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https://support.bioconductor.org/p/80566/
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Not able to display actual grayscale image using EBImage
1
1
Entering edit mode
Kpurva ▴ 10
@kpurva-10058
Last seen 5.6 years ago
Germany
I am using the EBImage package for image processing in R. I have a 260 by 134 Matrix which I converted to an image using
> image1 <- as.Image(matrix1)
And, here is the image object summary
> image1
colorMode : Grayscale
storage.mode : double
dim : 260 134
frames.total : 1
frames.render: 1
imageData(object)[1:5,1:6]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0 0 0 0 0 0
[2,] 0 0 0 0 0 0
[3,] 0 0 0 0 0 0
[4,] 0 0 0 0 0 0
[5,] 0 0 0 0 0 0
A value greater than zero for a specific cell in the image object looks like this:
> imageData(image1)[9,2]
[1] 3686.308
I then use the display function in the EBImage package to view the image which is build from the matrix data.
> display(image1, method = "raster")
However, I get a binary image i.e just black and white pixels. I have shown it below. My data does not just consist of 0 and 1. The background values are zero but the regions with the actual image pattern have values higher than 1. How do I display the image using grayscale and using the functions in this package? Did someone face a similar issue? I also could not find a parameter to specify a gradient level.
ebimage image processing image display • 1.3k views
ADD COMMENT
2
Entering edit mode
Andrzej Oleś ▴ 750
@andrzej-oles-5540
Last seen 18 months ago
Heidelberg, Germany
The reason why the image renders only as black and white is because the display function expects values in the range [0, 1]. Values higher than 1 are clipped and effectively display as 1 (white).
To properly visualize your data you first need to scale the pixel intensity values to the [0, 1] range. This can be done with the help of the normalize function:
image1 <- normalize(image1)
display(image1, method = "raster")
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2022-05-20 12:03:49
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https://sciencehouse.wordpress.com/2012/05/25/nonlinearity-in-your-wallet/
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# Nonlinearity in your wallet
Many human traits like height, IQ, and 50 metre dash times are very close to being normally distributed. The normal distribution (more technically the normal probability density function) or Gaussian function
$f(x) = \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/2\sigma^2}$
is the famous bell shaped curve that the histogram of class grades fall on. The shape of the Gaussian is specified by two parameters the mean $\mu$, which coincides with the peak of the bell, and the standard deviation $\sigma$, which is a measure of how wide the Gaussian is. Let’s take height as an example. There is a 68% chance that any person will be within one standard deviation of the mean and a little more than 95% that you will be within two standard deviations. The tallest one percent are about 2.3 standard deviations from the mean.
The fact that lots of things are normally distributed is not an accident but a consequence of the central limit theorem (CLT), which may be the most important mathematical law in your life. The theorem says that the probability distribution of a sum of a large number of random things will be normal (i.e. a Gaussian). In the example of height, it suggests that there are perhaps hundreds or thousands of genetic and environmental factors that determine your height, each contributing a little amount. When you add them together you get your height and the distribution is normal.
Now, the one major thing in your life that bucks the normal trend is income and especially wealth distribution. Incomes are extremely non-normal. They have what are called fat tails, meaning that the income of the top earners are much higher than would be expected by a normal distribution. A general rule of thumb called the Pareto Principle is that 20% of the population controls 80% of the wealth. It may even be more skewed these days.
There are many theories as to why income and wealth is distributed the way it is and I won’t go into any of these. What I want to point out is that whatever it is that governs income and wealth, it is definitely nonlinear. The key ingredient in the CLT is that the factors add linearly. If there were some nonlinear combination of the variables then the result need not be normal. It has been argued that some amount of inequality is unavoidable given that we are born with unequal innate traits but the translation of those differences into income inequality is a social choice to some degree. If we rewarded the contributors to income more linearly, then incomes would be distributed more normally (there would be some inherent skew because incomes must be positive). In some sense, the fact that some sectors of the economy seem to have much higher incomes than other sectors implies a market failure.
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## 3 thoughts on “Nonlinearity in your wallet”
1. Andy says:
Income looks pretty close to lognormal to me. See figure 1 here: http://fmwww.bc.edu/ec-p/wp671.pdf. With maybe a power-law tail. Couldn’t this just mean that the individual factors which contribute to income (conscientiousness, drive, fluid intelligence, etc) are normally distributed and that they act multiplicatively? It’s worth pointing out that others think that achievement is also log-normally distributed: http://www.johndcook.com/blog/2009/09/29/achievement-is-log-normal/.
Like
2. Lognormal is already a fat tailed distribution and the power law tail makes it even fatter. Multiplicative is nonlinear.
Like
3. […] Scientific Clearing House Carson C. Chow « Nonlinearity in your wallet […]
Like
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2017-12-11 07:46:14
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https://seminars.math.toronto.edu/seminars/list/events.py/process?action=display&file=531079dad1b512b781d7739bf22ac2bf-submission-pkl-1507129742.3022101
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# Dynamics mini-course
Event Information Polynomial foliations of $\mathbb{C}^2$
15:05 on Wednesday November 22, 2017
16:00 on Wednesday November 22, 2017
HU1018, 215 Huron St.
Yury Kudryashov
Cornell University
Consider a vector field $v$ in $ℂ^2$ given by $\dot x=P(x, y)$, $\dot y=Q(x, y)$, where $P$ and $Q$ are polynomials of degree $n$. The geometric properties of $v$ for generic $P$, $Q$ are very different from those of a generic polynomial vector field in $\mathbb{R}^2$. I shall give a very brief overview of these differences, and discuss some recent result.
1. (in collaboration with V. Ramirez) For a generic quadratic vector field, consider the spectra of its finite zeros, and the Camacho-Sad indices of its singularities at the infinite line. Well-known index theorems imply four relations on these numbers, while a simple dimension count shows that there should be at least five relations.
We have found the missing relation, and proved that it can not be represented as an index theorem.
2. (in collaboration with N. Goncharuk) The foliation defined by a generic vector field $v$ possesses infinitely many homologically independent complex limit cycles. Though this result is known for at least 39 years, the proof involved rather long estimates on some integrals.
We provide a very short geometric proof of this statement, that works in some open subsets of a more natural class of polynomial foliations of $\mathbb{CP}^2$ with prescribed $\textbf{projective}$ degree.
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2018-02-25 00:03:51
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https://www.e-olymp.com/en/contests/16667/problems/172578
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favorite We need a little bit of your help to keep things running, click on this banner to learn more
Competitions
# Beehives
Bees are one of the most industrious insects. Since they collect nectar and pollen from flowers, they have to rely on the trees in the forest. For simplicity they numbered the n trees from 0 to n - 1. Instead of roaming around all over the forest, they use a particular list of paths. A path is based on two trees, and they can move either way i.e. from one tree to another in straight line. They don’t use paths that are not in their list.
As technology has been improved a lot, they also changed their working strategy. Instead of hovering over all the trees in the forest, they are targeting particular trees, mainly trees with lots of flowers. So, they planned that they will build some new hives in some targeted trees. After that they will only collect their foods from these trees. They will also remove some paths from their list so that they don’t have to go to a tree with no hive in it.
Now, they want to build the hives such that if one of the paths in their new list go down (some birds or animals disturbs them in that path) it’s still possible to go from any hive to another using the existing paths.
They don't want to choose less than three trees and as hive-building requires a lot of work, they need to keep the number of hives as low as possible. Now you are given the trees with the paths they use, your task is to propose a new bee hive colony for them.
#### Input
Starts with the number of test cases t (t50). Each case starts with a blank line. Next line contains two integers n (2n500) and m(0m20000), where n denotes the number of trees and m denotes the number of paths. Each of the next m lines contains two integers u v (0u, v < n, uv) meaning that there is a path between tree u and v. Assume that there can be at most one path between tree u to v, and needless to say that a path will not be given more than once in the input.
#### Output
For each case, print the case number and the number of beehives in the proposed colony or 'impossible' if its impossible to find such a colony.
#### Note:
Dataset is huge. Use faster I/O methods.
Time limit 5 seconds
Memory limit 128 MiB
Input example #1
3
3 3
0 1
1 2
2 0
2 1
0 1
5 6
0 1
1 2
1 3
2 3
0 4
3 4
Output example #1
Case 1: 3
Case 2: impossible
Case 3: 3
Source 2012 ACM Southwestern European Regional Programming Contest (SWERC), Problem A
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2021-03-07 12:31:58
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https://www.business-science.io/business/2017/11/28/customer_churn_analysis_keras.html
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# Customer Analytics: Using Deep Learning With Keras To Predict Customer Churn
Written by Matt Dancho on November 28, 2017
Customer churn is a problem that all companies need to monitor, especially those that depend on subscription-based revenue streams. The simple fact is that most organizations have data that can be used to target these individuals and to understand the key drivers of churn, and we now have Keras for Deep Learning available in R (Yes, in R!!), which predicted customer churn with 82% accuracy. We’re super excited for this article because we are using the new keras package to produce an Artificial Neural Network (ANN) model on the IBM Watson Telco Customer Churn Data Set! As for most business problems, it’s equally important to explain what features drive the model, which is why we’ll use the lime package for explainability. We cross-checked the LIME results with a Correlation Analysis using the corrr package. We’re not done yet. In addition, we use three new packages to assist with Machine Learning (ML): recipes for preprocessing, rsample for sampling data and yardstick for model metrics. These are relatively new additions to CRAN developed by Max Kuhn at RStudio (creator of the caret package). It seems that R is quickly developing ML tools that rival Python. Good news if you’re interested in applying Deep Learning in R! We are so let’s get going!!
## Customer Churn: Hurts Sales, Hurts Company
Customer churn refers to the situation when a customer ends their relationship with a company, and it’s a costly problem. Customers are the fuel that powers a business. Loss of customers impacts sales. Further, it’s much more difficult and costly to gain new customers than it is to retain existing customers. As a result, organizations need to focus on reducing customer churn.
The good news is that machine learning can help. For many businesses that offer subscription based services, it’s critical to both predict customer churn and explain what features relate to customer churn. Older techniques such as logistic regression can be less accurate than newer techniques such as deep learning, which is why we are going to show you how to model an ANN in R with the keras package.
## Churn Modeling With Artificial Neural Networks (Keras)
Artificial Neural Networks (ANN) are now a staple within the sub-field of Machine Learning called Deep Learning. Deep learning algorithms can be vastly superior to traditional regression and classification methods (e.g. linear and logistic regression) because of the ability to model interactions between features that would otherwise go undetected. The challenge becomes explainability, which is often needed to support the business case. The good news is we get the best of both worlds with keras and lime.
### IBM Watson Dataset (Where We Got The Data)
The dataset used for this tutorial is IBM Watson Telco Dataset. According to IBM, the business challenge is…
A telecommunications company [Telco] is concerned about the number of customers leaving their landline business for cable competitors. They need to understand who is leaving. Imagine that you’re an analyst at this company and you have to find out who is leaving and why.
• Customers who left within the last month: The column is called Churn
• Services that each customer has signed up for: phone, multiple lines, internet, online security, online backup, device protection, tech support, and streaming TV and movies
• Customer account information: how long they’ve been a customer, contract, payment method, paperless billing, monthly charges, and total charges
• Demographic info about customers: gender, age range, and if they have partners and dependents
### Deep Learning With Keras (What We Did With The Data)
In this example we show you how to use keras to develop a sophisticated and highly accurate deep learning model in R. We walk you through the preprocessing steps, investing time into how to format the data for Keras. We inspect the various classification metrics, and show that an un-tuned ANN model can easily get 82% accuracy on the unseen data. Here’s the deep learning training history visualization.
We have some fun with preprocessing the data (yes, preprocessing can actually be fun and easy!). We use the new recipes package to simplify the preprocessing workflow.
We end by showing you how to explain the ANN with the lime package. Neural networks used to be frowned upon because of the “black box” nature meaning these sophisticated models (ANNs are highly accurate) are difficult to explain using traditional methods. Not any more with LIME! Here’s the feature importance visualization.
We also cross-checked the LIME results with a Correlation Analysis using the corrr package. Here’s the correlation visualization.
We even built an ML-Powered Interactive PowerBI Web Application with a Customer Scorecard to monitor customer churn risk and to make recommendations on how to improve customer health! Feel free to take it for a spin.
### Credits
We saw that just last week the same Telco customer churn dataset was used in the article, Predict Customer Churn – Logistic Regression, Decision Tree and Random Forest. We thought the article was excellent.
This article takes a different approach with Keras, LIME, Correlation Analysis, and a few other cutting edge packages. We encourage the readers to check out both articles because, although the problem is the same, both solutions are beneficial to those learning data science and advanced modeling.
## Prerequisites
We use the following libraries in this tutorial:
Install the following packages with install.packages().
If you have not previously run Keras in R, you will need to install Keras using the install_keras() function.
## Import Data
Download the IBM Watson Telco Data Set here. Next, use read_csv() to import the data into a nice tidy data frame. We use the glimpse() function to quickly inspect the data. We have the target “Churn” and all other variables are potential predictors. The raw data set needs to be cleaned and preprocessed for ML.
## Preprocess Data
We’ll go through a few steps to preprocess the data for ML. First, we “prune” the data, which is nothing more than removing unnecessary columns and rows. Then we split into training and testing sets. After that we explore the training set to uncover transformations that will be needed for deep learning. We save the best for last. We end by preprocessing the data with the new recipes package.
### Prune The Data
The data has a few columns and rows we’d like to remove:
• The “customerID” column is a unique identifier for each observation that isn’t needed for modeling. We can de-select this column.
• The data has 11 NA values all in the “TotalCharges” column. Because it’s such a small percentage of the total population (99.8% complete cases), we can drop these observations with the drop_na() function from tidyr. Note that these may be customers that have not yet been charged, and therefore an alternative is to replace with zero or -99 to segregate this population from the rest.
• My preference is to have the target in the first column so we’ll include a final select operation to do so.
We’ll perform the cleaning operation with one tidyverse pipe (%>%) chain.
### Split Into Train/Test Sets
We have a new package, rsample, which is very useful for sampling methods. It has the initial_split() function for splitting data sets into training and testing sets. The return is a special rsplit object.
We can retrieve our training and testing sets using training() and testing() functions.
### Exploration: What Transformation Steps Are Needed For ML?
This phase of the analysis is often called exploratory analysis, but basically we are trying to answer the question, “What steps are needed to prepare for ML?” The key concept is knowing what transformations are needed to run the algorithm most effectively. Artificial Neural Networks are best when the data is one-hot encoded, scaled and centered. In addition, other transformations may be beneficial as well to make relationships easier for the algorithm to identify. A full exploratory analysis is not practical in this article. With that said we’ll cover a few tips on transformations that can help as they relate to this dataset. In the next section, we will implement the preprocessing techniques.
#### Discretize The “tenure” Feature
Numeric features like age, years worked, length of time in a position can generalize a group (or cohort). We see this in marketing a lot (think “millennials”, which identifies a group born in a certain timeframe). The “tenure” feature falls into this category of numeric features that can be discretized into groups.
We can split into six cohorts that divide up the user base by tenure in roughly one year (12 month) increments. This should help the ML algorithm detect if a group is more/less susceptible to customer churn.
#### Transform The “TotalCharges” Feature
What we don’t like to see is when a lot of observations are bunched within a small part of the range.
We can use a log transformation to even out the data into more of a normal distribution. It’s not perfect, but it’s quick and easy to get our data spread out a bit more.
Pro Tip: A quick test is to see if the log transformation increases the magnitude of the correlation between “TotalCharges” and “Churn”. We’ll use a few dplyr operations along with the corrr package to perform a quick correlation.
• correlate(): Performs tidy correlations on numeric data
• focus(): Similar to select(). Takes columns and focuses on only the rows/columns of importance.
• fashion(): Makes the formatting aesthetically easier to read.
The correlation between “Churn” and “LogTotalCharges” is greatest in magnitude indicating the log transformation should improve the accuracy of the ANN model we build. Therefore, we should perform the log transformation.
#### One-Hot Encoding
One-hot encoding is the process of converting categorical data to sparse data, which has columns of only zeros and ones (this is also called creating “dummy variables” or a “design matrix”). All non-numeric data will need to be converted to dummy variables. This is simple for binary Yes/No data because we can simply convert to 1’s and 0’s. It becomes slightly more complicated with multiple categories, which requires creating new columns of 1’s and 0s for each category (actually one less). We have four features that are multi-category: Contract, Internet Service, Multiple Lines, and Payment Method.
#### Feature Scaling
ANN’s typically perform faster and often times with higher accuracy when the features are scaled and/or normalized (aka centered and scaled, also known as standardizing). Because ANNs use gradient descent, weights tend to update faster. According to Sebastian Raschka, an expert in the field of Deep Learning, several examples when feature scaling is important are:
• k-nearest neighbors with an Euclidean distance measure if want all features to contribute equally
• k-means (see k-nearest neighbors)
• logistic regression, SVMs, perceptrons, neural networks etc. if you are using gradient descent/ascent-based optimization, otherwise some weights will update much faster than others
• linear discriminant analysis, principal component analysis, kernel principal component analysis since you want to find directions of maximizing the variance (under the constraints that those directions/eigenvectors/principal components are orthogonal); you want to have features on the same scale since you’d emphasize variables on “larger measurement scales” more. There are many more cases than I can possibly list here … I always recommend you to think about the algorithm and what it’s doing, and then it typically becomes obvious whether we want to scale your features or not.
The interested reader can read Sebastian Raschka’s article for a full discussion on the scaling/normalization topic. Pro Tip: When in doubt, standardize the data.
### Preprocessing With Recipes
Let’s implement the preprocessing steps/transformations uncovered during our exploration. Max Kuhn (creator of caret) has been putting some work into Rlang ML tools lately, and the payoff is beginning to take shape. A new package, recipes, makes creating ML data preprocessing workflows a breeze! It takes a little getting used to, but I’ve found that it really helps manage the preprocessing steps. We’ll go over the nitty gritty as it applies to this problem.
#### Step 1: Create A Recipe
A “recipe” is nothing more than a series of steps you would like to perform on the training, testing and/or validation sets. Think of preprocessing data like baking a cake (I’m not a baker but stay with me). The recipe is our steps to make the cake. It doesn’t do anything other than create the playbook for baking.
We use the recipe() function to implement our preprocessing steps. The function takes a familiar object argument, which is a modeling function such as object = Churn ~ . meaning “Churn” is the outcome (aka response, predictor, target) and all other features are predictors. The function also takes the data argument, which gives the “recipe steps” perspective on how to apply during baking (next).
A recipe is not very useful until we add “steps”, which are used to transform the data during baking. The package contains a number of useful “step functions” that can be applied. The entire list of Step Functions can be viewed here. For our model, we use:
1. step_discretize() with the option = list(cuts = 6) to cut the continuous variable for “tenure” (number of years as a customer) to group customers into cohorts.
2. step_log() to log transform “TotalCharges”.
3. step_dummy() to one-hot encode the categorical data. Note that this adds columns of one/zero for categorical data with three or more categories.
4. step_center() to mean-center the data.
5. step_scale() to scale the data.
The last step is to prepare the recipe with the prep() function. This step is used to “estimate the required parameters from a training set that can later be applied to other data sets”. This is important for centering and scaling and other functions that use parameters defined from the training set.
Here’s how simple it is to implement the preprocessing steps that we went over!
We can print the recipe object if we ever forget what steps were used to prepare the data. Pro Tip: We can save the recipe object as an RDS file using saveRDS(), and then use it to bake() (discussed next) future raw data into ML-ready data in production!
#### Step 2: Baking With Your Recipe
Now for the fun part! We can apply the “recipe” to any data set with the bake() function, and it processes the data following our recipe steps. We’ll apply to our training and testing data to convert from raw data to a machine learning dataset. Check our training set out with glimpse(). Now that’s an ML-ready dataset prepared for ANN modeling!!
#### Step 3: Don’t Forget The Target
One last step, we need to store the actual values (truth) as y_train_vec and y_test_vec, which are needed for modeling our ANN. We convert to a series of numeric ones and zeros which can be accepted by the Keras ANN modeling functions. We add “vec” to the name so we can easily remember the class of the object (it’s easy to get confused when working with tibbles, vectors, and matrix data types).
## Model Customer Churn With Keras (Deep Learning)
This is super exciting!! Finally, Deep Learning with Keras in R! The team at RStudio has done fantastic work recently to create the keras package, which implements Keras in R. Very cool!
### Background On Artifical Neural Networks
For those unfamiliar with Neural Networks (and those that need a refresher), read this article. It’s very comprehensive, and you’ll leave with a general understanding of the types of deep learning and how they work.
Source: Xenon Stack
Deep Learning has been available in R for some time, but the primary packages used in the wild have not (this includes Keras, Tensor Flow, Theano, etc, which are all Python libraries). It’s worth mentioning that a number of other Deep Learning packages exist in R including h2o, mxnet, and others. The interested reader can check out this blog post for a comparison of deep learning packages in R.
### Building A Deep Learning Model
We’re going to build a special class of ANN called a Multi-Layer Perceptron (MLP). MLPs are one of the simplest forms of deep learning, but they are both highly accurate and serve as a jumping-off point for more complex algorithms. MLPs are quite versatile as they can be used for regression, binary and multi classification (and are typically quite good at classification problems).
We’ll build a three layer MLP with Keras. Let’s walk-through the steps before we implement in R.
1. Initialize a sequential model: The first step is to initialize a sequential model with keras_model_sequential(), which is the beginning of our Keras model. The sequential model is composed of a linear stack of layers.
2. Apply layers to the sequential model: Layers consist of the input layer, hidden layers and an output layer. The input layer is the data and provided it’s formatted correctly there’s nothing more to discuss. The hidden layers and output layers are what controls the ANN inner workings.
• Hidden Layers: Hidden layers form the neural network nodes that enable non-linear activation using weights. The hidden layers are created using layer_dense(). We’ll add two hidden layers. We’ll apply units = 16, which is the number of nodes. We’ll select kernel_initializer = "uniform" and activation = "relu" for both layers. The first layer needs to have the input_shape = 35, which is the number of columns in the training set. Key Point: While we are arbitrarily selecting the number of hidden layers, units, kernel initializers and activation functions, these parameters can be optimized through a process called hyperparameter tuning that is discussed in Next Steps.
• Dropout Layers: Dropout layers are used to control overfitting. This eliminates weights below a cutoff threshold to prevent low weights from overfitting the layers. We use the layer_dropout() function add two drop out layers with rate = 0.10 to remove weights below 10%.
• Output Layer: The output layer specifies the shape of the output and the method of assimilating the learned information. The output layer is applied using the layer_dense(). For binary values, the shape should be units = 1. For multi-classification, the units should correspond to the number of classes. We set the kernel_initializer = "uniform" and the activation = "sigmoid" (common for binary classification).
3. Compile the model: The last step is to compile the model with compile(). We’ll use optimizer = "adam", which is one of the most popular optimization algorithms. We select loss = "binary_crossentropy" since this is a binary classification problem. We’ll select metrics = c("accuracy") to be evaluated during training and testing. Key Point: The optimizer is often included in the tuning process.
Let’s codify the discussion above to build our Keras MLP-flavored ANN model.
We use the fit() function to run the ANN on our training data. The object is our model, and x and y are our training data in matrix and numeric vector forms, respectively. The batch_size = 50 sets the number samples per gradient update within each epoch. We set epochs = 35 to control the number training cycles. Typically we want to keep the batch size high since this decreases the error within each training cycle (epoch). We also want epochs to be large, which is important in visualizing the training history (discussed below). We set validation_split = 0.30 to include 30% of the data for model validation, which prevents overfitting. The training process should complete in 15 seconds or so.
We can inspect the final model. We want to make sure there is minimal difference between the validation accuracy and the training accuracy.
We can visualize the Keras training history using the plot() function. What we want to see is the validation accuracy and loss leveling off, which means the model has completed training. We see that there is some divergence between training loss/accuracy and validation loss/accuracy. This model indicates we can possibly stop training at an earlier epoch. Pro Tip: Only use enough epochs to get a high validation accuracy. Once validation accuracy curve begins to flatten or decrease, it’s time to stop training.
### Making Predictions
We’ve got a good model based on the validation accuracy. Now let’s make some predictions from our keras model on the test data set, which was unseen during modeling (we use this for the true performance assessment). We have two functions to generate predictions:
• predict_classes: Generates class values as a matrix of ones and zeros. Since we are dealing with binary classification, we’ll convert the output to a vector.
• predict_proba: Generates the class probabilities as a numeric matrix indicating the probability of being a class. Again, we convert to a numeric vector because there is only one column output.
## Inspect Performance With Yardstick
The yardstick package has a collection of handy functions for measuring performance of machine learning models. We’ll overview some metrics we can use to understand the performance of our model.
First, let’s get the data formatted for yardstick. We create a data frame with the truth (actual values as factors), estimate (predicted values as factors), and the class probability (probability of yes as numeric). We use the fct_recode() function from the forcats package to assist with recoding as Yes/No values.
Now that we have the data formatted, we can take advantage of the yardstick package. The only other thing we need to do is to set options(yardstick.event_first = FALSE). As pointed out by ad1729 in GitHub Issue 13, the default is to classify 0 as the positive class instead of 1.
#### Confusion Table
We can use the conf_mat() function to get the confusion table. We see that the model was by no means perfect, but it did a decent job of identifying customers likely to churn.
#### Accuracy
We can use the metrics() function to get an accuracy measurement from the test set. We are getting roughly 82% accuracy.
#### AUC
We can also get the ROC Area Under the Curve (AUC) measurement. AUC is often a good metric used to compare different classifiers and to compare to randomly guessing (AUC_random = 0.50). Our model has AUC = 0.85, which is much better than randomly guessing. Tuning and testing different classification algorithms may yield even better results.
#### Precision And Recall
Precision is when the model predicts “yes”, how often is it actually “yes”. Recall (also true positive rate or specificity) is when the actual value is “yes” how often is the model correct. We can get precision() and recall() measurements using yardstick.
Precision and recall are very important to the business case: The organization is concerned with balancing the cost of targeting and retaining customers at risk of leaving with the cost of inadvertently targeting customers that are not planning to leave (and potentially decreasing revenue from this group). The threshold above which to predict Churn = “Yes” can be adjusted to optimize for the business problem. This becomes an Customer Lifetime Value optimization problem that is discussed further in Next Steps.
#### F1 Score
We can also get the F1-score, which is a weighted average between the precision and recall. Machine learning classifier thresholds are often adjusted to maximize the F1-score. However, this is often not the optimal solution to the business problem.
## Explain The Model With LIME
LIME stands for Local Interpretable Model-agnostic Explanations, and is a method for explaining black-box machine learning model classifiers. For those new to LIME, this YouTube video does a really nice job explaining how LIME helps to identify feature importance with black box machine learning models (e.g. deep learning, stacked ensembles, random forest).
#### Setup
The lime package implements LIME in R. One thing to note is that it’s not setup out-of-the-box to work with keras. The good news is with a few functions we can get everything working properly. We’ll need to make two custom functions:
• model_type: Used to tell lime what type of model we are dealing with. It could be classification, regression, survival, etc.
• predict_model: Used to allow lime to perform predictions that its algorithm can interpret.
The first thing we need to do is identify the class of our model object. We do this with the class() function.
Next we create our model_type() function. It’s only input is x the keras model. The function simply returns “classification”, which tells LIME we are classifying.
Now we can create our predict_model() function, which wraps keras::predict_proba(). The trick here is to realize that it’s inputs must be x a model, newdata a dataframe object (this is important), and type which is not used but can be use to switch the output type. The output is also a little tricky because it must be in the format of probabilities by classification (this is important; shown next).
Run this next script to show you what the output looks like and to test our predict_model() function. See how it’s the probabilities by classification. It must be in this form for model_type = "classification".
Now the fun part, we create an explainer using the lime() function. Just pass the training data set without the “Attribution column”. The form must be a data frame, which is OK since our predict_model function will switch it to an keras object. Set model = automl_leader our leader model, and bin_continuous = FALSE. We could tell the algorithm to bin continuous variables, but this may not make sense for categorical numeric data that we didn’t change to factors.
Now we run the explain() function, which returns our explanation. This can take a minute to run so we limit it to just the first ten rows of the test data set. We set n_labels = 1 because we care about explaining a single class. Setting n_features = 4 returns the top four features that are critical to each case. Finally, setting kernel_width = 0.5 allows us to increase the “model_r2” value by shrinking the localized evaluation.
#### Feature Importance Visualization
The payoff for the work we put in using LIME is this feature importance plot. This allows us to visualize each of the first ten cases (observations) from the test data. The top four features for each case are shown. Note that they are not the same for each case. The green bars mean that the feature supports the model conclusion, and the red bars contradict. A few important features based on frequency in first ten cases:
• Tenure (7 cases)
• Senior Citizen (5 cases)
• Online Security (4 cases)
Another excellent visualization can be performed using plot_explanations(), which produces a facetted heatmap of all case/label/feature combinations. It’s a more condensed version of plot_features(), but we need to be careful because it does not provide exact statistics and it makes it less easy to investigate binned features (Notice that “tenure” would not be identified as a contributor even though it shows up as a top feature in 7 of 10 cases).
## Check Explanations With Correlation Analysis
One thing we need to be careful with the LIME visualization is that we are only doing a sample of the data, in our case the first 10 test observations. Therefore, we are gaining a very localized understanding of how the ANN works. However, we also want to know on from a global perspective what drives feature importance.
We can perform a correlation analysis on the training set as well to help glean what features correlate globally to “Churn”. We’ll use the corrr package, which performs tidy correlations with the function correlate(). We can get the correlations as follows.
The correlation visualization helps in distinguishing which features are relavant to Churn.
The correlation analysis helps us quickly disseminate which features that the LIME analysis may be excluding. We can see that the following features are highly correlated (magnitude > 0.25):
Increases Likelihood of Churn (Red):
• Tenure = Bin 1 (<12 Months)
• Internet Service = “Fiber Optic”
• Payment Method = “Electronic Check”
Decreases Likelihood of Churn (Blue):
• Contract = “Two Year”
• Total Charges (Note that this may be a biproduct of additional services such as Online Security)
## Feature Investigation
We can investigate features that are most frequent in the LIME feature importance visualization along with those that the correlation analysis shows an above normal magnitude. We’ll investigate:
• Tenure (7/10 LIME Cases, Highly Correlated)
• Contract (Highly Correlated)
• Internet Service (Highly Correlated)
• Payment Method (Highly Correlated)
• Senior Citizen (5/10 LIME Cases)
• Online Security (4/10 LIME Cases)
#### Tenure (7/10 LIME Cases, Highly Correlated)
LIME cases indicate that the ANN model is using this feature frequently and high correlation agrees that this is important. Investigating the feature distribution, it appears that customers with lower tenure (bin 1) are more likely to leave. Opportunity: Target customers with less than 12 month tenure.
#### Contract (Highly Correlated)
While LIME did not indicate this as a primary feature in the first 10 cases, the feature is clearly correlated with those electing to stay. Customers with one and two year contracts are much less likely to churn. Opportunity: Offer promotion to switch to long term contracts.
#### Internet Service (Highly Correlated)
While LIME did not indicate this as a primary feature in the first 10 cases, the feature is clearly correlated with those electing to stay. Customers with fiber optic service are more likely to churn while those with no internet service are less likely to churn. Improvement Area: Customers may be dissatisfied with fiber optic service.
#### Payment Method (Highly Correlated)
While LIME did not indicate this as a primary feature in the first 10 cases, the feature is clearly correlated with those electing to stay. Customers with electronic check are more likely to leave. Opportunity: Offer customers a promotion to switch to automatic payments.
#### Senior Citizen (5/10 LIME Cases)
Senior citizen appeared in several of the LIME cases indicating it was important to the ANN for the 10 samples. However, it was not highly correlated to Churn, which may indicate that the ANN is using in an more sophisticated manner (e.g. as an interaction). It’s difficult to say that senior citizens are more likely to leave, but non-senior citizens appear less at risk of churning. Opportunity: Target users in the lower age demographic.
#### Online Security (4/10 LIME Cases)
Customers that did not sign up for online security were more likely to leave while customers with no internet service or online security were less likely to leave. Opportunity: Promote online security and other packages that increase retention rates.
## Next Steps: Business Science University
We’ve just scratched the surface with the solution to this problem, but unfortunately there’s only so much ground we can cover in an article. Here are a few next steps that I’m pleased to announce will be covered in a Business Science University course coming in 2018!
Your organization needs to see the financial benefit so always tie your analysis to sales, profitability or ROI. Customer Lifetime Value (CLV) is a methodology that ties the business profitability to the retention rate. While we did not implement the CLV methodology herein, a full customer churn analysis would tie the churn to an classification cutoff (threshold) optimization to maximize the CLV with the predictive ANN model.
The simplified CLV model is:
Where,
• GC is the gross contribution per customer
• d is the annual discount rate
• r is the retention rate
### ANN Performance Evaluation and Improvement
The ANN model we built is good, but it could be better. How we understand our model accuracy and improve on it is through the combination of two techniques:
• K-Fold Cross-Fold Validation: Used to obtain bounds for accuracy estimates.
• Hyper Parameter Tuning: Used to improve model performance by searching for the best parameters possible.
We need to implement K-Fold Cross Validation and Hyper Parameter Tuning if we want a best-in-class model.
### Distributing Analytics
It’s critical to communicate data science insights to decision makers in the organization. Most decision makers in organizations are not data scientists, but these individuals make important decisions on a day-to-day basis. The PowerBI application below includes a Customer Scorecard to monitor customer health (risk of churn). The application walks the user through the machine learning journey for how the model was developed, what it means to stakeholders, and how it can be used in production.
For those seeking options for distributing analytics, two good options are:
• Shiny Apps for rapid prototyping: Shiny web applications offer the maximum flexibility with R algorithms built in. Shiny is more complex to learn, but Shiny applications are incredible / limitless.
• Microsoft PowerBI and Tableau for Visualization: Enable distributed analytics with the advantage of intuitive structure but with some flexibilty sacrificed. Can be difficult to build ML into.
You’re probably wondering why we are going into so much detail on next steps. Business Science University, an online school dedicated to helping data science learners.
Benefits to learners:
• Build your own online GitHub portfolio of data science projects to market your skills to future employers!
• Learn real-world applications in People Analytics (HR), Customer Analytics, Marketing Analytics, Social Media Analytics, Text Mining and Natural Language Processing (NLP), Financial and Time Series Analytics, and more!
• Use advanced machine learning techniques for both high accuracy modeling and explaining features that have an effect on the outcome!
• Create ML-powered web-applications that can be distributed throughout an organization, enabling non-data scientists to benefit from algorithms in a user-friendly way!
Enrollment is open so please signup for special perks. Just go to Business Science University and select enroll.
## Conclusions
Customer churn is a costly problem. The good news is that machine learning can solve churn problems, making the organization more profitable in the process. In this article, we saw how Deep Learning can be used to predict customer churn. We built an ANN model using the new keras package that achieved 82% predictive accuracy (without tuning)! We used three new machine learning packages to help with preprocessing and measuring performance: recipes, rsample and yardstick. Finally we used lime` to explain the Deep Learning model, which traditionally was impossible! We checked the LIME results with a Correlation Analysis, which brought to light other features to investigate. For the IBM Telco dataset, tenure, contract type, internet service type, payment menthod, senior citizen status, and online security status were useful in diagnosing customer churn. We hope you enjoyed this article!
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2020-01-25 23:52:46
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https://tex.stackexchange.com/questions/407276/hidden-chapter-in-header-using-scrpage2
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# Hidden \chapter* in header using scrpage2
How can I include a "hidden" \section*{} or \chapter*{} in the header (centered) using scrpage2?
\documentclass[a4paper,12pt]{scrbook}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrpage2}
\automark[chapter*]{section*} %%leftpageheader: chapter* / rightpagehaeder: section*
\begin{document}
\chapter*{Bla}
\pagebreak
\section*{Blub}
\end{document}
• Do you just want unnumbered chapers and sections? Dec 22 '17 at 14:10
• Yes. unnumbered Dec 22 '17 at 14:14
• Do you use \chapter* here to have an unnumbered chapter or to prevent the chapter from being in toc (or both)? Dec 22 '17 at 15:05
• @TeXnician I just don't want any numbers. (but I want other parts of the book with numbered chapters as well) Dec 22 '17 at 15:08
• Please note: scrpage2 is obsolete. You should use scrlayer-scrpage instead of scrpage2. Dec 22 '17 at 15:22
scrbook like book does not provide automatic running head for \chapter* and \section*. If everything should be unnumbered, set \secnumdepth to -2 and use \chapter and \section instead of \chapter* and \section*:
\documentclass[a4paper,12pt]{book}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrpage2}
\setcounter{secnumdepth}{-2}%
\usepackage{blindtext}
\begin{document}
\chapter{Bla}
\section{Blub}
\Blindtext[10]
\end{document}
If you have numbered and not numbered chapters and section, you can use:
\documentclass[a4paper,12pt]{book}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrpage2}
\usepackage{blindtext}
\begin{document}
\chapter*{Bla}
\markboth{\MakeUppercase{Bla}}{}
\section{Blub}
\markright{\MakeUppercase{Blub}}
\Blindtext[10]
\end{document}
But using scrbook it is more easy to use \addchap and \addsec:
\documentclass[a4paper,12pt]{scrbook}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrpage2}
\usepackage{blindtext}
\begin{document}
\Blindtext[10]
\end{document}
If you want to avoid the ToC entries:
\documentclass[a4paper,12pt,headings=optiontotoc]{scrbook}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrpage2}
\usepackage{blindtext}
\begin{document}
\tableofcontents
\Blindtext[10]
\end{document}
And please note, that scrpage2 is obsolete. You should use scrlayer-scrpage:
\documentclass[a4paper,12pt,headings=optiontotoc]{scrbook}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrlayer-scrpage}
\usepackage{blindtext}
\begin{document}
\tableofcontents
\Blindtext[10]
\end{document}
See the KOMA-Script manual or the German KOMA-Script manual or the German KOMA-Script book for more information about scrlayer-scrpage, \addchap, \addsec and option headings.
• Thank you @Schweinebacke – but what if I want numbered and unnumbered \chapterin the same document (like an unnumbered appendix)? Dec 22 '17 at 15:10
• @לאהפּאַסטעך See the additional examples. Dec 22 '17 at 15:20
• Thank you both! the addchap / addsec command is new to me, but much better than the "old" \chapter* And thanks for the information that scrpage2is obsolete! Dec 22 '17 at 15:30
As your intention is to get unnumbered sections or chapters you can use KOMA's commands \addchap and \addsec which will produce unnumbered headings, but set the head marks and add the title to toc.
\documentclass[a4paper,12pt]{scrbook}
\usepackage[T1]{fontenc}
\usepackage[automark]{scrpage2}
• Thank you! I think using \addchapis the best idea here! Dec 22 '17 at 15:31
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2022-01-24 19:56:49
|
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|
http://mathhelpforum.com/trigonometry/101529-determining-quadrants-pi.html
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# Math Help - Determining quadrants from pi
1. ## Determining quadrants from pi
Im currently studying to take the ACT for the third time and am confused about how to figure out what quadrant different triangles will get zoned into based on pi. When looking up a missed question it says since pi/2 < x < pi; then x must be in quadrant II. Are there formulas like this for finding out all the quadrants? If so I would appreciate if someone could tell me about them. Thanks
2. Originally Posted by Jesse 8620
Im currently studying to take the ACT for the third time and am confused about how to figure out what quadrant different triangles will get zoned into based on pi. When looking up a missed question it says since pi/2 < x < pi; then x must be in quadrant II. Are there formulas like this for finding out all the quadrants? If so I would appreciate if someone could tell me about them. Thanks
In radians, $\pi$ corresponds to 180 degrees, so $2\pi$ is 360 degrees. That should help you figure out how to write the four quadrants in terms of pi.
|
2014-03-11 06:31:23
|
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https://socratic.org/questions/how-do-you-find-the-local-max-and-min-for-f-x-x-4-2x-2-1
|
How do you find the local max and min for f(x) = x^4 - 2x^2 + 1?
Sep 26, 2016
max at (0 ,1), mins at (-1 ,0) and (1 ,0)
Explanation:
Differentiate f(x) and equate to zero to find $\textcolor{b l u e}{\text{critical points}}$
$\Rightarrow f ' \left(x\right) = 4 {x}^{3} - 4 x$
$4 {x}^{3} - 4 x = 0 \Rightarrow 4 x \left({x}^{2} - 1\right) = 0 \Rightarrow 4 x \left(x - 1\right) \left(x + 1\right) = 0$
$\Rightarrow x = 0 , x = - 1 , x = 1$
Find the y coordinates.
$f \left(0\right) = 0 - 0 + 1 = 1 \Rightarrow \left(0 , 1\right)$
$f \left(- 1\right) = 1 - 2 + 1 = 0 \Rightarrow \left(- 1 , 0\right)$
$f \left(1\right) = 1 - 2 + 1 = 0 \Rightarrow \left(1 , 0\right)$
To find if max/min use the $\textcolor{b l u e}{\text{second derivative test}}$
• If f''(a) > 0 , then minimum
• If f''(a) < 0 , then maximum
The second derivative is.
$f ' ' \left(x\right) = 12 {x}^{2} - 4$
and $f ' ' \left(0\right) = - 4 < 0 \Rightarrow \left(0 , 1\right) \text{ is a maximum}$
$f ' ' \left(- 1\right) = 8 > 0 \Rightarrow \left(- 1 , 0\right) \text{ is a minimum}$
$f ' ' \left(1\right) = 8 > 0 \Rightarrow \left(1 , 0\right) \text{ is a minimum}$
graph{x^4-2x^2+1 [-10, 10, -5, 5]}
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2021-07-24 19:35:00
|
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https://www.gap-system.org/Manuals/pkg/semigroups-3.0.20/doc/chap13.html
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Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Bib Ind
### 13 Attributes and operations for semigroups
In this chapter we decribe the methods that are available in Semigroups for determining the attributes of a semigroup, and the operations which can be applied to a semigroup.
#### 13.1 Accessing the elements of a semigroup
##### 13.1-1 AsListCanonical
‣ AsListCanonical( S ) ( attribute )
‣ EnumeratorCanonical( S ) ( attribute )
‣ IteratorCanonical( S ) ( operation )
Returns: A list, enumerator, or iterator.
When the argument S is a semigroup in the representation IsEnumerableSemigroupRep (6.1-4), AsListCanonical returns a list of the elements of S in the order they are enumerated by the Froidure-Pin Algorithm. This is the same as the order used to index the elements of S in RightCayleyGraphSemigroup (13.2-1) and LeftCayleyGraphSemigroup (13.2-1).
EnumeratorCanonical and IteratorCanonical return an enumerator and an iterator where the elements are ordered in the same way as AsListCanonical. Using EnumeratorCanonical or IteratorCanonical will often use less memory than AsListCanonical, but may have slightly worse performance if the elements of the semigroup are looped over repeatedly. EnumeratorCanonical returns the same list as AsListCanonical if AsListCanonical has ever been called for S.
If S is an acting semigroup, then the value returned by AsList may not equal the value returned by AsListCanonical. AsListCanonical exists so that there is a method for obtaining the elements of S in the particular order used by RightCayleyGraphSemigroup (13.2-1) and LeftCayleyGraphSemigroup (13.2-1).
See also PositionCanonical (13.1-2).
gap> S := Semigroup(Transformation([1, 3, 2]));;
gap> AsListCanonical(S);
[ Transformation( [ 1, 3, 2 ] ), IdentityTransformation ]
gap> IteratorCanonical(S);
<iterator>
gap> EnumeratorCanonical(S);
[ Transformation( [ 1, 3, 2 ] ), IdentityTransformation ]
gap> S := Monoid([Matrix(IsBooleanMat, [[1, 0, 0],
> [0, 1, 0],
> [0, 1, 0]])]);
<commutative monoid of 3x3 boolean matrices with 1 generator>
gap> it := IteratorCanonical(S);
<iterator>
gap> NextIterator(it);
Matrix(IsBooleanMat, [[1, 0, 0], [0, 1, 0], [0, 0, 1]])
gap> en := EnumeratorCanonical(S);
<enumerator of <commutative monoid of 3x3 boolean matrices with 1
generator>>
gap> en[1];
Matrix(IsBooleanMat, [[1, 0, 0], [0, 1, 0], [0, 0, 1]])
gap> Position(en, en[1]);
1
gap> Length(en);
2
##### 13.1-2 PositionCanonical
‣ PositionCanonical( S, x ) ( operation )
Returns: true or false.
When the argument S is a semigroup in the representation IsEnumerableSemigroupRep (6.1-4) and x is an element of S, PositionCanonical returns the position of x in AsListCanonical(S) or equivalently the position of x in EnumeratorCanonical(S).
See also AsListCanonical (13.1-1) and EnumeratorCanonical (13.1-1).
gap> S := FullTropicalMaxPlusMonoid(2, 3);
<monoid of 2x2 tropical max-plus matrices with 13 generators>
gap> x := Matrix(IsTropicalMaxPlusMatrix, [[1, 3], [2, 1]], 3);
Matrix(IsTropicalMaxPlusMatrix, [[1, 3], [2, 1]], 3)
gap> PositionCanonical(S, x);
234
gap> EnumeratorCanonical(S)[234] = x;
true
##### 13.1-3 Enumerate
‣ Enumerate( S[, limit] ) ( operation )
Returns: A semigroup (the argument).
If S is a semigroup with representation IsEnumerableSemigroupRep (6.1-4) and limit is a positive integer, then this operation can be used to enumerate at least limit elements of S, or Size(S) elements if this is less than limit, using the Froidure-Pin Algorithm.
If the optional second argument limit is not given, then the semigroup is enumerated until all of its elements have been found.
For reasons of performance, S is enumerated in batches according to the option batch_size, which can be specified when S is created; see Section 6.3.
gap> S := FullTransformationMonoid(7);
<full transformation monoid of degree 7>
gap> Enumerate(S, 1000);
<full transformation monoid of degree 7>
gap> Display(S);
<partially enumerated semigroup with 8197 elements,
224 rules, max word length 11>
##### 13.1-4 IsFullyEnumerated
‣ IsFullyEnumerated( S ) ( operation )
Returns: true or false.
If S is a semigroup with representation IsEnumerableSemigroupRep (6.1-4), then this operation returns true if the Froidure-Pin Algorithm has been run to completion (i.e. all of the elements of S have been found) and false if S has not been fully enumerated.
gap> S := FullBooleanMatMonoid(4);
<monoid of 4x4 boolean matrices with 7 generators>
gap> Enumerate(S, 1000);
<monoid of 4x4 boolean matrices with 7 generators>
gap> IsFullyEnumerated(S);
false
gap> Size(S);
65536
gap> IsFullyEnumerated(S);
true
#### 13.2 Cayley graphs
##### 13.2-1 RightCayleyGraphSemigroup
‣ RightCayleyGraphSemigroup( S ) ( attribute )
‣ LeftCayleyGraphSemigroup( S ) ( attribute )
Returns: A list of lists of positive integers.
When the argument S is a semigroup in the representation IsEnumerableSemigroupRep (6.1-4), RightCayleyGraphSemigroup returns the right Cayley graphs of S, as a list graph where graph[i][j] is equal to PositionCanonical(S, AsListCanonical(S)[i] * GeneratorsOfSemigroup(S)[j]). The list returned by LeftCayleyGraphSemigroup is defined analogously.
gap> S := FullTransformationMonoid(2);
<full transformation monoid of degree 2>
gap> RightCayleyGraphSemigroup(S);
[ [ 1, 2, 3 ], [ 2, 1, 3 ], [ 3, 4, 3 ], [ 4, 3, 3 ] ]
gap> LeftCayleyGraphSemigroup(S);
[ [ 1, 2, 3 ], [ 2, 1, 4 ], [ 3, 3, 3 ], [ 4, 4, 4 ] ]
#### 13.3 Random elements of a semigroup
##### 13.3-1 Random
‣ Random( S ) ( method )
Returns: A random element.
This function returns a random element of the semigroup S. If the elements of S have been calculated, then one of these is chosen randomly. Otherwise, if the data structure for S is known, then a random element of a randomly chosen $$\mathscr{R}$$-class is returned. If the data structure for S has not been calculated, then a short product (at most 2 * Length(GeneratorsOfSemigroup(S))) of generators is returned.
#### 13.4 Properties of elements in a semigroup
##### 13.4-1 IndexPeriodOfSemigroupElement
‣ IndexPeriodOfSemigroupElement( x ) ( operation )
Returns: A list of two positive integers.
If x is a semigroup element, then IndexPeriodOfSemigroupElement(x) returns the pair [m, r], where m and r are the least positive integers such that x^(m + r) = x ^ m. The number m is known as the index of x, and the numberr is known as the period of x.
gap> x := Transformation([2, 6, 3, 5, 6, 1]);;
gap> IndexPeriodOfSemigroupElement(x);
[ 2, 3 ]
gap> m := IndexPeriodOfSemigroupElement(x)[1];;
gap> r := IndexPeriodOfSemigroupElement(x)[2];;
gap> x ^ (m + r) = x ^ m;
true
gap> x := PartialPerm([0, 2, 3, 0, 5]);
<identity partial perm on [ 2, 3, 5 ]>
gap> IsIdempotent(x);
true
gap> IndexPeriodOfSemigroupElement(x);
[ 1, 1 ]
##### 13.4-2 SmallestIdempotentPower
‣ SmallestIdempotentPower( x ) ( attribute )
Returns: A positive integer.
If x is a semigroup element, then SmallestIdempotentPower(x) returns the least positive integer n such that x^n is an idempotent. The smallest idempotent power of x is the least multiple of the period of x that is greater than or equal to the index of x; see IndexPeriodOfSemigroupElement (13.4-1).
gap> x := Transformation([4, 1, 4, 5, 1]);
Transformation( [ 4, 1, 4, 5, 1 ] )
gap> SmallestIdempotentPower(x);
3
gap> ForAll([1 .. 2], i -> not IsIdempotent(x ^ i));
true
gap> IsIdempotent(x ^ 3);
true
gap> x := Bipartition([[1, 2, -3, -4], [3, -5], [4, -1], [5, -2]]);
<block bijection: [ 1, 2, -3, -4 ], [ 3, -5 ], [ 4, -1 ], [ 5, -2 ]>
gap> SmallestIdempotentPower(x);
4
gap> ForAll([1 .. 3], i -> not IsIdempotent(x ^ i));
true
gap> x := PartialPerm([]);
<empty partial perm>
gap> SmallestIdempotentPower(x);
1
gap> IsIdempotent(x);
true
#### 13.5 Expressing semigroup elements as words in generators
It is possible to express an element of a semigroup as a word in the generators of that semigroup. This section describes how to accomplish this in Semigroups.
##### 13.5-1 EvaluateWord
‣ EvaluateWord( gens, w ) ( operation )
Returns: A semigroup element.
The argument gens should be a collection of generators of a semigroup and the argument w should be a list of positive integers less than or equal to the length of gens. This operation evaluates the word w in the generators gens. More precisely, EvaluateWord(gens, w) returns the equivalent of:
Product(List(w, i -> gens[i]));
see also Factorization (13.5-2).
for elements of a semigroup
When gens is a list of elements of a semigroup and w is a list of positive integers less than or equal to the length of gens, this operation returns the product gens[w[1]] * gens[w[2]] * .. . * gens[w[n]] when the length of w is n.
for elements of an inverse semigroup
When gens is a list of elements with a semigroup inverse and w is a list of non-zero integers whose absolute value does not exceed the length of gens, this operation returns the product gens[AbsInt(w[1])] ^ SignInt(w[1]) * .. . * gens[AbsInt(w[n])] ^ SignInt(w[n]) where n is the length of w.
Note that EvaluateWord(gens, []) returns One(gens) if gens belongs to the category IsMultiplicativeElementWithOne (Reference: IsMultiplicativeElementWithOne).
gap> gens := [
> Transformation([2, 4, 4, 6, 8, 8, 6, 6]),
> Transformation([2, 7, 4, 1, 4, 6, 5, 2]),
> Transformation([3, 6, 2, 4, 2, 2, 2, 8]),
> Transformation([4, 3, 6, 4, 2, 1, 2, 6]),
> Transformation([4, 5, 1, 3, 8, 5, 8, 2])];;
gap> S := Semigroup(gens);;
gap> x := Transformation([1, 4, 6, 1, 7, 2, 7, 6]);;
gap> word := Factorization(S, x);
[ 4, 2 ]
gap> EvaluateWord(gens, word);
Transformation( [ 1, 4, 6, 1, 7, 2, 7, 6 ] )
gap> S := SymmetricInverseMonoid(10);;
gap> x := PartialPerm([2, 6, 7, 0, 0, 9, 0, 1, 0, 5]);
[3,7][8,1,2,6,9][10,5]
gap> word := Factorization(S, x);
[ -2, -2, -2, -2, -3, -2, -2, -2, -2, -2, 5, 2, 5, 5, 2, 5, 2, 2, 2,
2, -3, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2 ]
gap> EvaluateWord(GeneratorsOfSemigroup(S), word);
[3,7][8,1,2,6,9][10,5]
##### 13.5-2 Factorization
‣ Factorization( S, x ) ( operation )
Returns: A word in the generators.
for semigroups
When S is a semigroup and x belongs to S, Factorization return a word in the generators of S that is equal to x. In this case, a word is a list of positive integers where an entry i corresponds to GeneratorsOfSemigroups(S)[i]. More specifically,
EvaluateWord(GeneratorsOfSemigroup(S), Factorization(S, x)) = x;
for inverse semigroups
When S is an inverse semigroup and x belongs to S, Factorization return a word in the generators of S that is equal to x. In this case, a word is a list of non-zero integers where an entry i corresponds to GeneratorsOfSemigroup(S)[i] and -i corresponds to GeneratorsOfSemigroup(S)[i] ^ -1. As in the previous case,
EvaluateWord(GeneratorsOfSemigroup(S), Factorization(S, x)) = x;
Note that Factorization does not always return a word of minimum length; see MinimalFactorization (13.5-3).
See also EvaluateWord (13.5-1) and GeneratorsOfSemigroup (Reference: GeneratorsOfSemigroup).
gap> gens := [Transformation([2, 2, 9, 7, 4, 9, 5, 5, 4, 8]),
> Transformation([4, 10, 5, 6, 4, 1, 2, 7, 1, 2])];;
gap> S := Semigroup(gens);;
gap> x := Transformation([1, 10, 2, 10, 1, 2, 7, 10, 2, 7]);;
gap> word := Factorization(S, x);
[ 2, 2, 1, 2 ]
gap> EvaluateWord(gens, word);
Transformation( [ 1, 10, 2, 10, 1, 2, 7, 10, 2, 7 ] )
gap> S := SymmetricInverseMonoid(8);
<symmetric inverse monoid of degree 8>
gap> x := PartialPerm([1, 2, 3, 4, 5, 8], [7, 1, 4, 3, 2, 6]);
[5,2,1,7][8,6](3,4)
gap> word := Factorization(S, x);
[ -2, -2, -2, -2, -2, -2, 2, 4, 4, 2, 3, 2, -3, -2, -2, 3, 2, -3, -2,
-2, 4, -3, -4, 2, 2, 3, -2, -3, 4, -3, -4, 2, 2, 3, -2, -3, 2, 2,
3, -2, -3, 2, 2, 3, -2, -3, 4, -3, -4, 3, 2, -3, -2, -2, 3, 2, -3,
-2, -2, 4, 3, -4, 3, 2, -3, -2, -2, 3, 2, -3, -2, -2, 3, 2, 2, 3,
2, 2, 2, 2 ]
gap> EvaluateWord(GeneratorsOfSemigroup(S), word);
[5,2,1,7][8,6](3,4)
gap> S := DualSymmetricInverseMonoid(6);;
gap> x := S.1 * S.2 * S.3 * S.2 * S.1;
<block bijection: [ 1, 6, -4 ], [ 2, -2, -3 ], [ 3, -5 ], [ 4, -6 ],
[ 5, -1 ]>
gap> word := Factorization(S, x);
[ -2, -2, -2, -2, -2, 4, 2 ]
gap> EvaluateWord(GeneratorsOfSemigroup(S), word);
<block bijection: [ 1, 6, -4 ], [ 2, -2, -3 ], [ 3, -5 ], [ 4, -6 ],
[ 5, -1 ]>
##### 13.5-3 MinimalFactorization
‣ MinimalFactorization( S, x ) ( operation )
Returns: A minimal word in the generators.
This operation returns a minimal length word in the generators of the semigroup S that equals the element x. In this case, a word is a list of positive integers where an entry i corresponds to GeneratorsOfSemigroups(S)[i]. More specifically,
EvaluateWord(GeneratorsOfSemigroup(S), MinimalFactorization(S, x)) = x;
MinimalFactorization involves exhaustively enumerating S until the element x is found, and so MinimalFactorization may be less efficient than Factorization (13.5-2) for some semigroups.
Unlike Factorization (13.5-2) this operation does not distinguish between semigroups and inverse semigroups. See also EvaluateWord (13.5-1) and GeneratorsOfSemigroup (Reference: GeneratorsOfSemigroup).
gap> S := Semigroup(Transformation([2, 2, 9, 7, 4, 9, 5, 5, 4, 8]),
> Transformation([4, 10, 5, 6, 4, 1, 2, 7, 1, 2]));
<transformation semigroup of degree 10 with 2 generators>
gap> x := Transformation([8, 8, 2, 2, 9, 2, 8, 8, 9, 9]);
Transformation( [ 8, 8, 2, 2, 9, 2, 8, 8, 9, 9 ] )
gap> Factorization(S, x);
[ 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1 ]
gap> MinimalFactorization(S, x);
[ 1, 2, 1, 1, 1, 1, 2, 2, 1 ]
#### 13.6 Generating sets
##### 13.6-1 Generators
‣ Generators( S ) ( attribute )
Returns: A list of generators.
Generators returns a generating set that can be used to define the semigroup S. The generators of a monoid or inverse semigroup S, say, can be defined in several ways, for example, including or excluding the identity element, including or not the inverses of the generators. Generators uses the definition that returns the least number of generators. If no generating set for S is known, then GeneratorsOfSemigroup is used by default.
for a group
Generators(S) is a synonym for GeneratorsOfGroup (Reference: GeneratorsOfGroup).
for an ideal of semigroup
Generators(S) is a synonym for GeneratorsOfSemigroupIdeal (7.2-1).
for a semigroup
Generators(S) is a synonym for GeneratorsOfSemigroup (Reference: GeneratorsOfSemigroup).
for a monoid
Generators(S) is a synonym for GeneratorsOfMonoid (Reference: GeneratorsOfMonoid).
for an inverse semigroup
Generators(S) is a synonym for GeneratorsOfInverseSemigroup (Reference: GeneratorsOfInverseSemigroup).
for an inverse monoid
Generators(S) is a synonym for GeneratorsOfInverseMonoid (Reference: GeneratorsOfInverseMonoid).
gap> M := Monoid([
> Transformation([1, 4, 6, 2, 5, 3, 7, 8, 9, 9]),
> Transformation([6, 3, 2, 7, 5, 1, 8, 8, 9, 9])]);;
gap> GeneratorsOfSemigroup(M);
[ IdentityTransformation,
Transformation( [ 1, 4, 6, 2, 5, 3, 7, 8, 9, 9 ] ),
Transformation( [ 6, 3, 2, 7, 5, 1, 8, 8, 9, 9 ] ) ]
gap> GeneratorsOfMonoid(M);
[ Transformation( [ 1, 4, 6, 2, 5, 3, 7, 8, 9, 9 ] ),
Transformation( [ 6, 3, 2, 7, 5, 1, 8, 8, 9, 9 ] ) ]
gap> Generators(M);
[ Transformation( [ 1, 4, 6, 2, 5, 3, 7, 8, 9, 9 ] ),
Transformation( [ 6, 3, 2, 7, 5, 1, 8, 8, 9, 9 ] ) ]
gap> S := Semigroup(Generators(M));;
gap> Generators(S);
[ Transformation( [ 1, 4, 6, 2, 5, 3, 7, 8, 9, 9 ] ),
Transformation( [ 6, 3, 2, 7, 5, 1, 8, 8, 9, 9 ] ) ]
gap> GeneratorsOfSemigroup(S);
[ Transformation( [ 1, 4, 6, 2, 5, 3, 7, 8, 9, 9 ] ),
Transformation( [ 6, 3, 2, 7, 5, 1, 8, 8, 9, 9 ] ) ]
##### 13.6-2 SmallGeneratingSet
‣ SmallGeneratingSet( coll ) ( attribute )
‣ SmallSemigroupGeneratingSet( coll ) ( attribute )
‣ SmallMonoidGeneratingSet( coll ) ( attribute )
‣ SmallInverseSemigroupGeneratingSet( coll ) ( attribute )
‣ SmallInverseMonoidGeneratingSet( coll ) ( attribute )
Returns: A small generating set for a semigroup.
The attributes SmallXGeneratingSet return a relatively small generating subset of the collection of elements coll, which can also be a semigroup. The returned value of SmallXGeneratingSet, where applicable, has the property that
X(SmallXGeneratingSet(coll)) = X(coll);
where X is any of Semigroup (Reference: Semigroup), Monoid (Reference: Monoid), InverseSemigroup (Reference: InverseSemigroup), or InverseMonoid (Reference: InverseMonoid).
If the number of generators for S is already relatively small, then these functions will often return the original generating set. These functions may return different results in different GAP sessions.
SmallGeneratingSet returns the smallest of the returned values of SmallXGeneratingSet which is applicable to coll; see Generators (13.6-1).
As neither irredundancy, nor minimal length are proven, these functions usually return an answer much more quickly than IrredundantGeneratingSubset (13.6-3). These functions can be used whenever a small generating set is desired which does not necessarily needs to be minimal.
gap> S := Semigroup([
> Transformation([1, 2, 3, 2, 4]),
> Transformation([1, 5, 4, 3, 2]),
> Transformation([2, 1, 4, 2, 2]),
> Transformation([2, 4, 4, 2, 1]),
> Transformation([3, 1, 4, 3, 2]),
> Transformation([3, 2, 3, 4, 1]),
> Transformation([4, 4, 3, 3, 5]),
> Transformation([5, 1, 5, 5, 3]),
> Transformation([5, 4, 3, 5, 2]),
> Transformation([5, 5, 4, 5, 5])]);;
gap> SmallGeneratingSet(S);
[ Transformation( [ 1, 5, 4, 3, 2 ] ), Transformation( [ 3, 2, 3, 4, 1 ] ),
Transformation( [ 5, 4, 3, 5, 2 ] ), Transformation( [ 1, 2, 3, 2, 4 ] ),
Transformation( [ 4, 4, 3, 3, 5 ] ) ]
gap> S := RandomInverseMonoid(IsPartialPermMonoid, 10000, 10);;
gap> SmallGeneratingSet(S);
[ [ 1 .. 10 ] -> [ 3, 2, 4, 5, 6, 1, 7, 10, 9, 8 ],
[ 1 .. 10 ] -> [ 5, 10, 8, 9, 3, 2, 4, 7, 6, 1 ],
[ 1, 3, 4, 5, 6, 7, 8, 9, 10 ] -> [ 1, 6, 4, 8, 2, 10, 7, 3, 9 ] ]
gap> M := MathieuGroup(24);;
gap> mat := List([1 .. 1000], x -> Random(M));;
gap> Append(mat, [1 .. 1000] * 0);
gap> mat := List([1 .. 138], x -> List([1 .. 57], x -> Random(mat)));;
gap> R := ReesZeroMatrixSemigroup(M, mat);;
gap> U := Semigroup(List([1 .. 200], x -> Random(R)));
<subsemigroup of 57x138 Rees 0-matrix semigroup with 100 generators>
gap> Length(SmallGeneratingSet(U));
84
gap> S := RandomSemigroup(IsBipartitionSemigroup, 100, 4);
<bipartition semigroup of degree 4 with 96 generators>
gap> Length(SmallGeneratingSet(S));
13
##### 13.6-3 IrredundantGeneratingSubset
‣ IrredundantGeneratingSubset( coll ) ( operation )
Returns: A list of irredundant generators.
If coll is a collection of elements of a semigroup, then this function returns a subset U of coll such that no element of U is generated by the other elements of U.
gap> S := Semigroup([
> Transformation([5, 1, 4, 6, 2, 3]),
> Transformation([1, 2, 3, 4, 5, 6]),
> Transformation([4, 6, 3, 4, 2, 5]),
> Transformation([5, 4, 6, 3, 1, 3]),
> Transformation([2, 2, 6, 5, 4, 3]),
> Transformation([3, 5, 5, 1, 2, 4]),
> Transformation([6, 5, 1, 3, 3, 4]),
> Transformation([1, 3, 4, 3, 2, 1])]);;
gap> IrredundantGeneratingSubset(S);
[ Transformation( [ 1, 3, 4, 3, 2, 1 ] ),
Transformation( [ 2, 2, 6, 5, 4, 3 ] ),
Transformation( [ 3, 5, 5, 1, 2, 4 ] ),
Transformation( [ 5, 1, 4, 6, 2, 3 ] ),
Transformation( [ 5, 4, 6, 3, 1, 3 ] ),
Transformation( [ 6, 5, 1, 3, 3, 4 ] ) ]
gap> S := RandomInverseMonoid(IsPartialPermMonoid, 1000, 10);
<inverse partial perm monoid of degree 10 with 1000 generators>
gap> SmallGeneratingSet(S);
[ [ 1 .. 10 ] -> [ 6, 5, 1, 9, 8, 3, 10, 4, 7, 2 ],
[ 1 .. 10 ] -> [ 1, 4, 6, 2, 8, 5, 7, 10, 3, 9 ],
[ 1, 2, 3, 4, 6, 7, 8, 9 ] -> [ 7, 5, 10, 1, 8, 4, 9, 6 ]
[ 1 .. 9 ] -> [ 4, 3, 5, 7, 10, 9, 1, 6, 8 ] ]
gap> IrredundantGeneratingSubset(last);
[ [ 1 .. 9 ] -> [ 4, 3, 5, 7, 10, 9, 1, 6, 8 ],
[ 1 .. 10 ] -> [ 1, 4, 6, 2, 8, 5, 7, 10, 3, 9 ],
[ 1 .. 10 ] -> [ 6, 5, 1, 9, 8, 3, 10, 4, 7, 2 ] ]
gap> S := RandomSemigroup(IsBipartitionSemigroup, 1000, 4);
<bipartition semigroup of degree 4 with 749 generators>
gap> SmallGeneratingSet(S);
[ <bipartition: [ 1, -3 ], [ 2, -2 ], [ 3, -1 ], [ 4, -4 ]>,
<bipartition: [ 1, 3, -2 ], [ 2, -1, -3 ], [ 4, -4 ]>,
<bipartition: [ 1, -4 ], [ 2, 4, -1, -3 ], [ 3, -2 ]>,
<bipartition: [ 1, -1, -3 ], [ 2, -4 ], [ 3, 4, -2 ]>,
<bipartition: [ 1, -2, -4 ], [ 2 ], [ 3, -3 ], [ 4, -1 ]>,
<bipartition: [ 1, -2 ], [ 2, -1, -3 ], [ 3, 4, -4 ]>,
<bipartition: [ 1, 3, -1 ], [ 2, -3 ], [ 4, -2, -4 ]>,
<bipartition: [ 1, -1 ], [ 2, 4, -4 ], [ 3, -2, -3 ]>,
<bipartition: [ 1, 3, -1 ], [ 2, -2 ], [ 4, -3, -4 ]>,
<bipartition: [ 1, 2, -2 ], [ 3, -1, -4 ], [ 4, -3 ]>,
<bipartition: [ 1, -2, -3 ], [ 2, -4 ], [ 3 ], [ 4, -1 ]>,
<bipartition: [ 1, -1 ], [ 2, 4, -3 ], [ 3, -2 ], [ -4 ]>,
<bipartition: [ 1, -3 ], [ 2, -1 ], [ 3, 4, -4 ], [ -2 ]>,
<bipartition: [ 1, 2, -4 ], [ 3, -1 ], [ 4, -2 ], [ -3 ]>,
<bipartition: [ 1, -3 ], [ 2, -4 ], [ 3, -1, -2 ], [ 4 ]> ]
gap> IrredundantGeneratingSubset(last);
[ <bipartition: [ 1, 2, -4 ], [ 3, -1 ], [ 4, -2 ], [ -3 ]>,
<bipartition: [ 1, 3, -1 ], [ 2, -2 ], [ 4, -3, -4 ]>,
<bipartition: [ 1, 3, -2 ], [ 2, -1, -3 ], [ 4, -4 ]>,
<bipartition: [ 1, -1 ], [ 2, 4, -3 ], [ 3, -2 ], [ -4 ]>,
<bipartition: [ 1, -3 ], [ 2, -1 ], [ 3, 4, -4 ], [ -2 ]>,
<bipartition: [ 1, -3 ], [ 2, -2 ], [ 3, -1 ], [ 4, -4 ]>,
<bipartition: [ 1, -3 ], [ 2, -4 ], [ 3, -1, -2 ], [ 4 ]>,
<bipartition: [ 1, -2, -3 ], [ 2, -4 ], [ 3 ], [ 4, -1 ]>,
<bipartition: [ 1, -2, -4 ], [ 2 ], [ 3, -3 ], [ 4, -1 ]> ]
##### 13.6-4 MinimalSemigroupGeneratingSet
‣ MinimalSemigroupGeneratingSet( S ) ( attribute )
‣ MinimalMonoidGeneratingSet( S ) ( attribute )
‣ MinimalInverseSemigroupGeneratingSet( S ) ( attribute )
‣ MinimalInverseMonoidGeneratingSet( S ) ( attribute )
Returns: A minimal generating set for a semigroup.
currently, no methods are installed to compute these attributes.
The attributes MinimalXGeneratingSet return a minimal generating set for the semigroup S, with respect to length. The returned value of MinimalXGeneratingSet, where applicable, is a minimal-length list of elements of S with the property that
X(MinimalXGeneratingSet(S)) = S;
where X is any of Semigroup (Reference: Semigroup), Monoid (Reference: Monoid), InverseSemigroup (Reference: InverseSemigroup), or InverseMonoid (Reference: InverseMonoid).
For certain types of semigroup, for example monogenic semigroups, a MinimalXGeneratingSet may be known a priori, or may be deduced as a by-product of other functions. However, since there are no methods installed to compute these attributes directly, for most semigroups it is not currently possible to find a MinimalXGeneratingSet with the Semigroups package.
See also SmallGeneratingSet (13.6-2) and IrredundantGeneratingSubset (13.6-3).
gap> S := MonogenicSemigroup(3, 6);;
gap> MinimalSemigroupGeneratingSet(S);
[ Transformation( [ 2, 3, 4, 5, 6, 1, 6, 7, 8 ] ) ]
gap> S := Semigroup([
> PartialPerm([1, 2, 3, 4, 5], [1, 2, 3, 4, 5]),
> PartialPerm([1, 2, 3, 4], [5, 2, 4, 1]),
> PartialPerm([1, 2, 4, 5], [4, 2, 3, 1])]);
<partial perm monoid of rank 5 with 2 generators>
gap> IsMonogenicInverseMonoid(S);
true
gap> MinimalInverseMonoidGeneratingSet(S);
[ [3,4,1,5](2) ]
‣ GeneratorsSmallest( S ) ( attribute )
Returns: A set of elements.
For a semigroup S, GeneratorsSmallest returns the lexicographically least set of elements X such that X generates S as a semigroup, and such that X is lexicographically ordered and has the property that each X[i] is not generated by X[1], X[2], ..., X[i-1].
It can be difficult to find the set of generators X, and it might contain a substantial proportion of the elements of S.
Two semigroups have the same set of elements if and only if their smallest generating sets are equal. However, due to the complexity of determining the GeneratorsSmallest, this is not the method used by the Semigroups package when comparing semigroups.
gap> S := Monoid([
> Transformation([1, 3, 4, 1]),
> Transformation([2, 4, 1, 2]),
> Transformation([3, 1, 1, 3]),
> Transformation([3, 3, 4, 1])]);
<transformation monoid of degree 4 with 4 generators>
[ Transformation( [ 1, 1, 1, 1 ] ), Transformation( [ 1, 1, 1, 2 ] ),
Transformation( [ 1, 1, 1, 3 ] ), Transformation( [ 1, 1, 1 ] ),
Transformation( [ 1, 1, 2, 1 ] ), Transformation( [ 1, 1, 2, 2 ] ),
Transformation( [ 1, 1, 3, 1 ] ), Transformation( [ 1, 1, 3, 3 ] ),
Transformation( [ 1, 1 ] ), Transformation( [ 1, 1, 4, 1 ] ),
Transformation( [ 1, 2, 1, 1 ] ), Transformation( [ 1, 2, 2, 1 ] ),
IdentityTransformation, Transformation( [ 1, 3, 1, 1 ] ),
Transformation( [ 1, 3, 4, 1 ] ), Transformation( [ 2, 1, 1, 2 ] ),
Transformation( [ 2, 2, 2 ] ), Transformation( [ 2, 4, 1, 2 ] ),
Transformation( [ 3, 3, 3 ] ), Transformation( [ 3, 3, 4, 1 ] ) ]
gap> T := Semigroup(Bipartition([[1, 2, 3], [4, -1], [-2], [-3], [-4]]),
> Bipartition([[1, -3, -4], [2, 3, 4, -2], [-1]]),
> Bipartition([[1, 2, 3, 4, -2], [-1, -4], [-3]]),
> Bipartition([[1, 2, 3, 4], [-1], [-2], [-3, -4]]),
> Bipartition([[1, 2, -1, -2], [3, 4, -3], [-4]]));
<bipartition semigroup of degree 4 with 5 generators>
[ <bipartition: [ 1, 2, 3, 4, -1, -2, -3 ], [ -4 ]>,
<bipartition: [ 1, 2, 3, 4, -1, -2 ], [ -3 ], [ -4 ]>,
<bipartition: [ 1, 2, 3, 4, -1 ], [ -2 ], [ -3 ], [ -4 ]>,
<bipartition: [ 1, 2, 3, 4, -2, -3, -4 ], [ -1 ]>,
<bipartition: [ 1, 2, 3, 4, -2 ], [ -1, -4 ], [ -3 ]>,
<bipartition: [ 1, 2, 3, 4, -2 ], [ -1 ], [ -3, -4 ]>,
<bipartition: [ 1, 2, 3, 4, -3 ], [ -1, -2 ], [ -4 ]>,
<bipartition: [ 1, 2, 3, 4 ], [ -1, -2, -3 ], [ -4 ]>,
<bipartition: [ 1, 2, 3, 4, -3, -4 ], [ -1 ], [ -2 ]>,
<bipartition: [ 1, 2, 3 ], [ 4, -1, -2, -3 ], [ -4 ]>,
<bipartition: [ 1, 2, -1, -2 ], [ 3, 4, -3 ], [ -4 ]>,
<bipartition: [ 1, -3 ], [ 2, 3, 4, -1, -2 ], [ -4 ]>,
<bipartition: [ 1, -3, -4 ], [ 2, 3, 4, -2 ], [ -1 ]> ]
#### 13.7 Minimal ideals and multiplicative zeros
In this section we describe the attributes of a semigroup that can be found using the Semigroups package.
##### 13.7-1 MinimalIdeal
‣ MinimalIdeal( S ) ( attribute )
Returns: The minimal ideal of a semigroup.
The minimal ideal of a semigroup is the least ideal with respect to containment.
It is significantly easier to find the minimal $$\mathscr{D}$$-class of a semigroup, than to find its $$\mathscr{D}$$-classes.
See also RepresentativeOfMinimalIdeal (13.7-2), PartialOrderOfDClasses (12.1-10), IsGreensLessThanOrEqual (Reference: IsGreensLessThanOrEqual), and MinimalDClass (12.1-6).
gap> S := Semigroup(
> Transformation([3, 4, 1, 3, 6, 3, 4, 6, 10, 1]),
> Transformation([8, 2, 3, 8, 4, 1, 3, 4, 9, 7]));;
gap> MinimalIdeal(S);
<simple transformation semigroup ideal of degree 10 with 1 generator>
gap> Elements(MinimalIdeal(S));
[ Transformation( [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ] ),
Transformation( [ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 ] ),
Transformation( [ 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 ] ),
Transformation( [ 6, 6, 6, 6, 6, 6, 6, 6, 6, 6 ] ),
Transformation( [ 8, 8, 8, 8, 8, 8, 8, 8, 8, 8 ] ) ]
gap> x := Transformation([8, 8, 8, 8, 8, 8, 8, 8, 8, 8]);;
gap> D := DClass(S, x);;
gap> ForAll(GreensDClasses(S), x -> IsGreensLessThanOrEqual(D, x));
true
gap> MinimalIdeal(POI(10));
<partial perm group of rank 10>
gap> MinimalIdeal(BrauerMonoid(6));
<simple bipartition *-semigroup ideal of degree 6 with 1 generator>
##### 13.7-2 RepresentativeOfMinimalIdeal
‣ RepresentativeOfMinimalIdeal( S ) ( attribute )
‣ RepresentativeOfMinimalDClass( S ) ( attribute )
Returns: An element of the minimal ideal of a semigroup.
The minimal ideal of a semigroup is the least ideal with respect to containment.
This method returns a representative element of the minimal ideal of S without having to create the minimal ideal itself. In general, beyond being a member of the minimal ideal, the returned element is not guaranteed to have any special properties. However, the element will coincide with the zero element of S if one exists.
This method works particularly well if S is a semigroup of transformations or partial permutations.
See also MinimalIdeal (13.7-1) and MinimalDClass (12.1-6).
gap> S := SymmetricInverseSemigroup(10);;
gap> RepresentativeOfMinimalIdeal(S);
<empty partial perm>
gap> B := Semigroup([
> Bipartition([[1, 2], [3, 6, -2], [4, 5, -3, -4], [-1, -6], [-5]]),
> Bipartition([[1, -1], [2], [3], [4, -3], [5, 6, -5, -6],
> [-2, -4]])]);;
gap> RepresentativeOfMinimalIdeal(B);
<bipartition: [ 1, 2 ], [ 3, 6 ], [ 4, 5 ], [ -1, -5, -6 ],
[ -2, -4 ], [ -3 ]>
gap> S := Semigroup(Transformation([5, 1, 6, 2, 2, 4]),
> Transformation([3, 5, 5, 1, 6, 2]));;
gap> RepresentativeOfMinimalDClass(S);
Transformation( [ 1, 2, 2, 5, 5, 1 ] )
gap> MinimalDClass(S);
<Green's D-class: Transformation( [ 1, 2, 2, 5, 5, 1 ] )>
##### 13.7-3 MultiplicativeZero
‣ MultiplicativeZero( S ) ( attribute )
Returns: The zero element of a semigroup.
MultiplicativeZero returns the zero element of the semigroup S if it exists and fail if it does not. See also MultiplicativeZero (Reference: MultiplicativeZero).
gap> S := Semigroup(Transformation([1, 4, 2, 6, 6, 5, 2]),
> Transformation([1, 6, 3, 6, 2, 1, 6]));;
gap> MultiplicativeZero(S);
Transformation( [ 1, 1, 1, 1, 1, 1, 1 ] )
gap> S := Semigroup(Transformation([2, 8, 3, 7, 1, 5, 2, 6]),
> Transformation([3, 5, 7, 2, 5, 6, 3, 8]),
> Transformation([6, 7, 4, 1, 4, 1, 6, 2]),
> Transformation([8, 8, 5, 1, 7, 5, 2, 8]));;
gap> MultiplicativeZero(S);
fail
gap> S := InverseSemigroup(
> PartialPerm([1, 3, 4], [5, 3, 1]),
> PartialPerm([1, 2, 3, 4], [4, 3, 1, 2]),
> PartialPerm([1, 3, 4, 5], [2, 4, 5, 3]));;
gap> MultiplicativeZero(S);
<empty partial perm>
gap> S := PartitionMonoid(6);
<regular bipartition *-monoid of size 4213597, degree 6 with 4
generators>
gap> MultiplicativeZero(S);
fail
gap> S := DualSymmetricInverseMonoid(6);
<inverse block bijection monoid of degree 6 with 3 generators>
gap> MultiplicativeZero(S);
<block bijection: [ 1, 2, 3, 4, 5, 6, -1, -2, -3, -4, -5, -6 ]>
‣ UnderlyingSemigroupOfSemigroupWithAdjoinedZero( S ) ( attribute )
Returns: A semigroup, or fail.
If S is a semigroup for which the property IsSemigroupWithAdjoinedZero (14.1-19) is true, (i.e. S has a MultiplicativeZero (13.7-3) and the set S ∖ { 0 } is a subsemigroup of S), then this method returns the semigroup S ∖ { 0 }.
Otherwise, if S is a semigroup for which the property IsSemigroupWithAdjoinedZero (14.1-19) is false, then this method returns fail.
gap> S := Semigroup([
> Transformation([2, 3, 4, 5, 1, 6]),
> Transformation([2, 1, 3, 4, 5, 6]),
> Transformation([6, 6, 6, 6, 6, 6])]);
<transformation semigroup of degree 6 with 3 generators>
gap> MultiplicativeZero(S);
Transformation( [ 6, 6, 6, 6, 6, 6 ] )
<transformation semigroup of degree 5 with 2 generators>
gap> IsGroupAsSemigroup(G);
true
gap> IsZeroGroup(S);
true
gap> S := SymmetricInverseMonoid(6);;
gap> MultiplicativeZero(S);
<empty partial perm>
fail
#### 13.8 Group of units and identity elements
##### 13.8-1 GroupOfUnits
‣ GroupOfUnits( S ) ( attribute )
Returns: The group of units of a semigroup or fail.
GroupOfUnits returns the group of units of the semigroup S as a subsemigroup of S if it exists and returns fail if it does not. Use IsomorphismPermGroup (6.5-5) if you require a permutation representation of the group of units.
If a semigroup S has an identity e, then the group of units of S is the set of those s in S such that there exists t in S where s*t=t*s=e. Equivalently, the group of units is the $$\mathscr{H}$$-class of the identity of S.
See also GreensHClassOfElement (Reference: GreensHClassOfElement), IsMonoidAsSemigroup (14.1-12), and MultiplicativeNeutralElement (Reference: MultiplicativeNeutralElement).
gap> S := Semigroup(
> Transformation([1, 2, 5, 4, 3, 8, 7, 6]),
> Transformation([1, 6, 3, 4, 7, 2, 5, 8]),
> Transformation([2, 1, 6, 7, 8, 3, 4, 5]),
> Transformation([3, 2, 3, 6, 1, 6, 1, 2]),
> Transformation([5, 2, 3, 6, 3, 4, 7, 4]));;
gap> Size(S);
5304
gap> StructureDescription(GroupOfUnits(S));
"C2 x S4"
gap> S := InverseSemigroup(
> PartialPerm([1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
> [2, 4, 5, 3, 6, 7, 10, 9, 8, 1]),
> PartialPerm([1, 2, 3, 4, 5, 6, 7, 8, 10],
> [8, 2, 3, 1, 4, 5, 10, 6, 9]));;
gap> StructureDescription(GroupOfUnits(S));
"C8"
gap> S := InverseSemigroup(
> PartialPerm([1, 3, 4], [4, 3, 5]),
> PartialPerm([1, 2, 3, 5], [3, 1, 5, 2]));;
gap> GroupOfUnits(S);
fail
gap> S := Semigroup(
> Bipartition([[1, 2, 3, -1, -3], [-2]]),
> Bipartition([[1, -1], [2, 3, -2, -3]]),
> Bipartition([[1, -2], [2, -3], [3, -1]]),
> Bipartition([[1], [2, 3, -2], [-1, -3]]));;
gap> StructureDescription(GroupOfUnits(S));
"C3"
#### 13.9 Idempotents
##### 13.9-1 Idempotents
‣ Idempotents( obj[, n] ) ( attribute )
Returns: A list of idempotents.
The argument obj should be a semigroup, $$\mathscr{D}$$-class, $$\mathscr{H}$$-class, $$\mathscr{L}$$-class, or $$\mathscr{R}$$-class.
If the optional second argument n is present and obj is a semigroup, then a list of the idempotents in obj of rank n is returned. If you are only interested in the idempotents of a given rank, then the second version of the function will probably be faster. However, if the optional second argument is present, then nothing is stored in obj and so every time the function is called the computation must be repeated.
This functions produce essentially the same output as the GAP library function with the same name; see Idempotents (Reference: Idempotents). The main difference is that this function can be applied to a wider class of objects as described above.
See also IsRegularDClass (Reference: IsRegularDClass), IsRegularGreensClass (12.3-2) IsGroupHClass (Reference: IsGroupHClass), NrIdempotents (13.9-2), and GroupHClass (12.4-1).
gap> S := Semigroup(Transformation([2, 3, 4, 1]),
> Transformation([3, 3, 1, 1]));;
gap> Idempotents(S, 1);
[ ]
gap> AsSet(Idempotents(S, 2));
[ Transformation( [ 1, 1, 3, 3 ] ), Transformation( [ 1, 3, 3, 1 ] ),
Transformation( [ 2, 2, 4, 4 ] ), Transformation( [ 4, 2, 2, 4 ] ) ]
gap> AsSet(Idempotents(S));
[ Transformation( [ 1, 1, 3, 3 ] ), IdentityTransformation,
Transformation( [ 1, 3, 3, 1 ] ), Transformation( [ 2, 2, 4, 4 ] ),
Transformation( [ 4, 2, 2, 4 ] ) ]
gap> x := Transformation([2, 2, 4, 4]);;
gap> R := GreensRClassOfElement(S, x);;
gap> Idempotents(R);
[ Transformation( [ 1, 1, 3, 3 ] ), Transformation( [ 2, 2, 4, 4 ] ) ]
gap> x := Transformation([4, 2, 2, 4]);;
gap> L := GreensLClassOfElement(S, x);;
gap> AsSet(Idempotents(L));
[ Transformation( [ 2, 2, 4, 4 ] ), Transformation( [ 4, 2, 2, 4 ] ) ]
gap> D := DClassOfLClass(L);;
gap> AsSet(Idempotents(D));
[ Transformation( [ 1, 1, 3, 3 ] ), Transformation( [ 1, 3, 3, 1 ] ),
Transformation( [ 2, 2, 4, 4 ] ), Transformation( [ 4, 2, 2, 4 ] ) ]
gap> L := GreensLClassOfElement(S, Transformation([3, 1, 1, 3]));;
gap> AsSet(Idempotents(L));
[ Transformation( [ 1, 1, 3, 3 ] ), Transformation( [ 1, 3, 3, 1 ] ) ]
gap> H := GroupHClass(D);
<Green's H-class: Transformation( [ 1, 1, 3, 3 ] )>
gap> Idempotents(H);
[ Transformation( [ 1, 1, 3, 3 ] ) ]
gap> S := InverseSemigroup(
> PartialPerm([10, 6, 3, 4, 9, 0, 1]),
> PartialPerm([6, 10, 7, 4, 8, 2, 9, 1]));;
gap> Idempotents(S, 1);
[ <identity partial perm on [ 4 ]> ]
gap> Idempotents(S, 0);
[ ]
##### 13.9-2 NrIdempotents
‣ NrIdempotents( obj ) ( attribute )
Returns: A positive integer.
