, that is to say:
${F}_{Qp}=\frac{{\sum }_{i=1}^{n}\frac{{Q}_{M}{}_{i}}{{Q}_{m}{}_{i}}}{n}$ (4)
${F}_{Qp}$ Majoration factor
${Q}_{M}{}_{i}$ Instantaneous maximum flow, m^3/s
${Q}_{m}{}_{i}$ Maximum daily averages flow, m^3/s
$n$ Duration in days
$i$ Counter day that begins the period of time $n$ .
With the obtained factor, the maximum average daily flow of the design flood is multiplied, trying to conserve the same total volume of the flood.
Optimal operation
The operation policy of the dam under study, is optimized by assigning different types of restrictions and penalties, either by spills or water deficit; in order to perform the long-term policies
analysis, the records are integrated into fortnightly intervals, optimizing the dispatch of the water demanded for a certain purpose (in this case electric power generation), dividing the volume in
such intervals throughout the year considering the flood and dry seasons. The groups are defined starting from dam’s capacity. With these groups, maximum expected benefits are determined on a
planning horizon of N stages with the definition of optimum operating policies for all possible initial states and dam’s water extraction, by stochastic dynamic programming. To perform optimization,
the maximum expected value of the total benefit generation is used as an objective function, imposing penalties for deficit or spillages in the dam (5).
${F}_{Obj}=Max E\left(GE-{C}_{1}DERR-{C}_{2}DEF\right)$ (5)
${F}_{Obj}$ Objective function
$E$ Expectation operator
$GE$ Energy generated
$DERR$ Spill
$DEF$ Deficit
${C}_{1}$ Spill penalty coefficient
${C}_{2}$ Deficit penalty coefficient
Once the optimal operation policy is obtained, a reservoir analytical simulation is performed in order to compare obtained levels respect historic ones.
Flood routing in dams
As result of optimized reservoir operation, the most frequent levels in the reservoir (middle elevations histogram) are obtained. Those elevations will become the initial conditions in the simulation
of design flood, generating a level analysis associated with their occurrence probability $\left({P}_{Elev}\right)$ . Design flood for 2, 5, 10, 20, 50, 100, 200, 500, 1 000, 2 000, 5 000 and 10 000
years return period are analyzed and for each initial elevation obtained from the optimal operation policy. In other hand, another scenario is simulated considering the NWL as initial condition in
the reservoir. From simulation results, discharge peak flows are associated with the exceedance probability $\left({P}_{Q}\right)$ . In order to estimate maximum discharge flow in closed digits (1
000, 2 000, etc.), outcomes are plotted describing the trend, associating them to theirs corresponding exceedance probability and return period.
Joint exceedance probability analysis
The joint exceedance probability associated to the maximum discharge rate is calculated by the total sum of the products of each probability of occurrence associated to the initial elevation ${P}_
{Elev}$ by the exceedance probability $\left({P}_{Q}\right)$ , using the formula:
${P}_{Ti}=\sum _{k=1}^{n}{P}_{Q}{}_{i}{P}_{Elev}{}_{k}$ (6)
${P}_{Ti}$ Joint exceedance probability for a ${Q}_{i}$ flow
${P}_{Q}{}_{i}$ Exceedance probability for a ${Q}_{i}$ flow, given an initial elevation k
${P}_{Elev}{}_{k}$ Probability of occurrence associated to initial elevation k.
From equation 5, there will be as many joint exceedance probabilities $\left({P}_{T}\right)$ as flows analyzed, which should be compared with results obtained considering as initial condition the
Hydraulic flooding through the river and inundation costs
Using selected discharge peak flows, hydraulic flooding is analyzed downstream of the dam, determining the affected areas by the flood. Defined areas are multiplied by a unit cost of affectation,
determining the cost associated with different probabilities of exceedance. Finally, costs - joint probabilities of exceedance curve define the expected risk, which is calculated by the area under
the curve. Expected risk is compared with the one obtained using the NWL as initial condition.
Analysis of the monthly average flows
In order to validate the design floods of the Cerro Prieto dam, it was performed a flow rates updating with the daily records provided by the National Water Commission (CONAGUA). For this purpose,
the contributions to the storage reservoir were conformed considering the own basin revenues as the sum of the increase (or decrease) in storage (DELTA V) + measured outputs (OUTPUTS) from 1963 to
2009, despising the years between 1986 and 1994 due to their incomplete data.
Update floods
A probabilistic analysis of annual maximum daily average values was performed by probability distribution functions, obtaining extrapolated values for different return periods. To perform adjustment
calculations for different probability distribution functions, the AX © (1996) program was used. The maximum daily average flows for durations from 1 to 15 days (Dominguez, 1981), were statistically
analyzed for different distribution functions, selecting the Double Gumbel function^15 as the best fit. For each duration, they were defined the maximum flow rates associated to different return
periods (2, 5, 10, 20, 50, 100, 200, 500, 1 000, 2 000, 5 000 and 10 000 years). From these results, the synthetic flows were transformed into real ones, determining flood’s shape by the alternating
blocks method, Dominguez et al.^16 in which at half of total time, the maximum value ${Q}_{1}$ is placed, forward the flow $Q2$ is placed, back flow $Q3$ , and so on.
Flood majoration
Considering the low regulation capacity of the reservoir, being more sensible to peak flows that volumes, it proceed to increase the design flood. For this purpose, the average of relation between
the daily instantaneous maximum flow rates and the average daily flows of maximum recorded flow from June 30 to July 20 2010 was determined, resulting a scale factor of 1.42 (Table 1). Based on this
factor and the peak time of the 2010 flood, the maximum daily mean flows extrapolated for the various return periods analyzed were adjusted (Figure 1). The peak day flow was affected by the average
factor obtained, keeping the volume of the original design flood.
Date Qmd Qmax Qmax/Qmd
dd/mm/yyyy m3/s m3/s
30/06/2010 34,89 61,67 1,77
1/7/2010 467,47 838,50 1,79
2/7/2010 288,44 727,47 2,52
3/7/2010 44,70 69,06 1,54
4/7/2010 35,88 45,15 1,26
5/7/2010 34,71 48,59 1,40
6/7/2010 85,47 147,13 1,72
7/7/2010 71,34 132,20 1,85
8/7/2010 33,91 59,04 1,74
9/7/2010 23,96 41,47 1,73
10/7/2010 29,40 32,22 1,10
11/7/2010 25,78 29,94 1,16
12/7/2010 21,28 23,34 1,10
13/07/2010 17,56 19,50 1,11
14/07/2010 14,88 16,60 1,12
15/07/2010 13,55 15,97 1,18
16/07/2010 11,79 13,24 1,12
17/07/2010 10,81 11,97 1,11
18/07/2010 10,17 11,97 1,18
19/07/2010 9,54 10,91 1,14
20/07/2010 8,64 10,51 1,22
Average 1,42
Table 1 Majoration factor. La Boca Dam
Optimal operation
Considering the average daily flows, as well as the characteristics of the reservoir presented in Table 2, the optimization of long-term operation policy was performed.^17–19 Runoff matrix for the
reservoir analytical simulation was built in to fortnightly intervals. To distribute the annual average volume in to intervals throughout the year (considering the flood and dry seasons), six monthly
groups were defined, which are conformed as showed in Table 3. Considering the useful capacity of the dam (39.00 hm3), the water volume has been distributed in 50 states by 0.7898 hm3 each one. The
absolute frequencies of runoff volumes, for each state and group of months, are defined in Table 3; the relative frequencies were obtained dividing the absolute frequencies between the total numbers
of years of the sample, this analysis has been considered as an approach to the probabilities associated with incomes of runoff matrix. Plotting the relative frequencies against the volume interval,
discontinuities appeared, so they were smoothed by distributing the volume maintaining the original shape form, forcing the sum to be equal to 1. From smoothed relative frequencies, it was developed
the reservoir optimal operation policy, by stochastic dynamic programming.^20 In Figure 2 the annual summary of levels obtained from the simulation of the optimal operation policy are presented, in
particular, the average biweekly levels in the reservoir that are compared with historical levels. As can be seen in Figure 2, the operation policy obtained from the simulation preserves the levels
above the historical average regime (increasing the generation), without reaching the maximum level of operation (NWL). Based on the levels obtained in the simulation, the relative frequencies
histogram of elevations for the months from August to November was elaborated (considered as the flood season). As a result, Figure 3 presents these relative frequencies for class intervals of 1.5 m
(the last interval to be considered is going to be the NWL, 448.50 m).
Basin area 266,00 km^2
Minimum annual runoff 20,08 hm^3
Average annual runoff 70,71 hm^3
Maximum annual runoff 213,24 hm^3
Design flow (Tr 10 000 years) – Spillway capacity 2 250,00 m^3/s
Silt level 425,74 m
NWL 448,05 m
MWL 449,20 m
Freeboard 0,84 m
Silt level capacity 1,62 hm^3
NWL capacity 39,49 hm^3
Regulation capacity (NWL – MWL) 3,17 hm^3
Total Capacity 42,66 hm^3
Area at NWL 465,75 ha
Area at MWL 498,36 ha
Table 2 General characteristics
Group Month
1 January + February + March + April
2 May + June + July
3 August
4 September
5 October
6 November + December
Flood routing
Flood routing simulation was analyzed considering different initial water levels in the reservoir, in order to obtain the maximum discharge flow magnitude. "Reservoir flood routing” Marengo et al.^21
program was used to analyze flood routing simulation. For this analysis two scenarios were considered:
a. Starting simulation at NWL (448.50 m),
b. Initial condition corresponding to the most frequently elevations obtained from the reservoir analytical simulation (Figure 3).
The income flows into the dam associated to different return periods are presented in Figure 1, (2, 5, 10, 20, 50, 100, 200, 500, 1 000, 2 000, 5 000 and 10 000 years).^22
Stage 1 - Initial Level 448.50 m (NWL)
From the flood routing simulation on Figure 4, the results for a 10 000 years return period are shown. Figure 4, represents that for floods up to 10 000 years return period, the highest elevation
obtained from analysis, 448,71 m, would be 0,49 m below MWL (449.20 m), so there is no danger for exceeding the maximum water level in the dam (MWL). The described analysis was performed in the same
way for each return period associated to different recurrence intervals. Figure 5 represents analysis results trending, return periods (years) are plotted in relation to the maximum reached elevation
in the reservoir.^23
Scenario 2 - Frequently average levels in the reservoir
The analysis was elaborated starting from several initial levels in the reservoir. For this purpose, the level histogram (Figure 3) obtained from the reservoir optimal operation policy was used,
resulting seven more frequent levels in the reservoir (Table 4), whose frequencies correspond to the Probabilities of occurrence to be analyzed. With initial levels defined in Table 4 and design
floods presented in Figure 3, routing through the reservoir is performed as described below.^24–26
Intervals @ 1,5 m Absolute Frequencies Relative Frequencies
438,0 439,5 15,00 0,125
439,5 441,0 3,00 0,025
441,0 442,5 16,00 0,130
442,5 444,0 19,00 0,160
444,0 445,5 13,00 0,110
445,5 447,0 17,00 0,140
447,0 448,5 37,00 0,310
Total 120 1
Table 4 Reservoir levels frequencies
Initial level 439.5 m
Routing trough the reservoir from initial elevation at 439,50 m, for return periods from 2 to 10 000 years were analyzed, determining its maximum discharge flow rate, maximum volume in the reservoir
and maximum elevation reached. The exceedance probabilities of maximum discharge flows considering values from 150 to 1 350 m3 /s, each 150 m3/s, were estimated by a linear interpolation between the
results using the reduced variable , as shown in Figure 6. The described procedure was applied the same way for the next initial levels: 441, 442.5, 444, 445.5, 447 and 448.5 m (NWL).^27–30
Overview results for different initial levels
Figure 7 shows the graphical analysis of the probability distributions of maximum discharge flows for the different initial levels studied
Joint exceedance probability analysis
On scenario number two, flood routing for different initial levels described are presented. The joint exceedance probability $\left({P}_{T}\right)$ was determined, with the exceedance probabilities
obtained for each maximum flow discharge rate and the probability of occurrence associated to each initial lift ${P}_{Elev}$ .^31 So, for each flow between 150 and 1 350 m3/s, it was performed the
sum of the product of the probability associated with each baseline $\left({P}_{Elev}\right)$ multiplied by the exceedance probability the corresponding flow $\left({P}_{Q/Elev}\right)$ , it means:
${P}_{Ti}={\sum }_{k=439,5}^{448,5}{P}_{Q}{}_{i/k}{P}_{Elev}{}_{k}$ (7)
Where ${P}_{Ti}$ is the joint exceedance probability associated to the maximum discharge flow rate ${Q}_{i}$ ; ${P}_{Elev}{}_{k}$ is the probability that the flood routing begins in $k$ elevation,
and ${P}_{Q}{}_{i/k}$ is the probability that the maximum discharge flow rate be greater than or equal to ${Q}_{i}$ given the initial elevation k. In Table 5, the joint exceedance probabilities
obtained for each analyzed flows are presented: From the analysis, the procedure presented in Table 5 was performed for flows between 150 and 1 350 m3/s, at each 150 m^3/s. Figure 8 presents the
trend for joint exceedance probability associated with each maximum discharge flow rate, compared with the one obtained considering at the NWL.^32–34
Initial level Flow (m3/s) 150
m Tr (years) PQ PELEV PQ PELEv
439.5 6.67 0.15 0.125 0.01875
441 6.67 0.15 0.025 0.00375
442.5 6.67 0.15 0.133333 0.02
444 6.25 0.16 0.158333 0.025333
445.5 6.06 0.165 0.108333 0.017875
447 5.88 0.17 0.141667 0.024083
448.5 5.26 0.19 0.308333 0.058583
PT 150 0.168375
Table 5 Joint exceedance probability
Hydraulic routing through the river
Considering maximum discharge flow, flood routing simulation through the river was performed. For this, the digital elevation model of the ASTER GDEM (2014) (ASTER Global Digital Elevation Model)
page was used, with a resolution of 20 meters in the vertical plane and 30 meters in the horizontal plan, analyzing the area between the spillway of La Boca Dam, and the reservoir of Marte R. Gomez
Dam, located downstream. This area was delimited using the Global Mapper ® platform (2014). The digital elevation model was exported to HEC-RAS ® program (2014) in order to carry out the simulation
of the flood along the river, considering the different flows analyzed, as well as the physiographic characteristics of the river. The results obtained from the HEC RAS simulation were incorporated
into the LAMINA ® program (2014), which determines the flooded area or surface from the water levels by each section of the river.^35 | {"url":"https://medcraveonline.com/OAJS/optimal-long-term-operation-policies-associated-with-the-generated-affectations--low-regulation-capacity-reservoirs.html","timestamp":"2024-11-11T04:55:58Z","content_type":"application/xhtml+xml","content_length":"229525","record_id":"<urn:uuid:f1c1c71e-8b84-4fea-8d49-69787a8e3266>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00403.warc.gz"} |
Brief summary of the NODE project
Being a language of nature, differential equations are ubiquitous in science and technology. Thus, solving them is a fundamental computational task, with renewed challenges due to the widespread
availability of HPC hardware. For applications, this task typically boils down to the numerical approximation of solutions. Most textbook algorithms focus on low order schemes, such as the popular
Runge–Kutta schemes, and a fixed single or double precision.
In the early days, many heuristics were invented to solve differential equations symbolically by hand. With the advent of computer algebra, systematic algorithms have been developed to compute closed
form solutions of differential equations, when possible. In theory, differential algebra even provides us with a complete elimination theory for non-linear equations. However, the complexity of these
methods is often prohibitive.
The NODE project aims at combining modern numerical and symbolic methods for solving differential equations. Our first main goal is to develop and implement new, more efficient, high-order numerical
schemes, together with efficient ways to control the error and certify the end-results. We expect this to be especially useful whenever traditional schemes become numerically unstable. We plan to
create a stand-alone open source HPC software library with a similar API as standard numerical libraries but with additional support for arbitrary precision and certification.
Our second main objective is to develop and implement differential counterparts of polynomial system solvers that are based on homotopy continuation. Such solvers benefit from more compact data
structures that avoid “intermediate expression swell”, a common evil in computer algebra. Therefore, they should be faster, both in theory and in practice. We will consider both numerical and
algebraic homotopies.
© 2022 Joris van der Hoeven
This webpage is part of the NODE project. Verbatim copying and distribution of it is permitted in any medium, provided this notice is preserved. For more information or questions, please contact
Joris van der Hoeven. | {"url":"http://magix.lix.polytechnique.fr/node/description.en.html","timestamp":"2024-11-04T15:24:10Z","content_type":"application/xhtml+xml","content_length":"5357","record_id":"<urn:uuid:d5cf5e6c-ee2d-491b-95ab-4e23681ccc3d>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00478.warc.gz"} |
What is a Mixed Stream Cash Flow? - Lunch Break Investing
What is a Mixed Stream Cash Flow?
What is a Mixed Stream Cash Flow?
Financial planning is a key part of any organization. Businesses need to see how their sales, production and cash flows will work out in the future. A key concept in financial planning is to examine
the cash flows linked to any investment decision. These refer to the movement of cash in and out of a business.
Mixed Stream cash flows have a combination of irregular inflows of cash. This is different from annuities which are simply outflows of cash. Mixed stream cash flows are slightly different to assess
as they have no specific pattern of cash receipt.
To understand cash flows, it is best to remember that they are the movement of cash into or out of a business after an action or an investment.
The movement of cash is directly due to the action or investment made. This is called a cash flow stream. These movements of cash are called mixed streams if they are going into the business. If the
business is making these payments to someone, they are called annuities.
Calculating the current value of the expected cash to come in can be done by applying some formulas usually called cash flow metrics.
The most common formulas are those of that help calculate what is known as Net Present Value (NPV), the profitability of an investment, how much returns it yields, and how many years it will take to
recover the cost of an investment.
These concepts are all used in evaluating cash flow streams to make investment decisions.
Defining Present Value of Cash Flows
The current worth (PV) of a future income stream is the sum of money or flow of cash expected in the future at a certain interest rate, which is called the rate of return.
It can be calculated by discounting these future income stream at the interest rate. The lesser the interest rate, the greater the worth of the expected income streams in the future.
Calculating the Value Of A Mixed Stream Cash Flow
The projected worth of a variable flow of income needs to be calculated for each individual stream and needs to be done on a table or preferably an Excel sheet.
To find out the projected worth of a variable or mixed stream cash flow we need to find out the interest factors for the future value expected. The Future Value Interest Factors (FVIF) are available
in interest factor tables.
People looking to calculate the future values can look up values against the expected interest rate and multiply the cash flow amount with the FVIF value for the given rate and the period of the cash
In terms of formula, the calculation of a future value of a variable income stream can be done in the following manner:
FV of a mixed stream cash flow = (I1 × FVIF[1)] + (!2 × FVIF[2] )+ (!3 × FVIF[3] +……. + In ×
The formula can be written as follows:
FV of a mixed stream cash flow = I1 × (1+i)^1 + I2 × (1+i)^2 + I3 × (1+i)^3+ …….. + I1 ×
I = Is the income expected for different periods (1,2,3,…n)
i = Interest rate
The projected interest factor is taken from the future value interest factors table (these are available online and in most financial management books).
The values can also be calculated for each year, provided the interest rate is given. The formula to calculate the FVIF is as below:
Mixed Stream Cash Flow Example
A business is evaluating an investment plan that will yield a variable flow of income over five periods as follows:
The business will get a rate of 7% on its invested capital. To find out how much the company will get at the after 5 years we can look up the value of the FVIF and then work out the variable stream:
in the working shown in the table, the FV of the variable stream cash flow is $130,110. This tells us that means that by year 5, the business will have earned total cash of $130,110 from this
investment plan.
Each cash flow carries its future value for the expected interest rate.
The total of these projected values of yearly cash inflow adds up to the total FV of mixed stream cash flow that the business will earn from its investment.
The result from using an Excel spreadsheet will be the same as the tabular calculation.
FV Interest Factors (FVIF) Table
The FVIF table is key in assessing any mixed cash flow scenario as it is useful to find the future value of a variable stream of income.
We can work out this table by using the FVIF formula discussed earlier, we can find out the future value interest factors just by using the calculation.
The formula is pasted into an excel sheet and dragging it so that you get the FVIF value.
Hope this explains mixed stream cash flow for you. You might be also interested in What is Base Case Cash Flow? and Cash Flow Adequacy Ratio.
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Understanding String Theory
Cosmology is the study of the universe's birth and evolution. The Standard Model of Cosmology, a widely accepted modern theory, states that some 15 billion years ago the universe emerged from the Big
Bang, an enormously energetic singular event that spewed forth all space and all matter. The universe's temperature at 10-43 seconds after the Big Bang, the so-called Planck time, is estimated to
have been 1032K, or some 10 trillion 10 trillion times hotter than the sun's interior. (1) In the first few picoseconds after the Big Bang, the universe expanded and cooled. About a
hundredth-thousandth of a second after the Big Bang, it was cool enough (10 trillion K) to produce protons and neutrons. About 300,000 years after Big Bang, electrically neutral atoms formed. A
billion years later, 100 billion galaxies and 100 billion stars (like our sun) were formed in each galaxy, and ultimately planets began to emerge.
Modern theories of creation are built upon quantum theory and Einstein's theory of gravity. The question is what happened before the Big Bang? Einstein's equations break down at the enormously small
distances and large energies found at the universe's origin. At distances 10-33cm, quantum effects take over from Einstein's theory. For questions involving the beginning of time, one must invoke the
ten-dimensional theory. The Big Bang probably originated in the breakdown of the original ten-dimensional universe into a four- and six-dimensional universe. Therefore, the history of the Big Bang
represents the breakup of previously unified symmetries, and the split universe was no longer symmetrical. Six dimensions have curled up.
Quantum physics abolishes time close to the Big Bang. How did the universe come into existence? Why does time vanish in the black hole? Did time exist before the universe came into being? These
questions and realities point to the existence of a Creator.
Unfortunately, quantum theory and Einstein's theory of gravity are mutually incompatible. In this new millennium, superstring theory, or simply string theory, resolves this tension. Three particle
theorists (Yoichiro Nambu, Leonard Susskind, and Holger Nielsen) independently realized that the dual theories developed in 1968 to describe the particle spectrum also describe the quantum mechanics
of an oscillating string. This marks the official birth of string theory in 1970, according to which the marriage of the laws of the large and the small is not only happy but also inevitable. Brian
Greene writes in his The Elegant Universe: String theory has the inherent capability to show that all of the astonishing happenings in the universe”from the frenzied dance of subatomic quarks
(components of protons or neutrons) to the stately dance of orbiting binary stars, from the primordial fireball of the big bang to the majestic whirl of celestial galaxies”are reflections of one
grand physical principle, one master equation.
Fundamental forces
During the past hundred years, physicists have proven the existence of four fundamental forces in nature: Gravitational force, electromagnetic force, the weak force, and the strong force. Gravity,
the most familiar force, keeps Earth revolving around the sun and our feet planted firmly on the ground. Electromagnetic force, the next most familiar force, is the driving force for such things as
lights, TVs, computers, and telephones.
The strong nuclear and weak nuclear forces are less familiar, because they operate in the atom's nucleus. The strong force (mediated by gluons) keeps quarks glued together inside protons and
neutrons, and keeps protons and neutrons tightly crammed together inside atomic nuclei. The weak force (mediated by W and Z particles) determines the radioactive decay of such radioactive materials
as uranium, plutonium, and tritium.
Gravitational force is mediated by graviton (the concept of graviton was introduced in 1974), photons for the electromagnetic force (a photon is the smallest EM force or the smallest packet of energy
for light). In Einstein's day, the strong and weak forces were unknown. For 30 years Einstein sought to unify the two distinct forces of gravity and electromagnetism.
String theory
Matter is composed of atoms, which in turn are made of nucleons (protons and neutrons in the nucleus) and electrons orbiting around the nucleus. Nucleons are made of three quarks each. Quarks are
made of string. According to the standard model of particle physics, the universe's elementary constituents are point-like ingredients with no internal structure. However, this standard model cannot
be a complete theory, for it does not include gravity. But according to string theory, atomic and subatomic particles are not point-like; rather, they consist of tiny one-dimensional filaments
somewhat like infinitely thin rubber bands. Physicists call these vibrating, oscillating, and dancing filaments strings.
String theory takes its name from this point of view. Unlike an ordinary piece of string, which itself is composed of molecules and atoms, the strings of string theory are alleged to lie deeply
within the heart of matter. They are so small”on average about as long the Planck length (10-33 cm, or about 100 billion billion [1020] times smaller than an atomic nucleus)”that they appear
point-like even when examined with our most powerful equipment. String theory offers a far fuller and more satisfying explanation than that of the standard model.
Moreover, this theory shows the harmonious union of general relativity and quantum mechanics”a major success. In this new millennium, the excitement in the physics community is that string theory may
provide the unified theory of all four forces and all matter. For this reason, string theory sometimes is described as possibly being the theory of everything.
String theory proclaims that the observed particle properties (i.e., mass, charge, and spin) are reflections of a string's various vibrations. Each preferred pattern of a string's vibration in string
theory appears as a particle whose mass and force charges are determined by the string's oscillatory pattern. All fundamental particles can be described as resonant patterns of these string
vibrations. There is even a mode describing the graviton. The same idea applies to the forces of nature as well. Hence everything, all matter and all forces, is unified under the microscopic string
oscillations”the notes that strings can play.
Extra Dimensions
Our universe has three spatial dimensions: length, width, and height. In formulating the general theory of relativity, Einstein showed that time is another dimension. According to the general theory
of relativity, space and time communicate the gravitational force through their curvature. The special theory of relativity is Einstein's law of space and time in the absence of gravity.
In 1919, the mathematician Theodor Kaluza unified Maxwell's electromagnetism and Einstein's theory of general relativity by adding a fifth dimension. Thus Kaluza was the one who suggested that the
universe might have more than three spatial dimensions.
For example, a garden hose viewed from a long distance looks like a one-dimensional object. When looked at closely, a second dimension, one shaped like a circle and curled around the hose, becomes
visible. The direction along the hose's length is long, extended, and easily visible. The direction circling around its thickness is short, curled up, and harder to see. Hence spatial dimensions are
of two types: large, extended, and therefore directly evident, or small, curled up, and far harder to detect. As for the garden hose, the curled-up dimension encircling its thickness is detected
either moving closer to the hose or using a pair of binoculars from a distance. If the hose is as thin as a hair or a capillary, its curled-up dimension is more difficult to detect.
In 1926, the mathematician Oskar Klein applied Kaluza's theory to quantum theory, which is used in modern string theory. Klein showed that our universe's spatial fabric may have both extended (the
three spatial dimensions of daily experience) and curled-up dimensions. The universe's additional dimensions are tightly curled up into a tiny space, a space so tiny that it has so far eluded
detection. These extra dimensions are believed to be minuscule, somewhere between 10-35 meters and 0.3 millimeters in size.
The equations of string theory show that the universe has nine space dimensions and one time dimension. At present, no one knows why the three space and one time dimensions are large and extended,
while all of the others are tiny and curled up.
Symmetry is a physical system property that does not change when the system is transformed. For example, a sphere is rotationally symmetrical, since its appearance does not change if it is rotated.
In 1971, supersymmetry was invented in two contexts at once: in ordinary particle field theory and as a consequence of introducing fermions into string theory. It holds the promise of resolving many
problems in particle theory, but requires equal numbers of fermions and bosons. Thus, it cannot be an exact symmetry of Nature.
Supersymmetry, a mathematical transformation, is a symmetry principle that relates a particle's properties of a whole number amount (integer) of spin (bosons) to those with half a whole (half-integer
or odd) number amount of spin (fermions). Bosons tend to be the mediators of fundamental forces, while fermions make up the matter experiencing these forces. Bosons can occupy the same space and have
an integral spin (0,1, .), while fermions cannot occupy the same space and have a half-integral spin ( 1/2, 3/2,.). Bosons transmit such forces as photons, gravitons, W and Z particles, mesons, and
gluons. Many bosons can occupy the same state at the same time. Fermions (e.g., electrons, muons, tau, protons, neutrons, quarks, and neutrinos) cannot share a given state at a given time with other
fermions. The fact that fermions make up matter explains why we cannot walk through walls: the inability of fermions (matter) to share the same space the way bosons (particles of force or energy)
Supersymmetry treats all particles of the same mass as different varieties of the same superparticle. This means that there is an equal matching between bosons and fermions. A supersymmetric string
theory is called a superstring theory. The original string theory only described bosons, and hence became known as bosonic string theory (BST). Thus it did not describe fermions or, for example,
include quarks and electrons.
Introducing supersymmetry to BST engendered a new theory that describes both the forces and the matter making up the universe: the theory of superstrings. String theorists have shown that all string
theories are different aspects of a string theory that has not 10 but 11 spatial dimensions. This was called M-theory. The M might stand for the mother of all theories or mystery, magic, matrix, or
membrane. The last two refer to mathematical techniques used in science. There is no space or time in M Theory. Furthermore, our space-time is not four-dimensional after all. M theory unites the four
forces of nature (e.g., gravity, quantum mechanics, strong force, and weak force) and, remarkably, is a mathematical and geometrical theory. It is attractive because it can explain gravity and an
atom's inside at the same time and thus resolve the contradiction between current theories.
Summary and Conclusions
String theory gives a theoretical description of elementary particles and treats them as one-dimensional curves (strings). Traditional models of interactions between elementary particles are based on
quantum field theory, which treats particles as dimensionless points. Theoretical physicists have not developed a workable theory of gravitation that is consistent with quantum mechanics' principles.
However, treating elementary particles as strings permits the derivation of a quantum theory that encompasses all four forces. Superstring theory, a combination of string theory and supersymmetry,
treats particles as very short (10-33 cm along its single dimension, which is 1020 smaller than a proton's diameter) closed strings (string loops). All of the masses, charges, and other properties of
elementary particles result from the vibration of these superstrings at different frequencies. The complex mathematical basis of superstrings involves 10 dimensions: 9 spatial dimensions, 6 of which
are invisible, and time. Since superstring theory provides a unified description of all elementary particles and fundamental forces, it is sometimes called the theory of everything.
Some major unsolved problems of string theory are how to condense, 10 dimensions to 6 (spatial) plus 4 (space and time) dimensions, and what is happening at distances smaller than 10-33 cm. In
addition, the experimental verification of the existence of strings in the near future poses quite a challenge. Since they are thought to be less than a billionth of a billionth the size of an atom,
we cannot use current technology to detect them directly. An indirect test, however, will be carried out within the next decade or so by the Large Hadron Collider, a huge atom smasher being built by
CERN (European Organization for Nuclear Research, located in Geneva, Switzerland). There also is an urgent need to develop new mathematics in areas of Riemann surfaces, algebraic geometry, singular
geometries, number theory, and other related fields.
1. K stands for Kelvin, a measurement of degree relating to, conforming to, or having a thermometric scale on which the unit of measurement equals the centigrade degree and according to which
absolute zero is 0A , the equivalent of “273.16A C.
• Adams, Steve. A Theory of Everything. New Scientist 161 (20 Feb. 1999).
• Arkani-Hamed, Nima et al. The Universe's Unseen Dimensions. Scientific American 283 (Aug. 2000): 62-69.
• Davies, P. C. W. and Julian Brown, Eds. Superstrings: A Theory of Everything? Cambridge, UK and New York: Cambridge University Press, 1988.
• Duff, Michael J. The Theory Formerly Known as Strings. Scientific American 278, (Feb. 1998): 64-69.
• Green, Michael M., John H. Schwarz, and Edward Witten. Superstring Theory. 2 vols. Cambridge, UK and New York: Cambridge University Press, 1987.
• Greene, Brian. The Elegant Universe. New York: W. W. Norton, 1999.
• Gribbin, John R. The Search for Superstrings, Symmetry, and the Theory of Everything. Boston: Little, Brown Co., 1998.
• Kaku, Michio. Hyperspace: A Scientific Odyssey through Parallel Universes, Time Warps, and the Tenth Dimension. New York: Oxford University Press, 1994.
• Mukhi, Sunil. The Theory of Strings: An Introduction. Current Science 77 (25 Dec. 1999): 1624-34.
• Peat, David F. Superstring and the Search for the Theory of Everything. Chicago: Contemporary Books, 1988.
• Polchinski, Joseph G. String Theory. 2 vols. Cambridge, UK and New York: Cambridge University Press, 1998. | {"url":"http://express.fountainmagazine.com/all-issues/2003/issue-41-january-march-2003/understanding-string-theory","timestamp":"2024-11-03T20:16:03Z","content_type":"text/html","content_length":"109095","record_id":"<urn:uuid:f37e3e2c-a0e3-4fda-8224-0caf8d0a84d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00009.warc.gz"} |
One Bit Comparator
1-Bit Magnitude Comparator - The Digital Comparator is another very usefulcombinational logic circuit used to compare the value of two binary digits. A magnitude digital Comparator is a
combinational circuit that compares two digital or binary numbers in order to find out whether one binary number is equal, less than or greater than the other binary number. We logically design a
circuit for which we will have two inputs one for A and another for B and have three output terminals, one for A > B condition, one for A = B condition and one for A < B condition.
A comparator used to compare two bits is called a single bit comparator. It consists of two inputs each for two single-bit numbers and three outputs to generate less than, equal to and greater than
between two binary numbers
Digital comparators actually use Exclusive-NOR gates within their design forcomparing their respective pairs of bits. When we are comparing two binary or BCD values or variables against each other,
we are comparing the “magnitude” of these values, a logic “0” against a logic “1” which is where the term Magnitude Comparator comes from. | {"url":"https://deldsim.com/study/material/14/one-bit-comparator/","timestamp":"2024-11-06T03:01:08Z","content_type":"text/html","content_length":"17170","record_id":"<urn:uuid:627cded3-dcbf-4a5f-aa74-8765db4b9706>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00218.warc.gz"} |
Frosty The Snowman's New Website & Giveaway
Have you seen the launch of the new FrostyTheSnowman.com website? Thanks to WarnerBros. Consumer Products the FrostyTheSnowman.com website has family fun activities to enjoy throughout the holiday
season. Interactive games, color pages, wallpapers, and more fun activities inspired by Frosty The Snowman. Have fun at Frosty's Winter Wonderland website http://bit.ly/WBFrosty and you can also
purchase the DVD @ http://bit.ly/DVD_Frosty WarnerBros. Consumer Products will also provide a copy of Frosty The Snowman’s Winter Wonderland on DVD to one lucky reader. United States winner, and no
PO Box address.
RULES: Contest closes 11:59pm Eastern December 22, 2011.
Open US winner. Leave your email in to post or available in profile, I will not search for it.
MUST complete Mandatory Entry. Winner will be chosen by random org. Entries that does not follow the rules will not win.
Mandatory Entry *Whats your favorite product at Frosty's store or game at his website? *******EXTRA ENTRIES***********
*Blog about the giveaway linking this post (5entries)
*Follow both on twitter & tweet "#Giveaway Frosty The Snowman’s Winter Wonderland on DVD @WB_Home_Ent @lifesandcastle http://goo.gl/aRrTd 12/22 #win #gift #Contest " (1 daily)
*Follow me Google + (5 entries)
*Give this post a google+1 (3 entries)
*Give me a Klout +K (1 k per day, daily)
*Grab Holiday Gift Guide Button (5 entries)
*Share this giveaway on your Google + wall, leave link(5entries)
*Stumble Life Is A SandCastle blog. (5 entries)
*Subscribe to my blog by email. (5 entries)
*(current giveaways)Enter The Night Owl Mama or giveaways on my site- one entry for each you enter.
*Add this giveaway to a contest linky & leave link (5 entries each link, up to 3contest listings)
"Disclosure: No sample was received to conduct review and company/pr providing the giveaway. Some photos used from sponsors website. All opinions are my own.
153 comments:
I like the wacky wobbler.
The Keepsake Ornament is really cute! :-)
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My favorite product is the Mini Animated Plush!
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Love the vintage shirts!
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I like the Frosty the Snowman throw looks cozy.
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My favorite is the Match Game.
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the hang man was fun!
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I love the Frosty the Snowman t-shirt:)
I like the Keepsake Ornament.
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That's a cute site for those of us that love Frosty. The match game looks like fun. Thanks and Happy Holidays! | {"url":"https://lifeisasandcastle.blogspot.com/2011/12/frosty-snowmans-new-website-giveaway.html","timestamp":"2024-11-06T20:13:22Z","content_type":"application/xhtml+xml","content_length":"300387","record_id":"<urn:uuid:6bd208f2-3fdb-4cc2-8184-824c04564c01>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00804.warc.gz"} |
sci.math FAQ: History of FLT
Archive-Name: sci-math-faq/FLT/history
Last-modified: December 8, 1994
Version: 6.2
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History of Fermat's Last Theorem
Pierre de Fermat (1601-1665) was a lawyer and amateur mathematician.
In about 1637, he annotated his copy (now lost) of Bachet's
translation of Diophantus' Arithmetika with the following statement:
Cubem autem in duos cubos, aut quadratoquadratum in duos
quadratoquadratos, et generaliter nullam in infinitum ultra
quadratum potestatem in duos ejusdem nominis fas est dividere:
cujus rei demonstrationem mirabilem sane detexi. Hanc marginis
exiguitas non caparet.
In English, and using modern terminology, the paragraph above reads
There are no positive integers such that x^n + y^n = z^n for n > 2 .
I've found a remarkable proof of this fact, but there is not enough
space in the margin [of the book] to write it.
Fermat never published a proof of this statement. It became to be
known as Fermat's Last Theorem (FLT) not because it was his last piece
of work, but because it is the last remaining statement in the
post-humous list of Fermat's works that needed to be proven or
independently verified. All others have either been shown to be true
or disproven long ago.
Tue Apr 04 17:26:57 EDT 1995
User Contributions:
Comment about this article, ask questions, or add new information about this topic: | {"url":"http://www.faqs.org/faqs/sci-math-faq/FLT/history/","timestamp":"2024-11-05T09:57:25Z","content_type":"application/xhtml+xml","content_length":"26800","record_id":"<urn:uuid:d7088c42-b0b3-4724-bed0-60f1d5c74066>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00072.warc.gz"} |
Swiss Olympiad in Informatics
Case #0: Stofl starts at the end of the left-hand L-shaped road and can walk to the southern exit.
Case #1: Stofl starts at the piazza in the north west, but it is not connected to the eastern road.
Tank Golf
To solve this task, you should first download the host server for the game, which allows you to test your bot locally.
Check out the Readme to the host server as well, there you find instructions on how to compile the host server. You can find some example bots in the directory bots/ to get you started.
If you encounter any problems while compiling the host server, don’t hesitate to ask questions to pascal@soi.ch.
See TankGolf on the Google Playstore for Android. The task is basically to write a bot for this game.
Two tanks are in a 2d physics simulation (upright), where a tank can shoot a bullet with a desired angle and intensity. The bullet causes an explotion wherever it impacts a surface, and the goal is
to push or throw the other tank into a hole in the map through the shockwave of the explosion. The player take turns at shooting, and one can only shoot when both tanks have come to a rest after the
last explosion (or the bullet shot by the opponent has fallen off the map). When a player falls through the hole, the opponent gets a point and the player respawns.
The task is to write a bot that, given the current positions and orientations of the tanks, determines an angle and intensity to shoot. The shot will be executed with a uniform random distribution
within a small range around the desired angle and intensity. The uncertainty radius depends on the square of the intensity, such that long range shots are more risky than short range.
The game server provides a function to look up the impact position of the bullet, and the positions where the tanks come to a rest after the explosion, given a game state and desired trajectory. The
bot can call this lookup function as often as you want, but please make sure that the bot doesn’t take longer than a few seconds for every turn (The time required to show the visualization isn’t
counted in this). The randomness will only be added when the bot commits to a final shot, and does not affect any lookups.
The description of the map can be found in geometry/maps/map1.json, or you can look at it using the provided visualization. The map used in the contest will be exactly the same.
To respawn, your bot can chose an x-coordinate to spawn at, then the tank will drop down at this x-coordinate from a fixed y-coordinate. Y is fixed to make sure you can’t spawn into the floor or the
other player. If you want to spawn right above the hole, you are very welcome to do so. There is also a query method available for respawning, similar to the query method for shots. A tank respawns
as soon as they fell off the map. If both tanks fall off the map, both spawn in first, then the bots continue in their alternating order of shooting turns.
There are two types of round, one where the bot can line up a shot, and one where the bot chooses a spawn position.
Your bot will always see itself as player A in the protocol, so player B is always the opponent in the data that you receive.
A game state has two representations, one for transmitting the full state and one for a subset of the full state:
full: playerA.x playerA.y playerA.angle playerB.x playerB.y playerB.angle last_impact.x last_impact.y scores.playerA scores.playerB
small: playerA.x playerA.y playerA.angle playerB.x playerB.y playerB.angle
A position that is not on the map is represented as -1, -1. For example a player position if they fell off the map or last_impact before the first shot are -1, -1.
Every round starts by the server sending the current type of round: ‘startrespawn’ or ‘startshoot’, followed by the full state (Initial spawn counts as respawn)
Server: startrespawn <full game state>
(player asks a query…)
Player: queryrespawn 12.345 where the player asks what would happen if I drop down at this x position
Server: state <full resulting state at rest>
(… zero or more queries in total)
Player: respawn 12.345 respawn at this location
(turn is over)
Server: startshoot <full game state>
(player asks a query..)
Player: queryrespawn 1.13 0.9 where the player asks what would happen if I shoot with this angle and intensity
Server: state <full resulting state at rest>
(.. zero or more queries in total)
Player: shoot 1.13 0.9 shoot with this angle and intensity
(turn is over)
Queries that are always possible while it’s your turn
Player: fullqueryshoot <small state> 1.13 0.9 where the player gives a game state and asks what would happen if I shoot with this angle and intensity
Server: state <full resulting state at rest>
Player: fullqueryrespawn <small state> 12.345 where the player gives a game state and asks what would happen if I drop down at this x position
Server: state <full resulting state at rest>
Note that this example only shows the communication for one of the players.
< is server to client and > is client to server
< startrespawn -1 -1 0 -1 -1 0 -1 -1 0 0 # We're player one and thus noone is on the map yet. Bullet hasn't impacted yet. Scores are at zero.
# read as: tank A: (-1, -1) 0 | tank B: (-1, -1) 0 | impact pos: (-1, -1) | scores: 0 0
> respawn 2 # Spawn at position 2
(other player respawns)
< startshoot 2 1 0 6 1 0 -1 -1 0 0 # Other player has spawned at coordinate 6. We shoot first
> queryshoot -1.13 0.6 # Simulate shot 1.13 radian to the right
< state 2 1 0 -1 -1 0 7.2 1 # Other player would fall in hole, we wouldn't move. We would get a point.
# read as: tank A: (2, 1) 0 | tank B: (-1, -1) 0 | impact pos: (7.2, 1) | scores: 1 0
> shoot -1.13 0.6 # Shoot as simulated before.
# What the player doesn't know immediately, but what we see in the next server command is, that the shot
# didn't work out because of the added randomness (impact was at 7.4, 1). We didn't get a point.
(other player shoots)
< startrespawn -1 -1 0 4.6 1 0 1.5 1 0 1 # The other player scored during their turn (last impact was at 1.5, 1). We need to select a respawn position.
> respawn 6
< startshoot 6 1 0 4.6 1 0 1.5 1 0 1 # We get to shoot immediately after respawning, because the other player was the last one to shoot.
Submit (100 points)
You play against the other participants of the SOI. At SOI-Day the programs will compete in a tournament to determine the best bot. You are allowed to submit up to three bots. Your best bot will be
used for the ranking. Bundle the files in a zip archive when submitting.
The contest is over, you can no longer submit.
Don’t hesitate to ask us any question about this task, programming or the website via email (info@soi.ch). | {"url":"https://soi.ch/contests/2019/round1/","timestamp":"2024-11-10T02:43:00Z","content_type":"text/html","content_length":"427520","record_id":"<urn:uuid:e9e1899d-3c84-4f89-88ae-51f695037f06>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00506.warc.gz"} |
A beginner's guide to interpreting odds ratios, confidence intervals and p-values - Students 4 Best Evidence
A beginner’s guide to interpreting odds ratios, confidence intervals and p-values
Posted on 13th August 2013 by Tim Hicks
Tutorials and Fundamentals
Who is it for?
Students of medicine or from the clinical sciences and professions allied to medicine wanting to enhance their understanding of medical literature they will encounter throughout their careers.
What will I learn and how?
How to interpret odds ratios, confidence intervals and p values with a stepwise progressive approach and a’concept check’ question as each new element is introduced.
How long will it take?
Approximately 20 minutes.
What it is not
A statistical textbook reworded or how to calculate any of these statistics.
Odds ratio
Confidence interval
P value
Bringing it all together – Real world example
Self test Answers
The first steps in learning to understand and appreciate evidence-based medicine are daunting to say the least, especially when confronted with the myriad of statistics in any paper. This short
tutorial aims to introduce healthcare students to the interpretation of some of the most commonly used statistics for reporting the results of medical research.
The scenario for this tutorial is centred around the diagram below, which outlines a fictional parallel two arm randomised controlled trial of a new cholesterol lowering medication against a placebo.
Odds ratio (OR)
An odds ratio is a relative measure of effect, which allows the comparison of the intervention group of a study relative to the comparison or placebo group.
So when researchers calculate an odds ratio they do it like this:
The numerator is the odds in the intervention arm
The denominator is the odds in the control or placebo arm = Odds Ratio (OR)
So if the outcome is the same in both groups the ratio will be 1, which implies there is no difference between the two arms of the study.
If the OR is > 1 the control is better than the intervention.
If the OR is < 1 the intervention is better than the control.
Concept check 1
If the trial comparing SuperStatin to placebo with the outcome of all cause mortality found the following:
Odds of all cause mortality for SuperStatin were 0.4
Odds of all cause mortality for placebo were 0.8
Odds ratio would equal 0.5
So if the trial comparing SuperStatin to placebo stated OR 0.5
What would it mean?
A) The odds of death in the SuperStatin arm are 50% less than in the placebo arm.
B) There is no difference between groups
C) The odds of death in the placebo arm are 50% less than in the SuperStatin arm.
Confidence interval (CI)
The confidence interval indicates the level of uncertainty around the measure of effect (precision of the effect estimate) which in this case is expressed as an OR. Confidence intervals are used
because a study recruits only a small sample of the overall population so by having an upper and lower confidence limit we can infer that the true population effect lies between these two points.
Most studies report the 95% confidence interval (95%CI).
If the confidence interval crosses 1 (e.g. 95%CI 0.9-1.1) this implies there is no difference between arms of the study.
Concept check 2
So if the trial comparing SuperStatin to placebo stated OR 0.5 95%CI 0.4-0.6
What would it mean?
A) The odds of death in the SuperStatin arm are 50% less than in the placebo arm with the true population effect between 20% and 80%.
B) The odds of death in the SuperStatin arm are 50% less than in the placebo arm with the true population effect between 60% and 40%.
C) The odds of death in the SuperStatin arm are 50% less than in the placebo arm with the true population effect between 60% and up to 10% worse.
P values
P < 0.05 indicates a statistically significant difference between groups. P>0.05 indicates there is not a statistically significant difference between groups.
Concept check 3
So if the trial comparing SuperStatin to placebo stated OR 0.5 95%CI 0.4-0.6 p<0.01
What would it mean?
A) The odds of death in the SuperStatin arm are 50% less than in the placebo arm with the true population effect between 60% and 40%. This result was statistically significant.
B) The odds of death in the SuperStatin arm are 50% less than in the placebo arm with the true population effect between 60% and 40%. This result was not statistically significant.
C) The odds of death in the SuperStatin arm are 50% less than in the placebo arm with the true population effect between 60% and 40%. This result was equivocal.
Bringing it all together – Real world example
A drug company-funded double blind randomised controlled trial evaluated the efficacy of an adenosine receptor antagonist Cangrelor vs Clopidogrel in patients undergoing urgent or elective
Percutaneous Coronary Intervention (PCI) who were followed up for specific complications for 48 hrs as outlined in the diagram below (Bhatt et al. 2009).
The results section reported “The rate of the primary efficacy end point was … (adjusted odds ratio with Cangrelor, 0.78; 95% confidence interval [CI], 0.66 to 0.93; P=0.005)
What does this mean?
A) The odds of death, myocardial infarction, ischemia-driven revascularization, or stent thrombosis at 48 hours after randomization in the Cangrelor arm were 22% less than in the Clopidogrel arm
with the true population effect between 34% and 7%. This result was not statistically significant.
B) The odds of death, myocardial infarction, ischemia-driven revascularization, or stent thrombosis at 48 hours after randomization in the Cangrelor arm were 34% less than in the Clopidogrel arm with
the true population effect between 7% and 22%. This result was statistically significant.
C) The odds of death, myocardial infarction, ischemia-driven revascularization, or stent thrombosis at 48 hours after randomization in the Cangrelor arm were 22% less than in the Clopidogrel arm with
the true population effect between 34% and 7%. This result was statistically significant.
This is a very basic introduction to interpreting odds ratios, confidence intervals and p values only and should help healthcare students begin to make sense of published research, which can
initially be a daunting prospect. However it should be stressed that any results are only valid if the study was well designed and conducted, which highlights the importance of critical appraisal as
a key feature of evidence based medicine.
I do hope you enjoyed working through this and would appreciate any feedback on the content, design and presentational aspects of this tutorial.
Self test Answers
Concept check 1. The correct answer is A.
Concept check 2. The correct answer is B.
Concept check 3. The correct answer is A.
Bringing it all together – Real world example. The correct answer is C.
You may also be interested in these blogs:
Why should students know about kappa value?
Efficacy of drugs: 3 examples to get you to truly understand Number Needed to Treat (NNT)
Key to statistical result interpretation: P-value in plain English
Surrogate endpoints: pitfalls of easier questions
How did they determine diagnostic thresholds: the stories of anemia and diabetes
Bhatt DL, Stone GW, Mahaffey KW, Gibson CM, Steg PG, Hamm CW, Price MJ, Leonardi S, Gallup D, Bramucci E, Radke PW, Widimský P, Tousek F, Tauth J, Spriggs D, McLaurin BT, Angiolillo DJ, Généreux
P, Liu T, Prats J, Todd M, Skerjanec S, White HD, Harrington RA. CHAMPION PHOENIX Investigators. (2013). Effect of platelet inhibition with cangrelor during PCI on ischemic events. N Engl J Med. Apr
No Comments on A beginner’s guide to interpreting odds ratios, confidence intervals and p-values
• DR. KAVITHA MOHANDAS
Very cogent explanation. Thanks. Kavitha
15th November 2018 at 9:37 am
Reply to DR.
• dr ashutosh
It would have been good if the etymology of terms were added.
what does 95% mean ? Is it the measure of confidence meaning ;that we are 95% confident of the applicable range’s(between 40% and 60%) dependability for its application for general population?
12th September 2018 at 12:23 pm
Reply to dr
• dr saima
very informative
29th August 2018 at 7:44 pm
Reply to dr
• ML
Needed clear, concise help in a hurry and you provided it. Thank you!
20th August 2018 at 2:26 am
Reply to ML
• Ritika Kaushal
Thank you ! This helped me understand aspect of biostats better than any book.
23rd July 2018 at 6:50 pm
Reply to Ritika
• Catherine Williams
Hi, I’m a little confused about the CI interval and your statement that even if it’s 95% if the interval “crosses” 1 it’s not statistically significant. I am reading a paper comparing the effects
of declawing of cats on various adverse out comes as compared with non declawed cats. In all of the comparisons (OR were calculated) the CI was 95% but the values listed under the CI were often
greater than 1. Examples include the following (1.99-7.84, 1.32 – 4.56)
14th February 2018 at 3:47 pm
Reply to Catherine
• P Koul
Hi Tim
I am running a Vaccine effectiveness study and wanted to calculate the Vaccine effectiveness from OR. The OR is 27 (CI 17-40). This translates to VE of 73 (calculated as 1-OR). Do the confidence
intervals also get changed to 60-83?
Thanks for your response.
21st December 2017 at 5:44 pm
Reply to P
• σχεσεις
I read this article completely concerning the resemblance of most up-to-date and earlier
technologies, it’s awesome article.
5th December 2017 at 5:09 pm
Reply to σχεσεις
• Mahesh
There is some nice introductory stuff here but I’m concerned that your explanation of p values feeds into popular misconceptions that are somewhat harmful (see comment 91 for a stark example).
The p value is the risk of obtaining the observed result, or a more extreme result, by chance if the null hypothesis were true. 0.05 was chosen by a researcher many decades ago and the entire
biomedical science community followed suit. There is nothing magical about 0.05 that makes a result “significant”.
I think it’s much more important that students learn about the scientific method, hypothesis testing, trial design, data types and distribution etc. before worrying about significance testing.
Otherwise the risk is that they go through their entire career mistakenly looking for the magical 0.05 like comment no.91
26th November 2017 at 7:52 pm
Reply to Mahesh
• Fredrik Nath
Hi Tim,
Felt I should just say I thought this little intro to OR and CI was excellent.
I’m a 67-year-old (full-time) neurosurgeon down south and for years, I’ve always gone by p levels in assessing stats presented in papers. Never worried much about the niceties of CI. I read your
very clear article and realise now I should have done this years ago! Well done and thanks.
30th September 2017 at 8:37 am
Reply to Fredrik
• Caitlin
Can somebody please help me? Can somebody explain this (AOR= 0,04 95% CI 0,02-0,09)? It would mean that an anterior and an lateral posterior episiotomie would be the best intervention to prevent
an intrapartumhemmorage. Kindly help.
3rd September 2017 at 4:00 pm
Reply to Caitlin
• Kira
Hi Tim,
I am a current post-graduate fellow working with the LA County Department of Public Health. While wrapping up my epidemiological research project I searched for a quick refresher or reference
guide for this very topic. You awesome data came up. Thank you so very much for compiling this information in a quick and straight forward manner. It helped me expedite my review of ORs essential
to completing the data analysis portion of my manuscript.
You’re so awesome!!!
21st March 2017 at 9:31 pm
Reply to Kira
• Sahana V
Let us consider the relationship between smoking and lung cancer. Suppose exposure to cigarette smoke increases the incidence of lung cancer by 20% (i.e. the relative risk is 1.2). Lung cancer
has a baseline incidence of 3% per year (in the non-exposed group). Suppose as well that baseline incidence in obese individuals is 1/3 less (i.e. 1%/yr.), and the relative risk associated with
the exposure is 1.2. You follow up 1000 non-obese and 1000 obese subjects with the exposure, and an equivalent number without the exposure. The study lasts 25 years. Work with 25-year cumulative
incidence and a denominator of 1000.