This function returns the number of idempotents in obj where obj can be a semigroup, $$\mathscr{D}$$-, $$\mathscr{L}$$-, $$\mathscr{H}$$-, or $$\mathscr{R}$$-class. If the actual idempotents are not required, then it is more efficient to use NrIdempotents(obj) than Length(Idempotents(obj)) since the idempotents themselves are not created when NrIdempotents is called.
See also Idempotents (Reference: Idempotents) and Idempotents (13.9-1), IsRegularDClass (Reference: IsRegularDClass), IsRegularGreensClass (12.3-2) IsGroupHClass (Reference: IsGroupHClass), and GroupHClass (12.4-1).
gap> S := Semigroup(Transformation([2, 3, 4, 1]),
> Transformation([3, 3, 1, 1]));;
gap> NrIdempotents(S);
5
gap> f := Transformation([2, 2, 4, 4]);;
gap> R := GreensRClassOfElement(S, f);;
gap> NrIdempotents(R);
2
gap> f := Transformation([4, 2, 2, 4]);;
gap> L := GreensLClassOfElement(S, f);;
gap> NrIdempotents(L);
2
gap> D := DClassOfLClass(L);;
gap> NrIdempotents(D);
4
gap> L := GreensLClassOfElement(S, Transformation([3, 1, 1, 3]));;
gap> NrIdempotents(L);
2
gap> H := GroupHClass(D);;
gap> NrIdempotents(H);
1
gap> S := InverseSemigroup(
> PartialPerm([1, 2, 3, 5, 7, 9, 10],
> [6, 7, 2, 9, 1, 5, 3]),
> PartialPerm([1, 2, 3, 5, 6, 7, 9, 10],
> [8, 1, 9, 4, 10, 5, 6, 7]));;
gap> NrIdempotents(S);
236
gap> f := PartialPerm([2, 3, 7, 9, 10],
> [7, 2, 1, 5, 3]);;
gap> D := DClassNC(S, f);;
gap> NrIdempotents(D);
13
##### 13.9-3 IdempotentGeneratedSubsemigroup
‣ IdempotentGeneratedSubsemigroup( S ) ( attribute )
Returns: A semigroup.
IdempotentGeneratedSubsemigroup returns the subsemigroup of the semigroup S generated by the idempotents of S.
See also Idempotents (13.9-1) and SmallGeneratingSet (13.6-2).
gap> S := Semigroup(Transformation([1, 1]),
> Transformation([2, 1]),
> Transformation([1, 2, 2]),
> Transformation([1, 2, 3, 4, 5, 1]),
> Transformation([1, 2, 3, 4, 5, 5]),
> Transformation([1, 2, 3, 4, 6, 5]),
> Transformation([1, 2, 3, 5, 4]),
> Transformation([1, 2, 3, 7, 4, 5, 7]),
> Transformation([1, 2, 4, 8, 8, 3, 8, 7]),
> Transformation([1, 2, 8, 4, 5, 6, 7, 8]),
> Transformation([7, 7, 7, 4, 5, 6, 1]));;
gap> IdempotentGeneratedSubsemigroup(S) =
> Monoid(Transformation([1, 1]),
> Transformation([1, 2, 1]),
> Transformation([1, 2, 2]),
> Transformation([1, 2, 3, 1]),
> Transformation([1, 2, 3, 2]),
> Transformation([1, 2, 3, 4, 1]),
> Transformation([1, 2, 3, 4, 2]),
> Transformation([1, 2, 3, 4, 4]),
> Transformation([1, 2, 3, 4, 5, 1]),
> Transformation([1, 2, 3, 4, 5, 2]),
> Transformation([1, 2, 3, 4, 5, 5]),
> Transformation([1, 2, 3, 4, 5, 7, 7]),
> Transformation([1, 2, 3, 4, 7, 6, 7]),
> Transformation([1, 2, 3, 6, 5, 6]),
> Transformation([1, 2, 3, 7, 5, 6, 7]),
> Transformation([1, 2, 8, 4, 5, 6, 7, 8]),
> Transformation([2, 2]));
true
gap> S := SymmetricInverseSemigroup(5);
<symmetric inverse monoid of degree 5>
gap> IdempotentGeneratedSubsemigroup(S);
<inverse partial perm monoid of rank 5 with 5 generators>
gap> S := DualSymmetricInverseSemigroup(5);
<inverse block bijection monoid of degree 5 with 3 generators>
gap> IdempotentGeneratedSubsemigroup(S);
<inverse block bijection monoid of degree 5 with 10 generators>
gap> IsSemilattice(last);
true
#### 13.10 Maximal subsemigroups
The Semigroups package provides methods to calculate the maximal subsemigroups of a finite semigroup, subject to various conditions. A maximal subsemigroup of a semigroup is a proper subsemigroup that is contained in no other proper subsemigroup of the semigroup.
When computing the maximal subsemigroups of a regular Rees (0-)matrix semigroup over a group, additional functionality is available. As described in [GGR68], a maximal subsemigroup of a finite regular Rees (0-)matrix semigroup over a group is one of 6 possible types. Using the Semigroups package, it is possible to search for only those maximal subsemigroups of certain types.
A maximal subsemigroup of such a Rees (0-)matrix semigroup R over a group G is either:
1. {0};
2. formed by removing 0;
3. formed by removing a column (a non-zero $$\mathscr{L}$$-class);
4. formed by removing a row (a non-zero $$\mathscr{R}$$-class);
5. formed by removing a set of both rows and columns;
6. isomorphic to a Rees (0-)matrix semigroup of the same dimensions over a maximal subgroup of G (in particular, the maximal subsemigroup intersects every $$\mathscr{H}$$-class of R).
Note that if R is a Rees matrix semigroup then it has no maximal subsemigroups of types 1, 2, or 5. Only types 3, 4, and 6 are relevant to a Rees matrix semigroup.
##### 13.10-1 MaximalSubsemigroups
‣ MaximalSubsemigroups( S ) ( attribute )
‣ MaximalSubsemigroups( S, opts ) ( operation )
Returns: The maximal subsemigroups of S.
If S is a finite semigroup, then the attribute MaximalSubsemigroups returns a list of the non-empty maximal subsemigroups of S. The methods used by MaximalSubsemigroups are based on [GGR68], and are described in [DMW16].
It is computationally expensive to search for the maximal subsemigroups of a semigroup, and so computations involving MaximalSubsemigroups may be very lengthy. A substantial amount of information on the progress of MaximalSubsemigroups is provided through the info class InfoSemigroups (2.6-1), with increasingly detailed information given at levels 1, 2, and 3.
The behaviour of MaximalSubsemigroups can be altered via the second argument opts, which should be a record. The optional components of opts are:
gens (a boolean)
If opts.gens is false or unspecified, then the maximal subsemigroups themselves are returned and not just generating sets for these subsemigroups.
It can be more computationally expensive to return the generating sets for the maximal subsemigroups, than to return the maximal subsemigroups themselves.
contain (a list)
If opts.contain is duplicate-free list of elements of S, then MaximalSubsemigroups will search for the maximal subsemigroups of S which contain those elements.
D (a $$\mathscr{D}$$-class)
For a maximal subsemigroup M of a finite semigroup S, there exists a unique $$\mathscr{D}$$-class which contains the complement of M in S. In other words, the elements of S which M lacks are contained in a unique $$\mathscr{D}$$-class.
If opts.D is a $$\mathscr{D}$$-class of S, then MaximalSubsemigroups will search exclusively for those maximal subsemigroups of S whose complement is contained in opts.D.
types (a list)
This option is relevant only if S is a regular Rees (0-)matrix semigroup over a group.
As described at the start of this subsection, 13.10, a maximal subsemigroup of a regular Rees (0-)matrix semigroup over a group is one of 6 possible types.
If S is a regular Rees (0-)matrix semigroup over a group and opts.types is a subset of [1 .. 6], then MaximalSubsemigroups will search for those maximal subsemigroups of S of the types enumerated by opts.types.
The default value for this option is [1 .. 6] (i.e. no restriction).
gap> S := FullTransformationSemigroup(3);
<full transformation monoid of degree 3>
gap> MaximalSubsemigroups(S);
[ <transformation semigroup of degree 3 with 7 generators>,
<transformation semigroup of degree 3 with 7 generators>,
<transformation semigroup of degree 3 with 7 generators>,
<transformation semigroup of degree 3 with 7 generators>,
<transformation monoid of degree 3 with 5 generators> ]
gap> MaximalSubsemigroups(S,
> rec(gens := true, D := DClass(S, Transformation([2, 2, 3]))));
[ [ Transformation( [ 1, 1, 1 ] ), Transformation( [ 3, 3, 3 ] ),
Transformation( [ 2, 2, 2 ] ), IdentityTransformation,
Transformation( [ 2, 3, 1 ] ), Transformation( [ 2, 1 ] ) ] ]
gap> MaximalSubsemigroups(S,
> rec(contain := [Transformation([2, 3, 1])]));
[ <transformation semigroup of degree 3 with 7 generators>,
<transformation monoid of degree 3 with 5 generators> ]
gap> R := PrincipalFactor(
> DClass(FullTransformationMonoid(4), Transformation([2, 2])));
<Rees 0-matrix semigroup 6x4 over Group([ (2,3,4), (2,4) ])>
gap> MaximalSubsemigroups(R, rec(types := [5],
> contain := [RMSElement(R, 1, (), 1),
> RMSElement(R, 1, (2, 3), 2)]));
[ <subsemigroup of 6x4 Rees 0-matrix semigroup with 10 generators>,
<subsemigroup of 6x4 Rees 0-matrix semigroup with 10 generators>,
<subsemigroup of 6x4 Rees 0-matrix semigroup with 10 generators>,
<subsemigroup of 6x4 Rees 0-matrix semigroup with 10 generators> ]
##### 13.10-2 NrMaximalSubsemigroups
‣ NrMaximalSubsemigroups( S ) ( attribute )
Returns: The number of maximal subsemigroups of S.
If S is a finite semigroup, then NrMaximalSubsemigroups returns the number of non-empty maximal subsemigroups of S. The methods used by MaximalSubsemigroups are based on [GGR68], and are described in [DMW16].
It can be significantly faster to find the number of maximal subsemigroups of a semigroup than to find the maximal subsemigroups themselves.
Unless the maximal subsemigroups of S are already known, the command NrMaximalSubsemigroups(S) simply calls the command MaximalSubsemigroups(S, rec(number := true)).
For more information about searching for maximal subsemigroups of a finite semigroup in the Semigroups package, and for information about the options available to alter the search, see MaximalSubsemigroups (13.10-1). By supplying the additional option opts.number := true, the number of maximal subsemigroups will be returned rather than the subsemigroups themselves.
gap> S := FullTransformationSemigroup(3);
<full transformation monoid of degree 3>
gap> NrMaximalSubsemigroups(S);
5
gap> S := RectangularBand(3, 4);;
gap> NrMaximalSubsemigroups(S);
7
gap> R := PrincipalFactor(
> DClass(FullTransformationMonoid(4), Transformation([2, 2])));
<Rees 0-matrix semigroup 6x4 over Group([ (2,3,4), (2,4) ])>
gap> MaximalSubsemigroups(R, rec(number := true, types := [3, 4]));
10
##### 13.10-3 IsMaximalSubsemigroup
‣ IsMaximalSubsemigroup( S, T ) ( operation )
Returns: true or false.
If S and T are semigroups, then IsMaximalSubsemigroup returns true if and only if T is a maximal subsemigroup of S.
A maximal subsemigroup of S is a proper subsemigroup of S which is contained in no other proper subsemigroup of S.
gap> S := ZeroSemigroup(2);;
gap> IsMaximalSubsemigroup(S, Semigroup(MultiplicativeZero(S)));
true
gap> S := FullTransformationSemigroup(4);
<full transformation monoid of degree 4>
gap> T := Semigroup(Transformation([3, 4, 1, 2]),
> Transformation([1, 4, 2, 3]),
> Transformation([2, 1, 1, 3]));
<transformation semigroup of degree 4 with 3 generators>
gap> IsMaximalSubsemigroup(S, T);
true
gap> R := Semigroup(Transformation([3, 4, 1, 2]),
> Transformation([1, 4, 2, 2]),
> Transformation([2, 1, 1, 3]));
<transformation semigroup of degree 4 with 3 generators>
gap> IsMaximalSubsemigroup(S, R);
false
#### 13.11 The normalizer of a semigroup
##### 13.11-1 Normalizer
‣ Normalizer( G, S[, opts] ) ( operation )
‣ Normalizer( S[, opts] ) ( operation )
Returns: A permutation group.
In its first form, this function returns the normalizer of the transformation, partial perm, or bipartition semigroup S in the permutation group G. In its second form, the normalizer of S in the symmetric group on [1 .. n] where n is the degree of S is returned.
The of a transformation semigroup S in a permutation group G in the subgroup H of G consisting of those elements in g in G conjugating S to S, i.e. S ^ g = S.
Analogous definitions can be given for a partial perm and bipartition semigroups.
The method used by this operation is based on Section 3 in [ABMN10].
The optional final argument opts allows you to specify various options, which determine how the normalizer is calculated. The values of these options can dramatically change the time it takes for this operation to complete. In different situations, different options give the best performance.
The argument opts should be a record, and the available options are:
random
If this option has the value true, then the non-deterministic algorithms in genss are used in Normalizer. So, there is some chance that Normalizer will return an incorrect result in this case, but these methods can also be much faster than the deterministic algorithms which are used if this option is false.
The default value for this option is false.
lambdastab
If this option has the value true, then Normalizer initially finds the setwise stabilizer of the images or right blocks of the semigroup S. Sometimes this improves the performance of Normalizer and sometimes it does not. If this option in false, then this setwise stabilizer is not found.
The default value for this option is true.
rhostab
If this option has the value true, then Normalizer initially finds the setwise stabilizer of the kernels, domains, or left blocks of the semigroup S. Sometimes this improves the performance of Normalizer and sometimes it does not. If this option is false, the this setwise stabilizer is not found.
If S is an inverse semigroup, then this option is ignored.
The default value for this option is true.
gap> S := BrauerMonoid(8);
<regular bipartition *-monoid of degree 8 with 3 generators>
gap> StructureDescription(Normalizer(S));
"S8"
gap> S := InverseSemigroup(PartialPerm([2, 5, 6, 3, 8]),
> PartialPerm([3, 6, 0, 2, 0, 0, 5, 7]));;
gap> Normalizer(S, rec(random := true, lambdastab := false));
#I Have 33389 points.
#I Have 40136 points in new orbit.
Group(())
#### 13.12 Attributes of transformations and transformation semigroups
##### 13.12-1 ComponentRepsOfTransformationSemigroup
‣ ComponentRepsOfTransformationSemigroup( S ) ( attribute )
Returns: The representatives of components of a transformation semigroup.
This function returns the representatives of the components of the action of the transformation semigroup S on the set of positive integers not greater than the degree of S.
The representatives are the least set of points such that every point can be reached from some representative under the action of S.
gap> S := Semigroup(
> Transformation([11, 11, 9, 6, 4, 1, 4, 1, 6, 7, 12, 5]),
> Transformation([12, 10, 7, 10, 4, 1, 12, 9, 11, 9, 1, 12]));;
gap> ComponentRepsOfTransformationSemigroup(S);
[ 2, 3, 8 ]
##### 13.12-2 ComponentsOfTransformationSemigroup
‣ ComponentsOfTransformationSemigroup( S ) ( attribute )
Returns: The components of a transformation semigroup.
This function returns the components of the action of the transformation semigroup S on the set of positive integers not greater than the degree of S; the components of S partition this set.
gap> S := Semigroup(
> Transformation([11, 11, 9, 6, 4, 1, 4, 1, 6, 7, 12, 5]),
> Transformation([12, 10, 7, 10, 4, 1, 12, 9, 11, 9, 1, 12]));;
gap> ComponentsOfTransformationSemigroup(S);
[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ] ]
##### 13.12-3 CyclesOfTransformationSemigroup
‣ CyclesOfTransformationSemigroup( S ) ( attribute )
Returns: The cycles of a transformation semigroup.
This function returns the cycles, or strongly connected components, of the action of the transformation semigroup S on the set of positive integers not greater than the degree of S.
gap> S := Semigroup(
> Transformation([11, 11, 9, 6, 4, 1, 4, 1, 6, 7, 12, 5]),
> Transformation([12, 10, 7, 10, 4, 1, 12, 9, 11, 9, 1, 12]));;
gap> CyclesOfTransformationSemigroup(S);
[ [ 12 ], [ 1, 11 ], [ 1, 11, 12, 5, 4, 6 ],
[ 1, 11, 12, 5, 4, 10, 9, 6 ], [ 1, 12, 5, 4, 6 ],
[ 1, 12, 5, 4, 10, 9, 6 ], [ 1, 12, 5, 4, 10, 9, 11 ],
[ 11, 12, 5, 4, 10, 9 ], [ 12, 5, 4, 10, 7 ], [ 4, 10, 7 ] ]
##### 13.12-4 DigraphOfActionOnPairs
‣ DigraphOfActionOnPairs( S ) ( attribute )
‣ DigraphOfActionOnPairs( S, n ) ( attribute )
Returns: A digraph, or fail.
If S is a transformation semigroup and n is a non-negative integer such that S acts on the points [1 .. n], then DigraphOfActionOnPairs(S, n) returns a digraph representing the OnSets (Reference: OnSets) action of S on the pairs of points in [1 .. n].
If the optional argument n is not specified, then by default the degree of S will be chosen for n; see DegreeOfTransformationSemigroup (Reference: DegreeOfTransformationSemigroup). If the semigroup S does not act on [1 .. n], then DigraphOfActionOnPairs(S, n) returns fail.
The digraph returned by DigraphOfActionOnPairs has n + n choose 2 vertices: the vertices [1 .. n] correspond to the points in [1 .. n], and the remaining vertices correspond to the pairs of points in [1 .. n]. This correspondence is stored in the vertex labels of the digraph; see DigraphVertexLabels (Digraphs: DigraphVertexLabels).
The edges of the digraph are defined as follows. For each pair {i, j} in [1 .. n], and for each generator f in GeneratorsOfSemigroup(S), there is an edge from the vertex corresponding to {i, j} to the vertex corresponding to {i ^ f, j ^ f}. Since f is a transformation, the set {i ^ f, j ^ f} may correspond to a pair (in the case that i ^ f <> j ^ f), or to a point (in the case that i ^ f = j ^ f). The label of an edge is the position of the first transformation within GeneratorsOfSemigroup(S) that maps the pair corresponding to the source vertex to the pair/point corresponding to the range vertex. See GeneratorsOfSemigroup (Reference: GeneratorsOfSemigroup) and DigraphEdgeLabels (Digraphs: DigraphEdgeLabels) for further information.
Note that the digraph returned by DigraphOfActionOnPairs has no multiple edges; see IsMultiDigraph (Digraphs: IsMultiDigraph).
gap> x := Transformation([2, 4, 3, 4, 7, 1, 6]);;
gap> y := Transformation([3, 3, 2, 3, 5, 1, 5]);;
gap> S := Semigroup(x, y);
<transformation semigroup of degree 7 with 2 generators>
gap> gr := DigraphOfActionOnPairs(S);
<digraph with 28 vertices, 41 edges>
gap> OnSets([2, 5], x);
[ 4, 7 ]
gap> DigraphVertexLabel(gr, 16);
[ 2, 5 ]
gap> DigraphVertexLabel(gr, 25);
[ 4, 7 ]
gap> DigraphEdgeLabel(gr, 16, 25);
1
gap> gr := DigraphOfActionOnPairs(S, 4);
<digraph with 10 vertices, 11 edges>
gap> DigraphVertexLabels(gr);
[ 1, 2, 3, 4, [ 1, 2 ], [ 1, 3 ], [ 1, 4 ], [ 2, 3 ], [ 2, 4 ],
[ 3, 4 ] ]
gap> DigraphOfActionOnPairs(S, 5);
fail
##### 13.12-5 DigraphOfActionOnPoints
‣ DigraphOfActionOnPoints( S ) ( attribute )
‣ DigraphOfActionOnPoints( S, n ) ( attribute )
Returns: A digraph, or fail.
If S is a transformation semigroup and n is a non-negative integer such that S acts on the points [1 .. n], then DigraphOfActionOnPoints(S, n) returns a digraph representing the OnPoints (Reference: OnPoints) action of S on the set [1 .. n].
If the optional argument n is not specified, then by default the degree of S will be chosen for n; see DegreeOfTransformationSemigroup (Reference: DegreeOfTransformationSemigroup). If the semigroup S does not act on [1 .. n], then DigraphOfActionOnPairs(S, n) returns fail.
The digraph returned by DigraphOfActionOnPairs has n vertices, where the vertex i corresponds to the point i. For each point i in [1 .. n], and for each generator f in GeneratorsOfSemigroup(S), there is an edge from the vertex i to the vertex i ^ f. See GeneratorsOfSemigroup (Reference: GeneratorsOfSemigroup) for further information.
Note that the digraph returned by DigraphOfActionOnPoints has no multiple edges; see IsMultiDigraph (Digraphs: IsMultiDigraph).
gap> x := Transformation([2, 4, 2, 4, 7, 1, 6]);;
gap> y := Transformation([3, 3, 2, 3, 5, 1, 5]);;
gap> S := Semigroup(x, y);
<transformation semigroup of degree 7 with 2 generators>
gap> gr := DigraphOfActionOnPoints(S);
<digraph with 7 vertices, 12 edges>
gap> OnPoints(2, x);
4
gap> gr2 := DigraphOfActionOnPoints(S, 4);
<digraph with 4 vertices, 7 edges>
gap> gr2 = InducedSubdigraph(gr, [1 .. 4]);
true
gap> DigraphOfActionOnPoints(S, 5);
fail
##### 13.12-6 FixedPointsOfTransformationSemigroup
‣ FixedPointsOfTransformationSemigroup( S ) ( attribute )
Returns: A set of positive integers.
If S is a transformation semigroup, then FixedPointsOfTransformationSemigroup(S) returns the set of points i in [1 .. DegreeOfTransformationSemigroup(S)] such that i ^ f = i for all f in S.
gap> f := Transformation([1, 4, 2, 4, 3, 7, 7]);
Transformation( [ 1, 4, 2, 4, 3, 7, 7 ] )
gap> S := Semigroup(f);
<commutative transformation semigroup of degree 7 with 1 generator>
gap> FixedPointsOfTransformationSemigroup(S);
[ 1, 4, 7 ]
##### 13.12-7 IsTransitive
‣ IsTransitive( S[, X] ) ( property )
‣ IsTransitive( S[, n] ) ( property )
Returns: true or false.
A transformation semigroup S is transitive or strongly connected on the set X if for every i, j in X there is an element s in S such that i ^ s = j.
If the optional second argument is a positive integer n, then IsTransitive returns true if S is transitive on [1 .. n], and false if it is not.
If the optional second argument is not provided, then the degree of S is used by default; see DegreeOfTransformationSemigroup (Reference: DegreeOfTransformationSemigroup).
gap> S := Semigroup([
> Bipartition([
> [1, 2], [3, 6, -2], [4, 5, -3, -4], [-1, -6], [-5]]),
> Bipartition([
> [1, -4], [2, 3, 4, 5], [6], [-1, -6], [-2, -3], [-5]])]);
<bipartition semigroup of degree 6 with 2 generators>
gap> AsSemigroup(IsTransformationSemigroup, S);
<transformation semigroup of size 11, degree 12 with 2 generators>
gap> IsTransitive(last);
false
gap> IsTransitive(AsSemigroup(Group((1, 2, 3))));
true
##### 13.12-8 SmallestElementSemigroup
‣ SmallestElementSemigroup( S ) ( attribute )
‣ LargestElementSemigroup( S ) ( attribute )
Returns: A transformation.
These attributes return the smallest and largest element of the transformation semigroup S, respectively. Smallest means the first element in the sorted set of elements of S and largest means the last element in the set of elements.
It is not necessary to find the elements of the semigroup to determine the smallest or largest element, and this function has considerable better performance than the equivalent Elements(S)[1] and Elements(S)[Size(S)].
gap> S := Monoid(
> Transformation([1, 4, 11, 11, 7, 2, 6, 2, 5, 5, 10]),
> Transformation([2, 4, 4, 2, 10, 5, 11, 11, 11, 6, 7]));
<transformation monoid of degree 11 with 2 generators>
gap> SmallestElementSemigroup(S);
IdentityTransformation
gap> LargestElementSemigroup(S);
Transformation( [ 11, 11, 10, 10, 7, 6, 5, 6, 2, 2, 4 ] )
##### 13.12-9 CanonicalTransformation
‣ CanonicalTransformation( trans[, n] ) ( function )
Returns: A transformation.
If trans is a transformation, and n is a non-negative integer such that the restriction of trans to [1 .. n] defines a transformation of [1 .. n], then CanonicalTransformation returns a canonical representative of the transformation trans restricted to [1 .. n].
More specifically, let C(n) be a class of transformations of degree n such that AsDigraph returns isomorphic digraphs for every pair of element elements in C(n). Recall that for a transformation trans and integer n the function AsDigraph returns a digraph with n vertices and an edge with source x and range x^trans for every x in [1 .. n]. See AsDigraph (Digraphs: AsDigraph). Then CanonicalTransformation returns a canonical representative of the class C(n) that contains trans.
gap> x := Transformation([5, 1, 4, 1, 1]);
Transformation( [ 5, 1, 4, 1, 1 ] )
gap> y := Transformation([3, 3, 2, 3, 1]);
Transformation( [ 3, 3, 2, 3, 1 ] )
gap> CanonicalTransformation(x);
Transformation( [ 5, 5, 1, 5, 4 ] )
gap> CanonicalTransformation(y);
Transformation( [ 5, 5, 1, 5, 4 ] )
##### 13.12-10 IsConnectedTransformationSemigroup
‣ IsConnectedTransformationSemigroup( S ) ( property )
Returns: true or false.
A transformation semigroup S is connected if the digraph returned by the function DigraphOfActionOnPoints is connected. See IsConnectedDigraph (Digraphs: IsConnectedDigraph) and DigraphOfActionOnPoints (13.12-5). The function IsConnectedTransformationSemigroup returns true if the semigroup S is connected and false otherwise.
gap> S := Semigroup([
> Transformation([2, 4, 3, 4]),
> Transformation([3, 3, 2, 3, 3])]);
<transformation semigroup of degree 5 with 2 generators>
gap> IsConnectedTransformationSemigroup(S);
true
#### 13.13 Attributes of partial perm semigroups
##### 13.13-1 ComponentRepsOfPartialPermSemigroup
‣ ComponentRepsOfPartialPermSemigroup( S ) ( attribute )
Returns: The representatives of components of a partial perm semigroup.
This function returns the representatives of the components of the action of the partial perm semigroup S on the set of positive integers where it is defined.
The representatives are the least set of points such that every point can be reached from some representative under the action of S.
gap> S := Semigroup([
> PartialPerm([1, 2, 3, 5, 6, 7, 8, 11, 12, 16, 19],
> [9, 18, 20, 11, 5, 16, 8, 19, 14, 13, 1]),
> PartialPerm([1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 16, 18, 19, 20],
> [13, 1, 8, 5, 4, 14, 11, 12, 9, 20, 2, 18, 7, 3, 19])]);;
gap> ComponentRepsOfPartialPermSemigroup(S);
[ 1, 4, 6, 10, 15, 17 ]
##### 13.13-2 ComponentsOfPartialPermSemigroup
‣ ComponentsOfPartialPermSemigroup( S ) ( attribute )
Returns: The components of a partial perm semigroup.
This function returns the components of the action of the partial perm semigroup S on the set of positive integers where it is defined; the components of S partition this set.
gap> S := Semigroup([
> PartialPerm([1, 2, 3, 5, 6, 7, 8, 11, 12, 16, 19],
> [9, 18, 20, 11, 5, 16, 8, 19, 14, 13, 1]),
> PartialPerm([1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 16, 18, 19, 20],
> [13, 1, 8, 5, 4, 14, 11, 12, 9, 20, 2, 18, 7, 3, 19])]);;
gap> ComponentsOfPartialPermSemigroup(S);
[ [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 18, 19, 20 ],
[ 15 ], [ 17 ] ]
##### 13.13-3 CyclesOfPartialPerm
‣ CyclesOfPartialPerm( x ) ( attribute )
Returns: The cycles of a partial perm.
This function returns the cycles, or strongly connected components, of the action of the partial perm x on the set of positive integers where it is defined.
gap> x := PartialPerm([3, 1, 4, 2, 5, 0, 0, 6, 0, 7]);
[8,6][10,7](1,3,4,2)(5)
gap> CyclesOfPartialPerm(x);
[ [ 3, 4, 2, 1 ], [ 5 ] ]
##### 13.13-4 CyclesOfPartialPermSemigroup
‣ CyclesOfPartialPermSemigroup( S ) ( attribute )
Returns: The cycles of a partial perm semigroup.
This function returns the cycles, or strongly connected components, of the action of the partial perm semigroup S on the set of positive integers where it is defined.
gap> S := Semigroup([
> PartialPerm([1, 2, 3, 5, 6, 7, 8, 11, 12, 16, 19],
> [9, 18, 20, 11, 5, 16, 8, 19, 14, 13, 1]),
> PartialPerm([1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 16, 18, 19, 20],
> [13, 1, 8, 5, 4, 14, 11, 12, 9, 20, 2, 18, 7, 3, 19])]);;
gap> CyclesOfPartialPermSemigroup(S);
[ [ 1, 9, 12, 14, 2, 20, 19, 3, 8, 11 ] ]
The content in this chapter is based partly on work by Zachary Mesyan. A full description of the objects described can be found in [MM16].
#### 13.14 Attributes of Rees (0-)matrix semigroups
##### 13.14-1 RZMSDigraph
‣ RZMSDigraph( R ) ( attribute )
Returns: A digraph.
If R is an n by m Rees 0-matrix semigroup M^0[I, T, Λ; P] (so that I = {1,2,...,n} and Λ = {1,2,...,m}) then RZMSDigraph returns a symmetric bipartite digraph with n+m vertices. An index i ∈ I corresponds to the vertex i and an index j ∈ Λ corresponds to the vertex j + n.
Two vertices v and w in RZMSDigraph(R) are adjacent if and only if v∈ I, w - n∈ Λ, and P[w - n][v] ≠ 0.
This digraph is commonly called the Graham-Houghton graph of R.
gap> R := PrincipalFactor(
> DClass(FullTransformationMonoid(5),
> Transformation([2, 4, 1, 5, 5])));
<Rees 0-matrix semigroup 10x5 over Group([ (1,2,3,4), (1,2) ])>
gap> gr := RZMSDigraph(R);
<digraph with 15 vertices, 40 edges>
gap> e := DigraphEdges(gr)[1];
[ 1, 11 ]
gap> Matrix(R)[e[2] - 10][e[1]] <> 0;
true
##### 13.14-2 RZMSConnectedComponents
‣ RZMSConnectedComponents( R ) ( attribute )
Returns: The connected components of a Rees 0-matrix semigroup.
If R is an n by m Rees 0-matrix semigroup M^0[I, T, Λ; P] (so that I = {1,2,...,n} and Λ = {1,2,...,m}) then RZMSConnectedComponents returns the connected components of R.
Connectedness is an equivalence relation on the indices of R: the equivalence classes of the relation are called the connected components of R, and two indices in I ∪ Λ are connected if and only if their corresponding vertices in RZMSDigraph(R) are connected (see RZMSDigraph (13.14-1)). If R has n connected components, then RZMSConnectedComponents will return a list of pairs:
[ [ I_1, Λ_1 ], ..., [ I_k, Λ_k ] ]
where I = I_1 ⊔ ⋯ ⊔ I_k, Λ = Λ_1 ⊔ ⋯ ⊔ Λ_k, and for each l the set I_l∪Λ_l is a connected component of R. Note that at most one of I_l and Λ_l is possibly empty. The ordering of the connected components in the result in unspecified.
gap> R := ReesZeroMatrixSemigroup(SymmetricGroup(5),
> [[(), 0, (1, 3), (4, 5), 0],
> [0, (), 0, 0, (1, 3, 4, 5)],
> [0, 0, (1, 5)(2, 3), 0, 0],
> [0, (2, 3)(1, 4), 0, 0, 0]]);
<Rees 0-matrix semigroup 5x4 over Sym( [ 1 .. 5 ] )>
gap> RZMSConnectedComponents(R);
[ [ [ 1, 3, 4 ], [ 1, 3 ] ], [ [ 2, 5 ], [ 2, 4 ] ] ]
#### 13.15 Changing the representation of a semigroup
##### 13.15-1 IsomorphismReesMatrixSemigroup
‣ IsomorphismReesMatrixSemigroup( S ) ( attribute )
‣ IsomorphismReesZeroMatrixSemigroup( S ) ( attribute )
‣ IsomorphismReesMatrixSemigroupOverPermGroup( S ) ( attribute )
‣ IsomorphismReesZeroMatrixSemigroupOverPermGroup( S ) ( attribute )
Returns: An isomorphism.
If the semigroup S is finite and simple, then IsomorphismReesMatrixSemigroup returns an isomorphism to a Rees matrix semigroup over some group (usually a permutation group), and IsomorphismReesMatrixSemigroupOverPermGroup returns an isomorphism to a Rees matrix semigroup over a permutation group.
If S is finite and 0-simple, then IsomorphismReesZeroMatrixSemigroup returns an isomorphism to a Rees 0-matrix semigroup over some group (usually a permutation group), and IsomorphismReesZeroMatrixSemigroupOverPermGroup returns an isomorphism to a Rees 0-matrix semigroup over a permutation group.
See also InjectionPrincipalFactor (12.4-7).
gap> S := Semigroup(PartialPerm([1]));
<trivial partial perm group of rank 1 with 1 generator>
gap> iso := IsomorphismReesMatrixSemigroup(S);;
gap> Source(iso) = S;
true
gap> Range(iso);
<Rees matrix semigroup 1x1 over Group(())>
gap> S := Semigroup(PartialPerm([1]), PartialPerm([]));
<partial perm monoid of rank 1 with 2 generators>
gap> Range(IsomorphismReesZeroMatrixSemigroup(S));
<Rees 0-matrix semigroup 1x1 over Group(())>
Goto Chapter: Top 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 Bib Ind
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2019-08-17 16:59:44
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https://engineering.stackexchange.com/tags/electrical/hot
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# Tag Info
10
Beryllium oxide is a very good electrical insulator but at the same time the best non-metal (except diamond) thermal conductor. So to summarize. In general, good thermal conductivity is correlated with good electrical conductivity, but it is not a strict relationship. For example, there is the empirical Wiedemann-Franz law for metals which states that the ...
8
Works similarly in the US (compared to UK responses). Natural monopolies are allowed to exist for utilities as it wouldn't make sense to have 3 or 4 different sewer lines running to everyone's home. The distribution company that owns the grid may or may not produce electricity itself. But there is (usually) only one power line that connects to a user.
7
ICEs (internal combustion engines) take some time to ramp up their output torque. Various mechanical systems have to react before more mixture is injected into cylinders, and that then makes more pressure upon combustion. Think of a throttle-body engine as example. You push on the gas pedal, which opens the throttle valve more. That causes more air to ...
6
The current generated by a conductor moving through a magnetic field is related to the length of the conductor and the magnetic field strength. A coil is simply a convenient way to get a longer length of conductor within a given space. The cross sectional area doesn't actually matter in terms of generating power except insofar as it determines the current ...
6
For metals, good electrical conductivity does indeed imply good thermal conductivity. This is known from the Wiedemann–Franz law, which gives the ratio between electronic contribution of thermal conductivity ($\lambda$) and electrical conductivity ($\sigma$) and is proportional to the temperature ($T$). $$\frac{\lambda}{\sigma} = LT$$ This gives the ...
6
In the UK there are three different parts of the complete system: generation, transmission and distribution, and supply (sales) to customers. There are many companies that generate electricity, varying in size from national (and even international) companies down to small organizations that operate a few wind turbines at one location. There are two parts ...
5
Unless they have hearing loss, young people tend to be able to hear a larger range of sound frequencies than older people, particularly higher frequencies: Some security companies recently began manufacturing machines designed to emit an annoying sound that prevents teenagers from loitering outside stores and shops. Teens are effectively driven away, but ...
4
I think the instant torque claim mostly applies to "off the line" acceleration. That is from a standstill and electric motor has 100% of its available torque available at 0 rpm (mostly). The tradeoff is the an electric motor is always going to see a drop in torque with speed. Generally when you compare wheel torque availability as a function of speed between ...
4
The sun is heating mode. When the room temperature reaches the set temperature, the air conditioner stops operating until the temperature falls below the set temperature and the starts operating again. When in heating mode, the air conditioner does not cool. This setting is used during cold weather periods, such as in winter. The snowflake is cooling mode. ...
4
Yes, but... There's not great data available on the Fokotek products, but in general switching HP sized motor loads with solid state relays is a fairly expensive proposition. Because the voltage and current values don't cross zero at the same time, motor loads (inductive loads) are harder to switch than resistive loads. 3HP is just on the cusp where SSRs ...
4
We really need more info on the batteries and type of connection. Will the connection be permanent? In that case soldering or welding are the best choices. If you're connecting to wire, you'll need good strain relief. Any repetitive motion at the solder joint / weld will cause failure eventually. In a vehicle the battery may have a tendency to jostle around....
4
New homes are inspected by local authorities to comply with local building requlations. As Mike says, give them a call. The National Electrical Code (NEC) are standardized guidelines for electrical construction created by the National Fire Protection Association (NFPA). The NEC are guidelines, not laws. The NEC is adopted into law by states and local ...
3
Interesting problem. Let’s see, if we have a cylinder 12 inches in diameter and 8 feet high that would be (pi*(0.5f)^2*8f)*7.48 g/ft3=47 gallons per operation. At 3 GPM, that would be about 15 minutes to fill the cylinder all the way. If it were used to lift a 6780 pound weight, 8 feet up, that would be 8*6780 foot pounds of work, in 15 minutes. That ...
3
Static electricity is a major issue when handling or dealing with explosives. When blast holes are loaded with ANFO (ammonium nitrate fuel oil) as the main explosive, and a specific density of ANFO is required in the holes, the ANFO is blown into the holes via compressed air and a tube. For short holes drilled with hand held machines, the tube is made of ...
3
I had to draw electrical and hydraulic schematics in Microsoft Visio at a past job. I'm glad that is behind me because that software is barely good enough for making flow charts. I would never recommend anyone use it for engineering diagrams. There are expensive professional software programs that solve this problem much better, but if you are on a budget I ...
3
I'd call the blue part the transmitter housing. This usually contains electronics for sensor signal conversion, which converts the sensor (or probe) low level signal to a standard 4-20 mA signal. The sensor (probe) is the metallic cylinder. And the sensor head (containing the sensing element, membrane/coil) is usually located at the end surface of the ...
3
The reason for the humming noise is improper grounding. Audio signals are low voltage level AC signals (over simplified). An audio signal could be for example 1V, 1 KHz signal. When there is improper grounding, a low level noise signal example 10mV 50Hz can get coupled to the audio signal causing humming. (Values are made up) I would suggest installing ...
3
Ever since 1975 use of at least one GFCI outlet and preferably more is mandatory in bathrooms or anywhere where there is the likelihood of moisture being near the outlet by NEC (National Electric Code). If you have only one GFCI in any bathroom, all other outlets in that bathroom must be downstream to that GFCI and controlled by it. You can ask your local ...
2
For pretty fundamental physical reasons any power generator is only a way of converting one sort of energy to another ie converting the energy you have available to the sort of energy that you want. For example : Chemical energy to mechanical power eg an internal combustion engine Mechanical power to electrical power eg a generator Chemical energy to ...
2
The choice of cable is down to the environment that the cable is working in. Designers will be looking at mechanical abuse - by accident or not, temperature, chemical hazards etc. So, taking your some of your examples: Computer power -assuming you mean desktop little movement ambient temperature low chemical hazard. Kettles / toasters - temperature risk, ...
2
I answered a similar question on SE Sustainable Living recently. Before anyone (including yourself) can answer your question you need to supply some more information. The first thing you need to find out & provide is, what is the heating value of the gas. In metric units this would be the mega joules (MJ) of heat that one cubic metre of gas produces. ...
2
No, it's a characteristic that electric motors produce most, if not all, of their torque from zero rpm - which is why the electric cars are so good at getting off the line and how electric motors don't always need gearing to start heavy machinery - just lots of energy...
2
If you couple together two motors of unequal speed through a gear system which allows the faster motor to match speed with the slower one, then the rotating speed will equal that of the slower one and their torque outputs will add. If you couple them in a manner where their output speeds do not match, then the more powerful motor will drive the other one ...
2
1) You could if you wanted to, but it's probably not necessary. Drawing less than the rated power should not hurt it. 2) yes and yes.
2
Voltage and Current multiply to give power. 15kV times 60mA is the same as 60Volt times 15 Amps (=900 Watts) You would not expect that to be a tiny thing, would you? And that power is without transformation losses. Keep in mind that 900 watts a common size for an electric space heater. This is not trivial power you are dealing with.
2
They would accomodate them by adding the necessary amount of either capacitance or inductance to the transmission lines to cancel the reactance. Note that this is commonly done today in order to trim the impedance of a large generator to match that of a high-voltage transmission line. It is also done to cancel the reactive component of a large electrical ...
2
DC motors make fine generators. You can test them by connecting the shaft of the motor to the chuck of a variable-speed electric drill and connecting the motor wires to a load resistor; you then measure output current and voltage as functions of drill speed.
2
The tweezers that you are describing have what's known as a Static Dissipative coating. This coating is conductive, but it has a very high resistance. What this means is that they are able to dissipate any differential charge slowly/safely, without there being a very low impedance path to ground that might produce an arc. Put simply, the tweezers will not ...
2
Yes! The logic gates from a RS485/422. Most likely the signal are level shifted. Below is an except form Low power RS-485/RS-422 transceiver ST485B, ST485C datasheet. Below is an another similar part. PROFIBUS Compliant 6kV VDE-Reinforced Isolated RS-485 Transceiver ISL32741E. The example circuit is converting signal 3.3V-5V and 5V-3.3V.
Only top voted, non community-wiki answers of a minimum length are eligible
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2021-01-20 23:42:20
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https://www.os2world.com/wiki/index.php/Known_Issues_with_the_eCS_Preview_1
|
# Known Issues with the eCS Preview 1
By Jim Burke
## When doing a CD boot, after the IBM506S driver finishes displaying the drive configurations, the next few lines crawl along on some machines. (pre 10/7/00)
With these machines, if we turn off the /V option with the IBM506S.ADD driver, the problem goes away. However, we felt that it's important to leave the /V on - at least for the preview version.
Workaround: Relax and have a cup of coffee or something. Go make a phone call.
## After the last phase of the installation, upon reboot, you might end up with a black screen and nothing happens after that. (pre 10/7/00)
Workaround: Simply hit Ctrl-Alt-Delete to reboot the computer.
Note: Appears to have been fixed in newer version of MCP (10/29/00)
## WarpCenter Display (pre 10/7/00)
When you click on the Disk Space Monitor of the WarpCenter, if you have an unformatted drive, it screws up the WarpCenter display to the right of the disk space monitor.
Work around: remove unformatted drives from the Disk Space Monitor
BTW: WarpCenter still does not display the "Connections" properly when you click on the OS/2 Warp image on the left hand side. The entire network folder does not get displayed (same problem in Warp 4 - looks like it was never fixed).
## CRC error with floppy causes a trap D if you ask eCS to retry (pre 10/7/00)
Workaround: Don't select retry. Select stop.
## pnpisa.snp hangs certain systems - particularly OEM notebooks (pre 10/7/00)
Workaround: Copy SNDBLAST.SNP on top of PNPISA.SNP. If you're booting from CD, you will have to generate the diskettes and do that to the diskette.
Workaround: Ignore the message
Note: This message no longer appears with later drop of the Convenience Pak (10/29/00)
## Object Desktop Installation program (pre 10/7/00) and Navigation problem (10/7/00)
You need to read http://www.stardock.com/support/os2/od/odbug20/sec1.htm before attempting to run the Object Desktop installation program.
The Object Desktop Navigation function does not work properly under eCS (and reportedly has not been working since certain Warp 4 fixpak). We will report this to Stardock and IBM.
Workaround: None
## Peer to peer doesn't remember userid password (10/7/00)
This is an old old bug. We will report that to IBM this time.
Workaround: Put in USERID as the userid and PASSWORD as the password. Change them after installation.
## Peer to peer service start troubles (10/18/00)
Peer services do NOT start automatically in eCS.
Workaround: Edit the IBMLAN.INI file, the line that now states:
wrkservices = ,MESSENGER
wrkservices = MESSENGER,PEER
Or
You can edit the STARTUP.CMD file, remove the line that reads:
NET START REQ
and replace it with:
NET START PEER
## Connections don't start automatically (10/18/00)
The peer to peer CONNECTIONS defined do *NOT* start automatically on LOGON, regardless of what you specify in the CONNECTIONS settings notebook.
Workaround: Include the appropriate NET USE commands in your startup.cmd.
## Peer can't see other eCS stations (10/18/00)
Yet another issue is the ability of the eCS machine(s) to see each other (and themselves) on a LAN. This is evident in both the "Create Connections" and the command line "NET VIEW" commands.
Workaround:
Edit the IBMLAN.INI file, and change the line that now reads:
srvhidden=yes to: srvhidden=no
## DHCPCD.CFG Problem (Prev 1 5/30/2001)
The problem is the dhcpcd.cfg file (in \mptn\etc).
I found it helpful to have the following lines in there:
option 1 # Subnet Mask
option 3 # Router
option 6 # Domain Name Server
option 15 # Domain Name
option 33 # Static Route
option 60 "IBMWARP_V3.1" # Vendor Class
option 77 "IBMWARP_V3.1" # User Class
If you don't have these, DHCP will not set up DNS and routing automatically.
Suggestion courtesy of Michal Necasek of Scitechsoft.com
## Can not install eCS on system with Tekram controller (10/22/00)
The Tekram driver runs in conflict with the sym8xx.add driver that's in the config.sys of the CD boot - causing various strange behavior.
Workaround:
These are the steps you need to follow for a successful install:
You will need to install from diskette. When generating the install diskettes, note the sequence. Makedisk creates disk #1, and #2 first - and then disk #0.
Change the label on the install diskette from VFDISK to DISK 0 by issuing the label command.
Modify config.sys (on diskette 1) as follows:
basedev = TMSCSIW.ADD
set copyfromfloppy=1
rem out:
rem basedev = ql510.add
Reboot system with installation diskette #0 in your floppy drive and turn off CD boot
Check disk zero's label. If it says VFDISK, change it to read DISK 0 with this command:
a:\label DISK 0
Note: Fixed. (10/29/00)
## Various WiseMachine errors (10/23/00)
Workaround:
\TVoice\WiseProg\Unzip\Unzip \TVoice\WiseMachine.zip
from the \TVoice directory.
Note: there is a space between Unzip and \TVoice\WiseMachine.zip
I cannot install without the Dani drivers (10/29/00)
Workaround:
After the screen where the slash and backslash are walking over the dots up and down (during the final installation phase from eCS diskette 2) when you see a screen with the text "If you booted from the eCS CD... etc, ... Press F3", press F3 twice to get the command prompt. Then insert eCS diskette one and type: "copy a:\DaniS506.add x:\os2\boot" and "exit" and "Enter" (for reboot).
If the above doesn't work, wait for the machine to reboot and when you see the eCS Prev1 boot blob on the upper left corner of your screen, hit ALT-F1 and then F2 to go to the command line. Perform the above copy and reboot the machine.
The config.sys file at the root of the installation partition x: includes the "basedev=DaniS506.add /a:2 /P:1F0 /a:3 /P:170" line from my eCS diskette 1 already. The installation of eComStation will now continue using the Dani-driver !
## 3Com ISA 3C509B card causes eCS to trap (10/29/00)
System traps during network install with the following message:
c0000005
DOSCALL1.DLL 0002:0000a459
Workaround:
This is a known bug. Most 3Com NIC drivers trap when used with a fast CPU (about 500MHz or more). Go to the 3Com web site and grab the very latest drivers which will solve the problem for *some* NIC models (the 3C509 is among them AFAIK).
==Chuck McKinnis' page NicPak has a good list of drivers and you can get some help there, too. Trap 6 during install (10/25/00)==
During installation got strange TRAP 006 errors.
Workaround:
Turn off second level cache during install.
## OS/2 cannot operate your hard drive or floppy diskette (10/29/00)
That message is put out when the country.sys file cannot be read for some reason.. The following is according to Daniela Engert:
Let's assume the hardware and the drivers are working properly (all adapter channels and all attached devices are operated by an eCS driver). Then there are two common scenarios which cause the mentioned error:
1) The bootmanager "sees" partitions differently from OS2DASD.DMD and propagates a bad boot drive letter up to OS2KRNL. In most cases this is caused by unsupported partition types (B, C, and F) located in front of the OS/2 boot partition (in partition scan order). [Note by Kim Cheung: this would happen, for instance, if you have a FAT32 partition in front of your eCS boot partition. Installing OS/2 with FAT32 around is an article from VOICE about this.]
2) The adapter channels or units are ordered differently in BIOS and OS/2. This usually happens when more than one adapter is in the system (e.g. an on-board IDE controller plus a SCSI host adapter or an additional EIDE controller, either on-board or a PCI plug-in, more than one of them equipped with DASD units). In this case, their order of appearance in CONFIG.SYS is important! Most BIOSes have options to reorder the adapter channels, and the OS/2 drivers are required to match that (either by reordering the ADDs in CONFIG.SYS or - in rare cases - by some ADD options trickery). Please note: this is not an absolute requirement, but it is hard to get away with a non-matching adapter/unit order.
Unfortunately resolving such a problem requires major knowledge and can't be done automagically.
## DFSEE and partitioning problems
To see what partitioning information you have on your drives, get Jan van Wijk's program, DFS.EXE from here: http://www.fsys.demon.nl/dfsee.htm It is almost an essential tool for some of the problems you might encounter.
## LVM can't access old boot manager partition (10/29/00)
There have been report that LVM fails to work with old boot manager partition left there from previous versions of OS/2. However, others had indicated that they have no problem.
Workaround:
If it affects you, remove the old boot manager and reinstall it under LVM. Reading material: LVM, FDISK and Partition Magic ; A Short Introduction to LVM and JFS (both from VOICE)
## TRAP E from In-Joy and DIOP dialer (pre 10/31/00)
{Could be trap D, too.}
TRAP 000e ERRCD=0002 ERACC=**** ERLIM=********
EAX=3b81939c EBA=f7161c14 ECX=00000000 EDXX=3b81939c
CS:EIP=0168:f77aa125d CSACC=c09b CSLIM=ffffffff
SS:ESP=0030:00004ab0 SSACC=1097 SSLIM=0000401f
DS=0160 DSACC=c093 DSLIM=ffffffff CRO=80010013
ES=0160 ESACC=c093 ESLIM=ffffffff CR2=3b81930a0
FFS=0000 FSACC **** FSLIM=********
GS=0000 GSACC=**** GSLIM=********
The system detected an internal processing error at location
##0168:fff1f322 - 000e:c322. 60000, 9084
07860652
Internal revision 14.062_W4
Workaround:
Try the new stack fix from ftp.service.boulder.ibm.com/ps/products/tcpip/fixes/v4.3os2 and get the file dated Jan 29 10:26, called ic27649.exe, and which is 844,651 bytes or you can get it from the In-Joy web site.
You can also try the WSeB UNI kernel at ftp://testcase.boulder.ibm.com/ps/fromibm/os2/. This is for single processor machines. Use the smp kernels for multi-processor ones. Don't use the W4 kernels. They are for Warp 4 based machines and eCS is based on the warp server for e-business kernel.
NOTE: Try these kernels after you try the stack fix and ONLY if you are experiencing trouble.
## Newer JFS version (pre 10/31/00)
There is a newer JFS version we have been told that ftp://testcase.boulder.ibm.com/ps/fromibm/os2/jfs1009.zip is a good fix to have and ftp://testcase.boulder.ibm.com/ps/fromibm/os2/jfs_miss.zip has some missing jfs trace files.
## I can't install eCS past 1024 cylinders (pre 10/31/00)
Depends on your hardware. If you hae a new bios that understands Int13 extensions, then you can install past the first 1024 cylinders. Older Pentiums don't have this, BTW.
Depending on your hardware, this can mean 512 MB, 1 GB, or 8 GB. With Logical Block Address (LBA), you can boot from up to 8 GB. This includes most Pentium class systems. With 486s and earlier, you're likely to be restricted to 512 MB or 1 GB, depending. [Note that this isn't a CPU restriction; it's a restriction due to the BIOS. For example, a new BIOS for a mainboard may allow it to boot from < 8 GB to anywhere on a large hard drive.]
## Can't install eCS to certain partitions (11/03/00)
I created a partition on my secondary master IDE drive and tried to install eCS to that partition. The installation seems to go perfectly until I remove the diskette and reboot, I get the message, "operating system missing, system halted." If I reboot I get my boot manager choices from before the installation attempt. It is as though LVM changes made in the install don't "stick."
Workaround:
You will need to replace the Warp 4 boot manager with the one created by LVM. LVMGUI has a menu item which says Bootmanager - use that. Click here for an image.
## 1608 error during network install (11/03/00)
This is a long standing issue dating as far back as Warp 3. Some NIC cards don't like to be poked by the MPTS sniffer. Warp Server install used to have a /NS option to deal with this but that option is not available with eCS.
Workaround:
Unplug the NIC card from your computer during install and reinsert it afterwords.
SmartSuite Install error: " Noncurrent OS Detection " Make sure 4os2 isn't used in config.sys.
## Unpack error when installing SmartSuite 1.5 (11/29/00)
> >> EPFIE196: An error occurred while executing M:\INSTTEMP.EPF\EPFIUPK2.EXE O:\ss\english\WORDPRO\LWPWK6O.DL_ L:\LOTUSW4\WORDPRO.
Workaround:
2. run LWPWK6O.EXE to extract LWPWK6O.DL_
3. xcopy the smartsuite directory to a harddrive example: XCOPY G:\SS C:\SS\ /S/E/V
4. Copy the extraced LWPWK6O.DL_ file into the \ss\english\wordpro directory.
5. From the \ss directory, run INSTALL.CMD
## SYS3175 when installing SmartSuite 1.5 (pre 11/29/00)
Installation stopped.
Workaround:
(b) Make sure WiseMachine is not running.
(c) From the \TVoice directory, unzip wisemachine.zip (with the overwrite and directory options if your unzip doesn't do it automatically). You get a new \TVoice\WiseProg\WiseManager\WiseMachineLocal.exe, \TVoice\WiseProg\WiseManager\SmartSuite\*.*, and \TVoice\WiseProg\DeskTop\CreateSmartSuite.cmd.
(d) Start up WiseMachine, Drag-n-drop SmartSuite to your destination drive. Modify the destination top-directory to say lotusw4
(e) Check the box "Use Unpack..." and click Yes at the warning message screen
(f) Insert the Lotus CD into your CD-ROM and point Source to one of the files in the \english directory of the CD (or whatever language directory there is).
(g) Click Install
This will unpack and transfer the ENTIRE Lotus CD to your hard drive.
(h) Close down WiseMachine.
(i) DON'T REBOOT THE MACHINE YET. Run the Lotus supplied install program. For my case, the install program doesn't fail with SYS3175 any more.
I wasn't able to get rid of this step because I think Lotus is putting some registration and path information into the emulated Windows registry files and I don't have time to investigate more details there to be able to get around it with WiseMachine. I should be able to do so in the future.
(j) Restart WiseMachine again, go to Installed Applications, right click on SmartSuite and do a Create ICON.
Using this method, you should be able to get all but Organizer to work.
## Smart Suite Launch Problem
Error opening ole
>It looks to me as if you have to physically delete those BASEDEV lines, >rather than just REM them out. >... >Peter
Thanks for your help ! I deleted the unwanted lines from config.sys and it works now.
## Removable Media Problems
because of broken removable media support (no matter if ZIP, ORB, or PCMCIA attached PRM devices). Almost any time I inserted/replaced media I got either
• a TRAP 3 in OS2DASD or OS2LVM (I don't know which one)
or
• no reaction on "refresh removable media"
no matter if FAT or HPFS. I couldn't find any reproducable pattern.
There is a new driver available for these removable drives.
There are two new OS/2 device drivers now available on the OS/2 DDPak (http://service.software.ibm.com/os2ddpak/) that may be of interest to some of you. They're currently beta, but will become "IBM Supported" when FVT completes.
• Driver for the internal 56K modem on the ThinkPad T20 family. It's listed under MODEMS/IBM. Yes, I know this is a WinModem, but it does work under OS/2 with this driver.
• Driver for Parallel Port-attached ZIP drives, ZIP 100, ZIP Plus, and ZIP 250. It's listed under REMOVABLE DISKS/IOMEGA.
## icmp: unknown protocol error
When you run ping or ftp and other programs you get an ICMP* error. From prev 2 and above, some files are missing from the installation cd. In this case the solution is to use another version of Warp or eCS and copy these files to the new installation mptn\etc directory.
x:\mptn\etc\protocol 6K
x:\mptn\etc\services 213K (missing but doesn't seem to be necessary to fix this)
• Internet Control Message Protocol
## Dos Lan Bug
Problem has been identified. It is an INSTALL problem (MPTS/TCPIP) where CONFIG.SYS is not updated. Actually, the MPTSCFG.INF file (VIEW it) had the answer as well. See the LAN Virtual DD section.
Two lines are not inserted into CONFIG.SYS during Install, the required Virtual drivers for DOS.
DEVICE=drive:\IBMCOM\PROTOCOL\LANVDD.OS2
DEVICE=drive:\IBMCOM\PROTOCOL\LANPDD.OS2
"Empty device table (no devices defined). SIO2K.SYS will not load."
CONFIG.SYS contained, very close to the end of the file:
device=c:\sio2k\sio2k.sys logfile=c:\sio2k\sio2k.log
device=c:\sio2k\uart.sys logfile=c:\sio2k\sio2k.log
device=c:\sio2k\vsio2k.sys logfile=c:\sio2k\vsio2k.log vIrqList(3,4)
By changing the order to:
DEVICE=E:\SIO2K\UART.SYS LOGFILE=E:\SIO2K\SIO2K.LOG
DEVICE=E:\SIO2K\VSIO2K.SYS LOGFILE=E:\SIO2K\SIO2K.LOG
DEVICE=E:\SIO2K\SIO2K.SYS LOGFILE=E:\SIO2K\SIO2K.LOG
the problem was fixed.
NB: This has nothing to do with eCS unless you are planning on using the Ray Gwinn SIO drivers.
## In WiseMachine, when you first click on "Installed Applications", there is nothing under the "Applications" branch. (pre 10/7/00)
Workaround: Right click on the "Applications" folder and do a refresh.
Note: Fixed (10/29/00)
## Not every Rexx script under WiseMachine has been fully tested. (pre 10/7/00)
When you do a drag-n-drop deployment of an application, WiseMachine executes a Rexx script for that application. These scripts were adopted from the network version and have not been fully tested under a local non-diskless station.
Workaround: If you have problem with certain application, be sure to go to our web site at www.ecomstation.com and follow the link for more recent instructions.
Note: Over 2/3 have been tested (10/29/00)
## WiseMachine is not WPS aware (pre 10/7/00)
The preview version of WiseMachine is not a SOM application. What that means is that if you moved an installed application from its original location to a new location, WiseMachine is not aware of that automatically.
Work around: What you need to do is to bring up WiseMachine again, do a drag-n-drop to the new location, and uncheck the "Copy program to destination location" option, and do an Install. This would force WiseMachine to update all of its record of that application. In the future, WiseMachine will be updated to include the ability to do this automatically.
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2018-02-20 19:18:31
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https://questions.examside.com/past-years/jee/question/the-sum-of-all-the-real-roots-of-the-equation-left-x-2-right-jee-advanced-1997-marks-2-fc5nxvmoehqxgvem.htm
|
NEW
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Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1
### IIT-JEE 1997
Fill in the Blanks
The sum of all the real roots of the equation $${\left| {x - 2} \right|^2} + \left| {x - 2} \right| - 2 = 0$$ is ............................
4
2
### IIT-JEE 1996
Fill in the Blanks
Let n and k be positive such that $$n \ge {{k(k + 1)} \over 2}$$ . The number of solutions $$\,({x_1},\,{x_2},\,.....{x_k}),\,{x_1}\,\, \ge \,1,\,{x_2}\, \ge \,2,.......,{x_k} \ge k$$, all integers, satisfying $${x_1} + {x_2} + \,..... + {x_k} = n,\,$$ is......................................
$${{\left[ {k + \left( {n - {{k(k + 1)} \over 2}} \right) - 1} \right]!} \over {\left[ {n - {{k(k + 1)} \over 2}} \right]!\,(k - 1)\,!}}$$
3
### IIT-JEE 1990
Fill in the Blanks
If $$\,x < 0,\,\,y < 0,\,\,x + y + {x \over y} = {1 \over 2}$$ and $$(x + y)\,{x \over y} = - {1 \over 2}$$, then x =..........and y =.........
$$- {\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}$$ ,$$- {\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 4}}$$
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[ Article ]
Journal of the Korean Astronomical Society - Vol. 51, No. 5, pp.129-142
ISSN: 1225-4614 (Print) 2288-890X (Online)
Print publication date 31 Oct 2018
Received 12 Mar 2018 Accepted 18 Sep 2018
# NEW PHOTOMETRIC PIPELINE TO EXPLORE TEMPORAL AND SPATIAL VARIABILITY WITH KMTNET DEEP-SOUTH OBSERVATIONS
Seo-Won Chang1, 2, 3 ; Yong-Ik Byun3 ; Min-Su Shin4 ; Hahn Yi3 ; Myung-Jin Kim4 ; Hong-Kyu Moon4 ; Young-Jun Choi4 ; Sang-Mok Cha4, 5 ; Yongseok Lee4, 5
1Research School of Astronomy and Astrophysics, The Australian National University, Canberra, ACT 2611 seowon.chang@anu.edu.au
2ARC Centre of Excellence for All-sky Astrophysics (CAASTRO)
3Department of Astronomy and University Observatory, Yonsei University, Seodaemun-gu, Seoul 03722
4Korea Astronomy and Space Science Institute, 776 Daedukdae-ro, Yuseong-gu, Daejeon 34055
5School of Space Research, Kyung Hee University, Giheung-gu, Yongin, Gyeonggi 17104
Correspondence to: S.-W. Chang
JKAS is published under Creative Commons license CC BY-SA 4.0.
## Abstract
The DEEP-South (the Deep Ecliptic Patrol of the Southern Sky) photometric census of small Solar System bodies produces massive time-series data of variable, transient or moving objects as a byproduct. To fully investigate unexplored variable phenomena, we present an application of multi-aperture photometry and FastBit indexing techniques for faster access to a portion of the DEEP-South year-one data. Our new pipeline is designed to perform automated point source detection, robust high-precision photometry and calibration of non-crowded fields which have overlap with previously surveyed areas. In this paper, we show some examples of catalog-based variability searches to find new variable stars and to recover targeted asteroids. We discover 21 new periodic variables with period ranging between 0.1 and 31 days, including four eclipsing binary systems (detached, over-contact, and ellipsoidal variables), one white dwarf/M dwarf pair candidate, and rotating variable stars. We also recover astrometry (< ±1–2 arcsec level accuracy) and photometry of two targeted near-earth asteroids, 2006 DZ169 and 1996 SK, along with the small- (~0.12 mag) and relatively large-amplitude (~0.5 mag) variations of their dominant rotational signals in R-band.
## Keywords:
methods: data analysis, techniques: photometric, stars: variables: general, asteroids: general
## 1. INTRODUCTION
The Deep Ecliptic Patrol of the Southern Sky (DEEPSouth: Moon et al. 2016) is a dedicated photometric study to physically characterize small bodies in our Solar System, as one of the secondary science projects of Korea Microlensing Telescope Network (KMTNet; Kim et al. 2016a). The DEEP-South employs a network of three identical 1.6-m telescopes located in Chile (CTIO), South Africa (SAAO) and Australia (SSO), allowing 24-hour monitoring of asteroids and comets. Light curves with BV RI-band colors have been acquired for more than two hundred Near-Earth Asteroids (NEAs) since late 2015. The major efforts to discover NEAs have been historically concentrated on planetary defense to ensure Earth’s safety from asteroid impacts. Several NEA discovery projects, such as the Catalina Sky Survey (CSS, Larson et al. 2003) and Panoramic Survey Telescope and Rapid Response System (PanSTARRS, Kaiser et al. 2002), have cataloged of NEAs more than 90 percent of the estimated population larger than 140 meters in regular scans of the sky. Comparing with other NEA surveys, the advantage of the DEEPSouth is the round-the-clock operation capability which is essential for either precision astrometry or time-series photometry with high temporal resolution.
As in the cases of other asteroid surveys (e.g., LONEOS: Bowell et al. 1995; LINEAR: Stokes et al. 2000; CSS: Larson et al. 2003), long-time baseline observations have the potential to advance our knowledge of variable and transient phenomena over different timescales from RR Lyraes (Miceli et al. 2008; Sesar et al. 2013; Torrealba et al. 2015) to periodic variables (Palaversa et al. 2013; Drake et al. 2014) to AGNs (Ruan et al. 2012). These time-domain data sets also increase our understanding of stellar structure and evolution: pulsating stars (e.g., δ Scuti, RR Lyrae, Cepheid, and Mira variables) are important probes of the internal structure in investigating different excitation mechanisms of oscillations in stars, eclipsing binary systems allow us to determine the masses of both stars in a direct way, and Type Ia supernovae are used as probes of cosmology. These new outcomes emphasize importance of accurate and homogeneous photometric measurements and calibrations in the entire survey data, in terms of further synergy with upcoming data releases by ongoing and future time-domain surveys.
Since the standard software packages of the DEEPSouth are designed for differential photometry of targeted moving objects (Yim et al. 2016), we have implemented an updated version of the source detection and time-series photometry pipeline to recover point sources which are not extracted in the original pipeline. First, the new pipeline is designed to conduct robust high-precision photometry and calibration of non-crowded fields with a varying point spread function (PSF). The PSF varies across images and changes with time due to focus, pointing jitter, optical distortion or atmospheric conditions, particularly in cases of wide field-of-view (FoV). Due to the large FoV (~4 deg2) of the KMTNet mosaic CCD images, we expect to see deformation of the image PSF as well as spatial variations in the pixel scale1. To mitigate this problem, we perform multi-aperture photometry in determining the optimal aperture for each point source at every epoch and correcting position-dependent variations in the PSF shape across the mosaic. In addition to this, the pipeline is designed to perform forced photometry, thus giving us the possibility to extract consistent measurement of explicit sources in every frame with respect to its reference (deep co-added) frame. This approach can reduce false detection rate at low flux levels.
Further, this paper addresses a multi-step calibration issue to tie all DEEP-South data coming from the three different telescopes to a consistent photometric system. As in the case of LINEAR and CSS surveys, standard stars are not available in every survey field for precision photometry, specially because of varying observation conditions. Due to the observing strategy of the program, however, we can determine relative calibration parameters (e.g., relative overall offset) by using multiply observed stars in any overlapping fields. Since some of these stars have been measured in other allsky multi-color surveys, we can transform our Johnson-Cousins filter system (only BV RI) to other photometric systems properly taking into account color terms. Similar calibration approaches have been successfully used in other asteroid surveys (Drake et al. 2009; Sesar et al. 2011).
Lastly, we have implemented a database management system for handling very large source catalogs. The source database is a key part of public data release enabling future scientific investigations. The data from most asteroid surveys are easily accessible in the form of an SQL-based relational database with an interactive web-based interface, containing all the epoch-based source photometry and metadata for more than tens to hundreds of million objects (e.g., LINEARdb: Sesar et al. 2011; Pan-STARRS1: Flewelling et al. 2016; ATLAS-VAR: Heinze et al. 2018). For convenience, all these systems provide internally linked tables by using a unique label for every object that was obtained by grouping individual source detections at a given matching radius. We can significantly increase speed of query execution for already linked objects in this approach. However, it does not allow us to investigate possible transients, variables or moving objects within the database itself in a flexible way. In order to accelerate data access and reduce query response time, we adopt an efficient data indexing technique called FastBit (Wu et al. 2009) that stores data in a column-oriented manner unlike traditional relational databases. In this paper, we focus on development and applications of catalog-based searches for variability in stars with fixed coordinates using this database. In the second paper of this series, we will present additional application example to search and identify moving objects with motion vectors.
This paper is organized as follows. In Section 2 we describe the DEEP-South survey and experimental time-series data. In Section 3 we outline the new photometry pipeline, calibration issues and the FastBit database system. In Section 4 we discuss a few example applications exploring temporal and spatial variability, especially periodic variable stars and targeted asteroids. We conclude with a view to utilize the pipeline for massive variability searches with a full set of the DEEPSouth survey data.
## 2. THE DEEP-SOUTH PHOTOMETRIC CENSUS FOR ASTEROIDS AND COMETS
The DEEP-South observations were made with three 1.6-m telescopes that have prime focus during the offseason of the Galactic bulge monitoring campaign of the KMTNet main survey. Survey observations comprise 135 distinct full nights per year between 2015 and 2019. The KMTNet camera consists of four 9k by 9k CCDs having a FoV of 2×2 degrees with a pixel scale of 0.4 arcsec per pixel (Kim et al. 2016a). The CCD chips are arranged with vertical (north-south direction) and horizontal (east-west direction) gaps of about 373 and 184 arcsec, respectively. The cameras have a 30 second readout time for the full mosaic with a readout noise of 10 electrons. The DEEP-South survey uses four standard Johnson-Cousins BV RI filters, but they used mostly R-band for time-series work. The quantum efficiency is 80–90% across the 4000–9000°A range, with a peak at R-band (~90%). About 50–100 science and associated calibration frames are produced in every nightly run (total 65–260 GB of data products). All the raw images are transferred to the KMTNet data center located at Daejeon in Korea for pre-processing.
The survey is designed to address a number of scientific goals that use five different observation modes (see Table 1 in Moon et al. 2016 for details). Opposition Census (OC) is the most frequently used mode for targeted photometry of NEAs in the regions on the sky around opposition. At that location, their color or brightness can be measured accurately. Each OC run consists of a series of exposures of more than two target fields that are visible in the opposition region. They normally employ exposure times short enough to avoid reducing the detection sensitivity for a moving object which may appear slightly streaked. From these OC observations, we can expect to obtain physical properties (e.g., rotation period and color indices), and to discover unknown moving objects. They also devote a small fraction of time to conduct both ecliptic survey and target-of-opportunity follow-up observations.
Summary of R-band time-series observations in the N02007-OC field
### 2.1. Experimental Data Sets
For the purpose of developing a new photometric pipeline, we use the DEEP-South first-year data collected from late-July 2015 through mid-April 2016. Here we present results for the most frequently observed field, N02007-OC, centered at RA=00:24:00, DEC=+05:00:00 (Figure 1). The first letter in the field designation indicates a sign of Ecliptic latitude: N and S for + and -, respectively. The next two digits are Ecliptic latitude in degrees; the last three digits are Ecliptic longitude in degrees. This field has full overlap with the Sloan Digital Sky Survey (SDSS) which provides colors of objects detected in the ugriz system. The observation logs are summarized in Table 1. We conducted time-series observations of the N02007-OC field only in the R-filter to measure periodic variability induced by rotation of the two targeted asteroids (see Section 4.3). These moving objects were detected in 200-second exposures with sufficient signal-to-noise ratio down to R ~ 24.
Number of epochs per target field in the DEEP-South year-one data obtained from the three KMTNet telescopes. The values are color-coded by the bar on the right (from 1 to 510 visits). The solid line shows the Galactic plane in equatorial coordinates using the Aitoff projection. The location of the selected field, N02007-OC, is recognized by its darkest color in the center.
Since early science commissioning observations were undertaken, there have been many updates and changes in CCD readout electronics and software for telescope operation. As part of quality assurance of the year-one data, we check problematic images (e.g., poor telescope tracking, readout error, or bad seeing) separately for each of the four CCD chips in mosaic, i.e., on a chip-by-chip basis. We define the individual amplifier images as the basic unit of our photometry pipeline, called a frame. After excluding 24 bad frames both automatically as well as manually through the pipeline (see Section 3), 16296 frames were used for this study.
## 3. DATA PROCESSING PIPELINE
The raw mosaic images are pre-processed by the KMTNet data reduction pipeline, including overscan, bias and flat-fielding reduction. The KMTNet camera produces strong crosstalk signals in multi-segment amplifiers for bright or saturated sources. The level of these electronic ghosts varies from frame to frame, but these can be removed by measuring crosstalk coefficients among frames within a single CCD image (see Kim et al. 2016b for details).
The components of our processing pipeline are separated into three basic units based on the format of data. Here we briefly summarize main procedures applied to the pre-processed data.
• 1. MEF unit (18432×18464 pixels): we obtain an accurate astrometric solution used SCAMP (Bertin 2006) after correcting for geometric field distortion that is well described by a polynomial with third order terms measured from the center of the mosaic. This distortion pattern appear to be stable over time at three different KMTNet sites. The astrometric solution was determined as follows: (i) we run SCAMP separately for each exposure, and then (ii) we run it again to find a final solution that is consistent across all input frames. This astrometric solution is verified against the 2MASS reference system (Skrutskie et al. 2006). The resultant internal and external astrometric accuracy on overlapping sources are less than 0.02±0.01 and 0.09±0.06 arcsec in both RA and Dec directions, respectively. In the future, we will use the second data release of Gaia as input to validate astrometric analysis.
• 2. CCD unit (9216×9232 pixels): we remove cosmic-ray hits or hot-pixels using the parallelized version of the L.A.Cosmic algorithm (van Dokkum 2001)2.
• 3. AMP unit (1152×9232 pixels): we adopt the strategy of treating all 32 frames separately in the photometry and calibration processes in order to reduce processing time.
### 3.1. Frame Catalog Generation
Source extraction is the first step in making a complete frame catalog for all point sources either in fixed or varying positions. We use the source detection algorithm implemented in the SExtractor (Bertin & Arnouts 1996) in which each source is regarded as as set of connected pixels that exceed threshold above the local background. In this work, all connected regions with more than 5 pixels above 2.5-σ of the local background were extracted. We also chose the following SExtractor parameters to optimize point source detection as well as to properly take into account blended sources: DETECT MINAREA=5, DETECT THRESH=2.5, DEBLEND NTHRESH=64, and DEBLEND MINCONT=0.0001. The average FWHM of each frame was used as a prior input for SEEING FWHM parameter. This process also produces several measurement characteristics such as FWHM, elongation, extraction flags, and star/galaxy classifier.
We utilize the multi-aperture photometry algorithm (Chang et al. 2015) in order to: (i) determine the optimal aperture for each object at each epoch, (ii) use an empirical aperture-growth-curve method for flux correction, and (iii) add a new tag that isolates peculiar situations where photometry returns improper measurements. For each frame, we measure photometry in a series of circular apertures (up to 12-pixel radius) without changing the sky annulus, and we use the last aperture (corresponding to 12-pixel radii) as the reference for aperture corrections. We refer the reader to Chang et al. 2015 for details of the photometric performance of the pipeline. The position-dependent PSF variation and pixel-scale variation across fields for given focus result in up to 10% changes in magnitude, depending on the selected aperture size for photometry. Figure 2 shows an example of position- and aperture-dependent magnitude offsets, with the reminder that the coverage of x-axis (~0.13 deg) is about eight times smaller than that of y-axis (~1.02 deg). This mayby expected to artificially enlarge the apparent size of the magnitude offset along the y-axis. The magnitude offsets vary significantly for the case of photometry performed with small-sized apertures. The magnitude differences between the reference aperture and the smaller apertures are well described by two-dimensional polynomial forms, so we can consider any spatial variation in the aperture corrections across a single frame.
Position- and aperture-dependent magnitude offsets (gray points) and that after the correction by a two-dimensional surface fitting (orange points) for an example frame. The dispersion of the magnitude offsets is much larger at smaller aperture sizes (top) than larger ones (bottom). The color maps show magnitude offsets in two-dimensional image frame.
Multiple visits to the same region of the sky enables us to achieve stable and uniform internal relative calibration in our instrumental photometric system. Bright, non-variable sources can be used as internal calibration stars, but it is necessary to check whether photometric quality of calibrators is sufficient and distributed homogeneously across the whole (x, y) plane in a given frame. For the latter, we calculate an additional quality-ensuring cut characterizing two-dimensional distribution of these internal calibrators by using two-dimensional Kolmogorov-Smirnov (K-S) test (e.g., Peacock 1983). The significance levels of flag statistic for the 2D K-S test can be summarized by the simple formula for the one-sample case (see also Equation 14.7.1 in Press et al. 1992). Using this probability PKS, we can figure out whether the source distribution is close to uniform across the frame. Figure 3 shows the spatial distributions of selected calibration stars for good and bad cases, in which relatively large value of probability (PKS > 0.15) indicates the presence of inhomogeneity.
Example of the spatial distribution of internal calibrators for good (gray points) and bad (red points) cases together with the calculated probability PKS. In the top panel, each frame location is indicated by frame numbers.
Following the suggestion of Ivezić et al. (2007), we also correct the spatial dependence of internal zeropoints around the photometric zeropoint separately for the 32 amplifiers in order to take into account atmospheric extinction gradients. We fit for linear gradients in both x and y coordinates simultaneously that can be expressed as:
$∆m=f\left(x,y\right)={C}_{0}+{C}_{1}x+{C}_{2}y$ (1)
where ∆m is the difference between reference and measured magnitudes for each star on every frame. The magnitude error is used as weighting factor for regression. C0 is a good diagnostic for discerning whether unrecognized temporal changes in atmospheric transparency have occurred during the observations. C0 should be close to zero when the quality of data is adequate. Figure 4 shows the histogram of C0 values for all frames along with cut-out images having three different quality cuts (good, intermediate, and bad). Finally, we apply the resulting zero-point surface to the instrumental magnitudes. A simple Boolean index is added to the database to be used as a photometric quality flag of our internal calibrations (see CALDEX in Appendix).
Histogram of C0 coefficients for all observed frames. The arrow indicates the range of photometric data quality observed under either perfectly (C0 ∼ 0) or partially nonphotometric conditions (-2 < C0 < -1). All insert images are presented as an example of different quality cuts.
### 3.2. Photometric Recalibration
To overcome the absence of observations used for photometric standardization, as mentioned in Introduction, we recalibrate our photometry to the SDSS Data Release 13 (SDSS DR13; Albareti et al. 2017) photometric catalog. The main reason being that there are no suitable photometric catalogs surveyed by other projects (e.g., no overlap with the first public data release of the Dark Energy Survey or poor photometric quality of the third data release of the Palomar Transient Factory). In addition to this, the well-studied SDSS color system is outstanding compared to other photometric systems (e.g., Pan-STARRS grizy filters or even for the Gaia broad bandpass) to aid in the study of point sources identified by our survey. Because only part of the DEEP-South field is covered by the SDSS DR13 photometric catalog, this approach is limited mainly by the declination limit. The recent SkyMapper Southern Survey provides the most comprehensive map of the Southern sky released to the public which contains well-calibrated magnitudes of point sources up to 18 (AB mag) in all uvgriz bands (Wolf et al. 2018), and thus it can serve as a calibration reference for the full data sets in the future.
From the SDSS DR13 database, we select suitable calibration stars with a series of clean flags (e.g., no edge, non-saturated, no cosmic-ray hits, primary object or no neighborhoods), magnitude errors below 0.05 in ri bands, and PSF magnitudes in the ranges 14≤ r ≤20 and 14≤ i ≤20. For the DEEP-South data, the mean magnitude error of the sources used becomes less than 0.05 mag in a given magnitude range. The transformation equation between the Johnson R and SDSS r system is simply derived as
$R=r+k\left(r-i\right)+C$ (2)
where k is the first-order color term and the possible zeropoint shift, C, is determined by iteratively rejecting outliers in the residual. The median absolute deviation of the residual of the fit for -0.3 < ri < 2.1 is 0.029. The color term k equals to -0.2837, which is similar to that found previously (k=-0.2936); please refer to the transformation equations derived by Lupton (2005).3 Figure 5 shows the transformation between SDSS ri and DEEP-South R magnitudes as defined by Equation (2). The scatter of magnitude difference between these two photometric systems is roughly constant with respect to magnitude. We see that the impact of a few photometric outliers is negligible at these color and magnitude levels.
Transformation between the DEEP-South R and SDSS r magnitudes as a function of r − i. The light points are stars with low photometric error (less than 0.1 magnitudes) in griz bands. Top: the solid line shows the best fitting with Equation (2); Bottom: corresponding difference between transformed DEEP-South and SDSS magnitudes as a function of the SDSS r magnitude.
As a final step, we ingest all frame catalogs into the FastBit database (see Appendix for technical details) that provides a set of compressed bitmap indexes to quickly retrieve the list of objects in a given sky region, observational parameters of selected sources or timeseries data (e.g., light curves).
## 4. SEARCHING FOR TEMPORAL AND SPATIAL VARIABILITY
### 4.1. Light Curve Production
Performing a cone search is the easiest way to construct light curves for stationary sources detected at least twice, called groups. We first build a master catalog of about 46,000 groups merged by matching positions (with a match radius of 0.5 arcsec) across different epochs and amplifiers, and we also generate a catalog of transient sources detected only once (see Section 4.3). Our smaller matching radius can reduce the number of spurious matches, but it causes us to lose a few real matches for faint sources. We find that about 80 new groups (mostly R > 20) can be associated by positional matching if we double the cutoff radius, thus the group sample considered here may not severely limit our variability results in the next section. Within each group, we determine mean positions by combining all astrometric measurements after excluding flagged data points. Now we only require the mean sky position and a selected search radius to produce a light curve of each group. We choose a loose cut for light curve production (i.e., matching radius of 0.8 arcsec). The following example command executes a query on the database and returns associated measurements:
ibis -d DeepSouth-DB-q "SELECT MJD, MAG_MAP, MAGERR_MAPWHERE (SQRT(POWER(XWIN_RA-RA_group, 2)+ POWER(YWIN_DEC-DEC_group, 2))BETWEEN 0 AND SEARCH_RADIUS)AND FLAGS < 4ORDER BY MJD".
In this paper, we limit the sample to ~13,000 stars which have reliable color information in the SDSS DR13 database. Note that there is overlapping regions with the Pan-STARRS1 data, but it is similar in depth to the SDSS in r-band for point sources.
In order to remove systematic noise caused by atmospheric variations or instrumental effects, we apply the photometric detrending algorithm (see Kim et al. 2009 for details)4 to the light curves. This algorithm finds position- and time-dependent systematics using a clustering technique, and then corrects the highly affected light curves by removing the systematics from individual light curves. Figure 6 shows the overall dispersion of light curves before and after the detrending procedure without any outlier clipping. Here, we use the root-mean-square (RMS) amplitude of light curves as a proxy of our photometric precision. The RMS values of DEEP-South light curves decrease to 0.3% precision level at the bright end (R ≤ 16). Applying this technique to the raw light curves gives good results in recovering the true brightness variations over the full magnitude range from 14 to 22.5.
Robust RMS of light curves before (left) and after (right) the photometric detrending as a function of magnitude. The solid lines are smoothed averages using a bin width of 0.5 magnitude. The horizontal lines indicate a photometric precision of 1% and 0.3%, respectively.
Figure 7 shows the detrended, final light curves of known variable stars from the AAVSO International Variable Star Index (VSX: Watson et al. 2017). We identify three out of five known variable stars in the N02007-OC field; two of them are W Ursae Majoristype eclipsing variables (EW) with periods shorter than 0.5 days (CRTS J002358.0+052711 and CRTS J002328.2+040635). The other one is a typical abtype RR Lyrae star, [MRS2008] 006.577957+03.925690, with heliocentric distance ~15.84 kpc. The photometric quality of folded light curves is good enough to conduct variability analysis. One star shown in Figure 7, CRTS J002358.0+052711, is close to the saturation limit (≈14–15 mag depending on seeing conditions) in the N02007-OC field causing the relatively large scatter of the light curve over all phase bins. Two of the unmatched variables (ASAS J002137+0513.8 and CD Psc) are bright (<12 mag) and saturated in our observations.
Final light curves of three known variable stars folded by their period. Both variable types and periods are noted above each phased light curve.
### 4.2. Periodogram Analysis for Variability Search
One explicit advantage of longitudinal network observations with the KMTNet facility is to alleviate aliasing signals due to the common occurrence of data gaps in unevenly spaced time-series data. Figure 8 compares the results of period analysis by various periodogram tools (e.g., VARTOOLS: Hartman & Bakos 2016 and MS Period: Shin & Byun 2004) for the CTIO sample only and for the combined data from three telescopes. The spectral peaks in the periodogram are severely affected by daily aliasing signals for the CTIO sample, but combining the time series data from the three sites greatly reduces the false peaks due to aliasing.
Period finding results by various periodogram tools for the CTIO sample only (left) and for the combined sample from three telescopes (right). Each periodogram power is normalized by the peak amplitude for comparison purpose. The arrows indicate a true period of 0.25288 days for one known EW variable star shown in Figure 7.