How to calculate this problem? Especially the construction of the table. Kindly help.
18th February 2017 at 3:59 am
Reply to Sahana
• noi noi παπουτσια
Do you mind if I quote a feww of your articles as long as I provide credit and
sources back to your webpage? My blog site is in the very same
niche as yours and my visitors would certainly benefit from some of
the information you proviude here. Pleae let mee know if this alright with you.
5th February 2017 at 5:59 am
Reply to noi
□ Selena Ryan-Vig
You’re welcome to quote any of the articles if you provide credit and link back to the original source. Many thanks!
6th February 2017 at 9:28 am
Reply to Selena
• Reena
fantastic resource, thanks so much!!
29th November 2016 at 4:08 am
Reply to Reena
• Oswald Pesh
Might you help me understand the following interpretation?:
“Group A reported significantly less difficulty in the instrumental activities of daily living (IADL) than the control group (effect size, 0.29; 99% confidence interval [CI], 0.03-0.55). Neither
Group B (effect size, 0.26; 99% CI, −0.002 to 0.51) nor Group C (effect size, 0.20; 99% CI, −0.06 to 0.46) had a significant effect on IADL”
What is the basis for interpreting no significant effect in groups B and C?
2nd October 2016 at 7:09 pm
Reply to Oswald
• SH
Hi Tim, your explanation is so much easy to understand. Just a question. Is Odds ratio the same as relative risk ratio?
Also I have difficulty understanding different study designs and ends up misinterpreting them.
Is there an easier way of understanding the difference between cohort studies, case control studys, retrospective cohort studies and cross-sectional studies
30th September 2016 at 5:55 am
Reply to SH
• Nancy
When doing a lit review, I find that results are frequently presented in different ways. I’d like to be able to convert them to read the same way so that I can compare them.
I know that I can convert OR1 by using the equation 1/x, where x=OR<1, which then reverses the factors being compared. For example,
FB is less likely in rural [OR=.26 (.12, .50)] than urban areas
converts to:
FB is more likely in urban [OR=3.85 (2.00, 8.33] vs rural areas
My question is how to convert:
HB is more likely in rural [OR=22.8 (10.6, 49.4)] than urban areas
Is it correct to say:
FB is more likely in urban [OR=22.8 (10.6, 49.4)] than rural areas
Or is there some calculation that needs to be done? FB and HB are dichotomous outcomes.
Thank you.
20th September 2016 at 7:17 pm
Reply to Nancy
• hamzah
Odds Ratio (OR) is a measure associations between exposure (risk factors) and the incidence of disease; calculated from the incidence of the disease in at risk groups (exposed to risk factors)
compared to the incidence of the disease in non-risk group (not exposed to a risk factor).
In this present study, by cross sectional study, We got OR 2.8 for variable keep livestock such as goats, sheep and pigs. Is this meaning Respondents or household who keep livestock such as
goats, sheep and pigs have a 2.8 times greater chance of contracting malaria compared to a respondents or household who do not raise cattle where the confidence interval [CI: 2.180 – 3492])?
What does different between OR and RR, for this case
Regards for your advice
22nd July 2016 at 9:12 am
Reply to hamzah
• eyasu
i know how to calculate crude odds ratio manually but how can i calculate adjusted odds ratio manually ?
10th June 2016 at 9:49 am
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Which of the following could be the quadratic equation of the g... | Filo
Which of the following could be the quadratic equation of the graph shown below?
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The parabola is "moving up," so eliminate choices A and E. Graph the remaining choices with your graphing calculator to find choice B is correct.
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Question Text Which of the following could be the quadratic equation of the graph shown below?
Topic Graph Equations
Subject Mathematics
Class Grade 12
Answer Type Text solution:1
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0 is even
もっと例文: 1 2
1. Official publications relating to the GRE tests both state that 0 is even.
2. Continuing the pattern of separation by two implies that 1 is odd and that 0 is even.
3. In the group of zero objects, there is no leftover object, so 0 is even.
4. For example, 1 is odd because and 0 is even because Making a table of these facts then reinforces the number line picture above.
5. The study findings showed that the risk of Apgar scores of 0 is even greater in first-born babies 14 times the risk of hospital births. | {"url":"https://ja.ichacha.net/mzj/0%20is%20even.html","timestamp":"2024-11-13T08:39:57Z","content_type":"text/html","content_length":"17604","record_id":"<urn:uuid:7016fe9c-d8a2-4145-9658-3e1f6150b5a3>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00137.warc.gz"} |
The Stacks project
Lemma 22.37.1. Let $R$ be a ring. Let $(A, \text{d}) \to (B, \text{d})$ be a homomorphism of differential graded algebras over $R$, which induces an isomorphism on cohomology algebras. Then
\[ - \otimes _ A^\mathbf {L} B : D(A, \text{d}) \to D(B, \text{d}) \]
gives an $R$-linear equivalence of triangulated categories with quasi-inverse the restriction functor $N \mapsto N_ A$.
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cogniDNA - What is Sir Isaac Newton's IQ?What is Sir Isaac Newton's IQ? - cogniDNA
Isaac Newton
IQ: 130 ± 1.5
Introduction to Isaac Newton
Sir Isaac Newton, born on December 25, 1642, in Woolsthorpe, England, is widely regarded as one of the greatest mathematicians, physicists, and astronomers in history. His pioneering work laid the
foundation for classical mechanics, including the laws of motion and universal gravitation, and had a profound impact on the development of modern science.
Newton attended Trinity College, Cambridge, where he made groundbreaking discoveries in mathematics, optics, and mechanics. His work on calculus, the nature of light, and the formulation of the laws
of motion cemented his reputation as a leading figure of the Scientific Revolution.
Newton's career is marked by numerous scientific achievements and contributions:
• Laws of Motion and Universal Gravitation: Formulated the three laws of motion and the law of universal gravitation, which became fundamental principles in physics.
• Mathematical Contributions: Developed calculus independently, providing powerful tools for mathematical analysis.
• Optics: Conducted pioneering experiments with light and prisms, establishing the nature of color and the composition of white light.
Newton's work revolutionized science and provided a new understanding of the natural world. His intellectual contributions continue to influence physics, mathematics, and many other fields to this
Isaac Newton's accomplishments demonstrate his profound intelligence and impact on science:
• Principia Mathematica: Authored "Philosophiæ Naturalis Principia Mathematica," considered one of the most important scientific works ever published.
• Development of Calculus: Independently formulated calculus, a fundamental mathematical framework.
• Optical Theories: Advanced the understanding of optics, light, and color through experimental and theoretical work.
Outcomes of Isaac Newton's Intelligence
Isaac Newton's estimated IQ reflects his ability to formulate revolutionary theories, solve complex mathematical problems, and develop foundational principles in physics. His contributions to science
have shaped the course of human knowledge and technological advancement.
🔍 How Did We Calculate This?
Note: Isaac Newton never took a formal IQ test, but his groundbreaking work in mathematics, physics, and astronomy, combined with his status as the foremost scientist of the 1600s, provides a solid
foundation for estimating his intellectual capability.
1️⃣ The Flynn Effect:
The Flynn Effect suggests that IQ scores tend to increase over time. For historical figures like Newton, an adjustment stabilizes the estimate, reflecting changes in average IQ levels since his era.
2️⃣ Modern Baseline:
A peer-reviewed source indicates that the modern average IQ for physicists is 133, which serves as a reference for recalibrating Newton's IQ estimate.
3️⃣ Flynn Effect Adjustment:
Applying the Flynn Effect adjustment, we estimate the average IQ for physicists in the 1600s to be around 89.
4️⃣ Estimating the Number of Physicists in the 1600s:
Approximately 300 physicists were estimated to be active worldwide during Newton's era, making his standing as a leading figure significant.
5️⃣ Newton's IQ:
Given his status as the top physicist of his time, we estimate Isaac Newton's IQ to be approximately 130. This estimate considers the historical context and Newton's extraordinary contributions to
science and mathematics. | {"url":"https://www.cognidna.com/celebrity-iq-scores/isaac-newton/","timestamp":"2024-11-05T03:30:43Z","content_type":"text/html","content_length":"22025","record_id":"<urn:uuid:253047ea-bf21-4fc0-bc48-efea11430a34>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00037.warc.gz"} |
Unimodular hyperbolic triangulations: circle packing and random walk
We show that the circle packing type of a unimodular random plane triangulation is parabolic if and only if the expected degree of the root is six, if and only if the triangulation is amenable in the
sense of Aldous and Lyons [1]. As a part of this, we obtain an alternative proof of the Benjamini–Schramm Recurrence Theorem [19]. Secondly, in the hyperbolic case, we prove that the random walk
almost surely converges to a point in the unit circle, that the law of this limiting point has full support and no atoms, and that the unit circle is a realisation of the Poisson boundary. Finally,
we show that the simple random walk has positive speed in the hyperbolic metric.
Funders Funder number
National Science Foundation
Horizon 2020 Framework Programme 676970
Natural Sciences and Engineering Research Council of Canada
Engineering and Physical Sciences Research Council EP/103372X/1
Israel Science Foundation 1207/15
Dive into the research topics of 'Unimodular hyperbolic triangulations: circle packing and random walk'. Together they form a unique fingerprint. | {"url":"https://cris.tau.ac.il/en/publications/unimodular-hyperbolic-triangulations-circle-packing-and-random-wa","timestamp":"2024-11-11T15:07:04Z","content_type":"text/html","content_length":"49049","record_id":"<urn:uuid:6a23f2f0-323c-496c-88e2-0bd5c7fb4f60>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00304.warc.gz"} |
Summaries, predictions, intervals, and tests for emmGrid objects — summary.emmGrid
Summaries, predictions, intervals, and tests for emmGrid objects
These are the primary methods for obtaining numerical or tabular results from an emmGrid object. summary.emmGrid is the general function for summarizing emmGrid objects. It also serves as the print
method for these objects; so for convenience, summary() arguments may be included in calls to functions such as emmeans and contrast that construct emmGrid objects. Note that by default, summaries
for Bayesian models are diverted to hpd.summary.
# S3 method for class 'emmGrid'
summary(object, infer, level, adjust, by,
cross.adjust = "none", type, df, calc, null, delta, side, frequentist,
bias.adjust = get_emm_option("back.bias.adj"), sigma, ...)
# S3 method for class 'emmGrid'
confint(object, parm, level = 0.95, ...)
test(object, null, ...)
# S3 method for class 'emmGrid'
test(object, null = 0, joint = FALSE, verbose = FALSE,
rows, by, status = FALSE, ...)
# S3 method for class 'emmGrid'
predict(object, type, interval = c("none", "confidence",
"prediction"), level = 0.95,
bias.adjust = get_emm_option("back.bias.adj"), sigma, ...)
# S3 method for class 'emmGrid'
as.data.frame(x, row.names = NULL, optional,
check.names = TRUE, destroy.annotations = FALSE, ...)
# S3 method for class 'summary_emm'
x[..., as.df = FALSE]
An object of class "emmGrid" (see emmGrid-class)
A vector of one or two logical values. The first determines whether confidence intervals are displayed, and the second determines whether t tests and P values are displayed. If only one value is
provided, it is used for both.
Numerical value between 0 and 1. Confidence level for confidence intervals, if infer[1] is TRUE.
Character value naming the method used to adjust \(p\) values or confidence limits; or to adjust comparison arrows in plot. See the P-value adjustments section below.
Character name(s) of variables to use for grouping into separate tables. This affects the family of tests considered in adjusted P values.
Character: \(p\)-value adjustment method to additionally apply across the by groups. See the section on P-value adjustments for details.
Character: type of prediction desired. This only has an effect if there is a known transformation or link function. "response" specifies that the inverse transformation be applied. "mu" (or
equivalently, "unlink") is usually the same as "response", but in the case where the model has both a link function and a response transformation, only the link part is back-transformed. Other
valid values are "link", "lp", and "linear.predictor"; these are equivalent, and request that results be shown for the linear predictor, with no back-transformation. The default is "link", unless
the "predict.type" option is in force; see emm_options, and also the section below on transformations and links.
Numeric. If non-missing, a constant number of degrees of freedom to use in constructing confidence intervals and P values (NA specifies asymptotic results).
Named list of character value(s) or formula(s). The expressions in char are evaluated and appended to the summary, just after the df column. The expression may include any names up through df in
the summary, any additional names in object@grid (such as .wgt. or .offset.), or any earlier elements of calc.
Numeric. Null hypothesis value(s), on the linear-predictor scale, against which estimates are tested. May be a single value used for all, or a numeric vector of length equal to the number of
tests in each family (i.e., by group in the displayed table).
Numeric value (on the linear-predictor scale). If zero, ordinary tests of significance are performed. If positive, this specifies a threshold for testing equivalence (using the TOST or
two-one-sided-test method), non-inferiority, or non-superiority, depending on side. See Details for how the test statistics are defined.
Numeric or character value specifying whether the test is left-tailed (-1, "-", "<", "left", or "nonsuperiority"); right-tailed (1, "+", ">", "right", or "noninferiority"); or two-sided (0, 2, "!
=", "two-sided", "both", "equivalence", or "="). See the special section below for more details.
Ignored except if a Bayesian model was fitted. If missing or FALSE, the object is passed to hpd.summary. Otherwise, a logical value of TRUE will have it return a frequentist summary.
Logical value for whether to adjust for bias in back-transforming (type = "response"). This requires a valid value of sigma to exist in the object or be specified.
Error SD assumed for bias correction (when type = "response" and a transformation is in effect), or for constructing prediction intervals. If not specified, object@misc$sigma is used, and a
warning is issued if it is not found or not valid. Note: sigma may be a vector, but be careful that it correctly corresponds (perhaps after recycling) to the order of the reference grid.
Optional arguments such as scheffe.rank (see “P-value adjustments”). In confint.emmGrid, predict.emmGrid, and test.emmGrid, these arguments are passed to summary.emmGrid.
(Required argument for confint methods, but not used)
Logical value. If FALSE, the arguments are passed to summary.emmGrid with infer=c(FALSE, TRUE). If joint = TRUE, a joint test of the hypothesis L beta = null is performed, where L is
object@linfct and beta is the vector of fixed effects estimated by object@betahat. This will be either an F test or a chi-square (Wald) test depending on whether degrees of freedom are available.
See also joint_tests.
Logical value. If TRUE and joint = TRUE, a table of the effects being tested is printed.
Integer values. The rows of L to be tested in the joint test. If missing, all rows of L are used. If not missing, by variables are ignored.
logical. If TRUE, a note column showing status flags (for rank deficiencies and estimability issues) is displayed even when empty. If FALSE, the column is included only if there are such issues.
Type of interval desired (partial matching is allowed): "none" for no intervals, otherwise confidence or prediction intervals with given arguments, via confint.emmGrid. Note: prediction intervals
are not available unless the model family is "gaussian".
object of the given class
passed to as.data.frame
required argument, but ignored in as.data.frame.emmGrid
passed to data.frame
Logical value. If FALSE, an object of class summary_emm is returned (which inherits from data.frame), but if displayed, details like confidence levels, P-value adjustments, transformations, etc.
are also shown. But unlike the result of summary, the number of digits displayed is obtained from getOption("digits") rather than using the optimal digits algorithm we usually use. Thus, it is
formatted more like a regular data frame, but with any annotations and groupings still intact. If TRUE (not recommended), a “plain vanilla” data frame is returned, based on row.names and
Logical value. With x[..., as.df = TRUE], the result is object is coerced to a data.frame before the subscripting is applied. With as.df = FALSE, the result is returned as a summary_emm object
when possible.
summary.emmGrid, confint.emmGrid, and test.emmGrid return an object of class "summary_emm", which is an extension of data.frame but with a special print method that displays it with custom
formatting. For models fitted using MCMC methods, the call is diverted to hpd.summary (with prob set to level, if specified); one may alternatively use general MCMC summarization tools with the
results of as.mcmc.
predict returns a vector of predictions for each row of object@grid.
The as.data.frame method returns an object that inherits from "data.frame".
confint.emmGrid is equivalent to summary.emmGrid with infer = c(TRUE, FALSE). The function test.emmGrid, when called with joint = FALSE, is equivalent to summary.emmGrid with infer = c(FALSE, TRUE).
With joint = TRUE, test.emmGrid calculates the Wald test of the hypothesis linfct %*% bhat = null, where linfct and bhat refer to slots in object (possibly subsetted according to by or rows). An
error is thrown if any row of linfct is non-estimable. It is permissible for the rows of linfct to be linearly dependent, as long as null == 0, in which case a reduced set of contrasts is tested.
Linear dependence and nonzero null cause an error. The returned object has an additional "est.fcns" attribute, which is a list of the linear functions associated with the joint test.
In doing testing and a transformation and/or link is in force, any null and/or delta values specified must always be on the scale of the linear predictor, regardless of the setting for `type`. If
type = "response", the null value displayed in the summary table will be back-transformed from the value supplied by the user. But the displayed delta will not be changed, because there (often) is
not a natural way to back-transform it.
When we have type = "response", and bias.adj = TRUE, the null value displayed in the output is both back-transformed and bias-adjusted, leading to a rather non-intuitive-looking null value. However,
since the tests themselves are performed on the link scale, this is the response value at which a *P* value of 1 would be obtained.
The default show method for emmGrid objects (with the exception of newly created reference grids) is print(summary()). Thus, with ordinary usage of emmeans and such, it is unnecessary to call summary
unless there is a need to specify other than its default options.
If a data frame is needed, summary, confint, and test serve this need. as.data.frame routes to summary by default; calling it with destroy.annotations = TRUE is not recommended for exactly that
reason. If you want to see more digits in the output, use print(summary(object), digits = ...); and if you always want to see more digits, use emm_options(opt.digits = FALSE).
The misc slot in object may contain default values for by, calc, infer, level, adjust, type, null, side, and delta. These defaults vary depending on the code that created the object. The update
method may be used to change these defaults. In addition, any options set using emm_options(summary = ...) will trump those stored in the object's misc slot.
Transformations and links
With type = "response", the transformation assumed can be found in object@misc$tran, and its label, for the summary is in object@misc$inv.lbl. Any \(t\) or \(z\) tests are still performed on the
scale of the linear predictor, not the inverse-transformed one. Similarly, confidence intervals are computed on the linear-predictor scale, then inverse-transformed.
Be aware that only univariate transformations and links are supported in this way. Some multivariate transformations are supported by mvregrid.
Bias adjustment when back-transforming
When bias.adjust is TRUE, then back-transformed estimates are adjusted by adding \(0.5 h''(u)\sigma^2\), where \(h\) is the inverse transformation and \(u\) is the linear predictor. This is based on
a second-order Taylor expansion. There are better or exact adjustments for certain specific cases, and these may be incorporated in future updates.
Note: In certain models, e.g., those with non-gaussian families, sigma is initialized as NA, and so by default, bias adjustment is skipped and a warning is issued. You may override this by specifying
a value for sigma. However, with ordinary generalized linear models, bias adjustment is inappropriate and you should not try to do it. With GEEs and GLMMs, you probably should not use sigma(model),
and instead you should create an appropriate value using the estimated random effects, e.g., from VarCorr(model). An example is provided in the “transformations” vignette.
P-value adjustments
The adjust argument specifies a multiplicity adjustment for tests or confidence intervals. This adjustment always is applied separately to each table or sub-table that you see in the printed output
(see rbind.emmGrid for how to combine tables). If there are non-estimable cases in a by group, those cases are excluded before determining the adjustment; that means there could be different
adjustments in different groups.
The valid values of adjust are as follows:
Uses the Studentized range distribution with the number of means in the family. (Available for two-sided cases only.)
Computes \(p\) values from the \(F\) distribution, according to the Scheffe critical value of \(\sqrt{rF(\alpha; r, d)}\), where \(d\) is the error degrees of freedom and \(r\) is the rank of the
set of linear functions under consideration. By default, the value of r is computed from object@linfct for each by group; however, if the user specifies an argument matching scheffe.rank, its
value will be used instead. Ordinarily, if there are \(k\) means involved, then \(r = k - 1\) for a full set of contrasts involving all \(k\) means, and \(r = k\) for the means themselves. (The
Scheffe adjustment is available for two-sided cases only.)
Makes adjustments as if the estimates were independent (a conservative adjustment in many cases).
Multiplies \(p\) values, or divides significance levels by the number of estimates. This is a conservative adjustment.
Uses our ownad hoc approximation to the Dunnett distribution for a family of estimates having pairwise correlations of \(0.5\) (as is true when comparing treatments with a control with equal
sample sizes). The accuracy of the approximation improves with the number of simultaneous estimates, and is much faster than "mvt". (Available for two-sided cases only.)
Uses the multivariate \(t\) distribution to assess the probability or critical value for the maximum of \(k\) estimates. This method produces the same \(p\) values and intervals as the default
summary or confint methods to the results of as.glht. In the context of pairwise comparisons or comparisons with a control, this produces “exact” Tukey or Dunnett adjustments, respectively.
However, the algorithm (from the mvtnorm package) uses a Monte Carlo method, so results are not exactly repeatable unless the same random-number seed is used (see set.seed). As the family size
increases, the required computation time will become noticeable or even intolerable, making the "tukey", "dunnettx", or others more attractive.
Makes no adjustments to the \(p\) values.
For tests, not confidence intervals, the Bonferroni-inequality-based adjustment methods in p.adjust are also available (currently, these include "holm", "hochberg", "hommel", "bonferroni", "BH",
"BY", "fdr", and "none"). If a p.adjust.methods method other than "bonferroni" or "none" is specified for confidence limits, the straight Bonferroni adjustment is used instead. Also, if an adjustment
method is not appropriate (e.g., using "tukey" with one-sided tests, or with results that are not pairwise comparisons), a more appropriate method (usually "sidak") is substituted.
In some cases, confidence and \(p\)-value adjustments are only approximate – especially when the degrees of freedom or standard errors vary greatly within the family of tests. The "mvt" method is
always the correct one-step adjustment, but it can be very slow. One may use as.glht with methods in the multcomp package to obtain non-conservative multi-step adjustments to tests.
Warning: Non-estimable cases are included in the family to which adjustments are applied. You may wish to subset the object using the [] operator to work around this problem.
The cross.adjust argument is a way of specifying a multiplicity adjustment across the by groups (otherwise by default, each group is treated as a separate family in regards to multiplicity
adjustments). It applies only to \(p\) values. Valid options are one of the p.adjust.methods or "sidak". This argument is ignored unless it is other than "none", there is more than one by group, and
they are all the same size. Under those conditions, we first use adjust to determine the within-group adjusted \(p\) values. Imagine each group's adjusted \(p\) values arranged in side-by-side
columns, thus forming a matrix with the number of columns equal to the number of by groups. Then we use the cross.adjust method to further adjust the adjusted \(p\) values in each row of this matrix.
Note that an overall Bonferroni (or Sidak) adjustment is obtainable by specifying both adjust and cross.adjust as "bonferroni" (or "sidak"). However, less conservative (but yet conservative) overall
adjustments are available when it is possible to use an “exact” within-group method (e.g., adjust = "tukey" for pairwise comparisons) and cross.adjust as a conservative adjustment. [cross.adjust
methods other than "none", "bonferroni", or "sidak" do not seem advisable, but other p.adjust methods are available if you can make sense of them.]
Tests of significance, nonsuperiority, noninferiority, or equivalence
When delta = 0, test statistics are the usual tests of significance. They are of the form (estimate - null)/SE. Notationally:
\(H_0: \theta = \theta_0\) versus
\(H_1: \theta < \theta_0\) (left-sided), or
\(H_1 \theta > \theta_0\) (right-sided), or
\(H_1: \theta \ne \theta_0\) (two-sided)
The test statistic is
\(t = (Q - \theta_0)/SE\)
where \(Q\) is our estimate of \(\theta\); then left, right, or two-sided \(p\) values are produced, depending on side.
When delta is positive, the test statistic depends on side as follows.
Left-sided (nonsuperiority)
\(H_0: \theta \ge \theta_0 + \delta\) versus \(H_1: \theta < \theta_0 + \delta\)
\(t = (Q - \theta_0 - \delta)/SE\)
The \(p\) value is the lower-tail probability.
Right-sided (noninferiority)
\(H_0: \theta \le \theta_0 - \delta\) versus \(H_1: \theta > \theta_0 - \delta\)
\(t = (Q - \theta_0 + \delta)/SE\)
The \(p\) value is the upper-tail probability.
Two-sided (equivalence)
\(H_0: |\theta - \theta_0| \ge \delta\) versus \(H_1: |\theta - \theta_0| < \delta\)
\(t = (|Q - \theta_0| - \delta)/SE\)
The \(p\) value is the lower-tail probability.
Note that \(t\) is the maximum of \(t_{nonsup}\) and \(-t_{noninf}\). This is equivalent to choosing the less significant result in the two-one-sided-test (TOST) procedure.
Non-estimable cases
When the model is rank-deficient, each row x of object's linfct slot is checked for estimability. If sum(x*bhat) is found to be non-estimable, then the string NonEst is displayed for the estimate,
and associated statistics are set to NA. The estimability check is performed using the orthonormal basis N in the nbasis slot for the null space of the rows of the model matrix. Estimability fails
when \(||Nx||^2 / ||x||^2\) exceeds tol, which by default is 1e-8. You may change it via emm_options by setting estble.tol to the desired value.
See the warning above that non-estimable cases are still included when determining the family size for P-value adjustments.
Warning about potential misuse of P values
Some in the statistical and scientific community argue that the term “statistical significance” should be completely abandoned, and that criteria such as “p < 0.05” never be used to assess the
importance of an effect. These practices can be too misleading and are prone to abuse. See the “basics” vignette for more discussion.
warp.lm <- lm(breaks ~ wool * tension, data = warpbreaks)
warp.emm <- emmeans(warp.lm, ~ tension | wool)
warp.emm # implicitly runs 'summary'
#> wool = A:
#> tension emmean SE df lower.CL upper.CL
#> L 44.6 3.65 48 37.2 51.9
#> M 24.0 3.65 48 16.7 31.3
#> H 24.6 3.65 48 17.2 31.9
#> wool = B:
#> tension emmean SE df lower.CL upper.CL
#> L 28.2 3.65 48 20.9 35.6
#> M 28.8 3.65 48 21.4 36.1
#> H 18.8 3.65 48 11.4 26.1
#> Confidence level used: 0.95
confint(warp.emm, by = NULL, level = .90)
#> tension wool emmean SE df lower.CL upper.CL
#> L A 44.6 3.65 48 38.4 50.7
#> M A 24.0 3.65 48 17.9 30.1
#> H A 24.6 3.65 48 18.4 30.7
#> L B 28.2 3.65 48 22.1 34.3
#> M B 28.8 3.65 48 22.7 34.9
#> H B 18.8 3.65 48 12.7 24.9
#> Confidence level used: 0.9
# --------------------------------------------------------------
pigs.lm <- lm(log(conc) ~ source + factor(percent), data = pigs)
pigs.emm <- emmeans(pigs.lm, "percent", type = "response")
summary(pigs.emm) # (inherits type = "response")
#> percent response SE df lower.CL upper.CL
#> 9 31.4 1.28 23 28.8 34.1
#> 12 37.5 1.44 23 34.7 40.6
#> 15 39.0 1.70 23 35.6 42.7
#> 18 42.3 2.24 23 37.9 47.2
#> Results are averaged over the levels of: source
#> Confidence level used: 0.95
#> Intervals are back-transformed from the log scale
summary(pigs.emm, calc = c(n = ".wgt.")) # Show sample size
#> percent response SE df n lower.CL upper.CL
#> 9 31.4 1.28 23 8 28.8 34.1
#> 12 37.5 1.44 23 9 34.7 40.6
#> 15 39.0 1.70 23 7 35.6 42.7
#> 18 42.3 2.24 23 5 37.9 47.2
#> Results are averaged over the levels of: source
#> Confidence level used: 0.95
#> Intervals are back-transformed from the log scale
# For which percents is EMM non-inferior to 35, based on a 10% threshold?
# Note the test is done on the log scale even though we have type = "response"
test(pigs.emm, null = log(35), delta = log(1.10), side = ">")
#> percent response SE df null t.ratio p.value
#> 9 31.4 1.28 23 35 -0.360 0.6390
#> 12 37.5 1.44 23 35 4.295 0.0001
#> 15 39.0 1.70 23 35 4.635 0.0001
#> 18 42.3 2.24 23 35 5.384 <.0001
#> Results are averaged over the levels of: source
#> Statistics are tests of noninferiority with a threshold of 0.09531
#> P values are right-tailed
#> Tests are performed on the log scale
con <- contrast(pigs.emm, "consec")
#> contrast ratio SE df null t.ratio p.value
#> percent12 / percent9 1.20 0.0671 23 1 3.202 0.0109
#> percent15 / percent12 1.04 0.0604 23 1 0.650 0.8613
#> percent18 / percent15 1.09 0.0750 23 1 1.194 0.5200
#> Results are averaged over the levels of: source
#> P value adjustment: mvt method for 3 tests
#> Tests are performed on the log scale
test(con, joint = TRUE)
#> df1 df2 F.ratio p.value
#> 3 23 7.981 0.0008
# default Scheffe adjustment - rank = 3
summary(con, infer = c(TRUE, TRUE), adjust = "scheffe")
#> contrast ratio SE df lower.CL upper.CL null t.ratio p.value
#> percent12 / percent9 1.20 0.0671 23 1.011 1.42 1 3.202 0.0343
#> percent15 / percent12 1.04 0.0604 23 0.872 1.24 1 0.650 0.9344
#> percent18 / percent15 1.09 0.0750 23 0.882 1.34 1 1.194 0.7027
#> Results are averaged over the levels of: source
#> Confidence level used: 0.95
#> Conf-level adjustment: scheffe method with rank 3
#> Intervals are back-transformed from the log scale
#> P value adjustment: scheffe method with rank 3
#> Tests are performed on the log scale
# Consider as some of many possible contrasts among the six cell means
summary(con, infer = c(TRUE, TRUE), adjust = "scheffe", scheffe.rank = 5)
#> contrast ratio SE df lower.CL upper.CL null t.ratio p.value
#> percent12 / percent9 1.20 0.0671 23 0.976 1.47 1 3.202 0.1090
#> percent15 / percent12 1.04 0.0604 23 0.841 1.28 1 0.650 0.9940
#> percent18 / percent15 1.09 0.0750 23 0.845 1.40 1 1.194 0.9165
#> Results are averaged over the levels of: source
#> Confidence level used: 0.95
#> Conf-level adjustment: scheffe method with rank 5
#> Intervals are back-transformed from the log scale
#> P value adjustment: scheffe method with rank 5
#> Tests are performed on the log scale
# Show estimates to more digits
print(test(con), digits = 7)
#> contrast ratio SE df null t.ratio p.value
#> percent12 / percent9 1.196684 0.06710564 23 1 3.202 0.0110
#> percent15 / percent12 1.038570 0.06042501 23 1 0.650 0.8613
#> percent18 / percent15 1.085945 0.07499759 23 1 1.194 0.5201
#> Results are averaged over the levels of: source
#> P value adjustment: mvt method for 3 tests
#> Tests are performed on the log scale
# --------------------------------------------------------------
# Cross-adjusting P values
prs <- pairs(warp.emm) # pairwise comparisons of tension, by wool
test(prs, adjust = "tukey", cross.adjust = "bonferroni")
#> wool = A:
#> contrast estimate SE df t.ratio p.value
#> L - M 20.556 5.16 48 3.986 0.0013
#> L - H 20.000 5.16 48 3.878 0.0018
#> M - H -0.556 5.16 48 -0.108 1.0000
#> wool = B:
#> contrast estimate SE df t.ratio p.value
#> L - M -0.556 5.16 48 -0.108 1.0000
#> L - H 9.444 5.16 48 1.831 0.3407
#> M - H 10.000 5.16 48 1.939 0.2777
#> P value adjustment: tukey method for comparing a family of 3 estimates
#> Cross-group P-value adjustment: bonferroni
# Same comparisons taken as one big family (more conservative)
test(prs, adjust = "bonferroni", by = NULL)
#> contrast wool estimate SE df t.ratio p.value
#> L - M A 20.556 5.16 48 3.986 0.0014
#> L - H A 20.000 5.16 48 3.878 0.0019
#> M - H A -0.556 5.16 48 -0.108 1.0000
#> L - M B -0.556 5.16 48 -0.108 1.0000
#> L - H B 9.444 5.16 48 1.831 0.4396
#> M - H B 10.000 5.16 48 1.939 0.3504
#> P value adjustment: bonferroni method for 6 tests | {"url":"https://rvlenth.github.io/emmeans/reference/summary.emmGrid.html","timestamp":"2024-11-14T06:58:01Z","content_type":"text/html","content_length":"62045","record_id":"<urn:uuid:783bbb0a-000b-4acd-8cac-2f9e3f423369>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00014.warc.gz"} |
Three Quantum Particles Hardy Entanglement from the Topology of Cantorian-Fractal Spacetime and the Casimir Effect as Dark Energy – A Great Opportunity for Nanotechnology
[1] L. Hardy, Nonlocality of two particles without inequalities for almost all entangled states. Phys. Rev. Lett. Vol. 71(11), 1993, pp. 1665-1668.
[2] J.S. Bell, Speakable and Unspeakable in Quantum Mechanics. Cambridge University Press, Cambridge 1991)
[3] Ji-Huan He et al, Quantum golden mean entanglement test as the signature of the fractality of micro spacetime. Nonlinear Sci. Lett B, Vol. 1(2), 2011, pp. 45-50.
[4] Mohamed S. El Naschie, Quantum Entanglement as a Consequence of a Cantorian Micro Spacetime Geometry. Journal of Quantum Information Science, Vol. 1(2), 2011, pp. 50-53.
[5] Mohamed S. El Naschie, Quantum Entanglement: Where Dark Energy and Negative, Accelerated Expansion of the Universe Comes from. Journal of Quantum Information Science, Vol. 3(2), 2013, pp 55-57.
[6] R. Penrose, The Road to Reality. J. Cape, London, UK. 2004
[7] M.S. El Naschie, M.A. Helal, L. Marek-Crnjac and Ji-Huan He, Transfinite corrections as a Hardy type quantum entanglement. Fractal Spacetime & Noncommutative Geometry in Quantum & High Energy
Physics, Vol. 2(1), 2012, pp. 98-101.
[8] M.S. El Naschie, Ji-Huan He, S. Nada, L. Marek-Crnjac and M. Helal, Golden mean computer for high energy physics. Fractal Spacetime and Noncommutative Geometry in Quantum and High Energy
Physics. Vol. 2(2), 2012, pp. 80-92.
[9] M.S. El Naschie and S.A. Olsen, When zero is equal one: A set theoretical resolution of quantum paradoxes. Fractal Spacetime & Noncommutative Geometry in Quantum High energy Physics, Vol. 1(1),
2011, pp. 11-24.
[10] M.S. El Naschie, Electroweak connection and universality of Hardy’s quantum entanglement. Fractal Spacetime & Noncommutative Geometry in Quantum High energy Physics, Vol. 1(1), 2011, pp. 25-30.
[11] L. Marek-Crnjac, Ji-Huan He and M.S. El Naschie, On the universal character of Hardy’s quantum entanglement and its geometrical-topological interpretation. Fractal Spacetime & Noncommutative
Geometry in Quantum & High Energy Phys. Vol. 2(2), 2012, pp. 118-12.
[12] M.S. El Naschie, L. Marek-Crnjac and Ji-Huan He, Using Hardy’s entanglement, Nash embedding and quantum groups to derive the four dimensionality of spacetime. Fractal Spacetime & Noncommutative
Geometry in Quantum & High Energy Phys. Vol. 2(2), 2012, pp. 107-112.
[13] Mohamed S. El Naschie, Experimentally Based Theoretical Arguments that Unruh's Temperature, Hawking's Vacuum Fluctuation and Rindler's Wedge Are Physically Real. American Journal of Modern
Physics, Vol. 2(6), 2013, pp. 357-361.
[14] Mohamed S. El Naschie, A Rindler-KAM Spacetime Geometry and Scaling the Planck Scale Solves Quantum Relativity and Explains Dark Energy. International Journal of Astronomy and Astrophysics, Vol.
3(4), 2013, pp. 483-493.
[15] Mohamed S. El Naschie, Topological-Geometrical and Physical Interpretation of the Dark Energy of the Cosmos as a “Halo” Energy of the Schrödinger Quantum Wave
Journal of Modern Physics, Vol. 4
(5), 2013, pp. 591-596.
[16] M.S. El Naschie, The quantum gravity Immirzi parameter – A general physical and topological interpretation. Gravity and Cosmology, Vol. 19(3), 2013, pp. 151-153.
[17] Mohamed S. El Naschie, Compactified dimensions as produced by quantum entanglement, the four dimensionality of Einstein’s smooth spacetime and ‘tHooft’s 4-ε fractal spacetime. American Journal
of Astronomy & Astrophysics, Vol. 2(3), 2014, pp. 34-37.
[18] Mohamed S. El Naschie, Electromagnetic—pure gravity connection via Hardy’s quantum entanglement.
Journal of Electromagnetic Analysis and Applications, Vol. 6(9), 2014, pp. 233-237.
[19] L. Marek-Crnjac and Ji-Huan He, An Invitation to El Naschie’s theory of Cantorian space-time and dark energy. International Journal of Astronomy and Astrophysics, Vol. 3(4), 2013, pp. 464-471.
[20] M.S. El Naschie, A review of E-infinity and the mass spectrum of high energy particle physics. Chaos, Solitons & Fractals, Vol. 19(1), 2004, pp. 209-236.
[21] M.S. El Naschie, Superstrings, knots and noncommutative geometry in E-infinity space. International Journal of Theoretical Physics, Vol. 37(12), 1998, pp. 2935-2951.
[22] Mohamed S. El Naschie, On a new elementary particle from the disintegration of the symplectic 't Hooft-Veltman-Wilson fractal spacetime. World Journal of Nuclear Science and ATechnology, Vol. 4
(4), 2014, pp. 216-221.
[23] W. Tan, Y. Li, H.Y. Kong and M.S. El Naschie, From nonlocal elasticity to nonlocal spacetime and nanoscience. Bubbfil Nano Technology, Vol. 1(1), 2014, pp. 3-12.
[24] D. Heiss (Editor):, Fundamentals of Quantum Information. Springer, Berlin 2002.
[25] I. Gengtsson and K. Zyczkowski, Geometry of Quantum States. Cambridge University Press, Cambridge 2006.
[26] M.A. Nielsen and I.L. Chuang, Quantum Computation and Quantum Information. Cambridge University Press, Cambridge, 2010.
[27] A. Furusawa and P. van Loock, Quantum Teleportation and Entanglement. Wiley-VCH, Weinheim, Germany, 2011.
[28] J.K. Pachos, Introduction to Topological Quantum Computation. Cambridge University Press, Cambridge, 2012.
[29] D.G. Marinescu and G.M. Marinescu, Classical and Quantum Information. Elsevier, Amsterdam, 2012.
[30] Mohamed S. El Naschie, Casimir-like energy as a double Eigenvalues of quantumly entangled system leading to the missing dark energy density of the cosmos. International Journal of High Energy
Physics, Vol. 1(5), 2014, pp. 55-63.
[31] J. Matsumoto, The Casimir effect as a candidate of dark energy. arXiv: 1303.4067[hep-th]26December 2013.
[32] K. Eric Drexler, Engines of Creation. Fourth Estate Ltd., London, 1990.
[33] P. Day (Editor), Unveiling The Microcosmos. Oxford University Press, Oxford, 1996.
[34] S.E. Lyshevski, Nano and Microelectromechanical Systems. CRC Press, Boca Ratan, 2001.
[35] L.E. Foster, Nanotechnology, Science, Innovation and Opportunity. Prentice Hall, Boston, 2006.
[36] M. Krummenacker and J. Lewis (Editors), Prospects in Nanotechnology. John Wiley, New York, 1995.
[37] M.S. El Naschie, Nanotechnology for the developing world. Chaos, Solitons & Fractals, Vol. 30(4), 2006, pp. 769-773.
[38] M.S. El Naschie, Chaos and fractals in nano and quantum technology. Chaos, Solitons & Fractals, Vol. 9(10), 1998, pp. 1793-1802.
[39] M.S. El Naschie, Nanotechnology and the political economy of the developing world. International, May 2007, pp. 7-12. (Periodical International Economic Magazine by AS&S Publishing Ltd, Camden
Town, London, UK, Registration No. 04761267).
[40] M.S. El Naschie, Can nanotechnology slow the aging process by inteferring with the arrow of time. International, June 2007, pp. 10-15. (Periodical International Economic Magazine by AS&S
Publishing Ltd, Camden Town, London, UK, Registration No. 04761267).
[41] M. Aboulanan, The making of the future via nanotechnology (in Arabic). International, July 2007, pp. 32-35. (Periodical International Economic Magazine by AS&S Publishing Ltd, Camden Town,
London, UK, Registration No. 04761267).
[42] M.S. El Naschie, From relativity to deterministic chaos in science and society. International, August 2007, pp. 11-17. (Periodical International Economic Magazine by AS&S Publishing Ltd, Camden
Town, London, UK, Registration No. 04761267).
[43] M.S. El Naschie, The political economy of nanotechnology and the developing world. International Journal of Electrospun Nanofibers and Application. I(I), 2007, pp. 41-50. Published by Research
Science Press, India.
[44] D. Brito and J. Rosellon, Energy and Nanotechnology: Prospects for solar energy in the 21st century. The James A. Baker III Institute for Public Policy of Rice University, December 2005, pp.
[45] O.E. Rössler and M.S. El Naschie, Interference through causality vacillation. In Symposium on the Foundations of Modern Physics, Helsinki, Finland, June 1994, pp. 13-16.
[46] O.E. Rössler and M.S. El Naschie: Interference is Exophysically Absent. In Endophysics – The World As An Interface. World Scientific, Singapore, 1998, pp. 159-160.
[47] M.S. El Naschie, A note on quantum gravity and Cantorian spacetime. Chaos, Solitons & Fractals, 8(1), p. 131-133 (1997).
[48] M.S. El Naschie, The symplictic vacuum exotic quasi particles and gravitational instantons. Chaos, Solitons & Fractals, 22(1), 2004, pp. 1-11.
[49] M. Agop, E-infinity Cantorian spacetime, polarization gravitational field and van der Waals-type forces. Chaos, Solitons & Fractals, 18(1), 2003, pp. 1-16.
[50] Mohamed S. El Naschie, A Rindler-KAM spacetime geometry and scaling the Planck scale solves quantum relativity and explains dark energy. International Journal of Astronomy and Astrophysics, Vol.
3(4), 2013, pp. 483-493.
[51] J. Cugnon, The Casimir effect and the vacuum energy. Few-body System, 53(1-2), 2012, pp. 181-188.
[52] K.A. Milton, Resource Letter VWCPF-1 van der Waals and Casimir-Polder forces. American Journal of Physics, 79, 2011, pp. 697.
[53] M. Ito, Gravity, higher dimensions, nanotechnology and particles physics. Journal of Physics, Conference Series, 89(1), 2007, p. 1-8.
[54] M.S. El Naschie, Casimir-like energy as a double Eigenvalue of quantumly entangled system leading to the missing dark energy density of the cosmos. International Journal of High Energy Physics,
1(5), 2014, pp. 55-63. | {"url":"https://sciencepublishinggroup.com/article/10.11648/j.nano.20150301.11","timestamp":"2024-11-11T23:08:37Z","content_type":"text/html","content_length":"90681","record_id":"<urn:uuid:e34a4395-47c4-4797-af8f-fee38d0bfa0d>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00369.warc.gz"} |
RE: st: definition of pseudo R^2 for dprobit or probit
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RE: st: definition of pseudo R^2 for dprobit or probit
From "Nick Cox" <[email protected]>
To <[email protected]>
Subject RE: st: definition of pseudo R^2 for dprobit or probit
Date Tue, 28 Oct 2003 10:20:38 -0000
I agreed strongly with Richard before his last paragraph.
My own bias is to try to steer the discussion in
the opposite direction, away from all ideas of "best":
* That discussion goes in a circle with a discussion of criteria
for "best", and there are lots, as everyone knows. After all,
we go round and round on preferred measures of location, scale,
shape, association in two-way tables, rank correlation, and so
* There are all sorts of theoretical and practical arguments for
saying that in many fields far too much emphasis is already placed
on single-number figures of merit (as compared with looking
at graphs, looking at residuals, detailed discussion of the
scientific and practical issues behind variable choice, model
structure, etc.). Sometimes it seems that researchers will
spend a very long time producing or collating data, formatting
it for software, writing programs, ..., and then expect to make a
quick decision on model virtues based on a few magic numbers!
* These questions of which measures to use
seem to arise primarily when response variables are categorical
(wide sense). The even wider context including measured responses
is, I hope everyone will agree, vital. After all, the history
presumably is that people wanted measures fulfilling the same
role as R^2 in (say) multiple regression -- even if that role
is often aggressive, not analytical, using R^2 to intimidate,
rather than to inform.
There are two simple ideals, it seems to me: that everyone
should state clearly what definition of R^2 they are
using; and that in principle enough information should be
provided to allow other measures to be calculated. Beyond
that, if measures fail to agree numerically, then choosing
one as best requires a special argument (which,
for all I know, could be "this is what people use in this
field, so I'll use it too").
There are more platitudes posing as homespun wisdom at
(and also some references and some code fragments).
[email protected]
> -----Original Message-----
> From: [email protected]
> [mailto:[email protected]]On Behalf Of Richard
> Williams
> Sent: 28 October 2003 02:32
> To: [email protected]
> Subject: Re: st: definition of pseudo R^2 for dprobit or probit
> At 08:04 PM 10/27/2003 -0600, Scott Merryman wrote:
> >[R] maximize, Methods and Formulas section
> >
> >Pseudo R2 = 1 - L1/L0, where L1 is the log likelihood of
> the full model and L0
> >is the log likelihood of the constant-only model.
> That is one of a couple of equivalent formulas but probably
> the simplest to
> write in an email message! Certainly clearer than what I
> wrote earlier.
> As a sidelight, this is one of many statistics that claims
> the name of
> "Pseudo R2". It would be nice if Stata explicitly labeled it as
> McFadden's R2, and perhaps reported a couple of the other
> alternatives in
> case anybody wants them.
> Of the various alternatives, McFadden's R2 seems to have
> emerged as the
> favorite and best. Anybody strongly disagree and think
> something else is
> better?
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2003-10/msg00735.html","timestamp":"2024-11-03T12:14:30Z","content_type":"text/html","content_length":"11355","record_id":"<urn:uuid:6ed44328-89c6-4164-84ff-4eb90681c90b>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00190.warc.gz"} |
Neophytos Charalambides receives Best Poster Award for research in the area of Data Science
The research can be applied to a wide range of big data applications that rely on the multiplication of two matrices in linear algebra.
Doctoral student Neophytos Charalambides received a Best Poster award from the 2022 SIAM Conference on Mathematics of Data Science (MDS22) for research describing a new approach for compressing
systems of linear equations. The resulting algorithm can be applied to a wide range of data science applications.
According to Neo, the work deals primarily with a fundamental problem and operation in linear algebra – that of multiplying two matrices (tables comprised of numbers), which appears in many
scientific and engineering disciplines. Though seemingly a relatively easy operation to carry out, with the advent of large data sets, it can take a lot of time to perform.
Since 1969, says Neo, when Volker Strassen first overcame the “cubic barrier” of matrix multiplication, there have been a lot of developments surrounding this problem. In this work, a randomized
approach is considered which compresses the matrices in such a way that high-quality approximations of their product are guaranteed; in terms of the Euclidean norm.
The approach draws ideas primarily from Randomized Numerical Linear Algebra, and is also related to graph compression and sparsification. Specifically, it is a direct generalization of a known
technique for compressing systems of linear equations, as well as the state-of-the-art approach for sparsifying large graphs through what is known as “effective resistances.”
The algorithm can be used in a variety of applications, such as extracting useful information from large networks, or even understanding the orbits of celestial bodies.
The research was conducted by Neo along with his advisor Prof. Alfred Hero, and Prof. Mert Pilanci at Stanford University. The poster was titled “Approximate Matrix Multiplication and Laplacian
Neo received two bachelor’s degrees (pure mathematics and electrical engineering) and two master’s degrees (mathematics and ECE) from Michigan, and a master’s degree in pure mathematics from Imperial
College London.
The Society for Industrial and Applied Mathematics (SIAM), in existence since 1952, has the goal of promoting “research that will lead to effective new mathematical and computational methods and
techniques for science, engineering, industry, and society.” | {"url":"https://eecsnews.engin.umich.edu/neophytos-charalambides-receives-best-poster-award-for-research-in-the-area-of-data-science/","timestamp":"2024-11-07T05:32:27Z","content_type":"text/html","content_length":"36905","record_id":"<urn:uuid:007bcd25-fb3f-4525-9932-354be1b5d4f4>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00866.warc.gz"} |
Urine Output Calculator
Last updated:
Urine Output Calculator
Our urine output calculator will show you an easy way of performing daily urine output calculations. Our tool will equip you with your patient's fluid balance and urine output in ml/ kg/ hr.
In the article below, we'll talk about the value of normal urine output per hour, dehydration, and the total body water volume. We'll also teach you how to calculate urine output in ml/ kg/ hr. 💧
We try our best to make our Omni Calculators as precise and reliable as possible. However, this tool can never replace a professional doctor's assessment. If any health condition bothers you, consult
a physician.
How to use the urine output calculator?
To calculate the urine output rate, you'll need the following data:
1. Your patient's age.
2. Your patient's weight.
3. The period of time over which the urine was collected.
4. The urine volume – the volume of urine collected during the given period of time.
5. Your patient's fluid intake during the given period of time.
6. It's ready! Our calculator will supply you with both the fluid balance and the urine output rate of your patient! 🚰
□ You will get a notification if your patient's urine output per hour is indicative of acute kidney injury.
□ Our calculator will let you know if your patient suffers from oliguria (peeing less than usual) or polyuria (peeing abnormally high volumes of urine).
Now it's time to go one step further. Check your patient's bladder volume and calculate the urine anion gap using the bladder volume calculator and the urine anion gap calculator, respectively.
How to calculate urine output in ml/kg/hr?
If you want to be better than our ml/ kg/ hr calculator, you need to practice! Follow our detailed instructions and the urine output calculation examples:
1. Collect your patient's weight, age, urine output, and the period over which the urine was collected. Our patient is 20 years old, weighs 80 kg, and we collected 3L (3000 mL) of urine during a
24-hours observation period.
2. Use the following equation to compute how much urine is output per hour:
Urine output (ml/kg/hr) = Collected urine / (Weight × Time),
□ Weight is given in kilograms (kg);
□ Collected urine is given in milliliters (mL); and
□ Time is given in hours.
Our patient's data:
x = 3000 / (80 × 24) = 3000 / 1920
x = 1.56 ml/kg/hr
3. Use the patient's age to determine if the urine output is within the normal range.
Our patient's over 18 years old – his urine output is 1.56 ml/kg/hr, which is within the normal range.
Hey, well done! 🎉
Discover the way to calculate the urine albumin creatinine ratio using the albumin creatinine ratio calculator.
Fluid balance
Fluid balance informs you whether your patient maintains their total body water volume. It allows you to correct the fluid intake, both orally and intravenously. In case of dehydration, the fluid
balance's value is negative. A regular person experiences symptoms of dehydration after the loss of around 7% of their total body water or 5% of their weight.
Fluid balance = Fluid intake - Collected urine
All of the variables are given in milliliters (mL).
You can also explore:
💡 Remember that your patient may lose significant amounts of water through their lungs, skin, and stool - especially when their body temperature is elevated.
What's the normal hourly urine output?
What's the minimum urine output per hour for healthy adults and children? Find out with one of the tables below!
For adults (≥18 years old)
Urine output (ml/kg/h) Meaning
<0.5 Oliguria
0.5-5 Healthy person
>5 Polyuria
For children (<18 years old)
Urine output (ml/kg/h) Meaning
<1 Oliguria
1-3 Healthy person
>3 Polyuria
How to calculate urine output for a 70 kg patient?
In order to do that, we need a bit more information:
• The amount of urine collected; and
• The time over which the urine was collected.
Then we may use the following equation:
Urine output (ml/kg/hr) = Collected urine / (Weight × Time)
Let's say that we gathered 300 ml of urine during a 6-hours observation.
Urine output (ml/kg/hr) = 300 ml / (70 kg × 6 hr)
Urine output (ml/kg/hr) = 0.71
What's the minimum urine output per hour?
A healthy urine output for an adult person should be greater than or equal to 0.5 ml/kg/hr. This minimum value is a bit different for children (<18 years old), which is 1 ml/kg/hr.
How do we measure urine output?
We can measure the urine output by inserting a Foley catheter into one's bladder. The catheter is a long tube that allows us to collect urine into the attached container. Then we can easily measure
the amount of urine collected during the given time.
We may also calculate the urine output per hour using the following equation:
Urine output (ml/kg/hr) = Collected urine / (Weight × Time) | {"url":"https://www.omnicalculator.com/health/urine-output","timestamp":"2024-11-13T07:49:09Z","content_type":"text/html","content_length":"560318","record_id":"<urn:uuid:2f4a7dbd-f832-4771-816c-37862aaca200>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00756.warc.gz"} |
Why to Encourage Your Child to Show Their Thinking | mathteacherbarbie.com
Why to Encourage Your Child to Show Their Thinking
“Show your work.” A common theme on math worksheets and tests. Yet, despite the frequent refrain, it’s actually quite difficult for teachers to describe what good examples of “showing work” actually
look like. It’s been called a “good and hard question to answer” by a college math instructor. Middle school math teacher Michelle Russell wrote an entire blog post about why this is difficult to
describe. However, if we go back to the “why” question, it might give us a glimpse into the “how.”
Students should show their work in math so that
• the teacher can identify understanding and misunderstanding
• teachers, students, parents, and tutors can identify errors and tailor tips accordingly
• some varieties of learning accommodations and needs can be unmasked and addressed at the most beneficial times
• students learn to communicate procedural steps while the stakes are lower and the problems are shorter
• students can identify and use patterns and efficiencies of thinking
• students can learn to correct errors along the way rather than having to start over each time
• students can demonstrate their brilliance to the adults around them
Why should students show their work in math?