We mainly use two different algorithms implemented in the VARTOOLS light curve analysis program to search for periodic variables with periods less than 40 days at an initial frequency resolution of 0.1/T (T is the time-span of the light curve): a Generalized Lomb-Scargle (LS) algorithm and an Analysis of Variance (AoV) algorithm using either phase binning or multi-harmonic model fitting (see Hartman & Bakos 2016 and references therein). To reduce erroneous results caused by outliers, we applied a typical sigma clipping (4-sigma) to each light curve before searching for semi-sinusoidal signals. Each algorithm gives a diagnostic parameter to test the significance of candidate periods. We choose different conservative selection cuts based on a false alarm probability (FAP) by visual inspection of example light curves at a range of values: e.g., LS FAP < -60, AoV FAP < -100, and AoVharm FAP < -140, respectively. Moreover, we remove the most obvious spurious detections having a value of peak frequency very close to 1 or 2 c/d (mostly around ~0.99 and ~2.02 days) due to the daily gaps even in the combined sample (see Table 1 for observation dates). Figure 9 shows the histograms of formal FAP values and initial periods from the LS period-search algorithm as an example after removing aliases. With these criteria, we check all the phase-folded light curves by eye and refined the initial estimate of the periods using fine grid searches (<0.01/T) around the highest peaks. Table 3 lists 21 new periodic variable stars in the period range of 0.1– 31 days. From most stars, the difference, ∆P, in the periods computed by using the LS and AoV algorithms is less than 0.5%. Table 2 summarizes the steps we used to identify a sample of periodic variable stars from our database for the N02007-OC field.
Periodic variable selection criteria
21 New Periodic Variable Stars in the N02007-OC field
Histograms of the FAP values (top) and initial periods (bottom) from the LS algorithm after filtering alias signals. Our conservative selection criterion is LS FAP < -60 indicated by dashed line. Example phased light curves (magnitude versus phase) are also shown for guiding purpose.
Figure 10 shows the locations of newly discovered variable stars in a color-color diagram, as well as all other stars with magnitude error less than 0.2 in gri bands. The effect of SDSS color errors induced by variability is negligible for our variable stars (error bars are smaller than the symbols). Most of them are close to the locus of stellar main sequence described by the SDSS gri colors (Covey et al. 2007, see dashed line in the same figure). We have only limited information to characterize their variability nature as their spectra are currently not available yet. However, the lightcurve properties and source colors suggest that these are mainly due to spot-induced rotational modulation. One color outlier that lies far outside the locus is a candidate white dwarf/M dwarf pair (WDMD: SDSS J002701.84+040710.5; V19), which is a likely short-period binary showing reflection from a close companion at the orbital period. The observed color is also consistent with those expected from genuine WDMD binaries (e.g., see Figure 2 of Rebassa-Mansergas et al. 2013). As shown in Figure 10, two of them show eclipsing binary (EB) signatures in their phase-folded light curves. The epochs of the primary and secondary eclipses are clearly seen over more than one complete cycle. SDSS J002201.40+053612.5 (V7) is identified as semi-detached EB with a 0.3808 d orbital period. SDSS J002247.39+045147.8 (V10) is an Algoltype EB with an orbital period of 2.953 days, having the detached component. We also show phase-folded light curves of all other discovered variables in Figure 11. SDSS J002215.91+055338.9 (V8) is a overcontact (OC) binary which is recognized by a continuous variation between eclipses such as W UMa-type systems. Lastly, SDSS J002051.80+042641.3 (V3) may be an ellipsoidal variable (ELV) showing double maxima and minima per orbital period due to tidal distortion.
Location of 21 new periodic variable stars (black dots) and all other stars (gray dots) in color-color diagram. The dashed line indicates the main-sequence stellar locus with solar metallicity (Covey et al. 2007). The locations of two eclipsing binaries and one WDMD candidate are indicated by circles and square with overplotted black dots, respectively. The bottom panels show their phase-folded light curves.
Phase-folded light curves of other 18 periodic variable stars. The small error bars indicate the interquartile range of the measurement errors for each light curve.
### 4.3. Recovery of Moving Objects
Another application is to retrieve trajectories of targeted asteroids and their light curves in the catalog of transient sources, i.e., those detected only once. It is not difficult to recover orbital paths of known asteroids on every frame by comparing with an ephemeris from the Minor Planet Center (MPC)5, the official organization that is responsible for the identification, designation and orbit computation for all types of moving objects (minor planets, comets and outer irregular natural satellites of the major planets). The names of the two objects we observed in this field are 2006 DZ169 and 1996 SK. The former asteroid was considered as a potential target for space missions due to its low rendezvous velocity (e.g., Mueller et al. 2011), while the latter is a potentially hazardous NEA (e.g., Lin et al. 2014). Their physical and dynamical properties are well summarized in the NEAs database, updated continuously by European Asteroid Research Node. There is no ambiguity in their measured rotation periods with a reliability code of 3 (i.e., secure result with full lightcurve coverage) based on the definition of Lagerkvist et al. (1989). Therefore, our observations allow us to confirm the results of variability analysis reported in previous works6.
We compute celestial coordinates of these NEAs at a given epoch based on the orbital data provided by the MPC database, and then we cross-identify single-epoch sources in the transient catalog with those precomputed positions. In this way, we recover 98% of the spatial locations of the NEAs after excluding flagged data. The coordinate difference between the MPC and our astrometric solution is less than ±1–2 arcsec in both RA and Dec direction. Figure 12 shows the projected path of each orbit (solid lines) over the observing span, as well as the single-epoch sources that are detected in this region of the sky. In order to highlight the possibility of searching for untargeted moving objects with moving speed similar to targeted ones, we only include all known moving objects within a given FoV (gray linked steaks or points) identified by the Virtual Observatory SkyBoT tool (Berthier et al. 2006).
Projected orbital paths (left) and light curves (right) of 2006 DZ169 and 1996 SK. Left: The background gray dots are single-epoch sources in the transient catalog, showing only known moving objects listed in the VO SkyBoT database. Right: The arrows indicate a subset of light curves zoomed in the inserted plots which show rapid (less than a few hours) brightness variations due to rotation.
Lastly, we present light curves of the two targeted NEAs which are representative in quality for other asteroids in the transient catalog because our experimental data were obtained in conditions optimal to measure their periods (see Figure 12). These light curves are a superposition of two components with different timescales; a long-term change in brightness with increasing or decreasing solar phase angle and a rapid periodic modulations in brightness due to rotation of the asteroid (see zoomed-in views of both light curves). After removing the long-term light variation and outliers, we measure rotation periods and full amplitudes for the NEAs using the periodogram tools. Their rotation periods are found to be similar to those in the NEAs database. The rotation period of 1996 SK is about 4.644 hours (Pnew) with a highly reliable result with full light curve coverage (c.f.,Pknown=4.645 hrs), while that of 2006 DZ169 is about Pknown=4.682 hours that is not clearly seen in the data. 1996 SK has a large amplitude of ~0.5 mag in R-band, while 2006 DZ169 exhibits a relatively low amplitude variation (~0.12 mag).
## 5. SUMMARY
We have described in detail the main algorithms of our new reduction pipeline to explore the temporal and spatial variability with DEEP-South observations. Our multi-aperture photometry technique produces a homogeneous set of photometric measurements for all point sources in non-crowded fields observed by a distributed network of three different telescopes. This is important as it will allow us to study the variability properties of targeted or untargeted objects in a large database of photometric time series without extra computational efforts. Taking into account spatial dependence of PSF variations, zeropoint variations, and systematic effects, we find the RMS scatter of the light curves for point sources to be reached down to 0.3% level at the bright end (R ≤ 16) and ~10% level at the faint end (R ~ 22) in the longest observing records of the DEEP-South year-one data. We also emphasize the use of the indexed database tools, such as FastBit, in both minimizing required storage space and fast query performance to produce large sets of light curves. In the future, we will apply this updated pipeline to process entire imaging data taken by DEEP-South survey to further explore stellar variability.
As applications of the photometric database, we first presented light curves of known variable stars to illustrate the overall quality of the photometric calibration. We find 21 new periodic variable stars with period between 0.1 and 31 days, including four EBs and one WDMD candidate which are evident by either (i) the shape of phase-folded light curve and/or (ii) colors to figure out variability classes. Since we limit the samples to sources only with the SDSS colors, we could miss either non-periodic variables or variable candidates without SDSS colors. To fully investigate properties of all remaining sources, we will attempt to use an infinite gaussian mixture model for detecting variable objects and suppressing false positives efficiently (e.g., Shin et al. 2009, 2012).
Additionally, we show the potential of database applications to retrieve the projected orbital paths and light curves of two targeted NEAs (2006 DZ169 and 1996 SK) in the experimental data. We expect to see more known asteroids with moving speed similar to targeted ones, and we will present results of exploring this possibility in the second paper of the series. All light curves of variable objects in this experiment can be accessed through the web site: http://stardb.yonsei.ac.kr/.
## Acknowledgments
We thank two anonymous referees for their constructive comments that improved this paper. This research was supported by the Korea Astronomy and Space Science Institute (KASI) under the R&D program (Project No.2015-1-320-18) supervised by the Ministry of Science, ICT and Future Planning. S.-W. C. acknowledges the support from KASI – Yonsei research collaboration program for the frontiers of astronomy and space science (2016-1-843-00). Parts of this research also were conducted by the Australian Research Council Centre of Excellence for All-sky Astrophysics (CAASTRO), through project number CE110001020. This research has made use of the KMTNet system operated by the KASI and the data were obtained at three host sites of CTIO in Chile, SAAO in South Africa, and SSO in Australia.
Notes
1 KMTNet (http://kmtnet.kasi.re.kr) Astrometric Calibration
3 SDSS (http://www.sdss.org) Photometric Transformations
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## APPENDIX A. DATABASE SYSTEM
The key technology of the FastBit is a bitmap-based indexing scheme that stores a list of row identifiers for each value of attribute column as sequences of bits (i.e., 0 or 1). It also reduces the size of index file significantly. Due to query-intensive nature of our experiments, this kind of indexing helps improve query performance. Liu et al. (2014) reported good performance in index creation, query and storage space of this algorithm by comparing with a typical relational database (e.g., MySQL or PostgreSQL). The reader is referred to Liu et al. (2014) for details of quantitative comparison between both database systems using high-resolution solar observations.
We check query response time for two cases: (i) a single one-thread application for different query options and (ii) multi-thread application for multiple query processing. In the former case, we compare performance of counting the number of hits returned without any options to that with only SELECT clause and that with an output option, respectively. For sizes up to 100 million rows, typical queries take less than one second on an Intel Xeon Processor with 2.66GHz clock speed (see top panel of Figure 4). FastBit database operation is limited by I/O performance that is dominated by data-accesses to disk than computations performed by the CPUs. In the multi-thread case, we made either 1000 or 10000 queries to be contained in two files where each SELECT statement is randomly defined. We can see improvement in speed by creating multiple query processes and executing these multiple queries in parallel, as shown in the bottom of Figure 13. We briefly present an example application in an astronomical context; therefore, we request the reader refer to Chou et al. (2011) for more details about FastBit-based parallel query processing.
Assessment of query execution performance using the FastBit database system for both single-thread (top) and multi-thread configurations (bottom). The dashed lines indicate the slope of perfect scaling with respect to single-node performance for reference.
We use command-line tools provided by FastBit for converting data format and building indexes. We hope this will make it easier for users of their own data sets to find relevant content. The following example command converts frame catalogs to the metadata tables in a raw binary form:
ardea -d DeepSouth-DB (directory-to-write-data)
-m "LOCAL_ID:uint, XWIN_IMAGE:double,
YWIN_IMAGE:double, XWIN_RA:double,
YWIN_DEC:double, MJD:double,
MAG_MAP:double, MAGERR_MAP:double,
FWHM_IMAGE:float, ELONGATION:float,
CLASS_STAR:float, FLAGS:short,
CALDEX:int, AMPS:int"
-v 5 (verbose_level)
-b ‘’(break/delimiters-in-text-data).
The below example command builds new indexes with basic bitmap options for binning, encoding, and compressing processes:
ibis -d DeepSouth-DB (target database)
-b "<binning nbins=B/><encoding equality/>"
-z (append-existing-indexes)
-v 5 (verbose_level).
We choose the binning option to reduce the number of bitmaps for attributes with (very) high cardinalities. The strategy of binning was discussed in detail in Section 2.4 of Wu et al. (2009), where it is shown that binning can improve the query response time based on order-preserving bin-based clustering. The maximum size of the index is primarily determined by three parameters: the number of rows N, the number of bins B, and the bitmap encoding. Under the equality encoding condition, our test database contains 15,629,598 rows each with 17 attribute columns, and its indexed size is about 2.4 gigabytes for small B (< 100).
Unfortunately, the use of FastBit scheme has intrinsic limitations as it is just a stand-alone data processing tool. It is not a database management system, so most SQL commands are not supported. It imposes a limit on the number of rows that can be stored in indexed tables (no more than 2 billion rows). Because it is also not well-optimized to run parallel processing in a cloud-like environment, we are now testing open source database systems, such as Redis7 and GeoMesa8, to overcome current difficulties.
### Figure 1.
Number of epochs per target field in the DEEP-South year-one data obtained from the three KMTNet telescopes. The values are color-coded by the bar on the right (from 1 to 510 visits). The solid line shows the Galactic plane in equatorial coordinates using the Aitoff projection. The location of the selected field, N02007-OC, is recognized by its darkest color in the center.
### Figure 2.
Position- and aperture-dependent magnitude offsets (gray points) and that after the correction by a two-dimensional surface fitting (orange points) for an example frame. The dispersion of the magnitude offsets is much larger at smaller aperture sizes (top) than larger ones (bottom). The color maps show magnitude offsets in two-dimensional image frame.
### Figure 3.
Example of the spatial distribution of internal calibrators for good (gray points) and bad (red points) cases together with the calculated probability PKS. In the top panel, each frame location is indicated by frame numbers.
### Figure 4.
Histogram of C0 coefficients for all observed frames. The arrow indicates the range of photometric data quality observed under either perfectly (C0 ∼ 0) or partially nonphotometric conditions (-2 < C0 < -1). All insert images are presented as an example of different quality cuts.
### Figure 5.
Transformation between the DEEP-South R and SDSS r magnitudes as a function of r − i. The light points are stars with low photometric error (less than 0.1 magnitudes) in griz bands. Top: the solid line shows the best fitting with Equation (2); Bottom: corresponding difference between transformed DEEP-South and SDSS magnitudes as a function of the SDSS r magnitude.
### Figure 6.
Robust RMS of light curves before (left) and after (right) the photometric detrending as a function of magnitude. The solid lines are smoothed averages using a bin width of 0.5 magnitude. The horizontal lines indicate a photometric precision of 1% and 0.3%, respectively.
### Figure 7.
Final light curves of three known variable stars folded by their period. Both variable types and periods are noted above each phased light curve.
### Figure 8.
Period finding results by various periodogram tools for the CTIO sample only (left) and for the combined sample from three telescopes (right). Each periodogram power is normalized by the peak amplitude for comparison purpose. The arrows indicate a true period of 0.25288 days for one known EW variable star shown in Figure 7.
### Figure 9.
Histograms of the FAP values (top) and initial periods (bottom) from the LS algorithm after filtering alias signals. Our conservative selection criterion is LS FAP < -60 indicated by dashed line. Example phased light curves (magnitude versus phase) are also shown for guiding purpose.
### Figure 10.
Location of 21 new periodic variable stars (black dots) and all other stars (gray dots) in color-color diagram. The dashed line indicates the main-sequence stellar locus with solar metallicity (Covey et al. 2007). The locations of two eclipsing binaries and one WDMD candidate are indicated by circles and square with overplotted black dots, respectively. The bottom panels show their phase-folded light curves.
### Figure 11.
Phase-folded light curves of other 18 periodic variable stars. The small error bars indicate the interquartile range of the measurement errors for each light curve.
### Figure 12.
Projected orbital paths (left) and light curves (right) of 2006 DZ169 and 1996 SK. Left: The background gray dots are single-epoch sources in the transient catalog, showing only known moving objects listed in the VO SkyBoT database. Right: The arrows indicate a subset of light curves zoomed in the inserted plots which show rapid (less than a few hours) brightness variations due to rotation.
### Figure 13.
Assessment of query execution performance using the FastBit database system for both single-thread (top) and multi-thread configurations (bottom). The dashed lines indicate the slope of perfect scaling with respect to single-node performance for reference.
### Table 1
Summary of R-band time-series observations in the N02007-OC field
Site Date FWHMa
(arcsec)
Airmass # Mosaic
Pointings
a Median over images on that date.
CTIO 2015-08-14 2.35±0.28 1.23–1.37 13
2015-08-16 1.37±0.17 1.22–1.31 15
2015-08-18 1.56±0.25 1.22–1.48 33
2015-08-22 3.52±0.44 1.22–1.55 35
2015-08-24 1.30±0.24 1.22–1.61 28
2015-11-01 1.38±0.16 1.22–1.47 26
2015-11-02 1.92±0.30 1.22–1.53 26
2015-11-03 1.47±0.24 1.22–1.62 25
2015-11-05 1.54±0.20 1.22–1.59 23
SAAO 2015-08-08 2.05±0.42 1.26–1.49 26
2015-08-10 1.28±0.44 1.26–1.49 32
2015-08-12 2.72±0.21 1.28–1.38 9
2015-08-14 2.68±0.33 1.26–1.39 23
2015-08-24 2.91±0.38 1.26–1.78 31
2015-11-09 2.79±0.49 1.35–1.58 12
SSO 2015-08-08 1.97±0.26 1.24–1.39 26
2015-08-10 2.38±0.33 1.24–1.46 29
2015-08-18 1.60±0.20 1.24–1.58 35
2015-08-20 1.40±0.26 1.24–1.61 32
2015-08-22 2.23±0.22 1.24–1.36 17
2015-11-07 3.02±0.52 1.31–1.62 14
Total 510
### Table 2
Periodic variable selection criteria
Selection criterion Number of objects
a Number of individual photometric measurements.
All 2.5-σ sources 15,629,598a
All possible groups (<0.5 arcsec) 46,054
SDSS stars with gri colors 13,261
Variable candidates by FAP cuts 309
Rejection of aliased signals 57
Final periodic variables 24
### Table 3
21 New Periodic Variable Stars in the N02007-OC field
VarID StarID
(J2000 coordinates)
R
(mag)
P
(mag)
AaR
(mag)
u
(mag)
g
(mag)
r
(mag)
i
(mag)
z
(mag)
Note
a Peak-to-through R-band variability amplitude of sinusoidal function fitted to phase-folded light curve.
V1 SDSS J002013.16+052242.2 16.24 2.1038 0.06 19.66 17.35 16.21 15.71 15.35
V2 SDSS J002031.74+055112.6 16.33 1.1568 0.04 20.26 17.72 16.38 15.78 15.43
V3 SDSS J002051.80+042641.3 17.50 0.3076 0.13 19.59 18.10 17.48 17.25 17.15 ELV
V4 SDSS J002111.05+054856.4 18.60 1.5046 0.04 22.12 20.04 18.61 17.23 16.46
V5 SDSS J002125.21+040357.0 16.79 31.5771 0.03 21.12 18.20 16.80 16.05 15.62
V6 SDSS J002141.37+051626.1 16.76 22.4097 0.01 20.83 18.14 16.71 15.89 15.47
V7 SDSS J002201.40+053612.5 18.35 0.3808 0.44 21.18 19.28 18.41 17.93 17.60 EB
V8 SDSS J002215.91+055338.9 20.19 0.2489 0.32 21.56 20.68 20.26 20.19 20.09 OC
V9 SDSS J002216.31+040439.5 19.59 23.4399 0.12 23.36 21.13 19.69 18.01 17.05
V10 SDSS J002247.39+045147.8 18.11 2.9532 >0.49 19.19 18.04 18.09 18.17 18.27 EB
V11 SDSS J002249.03+055356.6 18.33 16.0571 0.04 22.39 19.82 18.34 17.33 16.79
V12 SDSS J002314.27+042543.6 19.00 0.2673 0.16 23.05 20.33 18.97 17.58 16.75
V13 SDSS J002351.64+054553.7 15.89 21.7415 0.02 19.84 17.10 15.88 15.41 15.12
V14 SDSS J002425.52+044130.3 17.95 1.6484 0.10 20.22 18.64 17.89 17.54 17.38
V15 SDSS J002435.70+051914.1 16.64 7.5176 0.03 19.04 17.30 16.59 16.30 16.13
V16 SDSS J002516.12+053728.4 15.09 11.9412 0.02 18.82 16.14 15.07 14.68 14.46
V17 SDSS J002549.95+041608.6 20.60 0.1234 0.10 23.56 21.98 20.54 19.66 19.13
V18 SDSS J002610.07+043147.2 18.15 2.8856 0.02 22.81 19.64 18.12 16.73 15.95
V19 SDSS J002701.84+040710.5 17.69 0.3570 0.03 18.46 18.02 17.66 16.78 16.15 WDMD
V20 SDSS J002704.93+040519.4 15.32 20.4821 0.02 18.45 16.20 15.38 15.08 14.89
V21 SDSS J002733.09+043701.9 17.70 3.1778 0.04 21.41 19.15 17.66 16.57 15.99
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2019-09-17 01:28:04
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https://gmatclub.com/forum/six-congruent-circles-are-packed-into-an-equilateral-triangle-so-that-261037.html
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Six congruent circles are packed into an equilateral triangle so that
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Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?
(1) The radius of each circle is 2
(2) The area of the triangle is $$48+28\sqrt{3}$$
[Reveal] Spoiler:
Attachment:
six_circles_packed_1.png [ 16.87 KiB | Viewed 624 times ]
[Reveal] Spoiler: OA
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Re: Six congruent circles are packed into an equilateral triangle so that [#permalink]
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08 Mar 2018, 21:22
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Bunuel wrote:
Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?
(1) The radius of each circle is 2
(2) The area of the triangle is $$48+28\sqrt{3}$$
[Reveal] Spoiler:
Attachment:
six_circles_packed_1.png
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Six congruent circles are packed into an equilateral triangle so that [#permalink]
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09 Mar 2018, 01:07
Bunuel wrote:
Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?
(1) The radius of each circle is 2
(2) The area of the triangle is $$48+28\sqrt{3}$$
[Reveal] Spoiler:
Attachment:
The attachment six_circles_packed_1.png is no longer available
Question: Area of uncovered part of triangle = Area of triangle - Area of six circles = ?
Statement 1: The radius of each circle is 2
Using this information the side of the equilateral triangle can be calculated (using 30-60-90 property) as mentioned in attachment, Hence
SUFFICIENT
Statement 2: he area of the triangle is $$48+28\sqrt{3}$$
sing this information the side of the equilateral triangle can be calculated as mentioned in attachment, Hence
SUFFICIENT
Answer: option D
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Re: Six congruent circles are packed into an equilateral triangle so that [#permalink]
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10 Mar 2018, 20:42
1
This post received
KUDOS
There is a simpler approach to it, by trying to draw it on paper.
see the Sketch attached.
GMATinsight wrote:
Bunuel wrote:
Six congruent circles are packed into an equilateral triangle so that no circle is overlapping and such that circles are tangent to one another or the triangle at any point of contact, as shown above. What is the area of the part of the triangle that is NOT covered by circles?
(1) The radius of each circle is 2
(2) The area of the triangle is $$48+28\sqrt{3}$$
[Reveal] Spoiler:
Attachment:
The attachment six_circles_packed_1.png is no longer available
Question: Area of uncovered part of triangle = Area of triangle - Area of six circles = ?
Statement 1: The radius of each circle is 2
Using this information the side of the equilateral triangle can be calculated (using 30-60-90 property) as mentioned in attachment, Hence
SUFFICIENT
Statement 2: he area of the triangle is $$48+28\sqrt{3}$$
sing this information the side of the equilateral triangle can be calculated as mentioned in attachment, Hence
SUFFICIENT
Answer: option D
Attachments
WhatsApp Image 2018-03-11 at 09.09.28.jpeg [ 99.73 KiB | Viewed 271 times ]
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Re: Six congruent circles are packed into an equilateral triangle so that [#permalink] 10 Mar 2018, 20:42
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2018-03-18 21:32:13
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https://mathoverflow.net/questions/209933/generalization-of-sprague-grundy-theorem
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# Generalization of Sprague-Grundy Theorem
In my research on Combinatorial Game Theory, I used a certain theorem that is essentially a generalization of the Sprague-Grundy theorem. Because the result hinges too much on the work of others to be truly considered my own, I tried to find a source for citing it. However, I cannot find any source to cite it from.
Define the disjunctive $k$-sum of $n$ games for $n \geq k$ to be the game in which the $n$ games are played in parallel, with each player being allowed to move in no more than $k$ of the games per turn. We can determine the Sprague-Grundy value of a position in the following way:
1. Find the Sprague-Grundy values of each component game.
2. Write each value in base $k + 1$
3. Now add the values without carrying.
The resultant value is the Sprague-Grundy value of the cumulative position.
This theorem is just Sprague-Grundy using Moore's Nim rather than normal Nim. So how should I cite this? As I said previously, I am hesitant to claim the result because it relies so heavily on Moore's Nim. On the other hand, I have not found it anywhere.
• I didn't notice the difference between Moore's results and this generalization until you pointed it out. I'd just cite both Moore and Sprague-Grundy as I think it's an immediate consequence of these two theorems. – Douglas Zare Jun 23 '15 at 5:44
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2019-04-24 14:55:23
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https://tex.stackexchange.com/questions/224946/change-the-height-of-minipage-in-the-figure-environment
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# Change the height of minipage in the figure environment?
How can I change the height of the minipage in the figure envirment? Since I tried it so \includegraphics[width=.6, height=0.8\linewidth]{img/moovit_4} and I got this error ! Illegal unit of measure (pt inserted). How can I put code here? it is not like stackoverflow, I can not see the icon right to the image icon in the header. I used these quotation marks
\begin{figure}[H]
\centering
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[width=.5\linewidth]{img/moovit_2}
\captionof{figure}{A figure}
\label{fig:test1}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\includegraphics[width=.6, height=.8\linewidth]{img/moovit_4}
\captionof{figure}{Another figure}
\label{fig:test2}
\end{minipage}
\end{figure}
• You have to pass a unity to .6. For example .6\linewidth or 3cm. – Sigur Jan 25 '15 at 23:57
• also don't use both width and height or you will distort the figure, and did you intend height=.8\linewidth ie setting the height of the image based on the available width? – David Carlisle Jan 26 '15 at 0:03
As @sigur has already pointed out in a comment, you need to adjust the way you specify the width of the second image. You may also want to align the minipages in a way that both captions are at the same level; this can be done, e.g., by setting the location specifier [b] for the two minipage environments.
A separate comment: Since the two minipages occur inside a float of type figure, it's not necessary to write \captionof{figure}{...}; the simpler \caption{...} will do too. Finally, the first \centering statement (right after \begin{figure}[H]) is redundant since the two minipages occupy the full width of the textblock -- nothing there to center, really.
\documentclass{article}
\usepackage[demo]{graphicx} % omit 'demo' option in real document
\usepackage{caption}
\usepackage{float} % for "H" location specifier
\begin{document}
\begin{figure}[H]
%%\centering %% redundant
\begin{minipage}[b]{.5\textwidth}
\centering
\includegraphics[width=.5\linewidth]{img/moovit_2}
\caption{A figure}
\label{fig:test1}
\end{minipage}%
\begin{minipage}[b]{.5\textwidth}
\centering
\includegraphics[width=.6\linewidth, height=.8\linewidth]{img/moovit_4}
\caption{Another figure}
\label{fig:test2}
\end{minipage}
\end{figure}
\end{document}
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2021-06-21 17:26:03
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https://en.wikipedia.org/wiki/Electron_degeneracy_pressure
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# Electron degeneracy pressure
Electron degeneracy pressure is a particular manifestation of the more general phenomenon of quantum degeneracy pressure. The Pauli exclusion principle disallows two identical half-integer spin particles (electrons and all other fermions) from simultaneously occupying the same quantum state. The result is an emergent pressure against compression of matter into smaller volumes of space. Electron degeneracy pressure results from the same underlying mechanism that defines the electron orbital structure of elemental matter. For bulk matter with no net electric charge, the attraction between electrons and nuclei exceeds (at any scale) the mutual repulsion of electrons plus the mutual repulsion of nuclei; so absent electron degeneracy pressure, the matter would collapse into a single nucleus. In 1967, Freeman Dyson showed that solid matter is stabilized by quantum degeneracy pressure rather than electrostatic repulsion.[1][2][3] Because of this, electron degeneracy creates a barrier to the gravitational collapse of dying stars and is responsible for the formation of white dwarfs.
## From the Fermi gas theory
Pressure vs temperature curves of classical and quantum ideal gases (Fermi gas, Bose gas) in three dimensions. Pauli repulsion in fermions (such as electrons) gives them an additional pressure over an equivalent classical gas, most significantly at low temperature.
Electrons are part of a family of particles known as fermions. Fermions, like the proton or the neutron, follow Pauli's principle and Fermi–Dirac statistics. In general, for an ensemble of non-interacting fermions, also known as Fermi gas, each particle can be treated independently with a single-fermion energy given by the purely kinetic term,
${\displaystyle E={\frac {p^{2}}{2m}},}$
where p is the momentum of one particle and m its mass. Every possible momentum state of an electron within this volume up to the Fermi momentum pF being occupied.
The degeneracy pressure at zero temperature can be computed as[4]
${\displaystyle P={\frac {2}{3}}{\frac {E_{\text{tot}}}{V}}={\frac {2}{3}}{\frac {p_{\rm {F}}^{5}}{10\pi ^{2}m\hbar ^{3}}},}$
where V is the total volume of the system and Etot is the total energy of the ensemble. Specifically for the electron degeneracy pressure, m is substituted by the electron mass me and the Fermi momentum is obtained from the Fermi energy, so the electron degeneracy pressure is given by
${\displaystyle P_{e}={\frac {(3\pi ^{2})^{2/3}\hbar ^{2}}{5m_{e}}}\rho _{e}^{5/3}}$,
where ρe is the free electron density (the number of free electrons per unit volume). For the case of a metal, one can prove that this equation remains approximately true for temperatures lower than the Fermi temperature, about 106 kelvin.
When particle energies reach relativistic levels, a modified formula is required. The relativistic degeneracy pressure is proportional to ρe4/3.
## Examples
### Metals
For the case of electrons in crystalline solid, several approximations are carefully justified to treat the electrons as independent particles. Usual models are the free electron model and the nearly free electron model. In the appropriate systems, the electron degeneracy pressure can be calculated and can be shown that this pressure is an important contribution to the compressibility or bulk modulus of metals.[5]
### White dwarfs
Electron degeneracy pressure will halt the gravitational collapse of a star if its mass is below the Chandrasekhar limit (1.44 solar masses[6]). This is the pressure that prevents a white dwarf star from collapsing. A star exceeding this limit and without significant thermally generated pressure will continue to collapse to form either a neutron star or black hole, because the degeneracy pressure provided by the electrons is weaker than the inward pull of gravity.
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2021-04-12 20:28:00
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http://physics.stackexchange.com/questions/57936/does-dark-matter-have-more-space-time-or-particle-characteristics
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# Does Dark Matter have more space-time or particle characteristics?
Dark Matter appears to have more in common with phenomena related to spatial geometry then a particle. I thought in General Relativity, space can be curved without the presence of matter so gravitational lensing does not imply there is matter present but that the space in a region is curved. If Dark matter has more characteristics related to spatial geometry, why is it referred to as a kind of exotic particle (WIMP).
-
possible duplicate: physics.stackexchange.com/questions/29459/… – Michael Brown Mar 25 '13 at 3:26
There are 2 popular approaches to this puzzle called the dark matter problem. First is through a modification to general theory of relativity as you had suggested. The second is through postulating presence of an unseen particle. I would suspect an unseen particle is the more likely reason for observed experimental discrepancies like galaxy rotation curves being flatter than expected. – Prathyush Mar 25 '13 at 3:46
But spacetime, according to GR, is constrained by the mass, energy, momentum, pressure, and shear. The equation we write down is $$G_{\mu\nu} = 8\pi T_{\mu\nu},$$ which unfortunately means almost nothing if you don't already know GR, but it just feels so good to write down people like me can't help it sometimes. Basically, the left side encompasses curvature and all that, while the right deals with the "stuff." Inferences about curvature therefore tell us something about the "stuff" there.
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2014-09-30 14:25:39
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https://alxivolnser.firebaseapp.com/351.html
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# Chi square pdf derivation of e
Seven proofs of the pearson chisquared independence test and. Karl pearson chisquare test the dawn of statistical inference. Mathematically, the pdf of the central chisquared distribution with. Derivation of the chisquare distribution a direct relation exists between a chisquaredistributed random variable and a gaussian random variable. The curve approaches, but never quite touches, the horizontal axis. I discuss how the chisquare distribution arises, its pdf, mean, variance, and shape. We will see how to use calculus to determine the values mentioned above of both the maximum value of the chi square distribution, which corresponds to its mode, as well as find the inflection points of the distribution.
I see your point but i need a more mathematicly rigorious derivation im afraid. When used without further qualification, the term usually refers to pearsons chi squared test, which is used to test whether an observed distribution could have arisen from an expected distribution under some assumption, or whether that assumption is likely to be wrong. It often arises in the power analysis of statistical tests in which the null distribution is perhaps asymptotically a chisquare distribution. Mathematically, the pdf of the central chi squared distribution with.
Proof let the random variable x have the chisquare distribution with n degrees of. Derivation of the pdf for one degree of freedomedit. However, in a distributional modeling context as with other probability distributions, the chisquare distribution itself can be transformed with a location parameter. Draw a careful sketch of the chisquare probability density function in each of the following cases. How to derive the density of the square of a standard normal and chi squared density from the gamma density. Aug, 2018 mathematical statistics uses techniques from various branches of math to prove definitively that statements regarding statistics are true. I wanted to know what the proof for the variance term in a central chisquared distribution degree n is. This pdf is called a chisquare pdf with n degrees of freedom.
I know that the answer is 2n, but i was wondering how to derive it. Chisquared test of independence minhaz fahim zibran department of computer science university of calgary, alberta, canada. Cochran theorem the second proof relies on the cochran theorem. Ratio of two normal random variables if x1 and x2 are independent and both have the normal distribution n0. Derivation of the chi square distribution a direct relation exists between a chi square distributed random variable and a gaussian random variable. Proof of variance formula for central chisquared distribution. After some simplification, you will arrive at the desired distribution. The chi square random variable is in a certain form a transformation of the gaussian random variable. The chi square test is used in data consist of people distributed across categories, and to know whether that distribution is different from what would expect by chance. The end lets derive chi squared pdf from normal distribution intuitively is published by aerin kim. We also revisited the proof done by pearson in 1900 and show that this proof.
The standard normal and the chisquare stat 414 415. The chi square test is used to make a judgment whether a laboratory method is capable of detection of gross alpha and beta radioactivity in drinking water for regulatory monitoring to protect health of population. Distributions derived from normal random variables 2, t, and f distributions statistics from normal samples. The gamma distribution is useful in modeling skewed distributions for variables that are not. E j2 e j, where o j is the observed count in cell j and e j is the estimate of the expected count under the null hypothesis. I discuss how the chi square distribution arises, its pdf, mean, variance, and shape. Distributions related to the normal distribution three important distributions. The characteristics of chi square variate are also. Show that the chisquare distribution with n degrees of freedom has probability density function fx 1 2n2. The two most common instances are tests of goodness of fit using multinomial tables and tests of independence in contingency tables. The new derivations are compared with the established derivations, such as by convolution, moment generating function, and. Because chi square distributions are a type of gamma distribution, and variances are found by squaring deviations from the mean, it follows that a function of the sample variance will have a chi square. To define the chisquare distribution one has to first introduce the gamma. Theorem an exponential random variable with parameter.
How can i obtain the pdf of the logarithm of a chisquared. An introduction to the chisquare distribution youtube. The chi square distribution is a special case of the gamma distribution and is one of the most widely used probability. The chi squared test refers to a class of statistical tests in which the sampling distribution is a chi square distribution. Sums of chi square random variables printerfriendly version well now turn our attention towards applying the theorem and corollary of the previous page to the case in which we have a function involving a sum of independent chi square random variables.
Lets derive chisquared pdf from normal distribution. In probability theory and statistics, the chisquare distribution also chisquared or. Because chisquare distributions are a type of gamma distribution, and variances are found by squaring deviations from the mean, it follows that a function of the sample variance will have a chisquare. The curve reaches a peak to the right of 0, and then gradually declines in height, the larger the. The chisquare test is used in data consist of people distributed across categories, and to know whether that distribution is different from what would expect by chance. Chi square is one of the most useful nonparametric statistics. Chi square formula with solved solved examples and explanation. The chisquare or \\chi2\ distribution can be described in many ways for example as a special case of the gamma distribution, but it is most intuitively characterized in relation to the standard normal distribution, \n0,1\. Free derivative calculator differentiate functions with all the steps. Mathematical statistics uses techniques from various branches of math to prove definitively that statements regarding statistics are true. The chi square distribution is used in the common chi square tests for goodness of fit of an observed distribution to a theoretical one, the independence of two criteria of classification of qualitative data, and in confidence interval estimation for a population standard deviation of a normal distribution from a sample standard deviation. The chi square or \\ chi 2\ distribution can be described in many ways for example as a special case of the gamma distribution, but it is most intuitively characterized in relation to the standard normal distribution, \n0,1\. The importance of the chi square distribution stems from the fact that sums of this kind are encountered very often in statistics, especially in the estimation of variance and in hypothesis testing.
For derivation from more basic principles, see the derivation in. You use this test when you have categorical data for two independent variables, and you want to see if there is an association between them. The following two sections cover the most common statistical tests that make use of the chi square. Since each chisquare distribution has degrees of freedom as a parameter, the f distribution will have two parameters. Mar 18, 2020 this video lecture gives a detailed information about how the pdf of a chi square variate is derived by using the pdf of standard normal variate. In few words, replace x in your chisquare distribution with expyab and multiple the result with 1a expya. We will see how to use calculus to determine the values mentioned above of both the maximum value of the chisquare distribution, which corresponds to its mode, as well as find the inflection points of the distribution.
Derivation of chisquared pdf with one degree of freedom from normal distribution pdf. Since sample variances have chisquare distributions, the f distribution is therefore related to the ratio of two chisquare distributions. In probability theory and statistics, the chisquare distribution with k degrees of freedom is the. Proofs related to chisquared distribution wikipedia. As the following theorems illustrate, the moment generating function, mean and variance of the chisquare distributions are just straightforward extensions of those for the gamma distributions. When used without further qualification, the term usually refers to pearsons chisquared test, which is used to test whether an observed distribution could have arisen from an expected distribution under some assumption, or whether that assumption is. It is heavily used in the analysis of variance anova. In probability theory and statistics, the noncentral chisquare distribution or noncentral chisquared distribution, noncentral distribution is a generalization of the chisquare distribution. The formula for the probability density function of the chisquare distribution is. The chi square formula is used in the chi square test to compare two statistical data sets. The chisquare distribution is a special case of the gamma distribution and is one of the most widely used probability. Mathematically, a squared standard score squared zscore from a normal distribution has a chi square distribution with one degree of freedom. Type in any function derivative to get the solution, steps and graph. Derivation of the pdf for two degrees of freedom of chi squared.
Max and inflection points of chisquare distribution. If x and y are independent multinomialm,p and multinomial n,p. The chisquare random variable is in a certain form a transformation of the gaussian random variable. Seven proofs of the pearson chisquared independence test. Let x be a chisquare random variable with ndegrees of freedom. Derivation of chi squared pdf with one degree of freedom from normal distribution pdf. Lectures in mathematical statistics changed from z to x z22, and the. If p is a positive integer, then applying equation b. The chisquared test refers to a class of statistical tests in which the sampling distribution is a chisquare distribution. Oct 23, 2012 a brief introduction to the chi square distribution.
Exploring the underlying theory of the chisquare test. The first derivation uses the induction method, which requires only a single integral to calculate. Exercises chi square is a distribution that has proven to be particularly useful in statistics. Notes on the chisquared distribution october 19, 2005. Notes on the chi squared distribution october 19, 2005. And one gets the chi squared distribution, noting the property of the gamma function. Distributions derived from normal random variables distributions derived from. The importance of the chisquare distribution stems from the fact that sums of this kind are encountered very often in statistics, especially in the estimation of variance and in hypothesis testing. Compared to previous proofs as for instance in buonocore and pirozzi 2014, we are the rst one to provide seven proofs for this seminal results with the use of a wide range of tools, like not. Oct 17, 2019 in channel modeling, the central chi square distribution is related to rayleigh fading scenario and the noncentral chi square distribution is related to rician fading scenario. We describe two new derivations of the chisquare distribution.
How to derive the density of the square of a standard normal and chisquared density from the gamma density. Equivalently, we may set up the problem as follows. Suppose that a random variable j has a poisson distribution with mean. On the pearsonfisher chisquared tteorem 6735 2 the pearsons proof in this section, we give the essentials of the pearson proof in modern notations, pointing out original formulas, numbers and sentences, in squared brackets, to simplify comparisons with the mentioned pearsons paper. Since each chi square distribution has degrees of freedom as a parameter, the f distribution will have two parameters. Noncentral c2, t, and fdistributions the results on transformation lead to many useful results based on transformations of normal random variables. Show that the chi squared distribution with k degrees of freedom does indeed have a gamma distribution. The chisquare distribution is connected to a number of other special distributions. Show that the chisquared distribution with k degrees of freedom does indeed have a gamma distribution. In channel modeling, the central chisquare distribution is related to rayleigh fading scenario and the noncentral chisquare distribution is related to rician fading scenario. Chisquare test of association between two variables the second type of chi square test we will look at is the pearsons chisquare test of association. In few words, replace x in your chi square distribution with expyab and multiple the result with 1a expya.
The sum of independent chi square random variables. Parameters 100, 1 here mean that we generate a 100. As the following theorems illustrate, the moment generating function, mean and variance of the chi square distributions are just straightforward extensions of those for the gamma distributions. Derivation of chisquared pdf with one degree of freedom from. A case of a failure of the chi square test and its amelioration are described. The end lets derive chisquared pdf from normal distribution intuitively is published by aerin kim. A brief introduction to the chisquare distribution. Since e x i p, the central limit theorem implies p nx n p. In probability theory and statistics, the chi square distribution also chi squared or. Sums of chisquare random variables printerfriendly version well now turn our attention towards applying the theorem and corollary of the previous page to the case in which we have a function involving a sum of independent chisquare random variables. Here is one based on the distribution with 1 degree of freedom. Derivation of the pdf for two degrees of freedom of chi.
From this representation, the noncentral chisquare distribution is seen to be a poissonweighted mixture of central chisquare distributions. Since sample variances have chi square distributions, the f distribution is therefore related to the ratio of two chi square distributions. You use this test when you have categorical data for two independent variables, and you want to see if. There are several methods to derive chi squared distribution with 2 degrees of freedom. This video lecture gives a detailed information about how the pdf of a chisquare variate is derived by using the pdf of standard normal variate. E 2 e for each pair of observed and expected values then sum them all up.
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2022-10-07 15:27:23
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https://calendar.math.illinois.edu/?year=2015&month=03&day=12&interval=day
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Department of
# Mathematics
Seminar Calendar
for events the day of Thursday, March 12, 2015.
.
events for the
events containing
Questions regarding events or the calendar should be directed to Tori Corkery.
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Thursday, March 12, 2015
11:00 am in 241 Altgeld Hall,Thursday, March 12, 2015
#### Exact formulas for the smallest parts function
###### Scott Ahlgren (UIUC Math)
Abstract: Building on ground-breaking work of Hardy and Ramanujan, Rademacher proved an exact formula for the values of the ordinary partition function. More recently, Bruinier and Ono obtained an algebraic formula for these values. We study the smallest parts function. Introduced by Andrews, its generating function is a prototypical example of a mock modular form of weight 3/2. Using automorphic methods, we obtain an exact formula and an algebraic formula for its values. The convergence of the exact formula is not obvious, and requires power savings estimates for sums of Kloosterman sums attached to a multiplier. These are proved with spectral methods following an argument of Goldfeld-Sarnak. (Joint work with Nick Andersen)
1:00 pm in 347 Altgeld Hall,Thursday, March 12, 2015
#### An first exit time problem inspired by modeling of gene expression
###### Jay Newby (Ohio State/MBI)
Abstract: A general class of stochastic gene expression models with self regulation is considered. One or more genes randomly switch between regulatory states, each having a different mRNA transcription rate. The gene or genes are self regulating when the proteins they produce affect the rate of switching between regulatory states. Under weak noise conditions, the deterministic forces are much stronger than fluctuations from gene switching and protein synthesis. Metastable transitions, such as bistable switching, can occur under weak noise conditions, causing dramatic shifts in the expression of a gene. A general tool used to describe metastability is the quasi stationary analysis (QSA). A large deviation principle is derived so that the QSA can explicitly account for random gene switching without using an adiabatic limit or diffusion approximation, which are unreliable and inaccurate for metastable events.This allows the existing asymptotic and numerical methods that have been developed for continuous Markov processes to be used to analyze the full model.
1:00 pm in Altgeld Hall 243,Thursday, March 12, 2015
#### Word Maps and Measure Preservation
###### Doron Puder (IAS Princeton)
Abstract: We establish new characterizations of primitive elements and free factors in free groups, which are based on the distributions they induce on finite groups. More specifically, for every finite group G, a word w in the free group on k generators induces a word map from G^k to G. We say that w is measure preserving with respect to G if given uniform distribution on G^k, the image of this word map distributes uniformly on G. It is easy to see that primitive words (words which belong to some basis of the free group) are measure preserving w.r.t. all finite groups, and several authors have conjectured that the two properties are, in fact, equivalent. In a joint work with O. Parzanchevski, we prove this conjecture.
2:00 pm in 140 Henry Administration Bldg,Thursday, March 12, 2015
#### Modular Equations in Two Variables
###### Dan Schultz (Penn State Math)
Abstract: By adding certain equianharmonic elliptic sigma functions to the coefficients of the Borwein cubic theta functions, a set of two-variable theta functions may be derived. These theta functions invert the F1(13;13;13;1|x,y) case of Appell's hypergeometric function and satisfy several identities akin to those satisfied by the Borwein cubic theta functions. I will discuss how these identities arise, several results concerning modular equations satisfied by these functions, and their applicability to a new two-parameter family of solvable nonic equations.
4:00 pm in 245 Altgeld Hall,Thursday, March 12, 2015
#### Semirandom methods in combinatorics
###### Jacques Verstraete (Univ of California San Diego)
Abstract: The development of the probabilistic method in combinatorics since it's inception by papers of P. Erdos has led to groundbreaking results across a broad mathematical landscape. In this talk, I will survey a technique which has come to be known as the semirandom method, starting with the ideas of V. Rodl. Some of the highlights include applications to combinatorial and projective geometry, and most notably the recent proof by Keevash of the existence of combinatorial designs. The main ideas will be discussed, without delving too far into the technical details, and a number of open problems will be presented.
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2022-05-28 04:06:29
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https://cran.ma.ic.ac.uk/web/packages/jfa/vignettes/v2_bayesian_sampling_workflow.html
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# Workflow: Bayesian audit sampling
## Scenario
This vignette aims to show how the jfa package facilitates auditors in the standard audit sampling workflow (hereafter “audit workflow”). In this example of the audit workflow, we will consider the case of BuildIt. BuildIt is a fictional construction company in the United States that is being audited by Laura, an external auditor for a fictional audit firm. At the end of the year, BuildIt has provided a summary of its financial situation in the financial statements. Laura’s job as an auditor is to formulate an opinion about the fairness and correctness of BuildIt’s financial statements.
Therefore, Laura needs to obtain sufficient and appropriate evidence for the claim that the misstatement in the financial statements is lower than a certain amount: the materiality. If the financial statements contain errors that are considered material, this means that the errors in the financial statements are large enough that they might influence the decision of someone relying on these financial statements. For Laura, the materiality is set at 5% of the total value of the financial statements.
Since BuildIt is a small company, there is only a single population upon which the financial statements are based. Therefore, laura can use the materiality for the overall financial statements as the performance materiality for the single population. Specifically, BuildIt’s population consists of 3500 items. However, before assessing the details in the population, Laura has performed a test of BuildIt internal control systems and found that they were quite reliable.
In order to formulate an opinion about the misstatement in the population, Laura separates her audit workflow into four stages. First, she will plan the size of the subset she needs to inspect from the financial statements to make a well-substantiated inference about them as a whole. Second, she will select the required subset from the financial statements. Third, she will inspect the selected subset and determines the audit (true) value of the items. Fourth, she will use the information from her inspected subset to make an inference about the misstatement in the population (e.g., in this case also the financial statements as a whole). To start her workflow, Laura first loads BuildIt’s financial statements in R.
library(jfa)
data("BuildIt")
## Setting up the audit
Laura wants to make a statement that, with 95% confidence, the misstatement in the financial statements is lower than the performance materiality of 5%. Based on last year’s audit at BuildIt, where the upper bound of the misstatement turned out to be 2.5%, she wants to tolerate at most 2.5% errors in the intended sample. Laura can therefore re-formulate her statistical statement as that she wants to conclude that, when 2.5% errors are found in her sample, she can conclude with 95% confidence, that the misstatement in the population is lower than the materiality of 5%. Below, Laura defines the materiality, confidence level, and expected errors.
# Specify the confidence, materiality, and expected errors.
confidence <- 0.95 # 95%
materiality <- 0.05 # 5%
expected <- 0.025 # 2.5%
Many audits are performed according to the audit risk model (ARM), which determines that the uncertainty about Laura’s statement as a whole (1 - her confidence) is a factor of three terms: the inherent risk, the control risk, and the detection risk. Inherent risk is the risk posed by an error in BuildIt’s financial statement that could be material, before consideration of any related control systems (e.g., computer systems). Control risk is the risk that a material misstatement is not prevented or detected by BuildIt’s internal control systems. Detection risk is the risk that Laura will fail to find material misstatements that exist in an BuildIt’s financial statements. The ARM is practically useful because for a given level of audit risk, the tolerable detection risk bears an inverse relation to the other two risks. The ARM is useful for Laura because it enables her to incorporate prior knowledge on BuildIt’s organization to increase the required risk that she will fail to find material misstatements. According to the ARM, the audit risk will then be retained.
$\text{Audit risk} = \text{Inherent risk} \,\times\, \text{Control risk} \,\times\, \text{Detection risk}$
Usually the auditor judges inherent risk and control risk on a three-point scale consisting of low, medium, and high. Different audit firms handle different standard percentages for these categories. Laura’s firm defines the probabilities of low, medium, and high respectively as 50%, 60%, and 100%. Because Laura performed testing of BuildIt’s computer systems, she assesses the control risk as medium (60%).
# Specify the inherent risk (ir) and control risk (cr).
ir <- 1 # 100%
cr <- 0.6 # 60%
## Stage 1: Planning an audit sample
Laura can choose to either perform a frequentist analysis, where she uses the increased detection risk as her level of uncertainty, or perform a Bayesian analysis, where she captures the information in the control risk in a prior distribution. For this example, we will show how Laura performs a Bayesian analysis. A frequentist analysis can easily be done through the following functions by setting prior = FALSE. In a frequentist audit, Laura immediately starts at step 1 and uses the value adjustedConfidence as her new value for confidence.
# Adjust the required confidence for a frequentist analysis.
c.adj <- 1 - ((1 - confidence) / (ir * cr))
However, in a Bayesian audit, Laura starts at step 0 by defining the prior distribution that corresponds to her assessment of the control risk. She assumes the likelihood for a sample of $$n$$ observations, in which $$k$$ were in error, to be poisson. Using the auditPrior() function, she can create a prior distribution that incorporates the information in the risk assessments from the ARM. For more information on how this is done, see Derks et al. (2019).
# Step 0: Create a prior distribution according to the audit risk model.
prior <- auditPrior(method = "arm", likelihood = "poisson", expected = expected,
materiality = materiality, ir = ir, cr = cr)
Laura can inspect the resulting prior distribution with the summary() function.
summary(prior)
##
## Prior Distribution Summary
##
## Options:
## Likelihood: poisson
## Specifics: ir = 1; cr = 0.6; dr = 0.0833333
##
## Results:
## Functional form: gamma(α = 2.325, β = 53)
## Equivalent sample size: 53
## Equivalent errors: 1.325
## Mode: 0.025
## Mean: 0.043868
## Median: 0.037764
## Variance: 0.0008277
## Skewness: 1.3117
## 95 percent upper bound: 0.099289
## Precision: 0.074289
The prior distribution can be shown by using the plot() function.
plot(prior)
Now that the prior distribution is specified, Laura can calculate the required sample size for her desired statement by using the planning() function. She uses the prior object as input for the planning() function to use her prior distribution.
# Step 1: Calculate the required sample size.
stage1 <- planning(materiality = materiality, expected = expected, conf.level = confidence, prior = prior)
Laura can then inspect the result from her planning procedure by using the summary() function. Her result tells her that, given her prior distribution she needs to audit a sample of 178 transactions so that, when at most 4.45 errors are found, she can conclude with 95% confidence that the maximum error in BuildIt’s financial statements is lower the materiality of 5%.
summary(stage1)
##
## Bayesian Audit Sample Planning Summary
##
## Options:
## Confidence level: 0.95
## Materiality: 0.05
## Hypotheses: H₀: Θ > 0.05 vs. H₁: Θ < 0.05
## Expected: 0.025
## Likelihood: poisson
## Prior distribution: gamma(α = 2.325, β = 53)
##
## Results:
## Minimum sample size: 178
## Tolerable errors: 4.45
## Posterior distribution: gamma(α = 6.775, β = 231)
## Expected most likely error: 0.025
## Expected upper bound: 0.049981
## Expected precision: 0.024981
## Expected BF₁₀: 9.6614
Laura can inspect how the prior distribution compares to the expected posterior distribution by using the plot() function. The expected posterior distribution is the posterior distribution that would occur if Laura actually observed a sample of 178 transactions, from which 4.45 were in error.
plot(stage1)
## Stage 2: Selecting a sample
Laura is now ready to select the required 178 transactions from the financial statements. She can choose to do this according to one of two statistical methods. In record sampling (units = "items"), inclusion probabilities are assigned on the transaction level, treating transactions with a high value and a low value the same, a transaction of $5,000 is equally likely to be selected as a transaction of$1,000. In monetary unit sampling (units = "values"), inclusion probabilities are assigned on the level of individual monetary units (e.g., a dollar). When a dollar is selected to be in the sample, the transaction that includes that dollar is selected. This favors higher transactions, as a transaction of $5,000 is five times more likely to be selected than a transaction of$1,000.
Laura chooses to use monetary unit sampling, as she wants to include more high-valued transactions. The selection() function allows her to sample from the financial statements. She uses the stage1 object as an input for the selection() function.
# Step 2: Draw a sample from the financial statements.
stage2 <- selection(data = BuildIt, size = stage1, units = "values", values = "bookValue", method = 'interval')
Laura can inspect the outcomes of her sampling procedure by using the summary() function.
summary(stage2)
##
## Audit Sample Selection Summary
##
## Options:
## Requested sample size: 178
## Sampling units: monetary units
## Method: fixed interval sampling
## Starting point: 1
##
## Data:
## Population size: 3500
## Population value: 1403221
## Selection interval: 7883.3
##
## Results:
## Selected sampling units: 178
## Proportion of value: 0.0001269
## Selected items: 178
## Proportion of size: 0.050857
## Stage 3: Executing the audit
The selected sample can be isolated by indexing the sample object from the sampling result. Now Laura can execute her audit by annotating the sample with their audit value (for exampling by writing the sample to a .csv file using write.csv(). She can then load her annotated sample back into R for further evaluation.
# Step 3: Isolate the sample for execution of the audit.
sample <- stage2\$sample
# To write the sample to a .csv file:
# write.csv(x = sample, file = "auditSample.csv", row.names = FALSE)
# To load annotated sample back into R:
# sample <- read.csv(file = "auditSample.csv")
For this example, the audit values of the sample are already included in the auditValue column of the data set .
## Stage 4: Evaluating the sample
Using her annotated sample, Laura can perform her inference with the evaluation() function. By passing the priorResult object to the function, she automatically sets method = "binomial" to be consistent with her prior distribution.
# Step 4: Evaluate the sample.
stage4 <- evaluation(materiality = materiality, conf.level = confidence, data = sample,
values = "bookValue", values.audit = "auditValue", prior = prior)
Laura can inspect the outcomes of her inference by using the summary() function. Her resulting upper bound is 2.278%, which is lower than the materiality of 5%. The output tells Laura the correct conclusion immediately.
summary(stage4)
##
## Bayesian Audit Sample Evaluation Summary
##
## Options:
## Confidence level: 0.95
## Materiality: 0.05
## Materiality: 0.05
## Hypotheses: H₀: Θ > 0.05 vs. H₁: Θ < 0.05
## Method: poisson
## Prior distribution: gamma(α = 2.325, β = 53)
##
## Data:
## Sample size: 178
## Number of errors: 0
## Sum of taints: 0
##
## Results:
## Posterior distribution: gamma(α = 2.325, β = 231)
## Most likely error: 0.0057359
## 95 percent credible interval: [0, 0.022781]
## Precision: 0.017045
## BF₁₀: 2179.8
She can inspect the prior and posterior distribution by using the plot() function. The shaded area quantifies the area under the posterior distribution that contains 95% of the probability, which ends at 2.278%. Therefore, Laura can state that there is a 95% probability that the misstatement in BuildIt’s population is lower than 2.278%.
plot(stage4)
## Conclusion
Since the 95% upper confidence bound on the misstatement in population is lower than the performance materiality Laura has obtained sufficient evidence to conclude that there is less than 5% risk that the population contains material misstatement.
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2023-02-01 03:07:10
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https://projecteuclid.org/euclid.ade/1396558057
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### Critical growth elliptic problem in $\mathbb R^2$ with singular discontinuous nonlinearities
#### Abstract
Let $\varOmega$ be a bounded domain in $\mathbb R^{2}$ with smooth boundary, $a > 0, \lambda>0$ and $0 < \delta < 3$. We consider the following critical problem with singular and discontinuous nonlinearity: \begin{eqnarray*} \begin{array}{rl} -\Delta u & = \lambda ( {\chi_{\{u < a\}}}{u^{-{\delta}}} + h(u) e^{u^2})~~\text{in} ~~\Omega, \\ u & > 0 ~\text{ in }~ \Omega,\\ u & = 0 ~\text{ on }~ \partial \Omega, \end{array} \end{eqnarray*} where $\chi$ is the characteristic function and $h(u)$ is a smooth nonlinearity that is a "perturbation" of $e^{u^2}$ as $u \to \infty$ (for precise definitions, see hypotheses (H1)-(H5) in Section 1). With these assumptions we study the existence of multiple positive solutions to the above problem.
#### Article information
Source
Adv. Differential Equations Volume 19, Number 5/6 (2014), 409-440.
Dates
First available in Project Euclid: 3 April 2014
Dhanya, R.; Prashanth, S.; Sreenadh, K.; Tiwari, Sweta. Critical growth elliptic problem in $\mathbb R^2$ with singular discontinuous nonlinearities. Adv. Differential Equations 19 (2014), no. 5/6, 409--440. https://projecteuclid.org/euclid.ade/1396558057.
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2017-09-24 11:58:14
|
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|
https://codereview.stackexchange.com/questions/6369/html-table-of-contents-parser
|
I've written a table of contents parser for a FOSS wiki project I maintain in my spare time. The class takes a HTML string, injects anchor takes before each H1,H2 etc. and then generates the contents for the headers.
My main concern is boundary checks within the code. I haven't got any unit tests for it as it's a bit hard to test without a slightly contrived set of example text.
I'm after any glaring issues, or easier ways of doing the tree parsing the way I've done it. I'm reluctant to wrap the InsertToc method in a giant try/catch but instead have all all edge cases catered for.
The source is here as well.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using HtmlAgilityPack;
public class TocParser
{
/// <summary>
/// Replaces all {TOC} tokens with the HTML for the table of contents. This method also inserts
/// anchored name tags before each H1,H2,H3 etc. tag that the contents references.
/// </summary>
public string InsertToc(string html)
{
HtmlDocument document = new HtmlDocument();
HtmlNodeCollection elements = document.DocumentNode.ChildNodes;
// Try parsing all H2 headers (as H1 is technically the page title).
// Add a fake root for the tree
{
}
StringBuilder builder = new StringBuilder();
builder.AppendLine("<div class=\"toc\">");
builder.AppendLine("<div class=\"toc-title\">Contents [<a class=\"toc-showhide\" href=\"#\">hide</a>]</div>");
builder.AppendLine("<div class=\"toc-list\">");
builder.AppendLine("<ul>");
builder.AppendLine("</ul>");
builder.AppendLine("</div>");
builder.AppendLine("</div>");
return document.DocumentNode.InnerHtml.Replace("{TOC}",builder.ToString());
}
/// <summary>
/// Generates the ToC contents HTML for the using the StringBuilder.
/// </summary>
{
// Performs a level order traversal of the H1 (or H2) trees
{
htmlBuilder.AppendLine("<li>");
{
htmlBuilder.AppendLine("<ul>");
htmlBuilder.AppendLine("</ul>");
}
htmlBuilder.AppendLine("</li>");
}
}
/// <summary>
/// Parses the HTML for H1,H2, H3 etc. elements, and adds them as Header trees, where
/// rootHeaders contains the H1 root nodes.
/// </summary>
{
foreach (HtmlNode node in parentNode.ChildNodes)
{
if (node.Name.StartsWith("h"))
{
{
// Add as a new child
}
{
{
}
}
node.PrependChild(anchor);
if (node.Name == rootTag)
}
else if (node.HasChildNodes)
{
}
}
}
/// <summary>
/// Represents a header and its child headers, a tree with many branches.
/// </summary>
{
public string Id { get; set; }
public string Tag { get; set; }
public int Level { get; private set; }
public List<Header> Children { get; set; }
public Header Parent { get; set; }
public string Title { get; set; }
{
Title = title;
Tag = tag;
int level = 0;
int.TryParse(tag.Replace("h", ""),out level); // lazy (aka hacky) way of tracking the level using HTML H number
Level = level;
ShortGuid guid = ShortGuid.NewGuid();
Id = string.Format("{0}{1}", Title.EncodeTitle(), guid);
}
public string GetTocNumber()
{
string result = SiblingNumber().ToString();
if (Parent != null && Level > 1)
{
while (parent != null && parent.Level > 0)
{
result += "." + parent.SiblingNumber();
parent = parent.Parent;
}
}
return new String(result.ToArray().Reverse().ToArray<char>());
}
public int SiblingNumber()
{
if (Parent != null)
{
for (int i = 0; i < Parent.Children.Count; i++)
{
if (Parent.Children[i] == this)
return i +1;
}
}
return 1;
}
public override bool Equals(object obj)
{
return false;
}
public override int GetHashCode()
{
return Id.GetHashCode();
}
}
}
• The comment beginning Try parsing all H2 headers is at best misleading. A more useful comment there would say why h2 is parsed iff no h1 was found. – Peter Taylor Nov 29 '11 at 16:35
I'd be weary of solutions that builds the HTML by hand. Work with the DOM and let it write out the HTML for you. HAP can do that for you.
Don't create one-time use extension methods. It appears you created an extension method for strings to encode your titles. If you can't use it anywhere else in your code, it doesn't belong as an extension. I'd argue that it should be a regular static method of your Header class as it might be specific to how you want your headers encoded. In this context, it is confusing to see that call there.
Your logic in your headers to get the "sibling number" and TOC prefix is more complicated than it needs to be. Especially the GetTocNumber() method, the logic is very confusing to glance at. I was having a hard enough time trying to figure out what it was doing. The string reversal at the end really killed it. They both could be done simpler. In fact, they could be calculated at once on construction with some refactoring.
That leads in to the critical thing that's missing in most of these methods, comments... there's not a lot of useful ones in there. Your comments should be explaining what is happening in the code that couldn't be determined at first glance. The code really should be self-documenting. When it isn't, you need to say what it's doing in comments. But no one cares that the next line will add some item to a list. You should me saying things like, "we need to ensure we don't have an empty list because..." or at least explain why some actions are needed.
I did a lot more that I thought I would do but I would rewrite it more like this.
p.s., I don't know what your HTML would look like so I don't know how the nesting actually worked. But this should give you an idea how it could be better implemented (IMHO).
//Does this really need to create instances of this class?
public static class TocParserEx
{
//Does this really need to be an instance method?
public static string InsertToc(string html)
{
var doc = new HtmlDocument();
//only place the TOC if there is a TOC section labeled
var tocPlaceholder = doc.DocumentNode
.DescendantNodes()
.OfType<HtmlTextNode>()
.Where(t => t.Text == "{TOC}")
.FirstOrDefault();
if (tocPlaceholder != null)
{
var newToc = HtmlNode.CreateNode(@"<div class=""toc"">
<div class=""toc-title"">Contents [<a class=""toc-showhide"" href=""#"">hide</a>]</div>
<div class=""toc-list""></div>
</div>");
tocPlaceholder.ParentNode.ReplaceChild(newToc, tocPlaceholder);
}
return doc.DocumentNode.WriteTo();
}
/// <summary>
/// </summary>
/// <param name="root">The root node which contains the headers</param>
{
.ToList();
{
var replacement = HtmlNode.CreateNode(String.Format("<a name=\"{0}\"/>", header.Id));
//replace the found header with the wrapper
}
}
/// <summary>
/// </summary>
/// <param name="tocSection">The TOC section to add to</param>
{
var ul = tocSection.AppendChild(HtmlNode.CreateNode("<ul/>"));
{
{
}
}
}
{
return HtmlNode.CreateNode(String.Format(@"<li>
<a href=""#{0}"">{1} {2}</a>
}
}
//this class really should be lightweight
{
public static Header Root { get { return _root; } }
public string Title { get; private set; }
public string Tag { get; private set; }
public string Id { get; private set; }
public int Level { get; private set; }
public HtmlNode Node { get; private set; }
public Header Parent { get; private set; }
public int EntryNumber { get; private set; }
public string Section { get; private set; }
private Header() : this(HtmlNode.CreateNode("<h0/>"), null) { }
{
Title = node.InnerText;
Tag = node.Name;
Id = EncodeTitle(Title) + ShortGuid.NewGuid();
Level = Int32.Parse(Tag.Substring(1));
Node = node;
Parent = parent ?? _root;
if (parent == null)
{
EntryNumber = 1;
Section = "1";
}
else
{
EntryNumber = parent.Children.Count;
Section = parent.Section + "." + EntryNumber;
}
}
{
if (parent == null)
return _root;
if (node == null)
throw new ArgumentNullException("node");
}
{
return System.Text.RegularExpressions.Regex.IsMatch(node.Name, @"h\d");
}
private static string EncodeTitle(string title)
{
//encode the title (whatever your logic is)
return String.Concat(title.Where(Char.IsLetterOrDigit));
}
}
And the "HTML" I tested it on:
<p>{TOC}</p>
<h1>This is a title!!!</h1>
<h1>Here's another title!!!</h1>
Generates something like this:
<p><div class="toc">
<div class="toc-title">Contents [<a class="toc-showhide" href="#">hide</a>]</div>
<div class="toc-list"><ul><li>
<a href="#ThisisatitlefyxTAeHYp0y2KaQOvD89JA">1.1 This is a title!!!</a>
</li><li>
<a href="#HeresanothertitleD3FHM3IpO0OAuNRRtmS1vw">1.2 Here's another title!!!</a>
</li></ul></div>
</div></p>
<a name="ThisisatitlefyxTAeHYp0y2KaQOvD89JA"><h1>This is a title!!!</h1></a>
<a name="HeresanothertitleD3FHM3IpO0OAuNRRtmS1vw"><h1>Here's another title!!!</h1></a>
• Hi Jeff thanks for the feedback, a few things - the extension method is used elsewhere in the project quite a lot, the HAP I'm using is 1.x as 2 is quite buggy still. I was hoping the code would be fairly self documenting but obviously not. It's creating a BST and doing in-order searches for each header (leaf) in the tree, to find what level that header is. The string is reversed as it's built from the bottom up.I agree about the statics although I did want it to be threadsafe. I'll probably borrow bits of your code but I don't want to rewrite the whole thing, making it more readable. – Chris S Nov 30 '11 at 10:15
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2019-09-16 01:29:50
|
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|
https://zbmath.org/?q=an:1083.35149
|
## Semiclassical analysis for the Kramers-Fokker-Planck equation.(English)Zbl 1083.35149
The authors study accurate semiclassical estimates of the resolvent of a class of pseudodifferential operators that include the Kramers-Fokker-Planck operator $P=v\cdot h\partial_x-V'(x)\cdot h\partial_v+{\gamma\over{2}}(-(h\partial_v)^2+v^2-hn),$ in $${\mathbb R}^{2n},$$ where $$V$$ is a smooth potential, $$x,v\in{\mathbb R}^n$$ and $$h>0$$ is essentially the temperature (so that in this case the semiclassical limit is given by a low-temperature limit). The class they consider is made of pseudodifferential operators that are neither elliptic nor self-adjoint, but that satisfy certain subelliptic conditions: if $$p=p_1+ip_2$$ is the symbol of $$p^w(x,D)$$ (semiclassical Weyl-Hörmander quantization), it is assumed that $$p_1\geq 0$$ and that:
$$\bullet$$ $$p\in S(\lambda^2,g_0),$$ $$\partial p\in S(\lambda,g_0),$$ $$\partial^2p_1\in S(1,g_0)$$ and $$\partial H_{p_2}p_1\in S(\lambda,g_0)$$, where $$1\leq\lambda(x,\xi)=\lambda\in C^\infty$$ is a $$g_0$$-admissible weight such that $$\lambda\in S(\lambda,g_0)$$ and $$\partial\lambda\in S(1,g_0)$$, where $$g_0=| dx| ^2+\lambda^{-2}| d\xi| ^2$$ is an admissible Hörmander’s metric on $${\mathbb R}^{2n}$$, and where $$H_{p_2}$$ is the Hamilton vector-field associated with $$p_2$$;
$$\bullet$$ $$p$$ has finitely many critical points $$C=\{\rho_1,\ldots,\rho_N\}$$ (where, for simplicity only, $$p(\rho_j)=0$$) such that, with $$c_0>0$$ sufficiently small, in a fixed ball $$B$$ containing $$C$$ one has $p_1+c_0H_{p_2}^2p_1\approx\text{ dist}_C^2;$
$$\bullet$$ outside $$B$$ one has $p_1+c_0H_{p_2}^2p_1\approx\lambda^2.$ Under these assumptions, the authors obtain precise resolvent estimates inside the pseudo-spectrum and, when $$p^w+Ch^2$$ is $$m$$-accretive, a precise description of the spectrum in a ball centered at the origin and radius $$Ch.$$ They are then able to apply the resolvent estimates and the description of the eigenspaces to the large time behavior of the semigroup $$e^{-tp^w/h}$$, $$t\geq 0,$$ associated with the Cauchy problem $(h\partial_t+p^w(x,D))u=0,\,\,\,\, u\bigl| _{t=0}=u_0.$ Near the critical points of $$p$$, the main technical tool is the use of the FBI-transform and of suitable weighted $$L^2$$-spaces of holomorphic functions; whereas away from the critical points, it is the use of the semiclassical Weyl-Hörmander calculus.
### MSC:
35S10 Initial value problems for PDEs with pseudodifferential operators 35P20 Asymptotic distributions of eigenvalues in context of PDEs 47D06 One-parameter semigroups and linear evolution equations 81Q20 Semiclassical techniques, including WKB and Maslov methods applied to problems in quantum theory 47A10 Spectrum, resolvent 47G30 Pseudodifferential operators
Full Text:
### References:
[1] Burq N., Comm. Math. Phys. 223 pp 1– (2001) · Zbl 1042.81582 [2] Davies E. B., R. Soc. Lond. Proc. Ser. A Math. Phys. Eng. Sci. 455 pp 585– (1999) · Zbl 0931.70016 [3] Davies E. B., Comm. Math. Phys. 200 pp 35– (1999) · Zbl 0921.47060 [4] DOI: 10.1002/cpa.20004 · Zbl 1054.35035 [5] DOI: 10.1002/1097-0312(200101)54:1<1::AID-CPA1>3.0.CO;2-Q · Zbl 1029.82032 [6] Freidlin M. I., Random Perturbations of Dynamical Systems. (1984) · Zbl 0522.60055 [7] Hellfer B., Hypoelliptic Estimates and Spectral Theory for Fokker–Planck Operators and Witten Laplacians. (2005) · Zbl 1072.35006 [8] DOI: 10.1007/s00205-003-0276-3 · Zbl 1139.82323 [9] Hörmander L., The Analysis of Linear Partial Differential Operators. (1985) [10] Kolokoltsov V. N., Semiclassical Analysis for Diffusions and Stochastic Processes. (2000) · Zbl 0951.60001 [11] DOI: 10.1016/S0031-8914(40)90098-2 · Zbl 0061.46405 [12] Risken H., 2nd ed. 18, in: The Fokker-Planck Equation. (1989) · Zbl 0665.60084 [13] DOI: 10.1007/BF02384749 · Zbl 0317.35076 [14] DOI: 10.1215/S0012-7094-90-06001-6 · Zbl 0702.35188 [15] Sjöstrand J., Structure of Solutions of Differential Equations (Katata/Kyoto, 1995) pp 369– (1996) [16] Talay D., Progress in Stochastic Structural Dynamics. 152 pp 139– (1999) [17] Tang S.-H., Math. Res. Lett. 5 pp 261– (1998) · Zbl 0913.35101 [18] DOI: 10.1137/S0036144595295284 · Zbl 0896.15006 [19] Zworski M., Proc. Amer. Math. Soc. 129 pp 2955– (2001) · Zbl 0981.35107
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2022-05-27 14:43:21
|
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|
https://zenodo.org/record/4034438/export/xd
|
Software Open Access
# Revisiting Iso-Recursive Subtyping
Zhou, Yaoda; Dos Santos Oliveira, Bruno Cesar; Zhao, Jinxu
### Dublin Core Export
<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>Zhou, Yaoda</dc:creator>
<dc:creator>Dos Santos Oliveira, Bruno Cesar</dc:creator>
<dc:creator>Zhao, Jinxu</dc:creator>
<dc:date>2020-09-17</dc:date>
<dc:description>The Amber rules are well-known and widely used for subtyping
iso-recursive types. They were first briefly and informally introduced
in 1985 by Cardelli in a manuscript describing the Amber
language.
Despite their use over many years, important aspects of the metatheory of the iso-recursive
style Amber rules have not been studied in depth or turn out to be
quite challenging to formalize.
This paper aims to revisit the problem of subtyping iso-recursive
types. We start by introducing a novel declarative specification
that we believe captures the spirit'' of Amber-style
iso-recursive subtyping. Informally, the specification states that
two recursive types are subtypes \emph{if all their finite
unfoldings are subtypes}. The Amber rules are shown to be sound
with respect to this declarative specification. We then derive a
\emph{sound}, \emph{complete} and \emph{decidable} algorithmic
formulation of subtyping that employs a novel \emph{double
unfolding} rule. Compared to the Amber rules, the double
unfolding rule has the advantage of: 1) being modular; 2)
not requiring reflexivity to be built in; and 3) leading to
an easy proof of transitivity of subtyping. This work
sheds new insights on the theory of subtyping iso-recursive types,
and the new double unfolding rule has important advantages over
the original Amber rules for both implementations and
metatheoretical studies involving recursive types. All results
are mechanically formalized in the Coq theorem prover. As far as
we know, this is the first comprehensive treatment of iso-recursive
subtyping dealing with unrestricted recursive types in a theorem prover.</dc:description>
<dc:identifier>https://zenodo.org/record/4034438</dc:identifier>
<dc:identifier>10.5281/zenodo.4034438</dc:identifier>
<dc:identifier>oai:zenodo.org:4034438</dc:identifier>
<dc:language>eng</dc:language>
<dc:relation>doi:10.5281/zenodo.4034437</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:title>Revisiting Iso-Recursive Subtyping</dc:title>
<dc:type>info:eu-repo/semantics/other</dc:type>
<dc:type>software</dc:type>
</oai_dc:dc>
69
19
views
|
2023-03-22 12:10:50
|
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|
https://codeforces.com/problemset/problem/518/F
|
F. Pasha and Pipe
time limit per test
4 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
On a certain meeting of a ruling party "A" minister Pavel suggested to improve the sewer system and to create a new pipe in the city.
The city is an n × m rectangular squared field. Each square of the field is either empty (then the pipe can go in it), or occupied (the pipe cannot go in such square). Empty squares are denoted by character '.', occupied squares are denoted by character '#'.
The pipe must meet the following criteria:
• the pipe is a polyline of width 1,
• the pipe goes in empty squares,
• the pipe starts from the edge of the field, but not from a corner square,
• the pipe ends at the edge of the field but not in a corner square,
• the pipe has at most 2 turns (90 degrees),
• the border squares of the field must share exactly two squares with the pipe,
• if the pipe looks like a single segment, then the end points of the pipe must lie on distinct edges of the field,
• for each non-border square of the pipe there are exacly two side-adjacent squares that also belong to the pipe,
• for each border square of the pipe there is exactly one side-adjacent cell that also belongs to the pipe.
Here are some samples of allowed piping routes:
....# ....# .*..# ***** ****. .***. ..#.. ..#*. ..#*. #...# #..*# #..*# ..... ...*. ...*.
Here are some samples of forbidden piping routes:
.**.# *...# .*.*# ..... ****. .*.*. ..#.. ..#*. .*#*. #...# #..*# #*.*# ..... ...*. .***.
In these samples the pipes are represented by characters ' * '.
You were asked to write a program that calculates the number of distinct ways to make exactly one pipe in the city.
The two ways to make a pipe are considered distinct if they are distinct in at least one square.
Input
The first line of the input contains two integers n, m (2 ≤ n, m ≤ 2000) — the height and width of Berland map.
Each of the next n lines contains m characters — the map of the city.
If the square of the map is marked by character '.', then the square is empty and the pipe can through it.
If the square of the map is marked by character '#', then the square is full and the pipe can't through it.
Output
In the first line of the output print a single integer — the number of distinct ways to create a pipe.
Examples
Input
3 3.....#...
Output
3
Input
4 2........
Output
2
Input
4 5#...##...####.####.#
Output
4
Note
In the first sample there are 3 ways to make a pipe (the squares of the pipe are marked by characters ' * '):
.*. .*. ... .*# **# **# .*. ... .*.
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2021-09-29 00:02:14
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.300197571516037, "perplexity": 2419.848890180253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780060908.47/warc/CC-MAIN-20210928214438-20210929004438-00622.warc.gz"}
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https://www.gradesaver.com/textbooks/science/physics/essential-university-physics-volume-1-3rd-edition/chapter-4-exercises-and-problems-page-69/61
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Essential University Physics: Volume 1 (3rd Edition)
F-35A: Yes; $a=.8\ m/s^2$ A-380: No
We see if there is enough force to raise the F-35A: $F_g=(18,000)(9.81)=177 \ kN$ This is less than the engine's force, so we find: $a = \frac{191,000-177,000}{18,000}=.8 \ m/s^2$ We see if there is enough force to raise the A-380: $F_g=(560,000)(9.81)=5,493 \ kN$ There is not enough force, for this is greater than the thrust of the engine.
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2021-03-06 00:16:42
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https://aimacode.github.io/aima-exercises/game-playing-exercises/ex_9/
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Exercise 5.9
This problem exercises the basic concepts of game playing, using tic-tac-toe (noughts and crosses) as an example. We define $X_n$ as the number of rows, columns, or diagonals with exactly $n$ $X$’s and no $O$’s. Similarly, $O_n$ is the number of rows, columns, or diagonals with just $n$ $O$’s. The utility function assigns $+1$ to any position with $X_3=1$ and $-1$ to any position with $O_3 = 1$. All other terminal positions have utility 0. For nonterminal positions, we use a linear evaluation function defined as ${Eval}(s) = 3X_2(s) + X_1(s) - (3O_2(s) + O_1(s))$.
1. Approximately how many possible games of tic-tac-toe are there?
2. Show the whole game tree starting from an empty board down to depth 2 (i.e., one $X$ and one $O$ on the board), taking symmetry into account.
3. Mark on your tree the evaluations of all the positions at depth 2.
4. Using the minimax algorithm, mark on your tree the backed-up values for the positions at depths 1 and 0, and use those values to choose the best starting move.
5. Circle the nodes at depth 2 that would not be evaluated if alpha–beta pruning were applied, assuming the nodes are generated in the optimal order for alpha–beta pruning.
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2020-08-14 05:22:18
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http://www.mathjournals.org/jot/1998-040-002/1998-040-002-007.html
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Previous issue · Next issue · Most recent issue · All issues
# Journal of Operator Theory
Volume 40, Issue 2, Fall 1998 pp. 357-372.
Characteristic matrix of the tensor product of operators
Authors Hideki Kosaki
Author institution: Graduate School of Mathematics, Kyushu University, Fukuoka, 810, Japan
Summary: The characteristic matrix of the tensor product of two Hilbert space operators is analyzed. The case where operators are not necessarily closable is also considered, and we determine (i) the closure of the graph of the algebraic tensor product $T_1 \algt T_2$, (ii) the maximal closable part of $T_1 \algt T_2$ in the sense of P. J\o rgensen ([1]).
Contents Full-Text PDF
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2018-10-23 05:52:02
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http://www.wspnet.de/pymaxima/installation/index_en.html
|
Installation
• First of all: you'll need Maxima, pyMaxima works fine with Maxima version 5.10.0 to 5.28
Note: without Maxima, pyMaxima won't work!
• Python (vers. 2.4 to 2.7) including Tkinter might be useful.
• On win32 (Win 2000, Win XP), pyMaxima requires the Python for Windows Extensions
• If you like the 3D-pygeo drawings, you must install VPython and PyGeo (Numerical Python is bundled with VPython)
Note: The installation of PyGeo is limited up to Python 2.7 You have to install the following package, if you are using Python 2.5 to 2.7:
• Unzip pyMaxima.zip in a subdirectory
• On Linux: fix the entries in the configuration file linuxInit.conf in the subdirectory ./init.
Call python pyMaxima.py in the shell.
• On win32: fix the entries in the configuration file winInit.conf in the subdirectory ./init.
Click on pyMaxima.py and don't close the black window in the back!
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2019-01-23 12:10:08
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https://www5.in.tum.de/wiki/index.php/Pendulum_Project
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# Pendulum Project
Inverted pendulum is mainly used as benchmark for testing new control algorithms. In the scope of Bavarian Graduate School of Computational Engineering Honors Project (BGCE) we developed classical controller using mathematical description of the inverted pendulum system and a fuzzy logic based controller. We have successfully developed a simulation model and implemented the above mentioned controllers on both the simulation and real system. We have also developed easy-to-use and informative graphical user interface (GUI) with visualization of the simulated system, state variable response diagram, and fuzzy rule control. All the controller functionalities for both the simulation and real system have been integrated into the GUI. Also, we have been able to implement Backpropagation based learning algorithm for the adaptive neural controller and obtain satisfactory preliminary results.
## Introduction
An inverted pendulum is a pendulum which is attached to a cart. (Fig. 1.1)
Fig. 1.1 Inverted Pendulum
Even though a regular inverted pendulum has a point mass attached to its rod, our hardware consists of a cart which is free to move horizontally and a rod attached to it. An inverted pendulum is inherently unstable, and must be actively balanced in order to remain upright, either by applying a torque at the pivot point or by moving the pivot point horizontally as part of a feedback system. [1]
In control theory and classical dynamics the inverted pendulum is considered to be a classic problem which is used as benchmark for testing of the control algorithms. As some examples of the control algorithms we used the inverted pendulum system to experiment with PID controller, fuzzy controller and lastly neural networks.
Besides its theoretical importance the inverted pendulum system has also seen action in real life. An inverted pendulum-like system is used during the take-offs of the rocket to keep the rocket in balance. Moreover, the shipping cranes in the seaports use a similar idea. Last but not the least, a self-balancing transportation device called Segway[2] uses same working principle as it is in an inverted pendulum.
## BGCE Project Scope
The classical approach of control theory is to develop a mathematical model for the system and then to develop a corresponding controller so that the closed loop is stable (PID- controller). A second approach is the use of a so-called fuzzy-controller, which doesn’t need an explicit model of the system. Another interesting possibility is to use self-learning controllers that learn automatically how to control the system. Techniques such as neural networks can be used for this task. In the scope of this project, these different approaches will be studied. The focus is not only on the controllers themselves, but also on their visualization and demonstration via an interactive graphical user interface.
## Classical Controller
Classical Controller tries to solve the physical equations governing the system in order to find the trajectory of the rod from the stable equilibrium position at rest to the unstable equilibrium position.
To achieve this goal we first have to find the mathematical representation of the system, meaning , we will model the system at first and then we try to find the relation between the output of the system which is the current state as feed back and the next input which hopefully helps us get closer to the unstable equilibrium position.[3]
### Classical Model
The purpose of system modeling is to provide an accurate mathematical description of the system. Once this description is obtained, the numerical constants of the system will be substituted into the equations to provide a working representation of the system. From there this model can be simulated and manipulated to obtain gain values on which the state feedback control is based.
This section will derive two sets of mathematical models of the system. The first model is a set of nonlinear differential equations. The second model is a set of linearized differential state equations.
Due to the difficulty level of nonlinear control and the scope of this project, the nonlinear equations will not be used for control calculations. However, the linearized equations provide a good estimate of the system under conditions mentioned in this chapter, and will be used for all control calculations.
Since the inverted pendulum is a complex system, the more accurate one wants to be the more complex the model gets and the slower the controller.
So having this fact on mind , we tried to use the most important terms and leave away the parts where we thought wouldn't be much of importance but of course doing this we moved away from a complex model to an easier to solve one in cost of accuracy.
The physical model of the inverted pendulum on a moving cart is seen in figure 1. The inverted pendulum system is free to move only in the X-Y plane. The symbolic descriptions are shown in table 3.1.
Using the physical model of the inverted pendulum from Fig. 1, we can now derive the differential equations that describe the system as seen in the following section. [3]
Table 3.1.1 Parameters of the Model
Fig. 3.1.1 Inverted Pendulum Model
in progress...
### Controller Design
The self-erecting inverted pendulum has a control design for the swing up, and a separate control design for the stabilization. The open loop swing up controller brings the pendulum upright close to the unstable point of equilibrium. Once the angular position has reached a software adjustable capture range, the closed loop stabilization controller takes over.