Teacher’s Perspective
• Shown work helps the teacher know what concepts a student understands an what misunderstandings a student has. This allows the teacher to know what they might need to go over individually or,
when many students have the same misunderstandings, reteach.
• Shown work allows a teacher to know what types of errors students are making. This allows the teacher to give study, test-taking, and other tips targeted specifically to the types of errors the
student makes.
• Shown work can help a teacher identify when a student might benefit from learning accommodations or different types or learning strategies.
I just wanted to know that they understand the process behind the problem. If that looks like 7+7=14 or it looks like tally marks or an array, I didn’t care. But also once I saw that they
understood on a couple questions I never required it beyond that.
former teacher Letty camire
Student’s Perspective
Have you ever heard a list of directions or instructions that were either too short (leaving you with questions) or too long (leaving your eyes and brain swimming from too much information)? Learning
to show work well helps us learn what’s important and what’s unimportant to tell someone any time we have to communicate a step-by-step process to someone else.
Attempting to follow the “show work” instructions in a college engineering class, my friend Ward received the demoralizing feedback that the amount of minute detail he showed demonstrated that he
either “didn’t study enough” or “had no aptitude in mathematics.” To a young engineering student, you can imagine how this must have hit his heart and his dreams, especially since he only did this in
an attempt to follow the instructions given. I hope this is increasingly unusual among even university faculty, and I can not excuse this professor of this unfeeling feedback. However, it does
illustrate the idea of how learning the skill of balancing detail with big picture while young and while the problems are short can pay off by helping avoid awkward moments like my friend Ward faced.
• Showing work builds the student’s communication skills. Communicating mathematics can be one of the hardest parts of the subject. But the communication skills built in this process go beyond
• Showing work organizes the student’s thinking.
• Showing work allows the student to make corrections along the way rather than starting over each time.
• Showing work allows the student to use efficiencies in and notice patterns of similar problems.
Parent’s or Tutor’s Perspective
• Shown work allows a parent or tutor to help the student find any errors, correct them, and avoid them in the future.
• Shown work allows a parent or tutor to see into the processes and procedures students are learning and using in class. Thus, the parent or tutor can speak to the student at the level they’re
prepared to understand and build the student’s mathematical confidence.
• Shown work helps a parent or tutor recognize common errors the student makes. They can then strategize together how to avoid this general type of error in the future (eg, helping the student know
when to slow down, when to check arithmetic on a calculator (if allowed), etc.)
• Shown work can help a parent or tutor identify when a student might benefit from learning accommodations or different types or learning strategies.
• Shown work can help a parent or tutor get to know the student better by showing a glimpse into the student’s brain and how they think.
What happens when students don’t show their work?
• The teacher does not know what the student understands and doesn’t understand, only whether the student can recite rote facts.
• The student has to recreate the entire process from the beginning whenever a followup or similar question is asked.
• Student needs for reteaching, for accommodations, and other indicators are buried and/or hidden.
• Parents are frustrated with not understanding or knowing how math is being taught to their children in class.
• Students may make patterns of error types that are masked by the lack of work, but could be identified and corrected if work is shown.
• Students may make memory errors in the middle of a problem.
• Instead of learning the skills to explain processes in these smaller settings, students may struggle to know how to explain larger processes in situations where the results matter more.
• The student may miss out on partial credit if the final answer is wrong, or be penalized for not showing work.
• The student may struggle with alternative question types: “we know / we’re pretty sure the answer is _____. Can you prove it?”
What if my student is neurodivergent or has physical or fine-motor limitations?
You know your child best! Hopefully, you and your child’s teachers have access to the technology and support they may need. I encourage you to seek out solutions that do allow your student to show
their work. Many of the reasons above apply just as well to students with non-standard needs. In addition, many of these students are very smart, though it unfortunately takes the right communication
tools to be able to demonstrate that to teachers and others. Help them unmask the intelligence you know is there. Finding a way to allow your child to show their thinking, even when it’s difficult,
can show off their brilliance. Take advantage, as much as you can, of the advances in both learning research and in technology that allow students to communicate mathematically in ways that are
accessible to them.
You’ve Got This! | {"url":"https://mathteacherbarbie.com/why-to-encourage-your-child-to-show-their-thinking/","timestamp":"2024-11-01T19:30:09Z","content_type":"text/html","content_length":"83412","record_id":"<urn:uuid:a6619867-8eb2-45f2-b774-d9220ea89cc8>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00401.warc.gz"} |
WEIBULL.DIST - Excel docs, syntax and examples
The WEIBULL.DIST function in Excel calculates the Weibull distribution probability density function or the cumulative distribution function for a given value.
=WEIBULL.DIST(x, alpha, beta, cumulative)
x The value at which to evaluate the function.
alpha The shape parameter of the Weibull distribution.
beta The scale parameter of the Weibull distribution.
cumulative A logical value that determines the type of function to use. TRUE for the cumulative distribution function, and FALSE for the probability density function.
About WEIBULL.DIST 🔗
When dealing with distributions and assessing probabilities in Excel, turn to the WEIBULL.DIST function. This function aids in computing the probability density function or cumulative distribution
function of a Weibull distribution based on specified parameters. It proves valuable in various statistical analyses, reliability studies, and survival analysis where the Weibull distribution model
is applicable.
Examples 🔗
Let's say you want to find the probability density function value at x = 10 for a Weibull distribution with alpha = 2 and beta = 5. To calculate this, use the formula:
=WEIBULL.DIST(10, 2, 5, FALSE)
Suppose you need to determine the cumulative distribution function value at x = 12 for a Weibull distribution with alpha = 3 and beta = 10. The formula to compute this would be:
=WEIBULL.DIST(12, 3, 10, TRUE)
Ensure that the provided values for alpha and beta are greater than zero, as the Weibull distribution requires positive parameters. Additionally, remember to adhere to the logical values of TRUE or
FALSE for the cumulative argument to receive the intended type of output.
Questions 🔗
What does the alpha parameter represent in the WEIBULL.DIST function?
The alpha parameter in the WEIBULL.DIST function indicates the shape parameter of the Weibull distribution. It determines the shape of the distribution curve.
How is the choice between the probability density function and cumulative distribution function made in the WEIBULL.DIST function?
The decision between using the probability density function and cumulative distribution function is based on the logical value of the cumulative argument. TRUE selects the cumulative distribution
function, while FALSE chooses the probability density function.
Can the WEIBULL.DIST function handle negative values for the parameters?
No, the Weibull distribution requires positive parameters for alpha and beta. Negative values are not suitable inputs for this function.
In what type of statistical analyses is the WEIBULL.DIST function commonly used?
The WEIBULL.DIST function is frequently utilized in reliability studies, survival analysis, and various statistical analyses where the Weibull distribution serves as an appropriate model for the
Related functions 🔗
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Free Forum - 16 vs 10 at true count 0 HELP
16 vs 10 at true count 0 HELP
I have searched for hours and found no definitive answer. Alas, I post here.
I understand the index for 16 v. 10 is 0. Greater than 0 means stand, less than 0 means hit.
However, what about at a true count of exactly 0, when the running count itself is 0.
Illustrious 18 regulates its indices at a count greater than or equal to the index. Therefore, 16 v. 10 at an index of 0 means stand at a true count of 0. However, the Advanced Strategy Lesson 14 on
this very website state hit at 0 or lower.
I understand the margin on this decision is incredibly small. However, I feel there should be a definitive answer. I haven't found a simulator where I can run this scenario many times, or else I
would just figure it out myself.
Of course, this is when surrender is not available.
Definitive answer
"I understand the margin on this decision is incredibly small. However, I feel there should be a definitive answer."
You have your definitive answer*; you just failed to recognize it when you saw it. :-)
*"Illustrious 18 regulates its indices at a count greater than or equal to the index. Therefore, 16 v. 10 at an index of 0 means stand at a true count of 0."
Further Nitpicking
Ha thanks for your quick response and from somebody so prominent in blackjack nonetheless.
Naturally, I would expect your response to be stand, as you developed the Illustrious 18.
Why, then, does the GameMaster blackjack school here say to hit on count 0? Also, what gives with a discrepancy such as Illustrious 18 saying to double 9 v. 7 at 3 but GameMaster saying double 9 v. 7
at 6? I am being nitpicky I know, but I would like to understand everything before I start putting money on the line.
I agree with GameMaster for 16 vs 10
A spreadsheet that I wrote a long time ago shows that for an infinite number of decks at a true count of zero and a bet of $100, you would lose (on average) $57.52 by hitting and you would lose (on
average) $57.58 by standing. These numbers include the loss if the dealer has an ace in the hole. Thus, you would save a whopping 6 cents (per $100 bet) by hitting at a true count of zero. Hitting
also helps eat cards (which is desirable at neutral and negative counts).
My understanding is that Don�s numbers are based on computer simulations (done by others) which have been rounded to the nearest integer. There is nothing wrong with that. However, Don mistakenly
assumes that the numbers are fixed in stone and thus his answer is not quite correct for your question. In other words, the exact index for 16 vs. 10 is some small positive number (e.g. 0.01) which
has been rounded to 0. Treating the index as exactly 0 has negligible effect in actual play but it can not be used to answer your question precisely.
Dr 21
Not true
When is the last time you played BJ against an infinite number of decks?
Let me know when you write a spreadsheet for 16 v. 10 for a number of decks that we actually use to play blackjack, such as six, with a running count of exactly zero, taking into account every
possible holding of 16, and then let me know what you find.
I would welcome such a study. Naturally, when we make plays using indices, we are incorporating an entire bucket of a range over an entire integer, and we are averaging results. It is possible that
at the extreme left end of the zero bucket, i.e., at precisely zero, 16 v. 10 would be a hit. But, I don't think so.
Whole decks/half decks
"Also, what gives with a discrepancy such as Illustrious 18 saying to double 9 v. 7 at 3 but GameMaster saying double 9 v. 7 at 6?"
An obvious discrepancy such as this (one number double the other) can only be because the first divides by whole decks to ascertain TC while the second divides by half decks. But then, you should see
the difference for all such index plays.
The Hi-Lo TC for doubling 9 v. 7, dividing by whole decks, is not +6. So, you'll have to ask GaMaster what their problem is.
The indices are risk-averse. (nt)
Further observation
In the lesson you reference, they first state:
"The most common decision any player makes at Blackjack is whether to hit or stand, consequently this will be the most common basic strategy variation and you should learn all the important ones. The
first is with a hand of 16 against a dealer's up card of 10. You should stand if the count is over 0 and hit if it is 0 or lower. This means that if the running count is 1 or higher, stand. Since the
'decision' number is 0, it's not necessary to calculate the true count -- the running count will do in this situation. Don't get confused here. Almost all basic strategy variations rely on the true
count, but for those where the decision number is 0, the running count will suffice."
But then, later, below, they say:
"12 vs. 4 Stand at 0 or higher (Yes, if the running count is at all minus, you hit 12 against a 4. It drives the other players at the table crazy!!!)"
Now, you may think that they actually are trying to make a fine distinction between the two plays -- and that is your right -- but I'm not so sure. I think they are just being inconsistent in their
interpretation of what to do at exactly zero, but I may be wrong.
Bottom line: I think it's ridiculously close at exactly zero, with no practical importance whatsoever, but I understand the interest in knowing the theoretically "correct" answer.
Never thought of that. Should have.
Why don't you think so?
Don wrote:
It is possible that at the extreme left end of the zero bucket, i.e., at precisely zero, 16 v. 10 would be a hit. But, I don't think so.
It is interesting that your post was titled �Not true� but your response stated that it is possible that at precisely zero, 16 v. 10 would be a hit.
If a player should hit his 16 against a dealer 10 at a true count of zero with an infinite number of decks, then he should definitely hit with 8 or fewer decks! If he hits his 16, he is obviously
hoping to receive a 5 or less. If he gets a 6 or higher, he will lose regardless of the effect that his hitting has caused on the dealer�s hand. If he gets a 5 or less, not only has he improved his
hand, but he has increased the likelihood of the dealer busting (due to the removal of a small card). This increase in the likelihood of the dealer busting due to the removal of a single small card
would not show up in the �infinite deck� game. Thus, he would save even more than 6 cents (per $100 bet) by hitting with a finite number of decks.
In conclusion: If a player should hit his 16 against a dealer 10 with an infinite number of decks, then he should definitely hit with 8 or fewer decks!
Dr 21
Risk Averse
I guess I don't really understand the concept of "risk-averse".
At certain counts, shouldn't there be a most statistically favorable play just as basic strategy is the most statistically favorable play given the knowledge of only the hand dealt and the dealer's
upcard. I don't understand how doubling 9 v. 7 at 3 could be to "risky" as with 4 and 5 but at 6 it is "risk-averse."
I feel like either it is favorable to double 9 v. 7 at 3 or not. There isn't a "risk-averse" basic strategy and a "risky" basic strategy after all.
I appreciate the advice.
Dr 21
Would it be wise then to hit at exact index numbers when the count is unfavorable (say, less than 1) in order to eat cards?
For example, one source of Illustrious 18 (http://jay.purplewire.com/blackjack/ill18.html) says hit at less than or equal to 0 for 12 v. 4. Another source (http://wizardofodds.com/blackjack/count/
highlow1.html) implies stand at greater than or equal to 0 for 12 v. 4. So, at exactly 0, one says hit, the other stand. But, since the count is at 0, should we hit to eat up cards to hopefully raise
the count? Or have you done a spreadsheet for this and found it favorable one way or the other?
I think I'm being too meticulous but it does drive me a bit crazy when two source of the same figure (Illustrious 18) differ in its subtleties.
Also, is there some program that I can run "infinite" trials on to figure this out myself?
Yes, you are being too meticulous
What you should get from Don's and my posts is we are talking pennies on the $100 for very close plays. You do not win money in this game by knowing every index and hand dependent index. In fact, a
counter can win with basic strategy only. (I personally would recommend that you at least use Don's I18 except for splitting tens.)
The way to win money is to bet high when the count is in your favor and bet as small as possible (or not at all) when the house has the edge. That's the easy part. The hard part is getting away with
it. Of course, you cannot over bet your bankroll and expect to survive. I would recommend Don's book to help you to learn the meaning of risk averse and appropriate bet spreads based on the count
(and your bankroll).
When the house has the edge, you will generally have a very small bet (e.g. $10) and the cost of an incorrect hit or stand for close plays will generally be less than a couple pennies. Certainly in
single and double deck, seeing an extra card is worth several pennies and I agree that hitting (even when mathematically incorrect) is worth it.
Dr 21
Risk-averse indices seek to sacrifice a small amount of EV to decrease variance.
Check the glossary in the Blackjack Basics section of this page. There is a lot of good information available here, even on the free site.
Furthermore ...
"Risk-averse indices seek to sacrifice a small amount of EV to decrease variance."
The point, however, is that we actually win MORE using r-a indices. How is this possible? Because, if we bet optimally, our bet size is a function of bankroll times edge divided by risk. So, by
lowering the risk even more than we lower the e.v., we increase the optimal bet size and, eventually, the SCORE, or hourly win rate.
So, if we are able to bet optimally, we don't win less when we use r-a indices; we win more.
Finally, as you can see from the study on this topic in BJA3, the whole concept doesn't amount to a hill of beans! ;-)
Thanks for the clarification Don, i knew there would be more to it! (nt)
Gamemaster is batting a thousand
First, let me say your spreadsheet is apparantly quite accurate. Griffin lists 0.06 (in percent) for the infinite deck EV delta on hard 16 v T (pg. 231 ToB), so that matches your 6 cents on a $100
bet exactly.
But I can't agree with: "If a player should hit his 16 against a dealer 10 at a true count of zero with an infinite number of decks, then he should definitely hit with 8 or fewer decks! If he hits
his 16, he is obviously hoping to receive a 5 or less. If he gets a 6 or higher, he will lose regardless of the effect that his hitting has caused on the dealer�s hand."
The problem with this logic is that the strategy EoRs for 6 and higher are all over the map!
From Don's BJA3 pg. 515 ...
EoR 16 v T (6): +1.6446
EoR 16 v T (7): -0.7109
EoR 16 v T (8): -0.0567
EoR 16 v T (9): +0.5524
EoR 16 v T (T): +1.1151
It seems the fact a 7 might show up as dealer's hole card is more important than its role as a hit card.
There's a fairly simple procedure for calculating an index based purely on the EoRs. It's described here:
I carried out the procedure for 16 v T, based on removal of the T upcard, for 1, 2, 6 and 8 decks. They came out as follows...
1 deck index: -0.1238
2 deck index: -0.02502
6 deck index: +0.04428
8 deck index: +0.0532
These follow the interpolation by 1/decks fairly well. So apparantly, the index changes in sign somewhere between 2 and 6 decks.
I also did the 1 deck calculation with the player's first two cards removed along with the T up. That makes quite a difference for one deck...
1 deck index for T,6 v T: +3.72646
1 deck index for 9,7 v T: -0.20216
The gamemaster does say to hit h16 v T for 6 decks at TC=0, and I think he's correct about that, especially since hitting is a little more risk averse than standing (i.e. better chance for a push).
Then in his single deck matrix (lesson 19) he has an index of 0 for 9,7 v T (and says to stand right at 0) and an index of +4 for T,6 v T. Right again! The man is psychic.
High quality post! (nt)
True Count = 0 is not the same as Basic Strategy
ET Fan,
If I am reading your post correctly, you are confusing basic strategy with the correct strategy for a true count of zero. In the original post, jcbsbrd asked about the correct strategy for 16 v 10
for a true count of zero. For a finite number of cards at a TC = 0, it is assumed that you have already included your two (or more) cards and the dealers up card in the count and thus the EOR of the
dealer�s ten and the exact cards that you are holding to make your 16 do not affect the correct strategy for a true count of zero. If the question was about basic strategy, then the cards in your
hand and the dealer�s up card should be considered in the analysis.
Dr 21
Of course it isn't
"True Count = 0 is not the same as Basic Strategy"
I just reread my post and I can't fathom how you got the impression I didn't know that.
"For a finite number of cards at a TC = 0, it is assumed that you have already included your two (or more) cards and the dealers up card in the count and thus the EOR of the dealer�s ten and the
exact cards that you are holding to make your 16 do not affect the correct strategy for a true count of zero."
The cards were, in fact, removed, therefore the effects of removal have an effect. Those particular cards are no longer available for player or dealer hitting, or for the dealer's hole card. The
count is just a general indicator. It doesn't have that level of specificity. I'm sure you can see that. | {"url":"https://bj21.com/boards/free/sub_boards/free/topics/16-vs-10-at-true-count-0-help","timestamp":"2024-11-14T00:14:54Z","content_type":"text/html","content_length":"57147","record_id":"<urn:uuid:55b5578c-b5e8-4c15-a1a4-74c6328bba4d>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00882.warc.gz"} |
Dynamics of Electron in TEM Wave Field
Big amount of works deals with solution of differential equations, associated with electron motion in electromagnetic field, using methods of classical electrodynamics. Solution of equation of an
electron motion in TEM wave field is interesting task because this equation is mathematical model of big number of wave processes, which are used for researches of different physical processes. The
proposed work dedicated to finding the solution for the equation of an electron motion in TEM wave field in laboratory system of coordinates using the theory of almost periodic functions. The work
demonstrates that the projections of electron velocity on coordinate axis conform to the wave equation, and, consequently, could be expanded into generalized Fourier series at any value of the wave
and electron parameters. In the present work, the formulas received before for electron velocity projection on coordinate axis, are transformed to a well-behaved form, and are broken down into
non-perfect generalized Fourier series. Non-perfect Fourier series for projections of electron velocity on coordinate axis are found by means of plotting of complex series, which are called in the
theory of almost periodic functions as ”closure of set”. For approximate computation of electron velocity it is possible to restrict oneself to finite number of series harmonics. Application of
method of electron velocity components transformation into generalized Fourier series made it possible to find in electron velocity components series terms, which do not depend on time and are equal
to average magnitudes of the respective values. Electron velocity components present functions of initial magnitudes of electron velocity components, of generalized phase magnitude and of the wave
parameters. Initial magnitudes are not preset at random, but calculated from the equations, the type of which is specified in the work. Electron trajectory in coordinate space is calculated by
integrating of the respective expressions for velocity projections on coordinate axis. For demonstration purpose the work deals with the example of electron dynamics in wave polarization plane with
consideration of only permanent addends and first harmonics of Fourier series for electron velocity projections on coordinate axis. An approximate solution of the equations of electron dynamics in
the plane of polarization of the wave is given. Solution for the equation of electron motion in TEM wave field in the laboratory coordinate system using the theory of almost periodic functions made
it possible to solve the problem of dynamics of relativistic electron in the field of progressing TEM wave. It made it possible to demonstrate the availability of time-independent summands in the
value of the speed of the electron, which moves in TEM wave. A very important circumstance is also the fact, that the theory makes it possible to investigate electron dynamics depending on the
original wave intensity.
• generalized Fourier series
• TEM wave
• wave equation
Dive into the research topics of 'Dynamics of Electron in TEM Wave Field'. Together they form a unique fingerprint. | {"url":"https://research.aalto.fi/en/publications/dynamics-of-electron-in-tem-wave-field","timestamp":"2024-11-09T12:30:17Z","content_type":"text/html","content_length":"65569","record_id":"<urn:uuid:10015770-1fbc-4a54-a5a3-8c5869ba7597>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00781.warc.gz"} |
AL3: Graph kernels under uncertainty
An important and growing branch of machine learning deals with graph-structured data. Typical applications ask to classify data in areas like chemo–informatics, e.g., to predict the toxicity of a
molecule. In this area, graph kernels form a rapidly developing set of techniques that are well suited for such tasks. One reason for their popularity is that they allow the use of welldeveloped
kernel methods. Many efficient graph kernels have been developed over the past. A particularly efficient algorithm is the Weisfeiler–Lehman kernel that scales well with growing inputs. This kernel is
based on isomorphism testing and plays an important role in descriptive complexity. Recently, it was demonstrated that it can be exploited to perform static code analysis. An important shortcoming of
the kernel is, however, that it requires discrete data that is not allowed to contain any form of uncertainty. In fact, the kernel can neither handle noisy nor incomplete data.
This dissertation project aims to develop efficient robust kernels that allow for uncertain input data. Existing robust graph kernels that are applicable to uncertain data do not scale as well as the
Weisfeiler–Lehman kernel. We aim to develop a robust kernel that is as efficient as the Weisfeiler–Lehman kernel. In general, the issues arising when applying the kernel to uncertain data are related
to those occurring in the context of the robust graph isomorphism problem—given two graphs which are almost isomorphic, find an “almost- isomorphism”. This problem asks for maps between graphs that
preserve most of the graph structure. Two-dimensional versions related to matrix multiplication open a direction to combat uncertain data via the route of algebraic operations. The plan of this
research project is to develop new graph kernels that are robust under uncertain data by incorporating randomisation into the Weisfeiler–Lehman graph kernel framework, and generalising this framework
to the two- dimensional setting.
The second aim of the dissertation project is to develop robust kernels for continuous and uncertain data. There are means to handle continuous data and there exist generic methods to turn discrete
kernels into continuous ones. While at first sight it appears that there is an intrinsic need for discrete data in the Weisfeiler–Lehman kernel, recent work shows how randomisation and hashing can be
exploited to transform the seemingly intrinsic discrete kernel into a kernel suitable for continuous data. This dissertation project plans to improve randomisation techniques to allow continuous
inputs to graph kernels. | {"url":"https://moves.rwth-aachen.de/research/projects/unravel/algorithms-and-complexity/al3-graph-kernels-under-uncertainty/","timestamp":"2024-11-14T11:54:58Z","content_type":"text/html","content_length":"34107","record_id":"<urn:uuid:443cdf82-46d4-40b6-81f8-e716824da4f8>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00572.warc.gz"} |
What is the status of Cosmic Inflation given the latest cosmological data? In a 80 pages-long publication, we present the first ever data analysis using the third-order slow-roll power spectra
complemented by an unprecedented Bayesian model comparison in the landscape of nearly three-hundred models of single-field slow-roll inflation.
In its simplest incarnation, Cosmic Inflation can be realised by a single field rolling down a potential and hundred of different theoretical embeddings have been proposed since the advent of the
paradigm. Comparing these models to data could be viewed as Herculean tasks and we have been sharing these enjoyments with Jérôme Martin and Vincent Vennin in the past decade.
In Ref. [1], we present the most recent and strongest constraints on single-field inflation imposed by the latest cosmological data, namely the Planck satellite 2020 Cosmic Microwave Background data,
the BICEP/Keck array 2021 polarization measurements, the South Pole Telescope third generation measurements and the full compilation of Baryonic Acoustic Oscillations for the Sloan Digital Sky Survey
Our analysis incorporates the most accurate theoretical predictions for the quantum fluctuations generated during inflation, the so-called third-order slow-roll spectra for both primordial
gravitational waves and curvature fluctuations (see this post). The Bayesian model comparison analysis uses some machine learning tools implementing the methods developed in Ref. [2] and include all
the new models presented in the Opiparous Edition of the Encyclopædia Inflationaris.
The following figure shows the one-dimensional marginalised posterior distributions obtained for the inflationary cosmological parameters, i.e., when the cosmic structures are seeded by the
inflationary quantum fluctuations.
Notice that the third slow-roll parameter \(\epsilon_3\) is now constrained in a range perfectly consistent with slow-roll and we find
\(-0.44 < \epsilon_3 < 0.55 \quad (95\%\,\texttt{CL})\).
One may also notice the maximum probability for \(\epsilon_1\) at non-vanishing values, showing a weak statistical preference for the presence of primordial gravitational waves in the current data
(mostly driven by the BICEP/Keck data). We have
\[\log(\epsilon_1) > 4.9 \quad (95\%\,\texttt{CL}),\]
\[\log(\epsilon_1) < -2.6 \quad (98\%\,\texttt{CL}).\]
The Bayes’ factors and maximum likelihood ratios for all models of the Encyclopædia Inflationaris are:
The reference model is denoted as “SR3” and represents the pure slow-roll analysis assuming no specific potential, only the natural priors for the slow-roll parameters \(\epsilon_i\in[-0.2,0.2]\).
Bars extended to the left mean the models are models, they are favoured when the bar is extended to the right (with respect to agnostic slow-roll). The bottom labels give the Jeffreys’ scale of
Bayesian evidence with respect to the best model. We find that \(40\%\) of all scenarios can be considered ruled-out (strongly disfavoured according to the Jeffreys’ scale) whereas \(20\%\) of the
models are most probable given the current data.
Our approach also allows us to constrain the reheating epoch, the transition period between cosmic inflation and the hot Big-Bang phase in which the universe is a relativistic plasma.
The following figures are scattered plots of inflationary models positioned according to their Bayesian evidence (horizontal axis) and the information gain on the reheating epoch (vertical axis). The
colour scale traces the mean value (over its posterior) of the reheating parameter \(\ln R_\mathrm{reh}\).
Each model is also encircled in a gauge counting the number of unconstrained model parameters (derived using Bayesian dimensionality). Non-encircled models are models for which all parameters are
constrained by the data, models with a full circle around have all their parameters unconstrained (which are then superfluous to fit the data). As such, the most probable and most efficient models
are those on the right having no circle around.