#### Closed Loop Stabilization Control Design
The stabilization control design is based on linear quadratic regulator (LQR) design with a tracking controller. The objective of this controller is to stabilize the pendulum rod in the upright unstable point of equilibrium while maintaining a software adjustable linear set point position. The LQR design will effectively return the state feedback gains needed to ensure stability of the system. However, to bring the steady state error of the linear position to zero, a tracking controller is added by integrating the error of the cart position, relative to the linear set point, over time. The gain adjustment of the integration result allows control over the zero steady state error convergence time.
Fig. 3.5.2.1 Simulink Control Design Block Diagram
## Fuzzy Controller
### Introduction
Instead of the mathematical modeling and writing differential equations describing the characteristics of the inverted pendulum system another controller technique named "Fuzzy Controller" is introduced initially to mimic human behavior in robotics. Since the human body is perfect in keeping its balance, the scientists came up with the inverted pendulum system to represent the human-like balancing as closest as it could be. To actualize this idea the fuzzy logic and fuzzy controllers are used to stabilize the inverted pendulum with its rod kept upright.
### Fuzzy Logic
Fuzzy logic (FL) is a form of multi-valued logic derived from fuzzy set theory to deal with reasoning that is approximate rather than precise. Just as in fuzzy set theory the set membership values can range (inclusively) between 0 and 1, in fuzzy logic the degree of truth of a statement can range between 0 and 1 and is not constrained to the two truth values {true (1), false (0)} as in classic predicate logic.[3]
FL provides a simple way to arrive at a definite conclusion based upon vague, ambiguous, imprecise, noisy, or missing input information. FL's approach to control problems mimics how a person would make decisions, only much faster.
### Distinctive Properties of Fuzzy Logic
FL incorporates a simple, rule-based IF X AND Y THEN Z approach in order to a solve control problem rather than attempting to model a system mathematically. The FL model is empirically-based, relying on an operator's experience rather than their technical understanding of the system. For example, rather than dealing with terms such as exact values (x=100 or 50<x<100), terms like "IF (x is too big) AND (x is getting smaller) THEN (push the cart)" etc. These terms are imprecise and yet very descriptive of what must actually happen. Similar style is used while describing the inverted pendulum system.
### Fuzzy Logic in Inverted Pendulum
In the inverted pendulum system we defined four state variables: position of the cart (s), velocity of the cart (sp), angular velocity of the rod(phip) and angle of the rod(phi_oben). Those state variables play a big role when the subcontroller "stabilizer" kicks in to keep the pendulum stable in the neighborhood of the 0 degree. Each state variable is divided into several membership functions shown as in the Fig. 4.4.1.
$\textstyle phip$ and $\textstyle phi\_oben$ are divided into 7 states categorized as: $\textstyle negative\_big$, $\textstyle negative$, $\textstyle negative\_small$,$\textstyle zero$, $\textstyle positive_small$, $\textstyle positive$, $\textstyle positive\_big$. Each category is associated with a range of values. The ranges were:
Angle: [-0.3, 0.3]
Angular Velocity: [-4, 4]
Position: [-0.4,0.4]
Velocity: [-0.5, 0.5]
At each time step those four state variables are computed and a fuzzy rule corresponding to that specific state of the system will become active. Thus, it makes the cart move in one direction as it is specified in that fuzzy rule.
Fig. 4.4.1 Membership Function Example
In Fig. 4.4.1 above, we can see five membership functions made up by triangles (except the left-and rightmost areas).
Fig. 4.4.2 Simulink Model Diagram of the Stabilizer
The erector swings the rod like a sine-wave and aims to bring the rod close to the region ($\textstyle +/- 0.3^\circ$) so that the stabilizer can kick in. Fig. 4.5.4 above depicts the inside of the stabilizer. The membership functions are defined in "zugehörigkeiten" block whereas the fuzzy rules are defined inside the S-Function builder called "fuzzy_stabilisator".
### Results
The fuzzy controller (FC) has produced during our implementation better results than the initial implementation. When the project started, we had a version of a fuzzy controller. The main problem with that FC was that the swinger part was slow and it took too much time for till the stabilizer kicked in. Our main focus was to implement a better swing-up controller and to improve the current set of fuzzy rules. However, our experiments even with new fuzzy rules did not improve the stabilizer. Hence, we kept the old fuzzy rules defined in the beginning of the project.
In our trials, we implemented Gaussian membership functions to get a smoother fuzzy controller acting on the inverted pendulum. Yet, we did not see any significant gain in terms of speedup and robustness of the pendulum system. That is why we decide rolled back those membership functions to their initial states where they are represented as triangles.
## Neural Networks
The characteristics of the inverted pendulum make identification and control more challenging, as we have seen from the above sections. This can be circumvented by using a neural network based identification-control approach [4],[5],[6]. This would involve first, developing an accurate model of the inverted pendulum system using neural networks – System Identification. Then, develop a neural network controller which determines the correct control action to stabilize the system.
### Artificial Neural Networks
The science of artificial neural networks is based on the neuron. In order to understand the structure of artificial networks, the basic elements of the neuron should be understood. Neurons are the fundamental elements in the central nervous system. Fig. 5.1 below depicts the components of a neuron [7].
Fig. 5.1 Components of a Neuron
A neuron is made up of 3 main parts -dendrites, cell body and axon. The dendrites receive signals coming from the neighbouring neurons. The dendrites send their signals to the body of the cell. The cell body contains the nucleus of the neuron. If the sum of the received signals is greater than a threshold value, the neuron fires by sending an electrical pulse along the axon to the next neuron.
The following model is based on the components of the biological neuron. The inputs X0-X3 represent the dendrites. Each input is multiplied by weights W0, W3. The output of the neuron model, Y is a function, F of the summation of the input signals. (Fig. 5.2)
Fig. 5.2 Dentrides
The main advantage of ANN is they operate as black boxes. This avoids most of the complex modeling activities, and reduces the model to just a set of weights. But, this also would be a disadvantage since the rules of operation in neural networks are completely unknown. But the disadvantage is the amount of time taken to train networks. It can take considerable time to train an ANN for certain functions.
### Neural Network Structures
There are 3 main types of ANN structures -single layer feed-forward network, multi-layer feed-forward network and recurrent networks [8]. In feed-forward networks the direction of signals is from input to output, there is no feedback in the layers. Other types of single layer networks are based on the perceptron model. The details of the perceptron are shown in Fig. 5.2.1.
Fig. 5.2.1 Perceptron
In the current work, we employ a multi-layer perceptron model with a single hidden layer.
Fig. 5.2.2 Three Layer Neural Network Structure
Increasing the number of neurons in the hidden layer or adding more hidden layers to the network allows the network to deal with more complex functions. Cybenko's theorem states that, "A feed-forward neural network with a sufficiently large number of hidden neurons with continuous and differentiable transfer functions can approximate any continuous function over a closed interval.". [9] The weights in MLP's are updated using the back-propagation learning algorithm [6], which is discussed in detail in the coming sections.
Inputs to the perceptron are individually weighted and then summed. The perceptron computes the output as a function $\textstyle F$ of the sum. The activation function, $\textstyle F$ is needed to introduce nonlinearities into the network. This makes multi-layer networks powerful in representing nonlinear functions. There are 3 main types of activation function -tan-sigmoid, log-sigmoid and linear [10]. Different activation functions affect the performance of an ANN.
Fig. 5.2.3 Activation Functions
The output from the perceptron is
$\textstyle y[k] = f (w [k].x[k])$.
### ANN Learning
Neural networks have 3 main modes of operation – supervised, reinforced and unsupervised learning [10]. In supervised learning the output from the neural network is compared with a set of targets, the error signal is used to update the weights in the neural network. Reinforced learning is similar to supervised learning however there are no targets given, the algorithm is given a grade of the ANN performance. Unsupervised learning updates the weights based on the input data only. The ANN learns to cluster different input patterns into different classes.
### 5.4 Back-propagation Learning Algorithm
Back-propagation Learning Algorithm is a supervised learning method, and is an implementation of the Delta rule [6]. As the name itself suggests, the errors are propagate backwards from the output nodes to the inner nodes and therefore the learning. So back-propagation is used to calculate the gradient of the error of the network with respect to the network's modifiable weights. This gradient is almost always then used in a simple stochastic gradient descent algorithm to find weights that minimize the error. But, the term "back-propagation" is used in a more general sense, and refers to the entire procedure encompassing both the calculation of the gradient and its use in stochastic gradient descent. Back-propagation usually allows quick convergence on satisfactory local minima for error in the kind of networks to which it is suited. But, this also brings out its inherent drawback since it usually converges to the local minima, instead of the global minima.
There are two passes before the weights are updated. In the first pass (forward pass) the outputs of all neurons are calculated by multiplying the input vector by the weights. The error is calculated for each of the output layer neurons. In the backward pass, the error is passed back through the network layer by layer. The weights are adjusted according to the gradient decent rule, so that the actual output of the MLP moves closer to the desired output. A momentum term could be added which increases the learning rate with stability.
During the (second) backward pass, the difference between the target output and the actual output (error) is calculated
$\textstyle e[k]=T[k] - y[k]$.
The errors are back propagated through the layers and the weight changes are made. The formula for adjusting the weights is
$\textstyle w[k+1] = w[k]+\mu e[k]x[k].$
Once the weights are adjusted, the feed-forward process is repeated. The weights are adapted until the error between the target and actual output is low. The approximation of the function improves as the error decreases.
The Algorithm can be summarized as below:
1. Present a training sample to the neural network.
2. Compare the network's output to the desired output from that sample. Calculate the error in each output neuron.
3. For each neuron, calculate what the output should have been, and a scaling factor, how much lower or higher the output must be adjusted to match the desired output. This is the local error.
4. Adjust the weights of each neuron to lower the local error.
5. Assign an inertial value for the local error to neurons at the previous level, giving greater responsibility to neurons connected by stronger weights.
6. Repeat from step 3 on the neurons at the previous level, using each one's inertial value as its error.
### System Identification
System identification is the process of developing a mathematical model of a dynamic system based on the input and output data from the actual process [11]. This means it is possible to sample the input and output signals of a system and using this data generate a mathematical model. An important stage in control system design is the development of a mathematical model of the system to be controlled. In order to develop a controller, it must be possible to analyze the system to be controlled and this is done using a mathematical model. Another advantage of system identification is evident if the process is changed or modified. System identification allows the real system to be altered without having to calculate the dynamical equations and model the parameters again. This circumvents most of the mathematically rigorous modeling activities involved in case of complex real systems. The mathematical model in this case is the black box, it describes the relationship between the input and output signals. The inverted pendulum system is a non-linear process. To adequately model it,non-linear methods using neural networks must be used. It can be seen from the literature that neural networks have been successful used in modeling a plethora of nonlinear systems.
As universal approximators, neural networks have found widespread application in nonlinear dynamic system identification.[12][13]. The most common method of neural network identification is called forward modeling. During training both the process and ANN receive the same input, the outputs from the ANN and process are compared, this error signal is used to update the weights in the ANN. This is an example of supervised learning-the teacher (pendulum system) provides target values for the learner (the neural network).
\includegraphics[width=0.5\textwidth]{supervised_learning.PNG} \textbf{Fig. 5.5.1} Supervised Learning
This training/identification can also be done off-line, having collected the input-output data from the inverted pendulum system. In the current work, for the input-output data for the identification of the model the non-linear pendulum model along with a classical feedback LQR controller(discussed in previous sections), is used.
### Specifics of the ANN Training
The quality of the neural model is tested by calculating the MSE (mean squared error). The MSE gives a good indication of the accuracy of the model. The MSE between the model and the process should be low. A model could have a low MSE but not predict any of the dynamics of the pendulum system. The output from the model and process is plotted to compare the dynamics. Basically, we want to see whether the model predicts the movement of the inverted pendulum. Increasing the number of hidden layer neurons allows for more complex functions to be modeled. During testing, neural networks with a range of hidden layer neurons were simulated. It was expected that as the number of hidden neurons increased the more accurate the model would become.
In progress..
## Graphical User Interface
This Pendulum GUI is a friendly graphical user interface for the BGCE Inverted Pendulum Project. You can easily switch into Classical controller, Fuzzy controller or Neural network by choosing the different Tab of this GUI, also can change between Simulation or Real-time. By default, the Classical controller mode. (Fig. 6.1)
Fig. 6.1 Default Pendulum GUI (Classical controller mode)
### Classical Controller Tab
The "Classical" tab consists of Cart Visualization panel, Control panel and State Variables panel.
-Simulation
First, type the required simulation time into Sim Stop Time text box, then click the Run Simulation button to start the simulation. The visualization of the pendulum will be showed in the Visualization axes. After the simulation is done, click the Show Values button in the State Variables panel, the plot of 4 state variables will be showed in the State Variables axis.
-Real-time System
Click the \textbf{Run Real-time System} button to start the real-time system by the pre- complied .wcp file.
### Fuzzy Controller Tab
The layout of Fuzzy Tab based on the classical tab but added a Fuzzy rule table panel. (Fig. 6.2.1)
-Simulation
Same as Classical Tab. Because of the features of fuzzy controller, one can also control the fuzzy rules by press/depress the toggle buttons of the Fuzzy rule table. There are tool tips for all the toggle buttons, implies the rule of each button. All the rules are activated by default, deactivate the rule by pressing the related button(turn red), activate by depressing(turn green). The membership function of the fuzzy rules will also be plotted in the sub-GUI during the simulation.
-Real-time System
Same as Classical Tab.
Fig. 6.2.1 Fuzzy controller simulation
Fig. 6.2.2 State Variables
Fig. 6.2.3 Membership Function of Fuzzy Rules
### Neural Networks Controller Tab
-Simulation
Same as Classical Tab.
-Real-time System
Same as Classical Tab.
## References
[1] "Wikipedia, The Free Encyclopedia",(2008). Available:http://en.wikipedia.org/wiki/Inverted_pendulum
[2] "User Manual",(2008) Available:http://www.segway.de
[3] "Fuzzy Sets and Applications: Selected Papers by L.A. Zadeh", ed. R.R. Yager et al. John Wiley, New York,(1995).[4]
[5] "Neural Networks for Control",W. T. Miller, R. S. Sutton, and P. J. Werbos,Cambridge, MA: MIT Press,(1990).
[6] "Neural Networks for Identification, Prediction and Control",D.Pham,X. Liu, Springer Verlag,(1995).
[7] "Neural Controller based on back-propagation algorithm",Saerens M., Soquet A., IEEE Proceedings UF, Vol. 138, No.1, pp 55-62, (1991).
[8] "Neural Networks",E. Davalo, P. Naim, Palgrave Macmillan (1991).
[9] "Neural Network Toolbox Users Guide",The Mathworks Inc (1998).
[10] "Approximation by superposition of a Sigmoidal Function, Mathematics of Control, Signals and Systems",Cybenko,G,Vol 2, No. 4, pp 303-314, (1989).
[11] "System Identification-Theory for the user",Ljung. L, Prentice Hall (1999).
[12] "Non-linear system identification using neural networks"S. Chen, S. A. Billings, and P.M. Grant,Int. J. Contr., vol. 51, no. 6, (1990).
[13] "Identification and control of dynamical systems using neural networks",K. S. Narendra and K. Parthasarathy, IEEE Transactions on Neural Networks, vol. 1, no. 1, (1990).
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2021-09-24 00:09:36
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https://tex.stackexchange.com/questions/209848/how-to-make-side-label
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# How to make side label
How to make and control a side label similar like this? This example i made with code:
\multirow{1}{*}{\begin{sideways} {\small This is my very very very veryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryveryvery long message} \end{sideways}}
My method is ugly hack and works only for one page. And i want to control it similar like fancyhdr (example: turn on after title, and turn off in listoftables/listoffigures at the end)
You don't need fancyhdr for this. For example, you can use eso-pic to place content on the page arbitrarily. Below I've built an interface that places content on the left side of the page as needed:
\documentclass{report}
\usepackage{fancyhdr,lipsum,graphicx}
\usepackage{eso-pic}
\fancypagestyle{sidelabel}{% Not really needed
\fancyfoot[C]{--\thepage--}
}
\providecommand{\sidelabeltext}{}
\AddToShipoutPictureFG{% Place this content on the ForeGround of every page
\AtTextCenter{%
\hspace*{\dimexpr-.5\textwidth-\sidelabelsep}%
\rotatebox{90}{%
\makebox[0pt]{%
\sidelabeltext%
}%
}%
}%
}
\newlength{\sidelabelsep}
\setlength{\sidelabelsep}{\marginparsep}
\newcommand{\setsidelabelsep}[1]{\setlength{\sidelabelsep}{#1}}
\pagestyle{sidelabel}
\begin{document}
\setsidelabeltext{This is some side text}
\lipsum[1-6]
\setsidelabeltext{This is some very very very very very very very very very very very
very very very very very very very very very very very very very long side text}%
\setsidelabelsep{0.5in}
\lipsum[1-6]
\setsidelabeltext{}
\lipsum[1-6]
\setsidelabeltext{This is some side text}%
\setsidelabelsep{20pt}
\lipsum[1-6]
\end{document}
\setsidelabeltext{<stuff>} updates what is printed on the left of the page, while \setsidelabelsep{<len>} updates the length that <stuff> is printed from the text. Default is \marginparsep.
Here's a way to do it with fancyhdr. With \setsideheader you tell what text should be printed from the point on.
\documentclass[a4paper]{report}
\usepackage{fancyhdr}
\usepackage{graphicx}
\usepackage{lipsum}
\pagestyle{fancy}
\fancyhf{} % clear all fields
\fancyfoot[C]{\thepage}
\hspace*{-3em}%
\smash{%
\rotatebox[origin=Br]{90}{%
}%
}%
}
\begin{document}
\chapter{A title}
\setsideheader{Some long text that goes in the side margin}
\lipsum[1-20]
\chapter{A title}
\setsideheader{Some long text that goes in the side margin for chapter 2}
\lipsum[1-20]
\clearpage
\pagestyle{plain}
\tableofcontents
\end{document}
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2020-09-26 15:49:49
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http://dictionnaire.sensagent.leparisien.fr/Relativistic%20rocket/en-en/
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Publicité ▼
## définition - Relativistic rocket
voir la définition de Wikipedia
Wikipedia
# Relativistic rocket
A relativistic rocket is any spacecraft that is travelling at a velocity close enough to light speed for relativistic effects to become significant. What "significant" means is a matter of context, but generally speaking a velocity of at least 50% of the speed of light (0.5c) is required. The time dilation factor, mass factor, and length contraction factor (all these factors equal the Lorentz factor) are 1.15 at 0.5c. Above this speed Einsteinian physics are required to describe motion. Below this speed, motion is approximately described by Newtonian physics and the Tsiolkovsky rocket equation can be used.
We define a rocket as carrying all of its reaction mass, energy, and engines with it. Bussard ramjets, RAIRs,[1] light sails, and maser or laser-electric vehicles are not rockets.
Achieving relativistic velocities is difficult, requiring advanced forms of spacecraft propulsion that have not yet been adequately developed. Nuclear pulse propulsion could theoretically achieve 0.1c using current known technologies, but would still require many engineering advances to achieve this. The relativistic gamma factor ($\gamma$) at 10% of light velocity is 1.005. The time dilation factor of 1.005 which occurs at 10% of light velocity is too small to be of major significance. A 0.10c velocity interstellar rocket is thus considered to be a non-relativistic rocket because its motion is quite accurately described by Newtonian physics alone.
Relativistic rockets are usually seen discussed in the context of interstellar travel, since most would require a great deal of space to accelerate up to those velocities. They are also found in some thought experiments such as the twin paradox.
## Relativistic rocket equation
As with the classical rocket equation, one wants to calculate the velocity change $\Delta v$ that a rocket can achieve depending on the specific impulse $I_{sp}$ and the mass ratio, i. e. the ratio of starting mass $m_0$ and mass at the end of the acceleration phase (dry mass) $m_1$. Subsequently specific impulse means the momentum produced by the exhaust of a certain amount of rocket fuel divided by the mass of that amount of rocket fuel. Thus specific impulse is a velocity, as opposed to the common usage of the word as the ratio of momentum and weight (weight would not make much sense in this context).
### Specific impulse
The specific impulse of relativistic rockets is the same as the effective exhaust velocity, despite the fact that the nonlinear relationship of velocity and momentum as well as the conversion of matter to energy have to be taken into account; the two effects cancel each other. I.e.
$I_{sp} = v_e$
Of course this is only valid if the rocket does not have an external energy source (e. g. a laser beam from a space station; in this case the momentum carried by the laser beam also has to be taken into account). If all the energy to accelerate the fuel comes from an external source (and there is no additional momentum transfer), then the relationship between effective exhaust velocity and specific impulse is as follows:
$I_{sp} = \frac {v_e}{\sqrt{1 - \frac{v_e^2}{c^2}}} = \gamma_e \ v_e,$
where $\gamma$ is the Lorentz factor.
In the case of no external energy source, the relationship between $I_{sp}$ and the fraction of the fuel mass $\eta$ which is converted into energy might also be of interest; assuming no losses, is
$\eta = 1 - \sqrt{1 - \frac{I_{sp}^2}{c^2}} = 1 - \frac{1}{\gamma_{sp}}.$
The inverse relation is
$I_{sp} = c \cdot \sqrt{2 \eta - \eta^2}.$
Here are some examples of fuels, the energy conversion fractions and the corresponding specific impulses (assuming no losses if not specified otherwise):
Fuel $\eta$ $I_{sp} / c$
electron-positron annihilation 1 1
proton-antiproton annihilation, using only charged pions 0.56 0.60
electron-positron annihilation with simple hemispherical absorption of gamma rays 1 0.25
electron-positron annihilation with hemispherical Compton scattering 1 >0.25
nuclear fusion: H to He 0.00712 0.119
nuclear fission: 235U 0.001 0.04
### Delta-v
In order to make the calculations simpler, we assume that the acceleration is constant (in the rocket's reference frame) during the acceleration phase; however, the result is nonetheless valid if the acceleration varies, as long as $I_{sp}$ is constant.
In the nonrelativistic case, one knows from the (classical) Tsiolkovsky rocket equation that
$\Delta v = I_{sp} \ln \frac {m_0}{m_1}$
Assuming constant acceleration $a$, the time span $t$ during which the acceleration takes place is
$t = \frac {I_{sp}}{a} \ln \frac {m_0}{m_1}$
In the relativistic case, the equation still valid if $a$ is the acceleration in the rocket's reference frame and $t$ is the rocket's proper time because at velocity 0 the relationship between force and acceleration is the same as in the classical case.
By applying the Lorentz transformation on the acceleration, one can calculate the end velocity $\Delta v$ relative to the rest frame (i. e. the frame of the rocket before the acceleration phase) as a function of the rocket frame acceleration and the rest frame time $t'$; the result is
$\Delta v = \frac {a \cdot t'} {\sqrt{1 + \frac{(a \cdot t')^2}{c^2}}}$
The time in the rest frame relates to the proper time by the following equation:
$t' = \frac{c}{a} \sinh \left(\frac{a \cdot t}{c} \right)$
Substituting the proper time from the Tsiolkovsky equation and substituting the resulting rest frame time in the expression for $\Delta v$, one gets the desired formula:
$\Delta v = c \cdot \tanh \left(\frac {I_{sp}}{c} \ln \frac{m_0}{m_1} \right)$
The formula for the corresponding rapidity (the area hyperbolic tangent of the velocity divided by the speed of light) is simpler:
$\Delta r = \frac {I_{sp}}{c} \ln \frac{m_0}{m_1}$
Since rapidities, contrary to velocities, are additive, they are useful for computing the total $\Delta v$ of a multistage rocket.
## Matter-antimatter annihilation rockets
It is clear on the basis of the above calculations that a relativistic rocket would likely need to be a rocket that is fueled by antimatter. Other antimatter rockets in addition to the photon rocket that can provide a 0.6c specific impulse (studied for basic hydrogen-antihydrogen annihilation, no ionization, no recycling of the radiation[2]) needed for interstellar space flight include the "beam core" pion rocket. In a pion rocket, antimatter is stored inside electromagnetic bottles in the form of frozen antihydrogen. Antihydrogen, like regular hydrogen, is diamagnetic which allows it to be electromagnetically levitated when refrigerated. Temperature control of the storage volume is used to determine the rate of vaporization of the frozen antihydrogen, up to a few grams per second (amounting to several petawatts of power when annihilated with equal amounts of matter). It is then ionized into antiprotons which can be electromagnetically accelerated into the reaction chamber. The positrons are usually discarded since their annihilation only produces harmful gamma rays with negligible effect on thrust. However, non-relativistic rockets may exclusively rely on these gamma rays for propulsion[3]. This process is necessary because un-neutralized antiprotons repel one another, limiting the number that may be stored with current technology to less than a trillion [4].
### Design notes on a pion rocket
The pion rocket has been studied independently by Robert Frisbee[5] and Ulrich Walter, with similar results. Pions, short for pi-mesons, are produced by proton-antiproton annihilation. The antihydrogen or the antiprotons extracted from it will be mixed with a mass of regular protons pumped inside the magnetic confinement nozzle of a pion rocket engine, usually as part of hydrogen atoms. The resulting charged pions will have a velocity of 0.94c (i.e. $\beta$ = 0.94), placing the theoretical upper limit on the speed of the rocket, and a Lorentz factor $\gamma$ of 2.93 which extends their lifespan enough to travel 2.6 meters through the nozzle before decaying into muons. Sixty percent of the pions will have either a negative, or a positive electric charge. Forty percent of the pions will be neutral. The neutral pions will decay immediately into gamma rays. These can't be reflected by any known material at the energies involved, although they can undergo Compton scattering. They can be absorbed efficiently a shield of tungsten placed between the pion rocket engine reaction volume and the crew modules and various electromagnets to protect them from the gamma rays. The consequent heating of the shield will cause it to radiate visible light, which could then be collimated to increase the rocket's specific impulse[2]. The remaining heat will also require the shield to be refrigerated[5]. The charged pions would travel in helical spirals around the axial electromagnetic field lines inside the nozzle and in this way the charged pions could be collimated into an exhaust jet that is moving at 0.94c. In realistic matter/antimatter reactions, this jet only represents a fraction of the reaction's mass-energy : over 60% of it is lost as gamma-rays, collimation is not perfect, and some pions are not reflected backwards by the nozzle. Thus, the effective exhaust velocity for the entire reaction drops to just 0.58c[2]. Alternative propulsion schemes include physical confinement of hydrogen atoms in an antiproton and pion-transparent beryllium reaction chamber with collimation of the reaction products achieved with a single external electromagnet; see Project Valkyrie.
## Sources
1. The star flight handbook, Matloff & Mallove, 1989
2. Mirror matter: pioneering antimatter physics, Dr. Robert L Forward, 1986
## References
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2021-01-23 14:45:04
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https://www.physicsforums.com/threads/on-finding-lie-algebra-representations.468972/
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# On Finding Lie Algebra Representations
antibrane
What I am trying to do is start with a Dynkin diagram for a semi-simple Lie algebra, and construct the generators of the algebra in matrix form. To do this with su(3) I found the root vectors and wrote out the commutation relations in the Cartan-Weyl basis. This gave me the structure constants of the algebra and then I was able to write down the matrices corresponding to the regular (adjoint) representation.
What I am confused about (assuming the above is a correct method) is how to find different matrix representations than this. The representation I am familiar with for su(3) consists of the "Gell-Man matrices" which are 3 dimensional and obviously my matrices from my construction are 8 dimensional.
How would I find different representations starting from the commutation relations? I know how to find them starting with the Lie Group. Let me know if I need to clarify something--thanks a lot.
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2022-12-02 13:41:04
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https://proxieslive.com/if-anything-can-be-verified-efficiently-must-it-be-solvable-efficiently-on-a-non-deterministic-machine/
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# If anything can be verified efficiently, must it be solvable efficiently on a Non-Deterministic machine?
Suppose, I wanted to verify the solution to $$2$$^$$3$$. Which is $$8$$.
The $$powers~of~2$$ have only one 1-bit at the start of the binary-string.
## Verify Solution Efficently
``n = 8 N = 3 IF only ONE 1-bit at start of binary-string: IF total_0-bits == N: if n is a power_of_2: OUTPUT solution verified, 2^3 == 8 ``
A solution will always be approximately $$2$$^$$N$$ digits. Its not possible for even a non-deterministic machine to arrive to a solution with $$2$$^$$N$$ digits faster than $$2$$^$$N$$ time.
Question
Can this problem be solved efficently in non-deterministic poly-time? Why not if the solutions can be verified efficently?
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2020-10-26 21:54:53
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http://physics.stackexchange.com/questions/82041/how-much-time-does-it-takes-an-electron-to-tunnel-through-a-barrier
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How much time does it takes an electron to tunnel through a barrier?
I know that in quantum mechanics there is no "time operator", so such a question is ill-posed. Anyway if the tunneling is instantaneous, this would imply an information transmission faster than $c$. On the other hand, how could someone define such a "time"?
-
Possibly duplicate of: physics.stackexchange.com/q/38237 – Ruslan Dec 15 '13 at 20:07
Assume a particle moves in a potential, in one dimension, of the form $V(x) = \infty$ if $x^2 > a^2$ else $V(x) = \gamma \delta(x)$,where $\gamma >0$. Let $E_0$ be the energy of the particle. Then $\langle v \rangle = \sqrt{\frac{2E_0}{m}}$
Frequency of collisions = $\frac{\langle v \rangle}{a}$
Frequency of tunelling = $\frac{\langle v \rangle}{a} * T$
$T$ is the transmission probability which can be calculated by solving the Schrodinger equation.
Time required for a particle to tunnel = $\frac{a}{\langle v \rangle * T}$
Edit: Corrected Typo. $V(x)=\infty$
-
I'm more interested in a finite size potential, say with FWHM width $a$. Then I don't like your dimensionality argument because in the potential barrier the velocity is imaginary, thus $t = \langle v \rangle / a$ is an imaginary time – Mattia Jan 24 '14 at 13:53
I think what you're asking here is:
"If a particle is about to tunnel through a barrier, how long will it take to get from one side to the other?"
If so then you should consider how the particle is tunnelling through the barrier. It tunnels through the barrier because it's wavefunction "leaks" through the barrier - which means that it has a non-zero probability to be located outside of the barrier (i.e. it's already outside the barrier "when it starts to tunnel through it"). Because of this, the question unfortunately doesn't have a well defined answer.
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2015-08-29 07:22:42
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https://elmwealth.com/how-much-of-a-good-thing-is-best-for-you/
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# How much of a good thing is best for you?
By Victor Haghani and Andrew Morton 1
A thought experiment:
You can invest your wealth in only two assets: a risk-free one and a market portfolio of all public equities. Your investment choices, however, are limited to: A) put 100% in the risk-free asset, or B) 10% in the risk-free asset and 90% in equities. You cannot mix A) and B) – you must choose one or the other. What is the lowest return (above the risk-free rate) you would need to expect from equities for you to choose B? 2
Do you have your number in mind? Great. Now imagine you’re still in this two asset world and you wake up one day and find that the expected return of equities is, in fact, exactly equal to your answer above. But now you’re no longer limited to just options A and B; you’re completely free to invest however much you like in equities, from 0% to 100% or more. How would you invest now?
We’ve asked about a dozen friends this question, all financial professionals. If you’re like them, you’ve likely read the question twice, trying to understand exactly what we’re getting at. You may feel like you just answered that question, and isn’t 90% the answer?
Well, no, we don’t think it is. Please read on as we try to explain why we think 45% is about the right answer, and how we can use the perspective of this problem to answer some other interesting questions.3
You can get an immediate intuition for the problem and its solution by replacing our first question with: how much ketchup on your fries would be so much that you’d be indifferent between no ketchup and that much ketchup? 4 And then replace our follow-up question with: how much ketchup is your optimal amount? Our first question was calibrating your risk aversion via indifference points, and the follow-up question involved choosing an optimal point anywhere in between. The halfway point is a pretty good estimate, and exact in some investing models.
Putting a price on risk:
The properties of risk aversion are central to this problem, so let’s analyse the simplest type of risk, a 50/50 coin flip. A side payment is needed to make a typical, risk-averse person indifferent to risking a fraction $Latex formula$ of his wealth on such a flip. How should this required side payment vary with $Latex formula$ ? There is a very good reason to suggest that it should be proportional to $Latex formula$, meaning we should demand four times the side payment for twice the risk. To see this, compare a single flip risking $Latex formula$ of your wealth with two consecutive flips each risking 1%. In both cases the mean and variance of total wealth change are the same (0 and 2).5 It seems reasonable that the total side payment should be the same too, which is only the case if the side payment is proportional to $Latex formula$.
Thus the problem of side payments for coin flips boils down to choosing the specific multiple of $Latex formula$ as your indifference point. Although it’s a matter of individual preference, we suggest a reasonable range for this multiple is 1 to 2.6 Let’s use 1, meaning you are indifferent to an offer of 1% compensation to take on a $Latex formula$ 10% of wealth flip, or 4% to take on a $Latex formula$ 20% flip and so on. You would accept coin flip offers with higher side payments than this, and decline those with lower.
To connect the coin example to the stock market, let’s simplistically model investing 90% of savings in the stock market for one year as risking 18% of wealth on the flip of a coin, meaning over a year we’ll either lose 18% or gain 18%.7 The rule suggests that we’d require a side payment of 3.2% (i.e. 0.182) of our wealth, or 3.6% on the 90% we have invested in equities, to be indifferent.8 You can see that in answering the question about what expected return you would need to be indifferent between putting 90% of your savings into equities or 0% in equities, you were calibrating your function of risk aversion.
Let’s now turn to the question of how much you should invest given the freedom to choose any trade size you like. Consider what happens as you increase your investment from 0. Your expected gain goes up proportionally, but your risk – and hence, the required reward for bearing it – goes up with the square of the investment. Adding these two effects gives the diagram below, illustrating why the optimal point is halfway between the indifference points of 0% and 90%.9 So, someone who is indifferent to investing 90% in equities if they had a 3.6% expected return would optimally invest 45% of his or her wealth in equities at that expected return. In general, optimal size is half the indifference point size.
What kind of questions can we answer with this simple framework?
• What expected return on equities would you need for your optimal allocation to be 75%?
Recall that the required minimum return you said would make you indifferent to being 90% invested in equities, R90%, is also the expected return for which 45% is your optimal allocation. Optimal allocation varies in proportion to expected return, and so for it to be 75%, the expected return needs to be 75%/45% higher than your R90%. For our prototypical investor who has R90% of 3.6%, the expected return of equities at which a 75% allocation would be optimal is 6%.10
• What is the effect of over-investing?
As can be seen from the diagram, if you invest double your optimal allocation, you’ve thrown away all the benefit of the investment, and things get worse at an increasing rate from there. This means we should err on the side of taking less risk in the face of uncertainty about the probabilities of future returns.
• At the other extreme, what about small investments? For example: a \$100 coin flip for an investor whose net worth is \$100,000?
The $Latex formula$ rule of thumb suggests being indifferent at a side payment of about (0.0012) x \$100,000 = 10 cents! That seems crazy. But if that investor has 45% in the stock market and the rest in the riskless asset, over about one hour his or her wealth will fluctuate by about \$100 (with stock volatility of 20%). Using the same 3.6% excess return of equities as earlier, the expected gain in that hour is about 19 cents. So if optimally invested wealth is producing 19 cents for \$100 of risk, he or she is roughly indifferent to a side payment of half that. The framework therefore suggests (if needed!) accepting smaller rewards for small risks than might at first seem intuitive.
• Should a passive investor in the stock market have a static allocation to equities?
Investment advisers typically subject their clients to a risk evaluation to determine an appropriate allocation to equities, often taking the form of the kind of indifference questions we posed. If you believe that the expected return and/or risk of the equity market change over time, then your optimal allocation to equities should also change.11
• Does horizon matter?
To the extent investments are like coin flips, following a random walk, with the risk we care about (variance) and the compensation we’re being offered to accept that risk both growing proportionately with time, then, whether we choose a month, a year or a decade as the horizon for our investment won’t affect our choice of the optimal amount we should invest in it. Whether we’re flipping a biased coin once or 300 times, the amount we put at risk as a fraction of wealth should be the same.
• How much is it worth to be able to invest in equities?
We don’t need to believe that we can beat the market for us to put a positive value on the opportunity to invest in equities. Recall that we’re choosing an allocation to equities that maximizes the surplus of what we’re expecting to be paid over what we need to get paid to accept that risk. That surplus is equal to half of the risk premium multiplied by our optimal allocation. So, for example, if we see equities priced to deliver a 5% expected return, and our optimal allocation is 60%, then our surplus is $Latex formula$ x 60% x 5% = 1.5% pa. This tidy sum should make us feel pretty good, even grateful, about the existence of the equity market and our ability to freely choose how much of it we’d like. We’re being invited to play a game with favorable odds, like betting on the flip of a coin that has a 60% heads bias, or playing in a poker game where the other players have more money than skill.12
• What is ‘Kelly’ betting, and how does it relate? The Kelly criterion, named after the scientist at Bell Labs credited with formulating it in 1956, tells us how to bet to maximize the expected growth rate of our wealth. Implicit in the Kelly criterion is a level of risk aversion that is half of what we’ve been using in these examples. So a Kelly investor in the stock market would need only 1.8% excess return to be indifferent to holding 90% of wealth in the market versus nothing – equivalently, he or she would only need to expect 3.6% to optimally be 90% invested in equities.13 This is a much more aggressive posture than most investors we’ve met seem comfortable with.
Conclusion:
We hope this discussion has given you some simple but versatile tools for thinking about a broad range of investment-related questions. Seeing the risk-taking decision more like tuning the dial on a radio, rather than flipping an on-off switch, enables us to get the most out of any game where the odds are in our favor, including investing in the stock market. The trick is to tune your portfolio to the point where the cost of risk, which is increasing quadratically, is just about to grow faster than the expected return you’re being paid to take that risk, which is increasing linearly. We hope you’ve found some ‘utility’ in this brief summary, and that we haven’t taken too many liberties in attempting to distill what are some of the most valuable insights in finance from the last 400 years, from Daniel Bernoulli to Bob Merton, and many brilliant minds in between.
• Arrow, Kenneth J. “Alternative Approaches to the Theory of Choice in Risk-Taking Situations,” Econometrica, Oct 1951.
• Bernoulli, Daniel. “Exposition of a New Theory on the Measurement of Risk,” (1738). Translated in Econometrica, (1954).
• Haghani, Victor and Richard Dewey. “Rational Decision-Making under Uncertainty: Observed Betting Patterns on a Biased Coin,” SSRN, 2016.
• Merton, Robert C. “Continuous-Time Finance.” 1990.
You can also see our pieces here, here and here on estimating the long-term expected return of the stock market and its risk.
1. [1] This not is not an offer or solicitation to invest, nor should this be construed in any way as tax advice. Past returns are not indicative of future performance.
2. [2] We will always use ‘the expected return of equities’ to mean the expected return above the risk-free rate.
3. [3] If you answered 45%, you can pat yourself on the back and get back to teaching your finance students.
4. [4] Sorry, but we’re assuming you’re North American and do like ketchup on your fries.
5. [5] To be precise, if the second flip is for stakes of 1% of the new wealth, the variance of wealth change is 2.0001%.
6. [6] A multiple of 1 corresponds to that of an investor with power utility function with relative risk aversion parameter $Latex formula$. The power utility function is given by: $Latex formula$ for $Latex formula$, $Latex formula$ for $Latex formula$. With $Latex formula$, we won’t accept a fair coin flip where we lose $Latex formula$ of our wealth, no matter how big the upside if we win.
7. [7] Investing in the stock market isn’t the same as flipping a coin with known probability of outcomes. There is uncertainty in addition to risk concerning the distribution of outcomes. The setup of the problem in this note leaves out many real world practicalities, such as the value of one’s human capital, the correlation of future consumption with the market, and the existence of other assets, to name just a few. We also have modeled the stock market as normally, rather than log-normally, distributed to the one year horizon.
8. [8] By the same logic, the risk of the stock market is 18%/0.9 = 20%.
9. [9] The net value for any allocation, $Latex formula$, to equities is: $Latex formula$, where $Latex formula$ is the proportion committed to the risky investment, and $Latex formula$ is the coefficient of risk aversion. The objective is maximized with respect to $Latex formula$ when $Latex formula$.
10. [10] Let’s use the notation $Latex formula$ for earning a side payment of fraction $Latex formula$ of wealth in exchange for risking fraction $Latex formula$ on a fair coin flip. What is the optimal fraction of wealth, $Latex formula$, of this risk we should choose, assuming we are indifferent to flips of $Latex formula$? To be indifferent, we need to choose $Latex formula$ such that $Latex formula$, or $Latex formula$. The optimal choice is half of that, so $Latex formula$. You can see that the optimal allocation, $Latex formula$, moves in proportion to expected return, $Latex formula$, and in inverse proportion to the square of risk, $Latex formula$. So, if stock market risk dropped by 10% (say from 20% to 18% in our example), then your optimal allocation to equities would go up by $Latex formula$.
11. [11] Except if the changes in the return and risk of the market are such as to keep the ratio of return to variance constant.
12. [12] We could also view this 1.5% a year as the risk-free payment we’d need to receive to forego being able to invest in the equity market, and only be able to invest in the risk-free asset. We’ve asked a version of this question to about 30 of our friends. We’re still conducting this survey, so stay tuned for the results in a future note.
13. [13] As above, assuming 20% annual stock market volatility. The Kelly criterion in continuous time gives an optimal fraction to invest, $Latex formula$ where $Latex formula$ is the annual risk of the stock market.
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2022-07-04 16:01:04
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https://www.codecademy.com/courses/learn-statistics-with-r/lessons/standard-deviation-in-r/exercises/standard-deviation
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Learn
Variance is a tricky statistic to use because its units are different from both the mean and the data itself. For example, the mean of our NBA dataset is 77.98 inches. Because of this, we can say someone who is 80 inches tall is about two inches taller than the average NBA player.
However, because the formula for variance includes squaring the difference between the data and the mean, the variance is measured in units squared. This means that the variance for our NBA dataset is 13.32 inches squared.
This result is hard to interpret in context with the mean or the data because their units are different. This is where the statistic standard deviation is useful.
Standard deviation is computed by taking the square root of the variance. sigma is the symbol commonly used for standard deviation. Conveniently, sigma squared is the symbol commonly used for variance:
$\sigma = \sqrt{\sigma^2} = \sqrt{\frac{\sum_{i=1}^{N}{(X_i -\mu)^2}}{N}}$
In R, you can take the square root of a number using ^ 0.5 or sqrt(), up to you which one you prefer:
num <- 25
num_square_root <- num ^ 0.5
### Instructions
1.
We’ve written some code that calculates the variance of the NBA dataset and the OkCupid dataset.
The variances are stored in variables named nba_variance and okcupid_variance.
Calculate the standard deviation by taking the square root of nba_variance and store it in the variable nba_standard_deviation. Do the same for the variable okcupid_standard_deviation.
# Take this course for free
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2021-09-16 19:47:24
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https://learn.careers360.com/engineering/question-pleaseplease-help-me-which-of-these-can-be-a-diagonal-element-in-a-square-matrix-with-6-elements-and-6-rows/
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Q
# Please,please help me Which of these can be a diagonal element in a square matrix with 6 elements and 6 rows ?
Which of these can be a diagonal element in a square matrix with 6 elements and 6 rows ?
• Option 1)
$a_{31}$
• Option 2)
$a_{77}$
• Option 3)
$a_{22}$
• Option 4)
$a_{15 }$
102 Views
As we have learned
Leading / principal diagonal of a matrix -
Diagonal from left hand side upper corner to the lower most element.
- wherein
For a diagonal element $a_{ij}; i= j$
Option 1)
$a_{31}$
Option 2)
$a_{77}$
Option 3)
$a_{22}$
Option 4)
$a_{15 }$
Exams
Articles
Questions
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2020-02-24 20:54:25
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http://codeforces.com/blog/entry/1256
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### Urbanowicz's blog
By Urbanowicz, 7 years ago, ,
Let integer number N to be power of 2. Now consider an array of N elements numbered from 0 to N − 1. We will name that array with “a” letter and “ai will denote its i-th element. Imagine that we have some rule to modify single ai and we want to apply that rule to every element between ax and ay inclusive. We will call this range modification.
Consider binary operation which is associative, commutative and has identity element. Let it be denoted by × sign, for example. Imagine that we want to evaluate that operation over elements between ax and ay inclusive (ax × ax + 1 × … × ay − 1 × ay). We will call this range query.
I am to describe non-recursive algorithms for performing range queries and modifications within O(logN) using data structure that takes O(N) additional space.
This is schematic representation of the data structure built for array of 8 elements. It consists of 4 layers (log2N + 1 in general). Each layer contains a set of ranges that don't overlap and cover the whole array. Ranges of the same layer have equal size that grows exponentially from bottom to top. All the ranges are numbered in row-major order from left to right, from top to bottom. The bottom-most (numbered from 8 to 15 here, from N to 2N − 1 in general) layer is intended to hold ai array values. Other ranges hold an answer for their range query.
One may call this structure a segment or interval tree, but thinking that this structure is a tree will give you nothing useful here.
Range Query over Static Array
First of all we will discuss how to implement range query while ignoring modifications. Assume that we have already built the structure somehow. Then the sample C++ implementation for sum operation looks as follows. (Please note that zero is an identity element of sum operation.)
int staticQuery(int L, int R) {
L += N;
R += N;
int sum = 0;
for (; L <= R; L = (L + 1) >> 1, R = (R - 1) >> 1) {
if (L & 1) {
sum += a[L];
}
if (!(R & 1)) {
sum += a[R];
}
}
return sum;
}
Let's break the function's code down. As was mentioned before, ai element resides at i + N index in the structure. That explains what first two statements do: query's range bounds are turned into indices of the bottom layer. Next statement initializes sum variable with identity. This variable will accumulate the sum in the range to return the answer.
Now look at the for-loop and try to figure out how it changes L and R indices. In this picture colored ranges are ranges that were traversed while answering staticQuery(3, 7). Yellow ones were ignored, red ones were accounted. It's plain to see that red ranges don't overlap and their union is 3–7 range.
For simplicity I will explain the movement of L index only. Note that we must get into account a[L] value if L is odd because other ranges that cover L-range (1, 2, 5 ranges for L = 11) depend on ai values that don't appear in the query's range. That is what first if-statement checks. Of course we should ignore a[L] value if L is even because there is wider range that covers L-range and doesn't confront with query's range (3 range for L = 6).
As you can guess from the picture, the next range we will examine has L / 2 index (in case of even L). Generally, we will get a set of covering ranges if we divide L by powers of 2. In case of odd L we aren't interested in covering ranges, but we can notice that “interesting” ranges reside rightwards. It's easy to prove that L = (L + 1) >> 1 assignment takes into account both cases.
As for the R index, everything is the same. But because of this index is responsible for the right edge, everything will be right-to-left. That's why second if-statement and R = (R - 1) >> 1 assignment slightly differ from their counterparts. And of course, we must stop the for-loop when L > R because when L-range and R-range meet, the query's range is completely covered.
Note 1. It is possible to implement the algorithm in such a way that it won't require operation to be commutative. You can even avoid identity element requirement.
Note 2. If operation is idempotent (minimum, for example), you can omit both if-statements and rewrite body of the for-loop in a single line:
m = min(m, min(a[L], a[R]));
And don't forget that identity element of minimum is positive infinity, not zero.
Note 3. The algorithm can easily be adapted for multidimensional queries. Just look at the code for 2D-case.
int staticQuery2D(int x1, int x2, int y1, int y2) {
x1 += N;
x2 += N;
int sum = 0;
for (; x1 <= x2; x1 = (x1 + 1) >> 1, x2 = (x2 - 1) >> 1) {
int i1 = y1 + N;
int i2 = y2 + N;
for (; i1 <= i2; i1 = (i1 + 1) >> 1, i2 = (i2 - 1) >> 1) {
if (x1 & 1) {
if (i1 & 1) {
sum += a[x1][i1];
}
if (!(i2 & 1)) {
sum += a[x1][i2];
}
}
if (!(x2 & 1)) {
if (i1 & 1) {
sum += a[x2][i1];
}
if (!(i2 & 1)) {
sum += a[x2][i2];
}
}
}
}
return sum;
}
Modification of Single ai Element
Performing single element modification is two-step. First, you modify the value stored at i + N index in order to modify ai. Second, you reevaluate values for all covering ranges from bottom layer to the very top.
void pop(int i) {
a[i >> 1] = a[i] + a[i ^ 1];
}
void popUp(int i) {
for (; i > 1; i >> 1) {
pop(i);
}
}
We already know how popUp() enumerates covering ranges, so let's focus on its subroutine. You can see that operation is evaluated over ranges with indices i and i ^ 1. It's not hard to figure out that these ranges are subranges of the same i >> 1 range. This reevaluation is correct because operation is associative. Please note that this code is not suitable if operation is not commutative.
Clearly, we can use single element modifications to build the structure up. As each modification has logarithmic complexity, the build up will take O(NlogN). However, you can do this within O(N) by using the following code. (You still need to load up values in the bottom-most layer, of course.)
void buildUp() {
for (int i = 2 * N - 1; i > 1; i -= 2) {
pop(i);
}
}
Range Query within O(log2N) after Transparent Range Modifications
Imagine we have a set of range modifications with different parameters. Then we apply these modifications over array in some order. This will result in some state of the array. Now imagine we apply the same modifications over original array but in another order. The result may be the same as previous and may be not. If result doesn't depend on the order we will call such modifications transparent. For example, “increment all the values between first and third inclusive by five” is a transparent modification.
Assume that the range we want to modify is not arbitrary but one of the presented in the structure. We can simply apply modification to its value, but it will leave the structure in inconsistent state. Although we know how to update covering ranges, we won't do that for every such modification because there is more efficient way to update them when we will perform modification for arbitrary range. So we will assume that covering ranges became consistent somehow. But there are still inconsistent subranges. Of course, we cannot update them directly, because their number is O(N). Instead, we will attach a modifier that will store parameters of that modification virtually applied to every subrange. Here's the code for performing increment with parameter p:
void modifierHelper(int i, int p) {
a[i] += p;
m[i] += p;
}
Although both statements look the same, they are different in sense. First statement performs modification itself using parameter p. Second statement calculates a parameter for modification in such a way that this modification is equivalent to sequential appliance of modifications with parameters m[i] and p.
Now, how to perform arbitrary modification. Recall that we already know how to represent arbitrary ranges from staticQuery() function. So the code for increment modification will be:
void pop(int i) {
a[i >> 1] = (a[i] + m[i >> 1]) + (a[i ^ 1] + m[i >> 1]);
}
void modify(int L, int R, int p) {
L += N;
R += N;
int LCopy = L;
int RCopy = R;
for (; L <= R; L = (L + 1) >> 1, R = (R - 1) >> 1) {
if (L & 1) {
modifierHelper(L, p);
}
if (!(R & 1)) {
modifierHelper(R, p);
}
}
popUp(LCopy);
popUp(RCopy);
}
Two last statements is a more efficient way to update covering ranges that I mentioned before. It's enough to do only these pop ups. Please note that the pop() subroutine has been changed. That's because i and i ^ 1 ranges are under influence of covering range's modifier.
Now, how to answer range queries. Note that we cannot just use staticQuery() function, because traversed ranges may be affected by modifiers in higher layers. To find out real value stored in the range we will use this subroutine which is just an appliance of all modifiers of covering ranges:
int realValue(int i) {
int v = a[i];
for (i >>= 1; i > 0; i >>= 1) {
v += m[i];
}
}
Then this code will perform range query:
int log2Query(int L, int R) {
L += N;
R += N;
int sum = 0;
for (; L <= R; L = (L + 1) >> 1, R = (R - 1) >> 1) {
if (L & 1) {
sum += realValue(L);
}
if (!(R & 1)) {
sum += realValue(R);
}
}
return sum;
}
Range Query and Opaque Range Modification within O(logN)
As opposed to transparent, opaque modifications' result depends on the order of appliance. However, transparent modifications can be viewed as opaque, and it makes opaque case more general than transparent one. In this section we will learn as an example how to perform range assignment and range sum query. (Range assignment changes each value in the range to desired value p.)
Simple trick with realValue() won't work anymore because now modifications in lower layers can affect modifiers in higher layers. We need a way to detach modifiers from the ranges we need to update. The way is “push down”. It is so called because it removes modifier at the range, and applies that modifier down to both subranges.
void push(int i) {
if (h[i >> 1]) {
modifierHelper(i, m[i >> 1]);
modifierHelper(i ^ 1, m[i >> 1]);
h[i >> 1] = false;
}
}
void pushDown(int i) {
int k;
for (k = 0; (i >> k) > 0; k++);
for (k -= 2; k >= 0; k--) {
push(i >> k);
}
}
Let's look at pushDown() at first. All you need to know about this is that it enumerates the same indices as popUp() does, but in reverse order. This reverse order is required because push downs propagate from top to bottom.
Now look at push(). There is new value h[i >> 1] which is true if modifier is attached to i >> 1 and false otherwise. If modifier is present then it is attached to both subranges and i >> 1 is marked as free of modifier.
In order to support range assignments and sum queries we need new modifierHelper() and pop() subroutines. I believe it will not be hard to understand how they work.
void modifierHelper(int i, int p) {
a[i] = p * q[i];
m[i] = p;
h[i] = true;
}
void pop(int i) {
if (h[i >> 1]) {
a[i >> 1] = m[i >> 1] * q[i >> 1];
} else {
a[i >> 1] = a[i] + a[i ^ 1];
}
}
The only thing you couldn't know is what's the q[i] values. Answer is simple: this is number of array elements which covered by i range. Please note that these values are necessary for the problem of our example only, you don't need them in general.
The most direct way to calculate q[i] values is to use this routine which is similar to buildUp() function:
void makeQ() {
for (int i = N; i <= 2 * N - 1; i++) {
q[i] = 1;
}
for (int i = 2 * N - 1; i > 1; i -= 2) {
q[i >> 1] = q[i] + q[i ^ 1];
}
}
Clearly, after all necessary push downs have been made, we can modify range straightforwardly. There are only two push downs we need to do. So the code for range assignment is as follows:
void modify(int L, int R, int p) {
L += N;
R += N;
int LCopy = L;
int RCopy = R;
pushDown(LCopy);
pushDown(RCopy);
for (; L <= R; L = (L + 1) >> 1, R = (R - 1) >> 1) {
if (L & 1) {
modifierHelper(L, p);
}
if (!(R & 1)) {
modifierHelper(R, p);
}
}
popUp(LCopy);
popUp(RCopy);
}
Absolutely the same holds for range sum query:
int query(int L, int R) {
L += N;
R += N;
pushDown(L);
pushDown(R);
int sum = 0;
for (; L <= R; L = (L + 1) >> 1, R = (R - 1) >> 1) {
if (L & 1) {
sum += a[L];
}
if (!(R & 1)) {
sum += a[R];
}
}
return sum;
}
Note 4. Sometimes you can avoid use of h[i] values. You can simulate detachment by “undefined” modifier.
Note 5. If your modifications are transparent, you can omit push downs in modify(). Also you can avoid push downs in query(), but this won't be as simple as removing a pair of statements.
Surprise!
In the very beginning I have said that N should be integer power of 2. In fact, that's not true. All the algorithms described here work correctly for any positive integer N. So for any array of size N this structure takes just O(N) of additional space.
•
• +21
•
7 years ago, # | +3 Hi! Excellent tutorial. However, I don't get the last statement: I have seen non-recursive implementations of this data structure for sizes 1<
• 7 years ago, # ^ | 0 I didn't understand "what happens" section. First of all say what's the N value? And what is your complain?
• 7 years ago, # ^ | ← Rev. 2 → 0 In your last statement you said that what you described above works not only for N which is a power of 2 but for any integer value of N. I tried to describe the situation when N is equal to 7 dividing the interval 1-7 into subintervals. In case if N is a power of 2 all the leaves (intervals of length 1) have numbers from N+1 to 2*N ( 1 - based). However, if N is not power of two then intervals of length 1 don't necessarily have numbers in that range and the code you used above ( l+=N, r+=N) that put l and r into the leftmost and rightmost intervals of lenghts 1 inside the query interval doesn't work. As an example, if N is 7, then interval 1-1 has number 4, but interval 2-2 has number 8, at least if we divide each interval into two halfs where the first is not longer than the second.
• 7 years ago, # ^ | 0 1 2 3 4 5 6 7 8 9 10 11 12 13That's how the tree will look like. I will return in 0-base. Zero index of array is located at 7 in tree. First - at 8. Second - at 9... And Sixth in 13.If you trace the algorithms, you'll see that everything is ok. Obviously 3-rd node contains bogus value but its value is never used.
• 7 years ago, # ^ | 0 You're right. Thank you, it was my mistake - I got used to fact that leaves should go visually from left to right but it is not true in this case.
7 years ago, # | ← Rev. 2 → 0 I'm sorry, but I found mistakes in section "Range Query within O(log2N) after Transparent Range Modifications". Implementations for modifierHelper(int i, int p) and pop(int i) are wrong. They cannot be implemented without q[i] values as described in "Range Query and Opaque Range Modification within O(logN)" section.And, yes, realValue too.
• 7 years ago, # ^ | ← Rev. 3 → 0 Here's proper implementations (I hope :) Also there is little mistake in popUp.void modifierHelper(int i, int p) { a[i] += p * q[i]; m[i] += p; } void pop(int i) { a[i >> 1] = (a[i] + m[i >> 1] * q[i]) + (a[i ^ 1] + m[i >> 1] * q[i ^ 1]); } int realValue(int i) { int v = a[i]; for (int j = i >> 1; j > 0; j >>= 1) { v += m[j] * q[i]; } }void popUp(int i) { for (; i > 1; i >>= 1) { pop(i); } }
7 years ago, # | ← Rev. 2 → +1 I use this implementation, it is more clear for me. RSQ can sum all numbers from interval [L, R), and RevRSQ adds DX to each number in interval [L, R)
• 7 years ago, # ^ | 0 I was waiting to see your implementation for a long time :)But as I can see neither RSQ nor RevRSQ support both range query and modification. Am I right?
• 7 years ago, # ^ | 0 yes. you are right. but it is easy to remember, and if you don't need both, you can use it.
7 years ago, # | 0 Give some problems link for practicing.Thanks.........
• 7 years ago, # ^ | +5 There is no problems that require you to use non-recursive implementation only. Hence, just lurk for segment/interval tree problems. Here is topic about some of these problems:It's in russian, but this will not be an obstacle, because most of links presented in that topic are for English problemsets.
» 2 years ago, # | ← Rev. 5 → 0 5 elements. (Read as 'count of slot 0 is 5') 0->5 1->0 2->0 3->0 4->0. Therefore for query <1, the answer should be 4 cuz 4 slots have value <1 (1,2,3 and 4).according to this implementation when array is not of size 2^n, the tree is coming out to be wrong. n=5. so we start filling at n=5.We have 4 slots with value [0,1) and 1 slot with value [5-6). The array would be like: (started filling at 5 cuz n=5, representing [0,1))index:0 1 2 3 4 5 6 7 8 9 10value:0 0 0 0 0 4 0 0 0 0 1now when making segment for array index 10, the value at index 5 gets overwritten. How to handle this?
• » » 2 years ago, # ^ | 0 10 is out of bounds, The last of five elements has index 9, not 10.
• » » » 2 years ago, # ^ | ← Rev. 2 → 0 why? I have 5 elements but then the count can go upto 6,ie, the queries could be <1,<2,<3,<4,<5,<6. So, my segment tree contains the segments of counts and not the elements. I used two methods to do this. In one method, I created a full binary tree. That worked. Then I read that the standard iterative method of segment tree is generalized and works for any n. But its failing this case. If i start filling elements from power(2,height of tree) to power(2,height of tree)+n, its working fine. But a tree of size 2*N is not.
• » » » » 2 years ago, # ^ | 0 Implementation described in this post uses inclusive bounds for queries. So, if you have 5 elements, then allowed queries are of form [L, R], where 0 ≤ L ≤ R ≤ 4.
• » » » » » 2 years ago, # ^ | 0 I want to store the count as segments. Suppose I were to assign any number of these elements to certain number of slots at any point of time. Now i want "how many slots have <=3 elements"...such kind of queries. Then in that case, a slot could contain 0-5 of these elements, which makes the segment size of root as 6 (0-5). This generalized implementation is failing for the above mentioned case.
• » » » » » » 2 years ago, # ^ | 0 If indices range from 0 to 5 then you have six elements and therefore should use 6 as an offset. Power of two works because it is larger than 6.
» 4 months ago, # | ← Rev. 4 → -8 < DELETED >
|
2018-05-27 21:40:28
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|
http://math.stackexchange.com/tags/recurrence-relations/hot
|
# Tag Info
4
Hint: Check that $\displaystyle \forall n\geq 0, x_{n+1}-x_n=\frac{(-1)^n}{n+1}$ The series $\displaystyle \sum (x_{n+1}-x_n)$ is therefore convergent, and so is the sequence $(x_n)$, say $x_n\to l$ Furthermore, $\displaystyle l= \sum_{k=0}^\infty (x_{k+1}-x_k) =\sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \log 2$. Edit: from a simulation on Mathematica, ...
4
Hint. From $$a_{n+1} = \frac{1}{\frac{1}{a_n}+n}$$ you get $$\frac1{a_n}-\frac1{a_{n+1}}=-n$$ then use a telescoping sum and a standard sum.
2
Let $b_n=1/a_n$ and find the recurrence satisfied by it.
2
I just wish to contribute a "quicker" development of Ragib Zaman's derivation. Just as he had shown, $$\cos^k \theta = \left( \frac{ e^{i\theta} + e^{-i\theta} }{2} \right)^k = \frac{1}{2^k} \sum_{n=0}^k \binom{k}{n} (e^{i\theta} )^n (e^{-i\theta})^{k-n} = \frac{1}{2^k } \sum_{n=0}^k \binom{k}{n} e^{i(2n-k)\theta}$$ Now, assuming that $\theta$ is real, we ...
2
If the first two terms are $a$ and $b$ then you can add up terms until you get to the 7th. It will be: $$t_7=8a+13b$$ Similarly you can write down and add up the first 10 terms to get: $$s_{10}=\sum_{n=1}^{10}t_n=88a+143b$$ Solving the equality now gives: $$11(8a+13b)=88a+143b$$ As both sides are equal the first two numbers can be anything.
2
Hint: Let $a_n$ be the number of strings which satisfy the conditions of length $n$. If a string begins with $b$ or $c$, then you are left with a string of length $n-1$, which can start with any character. How many strings are there of this type (it is related to $a_{n-1}$)? If a string begins with $a$, then the second letter must be a $b$ or $c$, then ...
1
It's a second order differential equation, so there are $2$ arbitrary constants in the general solution. As you could see, there was a constraint on $c_2$, but there is none on $c_0$ and $c_1$, and there is no mutual constraint between the latter two constants. It seems most convenient to separate things out into $2$ series, one where all the exponents are ...
1
We can see that each solution is completed characterized by two numbers $c_0$ and $c_1$. Also, the terms that are influeced by $c_0$ do not overlap with the terms that are influenced by the terms influenced by $c_1$. In particular observe the relationship: $$y(c_0,c_1)=c_0y(1,0)+c_1y(0,1)$$
1
As a complement to the previous answers, observe that you can give initial conditions to select a unique solutions. The choice of initial data corresponds with the choice of the constanta $c_0$ and $c_1$: $c_0=1$ and $c_1=0$ corresponds to $y(0)=1$ and $y'(0)=0$. $c_0=0$ and $c_1=1$ corresponds to $y(0)=0$ and $y'(0)=1$.
1
The general theory about second order homogeneous differential equations says that (i) there is an open interval $I$ on the $x$-axis containing containing $0$ in its interior such that the given ODE has a two-dimensional vector space ${\cal L}$ of solutions $y(\cdot):\>I\to{\mathbb R}$, and (ii) these solutions are even real analytic, i.e. have convergent ...
1
If one denotes the first terms by $a_0$ and $a_1$, the $7$th term will be a linear combination of those, say $p_7 a_0+ q_7 a_1$. Same goes for the sum: $s_{10} =u_{10} a_0+ v_{10} a_1$. You do not really need to compute explicitely those linear terms. If you choose peculiar solutions: $a_0=0$ and $a_1=1$, the property works ($t^{0,1}_7=8$, ...
1
To gain intuition (and get the "right" solution), it is simpler to start consider the case where $n$ is a power of $2$, i.e. $n=2^k$ for some integer $k \geq 0$. (In what follows, all logarithms are in base $2$, because computer science.) We can write \begin{align} T(2^k) &= 16 T(2^{k-1}) + 2\cdot 2^{4k} = 2^4 T(2^{k-1}) + 2^{4k+1} \\ &= 2^4 ... 1 Suppose we seek to evaluateS(N) = \sum_{n=N}^\infty \sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j} (1+j)^n \frac{t^n}{n!}.$$This is$$S(N) = \sum_{n=N}^\infty \frac{t^n}{n!} \sum_{k=N}^n \sum_{j=0}^k {k\choose j} (-1)^{k-j} (1+j)^n.$$Now introduce$$(1+j)^n = \frac{n!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \exp((1+j)z) \; dz.$$We ... 1 Let x=A_\text{before};y=B_\text{before} x'=A; y'=B \lambda=C_{A\rightarrow B}; \mu=C_{B\rightarrow A} From equations given:$$\begin{align} \left. \begin{array}{1} x&=\dfrac{x'-(-\mu y)}{1-\lambda}\\ y&=\dfrac{y'-\lambda x}{1-\mu} \end{array} \right\}\\\\ \left. \begin{array}{1} (1-\lambda)x-\mu y&=x'\\ (1-\mu) y+\lambda ...
1
Hint: Use the *principle of superposition of solutions; find particular solutions of $$a_n=5a_{n-1} - 6_{n-2} + 4^n\quad (\text{resp. }{} + 2n, {}+ 3)$$ then add these particular solutions. Example for $4^n$: A particular solution will have the form $a_n=C 4^n$. It is indeed a solution if $$C4^n=5C4^{n-1}-6C4^{n-2}+4^n\iff 16C=20C-6C+16\iff C=8.$$
1
Note that the sequence satisfies the recurrence relation $x_{n+1} = x_n + \frac1{x_n} = \frac{x_n^2 + 1}{x_n}$ or $x_{n+1}x_n = x_n^2 + 1$. So if it converges, the limit $x$ must satisfy $$x^2 = x^2 + 1$$ which cannot be. Therefore the sequence cannot converge. Note that the sequence is always positive, so $1/x_n > 0$ and $x_{n+1} > x_n$, so the ...
Only top voted, non community-wiki answers of a minimum length are eligible
|
2016-05-01 18:22:49
|
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https://math.stackexchange.com/questions/3143471/how-many-permutations-are-there-of-m-m-a-a-a-t-t-e-i-k-so-that-no-two
|
# How many permutations are there of M, M, A, A, A, T, T, E, I, K, so that no two consecutive letters are the same?
How many permutations are there of $$M, M, A, A, A, T, T, E, I, K$$ so that there are no two consecutive letters are the same? I would use the Inclusion-exclusion principle where $$A_{i} = \{ \text{on} \ i-\text{th} \text{ and }(i+1)-\text{th} \text{ position, there are two same consecutive letters} \}.$$ So my answer would be $$\frac{10!}{2!3!2!} - 9 \cdot \left(\frac{9!}{3!2!} + \frac{9!}{3!2!} +\frac{9!}{2!2!}\right)+ 8 \cdot \left(\frac{8!}{2!2!}\right) +8 \cdot 7 \cdot \left(\frac{8!}{3!} + \frac{8!}{2!} +\frac{8!}{2!}\right) - 7 \cdot 6 \cdot 5 \cdot 7!$$
As you observed, there are $$10$$ letters, of which $$3$$ are $$A$$s, $$2$$ are $$M$$s, $$2$$ are $$T$$s, $$1$$ is an $$E$$, $$1$$ is an $$I$$, and $$1$$ is a $$K$$.
If there were no restrictions, we would choose three of the ten positions for the $$A$$s, two of the remaining seven positions for the $$M$$s, two of the remaining five positions for the $$T$$s, and arrange the $$E$$, $$I$$, $$K$$ in the remaining three positions in $$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! = \frac{10!}{3!7!} \cdot \frac{7!}{2!5!} \cdot \frac{5!}{2!3!} \cdot 3! = \frac{10!}{3!2!2!}$$ in agreement with your answer.
From these, we must subtract those arrangements in which or more pairs of identical letters are adjacent.
Arrangements with a pair of adjacent identical letters: We have to consider cases, depending on whether the identical letters are $$A$$s, $$M$$s, or $$T$$s.
A pair of $$A$$s are adjacent: We have nine objects to arrange: $$AA, A, M, M, T, T, E, I, K$$. Choose two of the nine positions for the $$M$$s, two of the remaining seven positions for the $$T$$s, and then arrange the five distinct objects $$AA$$, $$A$$, $$E$$, $$I$$, $$K$$ in the remaining five positions, which can be done in $$\binom{9}{2}\binom{7}{2}5!$$ ways.
A pair of $$M$$s are adjacent: We have nine objects to arrange: $$A, A, A, MM, T, T, E, I, K$$. Choose three of the nine positions for the $$A$$s, two of the remaining six positions for the $$T$$s, and arrange the four distinct objects $$MM, E, I, K$$ in the remaining four positions, which can be done in $$\binom{9}{3}\binom{6}{2}4!$$ ways.
A pair of $$T$$s are adjacent: By symmetry, there are $$\binom{9}{3}\binom{6}{2}4!$$ such arrangements.
Arrangements with two pairs of adjacent identical letters: This can occur in two ways. Either the pairs are disjoint or they overlap.
Two pairs of $$A$$s are adjacent: This can only occur if the three $$A$$s are consecutive. Thus, we have eight objects to arrange: $$AAA, M, M, T, T, E, I, K$$. Choose two of the eight positions for the $$M$$s, two of the remaining six positions for the $$T$$s, and arrange the four distinct objects $$AAA, E, I, K$$ in the remaining four positions, which can be done in $$\binom{8}{2}\binom{6}{2}4!$$ ways.
A pair of $$A$$s are adjacent and a pair of $$M$$s are adjacent: We have eight objects to arrange: $$AA, A, MM, T, T, E, I, K$$. Choose two of the eight positions for the $$T$$s and arrange the six distinct objects $$AA, A, MM, E, I, K$$ in the remaining six positions in $$\binom{8}{2}6!$$ ways.
A pair of $$A$$s are adjacent and a pair of $$T$$s are adjacent: By symmetry, there are $$\binom{8}{2}6!$$ such arrangements.
A pair of $$M$$s are adjacent and a pair of $$T$$s are adjacent: We have eight objects to arrange: $$A, A, A, MM, TT, E, I, K$$. Choose three of the eight positions for the $$A$$s and then arrange the remaining five distinct objects $$MM, TT, E, I, K$$ in the remaining five positions in $$\binom{8}{3}5!$$ ways.
Arrangements with three pairs of adjacent identical letters: We again consider cases.
Two pairs of $$A$$s are adjacent and a pair of $$M$$s are adjacent: We have seven objects to arrange, $$AAA, MM, T, T, E, I, K$$. Choose two of the seven positions for the $$T$$s and arrange the five distinct objects $$AAA, MM, E, I, K$$ in the remaining five positions in $$\binom{7}{2}5!$$ ways.
Two pairs of $$A$$s are adjacent and a pair of $$T$$s are adjacent: By symmetry, there are $$\binom{7}{2}5!$$ such arrangements.
A pair of $$A$$s are adjacent, a pair of $$M$$s are adjacent, and a pair of $$T$$s are adjacent: We have seven objects to arrange: $$AA, A, MM, TT, E, I, K$$. Since all the objects are distinct, there are $$7!$$ such arrangements.
Arrangements containing four pairs of adjacent identical letters: We have six objects to arrange: $$AAA, MM, TT, E, I, K$$. Since all the objects are distinct, there are $$6!$$ such arrangements.
By the Inclusion-Exclusion Principle, there are $$\binom{10}{3}\binom{7}{2}\binom{5}{2}3! - \binom{9}{2}\binom{7}{2}5! - \binom{9}{3}\binom{6}{2}4! - \binom{9}{3}\binom{6}{2}4! + \binom{8}{2}\binom{6}{2}4! + \binom{8}{2}6! + \binom{8}{2}6! + \binom{8}{3}5! - \binom{7}{2}5! - \binom{7}{2}5! - 7! + 6!=47760$$ arrangements in which no two adjacent letters are identical.
• I have added the final result, it matches perfectly with my computations. So +1 – Oldboy Mar 12 at 9:30
I used the following program to check it:
public class PermutationCounter {
public static int[] duplicate(int[] source) {
int[] dest = new int[source.length];
System.arraycopy(source, 0, dest, 0, source.length);
return dest;
}
public static boolean allLettersExhausted(int[] source) {
for(int i = 0; i < source.length; i++) {
if(source[i] != 0) {
return false;
}
}
return true;
}
private char[] letters;
private int[] count;
public PermutationCounter(char[] letters, int[] count) {
this.letters = letters;
this.count = count;
}
public int countPermutations() {
return countPermutations(-1, count, "");
}
private int countPermutations(int last, int[] alphabet, String permutation) {
if(allLettersExhausted(alphabet)) {
System.out.println(permutation);
return 1;
}
int sum = 0;
for(int i = 0; i < alphabet.length; i++) {
if(i != last && alphabet[i] > 0) {
int[] alphabetCopy = duplicate(alphabet);
alphabetCopy[i]--;
sum += countPermutations(i, alphabetCopy, permutation + letters[i]);
}
}
return sum;
}
public static void main(String[] args) {
char[] letters = new char[] {'M', 'A', 'T', 'E', 'I', 'K'};
int[] count = new int[] {2, 3, 2, 1, 1, 1};
PermutationCounter counter = new PermutationCounter(letters, count);
System.out.println("TOTAL PERMUTATIONS: " + counter.countPermutations());
}
}
My answer is 47760 (much lower number) and the full list of all permutations can be found here (yes, the word "MATEMATIKA" is also there).
The code starts with two lists. The first contains the list of letters, the second list contains the number of appearances of each letter in the final word. The code is recursive: it tracks the last letter and counts all shorter words starting with a different letter.
|
2019-06-19 11:42:51
|
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|
https://codereview.stackexchange.com/questions/73572/mapping-yielding-lots-of-warnings-in-use/73590
|
# Mapping yielding lots of warnings in use
Recently, I wrote a C header file that defines a new type called HashMap that allows for the storage of key/value pairs.
I am just learning about the wonders of void pointers and as I use them, I get lots and lots of warnings about making pointers without a cast. The header file itself does not yield any warnings, but when I use the header file, I get lots.
Here is the header file (I am leaving out all the guards and other preprocessor commands):
typedef struct {
int size;
void **k;
void **v;
} HashMap;
void hashmap_init(HashMap *hm, int size) {
hm->k = malloc(size);
hm->v = malloc(size);
hm->size = size;
}
void hashmap_push(HashMap *hm, void *k, void *v, int index) {
hm->[index] = k;
hm->[index] = v;
}
void *hashmap_get(HashMap *hm, void *k) {
int i;
for(i = 0; i < hm->size; i++)
if(hm->k[i] == k)
return hm->v[i];
}
And the file in which I am testing the header:
int main(void) {
HashMap hm; // a new HashMap
hashmap_init(&hm, 10); // Initialize the HashMap with 10 spots for memory
hashmap_push(&hm, 3, 7, 0); // In the first index, put 3 as the key and 7 as it's value
int v = (int)hashmap_get(&hm, 3); // Get the key dubbed 3 from the hashmap
printf("%d\n", v); // => 7
return 0;
}
I was thinking: In Java, when using a HashMap, you initialize it with 2 different values. Would it be better if I did that here? (this is just a second thought - you don't have to answer this too)
### You have a bug
This doesn't compile for me:
hm->[index] = k;
hm->[index] = v;
I think you meant this:
hm->k[index] = k;
hm->v[index] = v;
### Understanding the warnings and responding to them
Recently, I wrote a C header file that defines a new type called HashMap that allows for the storage of key/value pairs.
As comments pointed out, this is not a hash map. It's a map. This wiki page explains what is a hash map.
I am just learning about the wonders of void pointers and as I use them, I get lots and lots of warnings about making pointers without a cast.
Pay attention to the warnings. Any unhandled warnings are bugs waiting to happen.
Since you mentioned you're new with pointers, I'm getting the feeling that maybe you're just not comfortable reading warnings yet. They are not that hard to read though, and you don't really have a choice: you must read and understand them so you could eliminate them.
Here's an example:
t.c: In function 'main':
t.c:31:5: warning: passing argument 2 of 'hashmap_push' makes pointer from integer without a cast [enabled by default]
hashmap_push(&hm, 3, 7, 0);
^
As a reminder, here's the method signature:
void hashmap_push(HashMap *hm, void *k, void *v, int index);
The method takes as 2nd parameter a void*, but you're passing it the int value 3. Well, that just doesn't work: you're passing the wrong type. To make it work, the compiler casts this integer to void*, and warns you about it. Is this really what you wanted? No. Casting an int variable to void* can't be good.
The next warning is related to the first:
t.c:15:6: note: expected 'void *' but argument is of type 'int'
void hashmap_push(HashMap *hm, void *k, void *v, int index) {
^
The first warning was about calling hashmap_push with the wrong parameter types, this one is about the hashmap_push receiving the wrong parameter types.
Next warnings:
t.c:31:5: warning: passing argument 3 of 'hashmap_push' makes pointer from integer without a cast [enabled by default]
hashmap_push(&hm, 3, 7, 0);
^
t.c:15:6: note: expected 'void *' but argument is of type 'int'
void hashmap_push(HashMap *hm, void *k, void *v, int index) {
^
I hope you guessed, this warning is about the same line, but this time argument 3, which is the int value 7. Same problem: casting an int to void*.
t.c:33:5: warning: passing argument 2 of 'hashmap_get' makes pointer from integer without a cast [enabled by default]
int v = (int)hashmap_get(&hm, 3);
^
t.c:20:7: note: expected 'void *' but argument is of type 'int'
void *hashmap_get(HashMap *hm, void *k) {
^
Based on the explanations earlier, can you tell what this is trying to tell you?
The 2nd argument of the call to hashmap_get is invalid, as the number 3 is of type int instead of void*. The warning after that is about the receiving side: the hashmap_get is expecting a void* but it's receiving an int.
Finally, this one:
t.c:33:13: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
int v = (int)hashmap_get(&hm, 3);
^
Can you tell what the warning is trying to tell you?
The hashmap_get method returns a void*, but you're casting it to int. This is not valid.
In short, all these warnings were trying to tell you that you're calling methods with the wrong types. It's a fairly lucky accident that this code works at all and produces the expected output. Without clearing up these warnings by calling the methods with the appropriate types, you cannot trust this code.
The header file itself does not yield any warnings, but when I use the header file, I get lots.
When the header file compiles without warnings, but the source file using that header produces warnings, that practically means the source file is misusing the header. The header is like a contract: it specifies what you can do. Your source file twisted that contract. You got a result that sort of magically seemed to work, but it's in violation of the contract, and it wasn't really supposed to work.
### Using the map without warnings
This (by no means good) code eliminates the warnings and works:
int main(map) {
HashMap hm;
hashmap_init(&hm, 10);
int key = 3;
int value = 7;
hashmap_push(&hm, &key, &value, 0);
int v = * (int*) hashmap_get(&hm, &key);
int key2 = 3;
float value2 = 5.3;
hashmap_push(&hm, &key2, &value2, 0);
float v2 = * (float*) hashmap_get(&hm, &key2);
printf("%d\n", v);
printf("%f\n", v2);
return 0;
}
This code passes the methods the types they expect, and doesn't make invalid casts, namely:
• &key is a int*, which is a subtype of void*, therefore valid
• Same for &value
• Casting the 1st call to hashmap_get to int* is valid, because it corresponds to the type of &value that was passed in earlier for &key
• Casting the 2nd call to hashmap_get to float* is valid, because it corresponds to the type of &value2 that was passed in earlier for &key2
I used different value types for the sake of a demonstration, but keep in mind that ideally (and in 99% of practical cases) all values of a map should be of the same type. Among other things, one reason for that is given some key k whose value you don't know, you cannot know in advance the correct type of the value. In this specific example, I knew in advance that the value of key2 is value2.
Finally, what do you think the program outputs?
7 5.300000
Well, this is a bit unexpected from a map. Normally I would expect from a map that if I set the value of key 3 to x, and then again set the value of key 3 to y, the second call would update the existing entry and overwrite it with y. This map works this way because of this code in hashmap_get:
if (hm->k[i] == k)
return hm->v[i];
That is, you are comparing the pointers of keys, rather than their values. In a more common map implementation, two distinct key objects both having the value 3 would be treated as the same key. To make this code work that way, it will take a substantial rewrite, and rethinking of the types.
### Other coding style issues
I suggest to use braces with all single-line statements, for example instead of:
for (i = 0; i < hm->size; i++)
if (hm->k[i] == k)
return hm->v[i];
Write like this:
for (i = 0; i < hm->size; i++) {
if (hm->k[i] == k) {
return hm->v[i];
}
}
It's a bit longer, but it can save you from some simple but common mistakes.
• Thank you for taking the time to write all of that. I also appreciate the hidden boxes along with the questions tied to them to get me thinking about it myself. – SirPython Dec 14 '14 at 17:40
|
2020-01-19 17:42:20
|
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|
http://fans.j-npcs.org/index.php?page=12&person=208
|
Determination of the spin of the Higgs-like boson in diphoton decay channel with center-edge asymmetry at the LHC
A.A. Pankov, A. V. Tsytrinov
Tech. Univ. of Gomel, Belarus
We discuss the discrimination of a $\sim$ 126 GeV spin-parity $0^+$ Higgs-like boson decaying into two photons, $H\to\gamma\gamma$, against the hypothesis of a minimally coupled $J^{P}=2^+$ narrow diphoton resonance with same mass and giving the same total number of signal events under the peak observed at LHC. We apply, as the basic observable of the analysis, the center-edge asymmetry $A_{\rm CE}$ of the cosine of the polar angle of the produced photons in the diphoton rest frame to distinguish between the tested spin hypotheses. We show that center-edge asymmetry $A_{\rm CE}$ should provide strong discrimination between the possibilities of spin-0 and spin-2 with graviton-like couplings, namely with the confidence level up to $8\sigma$ depending of fraction of $q\bar{q}$ production of the spin-2 signal.
|
2021-06-21 12:16:37
|
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|
https://yi-ding.me/blog/MLfolds/basics/Bayesian-Learning-on-the-Candy-Example/
|
Bayesian Learning on the Candy Example
For the Bayesian learning, Professor Arindam Banerjee has a very good example of candy bages. In this post, we use the candy example to explain and summarize Bayesian learning. Related code can be found here.
Bayes Theorem
Before we reach out to Bayesian learning. We need first review the idea of Beyas Theorem.
$p(\theta|\mathcal X)= \frac{p(\mathcal X|\theta)p(\theta)}{p(\mathcal X)}$
$$p(\theta)=$$ prior probability of hypothesis $$\theta$$
$$p(\mathcal X)=$$ prior probability of training data $$\mathcal X$$
$$p(\theta \mid \mathcal X)=$$ posterior probability of $$\theta$$ given data $$\mathcal X$$
$$p(\mathcal X \mid \theta)=$$ probability of $$\mathcal X$$ given $$\theta$$
Bayesian Estimation
Usually, we have more than one hypotheses and we want the most probable hypothesis given the training data. That is, Bayesian Estimation can help us choose an appropriate hypothesis from multiple hypotheses (in discrete hypothesis space), or compute the apporprate parameter (in continuous hypothesis space).
Maximum a posteriori (MAP), as its name suggests, is to estimate the parameter $$\theta$$ by maximize the posteriori $$p(\theta \mid \mathcal X)$$.
\begin{align} \theta_{\text{MAP}} &= \text{argmax}_\theta \phantom{2} p(\theta|\mathcal X) \\ &= \text{argmax}_\theta \phantom{2} \frac{p(\mathcal X|\theta)p(\theta)}{p(\mathcal X)} \\ &= \text{argmax}_\theta \phantom{2} p(\mathcal X|\theta)p(\theta) \end{align}
Correspondingly, the Maximum Likelihood (ML) is as follows:
$\theta_{\text{ML}} = \text{argmax}_\theta \phantom{2} p(\mathcal X|\theta)$
Basic Idea
According to Professor Banerjee, “The Bayesian view of learning is updating probability distribution over the hypothesis space.”
Here we use the following notations: $$H$$ is the hypothesis variable, values $$h_1$$, $$h_2$$, …. $$P(h_i)$$ is the prior. We can assume $$P(h_i)=P(h_j)$$ when we have no further information about the prior.
Training data $$\mathcal X=\{x^1,x^2,...,x^N\}$$
Given data, each hypothesis has a posterior:
$P(h_i|\mathcal X)=\alpha P(\mathcal X|h_i)P(h_i)$
For prediction, we can use a weighted average over all the hypotheses instead of picking the best-guess one:
$P(x_\text{test}|\mathcal X)=\sum_i P(x_\text{test}|h_i)P(h_i|\mathcal X)$
The Candy Example
Suppose there are five kinds of bags of candies:
• 10% are $$h_1$$: 100% cherry candies
• 20% are $$h_2$$: 75% cherry candies + 25% lime candies
• 40% are $$h_3$$: 50% cherry candies + 50% lime candies
• 20% are $$h_4$$: 25% cherry candies + 75% lime candies
• 10% are $$h_5$$: 100% lime candies
(The figure is taken from Prof. Banerjee’s slides)
We take one candy at each time and put the candy back. Then we have the following observations:
We need to answer two questions:
1. What kind of bag is it?(Estimation)
2. What flavor will the next candy be? (Prediction)
Estimation (Bayesian posterior updating)
Here we compute the the posterior after each candy is drawn:
Before any candy is drawn, the posterior is the prior:
# of candies drawn $$p(h_1|d)$$ $$p(h_2|d)$$ $$p(h_3|d)$$ $$p(h_4|d)$$ $$p(h_5|d)$$
0 0.1 0.2 0.4 0.2 0.1
After the first candy is drawn, we have:
$p(d) = \sum_{i=1}^{5}p(h_i)p(d|h_i) = 0.1*0 + 0.2*0.25 + 0.4*0.5 + 0.2*0.75 + 0.1*1 = 0.5$ $p(h_1|d) = \frac{p(d|h_1)p(h_1)}{p(d)} = \frac{0.1*0}{0.5} = 0$ $p(h_2|d) = \frac{p(d|h_2)p(h_2)}{p(d)} = \frac{0.2*0.25}{0.5} = 0.1$ $p(h_3|d) = \frac{p(d|h_3)p(h_3)}{p(d)} = \frac{0.4*0.5}{0.5} = 0.4$ $p(h_4|d) = \frac{p(d|h_4)p(h_4)}{p(d)} = \frac{0.2*0.75}{0.5} = 0.3$ $p(h_5|d) = \frac{p(d|h_5)p(h_5)}{p(d)} = \frac{0.1*1}{0.5} = 0.2$
That is:
# of candies drawn $$p(h_1|d)$$ $$p(h_2|d)$$ $$p(h_3|d)$$ $$p(h_4|d)$$ $$p(h_5|d)$$
1 0 0.1 0.4 0.3 0.2
Similarly, we can have the following results:
# of candies drawn $$p(h_1|d)$$ $$p(h_2|d)$$ $$p(h_3|d)$$ $$p(h_4|d)$$ $$p(h_5|d)$$
2 0 0.038 0.308 0.346 0.308
3 0 0.013 0.211 0.355 0.421
4 0 0.004 0.132 0.335 0.529
5 0 0.001 0.078 0.296 0.624
6 0 0.000 0.044 0.251 0.705
7 0 0.000 0.024 0.206 0.770
8 0 0.000 0.013 0.165 0.822
9 0 0.000 0.007 0.130 0.864
10 0 0.000 0.003 0.101 0.896
The updating process can be shown in the figure below:
Prediction
For prediction, we use a weighted average over all the hypotheses. Before any candy is drawn,
\begin{align} P(\text{next candy is lime} | \mathbf{d}) &= \sum_{i=1}^{5} P(\text{next candy is lime} | h_i) P(h_i|\mathbf{d}) \\ &= 0*0.1 + 0.25*0.2 + 0.5*0.4 + 0.75*0.2 + 1*0.1 = 0.5 \end{align}
After the first lime candy is drawn, we have \begin{align} P(\text{next candy is lime} | \mathbf{d}) &= \sum_{i=1}^{5} P(\text{next candy is lime} | h_i) P(h_i|\mathbf{d}) \\ &= 0*0 + 0.25*0.1 + 0.5*0.4 + 0.75*0.3+ 1*0.2 = 0.65 \end{align}
Similarly, we have the following figure.
|
2022-10-07 21:20:41
|
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|
https://simpleitk.readthedocs.io/en/master/link_N4BiasFieldCorrection_docs.html
|
# N4 Bias Field Correction¶
## Overview¶
The N4 bias field correction algorithm is a popular method for correcting low frequency intensity non-uniformity present in MRI image data known as a bias or gain field. The method has also been successfully applied as flat-field correction in microscopy data. This method assumes a simple parametric model and does not require tissue classification.
This example demonstrates how to use the SimpleITK N4BiasFieldCorrectionImageFilter class. This filter has one required input image which is affected by a bias field we wish to correct. The primary input is required to have a “real” pixel type of either sitkFloat32 or sitkFloat64. Additionally, there is an optional “MaskImage” input which specifies which pixels are used to estimate the bias-field and suppress pixels close to zero. It is recommended that the mask image follows the common conventions for masked images, which is being of pixel type sitkUint8 and having values of 0 and 1 representing the mask. Additionally, the mask image and the main input image must occupy the same physical space to ensure pixel to pixel correspondence.
Some basic parameters for using this algorithm are parsed from the command line in this example. The shrink factor is used to reduce the size and complexity of the image. The N4 algorithm uses a multi-scale optimization approach to compute the bias field. The SetMaximumNumberOfIterations method takes an array used to set the limit of iterations per resolution level, thereby setting both the iterations and the number of scales ( from the length of the array ). The output of the filter is the bias corrected input image.
## Code¶
#!/usr/bin/env python
from __future__ import print_function
import SimpleITK as sitk
import sys
import os
if len(sys.argv) < 2:
print("Usage: N4BiasFieldCorrection inputImage " +
"outputImage [shrinkFactor] [maskImage] [numberOfIterations] " +
"[numberOfFittingLevels]")
sys.exit(1)
if len(sys.argv) > 4:
else:
maskImage = sitk.OtsuThreshold(inputImage, 0, 1, 200)
if len(sys.argv) > 3:
inputImage = sitk.Shrink(inputImage,
[int(sys.argv[3])] * inputImage.GetDimension())
[int(sys.argv[3])] * inputImage.GetDimension())
inputImage = sitk.Cast(inputImage, sitk.sitkFloat32)
corrector = sitk.N4BiasFieldCorrectionImageFilter()
numberFittingLevels = 4
if len(sys.argv) > 6:
numberFittingLevels = int(sys.argv[6])
if len(sys.argv) > 5:
corrector.SetMaximumNumberOfIterations([int(sys.argv[5])]
* numberFittingLevels)
output = corrector.Execute(inputImage, maskImage)
sitk.WriteImage(output, sys.argv[2])
if ("SITK_NOSHOW" not in os.environ):
sitk.Show(output, "N4 Corrected")
# Run with:
#
# Rscript --vanilla N4BiasFieldCorrection.R inputImage, outputImage, shrinkFactor, maskImage, numberOfIterations, numberOfFittingLevels
#
library(SimpleITK)
args <- commandArgs( TRUE )
if (length(args) < 2) {
stop("At least 2 arguments expected - inputImage, outputImage, [shrinkFactor], ",
}
if (length( args ) > 4) {
} else {
maskImage <- OtsuThreshold( inputImage, 0, 1, 200 )
}
if (length( args ) > 3) {
inputImage <- Shrink( inputImage, rep(strtoi(args[3]), inputImage$GetDimension()) ) maskImage <- Shrink( maskImage, rep(strtoi(args[3]), inputImage$GetDimension()) )
}
inputImage <- Cast( inputImage, 'sitkFloat32' )
corrector <- N4BiasFieldCorrectionImageFilter()
numberFittingLevels <- 4
if (length ( args ) > 6) {
numberFittingLevels <- strtoi( args[[6]] )
}
if (length ( args ) > 5) {
corrector$SetMaximumNumberOfIterations( rep(strtoi( args[[5]], numberFittingLevels)) ) } output <- corrector$Execute( inputImage, maskImage )
WriteImage(output, args[[2]])
|
2020-09-29 01:59:13
|
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|
http://encyclopedia.kids.net.au/page/he/Heron%27s_formula
|
## Encyclopedia > Heron's formula
Article Content
# Heron's formula
In geometry, Heron's formula states that the area of a triangle whose sides have lengths a, b, c is
$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$
where
$s=\frac{a+b+c}{2}$
This formula is credited to Heron of Alexandria, although it is possible that it may have been known long before Heron's time.
The formula is in fact a special case of Brahmagupta's formula for the area of a cyclic quadrilateral; both of which are special cases of Bretschneider's formula[?] for the area of a quadrilateral.
Expressing Heron's formula with a determinant in terms of the squares of the distances between the three given vertices,
$A = \sqrt{ \frac{1}{16} \begin{bmatrix} 0 & a^2 & b^2 & 1 \\ a^2 & 0 & c^2 & 1 \\ b^2 & c^2 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{bmatrix} }$, illustrates its similarity to Tartaglia's formula for the volume of a four-simplex.
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2022-12-02 07:04:27
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https://stats.stackexchange.com/questions/74808/determine-separation-between-two-modes-from-distribution
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# Determine separation between two modes from distribution
I’ve got a sample of pairwise distances between points in a 2D picture. Some of these points lie within the same object. Their distance to each other is thus smaller than some well-defined threshold (the object’s diameter). Points that lie in different objects (predominantly) have a pairwise distance greater than said threshold. Points that lie within the same object are however rare (<10%).
I would like to determine this distance threshold empirically from my sample.
For “appropriate” parameters (well, herein lies the rub, doesn’t it?) the threshold is visible in the density plot:
The threshold is marked by the arrow. This is the objectively right cut-off for my application: it is the dip after the first tall plateau which corresponds to the distribution of the few points lying within the same object, and it corresponds to the object diameter that can be individually verified in the original picture, but not easily automatically deduced from my data.
Unfortunately, I have no idea how to determine it in an automated fashion. Even the adjust argument / bandwidth for the density function has been found by trial and error, and a different input data set I’ve tried requires a different bandwidth.
Is there any hope? Or should I just give up?
• As a point of terminology, I wouldn't use the term "cut-off" for an interior point. As a point of statistical interpretation, your arrow points to a minor antimode, but just from (a) general data analysis experience (b) complete ignorance of your application, I can't see any reason to take it more seriously than any other such detail. I would want to know much more about the raw data (e.g. any granularity in recording), to see results of varying bandwidth and kernel in density estimation, and to see such a feature being repeatedly reproducible in different datasets. – Nick Cox Nov 7 '13 at 0:24
• @Nick Right, “cut-off” is a complete misnomer that slipped in because of the application I need it for. Regarding your (b), I’ll amend the question. In a nutshell, I expect a distribution with two peaks: one of points which lie closer to each other than my cut-off, and one distribution of all other pairwise distances. – Konrad Rudolph Nov 7 '13 at 0:35
• @Nick I’ve essentially rewritten most of the question – sorry for that; however, I’ve realised that the first description was complete crap. I hope this one’s better. – Konrad Rudolph Nov 7 '13 at 0:46
• Optimistically, it is the first local minimum on the density function, with higher densities on either side. That's computable. But to be convincing, it has to persist over a good range of possible bandwidths. Compare work on mode trees, e.g. citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.30.5736 – Nick Cox Nov 7 '13 at 0:51
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2019-07-20 22:29:24
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https://www.physicsforums.com/threads/computing-arctan-inverse-tan-function.912681/
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# Computing arctan (inverse tan) function
Tags:
1. Apr 26, 2017
### Vital
1. The problem statement, all variables and given/known data
Hello!
Surprisingly I get different results when I try to compute the inverse tangent function.
My goal is to compute it both manually and using calculator in radiant mode.
2. Relevant equations
My goal is to compute arctan(½) both manually and using calculator in radiant mode.
3. The attempt at a solution
(1) First problem with calculator:
When I try to compute arctan(½) using any online calculator, I stumble upon different results; some calculators even refuse to compute it.
- https://web2.0calc.com gives me the result I see in the textbook
arctan(½) = 0.463647609001
- http://calculator.tutorvista.com/arctan-calculator.html
arctan(½) = 0.785...
(2) Second problem - manual computation gives different answer from any of above ones:
arctan(½) means that tan(θ) = ½
This means that θ can be in Quadrant I or in Quadrant III.
tan(θ) = sin(θ) / cos(θ)
sin(θ) = √1 - cos2(θ) (the whole expression is under square root sign, sorry for not being able to show it correctly)
hence, tan(θ) = ( √1 - cos2(θ)) / cos(θ)
Let cos(θ) = x, then tan(θ) = ( √1 - x2) / x = ½
Squaring both sides I get:
(1 - x2) / x2 = ¼
x = +- 2/√5
Let's choose positive cos, then cos(θ) = 2/√5 ≈ 0.8944
sin(θ) = √1-cos2(θ) = √1-0.89442 ≈ 0.4472
Thus tan(θ) = 0.4472 / 0.8944 = 0.5 (without any rounding, it is strictly 0.5)
Which is not 0.463647609001 (this can be rounded to 0.5, but it is not the same).
Please, help me to find my mistake in manual computation.
Thank you!
2. Apr 26, 2017
### Staff: Mentor
You simply recalculated tan(θ), not arctan(1/2).
You used that site incorrectly. It cannot parse 1/2 and is returning arctan(1). Try inputting 0.5 instead.
WolframAlpha is always a good resource for these calculations: http://www.wolframalpha.com/input/?i=arctan(1/2)
3. Apr 26, 2017
### Vital
I see. Thank you very much!
Could you, please, also take a look at my manual computation and why I don't get a correct answer there? Thank you very much for your help.
4. Apr 26, 2017
### Staff: Mentor
Because you are not computing anything! Nowhere are you writing an equation for θ. The best you will come to is $\theta = \arccos(2/\sqrt{5})$, which doesn't bring you closer to a numerical result than $\arctan(1/2)$. You are better off using the series expansion
$$\arctan(x) = x -\frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \ldots$$
for $|x| < 1$.
5. Apr 26, 2017
### PeroK
If you have Excel spreadsheet, then simply typing $=ATAN(0.5)$ into a cell does the job.
6. Apr 26, 2017
### Vital
I do ) Thank you. But my main goal is to do the same manually - it is easy to use technical devices, but that doesn't help to learn. )
7. Apr 26, 2017
### Vital
Sorry, that is not the part I have leaned yet, so I have to use those easy tools which have been covered in the book thus far.
8. Apr 26, 2017
### PeroK
You were obviously struggling to find a technical device that you could successfully operate, so I suggested Excel and how to use it. If you already know how to compute this on a calculator, why are you posting your difficulties in doing so?
9. Apr 26, 2017
### Vital
When I posted the question, DrClaude has kindly helped me to see a mistake I did when trying to use an online calculator. So now that is clear to me.
If you take a look at my initial post, you will see that I am also struggling with a "manual" approach.
10. Apr 26, 2017
### Ray Vickson
As already pointed out, the manual method you used gave you $\cos(\theta)$, not $\theta$ itself.
In order to compute $\arctan(t)$ for any input $0 < t < \pi/2$ you need to be able to compute $\tan(\theta)$ for numerical inputs $\theta$, because what you want to do is solve the equation $\tan(\theta) = t$ numerically. Typical equation-solving methods obtain (approximate) solutions by, essentially, searching through different values of $\theta$ until finding one that "works". You can Google the following topics about solving single-variable equations numerically: bisecting search, secant method, regula falsi, Newton's method, and others.
11. Apr 26, 2017
### Vital
I see. Thank you very much for these suggestions. I will study them.
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2017-10-22 10:55:04
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http://blogs.jccc.edu/dpatter/2016/03/
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# GDC16 – Day 2
So yesterday was all math, and today was all education. …well sort of. Academic artists have a world view that is quite different from that of your typical scientist. I come from a very analytic background. Even my hobbies, playing poker and racing cars, and my artistic work in photography are a blend of the creative and the analytic. Once I wrapped my brain around the different way in which they were looking at the problem of pedagogy, there were some good take-away points from the morning and afternoon sessions.
One of the things that true of both artist and scientist is that we fail, and I mean we fail a lot. Most of our careers is spent creating one failure after another. This is NOT a bad thing, unless it’s the only thing you’re doing. The road to success is lined with a long sequence of failures, and those who do succeed accept this and even embrace it. When students finally arrive in a college classroom, they are absolutely terrified of failure. Personal note: This should not be a surprise given how frequently we subject our kids to high-stakes standardized testing, but that’s a topic for another article. Game development, much like the development of a new scientific model, is a process of constant iteration. You try, test, fail, tweak, test, fail, rinse and repeat. So many of our students are fearful of that initial failure that they don’t even attempt the work and fail in a far more destructive way. Mitigating, navigating, and learning from failure through iteration is paramount in game development just as it is in science, but it’s a tough thing to teach.
There were more ideas that I picked up throughout the day, but nothing really groundbreaking. I have some ideas related to the iteration process and managing failure I mentioned above that I will craft into some new exercises for the Astronomy courses. I’ll have more on that as I develop the idea into more than just thoughts in my head.
The last talk I attended was by Margaret Moser, “Teaching Designers to Code”. I thought that this would be useful not only because we occasionally get a more design-focused student in the Math and Physics for Games course, but also because even many of the programming-focused students are really just learning how to code efficiently. The two big take-away items here were not ones that were startling revelations, but rather stern reminder of what I already knew and should be deploying in my classroom. Something I could do better when describing algorithms in class is not to start writing line-items of pseudo code or equations, but instead to start with a block diagram of the process. This, of course, is the key to coding more complex solutions. Don’t think in terms of individual lines, but think in terms of functional blocks. That’s how I think of my scientific work. Think about the big blocks of the problem before you drill down into writing specific equations.
The evening ended brilliantly. Professor Hanna and I made our way down to a region where my good friend and fellow photographer, Bash Beard, took me when I was in town for the AGU Fall Meeting, Beldon Place. At first, it just looks like a sketchy back alley, but when you peek around the corner, it’s lined with one amazing restaurant after another with just about every type of food you can think to eat. We tried the last restaurant at the end of the alley, Brindisi Cucina Di Mare. Of course I was going to sniff out a seafood joint. What else did you expect?
Tomorrow is the first day of the main conference which means SWAG!!! The Expo Hall opens and it’s time to meander around the vendor booths and see what kind of cool loot I can drag back for my students, for Lillian, and for my officemate’s geocaches. It’s also the day of the Game Developers Choice Awards. Think the Oscars for video games, only with a bit more edge to it.
# GDC16 – Day 1
It’s Math Day!!! The first two days of the Game Developers Conference is always given to specialty tracks and summits. For the past twelve years, there has been a Math for Game Programmers summit on Monday and Physics for Game Programmers on Tuesday. They’ve combined the two into a one-day summit, with the majority of talks being focused on the mathematics, but often with the physics of the game play in mind. As with any conference, some of the talks are huge hits, some are misses, some are good talks but not directly pertinent to your work or interests. This was the case on Monday as well. Fortunately, there were far more hits than misses. On the balance, it was a fantastic summit.
First up was J. Kyle Pittman (@piratehearts) to discuss jump physics. In our course, Math and Physics for Game Programmers, we teach students the basics of uniformly accelerated motion and the Forward Euler Method for numerical integration. When doing this, we certainly talk about free-fall motion and the kinematics of a jump, but Mr. Pittman made some great arguments for why doing jumps physically correctly could often be incorrect for good game play. Often, the acceleration due to gravity, $g$, is preset to a given value, often 9.8 m/s2 downward, and the dynamics of a jump are then determined by the vertical velocity, $v_y$, of the jump. This can lead to, although realistic, jumps that feel too floaty or too heavy, and often jump physics that do not match the necessary trajectories demanded by the level design. Rather than pre-determining the value of $g$, Mr. Pittman suggests allowing it to be determined by the constraints of the level design. If a gap distance and maximum height are known, the value of $g$ and $v_y$ can be calculated so that it is possible for the character to complete the jump.
$g = - \frac {2 h} {t_h}$
$v_y = \frac {2 h v_x} {x_h}$
where $h$ is the maximum height of the jump, $t_h$ is the time to reach the maximum height, $v_x$ is the maximum horizontal speed of the player’s character, and $x_h$ is the horizontal distance covered to the reach the maximum height. Combining these yields a value for the acceleration due to gravity that relies solely on the level design parameters and ensures that a jump is makeable by a player.
$g = - \frac {2 h {v_x}^2} { {x_h}^2}$
Groovy!
The other technique Mr. Pittman explained was using a different value for gravity for ascent versus decent. Mario’s jump physics included increasing the acceleration due to gravity by a factor of three once the jump reached it’s maximum height. This, of course, is physically inaccurate, but in developing a game, you’re goal is to produce good gameplay with believable physics, but simulation-level accurate physics. It’s funny how our brains will sometimes see a simulation that is physically accurate, but believe it’s totally fake.
Next up was Squirrel Eiserloh’s talk on camera shake. I always think of this as the Star Trek effect. When there’s trauma to a player, either from being hit or landing on a surface after a jump or fall, a bit of camera shake can lend realism to the visuals, much in the same way as camera shake was used, perhaps overused, in the original Star Trek episodes to indicate trauma to the ship. The big take-away from this talk was to use both rotational and translational shake in 2D environments, but only rotational shake in 3D. If you jitter the translation of a camera, you can easily end up with the camera inside of a wall. In a VR environment, just don’t, unless it is your intent to cause motion sickness and nausea in your players.
Later in the day, Simon Strange discussed finding distances in the training simulator, Strike Group Defense, a serious game designed to train Navy personnel on the best anti-ship missile defense strategies and tactics. Since naval engagements happen over hundreds of miles, the curvature of the earth is significant and must be taken into account. Given this, the game makes use of spherical coordinates to determine distance. Rather than using the complex trigonometry of calculating arc length on a great circle between two points, Mr. Strange opted instead to rotate the world so that the player was at the north pole. This means that the polar angle is a direct measure of the distance between the player and any point on the map, $d = R_{Earth} \phi$. This is brilliant, and something that Prof. Grondahl and I have tried to explain to our students over the years. Invest in a bit of work up front in order to save a lot of work later on. This is exactly what this method does. You invest in a bit of work ahead of time to rotate the player position to the +z-axis. After you do this, calculating distances on the curved surface of the Earth is a simple and, more importantly, computationally cheap calculation.
The last talk that I’ll highlight is from the always brilliant Gino van der Bergen. This year’s talk was on enforcing rotational joint limits in quaternian space. Quaternians are awesome mathematical objects first developed by Hamilton in the late 1800s. In fact, Hamilton was so excited about his recognition of the key to quaternian on his way home from the pub that he etched the basic complex numbers rules into the stonework of a bridge in Dublin. You have to love mathematical graffiti. Mr. van der Bergen did a great job of explaining why and how quaternions eliminate the singularity and degeneracy issues inherent in traditional matrix-based rotation operators.
Keep an eye on the website Essential Math. You can find slides from the Math for Game Programmers summits of previous GDCs, and in time the 2016 slides, including Mr. van der Bergen’s presentation on quaternian space, will be posted for public consumption.
At the end of the day, I took my camera and set out to capture some images around the Powell and Market region. If you’ve ever been in this area of San Francisco, you know there is a significant homeless population. The reasons for persons finding themselves homeless are as various and unique and the people themselves, but one thing that is common among them is their effective invisibility. Of course, people physically see them, but they consistently choose to disregard their presence, treating them as essentially invisible and separate from the rest of the sea of humanity shopping, commuting to and from conferences and work, waiting to ride the cable cars, and doing all the other things we’re used to doing. One such individual, an amputee, was sitting in his wheelchair off to the side of the rest of the buzz of tourists and townies. The look on his face was one of resigned dejection as people walked by with not even a glimmer of recognition that they just passed a fellow human being.
# GDC16 – Day 0
So today was interesting. I’m all set to check my bag and fly out to San Francisco and I find out that my flight is overbooked. …great. The ticket agent asks if I would mind moving to another flight. I’m in the process of rejecting by default as I needed to be here by 1700h PDT. Before I get around to doing that and demanding the seat I purchased, he says, “How about a non-stop that leaves 22 minutes later than your flight, but arrives two hours earlier. …blink… Yup! That will work just fine, thank you. As it happens, it was the flight that the Game Development Program Chair, Prof. Russ Hanna was on as well.
The rest of the trip goes great. Flight in was smooth, BART into town was smooth, got my MUNI pass easy-peasy. The problem started when I got to hotel. Paperwork is such a fun thing. Turns out there was a mishandling of the paperwork between the college and the hotel so the new front desk clerk didn’t think that my room was paid. Great. There are worse places than San Francisco to be homeless for a day. Fortunately, the folk at the Hotel Stratford were amazing and very understanding! They knew that this would work itself out in the morning when the business offices opened, and they did.
What to do while in the Powell and Market area and your stressed out? Eat! Senior Scientist at Fundamental Technologies, Jerry Manweiler, introduced me to a great burger joint on Union Square, the Burger Bar. Wet, stressed, and hungry, That’s where Russ and I went. It’s an amazing place with an amazing view, although I will say the view during the Christmas season when I come here for the American Geophysical Union Fall Meeting is a bit more colourful.
One thing that did go right was registration at GDC. It used to be such a madhouse. They’ve really streamlined the process and have it working very smoothly. Russ and I were able to walk right up, sign in, and grab our badges. We were in and out in less than five minutes. Brilliant! That left time for some photography work. Before wrapping up the day with more food at an Irish-themed deli. I’m a sucker for corned beef sandwiches!
I’m part of a small group of photogs that post weekly photos on a theme in an effort to push each of us outside of our normal comfort zones and force us out of our photographic ruts. Last week’s theme was “Motion,” and what I wanted was a bus or streetcar moving through the steam rising from the storm grates. I picked the wrong side of the street. Although I didn’t get the steam in the shot like I wanted, I did get some panning practice in before racing season. The lens I used was a 28-mm prime thrift-store special. Auto nothing! It forced me to think about and set every aspect of the exposure and the focus. It’s not tack sharp, but by this time it was starting to pour down rain, and I wasn’t keen on sticking around to grab another.
EXIF
Device: Nikon D7000
Lens: Tokina 28mm f/2.8
Focal Length: 28mm
Focus Mode: Manual
AF-Area Mode: Single
Aperture: f/8
Shutter Speed: 1/15s
Exposure Mode: Manual
Exposure Comp.: 0EV
Metering: Matrix
ISO Sensitivity: ISO 100
Tomorrow, the Math for Game Programmers summit begins! All math, all day!
# Found a new planet? Pics or it doesn’t exist.
In January, right at the beginning of the Spring semester, Professor of Planetary Astronomy Michael Brown and Assistant Professor of Planetary Astronomy Konstantin Batygin, both from the California Institute of Technology, published a remarkable prediction in the Astronomical Journal, Evidence for a Distant Giant Planet in the Solar System (read the article here, http://web.gps.caltech.edu/~kbatygin/Publications_files/ms_planet9.pdf) In the article, Dr. Batygin, the theoretician of the pair, uses Dr. Brown’s observations of objects within the Kuiper Belt to argue for the existence of an object in a Sedna-like orbit and approximately ten times the mass of Earth. The basis for their claim is the ordered clustering of the perihelions of the orbits of multiple Kuiper Belt objects, a phenomenon that has a 0.007% of occurring randomly (yes, the significance of the number was not lost on me). After a significant amount of modelling and numerical analysis, Dr. Batygin predicts the likely orbital parameters of the Neptune-sized object as being inclined as much as 40 degrees to the ecliptic and having a semi-major axis of ~700 AU with an eccentricity of ~0.6. This places its perihelion around 280 AU. That’s really far out there, and places the proposed object as a member of the distant Kuiper Belt or inner Oort Cloud, or Hills Cloud, rather than a member of the inner Kuiper Belt wherein resides more familiar objects like Pluto and Eris. The mathematics are extremely compelling and the discussion and conclusions well-reasoned, but as the modern saying goes, “Pics or it didn’t happen.”
This certainly isn’t the first time that an object has been predicted to exist by mathematical analysis of the orbit of other objects. The most famous, and earliest, is the prediction by French astronomer and mathematician Urbain Le Verrier of the existence of an eighth planet beyond the orbit of Uranus that would account for the Uranus’ increase and subsequent decrease in orbital speed unrelated to its solar distance. Le Verrier worked on the problem in the summer of 1846 during his position at the Paris Observatory. Using Newton’s mechanics and Law of Gravity and the observed positions of Uranus, he calculated where a more distant planet would have to be and how massive it would have to be to produce the observed deviations. After completing his work, two astronomers, Johann Gottfried Galle and Heinrich Louis d’Arrest at the Berlin University, began searching in the vicinity of Le Verrier’s predicted position for the new planet. Galle, looking through the telescope, called out positions and brightnesses of the visible objects to d’Arrest who compared the observations to previously recorded charts until Galle called out an object that was not on the chart. They had found the planet that would later come to be called Neptune within a single degree of Le Verrier’s calculations. It was a remarkable piece of work from both Le Verrier and from Galle and d’Arrest, and it was a triumph for Newtonian mechanics.
This method of discovering new objects was later attempted by William H. Pickering, Professor of Physics at Harvard University. Based on his calculations, using the apparent discrepancies in the orbits of both Uranus and Neptune, he attempted to image the proposed trans-Neptunian object at the Mount Wilson Observatory outside of Pasadena, CA. His search was unsuccessful, but the hunt for “Planet X” was picked up by Percival Lowell who had founded the Lowell Observatory in Flagstaff, AZ. Lowell’s attempts were equally unsuccessful. After Lowell’s passing, the search was tasked to an amateur astronomer from Burdett, KS, 23-year old Clyde Tombaugh. Rather than relying on sophisticated calculations, Tombaugh was tasked with systematically searching the Zodiac for anything non-stellar. In late January 1930, he captured two images of the object that we now know as Pluto. As it turns out the position of Pluto did not in any way correlate to Pickering’s calculations. In this case, the discrepancies were due to the lack of precision in the measurement of the masses of the outer planets.
Today, modern astronomers use the periodic motion of stars to mathematically infer the presence of extrasolar planets. The first of these discoveries was 51 Pegasi b. As two bodies orbit each other, such as a planet around its host star, the two bodies both move about their common center of mass. The planet being significantly less massive moves far more noticeably than the star, but the star does still move. Its motion is detectable by analysis of its light spectrum’s becoming alternately slightly bluer and then slightly redder as the star moves towards us and then away from us, respectively. This method has been used to locate many such extrasolar planets that we’re still unable to image directly. These are generally accepted as exceptions to the “Pics or it doesn’t exist” rule in science. The mathematics and analysis are so strongly compelling and there is no other viable alternate explanation that it is accepted that orbiting planetary bodies are responsible for the variations in the radial velocity of 51 Peg and other stars.
So now if we’re willing to take the mathematical word of the existence of extrasolar planets such as 51 Peg b, then why not for this new object proposed by Dr. Batygin and Dr. Brown from Cal Tech? The difference lies in the complexity of the problem. The extra solar planet problem is by comparison a simple problem. The radial velocity curve for 51 Peg is very clean, and the analysis of the data use methods that have long been vetted and refined by astronomers studying binary stars for which one can see the two separate objects. In other words, there is precedent for the methodology. This is not to say that Dr. Batygin’s methods are controversial or that the mathematical tools are not well understood. The Hamiltonian mechanics he deploys in his paper are extremely well understood and have been for over a century, but the data with which Dr. Batygin is working and the significant complexity created by moving from a two-body problem to an n-body problem make the analysis more difficult and intricate as well as making the results of those analyses less precise. For this reason, while I am very excited about this new prediction, I want to see an image before I take it as fact.
The observational discovery this new object, if it exists, likely won’t happen soon. Even at its proposed closest approach to the Sun, 280 AU, the intensity of sunlight striking the object is 0.00128% that of what it is here at Earth. Not only is the light very dim at that distance, most Kuiper Belt and inner Oort Cloud objects are coated with carbonaceous dust making their surfaces very dark and non-reflective. As our observational tools and techniques improve, we may eventually be able to start imaging these remote sentinels of our solar system, but until then, we’re left with only the predictions.
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2020-05-31 03:55:18
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http://tailieu.vn/doc/zend-php-certification-study-guide-p11-269953.html
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# Zend PHP Certification Study Guide- P11
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## Zend PHP Certification Study Guide- P11
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Zend PHP Certification Study Guide- P11: Hãy thẳng thừng, Giả sử bạn đang thuê một ai đó để giám sát hệ thống và PHP của bạn có nó thu hẹp xuống để hai ứng cử viên. Một trong những ứng cử viên nói, "Oh yeah, tôi biết tất cả về PHP." Các ứng cử viên khác nói, "Oh yeah, tôi biết tất cả về PHP, tôi đã được thông qua kỳ thi chứng chỉ Zend." câu hỏi tiếp theo của bạn có thể sẽ là "Zend Chứng nhận là gì?" Và các ứng viên nói, "Một công ty chuyên về...
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## Nội dung Text: Zend PHP Certification Study Guide- P11
2. 12 Debugging and Performance M AKING MISTAKES IS HUMAN, and so is fixing them. In your day-to-day program- ming adventures, it’s inevitable to introduce bugs in your PHP code, especially when you’re writing very complex applications with tens of thousands of lines of code spread across tens of files. When you’re prototyping an application, being able to avoid common programming mistakes is important to ensure that your code will be well-written from the very start. In this chapter, we’ll provide you with some guidelines on writing efficient code, debug- ging faulty scripts, and identifying bottlenecks when performance becomes an issue for both you and your clients. Terms You’ll Need to Understand n Bug n Coding standard n Code readability n Comparison operators n Performance n Caching n Portability Techniques You’ll Need to Master n Writing readable code n Proper commenting n Comparing heterogeneous data n Debugging
3. 186 Chapter 12 Debugging and Performance n Identifying and preventing performance bottlenecks n Preventing performance issues n Improving database performance n Using content and bytecode caching Coding Standards Writing your code in a structured manner is, perhaps, the smartest decision you can make. Although there aren’t any predefined coding standards that everyone in the pro- gramming community recognizes as better than the rest, deciding from the very begin- ning on a set of conventions will go a long way toward helping you make fewer mistakes. Documenting your code is particularly important.To make this job—probably at the top of the Ten Most Hated Tasks of programmers worldwide—a bit easier, you can even use one of the many automated tools available on the market, such as PHPDocumentor, which can extract documentation directly from your code if you structure your com- ments in a particular way. Regardless of how you introduce them in your applications, good comments and documentation will make sharing your code with other members of your team easier, as well as make sure that you’ll remember what it does when you get back from that three- week vacation. Remember, preventing bugs is much better than hunting for them. Extra whitespace and empty lines, although unimportant as far as the functionality of your code is concerned, can be an extremely valuable tool for writing better code: if ($foo == ‘bar’) {$i = 0; /** * foreach loop, get the content out of it */ foreach ( …. ) { } } By separating your code into logical groups, your source will be cleaner and easier to read. Also, indenting each line according to the code block it belongs to helps you figure out immediately what the structure of your script is.
4. Coding Standards 187 Flattening if Statements Consider the following snippet of code: if ($is_allocated) { if ($has_been_mangled) { if ($foo == 5) { print “foo is 5”; } else { print “You entered the wrong data!”; } } else { return false; } } else { return false; } As you can see, the many nested if statements here cause the code to look very busy and difficult to read. An easy way to improve the situation consists of “flattening” your if statements so that you can achieve the minimum level of indentation without com- promising either the functionality of your code or its performance.The preceding script, for example, could be rewritten as follows: if (!$is_allocated) { return false; } if (!$has_been_mangled) { return false; } if ($foo == 5) { print “foo is 5”;
5. 188 Chapter 12 Debugging and Performance } else { print “You entered the wrong data!”; } This approach gives you a better structure with fewer levels of nesting so that your code is easier to understand. Note that the type of operations performed is pretty much the same as before—and the elimination of two else statements will make the code easier to parse for the interpreter. Splitting Single Commands Across Multiple Lines One of the great things about PHP is that it doesn’t require you to write a single state- ment all on one line of code. In fact, any statement can be split across an arbitrary num- ber of lines without any change in its functionality—provided, of course, that the split doesn’t take place in the middle of a text string.This is particularly useful when you have a complex line of code that spans a large number of characters: $db->query(“select foo, bar, mybar as foobar from tbl1 where tbl1.mybar=’foo’”); This database query is split over several lines.The main advantage here is that you can immediately see what the query does, which tables are involved, and which conditions you are placing in the where clause. If the same query had been placed all on the same line, understanding its purpose would have taken a lot more effort, and the risk of intro- ducing new bugs by modifying it would have been greater. Concatenation Versus Substitution If you are inserting data into a long string—such as a database query—you can use the concatenation operator, but doing so often limits your ability to read the query properly:$db->query (“insert into foo(id,bar) values(‘“.addslashes($id). “‘,’”.addslashes($bar).”’)”); On the other hand, you could just use one of the printf() functions to do the job for you: $db->query(sprintf(“insert into foo(id,bar) values(‘%s’,’%s’)”, addslashes($id), addslashes($bar) )); 6. One Equal, Two Equals, Three Equals 189 As you can see, the entire statement is now a lot easier to read, although you will lose some performance by switching to sprintf() from the concatenation operator, which is native to the PHP interpreter and doesn’t require the execution of any external libraries. The literals in the string passed to sprintf() will be substituted with the values of the parameters passed afterwards in the order in which they appear in the call. Combined with the ability to split your commands over several lines, this approach can enhance readability to a large degree. Choose Your Opening Tags Carefully Mixing PHP and HTML code is one of the characteristics of PHP that make it both easy to use and powerful, although it’s easy to abuse this capability and come up with code that is difficult to read. When writing code for an application that could run on heterogeneous systems, it’s always a good idea to be very careful about which opening tag styles you use. In Chapter 1, “The Basics of PHP,” we mentioned that there are several of them, but only the canonical tags are fully portable. Short tags (which include the echo tag 7. 190 Chapter 12 Debugging and Performance Clearly, the statement should have been written as follows: if ($a == 5 { print “a is 5”; } In this case, the condition is a comparison operator, and it will be evaluated as true only if the value of $a is 5. There is, luckily, a very easy way to avoid this mistake once and for all, without any possibility of ever slipping again: make sure that the condition is written in such a way that it cannot possibly be misinterpreted: if (5 ==$a) { print “a is 5”; } With this approach, if you mistakenly only use one equal sign instead of two, as in 5 = $a, the interpreter will print out an error because you can’t assign anything to an imme- diate value. If you make a habit of writing all your conditions this way, you will never fall in the assignment trap again! There’s Equal and Equal As we mentioned in Chapter 1, PHP is a loosely typed language.This means that, under the right circumstances, it will automatically juggle data types to perform its operations according to how programmers are most likely to want it to. There are scenarios, however, in which this is not a desirable approach, and you want, instead, PHP to be strict and literal in the way it compares data. Consider, for example, what would happen if you were dealing with information coming from a patient’s med- ical record. In this situation, you’ll want to make sure that nothing is left to chance and that PHP doesn’t attempt to interpret user input in too liberal a way. Generally speaking, it’s always a good idea to use the identity operators (=== and !==) whenever you know that a value has to be of a certain type: if ($a !== 0) { echo ‘$a is not an integer zero’; } Testing for Resource Allocation One of the most common mistakes that causes code to become unreliable consists of using external resources without ensuring that they are available. For example, look at the following code: 8. Ternary Operators and if Statements 191$res = mysql_query(“select foo from bar”); while ($row = mysql_fetch_array($res)) { print $row[‘foo’].””; } See what’s wrong? The author doesn’t test for the query’s failure before moving on to perform other tasks that use the resource returned by mysql_query().The query could fail for a number of reasons, even though it is syntactically correct—for example, the server might be unavailable, or there could be a network interruption.What’s worse in this particular case, the MySQL extension does not cause a fatal error if a query cannot be executed.Therefore, the script moves on, and a cascade of additional problems could be caused by this initial blunder. If, on the other end, error conditions are properly tested for, this issue doesn’t even present itself: if (!$res = mysql_query(“select foo from bar”)) { /** * no valid result, log/print error, mysql_error() will tell you */ } else { while ($row = mysql_fetch_array($res)) { print $row[‘foo’].””; } } It’s undoubtedly hard to write an if statement every time you execute a query—but also necessary if you are serious about error management.To make things a bit easier on yourself (and your entire team), you could adopt one of the many abstraction layers available or write one yourself.This way, the actual error management can be performed in a centralized location (the abstraction layer), and you won’t have to write too much code. It’s important to keep in mind that this process is required whenever you interact with an external resource, be it a database, a file, or a network connection. Starting with PHP 5, you can use other error-control structures known as exceptions. However, remember that these are not available in PHP 4 and, therefore, cannot be used to solve a problem that appears in the exam. Ternary Operators and if Statements if statements are necessary control structures for all but the simplest of PHP scripts. As a result, sometimes they will tend to be very complex, even if you nest them on various levels. 9. 192 Chapter 12 Debugging and Performance Luckily, the ternary conditional operator that you saw in Chapter 1 can be used to simplify the use of if statements by embedding them directly in a larger expression. For example, consider the following snippet of code: function is_my_country($country) { if (strlen($country) == 3) { return 1; } else { return 0; } } It could also be written as function is_my_country($country) { return (strlen($country)==3) ? 1 : 0; } As you can see, the function is much shorter than the if statement in the preceding example.This can be very valuable if you’re dealing with a complex piece of code such as the following:$db->query(sprintf(“insert into foo(f1,f2,f3) values(‘%s’,’%s’,’%s’)”, (isset($_SESSION[‘foobar’])) ? ‘yes’ : ‘no’, (isset($_POST[‘myfoo’]) && $_POST[‘myfoo’]!=’’) ?$_POST[‘myfoo’] : ‘no’, ‘foo’)); A call such as the preceding one would have been a lot more complex if it had been written using traditional if statements—not to mention that you would have needed either a number of new variables to hold the information, or a different set of function calls for each possible scenario. Logging and Debugging Displaying error messages to the browser is a problem from many points of view. First, you’re letting your visitors know that something in your code is broken, thus shaking their confidence in the solidity of your website. Second, you’re exposing yourself to potential security vulnerabilities because some of the information outputted might be used to hack into your system.Third, you’re preventing yourself from finding out what error occurred so that you can fix it. A good solution to this problem consists of changing your php.ini setting so that the errors are not displayed to the screen, but stored in a log file.This is done by turning display_errors to off and log_errors to on, as well as setting a log file where the
14. Exam Prep Questions 197 This is what “bytecode caches” do.They are usually installed as simple extensions to PHP that act in a completely transparent way, caching the bytecode versions of your script and skipping the parsing step unless it is necessary—either because the script has never been parsed before (and, therefore, can’t be in the cache yet) or because the origi- nal script has changed and the cache needs refreshing. A number of commercial and open-source bytecode caches (also called accelerators) are available on the market, such as the one contained in the Zend Performance Suite, or the open-source APC. Most often, they also modify the bytecode so as to optimize it by removing unnecessary instructions. Bytecode caching should always be the last step in your optimization process because no matter how efficient your code is, it’s always going to provide you with the same per- formance boost. And, as a result, it could trick you into a false sense of security that would prevent you from looking at the other performance optimization techniques available. Exam Prep Questions 1. How can the following line of code be improved? $db->query(“insert into foo values($id,$bar)”); A. Use addslashes and sprintf to avoid security holes and make the code cleaner B. Split the query over several lines C. Use mysql_query() instead of$db->query() D. Define the table fields that will be affected by the INSERT statement E. Use mysql_query() instead of \$db->query() and addslashes to avoid security holes Answers A, B, and D are correct. First of all, you need to ensure that the query is secure; this is done by executing addslashes (or the equivalent function for your DBMS of choice) to prevent scripting attacks. If your query is long, it’s not a bad idea to split it over several lines to get a better overview of your code. Use sprintf() where possible to make the code cleaner. Finally it’s always a good idea to define the table fields that will be filled by an INSERT statement to prevent unexpected errors if the table changes.
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2017-12-11 02:18:32
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https://www.lavoisier.fr/livre/autre/linear-partial-differential-operators/hormander/descriptif_2856540?combien=1
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# Linear Partial Differential Operators, Softcover reprint of the original 1st ed. 1963Grundlehren der mathematischen Wissenschaften Series, Vol. 116
Langue : Anglais
• ## Sommaire
The aim of this book is to give a systematic study of questions con cerning existence, uniqueness and regularity of solutions of linear partial differential equations and boundary problems. Let us note explicitly that this program does not contain such topics as eigenfunction expan sions, although we do give the main facts concerning differential operators which are required for their study. The restriction to linear equations also means that the trouble of achieving minimal assumptions concerning the smoothness of the coefficients of the differential equations studied would not be worth while; we usually assume that they are infinitely differenti able. Functional analysis and distribution theory form the framework for the theory developed here. However, only classical results of functional analysis are used. The terminology employed is that of BOURBAKI. To make the exposition self-contained we present in Chapter I the elements of distribution theory that are required. With the possible exception of section 1.8, this introductory chapter should be bypassed by a reader who is already familiar with distribution theory.
I: Functional analysis.- I. Distribution theory.- 1.0. Introduction.- 1.1. Weak derivatives.- 1.2. Test functions.- 1.3. Definitions and basic properties of distributions.- 1.4. Differentiation of distributions and multiplication by functions.- 1.5. Distributions with compact support.- 1.6. Convolution of distributions.- 1.7. Fourier transforms of distributions.- 1.8. Distributions on a manifold.- II. Some special spaces of distributions.- 2.0. Introduction.- 2.1. Temperate weight functions.- 2.2. The spaces ?p, k.- 2.3. The spaces $$\mathcal{B}_{p,k}^{loc}$$.- 2.4. The spaces ?(s).- 2.5. The spaces ?(m, s).- 2.6. The spaces $$\mathcal{H}_{(s)}^{loc}\left( \Omega \right)$$ when ? is a manifold.- II: Differential operators with constant coefficients.- III. Existence and approximation of solutions of differential equations.- 3.0. Introduction.- 3.1. Existence of fundamental solutions.- 3.2. The equation P (D) u = f when f ? ??.- 3.3. Comparison of differential operators.- 3.4. Approximation of solutions of homogeneous differential equations.- 3.5. The equation P (D) u = f when f is in a local space $$\subset {\mathcal{D}'_F}$$.- 3.6. The equation P (D) u = f when $$f \in \mathcal{D}'$$.- 3.7. The geometric meaning of P-convexity and strong P-convexity.- 3.8. Systems of differential operators.- IV. Interior regularity of solutions of differential equations.- 4.0. Introduction.- 4.1. Hypoelliptic operators.- 4.2. Partially hypoelliptic operators.- 4.3. Partial hypoellipticity at the boundary.- 4.4. Estimates for derivatives of high order.- V. The Cauchy problem (constant coefficients).- 5.0. Introduction.- 5.1. The classical existence theory for analytic data.- 5.2. The non-uniqueness of the characteristic Cauchy problem.- 5.3. Holmgren’s uniqueness theorem.- 5.4. The necessity of hyperbolicity for the existence of solutions to the noncharacteristic Cauchy problem.- 5.5. Algebraic properties of hyperbolic polynomials.- 5.6. The Cauchy problem for a hyperbolic equation.- 5.7. A global uniqueness theorem.- 5.8. The characteristic Cauchy problem.- III: Differential operators with variable coefficients.- VI. Differential equations which have no solutions.- 6.0. Introduction.- 6.1. Conditions for non-existence.- 6.2. Some properties of the range.- VII. Differential operators of constant strength.- 7.0. Introduction.- 7.1. Definitions and basic properties.- 7.2. Existence theorems when the coefficients are merely continuous.- 7 3 Existence theorems when the coefficients are in C?.- 7.4. Hypoellipticity.- 7.5. The analyticity of the solutions of elliptic equations.- VIII. Differential operators with simple characteristics.- 8.0. Introduction.- 8.1. Necessary conditions for the main estimates.- 8.2. Differential quadratic forms.- 8.3. Estimates for elliptic operators.- 8.4. Estimates for operators with real coefficients.- 8.5. Estimates for principally normal operators.- 8.6. Pseudo-convexity.- 8.7. Estimates, existence and approximation theorems in ?(s).- 8.8. The unique continuation of singularities.- 8.9. The uniqueness of the Cauchy problem.- IX. The Cauchy problem (variable coefficients).- 9.0. Introduction.- 9.1. Preliminary lemmas.- 9.2. The basic L2 estimate.- 9.3. Existence theory for the Cauchy problem.- X. Elliptic boundary problems.- 10.0. Introduction.- 10.1. Definition of elliptic boundary problems.- 10.2. Preliminaries concerning ordinary differential operators.- 10.3. Construction of a parametrix.- 10.4. Local theory of elliptic boundary problems.- 10.5. Elliptic boundary problems in a compact manifold with boundary.- 10.6. Various extensions and remarks.- Appendix. Some algebraic lemmas.- Index of notations.
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2019-08-20 21:22:43
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https://raja.readthedocs.io/en/develop/sphinx/user_guide/tutorial/vertexsum_coloring.html
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# Iteration Space Coloring: Mesh Vertex Sum¶
This section contains an exercise file RAJA/exercises/vertexsum-indexset.cpp for you to work through if you wish to get some practice with RAJA. The file RAJA/exercises/vertexsum-indexset_solution.cpp contains complete working code for the examples discussed in this section. You can use the solution file to check your work and for guidance if you get stuck. To build the exercises execute make vertexsum-indexset and make vertexsum-indexset_solution from the build directory.
Key RAJA features shown in this example are:
• RAJA::forall loop execution template method
• RAJA::TypedListSegment iteration space construct
• RAJA::TypedIndexSet iteration space segment container and associated execution policies
The example computes a sum at each vertex on a logically-Cartesian 2D mesh as shown in the figure.
Each sum is an average of the area of the four mesh elements that share the vertex. In many “staggered mesh” applications, an operation like this is common and is often written in a way that presents the algorithm clearly but prevents parallelization due to potential data races. That is, multiple loop iterates over mesh elements may attempt to write to the same shared vertex memory location at the same time. The example shows how RAJA constructs can be used to enable one to express such an algorithm in parallel and have it run correctly without fundamentally changing how it looks in source code.
We start by setting the size of the mesh, specifically, the total number of elements and vertices and the number of elements and vertices in each direction:
//
// 2D mesh has N^2 elements (N+1)^2 vertices.
//
constexpr int N = 1000;
constexpr int Nelem = N;
constexpr int Nelem_tot = Nelem * Nelem;
constexpr int Nvert = N + 1;
constexpr int Nvert_tot = Nvert * Nvert;
We also set up an array to map each element to its four surrounding vertices and set the area of each element:
//
// Define mesh spacing factor 'h' and set up elem to vertex mapping array.
//
constexpr double h = 0.1;
for (int ie = 0; ie < Nelem_tot; ++ie) {
int j = ie / Nelem;
int imap = 4 * ie ;
e2v_map[imap] = ie + j;
e2v_map[imap+1] = ie + j + 1;
e2v_map[imap+2] = ie + j + Nvert;
e2v_map[imap+3] = ie + j + 1 + Nvert;
}
//
// Initialize element areas so each element area
// depends on the i,j coordinates of the element.
//
std::memset(areae, 0, Nelem_tot * sizeof(double));
for (int ie = 0; ie < Nelem_tot; ++ie) {
int i = ie % Nelem;
int j = ie / Nelem;
areae[ie] = h*(i+1) * h*(j+1);
}
Then, a sequential C-style version of the vertex area calculation looks like this:
std::memset(areav_ref, 0, Nvert_tot * sizeof(double));
for (int ie = 0; ie < Nelem_tot; ++ie) {
int* iv = &(e2v_map[4*ie]);
areav_ref[ iv[0] ] += areae[ie] / 4.0 ;
areav_ref[ iv[1] ] += areae[ie] / 4.0 ;
areav_ref[ iv[2] ] += areae[ie] / 4.0 ;
areav_ref[ iv[3] ] += areae[ie] / 4.0 ;
}
We can’t parallelize the entire computation at once due to potential race conditions where multiple threads may attempt to sum to a shared element vertex simultaneously. However, we can parallelize the computation in parts. Here is a C-style OpenMP parallel implementation:
std::memset(areav, 0, Nvert_tot * sizeof(double));
for (int icol = 0; icol < 4; ++icol) {
const std::vector<int>& ievec = idx[icol];
const int len = static_cast<int>(ievec.size());
#pragma omp parallel for
for (int i = 0; i < len; ++i) {
int ie = ievec[i];
int* iv = &(e2v_map[4*ie]);
areav[ iv[0] ] += areae[ie] / 4.0 ;
areav[ iv[1] ] += areae[ie] / 4.0 ;
areav[ iv[2] ] += areae[ie] / 4.0 ;
areav[ iv[3] ] += areae[ie] / 4.0 ;
}
}
What we’ve done is broken up the computation into four parts, each of which can safely run in parallel because there are no overlapping writes to the same entry in the vertex area array in each parallel section. Note that there is an outer loop on length four, one iteration for each of the elements that share a vertex. Inside the loop, we iterate over a subset of elements in parallel using an indexing area that guarantees that we will have no data races. In other words, we have “colored” the elements as shown in the figure below.
We partition the mesh elements into four disjoint subsets shown by the colors and numbers so that within each subset no two elements share a vertex.
For completeness, the computation of the four element indexing arrays is:
//
// Gather the element indices for each color in a vector.
//
std::vector< std::vector<int> > idx(4);
for (int ie = 0; ie < Nelem_tot; ++ie) {
int i = ie % Nelem;
int j = ie / Nelem;
if ( i % 2 == 0 ) {
if ( j % 2 == 0 ) {
idx[0].push_back(ie);
} else {
idx[2].push_back(ie);
}
} else {
if ( j % 2 == 0 ) {
idx[1].push_back(ie);
} else {
idx[3].push_back(ie);
}
}
}
## RAJA Parallel Variants¶
To implement the vertex sum calculation using RAJA, we employ RAJA::TypedListSegment iteration space objects to enumerate the mesh elements for each color and put them in a RAJA::TypedIndexSet object. This allows us to execute the entire calculation using one RAJA::forall call.
We declare a type alias for the list segments to make the code more compact:
using SegmentType = RAJA::TypedListSegment<int>;
Then, we build the index set:
RAJA::TypedIndexSet<SegmentType> colorset;
colorset.push_back( SegmentType(&idx[0][0], idx[0].size(), host_res) );
colorset.push_back( SegmentType(&idx[1][0], idx[1].size(), host_res) );
colorset.push_back( SegmentType(&idx[2][0], idx[2].size(), host_res) );
colorset.push_back( SegmentType(&idx[3][0], idx[3].size(), host_res) );
Note that we construct the list segments using the arrays we made earlier to partition the elements. Then, we push them onto the index set.
Now, we can use a two-level index set execution policy that iterates over the segments sequentially and executes each segment in parallel using OpenMP multithreading to run the kernel:
using EXEC_POL1 = RAJA::ExecPolicy<RAJA::seq_segit,
RAJA::omp_parallel_for_exec>;
RAJA::forall<EXEC_POL1>(colorset, [=](int ie) {
int* iv = &(e2v_map[4*ie]);
areav[ iv[0] ] += areae[ie] / 4.0 ;
areav[ iv[1] ] += areae[ie] / 4.0 ;
areav[ iv[2] ] += areae[ie] / 4.0 ;
areav[ iv[3] ] += areae[ie] / 4.0 ;
});
The execution of the RAJA version is similar to the C-style OpenMP variant shown earlier, where we executed four OpenMP parallel loops in sequence, but the code is more concise. In particular, we execute four parallel OpenMP loops, one for each list segment in the index set. Also, note that we do not have to manually extract the element index from the segments like we did earlier since RAJA passes the segment entries directly to the lambda expression.
Here is the RAJA variant where we iterate over the segments sequentially, and execute each segment in parallel via a CUDA kernel launched on a GPU:
using EXEC_POL2 = RAJA::ExecPolicy<RAJA::seq_segit,
RAJA::cuda_exec<CUDA_BLOCK_SIZE>>;
RAJA::forall<EXEC_POL2>(cuda_colorset, [=] RAJA_DEVICE (int ie) {
int* iv = &(e2v_map[4*ie]);
areav[ iv[0] ] += areae[ie] / 4.0 ;
areav[ iv[1] ] += areae[ie] / 4.0 ;
areav[ iv[2] ] += areae[ie] / 4.0 ;
areav[ iv[3] ] += areae[ie] / 4.0 ;
});
The only differences here are that we have marked the lambda loop body with the RAJA_DEVICE macro, used a CUDA segment execution policy, and built a new index set with list segments created using a CUDA resource so that the indices live in device memory.
The RAJA HIP variant, which we show for completeness, is similar:
using EXEC_POL3 = RAJA::ExecPolicy<RAJA::seq_segit,
RAJA::hip_exec<HIP_BLOCK_SIZE>>;
RAJA::forall<EXEC_POL3>(hip_colorset, [=] RAJA_DEVICE (int ie) {
int* iv = &(d_e2v_map[4*ie]);
d_areav[ iv[0] ] += d_areae[ie] / 4.0 ;
d_areav[ iv[1] ] += d_areae[ie] / 4.0 ;
d_areav[ iv[2] ] += d_areae[ie] / 4.0 ;
d_areav[ iv[3] ] += d_areae[ie] / 4.0 ;
});
The main difference for the HIP variant is that we use explicit device memory allocation/deallocation and host-device memory copy operations.
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2023-04-01 07:14:34
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|
https://math.stackexchange.com/questions/664982/explicit-topological-conjugation-between-two-odes/665843
|
# Explicit topological conjugation between two ODEs
I am figuring out (trying) to find an explicit topological conjugation between two ODEs.
$$\frac{dx}{dt} = -x \quad (\text{flow: } e^{-t})$$
and
$$\frac{dx}{dt} = Ax, \qquad A = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}$$ (a $2 \times 2$ matrix with hyperbolic eigenvalues)
My question at hand is it is easy to find a homeomorphism between two matrices, but a matrix and a non-matrix has me a bit confused at the moment. I got stuck after finding the matrix exponential of the $2 \times 2$ matrix because I do not know what to do afterwards...
• I know very little about this subject, but I am assuming $x$ is a vector of length 2 here? In which case, $-x = -Ix$ where I is the identity matrix, so you really do have two matrices. – Steven Gubkin Feb 5 '14 at 20:44
• Got something from the answer below? – Did Mar 17 '14 at 8:17
$$\dot y=\begin{bmatrix} -2 & 1\\ 1 & -2 \end{bmatrix}y$$ to $$\dot z=Bz=\begin{bmatrix} -3 & 0\\ 0 & -1 \end{bmatrix}z.$$
The previous transformation is done by defining $z=Cy$ with $C=\begin{bmatrix} -1 & 1\\ 1 & 1\end{bmatrix}$, and noting that $CAC^{-1}=B$.
Next you already have the second equation as you want it, so $x_2=z_2$. For the first coordinate just define $x_1=z_1^{1/3}$ and perform the required computations.
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2019-09-16 08:43:26
|
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https://cds.cern.ch/collection/CMS%20Theses?ln=ca
|
The CDS website will be unavailable on Wednesday, June 29th, from 7am to 7.30am CEST, due to a planned intervention.
# CMS Theses
2022-06-26
20:23
Application of Deep Learning techniques in the search for BSM Higgs bosons in the $\mu\mu$ final state in CMS / Diotalevi, Tommaso The Standard Model (SM) of particle physics predicts the existence of a Higgs field responsible for the generation of the particles' mass [...] CERN-THESIS-2022-073 - 214 p.
2022-06-26
12:23
Search for high-mass resonances in final states with a boosted-dijet resonance in proton-proton collisions at sqrt{s} = 13 TeV with the CMS detector / Quaranta, Claudio The Standard Model (SM) of particle physics provides the best description of the fundamental constituents of the universe and their interactions [...] CERN-THESIS-2022-072 - 173 p.
2022-06-20
13:44
Search for a $HH$ production in the $b \bar{b} ZZ(4l)$ final state at $\sqrt{s} = 13$TeV with full Run II data / Margjeka, Ilirjan A search for resonant and non-resonant double Higgs boson production at the LHC with the CMS experiment is presented [...] CMS-TS-2022-009 ; CERN-THESIS-2022-068. - 2022. - 218 p.
2022-06-09
14:05
Prospects of Nonresonant Higgs Boson Pair Production Measurement in the $WW\gamma\gamma$ Channel at the HL-LHC with the Phase-II CMS Detector / Guzel, Oguz Since the discovery of the Higgs boson in 2012 by the CMS and ATLAS experiments at the CERN's Large Hadron Collider in Geneva, Switzerland, physicists have tried to measure accurately its properties and to understand better the underlying electroweak symmetry breaking mechanism [...] CERN-THESIS-2022-059 - Istanbul : Istanbul Technical University, 2022-06-02. - 129 p.
2022-06-06
22:39
Searches for undiscovered processes using the multilepton final state in proton-proton collisions at CMS / Verbeke, Willem The standard model (SM) of particle physics represents our most fundamental knowledge of nature [...] CERN-THESIS-2022-057 - 278 p.
2022-05-24
22:30
Search for the decay of Standard Model Higgs bosons to a charm quark-antiquark pair using 138 fb$^{-1}$ of CMS proton-proton collision data at $\sqrt{s}=13~$TeV / Burkle, Bjorn The Standard Model of particle physics describes the fundamental ways in which the particles of our universe interact [...] CERN-THESIS-2022-050 -
2022-05-24
11:50
Performance of High Granularity Calorimeter prototypes for the CMS HL-LHC upgrade in beam test experiments at CERN / Pandey, Shubham The CMS collaboration has opted for a High Granularity Calorimeter (HGCAL) to replace the current end cap electromagnetic and hadronic calorimeters in the view of high-luminosity phase of the LHC (HL-LHC) [...] CERN-THESIS-2022-048 CMS-TS-2022-008. - 2022. - 269 p.
2022-05-16
18:49
Top Quark Studies with the CMS Experiment: Rare Process and Precision Measurements / Zhang, Wenyu Top quark is the heaviest known fundamental particle [...] CERN-THESIS-2022-043 - 169 p.
2022-05-10
14:09
Performance of the GE1/1 detectors for the upgrade of the CMS muon forward system / Mocellin, Giovanni "The CERN Large Hadron Collider (LHC) will undergo major upgrades over the course of the decade that will lead to the High Luminosity LHC [...] CERN-THESIS-2021-327 10.18154/RWTH-2021-07066. -
2022-05-06
16:55
Studies of eco-friendly gas mixtures for RPC detectors at CERN LHC experiments / Verzeroli, Mattia Several gaseous detectors are operated at the CERN LHC experiments [...] CERN-THESIS-2022-040 - 161 p.
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2022-06-29 01:44:37
|
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https://zbmath.org/?q=an:0646.35072
|
# zbMATH — the first resource for mathematics
Absence of shocks in an initially dilute collisionless plasma. (English) Zbl 0646.35072
The Cauchy problem for the relativistic Vlasov-Maxwell equations $\partial_ tf_{\alpha}+\hat v_{\alpha}\cdot \nabla_ xf_{\alpha}+e_{\alpha}+e_{\alpha}[E+c^{-1}\hat v_{\alpha}\wedge B]\cdot \nabla_ vf_{\alpha}=0$
$E_ t=c Cur B-j,\quad \nabla \cdot E=\rho;\quad B_ t=-c Curl E,\quad \nabla \cdot B=0$ is studied in three dimensions. The authors prove: If the initial data satisfy the constraints $$(\nabla \cdot E_ 0=\rho_ 0\equiv 4\pi \int_{k^ 3}\sum_{\alpha}e_{\alpha}f_{\alpha_ 0}dv$$, $$\nabla \cdot B_ 0=0)$$ and have compact support and sufficiently small $$C^ 2$$ norm, then there exists a unique global $$C^ 1$$-solution. This is proved using the iteration method. This class of problems have been studied by the authors, Bardos and Degond [the authors, Math. Methods Appl. Sci. 9, 46- 52 (1987); C. Bardos and P. Degond, Ann. Inst. Henri Poincaré, Anal. Non Linéaire 2, 101-118 (1985; Zbl 0593.35076)].
Reviewer: B.Guo
##### MSC:
35Q99 Partial differential equations of mathematical physics and other areas of application 82C70 Transport processes in time-dependent statistical mechanics 35B40 Asymptotic behavior of solutions to PDEs
Full Text:
##### References:
[1] Bardos, C., Degond, P.: Global existence for the Vlasov-Poisson equation in 3 space variables with small initial data. Ann. Inst. Henri Poincaré, Analyse non linéaire2, 101-118 (1985) · Zbl 0593.35076 [2] Bardos, C., Degond, P., Ha, T.-N.: Existence globale des solutions des équations de Vlasov-Poisson relativistes en dimension 3. C. R. Acad. Sci. Paris301, 265-268 (1985) · Zbl 0598.35109 [3] Glassey, R., Strauss, W.: Singularity formation in a collisionless plasma could occur only at high velocities. Arch. Ration. Mech. Anal.92, 59-90 (1986) · Zbl 0595.35072 [4] Glassey, R., Strauss, W.: High velocity particles in a collisionless plasma. Math. Methods Appl. Sci.9, 46-52 (1987) · Zbl 0649.35079 [5] Horst, E.: Global solutions of the relativistic Vlasov-Maxwell system of plasma physics. Habilitationsschrift, Universität München 1986 [6] John, F.: Blow-up of solutions of nonlinear wave equations in three space dimensions. Manuscr. Math.28, 235-268 (1979) · Zbl 0406.35042
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-12-04 10:52:59
|
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https://socratic.org/questions/a-tank-contains-n-2-at-1-0-atm-and-o-2-at-2-0-atm-helium-is-added-to-this-tank-u
|
# A tank contains N_2 at 1.0 atm and O_2 at 2.0 atm. Helium is added to this tank until the total pressure is 6.0 atm. What is the partial pressure of the helium?
${p}_{H e}$ $=$ $3.0 \cdot a t m$
We know that the total pressure is $6.0 \cdot a t m$; i.e. $6.0 \cdot a t m$ $=$ ${p}_{{N}_{2}} + {p}_{{O}_{2}} + {p}_{H e}$.
Since we have been given the other partial pressures, calculation of ${p}_{H e}$ is trivial.
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2020-05-29 01:24:58
|
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https://tech.io/playgrounds/54906/rust-for-python-developers---unsigned-signed-integers-and-casting/sign-and-magnitude
|
# Rust for Python Developers - Unsigned, Signed Integers and Casting
Shin_O
1,491 views
### Open Source Your Knowledge, Become a Contributor
This story was originally published on Medium
## Signed, Ones' Complement and Two's Complement
In computing, signed number representations are required to encode negative numbers in binary number systems. Let’s examine sign-and-magnitude, ones’ complement, and two’s complement.
## Sign-and-Magnitude
Sign-and-Magnitude is also called Signed Magnitude. The first bit (called the most significant bit or MSB) tells if it is positive by 0 or a negative by 1. The rest is called magnitude bits.
As I mentioned it before that signed integer types have the min and the max from -(2ⁿ⁻¹) to 2ⁿ⁻¹-1 where n stands for the number of bits. Since we use the first bit for the positive and negative signs we have n-1 in the 2ⁿ⁻¹.
For 4-bit the min and max are from -(2³) to 2³–1, which is -8 to +7.
As you see in the diagram above, the positive and the negative have the same digits except for the sign bit.
The problem of the signed magnitude is that there are two zeros, 0000 and 1000.
Open Source Your Knowledge: become a Contributor and help others learn.
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2020-09-30 21:15:09
|
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https://indico.cern.ch/event/868940/contributions/3814843/
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# ICHEP 2020
July 28, 2020 to August 6, 2020
virtual conference
Europe/Prague timezone
## MADMAX: A Dielectric Haloscope Experiment
Jul 31, 2020, 9:30 AM
25m
virtual conference
#### virtual conference
Talk 09. Dark Matter Detection
### Speaker
Xiaoyue Li (Max Planck Institute for Physics)
### Description
Axions emerge naturally from the Peccei-Quinn (PQ) mechanism which addresses the absence of CP violation in QCD; they also turn out to be a good cold dark matter (CDM) candidate. If PQ symmetry breaking had occurred after inflation, the axion mass is likely to range from $\sim26 ~\mu$eV to $\sim1$ meV, which is yet to be explored experimentally.
We present a novel dielectric haloscope experiment dedicated to the direct detection of axion CDM in the mass range of 40 to 400 $\mu$eV — the MAgnetized Disc and Mirror Axion eXperiment (MADMAX). Multiple dielectric discs and a metal mirror are placed in a strong magnetic field ($\sim10$ T dipole) to utilize the axion-induced coherent electromagnetic waves emitted from each disc surface and their resonances within the discs-mirror system, such that the axion-induced signal can be boosted to a level detectable by state-of-the-art low noise amplifiers. The design and sensitivity of MADMAX, ongoing R&D activities and the project roadmap will be presented.
### Primary author
Xiaoyue Li (Max Planck Institute for Physics)
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2021-12-02 16:33:52
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http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:1196.33017
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Zbl 1196.33017
Yan, Zhenya
Extended Jacobian elliptic function algorithm with symbolic computation to construct new doubly-periodic solutions of nonlinear differential equations.
(English)
[J] Comput. Phys. Commun. 148, No. 1, 30-42 (2002). ISSN 0010-4655
Summary: With the aid of computerized symbolic computation, the extended Jacobian elliptic function expansion method and its algorithm are presented by using some relations among ten Jacobian elliptic functions and are very powerful to construct more new exact doubly-periodic solutions of nonlinear differential equations in mathematical physics. The new $(2+1)$-dimensional complex nonlinear evolution equations is chosen to illustrate our algorithm such that sixteen families of new doubly-periodic solutions are obtained. When the modulus $m\rightarrow 1$ or 0, these doubly-periodic solutions degenerate as solitonic solutions including bright solitons, dark solitons, new solitons as well as trigonometric function solutions.
MSC 2000:
*33E05 Elliptic functions and integrals
33F05 Numerical approximation of special functions
35B10 Periodic solutions of PDE
35Q53 KdV-like equations
37K40 Soliton theory, asymptotic behavior of solutions
Highlights
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2013-05-22 04:32:16
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https://www.physicsforums.com/threads/find-mass-and-center-of-mass-of-ice-cream-cone.403936/
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# Find mass and center of mass of ice cream cone
1. May 17, 2010
### Michels10
1. The problem statement, all variables and given/known data
A toy manufacturer wants to create a toy ice cream cone by fitting a sphere of radius 4 cm inside a cone with a height of 8 cm and radius of the base of 3 cm. The base of the cone is concave, but the rest of the cone is solid plastic so that with the sphere attached there is no hollow space inside. The sphere and cone are constructed from special materials so that the density of the sphere is proportional to the distance from the tip of the cone, with constant of proportionality 1.4, and the density of the cone is proportional to the distance from its vertical axis, with constant of proportionality 1.8. Find the mass and center of mass of the ice cream cone. (Hint: decide on a position for the whole ice cream cone. If the center of the sphere is the origin, then the tip of the cone cannot also be at the origin. It is a good idea to have the cone oriented vertically the way you would normally hold one.) Also, provide a plot showing the entire ice cream cone.
Note from professor: Instead of dealing with these
integrals, I will allow you to change the density functions given so
that instead the densities for both the cone and sphere are
proportional to the square of the distance described. With these
simpler density functions (without the square roots), the integrals
should be much simpler to solve.
2. Relevant equations
Sphere: z = x^2 + y^2 + z^2
Cone: z = sqrt(x^2 + y^2)
Vsphere = 4/3*pi*r^3
Vcone = (pi*h*r^2)/3
density = mass/volume
center of mass:
x(bar) = Myz/m
y(bar) = Mxz/m
z(bar) = Mxy/m
where M is the moment about a coordinate plane.
3. The attempt at a solution
Not exactly sure where to start... If I have no mass how can I begin to solve for this? Should I start by solving for the volume that the sphere+cone take up?
Any help is greatly appreciated!
Thanks,
Michels10
2. May 17, 2010
### Staff: Mentor
Here is the information you need to use to get the mass.
The mass of the sphere = density * volume, with the density as specified above. The mass of the cone can be computed similarly. Also, keep in mind the hint given by your professor.
In both, the density is not constant (i.e., is a function).
Your equation above is incorrect. You have z as a function of x, y, and z. What you should have is x^2 + y^2 + z^2 = r^2.
3. May 17, 2010
### Michels10
Dont I still have two unknowns? density and mass? I guess my problem now is finding the equation for density. Do I need to set the equations equal to and find out where they intersect?
I've solved for both volumes:
Vcone = 24*pi
Vsphere = (32/3)*pi
Thanks,
Michels10
Last edited: May 18, 2010
4. May 18, 2010
### Staff: Mentor
Your density functions will be in units of mass/volume. Due to the way the density functions are defined, the center of mass will be along the central axis of the cone+sphere, which I'm assuming you will orient so that the z-axis passes vertically through the center of the cone and sphere.
It's not enough to get the volume of each of the two solids. The formulas I gave before should be interpreted as incremental mass, $\Delta m$, which you will need to use in each integral.
Of course you need the equations of the cone and sphere. Hopefully you have already graphed them.
5. May 18, 2010
### Michels10
how do these equations look:
cone- z2 = (64/9)(x2+y2)
sphere- 4 = x2+y2+z2
Cone:
Int(Int(Int(1.4*(64/9)(r2)*r,dr),d(theta)),dz)
r=0 to 3
theta = 0 to 2pi
z = 0 to 8
Sphere:
Int(Int(Int(1.8*p2*sin(phi),d(p),d(theta)),d(phi))
p = 0..2
theta = 0..2pi
phi = 0..pi
Last edited: May 18, 2010
6. May 18, 2010
### Michels10
i've attached a picture of the cone/sphere graphed.
#### Attached Files:
• ###### cone_graphed.jpg
File size:
21.4 KB
Views:
241
7. May 18, 2010
### Staff: Mentor
The equations in post #5 don't go with the graphs in post #6. Your equation for the sphere has its center at the origin. The graph of the sphere has its center at somewhere around (0, 0, 11).
Also, your equation for the sphere has a radius of 2. In the problem description, the radius is supposed to be 4.
8. May 18, 2010
### Michels10
I updated one of my last posts with some integrals... But back to the graph..
The cone has a height of 8. The total height from the bottom of the cone to the center of the sphere is 8+sqrt(7). And the radius is still 4 for the sphere, and still 3 for the cone
9. May 18, 2010
### Staff: Mentor
Read what I said in post #7.
As for your integrals, I think you are jumping the gun. It doesn't look to me like you have the density functions figured out yet.
10. May 18, 2010
### Michels10
Okay, then how about I change them...
sphere- 16 = x2+y2+(z - (8+sqrt(7)))2
cone- x2 + (y-3)2 = (z-8)2
11. May 18, 2010
### Staff: Mentor
Where did the 8 + sqrt(7) come from?
Why is the y - 3 term in the equation for the cone?
12. May 18, 2010
### Michels10
I think i've about given up on this one. I clearly have no idea what I'm doing. I was trying to rearrange the equations to fit my graph. I saw my professor, he gave me an equation of
z^2 = 64/9 * x^2 + y^2 for the cone
and said that the height from the center of the sphere to the bottom of the cone was 8+ sqrt(7). the radius of the cone is 3, the radius of the ball is 4... To find the distance from the center of the sphere to the cone-- that leaves us with a triangle that has 4^2 = x^2 + 3^2.
x= +- sqrt(7). Add this to the height of the cone and we get 8+sqrt(7)
I've got an example in my book that calculates the density+mass+center of mass, so I tried adapting that to this problem and that is why I 'jumped the gun' with the integrals.
If the tripple integral of the constant of proportionality multiplied by the formula of the equation doesn't get me the mass, then im not sure what does.
p(x,y,z) = K*(x^2+y^2+z^2) for density.
where K is a constant
and
int(int(int( p(x,y,z) dx )dy)dz)
the tripple integral of the density equation for the mass.
13. May 18, 2010
### Staff: Mentor
This looks fine. The assumption here is that the tip of the cone is at the origin.
That's fine. I just didn't see where you got it, and now I do. This makes the equation of the sphere x2 + y2 + (z - (8 + sqrt(7))2 = 16.
But not for this problem. In this problem, the density varies in different ways for the cone and the sphere.
Before setting up the triple (one p in triple) integrals, see if you can find formulas for the density functions as described in your first post, and using your instructor's suggestion. You'll need one function for the cone density and another for the sphere density.
14. May 19, 2010
### Michels10
Okay, I'm assuming it has something to do with the slope of the cone, and the increasing radius/size as you move up from the bottom of the cone... Since the radius is 3 and the height is 8, this gives us a slope of 8/3.
I'm assuming that I need to use that 8/3 in my equation, I'm not sure where though...
Cone:
$$\int^{2pi}_{0}\int^{8}_{0}\int^{(8/3)x}_{0} (1.4*r^2) r dr dz dtheta$$
Not sure about the sphere, though...
15. May 19, 2010
### Staff: Mentor
Double integrals should suffice for both the cone and sphere.
For the cone, the region over which integration is performed is {{r, theta) | 0 <= r <= 3, 0 <= theta <= 2pi}.
The density function is as you have it, 1.4*r^2 (using the note from your professor). The integral for the mass of the cone should look like this:
$$m = \int_{\theta = 0}^{2\pi} \int_{r = 0}^3 (z_{top} - z_{cone}) 1.4 r^2 r dr d\theta$$
What I'm showing as ztop is the z value at the top of the cone, which is the z value on the lower half of the sphere. zcone is the z value at a point on the surface of the cone.
16. May 19, 2010
### Michels10
Okay very cool. I suppose now I have to find out how deep the cone is in the center (along the z axis) due to concavity.
Thanks you for your help thus far. I am going to keep going at it! I'll post back soon with an update.
17. May 19, 2010
### Staff: Mentor
Yes, that's basically it. The top surface of the cone is the bottom surface of the sphere. You don't have to find how deep the top surface of the cone is - you just need the z value, which will be the z value on the lower half of the sphere.
Still ahead, you need to find Mz, the moment about the xy plane, and the mass and moment for the sphere.
18. May 19, 2010
### Michels10
The center of mass of the cone is somewhere along the Z axis. This yields a problem when dealing with double integrals doesn't it? I cant substitute a Z into the equation when dealing with polar coordinates.
I believe the equation is: 1/mass Int(Int((z*density)rdr)dtheta)
edit: i've been reading that the center of mass of a cone is simply h/4... But since my cone is concave, I'm not sure how to go about it.
Last edited: May 19, 2010
19. May 19, 2010
### LCKurtz
Yes, it does. You need a triple integral.
Just add the dz integral between the lower and upper limits (cone and lower sphere) as the inside integral (r dz dr dtheta).
20. May 20, 2010
### Michels10
Okay, got that, now what should I do for the sphere? Finding the density of the sphere should be relatively easy... Its not just the triple integral of volume * density constant *radius?
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2017-10-20 23:36:32
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https://math.stackexchange.com/questions/3164169/circle-drawn-on-focal-chord-of-a-parabola
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# Circle drawn on focal chord of a parabola
Is it possible for a circle with diameter as any focal chord of a parabola to cut the parabola at 4 points (2 being the extremities of the focal chord)?
We were asked to find the product of the parameters(t) of the point cut by the circle on the parabola (other than the extremities of the focal chord).Doing some algebra we obtain this product as 3,but I feel that the parameters would be imaginary since I don't think that such a circle can exist.Am I correct?
• What do you mean by “the” focal chord? There is an infinite number of them. – amd Mar 27 at 19:22
• @amd Any one of the infinite focal chords can be chosen. The product of the parameters will turn out to be same irrespective of which one you take. – Vaishakh Sreekanth Menon Mar 28 at 2:08
It is certainly possible for a circle whose diameter is a focal chord to meet a parabola in four places, although not when that chord is the latus rectum.
The parabola $$4fy=x^2$$ has its focus at $$F=(0,f)$$. A line through $$F$$ with slope $$m$$ meets the parabola at points $$P_{\pm} := f\left(\;2 (m \pm \sqrt{1 + m^2}), 1 + 2 m^2 \pm 2 m \sqrt{1 + m^2}\;\right) \tag{1}$$ The circle with diameter $$\overline{P_{+}P_{-}}$$ has equation $$x^2+y^2- 2 f m x - 2 f y ( 1 + 2 m^2 ) - 3 f^2 = 0 \tag{2}$$ Substituting $$y=x^2/(4f)$$ gives $$\left(x^2-4 f m x -4 f^2\right) \left(x^2+4f m x + 12 f^2 \right) = 0 \tag{3}$$ The roots of the first factor are the $$x$$-coordinates of $$P_{+}$$ and $$P_{-}$$. The roots of the second factor are $$x = 2f \left(-m \pm \sqrt{-3 + m^2}\right) \tag{4}$$ which are real and distinct, leading to two more circle-parabola intersection points, whenever $$|m|>\sqrt{3}$$. (For $$m=\pm \sqrt{3}$$, the two points coincide.) This agrees with @amd's comment to @GReyes' answer. $$\square$$
We observe (either by direct computation from $$(1)$$ and $$(4)$$, or by invoking Vieta's formulas on the constant terms of the factors of $$(3)$$) that the product of the $$x$$-coordinates of $$P_{+}$$ and $$P_{-}$$ is $$-4f^2$$, which is independent of $$m$$; likewise for the product of the $$x$$-coordinates of the other points of intersection, $$12 f^2$$. There's probably a nice geometric interpretation of these facts.
You are right. There are no such real points. If the equation of the parabola is $$y^2=4px$$, the radius of your circle is $$|y(p)|=2p$$. But then, the distance from any point on the circle to the focus is $$2p$$, whereas the distance to the directrix, on the left side of the circle, is less than $$2p$$ (the $$x$$-coordinate of those points is less than $$p$$) and, on the right half, larger than $$2p$$. Therefore, there are no other real intersections with the parabola, since on the parabola the two distances are equal.
• What's described here is only true of the latus rectum. Counterexample: the parabola $y^2=x$, with focus at $(1/4,0)$ and the focal chord drawn through the point $(4,2)$. A circle on this focal chord cuts the parabola in $4$ points. – nickgard Mar 27 at 13:58
• Indeed. For the parabola $x^2=4y$, any focal chord with slope outside the interval $[-\sqrt3,\sqrt3]$ will generate four intersection points. – amd Mar 27 at 19:40
• I agree. For some reason I understood that the focal chord referred to the latus rectum, probably the use of "the" instead of "a". Not true for a general chord through the focus, of course. – GReyes Mar 27 at 22:19
• But even if we do not chose the latus rectum won't his logic of distances being greater on one side of the circle and less on the other still hold? – Vaishakh Sreekanth Menon Mar 28 at 2:10
• No, it does not work. For one thing, the focus is not the center of the circle anymore. – GReyes Mar 28 at 4:20
If $$A(at^2,2at)$$ is one of the extremities of the focal chord, we know that the other end will be $$B\left(\dfrac{a}{t^2}, -\dfrac{2a}{t}\right)$$
Let the circle with $$AB$$ as diameter intersect the parabola at $$C(t')$$. Since $$\angle ACB$$ is a right angle, we have $$\dfrac{2}{t+t'} \times \dfrac{2}{t'-\dfrac{1}{t}}=-1$$
or $$t'^2+ \left(t-\dfrac{1}{t} \right)t'+3 =0$$
We see that this will have two real roots when $$\left(t-\dfrac{1}{t} \right)^2 > 12$$
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2019-05-26 00:58:00
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https://datascience.stackexchange.com/tags/feature-selection/hot
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Tag Info
3
There seems to be two possible approaches to your problem : If they are just identification features that you know aren't informative, you should remove them yourself. SelectKBest - like almost any other EDA tools - works on all the features you provide it, there is no way it knows what features are supposedly uninformative identification features and which ...
2
Its best to remove such a variable. Reasons are following: Artificial imputation can add bias and result cannot be justified because 99% data for the particular variable was artificially created. The variables/features that you choose for building the predictive model should have low correlation with the target/outcome variable/feature. Because, variable ...
1
Every time you compress the feature space you are losing some information. The original feature engineering stage you outlined sounds like a meaningful compression & might make sense in the context of your problem. The second compression on the other hand might only serve to lose some information. I would only perform the second compression if the ...
1
Exogenous simply means a value that is determined outside the context of your model & is then imposed on your model. Endogenous means the model determines the value. I don't know about "external" as this word seems to depend on context. But you would be right to say these variables are exogenous.
1
There's a good chance that it's a sign of overfitting: the fact that the importance of the features is not stable can be considered as an indication that the model itself is not stable, and this typically happens when it doesn't have enough information in the data to be sure how to use the features. As a result minor variations in the features or data cause ...
1
No, RFE cannot guarantee that it finds the feature subset with optimal score. As with most greedy processes, the point of RFE is to reduce the computational cost (fitting a model for each of the $2^m$ feature subsets), at the cost of perhaps not finding the actual optimum (but hopefully "close enough"). See also https://stats.stackexchange.com/...
1
This important point is missing: SFS is suitable as it has no assumption for features to be categorical or numerical. However, one-hot encoding is redundant when you are planning to use SFS. You just make the process longer by one-hot encoding since by doing so SFS needs to check more number of features than what it actually is.
Only top voted, non community-wiki answers of a minimum length are eligible
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2020-10-20 22:56:16
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https://deepai.org/publication/distances-between-states-and-between-predicates
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# Distances between States and between Predicates
This paper gives a systematic account of various metrics on probability distributions (states) and on predicates. These metrics are described in a uniform manner using the validity relation between states and predicates. The standard adjunction between convex sets (of states) and effect modules (of predicates) is restricted to convex complete metric spaces and directed complete effect modules. This adjunction is used in two state-and-effect triangles, for classical (discrete) probability and for quantum probability.
## Authors
• 18 publications
• 4 publications
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• ### Wave function representation of probability distributions
Orthogonal decomposition of the square root of a probability density fun...
12/21/2017 ∙ by Madeleine B. Thompson, et al. ∙ 0
• ### Categorical Equivalences from State-Effect Adjunctions
From every pair of adjoint functors it is possible to produce a (possibl...
01/29/2019 ∙ by Robert Furber, et al. ∙ 0
• ### Modeling Sequences with Quantum States: A Look Under the Hood
Classical probability distributions on sets of sequences can be modeled ...
10/16/2019 ∙ by Tai-Danae Bradley, et al. ∙ 0
##### This week in AI
Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.
## 1 Introduction
Metric structures have a long history in program semantics, see the overview book [2]. They occur naturally, for instance on sequences, of inputs, outputs, or states. In complete metric spaces solutions of recursive (suitably contractive) equations exist via Banach’s fixed point theorem. The Hausdorff distance on subsets is used to model non-deterministic (possibilistic) computation. In general, metrics can be used to measure to what extent computations can be approximated, or are similar.
This paper looks at metrics on probability distributions, often called states. Various such metrics exist for measuring the (dis)similarity in behaviour between probabilistic computations, see e.g. [6, 11, 4]. This paper does not develop new applications, but contributes to the theory behind distances: how they arise in a uniform way and how they can be related. The paper covers standard distance functions on classical discrete probability distributions and also on quantum distributions. For discrete probability we use the total variation distance, which is a special case of the Kantorovich distance, see e.g. [13, 5, 30, 29]. For quantum probability we use the trace distance for states (quantum distributions) on Hilbert spaces, and the operator norm distance for states on von Neumann algebras. One contribution of this paper is a uniform description of all these distances on states as ‘validity’ distances.
In each of these cases we shall describe a validity relation between states and predicates , so that the validity is a number in the unit interval . This validity relation plays a central role in the definition of various distances. What we call the ‘validity’ distance on states is given by the supremum (join) over predicates in:
d(ω1,ω2)=⋁p∣∣ω1⊨p−ω2⊨p∣∣. (1)
In general, states are closed under convex combinations. We shall thus study combinations of convex and complete metric spaces, in a category .