Weighted over the landscape, we find that the current data constrain the kinematics of reheating by \(1.3\) bits. The precision reached by the current cosmological data is such that almost half of
the inflationary landscape is out of the game, but, also, for each of the favoured models, the way the universe reheated to become a plasma can now be inferred by astrophysical and cosmological
observations. We are talking here of an epoch of the universe being, at least, at redshift \(z > 10^{10}\)!
The future is bright and we are looking forward to the Euclid satellite measurements (see also this post). | {"url":"https://curl.group/feed.xml","timestamp":"2024-11-04T09:05:19Z","content_type":"application/atom+xml","content_length":"32412","record_id":"<urn:uuid:8340ff9c-049c-435d-892b-75e9ff11f351>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00821.warc.gz"} |
USACO 2016 February Contest, Silver
Problem 2. Load Balancing
Contest has ended.
Log in to allow submissions in analysis mode
Farmer John's $N$ cows are each standing at distinct locations $(x_1, y_1) \ldots (x_n, y_n)$ on his two-dimensional farm ($1 \leq N \leq 1000$, and the $x_i$'s and $y_i$'s are positive odd integers
of size at most $1,000,000$). FJ wants to partition his field by building a long (effectively infinite-length) north-south fence with equation $x=a$ ($a$ will be an even integer, thus ensuring that
he does not build the fence through the position of any cow). He also wants to build a long (effectively infinite-length) east-west fence with equation $y=b$, where $b$ is an even integer. These two
fences cross at the point $(a,b)$, and together they partition his field into four regions.
FJ wants to choose $a$ and $b$ so that the cows appearing in the four resulting regions are reasonably "balanced", with no region containing too many cows. Letting $M$ be the maximum number of cows
appearing in one of the four regions, FJ wants to make $M$ as small as possible. Please help him determine this smallest possible value for $M$.
INPUT FORMAT (file balancing.in):
The first line of the input contains a single integer, $N$. The next $N$ lines each contain the location of a single cow, specifying its $x$ and $y$ coordinates.
OUTPUT FORMAT (file balancing.out):
You should output the smallest possible value of $M$ that FJ can achieve by positioning his fences optimally.
Problem credits: Brian Dean
Contest has ended. No further submissions allowed. | {"url":"https://usaco.org/index.php?page=viewproblem2&cpid=619","timestamp":"2024-11-09T13:38:22Z","content_type":"text/html","content_length":"8638","record_id":"<urn:uuid:f19fedf8-ab2f-4c85-9bb5-c2450aa645b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00166.warc.gz"} |
unified field theory summary | Britannica
unified field theory, Attempt to describe all fundamental interactions between elementary particles in terms of a single theoretical framework (a “theory of everything”) based on quantum field
theory. So far, the weak force and the electromagnetic force have been successfully united in electroweak theory, and the strong force is described by a similar quantum field theory called quantum
chromodynamics. However, attempts to unite the strong and electroweak theories in a grand unified theory have failed, as have attempts at a self-consistent quantum field theory of gravitation. | {"url":"https://www.britannica.com/summary/unified-field-theory","timestamp":"2024-11-13T06:29:19Z","content_type":"text/html","content_length":"60739","record_id":"<urn:uuid:08a5ef3d-697c-4648-90e5-8aed147ea3cd>","cc-path":"CC-MAIN-2024-46/segments/1730477028326.66/warc/CC-MAIN-20241113040054-20241113070054-00516.warc.gz"} |
Cardinal characteristic
From Encyclopedia of Mathematics
2020 Mathematics Subject Classification: Primary: 54A25 [MSN][ZBL]
of a topological space
A function associating an infinite cardinal number to each space and taking the same value on homeomorphic spaces. Cardinal characteristics are also called cardinal invariants. The domain of
definition of a cardinal invariant is the class of all topological spaces or some subclass of it. The following cardinal invariants arose at the first stage of development of general topology. Let
$X$ be an arbitrary topological space. A trivial invariant is its cardinality $|X|$, i.e. the cardinality of the set of all its points. Its weight $w(X)$ is the smallest cardinality of a base of $X$.
The density $d(X)$ is the smallest cardinality of a dense subset of $X$. The Suslin number $c(X)$ is the least infinite cardinal number $\mathfrak{t}$ such that the cardinality of every family of
pairwise-disjoint non-empty open sets does not exceed $\mathfrak{t}$. The Lindelöf number $l(X)$ is the least infinite cardinal number $\mathfrak{t}$ such that every open covering of $X$ has a
subcovering of cardinality $\le\mathfrak{t}$. These simple notions immediately showed their importance by entering in a decisive way in fundamental theorems and problems. Examples: a regular space of
countable weight is metrizable (the Urysohn–Tikhonov theorem, 1925); a compact Hausdorff space is metrizable if and only if its weight is countable; the Suslin number of the space $X$ of a compact
group is countable; for every space $X$ of countable weight its Lindelöf number $l(X)$ is countable. The Suslin problem — is it true that every ordered connected compact Hausdorff space $X$ for which
$c(X)=\aleph_0$ is homeomorphic to the interval $[0,1]$ — leads to the question of the relationship between two cardinal invariants: the density and the Suslin number. For a positive solution of
Suslin's problem it is sufficient to show, under the above assumptions, that $d(X) \le c(X)$. The question of comparison of cardinal invariants — the solution of which, as is clear from the above
example, may be of key significance for a definitive conclusion on the structure of a space — is central in the theory of cardinal invariants. The reason for this lies in the very nature of the
concept of a cardinal invariant: the values of a cardinal invariant are cardinal numbers, the class of which is well-ordered by magnitude. Consequently, one can try to compare the values of any two
cardinal invariants $\phi_1$ and $\phi_2$. A series of mutually related questions arises. Is it true that for all $X$, $\phi_1(X) \le \phi_2(X)$; for which $X$ does $\phi_1(X) \le \phi_2(X)$ hold;
when is $\phi_1(X) = \phi_2(X)$, etc.
It is possible to do arithmetic with cardinals: to multiply and to add them, and to raise them to a power. Correspondingly, it is possible to do arithmetic with cardinal invariants — to multiply and
to add them as functions, etc. This opens up new possibilities for comparing cardinal invariants, using arithmetic. There always holds $$ c(X) \le d(X) \le w(X)\,;\ \ \ l(X) \le w(X) $$ that is, the
Suslin number does not exceed the density, the density does not exceed the weight, and the Lindelöf number does not exceed the weight. But the density and the Lindelöf number are not comparable in
this sense: There are spaces $X$,$Y$ and $Z$ for which $$ d(X) < l(X)\,,\ \ \ l(Y)<d(Y)=c(Y)\,,\ \ \ l(Z)=d(Z)=c(Z) \ . $$
The incomparability of cardinality and weight is unexpected at first sight: There are countable normal $T_1$-spaces of uncountable weight. But always $d(X) \le |X|$ and $l(X) \le |X|$. For every
$T_0$-space $X$, $|X| \le \exp(w(X))$ (one writes $\exp(\mathfrak{t})$ instead of $2^{\mathfrak{t}}$). For every Hausdorff space $X$, $|X| \le \exp(\exp(d(X)))$. Always $w(X) \le \exp(|X|)$.
In a comparison problem it may happen that not just two, but more cardinal characteristics are involved. In that direction, particularly subtle, beautiful and often unexpected results have been
obtained, striking in their generality: For each Hausdorff space $X$, $|X| \le \exp(c(X)\cdot\chi(X))$, where the character $\chi(X)$ is the least infinite cardinal number $\mathfrak{t}$ such that at
each point of $X$ there is a local base of cardinality $\le \mathfrak{t}$ (see [1], [2]). Much research in the theory of cardinal invariants was stimulated by the problem of estimating the
cardinality of a compact Hausdorff space satisfying the first axiom of countability, a problem which remained unsolved from 1923 to 1969. It then turned out that for each Hausdorff space $X$, $|X| \
le \exp(l(X)\cdot\chi(X))$ (Arkhangel'skii's theorem, see [2], [4]).
The computation of cardinal invariants takes place in all parts of general topology because of the set-theoretic nature of the latter. Therefore, the theory of cardinal invariants finds application
in practically all domains of general topology and in each approach to the investigation of spaces.
In particular, in the study of spaces by coverings, the Lindelöf number, the density and the Suslin number appeared from the very beginning. In the investigation and classification of spaces by
continuous mappings (in particular, in the development of the theory of dyadic compacta and of the theory of absolutes) new cardinal invariants arose and played a key role: the spread and the $\
pi$-weight. The spread $s(X)$ of a space $X$ is the least upper bound of the cardinalities of discrete subspaces of $X$, and the $\pi$-weight $\pi w(X)$ of a space $X$ is the minimum of the
cardinalities of families $\mathcal{V}$ (called $\pi$-bases) of non-empty open sets in $X$ such that for each non-empty open set $U$ in $X$ there is a $V \in \mathcal{V}$ such that $V \subset U$. In
the investigation of spaces by inverse spectra a major role is played by the Suslin number, the character and the weight.
Thus, there is an approach to general topology for which cardinal invariants appear both as the principal means of investigation of the structure of spaces, as a basic language in which the
properties of spaces from various classes can be expressed and, finally, as a means of classification and selection of new classes of topological spaces. Basic here is, again, the problem of the
comparison of cardinal characteristics. The fundamental question can be posed as follows. Given a class $\mathcal{P}$ of topological spaces to which the domains of definition of cardinal invariants
are restricted. What are the basic relations between the cardinal invariants under these restrictions? By developing the theory of cardinal invariants for a class $\mathcal{P}$ one obtains the
"cardinal portrait" of $\mathcal{P}$. A comparison of the cardinal portraits of two classes $\mathcal{P}_1$ and $\mathcal{P}_2$ allows one to judge the relationships between these classes and also to
give effective means of proving that a concrete space belongs to one class or other.
This approach can be demonstrated by the class of metrizable spaces. The characteristic feature here is that for this class a number of fundamental cardinal invariants coincide: the Suslin number is
equal to the density, to the weight and to the Lindelöf number. This fact is often applied; for example, to prove that some space is non-metrizable, it is enough to prove that at least two of the
invariants mentioned above differ.
In the class of metrizable spaces the theory of cardinal invariants distinguishes itself from the general theory mainly by its simplifications, whereas in the class of compact Hausdorff spaces it
changes its appearance completely and in a non-trivial way. Responsible for the particular appearance of this theory is the fact that for compact Hausdorff spaces the character and pseudo-character
coincide, as well as the weight and the network weight. The pseudo-character $\psi(x,X)$ of $X$ at $x$ is the smallest number of open sets whose intersection is the point, and the character $\chi
(x,X)$ of $X$ at $x$ is the least cardinality of a local base at $x$. The network weight $\mathrm{nw}(X)$ is the least cardinality of a family $\mathcal{S}$ of sets in $X$ satisfying the condition:
If $x \in U \subset X$, where $U$ is open in $X$, then there is a $P \in \mathcal{S}$ for which $x \in P \subset U$ (such families are called networks in $X$). For every compact Hausdorff space $X$
the following hold: 1) $\psi(,X) = \chi(x,X)$ for all $x \in X)$; and 2) $\mathrm{nw}(X) = w(X)$. Therefore, the weight cannot increase under a continuous mapping onto a compact Hausdorff space, and
if a compact Hausdorff space $X$ is the union of two subspaces $X_1$ and $X_2$, then the weight of $X$ does not exceed the maximum of the weights of $X_1$ and $X_2$ (the addition theorem for
weights). For the same reason, the weight of a compact Hausdorff space never exceeds its cardinality; in particular, every countable compact Hausdorff space is metrizable. None of these theorems of
the theory of cardinal invariants for the class of compact Hausdorff spaces can be extended to the class of completely-regular spaces. An important specific result is the following: If $X$ is a
compact Hausdorff space, $\mathfrak{t}$ is a cardinal number, $\mathfrak{t} \le \aleph_0$ and if $\chi(x,X) \ge \mathfrak{t}$ for all $x \in X$, then $|X| \ge \exp\mathfrak{t}$ (the Čech–Pospišil
theorem). Almost-all metrizability criteria for compact Hausdorff spaces are also theorems about cardinal invariants. Thus, metrizability of a compact Hausdorff space $X$ is equivalent to any of the
following conditions: a) $w(X) = \aleph_0$; b) $\mathrm{nw}(X) = \aleph_0$; c) the diagonal in $X \times X$ is a $G_\delta$-set; or d) $X$ has a point-countable base.
In the investigation of the structure of compact Hausdorff spaces $X$ the tightness $t(X)$ plays a major role. The tightness $t(X)$ (see [2], [4]) of $X$ is the least cardinal number $\mathfrak{t} \
ge \aleph_0$ such that if $x \in X$, $A \subset X$ and $x \in \bar A$, then there is a $B \subset A$ for which $x \in \bar B$ and $|B| \le \mathfrak{t}$. The tightness does not increase when a
compact Hausdorff space $X$ is raised to a finite power (in the class of completely-regular spaces this is not true).
If the tightness of a compact Hausdorff space $X$ does not exceed $\mathfrak{t}$, then for each $x \in X$ there is a family $\mathcal{V}$ of non-empty open sets in $X$ such that $|\mathcal{V}| \le \
mathfrak{t}$ and each neighbourhood $O_x$ of $x$ contains an element of $\mathcal{V}$. Therefore the $\pi$-weight of each separable compact Hausdorff space of countable tightness is equal to $\
aleph_0$. The spread of a compact Hausdorff space majorizes its tightness.
The fundamental properties of dyadic compacta are also, to a large extent, governed by theorems on cardinal characteristics. Thus, for each dyadic compactum the weight coincides with the spread and
the tightness. The class of dyadic compacta contains the class of compact Hausdorff topological groups, thus, in particular, every compact Hausdorff group of countable tightness is metrizable.
In the theory of dyadic compacta (cf. Dyadic compactum) and in other parts of the theory of cardinal invariants, the question of the behaviour of these invariants under multiplication is of major
importance. The following two theorems play an essential role here; the first of these implies the second. If $\mathcal{F}$ is a family of spaces such that $d(X) \le \mathfrak{t}$ for each $X \in \
mathcal{F}$ and if $|\mathcal{F}| \le \exp(\mathfrak{t})$, then the density of the product of the spaces from $\mathcal{F}$ does not exceed $\mathfrak{t}$ (see [1]–[4]). If $\mathcal{F}$ is the
product of any set of spaces of densities not exceeding $\mathfrak{t}$, then $c(X) \le \mathfrak{t}$. In the latter result there is no limit on the number of factors. In particular, the Suslin number
of any Tikhonov cube (the product of an arbitrary set of segments) is countable. Thus, the condition $c(X) = \aleph_0$ places no restriction on the cardinality of a space.
Many simply formulated questions on the behaviour of cardinal invariants under multiplication have turned out to be very delicate. For example, the question: Is it true that always $c(X \times X) = c
(X)$? turns out to be related to the Suslin hypothesis and the continuum hypothesis.
On the other hand, the behaviour of cardinal invariants when passing from a space $X$ to its image $Y$ under a continuous mapping $f : X \rightarrow Y$, is, on the whole, governed by simple general
For example, $c(Y) \le c(X)$, $d(Y) \le d(X)$, $\mathrm{nw}(Y) \le \mathrm{nw}(X)$, $l(Y) \le l(X)$. If $f$ is a quotient mapping onto, then $t(Y) \le t(X)$. The fact that the foundation of the
theory of cardinal invariants consists of a system of simple universal rules of this kind also can be considered as one of the reasons ensuring the broad applicability of the theory.
Significant information on the structure of spaces is obtained by consideration of the question: How do cardinal invariants behave on passing to a subspace? A cardinal invariant $\phi$ for which $Y \
subset X$ always implies $\phi(Y) \le(X)$ is called monotone. These include: weight, network weight, tightness, character, and spread. Non-monotone are the Suslin number, the density and the Lindelöf
number. The following questions arise: Which are the spaces $X$ for which $c(Y) \le \mathfrak{t}$ for all $Y \subset X$; which are the spaces $X$ for which $d(Y) \le \mathfrak{t}$ for all $Y \subset
X$; what effect on the topology of $X$ has the requirement: $l(Y) \le \mathfrak{t}$ for all $Y \subset X$? The answer to the first question is simple: the condition means that the spread of $X$ does
not exceed $\mathfrak{t}$. But the two subsequent conditions single out new classes of spaces. The investigation of these classes turns out to depend on special hypotheses of set theory, in
particular on Martin's axiom.
The theory of cardinal invariants has a peculiar character on the spaces of topological groups. E.g., the criterion for metrizability reduces here simply to the first axiom of countability. The major
properties of linear topological spaces, in particular, of the spaces $C(X)$ of continuous real-valued functions on a space $X$, can be formulated in the language of cardinal invariants. This refers
to the theorems on Eberlein compacta (each Eberlein compactum is a Fréchet–Urysohn space, the weight of an Eberlein compactum is equal to its Suslin number); and the following theorem: If $X$ is a
compact Hausdorff space, then the tightness of $C(X)$ in the topology of pointwise convergence is countable.
Between a number of cardinal invariants of the spaces $X$ and $C(X)$ there is a duality-type correspondence.
[1] I. Juhász, "Cardinal functions in topology" , North-Holland (1971)
[2] A.V. Arkhangel'skii, V.I. Ponomarev, "Fundamentals of general topology: problems and exercises" , Reidel (1984) (Translated from Russian)
[3] R. Engelking, "Outline of general topology" , North-Holland (1968) (Translated from Polish)
[4] A.V. Arkhangel'skii, "Structure and classification of topological spaces, and cardinal invariants" Russian Math. Surveys , 33 : 6 (1978) pp. 33–96 Uspekhi Mat. Nauk , 33 : 6 (1978) pp. 29–84
The usual terminology in the literature (cf. [1]–[4]) is cardinal function, or cardinal invariant.
The Suslin number of a space $X$ is also called its cellularity, and its Lindelöf number also its Lindelöf degree (the latter is often denoted by $L(X)$).
The Urysohn–Tikhonov theorem, mentioned above, is also called the Urysohn metrization theorem.
The fact that the class of dyadic compacta contains the class of compact Hausdorff topological groups is called Kuzminov's theorem (on compact groups).
For the notion of a Fréchet–Urysohn space see Sequential space.
The problem of whether $c(X) = c(X \times X)$ for every space $X$ was solved by S. Todorčević [a2], who found, without using extra set-theoretic hypotheses, spaces $X$ satisfying $c(X) < c(X \times
The problems whether every hereditarily separable space is Lindelöf and whether every hereditarily Lindelöf space is separable generated a lot of research. Many examples were constructed using
various extra set-theoretical assumptions, in particular the Continuum Hypothesis. Todorčević [a1] showed that the statement "every hereditarily separable space is Lindelöf" is consistent with the
usual axioms of set theory. For much more information and other recent developments see various articles in [a3], in particular Chapts. 1; 2, and [a4].
[a1] S. Todorčević, "Forcing positive partition relations" Trans. Amer. Math. Soc. , 280 (1983) pp. 703–720 Zbl 0532.03023
[a2] S. Todorčević, "Remarks on cellularity in products" Compos. Math. , 57 (1986) pp. 357–372 Zbl 0616.54002
[a3] K. Kunen (ed.) J.E. Vaughan (ed.) , Handbook of set-theoretic topology , North-Holland (1984) pp. Chapts. 1–2 Zbl 0546.00022
[a4] I. Juhász, "Cardinal functions. Ten years later" , MC Tracts , 123 , North-Holland (1980) Zbl 0479.54001
[b1] M.E. Rudin, "Lectures on set theoretic topology", Amer. Math. Soc. (1975) ISBN 0-8218-1673-X Zbl 0318.54001
How to Cite This Entry:
Cardinal characteristic. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Cardinal_characteristic&oldid=54749
This article was adapted from an original article by A.V. Arkhangel'skii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098.
See original article | {"url":"https://encyclopediaofmath.org/wiki/Cardinal_characteristic","timestamp":"2024-11-03T01:33:55Z","content_type":"text/html","content_length":"35317","record_id":"<urn:uuid:104fd7f7-f849-4d85-ad22-bbc6c01b8c94>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00477.warc.gz"} |
Final Temperature Calculator - Savvy Calculator
Final Temperature Calculator
About Final Temperature Calculator (Formula)
Final temperature is a key concept in thermodynamics, especially when studying how different substances interact in heat exchange. When two substances at different initial temperatures come into
contact, heat is transferred from the hotter substance to the cooler one until equilibrium is reached. The final temperature of the combined system depends on the masses, specific heat capacities,
and initial temperatures of the substances involved. Understanding how to calculate final temperature is essential in various scientific, engineering, and everyday applications.
The formula for calculating final temperature after heat exchange is:
Final Temperature (TF) = (m₁c₁t₁ + m₂c₂t₂) / (m₁c₁ + m₂c₂)
• m₁ and m₂ are the masses of the two substances.
• c₁ and c₂ are the specific heat capacities of the substances.
• t₁ and t₂ are the initial temperatures of the substances.
• TF is the final temperature after equilibrium is reached.
This formula helps calculate the equilibrium temperature after two substances, with different initial conditions, exchange heat until they reach thermal equilibrium.
How to Use
1. Measure Initial Conditions: Find the masses (m₁ and m₂), specific heat capacities (c₁ and c₂), and initial temperatures (t₁ and t₂) of both substances.
2. Apply the Formula: Use the formula to calculate the final temperature by substituting the known values.
3. Interpret the Result: The final temperature shows the thermal equilibrium after heat has been exchanged between the two substances.
Suppose you have two substances. The first substance has a mass of 2 kg, a specific heat capacity of 1000 J/kg°C, and an initial temperature of 80°C. The second substance has a mass of 3 kg, a
specific heat capacity of 500 J/kg°C, and an initial temperature of 20°C. Using the formula, you can calculate the final temperature (TF):
TF = (2 * 1000 * 80 + 3 * 500 * 20) / (2 * 1000 + 3 * 500)
TF = (160000 + 30000) / (2000 + 1500)
TF = 190000 / 3500
TF = 54.29°C
In this example, the final temperature after heat exchange is 54.29°C.
1. What is final temperature?
The final temperature is the equilibrium temperature reached when two substances exchange heat until they are at the same temperature.
2. How is final temperature calculated?
The final temperature is calculated using the formula TF = (m₁c₁t₁ + m₂c₂t₂) / (m₁c₁ + m₂c₂), which accounts for the masses, specific heat capacities, and initial temperatures of the substances.
3. What is specific heat capacity?
Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius.
4. Does mass affect the final temperature?
Yes, the mass of each substance plays a crucial role in determining the final temperature after heat exchange.
5. Can the final temperature be lower than both initial temperatures?
No, the final temperature will always lie between the initial temperatures of the two substances.
6. What happens if the masses or specific heat capacities are equal?
If the masses and specific heat capacities of both substances are equal, the final temperature will be the average of the two initial temperatures.
7. Why do we use specific heat capacity in the formula?
Specific heat capacity is used because different substances require different amounts of heat to change their temperature, affecting the final temperature.
8. Can this formula be used for gases?
Yes, the formula can be used for gases, as long as you know their specific heat capacities.
9. What if one substance loses heat while the other gains heat?
The formula takes this into account by balancing the heat gained and lost, ensuring thermal equilibrium.
10. What units should be used in the formula?
Ensure that mass is in kilograms (kg), specific heat capacity is in J/kg°C, and temperature is in degrees Celsius (°C) for consistency.
11. Can this formula be used for more than two substances?
For more than two substances, the formula can be extended by adding additional terms for each substance.
12. What is thermal equilibrium?
Thermal equilibrium is the state at which two or more substances in thermal contact no longer exchange heat and have the same temperature.
13. Why does the larger mass or specific heat capacity affect the final temperature more?
Substances with larger masses or higher specific heat capacities store and transfer more heat, influencing the final temperature more significantly.
14. How can I measure specific heat capacity?
Specific heat capacities are typically measured in a laboratory, but common values for substances like water, metals, and gases are widely available.
15. Can the final temperature be predicted without using the formula?
While estimates can be made, the formula ensures accuracy by accounting for all relevant variables.
16. Is the final temperature always an average of the initial temperatures?
No, the final temperature is weighted by the masses and specific heat capacities of the substances, so it is not a simple average.
17. Does the formula apply to phase changes?
This formula applies to heat exchange without phase changes. If phase changes occur (e.g., melting, boiling), additional calculations are needed.
18. What if the initial temperatures are the same?
If both substances have the same initial temperature, no heat is exchanged, and the final temperature remains the same.
19. Why is final temperature important in engineering?
Understanding final temperature is crucial for designing systems like engines, HVAC units, and industrial processes that involve heat exchange.
20. Can this formula be applied in daily life?
Yes, the final temperature formula can be used in various everyday scenarios, such as mixing hot and cold water or determining cooking times.
The final temperature calculation is an essential tool in thermodynamics, helping to determine the equilibrium temperature after heat exchange between two substances. By using the formula TF =
(m₁c₁t₁ + m₂c₂t₂) / (m₁c₁ + m₂c₂), you can accurately calculate the final temperature and better understand the principles of heat transfer. This knowledge can be applied in a wide range of
scientific and practical applications, from laboratory experiments to real-world engineering challenges.
Extinction Coefficient Calculator
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Glossary of Verbs Associated with Math Standards
Glossary of Verbs Associated with the New York State Next Generation Mathematics Learning Standards
Key vocabulary was identified to be defined in a glossary of verbs associated with the New York State Next Generation Mathematics Learning Standards. This glossary contains a list of verbs that
appear throughout the Mathematics Standards and are explained in the context in which they are used.
Downloadable Resource: PDF Version of this Glossary
Word Definition/context of use in the standards
Analyze Analyze requires students to examine carefully, take apart mathematically, and break down into components or essential characteristics to identify causes, key factors, and possible
Apply Apply requires a student to use mathematical knowledge in a variety of situations.
Calculate Calculate requires a student to determine an answer.
Classify Students classify by determining characteristics (attributes) that objects (numbers, shapes, etc.) share, and characteristics (attributes) they don't share.
Compare Students compare by examining two or more objects, numbers or mathematical situations in order to determine similarities and differences.
Compose Compose requires students to form or make something (numbers, functions, sets, etc.) by combining parts.
Convert Students convert by changing the form (e.g. measurement, different units) without a change in the size or amount.
Decompose Students decompose by separating into parts in terms of simpler components that allows for students to see groupings, relationships and patterns.
Demonstrate Students demonstrate understanding and application of the content in the standard through narrative (oral or written), modeling (including pictures, diagrams or technology), algebraic
work or any mathematically appropriate method that clearly communicates the steps leading to the solution or conclusion needed.
Derive Derive requires the student to utilize current or specified knowledge to formulate a “new” theorem, formula or relationship.
Describe Describe requires that students illustrate their thinking or justifications through verbal (oral or written) statements that may reference a drawing/diagram/model.
Determine To determine requires finding something out or establishing exactly, typically as a result of research or calculation.
Develop Develop requires a student to engage in experimentation or argumentation that leads to a mathematically appropriate conclusion.
Differentiate Differentiate requires a student to determine the difference between two or more things.
Distinguish Distinguish requires students to recognize distinct or different characteristics (attributes).
Evaluate Evaluate requires that a student find the value of a mathematical expression.
Explain Explain requires a student to provide verbal (oral or written) evidence to support a conclusion or solution.
Explore Explore requires the student to learn the concept in the standard through a variety of instructional activities. Repeated experiences with these concepts, with immersion in the
concrete, are vital.