We also study metrics on predicates. The algebraic structure of predicates will be described in terms of effect modules. Here we show that suitably order complete effect modules are Archimedean, and thus carry an induced metric, such that limits and joins of ascending Cauchy sequences coincide. In our main examples, we use fuzzy predicates on sets and effects of von Neumann algebras as predicates; their distance can also be formulated via validity , but now using a join over states in:
d(p1,p2)=⋁ω∣∣ω⊨p1−ω⊨p2∣∣. (2)
The ‘duality’ between the distance formulas (1) for states and (2) for predicates is a new insight.
A basic ‘dual’ adjunction in a probabilistic setting is of the form , between effect modules and convex sets. Effect modules are the probabilistic analogues of Boolean algebras, serving as ‘algebraic probabilistic logics’ (see below for details). Convex sets capture the algebraic structure of states. This adjunction thus expresses the essentials of a probabilistic duality between predicates and states. Since predicates are often called ‘effects’ in a quantum setting, one also speaks of a duality between states and effects.
This paper restricts the this adjunction to an adjunction between directed complete effect modules and convex complete metric spaces. This restriced adjunction is used in two ‘state-and-effect’ triangles, of the form:
@C-1.5pc DcEMod ^op@/^2ex/[rr] & ⊤& ConvCMet @/^1.5ex/[ll] & & DcEMod ^op@/^2ex/[rr] & ⊤& ConvCMet @/^1.5ex/[ll] & Kℓ_fin(D)[ul]^Pred[ur]_Stat & & & & vNA ^op[ul]^Pred[ur]_Stat &
Details will be provided in Section 4. Thus, the paper culminates in suitable order/metrically complete versions of the state-and-effect triangles that emerge in the effectus-theoretic [16, 9] description of state and predicate transformer semantics for probability (see also [18, 20]).
## 2 Distances between states
This section will describe distance functions (metrics) on various forms of probability distributions, which we collectively call ‘states’. In separate subsections it will introduce discrete probability distributions on sets and on metric spaces, and quantum distributions on Hilbert spaces and on von Neumann algebras. A unifying formulation will be identified, namely what we call a validity formulation of the metrics involved, where the distance between two states is expressed via a join over all predicates using the validities of these predicates in the two states, as in (1).
### 2.1 Discrete probability distributions on sets
A finite discrete probability distribution on a set is given by ‘probability mass’ function with finite support and . This support is the set . We sometimes simply say ‘distribution’ instead of ‘finite discrete probability distribution’. Often such a distribution is called a ‘state’. The ‘ket’ notation is useful to describe specific distributions. For instance, on a set we may write a distribution as . This corresponds to the probability mass function given by , and .
We write for the set of distributions on a set . The mapping forms (part of) a well-known monad on the category of sets, see e.g. [15, 17, 18] for additional information, using the same notation as used here. We write for the associated Kleisli category, and for the category of Eilenberg-Moore algebras. The latter may be identified with convex sets, that is, with sets in which formal convex sums can be interpreted as actual sums. Thus we often write ; morphisms in are ‘affine’ functions, that preserve convex sums. Convex sets have a rich history, going back to [33], see [27, Remark 2.9] for an extensive description.
###### Definition 1
Let be two distributions on the same set . Their total variation distance is the positive real number defined as:
tvd(ω1,ω2)=12∑x∈X∣∣ω1(x)−ω2(x)∣∣. (3)
The historical origin of this definition is not precisely clear. It is folklore that the total variation distance is a special case of the ‘Kantorovich distance’ (also known as ‘Wasserstein’ or ‘earth mover’s distance’) on distributions on metric spaces, when applied to discrete metric spaces (sets), see Subsection 2.2 below.
We leave it to the reader to verify that is a metric on sets of distributions , and that its values are in the unit interval .
###### Example 2
Consider the sets and
with ‘joint’ distribution
given by . The first and second marginal of , written as and , are: and . We immediately see that is not the same as the product of its marginals, since . This means is ‘entwined’, see [24, 18]. One way to associate a number with this entwinedness is to take the distance between and the product of its marginals. It can be computed as:
For a function there are two associated ‘transformation’ functions, namely state transformation (aka. Kleisli extension) and predicate transformation . They are defined as:
f∗(ω)(y)=∑xf(x)(y)⋅ω(x)andf∗(q)(x)=∑yf(x)(y)⋅q(y). (4)
Maps are called (fuzzy) predicates on . In the special case where the outcomes are in the (discrete) subset , the predicate is called sharp. These sharp predicates correspond to subsets , via the indicator function .
For a state we write for the validity of predicate in state , defined as the expected value in . Thus, ; the latter sum is commonly written as . Further, the fundamental validity transformation equality holds: .
We conclude this subsection with a standard redescription of the total variation distance, see e.g. [13, 34]. It uses validity , as described above. Such ‘validity’ based distances will form an important theme in this paper. The proof of the next result is standard but not trivial and is included in the appendix, for the convenience of the reader.
###### Proposition 3
Let be an arbitrary set, with states . Then:
tvd(ω1,ω2)=⋁p∈[0,1]X∣∣ω1⊨p−ω2⊨p∣∣=maxU⊆Xω1⊨1U−ω2⊨1U
We write maximum ‘’ instead of join to express that the supremum is actually reached by a subset (sharp predicate). Completeness of the Kantorovich metric is an extensive topic, but here we only need the following (standard) result. Since there is a short proof, it is included.
###### Lemma 4
If is a finite set, then , with the total variation distance , is a complete metric space.
• Let and be a Cauchy sequence. For each we have . Hence, the sequence is Cauchy too, say with limit . Take . This is the limit of the .
### 2.2 Discrete probability distributions on metric spaces
A metric on a set is called 1-bounded if it takes values in the unit interval , that is, if it has type . We write for the category with such 1-bounded metric spaces as objects, and with non-expansive functions between them, satisfying . From now on we assume that all metric spaces in this paper are 1-bounded. For example, each set carries a discrete metric, where points have distance if they are equal, and otherwise.
For a metric space and two functions from some set to there is the supremum distance given by:
A ‘metric predicate’ on a metric space is a non-expansive function . These predicates carry the above supremum distance . We use them in the following definition of Kantorovich distance, which transfers the validity description of Proposition 3 to the metric setting.
###### Definition 5
Let be two discrete distributions on (the underlying set of) a metric space . The Kantorovich distance between them is defined as:
kvd(ω1,ω2)=⋁p∈Met(X,[0,1])∣∣ω1⊨p−ω2⊨p∣∣. (6)
This makes a (1-bounded) metric space.
The Kantorovich-Wasserstein duality Theorem gives an equivalent description of this distance in terms of joint states and ‘couplings’, see [28, 34] for details. Here we concentrate on relating the Kantorovich distance to the monad structure of distributions. The next lemma collects some basic, folkore facts.
###### Lemma 6
Let be metric spaces.
1. The unit function given by is non-expansive.
2. For each non-expansive function the corresponding state transformer from (4) is non-expansive.
As special cases, the multiplication map of the monad is non-expansive, and validity in its first argument as well.
3. If and are non-expansive, then so is . Moreover, the function is itself non-expansive, wrt. the supremum distance (5).
As a result, validity is non-expansive in its second argument too.
4. Taking convex combinations of distributions satisfies: for ,
kvd(r⋅σ1+s⋅σ2,r⋅τ1+s⋅τ2)≤r⋅kvd(σ1,τ1)+s⋅kvd(σ2,τ2).
• We do points (1) and (4) and leave the others to the reader. The crucial point that we use to show for (1) is that the unit map is non-expansive is: . Hence we are done because the join in (6) is over non-expansive functions in:
For point (4) we first notice that for and ,
(μ(Ω)⊨p)=∑xμ(Ω)(x)⋅p(x)=∑x(∑ωΩ(ω)⋅ω(x))⋅p(x)=∑ωΩ(ω)⋅(∑xω(x)⋅p(x))=∑ωΩ(ω)⋅(ω⊨p)=(Ω⊨((−)⊨p)),
where is used as (non-expansive) predicate on . Hence for with ,
###### Corollary 7
The monad on lifts to a monad, also written as , on the category , and commutes with forgetful functors, as in:
@R-0.5pc Met [d][rr]^-D & & Met [d] Sets [rr]^-D & & Sets (7)
We write for the category of Eilenberg-Moore algebras of this lifted monad, with ‘convex metric spaces’ as objects, see below.
The lifting (7) can be seen as a finite version of a similar lifting result for the ‘Kantorovich’ functor in [5]. This captures the tight Borel probability measures on a metric space . The above lifting (7) is a special case of the generic lifting of functors on sets to functors on metric spaces described in [3] (see esp. Example 3.3).
The category of the monad contains convex metric spaces, consisting of:
1. a convex set , that is, a set with an Eilenberg-Moore algebra of the distribution monad on ;
2. a metric on ;
3. a connection between the convex and the metric structure, via the requirement that the algebra map is non-expansive: , for all distributions .
The maps in are both affine and non-expansive. We shall write for the full subcategory of convex complete metric spaces.
###### Example 8
The unit interval is a convex metric space, via its standard (Euclidean) metric, and its standard convex structure, given by the algebra map defined by the ‘expected value’ operation:
The identity map is a predicate on that satisfies:
(ω⊨id)=∑xω(x)⋅id(x)=∑xω(x)⋅x=α(ω).
This allows us to show that is non-expansive:
∣∣α(ω1)−α(ω2)∣∣=∣∣ω1⊨id−ω2⊨id∣∣≤⋁p∣∣ω1⊨p−ω2⊨p∣∣=kvd(ω1,ω2).
In fact, we can see this as a special case of non-expansiveness of multiplication maps from Lemma 6 (2): indeed, , for the two-element set , and the algebra corresponds to the multiplication .
### 2.3 Density matrices on Hilbert spaces
The analogue of a probability distribution in quantum theory is often simply called a state. We first consider states of Hilbert spaces (over ), and consider the more general (and abstract) situation of states on von Neumann algebras in subsection 2.5.
A state of a Hilbert space is a density operator, that is, it is a positive linear map whose trace is one: . Recall that the trace of a positive operator is given by , where is any orthonormal basis for ; this value does not depend on the choice of basis , but might equal [1, Def. 2.51]. The same formula also works for when is not necessarily positive, but bounded with — where and is the adjoint of and where the square root is determined as the unique positive operator with . Such , which are aptly called trace-class operators, always have finite trace: , see [1, Def. 2.5{4,6}]. When is finite dimensional, any operator is trace-class, and when represented as a matrix, its trace can be computed as the sum of all elements on the diagonal. If is a density operator, then the associated matrix is called a density matrix. We refer for more information to for instance [1], and to [31, 32, 35] for the finite-dimensional case.
A linear map is called self-adjoint if and positive if it is of the form . This yields a partial order, with iff is positive. A predicate on is a linear map with . It is called sharp (or a projection) if . Predicates are also called effects. We write for the set of effects of . For a state of the validity is defined as the trace . To make sense of this definition we should mention that the product of bounded operators is trace-class when either or is trace-class [1, Def. 2.54] — so is trace-class because is.
###### Definition 9
Let be two quantum states of the same Hilbert space. The trace distance between them is defined as:
trd(ϱ1,ϱ2)=12tr(∣∣ϱ1−ϱ2∣∣)=12tr(√(ϱ1−ϱ2)†(ϱ1−ϱ2)). (8)
This definition involves the square root of a positive operator . With the examples below in mind it is worth pointing out that in the finite-dimensional case — when is essentially a positive matrix — the square root of can be computed by first diagonalising the matrix , where is a diagonal matrix; then one forms the diagonal matrix by taking the square roots of the elements on the diagonal in ; finally the square root of is .
The trace distance is an extension of the total variation distance : given two discrete distributions on the same set, then the union of their supports is a finite set, say with elements. We can represent via diagonal matrices as density operators . They are states, by construction. Then .
###### Example 10
We describe the quantum analogue of Example 2
, involving the ‘Bell’ state. As a vector in
the Bell state is usually described as . The corresponding density matrix is the following matrix.
β=1√2(1001)1√2(1001)=12(1001000000001001)
Its two marginals (partial traces) are equal matrices, namely:
The product state is obtained as Kronecker product, see e.g. [31].
We can now ask the same question as in Example 2, namely what is the distance between the Bell state
and the product of its marginals. We recall that the Bell state is ‘maximally entangled’ and that the quantum theory allows, informally stated, higher levels of entanglement than in classical probability theory. Hence we expect an outcome that is higher than the value
obtained in Example 2 for the classical maximally entwined state.
The key steps are:
Hence:
trd(β,β1⊗β2)=12tr(∣∣β−β1⊗β2∣∣)=12(\nicefrac12+\nicefrac14+\nicefrac14+\nicefrac12)=34.
In the earlier version of this paper [19] these distance computations are generalised to -ary products, both for classical and for quantum states. Both distances then tend to , as goes to infinity, but the classical distance is one step behind, via formulas versus . Here we only consider .
The following result is a quantum analogue of Proposition 3. Our formulation generalises the standard formulation of e.g. [31, §9.2] and its proof to arbitrary, not necessarily finite-dimensional Hilbert spaces. We’ll see an even more general version involving von Neumann algebras later on.
###### Proposition 11
For states on the same Hilbert space ,
As before, the maximum means the supremum is actually reached by a sharp effect. The proof of this result is in the appendix.
### 2.4 Preliminaries on von Neumann algebras
Our final example of a distance function requires a short introduction to von Neumann algebras. We do not however pretend to explain the basics of the theory of von Neumann algebras here; for this we refer to [26]. We just recall some elementary definitions and facts which are relevant here.
To define von Neumann algebras we must speak about -algebras first.
###### Definition 12
A -algebra is a complex vector space endowed with:
1. an associative multiplication that is linear in both coordinates;
2. an element , called unit, such that for all ;
3. a unary operation , called involution, such that , , , and for all and ;
4. a complete norm, , with and for all .
N.B. In the literature the unit is usually not included as part of the definition of -algebra, and what we have defined above is called a unital -algebra instead.
Two types of elements deserve special mention: an element of a -algebra is called self-adjoint when , and positive when for some .111In [26] a different but in the end equivalent definition of “positive” is used, see Theorem 4.2.6 of [26]. Elementary matters relating to self-adjoint elements are usually easily established: the reader should have no trouble verifying, for example, that every element of a -algebra can be written as for unique self-adjoint (namely, and .) On the other hand, the everyday properties of the positive elements are often remarkably difficult to prove from basic principles, such as the facts that the sum of positive elements is positive, that the set of positive elements of is norm closed (see parts (iii) and (i) of Theorem 4.2.2 of [26]), that every positive element has a unique positive square root, (see Theorem 4.2.6(ii) of [26]), and that every self-adjoint element of may be written uniquely as where with (see Proposition 4.2.3(iii) of [26]).
The elements of a -algebra are ordered by when is positive. We write for subset of effects ; they will be used as quantum predicates. Such an effect is called sharp (or a projection) if .
###### Definition 13
A -algebra is a von Neumann algebra (aka. -algebra) if firstly the unit interval is a directed complete partial order (dcpo), and secondly the positive linear functionals that preserve these (directed) suprema separate the elements of . In the notation introduced below this means that follows if for all states .
There are several equivalent alternative definitions of the notion of ‘von Neumann algebra’, but this one, essentially due to Kadison (see [25]), is most convenient here.
We consider as morphisms between von Neumann algebras: linear maps which are unital (that is, ), positive ( implies ) and normal. The latter normality requirement means that the restriction preserves directed joins (i.e. is Scott continuous). This yields a category of von Neumann algebras. It occurs naturally in opposite form, as .
Each non-zero map in has operator norm equal to 1, i.e. , where . Below we apply the operator norm to a (pointwise) difference of parallel maps in . Using as distance, each homset of is a complete metric space.222Here’s a proof that is complete: We must show that a Cauchy sequence in converges. By Theorem 1.5.6 of [26] -converges to a bounded linear map . It’s clear that will be unital, and positive (since the norm-limit of positive elements of is again positive, see Theorem 4.2.2 of [26]), so it remains to be shown that is normal. Given directed we must show that . For this it suffices to show that for all positive normal linear functionals , which is the case when is normal. But since -converges to , this is indeed so (because the predual of is complete, see the text under Definition 7.4.1 of [26].)
### 2.5 States of von Neumann algebras
A state of a von Neumann algebra is a morphism in . We write for the set of states; it is easy to see that it is a convex set. For an effect we write for the value . When is the von Neumann algebra of bounded operators on a Hilbert space , then ‘effect’ has a consistent meaning, since . Moreover, density operators on are in one–one correspondence with states of , via ; in fact, this correspondence extends to a linear bipositive isometry between trace-class operators on and normal — but not necessarily positive — functionals on (see [1, Thm 2.68]).
For states of von Neumann algebras we use half of the operator norm as distance, since it coincides with the ‘validity’ distance whose formulation is by now familiar. The proof is again delegated to the appendix.
###### Proposition 14
Let be two states of a von Neumann algebra . Their validity distance , as defined on the left below, satisfies:
vld(ϱ1,ϱ2)\coloneqq⋁e∈[0,1]A∣∣ϱ1⊨e−ϱ2⊨e∣∣=maxs∈[0,1]A sharpϱ1⊨s−ϱ2⊨s=12∥ϱ1−ϱ2∥op
Via the last equation it is easy to see that is a complete metric.
###### Corollary 15
Let be a von Neumann algebra.
1. For each predicate the ‘evaluate at ’ map is both affine and non-expansive.
2. The convex map is non-expansive.
3. The ‘states’ functor restricts to a functor .
1. It is standard that the map is affine, so we concentrate on its non-expansiveness: for states we have:
∣∣eve(ϱ1)−eve(ϱ2)∣∣=∣∣ϱ1⊨e−ϱ2⊨e∣∣≤⋁a∈[0,1]A∣∣ϱ1⊨a−ϱ2⊨a∣∣=vld(ϱ1,ϱ2).
2. Suppose we have two formal convex combinations and in . The map is non-expansive since:
vld(α(Ω),α(Ψ))=⋁e∣∣(∑iri⋅ωi)⊨e−(∑jsj⋅ϱj)⊨e∣∣=⋁e∣∣∑iri⋅ωi(e)−∑jsj⋅ϱj(e)∣∣=⋁e∣∣∑iri⋅eve(ωi)−∑jsj⋅eve(ϱj)∣∣=⋁e∣∣Ω⊨eve−Ψ⊨eve∣∣≤⋁p∈Met(Stat(A),[0,1])∣∣Ω⊨p−Ψ⊨p∣∣kvd(Ω,Ψ).
3. We have to prove that for a positive unital map between von Neumann algebras the associated state transformer is affine and non-expansive. The former is standard, so we concentrate on non-expansiveness. Let be states of . Then:
vld(f∗(ϱ1),f∗(ϱ2))=⋁e∈[0,1]A∣∣f∗(ϱ1)(e)−f∗(ϱ2)(e)∣∣=⋁e∈[0,1]A∣∣ϱ1(f(e))−ϱ2(f(e))∣∣≤⋁d∈[0,1]B∣∣ϱ1(d)−ϱ2(d)∣∣=vld(ϱ1,ϱ2).
## 3 Distances between effects (predicates)
There are several closely connected views on what are predicates in a probabilistic setting. Informally, one can consider fuzzy predicates on a space , or only the sharp ones . Instead of restricting oneself to truth values in , one can use -valued predicates , which are often called ‘observables’. Alternatively, one can restrict to the non-negative ones . There are ways to translate between these views, by restriction, or by completion. The relevant underlying mathematical structures are: effect modules, order unit spaces, and ordered cones. Via suitable restrictions, see [22, Lem. 13, Thm. 14] for details, the categories of these structures are equivalent. Here we choose to use effect modules because they capture -valued predicates, which we consider to be most natural. Moreover, there is a standard adjunction between effect modules and the convex sets that we have been using in the previous section. This adjunction will be explored in the next section.
In this section we recall some basic facts from the theory of effect modules (see [16, 9, 23]), and add a few new ones, especially related to -joins and metric completeness, see Proposition 18. With these results in place, we observe that in our main examples — fuzzy predicates on a set and effects in a von Neumann algebras — the induced ‘Archimedean’ metric can also be expressed using validity , but now in dual form wrt. the previous section: for the distance between two predicates we now take a join over all states and use the validities of the two predicates in these states.
We briefly recall what an effect module is, and refer to [16] and its references for more details. This involves three steps.
1. A partial commutative monoid (PCM) is given by a set with an element and a partial binary operation which is commutative and associative, in a suitably partial sense,333That is: is defined iff is defined, and they’re equal in that case; and is defined iff is defined, and they’re equal in that case. and has has unit element.
2. An effect algebra is a PCM in which each element has a unique orthosupplement with , where . Moreover, if is defined, then . Each effect algebra carries a partial order given by: iff for some . It satisfies iff . For more information on effect algebras we refer to [12].
3. An effect module is an effect algebra with a (total) scalar multiplication operation which acts as a bihomomorphism: it preserves in each coordinate separately scalar multiplications and partial sums , when defined, and maps the pair to .
We write for the category of effect modules. A map in preserves , sums , when they exist, and scalar multiplication; such an then also preserves orthosupplements. There are (non-full) subcategories of directed complete and -complete effect modules, with joins of directed (or countable ascending) subsets, with respect to the existing order of effect algebras. The sum and scalar multiplication operations are required to preserve these joins in each argument separately444In fact, it can be shown that maps preserve joins automatically, see Lemma 17 (1i). Preservation by scalar multiplication can also be proved, but is outside the scope of this paper.. Since taking the orthosupplement is an order anti-isomorphism it sends joins to meets and vice-versa. In particular, /directed meets exist in -/directed complete effect modules. Morphisms in and are homomorphisms of effect modules that additionally preserve the relevant joins.
Below it is shown how this effect module structure arises naturally in our main examples. The predicate functors are special cases of constructions for ‘effectuses’, see [16].
###### Lemma 16
1. For the distribution monad on there is a ‘predicate’ functor on its Kleisli category:
This functor is faithful, and it is full (& faithful) if we restrict it to the subcategory with finite sets as objects.
2. There is also a ‘predicate’ functor:
This functor is full and faithful.
Writing on both sides in point (2) looks rather formal, but makes sense since the category of von Neumann algebras is naturally used in opposite form, see also the next section.
1. It is easy to see that the set of fuzzy predicate on a set is an effect module, in which a sum exists if for all , and in that case . Clearly, and for a scalar . The induced order on is the pointwise order, which is (directed) complete.
For a Kleisli map the predicate transformation map from (4) preserves the effect module structure. Moreover, it is Scott-continuous by the following argument. Let be a directed collection of predicates, and let . Write the support of as . Then:
f∗(⋁iqi)(x)f(x)(y1)⋅(⋁iqi)(y1)+⋯+f(x)(yn)⋅(⋁iqi)(yn)=(⋁if(x)(y1)⋅qi(y1))+⋯+(⋁if(x)(yn)⋅qi(yn))=⋁i1⋯⋁inf(x)(y1)⋅qi1(y1)+⋯+f(x)(yn)⋅qin(yn)=⋁if(x)(y1)⋅qi(y1)+⋯+f(x)(yn)⋅qi(yn)since (qi) is directed=⋁if∗(qi)(x)=(⋁if∗(qi))(x).
Assume for , and let , . Write for the singleton predicate that is on and zero everywhere else. Then . Hence , showing that is faithful.
Now let be finite sets and a map in . Define . We claim that is a distribution on , say, and that . This works as follows.
∑y∈Yf(x)(y)=∑ih(1{yi})(x)=(∑\vbox\Large\displaylimitsih(1{yi}))(x)=h(∑\vbox\Large\displaylimitsi1{yi})(x)=h(1
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2021-10-18 10:54:11
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http://gatkforums.broadinstitute.org/gatk/discussion/3468/depthofcoverage
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# DepthofCoverage
Member Posts: 3
Hi,
im trying to calculate the coverage of a exome sequencing project.
My Code is
java -jar $gatk_jar \ -T DepthOfCoverage \ -R$ref_genome \
-I "$sample_name".recal_reads.bam \ -o$result_dir/$sample_name/coverage/"$sample_name" \
-geneList $ref_refseq \ -L test.bed \ -ct 10 -ct 20 -ct 30 but i get an ERROR that i cant solve... ##### ERROR MESSAGE: Unknown file is malformed: Could not parse location from line: chr1 66999814 67000061 NM_032291_exon_0_10_chr1_66999825_f 0 + This error relates to my refseq_file but i cant find the problem with it. This is the first line from my geneList so something is messed up. I created it along http://www.broadinstitute.org/gatk/guide/article?id=1329 and used the script sortByRef.pl to sort it by the fai file from your bundle Can you give me any advise to check on ? Thank you very much for helping! Tagged: ## Answers • Administrator, Dev Posts: 10,995 admin Hi there, Try passing the refseq as -geneList:REFSEQ$ref_refseq to ensure that the file gets parsed using the correct format presets.
Geraldine Van der Auwera, PhD
• Member Posts: 3
edited November 2013
Yeah this solves the error!
But now I have a new one, which comes from the order of the file. I have tried to use our suggested perl script. My Command:
MESSAGE: Input file is not sorted by start position.
##### ERROR We saw a record with a start of chr1:48998527 after a record with a start of chr1:66999825, for input source: geneTrack.refSeq.sorted.txt
I know that message is clear but i dont know what is going wrong.
I called your perl script with:
./sortbyref.pl geneTrack.refSeq ucsc.hg19.fasta.fai > geneTrack.refSeq.sorted.txt
But i guess there is something wrong with that .fai file ?
• Member, Dev Posts: 543 ✭✭✭✭
It's been a while since I've used it, but I think that script has a default value (overridden on the command line) for which column contains the actual position values. I think that default must be wrong for your file
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2017-01-19 23:26:52
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http://www.chegg.com/homework-help/questions-and-answers/1-many-electrons-must-transferred-away-bee-produce-charge-930pc-2suppose-two-bees-charge-9-q1387804
|
1.How many electrons must be transferred away from a bee to produce a charge of 93.0pC ?
2.Suppose two bees, each with a charge of 93.0pC , are separated by a distance of 1.20cm . Treating the bees as point charges, what is the magnitude of the electrostatic force experienced by the bees? (In comparison, the weight of a 0.140-g bee is 1.37x10^-3N.
3.The Earth produces an electric field of magnitude 110N/C . What force does this electric field exert on a bee carrying a charge of 93.0pC ? (For comparison, the weight of a bee is approximately 1.37x10^-3.)
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2015-07-30 05:43:17
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https://kopavguldfrul.firebaseapp.com/20707/44233.html
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# En smart kalkylator för din iPhone och Apple Watch Magasin
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### Texas ti 84 solver, ever struggled with solving an algebraic
After you enter the expression, Algebra Calculator will solve the equation 4x+7=2x+1 for x to get x=-3. Here are more examples of how to solve equations in Algebra Calculator. Feel free to try them now. When you enter an equation into the calculator, the calculator will begin by expanding (simplifying) the problem. Then it will attempt to solve the equation by using one or more of the following: addition, subtraction, division, taking the square root of each side, factoring, and completing the square.
Det är en av de 10. The Equation Solver on your TI-84 Plus calculator is a great tool for Ö 43210 10,2a.
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Solution of equation f(x)=0 with iterative method Graphing Calculator Software Applications (APPS) are pieces of software that can be downloaded onto a TI and determinant and using matrices to solve a system of linear equations.
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2022-05-24 18:53:21
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http://mathhelpforum.com/math-software/139973-find-maximum-value.html
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# Math Help - Find maximum value
1. ## Find maximum value
Code:
function max()
A = [ 5 4 7 2 6 3 1 ];
max=0;
for j = 1:7,
if( A(j) > max ) max = A(j);
end %first end
% line A
end %second end
max
When i insert the max word at line A i get it printed all the time the loop is iterated, even though it's outside the for loop.
Why is that?
2. The above is Matlab code
3. Originally Posted by AUCC
Code:
function max()
A = [ 5 4 7 2 6 3 1 ];
max=0;
for j = 1:7,
if( A(j) > max ) max = A(j);
end %first end
% line A
end %second end
max
When i insert the max word at line A i get it printed all the time the loop is iterated, even though it's outside the for loop.
Why is that?
It is not outside the for-loop, that ends at the second end, the first end terminates the if-construct.
CB
4. Should there be an end statement to terminate the function ?
5. Originally Posted by AUCC
Should there be an end statement to terminate the function ?
It won't do any harm, but is redundant.
A single function in an .m file does not require and end (and as far as I recall a function definition in a script file is terminated by the eof marker or the next function statement - the same goes for functions and sub-functions in the same file)
Good practice would say put the end statement in to terminate the function, Matlab will catch up eventually and insist on it.
CB
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2014-03-16 19:12:30
|
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http://mathhelpforum.com/calculus/186482-solids-revolution-question.html
|
# Math Help - Solids of revolution question
1. ## Solids of revolution question
Can you help me to solve these 2 questions? They are very confusing and I cannot find the right way.
I have the answer from the book but I need the way to solve it.
Thanks a lot!!!!
1) A cylindrical hole of diameter 2 cm is bored through a sphere of radius 4 cm. The hole has been bored so that its main axis passes through the centre of the sphere. Calculate the volume of the remainder of the sphere.
2) Consider that part of the parabola y = (x-1)*(x+2) which lies below the x-axis. Calculate the volume of the solid formed when this part is rotated about the line y =4.
2. ## Re: Solids of revolution question
Originally Posted by tosucceed999
Can you help me to solve these 2 questions? They are very confusing and I cannot find the right way.
I have the answer from the book but I need the way to solve it.
Thanks a lot!!!!
1) A cylindrical hole of diameter 2 cm is bored through a sphere of radius 4 cm. The hole has been bored so that its main axis passes through the centre of the sphere. Calculate the volume of the remainder of the sphere.
The simplest way to do this is to first find the volume of the entire sphere, $(4/3)\pi(4)^3$, then subtract the volume removed. That will be a cylinder plus two "caps" at either end. The radius of the cylinder is, of course, 2 cm. To find the height, h, draw a circle with a thin rectangle inscribed in it. The line from the center of the circle to a corner of the rectangle has length the radius of the sphere, 4 cm. A line from the center of the sphere perpendicular to the side of the rectangle has length the radius of the cylinder, 2 cm. Use the Pythagorean theorem to find half the length of the cylinder. Two find the volume of a cap, think of it as a stack of disks. Setting up a coordinate system with the origin at the center of the sphere, the equation of the sphere's projection in the xy-plane is $x^2+ y^2= 16$ or [itex]x= \pm\sqrt{16- y^2}[/tex] and that is the radius of one of the disks. Its area is $\pi x^2= \pi(16- y^2)$. If its thickness is dx, the volume of one disk is $\pi(16- y^2)dy$ and so the volume of them all is [tex]\pi \int 16- y^2 dy[tex]. The integral is take from the top of the cylinder to the top of the sphere.
2) Consider that part of the parabola y = (x-1)*(x+2) which lies below the x-axis. Calculate the volume of the solid formed when this part is rotated about the line y =4.
I would do this in two parts.
First look at the parabola only. Imagine a line from $y= x^2+ x- 2$ to y= 4. Rotating around the axis y= 4 gives a disk of radius $4- (x^2+ x- 2)$ and so of area $\pi (6- x^2- x)^2$. Taking each disk to have "thickness" dx, The volume of each disk is $\pi (6- x^2- x)^2 dx$ and the volume of all put together is $\pi\int (6-x^2-x)^2 dx$. The integral is, of course, from x= -2 to x= 1 where the parabola crosses the x-axis.
Second, subtract the volume of the cylinder made up of the portion that is above the x-axis. That will be a cylinder with radius 4 and height 1-(-2)= 3.
Thank you.
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2015-05-24 21:44:45
|
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http://youlifereporter.it/dtsx/norm-in-latex-overleaf.html
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2020-07-10 20:04:52
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http://www.inf.ed.ac.uk/publications/report/0367.html
|
Informatics Report Series
Report
EDI-INF-RR-0367
Related Pages Report (by Number) Index Report (by Date) Index Author Index Institute Index Home
Title:Verifiable Agent Dialogues
Authors: Christopher Walton
Date: 2005
Publication Title:Journal of Applied Logic, Special Issue on Logic-Based Agent Verification
Publisher:Elsevier
Publication Type:Journal Article Publication Status:Pre-print
ISBN/ISSN:15708683
Abstract:
In this paper, we introduce the Multi-Agent Protocol (MAP) language which expresses dialogues in Multi-Agent Systems. MAP defines precisely the pattern of message exchange that occurs between the agents, though it is independent of the actual rational processes and message-content. This approach makes MAP applicable to a wide range of different agent architectures, e.g.\ reactive, proactive, and deductive agent systems. In the first half of the paper we specify the syntax of MAP, together with an operational semantics that defines an abstract implementation of the language. Our specification is derived from process calculus and thus forms a sound basis for the verification of our protocols. We also sketch a connection between MAP and temporal logic. In the latter half of the paper we define a translation from MAP into PROMELA, which is the specification language of the SPIN model checker. This translation allows us to prove properties of our protocols, such as termination, liveness, and correctness. The verification process is defined by $P \models \varphi$, where P is our translated agent protocol, and $\varphi$ is a linear temporal logic formula. By performing model checking on our protocols we can obtain a high degree of confidence in the resulting agent dialogues.
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2018-06-24 12:49:40
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|
https://www.talkstats.com/threads/how-to-combine-rows-of-paired-data-into-one-row.43820/
|
# how to combine rows of paired data into one row
#### statsguy20
##### New Member
Hi,
I have data that is structured like the following:
location team score
away ny 100
home la 80
away chi 95
home det 85
But I would like to transform the pairs of data to be structured like this:
away_team home_team away_score home_score
ny la 100 80
chi det 95 85
Does anyone know how to do this? Thanks!
#### chetan.apa
##### Member
Probably, you could do this easily in a data step..something like this
I have assumed that the data rows to be paired are in consecutive lines. Note that I have not tested this code.
Code:
data abc(keep=away_team home_team away_score home_score);
set abc;
retain away_team home_team away_score home_score;
if location="home" then do;
home_team=team;
home_score=score;
end;
else do;
away_team=team;
away_score=score;
end;
if mod(_n_,2)=0 then output;
run;
#### statsguy20
##### New Member
amazing, thank you! what operation is the retain statement performing in the code you wrote?
|
2022-08-10 07:35:18
|
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|
https://www.physicsforums.com/threads/a-simple-momentum-question.199146/
|
# A simple momentum question.
1. Nov 18, 2007
### ken62310
PHYSIC momentum QUESTION~
1. The problem statement, all variables and given/known data
A billiard ball moving with velocity 4i+3j m/s strikes a stationary ball of the same mass. After the collision the struck ball moves with a speed of 4m/s at an angle of 30 degree with respect to the line of original motion of the striking ball. Find velocity of the striking ball after collision.
2. Relevant equations
Pi=Pf
P=mv
3. The attempt at a solution
Pxi=mv1=4
Pxf=mv2cos(theta)+m(4)cos30
Pyi=mv1=3
Pyf=mv2sin(theta)+m(4)sin30
am i correct with that?? i am really confuse
Last edited: Nov 18, 2007
2. Nov 18, 2007
### Mindscrape
You have the right idea, however the problem says that the stricken ball goes at a 30 degree angle with respect to the stricker's original line of motion, so you will want to add the two angles.
3. Nov 18, 2007
### ken62310
i am not quite understand what you mean.
how do i add the two angles? could you be more detailed?
4. Nov 18, 2007
### Mindscrape
Well, let me preface this by saying that what I am telling you is only one approach. I can think at least one other reasonable way to do this problem, however, what I am telling you is pretty straightforward, or at least hopefully it will seem that way.
So if I am walking along some path at 50 degrees and then the path forks off so that one trail goes at 30 degrees with respect to the original direction I am walking, what is the total angle that I have changed by? It would be 80 degrees, right? The same is true for your ball. The first ball has some trajectory with an angle (that I'll let you find), and then it deviates from the angle by an additional 30 degrees.
Last edited: Nov 18, 2007
5. Nov 19, 2007
### ken62310
so that means i need to find the angle that formed by the 4i+3j right?
now i have another question.
I found that the angle of the 4i+3j would be 37 degree.
and for x i will add that to the 30 degree right. then what about for the y?
do i need to add the different angle? should it be respect to the y-axis?
6. Nov 19, 2007
### ken62310
for x: mv1=mv2 cos67+ mv3 cos theta.
for y: mv1=mv2 sin30+ mv3(37-theta).
...
please tell me if i am correct..
Last edited: Nov 19, 2007
7. Nov 19, 2007
### katchum
Why mv1? If you projected it to the x-axis, it should have a cosinus in it. Am I missing something?
8. Nov 19, 2007
### ken62310
oh...so for x: mv1 cos 37 and for y: mv1 cos 53 ? am i right?
9. Nov 19, 2007
### rl.bhat
Initial velocity of the striking ball is 5 m/s. If the collision is perfectly elastic KE is conserved. Since masses of the balls are same V^2 = V1^2 + V2^2. Therefore the velocity of the strking ball after collision will be 3 m/s. And the two balls will move perpendicular to each other.
10. Nov 19, 2007
### ken62310
so all the angle stuff are useless in this case??
11. Nov 19, 2007
### Mindscrape
It depends on what you have learned so far in physics. If you don't know about kinetic energy and perfectly elastic collisions then conservation of momentum is the only choice you have. Anyway, for the momentum case you should get something to the effect of
$$v_{xi1} = 4 = v_{xf1} + 4cos(arctan(3/4) + 30)$$
and
$$v_{yi1} = 3 = v_{yf1} + 4sin(arctan(3/4) + 30)$$
so
$$||\vec{v}|| = (v_{yf1}^2 + v_{xf1}^2)^{(1/2)}$$
12. Nov 19, 2007
### katchum
I have a question of my own.
How do you find the angle of the first ball after the collision?
Here is my attempt:
x: m.vi1.cos37 = 4 = m.vf1.cosx + m.vf2.cos7 and vf2 = 4
y: m.vi1.sin37 = 3 = m.vf1.sinx + m.vf2.sin7
Two parameters vf1 and x.
vf1 is the final velocity. (absolute)
x is the angle between the first ball after collision and the x axis.
I find it odd that I didn't use the equation of kinetic energy, there has to be something wrong...
Last edited: Nov 19, 2007
13. Nov 19, 2007
### rl.bhat
Momentum
If you take the direction of the initial velocity as axis of reference, the problem becomes easy. If you shift the co-ordinate of the initial velocity to the point of impact with out any rotation, the components of final velocities of two balls involve several angles. Angle of V1 with new X-axis will be 37 +30 degree. But the angle of V2 with new X-axis is unknown.. If you use coservation of energy we can avoid all these complication.
I didn't understood mindscrape's equations. vxf1 is not in the direction of x-component of the initial velocity and y-component should be the difference of two components of v1 and v2 along y axis.
Last edited: Nov 19, 2007
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2018-02-20 23:49:43
|
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http://physics.stackexchange.com/questions/230336/height-issues-of-the-time-traveler
|
# Height Issues of the Time Traveler [duplicate]
A person from the year 2250 goes back in time. They go back 60 Million years, because they want to observe dinosaurs. Imagine their surprise when they see T-Rex's running around like little chickens!!
That's because they neglected the Hubble Expansion that had occurred in the last 60 Million years. So they are much larger than anything that existed so far in the past.
Is this correct?
-
## marked as duplicate by John Rennie, David Z♦Jan 19 at 14:34
So it sounds like the time traveler would be larger, just not very much larger. Like only infinitesimally larger. – Jiminion Jan 18 at 19:19
exactly. However, this would be a non-trivial calculation to determine the exact expansion, and my guess is that it would be less than an atom's diameter. – Sam Blitz Jan 18 at 19:26
– rob Jan 18 at 23:51
I find some of the answers unclear. It is unclear whether matter (or galaxies, for that matter) do not expand, or they expand so little as to be impossible to discern. Subtle, but very different answers. – Jiminion Jan 19 at 15:19
I don't find the answers in the referenced question responsive. – Jiminion Jan 19 at 22:02
No, because Hubble expansion has negligible effects on very small systems (such as human beings).
Here is an answer which explains the maths behind it : Can the Hubble constant be measured locally??
-
Shortly, no, this is not correct.
Here's why. Hubble's law gives us that for a distance of one megaparsec, that space expands by approximately 70 km/s (the data varies, but it's somewhere between 60-80 km/s - it doesn't matter, and you'll see why). Now, how tall is your average human? Let's be generous and say your time traveler is 2m tall. Now, how many MPc is that? Oh, about $6.4 \times 10^{-23} \text{ MPc}$. So, even naively neglecting the fact that the expansion of the universe does not affect gravitationally, electromagnetically, chemically, or otherwise bound bodies (see mlg's answer above, i.e. the earth does not expand with the universe), if we assumed it did, then we would find that your 2m tall person has grown about a centimeter. Not dwarfing anything!
-
I think I should go from another direction.
Yes, obviously the Hubble constant refers to intergalactic motion, and cannot be properly applied to intragalactic effects. That does not necessarily mean that such effects do not exist (it just means that they are too minor to measure, and/or usually overshadowed by other effects; that said, I believe that GPS is precise enough that Hubble drift would've affected it if it worked on these scales, but I hadn't done the calculation, and for all I know perhaps it does and it's just brushed off as another easy correction).
But that aside, consider what the Hubble constant means: it is (roughly, due to complicated inflationary models, but it works as a first approximation) the inverse of the time since Big Bang. Everyone knows how much it had been since Big Bang: 13 billion years (give or take a bit).
That means that the fraction of the Hubble expansion that had occurred over the last X million years is about X/13000. (Well, more like 13600 really, but whatever.)
For x=70 (an appropriate value for seeing T-Rex - 60 would put the traveller in the early Paleogene*), this would be 70/13000, or about 1/200. Yes, Sam Blitz's estimation is correct: about half a percent, or, for typical human height, about a centimeter.
If they wanted to visit the Triassic period instead, the time gap would roughly triple, so they would instead be about three centimeters taller. Still not significant (and probably not noticeable).
*) Which might be the reason he's seeing chickens instead of dinosaurs: because all the big dinosaurs have gone extinct, and the few that still remain (mainly ancestors of modern birds) are tiny and look like chickens! Since they also happen to be fairly close relatives of T-Rex (and even closer, IIRC, of Velociraptor), the result is a lot similar to a chicken-sized T-Rex (except with feathers, obviously - though perhaps the actual T-Rex also had them).
-
GPS has been in operation less than 40 years, and has an accuracy of a few centimeters if you use really fancy techniques and equipment that didn't exist for most of those 40 years; more like 1m at best without them. Given the Hubble figure of 70km/s/MPc, that's a change in the Earth's diameter of 3.6 cm in 40 years. A close thing, but I'm going to call a "not quite" on your GPS verification theory. – hobbs Jan 19 at 14:07
A maybe more practical concern — plate tectonics adds a confounding motion of multiple cm/year to any given point on the Earth's surface, making it tricky to measure an effect on the order of 1 mm/year. – hobbs Jan 19 at 14:14
GPS works by triangulating distance of multiple satellites, I don't know how the Earth's diameter plays a role. Maybe you are referring to the tables involved. I suppose expanding space changes the freq. slightly of the radio waves used? (Again, not sure what expanding space really means.) – Jiminion Jan 19 at 19:37
just try an online calculator like Wolfam : 13.65 billion years after the big bang
• redshift = 0.00474
• time ago (lookback time) = 65.6 million years distance (comoving) = 65.7 million ly (light years) = 20.2 Mpc (megaparsecs) = 6.22×10^20 km (kilometers) = 3.86×10^20 miles
• fraction of total observable radius = 0.00141
• scale factor = 0.995 × current value
• epoch = dark energy dominated, post-recombination
• radiation temperature = 2.74 K (kelvins)
• Hubble parameter (expansion rate) = 70.6 km/s/Mpc (kilometers per second per megaparsec) (based on 5-year WMAP data and Lambda-CDM model;
• current universe age: 13.7 billion years
size of a T-Rex squeleton is now about 12 m. If it expands , it gains $12*(1-0.995) = 0.06 m = 6 cm$ , still too big to look like a chicken.
Moreover, it seems that material objects , like a squeleton which is bound by EM fields, doesn't expand.
-
When say that EM bound stuff does expand, is this because an atom is mostly empty space? (A nucleus is like a baseball in a football stadium; that sort of thing). But due to the same forces, wouldn't the atom still have to expand? [Space expansion is very unclear at the micro level....] – Jiminion Jan 19 at 15:24
@Jiminion When space expands, it happens a small perturbation, quickly corrected by the EM dynamics – igael Jan 19 at 15:33
What do you mean by 'correction' ? – Jiminion Jan 19 at 19:38
@Jiminion if space expands, the laws of EM being the same, the equilibrium distances remain "constant" in the atoms , lattices of matter, thermodynamic , etc . There is an issue here about the new equilibrium but it is another question which I don't have the answer. – igael Jan 19 at 20:31
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2016-07-28 10:40:01
|
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https://support.bioconductor.org/p/113687/
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Question: GCRMA not working inside tryCatch block in R
0
5 months ago by
Houston, TX
perdedorium0 wrote:
I've written a script with a for() loop to process the raw microarray CEL files of five different GEO data sets in R, but it screws up when it gets to the GCRMA processing part. I get a "Was unable to process microarray data" message in my log file, but no error message, despite having an error block. Also, it works just fine when I don't use a tryCatch block, which makes me think it's something with the way I have my error handling set up. But I have the exact same setup for reading the CEL files into an Affybatch, and that works just fine. What am I doing wrong here?
Here's the GCRMA processing step (which is not working):
eset <- tryCatch(
{
gcrma(affy.data)
}, warning = function(w) {
# For warnings, write them to the output file.
cat(paste("Warning in GEO data set", i, "when performing GCRMA processing:", conditionMessage(w)), file=logfile, append=TRUE, sep = "\n")
}, error = function(e) {
# For errors, write them to the output file and then skip to the next data set.
cat(paste("Error in GEO data set", i, "when performing GCRMA processing", conditionMessage(e)), file=logfile, append=TRUE, sep = "\n")
return(NULL)
}
)
if(is.null(eset)) {
cat(paste("Was unable to process microarray data for GEO data set", i, ". Skipping to next data set."), file=logfile, append=TRUE, sep = "\n")
next
} else {
# If everything went all right, make a note of that in the output file.
cat(paste("Successfully processed GEO data set", i), file=logfile, append=TRUE, sep = "\n")
}
... and here's the step for reading into an Affybatch, which is:
# Read the CEL files into an Affybatch object.
affy.data <- tryCatch(
{
}, warning = function(w) {
# For warnings, write them to the output file.
cat(paste("Warning in GEO data set", i, "when reading CEL files:", conditionMessage(w)), file=logfile, append=TRUE, sep = "\n")
}, error = function(e) {
# For errors, write them to the output file and then skip to the next data set.
cat(paste("Error in GEO data set", i, "when reading CEL files:", conditionMessage(e)), file=logfile, append=TRUE, sep = "\n")
return(NULL)
}
)
if(is.null(affy.data)) {
cat(paste("Was unable to read in CEL files for GEO data set", i, ". Skipping to next data set."), file=logfile, append=TRUE, sep = "\n")
next
} else {
cat(paste("Successfully read CEL files for GEO data set", i), file=logfile, append=TRUE, sep = "\n")
}
(Btw, i is the loop variable that holds the name of the data set.)
modified 5 months ago by James W. MacDonald49k • written 5 months ago by perdedorium0
Answer: GCRMA not working inside tryCatch block in R
2
5 months ago by
Martin Morgan ♦♦ 23k
United States
Martin Morgan ♦♦ 23k wrote:
I didn't take a close look but using tryCatch() to catch warnings is not usually the right thing to do -- when a warning is encountered control flow goes to the handler and then to the rest of the 'top level' code, so code after the warning is never executed. Compare
> tryCatch({ warning("oops"); 1 }, warning = function(...) {}); 2
NULL
[1] 2
> withCallingHandlers({ warning("oops"); 1 }, warning = function(...) invokeRestart("muffleWarning")); 2
[1] 1
[1] 2
That did the trick! And thanks, I did not know that about tryCatch().
Answer: GCRMA not working inside tryCatch block in R
2
5 months ago by
United States
James W. MacDonald49k wrote:
You might try Martin Maechler's error catching code, from not as long ago as I remember it...
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2019-03-26 04:38:59
|
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|
https://socratic.org/questions/how-do-you-write-an-equation-in-standard-form-of-the-parabola-that-has-vertex-8-
|
# How do you write an equation in standard form of the parabola that has vertex (-8,-3) and passes through the point (4,717)?
Feb 20, 2018
$y = 5 {x}^{2} + 80 x + 317$
#### Explanation:
The formula for a parabola can be written as $y = a {\left(x - h\right)}^{2} + k$, where $\left(h , k\right)$ is the vertex of the parabola.
Here, $h = - 8$ and $k = - 3$.
We can input the above into the equation:
$y = a {\left(x - \left(- 8\right)\right)}^{2} + \left(- 3\right)$, which simplifies to:
$y = a {\left(x + 8\right)}^{2} - 3$
We know that one of the coordinates is $\left(4 , 717\right)$. So when $x = 4 , y = 717$. We need to find $a$, and so we can input:
$717 = a {\left(4 + 8\right)}^{2} - 3$
$717 = a \left({12}^{2}\right) - 3$
$717 = 144 a - 3$
$144 a = 720$
$a = \frac{720}{144}$
$a = 5$
Since $a = 5$, we can input this into $y = a {\left(x + 8\right)}^{2} - 3$.
$y = 5 {\left(x + 8\right)}^{2} - 3$
$y = 5 \left({x}^{2} + 16 x + 64\right) - 3$
$y = 5 {x}^{2} + 80 x + 320 - 3$
$y = 5 {x}^{2} + 80 x + 317$, the formula for the parabola.
We can graph it:
graph{5x^2+80x+317 [-29.24, 35.7, -11.13, 21.33]}
Notice the vertex is at $\left(- 8 , - 3\right)$.
So the above is proved.
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2020-01-27 00:17:55
|
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https://www.statistics-lab.com/%E9%87%91%E8%9E%8D%E4%BB%A3%E5%86%99%E9%87%91%E8%9E%8D%E8%AE%A1%E9%87%8F%E7%BB%8F%E6%B5%8E%E5%AD%A6%E4%BB%A3%E5%86%99financial-econometrics%E4%BB%A3%E8%80%83random-walk-hypothesis/
|
### 金融代写|金融计量经济学代写Financial Econometrics代考|Random Walk Hypothesis
statistics-lab™ 为您的留学生涯保驾护航 在代写金融计量经济学Financial Econometrics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写金融计量经济学Financial Econometrics代写方面经验极为丰富，各种代写金融计量经济学Financial Econometrics相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 金融代写|金融计量经济学Financial Econometrics代考|Background
The basic idea is that movement of the stock prices are unpredictable and random. Jules Augustin Frédéric Regnault, a French stock broker’s subordinate primarily explained the rationale behind the stock prices stochastic movements using a random walk model. Jules in his book titled ” Calcul des chances et philosophie de la bourse” states that “l’écart des cours est en raison directe de la racine carrée des temps”. 1 Which means “the price difference is a direct result of the square root of the times” and it clearly indicates that the stock prices movement follows a stochastic process. Later it becomes the foundation of Louis Bachelier PhD thesis titled “Th’corie de la Sp’eculation”. Louis often credited as the first person to introduce advanced mathematics into the field of finance. He established a mathematical model of the stochastic process (known as Brownian motion) for valuing the stock options. For a longer period of time Louis’ contribution was overlooked or ignored as his study applied mathematics into the field of finance. Possible reason could be during the nineteenth century the field of interdisciplinary research was not developed or quite accustomed. Then in 1964, Paul H. Cootner, a professor of MIT Sloan School of Management in his book titled “The Random Character of Stock Market Prices” educes the ideas on the stochastic process of stock prices movements. Later on the same ideas are well established by the several known scholars namely Eugene Fama, Burton Malkiel, and others.
## 金融代写|金融计量经济学Financial Econometrics代考|What Is Random Walk Hypothesis and Its Implications
Random walk hypothesis assumes that price movements of individual securities in the stock markets follow a random walk and successive price movements are independent to each other. Therefore, the random walk hypothesis posits that it is impossible to forecast the stock prices movements. Random walk hypothesis also suggests that it is impossible to beat the market by the market participants in the long run. Outperformance of the market by an investor is only possible by taking an extra amount of risk. The Random Walk hypothesis is heavily criticized on several grounds such as market participants differ in terms of the amount of time they spend in the financial market. Then several numbers of the known and unknown factors are responsible for driving the stock prices (Maiti, 2020). It is often not likely to detect the associated trends or patterns that might exist in the stock prices movements due to the presence of several such distinct factors. To test the random walk hypothesis in practice, Wall Street Journal (WSJ) initiated the “Dart Throwing Investment Contest” in the year 1988. Two groups were formed: one group belongs to the professional investors whereas other belongs to the dummy. Professional investors group consists of the professionals working with NYSE whereas WSJ staff groups as the dummy. In other words professional investors represent “skill” whereas dummy represents “luck”. Professional investors selected stocks based on their skills whereas dummy made their selection of stocks based on the outcome of the dart throwing (luck). After 100 contests the outcomes come as following: professional investors won 61 times or skill wins 61 times versus luck. However, professional investors (skill) are able to beat the market (DJIA) 51 times out of 100 . Dart Throwing Investment Contest does not provide any ultimate consensus on “luck versus skills”. The current consensus is that the random walk hypothesis is linked to the efficient market hypothesis (EMH). Mathematically a simple random walk model with a drift is represented by following Eq. (2.1):
$$\text { Price }{t}=\text { Price }{t-1}+\alpha_{t}$$
$/ / \operatorname{Mean}(\mu)$ is zero and standard deviation $(\sigma)$ is constant where
Price $_{t}$ represents current stock price at time ( $\left.t\right)$
Price $_{t-1}-$ represents stock price at time $(t-1)$
$\alpha_{t}$ represents the drift term
## 金融代写|金融计量经济学Financial Econometrics代考|Efficient Market Hypothesis
Efficient market hypothesis assumes that the security prices reflect all available information and follow a random walk. However, the strength of these assumptions depends on the form of EMH as explained by Fama (1970). Thus, the direct implication of EMH is that it is impossible to beat the market steadily merely by ensuing a specific risk adjusted strategy. Fama (1970) labels efficient market hypothesis into three types based on the level of the relevant information as weak, semi-strong and strong forms of EMH, respectively. Weak forms of EMH assume that all historical stock prices information is impounded in the current price of the stock. Then a semi-strong form of EMH assumes that all publicly available information in addition to those historical stock prices information are well impounded in the current price of the stock. Finally, strong forms of EMH assume that all available information both private (insider information) and public is fully impounded in the current price of the stock. The joint hypothesis problem makes it difficult to test the EMH. The argument put forward as the EMH could only be tested with the help of a market equilibrium model (Asset pricing model). The EMH basically tests whether the properties of expected returns suggested by the assumed market equilibrium model (asset pricing model) are noticed in actual returns. If the tests reject, then it is difficult to interpret whether the rejection is made due to the inefficiency of the market or a bad market equilibrium model (asset pricing model). Fama (1970) stressed that the EMH is always tested jointly along with the market equilibrium model and vice versa. Thereafter significant number of studies are done on EMH and asset pricing. All of these studies unanimously emphasized that the $\mathrm{EMH}$ is very simple in principle but testing it proved to be difficult. Both the random walk hypothesis and efficient market hypothesis have significant importance in applied financial econometrics study as both of them could provide some relevant information on relative market efficiency. Still distinct views exist on whether markets are really efficient? There are several studies that evidence support for EMH such as on an average mutual funds do not able to outperformed the market. Then “Dart Throwing Investment Contest” results show that the performance of skills versus luck are quite similar in beating the market. On the other hand, there are group of studies that evidence against EMH is as follows: presence of stock market anomalies, presence of excessive volatility in the stock market, behavioural finance theories, and others.
## 金融代写|金融计量经济学Financial Econometrics代考|What Is Random Walk Hypothesis and Its Implications
价格 吨= 价格 吨−1+一个吨
//意思是(μ)是零和标准偏差(σ)是常数，其中
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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2023-02-02 15:16:58
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https://docs.rs/malachite-base/latest/malachite_base/num/arithmetic/traits/trait.DivRound.html
|
# Trait malachite_base::num::arithmetic::traits::DivRound
pub trait DivRound<RHS = Self> {
type Output;
fn div_round(self, other: RHS, rm: RoundingMode) -> Self::Output;
}
Expand description
Divides a number by another number and rounds according to a specified rounding mode.
## Implementations on Foreign Types
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Up}) = f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \N$.
##### Panics
Panics if other is zero, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Up}) = f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \N$.
##### Panics
Panics if other is zero, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Up}) = f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \N$.
##### Panics
Panics if other is zero, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Up}) = f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \N$.
##### Panics
Panics if other is zero, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Up}) = f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \N$.
##### Panics
Panics if other is zero, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Up}) = f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \N$.
##### Panics
Panics if other is zero, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = \operatorname{sgn}(q) \lfloor |q| \rfloor.$$
$$f(x, y, \mathrm{Up}) = \operatorname{sgn}(q) \lceil |q| \rceil.$$
$$f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \Z$.
##### Panics
Panics if other is zero, if self is Self::MIN and other is -1, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = \operatorname{sgn}(q) \lfloor |q| \rfloor.$$
$$f(x, y, \mathrm{Up}) = \operatorname{sgn}(q) \lceil |q| \rceil.$$
$$f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \Z$.
##### Panics
Panics if other is zero, if self is Self::MIN and other is -1, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = \operatorname{sgn}(q) \lfloor |q| \rfloor.$$
$$f(x, y, \mathrm{Up}) = \operatorname{sgn}(q) \lceil |q| \rceil.$$
$$f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \Z$.
##### Panics
Panics if other is zero, if self is Self::MIN and other is -1, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = \operatorname{sgn}(q) \lfloor |q| \rfloor.$$
$$f(x, y, \mathrm{Up}) = \operatorname{sgn}(q) \lceil |q| \rceil.$$
$$f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \Z$.
##### Panics
Panics if other is zero, if self is Self::MIN and other is -1, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = \operatorname{sgn}(q) \lfloor |q| \rfloor.$$
$$f(x, y, \mathrm{Up}) = \operatorname{sgn}(q) \lceil |q| \rceil.$$
$$f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \Z$.
##### Panics
Panics if other is zero, if self is Self::MIN and other is -1, or if rm is Exact but self is not divisible by other.
##### Examples
See here.
Divides a value by another value and rounds according to a specified rounding mode.
Let $q = \frac{x}{y}$:
$$f(x, y, \mathrm{Down}) = \operatorname{sgn}(q) \lfloor |q| \rfloor.$$
$$f(x, y, \mathrm{Up}) = \operatorname{sgn}(q) \lceil |q| \rceil.$$
$$f(x, y, \mathrm{Floor}) = \lfloor q \rfloor.$$
$$f(x, y, \mathrm{Ceiling}) = \lceil q \rceil.$$
$$f(x, y, \mathrm{Nearest}) = \begin{cases} \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor < \frac{1}{2}, \\ \lceil q \rceil & q - \lfloor q \rfloor > \frac{1}{2}, \\ \lfloor q \rfloor & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is even}, \\ \lceil q \rceil & \text{if} \quad q - \lfloor q \rfloor = \frac{1}{2} \ \text{and} \ \lfloor q \rfloor \ \text{is odd.} \end{cases}$$
$f(x, y, \mathrm{Exact}) = q$, but panics if $q \notin \Z$.
##### Worst-case complexity
Panics if other is zero, if self is Self::MIN and other is -1, or if rm is Exact but self is not divisible by other.
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2022-11-30 17:27:00
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|
https://courses.csail.mit.edu/6.856/21/Notes/n15-sampling.html
|
\documentclass[12pt]{article} \usepackage{wide,me} \parindent0pt
## Polling
Outline
• Set has size $u$, contains $s$ “special” elements
• goal: count number of special elements
• sample with probability $p=c(\log n)/\epsilon^2 s$
• with high probability, $(1\pm\epsilon)sp$ special elements
• if observe $k$ elements, deduce $s \in (1\pm\epsilon)k/p$.
• Problem: what is $p$?
Related idea: Monte Carlo simulation
• Probability space, event $A$
• easy to test for $A$
• goal: estimate $p=\Pr[A]$.
• Perform $k$ trials (sampling with replacement).
• expected outcome $pk$.
• estimator $\frac{1}{k}\sum I_i$
• prob outside $\epsilon < \exp(-\epsilon^2 kp/3)$ ($\epsilon < 1$)
• for prob. $\delta$, need $k=O\left(\frac{\log 1/\delta}{\epsilon^2 p}\right)$
• Define $(\epsilon,\delta)$-approximation scheme
• what if $p$ unknown? For now, assume confirming proposed $p$.
• What if $p$ is small?
More general estimation
• Have random variable $X$
• Want to estimate $E[X]$
• turns out can do this with few samples if $\mu < \sigma$.
• Proof in HW.
• What if want to estimate $\sigma$?
• Just need to estimate mean of random variable $X^2$.
• There are schemes that are “competitive” versus unknown $\mu,\sigma$
Another generalization: conditional sampling
• choose from some distribution, conditioned on some event
• if event common, just choose from distribution till event happens
• rare events hard to condition on.
• Choosing a random graph coloring is easy
• Choosing a random legal graph coloring is hard (and useful in physics)
For now we'll study “sampling for (better) algorithms.” Later, “algorithms for (better) sampling”
Handling unknown $p$
• Sample $n$ (unknown) times till get $\mu_{\epsilon,\delta}=O(\log \delta^{-1}/\epsilon^2)$ hits
• w.h.p, $p \in (1\pm\epsilon)\mu_{\epsilon,\delta}/n$
• let $k = \mu_{\epsilon,\delta}/p$
• so when take $k$ samples expect $\mu_{\epsilon,\delta}$ hits
• consider first $k/(1+\epsilon)$ samples
• expect $\mu_{\epsilon,\delta}/(1+\epsilon)$ hits
• Chernoff says w.h.p $< (1+\epsilon)\mu_{\epsilon,\delta}/(1+\epsilon)=\mu_{\epsilon,\delta}$ hits
• so won't have enough hits to stop before this point
• similar argument that won't stop too late
• instead will stop at some point in proper interval
• and this get good estimate
Discussion
• For error probability $\delta$, need $kp=\mu_{\epsilon,\delta}=(4/\epsilon^2)\ln 2/\delta$
• So, can get tiny probability of error cheaply, but sensitive to $\epsilon$.
• Note for many apps, want “with high probability:” $\delta=1/n^2$, so $kp \approx \log n$.
• Note also, to get good sample, need $k >> 1/p$.
• i.e., sample needed is larger as event we want to detect is rarer.
• Sampling without replacement gives lower probability of large deviation
### Transitive closure
Problem outline
• databases want size
• matrix multiply time
• compute reachibility set of each vertex, add
Sampling algorithm
• generate vertex samples until $\mu_{\epsilon\delta}$ reachable from $v$
• deduce size of $v$'s reachability set.
• reachability test: $O(m)$.
• number of sample: $n/$size.
• $O(mn)$ per vertex---ouch!
Can test for all vertices simultaneously
• increase mean to $O(1/\epsilon^2\log (n/\delta))$,
• so $\delta/n$ failure probability per vertex
• so $\delta$ overall (union bound)
• $O(mn)$ for all vertices (still ouch).
Avoid wasting work
• stop propagating to a vertex once it has seen enough
• if a vertex has seen $k$ samples, so have all its predecessors
• so no need to forward observations
• so send at most $\log n/\delta$ samples over an edge
• also, after $O(n\mu_{\epsilon,\delta/n})$ samples, every vertex has $\log n/\delta$ hits. No more needed.
|
2021-05-17 10:57:27
|
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|
https://codereview.stackexchange.com/questions/197681/pounds-to-tons-converter
|
# Pounds-to-tons converter
Finished my first lone project and I'm wondering if the execution is passable. This is a basic lbs to tons converter. The customer drives on the scale loaded, then dumps the load and returns unloaded to the scale again. We need to convert the leftover lbs to tons. I am wondering if this is a concise way to perform this task.
loaded weight - unloaded weight = Total lbs in scrap
total lbs * 0.0005 = weight in tons
function calc() {
let final = document.getElementById("result").value =
Math.round(${(loadedAmount - unloadedAmount) * 0.0005} * 100)/100; let resultInPounds = document.getElementById("result").value = Math.round(${(loadedAmount - unloadedAmount)});
} else {
}
} else {
}
}
console.log(loadedAmount: ${loadedAmount}); console.log(unloadedAmount:${unloadedAmount});
document.getElementById("result").innerHTML=final + " Tons or " +
resultInPounds + " lbs";
console.log(result);
}
<!DOCTYPE html>
<html>
<meta charset="utf-8">
<title></title>
<style>
</style>
<body>
<script src="app.js"></script>
<div >input loaded pounds
</div>
<div > input unloaded pounds
</div>
<div><label type="number" id="result" ></label></div>
<button type="submit" id="button" onclick="calc()">Click To
Convert</button>
</body>
</html>
• "total lbs * 0.0005 = weight in tons" That's quite a rounding error you produce there, almost 10% just with one action. Are you sure you want to do this?
– Mast
Jul 2 '18 at 21:21
• @Mast There are many kinds of tons, some of which are 2000 pounds. Jul 3 '18 at 19:17
• @200_success Interesting, you're right. Our ton is your metric ton. Other ton(ne)s are still in use it would seem.
– Mast
Jul 3 '18 at 20:01
Please don't take any of these comments personally, it's clear you are learning and attempting to write clean code and we all started right where you are. With that said I have gone through your code and added comments to outline some things I would change or do differently.
function calc() {
// You should do some basic checks to make sure this value is actually a number
// i.e.
// parse the flaot
// check if it's nan (Not A Number)
// this line is doing to much and makes it hard to read and will make it hard to maintain later in life.
// variables do have some overhead but not enough to warrant not using them to make your code cleaner.
let final = document.getElementById("result").value = Math.round(${(loadedAmount - unloadedAmount) * 0.0005} * 100)/100; let resultInPounds = document.getElementById("result").value = Math.round(${(loadedAmount - unloadedAmount)});
// try something along these lines, i have no idea if your calculations are correct
let final = Math.Round((((loadedAmount - unloadedAmount) * 0.0005) * 100)/100);
document.getElementById("result").value = final;
document.getElementById("result").value = resultInPounds;
// you can see this is much easier to follow and understand what it's intention is.
// You don't need to do these checks here if you've moved them above this so you can remove all of this code. Your variables will already have numbers in them and you would have already presented and errors to the user
} else {
}
} else {
}
}
// this is fine
console.log(loadedAmount: ${loadedAmount}); console.log(unloadedAmount:${unloadedAmount});
// this is fine although you are using templates before this and suddenly stopped, maybe use templates here or change the others to not use template.
document.getElementById("result").innerHTML=final + " Tons or " + resultInPounds + " lbs";
// the variable 'result' doesn't exist anywhere and will always log undefined
console.log(result);
}
<!DOCTYPE html>
<html>
<meta charset="utf-8">
<title></title>
<style>
</style>
<body>
<script src="app.js"></script>
<div >input loaded pounds
</div>
<div > input unloaded pounds
</div>
<div><label type="number" id="result" ></label></div>
<button type="submit" id="button" onclick="calc()">Click To
Convert</button>
</body>
</html>
Based on some feedback here is an example that implements all the changes suggested so far.
window.calc = function() {
var self = this;
// You should do some basic checks to make sure this value is actually a number
// i.e.
// parse the float
self.percentage = 0.0005;
// make this a more meaningful message
// end the function, no calculations can be done
return;
}
// do your calculations
self.final = Math.round((self.loadedAmount - self.unloadedAmount) * self.percentage * 100 / 100);
// display the results
document.getElementById("result").value = ${self.final} Tons or${self.resultInPounds} lbs;
// log the results
console.log(loadedAmount: ${self.loadedAmount}); console.log(unloadedAmount:${self.unloadedAmount});
// the variable 'result' doesn't exist anywhere and will always log undefined
console.log(${self.final} Tons or${self.resultInPounds} lbs);
}
<h3>Negaunee Iron and Metal</h3>
<br>
<br>
<button type="button" id="button" onclick="window.calc()">Convert</button>
<br>
<output for="loaded unloaded" id="result"></output>
• in my opinion, you shouldn't keep his code and your correction in the same file. It could be confusing, e.g. resultInPounds and final are both defined twice.
– Ivan
Jul 3 '18 at 18:13
• Good point however I was more or less trying to highlight the suggestions. I'll make a revision only containing my suggested changes. Jul 3 '18 at 18:23
The label element can be used for the labels of the input fields.
The placeholder attribute shouldn’t be used for labeling.
The output element can be used for the result. (You are using a label, which is not appropriate; and note that label can’t have a type attribute.)
Maybe a button with the button type is more appropriate here. (Button type “button” vs. “submit”)
This results in:
<label for="loaded">Loaded pounds</label>
|
2021-09-25 00:10:41
|
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|
http://math.stackexchange.com/questions/142556/number-of-3-digit-numbers-in-ap-or-gp?answertab=oldest
|
# Number of 3 digit numbers in AP or GP
How many three digit numbers have the property that their digits taken from left to right form an arithmetic or geometric progression?
-
1. If this is homework - please add the homework tag. 2. try to write what you have tried so far. – nbubis May 8 '12 at 9:11
@ nubis I believe in completing my homework myself. I am stuck at this problem and counted the answer as 54 and given answer is 42. – Arpit Bajpai May 8 '12 at 10:52
Let us count. The list of geometric progressions is short, though not as short as I first believed! There are the obvious $1, 2, 4$, and $2, 4, 8$, and $1, 3, 9$, and their reverses. Then there is the less obvious $4,6,9$ and its reverse, which I had missed. And of course $a, a, a$ where $a$ is any of $1$ to $9$. These last also happen to be arithmetic progressions. What about "common ratio" $0$? We will go along with Wikipedia's definition and not allow that.
Now let's count the arithmetic progressions. There are the $9$ with common difference $0$.
There are $7$ increasing ones with common difference $1$, and $8$ decreasing ones, since $0$ can be the final digit in that case.
There are $5$ increasing ones with common difference $2$, and $6$ decreasing ones.
There are $3$ increasing ones with common difference $3$, and $4$ decreasing ones.
There is $1$ increasing one with common difference $4$, and there are $2$ decreasing ones.
If we decide to forget about the sequences $a, a, a$ we get a count of $44$, for we listed $6$ geometric progressions and $36$ arithmetic progressions. Adding in the sequences $a, a, a$, which we definitely should, since they are indeed in both categories, gives us $53$.
For whatever it is worth, Wikipedia does not allow common ratio $0$. If we accept that, the correct count is $53$.
-
You've missed the geometric progressions 469 and 964. But then I don't understand how, starting with 42, and adding in the aaa-numbers, of which there are 9, you get 53. Did you really get the same answer I got, earlier, by making two cancelling mistakes? – Gerry Myerson May 8 '12 at 13:45
Yes, I did add $42$ and $9$, getting $53$. And did entirely miss common ratios $3/2$, $2/3$. Coffee had not yet kicked in. – André Nicolas May 8 '12 at 14:21
111 123 135 147 159 210 222 234 246
258 321 333 345 357 369 420 432 444
456 468 531 543 555 567 579 630 642
654 666 678 741 753 765 777 789 840
852 864 876 888 951 963 975 987 999
should be all the arithmetic progressions, that's 45 right there, so 42 can't be right.
124 139 248 421 469 842 931 964 so I get 53. What did I miss?
Are we counting 100 200 ... 900 as geometric progressions with constant ratio zero?
EDIT: For what it's worth, the number of 3-term geometric progressions with entries from $\{{1,2,\dots,n\}}$ is $(6/\pi^2)n\log n+O(n)$. There's a proof in my paper, Trifectas in geometric progression, Austral. Math. Soc. Gaz. 35 (2008) 189–194, available online at http://www.austms.org.au/Publ/Gazette/2008/Jul08/TechPaperMyerson.pdf. In the paper, I leave counting the 3-term arithmetic progressions as an exercise.
-
|
2015-01-28 20:21:21
|
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|
https://tex.stackexchange.com/questions/653888/how-to-draw-a-loop-passing-through-by-a-edge
|
# How to draw a loop passing through by a edge?
I wish draw the following graph:
Follows the my attempt:
\usetikzlibrary{positioning}
\begin{tikzpicture}
\tikzset{pointblack/.style={fill=black, circle, minimum width=3pt, scale=0.6}}
%%
\node at (2, 0.5) [pointblack] (xhw) {};
\node at (2, -0.5) [pointblack] (xwz) {};
\node at (6, 0) [pointblack] (out) {};
\node at (1,0) [pointblack] (x) {};
\node at (4, 0.5) [pointblack] (y) {};
%%% edges
\draw
(out) to [bend left = 10] (xhw)
(out) to [bend right = 25] (xhw)
(out) to (xwz);
\draw[dashed]
(y) to [bend left = 35] (x)
(y) to [bend right = 45] (x);
\end{tikzpicture}
Remain draw the dashed loop passing by edge above. How can I to do?
• With a screenshot (and some indication of „above“) it would be easier. However, you probably want to look for in and out, which specify angles of outgoing and incoming lines and bends the line as needed. Aug 13 at 4:14
Like this:
\documentclass[margin=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{positioning}
\begin{document}
\begin{tikzpicture}[
node distance = 7mm and 11mm,
pointblack/.style = {fill=black, circle, minimum width=2pt}
]
\begin{scope}[nodes=pointblack]
\node (x) {};
\node[above right=of x] (xhw) {};
\node[below right=of x] (xwz) {};
\node[right=of xhw] (y) {};
\node[below right=of y] (out) {};
\end{scope}
%%% edges
\draw[dashed]
(x) to (y) (y) to [bend right = 45] (x)
(x) to [out=-60,in=-0,distance=33mm] (x);
\draw
(xhw) to (out)
to [bend right = 45] (xhw)
(out) to (xwz);
\end{tikzpicture}
\end{document}
Addendum: in the case, that you like to have more narrow bounding box around picture, than use of the library bbox is thw way to go:
\documentclass[margin=3mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{bbox, % <---
positioning}
\begin{document}
\begin{tikzpicture}[
node distance = 7mm and 11mm,
bezier bounding box, % <---
pointblack/.style = {fill, circle, outer sep=0pt}
]
...
|
2022-10-04 03:38:53
|
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|
https://cs.stackexchange.com/questions/27790/policy-function-%CF%80-in-reinforcement-learning-unclear
|
# Policy function π in Reinforcement learning unclear
I have one question about policy function in Reinforcement learning.
in fact this function indicates which action should be done in each state? Or
this function indicate for get the specific reward in each state, which action shoud be selected?
or something else?
A policy $\pi$ maps each state to an action. Generalizations of this allow for a stochastic mapping (each state to an action with given probability). Given a policy you can estimate or compute the expected reward (or cost) at any state. An optimal policy is simply one that produces the most reward overall.
|
2019-10-24 04:25:49
|
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|
http://sirsmoke.blogspot.com/2010/12/
|
## Tuesday, December 21, 2010
### Grenze
I'm reaching the edge of my sanity
Music by: Pink Floyd [ The Great Gig in the Sky]
## Monday, December 13, 2010
### Gerüst
"... soon comes rain
frost or flames
skeleton me
Fall asleep
spin the sky
skeleton me
Love, don't cry..."
Music by: The YYY's [Skeletons]
## Thursday, December 09, 2010
### Ereignishorizont
An event horizon is a boundary in spacetime beyond which events cannot affect an outside observer. The most common case of an event horizon is that surrounding a black hole. Light emitted from beyond the horizon can never reach the observer. Likewise, any object approaching the horizon from the observer's side appears to slow down and never quite pass through the horizon, with its image becoming more and more redshifted as time elapses. The traveling object, however, experiences no strange effects and does, in fact, pass through the horizon in a finite amount of proper time.
The most commonly known example of an event horizon derives from general relativity's description of a black hole, a celestial object so dense that no nearby matter or radiation can escape its gravitational field. Often, this is described as the boundary within which the black hole's escape velocity is greater than the speed of light. However, a more accurate description is that within this horizon, all lightlike paths (paths that light could take) and hence all paths in the forward light cones of particles within the horizon, are warped so as to fall farther into the hole. Once a particle is inside the horizon, moving into the hole is as inevitable as moving forward in time, and can actually be thought of as equivalent to doing so, depending on the spacetime coordinate system used.
Here I must cite a proper font to describe a suitable definition:
"The criterion for determining whether an event horizon for the universe exists is as follows. Define a comoving distance dE by
$d_E=\int_{t_0}^\infty \frac{c}{a(t)}dt\ .$
In this equation, a is the scale factor, c is the speed of light, and t0 is the age of the universe. If $d_E \rightarrow \infty$ (i.e. points arbitrarily as far away as can be observed), then no event horizon exists. If $d_E \neq \infty$, a horizon is present... "
The description of event horizons given by general relativity is thought to be incomplete. When the conditions under which event horizons occur are modelled using a more comprehensive picture of the way the universe works, that includes both relativity and quantum mechanics, event horizons are expected to have properties that are different from those predicted using general relativity alone.
At present, it is expected that the primary impact of quantum effects is for event horizons to possess a temperature and so emit radiation. For black holes, this manifests as Hawking radiation, and the larger question of how the black hole possesses a temperature is part of the topic of black hole thermodynamics. For accelerating particles, this manifests as the Unruh effect, which causes space around the particle to appear to be filled with matter and radiation.
Duh....
|
2018-05-27 03:21:39
|
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|
https://sciencehouse.wordpress.com/2020/09/06/why-it-is-so-hard-to-forecast-covid-19/
|
# Why it is so hard to forecast COVID-19
I’ve been actively engaged in trying to model the COVID-19 pandemic since April and after 5 months I am pretty confident that models can estimate what is happening at this moment such as the number of people who are currently infected but not counted as a case. Back at the end of April our model predicted that the case ascertainment ratio ( total cases/total infected) was on the order of 1 in 10 that varied drastically between regions and that number has gone up with the advent of more testing so that it may now be on the order of 1 in 4 or possibly higher in some regions. These numbers more or less the anti-body test data.
However, I do not really trust my model to forecast what will happen a month from now much less six months. There are several reasons. One is that while the pandemic is global the dynamics are local and it is difficult if not impossible to get enough data for a detailed fine grained model that captures all the interactions between people. Another is that the data we do have is not completely reliable. Different regions define cases and deaths differently. There is no universally accepted definition for what constitutes a case or a death and the definition can change over time even for the same region. Thus, differences in death rates between regions or months could be due to differences in the biology of the virus, medical care, or how deaths are defined and when they are recorded. Depending on the region or time, a person with a SARS-CoV-2 infection who dies of a cardiac arrest may or may not be counted as a COVID-19 death. Deaths are sometimes not officially recorded for a week or two, particularly if the physician is overwhelmed with cases.
However, the most important reason models have difficulty forecasting the future is that modeling COVID-19 is as much if not more about modeling the behavior of people and government policy than modeling the biology of disease transmission and we are just not very good at predicting what people will do. This was pointed out by economist John Cochrane months ago, which I blogged about (see here). You can see why getting behavior correct is crucial to modeling a pandemic from the classic SIR model
$\frac{dS}{dt} = -\beta SI$
$\frac{dI}{dt} = \beta SI - \sigma I$
where $I$ and $S$ are the infected and susceptible fractions of the initial population, respectively. Behavior greatly affects the rate of infection $\beta$ and small errors in $\beta$ amplify exponentially. Suppression and mitigation measures such as social distancing, mask wearing, and vaccines reduce $\beta$, while super-spreading events increase $\beta$. The amplification of error is readily apparent near the onset of the pandemic where $I$ grows like $e^{\beta t}$. If you change $\beta$ by $\delta \beta$, then the $I$ will grow like $e^{\beta t+\delta \beta t}$ and thus the ratio is growing (or decaying) exponentially like $e^{\delta \beta t}$. The infection rate also appears in the initial reproduction number $R_0 = \sigma/\beta$. From a previous post, I derived approximate expressions for how long a pandemic would last and show that it scales as $1/(R_0-1)$ and thus errors in $\beta$ will produce errors $R_0$, which could result in errors in how long the pandemic will last, which could be very large if $R_0$ is near one.
The infection rate is different everywhere and constantly changing and while it may be possible to get an estimate of it from the existing data there is no guarantee that previous trends can be extrapolated into the future. So while some of the COVID-19 models do a pretty good job at forecasting out a month or even 6 weeks (e.g. see here), I doubt any will be able to give us a good sense of what things will be like in January.
## One thought on “Why it is so hard to forecast COVID-19”
1. Ishi Crew says:
Out of curiosity I skimmed Cochrane’s model because i remebered his name though forgot who it was—he’s at Hoover/U Chicago so now i remember. (birds of a feather flock together –a sort of informal and generalized quarantining or lockdown, or lockout—‘competetive exclusion’–human groups or tribes tend to share memes or ‘viruses of the mind’).
His model looks reasonable, intuitive —just makes the ‘constants’ beta, R_0 time dependent. I sort of call this ‘adding more nonlinearity’.
In a sense (or in some ranges) his slightly more complex model (he cites someone at stanford which i guess is related) doesn’t change anything. A few more wiggles, but basically same limit for most cases–and he points out to an extent one can choose the differences either through informal rules (persoal choice, education, nudges) or govt policy.
(Same issue about laws versus behavioral change arrived informally arises in discussions of climate change, economic growth, and population—-some people prefer controlling population via ‘fertility taxes’ (a mild form of china’s one child policy) while others promote using educattion instead (ie educated people can plan their own family sizes and likely this will have same effect as a law on sizes) .
Some people favor lots of carbon, pollution, value added, Pigouvian or sin taxes, to cut consumption of carbon and problem producing goods, and forms of wealth or progressive income taxes for basic income, health, and education services, to make the economy stable—less prone to unrest –which itself can affect population and climate . Others favor using ‘shaming’, persuasion by creating alternative models of consumption and distribution, etc. rather than laws.)
My ‘guess’ is that COVID-19 will go as ‘business as usual’ like other viruses and diseases, so its not the black plague (which i think killed off 1/3rd of europe). So January will be like an average of the last 6-10 months-which means the same, or else the virus will be gradually be wiped out with possibly periodic recurrences, or possibly it will just become a chronic condition at least for some (including me—something you live with–even tho i tested negative).
Actually, my interpretations of the idea that globally the epidemic is ‘flat’ is that is an application of the law of large numbers and/or CLT (i forget exactly how these differ–sometimes they may not, sometimes they may) —just take a random sample off the earth and one will find some ‘modes’ increasing while others decrease –a la the Fermi-Pasta-Ulam experiments. Eventually it will approach the stationary or equilibrium distribution (basically like an ‘old growth forest’–it doesn’t mean its totally static –trees fall, occasional small fires, new growth, etc. . ).
Whatever pattern of governmental regulation took—extreme lockdowns, or minor ones— it willl all sort of even out—possibly at different rates. (Actually the pprevious post on ‘why there is no herd immunity’ suggests that if different regions have very different rules and there is migration between them, that could mean reaching the equiblibirum or stationary distibution could take a very long time–and maybe never will.)
There are so parameters in these models in theory probably everything is possible. Cochran emphasizes the difference between epidimiology as done by people who study animals for whom R and B are constants, versus economists who study human animals which may be able to change behavior (there is some debate about this–eg how hard wired are people, if they even have any free will–they may think so, just as they may believe god doesn’t exist, but may be wrong. Of course if they do have free will then they should be able to create god. ).
So its quite possible ‘this doesn’t change everything’ .
————–
As an aside I looked at Cochran’s first blog which is on finance and mentions MMT modern monetary theory’ and S kelton who promotes it). I agree with his assessment of MMT–its not a theory (or at best warmed over keynsianism with a twist called ‘the tax break for the rich’—MMT says ‘let the the poor have their cake and eat it2’ so long as they work at their govt paid for guaranteed job (basically voluntary wage slavery —if you want to eat, work)) . ; and also dont take cake from the rich to pay for it. The poor can make and eat their own cake or whatever they are able to affrod to make –eg stone soup.
MMT people pass this off as an alternative to a UBI to help the poor; but i view it as a possible trojan horse —people are just going to turned into a proletarian caste like hamsters who make their own wheels. ) In a sense we already have this—its called the youtube/podcast culture.
Actually MMT and UBI are equivalent in some limits–just as are capitalism, marxism-communism , and anarcho-communism. This is my tendency to see all models as limits of the same model—same for COVID, gun regulations, voting versus non-electoral politics (ie personal change oor protesting.)
——————-
He also writes on inequality—says for example in his view its misplaced for people Like Krugman or Pietty and others to care about inequality. He says a poor laborer in California doesn’t care about the jet owned by a billionaire flying above the clouds above the farm or factory s/he’s working at–s/he’s worried about paying the rent. S pinker has the same view—says ‘envy’ or worry about income and wealth inequality are myths by ‘liberals’ and ‘leftists’ trying to make the poor angry.
my view is they aren’t myths—poor people do worry about those billionaires and jets because those are the people who own and run where they work, and determine whether they can pay rent.
if they dont it may also be a behavioral strategy —dont worry about things you can’t change because its unhealthy (R Aapolsky advises this from Stanford).
and as far as being grateful that the rich do provide some jobs, they may not be happy that it comes with all sorts of health hazards—some of which would be preventible by making fewer profits, etc.
——————-
If Cochran used the same kind of behavioral logic for COVID that he uses in his analyses of economics of inequality he probably would say a billionaire or poor person in Wuhang (sic?) China who has COVID has no effect on someone in the USA and hence should be ignored.
—— Most of the people sort of promoting less stringent controls and more ‘herd imunity’ emphasize economic issues —i wonder how these interact. (I discussed this a bit with people who volunteer on the ‘endcovid’ list maintained by NECSI but it didn’t go far—no modelers were in that group. Putting that in a model would add more complexity.)