Explore indicates that the topic is an important concept that builds the foundation for progression toward mastery in later grades. However, mastery at the current level is not
expected for that standard.
Express Express requires students to change an amount or quantity into a different form.
Fluent The word fluent is used in the Standards to mean “fast and accurate.” Fluency in each grade involves a mixture of just knowing some answers, knowing some answers from patterns and
knowing some answers from the use of strategies.
For additional information refer to pages 18-19 of Progressions for the Common Core State Standards in Mathematics.
Principles and Standards for School Mathematics states, “Computational fluency refers to having efficient and accurate methods for computing. Students exhibit computational fluency
when they demonstrate flexibility in the computational methods they choose, understand and can explain these methods, and produce accurate answers efficiently.
Required Grade Level Fluencies for Grades K-8:
Required grade level fluencies are available from EngageNY at Required Fluencies for Grades K-8 Standards for Mathematics.
Standards that are recommended fluencies at the High School level are identified in each set of standards for Algebra I, Algebra II and Geometry.
Generate Generate requires students to create something by the application of one or more mathematical rules or operations.
Identify Identify requires students to recognize a mathematical concept using prior knowledge.
Interpret Interpret requires students to make sense of and assign meaning to a mathematical task and explain the reasoning behind it.
Justify Justify requires a student to show evidence and/or steps that illustrate the mathematics leading to a solution or conclusion. Note: Words are acceptable but not necessary.
Know Know requires students have a firm mathematical understanding through awareness of situations, facts, information, and skills.
Make Make requires a student to create a picture, diagram or model to illustrate a mathematical concept.
Prove Prove requires students to demonstrate that an argument is universally true where each step and conclusion must be supported by evidence and/or reasoning. This can be shown through a
variety of strategies.
Recognize Recognize requires students to identify mathematical concepts based on previous facts or knowledge.
Reference Reference requires students to apply a specified mathematical concept.
Represent Represent requires students to communicate a mathematical concept through pictures, diagrams, models, symbols, or algebraic notation.
Solve Solve requires the students to find the answer to specified problem.
Specify Specify requires the student to clearly articulate or describe mathematical properties or procedures.
State State requires students to give an answer without calculations or underlying work.
Understand Understand requires a student to grasp sufficient knowledge of a mathematical concept in order to explain or apply it.
Use Use requires the student to apply designated processes, strategies or mathematical concepts.
Verify Verify requires students demonstrate that a mathematical concept is true or accurate.
Written Method A written method/representation is any way of representing a strategy using words, pictures or numbers. | {"url":"https://www.nysed.gov/curriculum-instruction/glossary-verbs-associated-new-york-state-next-generation-mathematics-learning","timestamp":"2024-11-14T07:47:58Z","content_type":"text/html","content_length":"72530","record_id":"<urn:uuid:2e1fa299-764f-48e5-a7e1-479d3167b1c1>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00134.warc.gz"} |
Building Debt With a Basement
Anyone who's been in this hobby for a while knows that a basement set-up is the way to go if you have one, and up until just recently, I was among the many who wished they had one.
Well, thanks to the combination of a marriage and a recent non-traveling job offer that landed me here in the Rocket City, I now have one! Plus, I have enough time and a little extra money to invest
in such a set-up and finally have that big beautiful SPS dominant coral reef tank in my living room that I always dreamed of. You know the one I'm talking about. Shock and Awe.
On the flip side of this of course, is the fact that I am married now and I can't go throwing money around like I used to. While spending large sums of cash on fish and corals is a much more
acceptable alternative to drugs and hookers, my other half still insists that I exercise some degree of restraint in funding this new project, in hopes that we can avoid rolling pennies for Ramen
noodles by the end of the month.
So this blog will serve as my daily... ish journal into the depths of my newly acquired basement and the saltwater aquarium hobby, while maintaining a budget-minded conscience and hopefully avoiding
starvation and/or a divorce.
To provide a little background, I got started in the coral reef hobby roughly 7-8 years ago while living near Clearwater Beach, Florida for a few years. That in no way, shape, or form, qualifies me
as a long term "expert", considering I never really had what I felt was a high quality thriving reef tank, certainly not by my own standards and obviously not by those of the "Tank of The Month"
judging outfit.
I started out with a standard 10 gallon tank after overhearing some people at work talking about this new "nano reef" phenomenon. It sounded like something that would be fun to try and wouldn't cost
a lot of money. Unfortunately, nobody was there to caution me that it had the same addictive qualities one might find in Warcraft, gambling, or crack cocaine.
After a 50 gallon and then up to a 90 gallon VHO lit euphyllia dominated tank, followed abruptly by 5 hurricanes, flooding, several power outages and a recession that put me out of a job, I decided
it was time to take a break from my hobbies and break out the suitcase once again.
Over the years, I'd settle in one place or another from six months to a few years and set up a little nano tank here and there but nothing too elaborate. My most recent tank was a little Finnex M30,
basically a 30 gallon tank with a built-in fuge and sump in the back.
This tank used to look half-way decent before several power outages last summer and a move to another state, which leads me to present day, circa fed up and on the verge of putting my hands in the
air and backing away slowly.
What I've decided to do in contrast, now that I have a basement and a reasonable budget to work with, is dedicate more time and effort into the hobby and expand my system. I'll be working with a
budget of roughly $500/month, give or take a little. I'm about two months into it and so far, I've purchased a few used tanks and built some shelves for the sumps in the basement.
As I'm writing this, my main display tank is nearly half-way full of RO/DI water.
This is a used 100 gallon tank that I snagged up on Craig's List. Along with the stand and canopy, two sumps, a 40-gallon rimless cube, the U-haul trailer and a tank of gas, I think I paid in the
neighborhood of $500-$600.
Luckily, I didn't have to do a lot of drilling. There just happened to be a vent on the wall where a 90-120 gallon tank would look nice that led right down into the basement.
All I had to do was disconnect the vent and cap it off. I also had a certified Electrician install new outlets in the basement on their own circuit to support the system.
I'm planning on having a 2-sump system, if it works. Theoretically, the first and upper sump will be for drainage, bio balls and a protein skimmer, sock filters, carbon, etc.
The other secondary (lower) sump will host the return pumps and (once again) theoretically, serve as a temporary holding area for water changes... more on that later.
The refugium is setting above the secondary sump and will drain directly into it so the pods don't get stuck in the filters. I'm hoping that once I turn the return pumps off, the water will rise
enough in the lower sump to give me the 30 or so gallons I'll need for water changes. I'm estimating a total volume of 300 gallons.
Total Comments 9
Here you can see the front of the refugium setting above a 72 gallon bow front that I also picked up off of Craig's List.
They also had some tanks that used to be in a Wal-Mart and I was able to purchase 5 of those and the bow front for roughly $400. One of those of course is being used as the refugium and the
other four will be frag tanks, or maybe I'll take a stab at breeding.
I'm planning on using the 72 bow front for softies and maybe the clown/anemone thing. The 40-gallon rimless cube will host a long awaited Black Angler and set on top of the wooden stand to the
right of the bow front.
On the other side, if facing the sumps, to my right will be my equipment rack where I'll have some Auto Top-Off, kalk dosing and other similar action going on.
Posted 12/27/2011 at 05:50 PM by speedpacer
I'll have a few Brute 33s behind the "equipment rack" if you will, for fresh and saltwater reserve. Right now, I just have some live rock and sand back there waiting to go into the upstairs
display and fuge, respectively.
On the other side to the left and facing out, if you're facing the sump, will be the frag tanks. Here's a sketch of what those will look like. I'm sure future photos will prove less confusing.
In order to build the shelving, I had to buy a circular saw and a power drill, which is cool because I'll have those for other projects in the future. Those and the lumber set me back about
$200 and I've spent a few hundred on a pump and some more live sand so I'm up to a little over $1000, which isn't bad considering everything I've gotten, especially compared to what all that
crap would've cost me new.
Here's a shot of the main display with the canopy. I'm planning on a DIY LED system.
And of course, overseeing the project while my wife is out of town...
That's about where I'm at right now. Since I've started this, I've gained another few inches of water in the tank. Hopefully, I'll be able to start the transferring process before the weekend
and weather permitting, I'll try to finish the shelving for the frag tanks.
By the end of April, I hope to have the entire system fully functional, automated, monitored and controlled via my iPhone. In order to save money on the monitoring, controlling and automation,
I'll be building and developing my own computer using a custom Arduino-based board/chipset. Basically, I'd like to automate everything but the feeding.
That's all for now. I'll post more pictures once the tank is full and I get a little aquascaping done.
Posted 12/27/2011 at 05:52 PM by speedpacer
The setup looks like its going to be great! Are you going to post any updates?
Posted 02/07/2012 at 06:39 PM by fpv930
Sorry, I just saw your comment. Thank you and yes! It's actually been coming along quite nicely. I'll try to take some pictures this weekend.
Posted 03/16/2012 at 04:53 AM by speedpacer
5. It looks pretty good. I'll have to tag along. Good Luck
Posted 04/07/2012 at 09:43 PM by Mel 2038
It's been a while since I've posted... been starting a new business. Everything has been doing really well.
The 100gal display tank upstairs has been running since the beginning, just a small cluster of zooanthids that I had left over from the other tank and I've gotten a few fish. The reason it
looks brighter on one side is because I'm using a compact T5 that I had over the smaller tank on the left side and then a standard 10gal light on the right side for now. I had a lot of
detritus on the bottom at first but I put a Tunze 6215 in there and it solved that problem.
My LED parts have been sitting in a box for 4 months but I finally broke those out and started to put it together yesterday, until I ran out of nylon washers. More of those on order.
I also built me a little workbench to help put all of this together.
So far, I just have the main display, sump, refugium and 72 gallon bowfront hooked up. I haven't put anything in the bowfront yet, will probably just be for LPS, and I'll probably do the
clownfish/anemone thing in the cube on the right, once I get everything stabilized. The QT/hospital tank is on the far left, which is where this all started from. It's still empty right now as
well. You can see the fuge up above the 72.
Posted 07/29/2012 at 09:26 AM by speedpacer
I've had the shelves for the frag tanks built for a while but they're still not hooked up yet.
And here's the backside...
What I'm working on now is the automation. It'll be controlled by an Arduino micro controller and 3 American DJ PC100A power strips with an iPad User Interface so I can control and monitor
everything at home, from work or while on vacation, etc.
Maybe next weekend, time permitting, I'll be able to finish the lights and start getting some corals! I also have a kalk reactor and a phosban reactor that have been sitting in the box for 4-5
months. I'll probably get around to hooking those up next weekend too.
Posted 07/29/2012 at 09:37 AM by speedpacer
Almost forgot... here's my plans for the LED lights (x2) over the main display upstairs. The daylights will be wired up in parallel with ELN-60-48D Mean Well drivers to control 24 LEDs (x4)
and then the 12 dawn/dusk (reds) will have their own driver, 12 moonlights will have their own and 12 high noons will have their own. The idea is that I'll be able to simulate sunrise/sunset
and lightning storms via the Arduino controller.
Posted 07/29/2012 at 10:08 AM by speedpacer
So, while I had a little downtime waiting on the nylon washers to finish my lights, and having to exchange my Mean Well 48D drivers for 48Ps since Arduino uses PWM, I went ahead and started
hacking my American DJ power strip. I've noticed a lot of guys are building small cases for their controller setups but since I have all of this room in the basement, I opted for a different
Since the American DJs are rack mountable, I decided to make the entire controller rack mountable. I'll be using 3 American DJs in all, so I ordered 3 1U cases, which fit the relays perfectly
(8 each):
And then I ordered a 2U case for the Arduino Mega 2560 and the Uno/Ethernet shield and the rest of the guts and glory.
There aren't enough PWM outputs on the Uno for my 7 channels so I decided to do a master/slave configuration and have the web server/client separate from the main I/O board. I still haven't
decided if I'm going to use the ethernet shield as a server or client, but I'd like to track historical data in a mysql database, so probably a client. I already have a LAMP (web) server in
the basement anyway.
Once it's all together, it should all fit snuggly into this rack:
I've got the prototype for the ATO working. It was a lot easier than I expected actually. Well, in part thanks to some other threads on here and reefprojects.com. Here you can see when the
float switch is up, the lamp is off:
And when the float switch is down, the lamp comes on:
Very cool! It's progress like this that drives me to move forward.
I just downloaded X-Code today and I may play around with the interface a little tomorrow. Then, maybe I'll have a better idea of how the web server/client will come into play. Everyone else
seems to be using LCD panels but since the bulk of my setup will be in the basement and the display is upstairs, I'm going for a iPhone/iPad display panel. Plus, I like the cool factor.
Posted 08/02/2012 at 06:26 PM by speedpacer | {"url":"https://reefcentral.com/forums/blog.php?s=f686724bd49f273a47b0ec0f4f76d40d&b=429","timestamp":"2024-11-11T17:52:16Z","content_type":"application/xhtml+xml","content_length":"70436","record_id":"<urn:uuid:34a9de1f-3c69-4b0d-8120-440320cdfe8c>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00122.warc.gz"} |
Financial Accounting
• Describe common types of investments
• Demonstrate an understanding of the time value of money
Generally speaking, to invest in stocks, bonds, and mutual funds, you need an investment account with a broker. You’ll want to evaluate brokers based on factors like costs (trading commissions,
account fees), investment selection, and investor research and tools.
Before selecting an investment broker, it is best to understand what types of investments are traditionally offered. We’ll take a look at the most commonly used types of investments:
• Stocks
• Bonds
• Mutual Funds
• Rental Real Estate
stock is a share of ownership in a single company. Stocks are also known as equities. Stocks are purchased for a share price, which can range from the single digits to a couple of thousand dollars,
depending on the company. Starbucks stock was trading at $60 per share in 2016, 2017, and 2018, showing very little growth, but in 2019 it increased to almost $100 per share. If you’d put $6,000 into
your brokerage account in 2018 and bought only Starbucks stock, you’d have 100 shares. By August of 2019, you could have sold those shares for $9,500, making a $3,500 profit in just one year.
In July of 2015, a single share of Amazon stock was trading at $500. By the end of 2018, the stock was up to almost $2,000 per share, but it dropped back down to about $1,500 in 2019. However, during
the coronavirus pandemic of 2020, the stock was trading at close to $2,500. A $5,000 investment in 2015 would only buy you 10 shares of stock, but by 2020, just five years later, your stock would
have been worth $25,000.
On the other hand, during that same time period, Ford Motor Company, a stock that had once traded at almost $40 per share in 1999, dropped from $15 per share to $5.
Picking winners in the stock market has been debated for as long as there have been publicly traded stocks, with as many different opinions and systems as there are stocks to pick from, which is
literally thousands. Most astute investors diversify, which is to say they buy a wide range of stocks in different industries in order to balance their portfolios.
The Need for Diversifying
One of the worst examples of investing in a single stock is Enron. When Enron bought Portland General Electric, all the employees who had stock in PGE suddenly found themselves owning stock in Enron
instead, and shortly after that, Enron went bankrupt and the stock went from $90 a share to $0 in the span of a few months. PGE employees lost their entire retirement savings.
A bond is essentially a loan to a company or government entity that agrees to pay you back in a certain number of years. In the meantime, you get interest payments. Bonds are generally less risky
than stocks. The trade-off is that the market value of bonds doesn’t fluctuate, so while they are safer, they offer less opportunity for growth.
Mutual Funds
Mutual funds are a mix of investments packaged together. Mutual funds are managed by professional investment firms, so they allow investors to skip the work of picking individual stocks and bonds
and instead purchase a diverse collection in one transaction. The inherent diversification of mutual funds makes them generally less risky than individual stocks.
Rental Real Estate
Another type of investments is purchasing rental real estate such as single-family homes, apartment complexes, condominiums, or even raw land to rent out to tenants—or lease properties to businesses
for use. Rental real estate is considered “passive” income, but buying and selling real estate can be complex, and a very specialized area of expertise is needed in order to minimize the investment
risks. You may, however, find you have an affinity for it and gain the experience and expertise to invest wisely and get great returns over the years.
Other Investment Options
There are other options, as well—such as cryptocurrency, puts and calls, precious metals, and a host of other less common alternatives that require a deeper understanding of investing and of each
specific type of investment. Although you may one day want to pursue some of these other investment options, it’s best to stick with a few solid stocks and mutual funds when you’re just getting
Your Investment Strategy
Your investment strategy depends on your saving goals, how much money you need to reach them, and your time horizon. If your savings goal is more than 20 years away (like retirement), almost all of
your money can be in equities because higher-risk equities can recover from market drops over the longer period of time until your retirement. If, however, you’re saving for a short-term goal like
emergency funds, you’re better off keeping your money in a savings account or another low-risk, easily accessible fund.
Later in this course, the accounting behind some of these investments will be covered more in detail. It is important for you to invest time to better understand your investment strategy and the
vehicles that move you in the direction of your financial goals.
Present Value vs. Future Value
The present value of an amount of money depends on several factors, but in its simplest form, it represents what a future amount of money is worth today. For example, you promise to give your
daughter $10,000 for college once she enrolls five years from now. If your investment account grows at about 8% every year, you would need to put $6,806 in your account today.
Let’s dive into how that growth works:
• Year 1: You put the $6,806 in your account, and it earns 8% during that first year. You now have the original $6,806 plus $544.48 (6,806 × .08) in earnings. These earnings are often called
interest, but they could also be growth, dividends, or rents. At the end of that first year, you have $7,350.48.
• Year 2: Your investment earns another 8%, but this time it’s on the balance of $7,350.48. This is compounding—where you earn a return on your original investment plus the growth. During the
second year, your $7,350.48 grows to $7,938.52, as you’ve earned $588.04 in interest.
□ Note: Compounding interest works for you when you are saving and investing, but conversely, it works against you when you are borrowing.
• Year 3: Your account grows to $8,573.60, given the same conditions as the previous year.
• Year 4: You have $9,259.49, given the same conditions as the previous year.
• Year 5: You have $10,000.25 to give to your daughter for her college tuition.
Compound Interest Example
Year Account balance, beginning of the year Earnings/growth at 8% Account balance, end of year
Year 1 $6,806.00 $544.48 $7,350.48
Year 2 $7,350.48 $588.04 $7,938.52
Year 3 $7,938.52 $635.08 $8,573.60
Year 4 $8,573.60 $685.89 $9,259.49
Year 5 $9,259.49 $740.76 $10,000.25
In other words, if someone offered you $7,000 today, or $10,000 five years from now, you’d be smart to take the $7,000 if you think you can sustain a return on your investments of 8% because the
future value of the $7,000 is more than $10,000 (it is, in fact, $10,285 and some change.)
In Summary
• The present value of $10,000 five years from now, discounted at 8%, with interest compounded annually, is $6,806.
• The future value of $6,806 at 8% compounded annually is $10,000.
You can calculate these amounts using a spreadsheet, a financial calculator, web-based calculators, commonly found tables, or even by hand if you are a math whiz. The trick to know is whether you are
looking for the present value of a future amount (often called discounting) or the future value of a present amount. In addition, if the compounding period is more often than yearly, depending on
what calculator you are using, you may have to do some quick math. For instance, if you are looking at tables for 8% compounded quarterly for 5 years, the number of periods (n) will be 20 (5 years
times 4 quarters per year) and the interest rate (r) will be 2% (8% per year divided by 4 quarters). In calculations, remember that 8% is actually 0.08 as a number.
The mathematical formula for Future Value (FV) is:
• [latex]C[/latex] = initial investment (present value)
• [latex]r[/latex] = rate of return
• [latex]n[/latex] = number of periods
Try it out: In our first example, the present value was [latex]6,806, n = 5, \text{ and } r = .08[/latex]
[latex]6806\times\left(1+.08\right)5 = 10000.25[/latex]
Future Value Tables
Because these calculations need to be done frequently, brokers and accountants create future value tables, which help people calculate future values without a financial calculator.
If you were using a future value table for this example, you would find the column for 8% and the row for n = 5, and you’d find a factor of 1.4693. That’s the future value of $1. So you would then
multiply the factor by your initial investment of $6,806, and you get $10,000.06 (the factor is rounded to the nearest ten-thousandth, making it slightly less accurate than using the actual
mathematical formula).
An annuity is a steady stream of monthly or annual income. There are tables and calculations for the future and present values of annuities as well. The calculations are more complex than those for a
single (lump) sum, but spreadsheets, calculators, and tables make the analysis possible. When it comes to periodic payments that are not all the same, or that have odd timing, a spreadsheet is going
to be your best bet.
Suppose for college, your sponsor, Yoshi Nakamura, offers you $6,806 today, or $2,000 for each of the next five years (a total of $10,000). The present value of an ordinary annuity (payments at the
end of each period) of $2,000 assuming an 8% investment rate for five years is $7,985.4 (2000 × 3.9927 factor from the present value of an annuity table). Based on this calculation, the $2,000
annuity is worth more than just $6,806, because you get some of the money each year, which you can then invest. In fact, using a future value analysis, by getting the $10,000 in installments and
investing it at 8%, you’ll end up with $11,733.20.
The important thing to know about the time value of money is that value is not fixed; it’s relative, which is another way of saying, “it depends.” It depends as well on factors other than just the
rate of return you think you could get from investing. It depends on how badly you need the money right now, and inflation, and taxes, and a host of other factors. Still, in essence, the thing to
remember is that a $1 bill right now is more valuable to you than that same $1 bill many years from now. | {"url":"https://www.dwmbeancounter.com/Lumen/content.one.lumenlearning.com/financialaccounting/chapter/investing/index.html","timestamp":"2024-11-08T12:21:56Z","content_type":"text/html","content_length":"153996","record_id":"<urn:uuid:5e580363-13b1-4698-81a7-e372a2912ea2>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00483.warc.gz"} |
Glossary » Units » Pressure » Microbar (barye, Barrie)
Microbar (barye, Barrie) (µbar) is a unit in the category of Pressure. It is also known as microbars. This unit is commonly used in the cgs unit system. Microbar (barye, Barrie) (µbar) has a
dimension of ML^-1T^-2 where M is mass, L is length, and T is time. It can be converted to the corresponding standard SI unit Pa by multiplying its value by a factor of 0.1.
Note that the seven base dimensions are M (Mass), L (Length), T (Time),
(Temperature), N (Aamount of Substance), I (Electric Current), and J (Luminous Intensity).
Other units in the category of Pressure include Atmosphere (metric) (at), Atmosphere (standard) (atm), Bar (bar), Barad (barad), Barye, CentiHg (0°C), Centimeter of Mercury (0°C) (cmHg (0 °C)),
Centimeter of Water (4°C) (cmH[2]O), Dyne Per Square Centimeter (dyn/cm^2), Foot of Water (4°C) (ft H[2]O), Gigapascal (GPa), Hectopascal (hPa), Inch of Mercury (0°C) (inHg (0 °C)), Inch of Mercury
(15.56°C) (inHg (15.56 °C)), Inch of Water (15.56°C) (inH[2]O (15.56 °C)), Inch of Water (4°C) (inH[2]O (4 °C)), Kilogram Force Per Square Centimeter (kgf/cm^2), Kilogram Force Per Square Decimeter
(kgf/dm^2), Kilogram Force Per Square Meter (kgf/m^2), Kilogram Force Per Square Millimeter (kgf/mm^2), Kilopascal (kPa), Kilopound Force Per Square Inch (kip/in^2, ksi, KSI), Megapascal (MPa), Meter
of Water (15.56°C) (mH[2]O, mCE (15.56 °C)), Meter of Water (4°C) (mH[2]O, mCE (4 °C)), Micron of Mercury (millitorr) (µHg (0 °C)), Millibar (mbar), Millimeter of Mercury (0°C) (mmHg, torr, Torr (0
°C)), Millimeter of Water (15.56°C) (mmH[2]O, mmCE (15.56 °C)), Millimeter of Water (4°C) (mmH[2]O, mmCE (4 °C)), Millitorr (mtorr), Newton Per Square Meter (N/m^2), Ounce Force (av.) Per Square Inch
(ozf/in^2, osi), Pascal (Pa, N/m^2), Pound Force Per Square Foot (lbf/ft^2), Pound Force Per Square Inch (psi, PSI, lbf/in^2), Poundal Per Square Foot (pdl/ft^2), Poundal Per Square Inch (pdl/in^2),
Standard Atmosphere (atm), Ton Force (long) Per Square Foot (tonf/ft^2 (UK)), Ton Force (long) Per Square Inch (tonf/in^2 (UK)), Ton Force (metric) Per Square Centimeter (tonf/cm^2 (metric)), Ton
Force (metric) Per Square Meter (tonf/m^2 (metric)), Ton Force (short) Per Square Foot (tonf/ft^2 (US)), Ton Force (short) Per Square Inch (tonf/in^2 (US)), and Torr (torr).
Additional Information
Related Glossary Pages
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m/s-K, 1m/s-K. ·, meter per square second · m/s, 1m/s2. ·, metric slug (hyl) · mug, 9.80665kg. ·, mho · mho, 1S. ·, microbar (barye, barrie) ...
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Unit Step and Delta Functions
Introduction to Heaviside unit step function and Dirac delta function. | {"url":"https://www.efunda.com/glossary/units/units--pressure--microbar_barye_barrie.cfm","timestamp":"2024-11-13T02:36:54Z","content_type":"text/html","content_length":"27488","record_id":"<urn:uuid:f989b752-61f9-483a-a84e-e0b6584439b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00194.warc.gz"} |
How do you find the derivative of $y=\
Hint: We solve the given equation using the identity formula of logarithm where the base of $\ln $ is always $e$. The first step would be to eliminate the logarithm function. Then we first define the
multiplication rule and how the differentiation of function works. We take multiplication of these two different differentiated values. We take the $\dfrac{dy}{dx}$ altogether.
Complete step-by-step solution:
We have $\ln a={{\log }_{e}}a$. So, $y=\ln \left( {{x}^{2}}y \right)$ becomes $y={{\log }_{e}}\left( {{x}^{2}}y \right)$.
We know ${{\log }_{e}}a=y\Rightarrow a={{e}^{y}}$. Applying the rule in case of $y={{\log }_{e}}\left( {{x}^{2}}y \right)$, we get
& y={{\log }_{e}}\left( {{x}^{2}}y \right) \\
& \Rightarrow {{x}^{2}}y={{e}^{y}} \\
We differentiate the given function ${{x}^{2}}y={{e}^{y}}$ with respect to $x$ using the chain rule.
We now discuss the multiplication process of two functions where \[f\left( x \right)=u\left( x \right)v\left( x \right)\]
Differentiating \[f\left( x \right)=uv\], we get \[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ uv \right]=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\].
The above-mentioned rule is the multiplication rule. We apply that on ${{x}^{2}}y$. We assume the functions where \[u\left( x \right)={{x}^{2}},v\left( x \right)=y\]
We know that differentiation of \[u\left( x \right)={{x}^{2}}\] is ${{u}^{'}}\left( x \right)=2x$ as $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and differentiation of $v\left( x \right)=y$
is \[{{v}^{'}}\left( x \right)=\dfrac{dy}{dx}\]. We apply the formula of \[\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}\dfrac{dy}{dx}\]. This followed the differential form of chain rule.
We now take differentiation on both parts of ${{x}^{2}}y={{e}^{y}}$ and get \[\dfrac{d}{dx}\left[ {{x}^{2}}y \right]=\dfrac{d}{dx}\left[ {{e}^{y}} \right]\].