Like
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2022-12-08 19:13:50
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https://eclipse.googlesource.com/virgo/org.eclipse.virgo.sample-greenpages/+/603f334f5e27c07a60e1d2357b849bf1ad1bfc01/guide/src/installing-greenpages.xml
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blob: 420f1276a9279a3e08946511669072fc18d88399 [file] [log] [blame]
Installing and Running @greenpages@ Installing and Running @greenpages@
Introduction @greenpages@ is a simple application that allows users to search an online email address directory. Each listing in the directory details the relevant email addresses and the name of the owner. @greenpages@ has only three screens: the search screen, the results screen and the listing detail screen. In the search screen, users can enter search criteria to be matched against the listings in the directory. The result screen displays any listings that match the criteria entered by the user. The listing detail screen shows all the data known about a given listing. Despite its simplicity, @greenpages@ is designed to demonstrate many different @webserv@ features and to act as a template from which other modular applications can be built. In particular, @greenpages@ demonstrates: bundle dependencies with Import-Package, load-time weaving with JPA and AspectJ, bundle classpath scanning, and service export, lookup and injection. In addition to demonstrating common @webserv@ features, @greenpages@ demonstrates integration with: Spring Framework 3.0; FreeMarker 2.3; EclipseLink 1.0.0; H2 1.0.71; and Commons DBCP 1.2.2. The @greenpages@ application is packaged as a PAR file containing four bundles. The greenpages.db bundle provides access to an external database and publishes a javax.sql.DataSource service. The greenpages.app bundle exports a greenpages package containing Directory and Listing interfaces. The greenpages.jpa bundle imports the greenpages package and uses the javax.sql.DataSource service to access the external database and publishes its contents as a greenpages.Directory service. The greenpages.web web application bundle imports the greenpages package and uses the greenpages.Directory service to respond to web requests.
Obtaining @greenpages@ This document provides instructions for building the complete @greenpages@ application and running it in @webserv@. To get the completed @greenpages@ application, including tests and explanatory skeleton parts: download the latest Zip file from @greenpages.download.url@ extract all the files from the Zip file to a convenient directory (preserving the directory structure). To extract the files on Windows: prompt> mkdir c:\springsource\samples prompt> cd c:\springsource\samples prompt> jar xf c:\path\to\@greenpages.zip.file@ prompt> set GREENPAGES_HOME=c:\springsource\samples\@greenpages.expanded.dir@ To extract the files on Unix systems: prompt$mkdir -p /opt/springsource/samples prompt$ cd /opt/springsource/samples prompt$unzip /path/to/@greenpages.zip.file@ prompt$ export GREENPAGES_HOME=/opt/springsource/samples/@greenpages.expanded.dir@ The environment variable GREENPAGES_HOME set here is not used by the projects, but is used as a shorthand in the instructions that follow. The @greenpages@ Zip file contains several directories with names that start greenpages. They contain the completed application which can be built and tested (as described in the next section).
Building and Installing @greenpages@
Building with @maven.full@ @greenpages@ uses @maven.full@ as its primary build system. Each bundle of the application can be built separately and the entire application can built and assembled into a PAR file from a single location. To build the application and assemble it into a PAR file: Make $GREENPAGES_HOME/ the current directory. Run the command mvn package. The first time this is run several files will be downloaded from @maven@ repositories. Subsequent runs will not need to do this. Verify that the greenpages-@greenpages.version@.par file exists in$GREENPAGES_HOME/greenpages/target.
Installing Dependencies into @webserv@ Unlike traditional Java EE applications, @greenpages@ does not package all of its dependencies inside its deployment unit. Instead, it relies on the mechanisms of OSGi to locate its dependencies at runtime. When running an OSGi application on @webserv@, these dependencies can be loaded into memory as needed, but first they must be made available to @webserv@. The Maven build included with @greenpages@ uses the dependency:copy-dependencies plugin to gather all the artifacts that @greenpages@ depends on that are not supplied by the @webserv@ runtime. These dependencies can then be installed into the @webserv@ repository. Dependencies are gathered automatically during the package phase. These dependencies can be found in $GREENPAGES_HOME/greenpages/target/par-provided. To install dependencies simply copy all the *.jar files from this directory into$SERVER_HOME/repository/usr (where $SERVER_HOME is the @webserv@ installation directory). Installing dependencies on Windows: prompt> cd %GREENPAGES_HOME%\greenpages prompt> copy target\par-provided\* %SERVER_HOME%\repository\usr Installing dependencies on UNIX: prompt$ cd $GREENPAGES_HOME/greenpages prompt$ cp target/par-provided/* $SERVER_HOME/repository/usr Notice that @webserv@ will not necessarily see these dependencies unless its repository indexes are rebuilt. Different repositories behave differently in this respect; some are passive (their indexes are built only once upon startup) and some are active (they can detect new files or files being removed dynamically). The usr repository is active so there is no need to restart @webserv@ when copying these files. The next time @webserv@ is started the -clean option will cause @webserv@ to re-scan the repository directories in any case. It is always safe to start @webserv@ with the -clean option. Starting and Configuring the Database @greenpages@ uses the H2 database to store all its data. Before starting the application, start the database server and populate the database with data. Change to the$GREENPAGES_HOME/db current directory. On Unix: prompt$cd$GREENPAGES_HOME/db On Windows: prompt> cd %GREENPAGES_HOME%\db Run the database startup script appropriate to the operating system. For Unix, this is run.sh, run in the background: prompt$sh run.sh & Press Return to continue. On Windows, run the run.bat command: prompt> run For both platforms, the command might invoke a browser window offering a connection to the database; close this window. Run the data population script appropriate to the operating system. For Unix, this is data.sh: prompt$ sh data.sh On Windows, run the data.bat command: prompt> data Run these commands once to start a database server for H2; the server will continue to run in the background.
Installing and Starting @greenpages@ PAR To install the @greenpages@ PAR into @webserv@ and start it: Copy the @greenpages@ PAR to the $SERVER_HOME/pickup directory. On Unix: prompt$ cd $SERVER_HOME prompt$ cp $GREENPAGES_HOME/greenpages/target/greenpages-@greenpages.version@.par pickup/ On Windows: prompt> cd %SERVER_HOME% prompt> copy %GREENPAGES_HOME%\greenpages\target\greenpages-@greenpages.version@.par pickup\ Start @webserv@ with the -clean option. On Unix: prompt$ $SERVER_HOME/bin/startup.sh -clean On Windows: prompt> "%SERVER_HOME%"\bin\startup.bat -clean Verify that @greenpages@ starts correctly by checking in the @webserv@ output for the log message: <DE0005I> Started par 'greenpages' version '@greenpages.version@'. Browsing the @greenpages@ Application Once installed and started, the @greenpages@ application can be accessed with a web browser using the address http://localhost:8080/greenpages. From the home page, a search query can be entered into the search box: After entering a query into the search box, the results page shows all the matches from the directory: Clicking on view, next to an entry in the search listing, displays the full details for that listing entry: Running @greenpages@ from Eclipse Using Eclipse and the @webserv@ tools, it is possible to run applications directly from the IDE. As changes are made to the application in the IDE, they can be automatically applied to the running application allowing for rapid feedback of changes in function. Importing the @greenpages@ Projects into Eclipse Before starting the @greenpages@ application from Eclipse, import the projects: Open the Import Wizard using FileImport. From the Import Wizard select GeneralExisting Projects into Workspace and click Next: Click Browse… and select$GREENPAGES_HOME/ as the root directory. In the Import Projects window, select all the projects which include greenpages in their name and click Finish: Validate that the imported projects appear in Package Explorer: There may be compilation errors at this stage.
Configuring @webserv@ Target Runtime Projects for @webserv@ are associated with a @virgo@ @webserv@ runtime environment in Eclipse. This is to allow launching and testing from within Eclipse, and also to allow classpath construction in Eclipse to mirror the dynamic classpath in the @webserv@ runtime. Compilation errors in the previous step will be resolved here. To configure a @webserv@ runtime environment: Open WindowShow ViewOther…. In the Show View dialog choose ServerServers to make the servers view visible: Right-click in the Servers (which may not be empty) view and select NewServer. In the New Server dialog, choose EclipseRTVirgo Web Server and click Next. Click Browse and select the $SERVER_HOME directory. Ensure that a JRE is selected supporting Java 1.6 or above. Click Finish to complete creation of the server: Select all projects (except Servers) in Package Explorer. Right-click on the projects and choose Close Project and then Open Project. It is possible that there remain spurious build errors from Eclipse (see the Problems view), in which case a project clean build may clear the problems. Select ProjectClean… from the main menu, and choose to Clean all projects. It may be necessary to repeat this on a few projects. (This process is sometimes known as the Eclipse dance.) Despite the dance steps outlined, there will remain some Warnings/Errors like this: It is safe to ignore these. When the @virgo@ Tooling starts the @webserv@ it uses a ‘warm start’ by default. It is useful to set the -clean option so that every server start is a clean one. This is done by an option on the @webserv@ Overview window, which is obtained by opening the @webserv@ entry in the Servers window. (Double-click, or right-click and choose Open.) The check box is labelled ‘Start server with -clean option’. Close the window before proceeding. Running @greenpages@ from Within Eclipse Now that @greenpages@ is successfully imported into Eclipse, run the project directly from within the IDE. If the @greenpages@ PAR file was previously copied to the pickup directory, be sure it is now removed so that it does not conflict with the deployment of the Eclipse project. On Unix: prompt$ cd $SERVER_HOME/pickup prompt$ rm greenpages-@greenpages.version@.par On Windows: prompt> cd %SERVER_HOME%\pickup prompt> del greenpages-@greenpages.version@.par Also, to prevent conflicts with the server configured in Eclipse, stop a currently-running @webserv@ by typing Control-C in the console window. To run @greenpages@ from within Eclipse: Right click on the @webserv@ instance in the Servers view and select the Add and Remove… menu item. Add greenpages (which is the containing project or PAR) to the server and finish. To start @webserv@ from within Eclipse right-click on the @webserv@ node in the Servers window and choose Start. The Servers view should now show the server and the added project: Verify that @greenpages@ is started correctly by checking for: <DE0005I> Started par 'greenpages' version '@greenpages.version@'. in the Console window. (If errors are shown implying that @greenpages@ failed to be installed, this may be because some dependencies were not copied to @webserv@, as described in . Check this.) Once installed and started @greenpages@ is again available from a web browser at the address http://localhost:8080/greenpages.
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2022-06-25 09:03:17
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https://www.physicsforums.com/threads/another-way-to-extend-the-complex-plane-insertsomethingthatgetsyourattention.499964/
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# Another way to extend the Complex Plane (Insertsomethingthatgetsyourattention)
#### Frogeyedpeas
Hey guys so I was thinking about how to extend the Complex Plane out to a third dimension and I started reading the whole tidbit about Quaternions and their mechanics when I realized that I want to propose a whole new question. Now please feel free to prove me wrong if you can answer it because I haven't found a whole lot.
Imagine a number J who satisfies the solution to the following equation
Logb(J) = -b
FOR ALL B:
There is no complex number that satisfies that solution and I believe (as uneducated as I might be in this subject) that there is no Quaternion, Octonion or any type of standard Algebraic extension of the number line that satisfies this equation. If this number J can be proposed as the new number extension to the complex plane, then,
We get numbers being described in the form of:
a + bi + cj. Now keeping in mind the ability for numbers to cross:
a + bi + cj + dji is what this number can look like...
Last edited:
#### Frogeyedpeas
Any replies, opinions, haikus, limericks, epic poetries, hard rock songs, and qwerty-piano recitals would be cool
#### coelho
Ok... if by Logb you mean log in the base b, then we have:
$$log_b(J)=-b$$
$$\frac{ln(J)}{ln(b)}=-b$$
$$ln(J)=-b ln(b)$$
$$J=e^{-b ln(b)}$$
$$J=(e^{ln(b)})^{-b}$$
$$J=b^{-b}$$
Which is, at worst, a complex number (if b were complex). If b were real, so is J.
#### disregardthat
If you want a number J to satisfy that equation, you are not extending the complex plane, you are reducing it by the equvalence relation $$b^{-b} = c^{-c}$$ for all complex b and c. If the function z^{-z} := e^{-z\log(z)} for some branch of the logarithm is surjective, you will have collapsed the complex plane to a single point.
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2019-06-18 16:37:41
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https://predictivehacks.com/numpy-hacks-for-data-cleansing/
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Predictive Hacks
# NumPy Hacks for Data Manipulation
The goal of this article is to provide some “Numpy Hacks” that will be quite useful during the Data Science Pipeline and especially during the Data Cleansing phase. As always, we will work with reproducible and practical examples. We will work with Pandas and NumPy libraries for our examples.
## random
### random.seed()
NumPy gives us the possibility to generate random numbers. However, when we work with reproducible examples, we want the “random numbers” to be identical whenever we run the code. For that reason, we can set a random seed with the random.seed() function which is similar to the random random_state of scikit-learn package.
### random.choice() | random.poisson() | random.rand()
With NumPy we can generate random numbers from distributions like poisson, normal, exponential etc, from the uniform distribution with the random.rand() and from a sample with the random.choice(). Let’s generate a Pandas data frame using the random module.
Example: We will create a pandas data frame of 20 rows and columns such as gender, age and score_a., score_b, score_c
import pandas as pd
import numpy as np
# set a random seed
np.random.seed(5)
# gender 60% male 40% female
# age from poisson distribution with lambda=25
# score a random integer from 0 to 100
df = pd.DataFrame({'gender':np.random.choice(a=['m','f'], size=20, p=[0.6,0.4]),
'age':np.random.poisson(lam=25, size=20),
'score_a':np.random.randint(100, size=20),
'score_b':np.random.randint(100, size=20),
'score_c':np.random.randint(100, size=20)})
df
### random.shuffle()
With the random.shuffle() we can shuffle randomly the numpy arrays.
# set a random seed
np.random.seed(5)
arr = df.values
np.random.shuffle(arr)
arr
## logical_and() | logical_or()
I have found the logical_and() and logical_or() to be very convenient when we dealing with multiple conditions. Let’s provide some simple examples.
x = np.arange(5)
np.logical_and(x>1, x<4)
And we get:
array([False, False, True, True, False])
np.logical_or(x < 1, x > 3)
And we get:
array([ True, False, False, False, True])
## where()
The where() function is very helpful when we want to apply an if else statement by assigning new values. Let’s say that we want to assign a value equal to “Pass” when the score is higher than 50 and “Fail” when the score is lower than 50. Let’s do it for the score_a column.
df['score_a_pass'] = np.where(df.score_a>=50,"Pass","Fail")
## select()
If we want to add more conditions, even across multiple columns then we should work with the select() function. Let’s that that I want to define the following column called demo as follows:
• if the gender is ‘m’ and the age is below 20 then ‘boy’
• if the gender is ‘m’ and the age is above 20 then ‘mister’
• if the gender is ‘f’ and the age is below 20 then ‘girl’
• if the gender is ‘f’ and the age is above 20 then ‘lady’
• else ‘null’
Let’s see how easily we can do it by using the select()
choices = ['Mister','Lady','Boy', 'Girl']
conditions = [
(df['gender'] == 'm') & (df['age']>20),
(df['gender'] == 'f') & (df['age']>20),
(df['gender'] == 'm') & (df['age']<=20),
(df['gender'] == 'f') & (df['age']<=20)
]
df['demo'] = np.select(conditions, choices, default=np.nan)
Note that we could have used the logical_and() in the conditions.
## digitize()
Many times, we want to bucketize our data into bins. We have explained how to create bins with Pandas. Let’s see how we can do it with NumPy. Let’s say that I can to create 5 bins from the score_a.
bins = np.array([0, 20, 40, 60, 80, 100])
df['Bins'] = np.digitize(df.score_a, bins)
## split()
You can also split the NumPy arrays into parts. Let’s say that you want to create a train (60%), validation (20%) and test (20%) datasets.
data_a, data_b, data_c = np.split(df.values, [int(0.6 * len(df.values)), int(0.8*len(df.values))])
data_a
data_b
data_c
## clip()
Sometimes we can set a range for the values and if they are outside this interval to get the minim and the maximum value respectively. Let’s assume that we want the data to take values from 0 to 100 and in our dataset, we have values below zero and values above zero. Let’s see the example below:
x = np.array([30, 20, 50, 70, 50, 100, 10, 130, -20, -10, 200])
np.clip(x,0,100)
As we can see below, the negative values became 0 and the values above 100 became 100:
array([ 30, 20, 50, 70, 50, 100, 10, 100, 0, 0, 100])
## extract()
Let’s say that we want to extract values that satisfy some conditions. Assume that on the previous example, we wanted to get the values which are less than 0 or greater than 100:
np.extract( (x>100) | (x<0), x )
And we get:
array([130, -20, -10, 200])
## unique()
The unique() function returns the unique values but we can use it to get a “value counts” of each element. For example:
# How to count the unique values of an array
x = np.array([0,0,0,1,1,1,0,0,2,2])
unique, counts = np.unique(x, return_counts=True)
dict(zip(unique, counts))
And we get:
{0: 5, 1: 3, 2: 2}
## argmax() | argmin() | argsort() | argpartition()
These functions are values useful. The argmax() and argmin() return the index of the max and min element respectively. Let’s say that we want to know the index of the data frame of the maximum score_a
np.argmax(np.array(df.score_a))
and we get 17. Now we can get the whole 17-th row of the data frame
df.iloc[np.argmax(np.array(df.score_a))]
The argsort() sorts the NumPy array and returns the indexes. Let’s say that I want to sort the score_a column:
df.iloc[np.argsort(np.array(df.score_a))]
If we want to get the N largest value index, then we can use the argpartition(). Let’s say that we want to get the top 5 elements of the following array:
x = np.array([30, 20, 50, 70, 50, 100, 10, 130, -20, -10, 200])
indexes = np.argpartition(x, -5)[-5:]
indexes
array([ 2, 3, 5, 7, 10], dtype=int64)
x[indexes]
array([ 50, 70, 100, 130, 200])
Let’s provide a final example by assuming the following scenario. Let’s say that we want to create three columns such as Top1, Top2 and Top3 for each row based on the scores of columns score_a, score_b and score_c. In other words, at which exam there was the highest score, then the second higher and finally the third higher. We can work with the argsort():
Tops =pd.DataFrame(df[['score_a','score_b','score_c']].\
apply(lambda x:list(df[['score_a','score_b','score_c']].\
columns[np.array(x).argsort()[::-1][:3]]), axis=1).\
to_list(), columns=['Top1', 'Top2', 'Top3'])
Tops
## Sum-Up
NumPy is a very popular and strong library. It is very fast and compatible with all AI and ML libraries like Scikit-Learn, TensorFlow etc. Thus, is it very important for every Data Scientist to be competent with NumPy. If you like this article, then you may like the tips about NumPy arrays.
### Get updates and learn from the best
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In Unix, there are three types of redirection such as: Standard Input (stdin) that is denoted by 0. Usually, it’s
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#### Content-Based Recommender Systems with TensorFlow Recommenders
In this post, we will consider as a reference point the “Building deep retrieval models” tutorial from TensorFlow and we
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2022-11-29 17:44:32
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https://www.projecteuclid.org/euclid.aop/1158673319
|
Annals of Probability
Second class particles and cube root asymptotics for Hammersley’s process
Abstract
We show that, for a stationary version of Hammersley’s process, with Poisson sources on the positive x-axis and Poisson sinks on the positive y-axis, the variance of the length of a longest weakly North–East path L(t,t) from (0,0) to (t,t) is equal to $2\mathbb{E}(t-X(t))_{+}$, where X(t) is the location of a second class particle at time t. This implies that both $\mathbb{E}(t-X(t))_{+}$ and the variance of L(t,t) are of order t2/3. Proofs are based on the relation between the flux and the path of a second class particle, continuing the approach of Cator and Groeneboom [Ann. Probab. 33 (2005) 879–903].
Article information
Source
Ann. Probab., Volume 34, Number 4 (2006), 1273-1295.
Dates
First available in Project Euclid: 19 September 2006
https://projecteuclid.org/euclid.aop/1158673319
Digital Object Identifier
doi:10.1214/009117906000000089
Mathematical Reviews number (MathSciNet)
MR2257647
Zentralblatt MATH identifier
1101.60076
Citation
Cator, Eric; Groeneboom, Piet. Second class particles and cube root asymptotics for Hammersley’s process. Ann. Probab. 34 (2006), no. 4, 1273--1295. doi:10.1214/009117906000000089. https://projecteuclid.org/euclid.aop/1158673319
References
• Aldous, D. and Diaconis, P. (1995). Hammersley's interacting particle process and longest increasing subsequences. Probab. Theory Related Fields 103 199–213.
• Baik, J., Deift, P. and Johansson, K. (1999). On the distribution of the length of the longest increasing subsequences of random permutations. J. Amer. Math. Soc. 12 1119–1178.
• Baik, J. and Rains, E. (2000). Limiting distributions for a polynuclear growth model with external sources. J. Statist. Phys. 100 523–541.
• Cator, E. and Groeneboom, P. (2005). Hammersley's process with sources and sinks. Ann. Probab. 33 879–903.
• Ferrari, P. A. and Fontes, L. R. G. (1994). Current fluctuations for the asymmetric simple exclusion process. Ann. Probab. 22 820–832.
• Groeneboom, P. (1989). Brownian motion with a parabolic drift and Airy functions. Probab. Theory Related Fields 81 79–109.
• Johansson, K. (2000). Transversal fluctuations for increasing subsequences on the plane. Probab. Theory Related Fields 116 445–456.
• Kim, J. and Pollard, D. (1990). Cube root asymptotics. Ann. Statist. 18 191–219.
• Seppäläinen, T. (2005). Second-order fluctuations and current across characteristic for a one-dimensional growth model of independent random walks. Ann. Probab. 33 759–797.
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2020-06-03 10:57:09
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https://chalkdustmagazine.com/tag/issue-11-puzzlehunt/
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# Chalkdust issue 11 puzzle hunt #5
The day is finally here: issue 11 is out now! To help you celebrate launch day in style, we’ve set a puzzle hunt: throughout the day, we are posting a series of puzzles. The answers to these puzzles form clues to the four-digit code for the door to let you into the Chalkdust issue 11 secret backstage lounge.
The fifth and final puzzle is set by Adam Townsend. You can see the puzzle below, or download it as a pdf.
Once you’ve solved this puzzle, head over to the Chalkdust issue 11 secret backstage lounge and unlock the door. See you inside!
Adam has been playing a game: he thinks of a secret word, tells you how many letters long it is, and lets you guess it. For each guess, he tells you how many letters in your guess are the correct letter in the correct position, and how many letters are the correct letter but in the wrong position.
For example, if Adam’s secret word is ‘GAME’, and you guess ‘ELMO’, he will tell you:
Correct letter,correct position Correct letter,wrong position E L M O 1 1 ? ? ? ?
as the ‘M’ is the correct letter in the correct position, and the ‘E’ is the correct letter in the wrong position. If Adam’s secret word is ‘AABB’ and you guess ‘AAAB’, he will tell you:
Correct letter,correct position Correct letter,wrong position A A A B 3 0 ? ? ? ?
as three letters are the correct letters in the correct position (AAAB), and there are no letters in the wrong positions as there is no third ‘A’ in Adam’s word. Adam doesn’t double count.
The secret words will spell out a clue to the code.
Correct letter,correct position Correct letter,wrong position H A T 0 2 E T A 0 2 A R C 0 0 ? ? ?
Correct letter,correct position Correct letter,wrong position M O D E 3 0 C U B E 2 0 ? ? ? ?
Correct letter,correct position Correct letter,wrong position G R A D I E N T 1 3 M A N I F O L D 1 3 C O N V E R G E 3 0 N O T A T I O N 2 4 S E Q U E N C E 0 3 M U R D E R E D 0 0 ? ? ? ? ? ? ? ?
Correct letter,correct position Correct letter,wrong position T A N 1 0 L A W 0 1 O D E 0 1 R O W 0 2 ? ? ?
Correct letter,correct position Correct letter,wrong position R A N G E 1 1 S H E A R 0 2 I N N E R 2 2 O U T E R 1 0 ? ? ? ? ?
# Chalkdust issue 11 puzzle hunt #4
The day is finally here: issue 11 is out now! To help you celebrate launch day in style, we’ve set a puzzle hunt: throughout the day, we are posting a series of puzzles. The answers to these puzzles form clues to the four-digit code for the door to let you into the Chalkdust issue 11 secret backstage lounge.
The fourth puzzle is set by Humbug. You can see the puzzle below, or download it as a pdf.
Be sure to come back at 5pm when the fifth puzzle will be posted.
## Humbug’s puzzle
Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0.
The number of copies of the digit 1 in the completed crossnumber is one more than a digit of the code.
#### Across
• 1 The sum of this number’s digits is 12. (5)
• 5 The middle digit of this number is equal to the first digit of 4D. (3)
• 6 The product of the digits of 5A. (2)
• 7 The sum of this number’s digits is 7. (2)
• 8 1A reversed. (5)
#### Down
• 1 The sum of this number’s digits is 14. (5)
• 2 This number is equal to 2 multiplied by the sum of its digits. (2)
• 3 This number is equal to 3 multiplied by the sum of its digits. (2)
• 4 1D reversed. (5)
• 7 The product of this number’s digits is 7. (2)
# Chalkdust issue 11 puzzle hunt #3
The day is finally here: issue 11 is out now! To help you celebrate launch day in style, we’ve set a puzzle hunt: throughout the day, we are posting a series of puzzles. The answers to these puzzles form clues to the four-digit code for the door to let you into the Chalkdust issue 11 secret backstage lounge.
The third puzzle is set by David Sheard. You can see the puzzle below, or download it as a pdf.
Be sure to come back at 3pm when the fourth puzzle will be posted.
## David’s puzzle
The digits 1 to 9 must be entered into the 9×9 grid below following normal sudoku rules: each digit must appear exactly once in each row, column, and 3×3 block. Additionally, the grid also contains thermometers and inequalities.
In each grey thermometer, the number in the circular bulb is the smallest number, and the numbers increase as you move away from the bulb. The sums of the digits on a diagonal indicated by an arrow must satisfy the inequality shown by the arrow (if the sum of the digits on a diagonal is written at the start of the arrow, then the resulting inequality must be true).
In this smaller example puzzle, the digits 1 to 4 must appear in each row, column, and 2×2 block exactly once. Starting from the grey bulbs, the numbers increase along each thermometer; and the numbers in the cells highlighted in blue in the solution add to more than or equal to 7. The sums of the digits on the other
indicated diagonals satisfy the inequalities written by their arrows.
The puzzle (click to enlarge)
The digit in the cell shaded blue is a factor of the code.
# Chalkdust issue 11 puzzle hunt #2
The day is finally here: issue 11 is out now! To help you celebrate launch day in style, we’ve set a puzzle hunt: throughout the day, we are posting a series of puzzles. The answers to these puzzles form clues to the code for the door to let you into the Chalkdust issue 11 secret backstage lounge.
The second puzzle is set by TD Dang. You can see the puzzle below, or download it as a pdf.
Be sure to come back at 1pm when the third puzzle will be posted.
## TD’s puzzle
The three circles below all meet at tangents. The area of the largest circle is $20\pi$.
The sum of the first and last digits of the code is less than the area of the blue quadrilateral.
# Chalkdust issue 11 puzzle hunt #1
The day is finally here: issue 11 is out now! To help you celebrate launch day in style, we’ve set a puzzle hunt: throughout the day, we are posting a series of puzzles. The answers to these puzzles form clues to the four-digit code for the door to let you into the Chalkdust issue 11 secret backstage lounge.
The first puzzle is set by Matthew Scroggs. You can see the puzzle below, or download it as a pdf.
Be sure to come back at 11am when the second puzzle will be posted.
## Scroggs’s puzzle
Each row and column of this grid contains the letters W, E, T, C, O, I, N, and S exactly once.
Each column and the first 7 rows contain the words given by the clues below in some position. For example, if a row contains the word `WET’, then the row could be WETCOINS,
NOWETSIC, ONSWETCI, or many other things.
Once the puzzle has been completed, the final row will contain a number. This number is one of the digits of the code.
#### Across
• 1 Conic ___. (7)
• 2 A charged atom or molecule. (3)
• 3 0 seconds in the future. (3)
• 4 315°. (2)
• 5 3D shape with volume $\pi r^2h/3$. (4)
• 6 225°. (2)
• 7 $\{2,3\}$ or $\emptyset$. (3)
#### Down
• 1 Can be found on page 65 of issue 11. (3)
• 2 $\in$. (2)
• 3 A baby’s favourite trig function (abbrev). (3)
• 4 adjacent ÷ hypotenuse (not abbreviated). (6)
• 5 135°. (2)
• 6 $\sqrt{100}$. (3)
• 7 ___ primes. (4)
• 8 adjacent ÷ hypotenuse (abbrev). (3)
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2021-07-28 22:32:48
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|
https://maharashtraboardsolutions.guru/category/class-9/
|
## Maharashtra Board 9th Class Maths Part 1 Practice Set 4.5 Solutions Chapter 4 Ratio and Proportion
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.
## Practice Set 4.5 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion
Question 1.
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution:
Let the number to be subtracted be x.
∴ (12 – x), (16 – x) and (21 – x) are in continued proportion.
∴ 84 – 4x = 80 – 5x
∴ 5x – 4x = 80 – 84
∴ x = -4
∴ -4 should be subtracted from 12,16 and 21 so that the resultant numbers in continued proportion.
Question 2.
If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.
Solution:
(28 – x) is the mean proportional of (23 – x) and (19-x). …[Given]
∴ -5(19 – x) = 9(28 – x)
∴ -95 + 5x = 252 – 9x
∴ 5x + 9x = 252 + 95
∴ 14x = 347
∴ x = $$\frac { 347 }{ 14 }$$
Question 3.
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution:
Let the first number be x.
∴ Third number = 26 – x
12 is the mean proportional of x and (26 – x).
∴ $$\frac { x }{ 12 }$$ = $$\frac { 12 }{ 26 – x }$$
∴ x(26 – x) = 12 x 12
∴ 26x – x2 = 144
∴ x2 – 26x + 144 = 0
∴ x2 – 18x – 8x + 144 = 0
∴ x(x – 18) – 8(x – 18) = 0
∴ (x – 18) (x – 8) = 0
∴ x = 18 or x = 8
∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18
∴ The numbers are 18, 12, 8 or 8, 12, 18.
Question 4.
If (a + b + c)(a – b + c) = a2 + b2 + c2, show that a, b, c are in continued proportion.
Solution:
(a + b + c)(a – b + c) = a2 + b2 + c2 …[Given]
∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2
∴ a2 – ab + ac + ab – b2 + be + ac – be + c2 = a2 + b2 + c2
∴ a2 + 2ac – b2 + c2 = a2 + b2 + c2
∴ 2ac – b2 = b2
∴ 2ac = 2b2
∴ ac = b2
∴ b2 = ac
∴ a, b, c are in continued proportion.
Question 5.
If $$\frac { a }{ b }$$ = $$\frac { b }{ c }$$ and a, b, c > 0, then show that,
i. (a + b + c)(b – c) = ab – c2
ii. (a2 + b2)(b2 + c2) = (ab + be)2
iii. $$\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}$$
Solution:
Let $$\frac { a }{ b }$$ = $$\frac { b }{ c }$$ = k
∴ b = ck
∴ a = bk =(ck)k
∴ a = ck2 …(ii)
i. (a + b + c)(b – c) = ab – c2
L.H.S = (a + b + c) (b – c)
= [ck2 + ck + c] [ck – c] … [From (i) and (ii)]
= c(k2 + k + 1) c (k – 1)
= c2 (k2 + k + 1) (k – 1)
R.H.S = ab – c2
= (ck2) (ck) – c2 … [From (i) and (ii)]
= c2k3 – c2
= c2(k3 – 1)
= c2 (k – 1) (k2 + k + 1) … [a3 – b3 = (a – b) (a2 + ab + b2]
∴ L.H.S = R.H.S
∴ (a + b + c) (b – c) = ab – c2
ii. (a2 + b2)(b2 + c2) = (ab + bc)2
b = ck; a = ck2
L.H.S = (a2 + b2) (b2 + c2)
= [(ck2) + (ck)2] [(ck)2 + c2] … [From (i) and (ii)]
= [c2k4 + c2k2] [c2k2 + c2]
= c2k2 (k2 + 1) c2 (k2 + 1)
= c4k2 (k2 + 1)2
R.H.S = (ab + bc)2
= [(ck2) (ck) + (ck)c]2 …[From (i) and (ii)]
= [c2k3 + c2k]2
= [c2k (k2 + 1)]2 = c4(k2 + 1)2
∴ L.H.S = R.H.S
∴ (a2 + b2) (b2 + c2) = (ab + bc)2
iii. $$\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}$$
9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of $$\frac{x+y}{x-y}, \frac{x^{2}-y^{2}}{x^{2} y^{2}}$$.
Solution:
Let a be the mean proportional of $$\frac{x+y}{x-y}$$ and $$\frac{x^{2}-y^{2}}{x^{2} y^{2}}$$
## Maharashtra Board 8th Class Maths Practice Set 2.3 Solutions Chapter 2 Parallel Lines and Transversals
Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 2.3 8th Std Maths Answers Solutions Chapter 2 Parallel Lines and Transversals.
## Practice Set 2.3 8th Std Maths Answers Chapter 2 Parallel Lines and Transversals
Question 1.
Draw a line l. Take a point A outside the line. Through point A draw a line parallel to line l.
Solution:
Steps of construction:
1. Draw a line l and take any point A outside the line.
2. Place a set-square, such that one arm of the right angle passes through A and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point A.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.
Line m is the required line parallel to line l and passing through point A.
Question 2.
Draw a line l. Take a point T outside the line. Through point T draw a line parallel to line l.
Solution:
Steps of construction:
1. Draw a line l and take any point T outside the line.
2. Place a set-square, such that one arm of the right angle passes through T and the other arm is on line l.
3. Place the second set-square as shown in the figure such that the vertex of the right angle is at point T.
4. Hold the two set-squares in place and draw a line parallel to line l through the edge of the second set-square. Name the line as m.
Line m is the required line parallel to line l and passing through point T.
Question 3.
Draw a line m. Draw a line n which is parallel to line m at a distance of 4 cm from it.
Solution:
Steps of construction:
1. Draw a line m and take any two points M and N on the line.
2. Draw perpendiculars to line m at points M and N.
3. On the perpendicular lines take points S and T at a distance 4 cm from points M and N respectively.
4. Draw a line through points S and T. Name the line as n.
Line n is parallel to line m at a distance of 4 cm from it.
## Maharashtra Board 8th Class Maths Practice Set 1.2 Solutions Chapter 1 Rational and Irrational Numbers
Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.2 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.
## Practice Set 1.2 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers
Question 1.
Compare the following numbers.
i. 7, -2
ii. 0, $$\frac { -9 }{ 5 }$$
iii. $$\frac { 8 }{ 7 }$$, 0
iv. $$-\frac{5}{4}, \frac{1}{4}$$
v. $$\frac{40}{29}, \frac{141}{29}$$
vi. $$-\frac{17}{20},-\frac{13}{20}$$
vii. $$\frac{15}{12}, \frac{7}{16}$$
viii. $$-\frac{25}{8},-\frac{9}{4}$$
ix. $$\frac{12}{15}, \frac{3}{5}$$
x. $$-\frac{7}{11},-\frac{3}{4}$$
Solution:
i. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7
ii. 0, $$\frac { -9 }{ 5 }$$
On a number line, $$\frac { -9 }{ 5 }$$ is to the left of zero.
∴ 0 > $$\frac { -9 }{ 5 }$$
iii. $$\frac { 8 }{ 7 }$$, 0
On a number line, zero is to the left of $$\frac { 8 }{ 7 }$$ .
∴ $$\frac { 8 }{ 7 }$$ > 0
iv. $$-\frac{5}{4}, \frac{1}{4}$$
We know that, a negative number is always less than a positive number.
∴ $$-\frac{5}{4}<\frac{1}{4}$$
v. $$\frac{40}{29}, \frac{141}{29}$$
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ $$\frac{40}{29}<\frac{141}{29}$$
vi. $$-\frac{17}{20},-\frac{13}{20}$$
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ $$-\frac{17}{20}<-\frac{13}{20}$$
vii. $$\frac{15}{12}, \frac{7}{16}$$
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48
Alternate method:
15 × 16 = 240
12 × 7 = 84
Since, 240 > 84
∴ 15 × 16 > 12 × 7
viii. $$-\frac{25}{8},-\frac{9}{4}$$
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8
ix. $$\frac{12}{15}, \frac{3}{5}$$
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15
x. $$-\frac{7}{11},-\frac{3}{4}$$
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44
#### Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational and Irrational Numbers Practice Set 1.2 Questions and Activities
Question 1.
Verify the following comparisons using a number line. (Textbook pg. no, .3)
i. 2 < 3 but – 2 > – 3
ii. $$\frac{5}{4}<\frac{7}{4}$$ but $$\frac{-5}{4}<\frac{-7}{4}$$
Solution:
We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3
## Maharashtra Board 9th Class Maths Part 1 Problem Set 7 Solutions Chapter 7 Statistics
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.
## Problem Set 7 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics
Question 1.
Write the correct alternative answer for each of the following questions.
i. Which of the following data is not primary ?
(A) By visiting a certain class, gathering information about attendance of students.
(B) By actual visit to homes, to find number of family members.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
(D) Review the cleanliness status of canals by actually visiting them.
(C) To get information regarding plantation of soyabean done by each farmer from the village Talathi.
ii. What is the upper class limit for the class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
(B) 35
iii. What is the class-mark of class 25 – 35?
(A) 25
(B) 35
(C) 60
(D) 30
(D) 30
iv. If the classes are 0 – 10, 10 – 20, 20 – 30, …, then in which class should the observation 10 be included?
(A) 0 – 10
(B) 10 – 20
(C) 0 – 10 and 10-20 in these 2 classes
(D) 20 – 30
(B) 10 – 20
v. If $$\overline { x }$$ is the mean of x1, x2, ……. , xn and $$\overline { y }$$ is the mean of y1, y2, ….. yn and $$\overline { z }$$ is the mean of x1,x2, …… , xn , y1, y2, …. yn , then z = ?
x1, x2, x3, ……. , xn
∴ $$\overline{x}=\frac{\sum x}{\mathrm{n}}$$
∴ n$$\overline{x}$$ = ∑x
Similarly, n$$\overline{y}$$ = ∑y
Now,
$$\text { (A) } \frac{\overline{x}+\overline{y}}{2}$$
vi. The mean of five numbers is 50, out of which mean of 4 numbers is 46, find the 5th number.
(A) 4
(B) 20
(C) 434
(D) 66
5th number = Sum of five numbers – Sum of four numbers
= (5 x 50) – (4 x 46)
= 250 – 184
= 66
(D) 66
vii. Mean of 100 observations is 40. The 9th observation is 30. If this is replaced by 70 keeping all other observations same, find the new mean.
(A) 40.6
(B) 40.4
(C) 40.3
(D) 40.7
New mean = $$\frac { 4000-30+70 }{ 100 }$$
= 40.4
(B) 40.4
viii. What is the mode of 19, 19, 15, 20, 25, 15, 20, 15?
(A) 15
(B) 20
(C) 19
(D) 25
(A) 15
ix. What is the median of 7, 10, 7, 5, 9, 10 ?
(A) 7
(B) 9
(C) 8
(D) 10
(C) 8
x. From following table, what is the cumulative frequency of less than type for the class 30 – 40?
(A) 13
(B) 15
(C) 35
(D) 22
Cumulative frequency of less than type for the class 30 – 40 = 7 + 3 + 12 + 13 = 35
(C) 35
Question 2.
The mean salary of 20 workers is ₹10,250. If the salary of office superintendent is added, the mean will increase by ₹ 750. Find the salary of the office superintendent.
Solution:
$$\text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$$
∴ The sum of all observations = Mean x Total number of observations
The mean salary of 20 workers is ₹ 10,250.
∴ Sum of the salaries of 20 workers
= 20 x 10,250
= ₹ 2,05,000 …(i)
If the superintendent’s salary is added, then mean increases by 750
new mean = 10, 250 + 750 = 11,000
Total number of people after adding superintendent = 20 + 1 = 21
∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii)
∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers
= 2, 31,00 – 2,05,000 …[From (i) and (ii)]
= 26,000
∴ The salary of the office superintendent is ₹ 26,000.
Question 3.
The mean of nine numbers is 77. If one more number is added to it, then the mean increases by 5. Find the number added in the data.
Solution:
∴ $$\text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$$
∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77
∴ sum of 9 numbers = 11 x 9 = 693 …(i)
If one more number is added, then mean increases by 5
mean of 10 numbers = 77 + 5 = 82
∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)
∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)]
= 127
∴ The number added in the data is 127.
Question 4.
The monthly maximum temperature of a city is given in degree Celsius in the following data. By taking suitable classes, prepare the grouped frequency distribution table
29.2, 29.0, 28.1, 28.5, 32.9, 29.2, 34.2, 36.8, 32.0, 31.0, 30.5, 30.0, 33, 32.5, 35.5, 34.0, 32.9, 31.5, 30.3, 31.4, 30.3, 34.7, 35.0, 32.5, 33.5.29.0. 29.5.29.9.33.2.30.2
From the table, answer the following questions.
i. For how many days the maximum temperature was less than 34°C?
ii. For how many days the maximum temperature was 34°C or more than 34°C?
Solution:
i. Number of days for which the maximum temperature was less than 34°C
= 8 + 8 + 8 = 24
ii. Number of days for which the maximum temperature was 34°C or more than 34°C
= 5 + 1 = 6
Question 5.
If the mean of the following data is 20.2, then find the value of p.
Solution:
∴ 20.2 (30 + p) = 610 + 20p
∴ 606 + 20.2p = 610 + 20p
∴ 20.2p – 20p = 610 – 606
∴ 0.2p = 4
∴ p = $$\frac { 4 }{ 0.2 }$$ = $$\frac { 40 }{ 2 }$$ = 20
∴ p = 20
Question 6.
There are 68 students of 9th standard from Model Highschool, Nandpur. They have scored following marks out of 80, in written exam of mathematics.
70, 50, 60, 66, 45, 46, 38, 30, 40, 47, 56, 68,
80, 79, 39, 43, 57, 61, 51, 32, 42, 43, 75, 43,
36, 37, 61, 71, 32, 40, 45, 32, 36, 42, 43, 55,
56, 62, 66, 72, 73, 78, 36, 46, 47, 52, 68, 78,
80, 49, 59, 69, 65, 35, 46, 56, 57, 60, 36, 37,
45, 42, 70, 37,45, 66, 56, 47
By taking classes 30 – 40, 40 – 50, …. prepare the less than type cumulative frequency table. Using the table, answer the following questions:
i. How many students have scored marks less than 80?
ii. How many students have scored marks less than 40?
iii. How many students have scored marks less than 60?
Solution:
Class
i. 66 students have scored marks less than 80.
ii. 14 students have scored marks less than 40.
iii. 45 students have scored marks less than 60.
Question 7.
By using data in example (6), and taking classes 30 – 40, 40 – 50,… prepare equal to or more than type cumulative frequency table and answer the following questions based on it.
i. How many students have scored marks 70 or more than 70?
ii. How many students have scored marks 30 or more than 30?
Solution:
i. 11 students have scored marks 70 or more than 70.
ii. 68 students have scored marks 30 or more than 30.
Question 8.
There are 10 observations arranged in ascending order as given below.
45, 47, 50, 52, JC, JC + 2, 60, 62, 63, 74. The median of these observations is 53.
Find the value of JC. Also find the mean and the mode of the data.
Solution:
i. Given data in ascending order:
45,47, 50, 52, x, JC+2, 60, 62, 63, 74.
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴ 106 = 2x + 2
∴ 106 – 2 = 2x
∴ 104 = 2x
∴ x = 52
∴ The given data becomes:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The mean of the given data is 55.9.
iii. Given data in ascending order:
45, 47, 50, 52, 52, 54, 60, 62, 63, 74.
∴ The observation repeated maximum number of times = 52
∴ The mode of the given data is 52.
Maharashtra Board Class 9 Maths Chapter 7 Statistics Problem Set 7 Intext Questions and Activities
Question 1.
To show following information diagrammatically, which type of bar- diagram is suitable?
i. Literacy percentage of four villages.
ii. The expenses of a family on various items.
iii. The numbers of girls and boys in each of five divisions.
iv. The number of people visiting a science exhibition on each of three days.
v. The maximum and minimum temperature of your town during the months from January to June.
vi. While driving a two-wheeler, number of people wearing helmets and not wearing helmet in 100 families.
(Textbook pg. no. 112)
Solution:
i. Percentage bar diagram
ii. Sub-divided bar diagram
iii. Sub-divided bar diagram
iv. Sub-divided bar diagram
v. Sub-divided bar diagram
vi. Sub-divided bar diagram
Question 2.
You gather information for several reasons. Take a few examples and discuss whether the data is primary or secondary.
(Textbook pg. no, 113)
[Students should attempt the above activity on their own.]
## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.5 Solutions Chapter 7 Statistics
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.
## Practice Set 7.5 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics
Question 1.
Yield of soyabean per acre in quintal in Mukund’s field for 7 years was 10, 7, 5,3, 9, 6, 9. Find the mean of yield per acre.
Solution:
Mean = 7
The mean of yield per acre is 7 quintals.
Question 2.
Find the median of the observations, 59, 75, 68, 70, 74, 75, 80.
Solution:
Given data in ascending order:
59, 68, 70, 74, 75, 75, 80
∴ Number of observations(n) = 7 (i.e., odd)
∴ Median is the middle most observation
Here, 4th number is at the middle position, which is = 74
∴ The median of the given data is 74.
Question 3.
The marks (out of 100) obtained by 7 students in Mathematics examination are given below. Find the mode for these marks.
99, 100, 95, 100, 100, 60, 90
Solution:
Given data in ascending order:
60, 90, 95, 99, 100, 100, 100
Here, the observation repeated maximum number of times = 100
∴ The mode of the given data is 100.
Question 4.
The monthly salaries in rupees of 30 workers in a factory are given below.
5000, 7000, 3000, 4000, 4000, 3000, 3000,
3000, 8000, 4000, 4000, 9000, 3000, 5000,
5000, 4000, 4000, 3000, 5000, 5000, 6000,
8000, 3000, 3000, 6000, 7000, 7000, 6000,
6000, 4000
From the above data find the mean of monthly salary.
Solution:
∴ The mean of monthly salary is ₹ 4900.
Question 5.
In a basket there are 10 tomatoes. The weight of each of these tomatoes in grams is as follows:
60, 70, 90, 95, 50, 65, 70, 80, 85, 95.
Find the median of the weights of tomatoes.
Solution:
Given data in ascending order:
50, 60, 65, 70, 70, 80 85, 90, 95, 95
∴ Number of observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, 5th and 6th numbers are in the middle position
∴ Median = $$\frac { 70+80 }{ 2 }$$
∴ Median = $$\frac { 150 }{ 2 }$$
∴ The median of the weights of tomatoes is 75 grams.
Question 6.
A hockey player has scored following number of goals in 9 matches: 5, 4, 0, 2, 2, 4, 4, 3,3.
Find the mean, median and mode of the data.
Solution:
i. Given data: 5, 4, 0, 2, 2, 4, 4, 3, 3.
Total number of observations = 9
∴ The mean of the given data is 3.
ii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
∴ Number of observations(n) = 9 (i.e., odd)
∴ Median is the middle most observation
Here, the 5th number is at the middle position, which is 3.
∴ The median of the given data is 3.
iii. Given data in ascending order:
0,2, 2, 3, 3, 4, 4, 4,5
Here, the observation repeated maximum number of times = 4
∴ The mode of the given data is 4.
Question 7.
The calculated mean of 50 observations was 80. It was later discovered that observation 19 was recorded by mistake as 91. What Was the correct mean?
Solution:
Here, mean = 80, number of observations = 50
$$\text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$$
∴ The sum of all observations = Mean x Total number of observations
∴ The sum of 50 observations = 80 x 50
= 4000
One of the observation was 19. However, by mistake it was recorded as 91.
Sum of observations after correction = sum of 50 observation + correct observation – incorrect observation
= 4000 + 19 – 91
= 3928
∴ Corrected mean
= 78.56
∴ The corrected mean is 78.56.
Question 8.
Following 10 observations are arranged in ascending order as follows. 2, 3 , 5 , 9, x + 1, x + 3, 14, 16, 19, 20. If the median of the data is 11, find the value of x.
Solution:
Given data in ascending order :
2, 3, 5, 9, x + 1, x + 3, 14, 16, 19, 20.
∴ Number if observations (n) = 10 (i.e., even)
∴ Median is the average of middle two observations
Here, the 5th and 6th numbers are in the middle position.
∴ $$\text { Median }=\frac{(x+1)+(x+3)}{2}$$
∴ 11 = $$\frac { 2x+4 }{ 2 }$$
∴ 22 = 2x + 4
∴ 22 – 4 = 2x
∴ 18 = 2x
∴ x = 9
Question 9.
The mean of 35 observations is 20, out of which mean of first 18 observations is 15 and mean of last 18 observations is 25. Find the 18th observation.
Solution:
$$\text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$$
∴ The sum of all observations
= Mean x Total number of observations
The mean of 35 observations is 20
∴ Sum of 35 observations = 20 x 35 = 700 ,..(i)
The mean of first 18 observations is 15
Sum of first 18 observations =15 x 18
= 270 …(ii)
The mean of last 18 observations is 25 Sum of last 18 observations = 25 x 18
= 450 …(iii)
∴ 18th observation = (Sum of first 18 observations + Sum of last 18 observations) – (Sum of 35 observations)
= (270 + 450) – (700) … [From (i), (ii) and (iii)]
= 720 – 700 = 20
The 18th observation is 20.
Question 10.
The mean of 5 observations is 50. One of the observations was removed from the data, hence the mean became 45. Find the observation which was removed.
Solution:
$$\text { Mean }=\frac{\text { The sum of all observations }}{\text { Total number of observations }}$$
∴ The sum of all observations = Mean x Total number of observations
The mean of 5 observations is 50
Sum of 5 observations = 50 x 5 = 250 …(i)
One observation was removed and mean of remaining data is 45.
Total number of observations after removing one observation = 5 – 1 = 4
Now, mean of 4 observations is 45.
∴ Sum of 4 observations = 45 x 4 = 180 …(ii)
∴ Observation which was removed
= Sum of 5 observations – Sum of 4 observations = 250 – 180 … [From (i) and (ii)]
= 70
∴ The observation which was removed is 70.
Question 11.
There are 40 students in a class, out of them 15 are boys. The mean of marks obtained by boys is 33 and that for girls is 35. Find out the mean of all students in the class.
Solution:
Total number of students = 40
Number of boys =15
∴ Number of girls = 40 – 15 = 25
The mean of marks obtained by 15 boys is 33
Here, sum of the marks obtained by boys
= 33 x 15
= 495 …(i)
The mean of marks obtained by 25 girls is 35 Sum of the marks obtained by girls = 35 x 25
= 875 …(ii)
Sum of the marks obtained by boys and girls = 495 + 875 … [From (i) and (ii)]
= 1370
∴ Mean of all the students
= 34.25
∴ The mean of all the students in the class is 34.25.
Question 12.
The weights of 10 students (in kg) are given below:
40, 35, 42, 43, 37, 35, 37, 37, 42, 37. Find the mode of the data.
Solution:
Given data in ascending order:
35, 35, 37, 37, 37, 37, 40, 42, 42, 43
∴ The observation repeated maximum number of times = 37
∴ Mode of the given data is 37 kg
Question 13.
In the following table, the information is given about the number of families and the siblings in the families less than 14 years of age. Find the mode of the data.
Solution:
Here, the maximum frequency is 25.
Since, Mode = observations having maximum frequency
∴ The mode of the given data is 2.
Question 14.
Find the mode of the following data.
Solution:
Here, the maximum frequency is 9.
Since, Mode = observations having maximum frequency
But, this is the frequency of two observations.
∴ Mode = 35 and 37
Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.5 Intext Questions and Activities
Question 1.
The first unit test of 40 marks was conducted for a class of 35 students. The marks obtained by the students were as follows. Find the mean of the marks.
40, 35, 30, 25, 23, 20, 14, 15, 16, 20, 17, 37, 37, 20, 36, 16, 30, 25, 25, 36, 37, 39, 39, 40, 15, 16, 17, 30, 16, 39, 40, 35, 37, 23, 16.
(Textbook pg, no. 123)
Solution:
Here, we can add all observations, but it will be a tedious job. It is easy to make frequency distribution table to calculate mean.
= 27.31 marks (approximately)
∴ The mean of the mark is 27.31.
## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.4 Solutions Chapter 7 Statistics
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.4 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.
## Practice Set 7.4 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics
Question 1.
Complete the following cumulative frequency table:
Solution:
Question 2.
Complete the following Cumulative Frequency Table:
Solution:
Question 3.
The data is given for 62 students in a certain class regarding their mathematics marks out of 100. Take the classes 0 – 10, 10 – 20,… and prepare frequency distribution
table and cumulative frequency table more than or equal to type.
55. 60, 81, 90, 45, 65, 45, 52, 30, 85, 20, 10,
75, 95, 09, 20, 25, 39, 45, 50, 78, 70, 46, 64,
42. 58. 31, 82, 27, 11, 78, 97, 07, 22, 27, 36,
35, 40, 75, 80, 47, 69, 48, 59, 32, 83, 23, 17,
77, 45, 05, 23, 37, 38, 35, 25, 46, 57, 68, 45.
47,49
From the prcparcd table, answer the following questions :
i. How many students obtained marks 40 or above 40?
ii. How many students obtained marks 90 or above 90?
iii. How many students obtained marks 60 or above 60?
iv. What is the cumulative frequency of equal to or more than type of the class 0 – 10?
Solution:
i. 38 students obtained marks 40 or above 40.
ii. 3 students obtained marks 90 or above 90.
iii. 19 students obtained marks 60 or above 60.
iv. Cumulative frequency of equal to or more than type of the class 0 – 10 is 62.
Question 4.
Using the data In example (3) above, prepare less than type cumulative frequency table and answer the following questions.
i. How many students obtained less than 40 marks?
ii. How many students obtained less than 10 marks?
iii. How many students obtained less than 60 marks?
iv. Find the cumulative frequency of the class 50 – 60.
Solution:
i. 24 students obtained less than 40 marks.
ii. 3 students obtained less than 10 marks.
iii. 43 students obtained less than 60 marks.
iv. Cumulative frequency of the class 50 – 60 is 43.
Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.4 Intext Questions and Activities
Question 1.
The following information is regarding marks in mathematics, obtained out of 40, scored by 50 students of 9th std. ¡n the first unit test. (Textbook pg. no. 120)
From the table, fill in the blanks in the following statements.
i. For class interval 10 – 20 the lower class limit is _____ and upper class limit is _____
ii. How many students obtained marks less than 10? 2
iii. How many students obtained marks less than 20? 2 + ____ = 14
iv. How many students obtained marks less than 30? ______ + _____ = 34
v. How many students obtained marks less than 40? ______ + ______ =50
Solution:
i. 10, 20
iii. 12
iv. 14 + 20
v. 34 + 16
Question 2.
A sports club has organised a table-tennis tournaments. The following table gives the distribution of players ages. Find the cumulative frequencies equal to or more than the lower class limit and complete the table (Textbook pg. no. 121)
Solution:
Equal to lower limit or more than lower limit type of cumulative table.
## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.3 Solutions Chapter 7 Statistics
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.
## Practice Set 7.3 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics
Question 1.
For class interval 20 – 25 write the lower class limit and the upper class limit.
Lower class limit = 20
Upper class limit = 25
Question 2.
Find the class-mark of the class 35-40.
Solution:
Class-mark
∴ Class-mark of the class 35 – 40 is 37.5
Question 3.
If class-mark is 10 and class width is 6, then find the class.
Solution:
Let the upper class limit be x and the lower class limit be y.
Class mark = 10 …[Given]
Class-mark
∴ x + y = 20 …(i)
Class width = 6 … [Given]
Class width = Upper class limit – Lower class limit
∴ x – y = 6 …(ii)
x + y = 20
x – y = 6
2x = 26
∴ x = 13
Substituting x = 13 in equation (i),
13 + y = 20
∴ y = 20 – 13
∴ y = 7
∴ The required class is 7 – 13.
Question 4.
Complete the following table.
Solution:
Let frequency of the class 14 – 15 be x then, from table,
5 + 14 + x + 4 = 35
∴ 23 + x = 35
∴ x = 35 – 23
∴ x = 12
Question 5.
In a ‘tree plantation’ project of a certain school there are 45 students of ‘Harit Sena.’ The record of trees planted by each student is given below:
3, 5, 7, 6, 4, 3, 5, 4, 3, 5, 4, 7, 5, 3, 6, 6, 5, 3, 4, 5, 7, 3, 5, 6, 4, 4, 3, 5, 6, 6, 4, 3, 5 ,7, 3, 4, 5, 7, 6, 4, 3, 5, 4, 4, 7.
Prepare a frequency distribution table of the data.
Solution:
Question 6.
The value of n upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
From this information prepare an ungrouped frequency distribution table of digits appearing after the decimal point.
Solution:
Question 7.
In the tables given below, class-mark and frequencies is given. Construct the frequency tables taking inclusive and exclusive classes.
i.
ii.
Solution:
i. Let the Lower class limit and upper class limit of the class mark 5 be x and y respectively.
∴x + y = 10
Here, class width = 15 – 5 = 10
But, Class width = Upper class limit – Lower class limit
∴ y – x = 10
∴ -x + y = 10 …(ii)
x+ y = 10
-x + y = 10
∴ 2y = 20
∴ y = 10
Substituting y = 10 in equation (i),
∴ x + 10 = 10
∴ x = 0
∴ class with class-mark 5 is 0 – 10
Similarly, we can find the remaining classes.
∴ frequency table taking inclusive and exclusive classes.
ii. Let the lower class limit and upper class limit of the class mark 22 be x andy respectively.
∴ x + y = 44 …(i)
Here, class width = 24 – 22 = 2
But, Class width = Upper class limit – Lower class limit
∴ y – x = 2
∴ -x + y = 2 …. (ii)
x + y = 44
– x + y= 2
2y = 46
∴ y = 23
Substituting y = 23 in equation (i),
∴ x + 23 = 44
∴ x = 21
∴ class with class-mark 22 is 21 – 23
Similarly, we can find the remaining classes
∴ frequency table taking inclusive and exclusive classes.
Question 8.
In a school, 46 students of 9th standard, were told to measure the lengths of the pencils in their compass-boxes in Centimetres. The data collected was as follows:
16, 15, 7, 4.5, 8.5, 5.5, 5, 6.5, 6, 10, 12, 13,
4.5, 4.9, 16, 11, 9.2, 7.3, 11.4, 12.7, 13.9, 16,
5.5, 9.9, 8.4, 11.4, 13.1, 15, 4.8, 10, 7.5, 8.5,
6.5, 7.2, 4.5, 5.7, 16, 5.7, 6.9, 8.9, 9.2, 10.2, 12.3, 13.7, 14.5, 10
By taking exclusive classes 0-5, 5-10, 10-15,…. prepare a grouped frequency distribution table.
Solution:
Question 9.
In a village, the milk was collected from 50 milkmen at a collection center in litres as given below:
27, 75, 5, 99, 70, 12, 15, 20, 30, 35, 45, 80, 77,
90, 92, 72, 4, 33, 22, 15, 20, 28, 29, 14, 16, 20,
72, 81, 85, 10, 16, 9, 25, 23, 26, 46, 55, 56, 66,
67, 51, 57, 44, 43, 6, 65, 42, 36, 7, 35
By taking suitable classes, prepare grouped frequency distribution table.
Solution:
Question 10.
38 people donated to an organisation working for differently abled persons. The amount in rupees were as follows:
101, 500, 401, 201, 301, 160, 210, 125, 175,
190, 450, 151, 101, 351, 251, 451, 151, 260,
360, 410, 150, 125, 161, 195, 351, 170, 225,
260, 290, 310, 360, 425, 420, 100, 105, 170, 250, 100
i. By taking classes 100 – 149, 150 – 199, 200 – 249… prepare grouped frequency distribution table.
ii. From the table, find the number of people who donated ₹350 or more.
Solution:
i.
ii. Number of people who donated ₹ 350 or more = 4 + 4 + 2 + 1 = 11
Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.3 Intext Questions and Activities
Question 1.
The record of marks out of 20 in Mathematics in the first unit test is as follows:
20,6, 14, 10, 13, 15, 12, 14, 17. 17, 18, 1119,
9, 16. 18, 14, 7, 17, 20, 8, 15, 16, 10, 15, 12.
18, 17, 12, 11, 11, 10, 16, 14, 16, 18, 10, 7, 17,
14, 20, 17, 13, 15, 18, 20, 12, 12, 15, 10
Answer the following questions, from the above information.
a. How many students scored 15 marks?
b. How many students scored more than 15 marks?
c. How many students scored less than 15 marks?
d. What is the lowest score of the group?
e. What is the highest score of the group? (Textbook pg. no. 114)
Solution:
a. 5 students scored 15 marks.
b. 20 students scored more than 15 marks.
c. 25 students scored less than 15 marks.
d. 6 is the lowest score of the group.
e. 20 is the highest score of the group.
Question 2.
For the above Question prepare Frequency Distribution Table. (Textbook pg. no. 115)
Solution:
## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.2 Solutions Chapter 7 Statistics
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.
## Practice Set 7.2 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics
Question 1.
Classify following information as primary or secondary data.
i. Information of attendance of every student collected by visiting every class in a school
ii. The information of heights of students was gathered from school records and sent to the head office, as it was to be sent urgently.
iii. In the village Nandpur, the information collected from every house regarding students not attending school.
iv. For science project, information of trees gathered by visiting a forest.
i. Primary data
ii. Secondary data
iii. Primary data
iv. Primary data
## Maharashtra Board 9th Class Maths Part 1 Practice Set 7.1 Solutions Chapter 7 Statistics
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 7 Statistics.
## Practice Set 7.1 Algebra 9th Std Maths Part 1 Answers Chapter 7 Statistics
Question 1.
The following table shows the number of Buses and Trucks in nearest lakh units. Draw percentage bar diagram. (Approximate the percentages to the nearest integer)
Solution:
Question 2.
In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar diagram (Approximate the percentages to the nearest integer)
Solution:
i. Sub-divided bar diagram:
ii. Percentage bar diagram:
Maharashtra Board Class 9 Maths Chapter 7 Statistics Practice Set 7.1 Intext Questions and Activities
Question 1.
A farmer has produced Wheat and Jowar in his field. The following joint bar diagram shows the production of Wheat and Jowar. From the gken diagram answer the following questions: (Textbook pg. no. 108)
i. Which crop production has increased consistently in 3 years?
ii. By how many quintals the production ofjowar has reduced in 2012 as compared to 2011?
iii. What is the difference between the production of wheat in 2010 and 2012 ?
iv. Complete the following table using this diagram.
Solution:
i. The crop production of wheat has increased consistently in 3 years.
ii. The production of jowar has reduced by 3 quintals in 2012 as compared to 2011.
iii. The difference between the production of wheat in 2010 and 2012 = 48 – 30 = 18 quintals
iv.
Question 2.
In the following table, the information of number of girls per 1000 boys is given for different states. Fill In the blanks and complete the table. (Textbook pg. no. 111)
Solution:
Draw percentage bar diagram from this information and discuss the findings from the diagram.
Question 3.
For the above given activity, the information of number of girls per 1000 boys is given for five states. The literacy percentage of these five states is given below. Assam (73%), Bihar (64%), Punjab (77%), Kerala (94%), Maharashtra (83%). Think of the number of girls and the literacy percentages in the respective states. Can you draw any conclusions from it? (Textbook pg. no. 112)
Solution:
By observing the number of girls per 1000 boys and literacy percentages in the given respective states, we can conclude that the literacy rate of girls is least in Bihar and is highest in Kerala.
## Maharashtra Board 9th Class Maths Part 1 Problem Set 5 Solutions Chapter 5 Linear Equations in Two Variables
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 5 Linear Equations in Two Variables.
## Problem Set 5 Algebra 9th Std Maths Part 1 Answers Chapter 5 Linear Equations in Two Variables
Question 1.
Choose the correct alternative answers for the following questions.
i. If 3x + 5y = 9 and 5x + 3y = 7, then what is the value of x + y ?
(A) 2
(B) 16
(C) 9
(D) 7
(A) 2
ii. ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26’. What is the mathematical form of the statement ?
(A) x – y = 8
(B) x + y = 8
(C) x + y = 23
(D) 2x + y = 21
(C) x + y = 23
iii. Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age?
(A) 20 years
(B) 15 years
(C) 10 years
(D) 5 years
(C) 10 years
Hints:
3x+ 5y = 9
5x + 3y = 7
8x + 8y = 16
∴ x + y = 2 .. [Dividing both sides by 8]
ii. Let the length of the rectangle be ‘x’ and that of breadth be ‘y’.
Perimeter of rectangle = 2[(x – 5) + (y – 5)]
∴ 26 = 2(x + y – 10)
∴ x + y – 10 = 13
∴ x + y = 23
iii. Let the age of Ajay bex years.
∴ x + (x + 5) = 25
∴ 2x = 20
∴ x = 10 years
Question 2.
Solve the following simultaneous equations.
i. 2x + y = 5 ; 3x – y = 5
ii. x – 2y = -1 ; 2x – y = 7
iii. x + y = 11 ; 2x – 3y = 7
iv. 2x + y = -2 ; 3x – y = 7
V. 2x – y = 5 ; 3x + 2y = 11
vi. x – 2y – 2 ; x + 2y = 10
Solution:
ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = 10/5
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.
ii. x – 2y = -1
∴x = 2y – 1 … .(i)
∴ 2x – y = 7 ….(ii)
Substituting x = 2y – 1 in equation (ii),
2(2y – 1) – y = 7
∴ 4y – 2 – y = 7
∴ 3y = 7 + 2
∴ 3y = 9
∴ y = 9/3
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 1
∴ x = 2(3) – 1
∴ x = 6 – 1 = 5
∴ (5, 3) is the solution of the given equations.
iii. x + y = 11
∴ x = 11 – y …(i)
2x – 3y = 7 …….(ii)
Substituting x = 11 -y in equation (ii),
2(11 – y) – 3y = 7
∴ 22 – 2y – 3y = 1
∴ 22 – 5y = 7
∴ 22 – 7 = 5y
∴ 15 = 5y
∴ y = $$\frac { 15 }{ 5 }$$
∴ y = 3
Substituting y = 3 in equation (i),
x = 11 – y
∴ x = 11 – 3 = 8
∴ (8, 3) is the solution of the given equations.
iv. 2x + y = -2 …(i)
3x – y = 7 …(ii)
2x + y = -2
+ 3x – y = l
5x = 5
∴ x = $$\frac { 5 }{ 5 }$$
∴ x = 1
Substituting x = 1 in equation (i),
2x + y = -2
∴ 2(1) +y = -2
2 + y = -2
∴ y = – 2 – 2
∴ y = -4
∴ (1, -4) is the solution of the given equations.
v. 2x – y = 5
∴ -y = 5 – 2x
∴ y = 2x – 5 …(i)
3x + 2y = 11 ……(ii)
Substituting y = 2x – 5 in equation (ii),
3x + 2(2x – 5) = 11
∴ 3x + 4x- 10= 11
∴ 7x = 11 + 10
∴ 7x = 21
∴ x = $$\frac { 21 }{ 7 }$$
∴ x = 3
Substituting x = 3 in equation (i),
y = 2x – 5
∴ y = 2(3) – 5
∴ y = 6 – 5 = 1
∴(3,1) is the solution of the given equations.
vi. x – 2y = -2
∴ x = 2y – 2 …(i)
x + 2y = 10 …..(ii)
Substituting x = 2y – 2 in equation (ii),
2y – 2 + 2y = 10
∴ 4y = 10 + 2
∴ 4y= 12
∴ y = $$\frac { 12 }{ 7 }$$
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 2
∴ x = 2(3) – 2
∴ x = 6 – 2 = 4
∴ (4, 3) is the solution of the given equations.
Question 3.
By equating coefficients of variables, solve the following equations. [3 Marks each]
i. 3x – 4y = 7 ; 5x + 2y = 3
ii. 5x + ly= 17 ; 3x – 2y = 4
iii. x – 2y = -10 ; 3x – 3y = -12
iv. 4x+y = 34 ; x + 4y = 16
Solution:
i. 3x – 4y = 7 …(i)
5x + 2y = 3 ….(ii)
Multiplying equation (ii) by 2,
10x + 4y = 6 …(iii)
∴ x = 1
Substituting x = 1 in equation (i),
3x – 4y = 7
∴ 3(1) – 4y = 7
∴ 3 – 4y = 7
∴ 3 – 7 = 4y
∴ -4 = 4y
∴ y = $$\frac { -4 }{ 4 }$$
∴ y = -1
∴ (1, -1) is the solution of the given equations.
ii. 5x + 7y = 17 …(i)
3x – 2y = 4 ….(ii)
Multiplying equation (i) by 2,
10x + 14y = 34 …(iii)
Multiplying equation (ii) by 7,
21x – 14y = 28 …..(iv)
∴ x = 2
Substituting x = 2 in equation (ii),
3x – 2y = 4
∴ 3(2) – 2y = 4
∴ 6 – 2y = 4
∴ 6 – 4 = 2y
∴ 2 = 2y
∴ y = $$\frac { 2 }{ 2 }$$
∴ y = 1
∴ (2,1) is the solution of the given equations.
iii. x – 2y = -10 ….(i)
3x – 5y = -12 …….(ii)
Multiplying equation (i) by 3,
3x – 6y = -30 …(iii)
Subtracting equation (ii) from (iii),
∴ y = 18
Substituting y = 18 in equation (i),
x – 2y = -10
∴ x – 2(18) = -10
∴ x – 36 = -10
∴ x = -10 + 36 = 26
∴ (26, 18) is the solution of the given equations.
iv. 4x + y = 34 …(i)
x + 4y = 16 …… (ii)
Multiplying equation (i) by 4,
16x + 4y = 136 …(iii)
Subtracting equation (ii) from (iii),
x = 8
Substituting x = 8 in equation (i),
4x + y = 34
∴ 4(8) + y = 34
∴ 32 + y = 34
∴ y = 34 – 32 = 2
∴ (8, 2) is the solution of the given equations.
Question 4.
Solve the following simultaneous equations.
Solution:
i. $$\frac{x}{3}+\frac{y}{4}=4$$
Multiplying both sides by 12,
4x + 3y = 48 …(i)
$$\frac{x}{2}-\frac{y}{4}=1$$
Multiplying both sides by 8,
4x – 2y = 8 …..(ii)
Subtracting equation (ii) from (i),
∴ y = 8
Substituting y = 8 in equation (ii),
4x – 2y = 8
∴ 4x – 2(8) = 8
∴ 4x – 16 = 8
∴ 4x = 8+ 16
∴ 4x = 24
∴ x = $$\frac { 24 }{ 4 }$$
∴ x = 6
∴ (6, 8) is the solution of the given equations.
ii. $$\frac { x }{ 3 }$$ + 5y = 13
Multiplying both sides by 3,
x + 15y = 39 …(i)
2x + $$\frac { y }{ 2 }$$ =19
Multiplying both sides by 2,
4x + y = 38 …….(ii)
Multiplying equation (i) by 4,
4x + 60y = 156 …(iii)
Subtracting equation (ii) from (iii),
4x + 60y =156 4x + y= 38
∴ y = 2
Substituting y = 2 in equation (i),
x + 15y = 39
∴ x+ 15(2) = 39
∴ x + 30 = 39
∴ x = 39 – 30 = 9
∴ (9,2) is the solution of the given equations.
iii. $$\frac { 2 }{ x }$$ + $$\frac { 3 }{ y }$$ = 13
Multiplying both sides by 5,
Question 5.
A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
According to the first condition,
a two digit number is 3 more than 4 times the sum of its digits.
10y + x = 4(x + y) + 3
∴ 10y + x = 4x + 4y + 3
∴ x – 4x + 10y – 4y = 3
∴ – 3x + 6y = 3
Dividing both sides by -3,
x – 2y = -1 …(i)
According to the second condition,
if 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18= 10x + y
∴ x – 10x + 10y – y = -18
∴ – 9x + 9y = -18
Dividing both sides by – 9,
x – y = 2 ……(ii)
Subtracting equation (ii) from (i),
∴ y = 3
Substituting y = 3 in equation (ii),
x – y = 2
∴ x – 3 = 2
∴ x = 2 + 3 = 5
∴ Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
The required number is 35.
Question 6.
The total cost of 8 books and 5 pens is ₹ 420 and the total cost of 5 books and 8 pens is ₹321. Find the cost of 1 book and 2 pens.
Solution:
Let the cost of one book be ₹ x and the cost of one pen be ₹ y.
According to the first condition,
the total cost of 8 books and 5 pens is ₹ 420.
∴ 8x + 5y = 420 …(i)
According to the second condition, the total cost of 5 books and 8 pens is ₹ 321.
5x + 8y = 321 ….(ii)
Multiplying equation (i) by 5,
40x + 25y = 2100 …(iii)
Multiplying equation (ii) by 8,
40x + 64y = 2568 … (iv)
Subtracting equation (iii) from (iv),
∴ y = 12
Substituting y = 12 in equation (i),
8x + 5y = 420
∴ 8x + 5(12) = 420
∴ 8x + 60 = 420
∴ 8x = 420 – 60
∴ 8x = 360
∴ x = $$\frac { 360 }{ 8 }$$
∴ x = 45
Cost of 1 book and 2 pens = x + 2y
= 45 + 2(12)
= 45 + 24
= ₹69
∴ The cost of 1 book and 2 pens is ₹69.
Question 7.
The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves ₹ 200, find the income of each.
Solution:
Let the income of first person be ₹ x and that of second person be ₹ y.
According to the first condition,
the ratio of their incomes is 9 : 7.
∴ $$\frac { x }{ y }$$ = $$\frac { 9 }{ 7 }$$
∴ 7x = 9y
∴ 7x – 9y = 0 …….(i)
Each person saves ₹ 200.
Expenses of first person = Income – Saving = x – 200
Expenses of second person = y – 200
According to the second condition,
the ratio of their expenses is 4 : 3
∴ $$\frac { x – 200 }{ y – 200 }$$ = $$\frac { 4 }{ 3 }$$
∴ 3(x – 200) = 4(y – 200)
∴ 3x – 600 = 4y – 800
∴ 3x – 4y = – 800 + 600
∴ 3x – 4y = -200 …(ii)
Multiplying equation (i) by 4,
28x-36y =0 …(iii)
Multiplying equation (ii) by 9,
27x-36y = -1800 …(iv)
Subtracting equation (iv) from (iii),
Substituting x = 1800 in equation (i),
7x – 9y = 0
∴ 7(1800) – 9y = 0
∴ 9y = 7 x 1800
∴ y = $$\frac { 7 \times 1800 }{ 9 }$$
y = 7 x 200
∴ y = 1400
∴ The income of first person is ₹ 1800 and that of second person is ₹ 1400.
Question 8.
If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 9 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Solution:
Let the length of the rectangle be ‘x’ units and the breadth of the rectangle be ‘y’ units.
Area of the rectangle = xy sq. units
length of the rectangle is reduced by 5 units
∴ length = x – 5
breadth of the rectangle is increased by 3 units
∴ breadth = y + 3
area of the rectangle is reduced by 9 square units
∴ area of the rectangle = xy – 9
According to the first condition,
(x – 5)(y + 3) = xy – 9
∴ xy + 3x – 5y – 15 = xy – 9
∴ 3x – 5y = -9 + 15
∴ 3x – 5y = 6 …(i)
length of the rectangle is reduced by 3 units
∴ length = x – 3
breadth of the rectangle is increased by 2 units
∴ breadth = y + 2
area of the rectangle is increased by 67 square units
∴ area of the rectangle = xy + 61
According to the second condition,
(x – 3)(y + 2) = xy + 67
∴ xy + 2x – 3y – 6 = xy + 67
∴ 2x – 3y = 67 + 6
∴ 2x – 3y = 73 …(ii)
Multiplying equation (i) by 3,
9x – 15y = 18 . ..(iii)
Multiplying equation (ii) by 5,
10x – 15y = 365 …(iv)
Subtracting equation (iii) from (iv), 10x- 15y= 365 9x-15y= 18
Substituting x = 347 in equation (ii),
2x – 3y = 73
∴ 2(347) – 3y = 73
∴ 694 – 73 = 3y
∴ 621 = 3y
∴ y = $$\frac { 621 }{ 3 }$$
∴ y = 207
∴ The length and breadth of rectangle are 347 units and 207 units respectively.
Question 9.
The distance between two places A and B on a road is 70 kilometres. A car starts from A and the other from B. If they travel in the same direction, they will meet in 7 hours. If they travel towards each other they will meet in 1 hour, then find their speeds.
Solution:
Let the speed of the car starting from A (first car) be ‘x’ km/hr and that starting from B (second car) be ‘y’ km/hr. (x > y)
According to the first condition,
Distance covered by the first car in 7 hours = 7x km
Distance covered by the second car in 7 hours = 7y km
If the cars are travelling in the same direction, 7x – 7y = 70
Dividing both sides by 7,
x – y = 10 …(i)
According to the second condition,
Distance covered by the first car in
1 hour = x km
Distance covered by the second car in 1 hour = y km
If the cars are travelling in the opposite direction
x + y = 70 …(ii)
∴ x = 40
Substituting x = 40 in equation (ii), x + y = 70
∴ 40 +y = 70
∴ y = 70 – 40 = 30
∴ The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.
Question 10.
The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.
According to the given condition,
the sum of a two digit number and the number
obtained by interchanging its digits is 99.
∴ 10y + x + 10x +y = 99
∴ 11x + 11y = 99
Dividing both sides by 11,
x + y = 9
If y = 1, then x = 8
If y = 2, then x = 7
If y = 3, then x = 6 and so on.
∴ The number can be 18, 27, 36, … etc.
Maharashtra Board Class 9 Maths Chapter 5 Linear Equations in Two Variables Practice Set 5 Intext Questions and Activities
Question 1.
On the glasses of following spectacles, write numbers such that (Textbook pg. no. 82)
i. Their sum is 42 and difference is 16.
ii. Their sum is 37 and difference is 11.
iii. Their sum is 54 and difference is 20.
iv. Their sum is … and difference is … .
ii. x + y = 37 and x – y = 11
∴ x = 24, y = 13
iii. x +y = 54 and x – y = 20
∴ x = 37, y =17
Question 2.
There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs. (Textbook pg. no. 92)
Here, if we take a pair of any two equations, we get following 6 pairs.
1. equation (i) and (ii)
2. equation (i) and (iii)
3. equation (i) and (iv)
4. equation (ii) and (iii)
5. equation (ii) and (iv)
6. equation (iii) and (iv)
Solution of each pair given above is (21, 15).
Here, all four equations are of same rectangle. By solving any two equations simultaneously, we get length and breadth of the rectangle.
Question 3.
Find the function.
∴ Given function = $$\frac { 6 }{ 14 }$$
Verify the answer obtained. (Textbook pg. no. 92)
For the fraction$$\frac { 6 }{ 14 }$$, if the numerator is multiplied by 3 and 3 is subtracted from the denominator, we get fraction $$\frac { 18 }{ 11 }$$.
Similarly, for the fraction $$\frac { 6 }{ 14 }$$, if the numerator is increased by 8 and the denominator is doubled, we get fraction $$\frac { 1 }{ 2 }$$.
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2022-09-29 08:56:26
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http://rspa.royalsocietypublishing.org/content/455/1987/2471
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# Growth and decay of random Fibonacci sequences
Mark Embree, Lloyd N. Trefethen
## Abstract
For 0 < β < β* ≈ 0.70258, solutions to the random recurrencexn+1 =xn±βxn−1 decay exponentially asn→ ∞ with probability one, whereas for β > β*, they grow exponentially. By formulating the problem as a Markov chain involving random matrix products and computing its invariant measure (a fractal) the Lyapunov constant σ(β) = limn→ ∞ |xn|1/n is determined numerically for a wide range of values β, and its dependence on β is observed to be non-smooth. (The limit is defined in the almost sure sense.) This generalizes recent work of Viswanath, who proved σ (1) = 1.131 988 24.... By a simple rescaling, these results also apply to the more general random recurrencexn+1 = αxn± βxn−1 for fixed α and β. These random recurrence relations have links with many fields, including ergodic theory, dynamical systems, heavy–tailed statistics, spectral theory, continued fractions, and condensed matter physics.
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2015-12-01 22:09:47
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https://blueollie.wordpress.com/2008/12/19/end-of-the-semester-obamas-picks-etc/
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# blueollie
## End of the Semester; Obama’s picks, etc.
Workout notes: yoga class then 6 miles on the track (6.38) in 58:11; 8 lap splits (7.5 laps per mile in lane 3): 9:58, 9:43, 9:30, 9:35, 9:48, 9:35. This was similar to my December 2 workout though 43 seconds slower though this run was much, much easier.
What I am noticing is that I feel like death during the first mile (out of breath, etc. and don’t really start feeling good until 20 minutes into it).
Academia Yep, the final grades are in. I decided to post them on blackboard; but this program is set up so that you load one column of data at a time. One of my snowflakes panicked when he saw his numerical course score and final exam score (both not-so-hot) but I hadn’t loaded the final grade column yet. He got a big “benefit of the doubt” for the final grade, though given his “when can I come in and see you” e-mail…well I am very tempted to change the grade that I assigned him thereby removing the “benefit of the doubt”. (yeah, I am going to come in over break to listen to whining…NOT!!!!)
Mathematics and Science
Interestingly enough, integrals came up in a blog that I read:
Having recently slogged through grading an enormous pile of graduate-level problem sets, I am compelled to share one of the most useful tricks I learned in graduate school.
This probably seems silly to the theoretical physicists in the audience, who have a habit of changing variables and units to the point where everything is dimensionless and equals one. However, in astrophysics, you frequently are integrating over real physical quantities (numbers of photons, masses of stars, luminosities of galaxies, etc) that still have units attached. While students typically do an admirable job of setting up the necessary integrals, they frequently go off the rails when actually evaluating the integrals, as they valiantly try to propagate all those extra factors.
Here’s an example of what I mean. Suppose you want to calculate some sort of rate constant for photoionization, that when multiplied by the density of atoms, will give you the rate of photo-ionizations per volume. These sorts of rates are always density times velocity times cross section:
Go ahead and read the column; it has some interesting mathematics in it. Finally, they boil down one integral to:
$\frac{8\pi}{c^2} \, \frac{(x_0\,kT)^3}{h^3} \, \sigma_0 \int_{x_0}^\infty \, \frac{1}{e^x-1} \, \frac{{\rm d}x}{x}$
Have fun! Ok, I haven’t tried this yet, but I strongly suspect that this is a residue integral or perhaps could be changed into a probability function by a clever change of variable.
Algebra
I also finished my abstract algebra grades. Here is an interesting little problem: roughly speaking, a field is a system of “numbers” in which every non zero element has a multiplicative inverse. For example, the integers don’t form a field (only 1 and -1 have a multiplicative inverse in the integers) but the rational numbers do form a field.
It is a known fact that the integers mod p (p a prime) do form a field.
So what about this question: consider the set $S$ consisting of $a + bi$ where $a$ and $b$ are mod p integers (p a prime) and $i$ is the square root of $-1$ and multiplication is defined as it is with complex numbers. What conditions must be met for $S$ to be a field? Hint: p = 3 and p = 7 yields fields whereas p = 5 does not!
Note: this problem is not at all difficult but you do have to play around a bit.
Science and Politics
I’ve been critical of some of Obama’s choices. But I’ve been ecstatic over others; from Science Debate 2008:
We want to congratulate President-elect Obama on continuing to assemble an outstanding science team.
A few days ago we told you about the appointment of Steven Chu as Energy Secretary.
Today we have two more outstanding appointments to announce:
1. We have learned that John Holdren will be President Obama’s Science Advisor. John has an excellent knowledge of science policy, and a deep understanding of how the public needs the government to engage on science policy issues. He is a recent past president of the AAAS and an early and ardent Science Debate 2008 supporter. You can watch a 1-minute video he did for us last February, promoting a primary science debate at the Franklin Institute in Philadelphia.
2. Jane Lubchenco, we’re told, will head up President Obama’s National Oceanic and Atmospheric Admninistration (NOAA). She is an outstanding choice with a deep background in marine biology. Jane is also a past AAAS president, and also an early supporter of Science Debate 2008. When we issued candidate invitations to an Oregon debate, Jane was a close advisor. Here’s a 1-minute video of her.
Why are these choices so important? Here is a 1-minute video President-elect Obama’s transition chief, John Podesta, did for us earlier this year that answers that. Clearly, this is a man who gets it, working, it seems, for a president who gets it.
There is widespread hope that these excellent picks (in science) will continue.
In all, I haven’t educated myself on all of his picks, though I’ll do so over the weekend. Here are my initial impressions:
* Department of Agriculture
Tom Vilsack: I am leaning favorable here.
* Department of Commerce
Bill Richardson: I am very happy; smart and multi-talented
* Department of Defense
Robert Gates: Yes, a Bush holdover but he appears to be smart and competent.
* Department of Education
Arne Duncan: Chicago area superintendent; I don’t know much.
* Department of Energy
Steven Chu: Nobel Laureate in Physics; head of the Livermore lab, passionate and out of this world brilliant.
* Department of Health and Human Services
Tom Daschle: lean favorable but I’ll have to learn more; I didn’t follow his Senate career all that closely though he was minority leader at one time.
* Department of Homeland Security
* Department of Housing and Urban Development
Shaun Donovan: really don’t know other than he is from New York.
* Department of the Interior
Ken Salazar: mixed; he has political skill and is the Colorado Senator; from reading the stuff he wrote I don’t get the impression that he is all that intellectual. Here is an example.
* Department of Justice
Eric Holder: I don’t know all that much about him.
* Department of Labor
* Department of State
Hillary Clinton: Love the pick. She is sharp, knowledgeable and well respected around the world.
* Department of Transportation
Ray LaHood: Barf; this guy is a mediocrity where I think excellence is needed. But one commenter at Prairie State Blue said:
Politically speaking. LaHood will help deflect some of the inevitable Republican criticism of these massive spending projects. And it’s not the job of a Cabinet Secretary to come up with fresh, imaginative solutions to complex problems. In fact, someone too in love with his own ideas could easily become a liability.
That is probably the best defense of this pick that I’ve seen.
* Department of the Treasury
Tim Geithner: seems sharp, creative and full of energy. I like the pick, but we’ll wait and see.
* Department of Veterans Affairs
Eric Shinseki: love this pick; he is best noted for falling from the grace of the Bush administration by telling the truth about how hard the Iraq occupation would be.
Ok, the Rick Warren thing
No, I don’t like the fact that is @sshole was chosen at all; I think that this wooish bigot sets a terrible example for the country. But remember this: politically speaking, this is one way to throw a bone to the yahoos (e. g., right wing evangelicals) without giving them one tiny bit of policy.
Some takes: Friendly Atheist (aka Hemant Mehta ) says to “chill out” (though he doesn’t like the pick either). He asks: “who gave Bush’s invocations? ” Of course, few really remember that.
Markos Moulitsas (founder of the Daily Kos) is disgusted but sees a silver lining:
I’m with Aravosis on this one.
I’m reading a lot about how Obama “reaches out” to his adversaries, and that’s why he’s building a track record of inviting avowed homophobes to stand front and center at his campaign events and now his inauguration.
Okay, I’m game. So we know being a gay-basher doesn’t disqualify you from a seat at the Obama table – in fact, it seems to be an outright qualification for proving Obama’s post-partisanship. If Obama prides himself on reaching out to all sides of every debate, then why is it that Obama has never sat down with, or promoted at his events, an avowed racist or anti-Semite?
Yeah. Where is David Duke’s invitation? Or as Blue Texan notes, when do Phelps and Hagee get their invitations? Heck, throw up Tom Tancredo up there for good measure, so us Latinos can feel some of the hate!
On the other hand, John Cole also has a point:
You would think that folks would be ecstatic that they have a President-elect who, for the first time I can remember, is publicly, openly, and repeatedly stating that he supports equal rights for gays and lesbians and that the Christian right is wrong about these issues […]
I think the Warren choice is bullshit, but if we want a silver lining, it’s that the President of the United States has just said:
I am fierce advocate for equality for gay and—well, let me start by talking about my own views. I think it is no secret that I am a fierce advocate for equality for gay and lesbian Americans. It is something I have been consistent on and something I intend to continue to be consistent on during my presidency.
That’s not a bad silver lining. But let me add this to John — Obama wouldn’t be out there making perhaps the strongest statement in support of gays and lesbians by a president (though he’s still not technically one, I know) if it wasn’t for the sturm and drang this choice generated. It is precisely this backlash that has forced Obama to clearly affirm his commitment to equality. And it will be continued pressure that will force him to do the right thing on the issue.
If we shut up, he’ll take the path of least resistance. And that path of least resistance is kowtowing to the conservative media, the clueless punditocracy, and bigots like Warren.
And let’s face it: it is Barack Obama’s style to blow off his strongest supporters and reach out to the other side.
Let me start by talking about my own views. I think that it is no secret that I am a fierce advocate for equality for gay and lesbian Americans. It is something that I have been consistent on and something that I intend to continue to be consistent on during my presidency. What I’ve also said is that it is important for America to come together, even though we may have disagreements on certain social issues, and I would note that a couple of years ago, I was invited to Rick Warren’s church to speak despite his awareness that I held views that were entirely contrary to his when it came to gay and lesbian rights, when it came to issues like abortion. Nevertheless I had an opportunity to speak, and that dialogue I think is part of what my campaign’s been all about, that we’re not going to agree on every single issue, but what we have to do is to be able to create an atmosphere where we can disagree without being disagreeable, and then focus on those things that we hold in common as Americans.
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2017-04-28 04:17:25
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https://www.zbmath.org/?q=ra%3Aackermann.nils+se%3A37
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## Localized concentration of semi-classical states for nonlinear Dirac equations.(English)Zbl 1309.35102
Summary: The present paper studies concentration phenomena of the semiclassical approximation of a massive Dirac equation with general nonlinear self-coupling: $-i \hbar \alpha \cdot \nabla w+a \beta w + V (x) w = g (| w| ) w.$ Compared with some existing issues, the most interesting results obtained here are twofold: the solutions concentrating around local minima of the external potential; and the nonlinearities assumed to be either super-linear or asymptotically linear at the infinity. As a consequence one sees that, if there are $$k$$ bounded domains $$\Lambda_j \subset \mathbb{R}^3$$ such that $$-a < \min_{\Lambda_j} V=V(x_j) < \min_{\partial \Lambda_j}V$$, $$x_j\in\Lambda_j$$, then the $$k$$-families of solutions $$w_\hbar^j$$ concentrate around $$x_j$$ as $$\hbar\to 0$$, respectively. The proof relies on variational arguments: the solutions are found as critical points of an energy functional. The Dirac operator has a continuous spectrum which is not bounded from below and above, hence the energy functional is strongly indefinite. A penalization technique is developed here to obtain the desired solutions.
### MSC:
35Q41 Time-dependent Schrödinger equations and Dirac equations 81V10 Electromagnetic interaction; quantum electrodynamics 35P15 Estimates of eigenvalues in context of PDEs 35B38 Critical points of functionals in context of PDEs (e.g., energy functionals)
Full Text:
### References:
[1] Ackermann, N., A nonlinear superposition principle and multibump solution of periodic Schrödinger equations, J. Funct. Anal., 234, 423-443, (2006) · Zbl 1126.35057 [2] Ambrosetti, A.; Badiale, M.; Cignolani, S., Semi-classical states of nonlinear shrödinger equations. arch, Rational Mech. Anal., 140, 285-300, (1997) · Zbl 0896.35042 [3] Ambrosetti, A.; Felli, V.; Malchiodi, A., Ground states of nonlinear Schrödinger equations with potentials vanishing at infinity, J. Eur. Math. Soc., 7, 117-144, (2005) · Zbl 1064.35175 [4] Byeon, J.; Jeanjean, L., Standing waves for nonlinear Schrödinger equations with a general nonlinearity, Arch. Rational Mech. Anal., 185, 185-200, (2007) · Zbl 1132.35078 [5] Byeon, J.; Wang, Z.Q., Standing waves with a critical frequency for nonlinear Schrödinger equations, Arch. Rational Mech. Anal., 165, 295-316, (2002) · Zbl 1022.35064 [6] Dautray R., Lions J.L.: Mathematical Analysis and Numerical Methods for Science and Technology, vol. 3. Springer, Berlin (1990) · Zbl 0683.35001 [7] Del Pino, M.; Felmer, P., Local mountain passes for semilinear ellipitc problems in unbounded domains, Calc. Var. Partial Differential Equations, 4, 121-137, (1996) · Zbl 0844.35032 [8] Del Pino, M., Felmer, P.: Multi-peak bound states for nonlinear Schrödinger equations, Annales de l’Institut Henri Poincare (C) Non Linear Analysis, vol. 15. No. 2. Elsevier, Masson, pp. 127-149, (1998) · Zbl 0524.35002 [9] Ding Y.H.: Variational Methods for Strongly Indefinite Problems, Interdiscip. Math. Sci., vol. 7. World Scientific Publ, Singapore (2007) · Zbl 1133.49001 [10] Ding, Y.H., Semi-classical ground states concentrating on the nonlinear potentical for a Dirac equation, J. Differ. Equ., 249, 1015-1034, (2010) · Zbl 1193.35161 [11] Ding, Y.H., Lee, C., Ruf, B.: On semiclassical states of a nonlinear Dirac equation. Proc. R. Soc. Edinburgh Sect. A Math. 143.04, 765-790 (2013) · Zbl 1304.35590 [12] Ding, Y.H.; Liu, Xiaoying, Semi-classical limits of ground states of a nonlinear Dirac equation, J. Differ. Equ., 252, 4962-4987, (2012) · Zbl 1236.35133 [13] Ding, Y.H.; Ruf, B., Existence and concentration of semi-classical solutions for Dirac equations with critical nonlinearities, SIAM J. Math. Anal., 44, 3755-3785, (2012) · Zbl 1259.35171 [14] Ding, Y.H.; Wei, J.C., Stationary states of nonlinear Dirac equations with general potentials, Rev. Math. Phys., 20, 1007-1032, (2008) · Zbl 1170.35082 [15] Ding, Y.H.; Wei, J.C.; Xu, T., Existence and concentration of semi-classical solutions for a nonlinear Maxwell-Dirac system, J. Math. Phys., 54, 061505, (2013) · Zbl 1282.81073 [16] Ding, Y.H.; Xu, T., On semi-classical limits of ground states of a nonlinear Maxwell-Dirac system, Calc. Var. Partial Differ. Equ., 51, 17-44, (2014) · Zbl 1297.35197 [17] Ding, Y.H.; Xu, T., On the concentration of semi-classical states for a nonlinear Dirac-Klein-Gordon system, J. Differ. Equ., 256, 1264-1294, (2014) · Zbl 1283.35099 [18] Esteban, M.J.; Séré, E., Stationary states of the nonlinear Dirac equation: a variational approach, Commun. Math. Phys., 171, 323-350, (1995) · Zbl 0843.35114 [19] Finkelstein, R.; LeLevier, R.; Ruderman, M., Nonlinear spinor fields, Phys. Rev., 83, 326-332, (1951) · Zbl 0043.21603 [20] Finkelstein, R.; Fronsdal, C.; Kaus, P., Nonlinear spinor field, Phys. Rev., 103, 1571-1579, (1956) · Zbl 0073.44705 [21] Floer, A.; Weinstein, A., Nonspreading wave packets for the cubic Schrödinger equation with a bounded potential, J. Funct. Anal., 69, 397-408, (1986) · Zbl 0613.35076 [22] Gilbarg D., Trudinger N.S.: Elliptic Partial Differential Equations of Second Order. Springer, Berlin (1998) · Zbl 0361.35003 [23] Ivanenko, D.D., Notes to the theory of interaction via particles, Zh.Éksp. Teor. Fiz., 8, 260-266, (1938) · Zbl 0021.27604 [24] Jeanjean, L.; Tanaka, K., Singularly perturbed elliptic problems with superlinear or asymptotically linear nonlinearities, Calc. Var. Partial Differ. Equ., 21, 287-318, (2004) · Zbl 1060.35012 [25] Lions, P.L.: The concentration-compactness principle in the calculus of variations: the locally compact case, Part II. AIP Anal. non linéaire1, 223-283 · Zbl 0704.49004 [26] Miguel, R.; Hugo, T., Solutions with multiple spike patterns for an elliptic system, Calc. Var. Partial Differ. Equ., 31, 1-25, (2008) · Zbl 1143.35027 [27] Miguel, R.; Yang, J.F., Spike-layered solutions for an elliptic system with Neumann boundary conditions, Trans. Am. Math. Soc., 357, 3265-3284, (2005) · Zbl 1136.35046 [28] Oh, Y.G., Existence of semi-classical bound states of nonlinear Schrödinger equations with potentials of the class ($$V$$)_{$$a$$}, Commun. Partial Differ. Equ., 13, 1499-1519, (1988) · Zbl 0702.35228 [29] Oh, Y.G., On positive multi-lump bound states of nonlinear Schrödinger equations under multiple well potential, Commun. Math. Phys., 131, 223-253, (1990) · Zbl 0753.35097 [30] Rabinowitz, P.H.: Minimax Methods in Critical Point Theory with Applications to Differential Equations. Amer. Math. Soc., Providence, 1986 · Zbl 0609.58002 [31] Rabinowitz, P.H., On a class of nonlinear Schrödinger equations, Z. Angew. Math. Phys., 43, 270-291, (1992) · Zbl 0763.35087 [32] Stein E.M.: Singular integrals and differentiability properties of functions, vol. 2. Princeton University Press, Princeton (1970) · Zbl 0207.13501 [33] Simon, B., Schrödinger semigroups, Bull. Am. Math. Soc. (N.S.), 7, 447-526, (1982) · Zbl 0524.35002 [34] Wang, X., On concentration of positive bound states of nonlinear Schrödinger equations, Commun. Math. Phys., 153, 229-244, (1993) · Zbl 0795.35118
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2022-05-24 03:25:46
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https://www.cryptologie.net/home/all/13
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Hey! I'm David, a security engineer at the Blockchain team of Facebook, previously a security consultant for the Cryptography Services of NCC Group. I'm also the author of the Real World Cryptography book. This is my blog about cryptography and security and other related topics that I find interesting.
# SHA-3 vs the world: slides posted July 2017
I just gave a talk at Defcon on SHA-3, derived functions and Strobe. The talk was recorded but I don't expect it to be online in the next 6 months, in the mean time the slides are available here .
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# BEAST: An Explanation of the CBC Attack on TLS posted July 2017
I made a video explaining the BEAST attack. As usual it's more of an overview so head over to something like this for more details.
# Defcon: SHA-3 vs the world posted July 2017
I'll be speaking at the Defcon Crypto village again this year (my talk of last year is here).
It will be about recent hash functions, it will focus a lot on SHA-3 and it will try to avoid any of the recent controversy on which hash function is better (it will be hard but I will try to be neutral and fair).
It'll be recorded if you can't make it. If you can make it, head to the crypto village at 11am on Friday. See the Defcon Crypto Village schedule here. And here is the abstract:
Since Keccak has been selected as the winner of the SHA-3 competition in 2012, a myriad of different hash functions have been trending. From BLAKE2 to KangarooTwelve we'll cover what hash functions are out there, what is being used, and what you should use. Extending hash functions, we’ll also discover STROBE, a symmetric protocol framework derived from SHA-3.
# How big are TLS records during a handshake? posted June 2017
I've asked some TLS size questions on Twitter and got some nice results :]
People mostly got it right for the Client Hello. But it wasn't as easy for the Server Hello.
Client Hello → 212 bytes
Server Hello → 66 bytes
These are just numbers I got from a TLS 1.2 handshake with a random website. These numbers are influenced by the browser I use and the configuration of the server. But they should be close to that range anyway as the structure of a Client Hello or a Server Hello are quite simple.
A better question would be: what is the bigger message? and the Client Hello would always win. This is because the Server Hello only replies with one choice from the list of choices the client proposed. For example the server will choose only one ciphersuite from the 13 suites the client proposed. The server will choose one curve from the 3 different curves proposed by the client. The server will choose a single signature algorithm from the client's 10 propositions. And on and on...
Everyone (mostly) got this one!
Certificate → 2540 bytes
Obviously, this is the biggest message of the handshake by far. The number I got is from receiving two different certificates where each certificate is about a thousand bytes. This is because servers tend to send the full chain of certificates to the client, longer chains will increase the size of this message. Probably why there are propositions for a certification compression extension.
ServerKeyExchange → 338 bytes
ClientKeyExchange → 75 bytes
Both of these messages include the peer's public key during ephemeral key exchanges. But the ServerKeyExchange additionally contains the parameters of the key exchange algorithm and a signature of the server's public key. In my case, the signature was done with RSA-2048 and of size 256 bytes, while the NIST p256 public keys were of size 65 bytes.
Using ECDSA for signing, signatures could have been smaller. Using FFDH for the key agreement, public keys could have been bigger.
Tim Dierks also mentioned that using RSA-10000 would have drastically increased the size of the ServerKeyExchange.
Maybe a better question, again, would be which one is the bigger message.
People mostly got it right here!
The rest of the handshake is negligible:
ChangeCipherSpec is just 6 bytes indicating a switch to encryption, it will always be the same size no matter what kind of handshake you went through, most of its length comes from the record's header.
Finished is 45 bytes. Its content is a MAC of the handshake transcript, but an additional MAC is added to protect the integrity of the ciphertext (ciphertext expansion). Remember, Finished is the first (and only) encrypted message in a handshake.
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# Crypto training at Black Hat USA posted June 2017
I'll be back in Vegas this year to give the crypto training of Black Hat. The class is not full yet so hurry up if that is something that interests you.
It will be a blend of culture, exercises and technical dives. For 2 days, students get to learn all the cool crypto attacks, get to dive into some of them deeply, and get to interact via numerous exercises.
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# Noise+Strobe=Disco posted June 2017
Noise is a protocol framework allowing you to build different lightweight TLS-like handshakes depending on your use case. Benefits are a short code size, very few dependencies, simplicity of the security guarantees and analysis. It focuses primarily on the initial asymmetric phase of the setup of a secure channel, but does leave you with two ciphers that you can use to read and write on both sides of the connection. If you want to know more, I wrote a readable implementation, and have a tutorial video.
Strobe is a protocol framework as well, focusing on the symmetric part of the protocol. Its simplicity boils down to only using one cryptographic primitive: the duplex construction. Which allows developers to benefit from an ultra short cryptographic code base supporting their custom-made symmetric protocols as well as their different needs of cryptographic functions. Indeed, Strobe can be used as well to instantiate a hash function, a key derivation function, a pseudo-random number generator, a message authentication code, an authenticated encryption with associated data cipher, etc... If you want to know more, I wrote a readable implementation and Mike Hamburg gave a talk at RWC.
Noise+Strobe=Disco. One of Noise's major character is that it keeps a running hash, digesting every message and allowing every new handshake message to mix the transcript in its encryption while authenticating previous messages received and sent. Strobe works like that naturally. Its duplex function absorbs every calls being made to the underlying primitive (the Keccak permutation), to the extent that every new operation is influenced by any operation that happened previously. These two common traits in Strobe and Noise led me to pursue a merge between the two: what if that running hash and symmetric state in Noise was simply Strobe's primitive? And what if at the end of a handshake Noise would just spew out two Strobe's objects also depending on the handshake transcript? I talked to Trevor Perrin about it and his elegant suggestion for a name (Disco) and my curiosity led to an implementation of what it would look like.
This is of course highly experimental. I modified the Noise's specification to see how much I could remove/simplify from it and the result is already enjoyable.
I've discussed the changes on the mailing list. But simply put: the CipherState has been removed, the SymmetricState has been replaced by calls to Strobe. This leaves us only with one object: the HandshakeState. Every symmetric algorithm has been removed (HDKF, HMAC, HASH, AEAD). The specification looks way shorter, while the Disco implementation is more than half the size of the Noise implementation.
The Strobe's calls naturally absorbs every operation, and can encrypt/decrypt the handshake messages even if no shared secret has been negotiated (with a non-keyed duplex construction), which simplifies corner cases where you would have to test if you have already negotiated a shared secret or not.
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# Readable implementation of the Noise protocol framework posted June 2017
I wrote an implementation of the Noise Protocol Framework. If you don't know what that is, it is a framework to create lightweight TLS-like protocols. If you do not want to use TLS because it is unnecessarily complicated, and you know what you're doing, Noise is the solution. You have different patterns for different usecase and everything is well explained for you to implement it smoothly.
To learn more about Noise you can also check this screencast I shot last year:
My current research includes merging this framework with the Strobe protocol framework I've talked about previously.
This led me to first implement a readable and understandable version of Noise here.
Note that this is highly experimental and it has not been thoroughly tested.
I also had to deviate from the specification when naming things because Golang:
• doesn't use snake_case, but Noise does.
• capitalizes function names to make them public, Noise does it for different reasons.
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# SIMD instructions in Go posted June 2017
One awesome feature of Go is cross-compilation. One limitation is that we can only choose to build for some pre-defined architectures and OS, but we can't build per CPU-model. In the previous post I was talking about C programs, where the user actually chooses the CPU model when calling the Make. Go could probably have something like that but it wouldn't be gooy. One solution is to build for every CPU models anyway, and decide later what is good to be used. So one assembly code for SSE2, one code for AVX, one code for AVX-512.
Note that we do not need to use SSE3/SSE4 (or AVX2) as the interesting functions are contained in SSE2 (respectively AVX) which will have more support and be contained in greater versions of SSE (respectively AVX) anyway.
The official Blake2 implementation in Go actually uses SIMD instructions. Looking at it is a good way to see how SIMD coding works in Go.
In _amd64.go, they use the builtin init() function to figure out at runtime what is supported by the host architecture:
func init() {
useAVX2 = supportsAVX2()
useAVX = supportsAVX()
useSSE4 = supportsSSE4()
}
Which are calls to assembly functions detecting what is supported either via:
1. a CPUID call directly for SSE4.
2. calls to Golang's runtime library for AVX and AVX2.
In the second solution, the runtime variables seems to be undocumented and only available since go1.7, they are probably filled via cpuid calls as well. Surprisingly, the internal/cpu package already has all the necessary functions to detect flavors of SIMD. See an example of use in the bytes package.
And that's it! Blake2's hashBlocks() function then dynamically decides which function to use at runtime:
func hashBlocks(h *[8]uint64, c *[2]uint64, flag uint64, blocks []byte) {
if useAVX2 {
hashBlocksAVX2(h, c, flag, blocks)
} else if useAVX {
hashBlocksAVX(h, c, flag, blocks)
} else if useSSE4 {
hashBlocksSSE4(h, c, flag, blocks)
} else {
hashBlocksGeneric(h, c, flag, blocks)
}
}
Because Go does not have intrisic functions for SIMD, these are implemented directly in assembly. You can look at the code in the relevant _amd64.s file. Now it's kind of tricky because Go has invented its own assembly language (based on Plan9) and you have to find out things the hard way. Instructions like VINSERTI128 and VPSHUFD are the SIMD instructions. MMX registers are M0...M7, SSE registers are X0...X15, AVX registers are Y0, ..., Y15. MOVDQA is called MOVO (or MOVOA) and MOVDQU is called MOVOU. Things like that.
As for AVX-512, Go probably still doesn't have instructions for that. So you'll need to write the raw opcodes yourself using BYTE (like here) and as explained here.
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2021-03-06 17:19:43
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http://davidwalsh.name/git-aliases
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Creating Git Aliases
By on
I create shortcuts for everything. I create variables which act as text shortcuts, shortcuts in TextMate to generate CSS/JS/HTML, and bash scripts so I don't have to type in the same commands over and over again. So why should version control software be any different? I commit and push religiously so I create Git aliases to save myself a few keystrokes.
Sample Aliases
#make "com" alias for "commit"
git config alias.com commit
#make "co" alias for checkout
git config alias.co checkout
#make "br" alias for branch
git config alias.br branch
# When you want to see just the differences of one function in one file in two different commits
git config alias.funcdiff '!sh -c "git show \"\$1:\$3\" | sed -n \"/^[^ \t].*\$4(/,/^}/p\" > .tmp1 && git show \"\$2:\$3\" | sed -n \"/^[^ \t].*\$4(/,/^}/p\" > .tmp2 &&
git diff --no-index .tmp1 .tmp2"' -
These are just some sample Git aliases. You can view more detailed (and by detailed I mean brain-numbing) examples of git aliases at the Git Wiki.
Recent Features
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I was inspired when I first saw Addy Osmani's original ShineTime blog post. The hover sheen effect is simple but awesome. When I started my blog redesign, I really wanted to use a sheen effect with my logo. Using two HTML elements and...
• Write Better JavaScript with Promises
You've probably heard the talk around the water cooler about how promises are the future. All of the cool kids are using them, but you don't see what makes them so special. Can't you just use a callback? What's the big deal? In this article, we'll...
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• Create a Clearable TextBox with the Dojo Toolkit
Usability is a key feature when creating user interfaces; it's all in the details. I was recently using my iPhone and it dawned on my how awesome the "x" icon is in its input elements. No holding the delete key down. No pressing it a...
• Truly Responsive Images with responsive-images.js
Responsive web design is something you hear a lot about these days. The moment I really started to get into responsive design was a few months ago when I started to realise that 'responsive' is not just about scaling your websites to the size of your...
Discussion
1. Nice! I use ci for commit though, more like Subversion. I also have di for diff, st for status and sta for stash.
2. Terry
I’m trying to get the funcdiff alias created. When I try to call it with something like:
git funcdiff “rel\path\to\file” functionName
or
both give me an error like:
fatal: Path ‘funcitonName’ does not exist in ‘sha1-2′.
fatal: Invalid object name ‘rel\path\to\file’.
I’ve Googled everywhere for anyone using this alias to see a sample syntax, but no one has one. Have you used it successfully?
3. Dan
You can make aliases for branch name, too. For example:
4. Peter
Dan, command
ruins git repo in windows. Because it overwrites “HEAD” with “head” in “.git” directory
For ones, that tried this: just rename the file back
5. Shahzeb
Very useful, thanks!
6. Thanks for the commands,
I need to know how to create an alias for paths.
For example if I need to write “git push PATH” instead of “git push gitolite@codex…”
Regards
7. [...] Creating Git Aliases [...]
Wrap your code in <pre class="{language}"></pre> tags, link to a GitHub gist, JSFiddle fiddle, or CodePen pen to embed!
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2014-10-26 05:57:05
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https://socratic.org/questions/how-do-i-find-the-sides-of-a-right-triangle-using-only-the-trig-functions-sin-co
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# How do I find the sides of a right triangle using only the trig functions sin cos and tan?
For example the $\left(3 , 4 , 5\right)$ right triangle has the same $\sin , \cos , \mathmr{and} \tan$ values as the $\left(6 , 8 , 10\right)$ right triangle (but obviously the lengths of the sides are different).
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2020-03-31 08:13:24
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https://howlingpixel.com/i-en/Ephemeris_time
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# Ephemeris time
The term ephemeris time (often abbreviated ET) can in principle refer to time in connection with any astronomical ephemeris. In practice it has been used more specifically to refer to:
1. a former standard astronomical time scale adopted in 1952 by the IAU,[1] and superseded in the 1970s.[2] This time scale was proposed in 1948, to overcome the drawbacks of irregularly fluctuating mean solar time. The intent was to define a uniform time (as far as was then feasible) based on Newtonian theory (see below: Definition of ephemeris time (1952)). Ephemeris time was a first application of the concept of a dynamical time scale, in which the time and time scale are defined implicitly, inferred from the observed position of an astronomical object via the dynamical theory of its motion.[3]
2. a modern relativistic coordinate time scale, implemented by the JPL ephemeris time argument Teph, in a series of numerically integrated Development Ephemerides. Among them is the DE405 ephemeris in widespread current use. The time scale represented by Teph is closely related to, but distinct (by an offset and constant rate) from, the TCB time scale currently adopted as a standard by the IAU (see below: JPL ephemeris time argument Teph).[4]
Most of the following sections relate to the ephemeris time of the 1952 standard.
An impression has sometimes arisen that ephemeris time was in use from 1900: this probably arose because ET, though proposed and adopted in the period 1948–1952, was defined in detail using formulae that made retrospective use of the epoch date of 1900 January 0 and of Newcomb's Tables of the Sun.[5][6]
The ephemeris time of the 1952 standard leaves a continuing legacy, through its ephemeris second which became closely duplicated in the length of the current standard SI second (see below: Redefinition of the second).
## History (1952 standard)
Ephemeris time (ET), adopted as standard in 1952, was originally designed as an approach to a uniform time scale, to be freed from the effects of irregularity in the rotation of the earth, "for the convenience of astronomers and other scientists", for example for use in ephemerides of the Sun (as observed from the Earth), the Moon, and the planets. It was proposed in 1948 by G M Clemence.[7]
From the time of John Flamsteed (1646–1719) it had been believed that the Earth's daily rotation was uniform. But in the later nineteenth and early twentieth centuries, with increasing precision of astronomical measurements, it began to be suspected, and was eventually established, that the rotation of the Earth (i.e. the length of the day) showed irregularities on short time scales, and was slowing down on longer time scales. The evidence was compiled by W de Sitter (1927)[8] who wrote "If we accept this hypothesis, then the 'astronomical time', given by the earth's rotation, and used in all practical astronomical computations, differs from the 'uniform' or 'Newtonian' time, which is defined as the independent variable of the equations of celestial mechanics". De Sitter offered a correction to be applied to the mean solar time given by the Earth's rotation to get uniform time.
Other astronomers of the period also made suggestions for obtaining uniform time, including A Danjon (1929), who suggested in effect that observed positions of the Moon, Sun and planets, when compared with their well-established gravitational ephemerides, could better and more uniformly define and determine time.[9]
Thus the aim developed, to provide a new time scale for astronomical and scientific purposes, to avoid the unpredictable irregularities of the mean solar time scale, and to replace for these purposes Universal Time (UT) and any other time scale based on the rotation of the Earth around its axis, such as sidereal time.
The American astronomer G M Clemence (1948)[7] made a detailed proposal of this type based on the results of the English Astronomer Royal H Spencer Jones (1939).[10] Clemence (1948) made it clear that his proposal was intended "for the convenience of astronomers and other scientists only" and that it was "logical to continue the use of mean solar time for civil purposes".[11]
De Sitter and Clemence both referred to the proposal as 'Newtonian' or 'uniform' time. D Brouwer suggested the name 'ephemeris time'.[12]
Following this, an astronomical conference held in Paris in 1950 recommended "that in all cases where the mean solar second is unsatisfactory as a unit of time by reason of its variability, the unit adopted should be the sidereal year at 1900.0, that the time reckoned in this unit be designated ephemeris time", and gave Clemence's formula (see Definition of ephemeris time (1952)) for translating mean solar time to ephemeris time.
The International Astronomical Union approved this recommendation at its 1952 general assembly.[12][13] Practical introduction took some time (see Use of ephemeris time in official almanacs and ephemerides); ephemeris time (ET) remained a standard until superseded in the 1970s by further time scales (see Revision).
During the currency of ephemeris time as a standard, the details were revised a little. The unit was redefined in terms of the tropical year at 1900.0 instead of the sidereal year;[12] and the standard second was defined first as 1/31556925.975 of the tropical year at 1900.0,[12][14] and then as the slightly modified fraction 1/31556925.9747 instead,[15] finally being redefined in 1967/8 in terms of the cesium atomic clock standard (see below).
Although ET is no longer directly in use, it leaves a continuing legacy. Its successor time scales, such as TDT, as well as the atomic time scale IAT (TAI), were designed with a relationship that "provides continuity with ephemeris time".[16] ET was used for the calibration of atomic clocks in the 1950s.[17] Close equality between the ET second with the later SI second (as defined with reference to the cesium atomic clock) has been verified to within 1 part in 1010.[18]
In this way, decisions made by the original designers of ephemeris time influenced the length of today's standard SI second, and in turn, this has a continuing influence on the number of leap seconds which have been needed for insertion into current broadcast time scales, to keep them approximately in step with mean solar time.
## Definition (1952)
Ephemeris time was defined in principle by the orbital motion of the Earth around the Sun,[12] (but its practical implementation was usually achieved in another way, see below).
Its detailed definition depended on Simon Newcomb's Tables of the Sun (1895),[5] interpreted in a new way to accommodate certain observed discrepancies:
In the introduction to Tables of the Sun the basis of the tables (p. 9) includes a formula for the Sun's mean longitude, at a time indicated by interval T (in Julian centuries of 36525 mean solar days[19]) reckoned from Greenwich Mean Noon on 0 January 1900:
Ls = 279° 41' 48".04 + 129,602,768".13T +1".089T2 . . . . . (1)
Spencer Jones' work of 1939[10] showed that the positions of the Sun actually observed, when compared with those obtained from Newcomb's formula, show the need for the following correction to the formula to represent the observations:
ΔLs = + 1".00 + 2".97T + 1".23T2 + 0.0748B
(where "the times of observation are in Universal time, not corrected to Newtonian time", and 0.0748B represents an irregular fluctuation calculated from lunar observations[20]).
Thus a conventionally corrected form of Newcomb's formula, to incorporate the corrections on the basis of mean solar time, would be the sum of the two preceding expressions:
Ls = 279° 41' 49".04 + 129,602,771".10T +2".32T2 +0.0748B . . . . . (2)
Clemence's 1948 proposal did not adopt a correction of this kind in terms of mean solar time: instead, the same numbers were used as in Newcomb's original uncorrected formula (1), but now in a reverse sense, to define the time and time scale implicitly, based on the real position of the Sun:
Ls = 279° 41' 48".04 + 129,602,768".13E +1".089E2 . . . . . (3)
where the time variable, here represented as E, now represents time in ephemeris centuries of 36525 ephemeris days of 86400 ephemeris seconds. The 1961 official reference put it this way: "The origin and rate of ephemeris time are defined to make the Sun's mean longitude agree with Newcomb's expression"[21]
From the comparison of formulae (2) and (3), both of which express the same real solar motion in the same real time but on different time scales, Clemence arrived at an explicit expression, estimating the difference in seconds of time between ephemeris time and mean solar time, in the sense (ET-UT):
${\displaystyle \delta t=+24^{s}.349+72^{s}.3165T+29^{s}.949T^{2}+1.821B}$ . . . . . (4)[20]
Clemence's formula, now superseded by more modern estimations, was included in the original conference decision on ephemeris time. In view of the fluctuation term, practical determination of the difference between ephemeris time and UT depended on observation. Inspection of the formulae above shows that the (ideally constant) unit of ephemeris time such as the ephemeris second has been for the whole of the twentieth century very slightly shorter than the corresponding (but not precisely constant) unit of mean solar time (which besides its irregular fluctuations tends gradually to increase), consistently also with the modern results of Morrison and Stephenson[22] (see article ΔT).
## Implementations
### Secondary realizations by lunar observations
Although ephemeris time was defined in principle by the orbital motion of the Earth around the Sun,[23] it was usually measured in practice by the orbital motion of the Moon around the Earth.[24] These measurements can be considered as secondary realizations (in a metrological sense) of the primary definition of ET in terms of the solar motion, after a calibration of the mean motion of the Moon with respect to the mean motion of the Sun.[25]
Reasons for the use of lunar measurements were practically based: the Moon moves against the background of stars about 13 times as fast as the Sun's corresponding rate of motion, and the accuracy of time determinations from lunar measurements is correspondingly greater.
When ephemeris time was first adopted, time scales were still based on astronomical observation, as they always had been. The accuracy was limited by the accuracy of optical observation, and corrections of clocks and time signals were published in arrear.
### Secondary realizations by atomic clocks
A few years later, with the invention of the cesium atomic clock, an alternative offered itself. Increasingly, after the calibration in 1958 of the cesium atomic clock by reference to ephemeris time,[17] cesium atomic clocks running on the basis of ephemeris seconds began to be used and kept in step with ephemeris time. The atomic clocks offered a further secondary realization of ET, on a quasi-real time basis[25] that soon proved to be more useful than the primary ET standard: not only more convenient, but also more precisely uniform than the primary standard itself. Such secondary realizations were used and described as 'ET', with an awareness that the time scales based on the atomic clocks were not identical to that defined by the primary ephemeris time standard, but rather, an improvement over it on account of their closer approximation to uniformity.[26] The atomic clocks gave rise to the atomic time scale, and to what was first called Terrestrial Dynamical Time and is now Terrestrial Time, defined to provide continuity with ET.[16]
The availability of atomic clocks, together with the increasing accuracy of astronomical observations (which meant that relativistic corrections were at least in the foreseeable future no longer going to be small enough to be neglected),[27] led to the eventual replacement of the ephemeris time standard by more refined time scales including terrestrial time and barycentric dynamical time, to which ET can be seen as an approximation.
## Revision of time scales
In 1976 the IAU resolved that the theoretical basis for its current (1952) standard of Ephemeris Time was non-relativistic, and that therefore, beginning in 1984, Ephemeris Time would be replaced by two relativistic timescales intended to constitute dynamical timescales: Terrestrial Dynamical Time (TDT) and Barycentric Dynamical Time (TDB).[28] Difficulties were recognized, which led to these being in turn superseded in the 1990s by time scales Terrestrial Time (TT), Geocentric Coordinate Time GCT(TCG) and Barycentric Coordinate Time BCT(TCB).[16]
## JPL ephemeris time argument Teph
High-precision ephemerides of sun, moon and planets were developed and calculated at the Jet Propulsion Laboratory (JPL) over a long period, and the latest available were adopted for the ephemerides in the Astronomical Almanac starting in 1984. Although not an IAU standard, the ephemeris time argument Teph has been in use at that institution since the 1960s. The time scale represented by Teph has been characterized as a relativistic coordinate time that differs from Terrestrial Time only by small periodic terms with an amplitude not exceeding 2 milliseconds of time: it is linearly related to, but distinct (by an offset and constant rate which is of the order of 0.5 s/a) from the TCB time scale adopted in 1991 as a standard by the IAU. Thus for clocks on or near the geoid, Teph (within 2 milliseconds), but not so closely TCB, can be used as approximations to Terrestrial Time, and via the standard ephemerides Teph is in widespread use.[4]
Partly in acknowledgement of the widespread use of Teph via the JPL ephemerides, IAU resolution 3 of 2006[29] (re-)defined Barycentric Dynamical Time (TDB) as a current standard. As re-defined in 2006, TDB is a linear transformation of TCB. The same IAU resolution also stated (in note 4) that the "independent time argument of the JPL ephemeris DE405, which is called Teph" (here the IAU source cites[4]), "is for practical purposes the same as TDB defined in this Resolution". Thus the new TDB, like Teph, is essentially a more refined continuation of the older ephemeris time ET and (apart from the < 2 ms periodic fluctuations) has the same mean rate as that established for ET in the 1950s.
## Use in official almanacs and ephemerides
Ephemeris time based on the standard adopted in 1952 was introduced into the Astronomical Ephemeris (UK) and the American Ephemeris and Nautical Almanac, replacing UT in the main ephemerides in the issues for 1960 and after.[30] (But the ephemerides in the Nautical Almanac, by then a separate publication for the use of navigators, continued to be expressed in terms of UT.) The ephemerides continued on this basis through 1983 (with some changes due to adoption of improved values of astronomical constants), after which, for 1984 onwards, they adopted the JPL ephemerides.
Previous to the 1960 change, the 'Improved Lunar Ephemeris' had already been made available in terms of ephemeris time for the years 1952-1959[31] (computed by W J Eckert from Brown's theory with modifications recommended by Clemence (1948)).
## Redefinition of the second
Successive definitions of the unit of ephemeris time are mentioned above (History). The value adopted for the 1956/1960 standard second:
the fraction 1/31 556 925.9747 of the tropical year for 1900 January 0 at 12 hours ephemeris time.
was obtained from the linear time-coefficient in Newcomb's expression for the solar mean longitude (above), taken and applied with the same meaning for the time as in formula (3) above. The relation with Newcomb's coefficient can be seen from:
1/31 556 925.9747 = 129 602 768.13 / (360×60×60×36 525×86 400).
Caesium atomic clocks became operational in 1955, and quickly confirmed the evidence that the rotation of the earth fluctuated randomly. This confirmed the unsuitability of the mean solar second of Universal Time as a measure of time interval for the most precise purposes. After three years of comparisons with lunar observations, Markowitz et al. (1958) determined that the ephemeris second corresponded to 9 192 631 770 ± 20 cycles of the chosen cesium resonance.[17]
Following this, in 1967/68, the General Conference on Weights and Measures (CGPM) replaced the definition of the SI second by the following:
The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.
Although this is an independent definition that does not refer to the older basis of ephemeris time, it uses the same quantity as the value of the ephemeris second measured by the cesium clock in 1958. This SI second referred to atomic time was later verified by Markowitz (1988) to be in agreement, within 1 part in 1010, with the second of ephemeris time as determined from lunar observations.[18]
For practical purposes the length of the ephemeris second can be taken as equal to the length of the second of Barycentric Dynamical Time (TDB) or Terrestrial Time (TT) or its predecessor TDT.
The difference between ET and UT is called ΔT; it changes irregularly, but the long-term trend is parabolic, decreasing from ancient times until the nineteenth century,[22] and increasing since then at a rate corresponding to an increase in the solar day length of 1.7 ms per century (see leap seconds).
International Atomic Time (TAI) was set equal to UT2 at 1 January 1958 0:00:00 . At that time, ΔT was already about 32.18 seconds. The difference between Terrestrial Time (TT) (the successor to ephemeris time) and atomic time was later defined as follows:
1977 January 1.000 3725 TT = 1977 January 1.000 0000 TAI, i.e.
TT − TAI = 32.184 seconds
This difference may be assumed constant—the rates of TT and TAI are designed to be identical.
## Notes and references
1. ^ 'ESAE 1961': 'Explanatory Supplement (1961), esp. p.9.
2. ^ 'ESAA (1992)': P K Seidelmann (ed)., especially at pp.41-42 and at p.79.
3. ^ B Guinot and P K Seidelmann (1988), at p.304-5.
4. ^ a b c
5. ^ a b
6. ^ For the components of the definition including its retrospective aspect, see G M Clemence (1948), esp. p.172, and 'ESAE 1961': 'Explanatory Supplement (1961), esp. pages 69 and 87.
7. ^ a b
8. ^
9. ^
10. ^ a b
11. ^ Clemence (1948), at p. 171.
12. ESAA (1992), see page 79.
13. ^ At the IAU meeting in Rome 1952: see ESAE (1961) at sect.1C, p. 9; also Clemence (1971).
14. ^ ESAA 1992, p. 79: citing decision of International Committee for Weights and Measures (CIPM), Sept 1954.
15. ^ ESAA (1992), see page 80, citing CIPM recommendation Oct 1956, adopted 1960 by the General Conference on Weights and Measures (CGPM).
16. ^ a b c ESAA (1992), at page 42.
17. ^ a b c W Markowitz, R G Hall, L Essen, J V L Parry (1958)
18. ^ a b
19. ^ The unit of _mean solar_ day is left implicit on p.9 but made explicit on p.20 of Newcomb (1895).
20. ^ a b Clemence (1948), p.172, following Spencer Jones (1939).
21. ^ ESAE (1961) at p.70.
22. ^ a b
23. ^ Clemence (1948), at pp.171-3.
24. ^
25. ^ a b B Guinot & P K Seidelmann (1988), at p.305.
26. ^ W G Melbourne & others, 1968, section II.E.4-5, pages 15-16, including footnote 7, noted that the Jet Propulsion Laboratory spacecraft tracking and data reduction programs of that time (including the Single Precision Orbit Determination Program) used, as ET, the current US atomic clock time A.1 offset by 32.25 seconds. The discussion also noted that the usage was "inaccurate" (the quantity indicated was not identical with any of the other realizations of ET such as ET0, ET1), and that while A.1 gave "certainly a closer approximation to uniform time than ET1" there were no grounds for considering either the atomic clocks or any other measures of ET as (perfectly) uniform. Section II.F, pages 18-19, indicates that an improved time measure of (A.1 + 32.15 seconds), applied in the JPL Double Precision Orbit Determination Program, was also designated ET.
27. ^
29. ^ IAU 2006 resolution 3
30. ^
31. ^ "Improved Lunar Ephemeris", US Government Printing Office, 1954.
## Bibliography
1972
1972 (MCMLXXII)
was a leap year starting on Saturday of the Gregorian calendar, the 1972nd year of the Common Era (CE) and Anno Domini (AD) designations, the 972nd year of the 2nd millennium, the 72nd year of the 20th century, and the 3rd year of the 1970s decade.
Within the context of Coordinated Universal Time (UTC) it was the longest year ever, as two leap seconds were added during this 366-day year, an event which has not since been repeated. (If its start and end are defined using mean solar time [the legal time scale], its duration was 31622401.141 seconds of Terrestrial Time (or Ephemeris Time), which is slightly shorter than 1908).
Atomic clock
An atomic clock is a clock device that uses an electron transition frequency in the microwave, optical, or ultraviolet region of the electromagnetic spectrum of atoms as a frequency standard for its timekeeping element. Atomic clocks are the most accurate time and frequency standards known, and are used as primary standards for international time distribution services, to control the wave frequency of television broadcasts, and in global navigation satellite systems such as GPS.
The principle of operation of an atomic clock is based on atomic physics; it measures the electromagnetic signal that electrons in atoms emit when they change energy levels. Early atomic clocks were based on masers at room temperature. Since 2004, more accurate atomic clocks first cool the atoms to near absolute zero temperature by slowing them with lasers and probing them in atomic fountains in a microwave-filled cavity. An example of this is the NIST-F1 atomic clock, one of the national primary time and frequency standards of the United States.
The accuracy of an atomic clock depends on two factors. The first factor is temperature of the sample atoms—colder atoms move much more slowly, allowing longer probe times. The second factor is the frequency and intrinsic width of the electronic transition. Higher frequencies and narrow lines increase the precision.
National standards agencies in many countries maintain a network of atomic clocks which are intercompared and kept synchronized to an accuracy of 10−9 seconds per day (approximately 1 part in 1014). These clocks collectively define a continuous and stable time scale, the International Atomic Time (TAI). For civil time, another time scale is disseminated, Coordinated Universal Time (UTC). UTC is derived from TAI, but has added leap seconds from UT1, to account for variations in the rotation of the Earth with respect to the solar time.
Barycentric Coordinate Time
Barycentric Coordinate Time (TCB, from the French Temps-coordonnée barycentrique) is a coordinate time standard intended to be used as the independent variable of time for all calculations pertaining to orbits of planets, asteroids, comets, and interplanetary spacecraft in the Solar system. It is equivalent to the proper time experienced by a clock at rest in a coordinate frame co-moving with the barycenter of the Solar system: that is, a clock that performs exactly the same movements as the Solar system but is outside the system's gravity well. It is therefore not influenced by the gravitational time dilation caused by the Sun and the rest of the system.
TCB was defined in 1991 by the International Astronomical Union, in Recommendation III of the XXIst General Assembly. It was intended as one of the replacements for the problematic 1976 definition of Barycentric Dynamical Time (TDB). Unlike former astronomical time scales, TCB is defined in the context of the general theory of relativity. The relationships between TCB and other relativistic time scales are defined with fully general relativistic metrics.
Because the reference frame for TCB is not influenced by the gravitational potential caused by the Solar system, TCB ticks faster than clocks on the surface of the Earth by 1.550505 × 10−8 (about 490 milliseconds per year). Consequently, the values of physical constants to be used with calculations using TCB differ from the traditional values of physical constants (The traditional values were in a sense wrong, incorporating corrections for the difference in time scales). Adapting the large body of existing software to change from TDB to TCB is an ongoing task, and as of 2002 many calculations continue to use TDB in some form.
Time coordinates on the TCB scale are conventionally specified using traditional means of specifying days, carried over from non-uniform time standards based on the rotation of the Earth. Specifically, both Julian Dates and the Gregorian calendar are used. For continuity with its predecessor Ephemeris Time, TCB was set to match ET at around Julian Date 2443144.5 (1977-01-01T00Z). More precisely, it was defined that TCB instant 1977-01-01T00:00:32.184 exactly corresponds to the International Atomic Time (TAI) instant 1977-01-01T00:00:00.000 exactly, at the geocenter. This is also the instant at which TAI introduced corrections for gravitational time dilation.
Barycentric Dynamical Time
Barycentric Dynamical Time (TDB, from the French Temps Dynamique Barycentrique) is a relativistic coordinate time scale, intended for astronomical use as a time standard to take account of time dilation when calculating orbits and astronomical ephemerides of planets, asteroids, comets and interplanetary spacecraft in the Solar System. TDB is now (since 2006) defined as a linear scaling of Barycentric Coordinate Time (TCB). A feature that distinguishes TDB from TCB is that TDB, when observed from the Earth's surface, has a difference from Terrestrial Time (TT) that is about as small as can be practically arranged with consistent definition: the differences are mainly periodic, and overall will remain at less than 2 milliseconds for several millennia.TDB applies to the Solar-System-barycentric reference frame, and was first defined in 1976 as a successor to the (non-relativistic) former standard of ephemeris time (adopted by the IAU in 1952 and superseded 1976). In 2006, after a history of multiple time-scale definitions and deprecation since the 1970s, a redefinition of TDB was approved by the IAU. The 2006 IAU redefinition of TDB as an international standard expressly acknowledged that the long-established JPL ephemeris time argument Teph, as implemented in JPL Development Ephemeris DE405, "is for practical purposes the same as TDB defined in this Resolution" (By 2006, ephemeris DE405 had already been in use for a few years as the official basis for planetary and lunar ephemerides in the Astronomical Almanac; it was the basis for editions for 2003 through 2014; in the edition for 2015 it is superseded by DE430).
Coordinated Universal Time
DUT1
The time correction DUT1 (sometimes also written DUT) is the difference between Universal Time (UT1), which is defined by Earth's rotation, and Coordinated Universal Time (UTC), which is defined by a network of precision atomic clocks.
DUT1 = UT1 − UTCUTC is maintained via leap seconds, such that DUT1 remains within the range −0.9 s < DUT1 < +0.9 s. The reason for this correction is partly that the rate of rotation of the Earth is not constant, due to tidal braking and the redistribution of mass within the Earth, including its oceans and atmosphere, and partly because the SI second (as now used for UTC) was already, when adopted, a little shorter than the current value of the second of mean solar time.Forecast values of DUT1 are published by IERS Bulletin A.
Weekly updated values of DUT1 with 0.1 s precision are broadcast by several time signal services, including WWV and MSF. These services transmit one pulse per second of some sort. To represent positive DUT1 values from +0.1 to +0.8 seconds, the pulses sent during seconds 1 through 8 are "emphasized" in some way, generally by transmitting a double pulse. The number of emphasized pulses gives the value of DUT1. Negative DUT1 values, from −0.1 to −0.8 seconds, are similarly represented by emphasizing pulses 9 through 16. For example, a DUT1 value of −0.4 would be transmitted by emphasizing pulses 9 through 12.
The Russian time signal RWM transmits an additional correction dUT1 in 0.02 s increments. Positive values of dUT1 from +0.02 to +0.08 s are encoded by emphasizing pulses 21 through 24; negative values are encoded by emphasizing pulses 31 through 34. The actual value of DUT1 is approximated by the sum of the transmitted DUT1 + dUT1.
The longwave RBU time signal also transmits dUT1.
Dynamical time scale
In time standards, dynamical time is the time-like argument of a dynamical theory; and a dynamical time scale in this sense is the realization of a time-like argument based on a dynamical theory: that is, the time and time scale are defined implicitly, inferred from the observed position of an astronomical object via a theory of its motion. A first application of this concept of dynamical time was the definition of the ephemeris time scale (ET).In the late 19th century it was suspected, and in the early 20th century it was established, that the rotation of the Earth (i.e. the length of the day) was both irregular on short time scales, and was slowing down on longer time scales. The suggestion was made, that observation of the position of the Moon, Sun and planets and comparison of the observations with their gravitational ephemerides would be a better way to determine a uniform time scale. A detailed proposal of this kind was published in 1948 and adopted by the IAU in 1952 (see Ephemeris time - history).
Using data from Newcomb's Tables of the Sun (based on the theory of the apparent motion of the Sun by Simon Newcomb, 1895, as retrospectively used in the definition of ephemeris time), the SI second was defined in 1960 as:
the fraction 1/31,556,925.9747 of the tropical year for 1900 January 0 at 12 hours ephemeris time.Caesium atomic clocks became operational in 1955, and their use provided further confirmation that the rotation of the earth fluctuated randomly. This confirmed the unsuitability of the mean solar second of Universal Time as a precision measure of time interval. After three years of comparisons with lunar observations it was determined that the ephemeris second corresponded to 9,192,631,770 ± 20 cycles of the caesium resonance. In 1967/68 the length of the SI second was redefined to be 9,192,631,770 cycles of the caesium resonance, equal to the previous measurement result for the ephemeris second (see Ephemeris time - redefinition of the second).
In 1976, however, the IAU resolved that the theoretical basis for ephemeris time was wholly non-relativistic, and therefore, beginning in 1984 ephemeris time would be replaced by two further time scales with allowance for relativistic corrections. Their names, assigned in 1979, emphasized their dynamical nature or origin, Barycentric Dynamical Time (TDB) and Terrestrial Dynamical Time (TDT). Both were defined for continuity with ET and were based on what had become the standard SI second, which in turn had been derived from the measured second of ET.
During the period 1991–2006, the TDB and TDT time scales were both redefined and replaced, owing to difficulties or inconsistencies in their original definitions. The current fundamental relativistic time scales are Geocentric Coordinate Time (TCG) and Barycentric Coordinate Time (TCB); both of these have rates that are based on the SI second in respective reference frames (and hypothetically outside the relevant gravity well), but on account of relativistic effects, their rates would appear slightly faster when observed at the Earth's surface, and therefore diverge from local earth-based time scales based on the SI second at the Earth's surface. Therefore, the currently defined IAU time scales also include Terrestrial Time (TT) (replacing TDT, and now defined as a re-scaling of TCG, chosen to give TT a rate that matches the SI second when observed at the Earth's surface), and a redefined Barycentric Dynamical Time (TDB), a re-scaling of TCB to give TDB a rate that matches the SI second at the Earth's surface.
Geocentric Coordinate Time
Geocentric Coordinate Time (TCG - Temps-coordonnée géocentrique) is a coordinate time standard intended to be used as the independent variable of time for all calculations pertaining to precession, nutation, the Moon, and artificial satellites of the Earth. It is equivalent to the proper time experienced by a clock at rest in a coordinate frame co-moving with the center of the Earth: that is, a clock that performs exactly the same movements as the Earth but is outside the Earth's gravity well. It is therefore not influenced by the gravitational time dilation caused by the Earth.
TCG was defined in 1991 by the International Astronomical Union, in Recommendation III of the XXIst General Assembly. It was intended as one of the replacements for the ill-defined Barycentric Dynamical Time (TDB). Unlike former astronomical time scales, TCG is defined in the context of the general theory of relativity. The relationships between TCG and other relativistic time scales are defined with fully general relativistic metrics.
Because the reference frame for TCG is not rotating with the surface of the Earth and not in the gravitational potential of the Earth, TCG ticks faster than clocks on the surface of the Earth by a factor of about 7.0 × 10−10 (about 22 milliseconds per year). Consequently, the values of physical constants to be used with calculations using TCG differ from the traditional values of physical constants. (The traditional values were in a sense wrong, incorporating corrections for the difference in time scales.) Adapting the large body of existing software to change from TDB to TCG is a formidable task, and as of 2002 many calculations continue to use TDB in some form.
Time coordinates on the TCG scale are conventionally specified using traditional means of specifying days, carried over from non-uniform time standards based on the rotation of the Earth. Specifically, both Julian Dates and the Gregorian calendar are used. For continuity with its predecessor Ephemeris Time, TCG was set to match ET at around Julian Date 2443144.5 (1977-01-01T00Z). More precisely, it was defined that TCG instant 1977-01-01T00:00:32.184 exactly corresponds to TAI instant 1977-01-01T00:00:00.000 exactly. This is also the instant at which TAI introduced corrections for gravitational time dilation.
TCG is a Platonic time scale: a theoretical ideal, not dependent on a particular realisation. For practical purposes, TCG must be realised by actual clocks in the Earth system. Because of the linear relationship between Terrestrial Time (TT) and TCG, the same clocks that realise TT also serve for TCG. See the article on TT for details of the relationship and how TT is realised.
Barycentric Coordinate Time (TCB) is the analog of TCG, used for calculations relating to the solar system beyond Earth orbit. TCG is defined by a different reference frame from TCB, such that they are not linearly related. Over the long term, TCG ticks more slowly than TCB by about 1.6 × 10−8 (about 0.5 seconds per year). In addition there are periodic variations, as Earth moves within the Solar system. When the Earth is at perihelion in January, TCG ticks even more slowly than it does on average, due to gravitational time dilation from being deeper in the Sun's gravity well and also velocity time dilation from moving faster relative to the Sun. At aphelion in July the opposite holds, with TCG ticking faster than it does on average.
List of cycles
This is a list of recurring cycles. See also Index of wave articles, Time, and Pattern.
List of non-standard dates
There are several non-standard dates that are used in calendars. Some are used sarcastically, some for scientific or mathematical purposes, and some for exceptional or fictional calendars.
Lunar month
In lunar calendars, a lunar month is the time between two successive syzygies (new moons or full moons). The precise definition varies, especially for the beginning of the month.
This article deals with the definitions of a 'month' that are mainly of significance in astronomy. For other definitions, including a description of a month in the calendars of different cultures around the world, see: month.
National Physical Laboratory (United Kingdom)
The National Physical Laboratory (NPL) is the national measurement standards laboratory for the United Kingdom, based at Bushy Park in Teddington, London, England. It comes under the management of the Department for Business, Energy and Industrial Strategy.
Second
The second is the base unit of time in the International System of Units (SI), commonly understood and historically defined as 1⁄86400 of a day – this factor derived from the division of the day first into 24 hours, then to 60 minutes and finally to 60 seconds each. Analog clocks and watches often have sixty tick marks on their faces, representing seconds, and a "second hand" to mark the passage of time in seconds. Digital clocks and watches often have a two-digit seconds counter. The second is also part of several other units of measurement like meters per second for velocity, meters per second per second for acceleration, and per second for frequency.
Although the historical definition of the unit was based on this division of the Earth's rotation cycle, the formal definition in the International System of Units (SI) is a much steadier timekeeper: 1 second is defined to be exactly "the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom" (at a temperature of 0 K).
Because the Earth's rotation varies and is also slowing ever so slightly, a leap second is periodically added to clock time to keep clocks in sync with Earth's rotation.
Multiples of seconds are usually counted in hours and minutes. Fractions of a second are usually counted in tenths or hundredths. In scientific work, small fractions of a second are counted in milliseconds (thousandths), microseconds (millionths), nanoseconds (billionths), and sometimes smaller units of a second.
An everyday experience with small fractions of a second is a 1-gigahertz microprocessor which has a cycle time of 1 nanosecond. Camera shutter speeds usually range from 1⁄60 second to 1⁄250 second.
Sexagesimal divisions of the day from a calendar based on astronomical observation have existed since the third millennium BC, though they were not seconds as we know them today. Small divisions of time could not be counted back then, so such divisions were figurative. The first timekeepers that could count seconds accurately were pendulum clocks invented in the 17th century. Starting in the 1950s, atomic clocks became better timekeepers than earth's rotation, and they continue to set the standard today.
Solar time
Solar time is a calculation of the passage of time based on the position of the Sun in the sky. The fundamental unit of solar time is the day. Two types of solar time are apparent solar time (sundial time) and mean solar time (clock time).
Terrestrial Time
Terrestrial Time (TT) is a modern astronomical time standard defined by the International Astronomical Union, primarily for time-measurements of astronomical observations made from the surface of Earth.
For example, the Astronomical Almanac uses TT for its tables of positions (ephemerides) of the Sun, Moon and planets as seen from Earth. In this role, TT continues Terrestrial Dynamical Time (TDT or TD), which in turn succeeded ephemeris time (ET). TT shares the original purpose for which ET was designed, to be free of the irregularities in the rotation of Earth.
The unit of TT is the SI second, the definition of which is currently based on the caesium atomic clock, but TT is not itself defined by atomic clocks. It is a theoretical ideal, and real clocks can only approximate it.
TT is distinct from the time scale often used as a basis for civil purposes, Coordinated Universal Time (UTC). TT indirectly underlies UTC, via International Atomic Time (TAI). Because of the historical difference between TAI and ET when TT was introduced, TT is approximately 32.184 s ahead of TAI.
Theoretical astronomy
Theoretical astronomy is the use of the analytical models of physics and chemistry to describe astronomical objects and astronomical phenomena.
Ptolemy's Almagest, although a brilliant treatise on theoretical astronomy combined with a practical handbook for computation, nevertheless includes many compromises to reconcile discordant observations. Theoretical astronomy is usually assumed to have begun with Johannes Kepler (1571–1630), and Kepler's laws. It is co-equal with observation. The general history of astronomy deals with the history of the descriptive and theoretical astronomy of the Solar System, from the late sixteenth century to the end of the nineteenth century. The major categories of works on the history of modern astronomy include general histories, national and institutional histories, instrumentation, descriptive astronomy, theoretical astronomy, positional astronomy, and astrophysics. Astronomy was early to adopt computational techniques to model stellar and galactic formation and celestial mechanics. From the point of view of theoretical astronomy, not only must the mathematical expression be reasonably accurate but it should preferably exist in a form which is amenable to further mathematical analysis when used in specific problems. Most of theoretical astronomy uses Newtonian theory of gravitation, considering that the effects of general relativity are weak for most celestial objects. The obvious fact is that theoretical astronomy cannot (and does not try to) predict the position, size and temperature of every star in the heavens. Theoretical astronomy by and large has concentrated upon analyzing the apparently complex but periodic motions of celestial objects.
Time standard
A time standard is a specification for measuring time: either the rate at which time passes; or points in time; or both. In modern times, several time specifications have been officially recognized as standards, where formerly they were matters of custom and practice. An example of a kind of time standard can be a time scale, specifying a method for measuring divisions of time. A standard for civil time can specify both time intervals and time-of-day.
Standardized time measurements are made using a clock to count periods of some period changes, which may be either the changes of a natural phenomenon or of an artificial machine.
Historically, time standards were often based on the Earth's rotational period. From the late 18 century to the 19th century it was assumed that the Earth's daily rotational rate was constant. Astronomical observations of several kinds, including eclipse records, studied in the 19th century, raised suspicions that the rate at which Earth rotates is gradually slowing and also shows small-scale irregularities, and this was confirmed in the early twentieth century. Time standards based on Earth rotation were replaced (or initially supplemented) for astronomical use from 1952 onwards by an ephemeris time standard based on the Earth's orbital period and in practice on the motion of the Moon. The invention in 1955 of the caesium atomic clock has led to the replacement of older and purely astronomical time standards, for most practical purposes, by newer time standards based wholly or partly on atomic time.
Various types of second and day are used as the basic time interval for most time scales. Other intervals of time (minutes, hours, and years) are usually defined in terms of these two.
William Markowitz
William Markowitz (February 8, 1907 in Vítkov, Austrian Silesia – October 10, 1998 in Pompano Beach, Florida) was an American astronomer, principally known for his work on the standardization of time.
His mother was visiting Melč, Vítkov in Austrian Silesia (now cs:Melč, Czech Republic) when William was born. The Polish family emigrated to the U.S. in 1910 and settled in Chicago.
William earned his doctorate from the university in 1931, under W.D. MacMillan. He taught at Pennsylvania State College before joining the United States Naval Observatory in 1936, working under Paul Sollenberger and Gerald Clemence in the time service department.
After having married Rosalyn Shulemson in 1943, Markowitz eventually became director of the department. He developed the ephemeris time scale, which had been adopted by the IAU in 1952 on a proposal formulated by Clemence in 1948, as an international time standard. He subsequently worked with Louis Essen in England to calibrate the newly developed atomic clocks in terms of the ephemeris second. The fundamental frequency of caesium atomic clocks, which they determined as 9,192,631,770 ± 20 Hz, was used to define the second internationally since 1967. At the International Astronomical Union (IAU) meeting in Dublin in 1955, he had proposed the system of distinguishing between variants of Universal Time, as UT0 (UT as directly observed), UT1 (reduced to invariable meridian by correcting to remove effect of polar motion) and UT2 (further corrected to remove (extrapolated) seasonal variation in earth rotation rate), a system which remains in some use today.
He served as President of the IAU commission on time from 1955 to 1961, and was active in the International Union of Geodesy and Geophysics, the American Geophysical Union, and the International Consultative Committee for the Definition of the Second.
After retirement in 1966, Markowitz served as professor of physics at Marquette University until 1972, and also held a post at Nova Southeastern University.
ΔT
In precise timekeeping, ΔT (Delta T, delta-T, deltaT, or DT) is a measure of the cumulative effect of the departure of the Earth's rotation period from the fixed-length day of atomic time. Formally it is the time difference obtained by subtracting Universal Time (UT, defined by the Earth's rotation) from Terrestrial Time (TT, independent of the Earth's rotation): ΔT = TT − UT. The value of ΔT for the start of 1902 is approximately zero; for 2002 it is about 64 seconds. So the Earth's rotations over that century took about 64 seconds longer than would be required for days of atomic time.
International standards
Obsolete standards
Time in physics
Horology
Calendar
Archaeology and geology
Astronomical chronology
Other units of time
Related topics
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https://crpgaddict.blogspot.com/2013/06/mines-of-titan-ship-of-theseus.html
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## Sunday, June 30, 2013
### Mines of Titan, Ship of Theseus
Dumping off a dead body in a nightclub.
Mines of Titan has a perhaps unique approach to the idea of a "party" and associated character development, though I don't necessarily mean in a good way. You get so much experience from combat, and combat is so plentiful, that it's relatively easy to train each character to the max in skills and attributes in short order. At the same time, combat is so lethal that you're bound to lose even the most experienced characters occasionally, requiring you to recruit and train another if you want to avoid too much reloading. The game seems specifically designed for this dynamic, as without it, your characters spend most of the game not undergoing any development.
Quaid is eager to learn; there's just not much I need him to learn.
Oddly, the existence of the "party" seems to continue despite the replacement of the original members. If a previous character broke into a computer terminal and saved $5,000 by hacking the transport reservations system, the party will still have the reservations even if none of the same party members are with you when you arrive at the speeder terminal. The same goes for access to the "war room" (the only place to train on automatic weapons, arc weapons, and battle armor). Quest points that you've already experienced remain experienced. It's very odd. Sneaking into the war room. I've rotated through about 9 characters since the first posting. Gideon is still going strong, and is trained to the max in everything I would possibly want to train him in. Everyone else has been replaced a couple of times. I've found it impossible to keep all six characters alive at any given time. Eventually, I settled on four as the optimal number; there's less distribution of experience that way, and it cuts down on the length of combat (I'm 95% certain that the number of enemies that appear is based on the number of members in the party). Increasing stamina at a "personal development institute." Even from the beginning, I allowed myself to reload if my entire party was slaughtered--you lose all your money if that happens, and money is a precious resource for training and equipment. Also, towards the end of this session, I started playing a part of the game that takes place far away from the cities, and I decided to allow myself to reload during character deaths in this area; it would simply take too long to walk back to a city every time I want to replace a character. Plot-wise, a lot has happened since my first posting. I've been to three of the four cities on the moon and I've solved (I think) most of the quests. Here are the higlights: 1. Computer hacking. I got Gideon's "Programming" skill up high enough that I was able to hack into a computer terminal and accomplish a bunch of things, including: • Access to the "war room" for training. • A bunch of classified documents that suggest intelligent alien life forms have attacked the city of Procenium and have infiltrated the mines; hence, the lost of contact with the city and the announced shut-down of the mines. This attack might have been provoked by an agent's killing of one of the creatures. The creature itself is reported to be a "small reddish-brown sack that resembles a partially-filled rubber balloon." • Information about how to reprogram "golum armor," which is normally customized for each individual. I haven't found any golum armor yet, though. • The ability to modify police records to remove fines and warrants. • A back door into the "speeder reservation system" where I can make reservations without having to pay the$5,000 it normally costs.
• E-mail exchanges between systems operators at different cities, arguing about one's hoarding of an "interface card."
2. The missing specimen. I traced some pink footprints in one part of Primus and found a group of dirty cops who had stolen the specimen from the library.
They fail to kill the witnesses.
In a long combat that left Benny dead, I defeated them and recovered the specimen, which had fallen from its bottle and broken during the fight. The specimen--clearly one of the sentient creatures, or at least a part of it--absorbed itself into my hand and gave me a vision of what had befallen it, with a group of "incompetent scientists" vivisecting it. I recognized the leader as the Surgeon General of Primus.
3. Cybil Graves. The owner of the munitions shop, Cybil Graves, had asked me to go to the surface and find a group of Nomads who were supposed to give her a box. The Nomads, you may recall, are ex-convicts from the cities who have somehow found a way to survive on the surface. I found them with only a few minutes of wandering around.
The Nomads gave me a device that directed me to Cybil's box, as well as the entrance to some underground caves. I explored them a little but found the monsters fairly tough, so I returned to Cybil and got my $4,000 reward. That seems ominous. She soon appeared among the police bounties, "wanted dead or alive," for having "aided the outlaw nomads by running guns and ammunition" and "attempting to release a toxic gas into the Parallax air ducts." Much later, I found Cybil in an armor shop in Parallax, trying to buy armor for her Nomad friends. I got annoyed with the game in this section. There was a scripted set of screens in which Cybil offered me$4,000 to let her go and I weighed the offer in comparison to her $3,000 bounty. See what the screen says: Would it have been that hard for the developers to allow me to decide whether to accept her offer, turn her in, or "do both"? This seems like a role-playing choice that shouldn't have been taken away from me. In any event, I shot her, turned her in, and got$7,000 from the whole deal.
4. Warring SysAdmins. Okay, the game calls them "SysOps," but that was before it knew any better. In accordance with the e-mail exchanges I hacked, I got a quest to Progeny SysOp to steal a prototype interface card "devised to crack any computer system" from the Primus SysOp.
A rare role-playing choice in the gmae.
I went to the Primus office, and I had three choices: try to buy the plans, take them forcefully, or talk him into giving it to me. I suspect the first option would have worked with more administration skill and the third with more charisma, but I didn't have either, so I just robbed the guy. It got me wanted by the police, but I hacked a computer terminal and removed the bounty. When I returned the card to the SysOp in Progeny, he gave me a "prototype terminal," but I'm not sure what that did for me. Perhaps it would have allowed me to hack stuff if my hacking skill wasn't already so high.
5. Miscellaneous Progeny Quests. There were two quests on Progeny: helping an armor shop owner named Herb figure out who'd been breaking into his shop, and finding a wanted criminal who was hiding in the mines. Neither required any role-playing choices; I just wandered into the right places and watched as a series of narration and cut scenes left me victorious. I did get some good money and a couple nice sets of battle armor for my troubles, though.
6. The Alien Conspiracy. A few events came together to depict what I assume is the main quest. When I visited a hospital, a grotesquely burned man somehow divined that I knew of the aliens' existence and shouted at me to "stop these fools before it's too late!" He said that someone named Clinton Cain knew how to find the entrance to their home.
I found Cain by accident in the university in Progeny. He related that while collecting specimens on the surface, he came across a creature that had been wounded by laser fire. The creature "touched his mind" and revealed itself as a "member of an ancient race that does not have any written language or tools," using mental powers for everything. He further said that humanity's actions had "caused them to think of our race as a hostile force that must be eliminated."
Getting the main quest.
He suggested that if I find out what happened to Proscenium, I might be able to stop it at the other colonies and "save Titan." He set my "finder" (I guess the box I got from the Nomads) to direct me to the relevant caves and bade me good luck. I suspect it's pretty obvious what happened at Proscenium, so I don't know how going there is going to help, but whatever wins the game.
Solving these various quests involved visiting all three primary cities: Primus, Progeny, and Parallax. They were relatively indistinguishable from each other, all having the same look, services, and enemies.
I have no idea why the game felt it was necessary to put so many computer terminals near each other. You have to use them basically once in the game to get the hacking information, plus maybe a few other times to get speeder reservations. One per city would have been fine.
I moved between them on speeders, which normally cost $5,000 per trip but are free with my hacking abilities. When you go from one city to another, you have the option to take a "window seat" and see the terrain fly by as you zoom along. All cities delivered up plenty of opportunities for combat, though later as I developed my "administration" skill, I was able to avoid a lot of them. Choosing between tactical combat and computer-controlled combat is tough. 95% of the time, the computer does fine and all my characters survive, but that means 1 in 20 times, someone dies. To ensure survival, I viscerally want to be in control of the battle, but to be honest, I lose characters at a greater frequency than the computer does. Thus, I either have to watch my guys die with no control over it, or I have to take control and more thoroughly ensure their deaths. Here, I'm trying to use the terrain to bring them to me one by one, but it never really works. I keep getting into situations where they can shoot me from around the corner, but I can't hit them. More random notes: • It appears that your skill level with weapons and medical equipment defines what you're allowed to buy. As my skill with handguns, rifles, and automatic weapons rose, so did the available choices in the stores. More important, as my skill in medicine rose, so did the medical devices I could buy. Characters with a high enough skill can by "Medkits" and automatically heal the entire party after each combat. Very handy. • The game has an infrequent "copy protect" mechanism. When you try to train in advanced weapons in the war room, it occasionally asks you a question that you can only answer by looking at diagrams in the manual. I like to imagine my party members frantically paging through their notes at times like this. • I couldn't role-play any of the quest-related stuff, as above, but this is apparently the game's idea of a "role-playing choice." What kind of deviant says "yes" here? • The caves are full of steam and lava traps, but it's mostly just a waste of time, since my characters with medical skill immediately and automatically heal everyone anyway. Basically, this does nothing but slow my progress through the caves as I have to pause and read the message. • The screens in which your party moves as a unit are essentially identical to BattleTech: The Crescent Hawk's Inception, the previous Westwood/Infocom offering. In fact, many of the interface elements of this game are identical to BattleTech. As I close this posting, I'm stuck in the caves. Every few steps, I have a tough combat with a series of alien creatures capable of inflicting mass damage. My party bathes in a sea of chemicals shot by the creatures in the lower left. Throughout the caves, I've encountered those red balloon creatures. Some of them have granted me a vision of unlocking a door. Others grant me a "mental ability" but these manifest themselves in items that I can pick up, including a "Render," a "Synapse Beam," a "Mind Melt," and a "Reaver Rifle." I assume some of these are weapons, but other than the rifle, I'm not sure what classes. I'll have to experiment. I was playing part of this game in an airport. You don't want to know what "small, reddish brown, partially inflated rubber balloon" made me think about. But I'm stuck. The only way out of the caves, other than the way I came in, is a passage surrounded by four chemical vents that he game won't let me pass. I've tried all of my items here to see if any of them will work, but none of them do. It's possible that I missed something in the cities or I need to be wearing some special kind of armor to pass. Either way, it's a long trek back to the cities, and I'm not looking forward to it. I wouldn't mind a little hint here if there's some way to move forward without returning to civilization first. As you may have noticed, there are plot elements here that are similar to Total Recall (the REAL Total Recall), especially given that the original game, Mars Saga, was set on Mars. We have a shadowy, profit-motivated corporation running things through corrupt "agents"; an outcast class (mutants in the film, Nomads here); a hostile exterior environment; and rumors of ancient civilizations. But the game precedes the film by a couple of years. It's possible that the developers were influenced by Philip K. Dick's We Can Remember It for You Wholesale, on which Total Recall was based, but the story doesn't really have all of these elements. It's much more focused on the "hidden memory" aspect. It may all just be a coincidence. I'm looking forward to seeing what I find in Proscenium, if I ever get there. #### 38 comments: 1. It's been over a decade since I last played this, but I believe you can pick up golum armor by killing police officers (the highest ranked ones) in the third city, possibly in the other two cities too. It's the best armor in the game and there is no real consequence that I recall, other than officers being one of the hardest enemies in the cities. And if you don't have Medkit Cs yet, they are capable of healing party members completely. I never did beat the game however, so I'm afraid I don't recall how to get past that part. I think you may be able to enter one of those vents, or be able to rearrange your party formation somehow (might be thinking of magic candle though). I'm looking forward to you beating this game though :) 1. I have a bit of a role-playing objection to killing police officers. Fortunately, I got through the game without the armor. 2. In Deus Ex, I had a role-playing objection to killing the typical soldiers on either side of the conflict. 3. Luckily Deus Ex gives you lots of non-leathal combat options. Can't you do the whole game without killing anyone? 4. I have the same objection to killing police officers in video games (I like being a good guy). I remember a few years back, I was playing GTA 4 with a friend and he literally grabbed the controller out of my hands in a fit of nerd rage because I refused to shoot the police! 2. "I have no idea why the game felt it was necessary to put so many computer terminals near each other." If nothing else, it's probably realistic. Bizarre that the game forces you to stab Cybil in the back. Apparently it's a real pain to find what you're looking for in the vents. 1. (check the ground) 2. While I was about to write you to ask you to be more explicit, I just tried screwing around, and I finally found a way to get bumped around by the "trap" vents until it put me on top of the access vent. What a dumb mechanic. 3. You'd think there would have been ways to implement that 'puzzle' that weren't so immersion breaking. 4. Pretty sure you got through via a bug. You never found the remote detector, right? 5. After I won, I looked at a walkthrough, and it looks like the "remote detector" is the same thing as the "prototype terminal" that I got in Quest #4 above. 6. The Cybil scenario seems prime for choice, but drops the ball. Do you have the option to remove her bounty and warrant, or is that skill only for yourself? If you can remove it then I wonder how the later scenario plays out. 7. Unfortunately, I didn't check, but I'm 99% sure the option only works for the party. Being able to remove bounties for NPCs would be a level of roleplaying and choice this game just isn't capable of. 3. 'Oddly, the existence of the "party" seems to continue despite the replacement of the original members.' - Not really, lots of games do this. Key inventory items and quest flags are stored centrally to prevent players losing them or repeating quests for addition rewards. Some of the Gold box\M&M titles allow you to replace the entire roster, it's just more apparent in Mines of titan due to permadeath. A lot of older titles worked on very strict memory requirements, so it made sense to store quest flags centrally rather than 4-6 times. 1. The quest flags make sense. After all, SOMEONE completed them, even if it wasn't any of the current party members. And frankly, they're still spending the reward money. I have more of a problem with things that should have been specific to the characters somehow applied to the whole party. The same is true in, say, Curse of the Azure Bonds, when you can replace everyone after the game has started, but they still have the bonds. 2. Maybe they just found the dead guys wallet\diary\ID\reservations\severed hand or something and figured they'd play along. - I quess the designer could include an explanation, but would you really want to sit through that more than once? If you're replacing a character chances are it's because you've made a mistake (or were unlucky), the designer has to weigh up if it's worth punishing the player further for the sake of realism. It's kinda like playing a PnP game with a reasonable DM. He'll let a few things slide to keep the game flowing. 4. That "role-playing" picture comment makes me think about how Richard Garriott included hostile children in just about every game starting from Ultima IV, knowing that people would go to great lengths to avoid killing them even though the game itself would make no judgement (or in Ultima IV, it would even ding you points for valor for fleeing the battle!) He was clearly some kind of RPG Nietzche. 1. "What kind of deviant says 'yes' here?" I'm reminded of people on the GOG forums complaining that GOG sells the censored version of Fallout where you can't attack the children. What kind of deviants? Sadly, a significant percentage of gamers. :^/ 2. In their defense, the children steal from your inventory whenever you walk past them... Now, you would think "Doesn't that make the censored version better then?"... except that the censorship job was extremely lazy. They removed the children models... but the scripts are still there. In the original version, you could at least steal back your items. In the censored version, you have invisible children, "talk" and that take your items that you can't interact with, thus making the player wonder if it's some kind of bug that makes their inventory items disappear at random. Plus it's not like killing children is without consequence in Fallout. According to the game it's about one of the worst things you can do, making everyone hostile against you. 3. The 'Child killer' trait was awesome. - Well, not for the player, but it's a good example of a gameplay deterrent and a developer handing the concept in a mature fashion. As for the censorship, I thought they just used the patched UK version*? A bunch of quests are screwed up - Anna's bones, and the Mr Nixon doll one due to the missing characters, but I've never seen items go missing in over 20 plays. * I'm not sure about the rest of europe (GOG is Polish), but the UK is pretty sensitive about the idea of killing kids in video games. They banned handguns over a school shooting in the mid 90s, so no-one's going to risk getting a game banned for the sake of virtual kiddie murder. 4. I'm not sure whether it's strange that you can't kill kids or strange that you can kill people you gain nothing from killing. It seems like you should be able to kill everyone, or only those it makes sense to kill. ie If we're gonna have senseless violence, might as well go the whole hog. 5. "If a previous character broke into a computer terminal and saved$5,000 by hacking the transport reservations system, the party will still have the reservations even if none of the same party members are with you when you arrive at the speeder terminal. The same goes for access to the "war room" (the only place to train on automatic weapons, arc weapons, and battle armor). Quest points that you've already experienced remain experienced. It's very odd."
How's it odd? If the game made you do everything again, you'd be bitching about it. It's a computer game, you're "the party".
2. I imagine this is what the "Ship of Theseus" in the post title is referring to.
3. First of all, the CRPG Addict does not "bitch." He points out glaring flaws in games. In this case, the flaw is that the game specifically says that the war room program downloads data on your CHARACTERS, so it doesn't make sense that I can wander in later with completely different ones.
Also, this renders the programming skill somewhat useless after you've already hacked the computers once.
6. "5. Miscellaneous Progeny Quests."
"5. The Alien Conspiracy."
Clearly The Quest to Solve Chester's Problem with Counting was not one of the games you completed.
1. The problem is, I insert things later and forget to re-number. I need to remember to add the numbers only when I'm finished with the items.
2. This is why LaTeX rocks:
\begin{enumerate}
\item Miscellaneous Progeny Quests.
\item The Alien Conspiracy
\end{enumerate}
and then it does the counting for me.
While we are on this topic: 'Characters with a high enough skill can by "Medkits" and'
7. Not sure if you've done this yet (just going by your screen grabs) but you can, and will have to, rearrange your party positions when wandering around caverns in order to get past rocks in your path. It always annoyed me - not sure why they didn't go with the single icon representing the team in these situations.
I literally just commented on the previous post before this one appeared, but it looks like you're strolling through the game so won't need any help. I'm especially amazed how far you've got without golum armour, I always found that essential before venturing too far out of the cities.
1. Another quick thing. You've already started limiting your party to 4, and I think that's the optimum. We always used to play with a full party when we were kids, and I think that's another reason why we never got very far.
Also, I don't remember those gas vents being such an issue - either there's another exit, or I just blundered through it. (It's not mining skill related, is it? A decent mining skill helps avoid some of those evil cave-ins.)
2. Ben, I wish I'd read your comment a few hours earlier. It would have saved a lot of frustration. I agree with the difficulty navigating the caves, and I didn't figure out how to rearrange the party until very late (the option is unhelpfully called "combat order"), but it was one of those otherwise-unpassable rocks that made me hunt until I could find it. I agree that a single icon would have been a lot nicer.
8. For that hacking example it would help a lot if someone noticed the disappearance of your bounty (the guys that was robbed for example) and a day later it would reappear with a much high bounty.
9. The party replacement methods cinches what I thought earlier. Not only are they using the Traveller RPG rules as their basis, but they are running it like it was a table-top campaign.
There are ways to cheat death/ressurect those who died, but they are made expensive to keep people from just playing recklessly. There needs to be some consequences to big risks or completely stupid actions (I tell him just because you're a dragon doesn't mean you get to push me around). Or sometimes combat just goes badly. A player whose character is dead is stuck unable to play (usually), so the next best thing to resurrection is letting them create a new character and just jump back in.
It comes down to gameplay convenience trumping narrative:
http://www.leftoversoup.com/archive.php?num=315
---------
Secondly, the system is definitely Traveller. I think I yammered about it for a while back with Star Command; the most obvious reason is Education as a major statistic, plus your health being the combination of the 3 physical statistics.
And then character creation is starting with 18 year old blank slates and running them through 4 year segments of careers. It gets them a backstory and skills, and can provide a starting place for characterization.
The system is normally designed to allow 2 ways of skill improvement- either training by spending game-months learning on a skill, or from bonus points at the end of campaign sessions. Skill points can only be applied to skills that were used during the campaign session. But any computer version tends to end up focused more on combat (which computers can simulate easily) than skill checks/diplomacy interactions.
Since they aren't doing the long-term training (which would be pretty boring), and they used a lot of combat, it ends up back like most other DOS rpg's, where combat ability has to trump everything else in priority.
1. There were not many SF RPGs back in the day, and considering everyone stole from D&D, including stupid, stupid things like AC going down, I'm not really suprised that steal from whatever was around, even terrible, terrible systems like Traveller.
10. Just a quick comment to show everyone that, unlike AmyK, I am still alive here.
1. Hey, I'm not dead either! :p
11. PetrusOctavianusJuly 2, 2013 at 9:10 AM
Theseus, eh?
I just completed The Aethra Chronicles (a game you can look forward to when you reach 1994), and as I checked on Wikipedia if there was an article about it (there isn't) I found another article about an Aethra from Greek mythology, who was the mother of - you guessed (or knew?) it - Theseus.
Small world.
12. The "personal development institute" where you can train up your stamina reminds me of the gym and aerobics room in Sentinel Worlds I. I wonder if you'll like this game any more than that one. I know you've finished the game by now, so I'll keep reading...
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2017-08-23 23:05:41
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http://m.rangers.mlb.com/news/article/2747942/
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Howe confident in Blalock's fielding
Howe confident in Blalock's fielding
MINNEAPOLIS -- Rangers dugout coach Art Howe knows what Hank Blalock will have to go through to become an accomplished first baseman, and he is confident that Blalock can make a quick transition.
"Hank has the ability to be an outstanding first baseman," Howe said. "He has as good of hands as anybody. He just needs to experience different things that come up over there."
Blalock is expected to be activated off the disabled list on Friday, when the Rangers open a three-game series with the Indians in Cleveland. He has been working out at first base in extended Spring Training, and that's where he'll be Friday night, less than a week after volunteering to move there from third base.
Howe had to make a similar transition in his playing career. Initially signed by the Pirates as a shortstop, Howe was traded to the Astros in 1976, where he played primarily second base and third base before incorporating first base in to his repertoire in 1978.
He has a checklist of points that Blalock will have to adapt to during his transition to first base. The list starts with the ground ball in the right-side hole between the first and second baseman.
It's a balance between two sides. When a ball is hit, the first baseman's natural instinct is to go cover the bag, but Howe says that the first baseman has to be a "fielder first, and a receiver second." That means his first move is to go after a grounder to his right and not automatically go cover the bag.
"If your first step is toward first, you're not going to be able to get to the ball, even if it's hit a foot-and-a-half to your right," Howe said.
That said, the rule is to let the second baseman get everything he can. That way, a team doesn't have to worry about the pitcher having to cover the bag.
"You want to keep the pitcher involved as little as possible, because he has to catch the ball on the run," Howe said. "I told Hank [to] always check with the second baseman to see where he is playing. If the ball is hit at a pretty good pace, the first baseman has to get it. But if it's a slow roller or bouncer, then let the second baseman get it."
Next on the list are cutoffs and relays. There are times when the third baseman is the cutoff man for plays at home, and times when it's the first baseman. The basic rule is, if there is a runner on first base, then the first baseman is the cutoff man at home because of the potential play at third base. If there is no runner at first base, the third baseman takes the cutoff.
"He just needs to communicate with the third baseman," Howe said.
Blalock will also have a trickier throw from first base on potential double-play balls.
"The 3-6-3 is the most difficult double play to make," Howe said. "I told him to concentrate on getting the lead runner, and the double play is a bonus. Besides, it's the pitcher's responsibility to get over and cover [first]."
Howe said handling pickoffs shouldn't be difficult, although Blalock will have to adjust to the different backgrounds in different stadiums.
"It's just like playing catch," Howe said. "Hank will be fine. I felt pretty confident. When you're an infielder, it's just like learning to use your left hand instead of your right hand. You're on the other side of the infield, but you're still an infielder. Being an infielder, you still have the instincts to play the position.
"Going from the outfield to the infield is not easy. Being in the outfield, you're used to having a lot of time when the ball is hit to you. But in the infield, the ball gets on you pretty fast. Anybody can play first base, but not everybody can be a good first baseman. If you look at the great infields, they have a solid first baseman on each and every one of those infields."
T.R. Sullivan is a reporter for MLB.com. This story was not subject to the approval of Major League Baseball or its clubs.
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2014-10-21 14:33:36
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