We place the chain rule and \[\dfrac{d}{dx}\left( {{e}^{y}} \right)={{e}^{y}}\dfrac{dy}{dx}\] to get \[y\times 2x+{{x}^{2}}\dfrac{dy}{dx}={{e}^{y}}\dfrac{dy}{dx}\].
We take all the $\dfrac{dy}{dx}$ forms altogether to get
& y\times 2x+{{x}^{2}}\dfrac{dy}{dx}={{e}^{y}}\dfrac{dy}{dx} \\
& \Rightarrow \dfrac{dy}{dx}\left( {{e}^{y}}-{{x}^{2}} \right)=2xy \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{2xy}{\left( {{e}^{y}}-{{x}^{2}} \right)} \\
We now replace the value of ${{x}^{2}}y={{e}^{y}}$ in the denominator and get
\[\dfrac{dy}{dx}=\dfrac{2xy}{\left( {{x}^{2}}y-{{x}^{2}} \right)}=\dfrac{2xy}{{{x}^{2}}\left( y-1 \right)}=\dfrac{2y}{x\left( y-1 \right)}\].
Therefore, differentiation of $y=\ln \left( {{x}^{2}}y \right)$ is \[\dfrac{2y}{x\left( y-1 \right)}\].
Note: We need to remember that in the chain rule \[\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}\], we aren’t
cancelling out the part \[d\left[ h\left( x \right) \right]\]. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to | {"url":"https://www.vedantu.com/question-answer/find-the-derivative-of-yln-left-x2y-rig-class-12-maths-cbse-6010bc30dfcfb40cf08a3ded","timestamp":"2024-11-02T14:39:07Z","content_type":"text/html","content_length":"184993","record_id":"<urn:uuid:183bfc80-13a8-4e6e-9839-4bb27e973201>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00828.warc.gz"} |
Bartlett's Test Calculator
Bartlett's test is used to test the assumption that variances are equal (homogeneous) across groups. For help in using this calculator, read the Frequently-Asked Questions or review the Sample
To learn more about Bartlett's test, read Stat Trek's tutorial on Bartlett's test.
@media (max-width: 699px) { div#GroupData {font-size:10pt;} }
Frequently-Asked Questions
Instructions: To find the answer to a frequently-asked question, simply click on the question.
What is Bartlett's test?
Bartlett's test is used to test the assumption that variances are equal (i.e., homogeneous) across groups. The test is easy to implement and produces valid results, assuming data points within groups
are randomly sampled from a normal distribution.
Because Bartlett's test is sensitive to departures from normality, a normality test is prudent. Several ways to check for departures from normality are described at: How to Test for Normality: Three
Simple Tests.
Note: Unlike Hartley's Fmax test, which also tests for homogeneity, Bartlett's test does not assume equal sample sizes across groups.
How does Bartlett's test work?
Bartlett's test is an actual hypothesis test, where we examine observed data to choose between two statistical hypotheses:
• Null hypothesis: Variance is equal across all groups.
H[0]: σ^2[i] = σ^2[j] for all groups
• Alternative hypothesis: Variance is not equal across all groups.
H[0]: σ^2[i] ≠ σ^2[j] for at least one pair of groups
Like many other techniques for testing hypotheses, Bartlett's test for homogeneity involves computing a test-statistic and finding the P-value for the test statistic, given degrees of freedom and
significance level. If the P-value is bigger than the significance level, we accept the null hypothesis; if it is smaller, we reject the null hypothesis.
What steps (computations) are required to execute Bartlett's test?
The steps required to conduct Bartlett's test for homogeneity are detailed below:
• Step 1. Specify the significance level ( α ).
• Step 2. Compute the sample variance ( s^2[j] ) for each group.
s^2[j] = [ Σ ( X[ i, j] - X[ j] )^ 2 ] / ( n[ j] - 1 )
where X[ i, j] is the score for observation i in Group j , X[ j] is the mean of Group j, n[ j] is the number of observations in Group j , and k is the number of groups.
• Step 3. Compute the pooled estimate of sample variance ( s^2[p] ).
N = Σ n[ i]
s^2[p] = [ Σ ( n[ j]-1 ) s^2[j] ] / ( N - K )
where n[ j] is the sample size in Group j , k is the number of groups, N is the total sample size, and s^2[j] is the sample variance in Group j.
• Step 4. Compute the test statistic (T).
A = ( N - k ) * ln( s^2[p] )
B = Σ [ ( n[ j] - 1 ) * ln( s^2[j] ) ]
C = 1 / [ 3 * ( k - 1 ) ]
D = Σ [ 1 / ( n[ j] - 1 ) - 1 / ( N - k ) ]
T = ( A - B ) / [ 1 + ( C * D ) ]
where A is the first term in the numerator of the test statistic, B is the second term in the numerator, C is the first term in the denominator , D is the second term in the denominator, and ln
is the natural logarithm.
• Step 5. Find the degrees of freedom ( df ), based on the number of groups ( k ).
df = k - 1
• Step 7. Find the P-value for the test statistic. The P-value is the probability of seeing a test statistic more extreme (bigger) than the observed T statistic from Step 4.
It turns out that the test statistic (T) is distributed much like a chi-square statistic with ( k-1 ) degrees of freedom. Knowing the value of T and the degrees of freedom associated with T, we
can use Stat Trek's Chi-Square Calculator to find the P-value - the probability of seeing a test statistic more extreme than T.
• Step 7. Accept or reject the null hypothesis, based on P-value and significance level. If the P-value is bigger than the significance level, we accept the null hypothesis that variances are equal
across groups. Otherwise, we reject the null hypothesis.
What should I enter in the field for number of groups?
Bartlett's is designed to test the hypothesis of homogeneity among nonoverlapping sets of data. The number of groups is the number of data sets under consideration.
What should I enter for significance level?
The significance level is the probability of rejecting the null hypothesis when it is true. Researchers often choose 0.05 or 0.01 for a significance level.
What should I enter for sample size?
Sample size refers to the number of observations in a group. For each group, enter the number of observations in the space provided.
Note: Unlike some other tests for homogeneity (e.g., Hartley's Fmax test), Bartlett's test does not require equal sample sizes across groups.
What should I enter for variance?
In the fields provided, enter an estimate of sample variance for each group. To compute sample variance ( s^2[j] ) for each group, use the following formula:
s^2[j] = [ Σ ( X[ i, j] - X[ j] )^ 2 ] / ( n[ j] - 1 )
where X[ i, j] is the score for observation i in Group j , X[ j] is the mean of Group j, and n[ j] is the number of observations in Group j .
What is degrees of freedom?
Bartlett's test computes a test statistic (T) to test for normality. The degrees of freedom ( df ) for a chi-square test of that statistic is:
df = N - k
where N is the total sample size across all groups, and k is the number of groups in the sample.
What is the test statistic (T)?
The test statistic (T) is the statistic used by Bartlett's test to make a decision about whether to accept or reject the null hypothesis of equal variances between groups. When T is very big, we
reject the null hypothesis; when T is small, we accept the null hypothesis.
The calculator computes a T statistic, based on user inputs. The formulas that the calculator uses to compute a T statistic are given at Bartlett's Test for Homogeneity of Variance.
What is the P-value?
If you assume that the null hypothesis of equal variance is true, the P-value is the probability of seeing a test statistic (T) that is more extreme (bigger) than the actual test statistic computed
from sample data.
How does the calculator test hypotheses?
Like many other techniques for testing hypotheses, Bartlett's test for homogeneity of variance involves computing a test-statistic and finding the P-value for the test statistic, given degrees of
freedom and significance level.
If the P-value is bigger than the significance level, the calculator accepts the null hypothesis. Otherwise, it rejects the null hypothesis.
Sample Problems
Problem 1
The table below shows sample data and variance for five groups. How would you test the assumption that variances are equal across groups?
Group 1 Group 2 Group 3 Group 4 Group 5
Sample Data
2.5 10 22.5 40 62.5
One option would be to use Stat Trek's Bartlett's Test Calculator. Simply, take the following steps:
• Enter the number of groups (5).
• Enter the significance level. For this problem, we'll use 0.05.
• For each group, enter sample size. In this example, the sample size is 5 for each group.
• For each group, enter a sample estimate of group variance.
Then, click the Calculate button to produce the output shown below:
From the calculator, we see that the test statistic (T) is 8.91505. Assuming equal variances in groups and given a significance level of 0.05, the probability of observing a test statistic (T) bigger
than 8.91505 is given by the P-value. Since the P-value (0.06326) is bigger than the significance level (0.05), we cannot reject the null hypothesis of equal variances across groups.
Note: To see the hand calculations required to solve this problem, go to Bartlett's Test for Homogeneity of Variance: Example 1. | {"url":"https://www.stattrek.com/online-calculator/bartletts-test","timestamp":"2024-11-12T22:18:52Z","content_type":"text/html","content_length":"65996","record_id":"<urn:uuid:d64ddb67-a36c-43e7-baa0-5a44008e843d>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00006.warc.gz"} |
Doxastic logic
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Doxastic logic is a modal logic concerned with reasoning about beliefs. The term doxastic derives from the ancient Greek δόξα, doxa, which means "belief." Typically, a doxastic logic uses 'Bx' to
mean "It is believed that x is the case," and the set ${\displaystyle \mathbb{B}}$ denotes a set of beliefs. In doxastic logic, belief is treated as a modal operator.
${\displaystyle \mathbb{B}}$: {${\displaystyle b_1, b_2, ..., b_n}$}
There is complete parallelism between a person who believes propositions and a formal system that derives propositions. Using doxastic logic, one can express the epistemic counterpart of Gödel's
incompleteness theorem of metalogic, as well as Löb's theorem, and other metalogical results in terms of belief.^[1]
Types of reasoners[]
To demonstrate the properties of sets of beliefs, Raymond Smullyan defines the following types of reasoners:
1. REDIRECT Template:Request quotation
1. REDIRECT Template:Request quotation
An accurate reasoner never believes any false proposition. (modal axiom T)
• Inaccurate reasoner:^[1]^[2]^[3]^[4] An inaccurate reasoner believes at least one false proposition.
• Conceited reasoner:^[1]^[4] A conceited reasoner believes his or her beliefs are never inaccurate. A conceited reasoner will necessarily lapse into an inaccuracy.
• Consistent reasoner:^[1]^[2]^[3]^[4] A consistent reasoner never simultaneously believes a proposition and its negation. (modal axiom D)
• Normal reasoner:^[1]^[2]^[3]^[4] A normal reasoner is one who, while believing p, also believes he or she believes p (modal axiom 4).
• Peculiar reasoner:^[1]^[4] A peculiar reasoner believes proposition p while also believing he or she does not believe p. Although a peculiar reasoner may seem like a strange psychological
phenomenon (see Moore's paradox), a peculiar reasoner is necessarily inaccurate but not necessarily inconsistent.
• Regular reasoner:^[1]^[2]^[3]^[4] A regular reasoner is one for whom all beliefs are distributive over logical operations. (modal axiom K)
• Reflexive reasoner:^[1]^[4] A reflexive reasoner is one for whom every proposition p has some q such that the reasoner believes q≡(Bq→p). So if a reflexive reasoner of type 4 [see below] believes
Bp→p, he or she will believe p. This is a parallelism of Löb's theorem for reasoners.
• Unstable reasoner:^[1]^[4] An unstable reasoner is one for whom there is some proposition p such that he or she believes he or she believes p, but who does not really believe p. This is just as
strange a psychological phenomenon as peculiarity; however, an unstable reasoner is not necessarily inconsistent.
• Stable reasoner:^[1]^[4] A stable reasoner is not unstable. That is, for every p, if he or she believes Bp then he or she believes p. Note that stability is the converse of normality. We will say
that a reasoner believes he or she is stable if for every proposition p, he or she believes BBp→Bp (believing: "If I should ever believe that I believe p, then I really will believe p").
• Modest reasoner:^[1]^[4] A modest reasoner is one for whom every believed proposition p, ${\displaystyle Bp\to p}$ only if he or she believes p. A modest reasoner never believes Bp→p unless he or
she believes p. Any reflexive reasoner of type 4 is modest. (Löb's Theorem)
• Queer reasoner:^[4] A queer reasoner is of type G and believes he or she is inconsistent—but is wrong in this belief.
• Timid reasoner:^[4] A timid reasoner is afraid to believe p [i.e., he or she does not believe p] if he or she believes ${\displaystyle Bp\to B\bot }$
Increasing levels of rationality[]
• Type 1 reasoner:^[1]^[2]^[3]^[4]^[5] A type 1 reasoner has a complete knowledge of propositional logic i.e., he or she sooner or later believes every tautology (any proposition provable by truth
tables) (modal axiom N). Also, his or her set of beliefs (past, present and future) is logically closed under modus ponens. If he or she ever believes p and believes p→q (p implies q) then he or
she will (sooner or later) believe q (modal axiom K). This is equivalent to modal system K.
□ p Template:Tee Bp
□ (BpTemplate:AndB(pTemplate:Impq))Template:ImpBq
• Type 1* reasoner:^[1]^[2]^[3]^[4] A type 1* reasoner believes all tautologies; his or her set of beliefs (past, present and future) is logically closed under modus ponens, and for any
propositions p and q, if he or she believes p→q, then he or she will believe that if he or she believes p then he or she will believe q. The type 1* reasoner has a shade more self awareness than
a type 1 reasoner.
□ B(pTemplate:Impq)Template:ImpB(BpTemplate:ImpBq)
• Type 2 reasoner:^[1]^[2]^[3]^[4] A reasoner is of type 2 if he or she is of type 1, and if for every p and q he or she (correctly) believes: "If I should ever believe both p and p→q, then I will
believe q." Being of type 1, he or she also believes the logically equivalent proposition: B(p→q)→(Bp→Bq). A type 2 reasoner knows his or her beliefs are closed under modus ponens.
□ B((BpTemplate:AndB(pTemplate:Impq))Template:ImpBq)
• Type 3 reasoner:^[1]^[2]^[3]^[4] A reasoner is of type 3 if he or she is a normal reasoner of type 2.
• Type 4 reasoner:^[1]^[2]^[3]^[4]^[5] A reasoner is of type 4 if he or she is of type 3 and also believes he or she is normal.
• Type G reasoner:^[1]^[4] A reasoner of type 4 who believes he or she is modest.
Gödel incompleteness and doxastic undecidability[]
Let us say an accurate reasoner is faced with the task of assigning a truth value to a statement posed to him or her. There exists a statement which the reasoner must either remain forever undecided
about or lose his or her accuracy. One solution is the statement:
S: "I will never believe this statement."
If the reasoner ever believes the statement S, it becomes falsified by that fact, making S an untrue belief and hence making the reasoner inaccurate in believing S.
Therefore, since the reasoner is accurate, he or she will never believe S. Hence the statement was true, because that is exactly what it claimed. It further follows that the reasoner will never have
the false belief that S is true. The reasoner cannot believe either that the statement is true or false without becoming inconsistent (i.e. holding two contradictory beliefs). And so the reasoner
must remain forever undecided as to whether the statement S is true or false.
The equivalent theorem is that for any formal system F, there exists a mathematical statement which can be interpreted as "This statement is not provable in formal system F". If the system F is
consistent, neither the statement nor its opposite will be provable in it.^[1]^[4]
Inconsistency and peculiarity of conceited reasoners[]
A reasoner of type 1 is faced with the statement "I will never believe this sentence." The interesting thing now is that if the reasoner believes he or she is always accurate, then he or she will
become inaccurate. Such a reasoner will reason: "If I believe the statement then it will be made false by that fact, which means that I will be inaccurate. This is impossible, since I'm always
accurate. Therefore I can't believe the statement: it must be false."
At this point the reasoner believes that the statement is false, which makes the statement true. Thus the reasoner is inaccurate in believing that the statement is false. If the reasoner hadn't
assumed his or her own accuracy, he or she would never have lapsed into an inaccuracy.
It can also be shown that a conceited reasoner is peculiar.^[1]^[4]
Self fulfilling beliefs[]
For systems, we define reflexivity to mean that for any p (in the language of the system) there is some q such that q≡(Bq→p) is provable in the system. Löb's theorem (in a general form) is that for
any reflexive system of type 4, if Bp→p is provable in the system, so is p.^[1]^[4]
Inconsistency of the belief in one's stability[]
If a consistent reflexive reasoner of type 4 believes that he or she is stable, then he or she will become unstable. Stated otherwise, if a stable reflexive reasoner of type 4 believes that he or she
is stable, then he or she will become inconsistent. Why is this? Suppose that a stable reflexive reasoner of type 4 believes that he or she is stable. We will show that he or she will (sooner or
later) believe every proposition p (and hence be inconsistent). Take any proposition p. The reasoner believes BBp→Bp, hence by Löb's theorem he or she will believe Bp (because he or she believes
Br→r, where r is the proposition Bp, and so he or she will believe r, which is the proposition Bp). Being stable, he or she will then believe p.^[1]^[4]
See also[]
• Raymond Smullyan
• Jaakko Hintikka
• George Boolos
• Common knowledge (logic)
Further reading[]
• Lindström, St. and Wl. Rabinowicz: DDL Unlimited. Dynamic Doxastic Logic for Introspective Agents. In: Erkenntnis 51, 1999, p. 353-385.
• Linski, L.: On Interpreting Doxastic Logic. In: The Journal of Philosophy 65, 1968, p. 500-502.
• Segerberg, Kr.: Default Logic as Dynamic Doxastic Logic. In: Erkenntnis 51, 1999, p. 333-352.
• Wansing,H.: A Reduction of Doxastic Logic to Action Logic. In: Erkenntnis 53, 2000, p. 267-283.
Metalogic and metamathematics
Non-classical logic
Modal logic
• Intuitionistic logic
• Constructive analysis
Intuitionism • Heyting arithmetic
• Intuitionistic type theory
• Constructive set theory
• Degree of truth
• Fuzzy rule
Fuzzy logic • Fuzzy set
• Fuzzy finite element
• Fuzzy set operations
• Structural rule
Substructural logic • Relevance logic
• Linear logic
Paraconsistent logic • Dialetheism
Description logic
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The BLS (Boneh-Lynn-Shacham) signature scheme is a cryptographic scheme that provides a type of digital signature with unique properties and is developed by Dan Boneh, Ben Lynn, and Hovav Shacham in
Overview of the BLS signature scheme:
Key Generation:
Select a pairing-friendly elliptic curve of prime order q that defines the additive prime order group G1 and a multiplicative prime order group G2.
Choose P, a generator point on the elliptic curve G1.
Choose a random secret key, x, from a large prime number space.
Compute the corresponding public key, Q = x * P.
The public key Q is shared, while the secret key x is kept private.
Compute the message hash, H = H_1(m) belongs to G1, where H_1 is a cryptographic hash function.
Multiply the H by the secret key x to generate the signature: S = x * H
To verify a signature S on a message m using a public key Q:
Compute the message hash, H = H_1(m).
Verify the signature by checking if e(S, P) = e(H, Q), where e() is the bilinear pairing function.
e(S, P) is the pairing of the signature S with the generator point P.
e(H, Q) is the pairing of the hashed message digest H and the public key.
If the equation holds, the signature is valid.
The BLS signature scheme has very important properties,
Deterministic: The same message always produces the same signature when signed with the same secret key. This property makes it useful in various applications.
Full Aggregation: One of the significant advantages of BLS signatures is their efficient aggregation property. Multiple signatures can be combined into a single signature that verifies against the
aggregated public key. Aggregation reduces the signature size and computational overhead of verification.
Taking into consideration the deterministic nature of BLS and aggregation property Insaanity protocol deploys much efficient and unique way to achieve selection mechanism and randomness generation
and for zero knowledge proof used in developing the Insaanity cryptographic library | {"url":"https://whitepaper.insaanity.io/insaanity-protocol-architecture/basic-architecture/threshold-cryptography-library/bls","timestamp":"2024-11-02T12:47:42Z","content_type":"text/html","content_length":"175076","record_id":"<urn:uuid:7c4285f1-0dd0-4cc2-b4c8-ee2cc7d74fdb>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00387.warc.gz"} |
Lectures on Disintegration of Measures
by L. Schwartz
Publisher: Tata Institute of Fundamental Research 1976
ISBN/ASIN: B000OK17R6
Number of pages: 139
The material in these Notes has been divided into two parts. In part I, disintegration of a measure with respect to a single sigma-algebra has been considered rather extensively and in part II,
measure valued supermartingales and regular disintegration of a measure with respect to an increasing right continuous family of sigma-algebras have been considered.
Download or read it online for free here:
Download link
(670KB, PDF)
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Example Resultant
[1, 2, 3, 5, 5, 6, 6, 6, 9, 10]
One additional thing to notice is that in the problem, they've asked to merge the given arrays so that it's stored in the first array. Please refer to the problem statement for a detailed
How to Solve?
An approach could be to put all the elements from arr2 to the back of arr1 and then sort arr1 using the in-built sort function.
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
for(int i = 0, j = m; i < n; ++i, ++j)
nums1[j] = nums2[i];
sort(nums1.begin(), nums1.end());
Time & Space Complexity
Space Complexity: O(1) [assuming space complexity for sort to be O(1)]
Time Complexity: This heavily depends on the time complexity of the sort function. The in-built sort function works in O(NlogN) time when sorting N elements.
Therefore, time complexity = O((N+M)*Log(N+M))
Let's improve Time Complexity
Whenever struggling to improve the time complexity, do think of this that have you taken all the given constraints into consideration?
Hint 1 In the earlier approach, we didn't take this fact into consideration that the given arrays are already sorted. Hint 2 The first element in the final array will be either arr1[0] or arr2[0]
(because these are the smallest ones).
Now, let's say the smallest one is arr1[0], now the 2nd element in the final array will be either arr1[1] or arr2[0] Hint 3 Try keeping 2 pointers, one for each of the input arrays. These will point
to the smallest elements in the respective arrays that have not been added to the final sorted array yet. Try to think in this direction? Explanation The idea is basically to have 2-pointers, i & j
(both equal to 0, initially), one for arr1 and one for arr2:
• arr1[i] represents the smallest element from arr1 which hasn't been pushed to the final array yet.
• Similarly, arr2[j] represents the smallest element from arr2 which hasn't been pushed to the final array yet.
Now, since arr1[i] & arr2[j] are always going to be the smallest of all the remaining elements, the next element in our final array will always be either arr1[i] or arr2[j], depending on which one of
the both is smaller.
The whole process will be as explained below:
1. Initialise an empty aux vector, 2 integer pointers i and j, both initialised to 0.
2. Then start iterating while i < len(arr1) and j < len(arr2):
□ If arr1[i] < arr2[j], then push back arr1[i] to aux and increment the value of i by 1.
□ Otherwise, push back arr2[j] to aux and increment the value of j by 1.
3. Now, after the above while loop, exactly 1 of the arrays out arr1 and arr2 will still have some remaining elements to be pushed to aux.
4. So, we can add the remaining elements of both the arrays to aux. (We're adding both just because we don't know which one is remaining). An alternate approach could also be to check which one has
remaining elements, and only add the elements from that array to the aux array.
5. Now that we have our final aux array, all that's needed is to copy the elements from aux array to arr1, as expected in the problem statement.
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
vector<int> aux;
int i = 0, j = 0;
while(i < m and j < n) {
if(nums1[i] < nums2[j])
while(i < m)
while(j < n)
nums1 = aux;
Time & Space Complexity
Space Complexity: O(N+M) [because the aux array will use space]
Time Complexity: O(N+M) [try to think on your own first]
Method 1 to understand Time Complexity
In each iteration of the 1st while loop, either i is incremented, or j is incremented.
This means that the number of times the 1st while loop runs will basically be equal to the value of i + j just after it ends. Now, we know that when it ends, either i will become equal to m or j will
be n.
Let's say, i becomes equal to m, and j is equal to some random number k. That means the first while loop ran M + K times.
Now, let's analyse the 2nd and 3rd while loop. The 2nd loop will run 0 times because i is already equal to m. The 3rd while loop will run N - K times, as we assumed the value of j to be equal to K
after the 1st while loop.
So, in total, no. of operations = (M + K) + (N - K).
Therefore, time complexity = O(N+M);
Method 1 (a crispier one)
Everytime any iteration is run, be it in while loop 1, 2 or 3, exactly 1 element is added to the auxiliary array.
We know that there are going to be exactly N + M elements in the aux array finally.
Therefore, time complexity = O(N + M)
Let's get rid of extra space as well
Please do read the above section before reading this one.
Here, we'll try to not use any auxiliary array, and put the merged elements in a sorted order directly to the arr1.
Hint 1
The problem is that we can't directly use arr1, because if we do so, then we'll end up overwriting the elements present in arr1, which will put is in a pickle.
But, what if instead of putting the elements from the start in increasing order, we put them from the end in decreasing order?
Hint 2 Instead of starting with i = 0 and j = 0 and incrementing them in each iteration, think of starting with i = m - 1 and j = n - 1, and decrementing them in each iteration. Hint 3 We may also
need a 3rd pointer k, initialised with m + n - 1, where we'll put our elements finally. Complete explanation
As already explained in the hints above, we'll basically put the elements in decreasing order in arr1, and since there is empty space for N elements in the right of arr1, there is definitely not
going to be a problem until then.
But what about the time when we've used up the empty space and we start overwriting the elements in arr1? We'll worry about that in a while, let's look at the complete approach first.
For this, we'll initialise 3 pointers: i = m - 1, j = n - 1 and k = n + m - 1, and start iterating while i >= 0 and j >= 0, doing the following in each iteration:
1. If arr1[i] > arr2[j], then arr1[k] = arr1[i] and decrement the values of i & k by 1.
2. Otherwise, then arr1[k] = arr2[j] and decrement the values of j & k by 1.
Kindly note that arr1[k] is used in both scenarios above because it's mentioned in the problem, that we need to merge the 2 arrays in-place in arr1.
Now, after the above while loop ends, as explained in the above section as well, exactly 1 of the 2 arrays will have some un-resolved elements. In a manner similar to the above section, we'll simply
add those elements appropriately.
Won't the overwriting be a problem? The answer is no. The thing is that by the time we over-write a particular index k in arr1, the value of index i would've already become smaller than or equal to k
, and that element arr1[k] would've already gone to its right place, making arr1[k] useless now. (1 specific case could be if i = k, here arr1[k] won't be useless, but it won't mess up the algorithm
for sure)
If you're really interested in understanding this thing, I urge you to take a few examples and try it yourself on pen and paper as this article has already reached almost 1500 words xD
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = m - 1, j = n - 1, k = n + m - 1;
while(i >= 0 and j >= 0) {
if(nums1[i] > nums2[j])
nums1[k--] = nums1[i--];
nums1[k--] = nums2[j--];
while(i >= 0)
nums1[k--] = nums1[i--];
while(j >= 0)
nums1[k--] = nums2[j--];
Time & Space Complexity
Space Complexity: O(1) [simply because no extra space is used xD]
Time Complexity: O(N+M) [please refer to the above section, it's explained in detail there] | {"url":"https://read.learnyard.com/dsa/2-pointers-2/","timestamp":"2024-11-04T01:36:24Z","content_type":"text/html","content_length":"223864","record_id":"<urn:uuid:f837eb6f-6ba0-466e-aedd-ffbde3c4ce24>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00410.warc.gz"} |
Poj 1182 food chain (Classic! Type and query set)
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